Visualizing the "Greenhouse Effect" – Emission Spectra

Ira
I will enjoy reading and considering your post. I have only quickly glanced at the first few paragraphs and my immediate brief observations are:
I consider your better analogy (ie., that of an electric blanket) to be a worse analogy. An electric blanket has a power source and consumes energy in order to provide heat. Where is this seperate power source in your model Earth?
It would also be useful if you would briefly explain how the Earth emits a smooth blackbody curve given that temperatures vary between about 325K (desserts) and 220 to 235K (Artic/Antartic). What are the emission frequencies at these varying temperatures.

The graphics have one (unfortunately very common) error. They are plotted as a wavenumber distribution but labeled with wavelength. You CANNOT just transform the wavenumber into wavelength as they have different peaks. Much like the difference between wavelength and frequency: http://commons.wikimedia.org/wiki/File:PlanckDist_ny_lambda_en.png
The peak of a distribution in wavelength for 280K is at 10.5 µm.
“The dashed line represents a “blackbody” curve characteristic of 300ºK (equivalent to 27ºC or 80ºF). Note how the ~7μ and ~15μ regions approximate that curve, while much of the ~10μ region is not re-emitted downward.”

Ira Glickstein, why don’t you explain to your readers why the various curves for blackbody radiation in your article peak in the range of 17μm to 19μm when the http://spectralcalc.com/blackbody_calculator/blackbody.php site comes up with a peak wavelength of 9.659μm for 300K and 11.828μm for 245K. This is also confirmed by the Wikipedia graph (http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png), showing a peak at about 9μm for 310K.
I look forward to your explanation.

John Marshall,
“Its temperature will fall below that of the surface, and the 2nd law of thermodynamics forbids heat flow from cold to hot ( this is heat flow by any means available) so this rising warm air, relative to the surrounding air though colder than the surface, cannot warm the surface.”
The second law says no such thing. The misconception that the second law says that heat cannot flow from a cooler to a warmer body is the old nineteenth century understanding which dealt with conducting bodies. It was then rewritten in terms of entropy – entropy must always increase in a closed system.
This is what the second law actually says:
“The second law of thermodynamics is an expression of the tendency that over time, differences in temperature, pressure, and chemical potential equilibrate in an isolated physical system (Wikipedia).”
The key point to note is that over time, differences in temperature will equilibriate. The idea of greenhouse gases radiating energy back to the Earth’s surface does not contradict this description as long as the temperatures will equilibriate over time.
The second law allows energy from a cooler body to radiate to a warmer body (in fact, allow is the wrong word – all bodies above absolute zero must radiate energy) because it knows that the warmer body must be radiating even more energy to the cooler body and that their temperatures would most definately equilibriate over time.

The reason I suspect we can’t use this energy is because it doesn’t exist – heat will not flow from the cold atmosphere to the warm surface of the Earth.

Yes it will. When the energy is emitted it is in a random direction and has no knowledge of it is heading towards a warmer or colder body.
Heat will flow from a cold to hot body but at the same time more energy is going the other way. The flow energy from the atmosphere to the surface does exist.

Ira
Further to my last comment, your article is interesting and I am still pondering upon its implications. Whilst ultimately at the very top of the atmospher all heat is radiated into space, I consider that below that boundary other forms of heat transfer are far more significant that radiation.
As I have commented numerous times, I do not consider the Earth to be a blackbody and the assumption that it behaves in this simplistic way distorts what is truely happening. The Earth is never at equilibrium. There are constant changes in albedo due to changes in cloud cover, ice, snow, vegetation., seasons etc These variations change the emission properties of the Earth. The oceans act as huge heat sinks such that diurnal changes over oceans is very different to changes in diurnal temperatures over land. There are peaks and valleys and all of this means that the Earth does not simplistically behave as a uniform blackbody.
You state: “the surface emits upward radiation in the ~7μ, ~10μ, and ~15μ regions of the longwave band. This radiation approximates a smooth “blackbody” curve that peaks at the wavelength corresponding to the surface temperature.” I accept that the curves set out in the “Bottom of the Atmosphere from Surface of Tropical Pacific (and, lower curve, from Alaska) tend to follow the 300K and 245k blackbody curve save for dips in the 8 to 14μ wavelengths, the effects of which you are seeking to highlight.
However, the “Top of the Atmosphere from Satellite Over Tropical Pacific” data is very different. This does not follow the 300k or 290k curve. Up to 7μ it seems to follow a 210/220k curve Between 7 and 8μ it takes a path crossing the 220 to 250K curve. Between 8 and 9μ it jumps up to about the 300K curve. Between 18 to 25μ it falls through the 270 down to the 250K curve. My view of this plot, ignoring the dips around the 9.5μ and the 13 to 18μ range is that it does not exhibit a smooth blackbody curve at any part through the 6 to 25μ wavelengths.
What is the explanation and how in the light of that explanation can it vbe asserted that the Earth (even in that snapshot in time) is exhibiting a smooth blackbody emission spectra?
I look forward to hearing your commenst so that I may better understand your article.

Other aspects that control the temperature range at the surface of the Earth are at least as important and they include convection (winds, storms, etc.) and precipitation (clouds, rain, snow, etc.) that transfer a great deal of energy from the surface to the higher levels of the Atmosphere.

Your other modes of heat transfer are SIGNIFICANT and can not be dismissed.
At altitude radiation is indeed the mode of heat leaving the planetary system.
As altitude gets lower radiation becomes less important and transports less and less of the total heat.

Water vapour is the main so called “greenhouse gas” yet it does not produce a “greenhouse effect” as in a warming effect because it is a fluid gas. Though it may absorb IR, having done so it obviously expands and convects up through the atmosphere taking the IR up to cloud level. It does so because it is a highly mobile fluid gas.
CO2 does exactly the same thing by absorbing heat at the surface and transferring up to cloud level where it is emitted as IR. The cloud level, which begins at around 5000 meters is the beginning if the emission hight of the atmosphere. We know that CO2 begins emit the same level as water vapour because rain water is acidic. The acidity is caused by the dilution of CO2 into the water vapour as it form into clouds on emission, becoming carbonic acid.
Water vapour has a cooling effect on the surface. In exactly the same way so too does CO2.
How is this a “greenhouse effect” ???
The only place that the “greenhouse effect” is actually observed is in the computer models, not in the real world. The only reason that computer models predict a “greenhouse effect” is because they use a parameter known as “convective parameterisation”. In a computer model, if you over estimate convection you will not see a “greenhouse effect”. But if you underestimate this parameter, it will obviously and naturally result in a “greenhouse effect”.
Greenhouses are designed predominantly to inhibit convection as every body knows.
So in the models we have a “greenhouse effect” but in the atmosphere we do not.

can you state how you generated the radiance curves for each temperature? Typically these type of figures have a fudge factor to correct for the black/white body fraction. I have never believed that one is allowed to do this from first principles.

“Well, if a piece of blueberry pie has gone missing, and little Johnny has blueberry juice dripping from his mouth and chin, and that is pretty good circumstantial evidence of who took it.”
This is why climate scientist ought not dabble in police work I think, for what if Charles was the one who nicked the piece and in passing Johnny smeared him blue with it?
But essentially the dissipation is 360˚ but mostly, naturally, concentrated at the nap of the earth, just like the stove which will burn if ones hand is too close but the farther away the hand gets the colder it gets, to the point of room temperature, the in between GHGs doing absolutely next to nothing to transfer the heat to that further away location due to it’s own dissipation rate even though the GHGs are pre-heated by being inside the house as well. But of course, eventually, the stove’ll raise the temperature in the kitchen, but only when the floor, ceiling, and the walls are more radiant, but only to a certain point (unless the stove doesn’t burst into flames the paint on the farthest wall ain’t gonna start cooking any time soon.) So whom is the construct of the thief that steals that heat? :p

tallbloke says:
March 10, 2011 at 4:26 am
“the hot-water bottle effect”
But what does heat the bottle, in the first place?
It is not “green-house effect”.
http://es.scribd.com/doc/28018819/Greenhouse-Niels-Bohr
Joe Lalonde says:
March 10, 2011 at 4:49 am
First, without planetary rotation, there is no convection
But….what does the Earth rotate…..an homopolar motor?
Then….heat by resistance?

We live in an upside down world.
Water vapor makes up anywhere from 1-4% of the atmosphere. I can’t find the average because the percentage is so variable. CO2 is .04% of atmosphere. So water vapor is about 100x more plentiful than CO2 in atmosphere. Do anyone know the difference between the strength of CO2 vs WV as a greenhouse gas. Also, we know CO2 gets saturated in its’ greenhouse gas ability. Does this happen to WV? Then we know that man only creates 3% of the atmosphere CO2, well we don’t really know that, it’s just another big guess/estimate. How could we know that when we have forests densities completely changing throughout our world. Here in the US our forests are growing at a very fast rate, still recovering from the clearcuts of >100 yrs ago.
Then we have a possible feedback between CO2 and WV. Is it positive or is it negative? IPCC and modern “science” says CO2 always causes a positive increase in WV, causing amplification of greenhouse effect. But measurement of humidity and water vapor apparently doesn’t show this to be reality.
http://www.climate4you.com/GreenhouseGasses.htm, in fact WV is decreasing the last few years.
Then of course we have the oceans and their cylces, the sun and planetary alignment, undersea volcanoes etc etc etc. But THEY know everything. The SCIENCE is settled.
Yet after 11 years of increasing CO2 and amplification models, their data shows we are back where we started from in temperature with no trend whatsoever.
http://www.woodfortrees.org/plot/hadcrut3vgl/from:1999
We don’t know nothing about climate!

william gray says:
March 10, 2011 at 5:16 am
How do we measure re-emittance of IR? I know absorbtion can be measured.
——-
You should be able to get pretty close by taking a spectrum at say 800 mbar in the tropics aimed sideways at night. ☺ That’s the only way I can think of doing it. Wouldn’t that be close?

Ira,
You should stop saying the greenhouse gases in the atmosphere warm the earth to this crowd. Too may engineers running around here. The atmosphere slows the rate at which the earth loses energy to deep space. Thus, the earth in the sun-earth-space system has a higher equilibrium temp. The “old” rules of thermodynamics still apply: a warm object cannot gain net thermal energy from a cold object.
Thank you for posting the good data though, especially of the IR toward and away from earth at the arctic. I am curious what the ground temperature was on the days that this data was collected, since at the arctic you often have an inverted lapse rate and the atmosphere could indeed “heat” the ground. I am curious if the ground temps are cold enough at the poles, and the lapse rate is positive instead of negative, so that the atmosphere could indeed “heat” the Earth, at least in these specific conditions and locations.

Sorry but this article is filled with misinformation.
“Perhaps a better analogy would be an electric blanket that, in addition to its insulating properties, also emits thermal radiation both down and up.”
no a much better analogy is a blanket. There is no additional heating coming from the atmosphere.
They seem to give out degrees and doctorates in corn flakes boxes these days.

As in virtually all such articles focusing on longwave emissions/apsorption/etc, no account is given for the effect of the oceans which are the main determinant of global temperatures on this planet. The oceans absorb the vast majority of solar radiation on this planet and they do so to a considerable depth. Water is a relatively slow conductor of heat and therefore tends to hold it for a good length of time. Ocean currents and temperature layering cause the heat to move from the equator to the poles. Atmospheric temperatures in the tropics are nearly what one would expect from a black body analysis. The so called “green house effect” is really only visible at higher latitudes. This is due almost entirely to the redistribution of heat by the oceans. Atmospheric effects are quite small in comparison. This is not a new idea. until a few years ago oceanic scientists such as Doctor Robert Stevenson could have described all of this in considerable detail. Unfortunately, “climate scientists” with a poor understanding of the influence of the oceans have completely ignored it.

tallbloke says: March 10, 2011 at 4:26 am
“….The greenhouse effect works by *SLOWING DOWN THE RATE THE EARTH COOLS AT*, by raising the altitude at which the atmosphere radiates to space . There is more than a semantic difference. Understanding it this way enables you to understand that it was reduced albedo 1979-1998 allowing more Solar energy to enter the oceans that caused the majority of the global warming at the end of the last millenium.
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/
////////////////////////////////////////////////////////
Thanks for the link to your article. I was one of those who was arguing similar points with Willis and I have not seen your article before today. It is an interesting read.
I too respect Willis’ views but he was unable to even begin to explain the physics involved in how heat could be entrained by the oceans given the wavelength of DLR and its penatrative depth and thus become well mixed.
The only point he came up with (which did not answer the question) was that but for GHGs, the oceans would freeze and he referred to a link on scienceofdoom which suggested that without GHGs, the oceans would freeze within about 4 years. The underlying data and codes were not attached to the scienceofdoom article so that that assessment could not be verified. However, as I tried to point out to Willis, it is too simplistic looking at average temperatures and average conditions. The oceans are extremely complex and act as both a huge storage reservoir and a huge heat pump. For example, if one looks at the Baltic, in late summer, the sea temp is 16 to 18C and yet within about 4 months, it freezes over notwithsanding GHGs. There are many parts of the oceans (and inland lakes/seas) that freeze within months and this will tend to give the impression that when viewed on an average basis the seas would freeze within years. However, of course, there are great swathes of the Pacific, Indian Ocean, Atlantic etc receiving immense amounts of solar energy which energy is then pumped around by currents etc. It is almost certainly the case that it is this input and distribution that stops the majority of oceans from freezing over within seasons. Further, one may enquire rhetorically as to what causes the ice to melt/recede on these frozen seas/lakes? It is not an increase in GHCs but rather an increase in solar energy either directly and/or indirectly (via currents/circulation patterns).
I consider it probable that the vast majority of recent warming is due to natural variations and one of the key contenders for this being changes in cloud cover and changes in albedo allowing more solar energy to have penatrated the oceans.

Scarlet Pumpernickel says:
March 10, 2011 at 3:30 am
So what concentration of CO2 saturates these wavelengths?
There is already proof from Venus that the Greenhouse effect is not exponential http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
——–
Thanks Scarlet, one impressive analysis. Also since NASA lists Venus’s *average* temperature at 464 C and if the surface was as high as 505 C then the natural dry lapse rate matches the graviation acceleration almost exactly, as it should. (778K-339K)/49.5km = ~8.87 °C/km (g=8.87 m/s2). Neat, 96.5% CO2 and no greenhouse effect at all. Now that’s some pure simple logic!

funny all this blanket stuff, we know that without greenhouse gases the surface can get very hot during the daytime – the moon, though at night leads to rapid cooling.
Seems to me that greenhouse gases lead to cooling during the day, and at night leads to a slow cooling , or in the desert, rapid due to less moisture.
After all if the sun didn’t come up next day how quick before the planet froze.
Such a simple elegant explanation.

I suggest we rid the atmosphere of o2 also. I mean it has this fat absorption/emission band at 50-70 GHz. One cannot be too careful you know.

To John Marshall and Vince Causey:
In regard to the second law discussion, I would add this way of looking at it, which is really just a simpler way of saying what Vince said. When you have energy being transfered between the ground and the air it will have a rate of flow based on the difference in temperature between them. If you make the air warmer and the ground stays the same temperature, then the rate of energy flow will slow down. It’s not a matter of whether energy flows in both directions or not. The question is relative rate of flow. Changing the temperature of either medium relative to the other will change the rate of flow in the form of conduction.
To Ira Glickstein:
Are there any sources that show what those bottom/top of atmosphere graphs would look like with double CO2? I wonder how much it would change and how?

to Harry Dale Huffman
Good post. Great analysis. Far more valid than the nonsense out there.
On earth, the data from the Antarctic interior regions, confirms exactly what you concluded. CO2 levels have no silly feedback effects.
It’s as if the climate scientists, ignorant of the basics of radiational heat transfer, have built an elaborate farce. They chase around bits and pieces, thousands of localized effects. Meanwhile they construct a farcical ‘greenhouse effect’, include in it what they think belongs there, blow them way out of proportion, and exclude or ignore what they ‘think’ doesn’t belong there.
Then they frighten themselves with their own farce of a nightmare. And they try to get others to join them in their self-created nightmare.
Incompetence abounds, not only as scientists, but as people.

I’m an engineer, not a physicist, but I think I may be able to clarify the misunderstandings with respect to heat transfer and the second law of thermodynamics.
Any matter that is not at zero Kelvin will emit energy.
Where a cold surface meets a hot surface, both emit energy but the hot surface will emit more than the cold.
Although there will be energy transfer in both directions, the net transfer must be from the hot body to the cold one, causing the hot body to cool and the cold body to warm.
So when the second law forbids the transfer of heat from the cold body to the warm, I think it refers to net heat transfer.
Disclaimer: my thermodynamics are pretty rusty so that might be all wrong.

mkelly says:
March 10, 2011 at 9:21 am
You are correct, all substances radiate according to their respective temperatures, there are no exceptions to that fact.

Phil. says:
March 10, 2011 at 9:31 am
“Venus has a bond albedo of 0.75 due to the sulfuric acid clouds whereas the earth has a bond albedo of 0.29, you bet the temperature depends on the albedo. Venus absorbs 25% of the light incident on it and Earth 71% but the ratio of temperatures still only depends on the distance from the sun? Even a physicist should see the problem with that.”
I think Phil has either missed the point here or he hasn’t read Harry’s article. http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
He clearly states that Earth and Venus are heated top down by incoming IR.
Quote: “This in fact indicates that the Venusian atmosphere is heated mainly by incident infrared radiation from the Sun, which is not reflected but absorbed by Venus’s clouds, rather than by warming first of the planetary surface. (It also indicates that the Earth atmosphere is substantially warmed the same way, during daylight hours, by direct solar infrared irradiation, and that the temperature profile, or lapse rate, for any planetary atmosphere is relatively oblivious to how the atmosphere is heated, whether from above or below.)”
Albedo applies to visible electromagnetic radiation not invisible IR.

I am enjoying the various views expressed in the various comments.
As regards the various arguments as to whether heat can flow from a cooler body to a warmer body, net heat flow, the rate of net heat flow etc, given the importance of this issue, it amazes me that there appears to be no experimental data on this. Whilst this is not my field, surely, it cannot be that difficult to devise a suitable experiment. Possibly along the lines:
1. A very larger insulated air chamber with a well mixed air temperature at say 290K.
2. Suspend within the chamber a modest size blackbody sphere at say 340K and measure the time taken to cool to 320K.
3. Repeat 2 above but this time additionally suspend within the chamber a significantly smaller blackbody sphere (say perhaps 1/10th surface area) at 300K say 2 metres away from the larger 340K blackbody. Measure the time taken for the 340K blackbody to cool to 320K. Measure the heat radiated on both sides of that blackbody to see whether there is a difference in the amount of heat being radiated on the side which is adjacent to the smaller cooler blackbody. Measure the heat being radiated by the smaller blackbody to see whether there is a difference in the amount of heat being radiated on the side adjacent to the warmer blackbody.
4. Repeat 3 above but with the smaller blackbody at 280K.
5. Repeat the experiment with different sizes of blackbodies, and different temperatures for each blackbody and different distances between the blackbodies.
6. Repeat the experiement with the chamber of air having 500ppm, 600ppm, 700 ppm and 800ppm of Co2.
7. Repeat the experiment but with a slow running fan placed under and sime distance away from the blackbody spheres.
8. Carry out a number of experimental runs as appropriate.
9. Collect data and analyse.
I am not suggesting that the above experiment but some experiment along those lines ought to establish what really happens in the real world.

John Marshall says:
March 10, 2011 at 2:04 am
May I respectfully suggest you go read a junior level physics text book.

Bryan,
“Vince Causey perhaps is confusing radiation with heat.”
Yes, I was using the two terms interchangably. If we take heat flow to mean something getting warmer, then it is certainly impossible for a cooler body to make a warmer body still warmer. What I was trying to say was that energy will flow from the cooler body but heat will not – ie, the warmer body will not heat up, but will still cool down and equilibriate with the cooler body.
I was attempting to respond to a poster who asserts that the GHG effect violates the 2nd law of thermodynamics which forbids heat flowing from a cooler body to a warmer body. A better worded response would be to say that GHG does not depend on ‘heat’ flowing in that direction, merely the flow of energy, and that energy flow will have the effect of slowing down heat loss from the warmer body.

GaryP says:
March 10, 2011 at 10:47 am
Nail, head. Radiative not kinetic.

Morris Minor,
“If this is true why isn’t this energy collected and used as an energy source (better than solar energy as no need for storage – this can be collected 24 hours /day).”
Actually heat is collected from the ground. Thermal heat collectors based on heat pumps are able to extract that heat from the 2 metres below the surface and can be used to warm houses.

GaryP says:
……”Here is an actual experiment. I was using a hot filament in high vacuum to heat a sample to high temperature (~1000°C). I could not get one sample hot enough so I put a shiny metal cylinder around but not touching the filament or sample. The reflected heat increased my sample temperature over 100°C. The added radiation from the relatively cold cylinder increased the temperature of the white hot filament. The net flow of heat of course was from hot to cold. This is a pure radiation example and some heat did radiate from the cold reflector to the hot filament.”…..
All your reflector did was to insulate the hot filament.
When you put on clothes you reduce heat loss from your body.
The clothes also radiate IR as well as reducing heat loss by conduction and convection.
You would not saythat you wear your own greenhouse effect clothing would you?

peter_ga says:
March 10, 2011 at 3:21 am
“Earth is warmer because of its oceans, through an indirect greenhouse mechanism, that has nothing to do with co2.”
Bingo!
see here:
http://climaterealists.com/index.php?id=1487&linkbox=true&position=4
“The Hot Water Bottle Effect.”
and while I’m about it the one most important thing I have realised from the advice of solar expert Leif Svalgaard is that it is not all about radiative physics (sorry Ira).
If we are going to explain movements of the Earth’s air circulation systems to fit observations then radiative physics just does not work.
We see a cooling stratosphere/mesosphere when the sun is active and a warming stratosphere/mesosphere when the sun is inactive. That is the opposite of conventional climatology.
Thus the air circulation systems move poleward when the sun is active and equatorward when it is inactive. That affects global albedo and energy input to the oceans.
What we are left with is atmospheric chemistry involving ozone overriding radiative processes to shift the air circulations, affect global cloudiness and albedo and thereby switch the oceans from net energy gain to net energy loss. The tropospheric air temperatures then follow in due course.
This is my latest effort on that issue:
http://www.irishweatheronline.com/irishweather/how-the-sun-could-control-earths-temperature.html
“How The Sun Could Control Earth’s Temperature”
Alternative suggestions are welcome but they need to fit real world observations at least as well as do my proposals.

“Since gases tend to re-emit most strongly at the same wavelength region where they absorb, the ~7μ and ~15μ are well-represented, while the ~10μ region is weaker.”
We went over this before. Kirchoff’s first of law radiation: solids, liquids, and dense gases emit continuous blackbody spectrums with the peak emission frequency determined by the temperature. The earth’s atmosphere below the thermosphere is dense in the context of Kirchoff’s law. When the gas molecules are densely packed they give up energy through collisions. Collisions don’t generate emission line spectra they generate continuous blackbody spectra. The “temperature” is a measure of the average speed of the molecules and hence the average energy involved in collisions between molecules.

I recently voted for this as being the best scientific blog but, in response to this article, we still get people who say that back radiation cannot happen or wish to divert the discussion to some strange theories about the temperature on Venus.
But there is a serious problem with Ira’s article!
As expressed before (6:10AM); his blackbody curves are showing a peak at about 18 micron when they should be peaking at about 10 micron for a 300K blackbody. So, the graphs don’t make sense
Now Joel Heinrich at 2.57 AM and Phil at 8.21AM, March 10, say this is because Ira has converted wave number to wavelength.
But this does NOT cut the mustard. This is a perfectly acceptable thing to do, albeit ending up with a non-linear scale rather than a linear scale. In fact, the
Source Data itself includes the wavelength scale, so it is nothing to do with Ira’s conversion.
On the ‘wave number’ scale the peak of the blackbody curves should be around a wave number of 1000 – which they are not! They peak around a wave number of 600 – which is WRONG!
So the whole article and its conclusions are WRONG!
Now, can any mathematician or physicist please explain to me why I am mistaken? What am I missing?
Come on Ira.

Since such articles are very comment-rich, I assume the science is far from settled, even the basic theory.
While the charts are ok, nobody yet calculated how much is this backward IR actually affecting the earth surface temperature. Remove oceans and 99% of atmosphere (oxygen and nitrogen) with their tremendous heat-keeping capacity and run the experiment with those GHGs again. Or look at the Mars with 6,000 ppm of CO2 and its temperature like blackbody, since oceans, nitrogen and oxygen are missing.

CommieBob,
Is that flashlight cooler than your eyes? I never said, nor have I seen anyone else on this thread say, that radiation from a hot source would somehow “block” the radiation from a cooler source. I don’t see how this example refutes anything anyone is stating. If you shine a flashlight in your eyes, it will warm you, not the other way around. The flashlight will not warm the sun. The filament in the flashlight is not a hot as the photosphere of the sun. The photons that represent the information being exchanged between the sun and the filament will result in a net warming of the filament, not the sun.
I would expect my students to state that the atmosphere (or any insulation layer), changes the rate at which energy is transmitted.
You say that wearing a coat warms the body. I say putting on a coat slows the rate at which your body transfers thermal energy to the surroundings. (and yes, this is obviously conduction, not radiation) There is not really much difference mathematically. It is that statement that greenhouse gases HEAT the earth” by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.

“”””” P. van der Meer says:
March 10, 2011 at 3:15 am
Ira Glickstein, why don’t you explain to your readers why the various curves for blackbody radiation in your article peak in the range of 17μm to 19μm when the http://spectralcalc.com/blackbody_calculator/blackbody.php site comes up with a peak wavelength of 9.659μm for 300K and 11.828μm for 245K. This is also confirmed by the Wikipedia graph (http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png), showing a peak at about 9μm for 310K.
I look forward to your explanation. “””””
“”””” Joel Heinrich says:
March 10, 2011 at 2:57 am
The graphics have one (unfortunately very common) error. They are plotted as a wavenumber distribution but labeled with wavelength. You CANNOT just transform the wavenumber into wavelength as they have different peaks. Much like the difference between wavelength and frequency: http://commons.wikimedia.org/wiki/File:PlanckDist_ny_lambda_en.png
The peak of a distribution in wavelength for 280K is at 10.5 µm. “””””
Well the explanations are quite simple:
For Joel Heinrich’s comment; you can in fact plot a graph of anything you like against anything else that you like, even if they aren’t in any way related.
But in this case they are. Wavelength and wave number also are related, and you can get from one to the other or from the other to one if you prefer.
And some of the Graphs Ira gives, have horizonatal axes, in both Wavelength and Wave number, as you can plainly see.
But as you can also plainly see the vertical scale has specific units. And ALL of the graphs have the wrong labels. The vertical axis is not RADIANCE. It IS “Spectral Radiance”, which is the contribution to total RADIANCE, per increment of either WAVE NUMBER as it is in these graphs, or WAVELENGTH which is also quite co0mmon. I actually am more used to seeing blackbody radiation plots using a Wavelength horizontal scale, or a logarithmic wavelength horizontal scale, and a per unit wavelength Spectral Radiance for the vertical scale; but chemists tend to use Wave Numbers and per wavenumber units, and the BB curves are different depending on which you use.
At 288 K , the BB Planck function has about a 10 micron peak spectral radiance; which is also 1000 cm^-1 BUT ONLY if the Spectral Radiance is specified per unit of WAVELENGTH increment. If the spectral radiance units are per wavenumber (cm^-1) then the peak is closer to 600 cm^-1 or around 18 microns.
Obviously the global warmists prefer the per wave number version, since the peak of the curve is right on the CO2 band, making CO2 seem more important; whereas the same data on a per micron wavelength plot peaks at 10 microns, or 1000 cm^-1 with the CO2 15 micron band now down lower on the long wavelength tail.
Although I am more used to the wavelength and the per micron spectral radiance scale, I do agree there is some merit in using frequency units, since the photon energies are proportional to frequency, and not to wavelength. But actually neither version is without problems.
On a per wavelength scale, you get high spectral radiance for solar UV; but remember that the range of wavelengths available for solar UV is just 2-3 hundred nanometres, versus the tens of microns available for the LWIR emissions from the surface.
Conversely, on a per wave number spectral radiance plot, the CO2 band gets a miserable 100 wave numbers, whereas that low amplitude water tail has many hundreds of wavenumbers bandwidth; so neither representation is without problems.
Some people graph black body radiation curves using numbers of photons, versus wavelength or wave number, and then you get a different peak again.
Now guffawingly, Peter Humbug in his now famous Physics Today special paper, emitted the per spectral units interval all together, giving a totally ridiculous plot, that implied an infinite amount of total energy; which interestingly he also called flux, rather than spectral radiance.
But we will cut him some slack, and put his error down to a simple typo; we know what you meant Peter. And incidently as he plotted it he did mean per wave number interval.
Perhaps if he had paid more attention to his subject and left out the political BS, about climate disruption, and 800,000 Kelvin earth Temperatures, then he might have caught his mistake. But as I said we will grant him some leeway. I personally never make typeos, because I review everything carefully.

@Ira
I don’t think it’s a good idea to go into quantum mechanics which is generally reserved for quantum scale phenomena. Wavelengths in the micron range in a dense gas means a single wavelength spans millions of molecules at once which is in the domain of classical thermodynamics. Temperature itself is not a quantum measure but rather an average speed of a great number of molecules. That said I’ll take a stab at a lay description of individual molecules and photons in the dense portion of the earth’s atmosphere.
I just finished reading an article in this month’s issue of Scientific American about a new technique for cooling matter down closer to absolute zero than anyone has obtained before and even better a way to do it with elements that previous techniques didn’t work with. Background information in the article talked about the speed of motion of room temperature gases which if I recall correctly ranges from about 2000 miles per hour to down near zero. The gadget they constructed used a pair of laser beams that act like Maxwell’s Demon allowing lower speed molecules to pass through the pair of beams in one direction only. The volume on one side is smaller than the other and when they trap all the atoms on the smaller side (which for some esoteric quantum reason I don’t really understand doesn’t raise their temperature by compression) they turn off the lasers and let the atoms expand back into the full volume and when they expand they cool down. By doing that over and over and over each time dropping the temperature a little bit they get to within a few millionth’s of a degree of absolute zero.
Preceding the laser stage they got most of their cooling by letting a gas of atoms out through a pinhole into a vacuum. This had the effect of both cooling the atoms and getting them all moving at close to the same speed. They aimed this beam of atoms at the receding edge of a fan blade moving at half the speed of the atoms which had the effect of slowing them down even more without disturbing their distribution or speed relative to each other. After having that done to them a few times they passed into the laser chamber for the final cooling cycles.
I just thought I’d throw that in because it’s way interesting, bleeding edge, and descriptive of how temperature is a measure of speed of motion.
Anyhow, certain gas molecules have a shape that is resonant with certain frequencies of thermal radiation. When photon of that frequency intersects the molecule it is absorbed. The absorbed energy raises the molecule to a higher excitation state. In a rarefied gas it would eventually emit a photon of the same frequency and fall back to the lower excitation state in a quantum jump. In a dense gas where it is rubbing shoulder to shoulder with other molecules however the higher excitation results in greater speed and means it’s going to collide with another molecule right away, before the average amount of time elapses for a quantum transition to occur. Don’t ask me how long it takes for a quantum jump to occur. All I know is that it isn’t a fixed amount of time but rather a probability of occuring where there is a small probability of being instantaneous, a small probability of taking forever, and a greater probability of some length of time in between those two. A quantum physicist should be able to give a more definitive answer. But the answer is moot because it has been experimentally demonstrated that in a dense gas collisions occur before any significant number of quantum emissions happen. Thus Kirchoff’s law predates quantum mechanics by over half a century.
Due to the absorption of a thermal photon at narrow resonant frequences in so-called GHGs the higher excitation means that molecule is going to have a more energetic collision than it would have otherwise. A collision will also cause a photon to be emitted but the photon frequency is determined by the collision energy not the energy added by the resonant photon. Since the molecules are all moving at anywhere from 0-2000mph to start with the collisions will cause photons to be emitted over continuously varying range of energies (or frequencies as photon energy and frequency are the same thing). So what we end up with is a continuous blackbody emission spectrum reflective of the individual collision energies in billions of atoms all moving at different speeds. A few of the collisions will be very low energy giving us lower frequency photons and some will be higher energy giving us higher energy photons. The closer we get to the average speed of the molecules the more collisions we get with that amount of energy in them. Thus we get a spectrum with a peak energy at a certain frequency that falls off in a smooth curve to either side of that frequency i.e. a continuous blackbody spectrum. Since temperature is a measure of the average speed of a great many individual molecules the peak emission frequency of the spectrum is a function of temperature.

What is incredible for me it’s that everyone who talk about the outgoing and the back scattered spectra, always considers the vertical paths only.
Does anybody out of here have graphs of the back scattered spectrum at ground for different angles at the same place? That is the azimuthal (90°), the 60°, the 45° and the 30° for example.
What I’m guessing here is that the graphs could be very different.
But the most interesting should be the outgoing spectrum at the TOA for the different angles, because the following links shows how the so called “limb radiation” is almost complementary to the nadir one (see the topmost graph of figure 3).
http://www.atmos-chem-phys-discuss.net/6/4061/2006/acpd-6-4061-2006-print.pdf
(The research pertain a radiosonde at abt. 34km, but how you can see from the radiance emitted from the deep space at that height there is almost no more energy back scattered).
What I’m arguing is that the increased absorption pit at 15um due to the doubling of the CO2 seen at the satellite nadir view doesn’t mean that the radiation is backscattered or held by the atmosphere, that energy just exits the atmosphere under different angles not seen by the satellites “eye”.

Fred Souder says:
March 10, 2011 at 12:32 pm
CommieBob,
… by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.

We are clearly talking at crossed purposes. One of the things I have expertise on is EM radiation. If the problem is vocabulary, so be it. If you are saying that a weak emitter can not radiate energy toward a stronger emitter, you are just plain, flat, dead, completely wrong.
What I can also tell you as an expert is this: If I build and energize a transmitter antenna, I can still measure the signals on it that are caused by adjacent transmitters. (The total voltage on the antenna will be the sum of the local transmitted signal and all the received signals.) In fact, a good example would be continuous wave radar. A signal leaves the transmitting antenna, strikes the sharp edge of an aircraft, re-radiates at the same frequency and, shifted by the doppler effect, returns to the sending antenna and is detected. http://www.tpub.com/content/fc/12404/css/12404_24.htm
For experimental evidence with gases and heat, Dave Springer has a post above where he describes the work of John Tyndall. http://en.wikipedia.org/wiki/John_Tyndall

Fred Souder says:
March 10, 2011 at 12:32 pm
“I would expect my students to state that the atmosphere (or any insulation layer), changes the rate at which energy is transmitted. You say that wearing a coat warms the body. I say putting on a coat slows the rate at which your body transfers thermal energy to the surroundings. (and yes, this is obviously conduction, not radiation)”
If I were in a physics class I’d say the coat slows down how fast your body loses heat. If I were at a soccer game on a cold day and my kid came off the field shivering I’d say put on this coat to warm yourself up.
“There is not really much difference mathematically. It is that statement that greenhouse gases HEAT the earth” by re-radiating IR that is not going to sit well with the thermodynamics vocabulary police.”
Agreed. But in the end the surface is warmer and it’s difficult to describe how the equilibrium temperature between the surface and the 3 Kelvin cosmic microwave background temperature rises such that the heat transfer rate through the additional insulation rises commensurately. Then you might have to explain that because it’s a dynamic system with a plethora of other things changing both more slowly and more rapidly than GHGs the equilibrium temperature is a moving target that is approached but never attained for very long but that when the system goes farther out of equilibrium the harder it tries to get back to equilibrium.

commieBob says:
March 10, 2011 at 1:52 pm
“What I can also tell you as an expert is this: If I build and energize a transmitter antenna, I can still measure the signals on it that are caused bay adjacent transmitters. (The total voltage on the antenna will be the sum of the local transmitted signal and all the received signals.) In fact, a good example would be continuous wave radar. A signal leaves the transmitting antenna, strikes the sharp edge of an aircraft, re-radiates at the same frequency and, shifted by the doppler effect, returns to the sending antenna and is detected.”
I only worked on a conventional pulsed (weather) radar 35 years ago. CW Doppler can’t do ranging without adding in frequency modulation. What a nightmare sorting out the return signal would have been with old analog electronics!
That said, constructive and destructive interference happens in all situations and really muddies up the picture. As long as the frequencies are different you sort it out using analog filters or a DSP running FFTs on it. In fact new fangled cars have active noise reduction where road & engine noise in the passenger compartment are drastically reduced by destructive interference – there are listening devices that generate sound of equal amplitude but 180 degrees out of phase. The energy in the sound waves has to go somewhere so in the car I reckon that heats the air a little bit.
But this raises a puzzling question for me when it comes to electromagnetic waves in a vacuum. Suppose we have two emitters at exactly the same frequency but 180 degrees out of phase. The wavefronts meet and perfectly cancel out. Say they are separated by a two light-seconds and each starts emitting at exactly the same time. Theoretically the wavefronts will meet halfway and cancel each other out so the radiation from either will never reach the other. The puzzling part is where does the energy go?
I had an interesting email conversation with a couple of academics who’d published an article in SciAm about common misconceptions about the Big Bang. They were describing the expansion of the universe and how the 3000K temperature of the early universe (the point at which it became transparent to radiation) had fallen to 3K today due the universe expanding by a factor of 1000 since that time. They explained it in quantum terms using photons and how the individual photons lost energy over time. I objected and said that photons propogating through a vacuum are immortal and unchanging. If the photon lost energy, I asked, where does the lost energy go and what form does it take? I went on further to say the particle description was inappropropriate. They should have explained it using waves which is appropriate for EM propogating through a vacuum – the fabric of space simply expanded and stretched out the electromagnetic waves along with it. That satisfies the law of conservation of energy. They came back with some vague BS about gravitational energy and refused to answer any further questions. There may very well be a gravitational answer but unfortunately there’s currently no theory of quantum gravity. Why do people insist on making things so complicated when there are simple answers staring them in the face?

Massimo PORZIO says:
March 10, 2011 at 1:24 pm
“What I’m arguing is that the increased absorption pit at 15um due to the doubling of the CO2 seen at the satellite nadir view doesn’t mean that the radiation is backscattered or held by the atmosphere, that energy just exits the atmosphere under different angles not seen by the satellites “eye”.”
Over the arctic looking down from 20km the bottom of the pit conforms perfectly to a 225K blackbody curve while outside the pit the curve fits perfectly to a 265K curve. Clear arctic air is very dry so water vapor has a minimal effect. 265K is -8C which is the surface temperature of the ice. Yet in the 15um window the ice appears to be -48C. What we are seeing in the 15um window is the so-called emission altitude. Dry adiabatic lapse rate is 1 Kelvin per 100 meters. The emission altitude is 4000 meters. This is the height at which the CO2 in the atmosphere has absorbed all the 15um radiation available upwelling from the ground and what remains is thermalized radiation from the atmosphere at that height.
Were we to reduce the 265K curve to a temperature such that the total reduction in volume underneath the lowered curve is the same volume as the hole then we would have the equilibrium surface temperature absent ALL atmospheric CO2. I believe that works out to about 15C. The thing of it is that the first 100ppm or so of CO2 does the lion’s share of the work so by the time we get to the surface equilibrium temperature difference between 280ppm and 560ppm CO2 (a doubling) the increase is only about 1.0C.
This takes us round to my hypothesis that the MOST IMPORTANT climate function of CO2 is in raising the average surface temperature of the earth from -23C (no greenhouse gases – the average surface temperature of the moon) to -8C. At -23C the earth is a snowball and positive feedback from the exceedingly high albedo of ice keeps it that way. Once CO2 is in the atmosphere to help melt some snow and ice this lowers the albedo dramatically where it matters the most (at low latitudes where the sun is the strongest) and additionally starts pumping water vapor into the atmosphere from the liquid water surface which accelerates the melt and we are back to a water planet instead of an ice planet. Once we have a water planet the low albedo of the global ocean and the higher absolute humidity keeps it a water planet (barely, with a few excursions to a snowball once every few hundred million years).
So CO2 serves as “kindling” to light the fire which melts the ice and turns what would otherwise be a frozen world into a liquid water planet. The water cycle, once activated takes over the show. And water limits the maximum temperature through negative feedbacks of evaporation and cloud formation.
During most of the earth’s history CO2 has been up around 2000ppm. The current 380ppm is dangerously low which probably has a lot to do with why we’ve been in an ice age for the past 3 million years.
Despite even 2000ppm CO2 the earth has still entered ice ages once in while and a few times severe enough to freeze the entire planet. It’s a bit of a puzzle what could possibly kick the earth out of a snowball episode but after about 10 million years it does indeed melt. The most popular hypothesis for the melt is that when the earth is frozen over all the normal CO2 sinks are gone. The global ocean doesn’t absorb any CO2 and there are no forests that grow on glaciers taking in CO2 and no chemistry between CO2 and ice forming carbonate compounds. But volcanoes don’t stop belching out CO2 so over the course of millions of years atmospheric CO2 rises and rises and rises until its greenhouse effect starts driving back the ice at an accelerating rate and viola – a water world is born again and the water cycle caps the maximum temperature at a point nice and comfy for living things from pole to pole.

John Marshall says:
March 10, 2011 at 2:04 am

Its temperature will fall below that of the surface, and the 2nd law of thermodynamics forbids heat flow from cold to hot ( this is heat flow by any means available) so this rising warm air, relative to the surrounding air though colder than the surface, cannot warm the surface.

I keep hearing this. It gets boring.
Cool Body radiates x in all directions, including towards Warm Body. Warm Body radiates 2x in all directions, including towards Cool Body.
Warm Body cools more slowly than if Cold Body were not present
Overall heat transfer is not from Cool Body to Warm Body, so that 2nd law has not been broken, and nobody (pun intended) goes to Thermodynamic Jail.

RJ says:
March 10, 2011 at 4:17 am

Would a human being cook if he or she was enclosed in a container of CO2
In theory a percentage of the heat given off would return and increase the body temperature if the GHG theory is correct. Something surely is seriously wrong with the GHG theory.

NO. They would just cool down more slowly, so the overall temperature of the human would increase slightly until equilibrium (energy gained = energy lost) is reached. This assumes they are living (which is not a given in a container of CO2, BTW), and is generating a constant amount of energy from that living process.
I really fail to see why something so obvious is so hard to get. When you add a blanket at night, do you ‘cook’? No. Your temperature just increases until equilibrium (energy gained = energy lost) is reached. Try it!

Joe Lalonde says:
March 10, 2011 at 4:49 am

First, without planetary rotation, there is no convection

I think a lot of convection is due to differing temperatures. This is why you get a lot of wind when clouds come over on a sunny day. Also why the wind (assuming few clouds) goes toward the sea (which is warmer) in the first part of the day, and way in the latter part. Simple temp.

as the planets own energy is the centrifugal force it generates at 1669.8Km/hr due to the vacuum of space.

No. 1. there is no such thing as centrifugal force. It is simple inertia. Centrifugal force, does exist, however, and that is what stops everything flying off. In this case, gravity. The rest of the concept you put forward makes no sense to me.

The atmosphere bends a great deal of light and solar energy with the suspended molecules in the atmosphere with the tilting of the planet to the sun. The hottest point of the sun is it’s equator that our planet drifts through due to proximity and size of the suns equator.

Well, between the tropics, anyway. It differs through the year.

Next very little consideration for the absorption and storage of heat that is then released at night.

That is a false assumption. Just because sunlight is shown does not mean the rest of it stops when the sun goes down.

“”””” Ira Glickstein, PhD says:
March 10, 2011 at 5:21 pm
Phil. says:
March 10, 2011 at 8:21 am
… the Petty data is spectral radiance (mW/m^2.sr.cm-1) plotted vs. cm-1 and is correct, the transformation to a plot in terms of wavelength is non-linear (and is why it’s misleading for Ira to have reversed the axis on his plots, should leave it in wavenumbers).
Thanks for your comment, Phil. The Petty plots have both wavenumber and wavelength, and I preserved his wavelengths in my simplified curves when I reversed the plots horizontally. The reason I think in wavelength is that my experience with IR is based on near-IR 1.06μ laser rangefinders and far-IR ~10μ Forward-Looking Infrared video sensors. However, the conversion between the two measures is simple:
Wavenumber = 10,000/Wavelength
The reason for the difference in peaks has nothing to do with the use of Wavenumber or Wavelength.
As Phil noted, Petty plots Radiance.
My earlier posting plots Spectral Intensity. (It is from http://en.wikipedia.org/wiki/File:Atmospheric_Transmission.png.
From Wikipedia:
The units of the spectral intensity are energy/time/area/solid angle/frequency. In MKS units this would be W·m-2·sr-1·Hz-1 (watts per square-metre-steradian-hertz). “”””
Notwithstanding ANYthing that you may find on wikipiddle, the units of “spectral Intensity” ARE NOT energy/time/area/solid angle/frequency.
“INTENSITY” (fancy word for candlepower) has NO MEANING for anything but a POINT SOURCE; so with /area/ in there the intensity (spectral or not) would be infinite.
The /area/ for a source of non zero area, converts Intensity (spectral or ot) into RADIANCE ore STEARANCE which some snobs like to use.
)
As to the per frequency for the “spectral Intensity” or “spectral Radiance”, it could be either per frequency (or wavenumber, or per wavelength; so long as it is stated; either /cm^-1 or /micron.
I like to use per wavelength; because I much prefer to use a normalized Black Body Radiation plot, which is totally universal; and acknowledges that the Planck Radiation Function is a function of the single independent variable lambda.T which immediately points to the Wien Displacement Law.
So a normalized plot plots on a horizontal scale (usually logarithmic) whose variable is (lambda.T)/|lambda.T|max; and the Y-axis, is spectral radiance/ |spectral radiance|max and usually is plotted on both a linear or a logarithmic vertical scale.
So the BB curve typically goes from say (0.1) t0 maybe (100) in X and (0) to (1) in Y or 1.0 down to 10^-5 on the logarithmic plot.
An excellent graph can be found in “Modern Optical Engineering” by the late Warren J Smith, who was for years with Infra-Red Industries in Santa Barbara California.
Now yes you can reformulate all of this in terms of frequency or wavenumbers; and then Wien’s Displacement law becomes T/f is constant.
And yes a frequency based plot acknowledges the Einstein relationship E = h.nu; but as I have already noted, neither a wavelength or wave number graph is very intuitive (on the spectral interval) since one form favors the UV, and the other favors the far IR, as far as viually indicating relative importance. There are simply huge wave number ranges associated with those miserable looking spectral radiance values in the H2O bands; but the very prominent looking CO2 band (in the wave number pictorial) has a measly 100 cm^-1 for spectral bandwidth.
Note that 25% (almost exactly) of BB radiation occurs at shorter than the spectral peak Wavelength on a per micron spectral radiance plot; I have no idea how much lies below the peak on a per wave number plot; but you can easily work it out.
As I said some people like their BB plots in numbers of Photons, not in direct energy or power units.
But nyet on the spectral Intensity; the earth surface or atmosphere is not a point source.
For the pedantic, the actual IRRADIANCE on a surface due to a finite area source (on axis) differs from that due to a perfect point source by less than 1% for a source to surface distance that is ten times the source diameter.
Also there is no such thing as a point source of anything (physical). For EM radiation the boundary conditions for Maxwell’s Equations cannot be satisfied for a point source; and for the Quantum Physicists, a point source by virtue of the Heisenberg uncertainty principle must have an infinite spectral bandwidth; so the energy or power in any finite wavelength or frequency range would be zero so it would be undetectable at any wavelength or frequency.
I don’t have a problem with Ira’s use of both the wavenumber and wave length horizontal scale, as Ira said, that makes no difference, except as Phil pointed out, the wavelength conversion scale from wave number would be non-linear. But just keep the per wave number or per micron units straight in the sepctral Radiance units, and then everything is kosher.
Peter Humbug solved that dilemma by simply leaving out the per frequency altogether; but I’ll give him a typo yellow card on that.

Ira
Thanks for your further comments. I now understand why you suggested an electric blanket as analogy. However, I remain unconvinced that it is in fact a superior analogy. In fact in some ways, I liked the comment by Alan McIntire March 10, 2011 at 6:08 am comparing the position with traffic flow over a section of road with accidents, road works whathaveyou. My response to that is that yes at peak times, there are snarl ups and the journey time is increased, however, late in the evenning and at night, there is less and less traffic indeed during the night no traffic such that all the cars eventually exit the highway and the road is completely free come morning. All the traffic has disippated akin to all the energy received by the Earth finally finding its way out to space so that there has been no effective build up of energy.
I digressed, reverting to your electric blanket. Lets for the sake of argument assume that you are right with respect to the power source (although I have concerns regarding the lack of independent power source), one of the problems I have is how this all works in the real world (as opposed to in isolation in a laboratory), by which I mean that other atmospheric processes are at play which may overwhelm the effective influence of effective heat transfer from the sky to the Earth. In particular, convection. Further, does the energy absorbed by say CO2 thermalise before it has had an opportunity to radiate.
Say that you normally make your bed by creating a sandwich of a sheet, electric blanket and ordinary blanket. When the electric blanket is plugged in, this keeps you nice and snug. Now consider that you have a 4 poster bed but without any curtains. Instead of your electric blanket being placed on top of the sheet, it is suspended at the top of the 4 poster frame say about 3 feet above the matress. You now only have the sheet and the blanket directly over you. When the electric blanket is switched on, the down radiated heat is not sufficient to warm you, nor does the down radiated heat sufficiently prevent heat loss being radiated from your body. With this set up, you are cold at night. Even in the stillness of your bedroom, convection overcomes you feeling any real benefit from the down radiated heat. I am not saying that the electric blanket has no effect, only that it is overcome by other processes.
The fact is that radiation is weak compared to convection. I have gas central heating running radiators. When the system is on, even if I place my hand within 1/2 inch of the face of the radiator (ie., place my hand parallel with the radiator close to the surface of the panel), I struggle to feel the heat. Yet, if I place my hand several feet above the radiator, I can still feel the heat. Of course, if I touch the radiators they are very hot to touch and unpleasant to keep your hand there for any length of time.
I must admit that I struggle to see that the amount of energy that the Earth is said to receive from back radiation is more than the energy it receives from the sun in circumstances where the source of that back radiation has ultimately all come from the sun (if one ignores geothermal energy being produced from the centre of the Earth). If that were true, on a sunny day, there would not be a substantial difference in temperature between the temperature in the sun and under a cloud. At night, our bodies could be warmed and bask in the radiant energy from the sky which would be safer than sun bathing, and the photo electric cell would not be tuned to capture sun light but rather research would be directed at tuning it to work on back radiation which is available 24 hours a day and whether it is sunny or not. It appears to me that the heat transfer from the Earth from the ground to lower atmosphere is dominated by other processes (particularly the water cycle) which completely overwhelm the effects of back radiation.
I am not saying that there is no ‘greenhouse’ effect (how I dislike that expression) but rather in the real world, the effect of back radiation may be considerably less than is speculated (and I use this word since there is no body of empirical observational data backing up the claims) and I consider that the effect of any change in CO2 concentratioons above present levels is likely to be extremely modest. Indeed, Miskolczi could well be right on how it all works.
PS. Whilst I do not necessarily agree with various points you make, I liked your post and it certainly has generated many interesting comments and views.
PPS. I think you should consider carefully whether you wish to run with the eklectric blanket since this in my opinion in many ways confues the issue. Would not a better analogy be a one way mirror which is losing some of its mirrored surface such that it allows most of the incoming sunlight to pass through but only allows some of the LWR emitted from the Earth to pass back through towards space and reflects some part of the LWR back down towards the Earth.

All these posts about backradiation towards quantum mechanics made me go back to the basics, looking at it using the old thermodynamics and I ended with a result where there is backradiation but without any resulting effect. I dissected the first basic steps where radiation hits the surface and this seems enough to get answers.
In the radiative solution, the non-GHG surface temperature is found by using E = sigma T^4 (SB) where E comes only from the sun. If then T could go up because of GHG, this would mean more energy input would be needed. So where does this come from, are GHG capable of generating energy? Many people found a justification for this, but think of the following?
If we look at the CO2 idea as a plate with T2 above earth, you have something like the
classic radiation between two plates here, where the sun would heat the lower one which is T1.
So sure radiation will go both ways, there is backradiation and I go with that.
But read that in an end state (equation 19.2) where both plates would be T1, Qnet will be zero. Surface1 will never get any warmer by the (back)radiation from above, despite the accumulated (heat)energy in plate 2 and radiation going everywhere. Radiation is energy but is never heat.
So you can do this calculation with multiple radiation layers (like CO2 layers) and the
result will be the same. T1 will not get any warmer with Qnet getting zero, and all the
layers acting as insulation (like some people think CO2 does) and all plates actually
permanently getting radiation from both sides.
(Earth atmospheer would also be forced to become T1 everywhere form the surface up if it was’nt for gravity)
So what are the tricks with radiation that many people don’t see?
For starters I want to say that the photons of EM radiation can be regarde as cold, only interaction with matter gives any thermal energy (to get a temperature) which is taken away (partly) when a(nother) photon leaves.
Ideas where the delay of escaping photons by GHG causes heat accumulation have no basis, heat comes from a surface (matter) and cold energy went with the photons that took off. It does’nt matter how long it takes before the photon gets out to space (in the sun it takes up to 170.000 years)
Most greenhouse effect statements/sites claim that the radiative balance of earth must be because of the First law, the greenhouse theory makes fame by using energy in-out balances etc. convincing 99.99 percent of the people. But this is all dead wrong, the Second law rules here and this means simple energy conservation is not the whole story.
Earth could conserve it’s energy by accumulating it until it would be as hot as the sun,
nothing against the First law this way.
Is’nt it strange to say at first: well, because of energy conservation earth must
radiate away what it receives and that way we can calculate what the temperature must be by using SB and so we know for sure it must be -18C (not mentioning atmosfeer, and what is actualy measured).
Then secondly: listen, we also have GHG and this makes earth +15C because bla bla. So in fact saying: it’s wrong, earth does not simply radiate away what it receives and SB does not apply, but we can calculate the greenhouse effect using it.
Or: GHG introduces new physics, if one places multiple layers between two radiative
surfaces the old laws of physics don’t work anymore. Extra energy will turn into heat and raise the original surface temperature (think of the plates, it won’t happen).
So then in the GHG physics T of earth surface is supposed to go up with still the same
amount of energy coming from the sun. Now we know the Second law wants to establish a dynamic radiative equilibrium at the surface with a certain temperature but also T = dE/dS, so this means that for T to be able to go up by backradiation up entropy must decrease?? Strictly looking at this step.
Looking closer it must look like this:
Why does the sun heat the earth to T? Because HQ radiation from the sun can leave earth as LQ IR radiation (to the cold space) in the process gaining entropy by leaving WASTE HEAT on the surface (and they don’t call this irreversible for nothing, the downgraded (photon)energy won’t be able to do the same trick at this place again).
Some LQ IR returns (backradiation), but can never get rid of more heat. Because in that case it would have to leave as even lower quality IR, and hey …… this can only happen at a LOWER surface temperature (think of the BB-spectra of earth next to the sun, all the energy flows from the high spectrum to the low one).
See the contradiction, that’s the mighty second law of thermodynamics in action that says: it’s all wrong with this heating by backradiation philosophy.
So backradiation will simply leave in a reversible proces (like reflection does) with NO
waste heat, and with the same frequency as it had and bounce whatever way it wants with no effect on the surface. It’s like in the Qnet = zero situation between the two plates (lotsa radiation, no more heating by the cold photons).
So what is ‘the heat’ that can’t go from low to high temperature, and warming the earth surface and what’s all the confusion?
Sadi Carnot who laid the foundations of the Second law meant it to be like this: heat =
entropy (read here), but Clausius later changed the definition.
Maxwell’s classic Theory of Heat states: heat is something which may be transferred from to another, according to the Second law of thermodynamics.
In any case, heat needs matter which has a temperature (as kinetic energy) and other
matter at lower T. Radiation has no temperature and can never ever represent heat. And so the photons of radiation can be regarded as cold.
Heat is the waste from the irreversible radiation phenomena happening due to what the Second law dictates and is equivalent to the entropy gained and what is called
dissipation.
Heat is released when a photon interacts with matter. Almost everybody thinks ‘the
heat’ has to do with all the energy (E= h x v) related to the cold travelling photon. So
we have seen that this is not the case, this is imaginary heat (which is not heat but
energy) that is never released as long as matter is’nt involved. Energy is only released
when the photon disappears in the matter and after that heat from this matter may flow if other colder matter is there, but this heat is not equivalent with the photon energy but is a function of the temperature difference between surface temperature and colder matter above it (Q = k*A*dT). If there is no colder matter, there is no heat (dT=0).
If earth had no atmosfeer, the photons would warm earth surface to SB temperature but no heat would ever emerge or be exchanged. Then if there was a layer of CO2 around it at 10 meters or 10 km, this would create backradiation, but the vacuum inbetween would be as cold as in space, and the backradiation would not make the surface warmer, and there would be no heat despite all the radiation.
Now because heat is flowing from earth surface we know we have an irreversible process (waste heat, making entropy rise) then we also know for sure the new lower energy photon can never release waste heat at this temperature where it came from ever again, so to think backradiation can go beyond that and even create a higher temperature is impossible by the laws of thermodynamics.
This IR photon can release energy to matter at a higher colder level, and this energy
could flow as heat to matter at an even higher colder level, but that than is in fact
where this heat is. At a higher colder level, and this is the only direction heat can go.
So the heat coming from the surface is the part of the solar radiation that did’nt get
radiated away (directly) as IR (and the total energy of this IR is thus lower than the LWR that hits the surface). Heat is the part that has nothing to do with radiation, and the IR that left has nothing to do with heat.
And so the balance becomes: IR = LWR – HEAT. So the IR that leaves the surface does’nt only have a lower frequency, but the total energy of it is also lower than the incoming LWR. This is the second blow for the backradiation.
It is this HEAT that warms the lower atmosfeer and gives the temperature’s measured (in air, so does not even have to be representing the surface), and it can only take place by conduction from surface to air(surface), (or the process in the top ocean layer) it must be transported from matter at T1 to matter at T2 and it finds its way up to space dissected from the photons that gave up their energy.
As I Googled on the photons in the sun, the radiation from core to surface passed by on many pages and since these are not infected with the greenhouse virus the discriptions go along the lines of my view. Photons jiggle in the ‘random walk’ for tenthousands of years from layer to layer, and they do not heat the core. And also found; this degradation of high quality X-ray photons to relatively low quality optical photons, is only to be expected from the Second law of thermodynamics.
And look at this: a photon can only travel a tiny distance before running into another
hydrogen nucleus. It gets absorbed by that nucleus and the re-emitted in a random
direction. If that direction is back towards the center of the Sun, the photon has LOST
GROUND! It will get re-absorbed, and then re-emitted, over and over, trillions of times. The path it follows is called a "random walk"
— Sir Arthur Stanley Eddington:
"The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations — then so much the worse for Maxwell’s equations. If it is found to be contradicted by observation — well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation."

Commie Bob (March 10, 2011@5:41am)
Your point no. 5 is completely wrong. Lets dissect. You say “For those who doubt that back radiation exists, consider this: Infra-red radiation is electromagnetic radiation. It is the same as radio waves and light. While infra-red, radio and light may propagate similarly and be subject to the same law (inverse square) you have forgotten about wavelength. The obscure rays from an incandescent solid are vastly superior in their ability to heat than the luminous rays. You go on “Here are two examples of radiation from a weaker source going toward a stronger one:” Radiation commieBob is not heat. Radiation produces heat, and heat produces radiation but the radiation in and of itself is not heat, it is a byproduct. Radiation propagates without difficulty in space where there is no effective temperature(forget the 3k). I can even generate and emit emr without heat as the cause simply by plugging a transmitter into the wall and hooking up an antenna. Simply put the 2nd law does not apply to radiation at all. If it did all electromagnetic radiation emitted from the antennas of radio, tv, etc. would automatically head for colder climes like the interior of antarctica. The coldest would win(net flow). I won’t bother with your flashlight as it’s only vague misdirection. Just like agw. Backradiation as claimed is an energy absorption(conversion to heat)/reemittance towards the surface from some fictitious emitting altitude and raising surface temperature through fake amplification is clearly Bogus.
5b is vacous nonsense. “If I stand near a strong radio transmitter, I can still tune in weaker transmitters. Nothing about the stronger transmitter’s signal prevents the weaker signal from getting to me. Of course not! A strong 30 meter wave doesn’t care what a weak 20m wave does! They don’t interact because their wavelength’s or frequencies are different so the stronger signal has no possible way to prevent the weaker signal from getting to you. “The net energy flux will still be from the stronger source toward the weaker one.” Here you are implying some form of energy transfer between a 30m wave and a 20m wave or vice versa. Do you have a magic wand?
“It is a net flux though, in other words, the net flux is the difference between the two signals.” Oh, please commieBob do not emit any more pseudoscience as the net result is I can only laugh out loud.

Phil.
you say :Yes provided that hν for the photon matched the energy difference between the two higher energy levels (in a Q branch for example).
Aha! Eureka moment, maybe? This makes perfect sense! The photon does not exist between points, but instantaneously communicates information (in its reference frame) between the point where it is emitted and the point where it is absorbed. Of course the photon energy must match perfectly where it is absorbed (and emitted). This also explains why photons are made up of the same “stuff” as matter. They merely communicate the state of the matter, which is governed by expectation probabilities that are predicted by quantum mechanics. When one “decay” or energy level drop happens that emits a photon, the other state shift elsewhere is necessary and already mapped by the photon. That would explain the “spooky” experiments we learn about in quantum mechanics that involve photons with equal energies and opposite momentum. By observing the polarization of one, we “influence” the polarization of the other instantaneously at any distance. Of course, the photons must “know” instantly the properties of the other because they are the vehicles of communication relaying the change of states, and the photons already “know” of their destination, as they are not under the influence of time. The properties of the photon, the energy and momentum, are the communications between states of matter separated by a distance. This feels like a logical connection between spec. relativity and quantum mechanics.
Now, how does gravity fit in with all this?…
Need more help, Phil.

Gary P. (March 10, 2011)
Here’s a better explanation for your “experiment”. The radiation being emitted by your filament where previously was allowed free escape is now being reflected by the interior surface of your metal cylinder. Reflection is not absorption/reemission. Energy reflected is not energy absorbed. To invoke backradiation your filament needs to heat the cylinder hot enough to radiate sensibly back to the object. This is clearly not happening. Your cylinder was “relatively cold”. In fact by placing a round reflector around a centered filament you have created a rudimentary cavity whereby radiation formerly lost is now harnessed and focused centrally. The metal of the cylinder did emit no sensible radiation of its own. There is technically no hot to cold flow at all since you have removed air. Oh, and your emergency reflective blanket does reflect the small amount of heat a body produces but it does a far more important thing to keep one warm. Normal blankets are porous and allow easy convection and consequent air exchange. A reflective blanket is far less porous and much better at trapping the air mass around the body reducing the level of air exchange. This allows your body to warm the water vapor in your cocoon. No backradiation needed thanks!

Ira,
I follow your logic, but the electric blanket analogy was still a bad idea.
On the other hand I’m amused that half the criticism comes from either complaints about nomenclature, or the “cold things can’t warm up warm things” argument.
For the cold things can’t heat up warm things crowd, please stop with the complicated explanations and torchered arguments. Just ask yourself, do igloos work or don’t they? You can quote as many definitions of laws of thermodynamics as you want, construct as many analogies as you wish, with plates, bowling balls, flashlights, coats….but the fact is that igloos work. They’re made of cold stuff like ice and snow, and a person can survive very severe cold weather inside of one because “back radiation” keeps them warm. Someone dumb enough to sleep outside the igloo instead of inside is just a cryogenic experiment.

nighttime says:
March 11, 2011 at 12:33 am
do the inuit build fires in their igloos.>>>
sigh. i suppose they just might sometimes which has what to do with anything?
If you insist I exclude the possibility of fire keeping them warm….
An igloo, constructed of only snow and ice, containing no additional heat sources of any sort other than what is generated by the human being(s) within them, works.
The dope sleeping outside next to a fire on the other hand will still be a cryogenic experiment with a cooked spot opposite the fire.

@ Dave Springer says:
March 10, 2011 at 4:30 pm
Hi Dave,
About your reply to my post, I agree with your explanation of what is seen at the arctic.
I would like to know your opinion about the small emission peak in the middle of the CO2 15um absorption pit.
It is my opinion that that could be the only real backscattered photon emission of CO2, and that the other emissions could be due just to temperature of the IR active gases in the atmosphere.
Maybe I’m wrong with that, but I believe that because the “peak” is always positive at ground and at the TOA, and looking down at the tropical atmosphere that peak appears only over the 19-20km where the CO2 became the predominant IR active gas.
I repeat, I could have wrote a very silly thing, I’m not an atmospheric physicist just an electronic engineer.

davidmhoffer says:
March 11, 2011 at 12:15 am
David, this is an interesting point but you are wrong about how igloos work. They do not work by back-radiation.
Igloos maintain a higher temperature than the outside environment in the same way that all enclosures, including greenhouses, do.
This is quite simple and holds for all internal 3 dimensional environments enclosed by two dimensional barriers.
It is the result of the reduced probability of all three modes of energy transfer from a three dimensional gaseous environment towards and absorption through the two dimensional solid barrier (wall, glass or in the case of igloos, ice). The further from the two dimensional surface, the lower the probability is that the energy will be able to leave the enclosed atmospheric environment.
Historically, traditional dwellings tended to be constructed around a central fireplace for this very reason.
I live in a house high on the Sussex Weald which dates back more than 200 years. The are 4 fireplaces on the ground floor built into one central brick stack in the centre of the house. The external walls are merely 25 mm weather boarded with a 50 mm cavity and the internal skin is just 15 mm plaster board. There is no cavity wall insulation whatsoever and no double glazing. Thats just 2″ inches of wood and plasterboard between me and the elements 600 feet above sea level high on the Sussex Weald.
In the winter it is evident that the above holds true in that the closer you are to the external walls on a cold night, the colder it feels. The same is true for greenhouses. Ask any horticulturalist which plants are his smallest and lowest yielding and he will tell you it is the plants nearest the glass. Ask any Eskimo which part of the igloo he prefers to sleep in and I guarantee he will tell you, “the middle”.
This is what the “greenhouse effect” is. The greenhouse effect is simply the reduced probability of energy loss from all three modes of transfer, towards and through an external two dimensional surface from a three dimensional gas.
This fact holds true for all internal environments from glass houses to timber clad houses and even to igloos.
It does not hold true for the open atmosphere and therefore there is no “greenhouse effect” in the atmosphere.
The two dimensional barrier, be it timber, brick, glass or even ice, simply inhibits all three modes of energy transfer. This is the so called “greenhouse effect”.
That is why humans have evolved to construct and reside in dwellings.
There is no “greenhouse effect” in the atmosphere.

RJ;
but it helps that posters like Hans and other explain why in this way.>>>
I am not just a skeptic, but a raging skeptic. There is so much hooey in IPCC AR4 that they should replace the 4 with an S. Which is why it drives me crazy when people go to great lengths to explain why there is no backradiation or it defies the laws of thermodynamics, or it creates energy… simple things like an igloo show that there IS backradiation, cold things CAN radiate heat to warm things, and NONE of it breaks any laws of physics.
Here’s my oversimplified explanation on a photon by photon basis. Frankly it p**sed me off to no end to figure out that the surface does in fact get warmer (thought by darn little and maybe even cooler when all interelated effects are included) but as for the concept of backradiation, there is nothing wrong with the basic ghg explanation other than presumed order of magnitude:
http://knowledgedrift.wordpress.com/2011/02/27/co2-exactly-how-does-it-warm-the-planet/

Phil and Ira
Phil, I found your comments (March 10, 2011 at 7:52 pm) very interesting. Is this based upon an actual experiment or only a theoretical calculation?
What I am getting at is whether this reflected (back) radiation has genuine power to do work.
Consider the attached power plant:. http://www.solar-green-wind.com/wp-content/uploads/2009/12/largest_solar_power_station-1528.jpg
Consider the Earth’s energy budget:
http://www.klimaatfraude.info/images/EnergyBudgetNew.jpg
The power station is using mirrors tuned to collect and focus 184 w/sqm (i.e., 161 + 23) of solar bandwidth energy. Of course, during a sunny day, these mirrors are also receiving some (or all) of the 333 m/sqm back radiation. However, they are not focused to collect that nor are they tuned to the wavelength of downwelling long wave radiation. They are of course, tracking the elevation of the sun and therefore are effectively tuned to collect and focus the 184 w/sqm solar bandwidth energy. I presume that this power station produces no power at night and presumably only modest amounts on cloudy days and may be none on heavy dark rain cloudy days. It can effectively only work in a sunny climate such as California, Spain, Affrica (although of course, the illustrated station appears to be in Germany but Germany are beginning to realise that the country is not well suited to solar power generation).
My question is: If there was real power in the 333m/sqm longwave back radiation, why isn’t the power station designed to collect and focus this longwave radiation? There is nearly twice as much energy (viz 333 cf 184 m/sqm) and this longwave energy is available 24 hours a day come rain or shine. This eliminated the storage issue which has always beset green energy projects.
If there was real power (ie, the ability to do work) in this backradiation it is inconceivable that someone would not be collecting it, or that there would not be mainstream research designed at collecting this. For the vast majority of countries solar radiation is a non starter (due to cloudiness and latitude) but these countries have plenty of backradiation.
I guess I am raising a question on the point made by Hans (Hans says:
March 10, 2011 at 8:00 pm) in that the radiation may be there but it does not have any energy/ability to do work.
Phil and Ira your further views/comments would be appreciated.

Phil. says:
March 10, 2011 at 4:32 pm
“If you use Wien’s Law to determine the maximum there are two forms, a frequency form and a wavelength form, the peak frequency does not correspond to the peak wavelength using c=λν.”
The superimposed dashed line blackbody curves don’t correspond to the frequency/wavenumber on the horizontal axis. If you plug in the peak frequency indicated on the horizontal axis to Wien’s Law formula you get a very different temperature in Kelvin from that labeled on the dashed curves. Interestingly the labeled values correspond to degrees Rankine!
You can find it with calculator below which gives results in F, C, K, and R.
http://www.ajdesigner.com/phpwien/wien_equation_t.php
Something is definitely whacked in those plots but I can’t quite figure out what it is.

davidm
I don’t think people have said there is no back radiation. What I think they have said is this back radiation does not result in a colder body warming a warmer body from this back radiation.
Cold things can radiate to hot things. But it will not warm the hotter thing when the radiation arrives is how I understand this process.
Its like a brilliant teacher passing knowledge to a student. The student can pass knowledge back to the teacher but this will not improve or add to the teachers knowledge (heat). But the student can improve the knowledge of younger child. The same with the sun earth CO2 relationship.
And I see you as a luke warmer not a true sceptic. You are sceptical of the overstated alarmists predictions not the GHG science itself.
Even though the GHG science could lead to the chicken in the oven situation. Or a person in a CO2 container cooking themselves due to back-radiation.

RJ says:
March 11, 2011 at 4:33 am
“I don’t think people have said there is no back radiation. What I think they have said is this back radiation does not result in a colder body warming a warmer body from this back radiation”
Correct. But it causes the warmer body to cool at a slower rate.
Radiative transfer between two objects of different temperature is a two way street.
Say the warmer body is radiating at 2 w/m^2 and the cooler body is radiating at 1 w/m^2. The net flow is, of course from warmer to colder, at exactly 1 w/m^2.
The warmer body is adding energy to the cooler body but the cooler body is also adding energy to the warmer body. The cooler body is simply adding less energy to the warmer body than the warmer body is adding to the cooler body. This will always be the case. In most situations we only talk about the net flow of energy which always goes from warmer to colder but in reality there is flow in both directions and one flow is larger than the other. This smaller flow is “back radiation” and it is real beyond dispute.

Dave Springer
……”In most situations we only talk about the net flow of energy which always goes from warmer to colder but in reality there is flow in both directions and one flow is larger than the other. This smaller flow is “back radiation” and it is real beyond dispute.”……
There is radiation both ways but the radiation from the colder surface has fewer photons of every wavelength than is leaving the hot surface.
Also the QUALITY of the radiation is less as Hans pointed out and is restricted to the temperature profile of the colder surface.
This means that the colder surface radiation from say 300K can only effect a cancellation of that part of the output profile of say a 400K warmer surface.
That why it’s a misuse of language to say that the colder surface “heats” or “warms” the hotter surface.
To say that the cold surface insulates the warmer surface is a more accurate description of the process.
It also agrees better with your two rocks illustration

Ira
RJ does not know re back radiation. I am just trying to understand this issue nothing more. Is the GHG theory flawed or not.
Re evidence of radiation. What are your views on this
http://slayingtheskydragon.com/Latest-News/climate-follies-encore.html
Lord: “Back radiation can be simply demonstrated by pointing a simple infrared detector at the underside of a cloud. Try it.”
Chorus: “My IR detector only cost $60! Simple! Agreed! Agreed!”
Slayer: “Clouds do not absorb and re-radiate heat back to Earth. Clouds add THERMAL MASS which takes longer to heat and cool. Warmists ‘support’ this false hypothesis with IR thermometer readings, but the IR readings of a hot Barbie is the same from any distance; ENERGY is not. Your $60 REMOTE thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the Barbie.”
We know that this was the final act of the follies for there has been no reply in 36 hours from the Lord or the Luke Chorus.
At this point from what I have read I do not believe that a colder body can heat a warmer body. Otherwise we get a ridiculous situation where a chicken above zero could be cooked by nothing more than back radiation.
My views at this point are
That most of the energy leaves earths surface by conduction not radiation
The amount of CO2 is far to small to heat earth even if back radiation does occur and does heat the earth. (Which I doubt)
Colder bodies can not heat a warmer one. So CO2 might slow the rate of cooling but can not warm earth.

Dave Springer says:
March 11, 2011 at 4:30 am
Dave Springer,
Your reply to wayne regarding the oneway insulating effect of GHG’s would only be valid if the Incoming EMR were entirely or even mostly SW.
But this is not the case as it is mostly LW. Some say 50-50 but that is not the whole picture.
Your point only has validity if incoming EMR is mostly SW. It is not and therefore you are as far from reality as it is possible to be.
The SW EMR (light) emitted by the sun is a by product of the suns extreme temperature, not the other way round. Which is why there is a 600 km high bulge in the atmosphere under the solar point covering 25% of the atmospheres surface area, tracking the solar point around the Earth 24/7 called the Diurnal Atmospheric Bulge.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1966SAOSR.207…..J&db_key=AST&page_ind=0&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES

” wayne says:
March 11, 2011 at 3:17 am
Hans:
Brian H:
Brian:
Yes, I totally agree with Hans explanation”
Thanks Wayne, Brian, and RJ. I hoped there would be at least some readers here making it to the end of my bad English analysis.
“Ira Glickstein, PhD says:
March 11, 2011 at 6:23 am
RJ and others seem to accept the existence of “back radiation” but doubt that it can warm the surface of the Earth because it comes from a cooler source, the bottom of the Atmosphere. Thanks for setting them straight by reiterating that radiation goes both ways, with the Earth heating the bottom of the Atmosphere and the bottom of the Atmosphere heating the Earth, with the net flow of energy going from the Earth to the Atmosphere.”
Ira,
radiation goes both ways but radiation is not heat. Radiation is cold and is able to create heat and that heat is not related to the photon energy. It happens only if there is matter, and there must be matter at Thot and at Tcold. Only then heat emerges as the energy flow from hot towards cold. This heat is a function of the surface temperature not of the photons as I pointed out here.

Small correction
The consequence for that square metre of having the hotter object there is the up-conversion of 100J of 15um centred radiation into 100J of 5um centred radiation.
Should read
The consequence for that square metre of having the hotter object there is the up-conversion of 100J of 15um centred radiation into 100J of 5um centred radiation to add to the effect of the hotter surface.

Ira Glickstein said:
“Thanks Dave Springer for clearly explaining how “back radiation” from the Atmosphere to the surface of the Earth reduces the net flow of energy from the Earth to Space. This flow reduction causes the Earth to warm until, at the resultant higher temperature, the surface emits more longwave radiation such that energy in = energy out, and average temperatures stabilize.”
So now we have a new, sceptic, version of greenhouse warming. Backradiation does not directly add heat to the Earth but by reducing the rate of radiative cooling “it causes the Earth to warm”. Forgive me for not being able to distinguish between these two fallacies.
A heated body’s molecules are vibrating in accordance with the energy it is absorbing. A cooler body emitting towards the hotter body cannot heat it up any more because the emissions (backradiation) are not at the energy level to make the molecules vibrate any faster. Likewise the radiative cooling version of greenhouse theory cannot make the hotter body’s molecules vibrate any faster. Of course there is such a thing as radiative cooling, but in a constant irradiance model (such as Kiehl & Ternberth’s) the Earth is not cooling down so cutting off the flow of radiant energy cannot add any heat as it cannot make the incoming energy vibrate the molecules any more. Reduction in radiative cooling may have an effect at night (and if so the Earth will be less cold than it otherwise would be) but this is not the basis of the ‘greenhouse effect’ which is about a build up of additional heat below a radiation barrier until it is breached and a fictitious radiative equilibrium is arrived at. Further, the net flow argument which is used to demonstrate compliance with the laws of thermodynamics in this theory is a fraud. On this argument you could have any amount of backradiation – a million watts for example – but as the flow out would still be greater than the flow in (1 million plus the 66 watts of solar) then there would be compliance.

Phil. says:
What on earth is the ‘quality’ of radiation?
The ‘quality’ of radiation or energy generally is its ability to do work or in other words turn into other energy forms.
If you google the appropriate words you will find plenty of hits particularly from solar energy capture systems.
In the meantime here are two sources
http://www.scribd.com/doc/31017446/The-Second-Law-of-Thermodynamics
http://www.ijoticat.com/index.php/IJoT/article/viewFile/185/171