## Direct Evidence of Earth’s Greenhouse Effect

April 10th, 2013 by Roy W. Spencer, Ph. D.

After yesterday’s post about what determines temperature, I thought I would revisit one of the most convincing evidences of Earth’s greenhouse effect.

As I’ve mentioned before, a handheld infrared thermometer is a great little tool to help gain physical insight into the thermal radiative (infrared) effect the atmosphere has on surface temperature.

Here I’m going to give an example of how the IR thermometer responds to a clear sky versus a cloud, and I invite alternative ideas of what is causing the resulting indicated temperature changes.

First, I want to demonstrate how the IR thermometer does indeed respond, remotely, to the temperature of any object it is pointed at. I made the following measurements inside our break room freezer (reading about 9 deg. F), and while pointed at the coffee pot (reading about 129 deg. F):

(I’m not going to address the absolute accuracy of the measurements, which is probably not better than a few degrees, since we will be dealing with temperature changes of ten of degrees or more. If you work with these things enough you will see they are sensitive to changes smaller than 1 deg. F.)

Then, I took measurements outside our UAH building while pointing the IR thermometer at the sky. For reference, the ambient air temperature at our weather station about 100 ft away was 78 deg. F, and the dewpoint was 63 deg. F.

The thermometer is pointed first at a clear patch of sky (reading 27 deg. F), and then an adjacent cloud (reading 41 deg. F):

Now, my question is this:

What caused the IR thermometer reading to warm up by 14 deg. when it went from clear sky to the cloud?

I’m especially interested to hear an answer from those who tell me there is no such thing as downwelling sky radiation (aka “back radiation”). No matter what you believe is happening, it is rather obvious that the cloud influences the temperature reading differently than clear sky. (If you are thinking it is a reflected sunlight effect, you can perform the experiment at night and see the same effect; furthermore, the highest cloud temperatures you will get are from the thickest, *blackest* clouds on the bottom…so it’s not a reflected sunlight effect).

What Does the IR Thermometer Actually Measure?

As I mentioned in my previous post “What Determines Temperature?”, temperature is an energy budget issue, the result of energy gain versus energy loss.

Inside the IR thermometer, there is a thermopile (an electronic circuit very sensitive to temperature differences) with thermistors measuring temperature at both ends. When you point the thermometer at an object with a different temperature than the thermopile, an IR lens (which has a beamwidth of about 5 deg.) allows IR radiation to flow between the lens end of the thermopile and the target object.

If the target object is warmer than the viewing end of the thermopile, the net IR flow is from the object toward the thermopile, which begins to warm. Circuitry measures how fast those temperature changes occur and extrapolates an estimate of the target object’s temperature. (The thermometer has no idea what the infrared emissivity of the object is, so my unit simply assumes an emissivity of 0.95).

If the target object is colder than the thermocouple, the net flow of IR radiation is from the thermocouple to the object, and the thermocouple cools.

In the cloud case, the cloud has a higher emitting temperature because it is at a lower altitude, and it is more opaque in the infrared than the clear sky is.

A similar effect can be achieved from just the clear sky, by pointing the IR thermometer up at different elevation angles. Increasing temperatures will be indicated as the elevation angle is lowered toward the horizon. Today, I measured about 15 deg. F pointing straight up to about 35 deg. F when pointed about 20 deg. above the horizon.

In this clear-sky case, infrared absorbers/emitters (aka “greenhouse gases”) in the atmosphere, which are partly (but not totally) transparent at the IR frequency band the thermometer is tuned to, become more opaque as the thermometer is pointed at lower elevation angles. As the elevation angle is lowered, the path length through the atmospheric absorbers increases, and the altitudes from which the IR emission is being received are lower and thus at higher emitting temperatures.

This is the most convincing, do-it-yourself, direct observational evidence of downwelling sky radiation I have been able to find, and it makes a great little science experiment for students. What makes it “direct” evidence is that it actually measures the surface temperature effect (at surface of the thermopile) of changing downwelling IR radiation from the sky. This is the same thing happening continuously at the surface of the Earth as the strength of the greenhouse effect changes from water vapor, clouds,..oh yeah, and carbon dioxide.

And if you STILL don’t see how this demonstrates the greenhouse effect, imagine what would happen if you suddenly removed all of that atmosphere and clouds: there would be a sudden increase in the rate of net IR flow from the surface of the Earth to outer space, and temperatures would drop. THAT is the greenhouse effect.

For those who do not believe the above explanation, give us your alternative answer to the question: what causes the IR thermometer indicated temperature to increase from (1) clear sky to cloud, and (2) zenith clear sky to low-elevation clear sky?

### 308 Responses to “Direct Evidence of Earth’s Greenhouse Effect”

1. Hi Roy: Yes an IR thermometer measures temperature and does not record any form of “back radiation” measured in W/m2 as heat transfer from cold to warm, as you very clearly point out yourself. Don’t you see that we are saying the same thing in this regard? Temperature is one thing, and heat transfer is something different, in particular heat transfer from cold to warm.

Further, describing the “greenhouse effect” as the total effect of the atmosphere is misleading and even dangerous, since in arguments by alarmists the “greenhouse effect” is quickly changed into the effect of CO2 alone. With this logic, CO2 has a (big) warming effect since the Earth would be a bit cooler without any atmosphere at all. I can’t believe that you consider this to be rational scientific logic, or am I wrong?

Best regards, Claes.

• Roy Spencer says:

I agree the effect of additional CO2, by itself, is very small.

• If the effect is so small that it cannot be observed, how do you that it is an “effect” at all.
Isn’t this like insisting that there are (or must be) ghosts, which cannot be observed, because there is no definite evidence that there are no ghosts? The burden of proof lies primarily on those claiming existence of an effect, not on those doubting its existence, right?

• Roy Spencer says:

Actually, it has been observed. The NASA AIRS instrument flying for over 10 years on Aqua has measured the decreased IR emission to space with increasing CO2 in certain spectral bands, and have retrieved the (gradually increasing) CO2 content based upon those measurements:

Yes, the signal is very small, and I have always maintained it quite possibly smaller than natural changes on the same time scale.

• You have to be careful when interpreting the reading of AIRS as I discuss in detail at

Evidencing warming of CO2 by these measurements are as questionable as evidencing “back radiation” by an IR-thermometer, which we have just agreed upon.

• Will Janoschka says:

The adabatic mostly O2 N2 and H2O. will
change slightly with increasing CO2.

Does this affect the outward flux
from the atmosphere?

• Nabil Swedan says:

The observed decrease in the IR to space can also be attributed to causes other than the greenhouse gas effect. An increase in the concentration of CO2 increases the molecular weight of air and lapse rate increases in absolute value. Temperature profile decreases and infrared radiations decreases as a result.

• DJC says:

Roy

Claes is right about the AIRS measurements. Some of the energy which CO2 may have absorbed (more on the way down than on the way up, anyway) can be transferred to molecules of water by diffusion. The water then radiates that same energy so it simply gets out another gate to space.

You could have learnt this from my paper published over a year ago – see Appendix FAQ 6 copied below from “Radiated Energy and the Second Law of Thermodynamics.” I hope you don’t take another year or more to learn about the content of this year’s paper “Planetary Core and Surface Temperatures.”

Q.6 What happens to the radiation which is absorbed by carbon dioxide?

When spectrometers near the top of the atmosphere (TOA) are pointed at a source of radiation on the surface they will detect rays which get straight through, but the rays with frequencies which can be absorbed by carbon dioxide are mostly missing, indicating that they have been absorbed by carbon dioxide molecules. When this happens some of the surplus radiated energy will be converted to thermal energy. This energy might or might not be shared with, for example, water vapour molecules. Whatever happens, subsequent spontaneous emission is more likely (because of the warming) but the new ray is highly unlikely to strike the spectrometer. If the new radiation heads towards warmer regions in the atmosphere, or to the surface itself, it will undergo resonant scattering. But if it heads upwards to cooler regions it will either get through to space or strike another molecule further up, where the process starts over again. One way or another, the energy gets out to space by another gate.

• DJC says:

Nabil

The effect of carbon dioxide on the thermal gradient works in two ways. Firstly, yes, it does make a very minute change in the mean specific heat of air. But its own specific heat has a mean of about 0.75 compared with air just over 1.0. Hence its effect is to reduce the gradient but only by about 1 part in 10,000. The greater effect is the reduction of the gradient by intra-molecular radiation working against the gravitationally induced thermal gradient. Water vapour appears to reduce surface temperatures by about 5C (from what would have been about 20C back to about 15C) but carbon dioxide’s effect could be as little as 1/1,0000th of this. So we have …

Cooling effect of carbon dioxide 0.001 x 5 = 0.005C
Warming effect of carbon dioxide 0.0001 x 30 = 0.003C
Net cooling effect of carbon dioxide 0.002C

So, it’s a close shave, but is minuscule.

• DJC says:

Sorry – typo 1/1,0000th should be 1/1,000th

• Jonas N says:

DCJ (and Claes & Roy)

I agree with that thought, that the from-space observed dips in the IR spectrum corresponding to the CO2 absorbtion bands just show that the atmosphere as a whole, has become a little more opaque at these frequency bands over the years. As expected and due to the increased CO2 content.

However, this does not iply that the atmosphere is heating by the same amout apparently missing (and increased lately) where the CO2-dips are.

The warming effect, if discernible, must be the change in total IR leaving the earth. Im not saying that this is not so, but think it remains to be established/measured properly.

A simple analogy would be: A small lake with constant level and water supply matching the flow through the outlet brook. If you put some scattered sticks in the brook (CO2-molecules) which redirect the waterflow there, you would notice that right behind them water levels are lower, however between the sticks the same water (as before) would flow through the brook, and there at a mincule higher level.

And the bigger question: Would it rise the water level in the lake?. Yes! At least if this is done exactly at the outlet. But to an extent you can observe/measure? Questionable …

• Jonas N says:

Further Roy,

Yes the handheld IR thermometer shows different temperatures, and you may call this information it detects ‘backradiation’. But you are also pointing at ‘objects’ at very different locations, and with different temperatures. So how could anybody expect something else?

The clouds are indeed warmer than whatever the signal pointed at the clear sky measures (probaly the scarse GHG molecules all the way up through the sky)

The experiment shows different temperatures at different altitudes, which I don’t think is disputed.

• Hops says:

Claes writes: “The burden of proof lies primarily on those claiming existence of an effect, not on those doubting its existence, right?”

This gets to one the great fallacies of the debate over climate. It is true that in science, the burden of proof is on those who make a claim. But when it comes to health and safety, the burden is on those who want to make a change.

The FDA doesn’t authorize a new drug until such time as it is proven unsafe. Likewise, we should not be changing the composition of the atmosphere unless proven safe beyond any reasonable doubt.

I think what we are doing to future generations is immoral. And that’s why this not just a dispassionate debate about a matter of science.

Hops

• John says:

Hops writes: “The FDA doesn’t authorize a new drug until such time as it is proven unsafe. Likewise, we should not be changing the composition of the atmosphere unless proven safe beyond any reasonable doubt.”

“I think what we are doing to future generations is immoral. And that’s why this not just a dispassionate debate about a matter of science.”

The general atmospheric composition remains approximately 78.09% nitrogen, 20.95% oxygen, .93% argon, .039% carbon dioxide, .0018% neon and other trace gasses. Carbon dioxide trace gas increased 10-15% since 1958 when measurements at Mona Loa began. What has been the immoral result? More terrestrial flora? What magic carbon dioxide level do you believe represents the moral ideal? What level represents moral failure? Mars and Venice have atmospheric carbon dioxide level in excess of 95%. Do they represent immoral planets? What action can anyone take that will prove safe beyond a reasonable doubt? One assumes the risk of death every time the drive their car. Reasonably safe based on who’s judgement? Please let us know your thoughts. What you’ve stated seems in-coherent.

• Guy Threepwood says:

Using passion instead of science is the problem.

Most plants and life on Earth evolved with far higher Co2 ‘pollution’- up to 7000 ppm in the Cambrian & >4000 in the Ordovician ice age, plants still prefer 1200-1500 ppm Co2 ‘pollution’ today- which you also breath today in public buildings with no ill effects, but they had consumed this resource down to a near starvation level of around 270 ppm by the industrial revolution.
We have helped restore a tiny fraction of this vital nutrient ‘immorally’ squandered by plants!

• Curt says:

Why, then, is the temperature of the sensor so different in the same location at virtually the same time when just pointed at something different?

• Christopher Game says:

Claes Johnson’s post http://www.drroyspencer.com/2013/04/direct-evidence-of-earths-greenhouse-effect/#comment-73731 is muddle headed. He is confusing himself by mathematically playing with words, and getting their physical meanings wrong.

He writes that “an IR thermometer measures temperature and does not record any form of “back radiation””. That is nonsense. The device directly responds to radiation, not to temperature directly. Indirectly, the temperature of the source of the radiation is inferred from the radiance of its radiation. Radiance works out so that it can tell temperature at a distance, taking into account the inverse square law. This basic physics of radiation is repeatedly and consistently denied by Claes, when he denies the existence of “back radiation”. Instead, Claes has his own privately invented mathematical scheme by which he cancels back radiation mathematically, but this mathematical operation does not abolish the physical fact.

Claes is here trying to make out that he basically in some respects agrees with Dr Spencer, but really at the heart of it he radically disagrees with Dr Spencer.

Dr Spencer is telling us of the almost universally accepted way of understanding atmospheric radiation, while Claes radically denies it. The weight of analysis and evidence in favour of the nearly universally accepted understanding is massively overwhelming. It would be impractical for me to try to support this statement in detail here, because of the massive amount of evidence and analysis. If you don’t like accepting this particular consensus, you will need to read a good range of textbooks and original research and do some experiments for yourself. I cannot do better than that.

Claes misleadingly tries to imply that Dr Spencer thinks that there is heat transfer from cold to warm. On a charitable interpretation, Claes has accidentally misread Dr Spencer. On a more likely interpretation, Claes is so blinded by his irrational belief in his own privately invented mathematical scheme that he is compelled by his inner workings to misread Dr Spencer.

Dr Spencer does not think that heat is transferred from cold to hot. Dr Spencer thinks that radiative energy can propagate from cold to hot. At the same time he thinks that radiative energy can propagate from hot to cold. Electromagnetic radiation can travel in opposite senses in the same spatial direction in the same place at the same time. It is not a material fluid that can travel in only one sense at one and the same place at the same time. Dr Spencer thinks that the heat transfer between the hot and cold bodies is the difference between the respective amounts of radiative energy emitted by one body and absorbed by the other. He also knows that it is fact of nature that this difference always works out so that heat is transferred from the hot to the cold body.

Claes, by implication in his post above, misrepresents Dr Spencer’s view.

This was all worked out long ago. Prevost 1791 is the main original source, and it has not been overthrown in over two centuries, not even by Claes. The idea was developed further experimentally by Balfour Stewart 1858 and more lucidly and theoretically by Gustav Kirchhoff 1860. Kirchhoff’s work led to Planck’s 1900 and Planck’s to quantum mechanics 1925. Claes thinks he can singlehandedly overthrow all this. No way, Claes.

• Claes Johnson says:

Independent heat transfer from cold to warm violates the 2nd Law of rhermodynamics. To believe that it anyway happens because it is conditional and always is coupled to a greater transfer from warm to cold, requires a physical mechanism for this coupling and accounting which however is lacking. When doing physics it is better to stick to principles of phyiscs than magics.

• Christopher Game says:

Claes, you are making the mistake of thinking that radiation by itself from one body to another is heat transfer. That is wrong in classical thermodynamics because it is not a process that takes the bodies from one state of thermodynamic equilibrium to another. Thermodynamic equilibrium usually obeys the principle of microscopic detailed balance, and one-way radiation of course is very far from that. You seem to be appealing to the second law of thermodynamics to provide an answer about how radiation propagates. That is a mistake. Thermodynamics does not provide information of that kind. Thermodynamics presupposes such information. You can check this in a reliable account of classical thermodynamics. You are asking for a coupling that does not exist and is not needed. The emission of thermal radiation is spontaneous. Remarkable and surprising perhaps, but not something that you will succeed in persuading rational and well-informed people does not happen. Given that, one also needs the Helmholtz reciprocity principle for radiative transfer, and Kirchhoff’s law that there is a unique and universal spectral radiance for black body radiation, with its corollary that for a body in thermodynamic equilibrium with its surroundings the ratio of the emissivity power to the absorptivity factor is equal to that universal spectral radiance. Of course Kirchhoff’s unique and universal spectral radiance is explicitly specified by Planck’s law, which arose from the investigation of Kirchhoff’s law.

You have long and conspicuously shown yourself to be impervious to reason on this matter, and I do not in the least expect to persuade you to see the light, but I thought it might be useful to post a warning to some readers that you do indeed hold to your own private doctrine that conflicts radically with the generally accepted and I think right understanding. You are on the wrong track to try to make out that the generally accepted understanding works by magic.

• Christopher: You write “The emission of thermal radiation is spontaneous. Remarkable and surprising perhaps”. Yes, it is so remarkable and surprising that it lacks physics rationale. It follows the unfortunate tendency of modern physics to believe that the more remarkable and surprising, the better.

What is “thermal radiation” as opposed to “radiation”. Isn’t “thermal radiation” = radiative heat transfer, which must satisfy the 2nd law? “Back radiation” like “back conduction” violates the 2nd law and to insist that it anyway is real, is magic and not physics. Why not be rational and give up a meaningless advocacy of a physical effect, which is not real and only serves to confuse the minds of both scientists andy people?

• Christopher Game says:

Claes, you still get the basic physics wrong. No, thermal it is not right to say simply “”thermal radiation” = radiative heat transfer”; in order to get the physics clear and right you need to think more systematically and express yourself more carefully than that. Your thinking here, or your expression of it, is muddled. That you write “”thermal radiation” = radiative heat transfer” shows right here and now that you are happy to speak so loosely and carelessly that I cannot expect to have a useful rational debate with you. I knew that already before I posted. I do not wish to spend further time trying to debate with you.

• Gordon Robertson says:

Claes…it’s amazing how many people cannot understand this basic tenet of thermodynamics. Clausius stated it in words, that infrared energy can flow both ways but heat can only be transferred from a warmer body to a cooler body without external input.

In their rebuttal to Gerlich and Tscheushner, Halpern et al indicated a complete misunderstanding of that basic principle. In a nutshell, they confused infrared energy with heat. They went so far as to attempt analyzing the 2nd law through entropy. There’s nothing wrong with that but they obviously pulled equations out of thin air and plugged in numbers.

They accused G&T of claiming that in a two body scenario, with a warmer and a cooler body radiating against each other, that one body was not radiating. Halpern et al were using a mathematical analysis based on Boltzmann without understanding why they were applying the math.

David Bohm once pointed out that equations without an understanding of their meaning were garbage. G&T had to point out to physicist Rahmstorf, a rabid alarmist, that the 2nd law is about heat, not infrared energy. I wish people would get that simple message. Thermodynamics is about heat, not infrared theory per se.

• Nabil Swedan says:

Christopher, There are no back-radiations, it is against the laws of physics. This experiment is a crude one and it measures back-radiations from the thermometer itself. If you cool the instrument to about zero absolute as they do in infrared astronomy, you get zero back-radiations. That is why there is infrared astronomy. In the presence of 333 w/m2 or even 1 w/m2 of back-radiations, there would have been no infrared astronomy, which is based on detecting minute infrared radiations from the universe.

2. Curt says:

From Roger Pielke Sr’s blog a while back, a similar post:

http://pielkeclimatesci.wordpress.com/2011/07/12/new-paper-measuring-total-column-water-vapor-by-pointing-an-infrared-thermometer-at-the-sky-by-forrest-m-mims-iii-et-al-2011/

referencing a paper by Mims et al, showing that a cheap infrared thermometer can provide a very good humidity measurement.

http://journals.ametsoc.org/doi/pdf/10.1175/2011BAMS3215.1

The abstract:

A 2-year study affirms that the temperature indicated by an inexpensive (\$20 to \$60) IR thermometer pointed at the cloud-free zenith sky (Tz) is a proxy for total column water vapor (precipitable water or PW). Tz was measured at or near solar noon, and occasionally at night, from 8 September 2008 to 18 October 2010 at a field in South-Central Texas. PW was measured by a MICROTOPS II sun photometer. The coefficient of correlation (r2) of PW and Tz was 0.90, and the rms difference was 3.2 mm. A comparison of Tz with PW from a GPS site 31 km NNE yielded an r2 of 0.79, and an rms difference of 5.8 mm. An expanded study compared Tz from eight IR thermometers with PW at various times during the day and night from 17 May to 18 October 2010, mainly at the Texas site and 10 days at Hawaii’s Mauna Loa Observatory. The best results were provided by two IR thermometers that yielded an r2 of 0.96 and an rms difference with PW of 2.7 mm. The results of both the ongoing 2-year study and the 5-month comparison show that IR thermometers can measure PW with an accuracy (rms difference/mean PW) approaching 10%, the accuracy typically ascribed to sun photometers. The simpler IR method, which works day and night, can be easily mastered by students, amateur scientists and cooperative weather observers.

Capsule: A \$20 infrared thermometer pointed at the cloud-free zenith sky can measure precipitable water vapor about as well as a sun photometer–and it can do so day or night.

• Roy Spencer says:

Yes, I exchanged e-mails with Forrest before he got that work published. Cool stuff. On clear summer mornings before I go to work I use my IR thermometer, just as Forrest suggests doing, to see how humid the atmosphere is. Here in Alabama, it gives a good idea of whether you will get thunderstorms later in the day.

• John Olson says:

The question should be what the IR instrument is actually measuring in the “clear sky” case? Since there is no “object” in the field of view, the IR averages the temperature of some nebulous mixture of “upper atmosphere”, or whatever altitude corresponds to the expected temporal response (what does the laser pointer “bounce” off of? In the case of your “black thunderstorm cloud” your altitude is lower than white, wispy cirrus clouds. Similar comment for pointing at zenith rather than horizon. How far (high) does the laser pointer get before being bounced back to the IR instrument? Should be some significant temperature differences at those different altitudes, no?

3. stephen richards says:

Measurement instrument bandwidth ? Exactly what object is it measuring (the coffee pot is reasonable because you are very close but after that ???

Roy this is not the way to demonstrate your thesis on backward radiation.

4. Norman says:

Roy,

Could the warmer temp of the cloud be the actual cloud temperature and not backraditaion? Low level clouds are water droplets and above the freezing point. Water vapor condenses and warms the air around it forming the water droplets. Wouldn’t this explain the warmer temp of the cloud?

Seems Prof. Nasif Nahle has conducted similar experiments but with a much different interpretation of results.

5. This “debate” has gone on far too long. Obviously, measuring the temperature of an “object” (especially, an undefined or ill-defined portion of the atmosphere) is not the same as changing the global mean surface temperature of a planet. The clear, definitive facts are:

1) The consensus greenhouse effect entails a gross violation of the conservation of energy, and

2) The proper Venus/Earth temperatures comparison (done only by me, in late 2010, but which should have been done over 20 years ago by some competent atmospheric or climate scientist, or undergraduate student, and made a part of freshman climate science ever since) shows there is no sign of a greenhouse effect in either of those two real, detailed atmospheres, and the CO2 climate sensitivity is zero.

• Curt says:

HDH:

I’m afraid you have absolutely no concept of what conservation of energy (1st law of thermodynamics means), as your post that you link proves. The Trenberth diagram of power flows has many limitations, but gross violation of conservation of energy is not one of them.

The Trenberth diagram shows averaged steady-state power flows in the earth’s climate system. If you examine any subsystem, or the whole system, these power flows balance, and energy is conserved.

Looking at the earth itself bounded at its surface, we have incoming solar radiation of 168 and incoming “back radiation” of 324 for a total of 492. We have outgoing 24 conductive/convective, 78 evaporative, and 390 radiative, for a total of 492. You could add 30 to each direction for the reflective. But there is no 1st law violation here.

Looking at the earth/atmosphere system at the “space” boundary, you have incoming solar of 342 and outgoing reflected solar of 107 and outgoing infrared of 235 for a total of 342. Again, these balance, no violation here.

You could do the same analysis for the atmosphere itself, with once again no violation.

It is an empirical fact that the earth’s surface emits far more power than it obtains from the sun. Yet the earth is not rapidly cooling. Where does the balancing power come from? (You seem to object to analyses that say that the earth’s surface radiates more than twice what it receives from the sun — I see no factor of two issue in the 1st law.)

I am not claiming that the Trenberth diagram gets the numbers exactly right, or even if it did, that it tells us much about how the system reacts to changes, but it is internally consistent without 1st law violations.

• John Olson says:

Curt: You might want to give Gerlich & Tscheuschner a careful read. If you aren’t persuaded to re-think Trenberth, then you might want to read Kramm (2009) responding to Smith (2008) in the ongoing scientific debate about AGW theory in the physics community (from first principles).

Total fossil fuel consumption in the post-industrial era has been no greater than 500 GTOE (gigatonnes oil equivalent). Half of this was taken up by nature, leaving just 250 GtCO2 added to atmospheric CO2 (3070 GtCO2). Therefore, total CO2 added by mankind over 160 yrs is about 34 ppmv, which can only be responsible for (at most) +0.5W/m2.

With such a small forcing I can’t help but side with those physicists who argue it isn’t possible to detect. Willson’s latest work is interpreted by many to suggest 1360 is a better baseline TSI TOA than 1365 or 1367 W/m2. If so, then this means the 20th century closed out with substantially greater solar forcing than at present, and solar cycle to cycle variation in solar forcing was about 10X greater than forcing from all fossil fuel CO2 summed from 1850 to 2013.

The tiny CO2 forcing from fossil fuel consumption (sum total over entire industrial era) cannot contribute more than about +0.4C (using dF = 5.35LN(C/C0) and dT = (0.8)dF) over 160 yrs. If 1850 was -0.3C, then 2013 should be +0.1C, all else being equal.

6. Fred Harwood says:

Roy, I’ve been using IR handheld meters in domestic heating situations from when they were readily available. They are unreliable, for example, when pointed at a shiny copper pipe, versus a black iron pipe. I’ve had to paint a flat white patch on copper to get anything near the actual temperature reading.
I don’t know why, just find it disturbing that IR ain’t IR in some situations.

7. Roy Spencer says:

Norman, yes it would be the temperature of the cloud surface.

8. Roy Spencer says:

Fred, yes the IR emissivity of some surfaces, especially shiny metallic ones, is well below 1. Paint almost always has a high emissivity, as does oxidized metal.

• Fred Harwood says:

Thanks, Roy. I guess I wonder how many other real-world emissivity differences exist for satellites, and sundry IR sensors.

No doubt, I’m not up to date.

9. Dr No says:

Where is Doug when we need him (for a laugh)?

• steveta_uk says:

Why would Doug contribute when his sensei is already present?

10. Stephen Wilde says:

I note that Roy confirms that the sensor measures the temperature of the cloud surface.

The cloud surface temperature is determined by the pressure related adiabatic lapse rate profile at the height of the cloud.

How does measurement of a temperature at the height of the cloud tell us anything about net radiative flows between the surface of the cloud and the location of the sensor ?

Isn’t the sensor actually confirming that there is a zero net flow between the two locations otherwise it could not accurately record the temperature of the cloud surface ?

A zero net flow confirms that the dominant factor in setting temperature at any given height is not net radiative energy flows at all but rather the pressure related adiabatic lapse rate profile.

If the sensor is recording the cloud surface temperature then all that is happening is that the substance of the cloud allows the sensor to focus at a lower warmer height than it can for a cloud free target area.

Nothing to do with radiative flows either way.

It is simply false to interpret the higher reading at the lower warmer cloud height as indicating a greater downward radiative energy flow from the cloud.

One can extend that principle to the optical depth of a cloud free atmosphere. The more opaque the atmosphere the lower the height at which the sensor will measure temperature and the higher the reading will be.

If the atmosphere were to be completely opaque the sensor would simply measure the temperature at the height of itself.

If the atmosphere were to be perfectly transparent the sensor would simply measure the top of atmosphere temperature.

The sensor is not measuring a downward flow of radiation at all, merely the temperature at the height at which opacity permits it to focus.

11. Stephen Wilde says:

Fred Harwood said:

“They are unreliable, for example, when pointed at a shiny copper pipe, versus a black iron pipe. I’ve had to paint a flat white patch on copper to get anything near the actual temperature reading.”

The reason would be that for a more complex target area the sensor would be less able to focus accurately.

Adding the flat white patch helps to improve focusing and so brings the sensor reading closer to reality.

12. Stephen Wilde says:

“what causes the IR thermometer indicated temperature to increase from (1) clear sky to cloud, and (2) zenith clear sky to low-elevation clear sky?”

The adiabatic lapse rate coupled with the tendency of the sensor to focus at different heights depending on opacity.

The important point after that is to realise that opacity might affect the height at which the sensor will measure temperature but opacity is not the sole determinant of that temperature because the vertical temperature profile is set by mass, gravity and insolation.

In so far as opacity tries to disturb the lapse rate profile set by mass, gravity and insolation then the effect will be countered by circulation changes which always seek to restore the vertical temperature profile set by mass, gravity and insolation.

The inevitable conclusion is that there can be zero net thermal effect from radiative energy flows within an atmosphere and the IR sensor is not measuring changes in net radiative flows at all, merely air temperatures at different heights depending on the height at which the sensor can most easily focus.

13. Mike Flynn says:

Dr Roy,

1. What was the altitude of the clouds you were targeting? What was the air temperature at that altitude in the location of the targeted cloud?

2. How does your IR measuring device respond to a heat source reflected in a mirror? Will it measure the temperature of the mirror surface, or the reflected IR emitted by the heat source?

I assume you have this information readily to hand.

Live well and prosper.

Mike Flynn.

14. Will Janoschka says:

The instrument you are using is call a flux meter
It measures both radiative flux and its direction.
You did describe how it works correctly but not
what it measures “flux” from the flux and its
own temperature it calculates radiance in one of
two forms:

When the object pointed at is warmer, it measures
emittance,that potential, for emitting the measured
radiant flux to this device from the object over
this fixed solid angle.

When the object pointed at is colder, it measures
absorbance. that potential’ for absorbing the measured
radiant flux from this device to the object at a
fixed solid angle.

The device then converts those terms to what is known
as “effective” black body temperature. this would
allow one to approximate the radiant flux to or
from any body at any temperature and known emissivity.
But notice that flux has always a magnitude and
direction. This is useful for getting some idea of”is”.

There is no device that can measure flux in both
directions let alone how much is in each direction
as indicated in the Trenberth view-graph.
Such spontaneous two way flux does not happen.
This would be a violation of 2LTD.
If your Climate Clowns can power source for the
refrigeration process yhey claim with back radiation
it may be interesting to consider, but still
cannot be demonstrated. It is at most a “claim”.

Radiation itself requires a knowledge of directional
emissivity and much of all geometry.

15. Will Janoschka says:

The instrument you are using is call a flux meter
It measures both radiative flux and its direction.
You did describe how it works correctly but not
what it measures “flux” from the flux and its
own temperature it calculates radiance in one of
two forms:

When the object pointed at is warmer, it measures
emittance,that potential, for emitting the measured
radiant flux to this device from the object over
this fixed solid angle.

When the object pointed at is colder, it measures
absorbance. that potential’ for absorbing the measured
radiant flux from this device to the object at a
fixed solid angle.

The device then converts those terms to what is known
as “effective” black body temperature. this would
allow one to approximate the radiant flux to or
from any body at any temperature and known emissivity.
But notice that flux has always a magnitude and
direction. This is useful for getting some idea of”is”.

There is no device that can measure flux in both
directions let alone how much is in each direction
as indicated in the Trenberth view-graph.
Such spontaneous two way flux does not happen.
This would be a violation of 2LTD.

If your Climate Clowns can show a power source for the
refrigeration process they claim with back radiation
it may be interesting to consider, but still
cannot be demonstrated. It is at most a “claim”.

Radiation itself requires a knowledge of directional
emissivity and much of all geometry.

16. KevinK says:

Dr. Spencer,

With respect, I agree that you have remotely measured the temperature of a cloud versus the temperature of a column of cloud free air. No doubt about that.

I also agree that “backradiation” exists, so does “back conduction”, “back EMF” (electric motors) and “back pressure” (steam engines).

Years ago they mounted differential steam pressure gauges in the front of railroad steam locomotives to measure the “back pressure”. This was done to minimize this pressure which reduces the overall efficiency of those machines. The size of the pipes used to transport steam inside the locomotive (and the sharpness of the bends) “pushed back” against the free flow of steam from the boiler to the cylinders, aka: gaseous pressure to mechanical motion conversion subsystem (aka: GPMMCS, sorry, I just had to throw in an acronym).

What you are measuring is no different than effects seen in other technologies. For example, in an anti-reflection multilayer optical interference coating (that purple looking film on the front of most camera lenses) some energy (visible light) is flowing “back” towards the lens (accurately shaped pieces of glass) AFTER being reflected from the glass due to Fresnel reflections (caused by refractive index mismatches at the air-glass interface).

This is very similar to mostly visible solar energy being absorbed and then later emitted as longer wavelength light from the surface of the Earth. Of course the absorption/reemission process takes a little longer than just a reflection. And then the longer wavelength light is absorbed and remitted back towards the surface (the “GHE” you have measured). Same as the first layer of dielectric material (almost totally transparent) in a multilayer optical interference filter “stack”.

It is only the presence of optical interference (constructive and destructive) that causes “more energy to stay” in the lens thus reducing the reflections that make your photos ”duller” (admittedly a subjective term, not enough space here to teach optical design).

There is a huge missing step between measuring this “back radiation” and DEMONSTRATING that it causes a higher “equilibrium” temperature at the surface of the Earth.

For it to do so at least two conditions must be met; 1) it must be shown that it slows the velocity of energy/heat relative to other materials in the system (the Oceans) OR 2) it must be shown that it delays (via many many passes) the progress of the energy/heat long enough so that some is left over after each cycle of incoming solar energy (ie: 24 hours).

Case 1) is what a thermal insulator does, that big bale of fiberglass stuffed in your wall SLOWS the velocity of the heat flowing through it. Since the “back radiation” is travelling at the speed of light it is unlikely to satisfy this case. Case 2) could occur if the distances travelled where long enough to delay the energy/heat by 24 hours.

Since light travels from the surface to the TOA in a few milliseconds (speed of light times 5 miles) and even after accounting for maybe 10-100-1000 bounces back and forth (surface to GHG) this delay is far less than 24 hours.

Now, some will certainly call me a lunatic when I admit the existence of “back conduction”. And perhaps I am, the spouse avers so at times. But consider two solid molecules in close contact, the one on the left is at 100 degrees, and the one on the right is at 99 degrees. Please remember, this is a thought experiment (much easier than real ones). So as argued by proponents of the “GHE”, each molecule transfers its vibrational energy without regard to the temperature of its neighbor. Just like a molecule of “GHG” emits IR in ALL directions. So the 99 degree module transfers some of its energy to its 100 degree neighbor, WITHOUT regard to its neighbor’s temperature. And of course, the 100 degree molecule transfers MORE of its energy to its 99 degree neighbor. Therefore “back conduction” exists, and yet the net flow of energy is STILL from the warmer location to the colder location.

I suggest you might want to study the science behind multi-layer optical interference coatings (it is not something I made up, there are companies that do nothing else but design these coatings).

And I also suggest that a quick study of “optical delay lines” might convince you that the “GHE” only acts to delay the flow of energy/light/heat from the Sun to the Earth to the Atmosphere and eventually to the Universe by perhaps a few tens of milliseconds. The “missing heat” is long “gone”, almost “with the wind” but in fact at a much higher speed.

Cheers, Kevin.

• Mike Flynn says:

Kevin,

You are, of course, correct. I wonder what a climatologist would make of back EMF in an unloaded motor? Do they really understand net effects of energy flows, of energy content of a body, and how insulators work?

Live well and prosper.

Mike Flynn.

• Will Janoschka says:

Back never happens.

All of your “backs” are some potential
for limiting the “forward”
It is the difference in potential that
determine what “happens” or “is”..

Consider one 12 Volt battery, another 6 Volt
battery A 1 Ohm resistor across 6 = 36 Watts
across 12 = 144 Watts. across (12-6) volts =
36 Watts, not 144-36 = 108 Watts! The
deference is What might be rather than “is”

It is always difference in potential
not some stupid absolute potential that
allows what “can be” to be.

With most thermodynamics the potential
is “T” so the difference is delta T.

With radiative transfer it is not T

The potential is Ta^4.
The difference in potential is Ta1^4-Ta2^4
Not ever (Ta1 – Ta2)^4 !!!

That is the ability to transfer radiative
energy, with the (-) showing the direction.

For small delta T it is 4T^3 delta T.
with the “delta” showing direction.

• Mike Flynn says:

Will,

You, of course, are also correct. The EMF generated by the motor is called “back” EMF by convention, and of course the power consumed by the motor is dependent on the difference in potentials, as you have pointed out.

Hence my wondering whether climatologists realise that “back” radiation is no different in character to “front” radiation, “sideways” radiation or “down a little bit, and slightly to the left” radiation. As far as I know, EMR has no intelligence, and does not know what produced it, or where it is heading.

A measuring instrument, likewise, has no knowledge of what it is being pointed at. I fear Dr Roy has made some erroneous assumptions in his fervour to “prove” the “greenhouse effect”.

I think some of us are in agreement, but nomenclature differences and definitions may need to be agreed on first. For starters, let us abandon things like “surface”, “forcings” and “back radiation”, until they are rigorously defined within the “real” physics community, rather than the ” climatological” physics community.

Live well and prosper.

Mike Flynn.

• Curt says:

You are correct that EMR has no intelligence and that it does not know where it is heading. Which is why there has to be back radiation. An object, even a CO2 molecule, has know way of knowing what it is radiating towards, and whether what it is radiating towards is warmer or cooler.

Greenhouse gas molecules up in the atmosphere radiate half net upwards, toward generally cooler sections of the upper atmosphere and space, and half net downwards, toward generally warmer sections of the lower atmosphere and the surface. The half that goes downward is called “back radiation”. That’s all.

Will talks about temperatures, or T^4 quantities in the case of radiative transfers, as potentials in thermal systems, comparable to voltages in an electrical system, which is a correct and standard systems analysis. Let’s see where that takes us.

Lets take Will’s two bodies, the first at absolute temperature Ta1 and the second at Ta2. For simplicity, we’ll say they both emit as “gray bodies” with emissivity “e”. The first one emits with a radiative power proportional to e*Ta1^4. The second emits with a radiative power proportional to e*Ta2^4.

The net power transfer from body 1 to body 2 is K*(e*Ta1^4-e*Ta2^4), where K incorporates the geometric relationships between the two bodies, and of course the net power transfer from body 2 to body 1 is K*(e*Ta2^4-e*Ta1^4). This net power transfer is heat transfer.

Note that, regardless of their temperatures, both bodies are radiating power towards the other. Of course, the body with the higher temperature is radiating more power than the one with the lower temperature, so the net power transfer, manifesting itself as heat transfer, is always from the warmer body toward the cooler body — no 2nd law violation.

But without some radiation from the cooler body to the warmer body, Will’s equations, which are regarded as not even controversial in the engineering heat transfer community, would not work. To assert otherwise would mean that bodies know what they are radiating towards, which you realize is nonsense.

• Mike Flynn says:

Kevin,

All molecules in the atmosphere emit EMR. All.

If you wish to define “back” radiation as that which is directed “downwards”, then I once again ask for yet another definition, as in general, “downwards” seems to be used as a synonym for “eventually contacts the Earth’s surface (another undefined term, naturally).

Only a comparatively small proportion of the emitted radiation emitted “downwards” intercepts the surface. The proportion depends on the angle of emission, and the height above the surface. To assume that half of the radiation emitted by a molecule within the atmosphere will eventually be intercepted by the surface is woolly, imprecise climatological thinking.

Just as ridiculous as saying that only “greenhouse gases” emit IR (or any other wavelength, some would assert). Just plain nonsensical. I know. I breathe a mixture of mainly oxygen and nitrogen. So far my lungs have not frozen, which would assuredly be the case if both gases were non EMR emitting ie at 0K.

So I agree with both you and Will. For me, “back radiation” is not useful. It serves merely to obfuscate, by introducing yet another term like “surface” which sounds nice, but is incapable of useful definition. Try it yourself. You are obviously smart enough to appreciate the futility of the exercise!

So, pardon me again, if I say warming due to CO2 in the atmosphere is preposterous rot. Even the fact deniers (the warmists), cannot deny that even using their woolly definitions, and their own figures, the heat is “missing”! Oh dear, where can it be?

I certainly don’t know, and neither it seems, does anybody else! This is science? Give me a break.

Live well and prosper,

Mike Flynn.

• Mike Flynn says:

Curt,

Sorry. I meant Curt, not Kevin.

Apologies.

Live well and prosper,

Mike Flynn.

• Curt says:

Wow… so many misconceptions in a single post!

“All molecules in the atmosphere emit EMR. All.”

The molecules that comprise 99% of the atmosphere do not emit (or absorb) any significant EMR in the spectrum emitted by the earth’s surface as thermal radiation. Since the absorption of a photon of a given wavelength and its emission are completely reversible processes, the percentage absorption and emission at every wavelength is identical.

The molecules that comprise 99.99% of the atmosphere do not emit (or absorb) any significant EMF in the spectrum emitted by the sun’s surface as thermal radiation, and to which our eyes are sensitive. This is why we call the atmosphere transparent.

So the N2, O2, and Ar in your lungs are transparent to the thermal radiation that your lungs emit. This radiation just passes from one surface to another in your lungs.

H2O and CO2 do absorb and emit EMR in the spectrum emitted by thermal radiation of things in the temperature range of the earth’s surface. This makes them fundamentally different from N2, O2, and Ar.

All of this can be (and has been) determined with controlled laboratory experiments. Consult a spectroscopy textbook. You are not understanding the basics. You would be flunked out of any university physical chemistry or engineering heat transfer course — nothing to do with climate science — if you persisted in your views through the course.

You also completely misunderstand the significance of the back radiation. It does not have to reach the earth for it to have an effect. It can be absorbed in lower levels of the atmosphere, thermalized there, and then re-emitted part up and part down, in a cascaded process. As long as the surface and the lower atmosphere are warmer than the upper atmosphere and space, there will be a net upward flux of this radiation, but any downward flux (and this is measurable) will serve to reduce the net upward flux.

As I said, this downward radiative flux is measurable. It can very precisely measured wavelength by wavelength with sophisticated (and expensive!) instrumentation. The point of Roy’s post is that you can also get a general sense for the effect with a cheap oven IR thermometer.

• Mike Flynn says:

Curt,

Are you a climatologist?

I can not see any other reason for your use of the term “thermal radiation”.

Your comment about completely reversible processes may have been accepted at some point in history.

The following is reasonably up to date –

” Absorption is the process by which radiant energy is absorbed and converted into other forms of energy. An absorption band is a range of wavelengths (or frequencies) in the electromagnetic spectrum within which radiant energy is absorbed by substances such as water (H2O), carbon dioxide (CO2), oxygen (O2), ozone (O3), and nitrous oxide (N2O). ”

This is from a presentation covering satellite remote sensing using different wavelengths for different purposes, and explaining the background to some of the theory involved in the design.

So yes, you are correct that the atmosphere is reasonably transparent to visible light, as any person who is not blind can plainly see.

Your statement relating to “transparent to thermal radiation” etc., is puzzling. I am aware of people measuring the temperature of the “air”, which appears to consist of mostly oxygen and nitrogen. Air conditioners seem to be able to cool this substance, and hair driers seem to be able to heat it. Adiabatic compression and expansion of the atmosphere seem to result in differences in energy content of an atmospheric parcel, or indeed of any gas.

Rapid expansion of high pressure nitrogen gas creates liquid nitrogen, used for freezing skin lesions amongst other things. So no, fact appears to outweigh your “conceptions”.

Your statement about back radiation seems to be in need of some clarification. You say that radiation can be absorbed, “thermalized” there . . . Did I read wrong, or did you say earlier that absorption is a completely reversible process?

It appears that you have created a new definition for “thermalize” – possibly a climatological dictionary contains a secret definition. of which i am unaware.

As at least one other poster has pointed out, your understanding of radiative flux appears to be some hundreds of years out of date.

Thank you for your advice to consult a spectroscopy textbook. I believe I’ll pass on that just now.

Live well and prosper,

Mike Flynn.

• Curt says:

Hi Mike,

I guess it’s time for some clarifications and responses. First of all, I am not a climate scientist. After a start in physical chemistry, I got my degrees in mechanical engineering. I studied thermodynamics (both in the chemistry and ME fields), heat transfer, and fluid dynamics, but my focus in grad school and professionally has been feedback control systems.

When I look at climate science, I am often reminded of the quote that is usually attributed to Lord Kelvin (after whom the absolute temperature scale is named) that “the steam engine did more for the science than the science did for the steam engine”. This is a reminder to me that the scientific disciplines used by climate scientists were developed for mechanical engineering purposes.

The term “thermal radiation” is still widely used in many disciplines to specify the electromagnetic radiation emitted by a substance simply as a function of its temperature. For example, incandescent light bulbs emit thermal radiation. Fluorescent light bulbs and LEDs emit electromagnetic radiation from other causes.

When I said the main components of the atmosphere were “transparent to thermal radiation”, I guess I should have been more specific. More strictly, these gases are transparent to the wavelengths emitted by substances on the earth’s surface just as a function of their temperature (whether tropical, temperate, or polar). I don’t think this is controversial at all, and it can easily be verified in the lab.

When I said that the absorption and emission of photons of any wavelength were reversible processes, I was stating the underlying quantum mechanical principle behind Kirchhoff’s Law of Thermal Radiation, one of whose corrolaries is that the absorption and emission coefficients of any substance at every individual wavelength are identical. I haven’t ever seen even a whisper that this law is considered controversial, let alone hundreds of years out of date.

I have no idea what point you are trying to make when you state that compressing a gas heats it, allowing it to expand cools it, or exposing it to a very hot surface (as the coils in a hair dryer) heats it. None of these effects have anything to do with radiative heat transfer, and would work equally well on gases with different absorption/emission spectra.

As to thermalizing absorbed radiative energy, I don’t think it is at all controversial that when a molecule absorbs a photon of EMR, its kinetic energy can increase due to vibration, rotation, or translation. On the macroscopic scale, when this happens to a lot of molecules, we say that the temperature increases, since the temperature is just a measure of the average kinetic energy of the molecules. This is how radiative energy can heat a substance.

The interesting thing is what happens next to such a gas molecule. One thing that could happen is that it could immediately re-emit a photon of the same wavelength, and its kinetic energy would reduce. But it turns out that it is far more likely that it will collide with another gas molecule first and transfer some of the kinetic energy to that molecule, even if that molecule cannot absorb the wavelengths of EMR. This is how the “non-greenhouse” gases can be heated by thermal radiation from the earth even though they do not absorb the radiation themselves.

I hope this clarifies my arguments.

• Mike Flynn says:

Hi Curt,

Lord Kelvin went to his deathbed believing that the Earth could not possibly be older than forty million years. The great man had done the calculations, and of course, he could not possibly be wrong. I happen to think Kelvin was totally wrong.

With whom do you agree? Kelvin or my good self?

Can I point out a few other nonsensical (in light of present knowledge) ideas that Lord Kelvin espoused. Caloric, the universal aether, and so on? Believe Kelvin if you wish.

I tend not be so gullible.

The scientific disciplines misused by supposed “climate scientists” are about as relevant as those used by Lord Kelvin to come up with a totally wrong answer.

With regard to thermal radiation, what distinguishes it from “non thermal radiation”? Precisely nothing. If you had defined “thermal radiation” as a portion of the EM spectrum defined by, say, frequency, it might have had some relevance. As it is, it is meaningless bit of quite unscientific handwaving.

Incandescent light bulbs emit a wide range of frequencies, depending on the temperature and composition of the filament. Is the visible light it emits part of your “thermal radiation” spectrum as well?

Can I point out that the cause of the light emitted from all your examples is caused by a flow of electrons? Efficiencies and colour temperatures differ, modes of emission differ, but the cause is the same – electricity.

I have no problem with you stating that the atmosphere is more transparent to some EMR wavelengths than others. You might care to define the Earths surface, while you are at it. Do you really mean to say that wavelengths emitted by objects not as a function of their temperature are actually absorbed by the atmosphere? I don’t think that’s the idea you intended to express.

With regard to Kirchoff, you appear to be a little behind the times. It can and has been shown that Kirchoff’s universality does not hold, and this becomes important in a number of engineering problems. But you probably know that. On the positive side, his laws relating to electric circuits have stood the test of time. Even I can’t find anything to disagree with in a practical sense.

My point about the gases is that you mention you accept that all gases can possess temperature. What are you measuring? Molecular motion? How do you do that from a distance through a vacuum? Use magic?

So no, your arguments are essentially more climatological handwaving and unsupported opinion.

But anyway,

Live well and prosper,

Mike Flynn.

• Will Janoschka says:

The instrument Dr.Spencer is using is call a flux meter
It measures both radiative flux and its direction.
You did describe how it works correctly but not
what it measures “flux” from the flux and its
own temperature it calculates radiance in one of
two forms:

When the object pointed at is warmer, it measures
emittance,that potential, for emitting the measured
radiant flux to this device from the object over
this fixed solid angle.

When the object pointed at is colder, it measures
absorbance. that potential’ for absorbing the measured
radiant flux from this device to the object at a
fixed solid angle.

The device then converts those terms to what is known
as “effective” black body temperature. this would
allow one to approximate the radiant flux to or
from any body at any temperature and known emissivity.
But notice that flux has always a magnitude and
direction. This is useful for getting some idea of”is”.

There is no device that can measure flux in both
directions let alone how much is in each direction
as indicated in the Trenberth view-graph.
Such spontaneous two way flux does not happen.
This would be a violation of 2LTD.

If your Climate Clowns can show a power source for the
refrigeration process they claim with back radiation
it may be interesting to consider, but still
cannot be demonstrated. It is at most a “claim”.

Radiation itself requires a knowledge of directional
emissivity and much of all geometry.

• KevinK says:

Will, with respect;

“back” does indeed happen.

Back EMF is an AC current flowing in the opposite direction to the forward AC current flowing to a motor. Of course since the direction of AC current changes 60 (or 50) times per second it is important to consider just one instant in time. At that instant there is current flowing to the motor and “back” EMF current flowing away from the motor. It is real, it can be measured, and it can be used to control the speed of a motor if desired.

It is not the same thing as the resistance in a DC circuit.

Back pressure in a plumbing setup is real and can be measured. Yes it is caused by the “resistance” to flow caused by the size of the pipe and bends, but it is manifested as a pressure in the “back” direction. This results in less pressure at the output of the plumbing system. This is why you use a larger pipe for longer runs to reduce the pressure drop at the destination.

Back radiation inside an AR multilayer optical interference filter heads back towards its source (the lens surface). It is not a “resistance” to forward flow. It is light traveling backwards relative to its initial source. Since you would destroy the optical interference by inserting a detector it cannot be measured. But very accurate models of interference filters predict the overall behavior of the
complete filter “stack”.

An optical delay line intentionally directs radiation in alternating backwards and forward directions to cause the light to traverse a longer optical distance in a smaller physical distance. This results in a delay to the optical radiation. If you input a “pulse” of laser light it will arrive delayed at a detector. Unlike the interference filter an optical delay line does not employ interference.

If you consider sunlight as a sine wave input (at the equator for this example) then the “back” radiation from GHGs is exactly the same as the “back” radiation from the dielectric layers of an optical interference filter.

Except, of course that there are some delays for absorption and re-emission. Since there is no optical interference present no energy “stays here” at the surface.
The GHE merely delays the flow of energy (alternating between visible light, absorbed thermal energy, emitted IR light, etc. etc.) through the system.

The missing heat is gone, travelling away at the speed of
light in a vacuum.

Cheers, Kevin

• Curt says:

Kevin:

Back EMF in a motor is a voltage, not a current. It exists in both AC and DC motors. And as far as the electrical circuit is concerned, it acts much like a resistor. In fact, “equivalent circuit” models of motors that electrical engineers like to use treat the back EMF phenomenon as a resistance.

In an electric circuit, the voltage drop across a resistance multiplied by the current through that resistance gives you the power that is transferred from the electrical to the thermal realm. Similarly, the back EMF voltage, multiplied by the current through the motor, gives you the power transferred from the electrical to the mechanical realm.

As the back EMF (voltage) increases with speed, it does serve to reduce the current that a given voltage supply can push through the motor, and therefore the torque that is generated by the motor. This is what makes motors self-regulating as to speed.

• KevinK says:

Curt, sorry, but back EMF in a motor is manifested as a current. In an AC motor is is out of phase with respect to the supply current.

DC motors are not “self regulating” with respect to speed, they are regulated by the amount of voltage and or current applied. They can be speed controlled in either voltage or current mode, or a combination of both. Or you can control them with more exotic schemes like “pulse width modulation” which is basically flipping the ON/OFF switch really really quickly to get the speed you want.

Most generic AC motors are self regulating with respect to speed (until overloaded) because of the relationship between the stator and rotor poles (specifically the number present in each rotation) and the applied frequency.

This is why just about ever AC motor on the common garden variety table saw runs at the same speed (3450 rpm at 60 Hz). With more complex control schemes (generating frequencies other than 60 Hz) speed control of synchronous AC motors has been more common in the last few decades.

Of course, there are many different flavors of motors and each has specific characteristics.

I have an MSEE degree with decades of experience so your helpful explanation of “equivalent circuit” is nice but unnecessary.

Cheers, Kevin.

• Curt says:

Kevin,

I too have graduate engineering degrees, and I have made my living in precision motor and generator control for 3 decades now. I also teach this stuff at the university level.

And I can say without hesitation that back EMF is voltage. Period. By definition. Of course, this voltage can affect the current through the motor, but it is a voltage. If I had a student that said on a test that it was a current, I would simply mark him wrong. (Actually, you have just given me a good idea for a test question for the next time I teach this!)

Take a motor that is unloaded electrically and spin it mechanically. Measure the voltage across the motor leads. That is the back EMF. There is no current flowing, since there is no electrical load. Back EMF is a voltage, period. (I am in the middle of a micro-hydro generation project now where we have to make sure that the back EMF of the unloaded motor/generator is properly synchronized to the grid before we can connect it. No current is flowing, but there is back EMF at this point.)

When I say a DC motor is self-regulating with respect to speed, I mean this. If you apply a constant DC voltage to a DC motor, it will reach a speed where the back EMF, which is proportional to the speed, and the voltage drop across the armature resistance, which is proportional to the current through it and therefore to the torque produced by it, add up to the supply voltage. When this generated torque equals the load torque, the motor will run at constant speed.

Now if the load torque increases, the motor will decelerate. This instantly reduces the back EMF voltage, causing more current flow through the armature resistance, generating more torque to counteract the load torque, all at a constant supply voltage. This is what I mean by self-regulating.

With the AC induction motors you cite, it is more complicated because the slip frequency is also involved, but the variation of back EMF with speed is a key part of the speed regulation when running “directly off line” from a constant-voltage constant-frequency AC source.

And I am well aware of the more sophisticated control possibilities. As an example, the controller I designed has been used by NASA for the last 20 years on the cranes that lift the Space Shuttle on and off the 747. This is field-oriented control of dual AC induction motors with pulse-width-modulated power transistors and digital current-loop closure. So yes, I am much more than casually acquainted with this field.

• Gordon Robertson says:

KevinK “And of course, the 100 degree molecule transfers MORE of its energy to its 99 degree neighbor”.

You are over-simplifying. You are loosely associating the transferred infrared energy with heat.

Think about it. When a body is heated, it’s atoms/molecules go to a higher energy level within the atom/molecule. What does it take to increase that energy level so that the atoms/molecules are at even higher energy level, hence warmer?

Ask yourself if IR from cooler GHGs have the requisite energy to raise the temperature in a warmer body. I don’t think they do, otherwise cooler bodies would be able to raise the temperature of warmer bodies through transmitting lower intensity IR to them.

There’s a reason why heat can only be transferred from warmer bodies to cooler bodies under normal means and I think it has something to do with the ability of atoms/molecules at a higher energy level to absorb lower intensity energy.

• Norman says:

Gordon Robertson

That is what the group at PSI believe. There argument is that at a given temperature all the possible modes of vibration for that temperature are taking place so an IR photon at the same temperature cannot change the temperature. It can only be absorbed when the body releases its own IR photon and drops in vibration, but it will not heat up, just stay the same.

It is no differnt than an atom with visible light. In an excited state an atom will not absorb a photon of the same energy it is already in. It would have to release a photon, drop to a lower energy level to then be able to absorb the incoming photon.

Not sure it works that way but I believe that is the theory they are working on. The IR coming back from the atmosphere is at a lower energy than the Earth’s surface so it could not be absrobed by any of the molecules within the surface until they released a photon and dropped to a lower state that allowed an absorption. That would explain why a colder body cannot warm a warmer one even though both are emitting radiation continuously.

• Gordon Robertson says:

Norman…the argument obviously has merit because as far back as 1850, Clausius stated that heat can only be transferred from a warmer body to a colder body. It makes sense that some mechanism is in place to prevent lower intensity IR from a cooler body from affecting the warmer body.

When you look at a refrigerator, they are using external power and a special refrigerant, along with a compressor, to extract heat from a colder body, then they radiate it to space via a radiator on the back of the fridge.

Heat can not normally be transferred from a cooler body to a warmer body and there is no mechanism in the atmosphere that would allow cooler gases to radiate the intensity of radiation required to raise surface temperatures.

I think that describes the greenhouse effect as well and that the oceans play a major role in warming the atmosphere.

• steveta_uk says:

“It makes sense that some mechanism is in place to prevent lower intensity IR from a cooler body from affecting the warmer body.”

It’s funny how something that “makes sense” can still be quite wrong.

You are all thinking of a cooler “body” and a warmer “body” exchanging “heat”. But that doesn’t happen at the molecular level.

A warm land surface does not consist are large number of individual molecules all of which are at 15C. The temperature of a molecule is almost a meaningless concept.

The temperature applies to the bulk material. In the case of a solid, most of the heat energy is contained within intermolecular bonds and as these are almost infinitely variable, a true black body spectrum is possible. But the specific energy value of a single IR photon emitted from the surface depends of the energy available as the precise location from which it was emitted. So just as it can emit a wide range of energies, it can absorb in exactly the same wide range, plus of course absorb much higher energies too.

From a cool gas, the above doesn’t apply. The intermolecular bonds of a solid don’t exist, and so the source of black-body radioatn doesn’t ewist. But a single CO2 molecule can absorb a photon from elsewhere, or gain energy from a colision, and emit this energy as an IR photon, in any direction (or possibly all, think of two-slit experiments) and if that IR photon hits the surface it can be absorbed and increase the energy levels at the point where it was absorbed. Temperature doesn’t come into it.

Of course, since far more energy is emitted from a warm surface than the cool air, the surface will in general transfer heat to the atmosphere, but this is just an averaging process – individual IR photon exchanges occur in all directions and this cannot violate any possible “laws” as it happens all the time.

• Nabil Swedan says:

Kevin, Have you ever experienced back-electrical current, back-water flow, or back-pneumatic flow? Unless the back pressure is greater than supply pressure, you will never experience these back-fluxes. The same is true for back-radiation flux. This can never happen based on the physics available. Only in fiction can it happen, and that is why the climate science and climate issue are in trouble.

17. jimbrock says:

wrt the cloud. It is at a lower altitude than the clear sky. Maybe the cloud IS warmer than space.

18. Alex says:

Beautiful discussion. Disagreement is what makes science.

“The growth of knowledge depends entirely on disagreement” — Karl Popper

Warmists would immediately block any comments that would put their catastrophism under discussion, killing the growth of knowledge, murdering science.

• Norman says:

Alex,

Very well said. I have been banned from both Skeptical Science and Open Mind and if you post anything on Real Climate not towing the party line you are boreholed. I guess they are so convinced of the corrctness of their beliefs that they view any contrary views as evil and people that are greed obsessed and want to destroy the Earth. They view themselves as the heroes fighting the forces of darkness.

19. Dave Jackson says:

Interesting, I have only ever played with one of these things and a thermal imaging camera to look for heat leaking out of my house. Intuitively, if you have two torches pointing at each other light goes in both directions. Why should infra red radiation be any different?

• Nabil Swedan says:

What makes you so sure that light goes both ways?

• KR says:

Perhaps because you can literally _measure it_?

• Nabil Swedan says:

Have you or anyone else measured it?

• KR says:

Yes.

* Google Scholar search on “radiation thermometry”: 38 thousand results in books, papers, citations, and patents.

This is _basic physics_ – and to be quite blunt, arguing against hundreds of years of physics measurements is not a strong (or in my opinion sensible) position to take.

• Curt says:

“What makes you so sure that light goes both ways?”

Because you can see it?

Seriously, take the example where you and I are in a dark room. We shine flashlights (torches) at each other. You have fresh batteries with high voltage in yours. I have old batteries with reduced voltage in mine.

Because of your higher voltage, more current flows through the resistance of your bulb’s filament, and that filament gets to a higher temperature than mine does. The EM radiation fields from our two flashlights pass through each other. Do you really expect that you would not be able to see any of the EMR from my flashlight?

• Nabil Swedan says:

It is not the same. Back-radiations imply radiation of cold, and cold does not radiate; it violates the law of thermodynamics. Radiation flows only from hot to cold, and there can be no back-radiations.

Back-radiations only exists in the mathematical formula for radiation-intensity calculations, similar to ohms-law. There is no physical back-electrical current, it only exists in the mathematical formula. The same is true in hydraulics, pneumatics, diffusion, etc.

• Curt says:

Now you are just being ridiculous. “Back-radiations imply radiation of cold” is a meaningless statement in a technical discussion. What is “cold”? What are its units? How do you measure it?

Substances emit thermal radiation as a function of their absolute temperature (which is always positive) and their emissivities within the range of wavelengths at which it is possible to emit at that temperature. The emissivity at a given wavelenghth can range from 0.0 (no emissions at that wavelength) to 1.0 (equivalent to a blackbody at that wavelength).

Substances have no way of knowing what they are radiating toward. A block of ice at 273K (0C) will radiate toward a block of dry ice at 193K (-80C), but it will also radiate equally toward objects at a room temperature of 296K (+23C). The difference is that it will radiate more toward the dry ice than the dry ice will radiate toward it, but the room temperature object will radiate more toward the ice than the ice radiates toward the object.

Even in the case where the ice is radiating toward the warmer object, its radiation is transferring energy to the warmer object, but less energy than the warmer object is radiating toward it.

The motor “back EMF” that KevinK brought up is a voltage, not a current. In systems analysis, it is a potential, like temperature (or temperature difference). And it is very real; you can measure it with a simple voltmeter. The same is true with pressures in hydraulic and pneumatic systems; these are potentials, and they are very real physical quantities.

• Nabil Swedan says:

Curt,

Ice block will radiate heat to a colder body at 193 K, but it cannot radiate “cold” to a warmer body at 296 K.

I have never heard or experienced back-electrical, pneumatic, hydraulic, or diffusive currents or fluxes. And certainly there can be no back radiation flux.

• Mike Flynn says:

Curt,

With a little effort on your part, you should be able to understand EMR in a useful fashion.

Couple of minor points, however. When you accuse somebody of “being ridiculous” rather than, say, “believing in the ridiculous”, you run the risk of being perceived as a person who indulges in personal attack, in lieu of presenting a cogent argument.

To then demand a definition of “cold” etc., would lead to your accusee demanding in like fashion, that you define “heat” and so on. Not a good attack, in my view.

However the other point is more serious. You state, apparently as a matter of fact, that
“Even in the case where the ice is radiating toward the warmer object, its radiation is transferring energy to the warmer object, . . . “.

Once again, this is simply not true, at least not in any useful sense. Your statement would imply that the warmer object’s energy content will increase. “Transferring” energy and all that. Now that the warmer object’s energy level has increased, will its temperature –

a) Rise
b) Fall
c) Remain the same.

And now you should understand why it is not possible to extract energy from the water surrounding a ship and use it to power the ship’s engines, leaving chunks of ice in its wake.

But maybe I am wrong. Maybe it is possible to raise the temperature of, say, water, merely by enveloping it with CO2. I think not.

Live well and prosper,

Mike Flynn.

• Curt says:

Hello Mike:

“Heat” has a specific thermodynamic definition — I have taken multiple classes in “heat transfer”; “cold” does not. I did get exasperated, I guess, but it’s virtually impossible to have a technical debate when meaningless terminology is used.

As to your bigger point, I agree (and have stated multiple times in this thread) that the warmer object will always radiate more to the cooler object than the cooler object will to the warmer object, so the NET heat transfer is always from warmer to cooler. (I regard “net heat transfer” as a redundancy, but never mind…) This is what the 2nd law requires, of course.

But I have several problems with the idea that cooler objects radiate nothing toward warmer objects. If this were so, would the magnitude of the heat transfer from warmer to cooler be dependent only on the sign of the difference in temperature, and not the magnitude? (We know experimentally that this is not the case.)

How does the cooler object know not to radiate in that direction, especially when it may be possible to radiate in other directions? What particles, waves, fields, decision-making algorithms are involved? (I went on at much more length elsewhere in the thread.)

In the case of the greenhouse effect itself, the argument is not that the (measurable) back radiation from the atmosphere transfers more energy from the atmosphere to the earth than earth to the atmosphere, but that it reduces the losses of heat the surface has received from the sun.

• KR says:

Nabil Swedan – Everything above absolute zero radiates EM (thermal radiation) as a function of temperature and emissivity. Everything. Even cold objects, even the background temperature of the universe (~3K in microwave energy).

The energy coming into an object (any object, in this case the ground) is the result of the absorptivity of that object and the EM impinging upon it. If a relatively colder object (say the atmosphere) is radiating EM towards that object, and it is radiating more than zero, or more than the background radiation of the universe over that angle, there is more incoming energy to the object than there would be without that object, warm or cold.

Energy leaving an object (the ground) is a function of emissivity and temperature. By the first law of thermodynamics, conservation of energy, incoming and outgoing energy are equal at equilibrium. And if a nearby object is colder than ground, yet warmer than absolute zero, that object will cause the ground to be warmer than it would be without it. Because incoming EM will be larger than it would in the absence of that (relatively cold, but warmer than outer space) object.

Not radiation of cold (what an odd term) but greater and lesser incoming EM radiation – dependent on nearby objects, their temperature and emissivity, and the solid angle they subtend from the object of interest.

This is conservation of energy, the first law of thermodynamics. I’m beginning to despair of you understanding or acknowledging the last few hundred years of physics (you have not so far) – but perhaps this conversation will be helpful to others…

• Mike Flynn says:

Curt,

You wrote : –

“In the case of the greenhouse effect itself, the argument is not that the (measurable) back radiation from the atmosphere transfers more energy from the atmosphere to the earth than earth to the atmosphere, but that it reduces the losses of heat the surface has received from the sun.”

I assume you meant what you wrote.

We agree. You rightly point out that the atmosphere does not cause the surface temperature to increase, but rather that it slows heat loss.

I point out that heat loss is accompanied by temperature drop, not increase. The greenhouse effect as you state, is merely a climatologically roundabout way of saying the atmosphere has a measurable insulating effect with respect to the solid Earth.

No “warming”. No increase in temperature. It works both ways, as you have previously stated in somewhat tortuous fashion.

The insulator reduces the amount of energy reaching the Earth, and also slows the rate at which it leaves the Earth.

Excellent. Sanity and rationality prevail at last. Thank you for the discussion.

Live well and prosper,

Mike Flynn.

• Gordon Robertson says:

Curt…you are talking about independent sources of light and that is the same mistake made by many climate scientists who should know better.

A better application of your flashlight would be shining it into a mirror that is designed to reflect only a fraction of the light it receives. You would still see some reflected light, and it would pass through your incident light, but would it make your flashlight glow brighter?

That’s the AGW argument, that back-radiated IR can be added to solar energy, as if the GHG-radiated energy is an independent radiator. It is not, it is dependent on surface IR, which is dependent on solar energy, and IR from the incoming solar energy. Physicist, Stephan Rahmstorf, advocates that theory. He claims the 2nd law is not violated because their is a positive balance of energy.

Gerlich and Tscheushner, two scientists skilled in thermodynamics, responded that the energy to which Rahmstorf is referring is not heat, but IR, and the 2nd law applies only to heat. They replied that you cannot summarily add up-dwelling and down-dwelling heat, as you can with IR.

Many climate scientists have failed to realize the difference between IR and heat and they have confused the 1st and 2nd laws of thermodynamics.

20. Stephen Wilde says:

To expand a little on my earlier comment I should have said that there is no CHANGE in the net radiative flow when the sensor is pointed at the cloud.

There is of course still a net upward radiative flow as evidenced by the adiabatic, pressure induced fall in temperature with height.

The fact is that all the air at the height of the cloud is at the same temperature as the cloud due to the vertical temperature profile.

All that happens when the cloud is targeted is that the temperature of the air AND the cloud at that height is measured and that is at a lower, warmer height than that which is measured when the cloud is not present.

THat is why one gets a change in the temperature reading. Nothing to do with a change in the net radiative flux at all.

The data from these sensors has been negligently interpreted to say something about the net flow of radiation between a cloud and the surface.

It measures nothing of the sort.

It simply measures the temperature of a specific height in the atmosphere and that temperature is set by mass, gravity and insolation.

Any radiative characteritics of constituent molecules have no effect whatever on that vertical temperature profile and the use of IR sensors to suggest that it does has been a perversion of science.

21. Massimo PORZIO says:

Nice essay Dr.Spencer,
You wrote:
“A similar effect can be achieved from just the clear sky, by pointing the IR thermometer up at different elevation angles. Increasing temperatures will be indicated as the elevation angle is lowered toward the horizon. Today, I measured about 15 deg. F pointing straight up to about 35 deg. F when pointed about 20 deg. above the horizon.”

Well, if you did it in a full clear sky assuming the same average pressure in both the field of views, this agrees with my opinion that flow measurements have nothing to do with the total energy which pass through a point (no matter if it is at the ground level or at the TOA).
I mean, the total down-welling radiation received by a single point is the integral of the full upward hemispheric field of view of an isotropic emitter/receiver placed in the measuring point, not that inferred by measuring only a narrow field of view in an arbitrary direction such as the zenith.
In case of ground measurements most of the energy come from angles different from the zenith indeed, because the open space is a very low source of LWIR and the optical path is shorter than the one of any other angle.
In case of TOA measurements instead, since the Earth surface is a good LWIR emitter, the current spectrographic measurements, taken from satellites, show the maximum outgoing energy at the nadir, but most of that energy is directly coming from that high emitting ground surface. That energy viewed from the ground is inversely proportional to the GHGs concentration and it is maximum at the TOA nadir view of course, while it is progressively switching to directly proportional to the GHGs concentration as the angle changes from the nadir FOV to the tangential FOVs.
In my opinion this implies that the current evaluation of the GHGs radiative forcing could be overestimated.
Have a nice day.

PS
He had a heart surgery, I hope he is recovering well.

22. Stephen Wilde says:

So the greenhouse effect is caused by total atmospheric mass not radiative characteristics of constituent molecules and the data from IR sensors does not indicate any variability in net radiative energy flows caused by the radiative characteristics of GHGs.

They simply measure the temperature at a given height, that height being determined by the optical depth of the atmosphere in front of the sensor and the temperature at that height being determined only by atmospheric mass, the strength of the gravitational field and the amount of energy being supplied to the atmosphere from outside (which could include geothermal as well as solar).

The current evaluation of GHG radiative forcing as derived from IR sensors is simply an artifact of wrongly interpreted data.

23. Rory says:

I observe, rather diffidently, and along the same lines as one of the comments, that your deep freeze and your coffee maker registered a temperature, probably independent of reflection of IR. Why assume that clouds are not honourable ir emitters, just like coffee pots.

Clearly low clouds would be warmer than a higher aggregate of clear air and 40F does not seem unreasonable for a cloud temperature, given a reasonable lapse rate.

• Nabil Swedan says:

Roy,
Clouds are not the same as carbon dioxide, they are droplets of water that obscure the line of sight between thermometer and outer space. They simply reduce radiations from the thermometer to the outer space and the thermometer records higher temperature. There are no back-radiations, its against the law of physics: Cold cannot be radiated, and radiations flows from hot to warm bodies period. The back-radiations recorded is from your thermometer itself. Cool it down close to zero, as they do in infrared astronomy, and you will get zero back-radiations.

• Will Janoschka says:

With IR instruments with detectors colder than the clouds
you do measure radiation “from” the clouds. but this “now”
radiation that does not violate 2LTD

• Nabil Swedan says:

No disagreement. But you will detect no back-radiations from a clear sky, which also means that CO2 forcing or back-radiations do not exist.

24. Gordon Robertson says:

Roy…IR is not heat. Heat is a measure of the excitedness of atoms and molecules, which is a measure of their kinetic energy. IR does not become heat until it contacts atoms or molecules and the energy is absorbed.

What you are measuring is electromagnetic energy, which is no different than light. Just as light contains no colour, IR contains no heat. Colour is a result of the eye responding to different frequencies.

I have no issue with the concept of back-radiation, but how much radiation can anthropogenic CO2 molecules radiate when they account for about 1/1000nds of 1% of atmospheric gases based on a density of about 390 ppmv? Trenberth-Kiehle are making the absurd claim that as much back-radiation is returned by GHGs as what is radiated through the entire flux of the surface.

Heat can only be transferred from a warmer body to a cooler body through ordinary means. To reverse that process, external energy is required. GHGs in the atmosphere are cooler than the surface, case closed. I know you have argued based on thought experiments that heat can flow from cooler objects to warmer objects. Sorry, you may be a brilliant meteorologist and an expert on AMSU units, but your understanding of thermodynamics is faulty.

I’;s a no-brainer. Since heat is a measure of the kinetic energy of a gas, which is the degree to which the gas molecules are excited, then transferring that energy to a cooler body requires emitting the warmer energy as infrared and having it absorbed by the cooler molecules. So far so good.

When the cooler gases re-emit whatever portion they intercepted, that energy is now at a lower intensity and in a reduced frequency range. How is that energy going to raise the temperature of the surface, as the AGW theory requires?

Please don’t talk about blankets. That is a naive theory that Craig Bohren has called plain silly. Heat is not trapped by GHGs and the 1% of GHGs in the atmosphere would be the equivalent of having 99 panes of glass removed from a greenhouse with 100 panes.

It’s not good enough to point a meter ta the sky and claim that as proof of a greenhouse effect. Where are the numbers? I think you pointed out before, with reference to a satellite, that no one has those numbers. Even Trenberth-Kiehle admitted their energy budget is based on a fictitious scenario.

So is the Greenhouse Effect Theory, which is based on a crude calculation of the temperature of a planet without an atmosphere and oceans and one with an atmosphere and oceans. It is far too complex to calculate either scenario given the number of parameters required.

• Roy Spencer says:

Gordon, no, the IR thermometer does NOT measure electromagnetic energy. It measures changes in the temperature of a thermistor on one side of a thermopile…that thermistor is responding to changes in radiative energy transfer.

This is exactly what the Earth’s surface does in response to changes in IR radiation…there are changes in temperature caused by changes in IR fluxes in and out of the surface.

That’s why the IR thermometer is a direct analog for the Earth’s surface in demonstrating how various levels of downwelling IR radiation can change surface temperature, which is what the greenhouse effect is.

• Leonard Weinstein says:

Roy,
I agree with most of you have said. However, to be technically correct, the IR thermometer is measuring the effect of received and absorbed electromagnetic energy by measuring how it changes the thermistor temperature. The IR input is electromagnetic energy. The response is a calibrated variation against known sources.

• Gordon Robertson says:

Roy…my point is that the heat source is remote and the meter is measuring a proxy for the heat. The meter requires IR to activate the thermistor and a thermopile is nothing more than a cascaded series of heat sensitive devices.

The only reason the meter can work is that someone has calibrated the incoming IR to what it SHOULD be from a legitimate heat source. Obviously, IR from the atmosphere, especially from an open sky, is a mix of various frequencies in the IR band of the EM spectrum. Unless there is a filter on the meter to reject certain frequencies, the meter is giving an average response to the entire IR spectrum.

I know you are talking about the greenhouse effect but AGW theory is an extended greenhouse theory. One basis of the theory is that GHGs act like a blanket to trap heat. The other, which I am using here, is that back-radiation from GHGs can raise the temperature of the surface, which is at a higher temperature than the GHGs.

Meteorologist, Craig Bohren, has dismissed the blanket theory as plain silly. I agree in the sense that ACO2 represents about 1/1000nds of 1% of atmospheric gases, based on a density of about 390 ppmv. ACO2 could only be said to trap a tiny amount of IR because we have no idea what becomes of the heat it acquires when it absorbs IR from the surface.

Bohren was kinder to the other theory, that back-radiation can be exchanged between bodies of different temperatures. He is a skeptic, however. My beef is not with you, who I support enthusiastically on your skeptical stance, it is with people like Trenberth-Kiehl, who have theorized a back-radiation in their energy budget diagram equivalent to all surface IR flux.

Bohren also addresses the cloud vs. open sky issue in his book on atmospheric radiation. He found a temperature close to -50 C in the open sky but only about -3 C when pointed at clouds. He explained clouds as the equivalent of a water surface due to the size of the water droplets. I take that to mean that clouds are equivalent to small lakes floating in the sky which have been warmed by solar radiation, and which likely take a long time to cool after sunset.

• Will Janoschka says:

Dr. Spencer Your instrument measured the difference.
in temperature on the two sides of the thermopile…
It is a measurement of thermal flux across a known
thermal conductivity It will show both magnitude and
direction of that flux. it indicates if that instrument
is emitting or absorbing that is all and always only
one way. Why do you not try to find out witch way the
Tour GHE claims but cannot demonstrate any back radiation.

• Nabil Swedan says:

Roy,
With this alleged back radiations or downwelling IR radiations, you inherently imply that cold is radiated back to the surface. First, cold cannot be radiated; second, how could cold warm the surface?

25. Stephen Wilde says:

Roy said:

“that thermistor is responding to changes in radiative energy transfer. ”

It is responding to the temperature at a targeted location which is dependent on optical depth.

The more opaque the air in front of it the lower the height at which it will record that temperature and the higher that temperature will be.

That temperature is set by mass, gravity and insolation with the radiant exchange being a consequence and not a cause of the vertical temperature profile.

I have answered your question clearly. The difference between the cloud affected measurement and the open sky measurement is due to the different optical depths causing the temperature to be measured at different heights.

That explanation removes any need to involve changes in radiative flux caused by clouds. Quite simply the clouds do not affect the net upward radiative flux and nor do GHGs.

Or, if you prefer, the IR sensor provides no evidence either way.

The greenhouse effect (or atmospheric thermal enhancement) is a result of atmospheric mass in its entirety held within a grvitational field and subjected to an energy source, not the radiative characteristics of constituent molecules.

• Leonard Weinstein says:

Stephen,
You are not correct. The clouds consist of water drops which emit nearly as black body radiators. The clear sky has select frequencies of radiation, which radiate less than an equivalent black body. Thus even at equivalent altitudes, the clouds would give a larger indicated temperature. The indicated temperature corresponds to received energy flux at the detector, and the presence of this back radiation always reduces the net radiation directly to space. The evaporation, convection, and condensation at some altitude carry some of the surface energy to higher altitudes, where it radiates to space. The less the direct NET radiation from the surface to space is, the larger the greenhouse effect is and the higher the average altitude of final radiation to space occurs. Combining this with the naturally occurring lapse rate then determines the temperature increase.

• Stephen Wilde says:

“Thus even at equivalent altitudes, the clouds would give a larger indicated temperature.”

At the same altitude the temperature of both air and cloud is the same due to the lapse rate.

Why then would the clouds give a larger indicated temperature ?

We are not comparing clear sky with cloudy sky. That does give a different reading because the heights at which the sensor focuses are different.

Instead we are comparing air at the same temperature and height as clouds and if both are the same temperature and if the sensor were to focus at the same height then the reading would be the same.

• Leonard Weinstein says:

Stephen,
The amount of radiation down from a given altitude is different due to the clouds acting close to a black body over the full spectral range, but gases only act like a black body in limited spectral ranges (the molecular radiation modes possible at the local temperatures). The atmospheric “window” of surface radiation directly to space is a result of the wavelengths not absorbed or radiated. Thus the integrated total energy radiated is less for the atmosphere, compared to clouds, even at the same temperature. Remember, the IR meter is not measuring temperature, but energy, and if the object measured is not a full black body, it gives a different reading.

• Stephen Wilde says:

“Remember, the IR meter is not measuring temperature”

Roy clearly stated that it measures the temperaure of the cloud surface.

Roy Spencer says:

April 10, 2013 at 3:28 PM

Norman, yes it would be the temperature of the cloud surface

26. Stephen Wilde says:

Roy said:

“there are changes in temperature caused by changes in IR fluxes in and out of the surface. ”

There are changes in temperature caused by fluctuations in the non radiative processes going on within the circulation of the atmosphere.

The IR flux at any given moment is a consequence of those non radiative processes superimposed on the basic upward radiative flux set by mass, gravity and insolation.

The non radiative processes change as necessary to offset any change in radiative flux that might otherwise have occurred from the radiative characteristics of constituent molecules.

The IR sensor is only recording the net outcome of both radiative and non radiative processes at a given height and that height depends on the optical depth between the sensor and the target. It does not record changes in the radiative flux between two given points.

• Roy Spencer says:

Stephen, I think you might be missing the main point. The experiment is not meant to explain why the atmospheric state is what it is. It’s to demonstrate that the temperature of a surface (the sky-viewing side of the thermopile in the IR thermometer) changes with changes in downwelling IR (clear sky vs. a cloud).

• Stephen Wilde says:

So are you measuring the temperature of the cloud or the temperature of the thermopile ?

The thermopile being lower in height than the cloud should be warmer than the cloud by virtue of the lapse rate.That temperature being set by the gravitationally induced lapse rate and not radiative exchange.

If it is warmer than the cloud then it is measuring its own gravitationally induced temperature and not that of the cloud.

If it is not warmer than the cloud then there is no warming effect from the cloud.

Of course the rules are different if the sensor points downward so that no part of the sky is in the field of view hence its greater usefulness with your freezer and coffee pot. No height or opacity complications intervene in that latter scenario.

• steveta_uk says:

Utter twaddle. Exactly how does this “gravitationally induced lapse rate” know whether the sensor is being pointed at a cloud or at the clear air adjacent to the cloud?

• AlecM says:

An IR sensor of this type measures the atmospheric window to avoid its signal varying with Relative Humidity.

Point it at the freezer and it will measure the AW part of the near black body IR emission in the 8-14 micron band, and it will be moderately accurate.

Point it to the sky and it will measure a temperature but the IR emission is mainly from trace gases offset by radiative energy interacting with the cosmic microwave background of space, so could be significantly lower than the average air temperature in its emission depth.

When pointed at the cloud, which provides a radiation field much closer to the real air temperature, the net difference in irradiance of the sensor minus that of the air will be closer to air temperature.

There is no ‘back radiation’. Lapse rate will control the temperature of the cloud. Its area in the view will establish by how much the signal rises.

• Curt says:

“An IR sensor of this type measures the atmospheric window to avoid its signal varying with Relative Humidity.”

Then how is this type of sensor a very effective humidity measurement?

http://journals.ametsoc.org/doi/pdf/10.1175/2011BAMS3215.1

• steveta_uk says:

AlecM, by comments was about the graviationally induced lapse rate and how the meter is apparently measuring only it’s own temperature, as stated by Stephen Wilde.

Your response seems to be utterly unrelated to this point.

27. Svend Ferdinandsen says:

Is it just a semantic discussion of the word “back”.
I believe it is common knowledge that all objects with a temperaturer above 0K radiates, and the radiation is independent of the surroundings (except eventual reflexion) and depends only of the temperature and the objects ability to radiate.

The use of an IR thermometer is a simpel and explaining method. I have used it myself, but at least in Denmark the sky is often below -20C, meaning no measurement, because most instruments stop below that temperature.
When pointing at clouds, you can estimate the hight of the clouds from the lapse rate and the ground temperature. It is anyway better that pure guessing.

• No, it is not a question about semantics but real physics. There is no “back radiation” in the same sense that there is no “back conduction” in heat transfer by conduction. Why? Because “back conduction” violates the second law and mathematically corresponds to solving the heat equation backwards which is an unstable process which has no physical realization. This is like unmixing the milk mixed into your coffee cup by a process of “back mixing”. Try it if you believe in “back mixing” and “back radiation”.

• Svend Ferdinandsen says:

You really stresses my understanding of electrical thermal noise and thermal radiation, which are related.
Every resistor produces thermal noise of itself independant of the connection to other resistors and there temperature, but of cause any net flow depends on the temperature differences.
In the same sense any object radiates energy by itself, independent of the surrounding.
Your way of thinking makes it very hard to calculate the energy tranfer beteen objects with very selective radiating surfaces like the Earth and the atmosphere.
It reminds me of traveling waves in a cable or waveguide, and it is possible to measure like you could do with a mirror in between the two bodies like a directional coupler.

• Curt says:

Actually Claes, there is “back conduction”. Put two bodies of slightly different temperatures in contact with each other. A commenter above used the example of one at 100C (373K) and one at 99C (372K). The molecules in each body have a statistical distribution of kinetic energies — the temperature is (by definition) representative of the average kinetic energy.

With these two bodies close in temperature, the two distributions are very largely overlapping. This means that in almost half of the collisions that form the basis of heat conduction, the cooler body will transfer energy to the warmer body. But more than half will transfer energy from the warmer to the cooler body, meaning that there will be a net energy transfer, manifesting itself as heat transfer, from the warmer to the cooler body.

If it were not the case that the cooler body could transfer some energy to the warmer body, the net conductive heat transfer from the warmer body to the cooler body would not be proportional to the difference in temperatures. And this proportionality of conductive heat transfer is widely known, and not the least bit controversial.

And with radiative transfer, we can actually measure the emissions of the cooler body toward the warmer body.

• Will Janoschka says:

“And with radiative transfer, we can actually measure the
emissions of the cooler body toward the warmer body.”
Toward something cooler only.there is no emission toward
the warmer, at the same temperature thermal radiative
heat transfer is zero and all emissions cease.

• Nabil Swedan says:

Good luck with the imaginative radiative transfer! There would have been already a practical and useful application for it in the market had it existed. Have you ever seen one?

• Nabil Swedan says:

“And with radiative transfer, we can actually measure the
emissions of the cooler body toward the warmer body.”

Is there a practical application?

• Curt says:

So tell me, Will: How does a body know what it is radiating toward and what that temperature is? What sensors does it use? What computational algorithms does it use to compare that temperature to its own temperature. How can it accomplish this at speeds faster than the speed of light to stop the photons it would otherwise emit?

You are asserting things that are completely contradicted by direct observations and measurements. Take the original subject of this post. By your logic, an IR thermometer should not be able to measure anything colder than itself because it will not receive any radiation from it. But you can take one of these thermometers at room temperature, point it at something in an open refrigerator, then at something in an open freezer. It will be able to tell the difference, and pretty accurately the temperatures of each. A more sophisticated sensor could tell you more accurately, but these cheap sensors are amazingly good. And they totally contradict your assertions.

• Nabil Swedan says:

Dear Curt,
Just like pneumatic flow requires difference in pressure, radiation flow requires difference in temperature. From higher values to lower values. If a body has equal or less temperature than the surroundings, it simply does not radiate anything.

IR thermometer is a non contact thermometer and it measures its own temperature at all times. This temperature is the net energy balance of conduction, convection, and radiation. The temperature is then calibrated to measure any of the following: radiation, temperature of the target, voltage, temperature of the thermocouple, amps, etc. All depends on the calibration curve and its intended translation.

IR thermometers do not measure back-radiations or greenhouse gas effect, they measure their own temperatures.

• Curt says:

“Is there a practical application?”

For measurements, of course! It is very nice to be able to determine the temperature of an object colder than the sensor. It is done all the time.

For capturing this energy (which is what I think you really mean), it can’t be done, because the warm device will always be radiating more away than it is receiving. That is the 2nd law limitation. (See “Maxwell’s Demon”.)

But radiative barriers for insulation to reduce heat loss from warm bodies (when the barrier is cooler than the warm body) are used all the time as well.

• Curt says:

“Just like pneumatic flow requires difference in pressure, radiation flow requires difference in temperature.”

Right here at the start you are completely wrong. Bodies radiate as a function of their temperature and emissivity alone. This is basic high school physics stuff.

Now if you are talking about the NET transfer of energy that occurs between two bodies as a function of what each of them radiates, you are correct. It does require a difference in temperature between the two bodies, and the bigger the difference, the greater the NET transfer of energy. I think we can agree that this is the case, as has been experimentally verified countless times.

But let’s compare my version of how this occurs to yours. In my version, each object emits electromagnetic radiation as a function of its own temperature and emissivity alone. Some of that radiation hits the other body. The radiation from Body A that hits Body B passes through the identical path “window” as the radiation from Body B that hits Body A.

If Body A has a higher temperature than Body B, it will radiate more power through this path “window” to Body B than Body B will radiate through this same path “window” to Body A. If the temperature difference is small, the difference will be small and there will be a small net heat transfer from A to B. If the temperature difference is larger, the difference in radiative transfer in the two directions will be larger and the net heat transfer will be larger.

If we now increase the temperature of Body B (while keeping the temperature of Body A constant) so that the temperature of Body B is now greater than that of Body A, nothing changes in the radiative emissions of Body A. But Body B is now radiating more toward Body A than the unchanged radiation of Body A toward Body B, so the net heat transfer is now from Body B to A.

In your version, each body must have complete knowledge of the other body’s temperature before it can emit a single photon toward the other. If Body B has a lower temperature than Body A, every atom of Body B must somehow “know” that it is not permitted to emit a thermally generated photon in the direction of Body A. Similarly, Body A must know the temperature of Body B so it can decide how much to radiate in the direction of Body B — the bigger the difference, the more it must radiate.

What is the mechanism by which each body knows the temperature of the other? What waves and/or particles are involved in this exchange? Assuming this exchange of information happens, what is the computational mechanism by which the atoms decide what they can emit, and in which direction?

And I have described an incredibly oversimplified situation, with just two bodies each of uniform temperature. In a more complete scenario, every point on the surface of every body would have to know the temperature in every direction in the hemisphere to which it could radiate, at every instant, and use this information instantaneously to decide how it could radiate in each possible direction of the hemisphere. How on earth could this possibly happen???

When we talk about “heat flow” or “heat flux” (with “flux” simply being the Latin word for flow), we are using these terms by analogy to fluid flow. It does not mean that the underlying mechanism is exactly the same. We still talk about “magnetic flux” even though we do not now believe that any particles are really flowing to create this flux. Even with fluid flow, we are talking about a macroscopic average flow — we do not believe that every fluid particle is moving at the exact same velocity in the exact same direction. When the net flow velocity is low, it is even possible that individual molecules could be moving in the opposite direction — that the rate of diffusion is higher than the average flow rate.

28. David Springer says:

Actually the presence of clouds has, on average, the opposite effect of greenhouse warming. Just like a beach umbrella intercepts sunlight to keep what’s beneath it cooler so too does the cloud. Average day/night temperature is cooler in wet climates than dry at similar latitudes. The climate type known as the tropical desert has the highest mean annual temperature of any climate type.

Be that as it may CO2 isn’t a cloud and more of it doesn’t shade the surface. Over dry land the effect of more CO2 is warming of the land and there’s no doubt of this. The radiative cooling path is restricted by additional CO2 and, being dry, there’s no evaporative cooling path that can respond, so the temperature of the land rises which pushes more energy through the atmospheric window and through conduction between air and ground.

Where I have a major bone of contention is over wet surfaces. I have yet to see anyone demonstrate that a body of water (or a wet solid surface for that matter) free to evaporate in response to mid-infrared illumination does anything other than simply evaporate more with no rise in temperature and the restriction in the radiative cooling channel is thus short-circuited by latent heat transport away from a wet surface.

Feel free, Dr. Spencer, to demonstrate that increasing mid-infrared illumination of the surface of a water body causes greenhouse warming. Good luck!

29. don penman says:

It is no surprise that water vapour gives off infrared radiation when it condenses.When a weather front causes water vapour to condense then there is a rise in temperature of the atmosphere ,this is not a greenhouse effect though a GH effect would be harder to seperate from the ambiant temperature.

30. Milton Hathaway says:

Dr. Spencer – I find your infrared thermometer experiment fairly convincing evidence of the atmospheric greenhouse effect. But to be honest, that’s probably because I’m already pretty convinced of the effect from my daily life. Here in the Coastal Pacific Northwest, where the temperature hovers above freezing for six months of the year and the humidity is high, if I have to stay late on a clear night, the windshield frosts up and I have to scrape it. This happens even if the temperature is above freezing. The other car windows, which are vertical, typically don’t ice up, but the hood does (but who cares about ice on the hood). We much prefer cloudy weather at night for this reason – scraping the windshield before you can head home is a pain. But a more typical pattern seems to be cloudy during the day and crystal clear and windless at night. This also causes thick fog to rise from low-lying areas, and (admittedly pretty) hoarfrost formations and interesting ice heaves rising from the mud, with the air temperature hovering above freezing. These occurrences are unique to clear nights. What causes this if not the greenhouse effect?

I did try your infrared thermometer experiment last fall. I got similar results, except that the reading pegged at <-40C pointed at clear sky, even on a sunny day.

I also find B&W infrared photography fairly convincing, too, but that might be even further from typical human experience than the infrared thermometer.

Having said all this, none of this intuition applies to CO2. As one poster mentioned, this seems like removing one pane of glass from a greenhouse with 100 panes. On the other hand, an earth-bound greenhouse loses heat through conduction and convection, whereas an atmospheric greenhouse (earth) can only lose heat through radiation. So the balance might be quite delicate. We were testing a product in a Thermotron oven, the temperature control tripped out for some reason, but the "product saver" feature failed to shut off power to the product. The oven was so well insulated that the product melted itself over the weekend from the small amount of heat given off from the product. Few things in my life have seemed as nonsensical as the resulting lump of molten dripping plastic. The relationship between heat and temperature can clearly be very counter-intuitive.

31. Mike Flynn says:

Dr Spencer,

I assume you are too busy to respond to my request for further information, in order that I can make a logical response to your question.

However, your use of a remote “temperature” measuring device may not be giving you appropriate advice. If you point your device towards outer space, what temperature should it indicate? If it is not indicating 4K or thereabouts, is it indicating the intensity and/or level of EMR from particulate matter in the atmosphere ( demonstrated by crepuscular rays etc), or gases emitting EMR as a consequence of their being above 0K?

In any case, noting the fact that water exhibits the property of total internal reflection, as well as normal Fresnel effects, explains the reflective properties of clouds.

This can be demonstrated by noting that clouds can, indeed, be seen, even though they are composed of “transparent” liquid. Furthermore, we perceive their colour to be white, in reflected sunlight. I can certify that clouds reflect UV, and also microwave radiation of far longer wavelengths than the infrared spectrum.

I will not suggest that you perform a simple experiment where you measure the “temperature” of an elevated horizontal sheet of white plastic (that thin white foam stuff is both highly reflective and easy to manoeuvre), which has been allowed to cool to ambient temperature at night. You would then have to come up wih a climatological explanation for the reason your remote sensing device was indicating a totally different temperature than a thermometer either attached to or embedded in the foam.

But hey, experimental verification has never been a tenet of climatology. As far as I am aware, here is precisely no experiment which shows an object having its temperature elevated purely by being surrounded by CO2. Not one. Nothing, in spite of tens of thousands of scientists sharing a “consensus”.

Do you not at least wonder why this should be?

Ah well, Newton, Lord Kelvin, Einstein, and many others went to their deathbeds avowing a firm belief in things that we now believe to be nonsensical at worst, and merely wrong, at best.

At least you have the potential to be rubbing shoulders with he best, if it should turn out you’re wrong.

No offence intended. I prefer science based on experiment and logic at a fairly basic level.

Live well and prosper.

Mike Flynn.

PS If you can find time to provide me with the additional information I requested, I believe I can convince you that there is a simpler explanation for the phenomenon you observed. The explanation does not involve the action of any “greenhouse gas”.

32. Stephen Wilde says:

There is gross confusion here as to how an IR sensor works and what it measures.

It certainly does not record the temperature of the thermopile itself because if it did then whenever it records a temperature of 0C or less the thermopile would ice up.

It doesn’t, even though it is designed to record temperatures as low as -20C according to one commenter.

What it does do is record the temperature of a ‘surface’. It records that temperature as Roy says and not the energy flow or flux as Leonard tried to assert.

So how does it define a surface to measure ?

It will only recognise a surface if sufficient light is reflected from that surface.

There is no problem with fully opaque surfaces that fill all or most of the field of view. That is why it works well for a coffee pot or a freezer.

There is a problem if there are multiple surfaces within the field of view but one commenter gets around that by putting white paint on the surface to be measured. The extra reflectivity of the target helps the sensor to focus on it.

There is another problem if the target is not sufficiently reflective for the sensor to identify the target as a surface and there we come to the nub of the issue as regards clouds compared to open sky.

Clouds are sufficiently reflective to be recognised as a surface so as long as there is sufficient cloud in the field of view the sensor will lock onto that cloud, recognise it as a surface and record the temperature. It seems to duly record a cloud surface temperature roughly comparable to what would be expected from the standard lapse rate.

Open sky is not sufficiently reflective for the sensor to effectively recognise a surface to measure. The sensor will therefore only recognise a surface when there is a deep enough layer of open sky in front of it for the cumulative reflectivity to build up to a level that the sensor can accept as a surface.

That means that the more transparent the air the higher the sensor must look to find a surface so as to measure the temperature of that surface.

So when there is no cloud in the ield of view the sensor will find a ‘surface’ at a higher level which will be colder and so of course the sensor will record a lower temperature from open sky as compared to a cloudy sky.

The fact is that the radiative flux has nothing whatever to do with the temperature recorded at the surface which it is measuring or with the temperature of the thermopile itself.

Both those temperatures are set by their respective heights in the vertical column and not by radiative fluxes.

As I aid before, the interpretation of IR sensor data has been grossly negligent.

• Stephen Wilde says:

The one remaining issue to be dealt with is that raised by Leonard who pointed out that even if air and cloud droplets at the same height might be the same temperature there is nonetheless a difference in the amount of energy that they radiate.

The implication is that the radiative difference allows the greater amount of radiation from the water droplets to alter the temperature of surrounding material (in particular the ground).

The logical problem with that is that if the water droplets are radiating energy away faster then they should become colder than the surrounding air but we see that they do not.

It follows therefore that the extra radiating capacity of the water droplets is being replaced as fast as it occurs and the only means by which that can be achieved is by direct solar heating from above or by conduction from the adjoining air molecules or the release of latent heat when the droplets condense out.

In any event the different radiative capability of the water droplets is clearly being offset by other processes because we see that the temperature of both the water droplets and the surrounding air molecules remains stable despite the different radiating capabilities.

The vertical temperature profile created by the pressure gradient remains stable despite the different radiative capabilities of constituent molecules.

It therefore follows that radiative characteristics have a zero net effect on the vertical temperature profile.

GHGs can however affect the vertical temperature profile but by different means.

Water vapour does it by latent heat absorption and release.

Ozone does it by reacting directly with incoming solar shortwave.

I have seen no evidence that radiative capabilities in themselves have any effect on the vertical temperature profile and as I have explained above we can no longer regard IR sensors as providing evidence in support of the proposition.

That leaves AGW radiative theory with a lot still to prove.

33. Thanks, Dr. Spencer.
Good experiment -> good understanding.

34. Kristian says:

In radiative terms, emission temperature of a surface (disregarding emissivity) is dictated by its heat gain from its heat source. And only that. The thermal energy of an object, defining its temperature, accumulates through the transfer and absorption of heat – warming the object (disregarding work done on the object). It does not accumulate by ‘cooling less’.

This is where the confusion arises. Spencer seems to be basing his argument on the strange, and frankly un-thermodynamic, notion that not only the radiative heat input to a surface (gain from hot reservoir, absorption) but also the radiative heat output (loss to cold reservoir, emission) from that surface, controls/regulates its temperature.

This is wrong.

Its temperature is set and constrained only by its absorption of heat from its hot reservoir.

The temperature of the surface in turn determines its emission to its cold reservoir.

This causal chain runs from left to right only. Not the opposite way:

Absorption of heat — temperature — emission of heat

This relates to pure radiative heat transfer.

The emitted flux does not in any way control the temperature of the body emitting it. The temperature controls it. Understand, without the absorbed heat, no temperature. Without temperature, no emitted heat. That’s how it works.

Only the radiative flux from a warmer body can raise the temperature of a cooler body. Because only a warmer body can transfer heat to that cooler body.

A black body in a vacuum at 290K emits 400 W/m^2. It does so no matter what. This flux is a function of its temperature. Based on heat gain from its heat source, a warmer body than itself. Its corresponding emission is dictated by the laws of physics.

If this body then receives a flux of 200 W/m^2 from another body adjacent to it at temperature 244K, then this flux will not be able to do anything in the way of increasing the level of thermal energy and thus the temperature of the 290K body. Its temperature is already set by its incoming flux from an even warmer body, say 6000K, its hot reservoir.

If you want to claim that the 290K body warms up to a higher steady-state temperature in the presence of the 244K body, then you are in effect saying that the 244K body operates as a second, independent heat source for the 290K body.

Why? Because nothing else has changed. The original heat source still sends but the 400 W/m^2 of 6000K spectrum radiation to the warm body. This will upon absorption in itself not warm it past the 290K it did before. The body also still emits its corresponding 400 W/m^2 flux of 290K spectrum radiation to its cold reservoir based on its temperature.

So if the temperature then still rises, then this must somehow be caused by something else. It must be caused by extra absorbed heat from somewhere. A positive transfer of (a gain in) thermal energy.

Well, the only other body in this system is the 244K one. The only thing that’s different is the 200 W/m^2 flux from the cool body to the warm.

Spencer claims this radiative flux slows the cooling rate of the warm body, thus heating it indirectly. But how specifically does it accomplish this without itself transferring HEAT to it and thereby raising the warm body’s kinetic energy level beyond what it would be otherwise – and hence its temperature? Does it somehow disallow half of the 400 W/m^2 of 290K spectrum radiation from leaving the surface of the warm body?

Remember now, if the warm body is provided with a constant energy/heat supply from its ultimate heat source, it will not cool in the meaning ‘temperature dropping’. Its emission temperature is kept up, sustained. So the cool body can do nothing to reduce its ‘cooling rate’. There is no cooling rate to be reduced.

What is reduced is ‘the heat transfer rate’ between the warmer body and the cooler. Q. This does not affect the surface temperature of the warmer body. Only the surface temperature of the cooler one.

The smaller the temperature difference between the two bodies, the smaller is Q. At T1 = T2, Q = 0. And the warm body can no longer heat the cool body. No more heat transfer.

35. Kristian says:

Sorry, I guess you can disregard the ‘(disregarding emissivity)’ in the first line of the post above.

36. Fulco says:

Dear Roy,

what you actualy showed is that a cloud acts like a blanket.
If you want to show that CO2, H2O or any other molecule with a dipole moment causes a greenhouse effect you need an athmosphere of pure N2 or something like that to compare with. To prove the greenhous effect exists we need to build a long container and fill it with different gasses and do the measurements.
Dipole gases are capable of capturing IR an store it in a rotation or vibration mode. This however does not contribute to the temperature of the gas. It can maintain this state for only a few microseconds. If an exited molecule collides with another molecule ik can transform this energy to translation energy which does contribute to the temperature.
Otherwise it wil reradiate IR in any direction. The reverse proces is however also true. dipole enhanced gasses cool faster.
Until then I stick with Miskolzci and put my bet on 0.48 C temperature rise if CO2 levels double.

Fulco

• Norman says:

Fulco

I have been asking the same thing. Get some measurements do some actual lab tests, spend some time doing experiments under where you can control variables to measure each different input effect.

When there are so many views and thoughts on a subject that are so divergent the only thing for a scientist do is to end the debate with experimental evidence from a lab. Earth system has too many variables. The “Yes Virgina” concept should be actaully tested in NASA vacuum chambers and measured precisely to see what that real outcome is. Thought experiments lead to endless debate. Actual experiments end debate and provide proof.

When are the actual experiments going to be run?

• Mike Flynn says:

Norman,

It doesn’t work, as anyone who tries quickly finds out.

No matter, faith doesn’t need fact. This too will pass – or is already in he process of passing!

Live well and prosper,

Mike Flynn.

• Nabil Swedan says:

Laboratory experiments were conducted decades ago. There is no evidence of the existence of greenhouse gas effect. Scientists still like these controversial experiments, because in science there is not a lot at stake. The climate issue, however, has become different just recently because our wallets are being on the line.

37. Fulco says:

By the way, This has nothing to do with Planck radiation.

38. Bill Hunter says:

I tend to think that in our rotating world greenhouse gases may be a condition necessary for a warmer surface but not a condition sufficient.

It seems more likely the lapse rate through the atmosphere is responsible for the warmer surface. That seems supported by Mars which has more CO2 per unit surface area than the earth 5.70mb compared to .35mb or earth. Mars has no greenhouse effect (or more likely a negative effect potentially arising from dust blocking incoming solar) and too thin of an atmosphere to support a pressure lapse rate.

We get a similar effect here where water in the atmosphere cools the surface. . . .its overridden by the heat pump capabilities of a thick atmosphere. In both scenarios radiation prevents the high emissivity surface from cooling to an average less than the atmosphere.

A close examination of radiation budgets strongly suggest the atmosphere warms the surface by being warmer than the surface at the point of contact with the surface. The atmosphere by virtue of a gravity heat pump driven by daily solar insolation oscillations (i.e. it gains heat by both conduction/convection and radiation, but loses it primarily by radiation only) if the solar radiation did not diurnally oscillate the atmosphere would have no way of warming above the average insolation.

Folks love to invoke the cool nearby object slowing the cooling of a warmer surface. But its also necessary to invoke a one way glass theory for the heat increase.

I would love to see something besides a thought experiment establishing that. . . .the thought being “what else could cause it?” How about a passive solar heater? Ever build one? I have. All I am saying is it seems likely the so-called and misnamed greenhouse effect is more complex than offered.

39. Rafael Molina Navas, Madrid says:

THE IMPORTANCE OF BEEING “BRAVE”

Congratulations, Dr. Spencer … You are a brave man explaining the issue so clearly, knowing how many viscerally skeptic “friends” you have.
But they are showing their erroneous ideas once more, in a way very similar to when you have previously explained it… I especially remember 3 o 4 years ago, after your article “Yes, Virginia, cold objects can …”
I really cant´understand their obstinacy!
I am seeing last year you dealt also with the issue: “Yes, Virginia, the “Vacuum” of Space Does have a “Temperature” … to no avail!

40. Kasuha says:

Dr. Spencer, thank you for the article. While I don’t have problems with understanding back radiation and thermal transport in general I wouldn’t expect the effect to be as strong as you demonstrated.

41. AlecM says:

The Pyrometer filters out >14 µm energy to eliminate main water vapour bands, so it’s mostly the ‘Atmospheric Window’.

The clear sky temperature is set by IR loss through the AW.

Putting a cloud in the view means that instead of the pyrometer sensor IR being offset by the cosmic microwave background, 2.7 °K, some or all of it is from the cloud at ~10 °C.

This is processed as a rise in temperature.

There is no ‘back radiation’, just a variation in the net IR signal difference.

Remember, heat generation rate per unit volume is the negative of the integrated divergence of all the monochromatic flux vectors at the volume of matter.

Most scientists get this sign wrong and imagine that a single S-B equation gives an energy hence heat flux, hence the false ‘back radiation’ claim.

42. Mike Flynn says:

To all,

Before getting too hung up on “IR”, remember it is just a definition of part of the literally infinite EMR spectrum. Bodies absorb EMR, as their nature dictates, and a temperature rise may ensue.

There is no (actually a tiny bit, due to inevitable energy transformation losses) IR emitted by the magnetron in a microwave oven. No need. Water in the oven heats nicely. UV lasers heat and cut nicely.

CO2 creates no energy. It increases the temperature of nothing. It absorbs EMR like every other matter in the known universe. It is transparent to some frequencies, and opaque to others in varying degree, like all known forms of matter.

Where’s the mystery?

Live well and prosper,

Mike Flynn.

43. Wayne2 says:

Dr. Spencer, several questions have come to mind as I thought through your experiment. I believe your overall thesis, but the experiment raises questions.

1. How much does the Sun heat clear air? As compared to something like a cloud, dust in the air, or the surface of the planet? I could posit that your experiments (both the cloud and the angle one) might simply be measuring the temperature of air closer to the warmer surface. Or also the temperature of things in the air (clouds, dust) that absorb more sunlight (visible and IR) than clean air.

2. To what degree are clouds actually sources of warmth? I’m thinking of things like the latent heat released by condensation, heat absorbed from the small amount of sunlight that the cloud actually absorbs (and the cloud tops will have 100% sunlight exposure during the day), etc.

3. To what degree is the atmosphere itself different when there are clouds in it — either because of the presence of the clouds, or because of the pre-conditions necessary for clouds to form? I’m thinking in terms of effects on vertical mixing, stratification, etc., and the resulting effects on temperature.

The last two may be trivially small, but I think the first one affects the applicability of your experiment.

At night, it makes intuitive sense to me that anything that reflects or absorbs outgoing radiation will keep us warmer than we’d otherwise be. And it’s something like the night case that your experiment is measuring. But during the day, it seems you’d need to stand on the ground with your sensor, and then also use your satellite in the sky looking down at the same cloud to see the net effect. Unfortunately, you can’t really do the same with CO2, since it’s much more diffuse than clouds, so you have to assume it works like clouds (at some wavelengths) but with weaker effect, I guess.

44. PeterB in Indianapolis says:

An IR thermometer doesn’t measure the TEMPERATURE OF ANYTHING.

http://www.omega.com/prodinfo/infraredthermometer.html

Now, anyone who is trying to claim that the “thermopile would ice up at temperatures of 0C” please read the information at the above link and make an actual attempt to understand it.

REMEMBER!!!

IR Energy is NOT HEAT!!!

Energy CAN be transferred from a cold body to a warm body, and vice versa. This does NOT violate ANY laws of Thermodynamics.

The only thing that the 2nd law of thermodynamics says is that the NET (say it again! NET! NET! NET!) transfer of HEAT (say it again, HEAT! HEAT! HEAT!) is from a warmer body to a colder body.

NOTICE that the 2nd law does NOT say that heat is NEVER transferred from a colder body to a warmer body, it merely says that ON NET, the HEAT TRANSFER must be from the warmer body to the colder body, until equilibrium is reached.

ALSO NOTE, ONCE EQUILIBRIUM IS REACHED, ENERY/HEAT TRANSFER DOES NOT SUDDENLY STOP (unless equilibrium is reached at a temperature of 0K). Once equilibrium is reached, the NET ENERGY TRANSFER IS ZERO, but energy is still being transferred BOTH WAYS, CONSTANTLY.

It is appalling how few people truly understand basic physics….

• Fulco says:

If an IR-photon is absorbed by a dipole-molecule temperature need not to rise as I explained. Only if this molecule colides with another molecule before is reemits the IR-photon there is a posibility to transfer rotation/vibration energy in translation energy thereby raising the temperature of the gas a bit.

• AlecM says:

‘ALSO NOTE, ONCE EQUILIBRIUM IS REACHED, ENERY/HEAT TRANSFER DOES NOT SUDDENLY STOP (unless equilibrium is reached at a temperature of 0K). Once equilibrium is reached, the NET ENERGY TRANSFER IS ZERO, but energy is still being transferred BOTH WAYS, CONSTANTLY.

It is appalling how few people truly understand basic physics….’

To get energy transfer, two radiation fields have to interact. Only net flux can do thermodynamic work. If there is none, there is no energy transfer, either way.

The monochromatic volume-specific rate of gain of heat, q, by matter is the negative of the divergence of the monochromatic flux density vector, a fancy way of stating that the radiation fields cannot transfer energy if they combine to zero!

If each body transferred equal energy to the other, it is easy to show there would be thermal runaway and we’d be a tenuous ball of cold gas.

The explanation is that if the radiation field consisted of ‘streams of photons’, as many imagine, these would fill empty surface vibrational states or in the case of GHG gas mixtures, the unactivated GHG molecules. However, this is set by the temperature of the matter. Maxwell-Boltzmannn statistics. Therefore if you claim there is energy from each body to the other, each body will heat up as the occupied density of states increases for each.

This does not happen and can’t.

45. PeterB in Indianapolis says:

It is possible that Rayleigh scattering is affecting the IR thermometer when you point it towards the horizon rather than pointing is straight up at the sky. This might account for the observed temperature differential on a clear day, and have nothing whatsoever to do with “Greenhouse Effect”.

• Fulco says:

Peter,

Rayleigh scattering is blue not red.

46. PeterB in Indianapolis says:

More information on the proper calibration and the sources of error in IR thermometers:

47. Stephen Wilde says:

PeterB said:

“anyone who is trying to claim that the “thermopile would ice up at temperatures of 0C” please read the information at the above link and make an actual attempt to understand it.”

Well I pointed out that it does not ice up at 0C so obviously the IR sensor does not measure the temperature of the thermopile which agrees with your comment.

However the IR sensor clearly displaya a temperature reading even if it is derived from the sensing of IR.

It gives reasonably accurate temperature for Roy’s freezer, and his coffee pot and for the temperature that would be expected from a cloud at a certain height given the lapse rate.

So I respectfully suggest that you have misunderstood something.

48. CC Squid says:

“What caused the IR thermometer reading to warm up by 14 deg. when it went from clear sky to the cloud?”

I live at 7000 ft. In Colorado and have a southern exposure. When the sky is clear and the outside temp. is 20f, my home is 73f. The next day when it is overcast, the house is kept at 64f by my furnace. The answer to your question is obviously that the heat that you measure has been stored in the cloud.

I do not know how the climate models are designed but if the programmers used the “black body radiation” theorem as the basis of their program I would ask if the radius of their “black body” changes with cloud height and thickness.

49. Bill Hunter says:

“Energy CAN be transferred from a cold body to a warm body, and vice versa. This does NOT violate ANY laws of Thermodynamics.”

thats just semantical gamesmanship, as is the claim that the greenhouse effect violates the 2nd law. Fact is the 2nd law in pure science only applies to what it has been empirically tested to apply to. Everything else is belief, not science.

And when it comes to beliefs there really is no difference between the belief of a priest and the belief of a scientists.

Both types of persons tend to belief their beliefs are in some way more valid than another’s belief.

But it is for the most part in this day and age, and it was not always this way, that priests seem to have a much better grasp that their beliefs are not science than scientists do.

Now my belief as I stated above is that it is the case that warm greenhouse gases in compliance with the tested laws of thermodynamics are allowed to provide energy for heat to offset in whole or in part the faster cooling of high emissivity planet surfaces. On the other side of the equation these gases may or may not absorb or reflect an equal amount of incoming light. At any rate its not good enough to claim an excess of outgoing absorption over incoming absorption (of all forms of energy transfer) without actually doing some calculations to demonstrate a net effect. Again here the argument for a greenhouse effect is a semantical argument not a scientific one as it has never been quantified.

One cannot claim a “greenhouse effect” out of far less than half a job of calculating all the forces at work. One must also quantify incoming radiation absorption by the atmosphere, heat gain and losses through conduction and convection, and other sources of potential energy sequesterization such as the passive solar heating model.

We know that the passive solar heating model requires that there be incoming heat sources that fluctuate around the average. Without the fluctuation the model does not work as the storage medium will have the exact same temperature as what is outside the storage medium. Its only when the incoming energy fluctuates that an opportunity arises to sequester additional heat. To sequester the additional heat you need to do it solely during times the fluctuation is on the high side as if heat losses occurred on the low side, again you could not make the model work.

In the passive solar heating model the passive solar heater is more efficiently the better the storage medium is insulated.

We know from the laws of emissivity that an atmosphere without radiative properties will transfer less heat to space so its in effect better insulated.

Finally, I have crudely calculated the emissivity of the atmosphere and have found it to be nearly identical in its current state to the emissivity of CO2 at a figure approximating about .13. Though I will admit I have not adequately verified that.

Thus it would appear from the standpoint of the passive solar model there is no ability to store more heat in the atmosphere by storing more CO2 there at least on a well mixed gas basis.

But there are significant opportunities to change the atmosphere emissivity with water vapor changes as it is not considered to be a well mixed gas and the emissivity of water vapor is considerably higher than CO2. And of course clouds can also have an effect.

There is much to do to verify such a model but at a minimum we know the model works as passive solar models do work.

50. kuhnkat says:

Roy,

those neat little IR thermometers were NOT designed to measure the temperature of atmospheric gases. They were NOT designed to measure anything past a few feet. I am appalled that a scientist of your caliber is falling for this kind of misrepresentation!! Why have you not done your homework using the phamphlets freely available from the manufacturers to determine how they should be appropriately be used!!!

Would you use the same equipment on your satellites to measure the temperature of the sun as you do the atmosphere without even an attempt at recalibration?!

You have NO IDEA where the radiation is coming from that the instrument is reading yet you make various claims based on your assumptions. Disgraceful.

51. KevinK says:

Bill Hunter wrote;

“as is the claim that the greenhouse effect violates the 2nd law. Fact is the 2nd law in pure science only applies to what it has been empirically tested to apply to. Everything else is belief, not science.”

Well…… Any claims that the GHE creates “extra energy”, or “net energy gains” do in fact FLY IN THE FACE OF THE 1st LAW. Engineers look inside our calculations for results that indicate either of these. This is a RED FLAG that we interpret as an indication that our calculations are WRONG. Then we ball up our paper (old times, replaced by spreadsheets now) and start over.

You might want to study the history of the laws of thermodynamics just a bit; they are in fact largely an offshoot of the study and perfection of steam engines. You know those awful fossil fuel burning machines that saved us from “Peak Wood”.

Sure, I suppose that there are some NEW FANGLED things (i.e. the GHE) that avoid the laws of thermodynamics, but is it STILL NOT WARMING………………….

Believing that climate science is not subject to the laws of thermodynamics while every other modern technology has followed them is, well, I’m actually at a loss for words……. I guess HUBRIS fits best.

Cheers, Kevin.

52. KevinK says:

PeterB wrote;

“NOTICE that the 2nd law does NOT say that heat is NEVER transferred from a colder body to a warmer body, it merely says that ON NET, the HEAT TRANSFER must be from the warmer body to the colder body, until equilibrium is reached.”

Well…. Almost exactly correct, except, for one small caveat, the 2nd law says nothing about equilibrium. The conditions present (energy flows, material temperatures, etc.) determine if equilibrium is in fact present. It (thermal equilibrium) is in fact a very rare creature that has only ever been spotted in textbooks; no trophies have ever been “bagged” in the wild.

The second law DOES NOT say; “EQUILIBRIUM WILL OCCUR”, ever…… It might occur, but real world observations show that it is a damn elusive little bugger, almost like a Sasquatch……

Cheers Kevin.

• Gordon Robertson says:

Kevin ““NOTICE that the 2nd law does NOT say that heat is NEVER transferred from a colder body to a warmer body, it merely says that ON NET, the HEAT TRANSFER must be from the warmer body to the colder body, until equilibrium is reached.”

You are confusing heat with infrared energy. There is no such thing as NET heat transfer between bodies of different temperatures. Heat will be transferred from the warmer body to the cooler body until equilibrium is reached, but heat cannot be transferred both ways.

IR can be radiated both ways because bodies of all temperatures radiate IR of different intensities. However, it has been incorrectly assumed by many climate scientists that lower intensity IR will raise the temperature of a body at a higher temperature.

If you read the treatise by Clausius on heat (he developed the 2nd law and coined the name ‘entropy’), he is emphatic that heat can only be transferred from a warmer body to a cooler body by ordinary means. He does state that IR can go both ways.

IR is electromagnetic energy and it has no heat content. It produces heat when it intercepts atoms or molecules in a body provided that the atoms/molecules are at a low enough energy state to absorb the energy.

53. KevinK says:

Whoops,my reference to a Sasquatch shouldhave probably referred to a “damn elusive BIG bugger”

Ha ha

Cheers Kevin.

54. KevinK says:

PeterB;

Perhaps a better phrase is;

“unless equilibrium is reached”

I agree totally with your point, just meant to point out that the 2nd law does not “enforce” equilibrium, that appears to be the job of the “climate cops”.

Just a little bit of sarcasm, I hope some can enjoy.

Cheers, Kevin.

55. KevinK says:

Curt;

“And I can say without hesitation that back EMF is voltage.”

Ok Curt, in the overall interest of keeping this thread “on topic” I will concede that you know FAR more than I ever will ever know about motors.

Just remember that EMF is an electromotive FORCE, it can be expressed as a voltage source or as a current source, remember that whole Thevinin’s theory thing. That would be EE 102 after we covered Ohm’s law in EE 101.

Cheers, Kevin.

56. Russell says:

Is the evil radiative behavior of clouds among the “blacklisted commentor memes” Anthony Watts new autocensorship software is set to destroy?

57. Bert Walker says:

Dr. Spenser,
I am sorry if this duplicates another post as I did not read all the comments (only ~10%) to see if this issue has already been brought up. I do enjoy your site, Thank you for hosting, authoring, and moderating it.

The answer to your question is obvious to any one who has gone outside on a cold clear night, and also later in the same night after clouds had moved in and experienced warmed air. Yet this is still only a regional effect.

I know you are the expert so please inform me where I am wrong but I do not believe the example you used is adequate to make your point.

To show the “greenhouse effect,” using only an IR thermometer one must show that IR is emitted from the earth, then re-emitted towards the earth.
It seems to me you could re-do your experiment on a partially cloudy night. Use your IR thermometer pointed at clear starlit sky shortly after evening has fallen, and then at the cloudy sky. Lastly point it at the ground to measure the earth temperature, the source of the IR. Presumably the Earth should show as the warmest of the three measurements, then the cloudy sky since it will be re-emitting more IR than the clear sky. If that works out, then “Greenhouse Effect is proved.

In each case of your example, the total radiant energy reaching the earth surface originating from solar radiation, hypothetically, could have been the same, and needs to be excluded. Failing to do so leaves the possibility that the increased H2O in the clouds simply converted a higher percentage of the visible (solar) spectrum (not measured by your IR thermometer) to IR, which you then demonstrated.

Thanks
Bert

58. PetterT says:

A (too?) simple explanation.
A CO2 molecule in the atmosphere will have a certain low temperature compared to the warmer earth surface. An object, like CO2, with a certain temperature emits electromagnetic radiation with a certain spectral radiance (power or energy quantity) as a function of frequency (or inverse wave length) according to Planck’s law. Cold CO2 in the atmosphere will have lower frequency and energy than radiation from warmer earth. Warm, high frequency and high energy radiation can increase the temperature of colder objects, no question about that. However, (back-) radiation from a cold object does not have the required frequency and power to increase the frequency and hence temperature of a warmer object. Low frequency radiation cannot create a higher frequency. The low frequency radiation will resonate and be reemitted (scattered) together with high frequency radiation from the warmer object. This contribution by the cold reemitted radiation will reduce the rate of cooling of the warmer object, but never increase its temperature.
The process is much like trying to excite an atom by using too low energy radiation; – it is not possible, because a sufficiently high energy is required.

• Ron C. says:

The reduced rate of cooling is another way of saying the flow of heat slows as the temperature of the cooler body nears that of the warmer body.

As Kristian points out, above, in the situation where the warmer body continues to be heated by a heat source, the cooling rate is unchanged. In fact, the temperature (and therefore the radiation) is set by the heat source, not the cooling. This is comparable to the earth and atmosphere, in that the surface continues to be heated by the sun.

• Fulco says:

PetterT,

First of all, the greenhouse effect has nothing to do with black body radiation and secondly black body radiation has a complete spectrum and not just one frequency.

• PetterT says:

Fulco,
if you look at the diagram here
http://en.wikipedia.org/wiki/Planck_law
you will see that low temperature radiation has a minimum wavelength and thus a maximum wavelength. Low temperature has lower max (and “cut off”) frequency compared to high temperature.
I am aware that radiation from CO2 in the atmosphere is not pure black body radiation, but rather some gray body radiation with some of the same properties.
If you, or somebody else, has a better explanation for why radiation from a cold body do not encrease the temperature of a warmer body, then I appreciate an explanation or a link/reference.
PetterT

• PetterT says:

correction:
…low temperature radiation has a minimum wavelength and thus a maximum frequency.

• PetterT says:

…and by the way,
a change of only a few ppm of CO2 in the atmosphere has such a small effect on decreasing the cooling rate of the earth surface that it is not measureable. In that I agree with Dr. Spencer. H2O as a “greenhouse gas” has many effects which are causing net cooling, as documented by Dr. Spencer and Dr. Lindzen.

59. Guenter Hess says:

@Christopher Game
you wrote:
“Claes, you are making the mistake of thinking that radiation by itself from one body to another is heat transfer. That is wrong in classical thermodynamics because it is not a process that takes the bodies from one state of thermodynamic equilibrium to another. Thermodynamic equilibrium usually obeys the principle of microscopic detailed balance, and one-way radiation of course is very far from that. You seem to be appealing to the second law of thermodynamics to provide an answer about how radiation propagates. That is a mistake. Thermodynamics does not provide information of that kind. Thermodynamics presupposes such information. You can check this in a reliable account of classical thermodynamics. You are asking for a coupling that does not exist and is not needed. The emission of thermal radiation is spontaneous.”

This is right on the money. Trying to refute the existence of “back radiation” with thermodynamics is futile and a misunderstanding of classical thermodynamics per se.
Best regards
Guenter Hess

60. JPC Lindstrom says:

Dear Dr Spencer,

My suggestion:

The difference between cloud and sky is the scattering properties of clouds. This in itself will increase the apparent flux (of light and IR) in all directions. Basic physics. Compare mirror vs. white walls in a room. Which room will appear brighter?

The scattered flux will originate from all directions but mainly from space and the ground (and reflected back to the ground).

A sheet of semi-transparent white perspex would probably give the same (qualitative) result.

61. Stephen Wilde says:

Has no one registered the point that the clear sky reading is lower as compared to the cloud reading simply because the lower reflectivity of clear air causes the sensor to measure temperature at a higher colder location ?

62. PhysicistPhillipe says:

Curt:

No, O^2 indeed absorbs in the UV spectrum…your claim that 99.99% of the atmosphere is transparent to incoming radiation is utter bunkum, we would fry to death otherwise. There is a reason we’re not bombarded with X/Gamma/UV/etc at the levels measured at the TOA boundary.

Yet we find the upper atmosphere is not searing at 5000 Kelvin. Want to guess why that is?

63. PhysicistPhillipe says:

I don’t understand the confusion here…compare cloudy nights to clear nights…see the difference in the rate of thermal loss..it is that simple. No one is arguing that downwelling LW increases the surface *warming* rate during daylight hours.

64. jjfox says:

Dr. Spencer, you need to understand that there is no such thing as a “greenhouse effect” because the IR activity that is exhibited by the atmosphere isn’t a warming affect at all because the upwelling IR is always greater then the downwelling IR, whether you are speaking of land or ocean surfaces. There is no net warming of the surface of the planet by downwelling IR. Why can’t you understand that?

And no, Virginia, cooling slower is not the same thing as warming, much the same as going broke slower isn’t the same thing as becoming wealthier. The world just doesn’t work that way.

Downwelling IR is real enough alright, it just doesn’t warm our planet’s surface.

The IR activity that is exhibited by the atmosphere is actually a cooling affect. It’s a “refrigeration effect”,not a “greenhouse effect”. It is how the atmosphere cools to space. It is driven by the latent energies of water changing phase in the atmosphere. That is the reason why the OLR to space rises and falls in an annual cycle in lockstep fashion with the water content of the atmosphere. You have access to that data, you can see it for yourself. It correlates very well.

If Earth was not a wet planet, with an ocean, our situation would be much the same as that on Venus.

Venus is a dry planet, lacking water. There is no condensation, fusion, sublimation or precipitation of water occurring in Venus’s atmosphere, and that is the reason why that planet cannot cool itself effectively to space

Dr. Spencer, you do need to understand that the IR activity that is exhibited by the atmosphere is not a warming “greenhouse effect” at all. It is a cooling “refrigeration effect” that cools our planet to space. Why can’t you understand that?

• Mike Flynn says:

jjfox and those of like mind.

Like you, i am unable to understand why anyone can think that returning only a portion of radiative energy to the surface that emitted it would result in anything but a drop in temperature. Even if 100% was somehow returned, the best that could happen is that the temperature would remain the same.

It doesn’t matter in the long run. The public in general are prepared to pay more attention to astrology than climatology, and this shows that everything is proceeding as it should.

Live well and prosper.

Mike Flynn.

65. coturnix19 says:

Dr. Spencer, if such an amazing, complex, self-sustaining chaotic process as life can and does exist on its own, without any help from creator or even deistic schrodinger’s “vital fotce”, what metaphysical arguments make you think it couldn’t apper on its own?

66. DJC says:

You write “if you suddenly removed all of that atmosphere and clouds: there would be a sudden increase in the rate of net IR flow from the surface of the Earth to outer space, and temperatures would drop. ”

Yes they might drop without any atmosphere at all, but what if you removed, say, 90% of the water vapour, leaving the atmosphere intact? Then temperatures would rise. We can see this in real world data where some regions are much drier than others. Studies such as that in the Appendix of my paper in the PSI/PROM menu (and a much larger one I am soon to publish) show that WATER VAPOUR COOLS – IT HAS NEGATIVE FEEDBACK

Inland tropical cities which are drier have both higher daily maximum and daily minimum temperatures than do those with higher rainfall in similar locations.

If ever there was proof that the radiative “greenhouse effect” is false pseudo science, this is it.

Do I have to write yet another article about your misunderstandings – that will be read by tens of thousands on PSI and other websites?

When are you going to stop misleading the public, Roy, with your misunderstandings of radiative heat transfer, the role of entropy in thermodynamics and Kinetic Theory – such as was used by Einstein but is largely overlooked by climatologists?

• Massimo PORZIO says:

-Off topic-
Doug, are you?
It’s everything went the right way, I hope.

Have a nice day.

Massimo

• DJC says:

Yes thanks Massimo – virtually no pain now and able to drive again.

• Massimo PORZIO says:

[email protected]

But having no reply I was worried.

I wish you all the best.

Massimo

67. DJC says:

(continued)

According to the original net energy diagram produced by NASA, there is about 19% of incident Solar radiation absorbed by the atmosphere, compared with only about 15% of it absorbed on the way back up from the surface. That should make you think!

The “warming” of the surface is not caused by radiative imbalance. The reason for it can be understood with the same Kinetic Theory which Einstein and others used about 100 years ago. It has to do with non-radiative diffusion processes in a gravitational field, because radiation can never transfer heat from cooler to warmer regions, whereas non-radiative “heat creep” can in fact transfer heat up the -g/Cp thermal gradient. This is the only possible process which can explain Venus surface temperatures where the heating of the surface has nothing to do with trapping energy that was first absorbed from Solar radiation by the surface.

The Venus surface only receives about 10% of the Solar insolation which Earth’s surface receives. Yet it is heated to well over 720K. Unless and until people understand the physical heat transfer processes involved, namely how the energy gets into the surface of Venus, then they will never understand what happens on Earth – because it is the same process of “heat creep” but just with different parameters.

Without an atmosphere the Sun could never heat the Earth’s surface to a mean of 288K because its radiation is only equivalent to about 255K. But you have to understand how and why the atmosphere causes the surface to be warmer, and yet not as warm as it would have been in the absence of most of the water vapour.

Water vapour has negative feedback – that is, a cooling effect, as temperature data indicates. That doesn’t sit well with any radiative “greenhouse effect” now does it?

That’s because there are many “old wives tales” circulating, such as …

(1) Dry desert regions are cooler at night (false)

(2) The atmosphere is transparent to incident insolation but opaque to IR from the surface (false)

(3) Water vapour traps heat and thus has positive feedback (false)

(4) Carbon dioxide does the same but is more effective than water vapour because it has a longer life span (laughable)

(5) High pressure produces and maintains high temperatures (false)

(6) The lapse rate exists because air expands as it rises and cools as a result. (false)

See my paper “Planetary Core and Surface Temperatures” on the Principia Scientific International Website (PROM menu) for support for what I am saying here.

Doug Cotton

68. DJC says:

I’m quite convinced now that it is the non-radiative process of diffusion of kinetic energy (at the molecular level) which sets the thermal gradient (badly named “dry adiabatic lapse rate”) which (after adjustment for intra-molecular radiation, mostly by H2O) then becomes less steep in moist regions. This gradient, together with the overall level set by Solar insolation, then determines the base surface temperature.

Now, suppose at some location that base temperature is 282K. Then, with temporary extra energy deposited by the Sun each day (flowing into and out of the outer crust and the first 100 metres or so of the atmosphere) we see the mean temperature raised to, say 288K.

Now, in the afternoon and early evening we see the temperature cooling at a reasonable rate, but the rate of cooling slows down as the base temperature is approached later in the night, because the rate of cooling slows when the temperature gap narrows. (After all, it might cool by 2 or 3 degrees per hour in the afternoon, but obviously doesn’t keep doing so throughout the night.)

What is happening is that all the energy in both the outer crust and the rest of the atmosphere is setting the base thermal plot which runs right through the surface and into the crust, mantle etc. Nothing will change that underlying support temperature in a hurry – in fact, nothing can but long-term changes in solar insolation levels as seen in the natural cycles.

The only role that radiation plays is in slowing this marginal cooling each day and thus extending the warmth of the day into the early evening. But non-radiative processes (conduction and evaporation) actually do about two thirds of this slowing anyway, and CO2 probably does only about one thousandth of the radiative slowing that water vapour does. All this has been happening through the life of the Earth and its atmosphere, so variations in carbon dioxide are insignificant, although more likely to be cooling effects anyway.

So, yes, we do observe slower cooling when low clouds are radiating overhead at night, but we should not infer that this will have any significant effect on mean temperatures and that base temperature. The base temperature is, in fact, affected the other way, as we see in the temperature data for inland tropical cities as in the Appendix of my paper. This data establishes that, as expected, it is the base temperature set by the thermal gradient and level of insolation which is the dominating factor in mean temperatures anywhere. That’s why water vapour has net negative feedback, despite its marginal slowing of daily cooling.

This physics will turn the climatology world upside down. They will never produce valid real world data that proves the opposite, because we have the fundamentals of physics backing what we say, whereas they have false ideas about radiative heat transfer and the relative significance of carbon dioxide. They also ignore the inconvenient truth that more Solar radiation is absorbed by the atmosphere on the way in than on the way out, and the fact that gravity sets up the thermal gradient, not rising air somehow “lapsing” down the tropospheric waterfall.

69. PhysicistPhillipe says:

DJC. You’re wrong, the 255K figure is derived using the Earth’s albedo which would be much lower if you remove the atmosphere. The insolated portion of the planetary surface right now recieves ~480W/m^2 in SWR..that would rise to almost 680W/m^2 in the absence of cloud cover. At the same, the radiative budget would decrease with the absence of the atmosphere’s nighttime insulation effect.

However, remember that the Earth is a swiftly rotating body with a massive thermal capacity given the planetary oceans. If we remove all GHGes from the atmosphere, while surface temps will initially drop, you’re now left with an atmosphere absorbing high frequency solar radiation (yes, this is why the most dangerous radiation does not fry us) but without a way to emit it. So the upper atmosphere would begin to heat up as tropospheric convection would cease. The planetary surface would then begin to warm after its initial cooling as this warming downwells..much like an actual greenhouse would warm, ironically.

70. PhysicistPhillipe says:

Also this argument that back radiation doesn’t exist is silly, again, monitor the difference in the cooling rate on cloudy nights and clear nights. No one is suggesting back radiation increases the warming *rate* during the daylight hours, but it certainly insulates the surface at night. Whether or not this radiative effect is significant on a planetary scale does not hinge on the direct effect of back radiation…it hinges on the significance of convection and the atmosphere’s ability to cool directly to space with GHGes present. The effect of back-radiation itself probably provides a 5-10K warming effect, in the real world. The Earth is not a zero-capacity greybody, rather it is a swiftly rotating wetbody with a massive thermal capacity.

• Rafael Molina Navas, Madrid says:

You say:
“No one is suggesting back radiation increases the warming *rate* during the daylight hours”
Beeing your statements much much more right than DJC´s, I must say that EVEN with a strong Sun, back radiation increases the warming rate.
Many of us have long discussed this point, particulary after:
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/
This was 2 or 3 years ago! To my surprise, I´m seeing same errors time and again.

Reference is also made to my post here four days ago.

71. DJC says:

PhysicistPhillipe

We at Principia Scientific International have been among the first to point out that the 255K figure is of course wrong itself because it assumes the Earth is a flat disk and no integration is involved to take account of varying levels of insolation all over the globe. (Having studied all this for many thousands of hours, I am quite aware of the adjustments for albedo.) So even your 680W/m^2 is inaccurate because it also belongs to flat Earth science. Furthermore, it is totally incorrect to assume surface emissivity is around 0.95, because more than half the net energy transferred from the surface to the atmosphere is not radiated at all, but is transferred by non-radiative processes. This immediately reduces emissivity below 0.5 and highlights a major error in the radiative models.

So I just used the oft-quoted 255K figure to keep the main point of the explanation simple. You missed the main point altogether.

So how about you address the rest of what I have written in my papers, articles and numerous comments here and on the PSI forum, because, whilst backradiation certainly exists, it cannot transfer thermal energy to a warmer surface – it can only slow that portion of surface cooling which is itself by radiation. (See my peer-reviewed paper Radiated Energy and the Second Law of Thermodynamics published back in March 2012.) And all that slowing of cooling is only a marginal effect, because the net effect of radiating molecules is to reduce the thermal gradient, and thus also reduce the surface temperature – as real world data shows.

And, remember, there is more incident Solar radiation absorbed on the way down than there is IR absorbed on the way back up. See my comment Your pot only boils at night, Roy in the thread before this.

Physicist Doug Cotton

72. DJC says:

PP The issue as to whether an atmosphere without any water vapour or CO2 etc would be able to radiate is not as simple as you make out. I have looked into this in some detail, even though such an atmosphere does not exist anywhere in our Solar system.

Firstly, if you compare measurements of emissivity for region of high humidity and with those for low humidity, it is very clear that the emissivity is not approaching a limit of zero.

Secondly, if you consider the thermosphere (without water vapour) it clearly absorbs SW radiation and must re-emit it. What happens, I suggest, is that a proportion of molecules that do absorb then have the ability to re-emit at the same wavelength. Now, extend this to the troposphere and there is no reason why a very small proportion of molecules there also act like those in the thermosphere, where temperatures reach hundreds of degrees, but the energy in some molecules could be far more than the mean.

However, it is far more relevant to discuss just a change in water vapour concentration and the effect of such on surface temperature. Contrary to what you say, real world data (such as in the study in the Appendix of my current paper) shows that moist regions have lower mean daily maximum and daily minimum temperatures, contrary to what you and climatologists assume.

73. David Socrates says:

Roy,

Thank you for being an empiricist – a rare beast in the climate change jungle. And thanks also for expressing your query about your findings in a succinct and crystal-clear way. I wish I could say the same for some of the commentators above in this thread who have waxed interminably on their various theories without actually addressing directly the subject of your article.

So here goes.

IR thermometer mechanism
First of all, your description of how the IR thermometer works is spot on. Ironically (in view of some of the wilder comments in this thread) it is a perfect, practical working example of the mutual exchange of radiative energy flow between two bodies. Any body at a temperature above absolute zero will exchange radiative energy with any other body within its field of view (Prevost’s Theory of
Exchanges, 1791). That is exactly what happens in this case between target and detector.

Pointing the IR thermometer at a clear sky

A clear sky (covering the IR thermometer’s whole field of view) obviously does NOT constitute “a solid target object, typically 1 metre away”!! So what is the IR Thermometer to make of such a target?

Well the device will certainly exchange infra red radiation with the GHG molecules in the atmosphere (mainly water vapour plus a much smaller proportion of CO2) and will certainly compute a ‘temperature’ value accordingly. But the figure you obtained of 27F (-2.8C) is not a meaningful atmospheric temperature because:

(i) The device is receiving radiation from molecules in a 3-dimensional volume, mainly within the first few hundred feet of the surface. This will of course increase the computed ‘temperature’ value significantly over what it would be if it were only receiving radiation from molecules actually adjacent to the surface! However…

(ii) It is not receiving radiation from ~98% of the atmospheric molecules within that 3 dimensional volume. So it is grossly under-recording the actual energy content of the atmosphere and therefore under-recording its true temperature.

I guess the effect of (ii) outweighs (i) by a large margin. That is why you get a temperature value of 27F rather than something closer to the 78F ambient that you might expect.

Pointing the IR thermometer at an extended cloudy sky

An extended cloud base (covering the IR thermometer’s whole field of view) also obviously does NOT constitute “a solid target object, typically 1 metre away”!! So what is the IR Thermometer to make of such an target within its field of view several thousand feet above the surface?

Well the device will certainly exchange infra red radiation with the water molecules in the cloud and it will certainly compute a ‘temperature’ value accordingly. The emissivity of low clouds is generally reported to be near 1.0 but, even so, clouds are not true ‘black bodies’ emitting exactly according to the Planck ‘black body’ spectrum. So the energy flow will be somewhat lower than from a true ‘black body’ spectrum and we would therefore expect the IR thermometer to under-record the actual temperature of the atmosphere at that height (as could be measured observationally by a weather balloon, for example).

With a ground temperature of 78F and a lapse rate of 3.5degF/1000ft, one would expect to find a cloud base having atemperature of 41F at a height of (78-41)/3.5 = 10,600ft. If the IR thermometer is under-recording for the reasons stated, the cloud’s true temperature would be higher and therefore the cloud would be at a lower altitude – which I suspect was probably so in your case.

But whatever the exact truth of its behaviour when an IR thermometer is pointed at the atmosphere, it surely puts to bed any idea that there is no such thing as radiation from atmosphere to surface.

74. So firstly I hope people would agree that the quanta of energy leaving a surface cannot depend on the final destination of the quanta i.e. its temperature, material and surface – it only depends on the source material and temperature.
I believe this describes Claes point of view:
The final destination of the radiation determines what happens to the quanta (rejected or absorbed)

where 100C- a very very very! small bit less than 100C 100C+ a very very very! small bit more than 100C
w greater than y
y greater than x
and x greater than z

oven at 101C transfers zero quanta to body at 10000C (equivalent to back radiation)
body at 10000C transfers w quanta to oven at 101C

body at 100C transfers zero quanta to oven at 101C (equivalent to back radiation)
oven at 101C transfers x quanta to body at 100C

oven at 100C- transfers zero quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 101C-

oven at 100C+ transfers x+1 quanta to body at 100C
body at 100C transfers zero quanta to oven at 100C+ (equivalent to back radiation)

body at 100C transfers y quanta to oven at 99C
oven at 99C transfers zero quanta to body at 100C (equivalent to back radiation)

oven at 10000C transfers w quanta to body at 100C body at 100C transfers zero quanta to oven at 10000C (equivalent to back radiation)

at 100C- to 100C+ oven temperature the body quanta changes from outputting x to receiving x+1 quanta
.
Somehow this does not seem to be a smooth or logical transition

Warmists would say quanta emitted from an object depends only on the object and its temperature. the final destination of the radiation is immaterial (well actually the quanta knows nothing until it hits the surface)
The sum of all quanta determines the rate of loss/gain of heat

oven at 101C transfers y quanta to body at 10000C (equivalent to back radiation)
body at 10000C transfers w quanta to oven at 101C

oven at 101C transfers y quanta to body at 100C
body at 100C transfers x quanta to oven at 101C (equivalent to back radiation)

oven at 100C- transfers x-1 quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 101C-

oven at 100C+ transfers x+1 quanta to body at 100C
body at 100C transfers x quanta to oven at 100C+ (equivalent to back radiation)

oven at 99C transfers z quanta to body at 100C (equivalent to back radiation)
body at 100C transfers x quanta to oven at 99C

oven at 10000C transfers w quanta to body at 100C
body at 100C transfers x quanta to oven at 10000C (equivalent to back radiation)

Consider 100C- to 100C+ oven temperature – the100C body quanta output is x and at 100C- it receives x-1 quanta and at 100C+ it receives x+1 quanta

A smooth and logical transition.
I assume that I have this wrong somehow so perhaps using x,y,z Claes could explain your position.

The energies:

http://1.bp.blogspot.com/-73PK0_vgbFg/UWTfiCNpjCI/AAAAAAAABEU/kT5cBjLFOMo/s1600/quanta+vs+temp.jpg

The Energy transfer as I believe Claes would say:

http://4.bp.blogspot.com/-7guLeJTj4U0/UWVPQPzXIkI/AAAAAAAABEs/LNqxJKnDIC0/s1600/slay+flow.jpg

75. David Socrates says:

thefordprefect says, April 15, 2013 at 11:24 AM

Pardon……???

76. Sky temperature and thermal imaging:
http://www.climateandstuff.blogspot.co.uk/2012/12/sky-temperature-and-thermal-imaging.html

Note thge camera expects BB radiation not discrete wavelengths so it gets the clear sky temperature wrong (only GHG radiation from clear sky)

The cloud temperature is more likely to be approx correct as the water droplets will emit grey body radiation at their temperature

Water Vapour and thermal imaging:
http://www.climateandstuff.blogspot.co.uk/2012/12/water-vapour-and-thermal-imaging.html

“Proof” that thermal imagers are calibrated to BB radiation and do not respond to water vapour (much!):

An uncooled bolometer array in a thermal imaging camera is at a temperature of approx 50C.
HOWEVER it can still measure the temperature of an object below -40C. How is this possible if the bolometer did not simply respond to the total flux balance of its IR plate

-273C does not radiate (zero to bolometer)
-40C object still radiates (to bolometer +).
1500C object certainly radiates (to bolometer +).
The bolometer radiates (away from bolometer -).
Just add the fluxes and you can get a 50C bolometer to measure -40C and 1500C (without the need to invent cold rays focussed on the bolometer through the lens!)

77. DJC says:

The Ford Prefect wrote I hope people would agree that the quanta of energy leaving a surface cannot depend on the final destination of the quanta i.e. its temperature, material and surface.

This is totally incorrect and I am getting fed up with people who think they know something about physics airing such nonsense.

Elementary physics tells us that the rate of all three processes, evaporative cooling, conduction and radiative cooling varies according to the temperature difference between the warmer source and the cooler absorbing target.

Back radiation slows the rate of radiative cooling of the surface. It does exist and that’s what it does as I have explained in my March 2012 paper and many comments since.

The processes of non-radiative cooling (which transfer about two thirds of all the energy that goes from the surface to the atmosphere) also slow down when the temperature gap at the boundary reduces, and speed up when that gap increases.

Show me any physics text book which says anything to the contrary!

78. Vincent Gray says:

This has nothing whatever to do with the greenhouse effect. There is a lapse rate. a reduction in atmospheric temperature with height. The atmosphere is heated by convection and latent heat loss, mainly when the sun shines. Each layer complies with the radiation laws, whetever is said by those who cannot understand thermodymanics. Every layer radiates heat upwards and downwards according to the Stefan/Boltzmann equation, modified by an emissivity coefficient.

Clouds obviously shelter part of this radiation.

79. DJC says:

(continued)

All the talk about whether surfaces “know” the temperature of an adjoining atmosphere, or whether photons “know” the temperature of their source is garbage and typical of the comments passed around in climatology discussions.

Consider a long metal rod with one end in boiling water at 100C and the other in water at 50C. A linear temperature gradient will develop in the rod from 100C down to 50C. Now, reduce the 50C to ice at 0C and (provided the rod is long enough) a new steeper thermal gradient develops. My point is that there is a feedback mechanism which causes the hot end to transfer energy at a slower rate. Just like “heat creep” (which explains Venus surface temperatures) all this can be easily explained with the same Kinetic Theory used by Einstein. At least Einstein recognised the importance of this, whereas climatologists completely ignore it and thus do not realise why an autonomous thermal gradient must develop in an atmosphere in a gravitational field.

Evaporative cooling also slows down as the temperature gap at the boundary reduces, because there is also a feedback mechanism at the molecular level.

The rate of transfer of thermal energy by radiation also slows down, but the mechanism is far more complex than the assumed two way heat transfer that dominates climatology energy diagrams etc. I am not going to try to summarise the physical process here, except to say that the vast majority of the radiation from the Earth’s surface is merely immediately re-emitting the energy that it received from the backradiation. As in my March 2012 paper Radiated Energy and the Second Law of Thermodynamics there is only a one way transfer of thermal energy which is quantified by the area between the Planck curves of the warmer source and cooler target. As the temperature gap reduces, so does this area.

80. DJC says:

No Vincent, the atmosphere is heated by a lot more than just convection, and the thermal gradient is not caused by convection as I explain in my February 2013 paper, Planetary Core and Surface Temperatures which is in the PROM menu on the Principia Scientific International website.

About 19% of incident Solar radiation is absorbed by the atmosphere (including clouds) and obviously this goes a long way towards warming the atmosphere. This process is even far more relevant on Venus where about 98% of incident Solar radiation is absorbed on the way down. With only 2% reaching the surface of Venus, how do you suppose convection (resulting from that 2%) is going to warm the whole Venus atmosphere on the way back up? How do you suppose the thermal gradient has developed in the Venus atmosphere when convection is obviously not the cause? How do you suppose sufficient energy gets into the surface of Venus to maintain it at a temperature of over 720K, even at the poles and during the 4 month long Venus night?

The only place you will find a valid explanation is in my latest paper, and I defy you or anyone to point to any alternative explanation that I could not pull to bits. If I sound angry because of all this pseudo physics that others keep airing, it is because I am.

81. DJC says:

And no Vincent every layer radiates electromagnetic energy, not “heat” because electromagnetic energy is not thermal energy: it has to be converted to thermal energy, and that does not happen in a warmer target.

Where I said above that “the vast majority of the radiation from the Earth’s surface is merely immediately re-emitting the energy that it received from the backradiation” I am of course referring to electromagnetic energy. The cooler atmosphere provides electromagnetic energy for most of the S-B “quota” of radiation which the warmer surface emits. Hence the warmer surface does not have to convert so much of its own thermal energy into EM energy. Hence the rate of radiative cooling is reduced.

Remember, however, that the thermal energy transferred by the surface to the atmosphere is only about 15% of the original incident Solar radiation. About 30% of that incident energy is transferred by non-radiative processes from the surface to the atmosphere. Since the 19% of incident radiation absorbed on the way down exceeds the 15% of upwelling radiation, we can easily see why water vapour and carbon dioxide etc have a net cooling effect. But there’s more to it than just that, and we need to consider the effect that they have in reducing the magnitude of the thermal gradient and, consequently the surface temperature – as real world data shows in the study in the Appendix of my latest paper.

• DJC says
because electromagnetic energy is not thermal energy: it has to be converted to thermal energy, and that does not happen in a warmer target.
————–

And what happens to all this non thermalized energy absorbed by the hotter object.

It is energy so there must be an effect when the hotter object absorbs it.

• DJC
If you suggest that a BB absorbing radiation from a cooler object immediately retransmits it then you will destroy BB radiation proportionality to temperature – you have categorically stated that the body will not be warmer when it is hit by radiation from a cooler object.

Yet you have it re-emitting all that “cold” radiation + all the radiation it would for its temperature as standard BB radiation.

i.e. the emitted energy will be greater than for a standard BB at its temperature.

• DJC says:

TFP writes “you will destroy BB radiation proportionality to temperature”

No I won’t and I don’t. Read my peer-reviewed Radiated Energy and the Second Law of Thermodynamics published on several websites (including Talkbloke’s Talkshop) in March, 2012.

When the Earth’s surface is losing two thirds of the thermal energy it transmits to the atmosphere by non-radiative processes, there is not enough thermal energy left for it to transfer as much by radiation as a blackbody would transfer to space.

About 50% of Solar insolation is absorbed by the Earth’s surface, and, then 30% goes back to the atmosphere by non-radiative processes, 15% by radiation and the remaining 5% or so goes through the atmospheric window direct to space. Most of the other 50% is reflected back to space, but 19% is absorbed by water vapour, carbon dioxide etc on the way down before it reaches the surface. Thus this 19% does not warm the surface and so there is a cooling effect. Notice that the 19% > the 15%.

Yes, you can say there is an amount of radiation from the surface which is similar to that of a blackbody, but that does not mean it is transferring thermal energy at that rate from within itself. Is a mirror cooling itself when it radiates reflected electro-magnetic energy? No. And neither is the surface when it “pseudo scatters” backradiation by immediately re-emitting it with exactly the same frequencies and intensities in a resonating process which never involves conversion of its energy to thermal energy.

And, in case you don’t know, radiative flux is not proportional to temperature and never has been. Try the fourth power of temperature.

• DJC says:

PS It’s obvious you didn’t comprehend the following which I quote from the comment to which you replied.

“The cooler atmosphere provides electromagnetic energy for most of the S-B “quota” of radiation which the warmer surface emits. Hence the warmer surface does not have to convert so much of its own thermal energy into EM energy. Hence the rate of radiative cooling is reduced.”

• DJC says:
“The cooler atmosphere provides electromagnetic energy for most of the S-B “quota” of radiation which the warmer surface emits. Hence the warmer surface does not have to convert so much of its own thermal energy into EM energy. Hence the rate of radiative cooling is reduced.”
=====
This is how I understand what you have written:

The low T atmosphere radiates quanta at wavelengths dependant on the GHGs upwards.
None reach the ground.
All eventually reach space at 0K (approx)

The grey body of the warmer ground radiates upwards
Because the GHGs in the atmosphere are radiating quanta at certain wavelengths these wavelengths will be excluded from the once greybody radiation (the spectrum from the ground will not be continuous).
The quanta radiated from the grey body ground will therefore be less by the amount being radiated by the atmosphere.
The ground will therefore cool more slowly.

The ground plus atmosphere therfore radiate upwards as a grey body of temperature T.

If this is not what you meant please have another go!

• DJC says:

The statement “the spectrum from the ground will not be continuous” is not correct. I referred to the full S-B “quota” as represented by the (full) Planck curve. I also referred to the fact that the one way effective transfer of thermal energy is related to the area between the Planck curves – a fcat which physicists and engineers have used for many years. It would be better if you just read the papers I have referred you to, as I really don’t have time to re-write here about 35 pages from the two of them.

• Ok you refuse to correct my statement.

Here you say
“If instead it meets a warmer target, possibly the Earth’s surface, it will just undergo resonant scattering without leaving any energy behind. Eventually it will get to space where it will travel on for who-knows-how-long until it strikes another target, which also may be warmer or cooler. The “temperature” of the radiation is really just the temperature of a blackbody for which the peak frequency would be the frequency of the radiation”

As far as I can see resonant scattering refers to an electron receiving sufficient energy to jump a level in an atom. When this falls back (usually “instantly”) a photon of the same wavelength is emitted. i.e. it will act abit like a mirror and the quantum of energy transported by the photon will simply reflect back from whence it came with no effect on the object it hit.

The energy transported by an 8um IR photon is
The smallest amount of energy (i.e. one quantum) that an object can absorb from IR light with a wavelength of 8µm
Energy quantum = hν
h is known as “Planck’s constant”, and has a value of 6.63 x 10^-34 Joule seconds (Js)
so we need to know the frequency ν
νλ= c
ν = c/λ
ν = (3.00 x 108 m/s)/(8*10-6 m) s-1
plugging into Planck’s equation:
E = (6.63 x 10^-34 Js)*( (3.00 x 10^8 m/s)/(8*10^-6 m) s-1)
E (1 quanta) =2.49×10^-20 J at 8μm IR
i.e. 0.155 electron volts.

In order to be exite an electron to higher orbit the current orbit to next orbit must be greater than or equal to 0.155eV

few atoms have a structure with such low eenergy levels between shells. What happens to the photons that do not hit one of these low energy shell atoms?

Correct me if I am wrong but I thought the energy level had to exactly match the shell to shell level to get absorbed.

• DJC says:

Have you ever considered what happens in reflection? Any frequency is re-emitted. Read the linked paper by Claes Johnson, Professor of Applied Mathematics. You need to think more in terms of the wave-nature of radiation and the concept of resonators. The cooler body’s Planck curve is always fully contained within that of the warmer body. This means that there are always matching frequencies and intensities being radiated by each body and these merely resonate between the bodies: some people suggest they do so as standing waves. If it were not the case and the EM energy were converted to thermal energy, then you could have one-way transfers of heat from cool to warm bodies, such as when radiation penetrates beneath the surface of water and there is no corresponding immediate return radiation that is in any way dependent upon the initial radiation.

Radiation each way is an independent process which must (on its own) obey the Second Law of Thermodynamics.

• DJC says:

Have you ever considered what happens in reflection? Any frequency is re-emitted. Read the linked paper by Claes Johnson, Professor of Applied Mathematics. You need to think more in terms of the wave-nature of radiation and the concept of resonators. The cooler body’s Planck curve is always fully contained within that of the warmer body. This means that there are always matching frequencies and intensities being radiated by each body and these merely resonate between the bodies: some people suggest they do so as standing waves. If it were not the case and the EM energy were converted to thermal energy, then you could have one-way transfers of heat from cool to warm bodies, such as when radiation penetrates beneath the surface of water and there is no corresponding immediate return radiation that is in any way dependent upon the initial radiation.

Radiation each way is an independent process which must (on its own) obey the Second Law of Thermodynamics.

Doug Cotton

• DJC says:

Roy – sorry about the duplication of the above comment. It did not appear to be doing anything when I first clicked ‘Submit’. Please delete one and this comment.

82. David Socrates says:

Keep at it Doug. ðŸ™‚

Your arguments make sense and I am pleased to see you and I are in broad agreement re. energy flow analyses. See my recent TB articles ATE Part I and ATE Part II and my assertion that Atmospheric Thermal Enhancement is a wholly non-radiative phenomenon, despite the fact that I also believe, as you do, that back radiation exists.

• DJC says:

Thanks for the support David. Unfortunately Roger (at Tallbloke’s) banned me last year and has never seen fit to entertain my explanations, such as in my latest paper. This is his loss, for I believe I could help steer his readers to the truth of the matter.

Unless and until someone shows me a more compelling explanation as to how sufficient thermal energy gets into the surface of Venus, then I will stick to my explanation of “heat creep” which has not been published elsewhere to the best of my knowledge. This has huge significance for understanding what happens on Earth and why water vapour has a net cooling effect, that is negative feedback. I have the empirical evidence in my favour on that one.

Until people understand the entropy conditions of the Second Law of Thermodynamics and the consequences, they will never understand “heat creep” and other material in my paper “Planetary Core and Surface Temperatures” and so they will fumble in the dark, quite unable to explain surface temperatures, let alone those in the crust, mantle and core of planets with atmospheres in our Solar system – including Earth. Heat creep explains all these.

83. pochas says:

DJC says:
April 16, 2013 at 5:48 AM

“Unless and until someone shows me a more compelling explanation as to how sufficient thermal energy gets into the surface of Venus, then I will stick to my explanation of “heat creep” which has not been published elsewhere to the best of my knowledge.”

You might consider the fact that convection follows the same physics on Venus as on earth. That is, low density parcels rise and cool, and high density parcels sink and are compressed. The difference is that on Venus the top of the atmosphere is heated vs. the earth where the surface is heated. Does this negate convection or defeat the adiabatic lapse rate? Not at all. The atmosphere still cools at the poles and sinks, to be replaced by warm parcels from the equator, which are replaced by rising parcels from below. This circulation establishes the lapse rate, same as on earth. Convection bands are clearly visible on photos of Venus.

• DJC says:

Contrary to what many climatologists think, compression does not cause a permanent increase in temperature. High pressure does not maintain high temperatures. There are high temperatures in the thermosphere, but very low pressure, so nor does expansion maintain low temperatures either.

The way that climatologists “explain” the thermal gradient (badly named “lapse rate”) is simply incorrect physics. It is established by diffusion of kinetic energy at the molecular level and can occur in a totally still atmosphere, as is almost the case on Venus.

Temperature is only a function of the mean kinetic energy of the molecules, whereas pressure is a function of both the mean KE and the density, as you can see in the ideal gas law. You need to study Kinetic Theory and to understand the entropy conditions of the Second Law of Thermodynamics. When you understand both, then you will understand “heat creep” as in my paper. An increase in temperature always requires additional net energy input.

A key point in my paper is that the same physics applies on all planets, not only in their atmospheres but also in their solid crusts and mantles where there is no convection of course. Diffusion of kinetic energy takes place in a still atmosphere and in solid matter. The lapse rate is not caused by convection, of which there is relatively little on Venus.

84. steveta_uk says:

DJC says “The only place you will find a valid explanation is in my latest paper”.

Can anyone think of a better example of a statement from an arrogant paranoid delusional nutcase than this?

Case closed.

• DJC says:

The challenge is for you or anyone to explain it correctly otherwise. So far no one has since the paper was published. I suggest you are being arrogant in suggesting you can it explain it otherwise, so I throw down the gauntlet to you to do so.

85. Rafael Molina Navas, Madrid says:

For PhysicistPhillipe (and DJC)
Please kindly note that hours ago I posted a repply related to your discussion two days ago where I made reference to Dr. Spencer´s work:
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

• DJC says:

Dr Spencer’s “work” in the linked article is nothing but seriously flawed physics.

86. coturnix19 says:

2DJC:

The heat creep thing is unphysical. When you have a column of an ideal gas at given temperature, its molecules do not move at certain speed, instead there is a disribution such that some ove slower, others faster. When those faster molecules from the bottom move higher, they indeed slow down, but those slow molecules from the bottom never make it to the top, so that average distribution and velocities stay the same. Temperature does not change, instead the pressure drops.

That is for an ideal gas, which behavior the air follows closely (except for water vapor component).

The situation you describe would indeed happen IF allof the molecules moved at the same speed. But that would be in a gas in strong diequillibrium.

• DJC says:

Your logic is flawed. You are assuming that there can exist an entropy gradient in a gravitational field, but the Second Law of Thermodynamics rules that out.

This is another major problem in climatology’s version of “physics” – namely that they totally ignore the entropy requirements of the Second Law of Thermodynamics.

I suggest you read the relevant sections of my paper, including the empirical evidence and the study in the Appendix which confirms the fact that radiating molecules (such as water vapour and CO2) decrease the thermal gradient by intra-molecular radiation, and thus lead to the observed lower daily maximum and minimum temperatures in regions which are more moist. If there were no gravitationally induced thermal gradient which is then reduced in magnitude by water vapour (ie wet adiabatic lapse rate is less) then you cannot explain observed surface temperatures with any other valid physics. Try if you wish and I’ll pick it apart.

87. Wagathon says:

We should bring this down to earth.

Point the infrared thermometer at a steak in the freezer. It’s colder than when pointed at the coffee pot.

Point the infrared thermometer at a spot on a granite counter top. Try it again with a steak held over the spot. No difference. Try it with the coffee pot held over the spot. Still no difference.

Drag the granite counter out of the house. Point the thermometer at the spot with the clear sky over it. Try it with the cloud over the spot. it’s colder! The warm cloud doesn’t seem to be heating the spot on the granite anymore than the coffee pot did.

What does any of this have to do with a greenhouse?

• DJC says:

Yes. It is hard to get across the point to Roy and climatologists in general, that, just because there is radiation from point A to point B, it does not mean there is a heat transfer. The electromagnetic energy in radiation is not the same as the kinetic energy which is related to temperature.

Radiation from a cooler source can do only one thing, namely slow the rate of radiativecooling of the warmer target.

It can never …

(a) slow the rate of non-radiative cooling which, in the case of Earth’s surface, is about two-thirds of all surface energy transferring to the atmosphere;

(b) transfer thermal energy into the warmer target, thus increasing its temperature.

The only valid physical phenomenon wherein a cooler source can transfer thermal energy to a warmer target is the non-radiative process of “heat creep” which explains how the required energy gets into the surface of Venus and all planets with atmospheres.

The explanation is in my paper Planetary Core and Surface Temperatures which is up for worldwide open review in the PROM menu on the Principia Scientific International website. Any one of you, Roy included, is welcome to submit any criticism thereof to our CEO John O’Sullivan. I’m still waiting.

Doug Cotton

• Wagathon says:

Do I need “a gravity effect” or a “radiative greenhouse effect” to understand why my face is cooled by putting my hand between it and the sun?

88. DJC says:April 16, 2013 at 4:57 PM
… You need to think more in terms of the wave-nature of radiation and the concept of resonators.
—————–
tfp: Please tell me what YOU believe resonant scattering is. I’m sure my very basic description was ok.
—————–

djc: … This means that there are always matching frequencies and intensities being radiated by each body and these merely resonate between the bodies:
——————
tfp: GHGs only radiate at narrow band frequencies these have to match the energy bands between shells for them to be absorbed and re-radiated. 0.155eV is a very low band gap
what molecules/atoms have matchingband gaps in water/ground?

You seem to be suggesting that the radiation emitted from the cooler body simply bounces(?) “resonates” between the hot and cold bodies.

Since the cold body continually emits quanta and you suggest this bounces without any effect (does not dissipate) then it seems to me that the space between cold and hot bodies must eventually contain an awful lot of quanta!

—————–
djc: to warm bodies, such as when radiation penetrates beneath the surface of water and there is no corresponding immediate return radiation that is in any way dependent upon the initial radiation.
—————–
tfp:
This is a problem with your theory!
I would suggest that quanta from a cold object entering the water adds to the energy content. Conduction evap. radiation subtract from the enerergy content – the energy remaining gets thermalised and the water warms. this sets the radiation/evap/cond losses.

• djc:
I do not think resonant scattering is really what you want to invoke for your theory. Perhaps there are other mechanisms for reflecting “cold” quanta?

http://www.dark-cosmology.dk/~pela/Encyclo/RT.html
Resonant scattering
If a photon’s energy matches (almost) exactly the energy difference between the ground state and the lowest (“first”) excited state of an atom, then when the atom de-excites, another photon with (almost) exactly the same energy is emitted, but in another direction. This is called resonant scattering, because the photons is in resonance with the atomic transition, or the line. If the photon has a slightly too short or long wavelength, it will simply pass right through the atom.

The most famous example of this is the hydrogen Lyman α line. This photon is abundantly emitted from neutral hydrogen, which comprises more than 90% of all atoms in the Universe. A photon initially emitted in the center of a cloud of atoms with which it is in resonance, cannot travel very far before it is absorbed and instantly re-emitted in another direction. Thus, it may take a very long time to escape the cloud.

However, the wavelength of the photon may slowly change after all: A photon with a shorter wavelength (blue in the figure on the right) is scattered, if the atom is moving away from the incoming photon with a velocity u such that in the reference frame of the atom, the Doppler shift has shifted the wavelength of the photon close to resonance. Being emitted with same wavelength in the reference frame of the atom, if the direction is opposite u, then in the reference frame of an external observer, the photon will seem to be emitted with a longer wavelength (red).

• Hmmm!

Photons carry no temperature information just wavelength. So how does your hot black body know that the long wavelength radiation from a hotter black body is not simply the short wavelengths from a cooler black body.

Your theory seems to suggest thnere is an awful lot of sorting of photons occurring in absorbtion/scattering/reflection. I cannot understand how this occurs.

Please can you provide a more scientific explanation than in your 2012 document. I am really getting confused!

89. Rafael Molina Navas, Madrid says:

DJC says:
“Dr Spencer’s “work” in the linked article is nothing but seriously flawed physics”, referring to
“Yes, Virginia, colder objects can make warmer …”
In another Dr. Spencer´s article in October 2010 (linked below) I wrote (slightly edited with details corrected after subsequent discussion):

“If we have only a warm object, it produces IR radiation at a rate proportional to its temperature (actually to its fourth power, but expresed in ºK …) and it will get cooler at a certain speed. Let us leave it getting cooler for an hour. If, starting again with the same object at the same initial temperature, we approach another colder object (but warmer than the air that was previously there), this will also produce radiation, at a lower rate though (but more than mentioned air). Part of this radiation will reach the first object, and part of its energy will be absorbed as heat. Nobody says that the temperature of the first object at that moment will increase. But in the first object, although emiting more radiation than the colder one, as it´s receiving additional heat, the cooling speed will diminish. And if we keep them so for an hour too, surprise surprise, the final temperature of the warmer object will have increased … in relation to its temperature after an hour when it was alone. And its mean temperature too. All this due to the sheer presence of the colder object”.
It´s naïve for me to hope you´ll understand my explanation rather than Dr. Spencer´s, but please kindly give it a go. When first posted, there were several well considered bloggers who saw “my” point (f.e.: Christopher Game)
By the way, my explanation was previous to my reading of “Yes, Virginia, …”

90. DJC says:

TheFordPrefect

“must eventually contain an awful lot of quanta”

No – there is no energy gain being claimed anywhere, just as there is no energy gain when light reflects back and forth between two mirrors. Go and study Prof Johnson’s “Computational Blackbody Radiation” paper as he was the pioneer in these discussions of resonance.

“Photons carry no temperature information just wavelength.”

Very wrong – Wien’s Displacement Law states temperature is proportional to the peak frequency in the Planck frequency distribution for any such spontaneous radiation.

GHGs only radiate at narrow band frequencies

Yes, which is the main reason they are like a picket fence with most of its pickets missing, standing up against the full spectrum radiation from the surface and thus having very little effect on the rate of radiative cooling. I should have made it clear that I was talking about true blackbodies radiating towards each other.

“I would suggest that quanta from a cold object entering the water adds to the energy content”

The whole point of my paper “Radiated Energy and the Second Law of Thermodynamics” is that your statement is false. The EM energy is not converted to thermal energy, and that is how and why the Second Law of Thermodynamics applies for radiation.

If the photon has a slightly too short or long wavelength, it will simply pass right through the atom.

Yes, but within the Earth’s surface it will resonate eventually with one of its own kind. Then random walk motion begins and probability theory says it will exit the Earth’s surface sooner or later, just as it will get to space sooner or later.

(blue in the figure on the right)

Might be best if you write your own stuff here rather than copying and pasting without acknowledgement.

“I am really getting confused!”

Yes I can see you are confused and that is because you need to stop thinking of radiation as a string of photons which, like little hand grenades, add thermal energy to everything they strike in violation of the Second Law of Thermodynamics. It doesn’t work like that. Try focussing the backradiation at night with a huge lens and see if it warms anything. After all, it has more photons than the sunlight which reaches us during the day – apparently /sarc.

Are you now going to try to tell me that on Venus (where the surface receives less than 10% of the incident Solar radiation that Earth’s surface receives) that backradiation somehow multiplies that energy by about a thousand so as to produce over 16,000W/m^2 – as would be necessary to maintain the Venus surface temperature?

I will continue discussion with you if and when you satisfactorily answer the question I have posed many times: “How does sufficient thermal energy enter the surface of Venus (even at the poles) to heat it and maintain it a temperatures above 720K everywhere? The answer is in my February 2013 paper “Planetary Core and Surface Temperatures” in the PROM menu at Principia Scientific International.

91. DJC says:

Rafael

You say “Part of this radiation will reach the first object, and part of its energy will be absorbed as heat.”

That is your assumption but it is incorrect for reasons explained in my March 2012 paper.

There may not be any immediate radiation the other way. If radiation from a cooler source were to penetrate beneath the surface of water, that layer of water does not immediately radiate back.

Hence, the first one-way “beam” of radiation is an independent, completed process which must thus obey the Second Law of Thermodynamics.

What you say violates that law and is thus incorrect.

Radiation cannot transfer thermal energy from a cooler source to a warmer target, such as a layer of water below the surface which cannot radiate back in any dependent way.

And, by the way, I can see you are not a physicist because you use the term “heat” incorrectly. Heat is a transfer of thermal energy, not energy itself.

• Rafael Molina Navas, Madrid says:

Dr. Spencer, as well as my example, say “objects”. It will take longer time to warm water surface, but eventually the same will happen, at least as an average if currents mix the water …
You are right I´m not a physicist, but my Spanish degree is equivalent to Naval Architect and Marine Enginneer (Engines, Boilers … and radiation, Thermodynamics…)
It´s rather a language matter: in Spanish we use “calor” mainly for both high whether temperature (“hace calor”=it´s hot) and “energía calorífica” o “térmica”, and I used “heat” …(to heat: “calentar”, with the sense of transfer, but not heat as noun)
I remember now Christopher Game told me the same in 2010.
Regarding rest of things in your posts, I have to read them more quitely, due to both facts of not beeing scientist and not beeing English my mother tongue.

92. DJC says:

The most important thing to understand about the Earth’s climate is that it is primarily controlled by the level of Solar radiation that is absorbed by the atmosphere and the surface.

The level of radiative flux varies slightly in natural cycles, and variations in albedo can also play a part.

What is not controlling surface temperature is any trapping of thermal energy. In fact, more energy is absorbed by the atmosphere on its way in from the Sun than on its way back from the surface. There is a strong propensity for radiative equilibrium to be maintained during extended periods of fairly level temperatures, such as from 1998 to the present. So I say that if anyone thinks there has been net trapping of energy in that period they have no valid data to back up their claim.

As explained in the above comments, radiation from the cooler atmosphere can only slow the rate of surface cooling which is itself by radiation. When two blackbodies radiate towards each other, the Planck curve for the cooler one is always contained within that for the warmer one. This means that there is the potential for resonating at all frequencies and intensities in the cooler Planck curve. But those frequencies and intensities which correspond to the area between the Planck curves do not resonate and, instead, have their electro-magnetic energy converted to thermal energy in a cooler target.

Now, about two-thirds of the thermal energy transferred from the Earth’s surface to the atmosphere is transferred by non-radiative processes. Radiation from the cooler atmosphere has no effect on these rates of cooling because its energy was never converted to thermal energy in the surface.

If radiative cooling is slowed then non-radiative cooling has a propensity to accelerate, because, as is well known in physics, it will do so if the temperature gap widens.

What sets the surface temperature, however, is non-radiative diffusion of kinetic energy setting the thermal gradient in the troposphere. Radiating molecules then reduce the magnitude of the gradient by about a third. Because radiative balance must still be maintained, the less steep gradient necessitates a lower surface temperature. So drier places on Earth are warmer than similar more moist places because the gradient above the dry places is steeper. This is confirmed by temperature data.

• Wagathon says:

The air in Antarctica is very dry…

• DJC says:

There is virtually no thermal gradient above the South Pole, temperatures being roughly -50C at the surface and at the tropopause. This is to be expected because of the “funnel effect” which drives winds downwards such that they then spread out in all directions at the Pole and start out back as Polar Easterlies. Strong wind will always dominate over diffusion and convection processes. This is covered in Section 13 of my 20 page paper “Planetary Core and Surface Temperatures” at PSI.

93. Wagathon says:

Abstract… Average annual balance of the thermal budget of the system Earth-atmosphere during long time period will reliably determine the course and value of both an energy excess accumulated by the Earth or the energy deficit in the thermal budget which, with account for data of the TSI forecast, can define and predict well in advance the direction and amplitude of the forthcoming climate changes. [Abdussamatov, HI. Bicentennial decrease of the total solar irradiance leads to unbalanced thermal budget of the earth and the little ice age. Applied Physics Research. Vol.4, No.1, pp. 178-184 (2012)]**

___________

94. @djc
tfp: – I referenced my quotation but here it is again:
http://www.dark-cosmology.dk/~pela/Encyclo/RT.html
————-
djc:
“must eventually contain an awful lot of quanta”
No – there is no energy gain being claimed anywhere, just as there is no energy gain when light reflects back and forth between two mirrors
————-
tfp: I thought lasers worked by inputting energy to a photon stream bouncing between mirrors.
A cold object will continue emitting quanta into your reflected light between mirrors

You state it cannot go anywhere so what happens to it?
————–
djc: “Photons carry no temperature information just wavelength.”

Very wrong – Wien’s Displacement Law states temperature is proportional to the peak frequency in the Planck frequency distribution for any such spontaneous radiation.
————–
A photon equates to a single wavelength (frequency) Wien does not come into this. A photon has an equivalent wavelength. a quantum from 8um wavelength of light is equivalent to 2.49×10^-20 J i.e. 0.155 electron volts its dimension is joules not “wavelength”.

However this is not what I want answered. You state categorically that a cool BB cannot heat a warm black body

Take 3 black bodies all emit a BB spectrum of radiation with wavelenths extending to infra red in ALL cases.
BB1 is warmer than BB2 is warmer than BB3

The warm BB1 will heat the cool BB2 and BB3 – agreed
But the warm BB1 radiates in the region 2 to 50um and of course much shorter peaking at 500nm say.
A cool BB2 radiates at 2 to 50um but has a peak at 1um
A cooler BB3 radiates at 2 to 50um but has a peak at 4um

In open space at 0K you have 3 sources of 2 to 50um IR some of this will be absorbed on 2 objects (from BB1) some will be absorbed on 1 object (from BB2) and some will be absorbed on none (from BB3).

How does each object know which quanta to absorb. What marks the BB3 quanta as cold and therefore ALL of it reflectable/resonatable.

What marks the radiation quanta with its temperature source?
What enables the boddies to reject the quanta from cooler sources?

95. Rafael Molina Navas, Madrid says:

DJC says:
(April 17, 2013 at 6:32 AM)
“Hence, the first one-way “beam” of radiation is an independent, completed process which must thus obey the Second Law of Thermodynamics.
What you say violates that law and is thus incorrect”.
You are a physicist, I presume. But with wrong basic ideas.
Imagine we have the colder object far from the warmer one, and from any other.
We know it emits radiation, the higher its temperature the more radiation emitted.
If we approach it to the warmer one, how and when does “he know” it? Is “he” able to “think”: “I have to stop emitting radiation towards that warmer object: my weaker radiation can´t compete with it …?”
Second Law of Thermodynamics refers to NET transferred thermal energy, not to any one-way beams of radiation …

96. DJC says:

My response is in my March 2012 paper. You completely misunderstand what I have explained there. Smaller distances increase the intensity – try getting close to the Sun! So the area between the Planck curves reduces and less thermal energy is thus transferred out of the warmer body.

The whole issue of the effect of radiation actually has little to do with surface temperatures which are set by non-radiative processes, just as are temperatures in the crust and mantle – all obeying the same physics and having thermal gradients calculated from the same formula, whether in the gaseous atmosphere or the solid regions below the surface.

When you can explain how sufficient energy gets into the surface of Venus, then you will have a correct understanding of the relevant physics relating to “heat creep.” For new readers, this is in my paper “Planetary Core and Surface Temperatures” in the PROM menu at Principia Scientifiic International. There is a brief summary in this comment on the latest thread here.

97. DJC says:

TheFordPrefect

Sorry – but your cited reference makes it clear that absorption is not always accompanied by electron excitation. You need to understand that it is a complex process converting electro-magnetic energy striking an atom into kinetic energy in an associated molecule, the latter having several degrees of freedom. My point is that when radiation from a cooler object strikes a warmer target there is no conversion of EM to KE because, if that happened, the KE could then transfer by non-radiative processes to other molecules and, bingo, you have a violation of the Second law of Thermodynamics with no dependent radiation the other way. You have to consider independent processes, and one-way radiation is just such a process.

In your three body example you must remember that a Planck curve for a warmer body always totally envelops that for a cooler body – that’s the mathematics of it. Then this means that there will always be matching frequencies and intensities between two perfect blackbodies. Just as temperature is a statistical mean, not the KE in one molecule, so too is the frequency distribution in radiation not the same as the frequency you somehow wish to allocate to an individual photon. As I keep saying, continually ignoring the wave nature of radiation gets you into confusing thoughts – as Prof Claes Johnson also says in his “Computational Blackbody Radiation” which it is time you read.

I am not going to discuss this further without your reading that key paper which was of course cited in my March paper. Reading a paper without reading important cited references therein is hardly valid study thereof. If you dispute Prof Johnson’s mathematics (he being a Professor of Applied Mathematics) then argue with him on his blog.

98. djc
Sorry – but your cited reference makes it clear that absorption is not always accompanied by electron excitation. You need to understand that it is a complex process converting electro-magnetic energy striking an atom into kinetic energy in an associated molecule
—————-
tfp:
you were claiming resonant scattering. this does not include kinetic energy it is [absorb photon – emit photon]
What are you now claiming reflects all those cold waves back the way they came?
———————
djc:
In your three body example …As I keep saying, continually ignoring the wave nature of radiation gets you into confusing thoughts
————————
tfp:
3 black bodies 3 temperatures and I talk soley about waves.
each bb radiates in the 2um to 50um range – I’m sure you cannot deny this?

3 bb equally spaced 1 light second apart each emmitting 2 to 50um. Space now has a mixture waves from 3 sources.
e.g.
1 2.5um wave can be absorbed by 2 bodies
1 2.5um wave can be absorbed by 1 body and “reflected” by 1 body
1 2.5um wave can be absorbed by none but reflected by all.

So I ask the simple question – what is the difference between each of these 2.5um waves?

The do not know where they came from BB or some generator – so the shape of the bb source is irrelevant. They are just waves of the same frequency travelling in a vacuum and therefore are tranferring the same quanta.

• DJC says:

The whole point about blackbody radiation is that it is not a wave with a single frequency. The radiation from the Sun has a huge range of frequencies, for example. The Planck curve also has intensities at each wavelength (frequency) and that plays a part as is explained in the first few sections of my paper Radiated Energy and the Second Law of Thermodynamics – so my response is in that peer-reviewed paper which my reviewers understood, even if you don’t because you haven’t really studied it and the linked references. There are also many comments of mine on the Tallbloke thread for the paper which might help your understanding.

When you can answer the question which my latest paper answers “How does sufficient energy get into the surface of Venus?” then you will have a far more accurate understanding of Planetary Core and Surface Temperatures.

If you only think in terms of radiation you will not get far with understanding the thermal gradient in the atmosphere, let alone those in the crust and mantle. And nor will you understand why water vapour cools rather than warms.

• Why are you talking water vapour, thermal gradients, crusts …
I have looked at CJ document which rambles on and on with many pictures of people with strange hairstyles. Would you care to summarise his science on my 3 body problem above.

My 3 body situation is what I want answers for.

I would modify this to make it more interesting by placing them 1 light second apart in the centre of a 100% reflective evacuated sphere 4 light seconds diameter at -273°C (this of course is not really relevant as 100% reflective sphere will not radiate)

Each BB radiates in the IR Range

The interior of the sphere will contain IR from all BBs being reflected around until absorbed. It will also have all the shorter wavelengths reflecting around, but lets just limit this to just the IR which all the bodies emit.

“As a corollary, the absorptivity of spontaneous radiation from a cooler source to a warmer
target must be zero” so this implies:
Some will not get absorbed because they originate from the coldest body
Some will get thermalized on the coldest BB but not on the Hottest BB.
some will get thermalized on warm and cold BBs

The source of the IR is not known (it has been bouncing around for over 1 second)it cannot be marked as originating from cold or hot (we are only looking at the long wave end of the BB radiation – not its peak) before hitting any object it would have been in transit for at least 1 second.

What tags it Cold IR, Warm IR, or Hot IR?
What physical property of the BB allows each to know the source of the IR and reflect or absorb it?

===============================
=========================
CJ’s paper.
The increase of the cut-off frequency with temperature can be under-stood as an increasing ability to emit coherent waves with increasing temperature/excitation or wave amplitude. At low temperature waves of small amplitude cannot carry a sharp signal.

tfp: I wonder how radio works!!! I wonder how older fibre optics (using IR) worked!!!!

CJ’s paper
An IR camera has sensors reacting to different frequencies of incoming infrared IR radiation outside the visible spectrum, and thus can produce a photographic image of an invisible object. By artificially coloring the image with red representing shorter IR wavelengths and blue longer wavelengths, and connecting frequency to temperature by Wien’s displacement law, the image can be read as a thermometer acting at distance

What the instrument measures is frequency which can be connected to temperature reflecting some reality, but the
connection to massive “downwelling” radiative energy, is just postulated through a certain formula and our analysis indicates that this is fiction rather than reality.

TFP:
What utter hogwash. Most IR cameras work by energy balance –
a bolometer (TOTALLY ignores frequency – it is just a thin black body receptor) reaches an equilibrium temperature from energy in and energy out per unit time. This temperature is measured and colourised by electronics.
Energy in comes from many sources – the image, the lens, the camera body, thermal conduction from back plate to sensor etc. (Anything above -273C!) energy out will be according to the BB radiation to surroundings and thermal conduction to anything at lesser temperature.
The image (IR focussed through the lens) will be the variable which adds energy to the sensor.

The camera expects a BB source and will not accurately measure air temperature as the only IR source is discrete bands from GHGs. A BB will add more energy than a discrete frequency.

For me to wade through mathematics I do not fully understand, I require the author have a basic knowledge of the real world!

• DJC says:

If you understood what I have been saying about the one way transfer of thermal energy by radiation, your three body “problem” is elementary.

I keep telling you – the energy transferred is represented by the area between the Planck curves and the thermal energy only ever transfers from the warmer BB to the cooler BB. Not all of it transfers – only that represented by the area between the Planck curves. The cooler BB keeps radiating its quota as per its Planck curve, but because it receives sufficient energy from the warmer BB, it does not use its own energy and so it does not cool, but in fact warms assuming there are no other non-radiative losses – ie assuming both are in space. That is the difference between true reflection and this “pseudo scattering” which just looks like scattering, but is really immediate re-emission of that radiation from the warmer BB which is represented by the area under the cooler BB’s Planck curve, that being totally enveloped by the warmer BB’s Planck curve. Because of this enveloping, it is possible for all frequencies (up to the intensities under the cooler Planck curve) to resonate.

Are you not aware that the quantification of heat transfer by radiation is common knowledge in physics and engineering? They work out heat transfers using the integration (expressed in the Stefan-Boltzmann Law) where that integration represents the area under the Planck curve. They then take the difference of the areas under the Planck curves for each body and of course that gives the area between the Planck curves.

• I take it you therefore disagree that each body is emitting radiation at IR wavelengths and that this radiation is just part of the BB spectrum?

If I use 3 diffraction gratings on the energy emitted from 3 BBs do I not see a spectrum of wavelengths from each?
Can I not then use a mirror on the longer than 2um IR end of the diffraction grating outputs?
Can I not then reflect all three IR only spectrums (all three being longer than 2um) onto a warm body?
How can this warm body then reject the longer than 2um that comes from a colder body but accept the longer than 2um radiation from the hotter body?

• DJC says:

Each true blackbody would be radiating its full spectrum of frequencies as determined by its Planck curve. I’m sure I have made that perfectly clear. The energy in that radiation gets “dropped off” in various portions when (and only when) the next target is cooler than the previous one. That energy which is not dropped off (and converted to thermal energy in the process) is passed on as part of the target’s quota of radiation. The process continues until the energy reaches something in outer space that is near the background radiation temperature in space – about 2.7K some say.

If you think you can violate the Second Law of Thermodynamics, then go ahead and make a fortune generating energy out of nothing. If, on the other hand, you actually agree with me and Clausius (circa 1850) that heat cannot be transferred from cold to hot bodies (in the same horizontal plane in a gravitational field) then I have one simple question for you …

If the electromagnetic energy in the incident radiation from a cooler source were in fact converted to thermal energy, then how would that additional kinetic energy “know” or “remember” that it came from radiation and it (supposedly) now has to be converted back to EM energy and go back as outward radiation so that the Second Law is not violated? If it warmed a layer of water beneath the surface of a lake the energy may well be released hours later by way of evaporation. Why do you have so much difficulty in understanding that the original one-way passage of radiation would be a completed independent process if its energy were converted to thermal energy? Clausius spoke about processes. Are you going to throw out over 160 years of physics since then?

What I am explaining is a dependent process whereby the incident radiation has no choice but to be re-emitted immediately in the same form, rather like diffuse reflection (in all directions) but not technically the same process.

Anyway, I’m sure other silent readers understand that the laws of physics must be obeyed. There’s no way of “excusing” the process based on “net” energy transfers, be they radiative or non-radiative.

99. Rafael Molina Navas, Madrid says:

DJC says:
(April 17, 2013 at 7:07 AM)
“There is a strong propensity for radiative equilibrium to be maintained during extended periods of fairly level temperatures, such as from 1998 to the present. So I say that if anyone thinks there has been net trapping of energy in that period they have no valid data to back up their claim”.
How could then be explained the huge increase in Global Oceanic Heat Content after 1998:
http://www.climate4you.com/images/NODC%20GlobalOceanicHeatContent0-700mSince1955%20With37monthRunningAverage.gif
(sorry: “Heat” is used by the author …)

100. DJC says:

The basis on which energy in the oceans is estimated is seriously flawed, and measurements only go to 700m depth and by no means cover all regions on the globe.

How do you explain increased energy in the oceans when sea surface temperatures have not risen at all since 1998? At least we know the energy in the very top layers of the ocean has remained static simply because the surface temperatures have done so. Perhaps some energy has risen in currents from depths below 700m, or perhaps the measurement are just insufficient to prove anything. Whatever has happened certainly has absolutely nothing to do with carbon dioxide levels.

101. Rafael Molina Navas, Madrid says:

It´s not true that “sea surface temperatures have not risen at all since 1998”: http://www.climate4you.com/images/HadSST3%20GlobalMonthlyTempSince1979%20With37monthRunningAverage.gif
And the continuous evaporation of sea surface water, especially when windy, keeps it normally cooler than just below, as least as an average …
And below 700 m is normally cooler …
In any case it is NOT proper of a physicist to say “Whatever has happened certainly has absolutely nothing to do with … ” (!!!)

• Doug Cotton says:

There is clearly a maximum around 1998 – maybe you are just looking at the 37 month running averages.

My paper and other investigations indicate that carbon dioxide does not have any net warming effect, so why should I not say that any warming cannot be due to carbon dioxide? You need to look for other reasons which have a sound basis in physics, because no theory in physics and no evidence in the real world shows any reason to believe that it is/was carbon dioxide causing the observed warming.

102. DJC says:April 18, 2013 at 10:01 PM
Each true blackbody would be radiating its full spectrum of frequencies as determined by its Planck curve.

If you think you can violate the Second Law of Thermodynamics, then go ahead and make a fortune generating energy out of nothing.
———-
tfp:
You have an understanding of the 2nd law most scientists would understand differently (see science of doom) so for me it is most important that you answer my restatement of an experiment:

Lets restate my question again
Take 4 black bodies at temperatures of 6000K, 6001K (two off), and 6002K.
put 3 different temperature black bodies in separate boxes each with with one opening
Put a difraction grating on each opening (to split the bb radiation into wavelengths), then separate out the IR using a mirror placed to reflect IR only.
Presumably no problem so far (IR was discovered by placing a thermometer in the spectrum beyond red after all)?

You now have 3 streams of IR (all the same frequencies/wavelengths) the stream from the hotter BB will contain more photons/quanta per second than cooler bbs but this will be the only difference.

Now join the IR streams together to converge on the 4th BB.
Again no problem?

What happens to the IR. Is it
absorbed (tfp=yes)
reflected (tfp=no if a true BB)
absorbed if it originated from a hotter body and “reflected” if it originated from a colder body i.e. it conforms to your theory.

If it conforms to your theory then please explain how the 4th body knows to reject some of the IR but not other IR of the same wavelength.

My theory is that all the IR gets absorbed on the 4th BB and the 4th BB gets warmer until the BB radiated equals the IR received per unit time. It will not thermally runaway.

So simple and it explains how thermal imaging cameras operate, how IR thermometers operate and does not destroy the 2nd law!

When water is penetrated by SW light (uv/blue but not IR) it simply warms according to the absorbed energy. The warmed water may at some point reach the surface and cause additional evapouration/conduction and IR emissions or it may stay in the depths conventional understanding of the second law allows this.

• DJC says:

(1) You have introduced a Maxwell demon which is refuted here.

(2) Furthermore you don’t define what you mean by “absorbed” – does it involve the conversion of electromagnetic energy to thermal (kinetic) energy or not in your mind?

(3) My discussion will only continue with you when you are the first in the world to find any error in my latest paper. See this comment and reply on that thread please.

103. DJC says:

You still don’t understand that intensity does matter. Hotter bodies can radiate the same frequencies as cooler bodies, but the intensity will always be greater, so not all of the radiation resonates – that above the Planck curve for the cooler body has its energy converted to kinetic energy. You still display absolutely no understanding of what I wrote in my paper last year.

What “explains how thermal imaging cameras work” ??? – and I assume you also mean bolometers. Perhaps you’d better read as far as the Appendix in my paper written over a year ago.

But as you won’t read it, I’ll copy and paste it here for the convenience of others to help them understand just how far from reality is your physics.

Q.4 How can an Infra-red thermometer measure cooler temperatures?

The original types of infra-red thermometers measure the frequency of the radiation, and then calculate the temperature using Wien’s Displacement Law. Infra-red cameras can do likewise to form an image by representing temperatures with different colours. However, the newer microbolometers have sensors which warm or cool at different rates, and these rates are used to determine temperature. As explained in the last paragraph of Section 5, radiation from another body at a slightly lower temperature can cause the rate of heat transfer from the warmer body to vary as the temperature difference between the two bodies varies. The instrument’s sensors are warmed (using electric input) but while they are warming they are also radiating energy to the object whose temperature is to be measured. Such radiation will reduce the rate of warming, so that net rate of warming will be affected by the temperature of the object because the energy transfer rate from the instrument to the object varies with thearea between the Planck curves.

104. DJC says:

I see that you actually agree with me when you write …

“When water is penetrated by SW light (uv/blue but not IR) it simply warms according to the absorbed energy.”

So at last you agree that the back radiation (obviously all in the IR) is not absorbed by warmer water on the surface because it doesn’t penetrate the water.

Full marks – it is “pseudo scattered” at the surface and thus does not penetrate or get absorbed in the sense of having its EM energy converted to kinetic energy in the water molecules.

105. djc:(1) You have introduced a Maxwell demon which is refuted here.
————
tfp: NO I have not.
There are 3 isolated sources of different but close temperatures.
They MUST create a spectum in the same IR band.
I have just used masks on the specra from difraction gratings that allows only 2um to 50um radiation to pass.

The hotter source will be emitting more quanta per unit time than the cooler.

————
djc: (2) Furthermore you don’t define what you mean by “absorbed” – does it involve the conversion of electromagnetic energy to thermal (kinetic) energy or not in your mind?
————
TFP: absorbed means that the quanta get converted to thermal energy in the solid body they hit
————
DJC: (3) My discussion will only continue with you when you are the first in the world to find any error in my latest paper. See this comment and reply on that thread please.
————
tfp: This is the way to do science??????
————
djc: You still don’t understand that intensity does matter. Hotter bodies can radiate the same frequencies as cooler bodies, but the intensity will always be greater, so not all of the radiation resonates
————

TFP: Are you suggesting that the wavicle at 10um can have different amplitudes. Please show a calculation where a quantum of energy from a wavelength of 10um has a modifier that assigns mor energy if the source was warmer (ignoring relativistic effect caused by vibration of the source atom)

I purposfully put the sources in separate chambers and filtered the IR separately and only combined them on the sink so you could not claim that the wavicles knew where they originated. However you still claim the know that their source was hot or cold!
—————

DJC:The original types of infra-red thermometers measure the frequency of the radiation, and then calculate the temperature using Wien’s Displacement Law. Infra-red cameras can do likewise to form an image by representing temperatures with different colours.
————–
tfp: I would love to see a reference to a camera than can distinguish frequency of maximum radiation and then on a pixel by pixel basis assing it a temperature. Please show a reference.
Using some semiconductors you can “see” wavelengths of a couple of microns but not much responds to longer wavelengths – too little energy to start knocking electrons around.
————
DJC . The instrument’s sensors are warmed (using electric input) but while they are warming they are also radiating energy to the object whose temperature is to be measured. Such radiation will reduce the rate of warming, so that net rate of warming will be affected by the temperature of the object because the energy transfer rate from the instrument to the object varies with thearea between the Planck curves.
————
tfp: this could be an explanation but a little logic shows it to be a shambles.
You are suggesting that something KNOWS the temperature difference between the 2 bodies (perhaps separated by a few microseconds in travel time) then miraculously adjusts the output up if cooler object is focussed.
Much easier to assume the bolometer IR plate temperature is the sum of all input radiation/conduction. If the focussed object is 0K then there will be no energy from the object so the plate will be T0, if the oject is above absolute zero then the sensor plate will be receiving energy from the focussed object and will be at temperature T2 and T2 will be bigger than T1.

106. djc: So at last you agree that the back radiation (obviously all in the IR) is not absorbed by warmer water on the surface because it doesn’t penetrate the water.

Full marks – it is “pseudo scattered” at the surface and thus does not penetrate or get absorbed in the sense of having its EM energy converted to kinetic energy in the water molecules.
————

The effect of LWIR on the water depends on the albedo – reflectivity.
that IR which reaches the molecules of the surface, warms (gets thermalised) the surface to greater than the temperature if there were no GHGs.
This means more emissions and more evaporation. It does not magically get pseudo scattered – which I assume is your new term that you use now you realise that solids do not do resonant scattering under normal circumstances.

107. Max™ says:

Two things:

1. I remain dubious that any bolometer which is capable of measuring the temperature of an object near 0 K is also capable of measuring the temperature of an object near 300 K and/or 1000 K or higher, without steps taken to recalibrate the instrument.

2. What is the emissivity of the air and cloud in the examples given in the top post?

• the thermal camer I have used has a number of ranges. covering -40C to 1500C. The AtoD converter used is limited to 8 bits (256 levels) so range switching is necessary. To go from range to range requires no recalibration.

O2 N2 etch emit no significant radiation. GHGs emit discrete bands of wavelengths broadened by doppler shift and pressure.

GHGs are NOT Black bodies so do not emit continuous spectrums. If the atmosphere were to emit bb radiation then it would pretty much scupper the use of thermal imaging cameras! (all you would see is a haze of air.

Clouds being water droplets would have a similar radiation pattern to water.

I did a video of steam being invisible until it heats a warm object:
http://climateandstuff.blogspot.co.uk/2012/12/water-vapour-and-thermal-imaging.html

108. It’s a pity that DJC has retired hurt!
I would realy like to know how he would explain the operation of a gas laser (eg. CO2)

A 1kw Co2 laser will cut stainless steel 2mm thick (wiki)
the melting point of stainleass steel is 1510C
The output wavelength of a CO2 laser is 10um approx

How does long wave IR at less than 1510C create a temperature of much more than 1510C. That’s an impressive back radiation!!!!

Is it perhaps that the sum of all the photon energy focussed in a small area adds and creates a high temperature?

109. Max™ says:

the thermal camer I have used has a number of ranges. covering -40C to 1500C. The AtoD converter used is limited to 8 bits (256 levels) so range switching is necessary. To go from range to range requires no recalibration.” ~tfp

233 to 1773 K broken down into how many steps?

I assume that it would need to be recalibrated to function down towards 0 K.

O2 N2 etch emit no significant radiation. GHGs emit discrete bands of wavelengths broadened by doppler shift and pressure.” ~tfp

Well, no, that’s not quite true.

There aren’t as many lines in the O2 and N2 spectra, but if you scale by concentration, nothing is very impressive outside of H2O, really.

As for lasers…

Light Amplication through Stimulated Emission of Radiation.

I’ve got a laser pointer sitting here,you can turn it on and stick it up your nose until the batteries die without noticing any increase in temperature, 650 nm wavelength at 5 milliWatts isn’t very impressive, neat looking though!

Simply bouncing photons around between mirrors won’t do anything, you have to excite the medium first (CO2 in your above example) and then stimulate emission of photons by feeding the appropriate wavelength light into the medium and encouraging it to transit to lower energy states after emitting more photons of your desired wavelengths.

A laser has nothing to do with any sort of “back-radiation”, so it isn’t the best example to support your position.

110. DJC says:

There is also discussion of lasers in the FAQ in the Appendix of my first paper, “Radiated Energy and the Second Law of Thermodynamics” so I am not going to get involved in another red herring which has nothing to do with spontaneous black body radiation. Microwave ovens prove that low frequency radiation is not absorbed by most substances – heating only occurs because whole water molecules are caused to flip in synchronisation with each passing wave.

These are the questions climatologists can’t answer …

(1) Why does real world data indicate that inland tropical cities with high rainfall have lower mean daily maximum and minimum temperatures if water vapour is supposed to have positive feedback, that is a warming effect?

(2) How does sufficient energy enter the surface of Venus to raise it to temperatures over 720K everywhere and maintain it at such temperatures when we know that radiation from the less hot atmosphere cannot raise the temperature above that from whence it came, and we also know that maintaining high pressure does not maintain high temperature?

(3) How do you explain the less steep thermal gradient in a moist troposphere when the release of latent heat is rarely likely to affect temperatures at the tropopause, and thus unlikely to affect the mean thermal gradient from the surface to the tropopause?

(4) Why does the thermal gradient in the outer crust often exceed 25C/Km when it could not possibly be maintained at anywhere near that gradient all the way through the mantle?

(5) How does sufficient energy get into the cores of planets such as Earth and Venus in order to maintain them at the estimated temperatures?

If you have read many of my previous comments, you will know where to find my answers in this year’s paper, “Planetary Core and Surface Temperatures” but don’t read it if you would be uncomfortable ridding yourself of naive greenhouse beliefs.

Doug Cotton

111. DJC says:

Roy I just posted the comment below with only a single link to my paper on PSI, but it did not appear. Now I’ll see what happens with the link removed. If it appears OK now without the link, I trust you can see why I thought PSI links were blocked somehow, because this has happened many times.

—————————

And thefordprefect has no valid response to my comments below his in italics below …

“O2 N2 each emit no significant radiation. “

Another old wives’ tale from climatologists. The fact is that they do, because empirical measurement shows that the limit of emissivity as water vapour is reduced towards zero is way above zero, meaning oxygen and nitrogen are radiating significantly.

“If the atmosphere were to emit bb radiation then it would pretty much scupper the use of thermal imaging cameras!

Firstly, this is countered in the FAQ’s in my March 2012 paper – our friend has no idea as to how such instruments measure the varying rate at which their sensors cool (or warm) depending upon whether the temperature of the source is cooler or warmer – all totally in accord with standard physics.

Clouds being water droplets would have a similar radiation pattern to water.

And they have similar absorption “patterns” with 19% of incident solar radiation being absorbed on the way in by clouds and the rest of the atmosphere, but only 15% being absorbed on the way back up – another problem for those believing the old wives’ tale that the atmosphere is transparent to incident Solar radiation and opaque to IR from the surface.

It seems the fordprefect has been totally sucked into believing the old wives tales of climatology and is probably afraid to change any such beliefs by reading contrary explanations of Planetary Core and Surface Temperatures.

112. @max
My comment about back radiation and lasers was not meant to be serious. What is important is that 10um CO2 laser output has a certain energy per quanta set by its frequency. It is generated by a hot but certainly not 1500C gas. Yet it manages to cut steel simply by concentratiun enough quanta in a small area. The nergy is additive.

@DJC
thefordprefect says: April 19, 2013 at 5:36 AM

• DJC says:

They are irrelevant loaded questions which make assumptions I simply don’t accept and are clearly at variance with what I have written. These are not issues which determine planetary core or surface temperatures. As I have said, my response is in my March 2012 paper which you have not made a serious attempt to understand, and nor have you considered what Prof Claes Johnson had to say on the subject in “Computational Blackbody Radiation” and also earlier in this thread.

I am not wasting my time on red herrings which have no significant effect on climate whatever happens.

If you don’t wish to be bound by the Second Law of Thermodynamics, that’s your choice. That law is all I need to answer your questions.

Furthermore, it’s time that you understood what a Planck curve is all about – namely a distribution of both frequencies and intensities. Engineering has been using the area between the Planck curves for calculating heat transfer between two parallel plates with plenty of empirical support. What’s your equation for such heat transfer if you have invented some new physics contrary to what I have quoted in my paper? The heat transfers between your black bodies will always be from warmer to cooler – that’s what the Second Law of Thermodynamics is telling you, and you can go and win a Nobel Prize if you can prove it wrong for radiation.

You will demonstrate a correct understanding of the laws of physics when and only when you can explain …

(1) how sufficient energy gets into the surface of Venus

(2) why water vapour (or rainfall) is negatively correlated with surface temperature

(3) why the thermal gradient in the outer crust and the atmosphere is a little less than the theoretical gravitationally induced thermal gradient.

If you want to make progress in your understanding of the real Solar system and planetary core and surface temperatures, read my paper by that name, because you have a lot to learn from it my friend.

Doug Cotton

• DJC says: April 21, 2013 at 5:15 AM

They are irrelevant loaded questions which make assumptions I simply don’t accept and are clearly at variance with what I have written. These are not issues which determine planetary core or surface temperatures. …
I am not wasting my time on red herrings which have no significant effect on climate whatever happens.
—————–
So you are saying that you are correct and a large percentage of scientist (including Roy Spencer) are totally wrong.

I suggested black bodies at different temperatures in isolating boxes.
shorter wavelengths than 2um are shunted off to space (use a prism if you like) the remaining IR radiation will all be the same frequencies (space got the peak frequency of radiation from the BBs). Hotter bodies will be emitting more quanta at each wavelength than cooler.
combine these IR sources onto another BB

You now refuse to explain how the receiver of this combined IR is capable of differentialing the source temperature in order to reject some and get warmed by other identical IR.

If you are unwilling to explain your theory perhaps CJ can be called in?

I would also like to know how cool CO2 can be stimulated to emit vast quanta of radiation at 10.6um which will burn steel.

113. Max™ says:

Serious or not, you’re presenting it in a horribly inaccurate manner.

If you can find me a laser where 1 kw input produces more than a 1 kw output with negative losses, then maybe it will resemble the magical effects necessary to compare it with GHG effects.

• Rafael Molina Navas, Madrid says:

ABSURD COMPARISON!
I´ll repeat what many times previously said by Dr. Spencer (and certainly by myself – humbly- and others):
GHGs effect is not any “invented” creation of energy from nothing …
For the sake of simplicity, let us consider some land Earth´s surface portion. During morning hours (and part of the afternoon) normally its temperature increases, needless to say that due to direct sunlight. But GHGs decrease the “bottom line” of energy radiated back to space in each hour. They don´t increase any instantaneous ground temperature, but they certainly increase mean temperature over the morning …
And during rest of the day, the ground cools, but at a lower pace due to the same (and starting from a temperature higher than if there were less GHGs in the atmosphere). Two facts that further increase mean temperatures.
WHERE DO YOU SEE A “MAGICAL EFFECT”?

• DJC says:

“But GHGs decrease the “bottom line” of energy radiated back to space in each hour. “

No they don’t. Over 24 hours, the atmosphere and the clouds absorb 19% of incident Solar Radiation (according to NASA energy diagrams) and then less than that, namely only 15% from upward radiation from the surface. In the sunlit hours that 19% is about double of course. So the atmosphere acts like a very effective umbrella, keeping the surface much cooler by day and a little warmer in the early hours of the night. Oxygen and nitrogen play the major role in this blanket effect absorbing at least two thirds of the thermal energy that transfers from the surface to the atmosphere.

When are people here going to stop trying to pull the wool over the eyes of silent readers by using totally false information and physics?

Radiation is not the key player in determining surface temperatures: if it were, then water vapour would have positive feedback, but the evidence is that it has negative feedback. Like to try to prove me wrong?

114. @Max™
This so far is not about the GHE it is simply the incorrect statement that radiation from a cool object cannot heat a hot object.

An excellent article on CO2 laser (old- 1960s?):

“For example, the output power from the 50 mm bore tube may decrease by only 10 per cent as the wall temperature increases 30° to 40° C.”
“Many applications have been suggested for CO2 lasers, but only a few of these have been given serious consideration and analysis. This is partly because little engineering has been done at 10 microns. Photon energy of about 0.1 eV is only about five times room temperature.”

tfp: so gas temperatures are less than 100C But it can cut steel at gt 1500C.
According to DJC you cannot transfer energy from a cooler to hotter body!
—————-

The action is all about power density.
http://www.miracoinc.com/PDFs/Laser_Design_Guide.pdf
shows a min spot size of 0.006 inches dia for a low power CO2 laser.
If one assumes a beam power of 1 watts
the area is of 1.8*10^-8 sq metres
Giving a power density is 548*10^5 watts /sq metre

I wouldn’t try pointing this up your nose!!

115. DJC: double of course. So the atmosphere acts like a very effective umbrella, keeping the surface much cooler by day and a little warmer in the early hours of the night. Oxygen and nitrogen play the major role in this blanket effect absorbing at least two thirds of the thermal energy that transfers from the surface to the atmosphere.
————
TFP: Clouds of course keep the ground significantly warmer during the night. Starry winters night = frost cloudy = no frost.
Thick clouds do cool by reflection during the day – high thin clouds tend to warm.
————–
DJC: Radiation is not the key player in determining surface temperatures: if it were, then water vapour would have positive feedback, but the evidence is that it has negative feedback. Like to try to prove me wrong?
————–
See this post
http://climateandstuff.blogspot.co.uk/2013/02/what-affects-dlwir.html
all data taken from one location.

Shows a definite
increase in DWLIR with Temperature
increase in DWLIR with ULWIR
increase in DWLIR with Absolute Humidity (called water vapour content on plot)

Please note that this is not cloud cover or rain it is humidity i.e.water vapour.

• DJC says:

I don’t dispute any of the content of this comment or the linked study. I refer to the effect of radiation from water vapour and how it has a marginal effect slowing that portion of surface cooling which is due to radiation. But water vapour sets a much lower base supporting temperature because it sets a much lower thermal gradient in the troposphere. That’s what you don’t understand because you refuse to read my latest paper “Planetary Core and Surface Temperatures” easily found on Google.

In the Appendix of that paper you will find my study of mean daily maximum and minimum temperatures for inland tropical cities. (The study is currently being expanded to many more cities.) If you read the whole paper you will find the physics which explains this.

116. Rafael Molina Navas, Madrid says:

DJC.- You yourself say:
(April 21, 2013 at 5:25 AM)
“ … So the atmosphere acts like a very effective umbrella, keeping the surface much cooler by day and a little warmer in the early hours of the night. Oxygen and nitrogen play the major role in this blanket effect absorbing at least two thirds of the thermal energy that transfers from the surface to the atmosphere”, and later:
“Radiation is not the key player in determining surface temperatures”
Of course, main players are Sun temperature, Earth´s distance from it, atmospheric incoming radiation absorbing properties, etc.
But if those properties don´t change as an annual average, or only oscilate around pretty constant values, they ARE NOT key players in determining surface temperature changes.
Anf if other “minor” players continuously change, always in the same sense, they become the key players.

• DJC says:

You are making an assertive statement without foundation. You must be aware of long term natural cycles – Roman warming, Dark Ages cooling, Medieval Warming, Little Ice Age, current warming. There are also shorter natural cycles, the 60 year one being the most obvious, as in the Appendix of my paper “Radiated Energy and the Second Law of Thermodynamics.”

These are all natural cycles and almost certainly governed by planetary orbits which can be shown astronomically to have an effect on Solar radiation levels and other factors (perhaps cosmic ray intensities) which can affect Earth’s albedo and hence the effective level of insolation from the Sun.

It’s all natural. Carbon dioxide has nothing to do with it – other than a net cooling effect (due to reducing the thermal gradient) which I estimate at about 0.002 degree.

Doug Cotton

• Rafael Molina Navas, Madrid says:

Most periods mentioned by you (not the 60 y.) should not be considered “cycles”, unless there were several interrelated maxima and minima. They could be due just to a couple of natural factors in conjunction, that later triggered some feedbacks.
F. e. : medieval warming (not global, actually) most probably due to a maxima of Sun radiation (undoubtly our main heater) and a low volcanic activity.
Can you find any concrete cause of the current warming that could be considered natural, cyclical or not?

• DJC says:

There’s been no “current warming” since the 60 year cycle peaked in 1998. The evidence for such is obvious in the Appendix of “Radiated Energy and the Second Law of Thermodynamics.”

There is absolutely no evidence for anything but natural warming and cooling throughout history. Carbon dioxide can only have a net cooling effect of about 0.002 degree, and water vapour cools by about 4 or 5 degrees. It’s all here and you can produce no evidence to the contrary.

Doug Cotton

117. Max™ says:

so gas temperatures are less than 100C But it can cut steel at gt 1500C.
According to DJC you cannot transfer energy from a cooler to hotter body
” ~tfp

>.>

You’re being deliberately dense, right?

Spontaneous transfers of energy from a cold to warm body are forbidden, a process where the only result is a transfer of energy from a cold to warm body is forbidden, a laser involves using work to excite molecules to a given energy level, then causing them drop back down to a lower level while emitting photons of a desired wavelength.

• A co2 laser emits radiation at 10.6um you cannot force it to emit at ultra short waves where the energy per quanta woud be more.
each quantum is fixed at 1.9*10^-20 joules
each quantum can only be 1.9*10^-20 joules no matter how it is pumped
a 10W beam laser outputs 5*10^20 quanta/second

the gas is below 100C (don’t have definite info but glass tube temp is maintained below 50C) the sum of all its output quanta allows cutting at 1500C

118. DJC says:

No thefordprefect – it is not about “radiation from a cool object cannot heat a warm object” it is about “spontaneously emitted (black body) radiation from a cool body cannot transfer thermal energy to a warmer body.”

Whilst it is true that the Clausius statement is not quite correct for non-radiative diffusion, conduction and convection between different altitudes in a gravitational field, it certainly is true for radiation between any altitudes.

It has never been proven incorrect, but for the above exception in a gravitational field. Gravity, however, has no effect on radiation, only on molecules which have mass moving by convection etc.

So what did the famous Clausius Statement say? I quote below and I consider it not in any way ambiguous …

Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.*

In regard to the exception in a gravitational field, this is of fundamental importance for climate considerations. Just Google “Planetary Core and Surface Temperatures” Cotton and read my February 2013 paper.

That latest paper contains all I have to say.

119. DJC says:

On the subject of lasers

If you were to plot a Planck curve of radiation from a laser it would have an area under it represented by a very thin and extremely high rectangle. The height of that rectangle protrudes well above the height of the Planck curve for the body it is heating. Hence, totally in accord with convectional physics, the heat transfer is represented by the area between the Planck curves which, in this artificially generated, non-spontaneous laser emission is clearly determined by the height (intensity) of the laser emission. This is all in accord with what I have said in my March 2012 paper, “Radiated Energy and the Second Law of Thermodynamics” which contains a comment on lasers in the Appendix. No surprises.

120. Mack says:

I was thinking a little bit about Roy pointng the little infared meter at the cloud and then clear sky. I’ll say yes, the little gizmo gives a true reading of 41F (~5C) for the cloud,and gives 27F (-2.8C) for blank sky. What I think gives that cloud heat is the incoming solar radiation. Roy says the same thing occurs at night, even the lower darker clouds click the meter to a higher temp. Yes, again those night clouds are warmer than clear sky. I think the warmth in these clouds is heat rising from the Earth surface.
So yes, yes, yes there appears to be a “greenhouse” effect at cloud covered night. You can feel it warmer, you can see the “blanket”. A blanket of clouds.
But daytime, cloud cover shields the surface from the sun causing a cooling,a blanket of clouds keeps us cool. Day equals night, ying equals yang, + greenhouse equals -greenhouse ie no greenhouse.
My tuppence worth as an internet trained armchair climatologist is that clouds in simple terms act as a thermostatic governor,a fine tuning, of the Earth’s temp. Their presence by day shielding the surface causing a “cooling” and by night slowing the rate of heat loss causing a “warming”.
As the song goes..clouds have to be looked at from both sides. Looking at them myself, esp the large cumulus, I marvel at this property of water.

121. DJC says: April 21, 2013 at 5:44 PM
On the subject of lasers

If you were to plot a Planck curve of radiation from a laser it would have an area under it represented by a very thin and extremely high rectangle. The height of that rectangle protrudes well above the height of the Planck curve for the body it is heating. Hence, totally in accord with convectional physics, the heat transfer is represented by the area between the Planck curves which, in this artificially generated, non-spontaneous laser emission is clearly determined by the height (intensity) of the laser emission. This is all in accord with what I have said in my March 2012 paper.
——————
Are you suggesting that a photon from a laser at 10.6um contain more energy than a 10.6um photon from a cube of ice and therefore will vaporise water at 100C?

If you are then this to me is wrong. A laser at 10.6um produces many photons at 10.6um but each photon only transports a quantum fixed at 1.9*10^-20 joules
each quantum can only be 1.9*10^-20 joules no matter how it is pumped.
If you suggest that the energy is more, then the wavelength will be less. This is established physics. but we know that the lase output is 10.6um wavelength.

The laser is able to vapourise water because of the energy density of the focussed spot which for a 10W beam laser outputs 5*10^20 quanta/second.

The ice will also output 10.6um photons. These are identical to the photons from the laser. They each still transport just one quanta of energy at 1.9*10^-20 joules

There are far fewer quanta at 10.6um being emitted from the ice. The ice will therefore not vaporise water.

Here’s another one I cannot understand using your theory:

Take a sun. Place a blackbody near it. the black body will warm
Take the same sun and attenuate the radiation (e.g. move it further away – The suns radiation spectrum at the BB will not change – just the number of quanta/second hitting the BB.
At some point the radiation from the sun will start to be reflected (according to your theory). this will suddenly occur when radiation energy from the sun falls below the BB radiation. Does this seem odd to you – the suns temperature has not changed?

I would still like you to provide, here, an answer to my 4 BB question.
http://www.drroyspencer.com/2013/04/direct-evidence-of-earths-greenhouse-effect/#comment-75417

• Doug Cotton (DJC) says:

No – there are many more photons (higher intensity) than would be emitted spontaneously. So the extra radiation cannot resonate. Many comments back I said “intensity matters.”

The radiating temperature of the Sun will always be over 5,000 degrees because nothing changes its frequency distribution. Distance only reduces its intensity, but not enough in the distance between Earth and the Sun. Perhaps you won’t experience any warming from a star in another galaxy though.

So now I have a question for you …

Why don’t low frequency photons in a microwave oven get absorbed in the usual way by any or all of the atoms in plastic and other substances? The radiation has to be at the right frequency to stimulate resonant “flipping” of whole water molecules in synchronisation with each passing wave, and heat is then generated by frictional processes when such flipping molecules collide with adjacent ones – not by normal atomic absorption. In other words, none of the low frequency radiation is absorbed (in the usual sense) by any substance in the microwave oven. Yet the same plastic bowls etc would easily be warmed by the Sun. Likewise, radio waves can travel around the world without much absorption by all the gases and solids they strike on the way. Why is it so?

• Doug Cotton (DJC) says:

I don’t know why you appear to have such difficulty understanding that all radiation can be represented by a plot of intensity against frequency (or wavelength) and that plot follows the Planck function for spontaneous true blackbody radiation. Any other radiation can be made up of just certain frequency bands, or even just one very narrow frequency band as in the case of a laser. But intensity matters. If the target cannot itself radiate the frequency it receives with an intensity as high as the incident radiation, then there can be no resonance for the extra intensity above the target’s Planck curve.

Suppose you stood close to the Sun but somehow filtered out all but a narrow band of frequencies such as in a laser. Yes, the Sun does radiate some at such low frequencies, and the intensity of even these frequencies (at the outside of its Planck curve) would still be far higher than that of our earth bound laser. So you’d also be cut in half by such Solar radiation. Fortunately Earth is at the right distance such that objects (like rocks on a beach) do not get unbearably hot, because the maximum temperature is of course determined by the Stefan-Boltzmann law – which uses intensity to determine such temperature.

122. bob paglee says:

Dr. Roy,

Thanks for the interesting lesson. But think of your radiation thermometer as a very small-aperture dish antenna receiving electromagnetic radiation with a big 5 degree beamwidth.

When you have pointed it toward a white cloud at a very high elevation angle, I believe you are receiving the reflected radiation emanating from the warm earth near your feet, even if you do this at night, because there is still some low-frequency radiation emanating from Earth also at night. A darker cloud will re-radiate even more — some as pure reflected signal, some as radiation emanating perhaps from the warmer droplets, and because there are more of them than in the white cloud.

But when you aim your radiation receiving antenna’s big 5 degree beam at low levels of elevation, you may be experiencing the electromagnetic radiation effect known to old radar engineers (like me) as “multipath”. Radiation bounces off any surface and is reradiated. This is a quite troublesome effect for narrow-band signals of limited spectrum because multipath radiation can add or subtract to the direct radiation, thereby varying the amplitude of the received signal. I don’t know if the broad spectrum of the radiation to which your thermometer responds would exhibit the same effect, but I suspect that it could at low angles of elevation.

You could try an experiment by putting your pot of hot coffee on a reflective surface, keeping your thermometer very close to the floor and moving it away from the pot to note if the reading varies slightly up and down in a roughly sinusoidal manner with increasing distance from the pot.

123. Max™ says:

“There are far fewer quanta at 10.6um being emitted from the ice. The ice will therefore not vaporise water.” ~tfp

Hmmm, you know this is why people say back radiation can’t do any work, right?

You seem to be disagreeing and agreeing at the same time.

Placing a block of ice next to a pot of water at nearly 373 K will not make the pot of water warmer, contrary to what Roy claims, the energy flow will be from the hot water to the ice, not the other way.

For every photon the block of ice emits which is absorbed by the water, far more travel the other way, thus the proper form of the equation is:

P = εσA(Th⁴ – Tc⁴)

When the SB law is expressed in this form there is no room to claim any warming due to back radiation.

124. Max™ says:April 23, 2013 at 3:37 PM
For every photon the block of ice emits which is absorbed by the water, far more travel the other way, …
—————
Yes you have that correct.
Now place the water next to a block at 0K – now you have NO photons from the block and the same number emitted from the water. SO the water cools more quickly!

Simple isn’t it – Thats all science says.

From this it is obvious that if the water is being gently heated then, when it reaches a stable temperature, next to a block of 0K then it emits n photons/sec and receives no photons /second

Now put it next to a 273K Ice block and let the temperature stabilise. It now receives m photons from the ice in addition to the energy from the heater.

It required to get rid of n photons to stabilise temperature without the additional m photons. It now has more energy to get rid of and it can only do that by becoming warmer and emitting n+k photons per second – where the energy of k photons = the energy of m photons (from the ice)

• Doug Cotton (DJC) says:

Where have you considered conduction due to molecular collisions at the boundary of your ice and water, or between your Earth’s solid surface and all the oxygen and nitrogen molecules striking it, acquiring two thirds of the energy the surface transfers to the atmosphere (as per NASA net energy diagram in my paper) thus providing a blanket effect?

125. Bob Ashworth says:

It is just the insulating affect of the cloud, less IR comes through to the earth from the sun. Direct evidence that water vapor cools the earth as do all gases and dust.

126. bob paglee says:

I forgot to add another comment to the rambling I made above (yesterday) regarding the effect on your radiation-measuring thermometer when you aim it at a low angle of elevation.

The 5-degree receiving beamwidth is probably stated for the half-power point (3 dB down from the peak of the response) but there is plenty of opportunity for signal to be picked up outside that 3 dB point due to the shape of the beam, and ALSO because it very likely HAS SOME BIG SIDELOBES. So I imagine you are measuring plenty of direct re-radiation from the warm Earth surface even if you believe you have the antenna aimed with a sufficiently high elevation.

Why don’t you run a beam-pattern measurement on that instrument with a small point source radiator located at a sufficient distance to assure far-field measurements? You may be surprised to see how far offset from the peak you must go before the signal drops down at least 10 dB.

127. Doug Cotton (DJC) says: April 23, 2013 at 6:18 PM
No – there are many more photons (higher intensity) than would be emitted spontaneously. …
April 23, 2013 at 6:45 PM…all radiation can be represented by a plot of intensity against frequency (or wavelength) and that plot follows the Planck function for spontaneous true blackbody radiation. Any other radiation can be made up of just certain frequency bands, or even just one very narrow frequency band as in the case of a laser. But intensity matters. If the target cannot itself radiate the frequency it receives with an intensity as high as the incident radiation, then there can be no resonance for the extra intensity above the target’s Planck curve.

I think I understand what you are saying.
The number of photons at any frequency from a source has to be above the number of photons possibly emitted at that frequency at the receiver before the source radiation can be thermalised by the receiver.

Is this it?

Trouble is that makes it even worse.

Use a lens and focus the radiation from a source at half the temperature into an area 1/15th the source area on the target then the source radiation will be reflected.
make the focussed spot 17th the size and the radiation from the source will be thermalised in the target.

Alternatively
A target at temp t emits n photons/second.
Take 15 sources at 1/2 the temperature and these will emit 15*1/(2^4)*n photons /second the intensity is less than those emitted from the target so these will not be thermalised on the target.

Take 17 sources at 1/2 the temperature and these will emit 17*1/(2^4)*n photons /second the intensity is greater than that of the target so these will be thermalised on the target.

This does not sound like logical physics to me.

• Doug Cotton (DJC) says:

You don’t understand the physics of optics. What a lens does to radiation travelling in one direction, it does the opposite to radiation going the other way. Try looking through a telescope the wrong way.

Now try using a huge lens to focus all that back radiation at night and cook a chook. After all, people do focus sunlight with a funnel and use it to cook. All those diagrams show similar amounts of back radiation to incident radiation.

There is a comment by another physicist which you should read on Roy’s latest thread (on insulation) where he talks about the same “pseudo scattering” that I do. You should read it. I’m not coming back to this old thread as I’m busy commenting on more recent threads.

128. Max™ says:

It required to get rid of n photons to stabilise temperature without the additional m photons. It now has more energy to get rid of and it can only do that by becoming warmer and emitting n+k photons per second – where the energy of k photons = the energy of m photons (from the ice)” ~tfp

Hang on, where does the “required to get rid of n photons” come from?

P = εσA(Th⁴ – Tc⁴)

For a black body (to simplify) and ignoring area, with Th = 373 K and Tc = 0 K you get P = 1097.64 W/m^2.

With Th = 373 K and Tc = 273 K you get

[email protected]:~\$ calgebra
>>> (5.6703*10^-8)*(373^4 – 273^4)
782.63220248

P = 782.63 W/m^2

129. Max™ says: April 24, 2013 at 11:58 AM
Hang on, where does the “required to get rid of n photons” come from?
—————-
I said energy but it doesn’t matter.
If more energy is entering the BB than leaving it the the temperature will not be stable – it will rise.
I was talking steady state and this only occurs when energy in/sec = energy out/sec.
It was emitting n photons before. It is receiving extra energy so to become stable again it must emit more energy/photons. the photon balance depends on the energy of each photon.

• Max™ says:

Ok, what happens if you have two black bodies at the same temperature radiating towards each other?

• Using the scientific expanation its easy – The same quantity of energy is transferred from a to b as from b to a

using slayer talk no radiation would be transferred as there is no area between the BB curves.

Where the radiation goes I leave up to you to explain.

The 2 bodies still cool if the environment is cooler

• Max™ says:

The energy from a is subtracted from that emitted by b, and the energy from b is subtracted from that emitted by a, if a=b, net=0!

Bring the objects into contact, what happens?

• Doug Cotton (DJC) says:

You are wrong in assuming thermal energy is transferred each way between such radiating bodies. Only electromagnetic energy travels each way. All this is covered in my peer-reviewed paper published on PSI and other websites in March 2012. If you disagree, then submit a formal review to the CEO of Principia Scientific International. But before you do, recognise the fact that at least one other physicist here on Roy’s forum is saying what I am saying. I really don’t care how many climatologists express different views – none of them really know much about modern physics.

130. Assume 2 identical half spheres 1m dia.
Both at same temperature in 0K environment
Face the flat sides to each other.

Assuming equal loss per unit area over all planes
The rounded plane will emit 2*the flat side
surface of curved plane of half sphere=(4πr^2)/2,
The flat surface of circle = πr^2.
When apart the surface area (per hemisphere) radiating to the environment will be
2πr^2+πr^2 = 3πr^2
For both hemispheres this will total 6πr^2
When approaching zero distance the area radiating to the environment will be 4πr^2 (the surface area of a sphere)
At zero = contact radiation will be zero but conduction will be high – both hemispheres are at the same temperature so energy transfer will be zero, net.
The completed sphere will radiate as one from an area of 4πr^2.
The difference in radiation will be proportional to the area of the radiating surface.
Together:apart is 2:3

• Doug Cotton (DJC) says:

At zero = contact radiation will be zero “,/i>

You don’t know that. Radiation continues between individual molecules no matter how close they are. How and why do you think it would switch off?

In fact, this is the reason why the thermal gradient in the solid outer crust of Earth is not as steep as the -g/Cp value – inter-molecular radiation reduces the gradient. You could have read all this in my February 2013 paper. What I say is backed up by real world data.

Now, answer the simple question about why higher levels of water vapour actually lead to cooler surface temperatures, not hotter temperatures as greenhouse conjectures would assume shoud happen. In fact, they assume sensitivity to water vapour would be doubling for each 30 degrees or so. In the real world, doubling reduces temperatures by about 3 to 4 degrees.

131. Doug Cotton (DJC) says: April 26, 2013 at 4:43 PM

Now, answer the simple question about why higher levels of water vapour actually lead to cooler surface temperatures, not hotter temperatures as greenhouse conjectures would assume shoud happen. …. In the real world, doubling reduces temperatures by about 3 to 4 degree
—————-
My analysis of data from http://www.nrel.gov/midc/srrl_bms/

Shows delta temperature flatlining up to 5g/cum water vapour then from 5 to 15g/cum temperature falls at 0.26 degC/ g/cum.

This does seem a bit odd atfirs sight however you seem to have forgotten that the replacement of o2/n2/ etc with h20 molecules actually makes the atmosphere require more energy to warm. If this energy is the same for 0g/cum water to 15g/com then the temperature will fall.

My calculations (need to be checked) show 1cum of air with 15g water vapour require 1.166kJ whilst the same volume of air with no watervapour require 1.005kJ for 1 deg C.

I think this would make the temp vs water mass just a bit positive.

Note I have not yet completed my analysis.

• Doug Cotton (DJC) says:

So you are trying to tell me that the specific heat of moist air is about 16% higher than that of dry air. Well the specific heat of water vapour at 225C in the atmosphere is 1.852 – let’s say 85% higher than that of dry air which is about 1.002. So, to get your 16% we would need air with nearly 19% water vapour. I would suggest trying about 3% at the most – meaning specific heat Cp would be about 2.5% higher. Thus the theoretical thermal gradient -g/Cp would be about 2.5% less steep, not your 16% I’m afraid.

But this does not explain the fact that the moist thermal gradient is about 6.5C/Km compared with the theoretical dry gradient of 9.8C/Km, a difference of about 33%. Obviously the percentage difference is far greater than that 2.5%. The reason is in my paper.

You seem to forget that the IPCC wants you to believe that mean water vapour levels raise temperatures by about 30 degrees. So presumably, if a dry region has only 20% of the mean water vapour level, then its surface temperature should have been raised only 6 degrees, and so it should have its mean temperatures about 24 degrees cooler than world means.

I hope I’ve at last made my point clear, and that you agree the IPCC is totally wrong in claiming such an effect supposedly due to a GHE..

132. Rafael Molina Navas, Madrid says:

DJC: You said (April 18):
“At least we know the energy in the very top layers of the ocean has remained static simply because the surface temperatures have done so”.

” …can atmospheric warming have “stalled” because of the enormous emission of reflective aerosols from coal burning in China and India in the last decade or so?”
said:
“In principle yes, but the evidence that more heat has gone into the ocean is very strong. As Rasmus wrote, warming has gone unabated; just the balance between ocean and atmosphere has changed”.

• Doug Cotton (DJC) says:

The autonomous thermal gradient in the atmosphere (due to diffusion of kinetic energy in a gravitational field) supports temperatures at the surface which, after all, should be what we are concerned about as human beings living at such altitudes.

Why should it worry you if the temperature of the deep ocean is 4 degrees rather than 3 degrees. Fo rthat matter, the temperature 9Km below the solid surface might be 270C or 275C. Who cares? If the surface is supported at a steady temperature, then it really doesn’t matter if there are natural variations at such depths. These regions won’t keep getting warmer indefinitely and are only experiencing natural up and down variations which we really can’t measire with much certainty anyway.

Now, please respect the fact that I really only wish to discuss the physics in my paper.

133. Rafael Molina Navas, Madrid says:

I agree with Dr. Spencer: GHG effect IS real, and IS slightly –but rather relentlessly- warming our planet, to say the least.
Even some Niñas had to “come out” -they weren´t actually Niños- due to the very regular continuous increase of the underlying average sea water temperatures at ENSO monitoring region:

• Doug Cotton (DJC) says:

Prove what you say in the light of counter arguments in my paper “Planetary Core and Surface Temperatures” and counter evidence based on actual temperature data, as in the study in the Appendix. If I had a million dollars I would safely offer it to anyone who could. No one will succeed, though. Try me!

• Rafael Molina Navas, Madrid says:

You do know our levels are different: I can´t argue at your level, and you can´t at mine …(or don´t want to)
In my last post, not a reply to any of yours, I just said I agree with Dr. Spencer and other scientists from NOAA … You´d better challenging them.
In the other hand, what could be the reason of the very regular increase of average underlying temperatures at ENSO monitoring area? Something like what you said in a post (April 18) –reply to another of mine- “Perhaps some energy has risen in currents from depths below 700m, or perhaps the measurement are just insufficient to prove anything. Whatever has happened certainly has absolutely nothing to do with …”?
By the way, I posted another reply (April 19). Did you see it?, or
Wasn´t it worth it because it wasn´t related to your paper?

• Doug Cotton says:

Sorry, I did miss your April 19 comment, so I’ve replied now in this comment.

134. Doug Cotton says:

I do challenge Roy Spencer and anyone else propagating similar pseudo physics. My comment on Roy’s latest post may also appear as an open letter on Principia Scientific International – or something similar, depending on the peer-review system to which it is already being subjected.

So my response to you and anyone is in my final comment here …

http://www.drroyspencer.com/2013/04/a-simple-model-of-global-average-surface-temperature/#comment-76950

135. Rafael Molina Navas, Madrid says:

Dr. Spencer and other scientists have been challenged by DJC. Let me carry on with my “engineer level” discussion.
According to DJC warming stopped in 1998 – maximum of a natural cycle-, and no more thermal energy has gone from atmosphere to seas since then.
What could be the reason of this pretty regular global mean sea level increase even after 1998?
We know ice melting has only been contributing to that rise in a mach smaller proportion than the warming of the seas.
By the way, melt ice is kind of haven to trapped energy -due to latent heat of fusion of that ice-, that not only doesn´t increase oceans´s temperatures but lowers them …

• Doug Cotton says:

According to DJC warming stopped in 1998

Gong! According to DJC, what has happened with climate, and will likely happen in the future, is explained in the Appendix of the paper “Radiated Energy and the Second Law of Thermodynamics” so silent readers can choose to accept incorrect “quotes” or get it straight from the horse’s mouth.

“fusion of ice” storing “latent heat” trapped in its haven? Mmmm! Interesting physics (LOL)

• Doug Cotton says:

Yes, and I quote from the same site that you got your plot from, namely http://www.climate4you.com/SeaTemperatures.htm#Global sea level

It is interesting that this simple empirical forecast has shown a steady trend towards lower values since about 2002.”

So what are you going to do about it? Reducing carbon dioxide emissions won’t help – so put your money into levy banks.

• Rafael Molina Navas, Madrid says:

” … a steady trend towards lower values”?
This is “cherry picking”, and, what is worse, you chose somethig relative not to what has happened, facts, but to “a simple extrapolation of the most recent observed annual sea level change” … until 2100 (¡!!).
And we should keep in mind that climate4you people are skeptics, and when they have to show “unwanted” figures, they frequently try and find something favorable to their ideas, whatever its reliability –the oldest trick in the book.
Let us look at 2 other global sea level graphs:

For me it´s clear that extrapolation must be erroneous and IS meaningless, to say the least.

136. Rafael Molina Navas, Madrid says:

Is it even WORST?:
The extrapolation is until 2100, but on the horizontal axis of the graph they write: 1995, 1996 … 2013 (¡!!)
A printing error? Difficult to believe. Most probably they are trying to DECEIVE. Not the first time I see something similar in a page of visceral skeptics.

137. Rafael Molina Navas, Madrid says:

What quoted by DJC comes from:
Note: Using the 3 year average shown in the diagram above, based on observed sea level changes, around 1999 the total sea level change from then until year 2100 would have been estimated to about 40 cm, in 2005 to about 30 cm (year 2005-2100), and in 2010 to about 22 cm (year 2010-2100). On July 14, 2012, the prognosis would be about 16 cm sea level increase until 2100. “It is interesting that this simple empirical forecast has shown a steady trend towards lower values since about 2002”.
And the graphs says:
“Extrapolated remaining sea level change until year 2100”
It seems that x-axis is correct, but what are they actually wanting to show?
Isn´t this a sophisticated decit?
If anybody wants to “investigate”, please kindly see the complete site:

It is the last graph but one.

138. Rafael Molina Navas, Madrid says:

UNCHAINED TRAGEDIES (CO2 to be blamed, arguably?)
“… Mercury levels in the world’s oceans have doubled over the past 100 years, according to the UN, with more mercury deposited in the Arctic than on any other part of the planet.
… Mercury levels there have for decades been linked to industrial pollution but recent research from Nasa has suggested that declining levels of sea ice in the region could be helping to push up levels of the substance”:

http://www.bbc.co.uk/news/science-environment-22425219

“The Arctic seas are being made rapidly more acidic by carbon-dioxide emissions, according to a new report.
Scientists from the Arctic Monitoring and Assessment Programme (AMAP) monitored widespread changes in ocean chemistry in the region.
… They say even if CO2 emissions stopped now, it would take tens of thousands of years for Arctic Ocean chemistry to revert to pre-industrial levels”.

http://www.bbc.co.uk/news/science-environment-22408341

139. Rafael Molina Navas, Madrid says:

Daily measurements of CO2 at the authoritative “Keeling lab” on Hawaii have topped 400 parts per million for the first time.

…The last time CO2 was regularly above 400ppm was about 3-5 million years ago – before modern humans existed.

http://www.bbc.co.uk/news/science-environment-22486153

140. Rafael Molina Navas, Madrid says:

Three species of migratory duck have shifted their wintering grounds northward in response to increasing temperatures, say scientists.
… At the northern end of their migratory flyway, in Sweden and Finland, there were approximately 130,000 more of the ducks in 2010 than in 1980.
… “Early winter temperature in south Finland,” Dr Hearn said, “increased by about 3.8C between 1980 and 2010.” (!!!)

If only we could do the same!

141. Rafael Molina Navas, Madrid says:

LIKELY IMPACT ON SEA LEVELS

Researchers have published their most advanced calculation
for the likely impact of melting ice on global sea levels.

The EU funded team say the ice sheets and glaciers could
add 36.8 centimetres to the oceans by 2100.

Adding in other factors, sea levels could rise by up to 69
centimetres, higher than previous predictions.

… the scientists stressed that sea level rise in line
with their projections could still make some islands in the Pacific
uninhabitable. And if global emissions of carbon dioxide are not curtailed then
the actual level of the sea by 2100 could be significantly higher than the
Ice2Sea (nombre del grupo de científicos) estimates….

http://www.bbc.co.uk/news/science-environment-22527273

142. John Nicol PhD Physics (Radiation Theory and experiment in gases for 30 years) says:

Dear Roy,

The effect of the cloud showing radiation back to the earth is NOT a greenhouse effect but is the reflection of the earth’s radiation back to earth. You do not need an IR thermometer to detect this effect which is one of two effects of clouds. The first occurs during daylight hours with the refection of sunlight, effectively cooling the earth. The second is that which you observe (reflection) but which will determine the return of radiation at ALL IR wavelengths (frequencies) characteristic of the radiation from the earth’s surface. To make useful measurements you really need an accurate spectrometer to measure the full range of individual frequencies – not the total as you have done.

As I understand it, the basis of the claims for global warming by carbon dioxide and other greenhouse gases, used by the modelers and the IPCC, is the falsely used calculation of the Effective Emission Temperature (EET) – see IPCC AR4 (2007) and AR5 (2013) – from an imaginary “model” of the earth in the form of a black, weightless shell without soil or atmosphere (N2 +O2) to represent a greenhouse-free earth. The night side of this model is at a temperature immediately after sun down of -273 C (0.0 K). The calculation of the EET for the sunlight side is about 303 K but the mean temperature is 288K, the same value as the measured mean temperature of the earth. The IPCC model EET power radiated, is then imagined to be spread uniformly over the whole of the earth which leads to a value of 254.9 K – this is impossible unless there is a thermally active material such as soil present to absorb half of the sun’s energy during the day and re-radiate it over night which is in fact the case for the real earth. The claim is then made, that in order for the earth to acquire the known “mean” temperature of 288 K, a further 33 K from greenhouse gases – mainly CO2 and H2O – is added from back radiation to provide the known mean temperature of 288 K.

However, the addition of the 255 K to 33 K in strict mathematical terms actually provides a constant temperature over the earth – not the varied temperature from the tropics of over 300 K and the much lower temperature at the poles from which the average of 288 K is obtained from measurements. Be that as it may, removing greenhouse gases from the model of Earth, should quite obviously leave us with both soil and most importantly, an inert atmosphere which neither absorbs nor radiates significant radiation energy at the temperatures concerned. During daylight hours, most of the warming of the atmosphere even now, is from contact between the lowest layer of air and the surface skin of the soil which reaches its highest temperatures at about midday for all latitudes.

Thus the air is warmed as has been shown by many simultaneous measurements of the surface skin temperature and that of the air at heights of between 1.2 m and 2 m as carried out by bureaux of meteorology all over the world. These show that at midday the temperature of the air is about equal to that of the skin, while at other times the air is warmer than that of the soil surface. Energy of course is also transmitted into the soil to be released at night. However, at night, the air cools against the soil’s skin, but without accompanying turbulence and convection, it cools only very slowly by varying amounts from about 5 C to 30 C again depending on latitude, wind and rainfall. The night time air thus MUST retain a temperature as shown by these measurements very similar to its daytime temperature of 288 K, providing for a mean temperature of the same value, but in this case varying according to the latitude, in contrast to the unrealistic estimates from the IPCC.

It is thus quite clear, that the known mean temperature of the earth arises because of the retention of energy in the thermally active air which has a heat capacity of about 1,000 Joules per kilogram. Since this will be the case whether or not greenhouse gases are present, these minor gases can have NO effect upon the temperature of the earth. This finding is perfectly consistent with those of my own work for carbon dioxide radiation transfer, the work of Jack Barrett in showing from the spectrum of CO2 that increases cannot produce significant warming and of the much earlier work by Gilbert Plass that carbon dioxide in the upper atmosphere provides significant cooling of the atmosphere. Much more recently, Will Happer from Princeton has shown also in a very complete and accurately calculated paper, that none of five greenhouse gases can have any significant effect on the earth’s temperature and that increases will provide more cooling then the associated back radiation will produce warming.

I would be very happy to engage in a discussion of this matter which is of such great importance to the future of civilisation in the world, even if eventually greater emphasis is placed on the use of renewable energy.

John Nicol, 60/50 Coriander Place, Forest Lake Queensland 4078 Australia. Ph: 07 3679 8848 Mob: 0409 761 503 Email: [email protected] .