Home/BlogThe Version 5.6 global average lower tropospheric temperature (LT) anomaly for February, 2015 is +0.30 deg. C, down a little from the January 2015 value of +0.35 deg. C (click for full size version):
The global, hemispheric, and tropical LT anomalies from the 30-year (1981-2010) average for the last 14 months are:
YR MON GLOBAL NH SH TROPICS
2014 01 +0.291 +0.387 +0.194 -0.029
2014 02 +0.170 +0.320 +0.020 -0.103
2014 03 +0.170 +0.338 +0.002 -0.001
2014 04 +0.190 +0.358 +0.022 +0.092
2014 05 +0.326 +0.325 +0.328 +0.175
2014 06 +0.305 +0.315 +0.295 +0.510
2014 07 +0.304 +0.289 +0.319 +0.451
2014 08 +0.199 +0.244 +0.153 +0.061
2014 09 +0.294 +0.187 +0.401 +0.181
2014 10 +0.365 +0.333 +0.396 +0.189
2014 11 +0.329 +0.354 +0.303 +0.247
2014 12 +0.322 +0.465 +0.178 +0.296
2015 01 +0.351 +0.553 +0.150 +0.126
2015 02 +0.296 +0.434 +0.157 +0.015
Note that the El Nino warmth in the tropics seems to have fizzled, falling about 0.25 deg C in the last few months to near the 1979-2010 average value, which is unusual since February has been the usual time of peak tropospheric warmth in response to previous El Nino events.
The global image for February, 2015 should be available in the next day or so here.
Popular monthly data files (these might take a few days to update):
uahncdc_lt_5.6.txt (Lower Troposphere)
uahncdc_mt_5.6.txt (Mid-Troposphere)
uahncdc_ls_5.6.txt (Lower Stratosphere)
What fits the global temperature trend data the best since the Holocene Optimum- Present is what I suggest below.
My thoughts on what drives the climate conform to what the data shows(present/past), unlike AGW theory which totally ignores the data both present and past.
AGW theory wants the data to conform to what it suggest, not the other way around.
https://climateaudit.files.wordpress.com/2015/01/md99-2275_jiang2015_1.png
More data which shows since the Holocene Optimum from around 8000BC , through the present day Modern Warm Period( which ended in 1998) the temperature trend throughout this time in the Holocene, has been in a slow gradual down trend(despite an overall increase in CO2, my first chart ), punctuated with periods of warmth. Each successive warm period being a little less warm then the one proceeding it.
My reasoning for the data showing this gradual cooling trend during the Holocene ,is Milankovitch Cycles were highly favorable for warming 10000 years ago or 8000 BC, and have since been in a cooling cycle. Superimposed on this gradual cooling cycle has been solar variability which has worked sometimes in concert and sometimes in opposition to the overall gradual cooling trend , Milankovitch Cycles have been promoting.
Then again this is only data which AGW enthusiast ignore if it does not fit into their scheme of things. I am going to send just one more item of data and rest my case.
http://www.murdoconline.net/wordpress/wp-content/uploads/2011/01/gisp2-ice-core-temperatures.jpg
“Temperatures in response to this will decline in the near future, in contrast to the steady state of temperature we presently have,or have been having for the past 15 years or so.”
– Salvatore Del Prete, 11/6/12
http://www.drroyspencer.com/2012/11/uah-v5-5-global-temp-update-for-october-2012-0-33-deg-c/#comment-64939
gotta admit, Salvatore has more guts than the IPCC when making verifiable predictions. 😉
Yes i agree Roy, but what i find even more interesting is the amount of times his shadow fails to refute him. Everytime Sal opens his mouth his shadow is not far behind failing again.
Now that takes guts.
@Salavatore “…unlike AGW theory which totally ignores the data both present and past”.
It not only ignores it, the IPCC creates it’s own data.
The quaint notion that the pre Industrial era had much lower levels of atmospheric CO2 is based on ice core proxies taken from Antarctic ice. According to ice core expert, Jaworowski, the content of CO2 in the cores varied widely in the same area, up to 2000 ppmv. The IPCC cherry picked a value that suited them.
In doing that, they completely ignored the effect of the Little Ice Age during which the Industrial era happened. Global temps were 1 C lower and one might expect global CO2 concentrations to be lower during such a cold spell since colder oceans absorb CO2.
The only other data they have is historical data measured by thermometers. Unfortunately for AGW theory, the data did not coincide with a gradual warming over the past century. So they went back and changed it.
They admitted that in the Climategate email scandal when IPCC Coordinating Lead Author Phil Jones, who keeps the official IPCC record, claimed he had applied Mike’s trick to 1940s temperature data to make it appear warmer.
NOAA and NCDC took care of the rest when they cut 4000 surface reporting stations from the 5000 available. It should be noted that the 5000, most of them in the US, only covered the 30% of the planet’s surface that is land. They barely covered the oceans at all.
After slashing the number of reporting stations, NOAA used interpolation and homogenization to fill in the gaps. Interpolation means filling in the temps of intermediate, missing stations with data from reporting sites up to 1200 miles apart. Homogenization means using climate models to smooth out irregularities left by such interpolation.
I might add that satellite telemetry is vastly superior to two a day surface thermometer readings averaged. The sat scanners cover billions and billions of data points (oxygen molecules) per instantaneous scanner position making their average much more accurate than thermometers and they cover 95% of the planet’s surface.
Dr. John Christy of UAH is on record as claiming there has been little or no true warming over the 33 year history of the sats.
Prete has immense faith in the gisp2 proxy. It ends in 1855. But why not check the new favorite Jiang 2015 against the Gisp2? Shift the gisp2 up 39 degrees and compare to the Jiang-proxy:
https://i.imgur.com/pAjt46C.png
Whatever happened to the Minoan warming in Jiang? Looks more like the Minoan cooling. And the Roman warming was gone before Caesar. And the medieval warming was not that impressing according to Jiang.
But of course: the main reason for believing in the correctness of gisp2 is that it ends in 1855. Get rid of the warm 30ies and 40ies in Greenland. And the warm noughties. Ingnore the modern warming. It does not fit into his scheme of things.
Prete has immense faith in the gisp2 proxy. Why not check the new favorite Jiang 2015 against the Gisp2? Shift the gisp2 up 39 degrees and compare to the Jiang-proxy:
https://i.imgur.com/pAjt46C.png
Whatever happened to the Minoan warming in Jiang? Looks more like the Minoan cooling. And the Roman warming was gone before Caesar. And the medieval warming was not that impressing according to Jiang.
But of course: the main reason for believing in the correctness of gisp2 is that it ends in 1855. Get rid of the warm 30ies and 40ies in Greenland. And the warm noughties. Ignore the modern warming. It does not fit into his scheme of things.
http://coacheshotseat.com/coacheshotseatblog/wp-content/uploads/2009/12/GreenlandIceCores15000.png
More data. I made my point.
Does the average from ’81 to ’10 have any scientific significance?
the tradition from meteorology is to use 30 year means for “normals”. The period 1981-2010 is the most recent 30-yr averaging period now in common use for computing departures from “normal”.
@Alick…”Does the average from ’81 to ’10 have any scientific significance?”
A better question might be whether the global average from 1950 – 1990 used by the surface record has any significance. The 1950s and 60s were periods of known cooling and that cooling makes up half the baseline. Such an average is bound to emphasize warming.
so by this measure, the running average is probably the highest it’s ever been for a non-El Nino phase, right?
…except that it is during an El Nino. If you corrected for El Nino/La Nina, I’m not sure what the answer to your question would be.
Dr. Spencer,
We are not technically in an El Nino by the definition of NOAA. Five consecutive values of the ONI have to be above 0.5 to be an El Nino. The last 3 have been 0.5, 0.7 and 0.7, so it is possible that this period will become classified as an El Nino, but NOAA gives it only a little better than 50% odds at this point. Even if it becomes an El Nino, it will be one of the weakest ever (you can’t have one less than 0.5 obviously). Of course the ONI could climb more this year.
If you do a multi-value regression of both Temperature and ONI, the results show that 1.0 of ONI corresponds to about 0.1 degrees on the 12 month average temperature. The value depends on how you smooth and lag the data and how much influence you include from the large volcanic events that you label on your graphs. As an example, the 2008 La Nina min ONI was -1.5 and the 2010 El Nino max ONI was 1.6. This 3.1 swing in ONI can be compared to the 0.4 swing in 12 month average temperature. Fitting the UAH data with a linear trend + 0.1*ONI (lagged a few months) gives a surprisingly good fit. Although perhaps not as entertaining as a cubic polynomial. Try it!
G. Loddy
We’ve not been technically in an El Nino, so better to refer to it as El Nino like conditions. At least until recently.
@Chris Brandow …”…so by this measure, the running average is probably the highest it’s ever been for a non-El Nino phase, right?”
If you look at the running average from 2001 – 2007, it looks quite similar and we still don’t know where it is going in the next few years.
All it will take is a couple of negative peaks to make the running average similar to that period.
You can’t claim that the relative flatness of the 2001 – 2007 period is all El Nino activity.
Thanks for the update, Roy.
Thanks, Dr. Spencer.
I have updated your graph my climate and meteorology pages.
Since GLOBAL is not the average of NH, SH, and TROPICS, they do not appear to be equally weighted. What is their relative weighting? And is there any significance to their relative weighting?
Thanks
Global is just the average of NH and SH.
Werner is correct.
Salvatore, another month has gone by….
“…here is my prediction for climate going forward, this decade will be the decade of cooling.”
– Salvatore del Prete, 11/23/10
http://ourchangingclimate.wordpress.com/2010/10/20/andrew-dessler-debating-richard-lindzen/#comment-8875
Thanks for reminding us how accurate Salvatore’s predictions are looking.
And how are the IPCC’s predictions doing?
see http://www.theclimatebet.com/
I meant that to support Sal, not detract from him.
The problem of wording….
but Sals predictions have been completely wrong?
The use of the word “normal” makes it sound like that is where the temperatures are “supposed” to be. I tend to believe the temperatures are always where they are “supposed” to be, given the inputs and how the mechanism works. I think science has a pretty good handle on the inputs but there is disagreement about how the mechanism works.
yes, it has always been a joke among meteorologists that weather is almost never “normal”. “Average” is a better term.
Living in south Texas I’m always annoyed in winter at the weatherman telling me it’s 20-25 degrees above/below normal. Winter here is cold then warm then cold then warm…it’s never the “normal” 60 degrees.
That still sounds like the “average” line is what the temperatures are “supposed” to be. I haven’t heard of any scientific reasoning as to why. To me, it is just a mathematical starting point. You could make that point anywhere. It shouldn’t change the basic shape of the graph/plot of points. It is the points relationship to each other that is important and not some arbitrarily placed horizontal line.
I feel like I’m not explaining myself very well. I want to put my finger on that horizontal line on that graph and ask, “why there”?
I would suggest a grid with an origin of ( year starting with, on the x-axis) and (the lowest monthly average temperature within the period of years to be displayed, -1 degree, on the y-axis). Example: 68 degrees F and 1972.
Otherwise I’ll just ignore the horizontal line, until told why I shouldn’t.
To give more refined explanation for past global temperature trends, ENSO, PDO,and AMO phases have to be incorporated and further superimposed upon the effects from Milankovitch Cycles and Solar Variability.
Volcanic Activity, and the Geo Magnetic Field having a place in this mix also.
The upshot being, an accurate temperature reconstruction/ verification should come about when each of the items mentioned in the above are evaluated and then combined in aggregate at any given time against the global temperature at any given time in accordance with the historical climatic global temperature record.
@Salvatore “…an accurate temperature reconstruction/ verification should come about when each of the items mentioned in the above are evaluated and then combined in aggregate…”
The IPCC has no interest in doing that, their mandate is to find proof of global warming. Remember, they are a political body run by a political body, the UN. The IPCC was initiated by UK PM Margaret Thatcher who had an axe to grind with coal miners who were defying her.
Thatcher was urged by an adviser to use her degree in chemistry to baffle the hoi polloi in the UN. She did a good job on them, convincing them that we have an impending disaster due to anthropogenic warming from sources like coal. That was her real agenda, to turn the world against British coal miners. From there, eco-alarmists got on board.
She had a hand in appointing one of the first IPCC directors, John Houghton, a climate modeler. Houghton was incredibly narrow-minded when it came to IPCC claims.
Hi Gordon Robertson,
Thank you for the political history info. While I had been aware Maggie Thatcher suffered carbon paranoia and wanted to stop Global Warming, I did not know she initiated the IPCC. It makes sense though, the British still want to rule the world even if they must resort to science fiction to help them. You may note similar attempts by Clinton/Gore in the 1990’s to dismantle U.S. coal in favor of Indonesian coal likely had similar commercial and territorial motives. However, the same dopey enviro-panic definitely helps sway gullible masses away from their legitimate claims to the Earth’s resources. The increasing costs, social dis-order and increasing violence as seen by ISIS, the Soviet Union Ukraine/Crimea situation and the ongoing British drama with Argentina over the Falkland Islands and other territorial thuggery largely due to attempt by some to grab access to the world’s hydrocarbons from others will only INCREASE as the reasoning capacity of the general population to see through obvious delusions becomes less evident due to an increasing population of the un-educated and an evident desire by leftist politicians to bring millions like them from third world countries into the first world. Crisis, antithesis, synthesis.
Have a great day!
Hello John. There is no ‘drama’ over the Falkland Islands. The people who live there don’t want to be Argentinian, and have voted. You know, democracy. The people who live there aren’t British – the vast majority were born on the island. Britain ‘owned’ the Falkland Islands before Argentina was even a country! British protection of the islands was based on their strategic position for service of the empire almost 200 years ago, not hydrocarbons. These were not known to exist.
Secondly, Britain does NOT still want to rule the world! We did, 200 years ago, but our time, like that of others, has passed. Britain has pulled back its military forces from all over the world, and it now has the smallest armed force since the Middle Ages. Britain is assessing its role in the world and is currently debating whether to leave the European Union – hardly a way of still ruling the world! Given what we have gifted the world, it is no wonder that we still play a part. And we are still a nuclear power – again, given our strategic geographical location.
Thirdly, Margaret Thatcher’s endorsement of global warming theory was based on short-term politics (coal mining union action) and advice from two trusted advisers. However, she completely changed her mind in later years, and saw it as a left-wing attempt at world domination…
“The doomsters’ favorite subject today is climate change. This has a number of attractions for them. First, the science is extremely obscure so they cannot easily be proved wrong. Second, we all have ideas about the weather: traditionally, the English on first acquaintance talk of little else. Third, since clearly no plan to alter climate could be considered on anything but a global scale, it provides a marvelous excuse for worldwide, supra-national socialism. All this suggests a degree of calculation…”
You’re welcome.
Gordon, I agree 100%.
In an earlier reply to a comment I mentioned Jaworowski, an ice core expert on ice cores proxies. Here’s one of his papers which reveals the scam behind IPCC claims about pre Industrial CO2 densities.
Note that the first part (Questioning the CO2 Ice Hockey Stick) is not Jaworowski. His paper follows part way down the page, titled “Do glaciers tell a true atmospheric CO2 story”?
http://www.greenworldtrust.org.uk/Science/Scientific/CO2-ice-HS.htm#ZJ
Well worth the read.
Consecutive months with monthly LT temperature at or above the following values (and the previous record). Records are marked with an asterisk.
0.00 36* (36)
0.05 35* (23)
0.10 21* (20)
0.15 21* (18)
0.20 10 (15)
0.25 6 (10)
0.30 6 (10)
**************************************************
Consecutive months with 13 month running average LT temperature at or above the following values (and the previous record). Records are marked with an asterisk.
0.00 74 (87)
0.05 72 (81)
0.10 69 (75)
0.15 28 (31)
0.20 25* (21)
0.25 7 (14)
0.30 0 (11)
It looks like several of these record-long streaks are in danger of falling. The running average is centered on August 2014, meaning that the last several months above 0.3 will be included in upcoming running averages. For example, even if the average drops to 0.00 for the upcoming months, the running averages for 0.00 – 0.20 will all set new records for longest streaks.
Alick, average is not what temperatures are supposed to be, it is an average of temperatures. My favorite example in meteorology is for rainfall. “Take one flood and one drought, add them together, divide by two and that is your average rainfall.” For your question about the time period, that is, as Dr. Spencer has said, somewhat arbitrary and conventional for those who follow these things. You may pick any time period you wish, if you’re so inclined. Then you’ll get to do your own averaging etc.
Personally, I don’t pay much attention to the horizontal line because it, like you say, doesn’t mean anything in itself. But that does not mean it doesn’t have political significance. If you’ll read David Appell, who seems to be an apologist for the alarmistas, he will find that the fact the temperature is above the line is great cause for alarm. Even more, if the recent measure had gone up 5 hundreths of a degree C instead of down 5 hundreths of a degree C, he would be all in a tizzy about how the sky if falling. My point is that the alarmistas watch the movement above the line, all 3 tenths of a degree of it, as if some great movement has occured.
Read some of Salvatore’s links. They will give you better perspective.
Have a snow free day.
Thanks. Very helpful.
If I look at the graph from about 2002 to 2015 one could create a new horizontal line at around 0.30 as the new normal. Also, it looks to me that from 1981 to 2015 there is a trend of about 0.01/year of warming. – if I’m interpreting the graph correctly. Of course, “past performance is no guarantee of future results” as they say in the finance world. But nonetheless 0.01/year does not seem like the “end of the world” either. 1 degree centigrade per century? Not sure what all the fuss is about. And not sure how government climate policy could move the tiny needle even if the sky were falling.
Hi Phyte On,
“… But nonetheless 0.01/year does not seem like the “end of the world” either.”
Your’ hunch proves correct. The Earth’s surface has been greatly warmer in the past. The eco-whackos, warmistas and the incurably credulous often worry that the “PERMAFROST” might thaw and all that assortment of frozen once living vegetation and animal life might be exposed to bacteria. Ask yourself this why don’t Mastadon’s, Whooly Mammoths, Dyre Wolves, tropical plants carpet the Polar regions like they once did? Well, because it’s far TOO COLD to support them now!!! Their permafrost remains from what I’ve read retain the carbon 14 isotope and therefore must have an age something less than 30-40,000 years ago. THE ONLY GEOLOGICAL EVIDENCE OF MASS EXTINCTION IS WHAT HAS BEEN CALLED THE ICE AGE!!! Warmer temperatures allow life to flourish cold temperatures tend to do the opposite. Let me ask you this what time of year does everyone run outside with little or no protective clothing to enjoy nature? If your normal it will likely prove to be Spring or Summer not Fall and Winter so much.
Have a great day!
1 degree centigrade per century is quite a lot, when you consider that the difference between fully glaciated ice age conditions and 20th century conditions is only about 6 degrees. Warming by 6K in only 600 years would be very significant indeed!
And of course Mastodons and Wooly Mammoths might be quite happy in a world that is much warmer and where the sea level is several hundred feet above its current level, but it may not be much fun trying to maintain a civilised human society as the sea swamps all of our major cities.
Hi Nigel Harris,
You state:
“…but it may not be much fun trying to maintain a civilised human society as the sea swamps all of our major cities.”
Ever heard of UNDERSEA TABLE MOUNTS?!!! Their islands that miss the ocean surface sometimes by about 1000 feet. They have surface (i.e., terrestrial vegetation) but now they’re under water. It seems sea levels have been rising for some time, I’ve read about 12-13 inches per century. Please provide any significant evidence this rate has accelerated, I haven’t seen it. Btw, the Maldives island in India sits only about a meter above the surface with it’s highest point being only 6 feet above the water’s surface. This island has had similar status for hundreds if not thousands of years, yet the eco-whackos constantly belie how it will soon be underwater. Apparently the natives living on the rather pleasant RESORT haven’t left yet and doubt anyone will soon.
Have a great day!
My statement should have read:
“This island has had similar status for hundreds if not thousands of years, yet the eco-whackos constantly bellow how it will soon be underwater.”
What I find frightening is that people as foolish as Nigel exist. The world economy has massive debt problems, the US government is on the verge of handing Iran the rights to build atomic bombs, there are countless other real problems in the world right now, yet he worries about some highly theoretical catastrophe 500 years in the future.
Hi Will Nitschke,
You stated:
“The world economy has massive debt problems, the US government is on the verge of handing Iran the rights to build atomic bombs, there are countless other real problems in the world right now, yet he worries about some highly theoretical catastrophe 500 years in the future.”
Iran has had the technology to build the A-bomb for decades. They can acquire the desired materials (including fissile material) with some effort but it doesn’t seem to difficult to me. A problem for us remains the apparent fact that they currently seek the development of ICBM’s capable of reaching the U.S., Europe and much of the planet. Last I checked the Iranians never signed the nuclear non proliferation agreements and are not obliged by any international treaty to care a fig about what the U.S. or anyone else thinks one way or another. The futile absurd agreement before the U.S. Congress does nothing substantial to halt their development (Netanyahu seems right about that). Besides who believes the Iranians won’t have the a-bombs when they want them anyways? They profess to live up to various nuke agreements they haven’t signed to portray the image they care about such things, but as Bibi pointed out they just got caught running two secret enrichment facilities. In reality, they don’t live up to these agreements and nothing can be done because they aren’t signatories anyways! Better hope the Suni Pakistani’s that have nukes don’t simply choose to share their know-how and material with their Shia/Iranian counterparts. In any case, from what I’ve read all the mid-East terrorists cells get their marching orders from the Iranian Mullahs. If the U.S. wants to prevent an attack we need good intelligence and the willingness to use effective force quickly.
However, the problem remains we have despots in our own government including the current administration that have provided arms to Isis, AlQueda (note Syrian rebels) and other terrorist organizations. Providing arms and assistance to groups that have attacked and declared war on the U.S. used to be called TREASON! Until this country faces it’s own problems, the international ones will seem all too insurmountable.
Have a great day!
Correction, I stated:
“In any case, from what I’ve read all the mid-East terrorists cells get their marching orders from the Iranian Mullahs.”
This statement pertains to coordinated attacks on Western targets not their own internal disputes.
Have a great day!
Alick, average is not what temperatures are supposed to be, it is an average of temperatures. My favorite example in meteorology is for rainfall. “Take one flood and one drought, add them together, divide by two and that is your average rainfall.” For your question about the time period, that is, as Dr. Spencer has said, somewhat arbitrary and conventional for those who follow these things. You may pick any time period you wish, if you’re so inclined. Then you’ll get to do your own averaging etc.
Personally, I don’t pay much attention to the horizontal line because it, like you say, doesn’t mean anything in itself. But that does not mean it doesn’t have political significance. If you’ll read David Appell, who seems to be an apologist for the alarmistas, he will find that the fact the temperature is above the line is great cause for alarm. Even more, if the recent measure had gone up 5 hundreths of a degree C instead of down 5 hundreths of a degree C, he would be all in a tizzy about how the sky if falling. My point is that the alarmistas watch the movement above the line, all 3 tenths of a degree of it, as if some great movement has occured.
Read some of Salvatore’s links. They will give you better perspective.
Have a snow free day.
Crakar says below, but does not seem to understand my solar criteria has yet to be reached therefore it is way to early to make a judgment as to how correct or not I might be.
I do not think these people read what is said and in what context.
March 4, 2015 at 10:15 PM
Yes i agree Roy, but what i find even more interesting is the amount of times his shadow fails to refute him. Everytime Sal opens his mouth his shadow is not far behind failing again.
Now that takes guts.
THE CRITERIA
Solar Flux avg. sub 90
Solar Wind avg. sub 350 km/sec
AP index avg. sub 5.0
Cosmic ray counts north of 6500 counts per minute
Total Solar Irradiance off .15% or more
EUV light average 0-105 nm sub 100 units (or off 100% or more) and longer UV light emissions around 300 nm off by several percent.
IMF around 4.0 nt or lower.
The above solar parameter averages following several years of sub solar activity in general which commenced in year 2005..
IF , these average solar parameters are the rule going forward for the remainder of this decade expect global average temperatures to fall by -.5C, with the largest global temperature declines occurring over the high latitudes of N.H. land areas.
https://i.imgur.com/pAjt46C.png
Some differences but the trend is essentially the same between both data sets.
Whatever happened to the Minoan warming Prete? The Roman warming was not so very Roman. And the MWP was nothing much.
The trend is not the same. The gisp2 proxy has LESS cooling from the climate optimum. The Jiang proxy has more in common with the better Greenland icecoreproxy from Vinther 2009.
All you manage to show is that you will only accept one proxy as correct. And you extrapolate this one proxy to the whole world. It does not even match proxies from the same area.
A human body was found thawing in the Alps in 1991. It was determined to have been there 5300 years and dubbed Oetzi, the Iceman. Other than the rise in atmospheric CO2, what has happened recently that accounts for this evidence? His body and belongings were fully intact in the rocky gully where he was found, so he couldn’t have been “rafted” down by the “river of ice” (glacier), which would have shredded him and scattered his belongings.
Norm, there can BE no other explanation. It has to man-made CO2. Obviously, man caused Oetzi to be buried in ice exactly 5300 years ago and man caused Oetzi to be thawed by burning CO2. There is NO possible natural explanations!
Dan, so glad you agree! Actually, it was a man who did Oetzi in; the snow buried him and became a glacier. No doubt there were others, but they had little chance of being found within a very short time before their remains rotted away. I was hoping you would offer a natural explanation – solar variation, water vapor, …, but you’re right – none of them can be shown to have hidden for so long, then appeared just in time to make us think we did it.
Hi Norm Rhett,
Great post! Imo, the glaciers have been thawing off and on for thousands of years. Please read my previous posts above. In fact, evidence of water/ice cut moraines in Europe helped Louis Agassiz convince Charles Lyell that an ICE AGE actually occurred. Remember it wasn’t that long ago (few thousand years) the polar ice caps probably didn’t even exist. Too many think like Charles Lyell that the “PRESENT IS THE KEY TO THE PAST.” Unfortunately for them, time is not merely periodic but linear. Events occur and don’t always re-occur in the same way. Some have the world view that time and events form a patter like a circle or an ever widening Gyre (T.S. Eliot). Personally, I don’t see it quite that way.
Have a great day!
Hi JohnKL,
No doubt the world has been warmer (and colder) than today. Ice caps grew and sometimes melted completely. Life, including human life, carried on. The problem now is that much of humanity lives in non relocatable cities below levels that , with significant melting in the foreseeable future, could be submerged. Nearly all of humanity relies on modern agriculture and distribution. It is uncertain, but clearly possible, that agriculture has already been adversely impacted by warming. If burning of fossil fuels is the primary cause of that change, and I have yet to see a valid alternative, we had better curb that unsustainable practice.
Thanks, I’ve had a good day so far. Same to you!
Norm the warm writes: “I have yet to see a valid alternative”
Well look no further than this 21st century new paradigm in climate science. In that you will most definitely not be able to refute it with valid physics (though you can take me on here if you wish to make a fool of yourself) and you will not be able to produce a centrifugal force field which acts differently and somehow does not form a temperature gradient in any empirical experiment such as in our group’s website, I would suggest your lack of finding an alternative rests somewhat in your lack of desire so to do.
Hi Norm Rhetts,
You state:
“It is uncertain, but clearly possible, that agriculture has already been adversely impacted by warming. If burning of fossil fuels is the primary cause of that change, and I have yet to see a valid alternative, we had better curb that unsustainable practice.”
This has been addressed in previous posts. Warm weather helps agricultural production, or do you think it an accident that growers use HOT HOUSES to grow tomatoes and other plants? In addition, since CO2 is a PLANT FOOD, the combination seems like a win/win since the planet still suffers ICE AGE conditions. If as you assert, the Earth’s hydrocarbons are “fossil fuels” and their consumptions cannot be sustained why do you worry if people continue to use them since we will soon run out of said hydrocarbons anyways? Unless of course you realize that Methane proves to be a VOLCANIC GAS and the Earth’s hydrocarbons likely will never run out (none of them have yet) you may wish to revise such seemingly dark vision.
Have a great day!
Norm,
I definitely do not agree with your pseudo-science blasphemy. Your statement that “snow buried him and became a glacier” is total fantasy. To imply that Oetzi was buried naturally is impossible. Those glaciers have been around for millions of years. To say that the climate today is the same as 5300 years ago is crazy. The climate today is warmer and changing faster than at any time in history and man-made CO2 is the cause. To make any statement about natural processes is just unfounded. Get with it, Norm.
5300 years ago was at the tail end of the Holocene Climatic Optimum. The HCO was a period of maximum NH heating due to orbital forcing. This is pretty well explained by Milankovitch cycles.
John, from what I’ve read, we should be in a cooling period of the Milankovitch cycles, contrary to the recent trend.
Norm Rhetts,
Depends on what data you rely on. Our current TREND will only appear as cogent as the data-set behind it. The satellite data doesn’t show any trend worth discussing the NASA, JMA, NOAA etc, do but the NASA (and thus other datasets that rely on it like JMA may be unreliable) data is being investigated for manipulation. The FACT remains it requires quite a lot of FAITH to BELIEVE anything reported about such data.
Have a great day!
John, as you well know, the periods of Milankovitch cycles are many thousands of years. No effect from them could be noticeable on the scale of a human lifetime. I was only making the point that what little effect there is would be in the wrong direction.
Norm Rhetts,
As you well know, I referred to current/recent trends.
Have a great day!
Milankovitch cycles do not regulate glacial periods. What does is the variation in the eccentricity of Earth’ orbit (due primarily to Jupiter’s gravity) and this alters the annual mean distance of the Earth from the Sun, thus affecting insolation levels and so the effective radiating temperature of Earth.
Temperature trends have nothing to do with carbon dioxide.
The reasons for the pause are all to do with natural cycles. Back radiation does not raise surface temperatures, so it has nothing to do with carbon dioxide levels. At Mauna where those levels are measured there has been no increase in temperature since 1959 as shown here.
To really understand what is happening you need an understanding of entropy and thermodynamic equilibrium. Without such, you are likely to be totally misled by those who likewise don’t understand thermodynamics.
When a ball is falling through a vacuum tube gravitational potential energy is converted to kinetic energy (just as happens to molecules in flight between collisions) and total energy remains constant. However, entropy increases and that is what the Second Law of Thermodynamics tells us.
As a result of that law, a stable density gradient forms in a planet’s troposphere. This happens when, in any horizontal plane, the pressure from below equals the pressure from above.
In Kinetic Theory we understand that pressure is caused by molecules striking a wall or imaginary boundary. Temperature is proportional to mean molecular kinetic energy and pressure is proportional to the product of temperature and density.
Hence, when we have thermodynamic equilibrium (with maximum entropy) we have equal numbers of molecules going upwards across a horizontal boundary as there are going downwards. In addition, when molecules are about to collide they have equal kinetic energy. This situation occurs when the sum (PE+KE) is constant even at different altitudes, and so that is why we have a temperature gradient.
As I said, this state of thermodynamic equilibrium also has a density gradient. This is because there is a slightly greater propensity for molecules to go downwards than upwards due to the fact that gravity curves their path if there is any horizontal component in their motion.
Now, because the temperature gradient is an equilibrium state, then we get downward convective heat transfer when that equilibrium is disturbed by newly absorbed solar radiation in the upper troposphere. This heat is what supplies that energy which James Hansen thought back radiation did. There’s evidence here
The graph on http://www.climate-change-theory.com/ is very neat and totally irrelevant. One might assume it is a graph of temperatures, but the title is “Angular momentum of the Sun and all planets”, perhaps interesting to an astronomer or spacecraft designer, but with no meaningful influence on terrestrial climate. The discussion of thermodynamics might be accurate; I didn’t bother to read it carefully.
The greenhouse effect is primarily a matter of radiant energy. Greenhouse gases are relatively transparent to visible light, so solar radiation can penetrate the atmosphere. That radiation is absorbed and reradiated at heat wavelengths, to which greenhouse gases are relatively absorbent.
Anyway, there is still no answer to the question, “If the gigatons of GHGs we have emitted burning fossil fuels are not the cause of warming that was absent for thousands of years, what is?”
Norm Rhett writes …
“no meaningful influence on terrestrial climate”
Who are you to know? The correlation with the well recognized ~1,000 year and superimposed 60-year natural climate cycles is compellingly statistically significant. Quite probably magnetic fields for the planets (which we know reach to the Sun) influence Sun spots, solar intensity and/or cosmic ray intensity which can then affect cloud formation. Oh, but you know that don’t you, because you read some of the text on the website.
“I didn’t bother to read it carefully”
No you wouldn’t deign to do that because I assume it doesn’t suit your agenda or pecuniary interests or something. Furthermore it’s probably well beyond your understanding due to a lack of knowledge or education in such thermodynamics.
“The greenhouse effect is primarily a matter of radiant energy.”
Well I’m offering a $5,000 reward to the first person in the world who can prove that using the laws of physics and backing it up with empirical studies that show the most prolific greenhouse gas water vapor causes the surface temperature to be about 30 degrees hotter, as claimed by the IPCC et al.
“What is?”
Well go back to the plot at the top of that website endorsed by our group of persons suitably qualified in physics. What qualifications in physics do you have?
PS Regarding the issue of radiation, not that it’s relevant to the 21st century new paradigm in climate science, I suggest you read the first of my peer reviewed papers on the Second Law of Thermodynamics (written three years ago) which is linked from the “Evidence” page on our group’s website. While there you might like to consider the significance of the fact that centrifugal force has been demonstrated empirically to form a temperature gradient, just as does the force of gravity in a planet’s troposphere, crust and mantle. This of course completely refutes the greenhouse radiative forcing conjecture.
Roy
Why do you do a 13-month running average instead of 12-months which would be far more logical. The 13-month one gives double weighting to one month of the year, and that’s a month furthest from the month in which you are plotting the mean anyway. There’s no reason why a 12 month mean could not be plotted half way between the 6th and 7th month.
I take it you’re not a scientist…
Well if you’re attempting to claim the $5,000 reward mentioned in my comment above (and not yet claimed since my book was published in March last year) then be sure that, when doing computations for entropy in a planet’s troposphere, you don’t inadvertently use the inappropriate equations for thermodynamic potentials which omit terms for gravitational potential energy. You may find it best to read first the website endorsed by my group of persons suitably qualified in physics. If in Australia some time, try to get along to one of my lectures on the 21st century new paradigm in climate science.
I’ll take that as a no then.
Think “Doc” Brown from Back To The Future, except a version that never invented anything that ever worked, and add a tin foil hat that he uses for protection against CIA infiltration into his brain.
Doug, I managed to get a ‘B’ in Thermodynamics at UC Berkeley as part of my physics major.
That reminds me, angular momentum is conserved.
You show insufficient understanding of the pre-requisites for momentum to be conserved. It may not be the case in an isolated system in a force field – only in a closed system.
Regarding conservation of angular momentum …
In the sum there are included 9 planets and the Sun, all with respect to Ssb. The sum is scalar. The planets are Mercury, Venus, Emb (Earth-Moon system), Mars, Jupiter, Saturn, Uranus, Neptune, Pluto. It does not include asteroids and spin momentum of bodies, namely of Sun.
The whole sum is almost constant, but there is a small difference of 8.11*10-7 of the whole. The difference from a constant value is mainly caused by performing the scalar sum instead of vector sum, and also it is divided between orbital angular momentum of asteroids, trans-neptunians and spin angular momentum of Sun. (but of these only asteroids were included in ephemerides calculation) (*2) – These high swings are at the times, when the Sun is approaching the solar-system barycenter too closely, and then the space curvature (not regarded by me), plays a significant role…? No, the Sun is moving retrograde at these times and its vector angular momentum is actually negative. There is no negative scalar angular momentum, which causes this difference.
On the scale of millennia, the Earth and Venus angular momentum (relative to Sun) grows, these planets move in concentric spirals(?) and are speeding up (as determined by huge Gaussian filtering), or rather by the change in orbit excentricities. The average approach to Sun after 5 millennia is still much smaller than annual approaching and receding due to excentricity… On this scale, the Mercury and Mars angular momentum (relative to Sun) shrinks.
[source]
Norm: a “B” hardly demonstrates exceptional understanding of thermodynamics, a topic I have studied extensively at post-graduate level, and I have also done so in regard to recent developments in understanding of maximum entropy production and the optimum paths taken to achieve such.
So a few questions:
(1) How do we determine from energy potentials when thermodynamic equilibrium is attained?
(2) Has entropy reached a maximum then within the constraints of the system?
(3) Explain how we can apply the Second Law of Thermodynamics and Kinetic Theory to show why a density gradient forms and stabilizes in a force field.
(4) Then do likewise for a temperature gradient.
(5) Express in words (using Kinetic Theory) the proportional relationship between pressure, density and temperature.
(6) Thus explain how the pressure gradient evolves as a corollary of the Second Law.
(7) Explain why we cannot assume sensible heat transfers are always from hot to cold in a vertical plane in an isolated system subjected to gravity.
Doug stay with the physics not the climate aspect. Both of which I disagree with you close to 100%.
Assertive statements are ignored. How about you explain your areas of specific disagreement, citing the laws of physics in support of your apparent misunderstandings. Then I can pinpoint for you where you go wrong.
Doug Cotton and Venus,
I am not sure your math works out with “heat creep” as an explanation of why Venus’s surface is so hot.
Your claim is the solar radiation is absorbed by the Venus atmosphere and then creeps down to the surface by means of gravity potential energy exchange. You claim carbon dioxide has no effect on this process.
The big huge problem with this conjecture and why it appears very flawed. Venus only absorbs (surface and atmosphere) 157 watts/meter^2 or so
https://books.google.com/books?id=3LEvFw_GIvoC&pg=PA35&lpg=PA35&dq=how+much+solar+energy+flux+is+absorbed+by+venus+atmosphere&source=bl&ots=9RlDIuVLvS&sig=B5cR8LR2S-S9hptFz1Au7lbsSNI&hl=en&sa=X&ei=rKP7VLj-GYecgwTWqYHADA&ved=0CDkQ6AEwBTgK#v=onepage&q=how%20much%20solar%20energy%20flux%20is%20absorbed%20by%20venus%20atmosphere&f=false
The surface is radiating at a rate of around 16,000 watts/meter^2
If the atmosphere was not doing anything to slow this high radiant flux your heat creep would do nothing to keep the temperature so high. Adding 157 watts/meter^2 to something losing 16,000 watts/meter^2 will do nothing to keep it warm.
The Sun’s direct radiation to the Venus surface supplies only about 20W/m^2, not the much greater and incorrect Science of Doom figure. The 20W/m^2 was estimated from measurements by Russian probes dropped to the surface.
Yes, for the temperature of a location on the surface to actually rise from 732K to 737K there would need to be over 16,000W/m^2 if radiation were what was raising the temperature. But it’s not. And radiation from a colder troposphere cannot raise the surface temperature one iota.
What does supply the necessary thermal energy to the Venus surface is downward convective heat transfer which is acting in accord with the process described in the Second Law of Thermodynamics whereby entropy is increasing and there is a propensity towards thermodynamic equilibrium.
The Sun’s radiation reaching Venus can only raise the temperature of regions that are at temperatures less than about 400K in the upper troposphere and above. From there the energy is transferred downwards, but not by radiation as that would violate the Second Law.
People like James Hansen had absolutely no concept of the new 21st century developments in our understanding of entropy. Nor have most readers here I suspect, so you could all start here from which I quote …
“The law of entropy, or the second law of thermodynamics, along with the first law of thermodynamics comprise the most fundamental laws of physics. Entropy (the subject of the second law) and energy (the subject of the first law) and their relationship are fundamental to an understanding not just of physics, but to life (biology, evolutionary theory, ecology), cognition (psychology). According to the old view, the second law was viewed as a ‘law of disorder’. The major revolution in the last decade is the recognition of the “law of maximum entropy production” or “MEP” and with it an expanded view of thermodynamics showing that the spontaneous production of order from disorder is the expected consequence of basic laws. This site provides basic texts, articles, links, and references that take the reader from the classical views of thermodynamics in simple terms, to today’s new and richer understanding.”
Doug I think you are missing my point of the Venus atmosphere. The surface of Venus is radiating away around 16000 watts/meter^2. The solar energy needed to replace this (incoming) is only around 157 watts/meter^2. I do not believe in your “heat creep” theory at all and I explained in detail why. Why you are wrong and why the atmosphere does not work in the simplistic fashion of a four molecule thought experiment.
Here is a simple test to demonstrate why you lack understanding that the atmosphere (in the region where weather dominates, troposphere) is so filled with molecules that they act the same as and actual surface and you can only measure the potential and kinetic energy relation based upon the height above a surface.
The simple test. Take two pieces of computer paper. Crumple one into a ball and the other leave as a sheet. Hold them the same height. They have the same mass so would possess the same potential energy. Drop them at the same time and see what happens. The crumpled one hits sooner and hence travels faster. It has a higher kinetic energy when it hits the surface than the sheet of paper. They start with the same potential energy but end with different kinetic energy. That is because the air molecules create a surface for the paper that it cannot move freely through.
Anyway, What happens with the 16000 watts/meter^2 that the surface of Venus is radiating away continuously?? Your heat creep theory (wrong as it is and will always be wrong even if you repeat it 10000 times more, Curt destroyed your theory with the ultra high centrifuge experiments) cannot provide the energy necessary to sustain this flux. The flux is leaving the surface, it is doing it continuously. You need to have this much energy returned in order for it to be maintained. Also where does this 16000 watts/meter^2 go? It is being emitted by the hot surface. It is not seen from the Earth so it is not making its way out. Where does it go?
Ultra high centrifuge machines create a temperature gradient with colder air at the center as shown in this comment and the following ones.
Where does it go? See this comment.
and this comment.
Go to this comment.
PS: My “maths” works out very well, thank you Norman, for all planetary tropospheres. For example, for the troposphere of Uranus my maths gave a temperature of 329K at the base thereof, whereas Wikipedia cites 320K. Pretty close I suggest.
Norman says “If the atmosphere was not doing anything to slow this high radiant flux “
The radiant flux is not “slowed” – only the flux of thermal energy is slowed, and in fact is reversed (by the atmosphere) during the 4-month-long sunlit period when a location on the equator of the Venus surface rises by 5 degrees, offsetting the inevitable cooling at night.
You may liken it to a ball bearing in a pot of water at 95°C. You turn on the stove (Sun) which boils the water and raises the ball bearing temperature to nearly 100°C. You turn off the stove (Sun) for a while (at night) until the ball-bearing temperature falls by 5 degrees. You turn it back on (daytime) and the ball bearing temperature rises again. Frankly I don’t care how much energy is transferred between the water and the ball bearing by radiation or otherwise. The point is, we know the ball bearing temperature will rise and fall accordingly.
Now it’s time for you to study my paper written two years ago on Planetary Core and Surface Temperatures” which is linked from our group’s website.
Of course the atmosphere slows cooling by both radiation and sensible heat transfer. Heat creep provides the energy required for balance at any internal interface. You have to consider the different directions day and night.
I’ll say no more at this point, because I’m trying to teach you to think for yourself, but I guess with your only getting a B in a first year course puts you at a significant disadvantage. Perhaps with a few hours of private tuition I could have helped you, but my fees for such are fairly high due to demand because of my success with students over the years since I gained my first degree with a physics major in the 1960’s.
Dr. Roy,
Please carry out a linear, least-squares fit to your data set. I think you will find that the trend line is positive, indicating a warming trend during the period for which you provide data.
Dr. Mark
There will be long term natural warming until the year 2058, after which nearly 500 years of long term cooling will commence, as explained here in The New 21st Century Paradigm in Climate Science.
We can learn a lot from the planet Venus as in my comment above.
We can also learn a lot from experiments with vortex cooling tubes which enable us to prove the greenhouse radiative forcing conjecture is false.
But you won’t understand why until you study the above-linked website now endorsed by our group of persons suitably qualified in physics.
Sorry – typo in the link:
http://climate-change-theory.com
(Over 6,300 visitors have viewed that site in two months.)
Angular momentum is conserved. The graph should be flat.
This is the scalar sum of the angular momentum of the Sun and nine planets you clot. Next time read the source, Norm the warm. This information has been around for years: http://semi.gurroa.cz/Astro/Orbital_Resonance_and_Solar_Cycles.pdf
Name calling lends nothing to the strength of an argument.
Because the Sun is the most massive body by three orders of magnitude, the barycenter of the solar system is essentially the center of the sun. Whatever happens between the planets is relatively tiny.
Quoting from http://arxiv.org/ftp/arxiv/papers/0903/0903.5009.pdf:
“The charts are self-relative, with minimum to maximum stretched to fill the image. Actually, if they displayed 0 also, the data would be one straight line at the top of the image, since the oscilation [sic] is incomparably smaller than the absolute, almost-constant value…”
What is the absolute vertical scale on your graph? Unless you can show how these tiny variations effect the energy balance of the world’s atmosphere (and oceans), I repeat, “the graph is totally irrelevant”.
No informed observer doubts that “there is a temperature gradient formed by gravity and not by back radiation from water vapor, carbon dioxide etc.” Worldwide average warming is not a gradient; it is a measure of the whole system. Except for the insignificant inflow of meteors and outflow of lighter gases, the world’s overall thermodynamic balance, i.e. temperature, is affected only by radiation.
Well what do you think the vertical scale is when related to temperature? Perhaps 1.5 degrees out of 287 degrees, that being about the last amount of warming between the Dark Ages and the Medieval Warming Period.
So, on the Kelvin scale, we have about 0.5% variation in mean world temperatures in the last 2000 years or so.
How flat would this long-term plot of temperatures in Ireland appear if it were correctly plotted with absolute zero (-273°C) as the axis? The issue is that there is a strong correlation for both the long term (934 year) and short term (60 year) cycles in the plot of angular momentum and the natural cycles in mean temperatures. I appreciate your attempted refutation, but I don’t buy it.
I’m glad you acknowledge the existence of the gravitationally induced temperature gradient which is the state of thermodynamic equilibrium with maximum entropy. Please then tell Roy, Pierrehumbert, the IPCC etc (who are thus not “informed observers”) that temperatures would not have been isothermal in the absence of IR-active molecules.
Your statement “the world’s overall thermodynamic balance, i.e. temperature, is affected only by radiation” is refuted in my peer-reviewed papers both linked from the “Evidence” page at http://climate-change-theory.com which is being visited by about 800 per week. The website itself provides a summary thereof.
“Because [sic] the sun is the most massive body of the solar system…[therefore] the barycenter of the solar system is essentially the center of the sun…”
It is a funny old center, then – sometimes 500,000 km outside the limb of the sun.
As an argument it is a non-sequitur; “Archimedes’ lever” ring a bell?
Astronomers routinely detect exo-planets by the wobbling of their stars around the barycenter of their systems.
At present the barycenter of our solar system is half way between the center of the sun and the surface*. In 2023 it will be way above the surface.
*defined as where the gas transitions from opaque to transparent.
It is gravity which “traps” thermal energy, not back radiation.
There is solid empirical evidence now in a vortex cooling tube that centrifugal force produces a temperature gradient, as does the force of gravity.
The evidence is here along with the explanation based on the laws of physics.
This is what climatologists need to understand ….
Entropy and the Second Law of Thermodynamics
The second law of thermodynamics (the entropy law or law of entropy) was formulated in the middle of the last century by Clausius and Thomson following Carnot’s earlier observation that, like the fall or flow of a stream that turns a mill wheel, it is the “fall” or flow of heat from higher to lower temperatures that motivates a steam engine. The key insight was that the world is inherently active, and that whenever an energy distribution is out of equilibrium a potential or thermodynamic “force” (the gradient of a potential) exists that the world acts spontaneously to dissipate or minimize. All real-world change or dynamics is seen to follow, or be motivated, by this law. So whereas the first law expresses that which remains the same, or is time-symmetric, in all real-world processes the second law expresses that which changes and motivates the change, the fundamental time-asymmetry, in all real-world process. Clausius coined the term “entropy” to refer to the dissipated potential and the second law, in its most general form, states that the world acts spontaneously to minimize potentials (or equivalently maximize entropy), and with this, active end-directedness or time-asymmetry was, for the first time, given a universal physical basis. The balance equation of the second law, expressed as S > 0, says that in all natural processes the entropy of the world always increases, and thus whereas with the first law there is no time, and the past, present, and future are indistinguishable, the second law, with its one-way flow, introduces the basis for telling the difference. [source]
That is why in a planet’s troposphere there is homogeneous molecular (PE+KE) when entropy is a maximum and so there are no unbalanced energy potentials. And that is why there is a temperature gradient formed by gravity and not by back radiation from water vapor, carbon dioxide etc.
Doug Cotton,
On Uranus. Have you actually calculated the energy released in a gravitational collapse? I have.
Uranus core is thought to be mostly iron. The core is 0.55 Earth mass or 3.284×10^24 kilograms. Heat capacity of iron is 450 joules/kg/K. The temperature of Uranus core is thought to be around 5000 Kelvin so it stores around 7.4×10^30 joules (give or take).
The derived calculation for energy from a cloud that collapses to a certain radius is 3/5XG(gravitational constant)xM^2(mass of the cloud in this case Uranus)/R(radius the planet is currently at).
If I did my math correctly
(6.67×10^-11 m^3/kgxsec^2)(86.81×10^24 kg)^2/(25,362,000 meters)=
1.189×10^34 joules.
So it is obvious that the energy from a gravitational collapse provides more than enough energy to explain Uranus’s 5000 Kelvin core and it cools only slowly because the thick atmosphere is an excellent insulator holding in most the heat.
Your theory greatly fails on Neptune and Jupiter. Both these planets are radiating away more energy than they receive from the sun. It would be impossible for your “heat creep” theory to explain that. Other processes are going on that you will not open up to to understand.
The reason all the planets have lapse rates based upon the gravitation potential is because lapse rates exist in areas of convection. Air cools as it rises because it expands and does work on its surroundings. The rate the air cools is based upon the density gradient which is determined by the gravitational field of the planet. No mystery. You just can’t understand why air cools as it expands. It is well established and experimental physics and it is why the vortex tube cools so drastically.
Norman
Have you read anything about Uranus and what makes it different from Neptune, let alone collapsing Jupiter? I also have explained that Uranus has a solid core 55% the size of Earth which, as with Earth and Venus, prevents collapsing such as happens with totally gaseous planets. In such collapsing gaseous planets it is obvious that more molecular gravitational potential energy is converted to kinetic energy than vice versa. Hence we observe net outward radiative flux, but there is no convincing evidence of significant net outward flux from Uranus, so it is neither generating internal energy nor cooling off. You could have read this on the group’s website.
Planets and moons are not still cooling off. Some may even have warmed to the existing temperatures. All exhibit close to the calculated temperature gradients based on -g/Cp and yet we find all are in radiative balance with the Sun at just the right altitude and just the right temperature. For example, the Uranus temperature plot goes from 5000K in the core down to the radiating temperature 59K at just the right altitude where there is a methane layer doing nearly all the absorbing and radiating in the stratosphere. Think about that. It is sheer coincidence – for all planets?
A location on the equator of Venus cools by about 5 degrees (from 737K to 732K) during the 4-month-long night. You might dispute the quantification, but it must cool by x degrees. There is no evidence of long term global warming or cooling in the period when measurements have been kept, so that location must warm back up by x degrees during the four months of sunlight, and thus the warming is due to the Sun. But it is not due to the Sun’s direct radiation to the surface. Instead it is due to downward convective heat transfers from the region where the Sun can do some warming because the temperatures are less than about 400K. You can confirm this with Stefan Boltzmann calculations.
If you had read the “Planetary Physics” group’s website you would realize that I do not claim that “heat creep” necessarily supplies all the required energy for the core, but it is not impossible for it to do so. If there is some internal energy generation, then heat creep makes up the extra that is required. Do you seriously imagine that the core of our Moon is kept at above 1300°C by some internal energy generation, especially when the surface can get as cold as -150°C?
My calculations for Neptune place its core temperature close to 7000°C and the difference between it and Uranus is correctly accounted for by the differences in gravity, height of atmosphere and weighted mean specific heat of the matter involved.
All your discussion about lapse rates being only due to air rising (supposedly in “parcels” that have nothing to hold them together) is thoroughly refuted in my websites, articles, papers, video and book. It is you who “just can’t understand” my hypothesis because you have not read it or studied the explanatory diagrams.
What a joke it is that you say I “just can’t understand” why air cools. It cools for one reason only – the mean molecular kinetic energy is reduced, whether that energy becomes gravitational potential energy or is dissipated by radiation or other heat transfers, phase changes or reactions. If we are considering an isolated system without wind or phase changes or chemical or nuclear reactions, then it cools because some of the kinetic energy is converted to gravitational potential energy. But there can be opposite conversions in downward convective heat transfers which are occurring because of the propensity for unbalanced energy potentials to be minimized, as the Law of Entropy (aka Second Law of Thermodynamics) says will happen.
If you understand the Law of Entropy, then you should be able to answer this question …
Suppose you construct a perfectly insulated thin sealed cylinder filled with pure nitrogen. The cylinder is 100 meters tall and subjected to Earth’s gravity at the base of the troposphere.
In that you agree that gravity forms a temperature gradient, then you will accept that there could be a state of thermodynamic equilibrium wherein the temperature at the top could be 0.98 degree cooler than that at the base of the cylinder. Let’s say it is 10.0°C at the base and 9.0°C at the top to the nearest tenth of a degree. Now suppose you initially installed electric heating units in the top quarter of the cylinder and so you are now able to raise the temperature of that top 25 meters to 9.5°C.
The question is: What happens when you turn off the heaters and wait until temperatures everywhere in the cylinder remain constant?
See also this comment.
Doug, I would encourage you to read up on Bernoulli’s principle (http://en.wikipedia.org/wiki/Bernoulli%27s_principle), which is pretty much what your are advocating (but nearly 400 years after he did).
“Bernoulli’s principle can be derived from the principle of conservation of energy. This states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy, potential energy and internal energy remains constant. (wikipedia)” In particular, in the limit where the bulk speed of the gas approaches zero, potential energy + internal energy = constant.
However, this principle assumes adiabatic conditions, where heat flow within the system is prohibited — ie true thermal equilibrium is not approached.
Adiabatic conditions do not prohibit heat transfers within the cylinder I am discussing when the additional thermal energy is added in the top quarter and then the heaters are turned off so that we once again have adiabatic conditions. I am not talking about “thermal equilibrium” – I am talking about thermodynamic equilibrium. In any event, you have not stated quantitatively what the temperatures would end up at when there is homogeneous (PE+KE) – and it has nothing to do with Bernoulli’s principle which you don’t need to “teach” to someone like myself whose first degree was in physics nearly 50 years ago.
Doug Cotton,
“cools for one reason only – the mean molecular kinetic energy is reduced, whether that energy becomes gravitational potential energy or is dissipated by radiation or other heat transfers, phase changes or reactions.”
I am certain you did not study physics in any known Universtiy or had a valid professor teach you. Energy is lost by a system when it does work.
Here is a simple explanation for you that even in your dense mental state you should be able to understand.
You have an perfectly elastic ball inside a container with fixed walls. The ball loses no energy in collision with this wall. The energy of the ball will not change within this system. Now let us have moving walls. Now the elastic ball meets with a wall and moves it outward (the wall has mass, moving mass a distance takes energy…did you learn this fact when you studied physics???). What happens to the energy of the ball now? The ball loses some energy because some of its internal energy went in pushing the wall out (moving the mass). The ball will have less energy. It cooled! Do you understand? It is a very well established experimentally proven physics. Expanding air in air will cool (in a vacuum it will not, it expends no energy when it expands then). You can test this many ways over and over with the same result. We have air pumps at my place of work. The use compressed air to move internal parts of the pump. The exhaust air expands and cools out of the vent. Take any compressed air source and blow it, it cools as it expands. Likewise with the box analogy. If an external energy moves the walls in and the elastic ball hits the wall it will pick up this energy of the wall moving in and will heat up. You need to understand this most simple and basic of physics before you conjecture and posutlate further. Until you get this your theory will sound astonishingly stupid! Understand please.
I studied physics for four years at Sydney University under Profs Harry Messel, Wernher von Braun and Julius Sumner Miller, all of whom you can read about in Wikipedia if you’re too young to remember the “Why is it so?” TV series.
I am not only quite aware of the Kinetic Theory of Gases (as used by Einstein and others) and I make use of it in the development of the hypothesis.
Get back to me when you have read the hypothesis and wish to ask any questions relating to such.
Define the boundaries of your expanding “parcel” of air when there are no walls to contain or restrict the random motion of molecules between collisions moving at about 1,800 kilometers per hour.
Doug
You have no clue what a mean free path means so do not state you understand how gases work. The wall is the surrounding air. An air molecule can only go around 93 nanometers (ground level increases as density goes down) before it hits another molecule and changes direction and energy. Do you know what a surface does? When an object hits a surface it changes direction and energy (if it interacts with that surface). The number of molecules in the atmosphere create the surfaces which restrict the motion of the molecules. You can see an air parcel every time you look up and see a cloud. The cloud forms when a moist air parcel rises, then cools from expanding and the moisture condenses into visible form. If you do not have turbulent mixing the parcel will exist for a period of time until conduction can equalize the temperature difference.
Doug do you know what restricts an explosion? The air itself exerts around 10000 newtons of force/meter^2 or 14.7 pounds/inch^2.
Doug read this:
http://fas.org/irp/imint/docs/rst/Sect14/Sect14_1b.html
It will help you understand what is going on and how these things work.
The motion of molecules is only “restricted” by collisions with other molecules. There is nothing else that can touch them, and certainly no imaginary wall. The fact that a molecule is moving away from some point proves nothing about the overall density in the region, or whether expansion or compression is happening at a macro level. The molecule’s direction may well be reversed at the next collision, so that it moves back where it came from.
(1) When molecules collide they share kinetic energy (not necessarily equally in each individual collision) though tending that way because the difference in momentum before the collision will always be greater than the difference in momentum after the collision. The leveling of energy (with an increase in entropy) is the irreversible process that leads towards thermodynamic equilibrium. The process is one of maximum entropy production, there being a maximum, which is thermodynamic equilibrium by definition.
(2) When molecules move between collisions their momentum is affected by gravity and thus can and will increase or decrease in downward or upward motion respectively. Momentum is not necessarily conserved in an isolated system that experiences an external force field. Momentum is only conserved in a closed system, of which the whole Solar System is a good approximation.
And regarding your linked article, air masses “form over extensive surfaces with light winds”
When will you get it into your head that I am talking about an “ideal” troposphere in calm conditions (no wind of any form) and no variation in composition, such as between dry air and clouds or rain or whatever. There can be no air parcels holding together in such instances without wind of any form. If there were hotter or colder air at the side, the temperatures would just level out in a horizontal plane. Likewise any variation in density would be leveled out horizontally. We are assuming all this type of leveling out has been done already in every horizontal plane. Otherwise we are just talking about weather and how it forms and changes. But we are not.
The Ideal Gas Law (which is derived from the same Kinetic Theory of Gases that I use anyway) tells us that pressure is proportional to the product of temperature and density. It tells us nothing more and nothing less. If pressure and density both reduce by, say, 10% then temperature remains the same. Pumping cold air into a car tire that has mostly somewhat hotter air inside it, may lead to a lower mean temperature even though there has been net compression.
“What happens to the energy of the ball now? The ball loses some energy because some of its internal energy went in pushing the wall out (moving the mass).”
All air comprises molecules. There are no walls, just other molecules to collide with.
When any molecule moves upwards in an isolated system it is doing work against gravity, and so it loses some kinetic energy, this having a cooling effect on a macro scale.
But when a molecule collides with another molecule it may lose KE if that molecule had less than it did, or it may gain KE if that molecule had more KE than it did.
From Kinetic Theory we understand that pressure is proportional to the product of temperature and density. It is the net effect of molecules striking a fixed wall. The more molecules there are the greater the pressure. The more mean KE there is the greater the pressure. Remove the wall and what then happens depends on the pressure, density and temperature of the air on the other side.
Doug Cotton,
Please explain where the 16000 watts/meter^2 energy emitted by the surface of Venus goes with your theory that carbon dioxide actually will cause slight cooling. Where does all this energy go? It is not showing up at the top of the atmopshere to be measured but it is being emitted nontheless.
You only concern yourself with how the surface warms but have zero explanation of where the energy goes. It does not just need to receive 16000 watts/meter^2 to maintain this surface temp, you have to explain where does the emitted radiation go?
The answer to your question pertaining to radiation may be deduced from my peer-reviewed paper “Radiated Energy and the Second Law of Thermodynamics” linked from the second page of the website endorsed by our group of persons suitably qualified in physics and visited by over 6,400 since January 8th.
Your problem is that you confuse the flux of (one-way) radiated electromagnetic energy with the actual transfer of thermal energy which is only ever from warmer sources of spontaneous radiation to cooler targets. And, before you ask “How does the target know the temperature of the source?” the answer is in the above paper.
Radiation into the surface is not the primary determinant of planetary surface temperatures, any more than it determines the temperature of that metal ball bearing in near-boiling water that is heated and cooled by 5 degrees.
The whole Venus+atmosphere system acts like a black body, not the surface which is only an internal boundary. The radiative flux that actually gets out to Space (matching the insolation) will in fact transfer almost the equivalent thermal energy to Space because the background radiation in space is equivalent to a temperature colder than 3K. All the rest of the radiation “bounces around” between (mostly) carbon dioxide molecules and the surface and this has a temperature leveling effect, reducing the magnitude of the temperature gradient from |g/Cp| to about 75% to 80% of that value. This causes the whole temperature plot to rotate downwards at the surface end (just as water vapor on Earth reduces the magnitude of the lapse rate) and so the carbon dioxide in the Venus atmosphere causes the surface temperature to be cooler than it would be with pure helium and hydrogen as in Uranus where the gradient is about 95% of |g/Cp| – all of which you could have read you know where – so stop wasting my time.
I’ll correct that and say “than it would be with a fairly dry air mixture (nitrogen and oxygen) for which Cp = ~1.0.”
Doug,
If Venus atmosphere were composed of helium and the surface was hot enough to emit 16000 watts/meter^2, nothing would stop the radiation from moving directly from surface to space and it would cool at a rapid rate. I think your explanation is a greenhouse effect.
” All the rest of the radiation “bounces around” between (mostly) carbon dioxide molecules and the surface and this has a temperature leveling effect”
It sounds like the carbon dioxide is acting to prevent the 16000 watt/meter^2 from leaving the planet.
Hi Norman,
You state:
“If Venus atmosphere were composed of helium and the surface was hot enough to emit 16000 watts/meter^2, nothing would stop the radiation from moving directly from surface to space and it would cool at a rapid rate. I think your explanation is a greenhouse effect.”
Like other diatomic gas compounds helium should not absorb/emit in the infrared, but in the much slower longer wavelength microwave region. So in fact helium would likely take much longer to radiate it’s energy to space and cool than carbon dioxide. Your statement appears false.
Have a great day!
Regarding the often-mentioned 16,000W/m^2 of radiative flux from the Venus surface, see this comment.
The Venus surface at night is not unlike the Earth’s surface in calm conditions in the early pre-dawn hours. When the excess “heat of the day” has nearly all dissipated on Earth, the rate of cooling becomes much slower, even though the environmental lapse rate is still present above that surface. Likewise on Venus there is a small net outward flux of thermal energy at night by both radiation and sensible heat transfers, that is, non-radiative processes. This could be quantified from the measurements which indicate there is about 5 degrees of cooling over the course of about four months for any location on the equator as it moves across the dark side.
But the opposite happens in the four-month Venus day and there is net input of thermal energy (equal to the loss the previous night) but none of that comes from the Sun’s direct radiation and none from radiation from less-hot regions in the troposphere. Virtually all of it thus comes from the non-radiative processes (conduction, diffusion, advection, convective heat transfer) which are restoring thermodynamic equilibrium with its associated temperature gradient. I have coined the term “heat creep” for such processes as they are slow and involve heat transfers from cooler to warmer regions (up the temperature gradient) subject to the condition that entropy is increasing. To understand “heat creep” you probably need to refer to the diagrams at the foot of this page.
Now, the main difference on Earth is that, unlike on Venus, on clear days (and mostly in non-polar regions) the Sun’s radiation can be strong enough to raise the Earth’s surface temperature for solid regions (though not much for oceans) and this temporary “heat of the day” is dissipated relatively quickly in the late afternoon and evening. You perhaps need to think about why the rate of cooling slows when the temperature gets down to that which is supported by the gravitationally induced temperature gradient and the resulting “heat creep” process. At that temperature the whole troposphere then has to cool by 1 degree for each degree of surface cooling, so a lot more energy has to get to Space. Hence the rate of cooling slows dramatically in those early pre-dawn hours as is well known.
However, for the whole Earth surface, the mean Solar radiation is estimated at about 168W/m^2 and most of that merely supplies non-radiative heat losses at the rate of about 102W/m^2 out of the surface. So the Sun’s direct radiation does not have a hope of explaining the mean surface temperature which would need an input of about 500W/m^2. James Hansen must have realized this, and so he added 324W/m^2 of back radiation to the 168W/m^2 of solar radiation absorbed by the surface. But unfortunately he was wrong in assuming that back radiation could help the Sun in this way. It most certainly does not make the Sun about three times as effective in regard to thermal energy that is actually added to the surface. Even Roy knows that back radiation does not penetrate water surfaces by more than a few nanometers, so how could it raise the temperature of the oceans? The only possible mechanism within the laws of physics is heat creep supplying the necessary thermal energy on Earth, just as it does on Venus – and Uranus, etc.
How is the UAH Global anomaly value calculated? it is definitely not the average of the NH, SH & Tropics values.
2015 02 +0.296 +0.434 +0.157 +0.015
Silent readers will note that Norman and Tim have avoided answering the three crucial questions …
(1) What happens in the 100m high cylinder when the heaters are turned off?
(2) How does the Venus surface actually obtain new additional thermal energy in order to rise in temperature by 5 degrees during its daytime?
(3) How do they explain the apparent huge coincidence that all planets exhibit the “right” temperature gradient (based on -g/Cp) from the core all the way to just the right altitude where there is radiative balance with insolation?
Can any of you silent readers who have not yet read the website answer these questions?
Can’ resist. Too much time on my hands. Silent readers rise up!
“(1) What happens in the 100m high cylinder when the heaters are turned off?”
No more energy is added or subtracted, integral variation Qp = integral m*Cp*dT = 0 forever since isolated.
Write an eqn. for this column’s entropy, maximize that entropy eqn. via calculus, find T(z) as Poisson did in the 1890s. This procedure is in the literature since 1998 and extended 2004, 2006. Of all linear potential temperature profiles, a constant potential temperature (Poisson eqn.) maximizes the entropy of an isolated layer of the atmosphere in hydrostatic equilibrium.
Why isn’t the equilibrium profile isothermal? Intuition from solids impedes the understanding of atmosphere convection. If a solid is isolated from its environment, and initially has non-uniform temperature, conduction eliminates all temperature gradients. However in an atmosphere, energy transfer in an isolated troposphere layer is dominated by convection not conduction (Lord Kelvin 1862). At entropy maximization, calculus requires the equilibrium temperature of an isolated atmospheric layer subjected to mixing with no condensation or evaporation of water to decrease with height at the dry adiabatic rate.
“(2) How does the Venus surface actually obtain new additional thermal energy in order to rise in temperature by 5 degrees during its daytime?”
The sun comes up on an enormous optical depth atmosphere. Actually no one really knows “by 5 degrees”.
“(3) How do they explain the apparent huge coincidence that all planets exhibit the “right” temperature gradient (based on -g/Cp) from the core all the way to just the right altitude where there is radiative balance with insolation?”
The derivation for -g/Cp (and Poisson eqn. potential temperature) applies the same everywhere in the universe there is atmosphere convection hydrostatic equilibrium.
You Ball4 need to catch up on recent advances in the understanding of thermodynamics, as explained here: http://www.entropylaw.com/ and the linked pages.
Maximum entropy production requires minimized energy potentials. So, rather than integrating with a wrong function anyway, you need to understand that the removal of all unbalanced energy potentials only happens when molecular (PE+KE) is homogeneous.
Hence, considering the exchange of PE and KE in molecular flight between collisions, we get, quite simply …
Your m*Cp*dT (representing the change in KE between two layers with height difference dH)
which we equate with the negative of the change in PE: m*g*dH and in one line derive the temperature gradient …
dT/dH = -g/Cp
Elementary my dear Ball4. You could have read it in the paper, the website or the book. It has nothing to do with fictitious rising, expanding, cooling “parcels” of air which have nothing to hold them together in the first place.
Doug 11:02pm: “Maximum entropy production requires minimized energy potentials.”
That’s actually correct Doug. Maxwell’s original column of gas max. entropy point as the entropy increases following the basic formula for your isolated gas column does occur when the energy potential in the column is minimized for which calculus is very useful tool. At that max. entropy point energy ceases to flow. This was worked out for Maxwell’s original arbitrary column of gas as of 1998 and extended in 2004, 2006 in the literature.
To formalize being correct, you will want to read up on those studies as silent readers have already done as well as Hamilton’s principle (well known law of least i.e.stationary action) for energy potential minimization as extended from D’Alembert, Lagrange, Laplace to learn more (don’t use wiki).
http://www.emis.de/classics/Hamilton/GenMeth.pdf
You have disregarded the molecular gravitational potential energy that is unbalanced when a sealed insulated cylinder is rotated from horizontal to vertical. If you were right about the vertical state still having maximum entropy then nothing would happen and no density gradient would form. So you are wrong.
Doug Cotton,
Heat of Earth. Doug you try to explain the Earth’s inner temperature caused by heat creep. You claim you have a physics background. Have you ever calculated the possibility that radioactive decay might be heating the Earth interior? Is it possible? Lets see.
http://www.ucl.ac.uk/EarthSci/people/lidunka/GEOL2014/Geophysics8%20-%20Thermal%20evolution/Heat.htm
The energy flux from Earth out is found to be around 0.08 watts/meter^2. The total surface area of Earth is 5.1×10^14 meter^2. So to replenish the energy lost you would need (0.08 watts/m^2)(5.1×10^14 m^2)=4.08×10^13 watts or 4.08×10^13 joules of energy/second must be generated to replace this outward flux.
Radioactive decay of Uranium 238 would provide the bulk of energy needed with the rest supplied by other radioactive materials in the Earth.
From this article:
http://arxiv.org/pdf/hep-ph/0501111.pdf
The total amount of Uranium in the Earth is estimated at around 10^17 kilograms
Using radioactive decay equations:
http://www.mpoweruk.com/nuclear_theory.htm
For Uranium 238 the decay rate for a gram/second is
(6.022×10^23)(0.693)/(1.41×10^17 seconds)(238 grams/mole)=12436 decays/second
Energy per gram =(12346 decays/second)(6.836×10^-13 joules/decay)=8.44×10-9 joules/gram/second for Uranium 238
That would make 8.44×10-6 joules/kilogram/second
Now multiply by the expected mass of uranium in the Earth at 10^17 kilograms you get 8.44×10^11 joules/second released by uranium 238 decay. Uranium 238 decay would be in the ballpark for energy needed to maintain the flux from the core.
You can read about the contribution of radioactive decay, fission etc in the paper, Norman.
Now answer my three questions as Ball4 tried to, but failed.
The flux from Earth’s surface, where the temperature is 288K, is about 390W/m^2 of radiation plus about 102W/m^2 of non-radiative heat transfers – nearly 500W/m^2 in total. You have no explanation as to what supplies most of the input required to match that output, and your 0.08W/m^2 from the core doesn’t help much.
In reality, the Sun’s direct radiation supplies about 168W/m^2 and the heat creep process supplies over 320W/m^2 some of which originally came from the surface, but most from where solar radiation was absorbed in the atmosphere.
Further info about radioactive decay: “Marone says, the vast majority of the heat in Earth’s interior—up to 90 percent—is fueled by the decaying of radioactive isotopes like Potassium 40, Uranium 238, 235, and Thorium 232 contained within the mantle. These isotopes radiate heat as they shed excess energy and move toward stability. “The amount of heat caused by this radiation is almost the same as the total heat measured emanating from the Earth.”
Read more at: http://phys.org/news62952904.html#jCp
However, firstly this is happening in the less hot mantle, not the core. Secondly, although it may appear to match net outward energy flow, the fact remains that a mere 0.08W/m^2 would be nowhere near enough to raise the core temperature to what it is, especially when the energy is coming from the less hot mantle.
In any event, the temperature gradient in the outer crust cannot be used to determine that there is necessarily an outward energy flow of the amount calculated, for the reasons in my paper.
It’s really not worth it, but briefly …
1) The cylinder would be isothermal to start with. If you briefly heat part of the gas and then turn off the heater, the gas will eventually return to an isothermal condition, just at a higher temperature.
3) Earth has about 4x the diameter and 6x the gravity. They ahve similar composition, so similar Cp. Thus the core temperature should be ~ 24x higher than the moon if -g/Cp governed the gradient. The earth’s core is several times warmer, but not anything close to 24 times higher.
Furthermore, if -g/Cp applied, then the bottom of the ocean would be much warmer than the surface, but the opposite is true.
2) Don’t have to for Venus now. And Doug wont listen anyway. 🙂
Regarding Tim Folkerts answers:
(1) Wrong. The cylinder would not be isothermal to start with, because I specified the initial temperatures. If you think isothermal conditions would then evolve you need to explain why there would be maximum entropy in such a state. You’ll be wrong anyway, as I may as well tell you in advance, because (PE+KE) must be homogeneous for there to be no unbalanced energy potentials.
(2) Elected not to answer
(3) Wrong because you did not realize that specific heat values rise to very high levels in very hot regions like Earth’s mantle where the temperature gradient is thus only about 1C°/Km. Also of course the net effect of gravity diminishes as you approach the center. Perhaps you might like to discuss just the nominal troposphere of Uranus and explain how it “knows” to be just the right temperature (320K) at the base thereof so that the temperature comes down to 59K in the methane layer in the stratosphere. Then you could discuss why the Venus surface temperature is so much higher than Earth’s surface temperature and try again to answer (2).
Footnote: Your argument, Tim, about the oceans was refuted in Section 14 of my February 2013 paper linked from the website. You fail to understand the role of intermolecular radiation that is explained in the paper, the website and the book, and which is far more prevalent in liquid water than in an atmosphere with perhaps less than 2% water vapor. In a nutshell, you have not read the paper and so you fumble around making such time-wasting points which are already refuted.
Doug Cotton,
Unless you provide valid and unbiased testing of #1 question about the state of cylinder. I believe it would be isothermal and I explained in detail why I believe this and all planetary atmospheres have the isothermal condition where there is no convection or energy absorption (why would that exist). You ignore these conditions but they exist. I state that the mean free path creates multiple surfaces in the cylinder of gas so the PE and KE only exchange in a small region. You say Tim Folkerts is wrong but you provide no evidence, we are just blindly to believe you because you are such an expert? This is science, provide the proof. You are the one rejecting established science, the burden of proof is only on you and no one else. Your vortex tube example is explained by expansion of gas leading to cooling (a proven fact which you can prove yourself at any time but choose not to for unknown reasons).
I think you are wrong about the surface of Venus changing. Everything I have read says it stays the same. Your source did not refer to the surface but up higher in the atmosphere I believe. I clicked on your link and read it. I am going by memory and may be wrong but no other sources makes such a claim. Where did they arrive at this information?
How could a particular location on the equator of Venus possibly not cool during 4 months of darkness? If it cools then it must warm again. The same happens on all planets with surfaces. Have you noticed Earth’s surface gets cooler at night? It can also warm the next morning even where thick cloud cover shades it from solar radiation. Think on that, because that’s heat creep in operation.
In any event, you have not explained what maintains the Venus surface temperature, whereas I have.
I don’t ignore anything. It is you who ignores the detailed explanation in my 18 page paper.
I provide ample evidence with a whole page of the website summarizing such.
I have provided details of an experiment with centrifugal force, plus a link to details of about 800 other experiments and a host of supporting evidence from other planets.
The vortex tube demonstrates the effect of centrifugal force establishing a temperature gradient in exactly the same way that gravity does. I don’t care how you think that happens. If you ignore the Ideal Gas Law and make up your own climatology “law” which the accepted Ideal Gas Law does not confirm, like temperature depending on density, good luck to you. All that the real law says is that pressure is proportional to the product of density and temperature.
Everything pertaining to sensible heat transfers happens originally at the molecular level through molecular collisions. You seem to be incapable of thinking like Einstein and Loschmidt at the molecular level, applying the well recognized and accepted Kinetic Theory of Gases.
I don’t care if you don’t read and understand my explanation of the physics – over 6,470 others have visited the Home page, over 1,780 the “Evidence” page and over 120 the German page of the website all just since January 8th. Over 100,000 have visited my first two climate websites in about four years. You are a drop in the bucket. Keep your personal beliefs if you wish, but stop killing people with your promulgation of the hoax.
In any event, you have never yet refuted the development of the hypothesis proven from the Second Law in the paper, so I rest my case thereon.
“You are the one rejecting established science”
I am not.
I have used established science. You are just brainwashed with the climatology garbage fissics. For example, the equations for thermodynamic potentials are developed with the very specific assumption that they are to be applied only where gravitational potential energy does not vary. When a stone falls, entropy increases. Do you get the significance of that statement? It is effectively denied by climatologists. Hence they use equations like Ball4 which have no term for gravitational potential energy, and so they “prove” there will be isothermal conditions because they started their “proof” with the assumption that gravity had no effect on entropy. That’s a circular argument if ever there was one, and yet it seem you don’t have the understanding of physics (such as I have gained over 50 years) to spot such errors in such arguments.
The Clausius statement only applies in a horizontal plane. It does not say there can be no sensible heat transfers from cooler to warmer regions in a gravitational field because it is derived from entropy equations which ignore gravity. I have proved beyond a shadow of a doubt that there are.
Of the thousands who have visited the website not a single person has written to the email address therein with any refutation or attempt at such. Not one person has quoted some statement in the website or paper and then claimed it is wrong because …. The only comments have been ones in general agreement and support, and in this way our group will grow. If there are any readers here who think they can write a refutation, I will guarantee to reply and show them where they are mistaken. If I cannot do so, I will publish their point and acknowledge they are right, and they can watch for my admission right here on this thread, and repeat their refutation here if they wish.
Now, read this comment below which was written before your last comment.
Doug 6:44am: “If there are any readers here who think they can write a refutation, I will guarantee to reply and show them where they are mistaken.”
Your work is shown faulty (pre-refuted) beyond reasonable doubt by the prior work accomplished published in 1998, 2004 and 2006. It is easy task to find where your work is faulty just by comparing it to the existing scientific method physics as I have pointed out in the past (this is formally called a literature search/review). Make your guarantee worth something, compare your work to the already published solutions on your own and show exactly where you are mistaken. I’ve already done so, you need to do it yourself.
And neither have you Norman thought about or answered Question (3) in this comment.
No Ball4, I have correctly refuted the false climatology physics which assumes isothermal conditions would exist in the absence of IR-active gases, contrary to the Second Law. I rely upon the Second Law of Thermodynamics to support what I say. They don’t. I also use that law to refute the claim that back radiation delivers more than twice the thermal energy to the surface than does the Sun, especially since back radiation does not even penetrate water by more than a few nanometers. I also dispute the implied claim that there is at least 15 degrees of surface warming for each 1% of water vapor above any particular region. I have supported that with a comprehensive study of real world data, proving that increasing water vapor leads to lower surface temperatures, as my hypothesis also shows to be the expected outcome.
You, like Tim and Norman, have not thought about (nor correctly answered) Question (3) here and until you, Tim and Norman do so, I rest my case on what is written in my two peer-reviewed papers about the Second Law of Thermodynamics which (as explained in other comments) is all about minimizing unbalanced energy potentials. The papers are linked at the foot of this page.
“No Ball4, I have correctly refuted the false climatology physics which assumes isothermal conditions would exist in the absence of IR-active gases, contrary to the Second Law.”
The published work in 1998, 2004, 2006 thoroughly refutes this prose statement Doug. Your prose cannot trump the physical laws and proper calculus. Every professional properly starts with a literature search for a thorough understanding of the state of the art. It is easy to show you fail to adequately perform this required step as, for example, I did previously with your incorrectly calculated “study of real world data.”
Your assertive prose says nothing, Ball4. Your cited “science” does not use a correct application of the Second Law to “prove” isothermal conditions would evolve in the plane of a force field, and they don’t evolve in a vortex tube or in the troposphere of Uranus, and that’s why you can’t answer question (3) as I predicted.
How could the Second Law of Thermodynamics be wrong?
The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations — then so much the worse for Maxwell’s equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.
—Sir Arthur Stanley Eddington, The Nature of the Physical World (1927)
Doug 8:05pm – That won’t work. What will work is you showing where exactly the existing published references do “not use a correct application of the Second Law” that Sir Eddington writes about. Absent that you have nothing but bluster.
Understanding the vortex tube is based on 1st and 2nd law same as the ref.s already in the literature, so the vortex tube supports the published accurate conclusions not your inaccurate conclusions.
Ball4: Read the experiment in which centrifugal force in a centrifuge machine creates a temperature gradient, just as gravity does in a planet’s troposphere. It’s summarized at http://climate-change-theory.com with a link as well.
Then help Curt answer this question because he’s stuck. See the comment above that one too.
Doug 2:57am: “..centrifugal force in a centrifuge machine creates a temperature gradient, just as gravity does in a planet’s troposphere.”
Thanks for keeping me out of bars at night Doug, you are singlehandedly cutting into my Fri. night bar bill so make this fun. I have read your link(s). Gravity is NOT a temporary gradient in a planet’s atmosphere. Gravity nowhere goes to zero in a planetary atmosphere like at the center of a standard centrifuge that Curt has used to sufficiently prove your theory of “heat creep” bogus: the scrap of paper would flame if sufficient oxygen available, the super centrifuge would melt under your “heat creep” theory. Neither are observed in nature just as your troposphere “heat creep” is science fiction 8:42pm when you incorrectly write: “It is gravity which “traps” thermal energy, not back radiation.” That’s bogus.
“What happens and why?”
Curt is not stuck, you are. The answer to your question was shown in the 1890s by Poisson’s eqn. which permits an understanding of the changes in temperature as air mixes through the isolated atmosphere column (your 100m isolated cylinder say). Your rod initial condition made hot at one end say (say 212F from boiling water at 1atm. & room temperature 70F at the other) will end up with this temperature distribution at thermodynamic equilibrium considering column PE + enthalpy conserved – note profile T(p) is non-isothermal as I made you aware 3/10 8:50am:
with To temperature at cylinder base z, p0 the pressure at cylinder base z:
T(p) = To * (p(z)/p0)^R/Cp
This result was shown to be the isolated column ideal T profile at max. entropy in 1998, the work was extended in 2004 and 2006. Have you finished your literature search? I thought not. That’s what happens; why it happens is as you write there are no further energy potentials to tap in the isolated column, Hamilton’s principle of least action has been applied (“the Hamiltonian”), 2LOT column entropy is at its maximum value with this T profile in your iron bar forever:
T(p) = To * (p(z)/p0)^R/Cp
Ball4 writes: “Gravity is NOT a temporary gradient in a planet’s atmosphere.”
I can hardly imagine a more totally incorrect understanding of the hypothesis that is in my website and linked paper on this.
The temperature gradient which is induced by a force field has nothing to do with there being any non-zero gradient in that force field itself. But, if there were no such development of a temperature gradient, then they could not produce gas at a very cold 1K (-272°C) by centrifugal force as in this experiment, and gas at around -50°C in a vortex cooling tube. All of which proves I’m right.
There is a wealth of evidence, as well as correct physics that proves that a density gradient and a temperature gradient represent the state of maximum entropy in a force field which, incidentally, can be a homogeneous or near-homogeneous force field.
Air does not need to “mix” due to rising advection. A planet like Uranus has no surface and no solar radiation at the base of its nominal troposphere where it’s hotter than Earth’s surface.
There is nothing that holds a “parcel” of air together in an ideal calm troposphere because molecules move randomly in all directions.
Only wind can keep a parcel of molecules reasonably contained and all moving in a given direction. But that then is not an adiabatic situation, and no significant warming or cooling takes place due to net downward or upward movement in such wind of any form.
For example, warm winds from a valley can be redirected up a mountain slope and will not cool as much as the “lapse rate” would indicate. Likewise, above the South Pole, winds blow air at -50°C down to the surface with no significant warming. That is why the surface is also -50°C. It gets its temperature from the upper troposphere where the winds converge from all longitudes and have to go downwards.
See “The properties of iron under core conditions from first
principles calculations” for results showing the specific heat of iron decreases rather than increases at high temperatures and pressures. Do you have reference to support your assertion?
http://www.homepages.ucl.ac.uk/~ucfbdxa/pubblicazioni/pepi03_2.pdf
It appears as though ‘total constant-volume specific heat per atom’ (Cv) for iron increases much more with rising temperature than it decreases with rising pressure … (Figure 7, p.113).
You have not thought about (let alone answered) Question (3) above.
Regarding Ball4’s answers
(1) Qp = integral m*Cp*dT = 0 forever since isolated.
Wrong because you omitted the necessary term for gravitational potential energy. You have not answered the question because it refers to a sealed insulated cylinder and so your reference to hydrostatic equilibrium is irrelevant. Try again and quantify the final temperature gradient, if any, and confirm whether or not the surface temperature would rise above 10°C. This will confirm (or otherwise) your understanding of your computations, which are not necessary anyway, as you would realize if you read the paper linked to our group’s website.
(2) I don’t care if you call it x degrees of warming. It has to warm because it has to cool at night. You have not answered the question. What form of heat transfer supplies the necessary additional thermal energy to raise the surface temperature? Whether the surface warmed or not, there would still need to be an input of thermal energy to maintain the temperature around 735K. That does not come from radiation absorbed by the surface.
(3) You completely miss the point of the question. Try reading it again. Let me put it another way: How does the core “know” what temperature it should be in order for the temperature gradient to be just right and the temperature itself just right as it is, for example, at 59K in the methane layer in the stratosphere of Uranus.
Your score 0/3.
Doug 10:20pm: “Wrong because you omitted the necessary term for gravitational potential energy.”
1) Not wrong Doug, what I wrote is the 1st law! I didn’t omit the term for gravitational potential energy Doug, PE+KE are in Q internal energy. You did omit part of the formula I wrote so your work is again faulty, apply constant volume in this case so 1st law is in form for your isolated column w/no heater:
integral variation Qv = integral m*Cv*dT = 0 forever since isolated.
“..your reference to hydrostatic equilibrium is irrelevant.”
Doug’s isolated gas column is in hydrostatic equilibrium so very relevant, very useful, see the prior work Doug.
“..quantify the final temperature gradient..”
As I wrote, of all possible T gradients, the one T gradient in Doug’s isolated column at max. entropy was worked out by Poisson non-isothermal in the 1890’s with To temperature at base z, p0 the pressure at base z:
T(p) = To * (p(z)/p0)^R/Cp
——
“2) What form of heat transfer supplies the necessary additional thermal energy to raise the surface temperature?”
Again, the sun comes up at dawn on Venus. Does Doug dispute that? Radiative, convective, conductive energy transfer are at work in the Venus extreme atmosphere optical depth.
——
“3) How does the core “know” what temperature..”
The core doesn’t “know” anything, the core is inanimate.
I’m glad I got 0/3 from Doug because if Doug ever gave me a score of 3/3 I would know my understanding is faulty for sure in each case. That would force me to have to go back and compare my answers to the correct existing literature and discover correct basic science which Doug has not yet done. After successfully doing that, Doug will learn 0/3 is correct for the answers given by Doug.
Ball4 says, March 10, 2015 at 8:50 AM:
“(…) what I wrote is the 1st law! I didn’t omit the term for gravitational potential energy Doug, PE+KE are in Q internal energy.”
For your information, Ball4, Q does not represent the ‘internal energy’ of the system. Q represents the ‘heat’ added to the system, as opposed to ‘work’ [W] done on it. Qv at constant volume (isometric), Qp at constant pressure (isobaric). ‘Internal energy’ is represented by the letter U, and is a very different entity.
I know using the term ‘heat’ for you is like uttering the true name of the Devil Himself, but you will need to come up with something better for Q than ‘internal energy’, I’m afraid.
Other than that, I agree with what you say.
Kristian! – ”Q represents the ‘heat’ added to the system”
There is no heat added to the system. Doug’s column heater was removed (“the heaters are turned off”) before asking his questions and column is constant volume, no mass input/output. For that case the variation in work on the contents is 0 so its resulting infinitesimal change in temperature is 0. Thus dU = variation in Qv. So to expand my 1st law equation for your consideration for Doug’s column:
delta U = integral variation Qv + 0 = integral m*Cv*dT = 0 forever since isolated, heater removed.
Because of the gravitational field the atmosphere possesses PE (per unit area assuming Doug’s cylinder is unit area). So Doug is correct to take into account PE which the previously correctly published formulations show explicitly by taking into account all forms of energy the layer can have, the sum of which must be constant.
PE = integral Zo to Ztop (density(z)*g*z) dz per unit area.
Note similarity to mgh.
Thus here Qv is the internal energy consisting of the total enthalpy + PE. For each infinitesimal layer of Doug’s 100m high column there is a variation in pressure, so the eqn.s already in the literature I’ve been mentioning for Doug take into account hydrostatic equilibrium (allowing for change from height to pressure in the integral) which is also beyond Doug but he doesn’t seem to have even absorbed that yet.
“Kristian! – ”Q represents the ‘heat’ added to the system”
There is no heat added to the system.”
Seriously!?
Whether heat [Q] is added to the system or not is immaterial to whether or not the letter Q in this particular case represents ‘heat added’ to the system. Q = 0. There. No heat added. Q is still ‘heat added’, Ball4.
Likewise, when there is no work [W] done on or by a system, δQ = dU. (In an adiabatic process, Q=0, and then work [W] (PdV) = dU.) That doesn’t mean Q (or W) is U. Q is ‘heat added to the system’, Ball4. And U is the ‘internal energy’ of that same system. Two different things. Always. Live with it.
Kristian: Where’s the physics beef exactly? There is none. Only semantic.
“..when there is no work [W] done on or by a system, δQ = dU.”
Ball 4: Since in this case W=0: Thus dU = variation in Qv.
No difference except I added a subscript for this constant volume process.
——
“That doesn’t mean Q (or W) is U.”
Q energy was added to Doug’s system in the past when the heaters were on, increasing the column internal energy U (i.e. delta U = integral variation in Qv).
Now, at the commencement of Doug’s 3 question challenge that Q energy has become part of internal energy U consisting of column enthalpy + PE. Afterwards there is 0 change in internal energy (delta U = 0) because 0 Q are added (integral variation in Qv =0) in constant volume process. Hence 1st law:
delta U = integral variation Qv + 0 = integral m*Cv*dT = 0 forever since isolated, heater removed.
There is no physics issue, only one of semantics wherein Kristian insists on using the unnecessary term heat which can end up confusing the physics discussion if not used synonymous with energy.
Simpler is better. Q is energy. W is energy. U is energy. No confusion.
Firstly, in physics, “heat is energy in transfer other than as work or by transfer of matter.”
You should use the term “thermal energy” which is not the only component of internal energy that varies in this scenario, because molecular gravitational potential energy varies as sure as a density gradient forms.
You cannot ignore the change in molecular gravitational energy in your computations maximizing entropy. When you rotate a sealed cylinder from horizontal to vertical some molecular PE is converted to KE until maximum entropy is attained and stable non-zero density and temperature gradients evolve.
An isothermal state does not have maximum entropy in this cylinder because, as is blatantly obvious, there are unbalanced energy potentials. The Second Law cannot be ignored. It is all about the propensity to minimize (unbalanced) energy potentials. Do you agree on that point? It is clearly explained at http://entropylaw.com
Doug 9:41pm: The change in molecular gravitational energy is not ignored in the computations maximizing entropy. PE is expressly contained in the existing literature since at least 1998. You would know that if you could do the correct calculus.
So Ball4, where is S for entropy in your calculations? How can you work out what the Second Law will tell you if you ignore entropy, which is what that law is all about? It’s a common oversight in climatology I find. Just ignore the Second Law – she’ll be right mate.
And, by the way, except for the application for steadily flowing fluids, the term “hydrostatic equilibrium” need not be used for the static situation wherein upward pressure balances the downward force of gravity. That state already has a term, namely thermodynamic equilibrium, because it is the one and only state of maximum entropy, and it has associated with it both a density gradient and a temperature gradient, even in an insulated sealed cylinder.
The term for molecular PE is not used in the entropy computations which are used to prove the Clausius “hot to cold” statement, so that applies only in a horizontal plane. Obviously climatologists apply it (incorrectly) in a vertical plane and deduced incorrectly that isothermal conditions would apply. It’s all there in Pierrehumbert’s “gold standard” climatology textbook.
Next time link me to whatever papers you rely on and I will pinpoint their errors in their application of the Second Law – that is, if they don’t ignore it altogether like you do, without a single mention of entropy normally designated with S would you not agree?
Of course I used PE it in my computations in the paper, and thus I got the right -g/Cp answer for the gradient, not zero. I’ve already told you that, even though you have never read the paper, nor discussed the content, nor answered question (3) about Uranus.
You still just don’t get it do you Ball4?
If molecular PE is included in entropy calculations, then it is blatantly obvious that you would have unbalanced energy potentials if the column were isothermal, simply because PE at the top is greater than PE at the bottom and yet KE would be the same.
So (PE+KE)top > (PE+KE)bottom
It’s just that simple Ball4 and not hard to understand. Now read http://entropylaw.com and try to understand why I am talking about unbalanced energy potentials.
Are you claiming that potential energy is not an energy potential Ball4?
“Are you claiming that potential energy is not an energy potential Ball4?”
No. As I wrote 9:54pm PE is included in the total energy conserved & entropy eqn. maximized in the already published literature that proves Doug is inept.
“..link me to whatever papers you rely on..”
Doug’s first job as a researcher is a literature search, Doug even fails to do that. Look for them in 1998, 2004, 2006 Doug – they are easy to find – that was my first step and I found they all pinpoint prove where Doug’s physics are faulty.
State the equation for entropy that is used in proving the Clausius “hot to cold” corollary which only applies in a horizontal plane because the equations for internal energy U (a thermodynamic potential) ignore changes in gravitational potential energy as explained here from which I quote …
A thermodynamic potential is a scalar quantity used to represent the thermodynamic state of a system. The concept of thermodynamic potentials was introduced by Pierre Duhem in 1886. Josiah Willard Gibbs in his papers used the term fundamental functions. One main thermodynamic potential that has a physical interpretation is the internal energy U. It is the energy of configuration of a given system of conservative forces (that is why it is a potential) and only has meaning with respect to a defined set of references (or data). Expressions for all other thermodynamic energy potentials are derivable via Legendre transforms from an expression for U. In thermodynamics, certain forces, such as gravity, are typically disregarded when formulating expressions for potentials. For example, while all the working fluid in a steam engine may have higher energy due to gravity while sitting on top of Mount Everest than it would at the bottom of the Mariana Trench, the gravitational potential energy term in the formula for the internal energy would usually be ignored because changes in gravitational potential within the engine during operation would be negligible.
(1) Well you should have referred to the Second Law as that is what I am talking about in the website http://climate-change-theory.com and the cylinder will approach thermodynamic equilibrium (as per the Second Law) which climatologists like to call hydrostatic equilibrium implying fluid flow of which there is none here.
(2) Good, I’m glad you seem to agree that there will be “heat creep” (downward diffusion, conduction and convective heat transfers) from the less-hot troposphere of Venus into the far hotter surface. Go and explain that to your climatology friends, because this lies at the heart of the new 21st century paradigm in my paper and book, and of course it could not happen if the state of thermodynamic equilibrium did not have an associated non-zero temperature gradient.
(3) Well, put it another way: Why is the temperature at just the right value of 320K? Now answer it, bearing in mind that similar happens on all planets with tropospheres, so it’s highly unlikely to be just a coincidence.
Answers:
(1) Some of the surplus thermal energy at the top will be transferred downwards by non-radiative processes even as far as the base of the cylinder, thus making the base of the cylinder hotter and restoring a 1 degree difference between the top and the base with each being slightly hotter than the initial state of thermodynamic equilibrium.
(2) The Venus surface receives thermal energy (original absorbed from insolation in the upper troposphere) by the “heat creep” mechanism explained here.
(3) The reason these temperatures are not coincidences is that the temperature profile builds up with the expected temperature gradient, and this is supported by thermal energy originally absorbed in the higher regions of the atmosphere and subsequently transferred downwards by the “heat creep” process.
Ball4, Norman and Tim – you all need to catch up with the 21st century …
“According to the old view, the second law was viewed as a ‘law of disorder’. The major revolution in the last decade is the recognition of the “law of maximum entropy production” or “MEP” and with it an expanded view of thermodynamics showing that the spontaneous production of order from disorder is the expected consequence of basic laws.”
Source: http://entropylaw.com
Now I suggest you all read the numerous comments on last month’s temperature data thread starting here.
And if you still don’t understand what I am explaining, then it really is time you read the detailed paper Planetary Core and Surface Temperatures.
Temperatures for planets and satellite moons are “anchored” by the fact that radiative balance with the Sun’s insolation must occur and be represented by a temperature usually found in the planet’s troposphere. If there is no atmosphere then it will be at the solid surface, as for the Moon. In the case of Uranus, it is in the methane layer in the stratosphere. That is where we find temperatures around the effective radiating temperature of (58±3)°K. [source]
The temperature profile builds up to warmer temperatures at lower altitudes and in the troposphere it follows closely the expected |g/Cp| temperature gradient. When we use the known data to calculate that gradient and then apply simple geometry to determine the temperature 350Km further down at the base of the nominal troposphere of Uranus we get about 329K. Estimates here say it is 320K and that makes sense because the gradient is always reduced a little by inter-molecular radiation.
Now, the point is that there must be a mechanism operating that allows the temperature profile to build up downwards, because otherwise it would be highly improbable that we would find the right temperature at the base of the troposphere. Furthermore, if all temperatures from the core outwards were supposedly due to the core still cooling off, then what would happen to the gradient calculations in the distant future when the base of the troposphere might be only, say 250K?
The mechanism (based on the Second Law of Thermodynamics) represents a major breakthrough in 21st century climate science, because it also functions on Venus, Earth etc and completely does away with any need to assume back radiation supplies the extra energy, which it simply cannot do anyway. You may read about it here.
Therein lies the death knell of the radiative greenhouse conjecture.
Gravity acts on molecules, not pressure or density or anything else. According to the Law of Entropy (Second Law of Thermodynamics) we can deduce that, when the final state with no unbalanced energy potentials is attained (that is, we have thermodynamic equilibrium) then (PE+KE) is homogeneous and so we have a temperature gradient.
We must also have equal numbers of molecules crossing each way over any horizontal boundary and, because gravity causes there to be a greater probability of downward molecular motion than upward motion, we thus also have a density gradient. The pressure gradient is merely a corollary of these two gradients, for which gravity is the cause of each.
Whilst there may appear to be some vague correlation between temperature and density (in a vertical plane) this does not imply that either is the cause of the other. Gravity is the cause of each gradient, as is obvious because both gradients do not exist in a horizontal plane.
There are no such things as “parcels” of air somehow clinging together in adiabatic (calm) conditions. Only wind holds air more or less together, but wind tends to destroy the temperature gradient.
For example, warm winds blowing over a plain may then be diverted up a sloping mountainside. The air in that wind will still be nearly as warm at the top of the mountain, and somewhat warmer than the ambient air was up there before the wind reached the top. In other words, the upward motion has not cooled the air because it was not an adiabatic process.
The temperature gradient only forms in the absence of wind of any form. The gradient develops by the very slow conduction-like process involving molecular collisions, just like the “car in garage” example in the website. In many cases one cannot detect any net air movement, but we know convective heat transfers (which include diffusion) are occurring.
So Maxwell was very wrong in his assumptions that the lapse rate requires overturning of the atmosphere with wind of any form or uneven surface heating, or imaginary (and impossible) parcels of air molecules somehow clinging together, rising, expanding and cooling. Wind destroys the very process he was imagining. But climatologists ever since have called on the authority of Maxwell, and ignored his teacher, Loschmidt (who knew better) because it suited their cause, which is ultimately to destroy capitalism, as now admitted publicly.
[Now it’s 12.30am and I’m signing out.]
If you look at individual molecules, then you have to consider that at higher altitudes, only molecules that started with above-average energy could even get there to begin with. The cooler, slower molecules never got there to be counted.
Put another way, look at ALL the molecules in a thin layer at some altitude (specifically all the ones that are moving upward). They all have the same initial PE(i) because they all have the same altitude. They have a distribution of KE, from zero to very large values.
IF ALL of them got up to some higher level with a larger final PE(f), then all would have to have a lower KE than they had before, and then your claim that PE+KE = constant would indeed be true! But some had zero KE, so they would have to have a NEGATIVE KE — which of course is physically impossible. Only SOME made it to the higher level — they would have to have a KE >= PE(f)-PE(i) to be in the group at the higher level.
So only SOME molecules get to the higher level — those that had above average KE to start with. Even after losing some KE, it turns out they would STILL have the same average KE as the larger group of molecules in the lower layer.
Until you can work through the detailed math of Kinetic Theory and sums over distributions of speed, you don’t have the tools to know how the actually evaluate this claim. Show that this claim is wrong mathematically, or stop claiming the KE+PE is constant.
****************************************
I think we are BOTH butting up against a common definition of insanity … repeating the same actions and arguments and expecting a different outcome. 🙂
Tim writes: “only molecules that started with above-average energy could even get there to begin with”
Absolute nonsense, Tim.
A molecule at sea level travelling upwards at a typical average speed of 1,800 Km/hour has sufficient momentum to get to the upper mesosphere at least if it did not collide with any others. Furthermore, it gets new energy on the way up from solar insolation of which about 20% is absorbed by the atmosphere. The fact that no region cools to absolute zero (-273°C) proves my point – all molecules still have plenty of KE. Those in a region at about 144K (-129°C) which would be colder than the upper mesosphere, still have half the kinetic energy of those at the surface at 288K.
The distribution of KE is irrelevant. In fact it evens out quite well in calm conditions in each horizontal plane. But we can argue without loss of generality that all have the same KE in any horizontal plane, because although some have more than the mean, and some less, the mean itself reduces as we go up. For every molecule 1 SD below the mean at one level there is one that much below the mean at another level.
What you and Norman and Ball4 don’t understand is the Second Law of Thermodynamics. You seem to think unbalanced energy potentials would evolve (with more PE at the top and yet the same KE) and the law tells us the exact opposite, namely that unbalanced energy potentials will dissipate, leaving us with homogeneous mean molecular (KE+PE) at all altitudes, and thus a temperature gradient. It’s not hard to understand.
Mathematically? Simple: we equate PE gain and KE loss and get -g/Cp as explained in my answer to Ball4 above.
You simply cannot ignore what the Second Law is telling you and it applies at a macro level, so your distribution issue is irrelevant and only the means matter. The “fictitious fissics” of climatology is false in many respects. That’s why I have written two comprehensive peer-reviewed papers on the Second Law of Thermodynamics, pointing this out. Study those papers, for they contain all I need to say and they are linked on the “Evidence” page of our group’s website.
Footnote:
And if you were right, Tim, about a force field like gravity or centrifugal force producing isothermal conditions then a vortex tube would not work, because all the air would remain isothermal along any radius of the cylinder. That does not happen, and room temperature air is separated into a hot stream (around 200°C) and a cold stream that can be around -50°C. That is absolute proof that a force field creates a temperature gradient.
Doug I contend that at least the cold stream is caused by expansion of compressed air doing work. You state this is wrong. You seem to own a vortex tube so instead of using compressed air source to drive the air through the tube use a fan to push the air molecules and then see if they cool. The temperature gradient should still form if the air is spinning.
Experiments and tests are not beneath you. Try different things, different configurations and see what happens.
Nonsense. The compressed air has already expanded before it enters the cylinder. Its temperature at that point has been measured in experiments. It then gets far colder in the center and far hotter at the circumference due to the force filed. Furthermore, the temperature gradient is that which may be calculated from the quotient of the acceleration due to that force field and the specific heat of the air, just as expected by the science. Yes a fan would produce a temperature gradient calculated in the same way by determining the magnitude of the centrifugal force. The temperature difference depends on the speed of the air that is injected by whatever means.
Some experiments with vortex tubes …
http://www.sciencedirect.com/science/article/pii/S0011227504001900
http://www.me.berkeley.edu/~gtdevera/notes/vortextube.pdf
http://wireilla.com/engg/ijmech/papers/2313ijmech01.pdf
“In this research numerical and experimental method will be employed to determine the significant factors on vortex tube behaviour with an incompressible flow in order to boost vortex tube energy (thermal) efficiency “
Doug Cotton,
That is not what I read about the vortex tube. The coldest temperature occurs right where the expanding gas enters the vortex tube and then it warms slightly as it progresses down the tube. You read the link I sent and it was a measured experiment. How do you claim different? What is your experimental evidence? Do you just make a claim and that is the case? If a fan would create the temperature gradient use it with your vortex tube and prove this is the case. Saying it does not prove it.
Please remember Galileo! When the question of falling objects came up and there was a controversy on the topic, Galileo went and ran and experiment by actually dropping balls of different mass. It ended the debate. So please do some experiments with your words. From what I read most posters do not accept your conclusions and understanding. Now you need to be a Galileo and actually advance science. What you have is conjecture, opinion and belief of what takes place (even your understanding of the vortex tube is controversial you claim it does this others claim it does that with no resolution in sight). Go out and experiment!
SIGH!
The vortex tube is NOT equilibrium! It is highly NON-equilibrium, with continued inputs of matter and energy to the system. As such, it tells us NOTHING about what would happen in equilibrium!
No one doubts that continued energy inputs can create temperature gradients in a vortex tube. Just like no one doubts that air convecting up and down will warm on the way down and cool on the way up. Heck, even carrying a balloon of air up an elevator will cool the air as it rises because it expands. None of this tells us what would happen in equilibrium, where no one is pushing or pulling or compressing or heating a system.
The Second Law is not about whether thermodynamic equilibrium is attained or not – it is about the propensity for systems to move towards such by minimizing unbalanced energy potentials. Are you going to argue that potential energy is not an energy potential? It seems you are, because it is ignored in the climatology “proof” that isothermal conditions would evolve.
It’s now over three years since BigWaveDave told you in this comment about how many PhD’s “forgot” about the well established knowledge that gravity forms a temperature gradient. And yet you still think you can ignore the Second Law of Thermodynamics and create a situation which has unbalanced energy potentials (more PE at the top) and which you claim is stable. (LOL)
Now run off and think about Question (3).
And just to rub it in, Tim, the Clausius “hot to cold” corollary of the Second Law is proved by using a simplified equation for entropy in which the term for gravitational potential energy is ignored. It is thus only true in a horizontal plane, whether you like it or not.
Obviously there would be isothermal conditions in a vertical plane if the Clausius statement applied to such. But, just in case you are not aware of it from your secondary school science classes, gravitational potential energy does vary with height, and so when molecules move between collisions with any vertical component in that motion, the change in PE causes a change in entropy which cannot be ignored. You choose to ignore it, but the gain in molecular KE is the one and only reason why there is warming in downward descending molecules and cooling in upward moving ones.
But the Second Law tells us which way things like temperature and density will vary as thermodynamic equilibrium is approached, and it tells us there will always be a propensity to increase entropy. In that entropy is a measure of progress towards the state with no unbalanced energy potentials, that is how we can determine which way temperature and density will move, and they do so towards a stable density gradient and a stable temperature gradient, because the Second Law defines those stable states as having no unbalanced energy potentials.
Footnote: The Second Law tells us that entropy increases towards a state of stability, not a state of disorder. The state of stability has maximum entropy and no unbalanced energy potentials within the constraints of the isolated system under consideration. The temperature gradient in a planet’s troposphere has no propensity to approach zero. You can read more at http://entropylaw.com which is not a site of mine.
Go back to this comment and explain the dilemma regarding flux for Earth’s surface, but don’t bother giving me the assertion that back radiation supplies twice as much thermal energy to the surface as does the Sun, thus helping the Sun to bring about the observed temperatures.
The warming and cooling of gas molecules – not imaginary parcels of air (which don’t exist anyway because they have nothing to keep the molecules all within any boundaries) is caused by the process explained in the Second Law of Thermodynamics. It cannot be by anything contrary to that law, because that law, like me when I use that law, is always right.
Doug why do you continue to make ignorant claims?
“he warming and cooling of gas molecules – not imaginary parcels of air (which don’t exist anyway because they have nothing to keep the molecules all within any boundaries)”
Do you have any clouds in Australia to look at forming?? What do you think a cloud is? It is a parcel of warm moist air that rose up expanded doing work against the surrounding mass of air that it can’t move into because of the surfaces created by the mean free path, the air surrounding the cloud does not have the same moisture content and may not even be rising at the time so no condensation takes place in that parcel. The parcels are real phenomena that can be visibly seen in a cloud formation. The parcel is kept together by surrounding air that is at a different temperature or moisture condition. Think about it. I sent you a link explaining it but I am not sure you read it. Don’t say something doesn’t exist because you can’t understand it and have a very very limited understanding of meteorology. You do not understand the weathermen use parcels concept to determine how severe a storm will be. A cold air mass is moving in and pushing up warm moist air over it (the two air masses do not mix one limit is the mean free path, the warmer air molecules can only go into the cold molecules a short distance at the interface and it is not enough to alter the density difference. Warm air expands and is lighter than denser cold air because the molecules are moving faster and push against the surrounding air more. Read up on it and learn before making ignorant comments please!
Well, what happens where there are no clouds? No temperature gradient? Does that make it colder where there are no clouds? /sarc
Why does empirical evidence prove that increasing water vapor leads to lower surface temperatures as in the study in my paper?
Why is it that you have so much trouble understanding that an isothermal state is obviously not the state of maximum entropy?
Blimey, you have absolutely not the faintest idea of what I have explained in the book and papers.
I am not talking about storms or wind of any form. When you open the doors of your hot car in the garage heat is transferred out the doors in all accessible directions, yet there’s no wind or storm or hurricane or whatever producing air that you can possibly detect moving, or maybe you can’t.
Read the four comments starting here.
Just to set you right, when we talk about the Second Law of Thermodynamics we are talking about a process which takes place in an isolated system.
All your moving air masses and whatever are some form of wind and they have nothing to do with isolated systems, adiabatic processes or the formation of the temperature gradient – in fact they tend to level out the temperatures. Warm air blowing up a mountainside, or cold air blowing downwards above the South Pole do not form a temperature gradient – they flatten it out.
Now stop wasting my time, because you could have learnt what I’m talking about by reading the website and the linked paper.
There were no “continued energy inputs” Tim in the sealed insulated cylinders in which temperature gradients evolved in over 99% of more than 850 experiments, about which you can read by following the link on our group’s website.
You, Tim Folkerts, continue to make assertive statements which have no foundation in the the laws of physics and are contradicted by empirical evidence. But still you bury your head in the sand for over three years, refusing to believe BigWaveDave, Loschmidt, myself and many others that the state of thermodynamic equilibrium in a planet’s troposphere has a density gradient and a temperature gradient.
Gaaa!
We were talking about the vortex tube, which DOES have inputs of gas and energy! Try not to change the subject!
Once again … systems which are NOT in equilibrium (eg a vortex tube) cannot be used to predict the conditions in a system that IS in equilibrium (eg the sealed insulated tubes).
The Second Law of Thermodynamics can be used to predict what will happen as thermodynamic equilibrium is approached. In a steady state of hydrostatic equilibrium (energy in = energy out) the vortex tube is shown to have the temperature gradient which we would expect from the science developed from the Second Law of Thermodynamics. The fact that it has a radial temperature gradient at all, no matter what, proves that isothermal conditions are not what would be approached, because it started out that way with isothermal air being fed into it.
You still have not learnt from BigWaveDave have you?
Give up, Tim! The Second Law is always right. It tells us in which direction things will have a propensity to evolve.
Now I am convinced that you are deliberately avoiding Question (3) because you know you can’t answer it with your isothermal conjecture which would violate the Second Law if it were the state of thermodynamic equilibrium. Remember too that you need to be able to explain the energy input that supports the 320K temperature (hotter than Earth’s surface) at the base of the nominal troposphere of Uranus that is 30 times further from the Sun than we are. The whole planet receives less than 4W/m^2 of insolation even at TOA and emits a similar amount. There’s no scope for imagining it’s all just cooling off.
“systems which are NOT in equilibrium (eg a vortex tube) cannot be used to predict the conditions in a system that IS in equilibrium”
Yes they can, because we can observe in which way they are tending. It’s damn obvious that the vortex tube is tending towards having a g(centrifugal)/Cp radial temperature gradient, and in fact it does achieve such.
And a vortex tube is in hydrostatic equilibrium anyway – there are no further changes in temperature when the air flow is staedy.
You just don’t understand the Second Law of Thermodynamics which can be used to predict such things as in my paper about which you have no clue because you have not studied it.
You have not proved me wrong on either point – you just throw around your assertive statements without one iota of understanding or proof. In fact you obviously still don’t understand why an isothermal state would not be maximum entropy, even though I have explained why in various comments.
You are unteachable Tim Folkerts, as BigWaveDave learnt three years ago.
Norman, Ball4 and Tim:
I am no longer going to be side tracked with all your red herrings.
(a) Answer those three questions (with reference to valid physics) before writing any more. I will ignore unsupported assertive statements.
(b) Prove your assertion that isothermal conditions in a vertical plane represent maximum entropy.
(c) Quote some statement in the website or paper and then …
(d) Refer to a reliable source of information about the Second Law of Thermodynamics, and use that to refute what is in the website or paper if you can.
Go to this comment.
Here is another “experiment” with a normal centrifuge machine. I quote:
“Molecules lose speed drastically when they are guided against the centrifugal force to the center of a rotating disk.”
[source]
This is all just as I would expect. The gain in centrifugal potential energy is offset by an equal loss in kinetic energy, meaning cooling.
You see (all of you) the cooling of these molecules (referred to above) as they are guided to the center of the centrifuge machine (using electrodes) comes about purely because their internal KE is being converted to PE. It has nothing to do with “parcels” of air expanding, or buoyancy forces. But it proves this happens for no other reason than that the molecules gain potential energy (because of the force field) just as they do in a gravitational force field, and just as I wrote in my paper two years ago, which none of you has yet deigned to read.
Here’s a bit more for those who are too lazy to click the above link …
“On the average, nitrogen molecules, for example, travel at a speed of more than 1,700 kilometers per hour at room temperature, or almost one-and-a-half times the speed of sound. … However, physicists at the Max Planck Institute of Quantum Optics in Garching have now found a rather simple way to slow down polar molecules to about 70 kilometers per hour. They let the molecules of various substances, such as fluoromethane, run up against the centrifugal force on a rotating disk, while being guided by electrodes. The speed of the decelerated molecules corresponds to a temperature of minus 272 degrees Celsius. The new method makes it possible to produce relatively large quantities of cold molecules in a continuous flow … “
That’s a temperature around 1K.
How much more evidence do you guys want?
“Here is another “experiment” with a normal centrifuge machine. “
1) that is NOT a “normal centrifuge machine.”
2) that is NOT thermodynamic equilibrium. There is a continuous input of particles; there is a continuous input of energy (used to continuously push backwards on the molecules as they spiral inward).
0 for 2 again. 🙂
And Tim, as here …
“… if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.”
Your isothermal conjecture, Tim, is blatantly obviously (to anyone with an understanding of entropy) contrary to the Second Law of Thermodynamics. You can raise all the red herrings you like, but until your acknowledge that “The law that entropy always increases holds … the supreme position among the laws of Nature“ you do nothing but make a fool of yourself arguing against what that law makes perfectly clear.
In a centrifuge machine, gas molecules in free flight between collisions “cool” if there is an inward component in their velocity because (and only because) some of their molecular kinetic energy associated with their speed of over 1,700Km/hour is converted to potential energy in the centrifugal force field and they are slowed down, even to as slow as 70Km/hour which represents a temperature of about 1°K in the centrifuge machine described in the article linked at the foot of the “Evidence” page here.
In a planet’s troposphere gas molecules in free flight between collisions “cool” if there is an upward vertical component in their velocity because (and only because) some of their molecular kinetic energy associated with their speed of over 1,700Km/hour is converted to potential energy in the force field of gravity and they are slowed down.
That is all we need to observe, because it’s pretty obvious that the opposite happens when molecules move downwards in a troposphere or outwards in a centrifuge machine. And thus we get a temperature gradient, as I have been telling you all along.
It happens at the molecular level, as you can see in the “experiment” in the article about the centrifuge machine with electrodes guiding molecules to the center.
It has nothing to do with buoyancy, expansion or imaginary parcels of air which cannot stay together in calm, cloudless conditions in an “ideal” atmosphere.
Read my four comments starting here.
“even carrying a balloon of air up an elevator will cool the air as it rises because it expands.”
No it won’t necessarily, because pressure and density each decrease, so temperature can remain constant. (Pressure is proportional to the product of temperature and density, as you should know.)
You could confirm what I say with a simple experiment. Fill a vacuum flask with cold moist air at 1°C. Take it up a mountain in your car to an altitude of 500 meters where the ambient temperature may be about 3 or 4 degrees cooler than that at the base of the mountain where you filled the flask. Has the water vapor in the air frozen and the air cooled to at least -2°C? I say no – but you try it.
Only when molecules move with their own kinetic energy against a force field (such as towards the center of a centrifuge machine) do they trade KE for PE and thus cool. When you supply the extra PE (using the fuel in your car) there is no change in temperature. Likewise when wind of any form supplies the necessary energy (such as when it blows up the side of a mountain) no significant cooling has time to occur in the way it does adiabatically in an isolated system.
Footnote 2:
Of course no individual molecule actually moves all the way up the troposphere. All molecules continually collide, almost certainly after moving less than a millionth of a meter. But KE is relayed from one to the next, and hence convective heat transfer can be observed at speeds of perhaps a few meters a minute in extreme cases such as the “hot car in garage” example when the convective heat transfer can be in all directions out of the car doors.
Remember too, that if a molecule has kinetic energy, say, two standard deviations below the mean at that altitude, then there is about a 97.5% probability that it will gain KE in its very next collision because 97.5% of the other molecules have more KE.
And finally a quote from here:
Clausius coined the term “entropy” to refer to the dissipated potential and the second law, in its most general form, states that the world acts spontaneously to minimize potentials (or equivalently maximize entropy), and with this, active end-directedness or time-asymmetry was, for the first time, given a universal physical basis.
Do you read those words “minimize [unbalanced] potentials” and realize that’s what I’ve been explaining happens when (PE+KE) is homogeneous? It’s not hard to understand.
And that’s what it’s all about.
And if you doubt that a force field causes molecules to cool as their KE is converted to PE whilst they move against that force field at over 1,700Km/hr, then read this comment and the linked article.
Doug Cotton,
I read your link to the atomic breaking system. The question for you is what is the number of molecules being decelerated by this device? I have stated many times in these long sessions of posts with you (not that you actually read what I say, maybe scan it to refute but no desire to understand) that with a few molecules in any test or thought experiment you would be completely correct. As a molecule moved up in a gravity field it would decelerate and lose energy. I do not dispute this. That is why I agree with your 4 molecule thought expermiment. I believe in the case of your link it is only a small amount of molecules and there are not enough to establish a surface based upon the mean free path so they would lose energy moving against a gradient. In thicker air you have millions of what would be Real surfaces. A surface is an object that will redirect or stop the motion of an object that strikes it. So to with many molecules. I have state repeatedly that the K.E. + P.E exchange in a normal atmopshere is only for a very small distance and that the molecules losing energy moving upward against the gravity field will regain this energy when struck by molecules moving down the gravity field. They will exchange energy and end with a zero increase or decrease based upon the gravity field.
Norman:
“In thicker air you have millions of what would be Real surfaces. A surface is an object that will redirect or stop the motion of an object that strikes it. So to with many molecules. I have state repeatedly that the K.E. + P.E exchange in a normal atmosphere is only for a very small distance and that the molecules losing energy moving upward against the gravity field will regain this energy when struck by molecules moving down the gravity field. They will exchange energy and end with a zero increase or decrease based upon the gravity field.”
Exactly so: and why Meteorologists consider air as being buoyant (ie vertical movement in the absence of forcing can be ignored). And why Mr Cotton’s “heat creep” theory is a non-starter.
I admire your fortitude in occasionally adding the the continued Doug Cotton threads here, but I gave up when it was obvious the man is not for logic (ie what happens in the real world).
When I explained how the atmosphere worked on Earth, his knowledge in response was woeful, yet, of course it was me an (ex) professional Meteorologist who was wrong (or alternatively something else – that got him banned).
He is far, far away from even where the Fairies live.
Yes well you answer the three questions above that no one else can, but which are answered in the website endorsed by our growing group of persons suitably qualified in physics. Meanwhile I have to prepare for the next climate lecture I’ll be giving here. Keep watching the hit counters on our website.
As I said, I will only discuss questions pertaining to (and quoting) specific points in that website or my peer-reviewed papers linked therein.
Douglas J. Cotton B.Sc.(physics), B.A.(econ), Dip.Bus.Admin.
Sydney & Macquarie Universities 1963 to 1972.
Retired physics educator and now climate science researcher and public speaker.
The thermodynamics of planetary tropospheres, crusts, mantles and cores is a matter of physics, not meteorology. Your understanding of physics is sadly lacking and I seriously doubt that you understand entropy maximization.
Toneb
I am hoping that Doug may learn to have a scientific mind at some point in his life.
“It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.”
Richard P. Feynman
I am encouraging Doug to do actual experiments. Everytime people engage with him the thread goes into the hunderds of posts with no progress. He does not change nor does he change anyone. If he would spend more time doing experiments and reporting the results he would do much more to advance science than endless repetition on his own pet theory. I think he has some kind of bet with someone on how many people will visit his websight. He includes a link to it in everyone of his posts and I have read all the material on it and still do not find my mind changed.
Your credentials would make you the correct person to ask.
There is a large section of stratosphere that is isothermal. Apart from the jetstreams (in only small regions of this space) I do not think there is any wind, energy absroption or vertical air motion going on. You would think in such conditions a lapse rate would form based upon the gravity field and yet none does. The only lapse rates seen in our atmosphere are in a highly convective zone or where the atmosphere actually absorbs lots of energy (upper stratosphere with ozone and UV light interaction). I still do not know where Doug has spent time answering why is there a few miles of isothermal atmosphere in the most calm region? I could be wrong, you would have the knowledge of this area much above my own internet research.
Regarding the stratosphere, see last paragraph of Section 13 and then Section 14 of my peer-reviewed paper “Planetary Core & Surface Temperatures” (February 2013.)
Doug Cotton,
You do confirm my suspicion that you do not read my posts but rather quick scan them responding to things not asked nor stated. Your Section 13 and 14 in NO WAY answer my question to you. The Stratosphere has an isothermal layer that extends upwards around 10 kilometers. You have not explained in any way why this nonconvective, non radiation absorbing, little wind, does not demonstrate “heat creep”. Why do the molecules 10 Kilometers above those at the bottom of the Stratosphere moving with the same average speed?
The only thing you say in these sections on the Stratosphere is:
“the tropopause ceiling, because the temperature inversion in the stratosphere stops further rising”
(from Section 13)
and
“Just as we see a temperature inversion in the stratosphere, where the Sun warms the top, so too do
we see it in the top layers of the ocean.”
It is not even reasonable to assume in any way that this is an answer to my question. I do not know who you are tying to fool with that.
The inversion in the Stratosphere starts after the isothermal layer which is several kilometers depth. Will you continue to ignore my question and point to items that do not relate in any way to what I am asking??
You have to explain 10 kilometers of isothermal Stratosphere and you have not and probably will not. With gravity acceleration of and your “heat creep” theory and a very dry Stratosphere the lapse rate should be 10 C/Kilometer. You have 10 kilometers of isothermal atmosphere so your “heat creep” theory should have the bottom of the Stratosphere to the layer before it is warmed by Solar UV and ozone interaction a 100 C colder than an isothermal state.
Please don’t point to your Section 13 and 14 as they answer absolutely NOTHING about the isothermal layer but please explain the nature of this isothermal layer. Why does it exist?
I am trying to encourage you to think. That’s how I always taught my physics students over the last five decades.
Convective heat transfer is downwards from the lower stratosphere and into the “valley” of the tropopause where there is actually a curved temperature profile not a level one. It’s like water flowing down each side of a valley into a lake at the bottom where it levels out.
This is obviously what must happen as the gradient changes from negative in the upper troposphere to positive in the stratosphere, as happens with all planets that have significant atmospheres. I have made it quite clear that the process that forms the gradient is slow, as with the “hot car in garage” example. Thus it is over-ridden when the rate of absorption of new solar energy is excessive and consistent, as it is in the stratosphere. So it is no surprise that there is a hump in the overall decline between the troposphere and the mesosphere. We see similar over-riding in the ocean thermoclines. So your temperatures in the tropopause in no way disprove the Second Law of Thermodynamics upon which my hypothesis is based and which is supported by this new centrifugal machine about which I have already posted on over 200 climate threads. So I’m busy and must get back to this task while my comment is in the clipboard.
Doug Cotton,
I wish you would apply your own system to yourself!! “I am trying to encourage you to think. That’s how I always taught my physics students over the last five decades.” You do not seem to do a very good job at thinking.
You are not explaining in the recent post why is 10 kilometers of the Stratosphere isothermal. That is a large distance and you ignore it like a disease.
“Convective heat transfer is downwards from the lower stratosphere and into the “valley” of the tropopause where there is actually a curved temperature profile not a level one.”
What the hell crap is that you are posting. Prove this downward convective heat transfer! And that has nothing to do with the 10 kilometers of isothermal Stratosphere. You dog the question with garbage and crap. Why? Do you use “If you can’t dazzle them with your brilliance, baffle them with your bullshit!”
Not one word in your post or in your website nor in your paper has any explanation of 10 kilometers of isothermal conditions in the Stratopshere!!
I have explained you “wonder” centrifuge clearly but you ignore this. Number of molecules.
See page 1733 here and note that tropopauses are not always isothermal – the temperature plots for the planets shown are all curved because there is a gradual change from a negative gradient in the upper troposphere to a positive gradient in the stratosphere where excessive heat is absorbed and thus causes the positive gradient. Why would you expect the slow diffusion process to work in the tropopause and create a negative gradient there? It is obviously dominated by the rush of heat down from the hotter stratosphere. I have no idea why you could not work that out for yourself. Besides, my paper is about what happens in tropospheres, crusts, mantles and cores.
Doug Cotton,
Other posters have commented on how you change the direction or topic when confronted with a specific question.
We are only talking about ONE THING here. We are not talking about Jupiter, Saturn or other planets. We are talking about 10 kilometers of the Earth’s lower Stratosphere that is isothermal.
10 kilometers Doug!! If your heat creep was even a remote reality that would require a change of 100 C rather than 0 C. Such a huge reality cannot be swept under the rug but you try to anyway. You ignore this elephant in the closet.
You understand physics you say. Your theory must cover all layers it can’t just magically work in some places and not others. If it fails in one region then it is bogus and other explanations for phenomena are accepted (like the lapse rate is caused by convection and expansion of air as it rises making a thermal gradient because gravity caused a density gradient so gases expand and cool based upon this).
Norman:
An atmosphere under instant switching on of gravity would develop an adiabatic atmosphere caused by greatest compression lower down. That is a one-time only event, and just as your bike tyre cools once you cease pumping so would an atmosphere as a BB. Under solar insulation a rotating planet would immediately develop differential heating and therefore a jet-stream, positioned by Coriolis. There would be convection, convergence and divergence both at the surface and aloft. The atmosphere would soon become incredibly turbulent. This process is a heat pump, naturally forming a DALR via vertical motion. Stop that vertical motion, as happens under nocturnal radiation conditions and the atmosphere tends to the isothermal. Look up a SkewlogT for a winter polar station. There is NO gravitational temperature effect. Full stop. it is NOT observed in real life and what’s more the world’s Met Centres do not model it.
Add GHG’s to the atmosphere and that just shifts the Earth’s BB temp to altitude from -18C at the surface to around 7km WITH THE SAME LAPSE, raising the ave surface temp 33C to 15C. The “Heat pump” mechanism is by far the dominant effect.
NWP models have the physics correct, otherwise how can they get amazing results globally for up to T+120??
The worlds Meteorologists (who like me, study the atmosphere 24/7) know those physics to be correct.
Fiddle with mind experiments all you like, but the atmosphere is where it happens in reality. And it staggers me there are people that deny it. It really does.
You are quite right, Toneb about the bike tire because you are right about high pressure not maintaining high temperatures as I have said many times.
You are not right about the troposphere tending to be isothermal at night. Even though upward convective heat transfer may stop altogether, the environmental temperature gradient remains intact because it is the state with maximum entropy and you will be well on your way to that $5,000 reward if you can prove physics tells us isothermal conditions would prevail.
You need to learn, Toneb, that the Second Law of Thermodynamics never fails to have a propensity towards minimizing unbalanced energy potentials. In short you need to understand thermodynamics far better than you do.
In this case, that means we have a propensity towards homogeneous mean molecular (PE+KE) and so a temperature gradient, reduced in magnitude by the temperature leveling effect of inter-molecular radiation, as happens in all planetary tropospheres.
If ever you want to discuss my hypothesis, feel free to read it here: http://climate-change-theory.com
Toneb,
I would tend to agree with your view. If Doug had any potential for being correct, unexplained anomalies would have arisen and the meteorological world would have started to research to figure these out. I sounds like most phenomena in the atmosphere can be explained by current understanding of physics and no “heat creep” is necessary to explain these conditions.
Best of Luck in trying to convince Doug. I am still trying to get him to explain to me why there is a 10 kilometer isothermal layer in the Stratosphere if “heat creep” is real.
He does not comprehend why a pressurized gas will cool when expanding in surrounding air. Even though he could grab a can of computer duster and spray it and see how it cools. He won’t do it though as it does not go with his Universe. His universe is real not the experimental one. The only sad thing is he has a group of people that think his physics is correct.
Of course there would be at least trillions of molecules in the centrifugal force machine experiment that produces a stream of gas at around 1°K. The vortex machine does likewise with a stream at about -50°C. There is no such thing as a Mean Free Path surface as molecules are randomly positioned. In general a molecule passes about 13 other molecules before colliding – but that’s just an average and it could be many more or fewer. You really are clutching at straws. Never mind, I’ll be busy now promoting this refutation of the greenhouse conjecture using this new machine. Curt, Tim, Ball4 and you can hide in your shells, because, as I told you many times, the Second Law of Thermodynamics is always right and cannot be ignored. It holds the supreme position over all laws of nature.
All of Doug’s examples are analogous to saying:
“Water naturally runs up hill. To prove it, I hooked a hose to the side of my house and put the other end up a hill. The water came out the top, so water runs up hill”.
The examples (vortex tube, centrifuge) have a “pump” to move things where they would not go on their own and do things they would not do spontaneously. As such, they cannot be used as examples of what things would do in thermodynamic equilibrium. Not that I think Doug will ‘get it’ …. ðŸ™
Mr. Cotton,
“There is no such thing as a Mean Free Path surface as molecules are randomly positioned.”
I have no clue what you are trying to say here. Did you read how I defined a surface? Do you ever read anything other than your own words? A surface is a place where an object (in this case molecules) will change direction and energy. When a molecule hits another it changes speed and energy, a molecule can only go a very tiny distance in the troposphere before it hits another molecule that acts like a surface to it.
I do not think you can grasp the concept even though it is quite real and it is what creates parcels of air which are also quite real.
Yes well molecules change direction and energy every single time they collide with other molecules. The new direction is seemingly random, though, as with a snooker ball, it can be determined from the angle of incidence and the relative velocities of the impacting balls or molecules. There is no reason why any particular molecule should be constrained within any imaginary boundary. Go and read up on Kinetic Theory – even the Wiki article will help you understand some of this important physics used successfully by Einstein and others, and used to derive the Ideal Gas Law. Here you will see a graphic of how molecules move.
Once again, you don’t realize that, in my website, book and papers I have discussed in great detail what happens with individual molecules both in flight between collisions and during collisions, and the consequent heat transfers that occur because of (and during) such collisions. In the absence of wind, and in a vertical insulated sealed cylinder in a force field like gravity a temperature gradient is the state of thermodynamic equilibrium because of the propensity for nature to move towards a state where KE of every pair of molecules about to collide is the same. That is when energy potentials vanish. That is what the Second Law of Thermodynamics is all about, and I explained all this over two years ago in the peer-reviewed paper “Planetary Core and Surface Temperatures” of which it seems you have read less than 5%.
Now I’m late for lunch and will be busy all afternoon and at a meeting in the evening. That gives you time to read and discuss any points regarding the physics explained in the website and paper.
Doug you have done nothing as you claim above. You have a 4 molecule thought experiment. You do not understand how this varies with many more molecules and why your theory falls apart at normal atmospheric conditions.
That is not content of my book, website or papers, so ignored.
Doug:
You claimed on last month’s thread: “When a ball bearing falls through a vertical vacuum cylinder entropy most certainly does increase even though (PE+KE) remains constant. What is reversible about such a process?”
Everything!!! Throw the ball up in a vacuum. As it goes up, it gains gravitational PE, loses KE. Eventually it comes down, losing gravitational PE, gaining PE. When it gets back to the original release height, it has as much KE as when it was released.
Therefore, by both commonsense and strict thermodynamic definition, these conversions are totally REVERSIBLE. No entropy is created in the conversion either direction.
Teachers expect students to understand these concepts in the first few weeks of an introductory undergraduate thermodynamics class. You cannot comprehend this after 50 years!
You need to figure this out. You are not The Smartest Man In The World, overturning two centuries of thermodynamic analysis. You simply cannot comprehend the most basic principles of the science.
You are totally wrong in saying “No entropy is created in the conversion either direction.”
The Second Law tells you entropy always increases in any natural process such as a ball falling. Once it falls it does not go back up. This is just so elementary you make an absolute fool of yourself.
Go and read http://entropylaw.com
Note Figure 4 with a ball rolling down a slope on this page.
Doug:
In last month’s thread, I pointed out that Maxwell’s analysis showed that if vertical columns tended toward a lapse rate, as you assert, an isolated system could turn thermal energy into work.
Your response was: “But no work can be extracted on any perpetual basis, because it would all cool down to absolute zero and then stop.”
You have just demonstrated that you don’t have the technical sophistication to realize that you have just admitted that your theories lead to BLATANT violations of the 2nd Law.
The 2nd Law does not just state that this type of energy conversion is impossible on a perpetual basis, it forbids this type of energy conversion AT ALL. If you had any comprehension of basic thermodynamic principles, of the type you should get early in your first course, you would understand this.
But after 50 years, you still don’t get it, so I will spell it out to you. The 2nd Law ABSOLUTELY PROHIBITS ANY of this type of conversion of thermal energy of an isolated system into work. You have ADMITTED that your theories are in COMPLETE VIOLATION of the laws of thermodynamics.
You need to go away and figure out how you could misunderstand such basic points so completely for so long.
Doug:
Your key idea is that you believe an ISOLATED column of gas in a gravitational field will tend toward the adiabatic lapse rate, whereas standard thermodynamic analysis, from the days of Maxwell to the present, says that it will tend toward isothermal conditions.
Yet the only real-world examples you put forward are not isolated systems, not even closed systems, but OPEN thermodynamic systems. You obviously don’t understand the difference between these types of systems, which are covered in the first few weeks of any introductory thermodynamics class.
You also have no concept of the difference between static thermodynamic equilibrium (which is what you are trying to assert conditions for) and dynamic steady-state conditions (which all of your examples exhibit). Now, this distinction may not be covered until the middle of an introductory thermodynamics course, but you would definitely be expected to understand it by the end of the course. 50 years later, you still do not get it, no matter how many times it is pointed out to you!
Doug:
Are you seriously trying to claim that Werhner von Braun was a physics professor at the University of Sydney in the 1960s, when he was leading America’s rocket program?
Doug also knows, apparently, that his idol Josef Loschmidt was somehow J.C. Maxwell’s teacher back in the 19th century.
Kristian,
I looked it up. Josef Loschmidt was working with Boltzmann. He was a mentor of him but nowhere is there mention of Maxwell.
It is what Curt has been pointing out about Doug. It is hard to believe he is a University trained physics major when he has so little rigor in his posts.
You, Curt, made a huge mistake assuming no temperature gradient evolves in a centrifuge machine. I have shown you an experiment proving you wrong here. The Second Law also proves you wrong.
Doug:
You still can’t understand it no matter how many times it is pointed out to you. The open thermodynamic systems you keep citing have nothing to do with your theory about thermodynamically isolated systems. Dynamic steady state conditions in these systems are completely different from thermodynamic equilibrium. These are beginner’s mistakes!
I have personally worked on systems that spin at 100,000 rpm in a high vacuum, so thermodynamically isolated. According to you, they should have melted down almost immediately. But there were absolutely no thermal issues of the type you assert!
You, Curt, have already proved you have no clue what entropy is.
Go and study http://entropylaw.com and note the Figure 4 in the second paper showing a ball rolling down a slope with entropy increasing.
You still can’t answer those three questions correctly.
I have used the Second Law of Thermodynamics and that is all I need. It has been around for two centuries, yes, but the understanding of it has come a long way in the last 15 to 20 years. That’s where http://entropylaw.com could help you if you are teachable, which I tend to doubt. I had not read that site when writing my papers and book, but I find it says exactly as I do about unbalanced energy potentials and the state of thermodynamic equilibrium. But you will dismiss it, no doubt, because it does not have enough integral signs to impress gullible minds. It just talks facts, as I do.
You claimed over and over again in comments above that a centrifuge machine would not produce a temperature gradient, but then you fail to acknowledge, let alone apologize about your error when I show you an experiment with one that does.
Yes he gave a series of lectures in the early 1960’s which I attended. Prof Harry Messel brought him out here, and also Prof Julius Sumner Miller.
Nice to see the return of Curt!
JohnKl
I missed your post way above:
Hi Norman,
You state:
“If Venus atmosphere were composed of helium and the surface was hot enough to emit 16000 watts/meter^2, nothing would stop the radiation from moving directly from surface to space and it would cool at a rapid rate. I think your explanation is a greenhouse effect.â€
Like other diatomic gas compounds helium should not absorb/emit in the infrared, but in the much slower longer wavelength microwave region. So in fact helium would likely take much longer to radiate it’s energy to space and cool than carbon dioxide. Your statement appears false.
Have a great day!
What would stop the surface from directly radiating into space at 16000 watts/meter^2 (lots of energy) and cooling Venus surface fairly rapidly. Helium does not absorb the IR energy. It would warm by conduction and convection and reach some higher temperature. Say you have a hot atmophere of helium that does not radiate IR. That is okay but it would not stop the surface from radiating away at all and that would still be 16000 watts/meter^2. As the surface cools the higher temperature helium gas will conduct this heat to the surface which will radiate it away and start cooling the whole system until it reaches an equilibrium temperature based upon how much solar energy it coming in and how much IR is leaving. It would be warmer than the Earth since the atmosphere would not stop any incoming solar radiation but not as hot as Venus is now.
So my question: Why does my statement appear false?
Thanks and have a nice day!
I agree 100%, Norman. GHGs warm the surface by blocking outgoing surface radiation (and replacing it with weaker radiation from the cooler top of atmosphere). With the helium atmosphere, the surface would never have gotten that hot to start with (and would rapidly cool if it did somehow warm up that much).
Hi Tim Folkerts,
You state:
“With the helium atmosphere, the surface would never have gotten that hot to start with (and would rapidly cool if it did somehow warm up that much).”
HILRIOUS!!! How does Helium “RAPIDLY COOL” by emitting in the MICRO-WAVE bandwidth while?!!! IT WOULD COOL MUCH FASTER IN THE INFRARED! Or why do you suppose it is that as CO2 and supposed GHG’s accumulate in the Earth’s upper troposphere and Stratosphere those atmospheric regions have been COOLING at a much faster rate than the surface temperatures have either been observed or can ADJUSTED to warm?!
Btw, as I mentioned below VENUS HAS MORE DI-ATOMIC ATMOSPHEIRC GASSES THAN EARTH! The Venusian atmosphere has 4 times the Nitrogen levels that Earth does. In fact, the large DI-ATOMIC level of Venusian atmospheric gas compounds likely helped CAUSE the enormous build up of CO2 and other tri-atomic gas compounds found on Venus NOW. See my reply to Norman below.
Have a great day!
Please not in all hilarity that the word above should have been typed as HILARIOUS!
Have a great day!
My question should have read:
“How does Helium “RAPIDLY COOL” by emitting in the MICRO-WAVE bandwidth?!!!
Yeah! especially when you incoming solar radiation arrives in the infrared, visible, UV and above spectrum.
Have a great day!
John,
A hint … before you declare that something is “HILARIOUS!”, you should first understand the science involved.
1) The statement about “rapidly cooling” dealt with the *surface*, not the *helium atmosphere*. The *surface* can and does emit IR quite freely. If the surface were ~ 730K emitting ~ 16000W/m^2, that is way more than the energy coming in (ie sunlight but no ‘back-radiantion’). Thus the surface would indeed rapidly cool until it was more like 300K (warmer on the sunny side; cooler on the night side) .
2) *Even if* we focus on the helium … the helium neither absorbs incoming solar radiation from the sun nor absorbs outgoing IR from the surface. As such, it only gains/loses energy by conduction with the surface. IF the surface is ~ 300 K (as established), then a 730 K atmosphere would conduct energy to the cooelr surface and rapidly cool.
Hi Tim Folkerts,
I apologize for being somewhat snarky and I did misread your statement regarding cooling, but that doesn’t help your case any.
You state:
“If the surface were ~ 730K emitting ~ 16000W/m^2, that is way more than the energy coming in (ie sunlight but no ‘back-radiation’).”
It’s also therefore way more energy from the sun than the atmosphere can re-radiate back to the surface. You’d do well to understand the science yourself Tim. Since the atmospheric particulate layer extends from the surface to an altitude of ~50km, supposedly reflects a little less ~95% of the incoming solar radiation (although I think the albedo gets over-stated) the amount absorbed by said particulate layer and that finally reaches the surface proves small. However, it’s enough to photograph the region as the Russians proved. In any case, you have yet to discover where the missing energy derives even assuming the GHG’s proved to be perfect absorbers/emitters. Indeed, when the CO2 gets combined with sulfur compounds and/or other GHG’s and particulates the atmosphere can indeed re-emit infrared-back to the surface, but the solar induced infrared proves far to inadequate to account for the energy you claim. So where does the energy come from. Btw, imo heat creep of solar radiation doesn’t account for it either.
You go on:
“Even if* we focus on the helium … the helium neither absorbs incoming solar radiation from the sun nor absorbs outgoing IR from the surface.”
Red Herring, I never claimed it did. Nor does Helium comprise much of the atmosphere. However, Nitrogen does and you have yet to account for my statement in it’s regard.
You then stated:
“IF the surface is ~ 300 K (as established), then a 730 K atmosphere would conduct energy to the cooler surface and rapidly cool.”
You never ESTABLISHED ~300K you merely asserted it. The lunar surface reaches well over 370K temps during the daytime. Venusian solar irradiance almost double that. If not for the particulate layer temps would be far in excess of that figure. Moreover, if as you claim CO2 traps so much IR why is the 1 bar atmospheric pressure Venusian atmospheric region at 50-65 km above the planets surface and above the particulate layer ( receiving full solar radation) have temps within normal earth range 120 deg F to 70-80 deg F?
Have a great day!
John,
The atmosphere radiates according to its temperature and emissivity. If the atmosphere is also 730K (a good approximation near the surface) and a blackbody (a less definite proposition, but not outlandish and in agreement with your own statements), then it radiates as much to the 730 K surface as the 730 K surface radiates to the atmosphere. In this case, no sunlight is needed to keep the surface @ 730 K. Ie, there will be 16,000 W/m^2 from the surface to the atmosphere, and 16,000 W/m^2 from the atmosphere to the surface = 0 W/m^2 net loss = 0 W/m^2 input needed.
Of course, the real atmosphere will be slightly cooler than the surface, and not a perfect blackbody, so there will be a little net conduction and a little net radiation, and some energy input will be needed from the sun to maintain the 730 K. Eg, if the IR is absorbed by the super-dense atmosphere in ~ 10m with a ~0.1 K temperature difference, there will be only ~ 9 W/m^2 of net radiation from surface to atmosphere. Only ~ 9 W/m^2 of input (eg from sunlight) would keep the temperature constant.
These numbers of course depend a lot on details of optical thickness, pressure broadening, what GHGs & particles are present, but this is completely in line with estimates you gave. Even if 95% of the sunlight is stopped before reaching the surface, that still leaves WAY more than 10 W/m^2 to slowly warm the surface during the day.
Corrections to my post:
“It’s also therefore way more energy from the sun than the atmosphere can re-radiate back to the surface.”
Should have read:
“It’s also therefore way more energy than impinges the surface from the sun, re-emitted by the surface and subsequently re-emitted again back to the surface from atmospheric gas molecules.”
My last full sentence in the last paragraph should read:
“Moreover, if as you claim CO2 traps so much IR radiation why does the Venusian atmospheric region at the 50-65 km region above the planets surface, at 1 bar of pressure and above the particulate layer ( receiving full solar radation) have temperatures within normal earth range from 70-80 deg F to about 100 deg F?”
My statements should have been clearer but I typed it quickly during lunch, my apology.
Have a great day!
Hi Tim Folkerts,
Thank you for the well stated reply. In and of itself your statement appears conservative and on points seems workable. The problem remains that it merely explains how Venusian temperatures can be maintained once you assume a set of parameters. You state:
“If the atmosphere is also 730K (a good approximation near the surface) and a blackbody (a less definite proposition, but not outlandish and in agreement with your own statements), then it radiates as much to the 730 K surface as the 730 K surface radiates to the atmosphere. In this case, no sunlight is needed to keep the surface @ 730 K. Ie, there will be 16,000 W/m^2 from the surface to the atmosphere, and 16,000 W/m^2 from the atmosphere to the surface = 0 W/m^2 net loss = 0 W/m^2 input needed.”
As you later admit:
“Of course, the real atmosphere will be slightly cooler than the surface, and not a perfect blackbody, so there will be a little net conduction and a little net radiation, and some energy input will be needed from the sun to maintain the 730 K.”
The problem remains that the energy required to generate the 16k w/m^2 flux hasn’t been accounted for by a long shot. You’ve only described how the flux can be maintained. Imo, neither the standard GHG theory nor the Doug’s heat creep theory begins to account for the energy required to generate and raise an atmosphere of tri-atomic gas compounds to a mass 93 times greater than the Earth’s requisite to generating such temperatures. It’s like a person orbiting the earth in a space capsule. It requires little energy to maintain the pace, perhaps a few retro-rocket thrusts to maintain orbit and altitude. However, this pales compared to the energy required to raise the capsule and set it in motion to begin with. Fortunately, empirical data provides the answer for the requisite energy but these theories have not.
Have a great day!
Hi Norman,
IMHO the quantity of back scattered radiation is quantized by the number of molecules present in the atmosphere. So, the effective quantity of energy “blocked” shouldn’t be so much more than the one returned at the ground by gravity, which is the result of the work previously e continuously done from the same ground to throw the very same molecules of gases up in the sky.
I’m by John side, that is, Helium should keep hotter the ground anyways. That, as the result of the work continuously done from the ground itself to keep up the atmosphere.
I’m not arguing that GHGs don’t work as supposed, I’ve no proof of that.
I’m just arguing that the mechanism of heating the ground by GHGs back radiation (radiated energy converted in molecular bending work and back converted in radiated energy), is not different than heating the ground by firing up the molecule which returning back by gravity release that energy back too (it’s heat converted to work by firing them up, which is converted back to heat once the very same molecules bump back to the surface).
Again IMHO, the last mechanism should be more efficient because all the energy is returned back to the ground, no energy escape to the outer space by that way, but I’ve no proof that it is more efficient, because I don’t have computed how much is this energy, compared to the GHGs back scattered energy.
Without disproving the above, I think that attributing all those 33K increase of temperature to the GHGs is at least not scientific for me.
Have a grat day.
Massimo
Hi Norman,
Thanks for the reply. You asked:
“So my question: Why does my statement appear false?”
Please note my comment simply referenced the ability of the gas molecules to radiate away acquired energy.
From what I understand 16000 Watts/meter^2 represents the energy required to result in the observed surface temp of + 870 °F (+ 465°C)on Venus not the total infrared irradiance from the surface. The Venusian atmosphere conveys most of that energy from the surface and since the mass of said atmosphere exceeds 93 times the Earth’s atmosphere ( thus having a pressure of 93 bar at the surface ) there appears to be plenty of gas molecules available to convey the energy away. If the entire atmosphere comprised Helium, Oxygen, Nitrogen or any other di-atomic gas compound and could only irradiate energy in the microwave range the temperature of atmospheric gas compounds must rise to enormous levels to rid itself of the enormous solar input ( Venusian solar irradiance appears to be ~2613.9 Watts/meter^2, which varies little since the planet follows an almost completely circular orbit about the sun ) received. A radiative heat trap results. Eventually, the accumulated energy in the atmosphere CONVEYED from the surface becomes so great and the resulting atmospheric temperature so enormous that CO2, sulfure dioxide and other tri-atomic gas compounds leach from the rocks and increasing volcanic action. Unlike Mercury the substantial Venusian gravitational pull ( 80% of the Earth’s ) traps the released gasses near the surface. The tri-atomic gas compounds then help radiate the energy away FASTER THAN COULD BE DONE BY THE DI-ATOMIC GASSES ALONE! In fact, that looks exactly like what happened. You may be aware that Earth’s atmosphere comprises 78% Nitrogen. Are you aware that Venus has four times the atmospheric Nitrogen that Earth has? Most of it appears NOW at high altitudes at this point in Venus’s history.
Have a great day!
JohnKl
“From what I understand 16000 Watts/meter^2 represents the energy required to result in the observed surface temp of + 870 °F (+ 465°C)on Venus not the total infrared irradiance from the surface.:
It is both John. The hot surface will emit energy based upon its temperature and emissitivity.
http://www.raytek.com/Raytek/en-r0/IREducation/Emissivity.htm?trck=emissivity
Here is a link on how radiation works. A hot blackbody object will emit radiation (leaving the surface) in a spectrum based upon its temperature. Venus surface is close to a blackbody.
http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html
The surface is emitting radiation at 16000 watts/meter^2. In order for it to stay at 465C it has to have this lost energy continuously replaced. If it receives more energy than this the surface will warm and radiate even more energy. If it receives less it will cool. It is not the gases in the atmosphere that are cooling the surface. It is the radiation being directly emitted by the surface.
Have a nice day!
Hi Norman,
Thank you for the reply and the links. You state:
“It is both John. The hot surface will emit energy based upon its temperature and emissitivity.”
Agreed! Which seems to me why the likely energy flux on Venus will prove to be even higher than 16k w/m^2 or the convective/conductive loss in any finite time period from the Venusian surface to the atmosphere must be extremely small due to similarities in terrestrial surface temperatures and the atmospheric gasses impinging. In the case of Venus you may well prove correct when you state:
“It is not the gases in the atmosphere that are cooling the surface. It is the radiation being directly emitted by the surface.”
Imo, you appear to be correct in this instance but it’s not universal. Our own planet does experience significant cooling due to atmospheric convection from the surface, otherwise surface temperatures should peak at lunar levels exceeding 100 deg centigrades. Such temps also appear on Earth orbiting manmade satellites.
Have a great day!
JohnKl
I agree with your post.
Have a nice day!
Hi Norman,
Thanks for the reply. As I mentioned to Tim Folkerts he proves absolutely correct that at this point in Venusian history only a few hundred w/m^2 can maintain present atmospheric conditions relatively well but it doesn’t even begin to explain the energy required for the system to exist and function at all. The complete solar irradiance of ~2613.9 Watts/meter^2 doesn’t begin to explain it unless it is coupled imo with planetary mass and atomic physics. Nor would it be explainable if all the energy impinging the planet was absorbed then emitted by the surface only to be captured by the atmosphere and fully directed back to the surface. The 16k w/m^2 flux supported by an atmosphere 93 times larger than Earth’s atmosphere must be accounted for and neither the Green House theory nor Doug’s heat creep applied merely to solar radiation acting physically and or chemically upon the Venusian surface accounts for it. Thanks and…
Have a great day!
P.S. Neither Venus nor any planet in this solar system can be separated from it’s history.
“What would stop the surface from directly radiating into space at 16000 watts/meter^2”
Nothing, and nothing would stop the upper troposphere cooling more (retaining the temperature gradient associated with helium) and collapsing to the surface as it liquefied and solidified.
As I have said all along, these planets are not just cooling off. Venus does cool 5 degrees in 4 months on the dark side, but the Sun’s energy warms it back up by the same amount the next Venus day. It just does not do so with direct radiation of less than 20W/m^2 into the surface: it does so by heating the region that is less than 400K in the upper troposphere, (which is the only region where it can raise the temperature as per Stefan Boltzmann) and then heat creep takes that energy down and into the surface.
Doug the surface of Venus does not heat or cool 5 degrees. Your source clearly states that is for the Upper Troposphere. Please reread your own source.
Doug Venus only receives 158 watts/meter^2 overall (because of the high reflectivity of its clouds). Your theory cannot replace a lost 16000 watts/meter^2 loss.
Re the oft-mentioned 16,000W/m^2 of radiative flux from the Venus surface, see this comment.
Massimo PORZIO
Always nice to hear from you. Even if energy would leave the surface and be tied up in the Helium that would not stop in any way the 16000 watts/meter^2 from leaving the surface. It is very true a hot surface will conduct heat to an atmopshere regardless of the type of gas. Yes helum, carbon dioxide, nitrogen will all get hotter.
But what you describe would in no way slow the loss of energy via radiation loss. If nothing blocks or stops that 16,000 watts/meter^2 from leaving it is gone. 16000 joules of energy per second/meter^2.
I have read lots of theories on Venus and non of them even want to deal with what is stopping the radiation from leaving. Some say the surface was heated by volcanism a few million years ago, others say compression of the atmopshere is what heats the surface, Doug has “heat creep”.
If carbon dioxide acted as a cooling agent (absorbing the energy of the hot surface and radiating it away), Venus surface would cool rapidly and would not stay warm and scientists would be able to see a very bright Infrared object. Venus would be one hot IR source.
In the IR what is seen from Venus is very cold.
http://www.dtic.mil/dtic/tr/fulltext/u2/772660.pdf
The conclusion is something is preventing the 16000 watts from making it to space and Earth based detectors. The radiation is being redirected.
I think the best explanation for Venus would be GHE since it would be the only one that would explain why a surface that is emitting 16000 watts/meter^2 is not showing this level of radiant energy to dectectors outside the planet.
Yes well Doug has been proven right with this experiment.
All planets have at least some IR-active gas, such as Uranus with its methane layer in the stratosphere which becomes its heat source at a cold (58±3)°K. It has mostly hydrogen and helium, but is hotter than Earth at the base of its nominal troposphere, all due to insolation of less than 4W/m^2 at TOA that has sent thermal energy down into the planet by “heat creep” over the life of the planet, there to remain “trapped” by gravity, not back radiation.
You all need to study thermodynamics because you all display huge misunderstandings of this physics in which I have specialized.
Yes well Doug has been proven right with this experiment.
See http://climate-change-theory.com as visited by over 6,666 since January 8th 2015.
Hi Norman,
“Even if energy would leave the surface and be tied up in the Helium that would not stop in any way the 16000 watts/meter^2 from leaving the surface. It is very true a hot surface will conduct heat to an atmopshere regardless of the type of gas. Yes helum, carbon dioxide, nitrogen will all get hotter.”
Maybe you missed my point, your assumption that GHGs “traps” the energy is a misunderstanding by my point of view. The just use some of that radiated energy to do a work and then they convert back that work to energy which only a part is scattered back to the ground.
In fact I could accept that the quantity of energy scattered back by the GHG molecules via the bending of their structure can raise the total radiated power at the ground to balance the incoming radiation from space, what I cannot understand is why you don’t apply the same process to the work done to keep all the molecules up in the sky.
Try to think the below.
Imagine a planet with a very negligible thermal capacity, and no atmosphere. As the sun radiation impinges on its ground it heat up to a value established by the SB adjusted for the geometry of the projection on the spherical body. As the sun stops to supply the energy to its ground, the planet temperature will fall immediately to almost 0K, because of its negligible thermal capacity (let me ignore the universe background temperature too).
Now imagine the very same planet but having an helium only atmosphere above its ground. After having heat the ground and its atmosphere as before, imagine the sun stops to supply the energy again, what I expect is that the ground remain to a temperature grater than 0K for the time needed to allow all the helium molecule to discharge their energy to the ground which dissipate it by radiation to the universe.
So, immediately after the sun energy has been removed the planet continued to radiate. Since only the sun energy has been removed, it’s obvious to me that that flux must exist also when the sun energy was still impinging the planet surface, so in that case the total radiated power at the ground must be the sum of the energy incoming from the sun plus the one which is continuously used to keep the molecules up in the sky.
No matter that the helium molecules don’t avoid the energy to exit to the space, what it matters is that the atmosphere is “charged” by bumping on the ground and that it discharges all that energy back to the ground, increasing the ground temperature exactly how GHGs do in the radiation domain, only more efficiently, because all the energy (and not only 50% as per the GHGs) is “trapped” on the “mechanical” domain.
I hope I’ve been clearer this time.
Do you get my point now?
Have a great day.
Massimo
Hi Norman,
by the way, thank you for the link on Venus spectral irradiance.
About that, I would highlight you that the temperature at the ground of Venus is about 735K and the equivalent radiating black body should be centered at 3.9um as seen here:
https://scholarsandrogues.files.wordpress.com/2011/05/venus-co2-spectrum-lg.jpg
The fact that in figure 1 of your link is not evidenced any radiation at about 8…9um is intriguing me. I suspect that there must be some other blocking shields (maybe the dense sulfuric acid clouds at the TOA) in fact no matter how much is dense the CO2 there, because it’s transparent on almost all the WL of that BB.
The site where I get the image above argues that the GHG is the culprit of Venus temperature:
http://scholarsandrogues.com/2011/05/06/venus-climate-v-co2-heating/
Note that in the 2nd energy flow diagram (the one they say it’s right), that graph say nothing about it’s all GHG effect or it is GHG effect plus the energy needed to keep the atmospheric gases (and the sulfuric clouds up there, as I suppose it is indeed.
The only way to evaluate the incidence of the two effects is compute the effective energy needed to do the two contemporary works. That is bending the CO2 molecular structure and keeping the atmosphere up there.
Now is time to go to sleep for me it almost 1 AM here.
Have a great day.
Massimo
“But what you describe would in no way slow the loss of energy via radiation loss.”
More assertive garbage. What stops the 5000K small solid core of Uranus cooling? There’s not much to block radiation in its atmosphere except the methane layer in its troposphere thousands of kilometers further up and only about 58K at that.
But we don’t measure any net outward radiation at TOA, and the only radiation in and out at TOA is less than 4W/m^2.
You have absolutely no idea of thermodynaics.
Massimo and others who want to learn what’s really happening, just read, study and inwardly digest all the information at http://climate-change-theory.com
Doug Cotton
I think it is you who have zero understanding to of thermodynamics and little math ability to go with your lack of understanding.
Here is some real math (something you shy away from).
From previous post:
Uranus core is thought to be mostly iron. The core is 0.55 Earth mass or 3.284×10^24 kilograms. Heat capacity of iron is 450 joules/kg/K. The temperature of Uranus core is thought to be around 5000 Kelvin so it stores around 7.4×10^30 joules (give or take)
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatcond.html
Use this equation.
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/thrcn.html#c1
Uranus is mostly Hydrogen which has a thermal conductivity of 0.172 watts/meter K.
So plug in the equation. This is high school physics but I did not graduate in college physics and the math is beyond my skill level at this time (use it or lose it). The goal is not a highly detailed precise energy profile just a ballpark estimate.
Uranus core is thought to be 1.2 times the size of Earth. That would give you a diameter of 15,300 km for Uranus core. Area of a sphere is 4(Pi)r^2. Radius of Uranus core would be 7650 km or 7,650,000 meters. Area of Uranus core is around 7.35×10^14 meter^2.
Uranus diameter is 51500 km. Radius is 25,750 km or 25,750,000 meters. Radius of Core is 7,650,000 meters. Distance core heat must travel to reach edge of Uranus is 18,100,000 meters.
Temperature of Uranus core estimated at 5000K while top of atmosphere is measured at 49 K.
Now we have all our variables to plug into our heat conduction formula (just a rough approximation not to Curt’s rigorous standards but should answer your question well enough).
Q/t (watts or joules/sec) = (0.172 w/meter K)(7.35×10^14 meter^2)(5000k-49K)/18,100,000 meters
Result=3.46×10^10 watts (joules/sec)
We have a core storing around 7.4×10^30 joules so divide
Time=7.4×10^30 joules/3.46×10^10 joules/sec
Time in seconds=2.14×10^20 seconds. Seconds in 1 year = 3.156×10^7.
So divide your time to cool Uranus core in seconds /seconds per year to get
2.14×10^20 seconds/ 3.156×10^7 seconds/year
=6.78×10^12 years. Uranus core could stay warm for 6 trillion years. That could explain why it is still warm and not cooled off yet from its heat of formation. I could have erred in the math. I am sure the insulating properties of the atmosphere can easily explain the reason the core is still warm after all these years. Good insulation.
But it does not explain why the temperature thousands of kilometers higher up at the base of the nominal troposphere of Uranus is 320°K which is close to the 329°K I calculated using my hypothesis. Nor does it explain why there is no convincing evidence of net radiative imbalance at TOA. Nor does it explain how the temperature gets down to just the right level 59°K at just the right altitude where there is a methane layer in radiative balance with the Sun at a temperature we can calculate to be just the right temperature of (58±3)°K – and the temperature gradient in the troposphere is at just the right -g/Cp gradient even though there is no surface receiving any solar radiation and creating turbulence which is supposed (incorrectly) to explain the temperature gradient that in fact forms at the molecular level as we would expect from a correct understanding of the Second Law of Thermodynamics which holds the supreme position in all the laws of nature. So, because your conjecture does not conform with that law, it crumbles.
And if the Uranus core cooled by even just 300 degrees then the base of the nominal troposphere would cool from 320°K to 20°K and, if the Sun were still radiating as at present, that is an impossible situation since the troposphere gradient would be nothing like what we calculate it to be using -g/Cp and the temperature for radiative balance with the Sun is (58±3)°K.
The formula for conduction rates is totally inapplicable for gases in a planet’s troposphere. If heat creep is operating, even the direction is wrong. If the temperature gradient equals that for the state of thermodynamic equilibrium then, by definition, there are no heat transfers across any internal boundary, but your calculations would show there to be.
Why on Earth you waste time doing such calculations and expecting me to accept them when my hypothesis clearly says something quite contrary, I really don’t know. But none the less I feel obliged to answer and to point out that, if you wish to challenge what I have explained, then you need to prove the physics wrong, not produce all these red herrings. In other words, you have to start talking about entropy, the Second Law, the dissipation of unbalanced energy potentials and the like. The website http://entropylaw.com will bring you up to date on developments since 1988.
Massimo PORZIO
Here is a way for you to think of Venus. The surface area of Venus is 4.6×10^14 meter^2. The entire surface is the same temperature so the whole surface is constantly emitting around 16000 watts/meter^2 (the emissitivity of Venus surface was calcualted to be very close to 1, almost a blackbody). The total solar energy that can be received by the entire surface (night and day included) is 158 watts/meter^2 (since Venus reflects 70% of the incoming solar radiation).
If you take the surface area of Venus times the emission rate, Venus surface is losing 7.36×10^18 joules/second. The total amount of available solar energy for Venus is 7.20×10^16 joules/second. Whatever Doug’s theory of “heat creep” the Sun provides only 1/100 of the energy needed to sustain the hot surface regardless of whatever path you want to formulate on how the energy gets to the surface.
If carbon dixoide is cooling the surface via radiation you would see a super high temperature emitting from Venus and it would cool very fast. Even if Venus was hit by an large object recently and was total molten (as Immanuel Velikovsky had proposed in the collision theory) if something did not redirect the radiant flux from the surface you would see it as a very bright IR object. If carbon dioxide cannot redirect the flow of radiation, what is doing it as it is more than obvious something is.
Carbon dioxide radiation from a cooler troposphere can neither cool nor warm a hotter planetary surface. It cannot even slow the overall rate of radiative and non-radiative cooling, because it only slows radiative cooling and the non-radiative cooling then accelerates to compensate. Nor can radiation from any IR-active molecules (including water vapor) affect surface temperatures. What happens with radiation was explained in my first paper on the Second Law of Thermodynamics in March 2012. It is linked from the ‘Evidence’ page but is not particularly relevant to the new 21st Century Paradigm in Climate Science.
What happens and what explains all temperatures in the tropospheres, crusts, mantles and cores of all planets and all satellite moons in all galaxies is explained at http://climate-change-theory.com and you people have not even got off Square One in your understanding of such thermodynamics because you refuse to read, study and inwardly digest the linked paper “Planetary Core and Surface Temperatures” (Feb 2013) at http://climate-change-theory.com
Doug Cotton,
Curt is 100% correct to highly question your physics training.
Curt: “Teachers expect students to understand these concepts in the first few weeks of an introductory undergraduate thermodynamics class. You cannot comprehend this after 50 years!
You need to figure this out. You are not The Smartest Man In The World, overturning two centuries of thermodynamic analysis. You simply cannot comprehend the most basic principles of the science.”
Doug you may have had a high school education in physics but not a college level. I have taken both (many years ago) high school and college level physics and there is a considerable difference. High School phyics is more concept orientated for the beginning students with some algebra math level (F=ma etc). At college level it is completely math oriented. College Physics describes the world in mathermatical precision not a simplistic conceptual world of High School physics.
Your three papers: “Planetary Surface Temperatures A Discussion of Alternate Mechanics” “Radiated Energy and the Second Law of Thermodynamics” and finally “Planetary Core and Surface Temperatures” are mostly conceptual works. They are high school level papers. A couple simple algerbra equations and that is all. Even if I don’t agree with any of your ideas you should have expressed them primarily in mathematical language with each variable and how it effects the outcome. Then others can determine the validity of your work. Having a genertic (no detailed math of energy flow) graph of Temperature with another (PE + KE) line and no math is worthless to science. Not one shread of intergal calculus showing how the energy in your system will move, precisely how wind will affect the flow, how convection will affect it how the energy is absorbed in the atmopshere and how it moves from there in explicit detailed math. You may fool some but if you actually went through a rigid 4 year physics course at Sydney you are not doing them favors by stating this. I would want my money back if your physics papers are what they consider adequate.
I also look at your sources. So much from Wikipedia. I would have failed my research papers if I had used this as a source. It is a very lazy researcher who would compose a paper with Wikepedia as the major source material. Where are the physics books. You are a very lazy scientist that won’t go visit a library and look at some really good physics books and use these as your source material.
Doug I think you are experiencing a mid-life crisis and need to feel important. Bad physics and insulting advanced educated types like Tim Folkerts or Curt.
My degrees are listed in each paper. Your calling me a liar is water off a duck’s back.
Submit your refutation and your claim for the $5,000 reward proving that the Second Law of Thermodynamics (on which the “heat creep” hypothesis is based) is apparently incorrect in your view. To qualify you also need to produce a study like mine but proving the opposite, namely that water vapor warms as the IPCC would like you to be gullible enough to believe.
I have shown you an experiment which proves Curt was wrong about his assumed and impossible isothermal conditions in a force field.
None of you can answer those three questions with any use of the laws of physics, whereas I can because I understand the thermodynamics involved.
And, by the way, I include in my papers and book all the computations that are needed to confirm the temperature gradient. The issue of determining surface temperatures from that gradient is straight forward geometry. To get the radiative temperature we just use Stefan Boltzmann on line calculators these days, and don’t need to spell out the calculations just to impress readers like yourself.
With convective heat transfers that are “trapped” by gravity we need not quantify the rate thereof, because there has been the life of the planet for such energy to be absorbed and thus trapped. You would have understood this if you had deigned to study what I have explained.
All you blurb about the lack of integral signs (or whatever it is that impresses you) is thus totally irrelevant, if you took the time to understand what the Second Law of Thermodynamics is all about. And in that regard, I only need to discuss the properties of the state in which unbalanced energy potentials have been fully dissipated. Once again, I do not have to put a time frame on this, though it is well known that diffusion and natural convective heat transfers are slow processes relative to even a slight breeze, let alone stronger winds.
I do not have to discuss wind of any form, and nor is such discussed in those greenhouse energy budgets. That is because weather tends to average out over the whole globe and have little if anything to do with annual climate trends.
I really don’t care if you are not interested in studying the hypothesis, but if I had a pecuniary interest in maintaining the hoax, I certainly would want to get my facts right and choose the right path for my future security – in other words, see if it would be best to find another vocation. I estimate you have until about the year 2023, perhaps 2026 before it all crumbles. I’m working on it.
You still have not woken up to the fact that my hypothesis is at the forefront of 21st century physics and not therefore to be found in outdated physics texts. Is anything up to date in the printed word? I only use Wikipedia references for the benefit of those, like yourself, not familiar with the terms used in physics such as entropy, convective heat transfer, kinetic theory, thermodynamic equilibrium etc. I bet you could not explain these without looking up Wikipedia or preferably http://entropylaw.com of which I was not aware at the time of writing the papers, but have referred you to above.
And you still don’t have a clue as to what the hypothesis is. You could not explain it in your own words, and nor could you reproduce the “heat creep” diagrams without referring back to them. That proves you have not studied what I have written.
And as for Tim Folkerts being “educated” in thermodynamics, you must be joking. Read this comment replying to Folkerts three years ago …
BigWaveDave March 1, 2012 at 4:19 pm
Tim Folkerts:
You asked …what qualifications do you have to judge a disagreement between PhD physicists on issues of fundamental thermodynamics?
I have been earning a living as an engineer specializing in cutting edge technology for very large scale thermal energy transfer processes and power systems for close to 40 years. My credentials include BS, JD and PE, and I have four patents.
As for my qualifications to engage in argument with PhD’s, I have many times been part of and have led teams with PhD team mates. I was also married to a PhD for 20 years.
Because the import of the consequence of the radial temperature gradient created by pressurizing a spherical body of gas by gravity, from the inside only, is that it obviates the need for concern over GHG’s. And, because this is based on long established fundamental principles that were apparently forgotten or never learned by many PhD’s, it is not something that can be left as an acceptable disagreement.
I still don’t know why this seems to be Doug’s favorite post of all times — it really doesn’t say much.
1) “Appeal to authority” is never a strong argument. Sure, Bigwave Dave has some knowledge of heat transfer. So do I. So do the other people in the original discussion. None of that per se will let us know who has the right answer.
2) Bigwave Dave claims education as an engineer and knowledge of “large scale thermal energy transfer processes and power systems”. This is very different from knowledge of the fundamental kinetic theory or statistical mechanics. Similarly, I know a lot of guys who know more about the engineering and operation of car engines than I do, but that doesn’t mean they know more about entropy or Otto cycles than I do.
So even if we are considering an “appeal to authority”, Bigwave Dave claims no authority on fundamental physics.
3) I agree when he predicts a “radial temperature gradient created by pressurizing a spherical body of gas by gravity”. This is, for example, what warmed the interior of the sun as it was forming. But note the word “pressurizing” — not “pressurized” (past tense). The action of doing work on a gas to pressurize it will indeed warm the gas “from the inside”.
His next sentence is a non sequitur. This one-time, non-equilibrium warming does not address what happens over the long-term once equilibrium is achieved. *IF* your “heat creep” is right, then he is right; if it is wrong, he is wrong.
To assume he is right is once again an appeal to authority (his and Doug’s). And neither of these are actually authorities on fundamental physics, anyway!
4) All of this is mostly an academic question. Both Doug’s theory and standard theory lead to the same conclusions in most cases. Both predict a lapse rate of g/Cp; both predict hot cores for planets. It then falls to “Occam’s Razor”. Doug’s theory requires rewriting large chucks of fairly basic thermodynamics. Standard theory –well — agrees with all the standard thermodynamics. Doug has one fairly bright scientist from 100+ years ago coming to his conclusion. Standard theory has lots of fairly bright scientists from 100+ years ago to the present coming to a different conclusion.
Even with 100’s of posts on dozens of websites, Doug apparently has failed to convince even one PhD engineer or physicist of his position. I suspect that 100’s of more posts are in the future, with pretty much the same impact. (if I am wrong, I would love to see the name of any engineering or physics professor who will voice support for Doug’s hypothesis).
Josef Loschmidt was a brilliant physicist in the 19th Century – he was Maxwell’s teacher and the first to estimate closely the actual size of air molecules. He understood kinetic theory which is all about molecules and their motion, separation etc.
From his correct understanding (and no doubt high intelligence) he postulated the gravito-themal effect on which my hypothesis is based of course.
As BigWaveDave pointed out, this “Loschmidt effect” obviates need for concern about greenhouse gases. In other words, it potentially puts climatologists out of a job. And it diminishes the value of certain well known domain names, which is why Anthony Watts ran an article trying to disprove it. Sadly for all of you the experiments with centrifugal force prove Loschmidt was right, and so am I.
Bad luck. You picked the wrong vocation. Take my advice and pursue some other source of income before it all crumbles, as it will. Try searching “climate hoax” on youtube.com and see just how many are pointing out just how wrong it all is. Many of these videos have each had hundreds of thousands of views. In the not too distant future, public opinion will vote out governments who back the hoax.
As I said, bad luck. I actually feel sorry for you as I relax in my semi-retirement (at age 69) in my million-dollar mansion here in the perfect climate of Sydney.
Doug Cotton,
What is your source that “Josef Loschmidt was a brilliant physicist in the 19th Century – he was Maxwell’s teacher” I found no such information. Need a source for this claim. Not saying it is not correct but I did not find such.
Hate to really pop your bubble on the new centrifuge device. It is not even close to trillions of molecules. I went the source of your article.
http://arxiv.org/pdf/1311.7119v1.pdf
Look at page 4. They claim 80000 molecules/second for this device. This is exactly what I have been saying but you do not read or comprehend the material. A small number of molecules will lose energy as they move up a gravity field and gain it when returning. NO ONE disputes this!! There is no mean-free path surface developed in 80,000 molecules. If you let 80,000 molecules free on the moon’s surface (at room temp) and could track each one’s velocity they would slow down as they moved up the moon’s gravity and speed up as they moved down and since they would most likely never meet they would not exchange energy.
Check out page 9 of this article:
http://ruc.noaa.gov/AMB_Publications_bj/2009%20Schlatter_Atmospheric%20Composition%20and%20Vertical%20Structure_eae319MS-1.pdf
It will show you how mean-free path changes with pressure. When you have less chance of particles exchanging energy you no longer have the very small distance molecules can move before encountering another one.
Go and play snooker. Slope the table. Play one ball up the slope for 1 meter where it strikes another ball that then carries on another 1 meter up the slope. It might help you to realize that two balls each moving one meter with a collision in the middle, are just as much slowed down in total as would be one ball travelling two meters. Someone else could have hit another ball in parallel with your two but going the two meters distance. It would arrive at the same moment as your second ball. So density of the molecules does not affect the loss in mean kinetic energy because the gain in potential energy is only dependent on the total distance traveled, no matter how many or how few molecules are involved.
Doug you will not even attempt to understand what I am saying about a pseudo surface (acts like a real surface). If you have a couple of snooker balls your ideas work fine, if you have thousands of snooker balls and no ball is close to being able to move more than a couple of inches before hitting another ball and you have some sort of vibrating sides to supply the balls with constant energy you would have a better picture of what I am describing. Your ideas will always work with a couple of molecules as long as they have long mean free paths. Your theory fails when path length is very short and I have explained why. You could see if on your snooker table if you set it up. Look again at the atomic simulator. Now put gravity at max but also put molecular number at max and see how the situation changes when the molecules can no longer freely move.
http://www.falstad.com/gas/
I’m not good enough at snooker Norman to spin the ball so that it bounces back towards me after a collision. Apparently you can play such trick shots with your molecules, ’cause I don’t see it happening all the time in the graphic here.
Yes well in a planet’s “ideal” troposphere such as we are considering (with dry air) we don’t get to the stage where the molecules can hardly move in a gas and thus start to liquefy as the bonds that we ignore in Kinetic Theory start to be strong enough to hold them in clusters. They are huge distances apart relative to their radius.
What happens is explained here which you have never quoted and thus have never refuted. Is it not “normal science” to try to understand what the other guy is explaining before trying to refute it? I understand what you’re trying to describe, but it just doesn’t happen in the real troposphere, and I know I’m right about that because my hypothesis is confirmed by real data from various planets.
“Your ideas will always work with a couple of molecules as long as they have long mean free paths.”
Nope. I used actual mean free paths in the first two iterations with 2 then 4 and then 6 molecules which I then use mathematical induction to extend to the whole troposphere.
Think back to your coordinate geometry from school days. You still get the same gradient for a line no matter how big or small the interval. So, when equating PE loss with KE gain it does not matter what dH is because there will be proportional variation in dT.
So m*g*dH = -m*Cp*dT gives us dT/dH=-g/Cp no matter what the MFP. That’s why there is no loss of generality in assuming vertical motion, because, for other directions all that matters is the variation in vertical height. That’s why (PE+KE)=constant represents thermodynamic equilibrium because, with no unbalanced energy potentials, the KE of each of a pair of molecules just before colliding should be the same, at least if we simplify it to assuming all molecules in a horizontal plane have the same KE, as we can also do without loss of generality.
Norman still doesn’t get it right when he says the mean free path changes with pressure. It is related to density, not pressure. If you increase the temperature whilst maintaining the density (in a sealed container for example) the pressure will increase but the mean free path is the same because it is related to density, not pressure. Gravity forms the density gradient which stabilizes when thermodynamic equilibrium is attained.
Doug Cotton,
Here is a sample of what a REAL science article might look like. Not the extensive math involved. That is the language of a physicist (which you are not, just a Wikipedia armchair researcher).
http://www.pmi.ou.edu/Biot2005/papers/FILES/076.PDF
I do not pretend to be a physics major. I studied chemistry but your work is very poor for a college level researcher.
Like I stated, if you actually went to Sydney University and majored in physics and you write articles that you do, then something is very wrong with the program out there or you were sleeping in all the classes.
I think I would give you an A for your work if you were a High School student. It is good work for that level. I would consider you a lazy unmotivated College student with such simplistic and only conceptual descriptions of a physics theory.
Curt has you pegged.
Go back to this comment about Curt’s enormous mistake regarding the centrifuge machine.
Doug:
I have repeatedly pointed out to you that your examples of thermodynamically open systems say nothing about what happens in a thermodynamically isolated system, which is what your claims are about.
The fact that you cannot understand the importance of this distinction simply demonstrates your fundamental incompetence in thermodynamics.
Yes, Norman, but on page 4 in their (3.5)equation for the change in entropy they show no term for the gravitational potential energy, because they are using such equations of thermodynamic potentials which are themselves derived with the assumption that gravitational PE does not vary – that is, their development does not relate to a planet’s troposphere where gravitational PE does vary. Climatologists make the same mistake because they plunge into their computations expecting to magically get the right answers when in fact they have ignored the prerequisites for those equations to be applicable.
The understanding of physics has deteriorated in many ways since my days when Prof Julius Sumner Miller (and Harry Messel) encouraged us to ask “Why is it so?” My own students over the years got could results because I encouraged them to understand thermodynamics, not just use the equations incorrectly. To understand entropy, we need to understand energy potentials, as you may realize if you study the material at http://entropylaw.com and that’s why I write as I do, emphasizing what’s happening, rather than trying to impress the gullible with triple integral signs etc.
Sorry typo: My own students got good results
Doug Cotton,
“emphasizing what’s happening, rather than trying to impress the gullible with triple integral signs etc.”
Triple integrals are not put in to impress the gullible, you moron. They are added to give a precise understanding of a system. The goal of physics is precise language to convey exacting ideas. They do not use math for the sake of impressing. I think it would only impress the phony physics people incapable of following precise logic. The rest would appreciate the precise language used to describe a physical system an completely as possible. Triple integrals just means it is for a 3-D space.
You do not use complex math because you do not know how or if you do you are just plain lazy. To correctly understand energy potentials you need to mathematically describe all the variables and how they relate in the most precise manner possible. You will never break into the science world with high school papers. Grow up and be a man about it. Do the work! Quit the lazy minded excuses. I really believe you are incapable of complex mathematical derivation (just as I myself am…I am not going to pretend I am this brilliant genius that does not know how to write an adult physics paper and make excuses why I don’t do it). My math is simplistic algebra at a high school level and only gives ballpark results and answers and would not satisfy an MIT trained engineer like Curt or any other of the higher level physics majors who read Roy Spencer’s blog).
Curt asserted I said (regarding entropy) “You even admit that it would keep decreasing until the gases reached absolute zero.” For the record I said no such thing relating to entropy which can never decrease in an isolated system, as I have said many times and as the Second Law says. I was not talking about entropy at all. I was talking about kinetic energy decreasing if work was being extracted from the system. In that explanation I showed why Maxwell was wrong in assuming a perpetual supply of energy could be drawn from the system because, if we attempted to do so, the system would cool right down. What Maxwell was claiming would violate the Second Law. In any event, the centrifuge experiment here proves Curt and Maxwell both wrong, as does all the other evidence that Curt hasn’t even read on the “Evidence” page of our group’s website.
Norman referred to “the higher level physics majors who read Roy Spencer’s blog)” to which I say let them come forward with their attempted refutations. Most physics PhD’s don’t really understand entropy I find. In any event, all the standard equations for thermodynamic potentials which they use are developed with the assumption that gravitational potential energy does not change. But it does in what we are talking about. Fortunately there is a neat, clear, unambiguous mathematical way to equate PE loss with KE gain and get quite simply the well known result -g/Cp which climatologists take about three times as long to derive because they introduce pressure only to find it cancels out again as here due to their lack of understanding of Kinetic Theory from which was derived the Ideal Gas Law that they used.
Doug:
Maxwell’s thought experiment showed that if you accepted the gravito-thermal effect, you could turn thermal energy of an isolated system into work without transferring any thermal energy to an external reservoir.
You agreed with this, but simply pointed out that this would only continue until the gases reached absolute zero.
But this very process FROM THE START results in decreasing entropy for the system, as there is a -Q/T entropy decrease in the isolated system as the thermal energy is converted to work. But this is a BLATANT VIOLATION of the 2nd Law, because the entropy of an isolated system cannot decrease (as you have agreed elsewhere).
The notion that you cannot convert thermal energy to work AT ALL without rejecting heat to a lower-temperature reservoir is the VERY HEART of the 2nd Law, understood even before the concept of entropy was solidified, and well before statistical mechanics quantified and explained it fully.
The fact that you cannot see this immediately signals that you have no real conceptual understanding of the fundamental principles of thermodynamics.
At first I missed this comment of Curt’s which proves he does not have a clue about entropy, and continues to confuse it with energy. See my reply above. I’m still laughing.
Doug:
You obviously have never taken (or at least understood) an actual thermodynamics course, or you would understand the analytical distinction between reversible processes, which do not increase entropy, and irreversible processes, which do.
And the conversion between gravitation potential energy and vertical (organized) kinetic energy is the classic, TEXTBOOK example of a reversible process.
I even spelled it out for you in terms that a high school student would understand. Start your ball bearing in a gravitational field with an upward velocity at a certain height in a vacuum. As it goes up (PE increases) and slows down (KE decreases). It gets to the top of its trajectory, then starts down (PE decreases) and accelerates (KE increases). When it gets back to the original height (PEend = PEstart), it has the same velocity magnitude as when it started (KEend = KEstart).
Now take another first year physics example by hanging this ball on the end of a spring. It will exchange between gravitational PE, spring PE, and KE in reversible conversions, producing undamped oscillations that don’t produce entropy. I remember deriving the equations for these in high school physics!
Now if there is any friction, there will be irreversible conversions that increase entropy. But this is a separate issue, which anyone that understood their first thermodynamics course would realize.
You are the one who has no clue as to what entropy is!
Yes well starting your ball by throwing it upwards is hardly a natural adiabatic process.
If and only if there is thermodynamic equilibrium do we see no further increase in entropy, because that is what the Second Law is all about (maximizing entropy) and that is the fact of life that I have used in the paper, the content of which you have neither read, understood, discussed or refuted. That also is why the ball falls when you let go of it – without pushing or throwing it, by the way. During flight total energy stays roughly the same – especially in a vacuum, but entropy is increasing as it falls, because otherwise it would not have fallen in the first place. Nothing (measurable) happens by itself in nature in such an isolated system unless entropy is increasing.
Go to this comment.
Doug:
I have read your confused rantings numerous times and pointed out your errors to you for years.
One of my amusements is getting you to make ever more ridiculous assertions in desperate defense of your wacky ideas.
Open up any introductory thermodynamics textbook. In a very early chapter, there will be a section on the distinction between reversible and irreversible processes. Try hard to understand the difference, because presently you have absolutely no concept of the distinction!
Oh, and your assertion that the act of throwing a ball is non-adiabatic shows that you do not even begin to comprehend the difference between work and heat. This is another concept from the opening weeks of introductory thermodynamics. Is there any basic concept in thermodynamics that you DO understand???
(Throwing a ball involves performing work on the ball — there is not heat transfer involved.)
Ignore the throwing action then. Suppose the system is isolated just after the throw. The ball eventually bounces to a stop. It no longer has the kinetic energy it started with and entropy has obviously increased.
In fact, it is almost inconceivable that there could be a reversible process in an isolated system in the real world. Such reversible processes only happen between two or more closed systems, so the Second Law is inapplicable and one or more of those systems can indeed lose entropy as it transfers heat to another system.
You are completely confused regarding entropy not changing in reversible processes, because that is talking about what can happen in two closed systems (where heat can still be exchanged) and it usually relates to an object being warmed by the environment and then subsequently cooling to the same temperature by transferring heat back to the environment which is the other system.
That process has nothing to do with the Second Law which only applies to an isolated system, not to a closed one.
In an isolated system only autonomous processes take place and such will only occur if there are unbalanced energy potentials on a macro scale. So this does not apply to individual molecule motion which is not measurable and is “averaged out” on a macro scale.
Doug, you say: “Suppose the system is isolated just after the throw. The ball eventually bounces to a stop. It no longer has the kinetic energy it started with and entropy has obviously increased.”
As I have pointed out elsewhere, it is inelastic collisions of the ball that convert kinetic energy into thermal energy, and thus increase entropy. It is NOT the conversion of gravitational potential energy to vertical potential energy, which produces work (force acting over a distance) and therefore does not increase entropy.
The problem for you is that the collisions of gas molecules are completely elastic, and therefore not entropy increasing.
A ball already traveling upwards is gaining entropy because the force of gravity is applying the acceleration due to gravity to its mass. Just because it returns to the same spot does not mean it has not gained entropy, as it does do both on the way up and the way down because the same force is applying the same acceleration each way. By the time it returns it has a totally different velocity because its direction is the exact opposite. There’s nothing reversible about that process. Even if it bounces it will eventually come to a stop (in an isolated system) because heat and sound will be generated during each bounce.
Read http://physics.stackexchange.com/questions/50160/entropy-change-during-reversible-processes
Now, the linked item here says …
Therefore S (which, by convention, represents only the entropy of some particular system) can either increase or decrease. Since we’re talking about a reversible process, the entropy of some other system must change by an equal and opposite amount, in order to keep the total constant.
One other thing: in thermodynamics, “closed” and “isolated” mean different things. “Isolated” means neither heat nor matter can be exchanged with the environment, whereas “closed” means that matter cannot be exchanged, but heat can. In your question you say the second law “prohibits a decrease in the entropy of a closed system,” but actually this only applies to isolated systems, not closed ones. When we apply the equations above, we’re not talking about an isolated system, which is why its entropy is allowed to change.
Now, because the Second Law applies to isolated systems (not closed ones which can still receive or lose heat) it amounts to saying entropy will increase if there are unbalanced energy potentials. Such “energy” applies to all forms of energy, and so, when the ball is up the top there sure is an unbalanced energy potential because of its gravitational potential energy. The same applies for molecules and we find by analysis and mathematical induction from small number examples, that we only get no unbalanced energy potentials in isentropic conditions – implying homogeneous mean molecular (PE+KE) – and that is why there is a temperature gradient in that maximum entropy state.
All this you could have read here: http://climate-change-theory.com so stop wasting my time with your incorrect understanding of entropy similar to that of the person who asked that question in the linked page. The answer given (quoted above) was correct and well explained.
Doug:
I had forgotten this one of your ludicrous errors (there are so many that it is hard to keep track): that the 2nd Law applies only to isolated systems. You have stated this repeatedly over the years.
NO! The 2nd Law is universal! That is why it is called a law. It’s application simplifies for closed systems compared to open systems (no mass transfers) and for isolated systems compared to closed systems (no heat transfer), so fewer terms are used in the equations.
But unless you understand the universality of the 2nd Law, you will never be able to perform any competent thermodynamic analysis!
Add this to the list of Doug’s fundamental thermodynamic errors!
The Second Law says entropy in an isolated system cannot decrease and will increase in that isolated system while ever there are unbalanced energy potentials.
Do you have any argument with the above?
I quote again from the linked source …
In your question you say the second law “prohibits a decrease in the entropy of a closed system,” but actually this only applies to isolated systems, not closed ones.
Go and argue with that physicist, because I’m sick of wasting my time teaching you correct physics when you display no desire to learn. Besides, I normally charge for my time.
I mean that, Curt. Go and argue with him because I can see that he has a correct understanding, whereas you are like that high school student asking him badly worded questions.
Silent readers can judge for themselves.
Curt now claims the Second Law also applies for closed systems in which entropy can either increase or decrease as heat is exchanged with the surroundings. That would make the law totally pointless, because we already have equations for changes in entropy as heat enters or leaves a closed system. The Second Law would be quite unnecessary and not tell us anything new in such instances. You would not know what would happen autonomously in any particular closed system.
As you can read in the linked page above, that physicist is quite specific about the Second Law only applying to isolated systems. It is only for isolated systems that we can say entropy never decreases. And there is clear cut explanation in the following quote that we are talking about the law that says entropy only increases, and the application of the law to isolated systems is all I am interested in doing and all I use in the development in the paper linked in http://climate-change-theory.com which explains all observed planetary temperatures.
The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations — then so much the worse for Maxwell’s equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.
—Sir Arthur Stanley Eddington, The Nature of the Physical World (1927)
Curt seems to overlook the fact that my hypothesis does in fact enable determination of planetary temperatures (above and below any surface) and also explains the required energy transfers. No other published hypothesis anywhere in world literature can make that claim.
Doug:
You still do not understand the difference in thermodynamics between work and heat, and between reversible and irreversible processes.
The entropy of an isolated system cannot decrease. I agree. (Which is why Maxwell’s disproof of your gravito-thermal heat-creep mechanism is so devastating. He shows how it would lead to an isolated system with decreasing entropy. You even admit that it would keep decreasing until the gases reached absolute zero.)
If there are temperature differences within the isolated system that are not fully insulated from each other, then the heat transfer from higher temperature to lower temperature will increase the overall isolated system.
But this is different from the conversion of gravitational potential energy to vertical kinetic energy. That is the performance of thermodynamic work — force acting over a distance. (Remember that from introductory physics???) And the performance of work DOES NOT increase entropy! (You probably never understood that in introductory thermodynamics.)
Now, if the falling ball has an INELASTIC collision at the bottom, that collision will convert organized kinetic energy into disorganized thermal energy, increasing the entropy of the system. So the isolated system with the ball at rest at a low elevation will have higher entropy that system starting with the ball at rest at a higher elevation.
But here is where your conceptual weakness leads you to improper conclusions. When you are talking about collisions involving gas molecules, these collisions are completely elastic, and therefore not entropy producing.
And because you do not have the fundamentals down to know what processes create entropy and what do not, you cannot tell the difference between these two cases.
Doug:
Stop! You are just embarrassing yourself now. Please consult a good textbook rather than random webpages.
Any textbook will tell you that the 2nd Law is universal, with isolated systems as a simplified special case.
Fundamentally, you have for a system:
dS (greater than or equal to) Integral of (dQ / T)
where dS is the change of entropy of the system, and the right side of the equation is the summation of heat flows across the boundary of the system and T is the temperature at which a heat transfer occurs at the boundary.
For an isolated system, there are no heat transfers across the boundary, so the right hand side is 0, with the result that the entropy of an isolated system cannot decrease.
But in the more general case, the 2nd Law still tells us much about the entropy of non-isolated systems.
Of course, the universe as a whole is by definition an isolated system.
On work and entropy, one of my undergraduate thermo texts says this:
“The view that work is ‘entropy-free’ is really the only way one can intelligently distinguish between heat and work. Work is energy transfer without associated entropy transfer, and heat is energy transfer with associated entropy transfer.”
These are basic, basic concepts in thermodynamics, but you don’t begin to understand them!
Footnote: Sometimes the Second Law is expressed as stating that the sum of the entropy of participating systems always increases. In practice these system must make up a composite isolated system comprising dependent sub-systems. If one or more of those sub-systems were only closed to the surroundings, then the surroundings could transfer heat into or out of that sub-system. Hence the surroundings would then become a participating system. However, it would not normally be dependent. This is why we cannot say water could flow up a hillside into a lake at the top provided that it subsequently flowed down further on the other side so that total entropy had a net increase. The fallacy lies in the fact that the two systems are independent, not dependent as would be the case with the two sides of a siphon.
There is nothing wrong with what Sir Arthur Stanley Eddington wrote back in 1927 (as quoted above) but I’m the first to agree that climatology text books (notably Pierrehumbert’s) have serious errors in physics and, as Curt claims, obviously incorrect “extensions” of the Second Law which, by the way, I have studied very extensively in writing two peer-reviewed papers on the subject.
In any event, in my hypothesis I don’t need to extend it to closed systems. I can apply it to a dry gas in a sealed insulated cylinder for example, and over 850 experiments (linked from the website) have confirmed with statistical significance that such is not isothermal. I can also apply it to that centrifuge and see that it agrees with observations there. And it agrees with observations in all planetary tropospheres.
Anyway, I’m leaving it to Curt to argue with that other guy now, because he is not talking about what is in my hypothesis at http://climate-change-theory.com .
From the days of Clausius entropy has always related to all forms of energy. Curt’s stupid text book has no answer to what happens when work autonomously creates heat, or heat creates work. This is the epitome of the false fissics in climatology and the extent they will go to in changing the laws of physics to do all they can to argue against the gravito-thermal effect which they know threatens their pecuniary and political interests. I repeat, there is nothing wrong with what Eddington wrote in 1927.
Climatology writers (who have brainwashed our young Curt) have overturned all this physics from the 19th century in order to achieve the UN goal to destroy capitalism at any cost.
One final note:
I just read Curt’s claim that “For an isolated system, there are no heat transfers across the boundary, so the right hand side is 0, with the result that the entropy of an isolated system cannot decrease.”
Blimey, all you are doing is effectively equating entropy with thermal energy. Whatever thermal energy does (increase, decrease or stay the same) you want entropy to do the same in any isolated system. And since you separate thermal energy from other forms of energy or work in your calculations and have no term for such in that equation, and since thermal energy cannot change in an isolated system if there is no reason for it to do work, entropy can never increase.
So, with the RHS = 0 you are also claiming entropy cannot increase in an isolated system.
The entropy equation relates to heat flow in and out of a closed system, not to what happens in an isolated system where thermal energy may not have to change at all.
In an isolated system, because it is perfectly insulated, everything would just stay the same. For example, you could start with warmer air at one end of a sealed horizontal cylinder and, according to you, there is no law of physics which will tell you that anything at all would happen. How would you explain conduction down a metal rod or diffusion or convective heat transfer in that cylinder? You have equated entropy with thermal energy, so the whole system would just stay static with warmer air up one end because you have no reason and no law to assume anything else would happen. Sadly for you it would.
So one final question for Curt, who makes the same mistake as the high school student here.
Heat a metal rod at one end and, while the other end is still cold, insert it into a perfectly insulated tightly fitting cylinder and then seal that cylinder. What happens and why? Prove your answer with your entropy equation and/or your climatology version of the Second Law wherein (because RHS=0) entropy cannot change in this isolated system comprising the inside of the cylinder and the enclosed rod that is hotter at one end.
Well come on Curt. Are you stumped with the above question about the rod?
I can show why entropy increases in the rod, even though total energy does not change. I can do it logically or computationally.
(a) Logically is easy when you understand that the Second Law operating in an isolated system tells you entropy will increase until all unbalanced energy potentials have been dissipated. So, logically, we obviously started with unbalanced energy potentials with a higher temperature (more mean KE per molecule) at the hot end, so we expect uniform temperatures to evolve at about the mean of the original end temperatures.
(b) Now I can also show why entropy increases computationally, but can you when you claimed above that dS=0? It’s not particularly hard and you should be able to do it. I just want to make the point that you must consider internal entropy changes which do not involve changes in total energy, because that is how the temperature gradient evolves in a force field, be it gravity or centrifugal force – as now demonstrated with a centrifuge machine as well as the vortex tube.
I’m still waiting Curt for your acknowledgement of your error regarding the centrifuge machine as discussed here.
Doug:
Good god, man! Do you have the IQ of a turnip???
I very carefully stated the equation/inequality for the 2nd Law above as:
dS (greater than or equal to) Integral of (dQ / T)
and pointed for for an isolated system, the right hand side is zero.
I also very clearly stated above that: “If there are temperature differences within the isolated system that are not fully insulated from each other, then the heat transfer from higher temperature to lower temperature will increase the overall isolated system [entropy].”
So for you to claim that I asserted that “entropy cannot increase in an isolated system” is an outright fabrication. You can’t even follow a simple logical argument!
I specifically stated that heat would flow from the hotter to colder parts of an isolated system, yet you claim I asserted the opposite!
By the way, that equation/inequality (do you even understand what “greater than” means?) is one of the most basic in all of thermodynamics, and you seem utterly unfamiliar with it!
Oh, and the textbook I quoted on the distinction between work and heat is an engineering thermodynamics text written in the 1960s. For you to claim that this was corrupted by climate science in the 1990s is just anothe example of how you are, as the kids would say, “reality challenged”.
Most of the book concerns the various types of conversions between work and heat, so for you to assert on the basis of one short quote that the “book has no answer to what happens when work autonomously creates heat, or heat creates work” is just another example of your logical impairment.
Work done on a whole system at the macro level is quite different from work done at the micro level by the force of gravity acting on molecules in flight between collisions. The work done by the engines of a plane does not have an impact on the internal temperature which (even without air conditioning) does not cool as much as the outside air just because the plane is climbing.
But there is no reason to disregard the effect of gravity upon those molecules, as Loschmidt realized in the 19th century and many PhD’s have overlooked ever since. This is an area of physics which is not covered in typical engineering courses, as I have discussed with friends who have done such and yet can now see that my hypothesis is correct. The main reason you keep disagreeing is that you have not yet read, studied and inwardly digested what is in the paper I wrote two years ago. If you had you would know that your statement about heat transfers always being from hot to cold is refuted with a correct understanding of entropy maximization, as explained in that paper.
The Clausius corollary of the Second Law only applies in a horizontal plane, as I have said many times, and proved in the paper, website and book.
And regarding your helium Venus, see this comment.
Massimo and those who want to learn what’s really happening, just read, study and inwardly digest all the information at http://climate-change-theory.com
Bye.
Roy and silent readers:
I told you Curt does not understand entropy. He jumps in and applies the entropy equation relating to heat transfers between two closed systems without realizing that it gives the overall change in entropy if and only if there is no other simultaneous process causing internal change in entropy within either system.
In other words, a prerequisite is that both systems have already reached thermodynamic equilibrium internally. That’s why he will not be able to explain what happens internally with the conduction down that metal rod in a sealed insulated cylinder in that isolated system. The entropy equation he quoted is totally irrelevant to what happens internally within an isolated system which starts out with unbalanced energy potentials, such as our rod that is hotter at one end.
To work out what happens internally within an isolated system that’s where you need the 19th century Second Law which is still the law in mainstream physics that has not been manipulated by the AGW cult to “prove” their claims, which totally ignore the valid version of the Second Law. I explained all this three years ago in my first peer-reviewed paper on the Second Law published on several websites in 2012.
To work out what happens within an isolated system (that, by definition, has no heat or matter transfers through its boundaries, and which is what the Second Law applies to) you have to recognize that entropy increases until the point where all unbalanced energy potentials are dissipated. It is well explained at http://entropylaw.com and that uses the same concept as in my 2013 paper here.
Footnote: Yes even Wikipedia confirms my second paragraph when they explain it here: Each system, by definition, must have its own absolute temperature applicable within all areas in each respective system in order to calculate the entropy transfer.”
There you go …
I would like to suggest that entropy could be defined as “a state function which increases autonomously in an isolated system if and only if there remain unbalanced energy potentials” because this embraces the close association with the Second Law of Thermodynamics and the state of maximum entropy (namely thermodynamic equilibrium) in which all unbalanced energy potentials have dissipated. Noting that such energy potentials include gravitational potential energy, we observe thermodynamic equilibrium when there are isentropic conditions in, for example, a planetary troposphere. Such conditions, in order to have no unbalanced energy potentials, would have a homogeneous sum of mean molecular kinetic energy and gravitational potential energy, and thus have opposite gradients in potential energy and kinetic energy, the latter being represented by temperature.
Curt,
Holy Cow! Doug can’t think or reason. I am reading through your posts. They are easy to follow and understand and he still gets them wrong and says you are making claims you are not. How he reasons is the greatest mystery of all on this thread. If any could figure this out!
You state dS>=dQ/t. You state energy cannot enter or leave an isolated system (by definition of an isolated system) therefore dQ=0 and the Right hand term will be zero which then becomes the change in entropy is > or equal to zero. You then state simply in an isolated system the entropy cannot DECREASE. It can still increase of not change. Then Doug launches into some bizzre attack on this most simple logic making the claim you can’t answer what happens if a rod heated at one end is then isolated. You clearly show the entropy can increase. Once all the energy is evened out there will be on change in entropy.
Amazing he can’t understand what you mean by a reversible process. An ideal elastic ball in a vacuum chamber will not change entropy as it moves up and down a gravity field. To change the entropy of this isolated system one would have to take energy away from the ball (opening the system up and no longer isolated) which would increase entropy for the system. Or add energy which would then decrease entropy.
Wow is all I can say. Wow!
Norman 10:51am: +1
Many posters have observed Doug long ago ceased citing correct fundamental eqn.s and principles in supporting a mere political view. It is sometimes humorous and a little fun to respond to his “new paradigm” posts when IMO there is little valid science value in them. It is absurd to view his AU$5,000 offer as possible since he cites his own non-experimental work in protecting from payout & ignoring a proper existing literature search response. You can reward yourself sufficiently by learning from any fumbles of the basic science in modern text books and existing relevant literature.
Ball4,
Thanks for the information. Curt has me convinced that Doug Cotton did not take University level physics classes. He knows nothing about thermodyanics and scrambles so much stuff it has to be literal pain to Curt and you.
I am completely convinced Doug is unable to derive his theory in mathematical terms. He argues that he can but thinks physics papers use math to impress readers. This might come close to the purest form of stupidity I have read. There are dirty forms of stupidity that might get some stuff correct, this is so pure it makes one applaud Doug Cotton’s reasoning. I think it is a rare gem to see such purified stupidity!
Math is used in physics because of it precise and calculateable nature. Stating “heat creep” exists and based upon the gravitation of a planet then stating it can be overcome by wind and other factors would require a precise detailed math to describe how the various variables will affect the “heat creep” so one can calculate the outcome.
Yes I have provided all the computations necessary to determine the temperature gradient, and the surface temperatures then follow from simple geometry. The computations using Stefan Boltzmann do not need to be presented, as there are on line calculators for such.
You can empirically measure the speed of heat creep in the “car in garage” example here (which is an extreme case) and compare it with typical wind speed. The diffusion and convective heat transfer process is obviously far slower than wind of any form. Anyone with a convection heater on one side of a room knows intuitively that it takes longer for the heat to reach the other side than it would for a cool breeze coming in a window on one side and out a window on the other side to affect temperatures.
I have successfully explained temperatures and associated necessary heat transfers for all planets in a way that no-one else in the world has done. Link me to any such research which you think correctly explains Uranus tropospheric temperatures and calculates the temperature at the base of that nominal troposphere to within a mere 9 degrees – as I did in computations that have been posted years ago on climate blogs for example.
Then spend half a day doing your own study of the correlation between temperature and precipitation, following the methodology in the Appendix of the paper and the book. Prove the IPCC right in their claim that water vapor of less than 2% raises the surface temperature about 30C° so regions with 4% should be at least 45° hotter than regions with 1% water vapor. I’m waiting.
Norman, the entropy equation (as here does not have a > sign. There could only be a greater entropy change if there were also simultaneous entropy increases within either of the two closed systems between which the energy is being transferred. That is why it is specifically stated that the two systems must be already in thermodynamic equilibrium within themselves and not be gaining or losing energy from a third system.
If you apply the entropy equation within an isolated system (as we were doing with the rod) you cannot just assume there is an increase in entropy because of a > sign. As I have outlined in an earlier comment (in which I supplied the computations which you failed to do when asked) I most certainly used the = sign, so study that comment.
Doug:
You will notice that the equation you cite uses “Qrev”, not just “Q”.
Come back when you understand the difference between these two. (It is a very basic concept in thermodynamics.)
Norman and Curt:
You cannot use the > sign in the entropy equation when considering an isolated system and what happens entirely within that system. That is why the equal sign is used in teh standard equation for entropy change. In assuming you can use a > sign you display a lack of understanding of the thermodynamics involved.
Curt failed to provide the computations I asked for which I have now provided about half way down this comment.
Doug:
Every time I think you cannot possibly get stupider, you do! Of course it is possible for the entropy of an isolated system to increase (if it is not already at maximum possible entropy), which is why every textbook I have ever seen uses the inequality!
A trivial example will suffice. You have a horizontal cylinder of gas isolated from the rest of the universe. To start, the two halves are thermodynamically isolated from each other, with the left half at 300K and the right half at 200K. Now the barrier is removed, and heat transfer starts from the left half to the right.
This increases the entropy of the system, as the entropy decrease of the left half, which starts at -Q/300, is less than the entropy increase of the right half, which starts at +Q/200.
This is one of the simplest problems you can have in an introductory thermo class. I am starting to think that Norman is correct that you have not ever actually formally studied thermodynamics.
When you get this trivial introductory problem COMPLETELY wrong, it is completely obvious to any observer that you are totally incapable of overturning the established paradigms of the science.
Yes that’s what I wrote here about three hours before you wrote this comment.
And you did not use you first argument that relied on a greater than (>) sign in the entropy equation. Instead, as I pointed out, you cannot use such a > sign in this situation, and indeed you did not in your calculations which follow the pattern of mine and are, of course, very standard ones you can even read in the Wikipedia Entropy article.
My whole point was to ensure that you understand that heat transfers are activated by unbalanced energy potentials which are always what is needed to start an isolated system moving towards the state of maximum entropy, which is thermodynamic equilibrium. So, when internal energy U (which is a thermodynamic potential) is greater in one region of an isolated system than in another, then there will be a propensity to dissipate that difference, and that is exactly what is happening in the “heat creep” process which I suspect you must be starting to understand. But stop wasting my time and read the detailed explanation and study the diagrams here.
Doug Cotton,
You might just want to admit you never studied any thermodynamics at the University level and have only taken it up as a retirement hobby by reading Wikipedia articles in your free time.
Curt knows what he is talking about and before I called him wrong I would really want to back it up with extensive research. I would give him the benefit of doubt that he took some very difficult courses at MIT.
Curt is 100% with the > sign. The Wikipedia article is only the GENERAL description of entropy. In an isolated or closed system the
>= sign is used and for exactly the reason stated by Curt. No energy can leave or enter a closed or isolated system so the entropy of this type of system can only increase (if it is isolated before all energy is balanced or at equilibrium state) or remain zero.
Check this out: Page 2. Read the description and look at the equation. Just what Curt used.
http://www.sfu.ca/~mbahrami/ENSC%20388/Notes/Entropy.pdf
Yet another laughable unforced error by Doug! (Is anyone keeping count?)
Internal energy U is NOT a thermodynamic potential. It is an extensive property. The intensive property of temperature T is the thermodynamic potential term for heat transfer.
If you have a mass 1*M of a substance at 300K and a mass 10*M of the same substance at 150K, the second part will have 5 times the internal energy of the first (with constant capacitance), but thermal energy will be transferred from the first part to the second.
Oh, and if you look at what goes into the internal energy U, you will see that it does not include terms like “organized” kinetic energy (wind, for a gas), gravitational potential energy, or chemical potential energy.
Well go and edit Wikipedia to bring it in line with “every text book” because, the equation dS = integral(dQ/T) is there in black and white.
You imply use of an = sign when you wrote “which starts at -Q/300, is less than the entropy increase …”
How would you know that? According to your > sign, Q/300 could be greater and perhaps exceed the entropy gain in the other half? You just don’t think>/i> young Curt and you could learn so much from someone like me who is “older and wiser” and who has specialized in (postgraduate) study of the Second Law of Thermodynamics and written two peer-reviewed papers on such.
Now start studying the website and the paper on Planetary Core and Surface Temperatures and stop wasting my time with questions you would not have to ask if and when you understand the hypothesis, which you don’t yet. Another useful site is http://entropylaw.com where you can learn of developments in the understanding of entropy and the Second Law since 1988.
Sorry about the failure to cancel italic – too much typing of that > sign (LOL)
Doug:
You continue to cite results of non-equilibrium open thermodynamic systems (your centrifuge examples) to argue about the equilibrium state of isolated thermodynamic systems.
This has been pointed out to you many times by multiple people, and you continue to ignore it. At this point, I can only conclude that you do not have the intellectual capability to understand the difference and its significance.
You also continue to insist that people who disagree with you must not have read your arguments.
NEWS FLASH! You are not such a brilliant writer that everyone who reads your arguments will instantly agree with you. More reality-disconnect issues…
It has nothing to do with whether thermodynamic equilibrium is actually reached or not. It has only to do with the direction temperatures take as entropy increases and unbalanced energy potentials start to dissipate.
As I thought, you don’t begin to understand the significance of the difference. In the dynamic case, you need to take into account all sorts of transient issues of conductivity and capacitances, and relative rates of different types of transfers, none of which apply to the equilibrium case.
Speaking of which, I asked you on a previous thread about the dynamics of your proposed “heat creep” mechanism, you ignored the question. So I will ask you again:
For “normal” conduction (which you at least seem to agree is valid in the horizontal direction), a substance has a conductivity property in Watts per meter per Kelvin (of difference). What is the equivalent parameter (with units) for your heat creep? What is the numerical value of this parameter for air?
Obviously the rate of heat transfer depends on the temperature difference above that which would be the case for the normal temperature gradient if the solid or gas is vertical in a gravitational field.
In the case of a planet’s troposphere the rates of entropy maximization are slow, but that does not prevent the general propensity towards such, and there has been the life of the planet in which the equilibrium density and temperature gradients formed. What is far more important to understand is the direction of heat transfer needed to maximize entropy and thus establish a new state of thermodynamic equilibrium. If there has been such a state, then new energy sends heat transfers in all accessible directions over the sloping thermal plane (because that plane represent thermodynamic equilibrium) rather like new rain water falling just in the center of a lake. All this is in my paper, so stop wasting my time.
Regarding the effect of centrifugal force in the centrifuge and the vortex tube, I find your wishy washy “argument” far from convincing, when it is obvious from observation that there has been a considerable change in temperature. Furthermore, the temperature gradient in the radius of the vortex tube agrees well with the quotient of the acceleration due to that centrifugal force and the specific heat of the gas. And furthermore, I have proven with The Second Law of Thermodynamics why this is so.
Doug Cotton,
“In the case of a planet’s troposphere the rates of entropy maximization are slow”
This is why you need elaborate mathematical expressions for your hypothesis (as with very few experimental evidence which have alternate explanations different from your own it will remain a hypothesis until overwhelming experimental evidence demonstrates it a reality. Since no one is excited to do experiments anymore then I guess we will not have a body of evidence to either reject your hypothesis or make it a valid theory with MANY experiments to back it up).
Without math you have no predictability. You have lots of wiggle room to explain anything right or wrong which makes it sound good to you but really does not determine anything.
With the Venus atmosphere you are in some ways agreeing with the GHE people. You state: ” So radiation from the surface warms the lower troposphere and some of that energy then returns to the surface by heat creep, as happens also on Earth and all planets.”
You are still accepting that CO2 absorbs outgoing IR and returns energy to the surface. The GHE hypothesis it that radiant energy returns and yours is that “heat creep” does it both are conceptually similar in that outgoing radiation is absorbed by CO2 and some form of energy returns and heats the surface.
But this is why math is so very important. You did not say all the radiation is absorbed so how much is? You need an equation that makes it calculable so it can be verified or rejected. Radiation loss is very rapid if nothing stops the flow. Heat creep is very slow. The surface is radiating very fast but can only receive heat slowly by heat creep. Now we need equations for the rate of heat creep and the rate of radiation loss.
Whereas I do not have enough evidence to reject your heat creep hypothesis (though I do not believe it currently) it would seem the GHE theory has slightly more validity than heat creep. It is empirical evidence that something it happening to Venus surface radiant energy and it is not leaving the planet. The GHE radiation redirection would be just as fast as the radiant energy leaving the surface. The Carbon dioxide or sulfur dioxide clouds would return the energy to the surface very rapidly as on remits and the other reflects. Heat creep would be a very slow process and I do not know how 16000 watts leaving could be replaced by a slow process that can only be worth so much energy per distance.
Need math Doug, I can calculate energy and see if your hypothesis has even a logic possibility of being workable. Thanks. If you write another paper detail all the effects in math equations so people can play with the idea.
Doug:
Once again you ignore the question. You say: “Obviously the rate of heat transfer depends on the temperature difference above that which would be the case for the normal temperature gradient if the solid or gas is vertical in a gravitational field.”
So obviously you have no idea. Look, I even primed you with the example of the standard conductivity parameter for a material, and gave you the units: Watts per meter per Kelvin. For air, the value of this parameter at STP is about 0.025 W/(m K).
Note the “Kelvin” term in the denominator. This gets multiplied by a temperature difference in K to compute the heat transfer. This is such basic, basic stuff, but you cannot understand it!
Since it is evident that you have never solved any problems of this type before, I will give you a very simple one and show you how to solve it:
You have a (horizontal, but it doesn’t really matter) sealed cylinder of air with a 0.1 m^2 cross section. Initially, it has a 5K per meter temperature gradient. What is the starting rate of heat transfer?
Answer:
Q = k * A * TempGradient
Q = 0.025 W/(m*K) * 0.1 m^2 * 5 K/m = 0.0125 W
So I ask you again, what is the equivalent “conductivity” parameter for your heat creep. What are its units?
If this same cylinder of air were vertical in earth’s gravitational field and starting in isothermal conditions, what would be the downward rate of heat transfer?
And your argument, Curtm about the system being “open” is pathetic, because the relative heat transfers to other systems are insignificant compared with what is happening within the centrifuge and the vortex tube. Each exhibits the effect of centrifugal force producing a temperature gradient which you cannot deny. You really are clutching at straws now because I have you on the ropes. And that is not surprising because the physics I present is correct.
Doug:
Yet again you demonstrate your abysmal ignorance of the most basic concepts in thermodynamics! Have you no shame?
Your vortex tube is the archetypal open system, with huge mass transfers in and out. (Put this on your study guide for the next thermo quiz: what makes a thermodynamic system “open” is mass transfers across the boundary.)
You have air forced into the tube at high pressure and velocity, and release to ambient at low pressure. Some of the air is dynamically heated; other parts are dynamically cooled. Because of the high-energy input, all of this is happening far faster than any conduction or even possible “heat creep” mechanisms can do anything significant.
Because of the highly open nature of this system, it says NOTHING about the behavior of an isolated system. By contrast an ultra-high-speed centrifuge or flywheel spinning in a vacuum is virtually perfectly isolated. You claim these systems would melt down very quickly. But they don’t! Why?
Doug,
If you respond to Curt maybe you might want to address a question I had asked that you did not answer. The surface of Venus emits 16000 watts/meter^2 of radiant energy (give or take). The TOA is quite cold and does not show this energy. What happens to it? Where does it go?
Also how does the 10 kilometer isothermal lower stratosphere exist with “heat creep”? Since the molecules at the 20 kilometer level have the same kinetic energy as those at the 10 kilometer level, what happens to the potenial enegry at top? Would the moleucules at the top have significant more potential energy than those at the bottom. Why does not energy exchange according your theory and create a lapse rate here?
Already answered here and very specifically here. Because of the excessive absorption rate in the stratosphere and its low density, the addition of new energy completely overpowers the much slower diffusion process that forms the gradient in the troposphere. Because the stratosphere is hotter than the tropopause (which has a curved temperature plot on most planets, not a linear one) there is obviously heat flow downwards into the tropopause region in the daytime as well as heat flow upwards from the troposphere at night and sometimes by day. If you can’t understand why the temperature gradient gradually changes from negative in the troposphere to positive in the lower stratosphere (below the maximum there) then I’m sure all the silent readers can.
Regarding the Venus surface, by far the majority of the inward thermal energy flow is by heat creep, as I have explained. So radiation from the surface warms the lower troposphere and some of that energy then returns to the surface by heat creep, as happens also on Earth and all planets.
The very fact that energy seems not to balance at the Venus surface is because of the oversight of this heat creep, so Venus provides substantial evidence for my hypothesis.
Now you might like to think about Uranus as I asked you in several previous comments but which you never answered correctly with any valid explanation based on physics and known data. Nor has anyone else in the world been able to answer if they neglect the downward diffusion and convective heat flows that must occur in all planetary tropospheres.
Regarding Uranus, go to this comment.
Curt has not shown the calculations for my question above. He incorrectly states “heat would flow from the hotter to colder parts of an isolated system” whereas the issue is what happens in a gravitational field in a planet’s troposphere where heat does not always flow from “hotter to colder” and in fact does not flow either way when there is thermodynamic equilibrium with its associated temperature gradient.
It is total unbalanced energy potentials which have to be considered as in my comment in italics above that I wrote in the Talk page on Wikipedia. When you include a term for molecular gravitational potential energy (which you must when it varies) then you find it is possible for convective heat transfer to “flow” from colder to hotter regions provided that entropy is increasing, which it will be if there remain unbalanced energy potentials.
There are such unbalanced energy potentials in an isolated isothermal column in an ideal atmosphere because molecular gravitational potential energy is greater at the top and yet mean molecular kinetic energy is not.
So the combination of (PE+KE) represents an unbalanced energy potential at the top. So maximum entropy is only reached when there is homogeneous (PE+KE) and that means a stable density gradient will evolve because more molecules will have a downward component of motion than will have an upward component. This happens because gravity curves the paths of molecules in any flight path that is not directly up or down. Gravity also accelerates molecules with its downward force, so there end up being molecules with greater KE at the bottom and less KE at the top when entropy reaches its maximum and we have a stable density gradient and a stable temperature gradient. And that’s why the centrifuge machine here produces 1°K gas at its center.
So the above is the “logical” explanation of what happens as entropy increases in a vertical column until thermodynamic equilibrium (the state of maximum entropy) evolves.
Computationally you imagine the rod divided in half and do two integrations. The top half is losing entropy and the bottom half gaining. Because the temperatures are higher in the top half, and T is in the denominator, the loss of entropy is less than the gain in the bottom half, so there is an overall gain in entropy until temperatures are equal.
But you have ignored the effect of gravitational potential energy changes at the molecular level which are more obvious in a gas, but do happen in solids as well because of minor net movement of molecules. So of course you get an incorrect result if you use only the “standard” equation for entropy which, in its derivation, specifically assumes gravitational potential energy is ignored, as explained here.
To get the right result computationally you have to express the change in potential energy in terms of thermal energy so that you can add that change to the change due to actual temperature changes. It is easier to use Kinetic Theory to do this, as I have in the paper linked to our group’s website. When you do so, you find the temperature gradient to be -g/Cp in the absence of radiation. Radiation then has a temperature leveling effect but it does not overpower the gravity effect in a planet’s troposphere. So water vapor reduces the magnitude of the gradient by up to about a third, thus rotating the thermal plot downwards at the surface end. All this is in the paper which Curt refuses to read, study and inwardly digest – to his loss.
As none of the perpetual commenters here have read, studied and inwardly digested my hypothesis, they are not qualified in my view to discuss same. Hence I will ask silent readers to come in on this if they have understood the paper. Those who have not will be ignored unless they show genuine interest and quote the text which they question. They should also include links that support any contrary statements they make.
Doug:
When will you get it through that thick skull of yours that we have read and digested your hypothesis, and rejected it as ridiculous, the product of someone who gets the most basic concepts of thermodynamics completely wrong, and not that of someone with new scientific insights.
Your hypothesis fails completely on both theoretical grounds, as it would lead to the entropy decrease of an isolated system, as Maxwell demonstrated 140 years ago, and on experimental grounds, as the successful operation of ultra-high-speed centrifuges and flywheels in a vacuum demonstrates.
Is there any silent reader who has studied my hypothesis here who thinks he/she can quote something from the February 2013 paper and then cite valid physics (with links) that refutes it? Does anyone know of a similar study of temperature and precipitation correlation which shows opposite results? Again, please link me to such a study.
My reply on the Wikipedia Talk page may help your understanding of the hypothesis …
Firstly “unbalanced energy potentials” are what drive the process of maximizing entropy that is described in the Second Law of Thermodynamics. If you read [[thermodynamic potentials]] you will see that they specifically ignore changes in gravitational potential energy because they assume that such are insignificant in an “engine” which they are discussing. Of course they are not insignificant in a planet’s troposphere. Then you incorrectly assume that I am talking about the macro potential energy associated with a whole object. I am not. I am talking about the interchange of molecular gravitational potential energy and molecular kinetic energy during the free path motion of each and every molecule between collisions. The Second Law is all about the fact that there is an autonomous propensity to increase entropy but only whilst there are still unbalanced energy potentials. When thermodynamic equilibrium is attained that is the maximum entropy state that we have. At the molecular level it means that any pair of molecules about to collide have the same kinetic energy on average at that height. Otherwise there would still be heat transfers by conduction or diffusion. But because molecules gain KE in downward motion and lose it in upward motion, we can deduce that PE loss = -KE gain which amounts to m*g*dH = m*Cp*dT and so we get the temperature gradient at thermodynamic equilibrium as dT/dH = -g/Cp. This temperature gradient is seen in all planetary tropospheres and even in Earth’s outer crust for example. It is overridden where there is wind or excessive heat absorption (as in the stratosphere) or excessive inter-molecular radiation as in water. But the inter-molecular radiation between water vapor molecules does have a temperature leveling effect that reduces the magnitude of the gradient by up to about a third, as also happens on Venus (due to CO2) and by about 5% to 10% in the nominal tropopshere of Uranus. The temperature gradient also evolves due to centrifugal force in a centrifuge and vortex tubes which rely on the same principle. There is overwhelming evidence of it everywhere, and over 850 experiments with sealed insulated cylinders have confirmed that isothermal conditions are not the state of thermodynamic equilibrium, simply because molecules have more PE at the top and yet the same KE. If you don’t consider such PE in entropy calculations then you have no explanation as to why anything falls anywhere under gravity, thus increasing entropy.
Chjoaygame had no valid response to the above comment I wrote in the Wikipedia talk page and he concluded as below, to which I replied as under …
I do not wish to discuss it further.
Chjoaygame (talk) 01:07, 14 March 2015 (UTC)
If that’s the best argument you can put forward whilst leaving all the other matters without refutation, I rest my case. You have no evidence that water vapor raises the surface temperature, let alone by about 15 degrees for each 1%, making rain forests 45 degrees hotter than deserts. You have no explanation as to how the required thermal energy gets to the base of the Uranus troposphere to make it hotter than Earth there and you have no counter arguments that in any way refute the fact that the state of thermodynamic equilibrium has a stable density gradient and temperature gradient. The definition I suggested for entropy is spot on, as is the (refined) first statement in my second comment above. This is not the place to discuss the climate debate, although the state of maximum entropy is very relevant. You can discuss such on Roy Spencer’s latest monthly temperature data thread, preferably after studying the new website linked therein and endorsed by our group of persons suitably qualified in physics.
Curt, Norman, Ball4 and anyone who wishes to challenge my hypothesis, note what I have just written in this comment above.
Doug 4:15am: “The formula for conduction rates is totally inapplicable…”
Which formula?
“If the temperature gradient equals that for the state of thermodynamic equilibrium..”
The universe entropy of earth’s troposphere increased between the time you wrote that and I write this so there is no state of thermodynamic equilibrium found. I have repeatedly & precisely shown (4:04pm) where your physics are wrong in your 100m cylindrical column & small bar by referring you to the published correct methods already in the literature since the 1890s and also shown precisely where your temperature and precipitation regional calculations go wrong.
Chew on this found from quick google search (which Doug could have easily done) using the 1976 committee determined avg. mid-latitude experimental lapse. Critically compare your incorrect method as previously noted on this blog when I showed you the right temperature & precipitation method for two city sized regions based on weather records.
http://www.engr.colostate.edu/~ramirez/ce_old/classes/cive322-Ramirez/CE322_Web/PrecipitableWaterExample.html
And in a nutshell, this is why it is pointless reasoning with Doug — and in his own words no less!
“Curt, Norman, Ball4 and anyone who wishes to challenge my hypothesis, note what I have just written in this comment above.”
Doug has a *hypothesis*. But it is a waste of his time to actually test the hypothesis with calculations — a waste to even consider alternate hypotheses. EVEN IF the calculations showed he was wrong, we cannot expect him to accept the results because his *hypothesis* says something contrary.
[And now I expect Doug to add 5 more posts — all of which will ultimately just serve to confirm this comment.]
Tim Folkerts,
You were most correct about Doug’s behavior.
Doug Cotton,
In the post above about Uranus.
“And if the Uranus core cooled by even just 300 degrees then the base of the nominal troposphere would cool from 320°K to 20°K and, if the Sun were still radiating as at present, that is an impossible situation since the troposphere gradient would be nothing like what we calculate it to be using -g/Cp and the temperature for radiative balance with the Sun is (58±3)°K.”
Why would this be the case? Why would the core cooling from 5000 kelvin to 4700 kelvin automatically reduce the nominal troposphere temperature from 320 K to 20 K (also you do not use the degree mark for K http://www.physlink.com/Education/AskExperts/ae129.cfm)?
Would a boiling pot of water cool 100 C if you lowered the temperature of the hot plate 100 C? If your hot plate was 500 C are you saying that if you turned it down to 400 C the water would stop boiling and would freeze solid? Your logic on this post makes so little sense I can’t follow how you arrived at this conclusion.
You would know the answer if you read, studied, inwardly digested and understood my hypothesis at http://climate-change-theory.com noting in particular the discussion of temperature gradients in sub-surface regions of planets and satellite moons.
Doug you need to change your name to Dodge Doug. You dodge anything that you can’t expalin. Reading your material will not explain in any way shape or form why if the core of Uranus goes down 300 K it would reduce the temperature of the nominal atmosphere by 300 K. You just make things up and when confronted you just say read my material. I have read your material it does not answer the question. You just dodge it with this silly post.
Doug Cotton,
Your thinking is a complete reversal of the conventional thought process on planetary atmospheres. I will not say you are wrong but I will say that if you understand the “accepted view” of atmospheres
it explains just as well why Uranus has a warm nominal troposphere and why it cools as it goes up based upon the gravitational constant of the planet.
The core of Uranus is hot (5000 K). This energy is moving outward from the surface of the core. In the gas just above the core convection would act as the primary heat mover (all gases and liquids support convection while solids do not). Convection exists in both liquids and gases in a gravity field because of hydrostatic equilibrium that develops in a gravity field. The gas will convect, as it rises it will push against its surroundings and cool leading to a lapse rate equal to the gravitational field as this is what develops the hydrostatic conditions (higher pressure density at the bottom with less pressure and density as you move up in a gravity field).
Figure 1 of this article:
http://www.aanda.org/articles/aa/pdf/2005/03/aa1695.pdf
The heat from the core initiates convection in the gas above and you get the standard lapse rate because the rising air cools as it expands. Sinking air warms, picks up more heat and rises again in convective circulation (which exists in tropospheres of planets with gas atmosphere). The upper layers of the atmosphere warm from solar radiation and create an inversion which stops the convective process and the atmosphere develops an isothermal state in the gravity field because there is no convection. The gas is no longer in vertical motion and the energy of the entire layer reaches the equilibrium state of isothermal conditions in such a gravity field and now the energy motion is only by conduction which is a very slow energy transfer in a gas so the core maintains its heat value for a very long time.
There are two ways to explain what is going on. You choose to blind yourself to the opposing opposite view even though it can completely explain the phenomena we see without “heat creep”.
I’m the first to agree that “my thinking” is in fact based on a whole different paradigm, as I wrote in this article about the 21st century new paradigm.
You have not addressed the issue of the “coincidences” in the temperatures for Uranus (and all other planets and moons) so go back to my comment on such.
The core of Uranus is not cooling off for the reasons explained in my hypothesis.
What I “choose” to do is based on the Second Law of Thermodynamics which you “choose” not to discuss because you have no understanding of this very specialized domain in thermodynamics in which I have done extensive post-graduate study and about which I have written two comprehensive peer-reviewed papers which members of our group (all suitably qualified in physics) agree are correct. You, young Norman, are way out of your depth arguing with a 69 year-old highly intelligent person like me who has been involved with physics most of his life after completing his first nine years of university education.
Doug Cotton,
If you could reason at all (which is apparent this ability is one you do not possess) you would have read my address to “coincidences” in temperature. Gases in convection rise. As they rise they expand. Expanding gas pushing against the mass of other gas molecules causes cooling because the molecules do work in expanding.
Gravity sets the pressure gradient and determines the pressure gradient which is what determines the expansion rate which determines the cooling rate which is based upon the gravitational field.
Doug Cotton,
You have been repeating your same ideas in several thousand posts on many blogs for years and it has changed nothing in the physics world. If you want to change established understanding it is up to YOU and no one else to come up with convincing experiments to prove a gravitational gradient will form in an isolated column and remain like this in equilibrium state (after all available energy has been exchanged). Would your time not be better spent actually doing some experiments. From above posts you seem to have sufficient funds to set up experiments (live in a million dollar mansion and are willing to give $5000 to anyone who can convince you your hypothesis is incorrect) save your $5000 and invest it in experminentational equipment and prove your hypothesis with experiments.
http://www.sciencebuddies.org/science-fair-projects/project_scientific_method.shtml
You have completed 1,2 and 3 of the scientific method. Now it is up to you to do 4,5 and 6. You have done much of communication but not of actual testing and results you have performed. If you can run conclusive experiments that a thermal gradient forms in a non rarefied gas (as used in your latest proof, the centrifuge which only has 80000 molecules/sec being decelerated).
I have drawn attention to “convincing experiments” (linked here) as well as convincing evidence from other planets and a convincing comprehensive study showing a negative correlation between Earth’s surface temperatures and precipitation levels, thus proving the opposite of what the IPCC claims about the most prolific greenhouse gas water vapor.
Do you get it yet, Norman? I have proved (with statistical significance) that water vapor cools, and so the AGW crowd are totally wrong in assuming it warms the Earth’s surface by a whopping 30 degrees or so. Hence I considered there to be a need to investigate the physics and discover where they went wrong, and why correct physics agrees with the evidence that water vapor does in fact cool. I have presented the world with such correct physics which you won’t yet find elsewhere.
Doug Cotton,
I refuted your water vapor cools argument with net radiation analysis. Radiation in vs radiation out. Tropical areas have much more energy entering than leaving. Sahara Desert actually has more radiation (over the year) leaving than entering. You take a few surface temperatures and falsely conclude that water vapor cools because the near ground temperature is lower than a desert temperature. You have to take the entire column of air to get a REAL study and it has already been done with net radiation studies which is all that really matters to determine heating and cooling effects. That you reject these studies does not make them wrong. If you do not agree with them why?
Doug Cotton,
Drawing attention to what you consider “convincing experiments” (I certainly do not) is not the same as doing your own experiments. I still can’t see why you are so reluctant to do experiments. If you have the time and money to perform them that is what makes science fun and challenging.
Doug Cotton,
You want this article:
https://tallbloke.files.wordpress.com/2012/01/graeff1.pdf
Where the precision of his measuring instruments is 0.1 K and he admits environmental factors can influence his setup.
Then you see in the REAL world of 1000 meters of water from all different regions (Tropics, midlatitudes and poles).
Mostly isothermal 500 meters and below for all locations. I believe you somehow manage to explain this with some intermolecular radiation preventing “heat creep”. If this is the case why would not the same intermolecular radiant energies that prevent heat creep in the ocean’s also stop a gradient from forming in Graeff’s experiment? Do laws of physics change? If there is no thermal gradient forming in ocean water then why does it form in ONE isolated experiment (maybe flaws in the insulation produced his measured effect, maybe other factors caused it).
Your theory is a complete fail on Venus which would also have intermolecular radiant energy exchanging between carbon dioxide molecules just like ocean water. If “heat creep” existed (and at this time without really solid experiments to prove it true, I do not accept it based upon your convictions) and was driving heat from the atmosphere down to the surface, this heat would warm the carbon dioxide molecules on the way down causing them to emit radiation and stopping the flow of heat downward. You have no logical argument that would stop thermal gradient forming in the Earth’s ocean via heat creep (which does NOT occur) but not also stop it from forming on Venus since both are composed of radiating material.
Whatever you, Norman, may think about Graeff’s experiments, a person who actually visited him (and appreciated the sophistication of his equipment) thinks differently. I linked you to her articles which you obviously did not study. Nearly all the 850 or so experiments detected that isothermal conditions did not exist in the inner cylinder, even when the outer cylinders were warmer at the top.
You and the AGW crowd are the ones presenting a new conjecture about isothermal conditions which are not the state of maximum entropy and thus are not what physics tells us would occur in a force field. So you go and produce a centrifuge machine which does not produce a temperature gradient, quite the opposite of what does happen in the real world and which is confirmed by the Second Law of Thermodynamics about which your understanding is barely school-boy level.
Graeff has done over 850 experiments (not “just one”) this century, most of them being left to settle for several months, hardly what happens in ocean currents, and none of them had a new source of energy heating them at the top as in ocean thermoclines which can cause heat transfers to any cooler regions below. But, none-the-less, in totally calm conditions in the dark of the winter night in the Arctic waters just north of Norway a gravito-thermal effect is seen in the ocean depths where it does get warmer.
The fact that carbon dioxide radiation in the Venus atmosphere does not completely level out the gradient provides evidence of what we expect, namely that carbon dioxide has so relatively few radiating frequencies that its effect is less than perhaps only 2% of its mass in water vapor molecules that have many more bands. This gives us all the more reason to understand that even the effect of CO2 in slowing radiative surface cooling is absolutely minuscule.
Now young Norman, I have had enough of your red herrings. Either put up valid physics (supposedly refuting the physics I present) or shut up.
“Heat creep” isn’t based on my “convictions.” It is based on the Second Law of Thermodynamics. Now prove the physics wrong smart young Norman with reference to what that law says about entropy, energy potentials etc – rather like at http://entropylaw.com
Doug Cotton,
I have already tried many times but you reject what I say so it does not seem possible to prove anything to you.
Doug Cotton
I neglected to link you to the Earth’s ocean thermal profile. Here look at this (Very isothermal at the poles where there is much less energy).
http://er.jsc.nasa.gov/seh/Ocean_Planet/activities/ts2ssac4.pdf
Yes well you also neglected to read even my most recent comment above (copied from the Wikipedia Talk page) which relates to this red herring and wherein I mentioned inter-molecular radiation in water, so go back to that comment. Better still, realize that if you were to read and understand the hypothesis you would be able to work out answers to your red herrings yourself.
Yes, Norman, and you neglected to think about all that thermal energy in the thermocline region between about 5°C and 20°C which is obviously going to send heat transfers to the colder (sub 5°C) regions below and destroy any slow formation of a gravito-thermal gradient, even if ocean currents have not already done so.
When I taught students physics I encouraged them to think, but I guess there are always people who are too narcissistic to be influenced by such encouragement.
In other words, the warmer ocean thermocline region with its positive temperature gradient acts very much like the warmer stratosphere with its positive temperature gradient, both regions sending thermal energy downwards at a rate which overpowers the slow formation of a negative temperature gradient in the regions below, namely the tropopause and the lower depths of the ocean.
Did you think that one through yourself, Norman?
Doug Cotton,
Please look again at this link.
http://er.jsc.nasa.gov/seh/Ocean_Planet/activities/ts2ssac4.pdf
Look at the polar ocean. There is NO THERMOCLINE region past 60 degrees North! It is a nearly isothermal ocean (less than a 5C temperature difference in 1500 meters of gravitational field. Looking at the graph the difference is probably less than 2 C in that whole depth and it is opposite of a “heat creep” so your hypothesis fails (only you can’t see it). It is a very subjective hypothesis. No detailed math and anything anyone says you give some vague and irrational response and pat yourself on the back like you are a genius.
Does it strike you Norman that you believe two totally different paradigms for Uranus and Venus?
(a) If you believe the AGW crap about runaway greenhouse effects on Venus, then you believe its surface is supplied with the necessary thermal energy primarily by radiation which is violating the Second Law by transferring thermal energy from a less-hot atmosphere into a hotter surface and raising the temperature of that surface a little while the Sun shines. The input to the surface you agree has to be over 16,000W/m^2 and yet, even at the top of the atmosphere (TOA) insolation from the Sun is at nowhere near that level.
(b) But Uranus (receiving less than 4W/m^2 of insolation at its TOA) apparently keeps the base of its nominal troposphere at temperatures hotter than Earth’s surface by somehow sending sufficient thermal energy from its “cooling” core that is 55% the mass of Earth and located thousands of kilometers further down. If you work out from your own calculations just how little per square meter is that energy from the core, you have a lot of explaining to do as to why convection would not cool the troposphere, taking its energy to the methane layer in the stratosphere where it would be easily radiated to space with a flux that would be orders of magnitude greater than 4W/m^2.
So which of your totally different paradigms (both of which are not supported by you with any of the laws of physics) shall we pick for Earth?
And, by the way, if you think that the Venus atmosphere somehow amplifies energy by sending energy back and forth to the surface and thus delivering in total more into the surface than it receives at TOA, then you should be able to set up parallel mirrors, place a lighted match in the middle and create all the light and heat you’ll ever need.
Doug Cotton
It is you who need to study physics. You do not understand at all what the GHE is supposed to do or why.
First and number 1. Not one climatologist is claiming the the atmosphere is getting colder and adding that energy to the surface. It is not how the system works and it would be well for you to at least understand the theory before you reject it.
Violation of the Second Law would be that the atmosphere gets colder and is adding its energy to the surface via radiation. This is what you have wrong about the GHE. Since you believe this which is not what is claimed.
You believe that the carbon dioxide has to be warm to emit and it does not (it will emit when warm but only certain wavelengths with an emissitivity of around 0.2 of black body). CO2 will absorb and resonant frequency radiation that reaches it. It will then reemit in any random direction. It works as an redirector of radiant energy like a mirror. That is all it has been stated to do no magic.
Visible light passes through carbon dioxide without being absorbed. It hits the surface and warms it. The warmed surface radiates at higher rates as it heats. The emitted IR is then absorbed by CO2. The CO2 can be cold or hot, it will absorb the IR photon and then reemit it in any random direction. If you have a large amount of CO2 50% radiant energy will be sent back to the surface.
Just as in a thermos bottle. The mirror inside redirects the radiant energy leaving the hot fluid and sends it back to the hot fluid keeping it warm for much longer than if no mirror were present (like a clear material so you still have a vacuum to prevent conduction and convection by nothing to stop the radiant energy from leaving).
Roy Spencer has tried to educate you but you do not want to understand how it works. It is like insulation. The energy can not leave. The planets are not closed systems. The GHE does not warm the surface above the surface temperature, it only returns energy that would be lost otherwise, it does not add any energy at all nor does it cool to add heat to the surface. Neither of these go on. Planets gain energy from the Sun and would continue to heat up if the energy did not leave at the same rate it comes in.
You’ll find an identical summary of the IPCC garbage in the opening pages of my book, which you can look inside free here. You don’t need to teach your grandmother how to suck eggs. But congratulations for memorizing what you were brainwashed with.
And regarding the Second Law of Thermodynamics, I have already referred you to an excellent website http://entropylaw.com which I only discovered a few weeks ago and which would help your understanding of the ramifications of that law.
Everyone should read that entropylaw.com website and at least the first few linked pages therein.
And what is even more annoying, Norman, is that you could have read what I explained two years ago (about the oceans) in Section 14 of the paper “Planetary Core and Surface Temperatures” for which the Contents page would have directed you where to look, assuming you are not motivated to read the whole paper yet …
1. Radiation and Heat Transfer
2. The Problems with the Greenhouse Conjecture
3. The Venus Dilemma
4. The Second Law of Thermodynamics
5. The State of Greatest Entropy
6. Quantification of the Thermal Gradient
7. Explanation at the Molecular Level
8. The Concept of “Heat Creep”
9. How Earth’s Surface Temperature is Supported
10. Laboratory Evidence for the Gradient
11. Planetary Evidence for the Gradient.
12. The “Pseudo” Lapse Rate.
13. Non-Radiative Heat Transfer Processes
14. Rebuttal of Counter Arguments
15. Support for the Mantle and Core Temperatures
16. Conclusions
17. Appendix – Study of Temperature / Rainfall Correlation
18. References
Doug Cotton,
Section 14 explains nothing really. Very poor explanation and it does not cover the isothermal water in polar regions.
Also in Section 12, regarding your criticism of Graeff’s 800 odd experiments, you could have read more, including this ..
“So, when Roderich Graeff included fine glass powder in one of his water cylinders, that would have reduced the mean specific heat, and thus increased the thermal gradient. The walls of the container would also have increased the gradient because of their much lower specific heat. Furthermore, the first cylinder would have had some interaction with the other one containing only water..
In short, Norman, all your attempted refutations were rebutted two years ago.
Sorry, that was of course Section 10. Laboratory Evidence for the Gradient
Doug Cotton,
Here goes wasted effort. On you it is a waste and it is likely no one else will read this post. But you are worth the attempt so here goes nothing.
Explanation of GHE with thermos bottles of two types. The current type.
http://offgridsurvival.com/stanleythermos/
Water stays warm at least 11 hours. I have read some keep liquid hot longer but this will do.
The other thermos still has a vacuum (no conduction or convection to cool the water) but a clear material that allows IR through, no mirrored surface to redirect thermal radiation.
Both have the same volume of boiling water 500 mL is a good number as this is what some thermoses are.
So you have 500 mL of boiling water poured into both thermoses. Heat capacity of water is 4.179 joules/gram C. 500 mL of water equals 500 grams. Room temperature we will use is 25 C.
Temperature difference between boiling and room temperature is 75 C.
Energy that will be lost to reach equilibrium with the room.
E=(4.179 joules/gram C)(500 grams)(75 C)
E=157,000 joules.
In the second flask with radiant energy loss (no mirror) the rate the fluid will emit radiation at peak is around 24 watts.
http://www.endmemo.com/physics/radenergy.php
325 cm^2 was a number given for 500 mL of water so it is good enough as long as both thermoses have the same radiating surface area.
the average radiation loss would be around the halfpoint between 24 and 10 watts or 17 watts (calculus would give you a much better watts but I do not need to be that specific for this post).
Time to lose 157,000 joules at the rate of 17 joules/second is around 2.5 hours.
The mirrored thermos will stay warm around 11 hours. The one with no mirror will last around 2.5 hours.
So maybe now you can see what is going on. The mirrored thermos is not adding heat to the liquid. It is not violating the Second Law of thermodynamics. It does not transfer energy from the cold mirror into the hot liquid. The mirror just redirects the radiation it does not add to it.
Now if you have a planetary system it is not a closed or isolated system. During the daylight hours entropy is decreasing as it gains energy. During the night the entropy increases. Planets are an open system and gaining energy on some cycle and losing it on another (day and night).
So if with the thermoses if you had a heat source that would add energy (joules) to the system every 4 hours what do you think happens to the water temperatures at this time?? Think and reason it through if you can.
We can pick some arbitrary value say 157,000 joules are added to each thermos every 4 hours.
So thermos 2 will go back to boiling then cool. Its temperature will never go above the boiling point as it is cooled to room temp before any additional heat is added. But what happens to Thermos 1? In 4 hours it is still quite hot. When 157,000 joules are added to this thermos the water will be much above boiling temperature and will continue to increase as energy is added every 4 hours. One does not heat above the boiling point, the other will heat much above the boiling point. The mirrored surface is not adding any energy to the water, it only keeps energy from leaving.
I think too many of the people that reject the GHE theory forget the planets are open systems.
Even on Venus. It does not matter if only 10 watts/meter hit the surface. That will not determine the equilibrium temperature.
If one surface receives 10 watts/meter^2 and another receives 1000 watts/meter^2 which will be hotter? The only thing energy in can tell you is the rate of heating but not what temperature the surface will finally reach.
If a surface is gaining 10 watts/meter^2 but only losing 9 watts/meter^2 that surface will heat up! If another surface is gaining 1000 watts/meter^2 but losing 1000 watts/meter^2 it will stay the same temperature. It is not energy in or out that matters. It is the NET energy. Energy in minus energy out. That is the only thing that will determine your surface temperature.
If a surface is gaining 10 watts/meter^2 but only losing 9 watts/meter^2 that surface will heat up!”
If it is receiving thermal energy at a rate that exceeds its rate of loss of thermal energy it will indeed rise in temperature to a limit which may be calculated depending upon whether it is radiation, conduction or convection that is supplying the thermal energy, and then depending on the temperature and distance of the source.
But note that a one-way pencil of radiation with low energy photons with low frequencies and/or insufficient intensity does not add thermal energy to a warmer target emitting higher energy photons with higher frequencies. That’s why the plastic bowl in your microwave oven does not get hot.
All this is explained in the first half of my 2012 paper linked from our group’s website and I have no intention of repeating more than 10 pages of explanation here.
No further correspondence will be entered into unless it pertains to something you don’t understand in either paper.
Doug Cotton,
You don’t need to respond because you are so goofy it is silly.
I read your article on the radiation and it is complete bogus.
It does not correlate to the REAL world. A thermos keeps liquids warm! Do you understand this? Radiant energy is leaving the liquid surface. Do you understand this? Radiant energy given off by the liquid is then returned to the liquid via mirrors to keep the liquid temperature the same. Do you understand what is going on. Energy is actually leaving the liquid and returning. There is no stupid lower energy photons scattering from the liquid. The energy leaves, it has lost those joules and now they return and it goes on like this for a long time.
Lack of reasoning again. GHE does not add lower energy photons to a higher energy surface. It returns photons that have been emitted, causing cooling that are then returned so the surface is at a similar state to when it emitted them.
Process: Surface emits IR, it cools as it loses energy. The IR is absorbed by CO2. It is then reemitted, some of this IR returns to the surface. The surface can receive it because it had just lost it and it will warm by the amount of the returning radiation, you totally neglect that it had cooled in emitting the radiant energy in the first place.
As written below by a Professor of Applied Mathematics: A blackbody “absorbs all radiation, emits absorbed frequencies below cut-off, and uses absorbed frequencies above cut-off to increase its temperature.” So some parts of the Earth’s surface sometimes receive from the Sun on a clear day sufficient energy at frequencies above their own peak radiating frequency (cut-off) to raise the temperature. All the frequencies below the frequency corresponding to the current temperature are re-emitted (without their electro-magnetic energy being converted to thermal energy) and all radiation back from colder regions in the atmosphere has frequencies below “cut-off” (the frequency the surface is emitting) and so all that back radiation is re-emitted without any of its electro-magnetic energy being converted to thermal energy.
And that, dear Norman, is precisely how, why, when and where every one-way pencil of radiation obeys the Second Law of Thermodynamics. Your IPCC conjecture with which you have been brainwashed in your education (which must have been after 1980 or so) is total garbage dreamed up initially by one man, James Hansen around the year 1981, because he was trying to explain the necessary energy input which we now know comes from the “heat creep” process, not from back radiation. As I said, the Sun cannot raise the mean surface temperature even above freezing point, so you can slow the cooling all you like, but from what temperature in the first place, young Norman? When it’s cloudy and also when the Sun’s radiation is not sufficient, even that solar radiation does not raise the surface temperature one iota.
Everything you have written about radiation is based on your wrong assumption that the electro-magnetic energy in every photon is all converted (100% of it) whenever it strikes even a warmer target. Well, Norman, even Roy Spencer will confirm that back radiation does not even penetrate water surfaces by more than a few nanometers if it came from a cooler region in the troposphere. Absolutely none of its electromagnetic energy gets converted to thermal energy in the water. You can read why in the paper I wrote three years ago, or, if computations impress you, then read this explanation by a Professor of Applied Mathematics …
“A blackbody thus can be seen as a system of resonators with different eigen frequencies which are excited by incoming radiation and then emit radiation. An ideal blackbody absorbs all incoming radiation and re-emits all absorbed radiation below cut-off. Conservation of energy requires absorbed frequencies above cut-off to be stored in some form, more precisely as heat energy thus increasing the temperature of the blackbody. As a transformer of radiation a blackbody thus acts in a very simple way: it absorbs all radiation, emits absorbed frequencies below cut-off, and uses absorbed frequencies above cut-off to increase its temperature.?
Radiation from a cooler source slows down that portion of the cooling of a warmer target which is itself by radiation.
If you want to learn about how radiation does this slowing of radiative cooling (due to a resonance process) and how radiation transfers thermal energy and when it does not, then read the paper I wrote three years ago on radiated energy.
It’s linked on the “Evidence” page here and you have no idea whatsoever as to what is explained there in in breakthrough 21st century science.
I will read your comments when they appear to be addressing the hypothesis in my 2013 paper that explains energy transfers and temperatures above and below any surface on any planet or moon. That hypothesis is based on physics which I know you don’t understand, but if my explanation has any shortcomings I will refine it and, if applicable, improve the explanation on the website endorsed by our group of persons suitably qualified in physics. I want to make sure the physics is understood by the “lowest common denominator” so you can help me there by displaying your misunderstandings.
Don’t forget that the mean radiation from the Sun which is absorbed by planetary surfaces is nowhere near sufficient to explain how those surfaces rise to the observed temperatures. And, as you will understand if you read the first paper (2012) above, back radiation from the colder troposphere does not help the Sun in attaining the mean surface temperatures observed.
That’s physics. That’s why I spent thousands of hours in research finding what does supply the energy and explaining it all with correct physics.
Doug you still do not understand the theory at all.
Carbon Dioxide and water vapor will emit radiant energy when heated but they will also emit after they absorb. They do not have to be warmed to redirect radiation. Will a super cooled mirror stop reflecting its energy back to its source? The temperature of the atmosphere does not effect the GHE. The Carbon dioxide molecules at the cooler regions of the atmosphere still absorb IR if the wavelengths are correct. The absorption is not based upon the K.E. the carbon dioxide molecules posses. That is a hurdle you need to figure out. Once they absorb the IR they will then reemit it even if still cold.
The energy of the sun can warm planets to many different temperatures. What determines how hot the surface gets is based upon how much energy leaves.
You just don’t understand. The surface loses energy when it radiates to carbon dioxide. It has cooled. The carbon dioxide only returns the same radiation that had been lost. It can then be reabsorbed by the surface.
Thermos bottles work Doug. They work by sending radiant energy back into the fluid. if the radiant energy did not return to the liquid but kept bouncing around from liquid surface to mirror then you would have a bunch of energy between the surfaces and the liquid would cool just as rapidly if there were no mirrors. The mirror is much cooler than the liquid emitting the IR in the thermos. But it still works you know. I guess in the hot climate of Australia you do not need a thermos to keep your drinks warm during a cold winter day while you are camping or hunting.
Go back to this comment and read my summary of the IPCC garbage including direct quotes from their website.
STOP Norman!
Before you write what I anticipate you are going to, it’s already refuted in one or both of my papers.
In so far as my hypothesis is concerned, everything to do with the rate of cooling of the surface (which is mostly non-radiative anyway) is irrelevant.
So read what that hypothesis is all about, because I know better than you all the standard arguments which climatologists teach each other and which you merely repeat here. Search the blogs and you’ll find I’ve responded to every single red herring you’ve ever raised over and over again in recent years. No one has ever proved me wrong. See if you can find anything anywhere that refutes the actual content of my papers, as distinct from what people like yourself assume is in the papers without reading them.
Doug Cotton,
You do not even use the term “Red Herring” correctly.
Here is what it is:
http://literarydevices.net/red-herring/
Here is an example of it. I ask a strong question and without even attempting to answer it you go somewhere else and ignore it completely. That is a true red herring tactic. I have not used red herring format. I reply to your posts directly. I disagree with your view and present and alternative view. THAT IS NOT A RED HERRING!
You use red herring tactic a lot.
Norman says:
March 14, 2015 at 11:34 AM
Doug Cotton,
In the post above about Uranus.
“And if the Uranus core cooled by even just 300 degrees then the base of the nominal troposphere would cool from 320°K to 20°K and, if the Sun were still radiating as at present, that is an impossible situation since the troposphere gradient would be nothing like what we calculate it to be using -g/Cp and the temperature for radiative balance with the Sun is (58±3)°K.”
Why would this be the case? Why would the core cooling from 5000 kelvin to 4700 kelvin automatically reduce the nominal troposphere temperature from 320 K to 20 K (also you do not use the degree mark for K http://www.physlink.com/Education/AskExperts/ae129.cfm)?
Would a boiling pot of water cool 100 C if you lowered the temperature of the hot plate 100 C? If your hot plate was 500 C are you saying that if you turned it down to 400 C the water would stop boiling and would freeze solid? Your logic on this post makes so little sense I can’t follow how you arrived at this conclusion.
Reply
Doug Cotton says:
March 14, 2015 at 3:24 PM
You would know the answer if you read, studied, inwardly digested and understood my hypothesis at http://climate-change-theory.com noting in particular the discussion of temperature gradients in sub-surface regions of planets and satellite moons.
Reply
Norman says:
March 14, 2015 at 8:38 PM
Doug you need to change your name to Dodge Doug. You dodge anything that you can’t expalin. Reading your material will not explain in any way shape or form why if the core of Uranus goes down 300 K it would reduce the temperature of the nominal atmosphere by 300 K. You just make things up and when confronted you just say read my material. I have read your material it does not answer the question. You just dodge it with this silly post.
You use a red herring here. I want you to explain why a 300 k loss of Uranus core would cool the nominal surface by the same amount. Rather than explain yourself you just ask me to read your material (which I have and think is weak and poorly defended and has no real math to work with of any type).
Norman – the answer is straight forward directly from my hypothesis – the temperature gradient does not change significantly because the mean specific heat does not change significantly at any location and nor does the force of gravity. It’s quite clear in my paper.
Now, see my final comment below, because the fact is you have no understanding of the 2013 paper and you should give me credit as the author for being able to detect your lack of understanding.
Here’s a simple test …
Draw the heat creep diagrams from memory, labeling them correctly also!
You can’t can you? Be honest! Try it and then compare with what is there in the paper. Because you can’t reproduce them from memory proves that you have not taken the time even to study the most important central issue in the whole hypothesis. I didn’t ask you to just read the papers. I asked you to study them and inwardly digest them. Then, instead of raving on about a thermos, you would have written acknowledging that I had stated that thermal energy is not transferred by one-way radiation from colder sources to hotter regions, but you could link me to some paper that you think proves me wrong, then I could pinpoint where the paper is wrong and link you to a paper like that cited just now, etc, in true scientific process. But true scientific process is obviously not to your liking. So just once more I will send you on your way with this genuine confirmation of my hypothesis …
“Doug Cotton shows how simple thermodynamic physics implies that the gravitational field of a planet will establish a thermal gradient in its atmosphere. The thermal gradient, a basic property of a planet, can be used to determine the temperatures of its atmosphere, surface and sub-surface regions. The interesting concept of “heat creep” applied to diagrams of the thermal gradient is used to explain the effect of solar radiation on the temperature of a planet. The thermal gradient shows that the observed temperatures of the Earth are determined by natural processes and not by back radiation warming from greenhouse gases. Evidence is presented to show that greenhouse gases cool the Earth and do not warm it.
John Turner B.Sc.;Dip.Ed.;M.Ed.(Hons);Grad.Dip.Ed.Studies (retired physics educator)”
Actually, Norman, I can’t give you a score of 100% for memorizing correctly all the IPCC garbage because you failed to understand how they calculate that mean surface temperature. I had already explained it in the text under the energy diagram on this page.
Unlike yourself, I make a point of really understanding the other side of the argument, and then I can pinpoint exactly where they go wrong in their science or logic. That’s the true scientific process you need to learn how to practice as you grow older and hopefully wiser.
So have you been a good student and learnt the figures in that energy diagram? You’ll see that they need to fudge enough downward radiation from the atmosphere (far more than goes upwards) in order to warm the surface with it because it is multiplying the effect of the Sun by about three. Take care you don’t get burnt by all that radiation at night. I’ll try a bit of backradiation baking at midnight as it is here now.
And never ever in any single comment of yours, Norman, have you addressed the hypothesis for the new 21st century paradigm in climate science, which is not based at all on radiation into the surface of Earth or any planet. You cannot refute that physics because you haven’t even bothered to study it and you most certainly could not explain it from memory, or draw the heat creep diagrams from memory, now could you? So you have just failed your exam on the new paradigm and I have wasted time on you, but not on silent readers, none of whom have come in to support you today.
Good bye.
“I have wasted time on you, but not on silent readers, none of whom have come in to support you today.”
The irony …
Tim/Norman/Curt/Ball4 and others who inhabit real world.
Like I said above – the man’s away with the Fairies. Whether it’s the same Fairies he was conversing with a couple of years back when I reported him to Roy for being abusive, is another question. This lot do seem slightly better mannered.
It is obvious to me that the Mark Twain quote is most apt where he is concerned.
PS: If you don’t know which – just ask.
No one is going to tell me the sky is green when I’ve observed it as blue all my profession life.
That’s the key my friend OBSERVATION.
Yes indeed. Observation such as the fact that the greenhouse gas water vapor cools in the real world (rather than warms in the imaginary IPCC world) is indeed very important for a correct understanding and for supporting the correct physics I have presented. I await your answer to the question here.
What is the sensitivity for each 1% of the most prolific “greenhouse gas” (namely water vapor) in Earth’s atmosphere?
To help any of you answer the question, here are some facts:
Fact 1: Water vapor absorbs a significant amount of incident solar radiation as shown here. The atmosphere absorbs about 20% of incident solar radiation and that absorbing is not by nitrogen, oxygen or argon. (Carbon dioxide also absorbs incident photons in the 2.1 micron range which each have about 5 times the energy of 10 micron photons coming up from the surface. On Venus over 97% of the energy from incident solar radiation is retained in carbon dioxide molecules.)
Fact 2: The concentration of water vapor varies between about 1% and 4%. (The concentration of carbon dioxide above Mauna Loa is 0.04% and, as this graph shows, temperatures there have not increased since 1959.)
Fact 3: The IPCC claims that water vapor does nearly all of “33 degrees of warming” of Earth’s surface. It must do most of it because it dominates CO2 in concentration and also in the number of frequency bands in which it absorbs and radiates, and water vapor lowers the “lapse rate” so that the temperature profile rotates downwards at the surface end, making the surface cooler. (In fact, as per my paper, there is no 33 degrees of warming being done by any back radiation because it is gravity which props up the surface end of the temperature profile.)
When you have answered the question, work out how much hotter the IPCC conjecture implies a region with 4% water vapor would be than a similar region with 1% water vapor at a similar altitude and latitude. Then look up the study in the Appendix of my paper and see what real world data tells us about how water vapor cools rather than warms. And if you don’t believe my study, then spend half a day doing your own.
Finally, note that it is quite clear in the energy diagram here and the text I wrote beneath it that they have certainly added 324W/m^2 of back radiation to 168W/m^2 of solar radiation in order to use this in Stefan Boltzmann calculations to determine the temperature of the surface. Obviously they worked out by difference what the back radiation figure had to be and made it 66% greater than the 195W/m^2 of upward radiation from the atmosphere to space. They need not have bothered, because their whole paradigm is wrong, because they ignored the fact that the Second Law of Thermodynamics tells us that gravity forms the temperature and density gradients – which represent the state of thermodynamic equilibrium.
Doug 4:16pm grows more distant, more detached from reality in the panic to post.
Fact. 1. Did you notice solar radiation absorbed by atm. includes O3, O2, CO2, N2 in addition to H2O? If on Venus CO2 holds 97% of the solar energy then ground and other atm. constituents at 3% must be mighty cold indeed. In reality carbon dioxide absorbs at EVERY wavelength not just ~2.1micron.
Fact 2. You link to a site that links to a paper that says: “we have found…there is an overall annual warming trend of temperatures dT/dt = 0.021±0.011C yr−1 at this observatory…Our basic hypothesis is that a large part of the temperature and (diurnal temperature range) trends at Mauna Loa can be attributed to changes in CO2….These trends are consistent with the observed increases in the concentrations of CO2 and its role as a greenhouse gas…”.
Discuss that. You don’t seem flustered by the paper’s disagreement with your political views.
Fact 3. Not “made it” 324 they measure the 324. There is no state of thermodynamic equilibrium ever, anywhere in earth’s real atmosphere. In reality its entropy increases & its energy is conserved. Since your regional calculations are known to be wrong, your conclusion(s) can not be correct.
What’s your answer to the question I asked?
I’ll respond to your points, and I expect you to answer my question in return.
Firstly, “a significant amount” does not imply all of that 20%. It’s a darn sight more absorption shown in that measured plot that is by water vapor than that by carbon dioxide.
It’s pretty obvious I meant “On Venus over 97% of the energy from incident solar radiation in the atmosphere” is retained in carbon dioxide molecules.” Less than 20W/m^2 of direct solar radiation gets through to the surface of Venus, and that’s been measured by Russian probes dropped there.
Regarding Mauna Lao, their claim of “overall rise” does not match with the temperature chart. They can blame warming on CO2 if they wish. I blame it on natural causes, because I have proved with thermodynamics that CO2 cannot raise the surface temperature, and nor can water vapor.
Oh I’ve got political views now have I? Well I do support the concept of capitalism which is what the UN has an agenda to destroy. All major parties here support capitalism.
Yep – I totally agree the atmosphere’s entropy increases. The Second Law tells us there is a propensity for that to happen until thermodynamic equilibrium is attained with, in this case, a stable density gradient and temperature gradient, as explained using the Kinetic Theory of Gases in my paper.
So I repeat: what’s your answer to the question I asked?
Doug says: “When you have answered the question, work out how much hotter the IPCC conjecture implies a region with 4% water vapor would be than a similar region with 1% water vapor at a similar altitude and latitude. “
Doug, you need to think about more than one idea at a time. The climate is a very complex systems with many interdependent factors.
In particular, water has several different effects. Some that come to mind include:
1) water vapor absorbs some outgoing IR (the ‘greenhouse effect’)
2) water vapor absorbs some incoming sunlight
3) water vapor condenses, changing the lapse rate
4) water forms clouds, reflecting away sunlight.
5) liquid water on the surface moderates temperatures via heat capacity
6) liquid water on the surface cools the surface via evaporation.
If you ONLY consider effect (1) you get a dramatically wrong answer. That does NOT mean that (1) is wrong. It simply means that other things besides (1) must be important. I know you have done some good work correlating humidity and temperature, taking into account latitude and altitude. Now take into account cloud cover and surface water heat capacity and difference in albedo and evaporative cooling. For example, high humidity would mean lots of water on the surface to evaporate, which would coll the surface. Do a multiple regression to find what impact each factor plays.
Don’t let yourself be fooled into thinking that “the IPCC conjecture” is ONLY (1). Any good climate model would include ALL of these. Your analysis also needs to include all of these before you can say that (1) is not significant.
Hi Tim Folkerts,
Your water list missed an important characteristic of water vapor. Water proves lighter than common air (i.e. oxygen and nitrogen molecules) allowing it rise quickly, cool and/or radiate away excess energy quickly and fully in a broad infra-red band. CO2 proves heavier than air and emits largely within a few narrow bands. As to the IPCC, it isn’t simply a matter of including facts in their analysis but also the comprehensiveness and weight given to them. Thanks for the post, and …
Have a great dsy!
John:
The atmosphere would have to be incredibly static for such separation to occur. In the real world, H2O concentration is far greater in the lower atmosphere than the upper. On the other hand, CO2, which as you point out has a higher molecular weight than O2 and N2, does not show this decrease in concentration with altitude.
Curt,
You claimed:
“The atmosphere would have to be incredibly static for such separation to occur.”
Really? Venus has a fairly turbulent atmsophere, yet nitrogen ( 4 times Earth’s atmospheric nitrogen mass ) with a molecular mass less than CO2 exists in large quantity in the UPPER Venusian atmosphere. So in fact SEPARATION CAN AND DOES OCCUR IN NATURE WITHIN TURBULENT ATMOSPHERES. Keep in mind the average wind speed 50km above the Venusian surface and 1 bar of pressure is ~100 mph.
As to water vapor, much of the water vapor rising from the Earth’s surface undergoes phase changes before it rises too high in the atmosphere preventing much of it from accumulating at high altitudes and I never asserted other-wise so I’m uncertain as to the purpose of your post in that respect. Btw, to an extent separation does occur since clouds do sometimes form in the stratosphere. Polar stratospheric clouds and others form in the region.
As to CO2 on Earth, you correctly point out that it remains fairly steady with altitude. Apparently true, but keep in mind it’s not perfect. CO2 levels in cities and locations with inversion layers can get as high as 1500 ppm. Which may concern some since at around 10000 ppm I’v read that breathing difficulties begin to appear. If true watching relative concentration levels in various locals may be wise.
Have a great day!
Doug 8:26pm: ”Regarding Mauna Lao, their claim of “overall rise” does not match with the temperature chart.”
Incorrect. Paper: “warming trend of temperatures dT/dt = 0.021±0.011C yr−1 at this observatory”
You wrote: ”The atmosphere absorbs about 20% of incident solar radiation and that absorbing is not by nitrogen, oxygen or argon. “
Incorrect. N2, O2, argon all absorb. All matter absorbs. At all frequencies. All temperatures. Nearly ALL of your views are political as they are known not science based.
”So I repeat: what’s your answer to the question I asked?”
There is only a political answer to “What is the sensitivity for each 1% of the most prolific “greenhouse gas” (namely water vapor) in Earth’s atmosphere?”
However, there are some simple analyses that can give basic understanding of the science by test but you demonstrate they are far beyond your level of comprehension as some calculus & radiation expertise is req’d. for which you do not have the earned degrees.
Ball4 and Norman
There are no noticeable notches in the plot linked in my comment above for nitrogen, oxygen or argon. But there certainly are for water vapor and I’m glad you’ve agreed that the atmosphere is nothing like “transparent” to incident solar radiation.
All of what I write is based on the Second Law of Thermodynamics and is correct physics. You cannot prove otherwise and never have even addressed the content of the hypothesis which neither of you could even summarize from memory, let alone draw the heat creep diagrams to demonstrate such understanding.
It sad actually that the post-baby boomer generation (and that includes two sons and three of my daughters) has been so brainwashed with the garbage about radiation because it sounds just so intuitive to the gullible who don’t understand thermodynamics – like blankets keeping the Earth warm – and it’s so easy to believe that the Sun can maintain the surface temperature on its own. It can’t do that with 168W/m^2 which supports a mean temperature of only -41°C.
What is the sensitivity to a 1% increase in water vapor?
You have not answered this based on mean WV percentage (less than 2%) and the warming of about 30 degrees supposedly caused by mostly water vapor delivering twice as much thermal energy to the surface as does the Sun’s direct radiation.
It’s not hard. Try about 15 degrees per 1%. That makes rain forests with 4% a whopping 45 degrees hotter than deserts with 1% water vapor. And, yes, water vapor does have regional impact and clouds do shade regions below.
Any answer which says water vapor causes there to be warmer mean surface temperatures is wrong. Water vapor lowers the thermal gradient. If it also jacked up the surface end of that profile, then the area under the thermal profile would be increased considerably and that would throw radiative balance with the Sun way out. That just doesn’t happen, and that variation rarely exceeds ±0.6%.
Radiation just does not work the way you think it does. Photons don’t explode like little hand grenades depositing thermal energy into everything they strike. I’ve given you links to my 2012 paper which cites the relevant material, notably the paper “Computational Blackbody Radiation.”
Radiation into a planet’s surface is not the primary determinant of the surface temperature. You are thinking within a totally incorrect paradigm, and you will never change until you throw out the radiation garbage from your mind and start from Square One with our group’s website and my linked papers.
Doug 5:54am: That’s a political statement, you merely have an opinion without experimental evidence. Unfortunately for you, well known correctly done experiments show precisely where your politics are drawn from faulty science. Entropy increases, energy is conserved. You demonstrate little understanding of these principles.
What is political is the United Nations’ plan to dominate world governments and destroy capitalism. My motive is altuistic, namely to save poverty, especially in developing countries, and to save lives that are being lost because humanitarian aid money (like $200 million from the Australia recently) is being diverted to carbon dioxide aid. I have “donated” thousands of dollars and thousands of hours of unpaid time (with an opportunity cost) to this cause. I put my money where my mouth is. You probably stand to lose money when the hoax is defeated. I don’t care which governments eventually toss out the hoax, just so long as that happens, the sooner the better. Try searching “climate hoax” on http://youtube.com and watch some of the hundreds of videos already seen by millions. There’s even one there by myself linked from the group’s website.
Now read, study and inwardly digest this comment below just written to Norman, and then the website endorsed by our group of persons suitably qualified in physics and my linked papers.
Doug 3:21pm – Your 3:11pm comment is just a political manifesto designed to support your political views. It lacks science method. Your regional calculation study is known to be incorrect. The basic science is not a hoax. Your $5,000 on offer is the hoax.
Doug Cotton,
I do want to compliment you on our website. It is a well done design with the excellent photo images.
What’s your answer to the question about water vapor sensitivity, Norman?
While you’re thinking about it, enjoy my photos at http://tasmania-holiday.com and several other linked websites.
Actually my wife’s a better photographer than me by far, with sales at the Royal Easter Show etc …
http://yocotton.com
Doug Cotton,
Thank you for the photo link. Some stunning photos in those collections. You and wife do have photo talent!
On your question about water vapor I will refer you to this web page which has 4 different Quick Time global maps of radiation. I have not completely thought them through but at least on clear sky it seems certain that water vapor is a strong forcing agent in keeping radiation from leaving the system. Oceans have much higher positive radiation levels when sun is shining on them then land areas and oceans would have the highest amount of water vapor over them on a long term basis. I will think on these and try to give you a valid answer.
http://cimss.ssec.wisc.edu/wxwise/homerbe.html
You may want to look at the global graphs yourself.
All that matters is temperatures, not radiation levels. We know the kind of temperatures found in non-polar oceans. The Sun’s direct radiation of 168W/m^2 cannot raise deep water to anywhere near such temperatures, because much of the radiation is absorbed at levels several meters below the surface where temperatures are already cooler. None of the back radiation, on the other hand, penetrates more than a few nanometers.
If the Earth were paved in black asphalt, had no atmosphere but received a mean of 168W/m^2 (about half what the Moon receives) then its mean temperature (using Stefan Boltzmann) would be around -41°C. You have to face up to this fact.
As you could have deduced from my hypothesis in the 2013 paper, thermal energy enters the ocean system primarily in non polar regions by conduction at the interface with the lower troposphere that has itself received thermal energy by heat creep and some from the solid surface on clear warm days. The Sun’s direct radiation (mean 168W/m^2 = -41°C) cannot warm the ocean surface to anywhere near the observed mean temperatures. The thermal energy (mostly from the atmosphere) then passes down through the ocean thermocline and into the depths. Any gravito-thermal gradient is obliterated by inter-molecular radiation. The thermal energy broadly speaking makes its way in the depths by currents and convection along the isothermals and that energy can only thus resurface in polar regions.
“Oceans have much higher positive radiation levels when sun is shining “
Of course. They reflect more and, in general, radiate more due to slightly higher temperatures usually. So what? That has nothing to do with the thermal energy they receive by heat creep. I am only interested in discussing my hypothesis now, not the debunked radiative greenhouse hoax. So please stop try to teach me what that hoax claims. I know what it claims better than you. I’ve read Pierrehumbert etc etc etc.
Thanks regarding the photos, Norman. There’s a certain amount of physics and math in photography. My aunt Olive Cotton was a famous Australian photographer and had an exhibition that was in the Sydney Art Gallery and the National Gallery in Canberra. She was of course a daughter of my grandfather, Prof. Leo Arthur Cotton (noted for his photographs in Antarctica) whose brother was Prof Frank Cotton who invented the anti-gravity suit. Mmmm – interesting that reference to gravity.
Doug in the past I was open to your thinking and doing some research on it.
http://www.drroyspencer.com/2013/04/global-warming-slowdown-the-view-from-space/
Couple years ago on Roy Spencer.
I do not think your conclusions are valid at this time. The more I thought and research it the less likely to me it seems you have a good or valid theory. In the past I asked for experimentation and still none has been done.
I really don’t care what a person like you, Norman, thinks. You are not qualified in physics and you make no attempt to study my website and papers, to understand what I have explained or to remember answers I have already given for the questions you ask several times.
Doug I know you are wrong and Roderich Graeff’s experiments are worthless and incorrect.
http://er.jsc.nasa.gov/seh/Ocean_Planet/activities/ts2ssac4.pdf
Graeff states the gradient formed by water would be 0.04 K/meter.
You have in the above graphs 1500 meters worth of data.
In 1500 meters a gravity gradient would be 60 C of hotter on bottom vs. top. Your claim is ocean currents and solar radiation mess up the effect. The solar energy at best is worth 25 C which is much smaller than the 60 C that should develop. In the polar water with little solar energy added you do not see any nothing of a thermal gradient. When it should be 60 C warmer at the 1500 meter level from the surface you should see something develop and YOU SEE NOTHING. Reality will easily claim that Graeff’s hypothesis is disproved and his experimental set up is flawed.
You really need to explain in great detail why a 60 C gradient does no show up, not even a 1 C gradient shows up in polar water, the opposite is the case. If you do not come up with really good reasons for this I am thinking you should reject your own hypothesis as invalid and incorrect.
I have already explained in the paper how and why Graeff got his calculations wrong. Gradient = 9.8 / 4.0 (from Cp here = 2.45 K/Km = 0.00245 K/meter.
You see what I mean? This is yet another example of how you forget what I have already explained and taught you. You are never going to prove teh Second Law of Thermodynamics wrong Norman, so give up.
Norman, you are not in any position to say Graeff’s experiments are useless. You have not spent many hours studying his apparatus and methodology, as I have. There is a whole section on his work it in my second paper. There is also the experiment with the centrifuge which gave the exact opposite results to what Curt assumed would be the case and assertively stated was the case. What the hell makes you think a new graduate in engineering would know more than myself with a degree in physics (far more physics than engineers do) and a total of nine years of university education (and a university scholarship in physics, and coming from a family with a mother Dr Marie Cotton who topped her university course and whose obituary, like my father’s, Dr Leo Frank Cotton also, was on the front page of The Sydney Morning Herald) and countless hours of post-graduate study in phyics, especially in the fields of radiation and thermodynamics.
You have absolutely no clue about this physics, Norman, and neither do Curt or 99.9% of climatologists know anything about the new 21st century breakthroughs (made by others, not myself) that are discussed in my papers and applied to atmospheric physics.
All you “know” Norman is what you have been brainwashed into believing. And because you don’t understand radiation and thermodynamics you have been like over 99.9% of the population who likewise don’t understand such, and thus you are gullible, there’s no other word for it. My first paper is about radiation and when it transfers thermal energy and when it does not. The second paper is about thermodynamics, entropy maximization and energy potentials about which you can also learn the basics at http://entropylaw.com which is a good site I only found last month.
My study showing (with statistical significance) that water vapor cools is repeatable. You can check the data which is all published. You could spend half a day selecting your own data along similar lines. One thing is for sure. You are never going to prove rain forests (with 4% WV) are 45C° hotter than dry regions with 1% WV (at similar latitudes and altitudes at similar times of the year) as the IPCC garbage implies they should be.
You write “The solar energy at best is worth 25 C which is much smaller than the 60 C that should develop.” but you haven’t a clue what you’re talking about and the statement is just so totally irrelevant to a process that has developed the observed temperature gradients slowly over the life of every planet and satellite moon. Inter-molecular radiation in water almost totally over-powers the gravito-thermal effect. Even in the atmosphere just 1% to 4% of water vapor, with molecules far more separated, over-powers about a third of the effect and, as you should know, reduces the magnitude of the temperature gradient by about a third. That is a fact.
The temperature plot in the troposphere rotates downward at the surface end as more water vapor is introduced, not upwards. Likewise for other radiating molecules like carbon dioxide.
Go back and read the comment where I explained how radiative balance with the Sun would be way out if that thermal plot rotated upwards and also raised the surface end. Sketch a graph yourself. Above all, come to an understanding that radiation between the surface and the colder atmosphere produces only a one-way transfer of thermal energy out of the surface, represented by the area between the Planck functions – as explained in my first paper.
The back radiation does not transfer thermal energy into the surface (as explained by that professor of applied mathematics) and all it can do is slow down about a third of the surface cooling which is itself by radiation. It does not help the Sun to make the surface hotter in the first place. Hold a mirror facing down above a sunny patch of ground (without blocking the insolation) and try yourself to make the ground hotter.
Back radiation does not slow down conduction, convection or evaporative cooling. That single fact demolishes the radiative greenhouse conjecture, as does the fact that the most prolific greenhouse gas, water vapor, cools rather than warms. There’s still a $5,000 reward for the first person in the world to prove otherwise and also prove my hypothesis substantially wrong, but no one has even tried, or if they have, they have found results like mine showing water vapor cools, as some others have also shown in less comprehensive but similar studies.
Doug:
You have a serious reality-disconnect problem on everyday matters as on technical matters. Where on earth did you get the idea that I am a recent graduate? I never said anything of the sort. The fact is that I have had 35 years of post-graduate technical experience, both in industry and teaching (at the University of California). If you want to argue academic pedigree — personal and family — mine is at least as illustrious as yours (not that it really matters).
In the 1980s, I worked for a prominent Silicon Valley company that had a policy of giving job applicants a technical grilling during interviews. The goal was to find out whether an applicant had a deep understanding of technical subjects. In each group I was in, the top managers quickly selected me to perform the grilling. I decided early on that radiative heat transfer was the best topic to separate the men from the boys.
At the time, I was surprised by how many people could play the academic game well enough to graduate, even with a decent grade point average, yet have no grasp whatsoever of the fundamentals. I place you firmly in this category. I would have rejected you immediately as a candidate.
The main reason I have bothered to engage in this dialog is basically the same reason some people are drawn to a train wreck — morbid fascination. I have not been disappointed. Here is a short list of the Doug Cotton “train wreck” egregious errors in introductory thermodynamics I have drawn out from you recently:
* No understanding of the difference between thermodynamically open, closed, and isolated systems. Shortly above this, you stated your belief that the difference between open and closed systems was heat transfer across the boundary, when it is really mass transfer.
* You continue to use open systems to argue about the behavior of isolated systems.
* You believe that thermodynamic work transfers increase entropy, when by definition they do not.
* You believe an isolated system that turns thermal energy into work until the gases reach absolute zero is not in violation of the 2nd Law of Thermodynamics.
* You believe that air at -50C from the upper troposphere to the surface would still be at -50C when it reached the surface, despite adiabatic compression.
* You cannot work out the heat flows in a simple two-cylinder isolated system, even when prodded to work it out step-by-step.
* You propose a new “heat creep” mechanism that would overturn 200 years of thermodynamic analysis, but when asked repeatedly for any quantitative details of this mechanism (what is the heat-creep equivalent of the conductivity constant for a material), you have no answer whatsoever).
(And I won’t even bother getting into your ridiculous ideas about radiative heat transfer that have no physical basis whatsoever.)
Thirty years ago, I learned about people like you who never really understood the material you studied. Your differences with standard thermodynamics (and I am talking about fields other than climate science) are because of your complete inability to understand the very basics of the field. You do NOT have ANY contributions to make to the science.
Doug Cotton,
Your own words “Inter-molecular radiation in water almost totally over-powers the gravito-thermal effect.” Not only must it overpower it but it must actually reverse it since the gradient is negative (temp goes down).
This would make Graeff’s experiments totally wrong. The same
Inter-molecular radiation must be acting in his column if it is acting in the ocean. How does this radiation work? You state IR can’t penetrate water yet somehow it can work through 1000 meters of water to equalize temperature?
Also Venus atmosphere is mostly carbon dioxide. If the small amount of water vapor can have this super profound effect (reducing your heat creep), why would NOT carbon dioxide totally overpower “heat creep” and prevent even a chance for a super hot surface. Think about it. You want to have your cake and eat it too. You can’t have it both ways. If inter-molecular radiation overpowers heat creep in the ocean and reduces heat creep in our atmosphere, then logical conclusion is that the same process would overpower heat creep completely and it could not exist on Venus. Venus should be as isothermal as the oceans of Earth!
the gradient is negative”
Yep, because thermal energy is being supplied at the top by conduction from the atmosphere, and it goes from hot to cold in the thermocline (downwards) by convective heat transfer.
I explained in the paper what Graeff did with his water experiments and why they showed a gradient.
It’s not IR “working through 1000 meters of water.” Blimey, you seem to think there is only one form of heat transfer in the universe.
I’ve already told you, carbon dioxide radiates in far fewer frequency bands than does water vapor, so its effect is far less per molecule.
What is the sensitivity to a 1% increase in water vapor, Norman? I’m not answering any more of your questions until you think about this question and answer it in your own words. That’s my teaching method, but so far you are not remembering what I have already taught you.
What is the sensitivity to a 1% increase in the concentration of water vapor, Curt?
Every single statement you make is assertive without so much as a quote from my website or papers, let alone even a link to some physics documentation to support your claims, or any empirical evidence that water vapor warms, for example.
All of what you say about what you think I believe is incorrect, as are your other statements also. In saying what you think I “believe” you have not quoted a single sentence from my website or the linked papers. For example, what I wrote about was a specific case of downward winds blowing air at -50°C from the upper troposphere to the South Pole That is not an adiabatic process in an isolated system, and that is why there is no significant change in temperature. You cannot quote any law of physics which says compression must maintain higher temperatures. You haven’t proved that :”heat creep” violates the Second Law and you never will because it is a corollary of that law. You cannot prove that variations in the gravitational potential energy of a molecule moving between collisions do not affect entropy. When a ball is falling in a vacuum tube the process is not reversible and entropy is increasing because potential energy potentials are decreasing. You need only to read http://entropylaw.com to learn about more recent understandings of entropy and the Second Law, and you will see examples involving gravitational potential energy.. I did explain what happens in a two cylinder experiment here and my paper on radiation is merely extending the work of a professor of applied mathematics, so you can argue with him about his computations. What the IPCC claims would violate the Second Law as I explain in that paper. You could read about how the concept of an ideal atmosphere being an isolated system is a useful one which gives a reasonable approximation of reality. It is after all what is used in all those energy budget diagrams. Without imagining a column of the troposphere as an isolated system you get nowhere. Likewise you get nowhere with understanding long term mean planetary temperatures if you introduce wind and weather conditions which may even reverse the Second Law process of increasing entropy. In short, Curt you have no understanding of the Kinetic Theory of Gases and the kind of information at http://entropylaw.com which I hope silent readers will study.
And you continue to demonstrate to silent readers that you are unable to answer the question “What is the sensitivity to a 1% increase in the concentration of water vapor, Curt?”
Doug 5:54am: Your answer is 42. Well documented.
Ball4,
Wrong Doug. Douglas Adams wrote The Hitchhiker’s Guide to the Galaxy and of course came up with the answer to life the universe and well everything “42.” The answer supposedly required the assistance of an enormous computer. Don’t count on Doug doing the math on this subject. Has he been providing a lot of math up til now?
You might try by learning to fly. According to the book all you have to do is throw yourself at the ground and miss!!!
Have a great day!
Then Ball4 produce links to such documentation and evidence from real world temperature data of your 42 degrees of warming for each 1% of water vapor. For example, if a region with 0.5% is warmed from 255K to (255+21)K find data to confirm. If a region with 4% water vapor is warmed by 168 degrees above 255K find evidence. I have produced evidence for my contention that it cools and never warms the surface, and I’ve produced the physics that supports this evidence, and which physics the evidence in Earth’s troposphere and in centrifugal force fields and in other planets all supports.
Poor Doug doesn’t get the joke…
Hi Curt,
How many people would? How many people have even read the book? Personally, I have but I don’t count myself among the majority on this one.
Have a great day!
No you don’t get the point that whatever calculated figure you enter (more realistically 10 to 25 degrees per 1% as in my comment below) you get absurd results with rain forests far hotter than deserts at similar latitude and altitude. If mean water vapor were just under 1% you could actually assume that, say, 0.8% produced 32 degrees of warming and thus get 40 degrees of warming per 1%. But I estimate 15 to 20 degrees per 1% is realistic if mean water vapor concentration were about 1.5% to 2% – assuming it is correct that it varies between about 1% and 4%.
How many have read the book? Well I banked a royalty cheque yesterday as it happens, but I have not recovered anywhere near the $3,000 initial outlay. I have personally sent out 96 out of my first 100 author’s copies, but (having made many millions from my software business, whilst doing part time tutoring in math and physics) I am not interested in profit on the book. The website has had nearly 7,000 hits since January 8th this year and the paper on which the book is based is now back on line there, so no one needs to buy the book now.
Doug:
I have a big enough target just quoting your misstatements here! I don’t quote your website or papers much because I am not a fan of fiction…
Let’s take your continued bizarre assertion that the conversion of gravitational potential energy to vertical velocity is an irreversible process that increases entropy.
Take a good hard rubber ball, like the “Super Ball” we had when we were kids. Drop it toward a hard surface. It will bounce back up virtually as high as the height from which it is released.
Now take an equivalently sized ball of putty and drop it toward the same surface. It will stick to the surface.
If you were correct that the conversion of graviational PE to downard KE was irreversible, increasing entropy, neither ball could bounce back up AT ALL, because that conversion would decrease entropy.
What’s the difference between the two cases? The collision of the hard rubber ball with the bottom surface is almost completely elastic, making the entire process almost completely reversible. The collision of the putty ball with the bottom surface in completely inelastic, converting all of the kinetic energy of the ball into thermal energy, increasing the entropy of the system due to this inelastic collision.
The comparison of the two cases shows that the irreversibility must come from the nature of the inelastic collision, not the falling itself. This is such a basic, basic point, and it has been explained to you repeatedly, but you don’t have the intellectual capability to grasp it.
And your fundamental error here completely invalidates the rest of your analysis, because the collisions of gas molecules are completely elastic, whether between molecules, or of a molecule and a surface.
I have shown, AND YOU HAVE AGREED, that if there were a “heat creep” mechanism, Maxwell’s thought experiment of putting a heat engine between two cylinders of different Cp would lead to a conversion in an isolated system of thermal energy to work until there is no more thermal energy. This is as blatant a 2nd Law violation as you can imagine. In a typical introductory thermodynamics class, the impossibility of this type of conversion is the first example of 2nd Law limitations given to students. Yet you cannot comprehend that this is a 2nd Law violation!
And in the bigger sense, you show complete ignorance of the process of scientific analysis. The reason we have thought experiments and laboratory experiments is to simplify things from the complexity of the real world/universe and to be able to isolate individual physical effects to test them. Until you thoroughly understand the underlying physical effects well, it is pretty pointless to try to untangle many interacting effects, especially when there are a lot of unkowns.
So when you ask, “What is the sensitivity to a 1% increase in the concentration of water vapor?”, you haven’t provided nearly enough information to answer that question. What exactly are the starting conditions? What assumptions should be made? What is the particulate load to seed condensation into clouds? What is the general albedo? Without answers to dozens of questions like this, it is an unanswerable question.
That is the key problem of climate science — in the real world, it is virtually impossible to untangle all of the different effects. But asserting things that are proved demonstrably false in controlled laboratory experiments, as you repeatedly do, does absolutely nothing to improve our understanding.
Yes Kinetic Theory assumes elastic collisions, and so do I when explaining that the KE of molecules about to collide must be equal for there to be no further diffusion of kinetic energy in such collisions. For this to be the case (PE+KE) must be homogeneous at all altitudes. It’s not hard to understand.
Entropy is determined at a macro level, not individual molecular motion. It is determined by mean net movement of molecules and mean net changes in energy potentials. When there is net downward movement of molecules in that cylinder rotated from horizontal to vertical, there is clearly a situation in which entropy is increasing until the system reaches thermodynamic equilibrium which, by definition, is the state of maximum entropy, and that state no longer has any unbalanced energy potentials at the macro level. Hence nothing changes at the macro level, and the density gradient and the temperature gradient that form simultaneously become stable simultaneously when molecules that are about to collide have (on average) equal kinetic energy. That’s why temperature becomes isothermal in a horizontal plane where (PE+KE) is still homogeneous, but PE is constant, so KE is likewise. But in a vertical plane, as you acknowledge, (PE+KE) is constant during all free flight of molecules, and so if there is a vertical component in that flight between collisions, and if the (PE+KE) was the same at the previous collision height and at the next collision height, then that interchange of PE and KE does indeed ensure that the new KE of the rising or falling molecule will, by the time it is about to collide, match the KE of the molecule already at the height where the collision is about to take place.
It’s not hard to understand.
To Roy and silent readers:
Regarding the sensitivity to the most prevalent greenhouse gas (water vapor) the real reason why Curt does not tender an answer (even within limits such as 10 to 25 degrees per 1%) is that he is stumped. He realizes that, if the Hansen-cum- IPCC conjecture about there being isothermal conditions in the troposphere were true (contrary to what the Second Law of Thermodynamics implies) then a mean of between 1% and 2% of water vapor would have to do at least 25 degrees (maybe 30 degrees) of that proverbial “33 degrees of warming” which is clearly stated in the IPCC website.
That leaves Curt with an inescapable conclusion that, if mean world water vapor were to increase by even 0.5% then there would be considerable warming, which would be contrary to the fact that cloud cover would increase, albedo would increase, the effective radiating temperature would be lowered (by 0.92 degree if albedo increased from 30% to 31% for example) and the magnitude of the temperature gradient would decrease, meaning the whole thermal profile would rotate and be lowered at the surface end. Everything points to cooling, not warming, including my study of empirical evidence over 30 years on three continents.
So Curt makes wishy washy excuses for not even trying to answer the question, but no one pulls wool over Cotton’s eyes.
Doug:
You don’t even have the intellectual ability to state the question you want to ask! You asked about the effect of a 1% increase in water vapor. That would be, for example, an increase from 10,000ppm to 10,100ppm. But that was not the question you were answering yourself.
If you want to play the scientific game, Doug, you must learn to express yourself precisely. As with your mathematical skills, you obviously don’t have the capability to do this!
And HOW DARE YOU state that your meaningless questions “leave [me] with the inescapable conclusion” or that I “am stumped”. You cannot know that! Go back to my earlier comment where I anticipated all of the confounding factors you just stated.
Roy and others:
Over recent years I have tied in knots numerous supporters of the AGW hoax, mostly those who have had an education after 1980 when institutions toed the “global warming” line with strong political agendas that were obvious in the propaganda from Al Gore (and now Obama) none of whom of course understood thermodynamics. Now we know the United Nations wants to control world governments and abolish capitalism.
The brilliant 19th century physicist (first to assess the size of air molecules) was decades ahead of his time when he postulated that gravity would form a temperature gradient at the molecular level, because molecules in free flight between collisions swap kinetic energy and gravitational potential energy. Maxwell did not take the trouble to understand his explanation, much like Curt, Ball4, Tim Folkerts and others do not either.
Maxwell’s thought experiments ignored the very process Loschmidt was describing because Maxwell failed to understand that a new state of thermodynamic equilibrium will always evolve (after temporary initial energy flows) when, with thermal connections, two systems become one with a new temperature gradient in between the original gradients.
Now with better understanding of the Second Law of Thermodynamics (especially since 1988 as explained here) the Loschmidt effect is easily understood and explained using the Kinetic Theory of Gases wherein the assumptions include the fact that “because they have mass, the gas molecules will be affected by gravity.”
Curt is tied in knots because, on a previous thread he assertively claimed that a radial temperature gradient would not form in a centrifuge machine with colder gas at its center. When I produced evidence of gas being cooled right down to 1K (-272°C) in this centrifuge machine he ignored such and of course did not admit his error.
As with others in the past, when they have their back against the wall, they avoid discussing the physics and evidence I present, avoid answering questions like “What is the sensitivity for a 1% increase in water vapor?” and, instead, misrepresent what I have said, make calls to “authority” or even their own qualifications from institutions promulgating the false physics that has been necessary to “explain” the hoax, which physics they assertively claim to be correct.
In Curt’s case, he claims that the equation usually used for entropy does not need to consider changes in gravitational potential energy at the molecular level, even though that energy is of course included in the internal energy, U and changes in the composition of internal energy affect entropy.
What Curt needs to understand is that the Second Law is about entropy increasing, and that happens only when (unbalanced) energy potentials are being reduced.
If we start with a sealed, insulated horizontal cylinder filled with gas that is isothermal, but rotate it to a vertical position, then there is unbalanced gravitational potential energy. We know that state is not stable because we know that a density gradient will form as entropy increases until that gradient becomes stable, and that is when we have the state of thermodynamic equilibrium.
I have explained with Kinetic Theory why Loschmidt was right about there also being a temperature gradient that evolves simultaneously as more molecules move downwards and gain kinetic energy. I have explained precisely how the density gradient forms (because gravity curves the paths of molecules) and I have explained that entropy is maximized when conditions are isentropic and the mean sum of molecular (PE+KE) is homogeneous. This directly implies that a temperature gradient exists. The rest follows, as shown in the heat creep diagrams here.
Without realizing that this slow downward diffusion and convective heat transfer up the temperature gradient happens due to excessive new energy at the higher altitudes, you have no explanation for any observed planetary surface temperatures anywhere in the Solar System, because the Sun’s direct radiation is insufficient, and back radiation does not help the Sun to raise the surface temperature.
Doug:
No matter how many times it is explained to you, you don’t have the intellectual horsepower to understand Maxwell’s argument. He ACCEPTED Loschmidt’s theories to see where they led. And they did NOT lead to a new equilibrium condition! You have even admitted that the gases would go all the way to absolute zero, turning all of their thermal energy into work without rejecting any heat to an external reservoir. This is a complete 2nd Law violation! So Loschmidt’s theories must be rejected.
And you absolutely refuse to come to grips with your erroneous use of completely open thermodynamic systems like your vortex tube to make statements about isolated systems. In trying to defend this, you have demonstrated that you don’t even understand what an open thermodynamic systems IS!
When it comes to real scientific understanding my money goes to Curt over Doug any day. Doug’s physics deals with no math so anything is then possible. Whatever he needs he makes up at the moment and calls it enlightened physics but has zero math to back it up so we might test his assertions with real calculations.
Doug Cotton,
Your twisted and distorted logic falls apart!!
“I have already explained in the paper how and why Graeff got his calculations wrong. Gradient = 9.8 / 4.0 (from Cp here = 2.45 K/Km = 0.00245 K/meter.
You see what I mean? This is yet another example of how you forget what I have already explained and taught you. You are never going to prove teh Second Law of Thermodynamics wrong Norman, so give up”
If the gradient for water is this small then why did Graeff find such a measurable amount of variation in temp with his test??? Something here is not right!
You claim you tied Curt in knots? No that is not the case. Your twisted logic has tied you up in so many knots of irrational thought that you are unable to allow an expert like Curt to help you untie them.
Don’t make Curt your enemy! He seem very intelligent and has a good understanding of thermodynamics. Why not listen to what he is saying and give his view a bit of respect? You may be 69 but that is not too old to learn. You have many faulty logical conclusions that do not make sense when merged.
With Venus you make very little sense. Your claim is that “heat creep” is a slow process. Your claim is that Carbon Dioxide is a cooling gas in an atmosphere (even if less than water vapor it still acts as a coolant according to your illogical system of thought process) makes it very difficult for your heat creep to warm the surface of Venus. You will not give a mathematical description of the rate of “heat creep” although Curt has asked for it more than once. You state it is slow but Venus surface radiation is very fast. Radiation is a super fast form of heat transfer! If the “heat creep” is slow how can it keep up with the tremendous loss of energy by the radiant outward flux?
Also you hate my bringing up a thermos bottle. A real physical object in the real world (not a thought experiment…far more real than your pseudo physics you peddle). The hot liquid inside a thermos is emitting radiation based upon its temperature and surface area but it takes several hours to cool, why? If the radiant energy is not returning to the hot fluid via reflection off the room temperature mirrors, what is keeping the fluid warm?? If radiation emitted from an object cannot be made to return to this object and maintain its heat how does it stay warm for hours and the better the thermos the longer the fluid stays warm. Why can this be based upon your unreal understanding of radiation and who you falsely perceive how things work (only in your world though, not in anyone else’s).
<I."why did Graeff find such a measurable amount of variation in temp with his test???"
Why do you keep asking the same question. I answered that by referring you to the explanation in my paper. You could just search the word “Graeff” in that pdf.
I really can’t be bothered reading more of your ill-informed crap, Norman. You need to read, study and inwardly digest the content of my Feb 2013 paper where, once you understand what I have explained, you’ll realize how totally and utterly irrelevant is your comment “Your claim is that Carbon Dioxide is a cooling gas … makes it very difficult for your heat creep to warm the surface of Venus.”
What is the sensitivity for each 1% of water vapor in Earth’s troposphere?
Go to this comment.
Hi Norman,
You seem to have as many issues as you claim Doug does. You state:
“Your claim is that Carbon Dioxide is a cooling gas in an atmosphere (even if less than water vapor it still acts as a coolant according to your illogical system of thought process) makes it very difficult for your heat creep to warm the surface of Venus.”
Well, a stronger absorber by definition means a stronger emitter. We’ve discussed this issue before and your sloppiness with terms and the laws of physics seems to cloud your thinking. REFLECTION IS SIMPLY A DIFFERENT PROCESS THAN ABSORPTION AND EMISSION. The coffee cup analogy fails because atmospheric gasses do not reflect all radiation back to the surface. Yes, CO2 absorbs IR and re-emits a portion back to the surface, but total emissions proceed at an overall FASTER RATE than their di-atomic neighbors. Which can only mean it helps cool the atmosphere in general faster than di-atomic gas compounds. Which is why the upper troposphere has a long history of significant temperature reduction as tri-atomic gas compounds accumulate there. The rate of decline far exceeds the supposed lower troposphere temperature increase which from just about every data-set available must be ADJUSTED to be taken seriously at all! In any case, the apparent differences between di-atomic and tri-atomic or greater gas compounds simply seem to be one of degree as Ball4 stated:
“N2, O2, argon all absorb. All matter absorbs. At all frequencies. All temperatures.”
That being the case whether one views a compound as COOLING or WARMING will prove to be a relative distinction at best. Thanks and…
Have a great day!
Correction:
“Coffee cup analogy” should read thermos bottle analogy. Pardon me!
Have a great day!
What gets absorbed from insolation and what does the absorbing are seen here.
An experiment for Norman to do:
Try holding a mirror above a sunny patch but a bit to the side so you don’t block the Sun some clear morning. Take care to reflect surface radiation from that patch back to itself. Keep temperature records each 10 minutes with spike thermometers in the ground, both in the patch and in similar ground at a short distance away. Adjust the mirror as required. I’m sure Norman will love to do this cheap experiment similar to ones I have done by day and also all through the night years ago.
Hi JohnKl
Ball4 is not correct with his statement that all gasses absorb all frequencies all temperatures. That is the thing with absorbing gases. They only absorb and emit specific frequencies based upon their bonding structure. It is a well established experimentally proven fact on that one.
I linked somewhere to a physics study of a radiant energy from surface to space and how long it would take to exist the atmosphere.
Rather than reflect radiation the better term is redirect (which surfaces do with visible light which allows you to see things). The white light hits surfaces and some redirect certain frequencies away from the surface and you see some color that that was not absorbed. What actually happens to the radiation at the surface is a complex process that requires some strong math. The basics is the light path is redirected.
I think the big thing Doug is missing about carbon dioxide absorption of radiant IR is that the carbon dioxide does not nee to speed up its motion after it receives the energy, just the atoms of the molecule start vibrating faster. It can exchange this energy with other molecules around it and heat them up or reemit in a random direction. When the energy is emitted it is redirecting the original path of the radiation from surface to space to then any random direction.
The whole total process is very complex and can’t easily be resolved. One should always keep their mind open and active on it.
What a joke!
Norman claims carbon dioxide does not need to “speed up its [a molecule’s] motion after it receives the energy, just the atoms of the molecule start vibrating faster.”
This PROVES your incompetence in physics. You cannot have the vibrational degrees of freedom increasing without the translational ones (x- y- and z-axis) each increasing and in fact being equal to each vibrational and rotational degree of freedom. Read about the Equipartition Theorem wherever you choose to do so.
Now, Roy and silent readers, with Norman making such a basic blunder in physics, are you going to take note of ANYTHING he says relating to physics, in which he is not qualified and about which he has next to nothing in the way of understanding?
This is pretty rich coming from a guy whose physics say a bouncing ball violates the laws of thermodynamics!
As I have pointed out above, if the conversion of gravitational potential energy to vertical kinetic energy is an irreversible, entropy-increasing process, as Doug has stated repeatedly, then the reverse process, that of converting vertical kinetic energy to gravitational potential energy would be an entropy-decreasing process, impossible in an isolated system. Yet this is exactly what happens when a ball (or gas molecule) bounces off the bottom surface of an isolated system!
So I, too, ask the question:
Roy and silent readers, with Doug making such a basic blunder in physics, are you going to take note of ANYTHING he says relating to physics, in which he is not qualified and about which he has next to nothing in the way of understanding?
Entropy is a macro property. If your “ball” is in fact an individual molecule that is experiencing random motion and collisions, the issue of entropy is irrelevant. But if you are determining whether there is any net movement of molecules in a given direction (detectable at the macro level) that is altogether different.
So when a horizontal sealed insulated cylinder is rotated to a vertical position in a gravitational field there will be an irreversible net movement of molecules downwards as entropy is increasing. As the Second Law tells us, when entropy then reaches a maximum we have thermodynamic equilibrium with its associated now-stable density gradient.
Simultaneously, the net movement downwards has also caused a net redistribution of kinetic energy, with a reduction in mean KE per molecule (temperature) at the top and a gain at the bottom, simply because more molecules have fallen than have risen. That’s how the temperature gradient evolves at the same time as does the density gradient and for the same reason. Because pressure is proportional to the product of density and temperature, we also get a pressure gradient as a corollary of the process described in the Second Law.
Norman 4:56am – “Ball4 is not correct with his statement that all gasses absorb all frequencies all temperatures.”
From Max Planck’s Nov. 1912 paper here is his formula (274) for the specific intensity of radiation from a volume of gas found from his experiments and those of many others:
v=incident beam frequency, T = gas temperature in kelvin
specific intensity absorbed by all gases tested = constant * v * (e^constant*v/T-1)^-1
By experiment, this specific intensity absorbed equation is never 0 at any frequency or any temperature. So all gases absorb all frequencies all temperatures.
Hi Norman,
You stated:
“Ball4 is not correct with his statement that all gasses absorb all frequencies all temperatures. That is the thing with absorbing gases. They only absorb and emit specific frequencies based upon their bonding structure. It is a well established experimentally proven fact on that one.”
As you state it’s Ball4’s claim and I’m quite happy and thank God the glass in my home windows do not absorb effectively in the visible light spectrum. Please note Ball4 did not just mention “gasses” but included all matter. In addition, I’m quite aware that different molecular shapes absorb energy differently. However, substances emit/absorb different energy frequencies with varying effectiveness. They may apparently also emit/absorb based on their given phase at any moment. For example, in a previous post I provided the broad but distinct emission spectrum for water vapor. Joel Shore quickly chimed in that in contradistinction liquid water absorbs at close to 100 percent. David Appel once lamented Angstrom’s saturation fallacy and pointed out the extent to which CO2 absorbs only a partially in certain bandwidths and thus it’s a fallacy to conclude that increasing atmospheric CO2 would not increase absorption.
Science concerns the facts and laws of nature and since I cannot rule out the possibility that all matter can be effected by and perhaps absorb/emit to some extent energy at any given wavelength even if not observed it seems rash to dismiss Ball4’s comment out of hand. Although, I take your point and certainly do understand that limits apply to absorption/emission of energy from matter at all levels.
You go on to state:
“The white light hits surfaces and some redirect certain frequencies away from the surface and you see some color that that was not absorbed. What actually happens to the radiation at the surface is a complex process that requires some strong math. The basics is the light path is redirected.”
Of course, Newton first used a prism to refract white light into the rainbow spectra of red, yellow, blue primary colors and intermediates we all now know and take for granted. As to the math, my schooling in physics may be limited but I’m quite trained since high school and beyond in higher math calculus included. The problem with relying too much on math is that any equation will only be as good as the assumptions behind it’s use in explaining and/or modeling phenomenon. Over your previous seemingly never ending discussions with Doug, I’ve noted you bring up Maxwell frequently and deride Doug’s math ability. Please note again some cautions in this regard. While I agree and have stated numerous times that Maxwell surpasses most any physicist I can think of in terms of achievements especially in regards to his equations, he has a track record of getting things wrong. Had the world including M Curie and others idolized him to much we likely never would have split the atoms of DECAYING matter (i.e. uranium) since he held to beliefs that prevented him from recognizing it at the time. His understanding of ABSORPTION and EMISSION also proved faulty and had to be corrected by Max Planck. That’s how science works. As to Doug’s ability to defend his position. Time will tell.
Have a great day!
JohnKl,
Thank you for the thoughtful and intelligent post. I think you have me confused with Curt on Maxwell. I mention him much less than Curt does.
I totally agree with you on the math. Wrong assumption will negate even the most perfect math derivations. The purpose of the math is for precision sake. It does not mean the equations are correct or will work but you need them as a starting point as they are useable for calculation purpose. If I was a NASA engineer trying to figure out obrital paths using Doug’s approach I would say gravity it stronger at the surface than higher above and that is is based upon the mass of the planet and that it always is and attractive force. Now all these statements may be true and correct but how will this help design a rocket or how much fuel I would need? Doug provides very little math with his ideas. Here is his crowning math achievement “T/H = -g/Cp” How can you verify his claims with this. He says heat creep is slow process. How slow. We know radiation is a very rapid method of heat exchange. In his Venus theory you have this super slow (which he provides no calculus to use, just it is slow wow that really helps) “heat creep” that keeps the surface Hot even though this hot surface is constantly radiating away from it 16000 watts/meter^2! You would need a very fast flow of energy to maintain the surface temperature since that 16000 joules of energy/second meter^2 is leaving at the speed of light.
I predict either Doug will stop communicating with me or Curt or now you (you have joined the unwanted because you did not blindly accept his hypothesis and can’t draw from memory his “heat creep” graph) and just go no posting randomly on Roy Spencer’s blog for the next several days.
I have provided all the computations I need to provide to support my hypothesis, once that hypothesis is understood. Maybe this new comment will help you understand better.
The thermal energy actually transferred by radiation out of the surface of Venus into (mostly) the very close regions in its lower troposphere is nothing like 16,000W/m^2 of thermal energy. That is the outward radiative flux of electromagnetic energy. I suggest you read my 2012 peer-reviewed paper on radiation and the Second Law of Thermodynamics and let me know if you ever publish an attempted refutation.
The input by day to the Venus surface by downward convective heat transfer must be at a comparable rate to the upward convective heat transfer at night, but I have not been up there to measure it, and neither have you. It can be more than the Sun delivers, because some thermal energy will cycle into the lower troposphere at night, and back to the surface by day. I have no intention of placing values on such rates, thanks, as I do not try to bluff people with guesswork.
If I stop fighting the hoax, you can probably assume I’m dead, but I’m into supplementation that slows the rate of human aging by about 30%, so let’s see.
Now it’s over to you to respond to the “coincidences issue” which you have not even understood yet.
2nd correction:
The upper atmosphere temp declined actually applies to the Stratosphere, not so much the upper troposphere.
I can see that you also, JohnKl, have not understood the new 21st century paradigm in climate science here because I see no reference to the effect of gravity which “obviates the need for concern over greenhouse gases” as BigWaveDave wrote here three years ago. It’s worth repeating one more time as his credentials might even impress Norman …
BigWaveDave March 1, 2012 at 4:19 pm
Tim Folkerts:
You asked …what qualifications do you have to judge a disagreement between PhD physicists on issues of fundamental thermodynamics?
I have been earning a living as an engineer specializing in cutting edge technology for very large scale thermal energy transfer processes and power systems for close to 40 years. My credentials include BS, JD and PE, and I have four patents.
As for my qualifications to engage in argument with PhD’s, I have many times been part of and have led teams with PhD team mates. I was also married to a PhD for 20 years.
Because the import of the consequence of the radial temperature gradient created by pressurizing a spherical body of gas by gravity, from the inside only, is that it obviates the need for concern over GHG’s. And, because this is based on long established fundamental principles that were apparently forgotten or never learned by many PhD’s, it is not something that can be left as an acceptable disagreement.
Hi Doug,
You stated:
“I can see that you also, JohnKl, have not understood the new 21st century paradigm in climate science here because I see no reference to the effect of gravity which “obviates the need for concern over greenhouse gases” as BigWaveDave wrote here three years ago.”
Doug, I didn’t deny that gravity acts centripetally to conserve mass and energy and don’t deny that the lapse rate helps establish temperature at any given altitude so I have no idea what your critique means other than you want to throw Big Wave Dave’s name in Norman’s face! In fact, I have asserted for above and at length in previous discussions that tri-atomic gas compounds help radiate to space faster and thus cool, that as Big Wave Dave suggests gravity acting upon the relative velocities of gas molecules can provide temperature differentials based on altitude and I’ve never asserted as Roy apparently does that the atmosphere would be completely iso-thermal and without temperature variance without tri-atomic and larger atmospheric gas molecules. So please explain your complaint opaque as it appears to be.
Have a great day!
And do you agree or not, JohnKl, that the density and temperature gradients represent the state of thermodynamic equilibrium (maximum entropy) and thus, as a consequence, do you agree with my explanation here (and in my linked paper) concerning the issue of how the required thermal energy gets into planetary surfaces in order to raise the surface temperature on the sunlit side?
If you don’t agree with my hypothesis, then explain in your own words how the necessary thermal energy gets down to the base of the nominal troposphere of Uranus in order to maintain temperatures hotter than Earth. Also explain in general terms how you would calculate (as I have) what that temperature ought to be, given that the methane layer in the Uranus stratosphere (about 350Km further up) is in radiative balance with the Sun at (58±3)K.
Doug:
You continue to make the egregious beginner’s error that the conversion of gravitational potential energy to vertical kinetic energy is an irreversible, entropy-increasing process. This is trivially refuted, for instance by the simple case of a bouncing ball.
This error completely invalidates all your subsequent analysis of entropy maximization at a non-zero lapse rate for an isolated system.
So I ask the question:
Roy and silent readers, with Doug making such a basic blunder in physics, are you going to take note of ANYTHING he says relating to physics, in which he is not qualified and about which he has next to nothing in the way of understanding?
Hi Doug,
You asked several questions. Allow me to pick one at a time.
You asked:
And do you agree or not, JohnKl, that the density and temperature gradients represent the state of thermodynamic equilibrium (maximum entropy)…?
No! As you yourself suggest in your own link they may tend toward INCREASING entropy but it would defy the 2nd law of thermodynamics if in any closed system entropy (amount of energy unavailable for useful work) stopped increasing!
You further ask:
“…do you agree with my explanation here (and in my linked paper) concerning the issue of how the required thermal energy gets into planetary surfaces in order to raise the surface temperature on the sunlit side?”
Largely, however a significant amount of direct sunlight (~10% of that found on Earth) reaches the surface and accounts in my opinion for most if not all of that increase. Moreover you have suggested in past posts that the solar energy including “heat creep” is sufficient to account completely for Venusian temperatures. Imo, unless one includes fissile geological action beneath the surface and the resulting energy and mass sufficient to form an atmosphere 93 times larger than the Earth’s the 16k w/m^2 remains woefully unaccounted for!
Have a great day!
No, JohnKl, the Second Law tells us entropy in an isolated system will eventually stop increasing. I very, very strongly recommend you read most of what is on this excellent site: http://entropylaw.com because, by definition, thermodynamic equilibrium has maximum entropy (within the constraints of an isolated system) and so entropy has stopped increasing. In really calm conditions in the early pre-dawn hours that can happen – close enough anyway – and advection ceases but the environmental lapse rate remains intact. See also this comment.
No, JohnKl, the sunlight reaching the Venus surface (say 20W/m^2 compared with 168W/m^2 for Earth) only supports a black body temperature of 137K which is about -136°C. At any existing temperature above that (actually about 735K) absolutely no thermal energy will be delivered to the surface by solar radiation. It would have to be over 16,548W/m^2 to raise the temperature even of a perfect black body in space that was already at 735K.
See my paper on radiation linked from the “Evidence” page at http://climate-change-theory.com and read about the Stefan-Boltzmann law.
See also this comment. There’s no problem regarding the 16,000W/m^2 when you understand the content of both my papers, but that requires two or three hours of studious attention to what is written therein.
Not at all Doug,
You assert without evidence:
“No, JohnKl, the Second Law tells us entropy in an isolated system will eventually stop increasing.”
The 2nd Law of Thermodynamics claims nothing of the sort. The 2nd Law of thermodynamics states:
In any CLOSED system the amount of energy unavailable for useful work must always INCREASE.
Your claims as to the existence of thermodynamic equilibrium are purely theoretical if not merely imaginative. Your link provides other links that make similar blunders like the following statement:
“The second law says only that entropy is maximized while the law of maximum entropy production says it is maximized (potentials minimized) at the fastest rate given the constraints. Like the active nature of the second law, the law of maximum entropy production is intuitively easy to grasp and empirically demonstrate.”
Of course imo they don’ demonstrate it by claims to thermodynamic equilibrium in OPEN systems.
Have a great day!
P.S. – Law always triumphs over theory.
I don’t use the over simplified version of the Second Law which climatologists love to quote, because it is incorrect.
See http://entropylaw.com and my two peer-reviewed papers on the Second Law of Thermodynamics.
If that law were not about entropy maximization then it would not lead to isothermal conditions in a horizontal plane. So take your garbage “law” to your unreal world.
From the above website …
The key insight was that the world is inherently active, and that whenever an energy distribution is out of equilibrium a potential or thermodynamic “force” (the gradient of a potential) exists that the world acts spontaneously to dissipate or minimize. All real-world change or dynamics is seen to follow, or be motivated, by this law. So whereas the first law expresses that which remains the same, or is time-symmetric, in all real-world processes the second law expresses that which changes and motivates the change, the fundamental time-asymmetry, in all real-world process. Clausius coined the term “entropy” to refer to the dissipated potential and the second law, in its most general form, states that the world acts spontaneously to minimize potentials (or equivalently maximize entropy), and with this, active end-directedness or time-asymmetry was, for the first time, given a universal physical basis. The balance equation of the second law, expressed as S > 0, says that in all natural processes the entropy of the world always increases, and thus whereas with the first law there is no time, and the past, present, and future are indistinguishable, the second law, with its one-way flow, introduces the basis for telling the difference. “
Even Wikipedia gets it right …
“According to the second law of thermodynamics the entropy of an isolated system never decreases; such a system will spontaneously evolve toward thermodynamic equilibrium, the configuration with maximum entropy.”
Note that it does not apply to a closed system: it applies to an isolated system.
So go and edit Wikipedia and tell me when your edits stick. William Connolly will probably love you for pointing out this sentence which he has failed to modify to suit the AGW agenda.
assert without evidence” ????
Indeed, JohnKl /sarc
Hi Doug,
You state:
“Note that it does not apply to a closed system: it applies to an isolated system.”
Fine I understood, apparently erroneously, that a CLOSED system meant one that was closed to both matter and energy. It turns out according to the modern vernacular and Wikipedia that such a definition applies to ISOLATED systems:
“In physical science, an isolated system is either (1) a thermodynamic system which is completely enclosed by walls through which can pass neither matter nor energy, though they can move around inside it; or (2) a physical system so far removed from others that it does not interact with them, though it is subject to its own gravity. Usually an isolated system is free from effects of long-range external forces such as gravity. The walls of an isolated thermodynamic system are adiabatic, rigid, and impermeable to matter.”
Frankly I’m not a professional Physicist and haven’t kept up with the modern linguistic gymnastics. The 2nd law definition I used came from a book several decades old before the AGW paranoia became a popular delusion that dealt with a completely different subject. Use the term ISOLATED if it makes you feel better in my definition YOU HAVE NOT PROVED THE EXISTENCE OF ANYTHING RESEMBLING MAXIMUM ENTROPY. I AGREE WITH YOU THAT IF ENERGY OR MATTER ENTERS AN ISOLATED (IF YOU WILL) SYSTEM THE CLAIM THEN THAT ENTROPY MUST INCREASE BECOMES ABSURD ON IT’S FACE. You also state:
“If that law were not about entropy maximization then it would not lead to isothermal conditions in a horizontal plane. So take your garbage “law” to your unreal world.”
Well Doug it’s not my law and my world works quite realistically, whether or not the 21st century paradigm disagrees or whether the definitions used for various terms mutate over time. However, let’s test just see how REALISTICALLY your bombast holds up. You claim above that the 2nd Law of Thermodynamics in some ill-defined way leads to isothermal conditions in a horizontal plane. Really!!! DOUG PLEASE ENLIGHTEN THE SILENT READERS WHERE YOU’VE OBSERVED AN ISOTHERMAL PLANE IN AND ISOLATED SYSTEM ( I.E. A SYSTEM THAT HAS NO INPUTS OF MATTER OR ENERGY)? OR EVEN HOW YOU MANAGED TO OBSERVE IT ASSUMING YOU WE’RE NOT LOCKED INTO SAID ISOLATED SYSTEM?
To be fair it’s dubious whether a truly ISOLATED system even exists. I don’t believe so. Frankly I’ve never seen one. However, we’ve all witnessed systems largely cut off from energy and matter inputs, but imo not absolutely. Personally, I AGREE WHOLEHEARTEDLY WITH THE QUOTE ABOUT CLAUSIUS YOU TOOK FROM YOUR OWN WEBSITE:
“The key insight was that the world is inherently active, and that whenever an energy distribution is out of equilibrium a potential or thermodynamic “force” (the gradient of a potential) exists that the world acts spontaneously to dissipate or minimize. All real-world change or dynamics is seen to follow, or be motivated, by this law. So whereas the first law expresses that which remains the same, or is time-symmetric, in all real-world processes the second law expresses that which changes and motivates the change, the fundamental time-asymmetry, in all real-world process. Clausius coined the term “entropy” to refer to the dissipated potential and the second law, in its most general form, states that the world acts spontaneously to minimize potentials (or equivalently maximize entropy), and with this, active end-directedness or time-asymmetry was, for the first time, given a universal physical basis. The balance equation of the second law, expressed as S > 0, says that in all natural processes the entropy of the world always increases, and thus whereas with the first law there is no time, and the past, present, and future are indistinguishable, the second law, with its one-way flow, introduces the basis for telling the difference. “
You do realize I hope that energy potentials exist everywhere and at all times whether or not we can detect them. Note how the vernacular stated at the beginning of the quote contradicts what Clausius later states. Which you in apparent FULL COGNITIVE DISSONANCE disagreed with when you stated:
“According to the second law of thermodynamics the entropy of an isolated system never decreases; such a system will spontaneously evolve toward thermodynamic equilibrium, the configuration with maximum entropy.”
WHICH IS IT DOUG?!!! YOUR FIRST QUOTE REFERENCING CLAUSIUS CLAIMED ENTROPY ALWAYS INCREASES!!! I AGREE WITH CLAUSIUS BY THE WAY AND NOT YOU!!! HOWEVER YOUR 2ND QUOTE CLAIMED IT MAXIMIZES!!! BOTH CANNOT BE TRUE DOUG!!! Times arrow is not on your side on this one I think.
Btw, Doug for a REAL WORLD example you may consider the universe itself. According to the 1st Law of Thermodynamics matter and energy cannot be destroyed. Over 2000 years ago the Greek philosophers Aristotle, Plato and probably others discovered that the universe proved FINITE and assuming no spiritual realm appears closed. Modern day astronomers using the Cobe satellite know that infrared energy now dissipates throughout the perceivable universe. In other words Doug we have unimpeachable evidence of universal increase in entropy throughout the cosmos and apparently irreversible. Btw, I say apparently because I do not believe the universe to be closed and agree with Isaac Newton ( truly my choice for most brilliant physicist of recorded time – and yes that includes Maxwell or Lochsmidt ) that LOWER men believe they can account for the universe and nature fully through it’s apparent laws and thus bind it with some simple explanation. In fact, God an entire spiritual realm including angels govern and guide, protect and maintain the ORDER we take for granted. The material universe cannot account for itself nor can any simple theory or even law of nature, including for example Einstein’s General Theory and Newton’s Laws of Gravitation and force equation F=MA. Indeed, modern physicists cocked-up the popular scientific fiction of DARK MATTER to explain why neither of the two afore mentioned scientific claims could explain the tight galactic formations, which held together better than either explanation could predict.
Without input of matter, energy or some force from outside itself any material system DIES!!!
Thanks Doug and have a great day!
My previous statement better reads:
“Without input of matter, energy or some force from outside itself any MERELY material system DIES!!!”
Have a great day!
Btw, one correction:
I stated:
“To be fair it’s dubious whether a truly ISOLATED system even exists. I don’t believe so.”
Actually, a largely ISOLATED system does come to mind, but it sees to me it would require some input from other realms to maintain it.
In physics an isolated system, can be subjected to an external force, whereas a closed system cannot. Hence conservation of momentum applies in a closed system, but not in an isolated system if there is an external force field like gravity.
The website http://entropylaw.com is not mine: I only came across it last month.
I suggest you refer to the statement of the Second Law that I quoted from Wikipedia. If you want more information, then there are two complete papers that I have written about the Second Law which has been an area of specialized post graduate study for me. The papers are linked from the “Evidence” page at our group’s website http://climate-change-theory.com .
In his 4:39 PM, March 17 comment, Doug repeats and confirms his basic idea upon which his whole gravito-thermal hypothesis rests:
“Yes Kinetic Theory assumes elastic collisions, and so do I when explaining that the KE of molecules about to collide must be equal for there to be no further diffusion of kinetic energy in such collisions. For this to be the case (PE+KE) must be homogeneous at all altitudes. It’s not hard to understand.”
He shares this peculiar notion with a certain Stephen Wilde.
They both appear to think that air molecules at the tropopause are cooler than the ones at the surface simply as a result of a direct, inversely coupled connection between their individual KE (velocity) and something conceived of as their molecular gravitational PE, in the sense that, as you move higher up into the troposphere, the amount of gravitational PE of each single air molecule grows progressively larger, and so the amount of KE of each single air molecule must grow equally smaller. Based on the idea that the sum of the two “(PE+KE) must be homogeneous at all altitudes.”
All you really need to point out, then, to properly address Doug’s hypothesis, is that gravitational PE of a volume of gas is irrelevant on a molecular level, meaning, it’s irrelevant to the temperature (the thermodynamics) of the gas. It is fundamentally a MACROscopic (Newtonian) function.
Microscopic PE in a gas is related to the attractive forces between its individual molecules, not to gravity. As Curt correctly noted on another thread, this PE from intermolecular forces can be safely ignored in a ‘thin’ gas like the atmosphere. But Doug’s problem is that he discards altogether even the idea of this actual form of microscopic PE and rather replaces it arbitrarily with a privately concocted idea of a molecular gravitational PE function. In his world it is this function, and only this, that eats away at the kinetic energy of the individual gas molecule as it moves higher up the atmospheric column.
But this would only be the case if the air molecules in our atmosphere never collided at all. If they did not interact. Then they would separately receive an initial conductive heating at the surface, from where they would consequently rush vertically upwards until gravity had slowed them down to a full stop, from which level they would then drop back down to the surface, ultimately reacquiring the velocity from when they were first heated in a fully reversible cycle, ideally bouncing back up from a perfectly elastic collision, basically moving up-and-down, up-and-down through Earth’s gravity field forever in a non-friction scenario.
We all know that this will not and does not happen in the real world.
Individual air molecules are not directly controlled by gravity in this way. The bulk air containing them is. But not the molecules themselves.
It is the elastic collisions between the air molecules that hold the atmosphere up, maintaining the hydrostatic equilibrium, and that similarly let the individual air molecules maintain their velocity at almost the same level (85% of surface velocity) even at the tropopause.
– – –
What is the average kinetic energy of an N2 (~air) molecule at the surface of the Earth?
KE = 3/2 kT
KE = 3/2 (1.38 x 10^-23 J/K)(288K) = 5.96 x 10^-21 J
This value is directly associated with an average molecular velocity of:
KE = 1/2 mv^2 [J = kg*(m/s)^2]
v = √(2KE/m)
v = √(2(5.96 x 10^-21 J)/4.65 x 10^-26 kg) = 506.3 m/s
So what is the average kinetic energy of an N2 molecule at the mean global tropopause height (12 km)?
KE = 3/2 (1.38 x 10^-23 J/K)(210K) = 4.35 x 10^-21 J
Which translates into a velocity of:
v = √(2(4.35 x 10^-21 J)/4.65 x 10^-26 kg) = 432.5 m/s
So there’s a difference in the average molecular KE between these two atmospheric levels of [5.96-4.35=] 1.61 x 10^-21 J, and a difference in the associated average molecular velocity of [506.3-432.5=] 73.8 m/s.
But what if we were to move the surface air N2 molecule straight up to the tropopause? How much would it gain in gravitational PE?
PE = mgh
At the surface, our reference point, its PE is 0, simply because h is zero.
At the tropopause, however, its PE relative to our surface reference point is:
PE = 4.65 x 10^-26 kg * 9.81 m/s^2 * 12000 m = 5.47 x 10^-21 J.
The difference in the inferred average molecular gravitational PE between these two atmospheric levels is therefore the same: 5.47 x 10^-21 J.
Compare this with the difference in average molecular KE between the two atmospheric levels: 1.61 x 10^-21 J.
Where is the direct PE/KE correspondence claimed by Doug (and Stephen)? The calculated difference in molecular gravitational PE (sfc-tropopause) is 3.4 times bigger than the equivalent difference in molecular KE, based on temperature/velocity …
– – –
If the N2 molecule started off upwards from the surface with a velocity of 503.6 m/s (at 288K), and if it didn’t collide with any other molecues on its way, how high would it get before gravity had slowed it down to a complete stop (0 m/s)?
KE at h_0: 5.96 x 10^-21 J
KE at h_1: 0 J
PE at h_0: 0 J
PE (PE_1) at h_1: 5.96 x 10^-21 J
h_1 = PE_1/mg = 5.96 x 10^-21 J/(4.65 x 10^26 kg * 9.81 m/s^2) = 13065 m.
13 km.
So what would the N2 molecule’s velocity be at the mean global tropopause height – 12 km?
v_sf – v_tr = √(2KE/m) – √2gh_tr = 506.3 m/s – √(2*9.81 m/s^2*12000m) = 506.3 – 485.2 = 21.1 m/s.
But what is the actual average molecular (N2) velocity at the mean global tropopause height?
We already know from above. It’s 432.5 m/s at 210K.
Not 21.1 m/s.
– – –
The reduction in air temperature (and hence the reduction in average molecular velocity) from surface to tropopause on Earth is fully accounted for by the gradual loss of disordered, internal (microscopic) kinetic energy as a result of the expansion of ascending air, and by the parallel gain in internal KE as a result of the compression of descending air, together making up the atmospheric adiabatic cycle.
As has been shown, there is no connection between the average molecular KE (directly associated with the temperature) of a volume of air in the atmosphere, and the notion of an average molecular gravitational PE of that same volume of air.
The environmental lapse rate, the tropospheric temperature profile, is fully the result of the convective movement of air up and down. The gravity-based ‘dry adiabatic lapse rate’ (DALR) could never be realised without such movement. It would remain a mere potential temperature gradient. Without surface heating and resulting convective movement, no lapse rate.
– – –
Internal (thermodynamic, temperature-related) energy of a system:
# Disordered MICROscopic (molecular) kinetic energy
# Disordered MICROscopic (intermolecular) potential energy
Newtonian (mechanical, motion/position/force field-related) energy of a system:
# Ordered MACROscopic (system) (bulk movement) kinetic energy
# Ordered MACROscopic (system) (gravitational) potential energy
Of course now, after having published the above comment, I see how this equation came out a bit confusing:
v_sf – v_tr = √(2KE/m) – √2gh_tr
Since the two velocity-terms are basically an (upward) KE expression (left one) vs. a (downward) PE expression (right one), they both describe the molecule’s velocity at the surface, only the latter (PE) one after having fallen from an initial v = 0 m/s at the tropopause (12000 m up).
Bear with me …
What Stephen Wilde claims with his wild conjecture is totally different and is similar to what is refuted in the right hand column on this page of the website endorsed by our group of persons suitably qualified in physics.
I am not talking about gravitational PE of a mass of gas. You need to study more carefully the detailed explanation in the Home page of our group’s website and my linked paper written two years ago, and you need to realize that consideration of two and then three layers of molecules (separated by about the mean free path) can be extended to the whole troposphere by mathematical induction.
You also need to realize that you cannot correctly answer the question about Uranus that I have posed yet again, this time to JohnKl. You need my hypothesis to do so, and absolutely every other explanation you attempt I will thoroughly debunk.
To all readers, this pinpoints where Kristian is wrong …
He writes “Individual air molecules are not directly controlled by gravity”
Of course they are. If that were not the case the density gradient would not form. It does form, and it forms entirely due to the fact that gravity curves the path of any molecule with a horizontal component in its velocity. This leads to a slightly greater probability of net downward motion than upward motion. Thus, because at thermodynamic equilibrium there must be equal numbers of molecules with equal KE passing upwards and downwards across any horizontal plane, we can understand that there will be more molecules in a plane below and less in a plane above because of the greater propensity to move downwards. Once again we can extend this by mathematical induction to the whole troposphere. There is no conjecture about a single molecule passing all the way up through the troposphere, or of any molecule running out of KE and then falling, because they don’t get down to absolute zero (0K) and, in any event, they gain more energy from absorbed insolation. Since the density gradient does exist (and is obviously due to gravity) I rest my case.
I will add, however, that our friend Kristian would do well to study the assumptions of Kinetic Theory (as used by Einstein and others) in which we read …
“Except during collisions, the interactions among molecules are negligible.
“Because of the above two, their dynamics can be treated classically.
“Because they have mass, the gas molecules will be affected by gravity.”
QED
Doug:
You continue to make the egregious beginner’s error that the conversion of gravitational potential energy to vertical kinetic energy is an irreversible, entropy-increasing process. This is trivially refuted, for instance by the simple case of a bouncing ball.
This error completely invalidates all your subsequent analysis of entropy maximization at a non-zero lapse rate for an isolated system.
So I ask the question:
Roy and silent readers, with Doug making such a basic blunder in physics, are you going to take note of ANYTHING he says relating to physics, in which he is not qualified and about which he has next to nothing in the way of understanding?
No I don’t. What is irreversible is the net change in position of some of the molecules as, for example, a stable density gradient forms. Likewise, if you have warmer air at one end of a horizontal sealed and perfectly insulated cylinder, entropy will increase, but only until all unbalanced energy potential are dissipated. That process is not reversible.
Kristian,
You are the scientist around here! Great job! You actually don’t mind getting into the math and seeing what is going on. Curt has asked Doug a bunch of times to provide a mathematical expression of his “heat creep” to give units and dimension to this aledged energy motion. To date? Crickets!
I really do not think Doug is capable of doing much math at all. Physics is mostly math, content is there to provide understanding of what the math is doing. If you can’t calculate it, don’t present it to the physics community. When you have some math, write up a mathematical description of the theory and present it to the physics community. Until then KEEP YOUR MOUTH SHUT and stop posting endlessly on websites until after you develop the math, please!!
Doug do what Kristian did above. He totally destroys the rational behind your idea! You are wrong and will soon need to admit it.
“The environmental lapse rate, the tropospheric temperature profile, is fully the result of the convective movement of air up and down. The gravity-based ‘dry adiabatic lapse rate’ (DALR) could never be realised without such movement. It would remain a mere potential temperature gradient. Without surface heating and resulting convective movement, no lapse rate.”
Kristian:
Exactly so …. and what I said at least once up thread.
The DALR is a natural consequence of air motion (turbulent/convective/Coriolis driven convergence & divergence) in an atmosphere. In Meteorology, air is considered buoyant (molecules are confined by their neighbours) at all altitudes unless forced to move (they most certainly do not creep – that process if did occur it would be many orders of magnitude weaker). When remaining ~stationary in the absence of insolation the vertical profile cools to an an isothermal. This is OBSERVED by the meteorological community and not up for discussion.
GHG’s have a small influence on the LR but it is primarily altered via convective LH release and radiative cooling to space.
It’s just empirical physics and I find it amazing that it being discussed here.
The Emperor’s NOT naked and Cotton certainly hasn’t spotted he is, when for ~150 yrs such great minds of the past and whilst under the current glare of technology hasn’t.
Toneb writes: “Without surface heating and resulting convective movement, no lapse rate.”
Totally incorrect, as I have explained to Roy Spencer many times, because isothermal conditions would not be the state of maximum entropy in a vertical plane in a force field. We see evidence in a centrifuge machine, here for example, and in every planetary troposphere, whether or not there is a surface.
Answer the above question about the nominal troposphere of Uranus which (over a distance of 350Km) exhibits a near perfect gravitationally induced -g/Cp temperature gradient, but which, at its base, has no surface and no direct solar radiation reaching down there, and yet is hotter than Earth despite receiving less than 4W/^2 of insolation at TOA.
The temperature and density gradients evolve at the molecular level as gravity acts on molecules moving between collisions. This is the state of maximum entropy (thermodynamic equilibrium) as I have proved beyond a shadow of a doubt.
The full, detailed and correct explanation of the thermodynamics involved is here supported by the linked peer-reviewed paper and endorsed by members of our group of persons suitably qualified in physics. Below is an example of such support …
“Doug Cotton shows how simple thermodynamic physics implies that the gravitational field of a planet will establish a thermal gradient in its atmosphere. The thermal gradient, a basic property of a planet, can be used to determine the temperatures of its atmosphere, surface and sub-surface regions. The interesting concept of “heat creep” applied to diagrams of the thermal gradient is used to explain the effect of solar radiation on the temperature of a planet. The thermal gradient shows that the observed temperatures of the Earth are determined by natural processes and not by back radiation warming from greenhouse gases. Evidence is presented to show that greenhouse gases cool the Earth and do not warm it.”
John Turner B.Sc.;Dip.Ed.;M.Ed.(Hons);Grad.Dip.Ed.Studies (retired physics educator)
Doug:
You continue to make the egregious beginner’s error that the conversion of gravitational potential energy to vertical kinetic energy is an irreversible, entropy-increasing process. This is trivially refuted, for instance by the simple case of a bouncing ball.
This error completely invalidates all your subsequent analysis of entropy maximization at a non-zero lapse rate for an isolated system.
So I ask the question:
Roy and silent readers, with Doug making such a basic blunder in physics, are you going to take note of ANYTHING he says relating to physics, in which he is not qualified and about which he has next to nothing in the way of understanding?
I make no such assumption. Find what you accuse me of doing in my paper and stop misquoting me. The analysis of the net movement of molecules (forming the density gradient) and the diffusion of kinetic energy in molecular collisions (forming the temperature gradient) so as to lead to a state of maximum entropy is a macro process and is not based on any such assumption.
Doug:
You can’t even keep track of your different scatterbrained arguments! In an above comment in this thread, you said:
“When a ball is falling through a vacuum tube gravitational potential energy is converted to kinetic energy (just as happens to molecules in flight between collisions) and total energy remains constant. However, entropy increases and that is what the Second Law of Thermodynamics tells us.”
http://www.drroyspencer.com/2015/03/uah-global-temperature-update-for-feb-2015-0-30-deg-c/#comment-184927
Roy and silent readers, with Doug making such a basic blunder in physics, are you going to take note of ANYTHING he says relating to physics, in which he is not qualified and about which he has next to nothing in the way of understanding?
Firstly I am qualified in physics and even gained a scholarship in such awarded by Sydney University in the 1960’s where I did my first degree (majoring in physics) from 1963 to 1966 inclusive. Since then I have helped numerous paying students understand physics and thus do well in their courses. Consequently I can easily detect misunderstandings in your physics, some of which I have already exposed in other comments, and no doubt there will be more in future.
There is an example of a ball rolling down a ramp and increasing entropy on this page.
The issue with molecules in free flight between collisions is nothing more nor less than the fact that (PE+KE) remains constant during such flight. Changes in entropy indisputably occur (at the macro level) when there are net movements of molecules to form a density gradient and net changes in the distribution of kinetic energy to form a temperature gradient, just as also happens in a centrifugal force field.
If your argument were correct, then you could have warm air in the “hot car in the garage” example here and there would be no heat transfers out of the car when the four doors were opened. You would just argue that all the molecules would bounce around in elastic collisions without there being any reason for any change in entropy.
Doug:
You are still completely unable to distinguish between reversible transfers (work) that do not increase entropy, like the conversion of gravitational potential energy to vertical kinetic energy, and irreversible transfers (heat) that do increase entropy, like an inelastic collision or mixing of hot and cold air.
This distinction is one of the first given to students in any introductory thermodynamics course, and they are expected to get it right by about the 3rd week of the course but you haven’t figured it out in 50 years! It’s absolutely pathetic!
If you had actually bothered to read and understand the page you linked, you would find that it said this: “the potential produces a flow from the falling weight through the moving paddle through the thermometer. This is precisely the one-way action of the second law and the experiment depends upon it entirely.”
It refers to the diagram on a separate page showing Joule’s famous experiment. The caption for that diagram says, “When a constraint is removed, potential energy in the form of a suspended weight is converted into the kinetic energy of a moving paddle wheel in a container of water sealed against other inflow or outflow of energy. The moving paddle wheel heats the water by a precise amount consistent with the falling weight.”
http://entropylaw.com/entropyenergy.html
It is NOT the conversion of gravitational potential energy to vertical kinetic energy that increases entropy. It is the conversion of that kinetic energy to thermal energy by the action of the paddlewheel in the fluid! This is basic, basic stuff that you do not have the intellectual horsepower to understand!
“the vertical profile cools to an an isothermal. This is OBSERVED by the meteorological community”
More garbage from Toneb – and totally ignoring the Second Law of Thermodynamics and how entropy is maximized.
Considering a 1Km high column of air, it is well known that, in calm conditions in the early pre-dawn hours, advection virtually stops, but the environmental lapse rate remains intact. Furthermore, if that temperature gradient becomes less steep (due to warm air blowing in horizontally at the top) then, even though the temperature is still lower at the top than at the surface, meteorologists call this an “inversion” and they know thermal energy is transferred downwards. That is heat creep up the temperature profile, maximizing entropy.
In meteorology, an inversion is a deviation from the normal change of an atmospheric property with altitude. It almost always refers to a “temperature inversion,”
You, Toneb, are way out of your depth when it comes to understanding the Second Law of Thermodynamics, about which I have written two comprehensive peer-reviewed papers based on extensive post-graduate research.
“The DALR is a natural consequence of air motion (turbulent/convective/Coriolis driven convergence & divergence) in an atmosphere.”
It is nothing of the kind. Wind of any form disrupts the slower process of forming the density and temperature gradients at the molecular level. You would know that I would respond thus if you understood my hypothesis, about which you have no idea.
The computation of the -g/Cp temperature gradient starts with an assumption of an “ideal” gas in an “ideal” isolated column of air in the troposphere, which column is not affected by intruding wind of any form, or any supply of new energy top or bottom or anywhere in that column.
See quote from Encyclopedia Britannica near the end of this thread. The lapse rate remains intact when there is no rising air.
You, Norman, have not yet apologized for criticizing my hypothesis based on your complete disregard for the Equipartition Theorem. Nor have you answered in your own words the question about the sensitivity to a 1% increase in water vapor.
Frankly, given that you have no qualifications in physics, and obvious serious misunderstandings, I am surprised that you have the hide to dabble in this discussion, Norman, as you are way out of your depth.
And, by the way, I have just proved where Kristian is wrong with reference to standard physics in the Kinetic Theory of Gases, one of the assumptions of which he specifically denied. Read my reply to him.
As with computer models, if the assumptions used in any math are incorrect it’s a case of garbage in, garbage out. I’ve done calculations like Kristian and got the same answer of 13Km. This is so elementary I would assume readers know how to do such school boy physics. But Kristian’s assumptions were wrong, because he completely ignored the fact that the atmosphere absorbs 20% of incident solar radiation which adds kinetic energy to molecules especially in the upper regions.
Regardless of what gas absorbed the insolation, that new thermal energy is dispersed by diffusion of kinetic energy in molecular collisions to all air molecules in such a way that the environmental temperature gradient (which has nothing to do with any lapsing process) evolves as entropy tends towards its maximum. This enables us to calculate the temperature at any point in the troposphere once we have calculated the effective radiating temperature and the location of the anchor point. That’s all the math I need (after calculating the gradient) and it’s such elementary school boy coordinate geometry that I can reasonably leave it to the reader to fill in the gaps.
But if you are impressed by computations then this physics by a professor of applied mathematics should convince you that electro-magnetic energy in radiation from the colder troposphere does not get converted to thermal (kinetic) energy in the warmer surface.
Those who cling to the false radiative forcing conjecture should read Mathematical Physics of BlackBody Radiation” where I’m sure Norman will be impressed by the math written by this professor of applied mathematics in December 2012, in agreement with my March 2012 paper linked at the foot of this page.
I also explained (as does this professor) that there is only a one-way transfer of thermal energy in a one-way pencil of radiation if the source of that spontaneous radiation is hotter than the target and the intensity (after attenuation due to distance) is still higher than that being emitted by the target. As I showed in my March 2012 paper, the thermal energy transfer is represented by the area between the Planck curves of the source and target (where the target’s curve is fully enveloped by that of the source) as is well known and established in engineering circles. That area represents the radiation which cannot resonate in the target.
So here are some quotes from the good professor’s paper where you will note on the first page examples of the Planck functions for different temperatures, and you will see what I mean about lower curves being totally under the ones for higher temperatures …
Hundred years later blackbody radiation is back at the center of discussion, now as the cornerstone of climate alarmism based on the idea of atmospheric “backradiation” from so-called “greenhouse gases” causing ”global
warming”. The weakness of this cornerstone is exposed …
but Planck now introduced the startling assumption that the vibrators did not emit energy in a continuous stream, but by a series of instantaneous gushes. Such an assumption was in flagrant opposition to Maxwell’s electromagnetic
laws ..
(My comment: Maxwell was not such a bright cookie, and he was also wrong in his attempted refutation of Loschmidt.)
We have seen that the Rayleigh-Jeans Radiation Law expresses near-resonance in a wave model with damping from a 3rd order time derivative and we now make a connection to acoustic resonance in string instruments modeled by a
wave equation with viscous damping from a 1st order time derivative.
(My comment: See the section on “Resonant Scattering” in my March 2012 paper linked at the foot of this page.)
Energy is transferred from the warmer to the colder body. We have thus proved a 2nd Law as an effect of dissipative radiation.”
Perhaps when you study the above 110 page on-line book you will realize that there is more to the Second Law than you thought, and photons are not like little hand grenades depositing thermal energy into everything they strike. That is why the radiative greenhouse conjecture is totally false. What does happen is explained here.
Go back to this comment Ball4 and others. I see no reason to have to continue to explain the content of my hypothesis here when you can all read, study and inwardly digest it here because none of you exhibits any evidence of having done so, especially Ball4 who seems to think that there has to be a gradient in gravity itself. (LOL)
Doug 7:10pm: I’ve read your stuff and easily found much of it flawed & already pointed out exactly where, as is your 7:05pm comment that merits no response. Actually you have failed to explain your stuff in this comment thread in a fundamental way. Your regional “study” calculations are known to be wrong. Once again, I point out you use a centrifuge analog to imagine & draw more flawed conclusions about nature. As Curt also correctly points out.
You present nothing but assertive statements. For a start, what you don’t understand is that entropy in a sealed insulated tall cylinder of gas that is isothermal will then increase (at the macro level) because of the unbalanced gravitational potential energy potential.
But entropy will only increase until the stable density gradient and temperature gradient form. That is the state of maximum entropy (called “thermodynamic equilibrium”) when mean molecular (PE+KE) is homogeneous at all heights.
Then, because entropy is at its maximum, there will indeed be no further increase in entropy. All the while in the whole process molecules are darting about (at about 1700 Km/hr) interchanging their own gravitational PE and KE in free flight and sharing KE in elastic collisions. What then happens is explained here.
Gravity does act on individual molecules as per the assumptions of Kinetic Theory, and as per plain common sense. Sometimes you would do well to admit your errors, but narcissistic people don’t deign to do so.
Everyone can go back to this comment where I prove Ball4 wrong and expose his huge mistake in which he claimed gravity does not act upon the mass of molecules. He will never ever admit this critical mistake because of his psychological nature. (LOL)
No such claim by me Doug. That approach won’t work, you need to deal with the existing eqn.s already in the literature.
Doug 9:16pm: “You present nothing but assertive statements.”
Incorrect Doug, I’ve provided the exact eqn.s pinpointing your faults, you ignore them.
“But entropy will only increase until the stable density gradient and temperature gradient form.”
Agree & I pinpointed the eqn. for the temperature gradient T(p) for you 3/13 4:04pm, worked out by Poisson in the 1890s – shown to be the max. entropy point in 1998. The existing literature like this can be easily compared to your stuff to pinpoint your flaws. Ignoring this established science will not improve your faulty papers and centrifuge based faulty imagination.
So you agree that entropy will increase when the sealed insulated isothermal cylinder is rotated from horizontal to vertical. Hence that isothermal state was not thermodynamic equilibrium, because by definition, thermodynamic equilibrium has maximum entropy which cannot thus increase. But you admit entropy would increase. So you are wrong.
Go back to this comment Ball4.
Sorry I got interrupted and forgot I posted the above. It is repeated below with more points …
And by the way, the equation for the Poisson gradient is nothing remotely like the accepted “dry adiabatic lapse rate” which is -g/Cp so the temperature gradient is the quotient of the acceleration due to gravity and the weighted mean specific heat of the gases.
An admission of error would be appropriate from you, Ball4, but predictably will not be forthcoming.
Doug 10:16pm: The T gradient -g/Cp is one possible gradient from among many, use it to compute the entropy in your column. Then compute the entropy in your column from the T(p) as shown by Poisson which will be higher and you will find no other profile with a still higher entropy than Poisson’s T(p), he was pretty good back in 1890s as he had the pre-req.s.
I am laughing because I know you cannot possibly compute the entropy in your column since you do not have the pre-req.s. If so, compute the isothermal T(p) = constant entropy, it will be less than Poisson T(p).
So you agree that entropy will increase when the sealed insulated isothermal cylinder is rotated from horizontal to vertical. Hence that isothermal state was not thermodynamic equilibrium, because by definition, thermodynamic equilibrium has maximum entropy which cannot thus increase. But you admit entropy would increase.
I replied here to your reference to the Poisson equation which is totally irrelevant because the temperature gradient is not caused by a variation in the density or by a variation in gravity or anything else such as controlled temperatures at the ends of a metal rod. It forms autonomously due to the force of gravity acting upon individual molecules. Gravity forms the density gradient and the temperature gradient, each being the same state of maximum entropy because there can only be one such state, and clearly the density gradient is stable, subject of course to perturbations due to wind and weather conditions. Likewise the temperature gradient is, because they are each one and the same state of maximum entropy towards which the system will tend to progress in calm conditions. We already have the correct term thermodynamic equilibrium. Hydrostatic equilibrium is only maintained by a continuous flow through a system, which there cannot be in a sealed cylinder.
You are still very confused Ball4 about what my hypothesis says and if you were face to face I would prove that by asking you to reproduce the heat creep diagrams from memory. Only you can test yourself on that one. If you can’t, then you don’t understand the hypothesis.
Go back to this comment.
Doug 10:04pm: “So you agree that entropy will increase when the sealed insulated isothermal cylinder is rotated from horizontal to vertical.”
Yes. If T(p) = constant to start then there is a higher entropy column T(p) as shown by Poisson which is max.
“Hence that isothermal state was not thermodynamic equilibrium..”
Yep.
“.by definition, thermodynamic equilibrium has maximum entropy which cannot thus increase..”
Yep.
“..the Poisson equation which is totally irrelevant…”
Nope, that T(p) profile I wrote is proven max. entropy for the isolated column in 1998. You will be able to compute higher entropy with no other profile. If you can, show it (laughter).
“Hydrostatic equilibrium is only maintained by a continuous flow..”
Makes no sense. Static is not flow.
“You are still very confused Ball4 about what my hypothesis says..”
Not at all. The correct existing literature is easy to access & understand for those have the pre-req.s and compare to easily find where faults exist in your hypothesis. I’ve pinpointed them for you. Ignoring them is not helping you improve.
You continue to ignore the prereqisites for the type of systems to which Poisson was referring. Prove if you can using the Poisson equation that the temperature gradient would be -g/Cp in a dry troposphere, because that sure is the accepted value and is proven in my paper and here albeit in a clumsy roundabout way that introduces pressure, only to find pressure than cancels out. I do it in two lines using Kinetic Theory.
In continuum mechanics, a fluid is said to be in hydrostatic equilibrium … when the flow velocity at each point is constant over time.
That my friend is the usual application of the term. There is no point in calling thermodynamic equilibrium “hydrostatic equilibrium” in a static case, though climatologists like to do so to blame the lapse rate on a continUous flow when it is not due to such. “Hydro” relates to water.
I have not ignored the pal-reviewed AGW literature. I have studied it extensively thanks and pinpointed, for example, where the “gold standard” Pierrehumbert textbook is wrong both in assuming isothermal conditions and using the albedo for the existing Earth (with clouds reflecting 20%) for their imaginary world without water vapor and thus without clouds.
Don’t bother with vague references to such literature or to the irrelevant Poisson equation which you obviously do not understand, whereas I have been explaining this stuff to my students for five decades. If you can’t express your argument in your own words referring specifically to quotes from my paper, then don’t bother because anything else from you will indeed be ignored.
Just to clarify one point: The use of the Poisson equation in considering hydrostatic equilibrium is inappropriate in a sealed cylinder. It is thermodynamic equilibrium which must evolve in that isolated system. That’s what the Second Law tells us. By definition, in such a state there are no unbalanced energy potentials (see http://entropylaw.com ) and no further net energy transfers across any external or internal boundary. This state has nothing to do with the scenarios worked with when applying the Poisson equation. Furthermore, the whole concept of parcels of air somehow rising and holding together for no reason is totally imaginary, even if it gives the right answer.
Whilst you agree the -g/Cp gradient is in all planetary tropospheres it is not formed by solar radiation heating a surface directly and causing upward flows by convective heat transfer which includes conduction and diffusion, but only heat transfers due to molecular collision which share KE.
The term “convective heat transfer” or “natural convection” (as distinct from “forced convection”) does not include wind of any form. It is only pure natural convective heat transfer which forms the temperature gradient, and it does so at the molecular level. Wind (or “forced convection” of any form, like a fan for example) does not form a vertical temperature gradient based on -g/Cp.
You completely missed the point about my question pertaining to other planetary tropospheres. The huge “coincidence” that the profiles get down to just the right temperature at just the right altitude is no coincidence at all with my hypothesis which explains that the profile builds up inwards from that anchor point (ie towards the core) and not outwards from a hot core cooling off.
Regarding the other planets, go to this comment.
Doug 11:01pm: You have a lot of learning to go. Your two line approach to deriving -g/Cp gradient makes the assumption of constant temperature* in a parcel. Poisson extended that by allowing the temperature to change. Therefore -g/Cp is an approximation and the final entropy will thus be lower than the T(p) from Poisson which is the exact solution and highest possible entropy at thermodynamic equilibrium of the isolated column. This applies to any air column.
*In the calculus, what this means is you took T outside the integral as a constant to get -g/Cp. Poisson did not do that, he allowed the more difficult calculus solution to vary the T(p) under the integral so as to integrate & obtain the ideal exact natural T gradient.
11:09pm: The atmosphere doesn’t flow with constant velocity like water in a pipe, you misunderstand wiki, as is usual. See the Fig. 2.3 wiki ref. given p. 62, 64:
“Thus eqn. (2.15) states that a fluid in hydrostatic equilibrium will align its constant-pressure surfaces everywhere normal to the local gravity vector.”
http://www.slideshare.net/mrceds/fluid-mechanics-frank-m-white-4-ed
11:30pm: “The use of the Poisson equation in considering hydrostatic equilibrium is inappropriate in a sealed cylinder.”
No. The Poisson T(p) IS the ideal exact temperature gradient achieved AT thermodynamic equilibrium of max. entropy when there are no unbalanced energy potentials in the sealed cylinder isolated air column.
Actually you relied on a definition of a parcel to get the two line -g/Cp, you just didn’t know to tell us. As usual.
In that case, the finite lifetime of a parcel was no concern to you. The derivation was based on the instantaneous rate of change of the parcel’s temperature (you assumed constant, Poisson used more generally not constant). Thus the parcel need live only a brief moment. Once its mathematical task was completed, the parcel was no longer of interest to you and can dissipate without damaging your approximate –g/Cp and Poisson’s exact derivation for T(p).
11:39pm: “Whilst you agree…”
I do not agree.
“You completely missed the point about my question pertaining to other planetary tropospheres”
I did not. I can easily see your point is faulty by comparing to existing literature. You are incapable of doing so. There is no solid evidence you are correct because it is well known your calculations are in error and your -g/Cp is only an approximation to nature’s exact ideal dry T(p) non-isothermal.
Ball4 writes: “But entropy will only increase until the stable density gradient and temperature gradient form.”
Agree,
So he agrees that the state of thermodynamic equilibrium (maximum entropy) is not isothermal. Hence he agrees the IPCC conjecture is wrong because it is built on an assumption that, in the absence of GHG, the surface temperature would have been the same as that 255K found nearly half way up the troposphere. Because the IPCC wrongly assumed isothermal conditions (which Ball4 now agrees would not be the case because such would not be the state of maximum entropy) they then assumed there had to be something (like back radiation perhaps?) raising the surface end of the temperature profile by rotating that profile up at that end and down at the tropopause end. However, what does more than enough rotating is gravity acting on molecules in flight between collisions. Then, the temperature leveling effect of radiation (involving mostly water vapor, carbon dioxide and methane molecules) reduces that gradient by about a third, thus reducing the supported surface temperature and cooling us all.
Doug 10:55pm: “So (Ball4) agrees that the state of thermodynamic equilibrium (maximum entropy) is not isothermal.”
Yep for an isolated air column, that’s what you will find in the specialist literature.
“Hence (Ball4) agrees..in the absence of GHG..255K”
No. I agree 255K results as earth atmosphere optical depth is decreased toward 0 asymptotically. I agree Venus 732K occurs as its optical depth increases WAY above 0.
The rest of the 10:55pm comment is gibberish. Devoid of fundamental science. Doug’s political view alone.
I am sick and tired of the anonymous few people here hiding their identity, qualifications etc and airing their views about what they think my hypothesis is saying, thus totally misquoting me and wasting my time in having to explain what they could have read in the first place in the website now visited by over 7,000 just since January 8th this year.
You will never, ever prove that a density and temperature gradient do not form in calm conditions (even without a planetary surface) at the molecular level, as the state of maximum entropy is approached, because that is why the density gradient is stable, and likewise the temperature gradient.
The reason you won’t prove me wrong is because the Second Law of Thermodynamics is not wrong.
And nor was BigWaveDave, three years ago, nor Dr Hans Jelbring when he studied this for his PhD in climatology and published in a peer-reviewed journal in 2003 here. And nor was Josef Loschmidt wrong in the late 19th century. Robert Brown’s pathetic attempt to refute the Loschmidt effect is proven wrong in the above linked website on the “WUWT errors” page.
We have solid evidence that my hypothesis is correct because the temperature profiles are physically built up inwards from the anchor point, which is at radiative equilibrium with the Sun. The heat is transferred inwards on the sunlit side by heat creep and outwards by radiation (globally) and outward convective heat transfer mostly on the dark side. It would be just far too great a coincidence for all planets to be just “cooling off” and yet having the right -g/Co temperature gradient all the way from the core to the pre-determined anchor point that depends on solar radiation and the location of gases that absorb such.
Now read the paper and stop wasting my time rewriting it all here.
Typo: -g/Cp gradient
OK, so let’s just quickly check Doug’s hypothesis that a ‘thermodynamic equilibrium’ in our atmosphere will necessitate a spontaneously generated, pure -g/Cp temperature gradient (‘dry adiabatic lapse rate’, DALR) to maximize entropy by having the sum of molecular KE and PE equal at all altitudes.
Let’s assume he’s right. Let’s posit a stable DALR from the surface to a tropopause at 12 km of altitude. Only gravity and the specific heat of the air matters. No convection, no release of latent heat, no atmospheric absorption of solar heat.
The temperature falloff rate in this case is 9.75K/km. If we assume a mean global surface temperature of 288K, then that would correspond to a temperature at the global mean tropopause height of 171K.
– – –
We will now find out whether or not PE+KE for each individual air molecule (N2 for convenience) balances between ground and tropopause level, 12 km up the atmospheric column. Is the sum of the two always the same?
PE+KE for an N2 molecule at the surface (288K)
PE + KE = 0 J + 5.96 x 10^-21 J = 5.96 x 10^-21 J
Its associated average velocity would be 506.3 m/s.
PE+KE for an N2 molecule at the tropopause (171K)
The KE of an N2 molecule at temperature 171K:
KE = 3/2 kT = 3/2 (1.38 x 10^-23 J/K)(171K) = 3.54 x 10^-21 J
This leaves us with a total sum of N2 molecular energy of:
PE + KE = 5.47 x 10^-21 J + 3.54 x 10^-21 = 9.01 x 10^-21 J
Its associated average velocity would be 390.2 m/s.
– – –
So, according to Doug’s hypothesis, (PE_sf + KE_sf) = (PE_tr + KE_tr) for individual air molecules.
It very clearly doesn’t, even in this ideal scenario:
PE_sf + KE_sf = 5.96 x 10^-21 J
PE_tr + KE_tr = 9.01 x 10^-21 J
For Doug’s claim to add up, gravity would have to make the air molecules at the tropopause move on average 485.2 m/s more slowly than their counterparts at the surface. So if the ones at the surface move at 506.3 m/s, then the ones at the tropopause could only move at 21.1 m/s.
Then and only then would (PE_sf + KE_sf) = (PE_tr + KE_tr).
In reality, N2 molecules at 171K fly around at an average velocity of 390.2 m/s. No matter where they are to be found in the gravity well.
– – –
With billions and trillions of air molecules careening through space much faster than the speed of sound, continually colliding with their neighbors, gravity has effectively lost its hold on each separate one of them.
It still affects them as a group, though, as ‘air’, but then only as an ordered, macroscopic, Newtonian function, not as a disordered, microscopic, thermodynamic (temperature-related) function.
Both Doug and Stephen Wilde seem unable to separate the two …
“gravity has effectively lost its hold on each separate one “
Nonsense. Does it loose its hold on a plane breaking the sound barrier, or a satellite crashing to Earth? These molecules are not travelling at the speed and altitude necessary to break out of the Earth system, except for a few in the exosphere. I have already mentioned the blatantly obvious fact that, especially in a planet’s upper troposphere and stratosphere, there is warming due to solar radiation. What happens to that energy is explained in the heat creep diagrams here.
And I have also mentioned that inter-molecular radiation (mostly between water vapor molecules) has a temperature leveling effect, reducing the magnitude of the gradient by up to about a third. So this process also affects the KE of molecules. The net result is a gradient of about 6.5K/Km at the equator, for example, where the troposphere is about 17Km to 18Km high and gets down to about -80°C to -90°C from about 25°C to 30°C at the surface. Checking using the means of these ranges, (85+27.5)/17.5 = 6.43.
You Kristian are unable to separate what my hypothesis says as distinct from Stephen Wilde’s conjectures which are similar to the error detailed in the second column here.
By the way, what’s your explanation as to how the necessary thermal energy gets down to the base of the nominal troposphere of Uranus from where it was initially absorbed in the stratosphere?
Go to this comment.
“Is the sum of the two always the same?”
Why should it be? You have completely overlooked the absorption of 20% of solar radiation by the atmosphere, this obviously increasing KE you clot! I have already explained this in an earlier comment and you would have worked it out yourself if you took the trouble to studythe paper and think. But some people just don’t think.
There are two possibilities:
(a) Heat creep happens
(b) Heat creep does not happen.
If (b) then …
the probability of the core temperatures of all planets and satellite moons being just right for the gradient to be just right to get down to just the right temperature at just the right altitude at the anchor point is less than 1 chance in 10^20.
Go back to my comment at 11:56pm.
Doug Cotton says, March 19, 2015 at 4:20 AM:
“”Is the sum of the two always the same?”
Why should it be?”
Because you claim it has to and will be when the -g/Cp temperature gradient is spontaneously established at ‘thermodynamic equilibrium’ in our atmosphere. That’s the scenario here. I’ve given it to you. You’re the one claiming that the PE+KE sum for individual air molecules at this point must be the same at all levels along the pure DALR, and you claim it over and over and over again. It’s a necessity. According to you. Because then apparently entropy is maximized, it cannot go any higher. You’re the one asserting this. Not me. Not us.
“You have completely overlooked the absorption of 20% of solar radiation by the atmosphere, this obviously increasing KE you clot!”
No, Doug. There specifically isn’t any 20% absorption of solar radiation going on here, Doug. None at all, in fact. Because that would’ve disrupted your -g/Cp (DALR) temperature gradient. There wouldn’t be a 9.75K/km falloff rate going up the tropospheric column if this were the case. Since there is also no convection going on at ‘thermodynamic equilibrium’. (And since the LR is dry, then there is also no release of latent heat of vaporization.)
This is a prerequisite of yours, Doug. If KE increases at some levels (like the higher ones), but not at others (like the lower ones), then your DALR temperature gradient could no longer be sustained. But it is. That’s the whole point.
Your objection is moot. You have nothing, Doug. We all see it. You’re just frantically waving your hands.
I’ve handed it to you, Doug. You have your ‘thermodynamic equilibrium’ with a temperature gradient exactly according to the DALR (-g/Cp) and nothing else. There are no external or internal disturbances here at all. Nothing to bring the state out of balance. Your ‘heat creep’ has made sure that the surface air is 117K warmer than the air at 12 km of altitude. As per your hypothesis that the equilibrated (max entropy) situation would require an atmospheric temperature gradient of 9.75K/km (-g/Cp).
“I have already explained this in an earlier comment and you would have worked it out yourself if you took the trouble to studythe paper and think. But some people just don’t think.”
There is nothing to work out here, Doug. Your hypothesis has been proven wrong. You claim something that couldn’t possibly be.
You cannot both have an atmospheric temperature gradient at ‘thermodynamic equilibrium’ equal to the DALR (9.75K/km) and a claimed state of max entropy where PE+KE for individual air molecules is exactly equal at all atmospheric levels. It simply doesn’t add up. Just look at the numbers, Doug. The math is exceedingly simple.
“There are two possibilities:
(a) Heat creep happens
(b) Heat creep does not happen.”
No, there is only one possibility, Doug: (b) heat creep does not happen.
Because it cannot happen …
Kristian:
Thanks for doing the math! You have just proved that Doug cannot even put together an internally consistent argument, even if you grant his flawed premises.
One of the things I’m finding highly amusing here is that whenever someone points out one of his flaws, Doug’s frantic attempts to counter always contradict some of his earlier assertions.
So I had to laugh when you pointed out that KE+PE is NOT constant over height at the DALR, as Doug insists would be the equilibrium case of an isolated system, he responded with an argument about solar absorption! Completely pathetic, but amusing as well.
Curt 1:50pm: +1.
While discussing Doug’s easy to find faulty science whether in paper, blog or book form, again the insights of Douglas Adams are most helpful:
“It startled (the faulty commenter) even more when just after (the faulty commenter) was awarded the Galactic Institute’s Prize for Extreme Cleverness (the faulty commenter) got lynched by a rampaging mob of respectable physicists who had finally realized that the one thing they really couldn’t stand was a smartass.”
Funny!
Then prove the physics wrong in my paper and do the study proving water vapor warms, and then you could be the one to win the $5,000 reward. You can start with the physics refutation here, where I will be able to expose your errors therein.
“Because that would’ve disrupted your -g/Cp (DALR) temperature gradient”
Not denied at all, in fact clearly spelled out in the paper as part of the explanation as to why water vapor cools rather than warms.
“I’ve handed it to you”
What you have “handed” is something totally removed from the full content of the paper, and specifically ignoring the key point which totally refutes what you are saying. The environmental temperature gradient is not -g/Cp for the reasons in the paper and because of the role of radiation that also plays its part in determining the final (environmental) state of overall thermodynamic equilibrium.
Now stop wasting my time and read, study and inwardly digest the paper before you even attempt to refute it. I will prove every such attempt wrong, so you’d better be careful in your explanation.
What I have stated in the hypothesis in the paper (and summarized here is correct and is a direct corollary of the Second Law of Thermodynamics.
You guys can try all you like (as with Kristian’s pathetic attempt above) but you will never prove the the Second Law wrong or prove that my hypothesis is in conflict with that law.
All along I have had my last card up my sleeve. I have asked the question about the “coincidence” in all planets several times on blogs and everyone avoids answering it, because those who think about it realize that the chance of the greenhouse paradigm being right rather than the heat creep hypothesis is less than one in 100000000000000000000.
Now who is to be first to challenge that “coincidence” issue?
Doug I have already answered your question and I think more than once. The only place you have lapse rates on planets is where you find convection (large motions of air). Rising and falling in a gravity field. Gravity creates a density gradient so that more molecules are found lower in the field. This density is based upon the hydrostatic equilibrium. Gravity wants to collapse the atmosphere while internal energy of the molecules holds it and the two opposing forces set up the density gradient. As energy is absorbed or moved from lower regions (like Uranus) to the higher areas and causes convection, this gas rises and expands in less dense air. The expansion requires internal energy from the gas to be used in expanding so the gas cools while rising and heats while sinking generating a lapse rate.
Toneb and Roy Spencer have explained this to you. I believe Toneb has explained this to you multiple times. Why should he continue to explain to you what you do not accept as reality? You do not accept parcels are real. You do not accept (even if you would test it yourself today) that expanding air will cool. Why should people waste time explaining what you do not accept as reality?
I’ll be all ears when you attempt to answer the coincidence issue, Norman. I’m not interested in reading yet another recap of the greenhouse conjecture and the incorrect physics promulgated by climatologists. Do you agree that the chance of that hypothesis being right is less than 1 in 10^20 because of the coincidence issue, or are you just not suitably endowed with sufficient understanding of probabilities to understand the point?
“The only place you have lapse rates on planets is where you find convection (large motions of air).>
Absolute nonsense. Upward large motions of air stop on clear calm nights in the early pre-dawn hours, but the environmental temperature gradient (lapse rate) is still evident. If it were not, then the mean surface temperature would be somewhat colder than 255K at night.
What would cause there to be large motions of upward air on the sunlit side in the 350Km high nominal troposphere of Uranus where we have a near perfect gravitationally induced temperature gradient very close to the calculated -g/Cp value? It is so close that when I used it to calculate the temperature at the base of that troposphere I got 329K whereas it is normally said to be 320K.
Norman does not understand how physicists define convection. “Although often discussed as a distinct method of heat transfer, convective heat transfer involves the combined processes of conduction (heat diffusion) and advection (heat transfer by bulk fluid flow)”
“Bulk fluid flow” does not mean wind which is “forced convection” that does not form a temperature gradient. It is what you may just be able to detect in the “hot car in garage” here.
Only the very slow process of “natural” convective heat transfer (including conduction and diffusion) causes the temperature gradient and it happens at the molecular level as the Second Law implies it will. This fact is easily proved with the Kinetic Theory of Gases, as was used successfully by Einstein and many others, but forgotten in climatology circles.
Now respond to the coincidence issue.
Doug Cotton
Are you sure? “Bulk fluid flow” does not mean wind which is “forced convection” that does not form a temperature gradient.”
I would say the wind is forming a temperature gradient.
Maybe I am wrong, you might have a valid counter point.
http://en.wikipedia.org/wiki/Chinook_wind
Ball4 and JohnKl
You may find this information useful.
http://www.meto.umd.edu/~zli/AOSC621/Lesson%205-6%20Absorption%20Spectroscopy.pdf
If I interpret this paper correctly it would indicate line intensity (S from Hitran data base) is directly proportional to absorption.
http://www.patarnott.com/atms749/pdf/TransmissionCalculation.pdf
Looking at Hitran database
N2
http://vpl.astro.washington.edu/spectra/n2hitranimages.htm
Too many links in posts. In moderation. Will repost in broken up fashion
Looking at Hitran database
N2
http://vpl.astro.washington.edu/spectra/n2hitranimages.htm
and CO2
http://vpl.astro.washington.edu/spectra/co2hitran2004imagesmicrons.htm
The Hitran values come from this database.
http://vpl.astro.washington.edu/spectra/allmoleculeslist.htm
Carbon Dioxide strongest line intensity (S) is 3×10^-18
While N2 is 3×10^-28 which would make it 10 billion times less absorbing when compared to carbon dioxide so for all practical purposes N2 is considered to be a nonabsorbing of IR radiation.
Go back to this comment about the coincidence issue which you deliberately ignore. It is repeated in a similar comment just a few up thread.
“N2 is considered to be a nonabsorbing of IR radiation.”
Wow! Have you just learnt that? All it does is store about 80% of the thermal energy in the atmosphere – which energy is trapped by gravity.
Doug Cotton,
I think you lose “heat creep” hypothesis with Venus for sure. As you stated Venus has a really slow day so you have a night side for months. That means the atmosphere above the surface is no longer receiving any solar radiation.
Your claim is that Venus atmosphere absorbs solar radiation at higher levels and then “heat creep” moves this energy down to the surface and keeps things quite hot.
Now you have a Venus night side atmosphere that receives zero radiation for months. It seems as if “heat creep” would exhaust the energy with no new supply to transmit down.
Radiation hypothesis can explain this. The large amount of radiating Carbon Dioxide absorbs and emits radiation all around in a raidation fog. The radiation moves very rapidlay and will move to all surfaces of the planet so the poles are the same temp as the equator and the night side. Some of the radiation trickles out of Venus upper atmosphere which is replaced by solar energy being absorbed at the same rate so the planet maintains and overall equilibrium temperature. I think radiation explains Venus condition far more realistically than your own.
Venus
Day:
Thermal Energy in by downward convective heat transfer
Thermal Energy out by radiation (not much because of back radiation)
Night:
Thermal Energy out by upward convective heat transfer
Thermal Energy out by radiation (not much because of back radiation)
“It would be just far too great a coincidence for all planets to be just “cooling off” and yet having the right -g/Co temperature gradient…
1) The earth’s oceans have the OPPOSITE gradient to what you predict.
2) Compared the the moon, the earth has 6x gravity and 4xradius and similar composition (ie similar Cp). The Earth’s core should be ~ 24x hotter than the moon’s by your hypothesis, when in fact it is not nearly this much hotter. (Cp may be a bit different, but not enough to fix your hypothesis.)
So your hypothesis fails for two obvious cases. And you haven’t shown any indication that it specifically works for any other terrestrial planet or solid moon. Where is your table of g, Cp, r, and T_core for the planets (and larger moon) showing how your predictions work?
All you really have is that all terrestrial planets have a temperature gradient — but that is to be expected due to initial accretion and gravitational differentiation and continued radioactive decay.
*********************************************
For Jovian planets, I would be surprised if the internal gradient was NOT ~ g/Cp. The gradient can’t rise above g/Cp or convection would start and heat would get carried off faster, inducing a return toward g/Cp. On the other hand, gravitational collapse and radioactive decay continue to add energy to the core, trying to warm it thus raise the gradient. The ‘natural’ solution is for a gradient of ~ g/Cp until the radioactive decay becomes small and gravity can’t compress the core further.
******************************************
RECAP
* There is no coincidence in jovian planets having a gradient of g/Cp.
* There is no evidence yet (and even some counter-evidence) that terrestrial planets follow your hypothesis.
The issue of ocean temperature gradients has been explained in earlier comments and in my paper linked here. It might surprise you to learn that the expected gradient of about -2K/Km is in fact found in the dark winter period in the calm waters north of Norway below the remains of the summer thermocline region.
I have also responded in the past to your comparison with the Moon. You have completely neglected the fact that the specific heat increases enormously at the hotter temperatures in Earth’s mantle, so the gradient is only around 1K/Km.
Your hypothesis, Tim Folkerts, is the one that fails with odds of less than one in 10^20 of being correct, as you should realize when you address the coincidences issue.
Sorry, Doug, but hand-waving assertions don’t cut it in science.
* you create an ad hoc reason why the ENTIRE UNIVERSE except for earth’s oceans should follow your rule.
* you assert that Cp values for the moon magically adjust to give your desired results (with no experimental backing).
* you assert that other terrestrial planets follow your rule with no calculations to back up the claim.
* you ignore that fact that gas giants should roughly follow the adiabatic lapse rate even without heat creep.
Heck, I would be impressed if you could simply show how the earth & moon follow your hypothesis. Find a source that gives heat capacity for mantle rock … heck for ANY rocks as a function of temperature. Show how these numbers give the correct core temperatures for earth an moon.
No hand-waving or unsupported assertions allowed. You are the one claiming it fits your hypothesis … put up or shit up.
You absolute clot, Tim Folkerts. Don’t you know that the top of the oceans in non-polar regions are warmed from above and some (probably most) of that extra energy must be transferred to cooler regions below, creating a thermocline and affecting the temperature gradient, over-powering the slow convection process that forms the gradient, just as in the stratosphere and tropopause region, even to the floor of the ocean. All this is explained in my paper. See also my reply to Norman regarding the region north of Norway in winter.
There is nothing magical about the well known fact that Cp varies with temperature. Even for the Venus troposphere this varies as in this table showing Cp of CO2 at 175K to be 0.709 but over 1.4 at temperatures above 2600K. Surely you know that the temperature gradient in the outer 10Km of Earth’s crust is about 25K/Km as seen in boreholes. Try extrapolating that to the core of the Earth!
There are stacks of calculations regarding other planets, including ones I have done and written up years ago on climate blogs. Read the series of articles linked in the second column on this page of our group’s website.
Heat creep does not cause the temperature gradient: gravity does, just as it causes a density gradient. Then the pressure gradient is a corollary.
When are you Tim Folkerts going to address the coincidences issue?
There is no “coincidence” to explain until your calculations show a good agreement between measurement and predictions.
So far you only CLAIM the cores match. The only “coincidence” is that Cp values coincidentally change in your world to exactly match your desired results.
Show values for ROCK, not gas. Show how Cp varies with temperature (and pressure ideally). Show how the actual numbers give the correct results.
“Surely you know that the temperature gradient in the outer 10Km of Earth’s crust is about 25K/Km as seen in boreholes. Try extrapolating that to the core of the Earth!”
So … you are saying that because your hypothesis ONLY works for the first ~ 10-100 km but fails badly after that — this is SUPPORT for your hypothesis. 🙂
You can work with just the tropospheres if you wish. Using the -g/Cp gradient we get quite accurate estimates of the temperatures at the base of the tropospheres of Earth, Venus, Uranus etc. How about doing some calculations yourself?
For example, you could try explaining the coincidence that the temperature at the base of the nominal troposphere of Uranus is 320K.
Q.1: Does the Uranus core “know” how it should be so that the temperature plot gets down to just the right 320K at the base of the Uranus troposphere?
Q.2: Does the region at the base of that troposphere “know” that about 350Km further up (in the stratosphere) the temperature has to get down to the radiating temperature of 58K but it must do so with a temperature gradient -g/Cp all the way through that 350Km?
Keep your focus on the questions I ask you Tim Folkerts – one at a time, as you feel the knot tightening.
oops … that should be “shut up” :-/
Footnote: Regarding the temperature data for several other planets and moons, Tim, see this well known paper.
Doug, I can’t fathom how you continue to miss points and go off on tangents.
You claim the INTERIOR of rocky planets and moons follows your hypothesis, yet ….
1) you admit your hypothesis only gives approximately the right result (25 K/km) in the very upper layers.
2) The gradient drops to ~ 1 K/km (~6000 K in ~6000 km) farther inside, yet there is no way Cp increases by 25x.
3) you quote a paper about *atmospheric* temperature to support a hypothesis about *internal* temperatures. Even if it is right (which is widely debated), it does not help out your argument.
SO ONCE AGAIN … since YOU claim that g/Cp explains the internal gradients, it is up to you to actually SHOW that g/Cp explains the internal gradients.
The questions to be answered Tim Folkerts are in this comment above. No red herrings will be entered into, especially matters relating to the old 20th century radiative greenhouse paradigm.
there is no way Cp increases by 25x.”
Yes there is actually – do some research before you make such assertive statements. In any event, the net force of gravity reduces because there is huge mass on the outer side of the inner mantle for example.
Regarding specific heat at high temperatures, the reason it increases is primarily because more degrees of freedom become unfrozen. See plots showing increases by factors of over 100x at typical temperatures found in the mantle.
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/debye.html
Hi Tim and Doug,
For what it’s worth, Tim has a good point and echoes one I’ve made before. You Doug frequently state EXTRAPOLATIONS like the temperature of the lunar core as FACT! Then use such as evidence for your claims. The term SCIENCE does not apply to such action.
Have a great day!
The fundamental point is that the state of thermodynamic equilibrium has a temperature gradient that enables conduction and convective heat transfers to warmer regions, as explained with the heat creep diagrams here.
The precise temperature of the core of a planet or moon is of course subject to significant error margins, because no one really knows accurate estimates of specific heat at such high temperatures. But relative differences in core temperatures (such as between Uranus and Neptune) do gel with the differences in the other parameters such as gravity, height of atmosphere and specific heat in the cooler regions at least.
If heat creep did not happen and if isothermal conditions were the state of thermodynamic equilibrium, then core temperatures would be probably less than a quarter of what they are.
Doug Cotton,
Evidence Curt is correct about you. In you mirror experiment you wanted me to try to prove some point clearly shows you do not look at the larger system to try and understand what is going on.
If the Earth’s surface is emitting 300+ watts/meter^2 and there are the same amount of watts returning a mirror would in NO WAY determine anything as it would only reflect back the same amount of energy the atmosphere would have returned had not the mirror you are holding been blocking its path. NOt a good expermient design at all. Day and night would not help. The energy reflected by the surface is not being absorbed and you would just return it and it still would not absorb. Bad thought process Doug.
I hope you were not as hostile to students as you are to anyone who would dare question your hypothesis or authority. I think you would be one of the worst teachers I could imagine for a student learning something. Agree 100% with what I say or I will call you stupid and declare myself the god of physics that none may dare Question!
Of course I know that the electro-magnetic energy in any radiation from any source which is cooler than the target is not converted to thermal energy in a warmer target. As soon as any distance is involved, a lower intensity is returned (as by the mirror) but the mirror will deliver more than the troposphere with its average height higher above the ground than the mirror. So, if anything is going to help the Sun to raise the surface temperature, the mirror has a better chance of doing so than back radiation. Of course both fail to do any warming at all, and I wrote a comprehensive peer-reviewed paper on this three years ago, based on the comprehensive computations by that professor of applied mathematics to which I have linked in a comment above. Such computations should surely impress you, Norman.
Maybe you’d like my computations of the probability of your being right.
Let’s say that the core of Uranus could reasonably have 100 different temperatures that could be distinguished within the limits of error, say from 0K to 9900K in intervals of 100K. Applying the calculated -g/Cp temperature gradient, only one of those temperatures gives the observed temperature at the anchor point (to the nearest 100K) which is the methane layer at about 58K in the stratosphere.
Apply this to say the relevant planets (with significant atmospheres) and similar satellite moons. We will say there are 10 of such. So the number of possibilities is
100^10 = 100,000,000,000,000,000,000.
As I keep saying, when will you address the coincidences issue?
Doug Cotton,
“Of course I know that the electro-magnetic energy in any radiation from any source which is cooler than the target is not converted to thermal energy in a warmer target.”
Really you think this is a correct statement? Then please explain why hot coffee poured into a thermos is still hot several hours later when you pour it out. The mirrors reflecting the radiation back into the hot liquid are most certainly much cooler than the radiating fluid however in the real world the coffee stay hot. Why?
Your statement is not a true statement. Does not matter what comprehensive computations by that professor of applied mathematics does. As JohnKl explained above, math is only as good as the assumptions it is based upon. Real world evidence suggests his math is wrong since the hot fluid must radiate energy and once it does it will be cooler than before and it will not stop radiating away its energy. Only possibility is the cold mirror is sending the energy back to the fluid, and the fluid must be absorbing the radiant energy or it would not remain hot (if radiation pseudo scattered rather than being reabsorbed and turned back into energy in the fluid). Sorry it is wrong and reality will not bend to your conjecture on the issue. You can ignore it but reality is what it is.
“Applying the calculated -g/Cp temperature gradient [to find the core temperature] …
And you have done this? Could you show us where you did this calculation? (And for that matter, could you tell us what value you used for the core and where you got that data?)
Doug Cotton,
I will offer an alternate explanation for why wet areas surface is cooler than dry areas that has nothing do do with water vapor cooling the atmosphere for allowing a gateway for Inrared emitted by Earth’s surface.
First use your own link.
http://www.waterencyclopedia.com/Ce-Cr/Climate-and-the-Ocean.html
Carbon Dioxide absorbs very little incoming solar raditation as compared with water vapor. Click on your link.
So higher water vapor areas will absrob more of the solar incoming IR spectrum than drier areas leaving less radiant energy from reaching the surface. This would be a neutral for water vapor, neither cooling nor warming. It would only determine what part of the system was warming or cooling as the energy will still be absorbed somewhere.
Point two:
http://www.waterencyclopedia.com/Ce-Cr/Climate-and-the-Ocean.html
If you look at this you will see warm moist air rises at the equator, condenses, rains out and then eventually sinks. Persistent desert areas are dry because the air above is dry and sinking and sinking air warms as it lowers. Persistent wet areas are such because air is rising. These rising areas are also pulling in surrounding air which would be cooler (receiving less solar energy than the equator relatively speaking).
The descending air over deserts may even limit convection not allowing the hotter air to move up as much kind of trapping the heat closer to the surface. Toneb would be able to answer that.
Anyway the two points can easily explain what you have observed in your research. Same information but totally different conclusion.
When doing science it is wise to come up with as many explanations you can for phenomena (wet areas cooler than dry areas in general) and also be open to other views. You concluded that this means water vapor is a coolant but that can be totally opposite of reality just depending upon your view and looking at the total picture and not a real over simplified one (like temperature with no other considerations as to cause of what you see other than water vapor is causing the cooling). NOT GOOD SCIENCE in any book.
Yes I explained how water vapor cools due to its radiation properties in the paper.
I also published a comprehensive study of real world data that proves increasing WV causes there to be lower mean daily maximum and minimum temperatures, as based on 30 years of temperature and precipitation records from 15 locations on three continents. When will you do your own study?
You have still not addressed the coincidences issue.
Doug Cotton,
Roy Spencer does not reject the radiation theory of GHE. He rejects the enhanced GHE. The positive feedbacks that will overwhelm the system and lead to catastrophe.
Only models predict a CAGW. Radiation alone from CO2 will cause a minor warming effect if CO2 is doubled.
Clouds are a really big factor that is ignored by models.
Your theory has been proven wrong by many intelligent people who have knowledge and experience I do not possess.
You have Curt who has extensively studied thermodyamics and makes clear arguements against your backward magic thinking and Toneb and actual trianed meteorologist that studies the atmosphere and the dynamics of it but you do not listen to these voices.
Well Roy should reject the radiation theory of GHE because the probability of it being correct is less than one in 10^20.
No, my hypothesis has not been proven wrong by anyone. No one has submitted what I requested for the $5,000 reward. Be the first, starting with a study such as mine in the paper. Remember that the IPCC wants you to be gullible enough to believe water vapor raises the surface temperature by about 30 of those “33 degrees of warming” which are a complete fiction because gravity props up the surface end of the thermal plot, not back radiation.
Roy also still believes that “33 degrees of warming” hoax because he, like you, is an Isothermalist and not a Loschmidtian.
Now, when will you address the coincidences question?
Doug 5:45pm: ”No, my hypothesis has not been proven wrong by anyone.”
Your hypothesis was proven wrong to start with in the specialist literature. As of 1998, 2004 and 2006, why bother. The proof is easy to find if you have the pre-req.s. Why not up the offer to $100^10 = $100,000,000,000,000,000,000. Sell the mansion to payoff the winner.
”No one has submitted what I requested for the $5,000 reward.”
They know Doug’s offer is a hoax, waste of time & not the 33K which is measured. The right physics is already in the literature.
”..gravity props up the surface end of the thermal plot, not back radiation….address the coincidences question?”
Baseless. Doug can’t even get the units right. Props? Gravity is a force field, radiation is energy. There are no coincidences. Your calculations are well known to be incorrect.
And where is your rebuttal of the coincidences issue? Where is your personal study of temperature/precipitation correlation that you could easily present us with here after perhaps half a day’s work? Where is you personal explanation in your own words proving the physics wrong, because your reference to those papers sure does not do so.
Nowhere, so I rest my case.