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## The Real Second Law of Thermodynamics

Many people often claim that the atmosphere cannot “heat” the earth’s surface and therefore the idea of the inappropriately-named “greenhouse” effect is scientifically impossible. The famous Second Law of Thermodynamics is invoked in support.

Let’s avoid a semantic argument about the correct or incorrect use of the word “heat”.

I claim that energy from the atmosphere is absorbed by the surface.

This absorbed energy has no magic properties. If the surface loses 100J of energy by other means and gains 100J of energy from the atmosphere then its temperature will stay constant. If the surface hasn’t lost or gained any energy by any other means, this 100J of energy from the atmosphere will increase the surface temperature.

I also claim that because the atmosphere is on average colder than the surface, more energy is transferred from the surface to the atmosphere compared with the reverse situation.

Let’s consider whether this violates the real second law of thermodynamics..

### The Conceptual Problem

In Heat Transfer Basics – Part Zero a slightly off-topic discussion about the “greenhouse” effect began. One of our most valiant defenders of the imaginary second law of thermodynamics said:

An irradiated object can never reach a higher temperature than the source causing the radiation

I have demonstrated previously in The First Law of Thermodynamics Meets the Imaginary Second Law that a colder body can increase the temperature of a hotter body (compared with the scenario when the colder body was not there).

In that example, there was more than one source of energy. So, with this recent exchange in Heat Transfer Basics it dawned on me what the conceptual problem was. So this article is written for the many people who find themselves agreeing with the comment above. As a paraphrased restatement by the same commenter:

If the atmosphere is at -30°C then it can’t have any effect on the surface if the surface is above -30°C

### Entropy Basics and The Special Case

Entropy is a difficult subject to understand. Heat and temperature are concepts we can understand quite easily. We all know what temperature is (in a non-precise way) and heat, although a little more abstract, is something most people can relate to.

Entropy appears to be an abstract concept with no real meaning – nothing you can get your hands around.

The second law of thermodynamics says:

Entropy of a “closed system” can never reduce

Before defining entropy, here is an important consequence of this second law:

Increasing entropy means that heat flows spontaneously from hotter to colder bodies and never in reverse

This fits everyone’s common experience.

• Ice melts in a glass of water
• A hot pan of water on the stove cools down to room temperature when the heat source is removed
• Put a hold and cold body together and they tend to come to the same temperature, not move apart in temperature

And all of these are easier to visualize than a mathematical formula.

What is entropy?  I will keep the maths to an absolute minimum, but we have to introduce a tiny amount of maths just to define entropy.

For a body absorbing a tiny amount of heat, δQ (note that δ is a symbol which means “tiny change”), the change in entropy, δS, is given by:

δS = δQ / T, where T is absolute temperature (see note 1)

It’s not easy to visualize – but take a look at a simple example. Suppose that a tiny amount of heat, 1000 J, moves from a body at 1000K to a body at 500K:

Example 1

The net change in energy in the system is zero because 1000J leaves the first body and is absorbed by the second body. This is the first law of thermodynamics – energy cannot be created or destroyed.

However, there is a change in entropy.

The change in total entropy of the system = δS1 + δS2 = -1 + 2 = 1 J/K.

This strange value called “entropy” has increased.

Notice that if the energy flow of 1000J was from the 500K body to the 1000K body the change in entropy would be -1 J/K. This would be a reduction in entropy – forbidden by the real second law of thermodynamics. This would be a spontaneous flow of heat from the colder body to the hotter body.

Updated note Sep 30th – this example is intended to clarify the absolute basics.

Think of the example above like this – If, for some reason, in a closed system, this was the only movement of energy taking place, we could calculate the entropy change and it has increased.

The example is not meant to be an example of only one half of a radiative energy exchange. Just a very very simply example to show how entropy is calculated. It could be conductive heat transfer through a liquid that is totally opaque to radiation.

### The Special Case

The simplest example demonstrating the second law of thermodynamics is with two bodies which are in a closed system.

Let’s say that we have a gas at 273K (Body 1) and a solid (Body 2) surrounded by the gas. The solid starts off much colder.

What is the maximum temperature that can be reached by the solid?

273K

Easy. In fact, depending on the starting temperature of the solid and the respective heat capacities of the gas and solid, the actual temperature that both end up (the same temperature eventually) might be a little lower or a lot lower.

But the temperature reached by the solid can never get to more than 273K. For the solid to get to a temperature higher than 273K the gas would have to cool down below 273K (otherwise energy would have been created). Heat does not spontaneously flow from a colder to a hotter body so this never happens.

This defining example is illuminating but no surprise to anyone.

It is important to note that this special case is not the second law of thermodynamics, it is an example that conforms to the second law of thermodynamics.

The second law of thermodynamics says that the entropy of a system cannot reduce. If we want to find out whether the second law of thermodynamics forbids some situation then we need to calculate the change in entropy – not use “insight” from this super-simple scenario.

So let’s consider some simple examples and see what happens to the entropy.

### Simple Examples

What I want to demonstrate is that the standard picture in heat transfer textbooks doesn’t violate the second law of thermodynamics.

What is the standard picture?

From "Fundamentals of Heat and Mass Transfer, 6th edition", Incropera and DeWitt (2007)

This says that two bodies separated in space both emit radiation. And both absorb radiation from the other body (see note 2).

The challenging concept for some is the idea that radiation from the colder body is absorbed by the hotter body.

We start with Example 1 above, but this time we consider an exchange of radiation and see what happens to the entropy of that system.

Example 2

What I have introduced here is thermal radiation from Body 2 incident on Body 1. We will assume all of it is absorbed, and vice-versa.

According to the Stefan-Boltzmann equation, energy radiated is proportional to the 4th power of temperature. Given that Body 2 is half the temperature of Body 1 it will radiate at a factor of 24 = 2 x 2 x 2 x 2 = 16 times less. Therefore, if 1000J from Body 1 reaches Body 2, then 62J (1000/16) will be transmitted in the reverse direction. However, the exact value doesn’t matter for the purposes of this example.

So with our example above, what is the change in entropy?

Body 1 loses energy, which is negative entropy. Body 2 gains energy, which is positive entropy.

δS1 = -(1000-62)/1000 = -938 / 1000 = -0.94 J/k
δS2 =(1000-62)/500 = 938 / 500 = 1.88 J/K

Total entropy change = -0.94 + 1.88 = 0.94 J/K.

So even though energy from the colder body has been absorbed by the hotter body, the entropy of the system has increased. This is because more energy has moved in the opposite direction.

There is no violation of the second law of thermodynamics with this example.

Now let’s consider an example with values closer to what we encounter near the earth’s surface:

Example 3

This isn’t intended to be the complete surface – atmosphere system, just values that are more familiar.

Surface:        δS1 = -(390-301)/288 = -89 / 288 = -0.31 J/k
Atmosphere: δS2 = (390-301)/270 = 89 / 270 = 0.33 J/K

Total entropy change = -0.31 + 0.33 = 0.02 J/K.

So even though the temperatures of the two bodies are much closer together, when they exchange energy, total entropy still increases.

Energy from the colder atmosphere has been absorbed by the hotter surface and yet entropy of the system has still increased.

Now, the example above (example 3) is an exchange of a fixed amount of energy (in Joules, J). Suppose this is the amount of energy per second (Watts, W) or the amount of energy per second per square meter (W/m²).

If the atmosphere keeps absorbing more energy than it is emitting it will heat up. If the earth keeps emitting more energy than it absorbs, it will cool down.

If example 3 was the complete system, then the atmosphere would heat up and the earth would cool down until they were in thermal equilibrium. This doesn’t happen because the sun continually provides energy.

### The Complete Climate System

The earth-atmosphere system is very complex. If we analyze a long term average scenario, like that painted by Kiehl and Trenberth there is an immediate problem in calculating the change in entropy:

From Kiehl & Trenberth (1997)

[Note from Sep 28th – This section is wrong, thanks Nick Stokes for highlighting it and so delicately! Preserved in italics for entertainment value only..] If we consider the surface, for example, it absorbs 492 W/m² (δQ = 492 per second per square meter) and it loses 492 W/m² (δQ = -492 per second per square meter).

Net energy change = 0. Net entropy change = 0.

Why isn’t entropy increasing? We haven’t considered the whole system – the sun is generating all the energy to power the climate system. If we do consider the sun, it is emitting a huge amount of energy and, therefore, losing entropy. But the energy generation inside the sun creates more entropy – that is, unless the second law of thermodynamics is flawed.

[Now the rewritten bit]

Previous sections explained that calculations of entropy “removed” (negative entropy) are based on energy emitted divided by the temperature of the source. And calculations of entropy “produced” are based on energy absorbed divided by the temperature of the absorber. In a closed system we can add these up and we find that entropy always increases.

So the calculation in italics above is incorrect. Change in entropy at the surface is not zero.

Change in entropy at the surface is a large negative value, because we have to consider the source temperature of the energy.

So as Nick Stokes points out (in a comment below), we can draw a line around the whole climate system, including the emission of radiation by the sun (see example 4 just below). This calculation produces a large negative entropy, because it isn’t a closed system. This is explained by the fact that the production of solar energy creates an even larger amount of positive entropy.

Example 4

### The Classic Energy Exchange by Radiation

I was in the university library recently and opened up a number of heat transfer textbooks. All of them had a similar picture to that from Incropera and DeWitt (above). And not a single one said, This doesn’t happen.

In any case, for someone to claim that an energy exchange violates the second law of thermodynamics they need to show there is a reduction in entropy of a closed system.

But one important point did occur to me when thinking about this subject. Let’s reconsider our commenter’s claim:

An irradiated object can never reach a higher temperature than the source causing the radiation

As I pointed out in the The Special Case section – this is true if this is the only source of energy. Yet the surface of the earth receives energy from both the sun and the atmosphere.

If the colder atmosphere cannot transfer energy to a warmer surface, and the second law of thermodynamics is the reason, the actual event that is forbidden is the emission of radiation by the colder atmosphere. When the colder atmosphere radiates energy it loses entropy.

After all, the entropy loss takes place when the atmosphere has given up its energy. Not when another body has absorbed the energy.

Our commenter has frequently agreed that the colder atmosphere does radiate. But he doesn’t believe that the surface can absorb it. He has never been able to explain what happens to the energy when it “reaches” the surface. Or why the surface doesn’t absorb it. Instead we have followed many enjoyable detours into attempts to undermine any of a number of fundamental physics laws in an attempt to defend “the imaginary second law of thermodynamics”.

### Conclusion

Entropy is a conceptually difficult subject, but all of us can see the example in “the special case” and agree that the picture is correct.

However, the atmosphere – surface interaction is more complex than that simple case. The surface of the earth receives energy from the sun and the atmosphere.

As we have seen, in simple examples of radiant heat exchange between two bodies, entropy is still positive even when the hotter body absorbs energy from the colder body. This is because more energy flows from the hotter to the colder than the reverse.

To prove that the second law of thermodynamics has been violated someone needs to demonstrate that a system is reducing entropy. So we would expect to see an entropy calculation.

Turgid undergraduate books about heat transfer in university libraries all write that radiation emitted by a colder body is absorbed by a hotter body.

That is because the first law of thermodynamics is still true – energy cannot be created, destroyed, or magically lost.

### Other Relevant Articles

The Amazing Case of “Back Radiation” – Part Three

The First Law of Thermodynamics Meets the Imaginary Second Law

Intelligent Materials and the Imaginary Second Law of Thermodynamics

Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics

### Notes

Note 1 – There are more fundamental ways to define entropy, but it won’t help to see this kind of detail. And for the purists, the equation as shown relies on the temperature not changing as a result of the small transfer of energy.

If the temperature did change then the correct formula is to integrate:

ΔS = ∫Cp/T. dT (with the integral from T1 to T2) and the result,

ΔS = Cp log (T2/T1),   this is log to the base e.

Note 2 – This assumes there is some “view factor” between the two bodies – that is, some portion of the radiation emitted by one body can “hit” the other. Just pointing out the obvious, just in case..

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### 158 Responses

1. Scienceofdoom:

Another excellent post! Much better than my take (which of course I’m unable to restrain myself from sharing).

As it can be shown that all bodies radiate heat, were it true that this fact violated the second law of thermodynamics, then the second law would be wrong, not the (provable) fact that all bodies radiate heat. Seeing as no one is contesting the validity of the second law, it is safe to say there is no violation.

Ok, I should have said EMR instead of heat, but the gist is correct.

Cheers!

JE

2. SoD,
I’m glad you have got into a proper definition of entropy. There’s a whole associated story of the flux of entropy through the atmosphere.

You can track entropy as with heat, cf Trenberth. It isn’t conserved, but once created cannot be destroyed, although it can be moved to somewhere else. And the earth story is how low-entropy SW comes in, how entropy is created when it is thermalised, how life processes etc create further entropy. In the end it is all exported (the system is stationary), and the greenhouse effect has a vital role.

You can mark off regions of space and do an entropy budget, as with energy. It won’t balance, but with proper accounting at boundaries, entropy within a box can only increase. So in a box including the surface, thge big term is the entropy increase resulting from SW, which could potentially heat something to 5700K, being reduced to thermal IR. It’s true that IR from cold air comes in too, and the heat associated with it undergoes an entropy decrease (free energy increase). But on balance entropy does not decrease.

Whenever heat does flow irreversibly from hot to cold, there is potential for a heat engine. This is what drives atmospheric motions, and indeed, life. But it doesn’t happen at the surface, where no useful work is done as SW is converted to heat at about 288K, and then to IR. Instead it happens as heat flows from the warm surface to the cold TOA. If the heat simply radiated to space, as it would without GHG, then no heat engine would be possible. The fact that GHG’s absorb “warm” IR and emit “cool” IR to sace from TOA makes life and weather possible.

To complete the entropy budget, you have to look at what goes out to space. This is negative entropy for the Earth (a gain by “out there”). The heat flux is fixed (about 235 W/m2), but to get the entropy flux you divide by the emission temperature. The fact that this, due to GHG and the lapse rate, is low, means that entropy is exported more efficiently. Without this, all the entropy-creating processes on earth (eg life) would not be possible.

The whole Earth bottom line is – 235 W/m2 comes in at 5700K and departs at about 255K. Net entropy created and exported – 235*(1/255-1/5700).

• Just for the record, 5700K is not a temperature for these heat transfer purposes. It is a color that approximates the spectral distribution of the 1350 w/m^2.

• Rigth,

If some temperature is chosen as the upper temperature for this kind of consideration, it’s the temperature at the point where the SW is absorbed or some effective average of those temperatures.

Radiation that reaches the Earth does not have any thermodynamic temperature, matter may have a temperature and radiation may be in equilibrium with matter, but radiation that reaches Earth is not in such an equilibrium.

3. Nick Stokes:

I think you have correctly pointed out that one part of what I wrote, “The Complete Climate System” is wrong.

The problem is all about drawing a line around a closed system. And I can’t draw one around the surface (I shouldn’t have) as it isn’t closed to the atmosphere. And I can’t draw one around the climate because it isn’t closed to solar energy..

But in any case I have written rubbish at that point and will fix it up.

Later note – now fixed up.

4. Quick reply before looking at post in more detail.

What has the Sun got to do with a calculation of “backradiation” at night “heating” the already warmer surface.
I think everyone on the Planet agrees that in daylight the Sun heats the Earth surface.
The example we discussed in Heat Transfer Basics – Part Zero was a night time atmosphere of -30C radiating 200J/s per metre squared onto an Earth surface at 15C.
In the example all 200J were accepted as being absorbed but when transformed into part of energy re radiated at 15C the average wavelength decreased showing that the “quality” of the radiation had improved contrary to the second law.

• on October 2, 2019 at 3:52 am | Reply LOL@Klimate Katastrophe Kooks

Nuclear fusion processes in the sun decrease entropy for the sun…. check.

Radiative emission from the sun increases entropy for the sun… check.

Radiative emission from the earth is “negative entropy for the Earth (a gain by ‘out there’).”, therefore radiative emission from the earth decreases Earth’s entropy?

Sun radiative emission increases entropy of the sun.
Earth radiative emission decreases entropy of the Earth?

Something’s not adding up.

5. @Bryan

You should read Nick Stokes comment above. When talking about thermodynamics and entropy in particular, great care must be used in defining your systems boundaries. You seem to forget that radiation is lost to space too.

Your night time atmosphere (243K) radiating 200W/m² toward the earths surface (288K) also radiates 200W/m² toward outer space (3K). Indeed entropy is lost when atmospheric radiation is absorbed by the earth’s surface but more entropy is created by the atmospheric radiation being absorbed by the interplanetary medium.

If you take ScienceOfDoom example 3 and reverse it to consider radiation emmited by the atmosphere and absorbed by the earth’s surface, you can see that entropy is lost when the radiation from the atmosphere is absorbed by the hotter surface :

Atmosphere: δS1 = -(390-301)/270 = -89 / 270 = -0.33 J/K
Surface: δS2 = (390-301)/288 = 89 / 288 = 0.31 J/K

This is right, net entropy equals -0.02J/K (as you say, “radiation quality” has increased since that energy will be re-emitted at a shorter wavelength)

Entropy is reduced. This looks wrong. And it is! Because in that calculation you forgot one term: the radiation from TOA to space :

If we suppose radiation from space to be j* = sigma * T^4 = 5.67*10^-8 * 3^4 = 4.59*10^-6 W/m²

Let’s consider the entropy budget for one square meter for a second :

Atmosphere: δS1 = -(390-0.00000459)/270 ~= -390/ 270 = -1.44 J/K
Space: δS2 = (390-0.00000459)/3 ~= 390/ 3= 130 J/K

Net entropy production from TOA-Space exchanges : 128.56 J/K

Net entropy production from atmosphere exchange with the earth’s surface and space : 128.56 J/K – 0.02 J/K = 128.54 J/K.

When you take all terms into account, no entropy is lost (even though some of it gets moved away.) In fact you see a massive amount of entropy created and, as Nick Stokes points out, it is the place where the entropy production is the largest (Sun-earth ~ 10 J/K.s.m², earth-atmosphere ~ 10^-2 J/Ksm² , atmosphere-outer space ~ 10² J/Ksm² )

I hope I got this right and did the calculation correctly. If I have, I think it should answer your objection.

6. propater

The question is a general one and is quite simple.
Does a colder surface raise the temperature of a surface at a higher temperature.
The night example was chosen deliberately to keep the problem as simple as possible.
Can 200J of radiation with a maximum centred around say 12um be increased in quality to 200J centred around 10um so as to increase the temperature of the hotter surface?
If allowed it would be consistent with the first law but IMHO it would violate the second law.
The reason I am reluctant to include the Sun in the calculation is that almost any perpetual motion device that satisfied the first law and failed the second could be rescued by such a sledgehammer argument.

• If you can harness energy from the sun without accounting for that energy, perpetual motion would appear to be possible. The sun provides the energy for a large number of processes here on earth and NET entropy increases during all of these processes.

7. @Bryan

Are you sure you are not confusing “heating” with “causing to cool more slowly than it would otherwise”?

8. Your 200J at 12um do not need to be “upgraded” to 10um to heat the surface. It is absorbed at 12um, the surface heat content has been increased by 200J (minus the part that has been reflected.) Temperature increases accordingly.

This is completly in accordance to the 1st law. There is no need to invoke the 2nd law here. The second law is not about energy. It is about information. And it postulates that the amount of information/certainty in/over a macroscopic system will always be stationnary or decrease in time. It says something about the way a system is ordered.

Here it is :

“In a system, a process that occurs will tend to increase the total entropy of the universe.”

It was first formulated by Clausius with regard to heat and temperature and the way heat flows in a macroscopic (that is for which the notion of temperature is well defined) system but ultimately it has more to do with information and disorder than about temperature and energy. For instance, you can calculate the entropy corresponding to one atom’s available electronic states (without taking heat or temperature into account.)

Here is a statistical definition of entropy :

S = Boltzmann’s constant * ln(Omega)

where Omega is the number states available to the system.

The “problem” that is addressed here with regard to the “greenhouse” effect is that at some point those 200J will be re-emitted at 10um. This correspond to an entropy decrease. That decrease would be a violation of the 2nd law if it was not compensated by the entropy increase resulting from TOA radiation to space.

You are right to say that at some place the entropy decreases but you are wrong to conclude that it constitutes a violation of the second law, because the entropy of the universe is increased when you take the whole process into account.

Once again this entropy decrease occurring when radiation at 12um “becomes” radiation at 10um can be considered in a way that does depend on information/disorder. Your notion of “radiation quality” is definitely about entropy and information (and not heat flows.) That is to say that the wave function of shortwave radiation is more information rich (more constrained, alows for less available states) than the wavefunction of longwave radiation.

To summarize, your opinion that it would violate the second law would be wrong. I think the article and the comments have shown how the phenomenon described does not violate the 2nd law (the total entropy of the universe is not decreased by it.)

I think your mistake is a misunderstanding and misaplication of the 2nd law based on Clausius’ statement (which is a special case) :

“Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature”

We do see (in this article, in others, in all the calculations) that more energy goes from the earth surface to the atmosphere than the other way around. So this is consistent with this formulation of the 2nd law. I, and others I guess, have a hard time understanding why you think otherwise.

I think the root of your misuderstanding and your disagreement is that you use the Clausius statement (which seems intuitive enough) but do not take care in its application. You do not define your systems boundaries correctly and do not account for all exchanges accross them (ie.: you forgot that radiation was also lost to space) or try to use it in non-macroscopic contexts (ie.: considering radiation wavelengths.)

One can not stress enough how important it is, when considering thermodynamics, to define precisely and correctly the systems considered, what their boundaries are and to correctly account for the energy transfers accross those boundaries.

That is how, I think, you end up with conclusions that seem intuitive but are, in fact, wrong.

9. @Bryan

I do not understand your last remark about perpetual motion machines and the sun.

I see no perpetual motion here and you can take the sun out of the picture if you like, it does not change anything.

In the end the earth’s surface is cooling toward the atmosphere and the atmosphere is cooling toward outerspace (the net energy flows from earth -> atmosphere -> space are positive all along the way, there is no way to build a perpetual motion on this.)

The main point here is that the rate at which the earth’s surface is cooling is slower in presence of an atmosphere compared to what it would be if the surface was cooling directly to outerspace. (It is logical: the cooler your cold source is, the faster your hot source cools down. If you raise your cold source temperature, your hot source cools down more slowly.)

• on October 2, 2019 at 5:25 am | Reply LOL@Klimate Katastrophe Kooks

A much-simplified alternative explanation:

In quantum field theory, every type of fundamental particle has its own quantum field ; fundamental particles are merely the spatially localized form of the smallest amount of a change in the quantum field… a quanta.

The universe is permeated by waves (although they’re not actually waves, they’re spirals, a consequence of the photon’s electronic and magnetic fields oscillating in quadrature, forming a circle. A circle translated through space-time is a spiral.

The image above shows the real (cosine… labeled ‘Re’ in the image) and imaginary (sine… labeled ‘Im’ in the image) components of an electromagnetic ‘wave’. When viewed in line with its direction of travel, it will appear to be a circle, and when viewed orthogonal to its direction of travel, it will appear to be a sinusoid, when in reality it’s a spiral.

This is because a sinusoid is a circular function.

You’ll note the peak amplitude of the sinusoid is analogous to the radius of the circle, the peak-to-peak amplitude is analogous to the diameter of the circle, and the frequency of the sinusoid is analogous to the rotational rate of the circle. You’ll further note the circumference of the circle is equal to 2 pi radians, and the wavelength of a sinusoid is equal to 2 pi radians, so the wavelength of the sinusoid is analogous to the circumference of the circle.

Thus the magnetic field and electric field (oscillating in quadrature) of a photon is a circle geometrically transformed into a spiral by the photon’s movement through space-time. This is why all singular photons are circularly polarized either parallel or antiparallel to their direction of motion. This is a feature of their being massless and hence having no rest frame, which precludes their exhibiting the third state expected of a spin-1 particle (for a spin-1 particle at rest, it has three spin eigenstates: +1, -1, 0, along the z axis… no rest frame means no 0-spin eigenstate). A macroscopic electromagnetic wave is the tensor product of many singular photons, and thus may be linearly or elliptically polarized if all singular photons comprising the macroscopic electromagnetic wave are not circularly polarized in the same direction.

All atoms and molecules (all comprised of fundamental particles, which are all comprised of quanta of waves) are harmonic oscillators.(with anharmonic force constants complicating the wave equation in the case of molecules).

This is why all matter has absorption peaks and valleys… wave resonance. Think in terms of everything in the universe composed of tiny antennae and waves.

Blackbodies absorb and emit at a wide range of wavelengths exemplified by the Planck curve, whereas a gas absorbs and emits at specific spectral lines. As a first approximation, atmospheric gases on Earth cannot and do not emit as a blackbody. The dominant source of blackbody radiation is transient oscillating dipoles induced by inter-molecular thermal vibrations within a material. Solids, liquids, plasma of sufficient density and gases of sufficient density can emit blackbody radiation, but our atmosphere cannot simply because the gas molecules spend the majority of their time relatively distant from each other, and thus they cannot sustain the inter-molecular oscillations necessary for blackbody radiation. As gas density increases, blackbody radiation production increases and eventually dominates the discrete emission spectra. Similarly, as gas density increases, blackbody absorption increases and eventually dominates the discrete absorption spectra.

So the universe is permeated by constantly-shifting energetic waves, with regions of higher and lower flux dependent upon the energetic processes occurring in that region of space-time.

Inherent to all invariant mass is a surrounding field (virtual photons) caused by Larmor radiation due to the bound electrons undergoing angular acceleration. This field has inherent to it field radiation pressure.

Photons, too, have field radiation pressure when they are incident upon invariant-mass matter… the angular momentum of the electronic and magnetic fields of a photon oscillating in quadrature is geometrically transformed to linear momentum upon incidence.

This field radiation pressure, for instance, is what prevents the sun from gravitationally collapsing. The process by which that field radiation pressure is generated also gives the sun its magnetic field.

Photons must be of sufficient energy (frequency) to overcome (electromagnetically cancel) this field radiation pressure in order to be emitted into the field or absorbed by an object.

As an analogy, think in terms of a bathtub filled with a fluid with special properties which will not allow any waves to form unless those waves are of greater energy than the energy of the bulk of the fluid. The waves, once formed, frictionlessly ‘ride along’ the surface of the fluid, but cannot be subsumed back into the bulk of the fluid until a region of the bulk fluid reduces its energy sufficiently to allow that wave back into the bulk fluid.

As two objects are coming into thermodynamic equilibrium, there will be photons emitted which are in transit between the two objects as the thermodynamic equilibrium state is passed through. Those photons are of insufficient energy to cancel the field radiation pressure of either object, and hence must be reflected, since they cannot be absorbed.

Simplistically, at exact thermodynamic equilibrium, the standing wave would preclude emission toward or absorption from either object due to the potential of each object being equal to the field radiation pressure of that standing wave.

Simplistically, if Object A is of higher temperature than Object B, Object A’s field radiation pressure is higher, thus the wave travels from A to B. And vice versa.

Of course, exact thermodynamic equilibrium can never really be achieved. It can be momentarily passed through, but never maintained. Heisenberg’s Uncertainty Principle (Fourier’s theorem for fields) ensures that there will always be at least a few quantum states which can be excited by absorption of (a few absorption resonances which will accept) the incident radiation.

We can move up one level to the fundamental particle level, with a higher abstraction of the concept.

We can move up another level to the atomic / molecular level, with yet more abstraction. This is the level you are discussing, in which it appears that radiation incident upon an object emitted by a lower potential object will be absorbed, and objects at a finite temperature emit under all circumstances.

The closer we get to the actual underlying physical reality, the less abstraction we deal with, and the more accurately we can describe energetic interaction on a per-interaction basis.

10. propater

……”The main point here is that the rate at which the earth’s surface is cooling is slower in presence of an atmosphere compared to what it would be if the surface was cooling directly to outerspace.”……

I agree with you and have repeated the same point many times.
The atmosphere acts like an insulator.
All 4 methods of heat exchange play their part.
The temperature time graphs of Earth with and without an atmosphere would show the temperature fall during nighttime.
The temperature of Earth with an atmosphere would not fall as quickly as without.
However;
At no time would the colder surface increase the temperature of the Earth Surface.
The maximum contribution of the colder surface would be limited to its own temperature(-30C lets say).

11. @Bryan

The point is not that the atmosphere “heats” actively the surface but that the atmosphere makes the surface hotter than it would be without an atmosphere (or an optically thinner atmosphere.)

If you increase the “insulating power” of the atmosphere, ie. by increasing its optical thickness, you end up with a hotter surface.

(by optical thickness, I mean the height, and so the temperature, at which the atmosphere becomes mostly transparent to longwave radiation and where radiation can be lost to space. I am not sure this is the right wording.)

The colder body does not increase the temperature of the hotter body by itself. It increases the hotter body’s temperature *compared to* what it would be otherwise.

Can you agree that, considering 2 scenarios :

A: a planet with an less insulating atmosphere
B: a planet with a more insulating atmosphere,

the surface temperature in scenario B is higher than in scenario A.

Then if you consider that the 2 planets are being provided each day with the same amount of energy from an outside source (to avoid their temperatures just going down to 3K.) I think you can agree that both planet’s temperatures will go up while we inject energy in their systems, then will go down when we stop.

I think you can also agree that planet A will see its temperature drop more rapidly than planet B’s.

Let’s consider we inject energy into our planets 12 hours a day then let them cool down 12 hours a day. If we do this for some time, we will see that the average temperature of both planet will stabalize around a mean temperature.

I think you can agree that this mean temperature will be lower for planet A than for planet B.

Then consider we change planet A’s atmosphere’s composition to make it as insulating as planet B’s atmosphere.

Do you not agree that, after some time, as we continue to heat them for 12 hours then let them cool down for 12 hours, planet’s A mean temperature will reach the same level as planet B?

Could you state at which point you disagree with me here?

Regarding my other post, do you still think the 2nd law of thermodynamics is violated when the earth’s surface absorbs 200J emitted by the atmosphere?

12. propater

……”Can you agree that, considering 2 scenarios :

A: a planet with an less insulating atmosphere
B: a planet with a more insulating atmosphere,”……

Yes but you cannot suddenly impose B on A and claim you have heated A
If B is suddenly imposed on A the rate of temperature drop will be reduced.
At no time will the temperature of A increase.

13. Bryan: You seem confused about the relation between Propater’s two planets. What do you mean by “you cannot suddenly impose B on A and claim you have heated A”? Propater’s contrast between A and B, which you seem to accept, implies that the subjunctive conditional, “if A’s atmosphere were to become more insulating (like B’s), then (barring other changes in the circumstances) A’s surface would become warmer than it is” is true. After all, the only difference supposed between A and B is this difference between their atmospheres. That is, the temperature of A would increase if A’s atmosphere came to resemble B’s. (Unless, of course, you simply ignore the fact that both continue to receive the same amounts of radiation energy from their suns.)

14. Bryan:

…”Yes but you cannot suddenly impose B on A and claim you have heated A
If B is suddenly imposed on A the rate of temperature drop will be reduced.
At no time will the temperature of A increase.”

If the two planets are identical. If their atmosphere are similar except for their radiative properties (which depends on their composition) If you alter A’s atmospheric composition to make it the same as B’s. You then have 2 identical planets with two identical atmospheres but A’s temperature is lower.

If A’s temperature does not rise to match B’s, how do you explain two identical planets receiving regularly the same amount of energy, loosing the same amount of energy to space (since their atmospheres are now identical) but having different temperatures?

You admit yourself that the rate at which A looses energy is now slower.

So, if we consider we instantly change the atmospheric composition of A at the end of a “day”, it now cools slower and at the end of the “night”, its temperature is higher than at the end of the previous night. And A will never be able to cool down to the low temperatures it reached before. That in itself is enough to change the mean temperature upward. But it gets worse.

A new “day” starts. The planet receives the usual amount of energy, but it starts at a higher temperature. At the end of that day, its temperature will be higher than the day before (because it started the day at a higher temperature and because the higher insulation also produce its effect during the day.)

Next “night” we start at a higher temperature than the first night with the new atmosphere.

You get the drift… A’s temperature will rise until a new equilibrium is reached. This equilibrium will be reached because energy radiated is proportional to T^4, so increasing radiative losses will eventually overwhelm the effect of increased insulation.

So, yes, increasing the insulation on A does ultimately lead to a higher temperature on A even though the atmosphere never “actively heated” the surface. What do you think is wrong with this?

And do you still think the 2nd law of thermodynamics is violated by the processes described in the article and the comments?

15. Bryan said:

Does a colder surface raise the temperature of a surface at a higher temperature..

..The reason I am reluctant to include the Sun in the calculation is that almost any perpetual motion device that satisfied the first law and failed the second could be rescued by such a sledgehammer argument.

Bryan indeed has a conceptual problem. A problem understanding what the second law is actually about.

Without the sun the earth and the atmosphere would move towards each other in temperature as both fell towards something approaching the cosmic background temperature of 3K (perhaps lifted slightly by any heat from the earth’s core).

What the second law of thermodynamics doesn’t say:

The entropy of a surface cannot decrease

How does a refrigeration system work?

What the second law of thermodynamics does say:

The entropy of a system cannot decrease

Bryan is taking one parcel of energy and “proving” with reference to an imaginary law of thermodynamics that a scenario can’t happen.

Negative entropy production is all around us. In each case, so far as anyone has found, it is more than compensated for by positive entropy production of that system.

Where in any thermodynamics textbook does it say that “No element of any system can ever have a reduction in entropy“?

16. Sod, in example 1 the direction of heat flow is the opposite of what it should be if you want to show that heat spontaneously flowing from cold to hot doesn’t violate the 2nd law of thermodynamics. Reversing the direction of the heat flow in the worked example results in a decrease in entropy, proving the violation.

17. @Bryan

You seem to have a problem with the concept of two-way exchanges.

Of course, when you consider things on a macroscopic scale, quite intuitively, a body A at a low temperature cannot bring a hotter body B to an even higher temperature. This is because if A can radiate heat/energy toward B, it means that B can radiate toward A.

Now the radiation from both bodies is a function of their own temperature and is not affected by outside factors. Since B’s temperature is higher than A’s, the flow of heat/energy from B to A is greater than from A to B. This is why A warms up while B cools down. Nobody disputes that.

Now, you seem to recognise the fact that the rate at which 2 bodies exchange temperatures depends on the temperature difference between them. Once again, it is something we can intuitively relate to. But why is it that way? It is not because of some magical properties of heat and temperature. It is because both bodies “send” heat/energy to the other one. (Even the colder body sends heat toward the hotter one. It is just that it sends less than what it receives in exchange from the hotter body.)

This two-way exchange allows us to explain why the cooling down/heating up happens more slowly between bodies with a smaller temperature difference.

Let’s take one of our previous example. We say a planet’s surface cools down more rapidly if it radiates directly toward space (3K) than if it radiates toward an atmosphere at 270K. You agree with that. Now what mechanism does explain this behaviour?

A two-way exchange between the two bodies explains it with no problem. That means that the hotter body *does* absorb heat/energy emitted by the colder body hence making the net exchange from hot to cold smaller.

Take a planet’s surface at 288K radiating directly to space :

The surface radiates 390W/m² toward space.
space radiates 4.6*10^-6W/m² toward the surface.
Net energy flow from surface to space 389,9999954W/m² : the surface is cooling down really fast.

Now consider a planet surface with an atmosphere at 270K radiating 301W/m² toward the surface :

the surface radiates 390W/m² toward the atmosphere (nothing really changed here, radiation depends on the body’s temperature)
the atmosphere radiates 301W/m² toward the surface
Net energy flow from surface to atmosphere 89W/m² : the surface is cooling down more slowly.

Does this makes sense to you? Can you agree that the atmosphere’s radiation toward the surface does have an effect on it? (you can clearly see the difference between the two scenarios)

When presented this way (we are not saying that the atmosphere is warming up the hotter surface, just that it emits energy that is absorbed by the surface, thus contributing to its heat content and affecting (reducing) the net energy transfert from the surface to the atmosphere), does it still strike you as incorrect?

When the 301J emitted by the atmosphere are absorbed (and there is no reason they would not be absorbed) by the surface, the surface’s heat content rises by 301J. This does not violate any law (in fact it is in accordance with the 1st law of thermo) because, at the same time it absorbed the 301J, it emitted 390J. The surface’s heat content is thus going down by 89J and the surface’s temperature decreases according to its specific heat. So the surface is still cooling down and the atmosphere heating up. Nothing is wrong here.

But you can not say the 301J from the atmosphere do not have an effect. The effect is that the surface is cooling down at the rate of 89W/m² rather than 390W/m². It is a significant effect. it does not contradict any law. It does not affect the direction of net heat exchange (it still goes from hot to cold, so Clausius’ statement is not broken by this phenomenon.)

18. on September 29, 2010 at 2:29 pm | Reply Leonard Weinstein

The Science of Doom,
I also have observed: “Many people often claim that the atmosphere cannot “heat” the earth’s surface and therefore the idea of the inappropriately-named “greenhouse” effect is scientifically impossible. The famous Second Law of Thermodynamics is invoked in support.”

These are an ignorant group that even I would class as just wrong rather than skeptics (I am a skeptic of CAGW). There are groups on all sides of the AGW issue, and letting a wrong headed subgroup bother you is not productive. If a basic error in understanding is shown by either side, please just ignore it. These errors decrease the strength of valid arguments on both sides by showing the errors some support.

You can argue logically as often as you wish, but the wrong headed people will never be convinced.

19. John Millett

in example 1 the direction of heat flow is the opposite of what it should be if you want to show that heat spontaneously flowing from cold to hot doesn’t violate the 2nd law of thermodynamics. Reversing the direction of the heat flow in the worked example results in a decrease in entropy, proving the violation.

In Example 1 the flow of heat is what we know to be true – from hotter to colder.

This causes an increase in entropy.

This is what I wanted to show and is in accordance with the second law of thermodynamics.

I can’t make any sense of your comment.

20. Yes, Sod, Example 1 demonstrated that heat flowing spontaneously from hot to cold results in an increase in entropy and, therefore, is in accordance with the 2nd law of thermodynamics. But, if I understand correctly, your purpose is to demonstrate that the reverse heat flow (from cold to hot) is also in accord with the 2nd law. Following the logic of the Example 1 demonstration, to do this you would show that the reverse flow also results in entropy increasing. My comment was to the effect that doing this would have demonstrated violation, not compliance.

21. Sod, on a re-reading I note that you have covered the point of the above post in the text if not in a specific graphic.

I can’t reconcile the following ststements:

Example 1: “Notice that if the energy flow of 1000J was from the 500K body to the 1000K body the change in entropy would be -1 J/K. This would be a reduction in entropy – forbidden by the real second law of thermodynamics. This would be a spontaneous flow of heat from the colder body to the hotter body”.

Example 3: “Energy from the colder atmosphere has been absorbed by the hotter surface and yet entropy of the system has still increased”.

Doesn’t the 2nd law apply separately to the two heat flows in the exchange as well as to the net flow? The first quote above seems to say that it does.

22. John Millett:

Doesn’t the 2nd law apply separately to the two heat flows in the exchange as well as to the net flow? The first quote above seems to say that it does.

No, the second law applies to a system.

I will try and rewrite the first example to make it crystal-clear. The first example was to demonstrate how the calculation was done. Thanks for identifying the misleading explanation.

The reason the second law of thermodynamics became “a law” was because no one could find a flaw in it.

And the formulation is definitely about the net movement of energy in a system. Otherwise the people who write about radiant heat transfer would be wrong because of the second law. Or the second law would be wrong because of the facts of radiant heat transfer.

The real second law anyway.

• Updated text added in article –

Think of the example above like this – If, for some reason, in a closed system, this was the only movement of energy taking place, we could calculate the entropy change and it has increased.

The example is not meant to be an example of only one half of a radiative energy exchange. Just a very very simply example to show how entropy is calculated. It could be conductive heat transfer through a liquid that is totally opaque to radiation.

23. Leonard Weinstein

…….”I also have observed: “Many people often claim that the atmosphere cannot “heat” the earth’s surface and therefore the idea of the inappropriately-named “greenhouse” effect is scientifically impossible. The famous Second Law of Thermodynamics is invoked in support.”……….

Consult any Physics Textbook.
Look up the word “heat” as used in the textbook.
You will find unanimous agreement that ;
Heat cannot flow from a colder body to an object at a higher temperature.
Further who gave you the monopoly on the use of the word climate sceptic?

24. Bryan:

SD has been very clear about this: it’s the net heat flow that behaves this way. Further, you haven’t explained what happens to the energy of the radiation that is emitted by the atmosphere when it reaches the surface. You have a choice here: either 1. Some of that energy is absorbed (depending on the albedo of the surface), and heat energy is thereby transmitted from the atmosphere to the surface (while, in accord with the second law, more heat energy goes the other way), or 2. Something entirely new and different occurs. If you choose the second option, you need to describe and explain what you think happens, and reconcile that account with observations of heat radiation and how surfaces interact with it.

25. Bryson Brown

I agree that SOD has been more careful with the use of the term heat.
It has quite a precise meaning when issued in a Physics Textbook.
Some people either do not care or do not know the difference say between IR radiation and heat.
This leads to all sorts of muddles and needless confusion.
If you have followed some of the previous threads SOD is particularly interested in what happens to the radiation from the colder atmosphere when it reaches a warmer surface.
I have tried to point out in the recent exchanges that the quality of the radiation would need to improve to to match or increase the temperature of the hotter surface.
SoD has replied with a reasonable argument that this is not completely prohibited by 2nd Law if we look at the greater flow of radiation from hot to cold.
However no actual calculations beyond this say that in fact it happens apart from being possible.

26. Bryan writes:

“I have tried to point out in the recent exchanges that the quality of the radiation would need to improve to to match or increase the temperature of the hotter surface….”

Well, now– frankly, I’m at a loss to understand this. Radiation impinging on a surface is either reflected, refracted or absorbed, depending on its albedo and transparency. There is no question of ‘matching quality’: if the radiation is absorbed, the energy of the radiation is taken up by the surface (which, of course, will simultaneously be radiating due to its own temperature and emissivity). The absorbed energy will make the surface hotter than it would be without that incoming radiation. In terms of the exchange with the colder atmosphere, more heat energy goes up than comes down– but heat still comes down, it adds to the heat energy of the surface, and so makes that surface hotter than it would be without the downward radiation.

Bryan does concede the ‘possibility’ of some heat flow from atmosphere to surface, but he complains that ‘no actual calculations…say that in fact it happens’. This is just wrong: the calculations tell us that the atmosphere is radiating (given its temperature and emissivity), and that some off that radiation is absorbed by the surface (given its albedo).

What makes no sense at all is the alternative suggestion, that the surface would or even could distinguish between a photon coming from a hotter source and a photon coming from a cooler source, absorb the first and somehow ignore, do away with or otherwise avoid taking up the energy of the second. Photons don’t come with ‘temperature of origin’ labels!

27. Bryson Brown

….”Photons don’t come with ‘temperature of origin’ labels!”…

Yes in fact they do.

The photon spectrum is centered around the characteristic temperature of the source.

• Bryan,

I think you are getting confused between the distribution of photons within a range of wavelengths that a black body emits and the specific wavenumber associated with an individual photon. An individual photon has a specific wavelength; there is no range associated with it.

28. Bryan said:

Bryson Brown

….”Photons don’t come with ‘temperature of origin’ labels!”…

Yes in fact they do.

The photon spectrum is centered around the characteristic temperature of the source.

Oh, the entertainment value!

First some background – previously on Science of Doom.. I tried to “lead the horse to water” with this question:

I don’t think you can claim clarity on the point about what happens to radiation when it reaches a warmer body.

Here is your chance to be clear.

1. When 10um radiation from a -10′C body reaches a 0′C surface how much of this 10um radiation is absorbed compared with the scenario of 10um radiation from a +10′C body.

In fact the whole article – Intelligent Materials and the Imaginary Second Law of Thermodynamics is quite instructive because it was based around the answer finally extracted to an even earlier question – “what happens to the radiation when it ‘reaches’ the ground?”. And you can read the entertaining responses there.

But I digress.

Finally, after many many many many attempts, I did get a response:

If the two photons have the same wavelength
The two photons would have equal chances of being absorbed of course.

I do not know where you are going with these self evident almost pointless comments.

And there we had it.

Because as regular readers of this blog will know – as well as anyone with a passing acquaintance of radiation – radiation is emitted with a spread of wavelengths.

And that was the point of “the pointless question”.

If +10’C radiation is absorbed by a 0’C surface then Bryan has admitted that the -10’C radiation will also be absorbed by a 0’C surface in almost exactly the same ratio. The reason for the slight difference is that the radiation of a -10’C source is very slightly different in wavelength composition to the radiation of a +10’C source – see the graph above..

Bryan claimed that the radiation was “mostly scattered”.

But he hasn’t yet explained how it is that the radiation from a +10’C body can be absorbed by a 0’C surface and yet this same 0’C surface has “zero” absorptivity for -10’C radiation.

In fact, he has “agreed” by his answer that the 0’C surface must have an almost identical absorptivity for -10’C radiation.

A conundrum if ever there was one.

That article was followed up by The First Law of Thermodynamics Meets the Imaginary Second Law and as expected Bryan couldn’t find the flaw in it.

The point of the article was that if a surface absorbs energy obviously it is going to have an effect on temperature.

Well, fascinating for the onlookers to see how an erroneous understanding of one part of physics causes so much stress on every other part.

29. […] the later article – The Real Second Law of Thermodynamics Possibly related posts: (automatically generated)The Amazing Case of “Back Radiation” […]

30. […] Update – New article – The Real Second Law of Thermodynamics […]

31. Hmm,
I once met a fellow engineering student who came across something like

“…no matter exits at absolute zero.”

Of course, this was in the context of an introduction to entropy and distribution of energy within the universe, and simply meant that there was no matter anywhere in the universe which had a temperature of absolute zero. However, this student became utterly convinced that, if you could somehow reduce some matter to zero temperature, it would quietly cease to exist. Some of us tried and failed to convince him that that was neither true, nor what the author meant. Not sure what became of him.

The point is that sometimes people can read completely correct information, and completely misunderstand it.

32. on October 1, 2010 at 2:26 am | Reply Leonard Weinstein

Bryan,
The atmosphere does not “heat” the ground as you pointed out. What it does is reduce the net radiation energy flux from the ground. I was stating a common usage rather than technical usage of the term “heat”, and appreciate your pointing that out. We should be more precise.

However, the greenhouse gases do absorb and radiate thermal energy. The net result of the atmospheric energy radiation is that the atmospheric greenhouse gases result in a higher ground temperature than otherwise. This arises from the fact that the radiated energy flux has a (T hot to the forth minus T cold to the forth) term which results in a lower energy flux than otherwise for a given ground temperature. The T cold is the effective temperature of a portion of the atmosphere. The only way the energy balance between input Solar energy and output to space is obtained is for some of the energy to be radiated to space to be radiated from the atmosphere, and this, combined with the lapse rate determines the surface temperature.

If the Earth’s temperature is increasing, there would not be an average balance, but some continual energy storage on Earth (land, oceans, and atmosphere) but these levels are smaller than the greenhouse effect levels and can be ignored for this argument.

As to my claim as climate skeptic, read my writeups on the Air Vent and my discussions with Science of Doom.

33. SoD

Certain gases such as CO2 and H2O have a significant radiative effect in the Infra Red.
The radiation from such gases from the colder atmosphere land on the warmer Earth.
Some think that there is 100% absorption (Earth surface is perfect black body)
I certainly do not think the Earth is a perfect Black body.

Anyway, what is the effect of this radiation?

1. Woods experiment still stands and the lesson for the wider atmosphere is that the radiative effect is very small.

2. Schack carried out experiments and came to the conclusion that the radiative effects of CO2 are significant at furnace temperatures but could be ignored at room temperatures.
3. Regular readers will remember Gords post on the Solar Heater(parabolic reflector).
When pointed to the night sky to collect as much of this radiation as possible the result was not an increase in temperature but the production of ice.
4. The statement that an irradiated surface cannot reach a higher temperature than the source of the radiation is accepted by all except certain people who for reasons best know to themselves want to ignore standard physics.
5. The Earth surface although not a perfect black body radiates a continuous spectrum centred around the average temperature of 15C(IPCC figure 390W/m2. )
The backradition from the -30C atmosphere is almost as large 324W/m2
Now if we take a circle of say 20Km radius the Earth surface and atmosphere would approximate to two flat planes facing each other at this scale.
Using SB equation, we find on calculation, an atmosphere with an emissivity of 1.6.
This is remarkable, coming from only 1% of the gases which only radiate specific wavelengths.
Further Climate Science tells us the radiation of the atmosphere is half up and half down so there must be 324W/m2 from the atmosphere going up.

I could go on and on but the above should convince any reasonable person that the case for AGW caused by increased CO2 emissions remains a fairly shaky conjecture.

• Brian:
“Regular readers will remember Gords post on the Solar Heater(parabolic reflector).
When pointed to the night sky to collect as much of this radiation as possible the result was not an increase in temperature but the production of ice.”

Maybe you need to read about parabolic reflectors.
IR radiation is emitted in random directions locally, a parabolic reflector is useful for collecting radiated energy from a distant source because the waves/photons travel near to parallel to each other. It wouldn’t do a very good job with a local noisy IR signal.

• Warmcast

The parabolic reflector will not focus nearly as well as focusing parallel rays.
However there should be some effect.
How would you explain the production of ice?
The colder atmosphere certainly is not raising the temperature of the Solar Heater.

• If ice formed, the local atmospheric temperature must have been cold.
I don’t think anyone is denying that temperatures are going to drop at night even with ‘warming’.

The other point is if the reflector was metal then you would get condensation build up if it was colder than the air. That may have turned to ice at some point later.

34. Bryan:

You could go on. And no doubt will.

But didn’t you agree that the absorptivity of a surface at wavelength, λ, was the same for a photon of wavelength λ, no matter what temperature atmosphere it was emitted from?

So why not go back and explain specifically why it is that if a surface at 0’C will absorb let’s say 0.9 (90%) of the incident radiation from an atmosphere of 10’C, why radiation from an atmosphere of -10’C will be not absorbed at all?

Otherwise people might think that you are dodging the important issue.

Surely you can put a bit of time into explaining this subject?

No more red herrings. Just explain the apparent inconsistency in your explanation. Pull it all together. Surely, it must be easy?

• scienceofdoom

……..”So why not go back and explain specifically why it is that if a surface at 0′C will absorb let’s say 0.9 (90%) of the incident radiation from an atmosphere of 10′C, why radiation from an atmosphere of -10′C will be not absorbed at all?”……..

I think that the bulk of the radiation in both cases is absorbed perhaps even the 90% that you quoted.

I don’t think that lets say 200J of photons centred around -10C is identically equivalent to 200J centred around +10C

35. Sod,

I wonder whether the radiative transfer folk have kept things as simple as they can be but no simpler – or words to that effect. Generally a system comprises the subject body, its surroundings/environement and a notional boundary separating them. A close examination of the Incropera and DeWitt extract reveals two systems, one the hot body surrounded by a vacuum (containing a photon gas) and the other the vacuum and its cool surroundings. A temperature potential exists across the vacuum from the boundary with the solid to that with the surroundings. Heat will flow to eliminate the potential. The body will lose entropy and the photon gas will gain a larger quantity of it all in accordance with the 2nd law of thermodynamics. Eliminating the temperature potential across the vacuum creates another one at the boundary with the surroundings. Again heat will flow to eliminate the potential. The photon gas will lose entropy, the surroundings will gain a larger quantity of it and the 2nd law will have been preserved. The process overall will have increased the entropy of the combined systems. Visually, only two heat-flow arrows were necessary, one crossing the boundary in each system, both in the same direction, hot to cold. Nowhere was it necessary to invoke heat flow in the direction cold to hot in violation of the Clausius corollary of the 2nd law.

36. […] See also: The Real Second Law of Thermodynamics […]

37. Bryan on October 1, 2010 at 12:25 pm:

scienceofdoom

……..”So why not go back and explain specifically why it is that if a surface at 0′C will absorb let’s say 0.9 (90%) of the incident radiation from an atmosphere of 10′C, why radiation from an atmosphere of -10′C will be not absorbed at all?”……..

I think that the bulk of the radiation in both cases is absorbed perhaps even the 90% that you quoted.

Cor blimey.

So you think that radiation from a colder atmosphere can be absorbed by a warmer surface?

I don’t think that lets say 200J of photons centred around -10C is identically equivalent to 200J centred around +10C

It’s not identical, just very similar in terms of proportion of total energy in each wavelength band.

Anyway back to my first question. You did realize what you wrote?

38. In the post “The Real Second Law of Thermodynamics” ScienceOfDoom aims at proving evidence for that the CO2 greenhouse effect is scientifically sound, that is, that the colder atmosphere will heat the (on average) warmer surface of the earth via so-called back-radiation from greenhouse gases. He initially sets the scene via the following statement:

“Let’s avoid a semantic argument about the correct or incorrect use of the word ‘heat’. … I claim that energy from the atmosphere is absorbed by the surface”.

The problem with this statement is that the difference between good science and bad science hides in the details. It is therefore of utmost importance to clearly define what ‘heat’ is and what it is not. ‘Heat’ is energy, but energy is not necessarily ‘heat’.

‘Heat’ is in thermodynamics defined as: “Thermal energy in transit”. Or, alternatively [http://en.wikipedia.org/wiki/Heat]: “The energy transferred from a high-temperature object to a lower-temperature object is called heat”. It is thus clear that ‘heat’ only refers to the net energy that is actually transferred from a warmer body to a cooler one. This is a consequence of the second law of thermodynamics, stating that the entropy of the system cannot reduce. It may either remain stationary (no change), in case of an equilibrium situation (e.g. no heat transfer between bodies of equal temperature), or increase for spontaneous ‘reactions’ (e.g. heat flows from hot to cold). The mysterious entropy concept is defined as follows [http://en.wikipedia.org/wiki/Entropy]: “The entropy change, δS, of a system at temperature T, absorbing an infinitesimal amount of heat, δq, in a reversible way, is given by δS=δq/T”. Please note here the use of the word ‘heat’ (and not ‘energy’).

SoD uses double accounting to reach his goal and treats ‘heat’ and ‘energy’ as if they were synonymous entities. This must be categorised as bad (and thus untrustworthy) science. It is, however, clear that all objects do emit thermal radiation energy (blackbody radiation). But this does not mean that this energy always is absorbed by (transferred to) other bodies that are exposed to this radiation (see also my comment: https://scienceofdoom.com/2010/09/12/heat-transfer-basics-part-zero/#comment-6546). Only a body that is cooler than the radiating body can absorb part of the incoming radiation energy (in relation to the temperature difference between the objects and to their geometry, etc.). This part of the radiated energy is thus actually transferred and is then called ‘heat’. Non-absorbed radiation energy is directly re-emitted and will not influence the temperature of the irradiated body.

A proper entropy calculation, using the true heat transfer in the equation above, will therefore show that the cooler atmosphere cannot heat the warmer surface of the earth. Such a hypothetical heat transfer would correspond to negative entropy and is thus forbidden by the laws of thermodynamics. This means that the CO2 greenhouse effect cannot exist.

Those of the readers who want to better understand atmospheric physics are recommended to visit Professor Claes Johnson’s blog, for example starting at: http://claesjohnson.blogspot.com/2010/09/lapse-rate-vs-radiative-forcing.html

39. JanS said:

..Only a body that is cooler than the radiating body can absorb part of the incoming radiation energy (in relation to the temperature difference between the objects and to their geometry, etc.)..

..A proper entropy calculation, using the true heat transfer in the equation above, will therefore show that the cooler atmosphere cannot heat the warmer surface of the earth. Such a hypothetical heat transfer would correspond to negative entropy and is thus forbidden by the laws of thermodynamics. This means that the CO2 greenhouse effect cannot exist..

All you have done is claim that something can’t happen (a colder body absorbing energy from a hotter body) and then used that claim to make another claim that “a proper entropy calculation” will demonstrate what you believe.

This article is about entropy. I demonstrate entropy increases. Therefore, the Second Law of Thermodynamics is not violated.

So why not actually provide the entropy calculation to show that my examples in the article are reducing entropy of a system?

Otherwise your claim is just a mantra.

40. John Millett on October 1, 2010 at 9:40 am:

I can’t make sense out of your comment other than you think that if you divide the system into two arbitrary systems it will help solve the imaginary problem of “heat flow from cold to hot”.

It’s a very simple system and just one system. With appropriate insulation around the outside of the system it will be a closed system.

An entropy calculation will reveal that entropy has increased – like in example 2 and example 3 in the article, and an energy calculation will reveal that energy has been conserved.

Both first and second law of thermodynamics are still intact. All clearly demonstrated and only confusing people who think that radiation from a colder surface can’t be absorbed by a hotter body.

However, you do have a fan.

41. Over on another blog someone who believes that energy from a colder body cannot be absorbed by a hotter body did stick their neck out about Incropera and Dewitt.

Apparently, as best I can tell, Incropera and DeWitt are right, but they aren’t showing energy from a colder body being absorbed by a hotter body.

You can read it in their words.

We did have one legendary commenter who thought that electromagnetic radiation in opposite directions cancelled out – his solution.. and he said Incropera and DeWitt were wrong because it would violate the second law of thermodynamics.

Of the many other people convinced that energy from a colder surface cannot be absorbed by a hotter body, none them have been bold enough to claim Incropera and DeWitt wrong (or comment that they are right!)

Now an apparent claim that they are right and we’ve just misunderstood what they are saying. (I have asked for clarification).

All fascinating stuff.

I can only see the plain fact that if Incropera and Dewitt (and the many other writers of heat transfer textbooks that I have checked) are correct – then energy from a colder surface is absorbed by a hotter body. And it’s not a violation of the second law of thermodynamics.

I offer the obvious point that that is why almost no one ever answers the question about Incropera and DeWitt..

• You have the patience o a saint… i would have smashed my monitor with my head by now! But i think where they get this from is the whole “heat can only flow from hot to cold”… without realizing, that “heat” is referring to the net flow of energy. But hell, the amount o times you have explained it, being very specific about it… :stares at the heavens shaking head:

42. On my earlier comment of October 3, 2010 at 10:27 pm, perhaps I spoke too soon.

Despite many questions, including my last I couldn’t get my “sparring partner” to actually say whether or not Incropera and Dewitt believed that energy from a colder source was absorbed by the warmer body.

He agreed that they were correct, was happy to answer every other question, but “I am refuted because they say that net heat flows from hot to cold”.

Fascinating stuff.

43. Fascinating stuff, indeed.

Sod, let’s step back a bit.

Which of the following qualifications to the Clausius corollary of the 2nd law of thermodynamics is the truer?

Heat may not flow from a cold region to a warmer one unless…

1….. accompanied by a larger heat flow in the opposite direction.

2……..work is done, say, by a refrigerator or an air conditioner.

44. John Millett

I don’t know how to pick “the truer”. I will answer your riddle with a riddle. Which is truer:

– that the earth rotates on its own axis;
– or the earth orbits the sun?

In any case, check out what the textbooks say:

Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics

• So, you’re version of the Clausius corollary of the entropic 2nd law would go like:

Heat may not flow from cold to hot unless either accompanied by a larger flow in the opposite direction or work is done by, say, a refrigerator or air conditioner.

This would imply that spontaneous flow and doing work are the same thing whereas they are opposites. Now to see what the textbooks say.

45. […] if anyone was saying that it would be ridiculous. I agree. To take one example from many, in The Real Second Law of Thermodynamics I said: Put a hold and cold body together and they tend to come to the same temperature, not move […]

46. […] Roads Less Travelled and especially Amazing Things we Find in Textbooks – The Real Second Law , The Real Second Law of Thermodynamics and The Three Body Problem. And for real measurements of back radiation, see The Amazing Case of […]

47. Late to the Party but

“The Fluctuation Theorem does much more than merely prove that in large systems observed for long periods of time, the Second Law is overwhelmingly likely to be valid. The Fluctuation Theorem quantifies the probability of observing Second Law violations in small systems observed for a short time.”

Click to access Review_37_with_figs.pdf

How does that work? Pretty much how SoD describes – the ‘hot’ particle can absorb energy from a ‘cold’ particle and therefore each becomes hotter or colder respectively. On a macro scale though these inequalities are generally insignificant and the second law holds true.

• FlipC

Its a long article but in a quick read it observes that for small numbers the second law sometimes fails.

So SoD may be occasionally correct if we are talking about very small numbers of molecules.

SoD will not be happy because he thinks his version of climate thermodynamics is always compatible with the second law.

• Yes, but when I read the contortions some commentators have gone through not to have a hot body ever made hotter using energy from a cold body at the molecular level and I know that this happens and has already been experimentally proved to be the case; I am compelled to point this out.

It doesn’t necessarily make SoD right when dealing with large numbers of molecules/atoms, but their point that the second law should refer to net heat applies at every level; it’s just that this reverse process with a large body is insignificant.

48. The Fluctuation Theorem is not really relevant to the discussion. It predict the probability of a decrease in the entropy of an isolated system to be inversely related to its size. For any macroscopic system this probability is so low you would have to wait longer than the age of the universe to observe such an entropy reduction.

Much of the confusion on this topic comes from the way heat is defined. If you define it as net flow of energy from one body to another the 2nd law implies the flow must be from hotter to colder. If it is regarded as a two way exchange of energy the 2nd law only requires the energy transferred from hotter to cooler is greater than the energy transferred from cooler to hotter. This is quite easy to demonstrate mathematically.

• David,

Much of the confusion on this topic comes from the way heat is defined.

Indeed!

A case can be made that the term should be banned from thermodynamics, and only energy and entropy, which don’t have multiple, mutually exclusive definitions, be used. But that’s not likely to ever happen.

• I quite agree that in the macroscopic terms that SoD is apparently using the Fluctuation Theory is insignificant, but at the microscopic level it does count which is what some commentators appear to be contesting.

To my mind the problem is that the term “heat can only travel from hot to cold” was coined prior to a true understanding of what heat was or to be more precise the forms in which it is stored and transmitted; and it’s this phrase that is constantly being repeated educationally; simply adding the tern “net” clears it up nicely.

Yes removing the term or phrase itself would be better, but as you say it’s unlikely to ever happen as it’s such a useful educational shorthand. Perhaps a better ‘lie to children’ would be a demonstration with two students facing each other – one with a lot of tennis balls (hot) and one with less (cold). The ‘hot’ student starts passing the tennis balls to the ‘cold’ student while occasionally the ‘cold’ one passes some back. The rate of passage slows or increases respectively until they each end up with the same number and simply pass one to the other at the same time. If you start with an uneven number of tennis balls they might just end up passing the same one back and forth 🙂

49. I am sure much of the confusion arises from the way thermodynamics is taught. Many textbooks at least still introduce the 2nd Law using heat engines and the Carnot cycle. Although this has the advantage of giving an historical perspective, as a chemist I was introduced to thermodynamics through the axiomatic approach of Gibbs and Guggenheim and I beleive this is a lot less likely to casue confusion. The fact that heat (in the sense of net energy transfer) cannot flow from hot to cold is then derived as a theorem and it is easily seen that in a two way exchange the same result applies to the difference between the energy transfers

50. ScienceofDoom states:

“As I pointed out in the The Special Case section – this is true if this is the only source of energy. Yet the surface of the earth receives energy from both the sun and the atmosphere.”

There is only one source of energy, the sun. You are double counting energy sources. The atmosphere is not its own energy source. That “energy” came from the sun. Same thing with your example 3.

There is no “imaginary” Second Law of Thermodynamics. A cooler body simply cannot make a warmer body warmer. There is no mention of a “net” effect.

Take your example Example 4. What happens to the sun when it receives radiation from the earth? Does the temperature of the sun increase minutely?

Or say with your example 2, with body 1 and 2. Assume those bodies have an internal source of energy that results in temperatures of 1000K and 500k, respectively. What happens to body 1 when it receive the radiated energy from body 2? Please explain your view.

• Jonathan,

There are at least two ways to avoid double counting for the surface energy balance:

1) Including three components: absorbed solar SW, IR emitted directly to space, and the net flux from the surface to the atmosphere.

2) Including more than three components: absorbed solar SW, all IR emitted directly from the surface, IR from the atmosphere to the surface, and other fluxes from the surface to the atmosphere. These other fluxes include convection and latent heat transfer.

Both approaches are correct (when applied correctly). From the fact that the sun is the original source of all energy considered does not follow that the second alternative would be false. It’s both correct and very useful in many ways.

The second law is always about the net effect. There’s no second law for anything but the net effect.

The existence of the the Earth makes the sun just a very little bit warmer than it would be without the Earth. The effect is too small to be noticeable, but it’s true.

The body of 1000 K is also a little warmer with the presence of the 500 K body than it would be without assuming that the power of the internal source is not changed by that.

• Jonathon,

You can see a worked example in The Three Body Problem where we see that a cold atmosphere increases the temperature of the warmer surface.

Entropy – the real measure of the second law of thermodynamics – is calculated and it’s clear that the second law of thermodynamics is not violated. Entropy increases in that example.

Check out Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics – where six heat transfer textbooks confirm the stuff that so many people dispute. Radiant energy moves in both directions, is absorbed by both the warmer and the colder surface, and changes the temperature of both surfaces.

On double counting..

The first law of thermodynamics describes the conservation of energy.

If the sun generates X Joules per second and doesn’t heat up, then it radiates out X Joules per second.

Let’s say that the earth’s climate system absorbs a fraction, Y Joules per second, from the sun.

From the conservation of energy – now that these Y Joules per second have arrived here – we have to account for them.

It’s very simple. The fact that the sun was the original source is irrelevant. Y Joules per second are absorbed in the climate system and need to be accounted for – that is, they are either absorbed (increasing internal energy, usually temperature) or emitted. Just as if they came up from geothermal sources or nuclear sources.

We just total up all energy absorbed by the climate system and then we know that energy emitted + energy retained = energy absorbed.

The same is true of any body no matter where we draw the boundary. Check out any heat transfer textbook. Check out any physics textbook. They all say the same thing.

Once you move 1 Joule of energy from the sun to the earth it is now in the climate system. Where is the double counting? It moved. It only exists in one place.

Take a banana. Move it from Jamaica to England. Now, someone in England sends the banana to France. So, for England, not a source EVER of bananas: 1 banana in = 1 banana out. But Jamaica is the source of all bananas!!! So I must be double counting.

Right, so let’s try instead:

0 bananas in = 1 banana out? No, that’s not correct.

Try again:

1 banana in = 0 bananas out? No, that’s not correct.

What is your point?

To help, please write a set of equations to describe what happens. I have already done so in both linked articles.

• SoD,

I did not initially see the comment tab to the right, so my response is posted below as a new comment.

51. Peka,

If the body of the the 1000K source warms up, you are creating energy. Per your thinking, the higher temperature 1000K source (body 1) would then in turn warm up the 500K body source (body 2). Where does the cycle end?

All versions of the Second Law do not mention a net effect because it leads to absurd possibilities. Per your point of view, two bodies at identical temperatures would warm each other up, because they are receiving IR energy from each other, right?

• The cycle ends without problems, because each additional warming is much less than the previous one.

Do you have an example of those absurd possibilities. Your example is just another example of inventing absurdities out of nothing. Here again the presence of the other body makes both a little warmer than they would be otherwise. There’s nothing absurd in it.

Have you heard about Aesop’s tale “The Tortoise and the Hare”? Think about that. Perhaps that helps you to understand that there’s nothing absurd in two bodies warming each other a little compared to the situation where the other is not there.

• Pekka,

If both bodies warm up a little, then you have created energy out of nothing, for it takes energy to warm up a body (the energy sources are fixed) That is the absurdity, and is in no way an invention.

• The energy comes from whatever keeps the bodies warm anyway. Without the other body they have only less to compensate for a fraction of what they will lose be emission to the outer space.

• on March 5, 2013 at 3:24 pm DeWitt Payne

Pekka,

I think you’re referring to one of Zeno’s paradoxes, Achilles and the tortoise, rather than Aesop’s fable. It’s all about convergent infinite series. Apparently Jonathan is unfamiliar with Calculus.

• Jonathan,

This example is explained in Radiation Basics and the Imaginary Second Law of Thermodynamics.

If you have an engineering background you should be able to understand the application of equilibrium equations for two bodies. It’s pretty simple.

Follow the maths in that problem and report back.

The fact that you have a conceptual problem doesn’t mean you are correct.

All we do in this simple equilibrium problem is solve 2 simultaneous equations for 2 unknowns – what is the new equilibrium temperature of the two bodies.

Maybe at some point during the next few hours I will think of a comparable mechanics problem that might aid in your conceptual problem.

Usually 2 equations in two unknowns where both affect each other does not lead to some kind of runaway condition.

How about a midge banging into an elephant. What is the momentum change for the midge? Clearly the momentum of the elephant doesn’t change – otherwise it would lead to a runaway momentum explosion as each one feeds into the other..?

Trouble is, your conceptual problem has no basis in physics so I can only guess at a problem which might be related closely enough to aid your thinking. That is, an analogy proves nothing, it is only for your possible illumination.

So, if that doesn’t help, please follow the simple maths in the linked article and comment on why you believe it is wrong (not including that fact that the answer isn’t what you were expecting).

And take a look at the article I already linked – Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics – where you can see the extracts from 6 heat transfer textbooks. Do professors in heat transfer not understand the subject either?

• SoD,

I would like you to look at the situation from a different angle, rather than your usual explanations, which I have read. How does the earth warm up with the addition of greenhouse gasses when the temperature of the sun remains the same? Where does that extra energy come from to warm the earth? There is no “magic” energy added when GHG’s are added.

Do you basically believe believe that if you have two bodies (A & B) in space radiating at two fixed (but different) temperatures, that body A warms B and B warms A?

I have not completely made up my mind on the issue, but there seems to be holes in your view. And when the 2nd Law appears to be violated, I take more notice.

Your argument about the textbooks is the classic “argument from authority”, which is not a valid scientific argument.

• Jonathan,

I would like you to look at the situation from a different angle, rather than your usual explanations, which I have read. How does the earth warm up with the addition of greenhouse gasses when the temperature of the sun remains the same? Where does that extra energy come from to warm the earth? There is no “magic” energy added when GHG’s are added.

I have already given an explanation below in my comment of March 5, 2013 at 1:00 am. Less energy leaves.

Let me make it even simpler.

1. Say Rin = 100 W/m2 is absorbed by the surface from the sun.
And Rout = 100 W/m2 is emitted by the climate system to space.

2. So the system is in equilibrium, at let’s say a surface temperature, Ts.

3. Rout reduces to 90 W/m2. Still Rin = 100 W/m2.

4. The first law of thermodynamics says that energy is conserved. More energy is entering than leaving so it must be accumulating.

5. Therefore, Ts increases.

6. No increase in solar radiation yet Ts has increased.

Which of these fundamental steps do you disagree with?

I have not resorted to “magic” energy.

Do you basically believe believe that if you have two bodies (A & B) in space radiating at two fixed (but different) temperatures, that body A warms B and B warms A?

I have given you an article which I wrote which explains this exact point and calculates the answer – Radiation Basics and the Imaginary Second Law of Thermodynamics.

To summarize, first we need to state the “before” case more specifically.

The “before” case is body A is in the vastness of space all by itself with only the background 3K of deep space to cool it.
Body A is at temperature Ta.

Likewise for body B, at temperature Tb.

Now we “wheel them in” into proximity with each other. So instead of having almost zero incident radiation on them, they now each have some incident radiation from the other.
The result is that post the “wheeling in”, Ta’ > Ta, and Tb’ > Tb.

In the specific example I calculated, Ta zoomed up from 1000 K to Ta’ = 1000.00006 K.

I have not completely made up my mind on the issue, but there seems to be holes in your view. And when the 2nd Law appears to be violated, I take more notice.

Then you must have done an entropy calculation. Please provide your calculations and results.

I have provided entropy calculations (and temperature changes) in The Three Body Problem – a simple example with three bodies to demonstrate how a “with atmosphere” earth vs a “without atmosphere earth” will generate increased equilibrium temperatures.

The transfer of heat is still from the hotter to the colder, just at a slower rate. The entropy is shown to have increased, proving that the second law of thermodynamics has not been violated.

Your argument about the textbooks is the classic “argument from authority”, which is not a valid scientific argument.

I’m just trying to help you see that heat transfer basics says something very different from your point of view.

Perhaps all heat transfer professors are wrong.

But they would expect the courtesy of at least an equation or two, rather than a vague hand-waving argument.

What is needed – at the very minimum – to overturn their stated theory is for you to identify which of their equations is wrong. Or, using these equations to demonstrate that a different result is obtained.

• Jonathan,
How does better insulation make the house warmer, when it’s heated by an electrical heater that keeps on running with a constant power? Where does that extra energy come from to warm the house? There is no “magic” energy added when insulation is added.

52. SoD,

Here is simplified view of what your explanation has accomplished. We have the sun at some temperature T-sun, and the earth is at T-earth. With the addition of greenhouse gasses, the earth has now increased in temperature even though the sun remains at the same temperature.

I would like your comments on the two examples I previously posted so I can understand your way of thinking.

(P.S. My background is that I am a licensed professional civil engineer. Civil engineering is mostly specialized physics, but deals mostly with Newtonian physics, although we are required to take a semester of Thermodynamics, in addition to the 3 semesters of general physics….but that was a long time ago)

• Jonathan,

Take a metal pipe of 0.3m outer diameter and 0.05m thick. The ambient air temperature, Ta= 15’C.

Attach a thermocouple to the outer wall of the pipe in one location and measure Tp.
Tp = Ta = 15’C

Pump X liter/second of 30’C water through the pipe, wait one hour, and measure the new temperature, Tp’.
15 < Tp’ < 30’C.

Add 0.1m thick insulation to the water pipe and wait 1 hour, then measure the new outer pipe temperature, Tp”.

Tp” > Tp’.

The temperature of the source of the heat (the water) has not changed so we must have created energy. Correct?

No. We have stored energy which has increased the temperature.

It is temperature difference that drives heat flow.

1. We are in equilibrium so Ein = Eout.
2. We change material properties to increase the resistance to heat flow – so Eout reduces.
3. Eout < Ein. Therefore, energy is accumulated in the body, therefore, temperature increases in the body.
4. Temperature keeps rising, increasing Eout.
5. Eventually a new equilibrium is reached where once again Ein = Eout, but the equilibrium temperature of the body has increased.

• SoD,

I have some company that arrived from Canada. So I will not be able to post for maybe a few days. But I will consider your above experiment. My visiting nephew is a Neurologist/psychiatrist. Maybe I’ll pick his brain!

• on April 28, 2013 at 3:16 am John Millett

Scienceofdoom,

This doesn’t look right, for two reasons. First, surface areas being greater on the outside than the inside and temperatures bearing the inverse relationship whilst maintaining the condition for equilibrium, Ein = Eout, Tp” is less than Tp’. Second, a side benefit of lagging a hot water pipe is avoiding the risk of accidental burning by reducing the exposed surface temperature.

• Jonathan: I’m glad to see that you are asking sensible questions and trying to make up your own mind. Perhaps I (a skeptical chemist who is used to thinking about how individual molecules behave) can help with possible contradictions of the Second Law of Thermodynamics (2LoT).

Imagine one billiard ball moving in the +y direction at 3 m/s colliding perpendicularly with a second ball moving in the +x direction at 4 m/s so that the first ball is motionless after the collision and the second is moving at 5 m/s at an angle (37 deg) compared with its initial path. Did we violate the 2LoT when kinetic energy was transfered from the slower-moving ball to the faster-moving one? Of course not. I am confusing kinetic energy with temperature – which is proportional to the mean kinetic energy of all of the molecules in the ball, not just their group translational kinetic energy. We need to be careful how we define temperature when we apply the 2LoT to various situations.

Now let’s explore what happens at the molecular level when a slower-moving molecule collides perpendicularly with a faster-moving molecule. Do we violate the 2LoT if the faster-moving molecule gains kinetic energy from the slower-moving one? Again the answer is no. However, it is easy to get fooled if you mistakenly equate a “faster-moving molecule” with a “hotter molecule”. Long after the 2LoT was accepted, physicists resolved the apparent contradiction between the 2LoT and the physics of molecules colliding on a microscopic scale by defining temperature as being proportional to the MEAN kinetic energy of a GROUP of molecules that are frequently colliding. The kinetic energy of a single molecule can change millions of times a second due to collisions, while the mean kinetic energy of a large group of colliding molecules remains constant. The 2LoT applies to heat transfer between two large groups of molecules, but not to collisions between individual molecules. Individual molecules have kinetic energy, but not “temperature”. A faster-moving molecule is more likely to lose than gain kinetic energy in a collision with a slower-moving molecule, so when we observe the NET result of 10^20 or more molecules colliding millions of times per second, heat always flows from hot to cold.

The rules governing heat transfer between molecules by photons are exactly the same as the rules governing transfer by collisions. Individual molecules can absorb a photon emitted by a molecule that is moving faster or slower than it is. Due to the Boltzmann distribution of molecular speed, you actually have no idea whether an emitting GHG molecule in the colder atmosphere is moving faster or slower than an absorbing molecule on the warmer ground. Neither has a well-defined temperature, so the 2LoT doesn’t prevent a molecule in the ground from absorbing a photon emitted by the atmosphere. However, we can be sure that the NET heat transfer will be FROM a large group of molecules on the warmer surface TO a large group of molecules in the cooler atmosphere. And this is precisely what the Stefan-Boltzmann law predicts will happen.

There is a one confusing difference between energy transfer by collisions and by photons. We can’t directly observe the individual collisions that transfer energy both ways between faster- and slower-moving molecules, but we can observe photons going each way. We point a photomultiplier at the ground and measure the upward flux of photons and at at the sky to measure the downward flux of photons. Then we can be CONFUSED into thinking that the 2LoT prevents the warmer earth from absorbing photons emitted by the colder atmosphere. The 2LoT applies only to the NET FLUX between these locations, not to the individual fluxes in each direction.

Here and elsewhere, a vocal group of skeptics try to apply the 2LoT without understanding temperature. They don’t recognize that individual molecules have no temperature, preventing the 2LoT from placing restrictions on the flow of individual photons between them. In my experience, they refuse to face the absurdities that develop when they try to apply the 2LoT to individual photons and molecules.

• Frank,

Your example of billiard balls is not valid as momentum is not conserved in that example. That’s, however, not essential for your point and your comment is mostly correct.

Thermodynamics was formulated (mainly during the first half of the 19th century) as a set of mathematical rules that explain many observations very well. One essential component was the Second law and the observation that conservation of entropy corresponds to the important limiting behavior of reversible processes. At that classical level nothing is said on what’s going on on more detailed level. Classical thermodynamics is a very powerful theory, but it’s based on postulates that are not derived from anything, they are just assumed and justified by the success of the resulting theory.

During the later 19th century Maxwell, Boltzmann, Gibbs and others made an important further step by deriving the second law from rather weak assumptions about microphysics and statistics. They observed that purely mathematical laws of statistics lead naturally to the Second law. Thus the Second law wasn’t anymore just postulated, it was derived from weak assumptions.

There were still some open questions in the statistical thermodynamics of 19th century as the classical physics did not explain fully the way the weak physical assumptions about the micro level should be made. Quantum mechanics was needed to fill the gaps. The first step in that direction was the derivation of the Planck’s law for radiation. Planck’s derivation involved effectively quanta, but his assumptions were ad hoc and lacked proper justification other than the success of the results. The development of QM provided that justification and it provided also a much better justification for the statistical thermodynamic theory.

What’s needed on the fundamental level for understanding the radiative mechanisms in the atmosphere was finalized by the Quantum Electrodynamics that Tomonaga, Schwinger, Feynman, and Dyson developed in late 1940’s and 1950’s. Fortunately the details and complex mathematics of QED are not needed, but only QED did fully explain, why the simpler formulation developed earlier works so well. (Someone might say that even the QED and more specifically the renormalization as part of it needs further justification that’s still lacking. The end of development of fundamental physics is not in sight.)

• Pekka: I chose the billiard balls to illustrate the dangers of mindlessly equating kinetic energy and temperature. I did intend to properly conserve momentum and kinetic energy in a ideal elastic collision, so let me know if I made a mistake.

I doubt we will get anywhere trying to convince skeptics that the 2LoT isn’t a “fundamental” law of nature, but I hoped to make it clear that it can’t be applied to energy transfer between molecules without running into absurdity. When physicists confronted those absurdities in the late 19th and early 20th centuries, they found they needed quantum mechanics and statistical mechanics to explain the physics of individual molecules. Unfortunately, many technical fields (such as engineering) only need to deal with the macroscopic world of classical mechanics, thermodynamics, and electromagnetic waves. When SOD tells them that the 2LoT applies only to the net flux of radiation between two locations, the statement seems totally arbitrary – they’ve never encountered a situation where the heat flux can be broken down into two components.

It’s sicken to hear the mindless zealots cite something Clausius said about the 2LoT in the 19th century as if it were gospel without recognizing that Clausius didn’t have a clear understanding of temperature at the time. Yet Clausius worked on the kinetic theory of gases and presumably wouldn’t have applied the 2LoT to the transfer of kinetic energy between two molecules of a gas.

• Frank,

First of all, I didn’t look carefully enough at your example. It does conserve momentum. While making it happen with billiard balls is impossible to the best trick players, the inverse case of hitting a ball at standstill with another with the speed of 5 and having as outcome the balls moving in perpendicular directions with speeds 3 and 4 is not that impossible. That kind of shots are common in billiard.

The main issue of finding arguments that could get skeptically minded understand what second law is about may, however, require different arguments, a special example of billiard balls is not likely to help much in that.

I have promised to think about, how to present in a systematic and accessible way those parts of fundamental physics that are most significant for understanding the atmosphere. I haven’t forgotten that, but doing that takes more time than I have had available for that task so far or will have in the immediate future (meaning a couple of weeks).

• It’s sometimes annoying that WordPress puts messages so eagerly to moderation. That has happened for a couple of my messages recently. One comment is presently in moderation in this subthread, but have no idea about what caused that this time. Neither have I been able to see the reason before.

That comment starts by noting that I made a wrong claim in my earlier comment concerning conservation of momentum.

• on April 28, 2013 at 2:46 am John Millett

Frank,

A reference to an actual field application of the photomultiplier observing simultaneous photon flows across a plane would be useful. Meanwhile, a though experiment:

Two enclosures equal in all respects, except temperature, connected by a frictionless pipe containing a blackbody slug which can be held fast or let loose, initially held fast. Radiative energy in the enclosures, incident on the blackbody surfaces, is absorbed and exerts pressure on it. The slug is therefore acted upon by opposing, unequal forces (pressure x area), the force from the warmer enclosure being the greater. Although in both enclosures the photons individually are bouncing around, the photon gas as a body is stationary, there is no flow. Free the slug and it moves in the direction of the greater force, inducing a flow of photons from the warm to the cool enclosure. Question: if instead of a physical barrier we had only the plane of its location, would photons from the cool enclosure move across the plane? We know that eventually the enclosures would come into equilibrium and that can occur only if the photon density is uniform within the enclosures, implying a relocation of photons from the warm enclosure across the plane to the cool one. What would induce a reverse relocation across the plane of a lesser number of photons?

• on April 28, 2013 at 3:41 pm DeWitt Payne

John Millet,

Question: if instead of a physical barrier we had only the plane of its location, would photons from the cool enclosure move across the plane?

Why wouldn’t they? Infrared photons don’t interact with each other. Pose the same problem with two different ideal gases at different pressures or temperatures. At equilibrium, the partial pressure of each gas in each container will be identical. The same thing would apply to photons if all the walls of the containers were perfectly reflective.

• John,

I cannot see the point of your question. I think I understand your first setup, and your description of it seems logical, but what’s the point in the second.

In absence of significant absorption the photon density in the enclosure reached its equilibrium state in the time it takes for light to traverse the full space. That equilibrium is not uniform, if the walls are at different temperatures. Photons do not form a gas that expands or contracts, because the photons have only negligible interaction with each other.

If the ordinary gas that fills the enclosure absorbs and emits strongly enough the photon density depends not only on the wall temperatures but also on the temperature of the gas. In that case we must study what happens to the gas. Photos affect that and that affects photons, but I still cannot see what’s the point of your question..

53. DeWitt Payne

“ Why wouldn’t they?”

The pressure gradient, which would first have to be eliminated.

At equilibrium the molecular gases are free to mix and eventually would become well mixed with uniform partial pressures. However, equilibrium is attained by eliminating the pressure difference between the two enclosures. This involves a flow of molecules down the pressure gradient from high to low pressure. This flow is one way. Two way flow may occur only with elimination of the pressure gradient. The photon gases would face the same pressure gradient as molecular ones. They would be free to mix only after its elimination.

• on April 29, 2013 at 3:34 pm | Reply DeWitt Payne

John Millet,

The pressure gradient, which would first have to be eliminated.

Wrong. It’s well known that gases diffuse from lower total pressure to higher total pressure. In fact, it can be a real problem when trying to keep oxygen out of a system. An inert dry box has to have a getter system to keep oxygen low even if it runs at above ambient pressure. It’s going to diffuse through the rubber gloves if nothing else. What counts is the partial pressure. A high flow rate can minimize this effect, but if the separator is, say, a membrane with small pore size that only allows diffusion, movement across the membrane will be immediate in both directions and depend only on the relative partial pressures, not the total pressures.

54. on April 30, 2013 at 8:04 am | Reply John Millett

Pekka Pirila

“I cannot see the point of your question”.

Rodrigo Caballero describes pressure in a fluid in terms of flux of momentum in a section of pipe between fast-moving molecules on its left and slow-moving ones on the right. He argues that more momentum enters from the left than leaves from the right; and that the section accumulates momentum and accelerates, causing flow from left to right. He makes no reference to a counter flow. He defines pressure as flux of momentum, hence pressure gradient.

I figured the same argument applied to a photon gas, since it exhibits properties of pressure and momentum. The first part of the comment was meant to show how one-way flow would come about under differential forces on a physical barrier. The second part premised that removing the barrier wouldn’t remove the pressure gradient and asked what would induce a counter flow.

At a deeper level, after some years following this site, almost exclusively, I have come to think that the conflict between back-radiation and 2LoT, perhaps the site’s most debated issue, is second order to that of radiative exchange. I am also inclined to the view that modeling radiative exchange as occurring between bodies surrounded by nothingness is unphysical. A realistic (and simpler) model, in my opinion, would be exchange between a body and its IR surrounding medium. The issue, then, is whether the exchange is one or two way. Is there an a priori physical imperative for 2-way exchange, other than the unproven hypothesis itself? If not, shouldn’t the simpler 1-way exchange be preferred? And couldn’t an energy density/pressure/temperature potential across the boundary between the body and medium permit that exchange and determine its direction?

• on April 30, 2013 at 1:16 pm | Reply DeWitt Payne

John Millet,

Rodrigo Caballero describes pressure in a fluid in terms of flux of momentum in a section of pipe between fast-moving molecules on its left and slow-moving ones on the right.

It’s usually more molecules per unit volume rather than faster moving molecules. A temperature gradient can cause a pressure gradient in a closed system, but it’s not required. That is, P is proportional to n as well as T.

I figured the same argument applied to a photon gas, since it exhibits properties of pressure and momentum.

Except that molecules can exchange momentum and energy with each other and photons, at least in the IR wavelength range, can’t. And even then, molecules can diffuse in any direction. Only in the case where the mass flow rate is significantly faster than the diffusion rate do you get flow only from high pressure to low. The concept of mass flow doesn’t apply to photons.

• John,
Radiation does not behave as a medium formed from molecules or atoms. Some properties are common, but many others are different. It’s possible to consider matter in absence radiation, but all interest in radiation comes from it’s interaction with matter. Radiative pressure is pressure on matter by radiation, not pressure on the radiation itself. There are many other similar differences.

55. John Millett wrote:

“A reference to an actual field application of the photomultiplier observing simultaneous photon flows across a plane would be useful.”

Roy Spenser discusses some experiments he has done and you could do with a \$50 hand-held IR thermometer at his website. If you really have an doubts about the two-way transfer of infra-red, buy one – or two if you insist on simultaneous measurements.

Alternatively, our host has written extensive about DLR and how accurately models of the atmosphere and radiation transfer theory predict the measured downward flux. See the multi-part series on the-amazing-case-of-back-radiation

https://scienceofdoom.com/2010/07/17/the-amazing-case-of-back-radiation/

Judith Curry has referred to a database where worldwide observations of DLR have been stored. These observations has been used by K&T to calculate that the average DLR globally is 333 W/m2 for their paper on the earth’s energy balance.

Since we have excellent measurements for the temperature and emissivity of the earth’s surface, we usually calculate the upward flux from the surface using oT^4 to be an average of 390 W/m2. (In an imaginary world where some of that upward flux cancelled, the resulting upward flux would be less than the measured downward flux, creating a real violation of the 2LoT).

56. on May 2, 2013 at 7:06 am | Reply John Millett

Frank,

Thanks for the reading list. The photomultiplier oversight is more than offset by the reference to Dr Spencer’s April, 2013 article proving DLR with his hand-held IR thermometer. It’s a while since I dropped by there.

Dr Spencer is to be commended for conducting his backyard experiment in 2010. It beats thought experiments. The summary of my 1500 word interpretation of it at the time reads “the experiment hasn’t proved that a cold body can make a warmer body warmer still. Nor has it demonstrated the fact of DLR; the temperature traces are explainable as reflecting energy imbalances between solar and surface fluxes only. The temperature differences inside and outside the box – the inside is hotter during the day and cooler at night – look to be a measure of the effect of air circulation, from which the box is insulated”.

The issue of a cold body making a warm body still warmer has had a bit of an airing here recently, Pekka putting it succinctly: a warmer cool body leads to a warmer warm body. When the cool body is warmed by the warm body, as is the case in the climate system, that should be modified to “a warmer cool body, warmed by the warm body, leads to a cooler warm body”, in my opinion.

The 300 responses to Dr Spencer’s article based on an IR thermometer made for fascinating, if at times tortuous, reading. This one dissuaded me from taking the bold step to acquire one:

“those neat little IR thermometers were NOT designed to measure the temperature of atmospheric gases. They were NOT designed to measure anything past a few feet. I am appalled that a scientist of your caliber is falling for this kind of misrepresentation!! Why have you not done your homework using the pamphlets freely available from the manufacturers to determine how they should appropriately be used!!!
Would you use the same equipment on your satellites to measure the temperature of the sun as you do the atmosphere without even an attempt at recalibration?!
You have NO IDEA where the radiation is coming from that the instrument is reading yet you make various claims based on your assumptions. Disgraceful”.

Still, Dr Spencer wasn’t stating temperatures of anything. His interest was in the temperature differences displayed when the instrument was variously pointed at clear sky zenith or low elevation or at a cloud. My hypothesis is that the change in temperature in the detector reflects a one-way flux in OR out of the instrument, the direction depending on the relative energy densities inside and outside the instrument. The orthodox view is of a two-way flux in AND out. The inside energy density is determined by the temperature of the instrument (which may be affected by the hand holding it?). The temperature gradient being small between the instrument and a low cloud warmed by both solar absorption and H2O phase change from vapour to liquid, the upwelling flux and, by extension, the flux out of the instrument is small and the detector cools only by a little. Away from the cloud, in clear skies, the temperature gradient increases and so does the amount the detector cools. The temperature gradient in clear sky at low elevation being smaller than at zenith, the amount the detector cools is also smaller.

• on May 2, 2013 at 11:40 am | Reply DeWitt Payne

John Millet,

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

Umm, IR thermometers are identical in fundamental design to pyrgeometers. They are designed to measure IR radiation in the thermal band from anything. You can, for example, use a simple IR thermometer to measure total column water vapor under clear sky conditions: http://journals.ametsoc.org/doi/pdf/10.1175/2011BAMS3215.1

The main difference between a pyrgeometer and a hand held IR thermometer is the field of view. A pyrgeometer also measures the dome temperature to correct for the small emissivity of the dome and the dome is transparent over a broader wavelength range.

One reason for the range restriction for measuring the temperature of a surface with an IR thermometer is precisely because atmospheric absorption/emission will interfere with the measurement since the air is likely not at the same temperature as the object being measured. There’s also the problem that even though the field of view is restricted, the effective area increases with distance so that it may include more than the object you’re trying to measure. And yes, if you were trying to measure total emission from the sky, you would have to measure at multiple angles from the horizon to zenith, convert the temperature to irradiance and integrate the measurements correcting for the measurement angle (multiply by cos(θ) IIRC if you define the zenith angle as 90 degrees) since the detector will be perpendicular to the direction measured rather than perpendicular to the zenith angle.

• Attempts to pick small parts of physical knowledge and to use them to prove a specific point are often bound to fail, because the method used in proving the point is equally dependent on the same or related theory.

Those who believe in the theory don’t need any proofs, while those who doubt it have no logical reason for taking the proof seriously.

The wide theory of radiation and its interaction with matter has been proven as essentially correct by a very large number of empirical observations over a period of more than 150 years. Some of these observations come from empirical tests designed to verify the validity of the theory, while by far most come from the practical application of the theory and the success of those applications.

We see in this thread again that John Millet does not accept any proofs and finds always holes in them. That’s because the approach cannot succeed in producing a proof. As long as he does not accept the whole theory, no proof is valid, when he does accept no proof is needed anymore.

All these simple empirical tests may be useful, as they help in learning about the phenomena. Observing radiation from various directions with an IR detector (“IR thermometer”) tells about the radiation. It tells about the radiation, when the principles of the IR detector are understood to a sufficient degree. The principles cannot be understood unless the underlying theory is accepted as valid.

It’s possible to figure out that the observations are in full agreement with the present understanding of IR radiation and the radiative processes in the atmosphere, but such simple observations by themselves are far too weak to prove anything. Again they are a confirmation for those who didn’t need any confirmation and worth very little to those who don’t know physics and prefer to keep their doubts.

• John: My point was simply that you personally – like Roy – could detect the existence of a TWO-WAY flux of photons between the atmosphere and the ground with an IR thermometer. Point the thermometer at the ground, then point it at the sky. Thermal IR is flowing in both directions. Right?

Yes, these primitive devises can’t detect all IR wavelengths accurately and therefore need to be calibrated against a source of known temperature (and emissivity) before measuring “temperature”. More accurate measurements show the same thing. (Even if the detector is technically not called a photomultiplier.)

Yes, the Sky Dragons who responded to Roy have some fairly dubious schemes for trying to avoid the consequences of two-way flux – mainly that photons somehow sense the temperature of their source and destination and decide whether to be absorbed or reflected. I’ve written above how totally absurd this seems to me. .

Worst of all, everyone who believes in the molecular theory of gases and the Boltzmann distribution of molecule speeds (not to mention the whole field of statistical mechanics) logically should recognize that the 2LoT does not applied to energy transfer between individual molecules – only to the net transfer of energy between groups of molecules large enough to have a defined temperature. Each photon transfers energy between two distinct individual molecules’ both of which have kinetic and internal energy, but not a temperature. The Sky Dragons at Roy’s and here always ignore this problem. They (you?) bloviate about the 2LoT without understanding the physics underlying the concept of temperature, hot and cold.

While misapplying the 2LoT, they also forget about the 1LoT – conservation of energy. The temperature of an object changes only when more or less energy flows into the object than flows out. It doesn’t make any difference where the energy is coming from or going to – the difference between these always results in a change in “internal energy”, thereby conserving energy. Internal energy includes kinetic energy of motion (which is proportional to absolute temperature) and energy stored in other vibrational, rotational, electronic, etc. energy states. (Heat capacity tells us how quickly temperature changes with an energy imbalance. When lots of vibrational and rotational energy states are available, more energy flows into or out of these modes and kinetic energy/temperature goes up or down more slowly.) You can babble about cold bodies warming warm bodies until you are blue in the face, but the blue color will be due to cooling only if the net energy flux from your face is greater than the net energy flux into your face.

• on May 4, 2013 at 8:26 am John Millett

Frank,

Radiation in the atmosphere is isotropic. The IR thermometer will give a reading whatever its orientation. Place two back to back horizontally with infinite clear sight and, presumably, they’ll read the same. Yes, in transit, 2-way fluxes every which way! That, however, tells us nothing about radiation’s interaction with matter which, as Pekka points out, is the only subject of interest in it.

At the boundary between a material body and its surroundings, IR exists in a bewildering variety of orientations and energies. Its constituent photons are orphans – we know not whence they came. Though a proportion of their total number will inevitably be incident upon the body, none can be assumed to have unconditional right of entry to it. To be absorbed into the body a photon must be of an energy sufficient to transition electrons to higher orbits around the atomic nucleus – consistent with the conservation law or 1LoT. The host body, not the visiting photon, determines the interaction between them.

Sky Dragons? No, not me. As mentioned to DeWitt some time ago, in this enterprise I am more lone ranger than groupie. Thanks to your cogent arguments, I have long understood that 2LoT applies only to energy transfer in macroscopic systems, not to molecule to molecule transfer. Less cogent ones have failed to satisfactorily explain that it applies only to net energy, implying 2-way, transfers between material bodies. That transfer requires a conversion from kinetic energy of the body to radiative energy as emission from the body. It also requires the reverse conversion as absorption into the other body. .My question: Why can’t it apply to two independent one-way transfers between a body and the radiation that surrounds it, one of emission, the other of absorption?

The problem, for me, with mutual transfer between bodies, as hypothesised by Prevost over 200 years ago, apart from it never having been seen, lies at the microscopic level. At the macro level at thermal equilibrium each body would be emitting and absorbing equal quantities of energy. But emission and absorption at the microscopic level are independent phenomena. What is the probability that these independent phenomena generate equal numbers of photons simultaneously entering and exiting the body? Or the probability that this occurs when the temperature gradient between them is zero? I would submit that the probability is maximum when the numbers of photons is minimum, i.e., zero. Then, the radiation reservoir surrounding the bodies would be at steady state; no flow would exist between the bodies, a flow about which, if it did exist, one would be entitled to ask: Why, apart from satisfying Prevost’s hypothesis? What physical function does it perform?

• John,

The modern formulation of Kirchhoff’s law states that emissivity of a body is equal to the absorptivity at the most detail level, i.e. when all potentially influential parameters of the radiation are the same. The most commonly discussed parameter is the wavelength, but for a full answer also direction of the radiation and its polarization must be the same.

The rate of absorption is determined by the product of absorptivity and intensity of incoming radiation, while the rate of emission is determined by the emissivity and temperature of the body according to Planck’s law. In thermal equilibrium of the full system the temperature and the incoming intensity match in such a way that the rate of emission is equal to the rate of absorption for every wavelength, every direction, and every polarization.

Kirchhoff’s original formulation stated that the overall level of radiation within a closed isothermal cavity has always the level determined by the temperature alone. That’s the level that in more detail satisfies the detailed balance I describe in the previous paragraph.

Kirchhoff’s law is fundamentally closely related to 2LoT, but it’s more detailed and tells, how radiation takes it’s part in the overall balance of the 2LoT.

The modern detailed formulation of Kirchhoff’s law is also very closely related to the quantum mechanical theory of interaction of radiation with matter. The quantum mechanical formula for absorption of a single photon is identical to the formula for the emission of an identical photon, when the material system does the reverse transition between two states. Temperature determines the likelihood that the material system is in the state that can emit the photon.

• John wrote: “But emission and absorption at the microscopic level are independent phenomena. What is the probability that these independent phenomena generate equal numbers of photons simultaneously entering and exiting the body?”

On a microscopic level, radiation’s interactions with matter are determined by the Schwartzschild equation:

dI/ds = emission – absorption = no*B(T) – no*I_0

where: I is the intensity of radiation at a particular wavelength, n is the number of molecules per unit volume, o is the absorption/emission cross-section, B(T) is Planck’s function for that wavelength and temperature and I_0 is the initial light intensity. After light passes through any material far enough, eventually emission and absorption will come into equilibrium and the intensity will not change with distance traveled. dI/ds = 0 and I becomes equal to B(T). In other words, after light travels through a homogeneous material for long enough, the material emits blackbody radiation. When it hasn’t traveled far enough, the intensity is changing from the incident intensity to blackbody intensity. (At wavelengths where o is zero, however, nothing changes.)

Absorption and emission are not really independent phenomena. Absorption is simply emission with time running backwards. When a photon of appropriate energy is near a molecule, the is some probability (expressed in quantum mechanics as an amplitude, a complex number) that the photon will disappear and the molecule will be in a higher vibrational, rotational or electronic energy state. The opposite is also true: There is some probability that a molecule a higher vibrational, rotational or electronic energy state will create a photon and find itself in a lower energy state. The quantum mechanical amplitude for emission is the complex conjugate of that for absorption, forcing the emission and absorption coefficients to be equal under most circumstance. (The fact that photons travel at the speed of light and don’t linger in the neighborhood of a potential absorbing molecule makes no difference.)

One can calculate how many IR photons are emitted per unit area from the surface of the earth at 288 degK (390 W/m2) where the peak emission is 10 um. You get about 20,000 photons per nm^2 per second. Distances between atoms in molecules are about 0.15 nm, so there are about 10 molecules (more than 1, less than 100) per nm^2 on the surface of a solid or liquid (where the molecular surfaces touch) and about 1% as many in a gas at 1 atm. On the earth and in the atmosphere, molecules exist in an intense, continuous shower of photons – about a thousand photons per second per molecule.

• on May 6, 2013 at 5:53 pm DeWitt Payne

John Millet,

What is the probability that these independent phenomena generate equal numbers of photons simultaneously entering and exiting the body

They’re not independent and they’re only random at the micro level. They’re linked by the First Law, conservation of energy and the Planck function. If more energy is emitted than absorbed by an object, its energy content must decrease which decreases its temperature and reduces emission and conversely for an excess of absorption vs. emission.

And finally, how does an excited state on the surface of an object or an excited gas molecule know whether the temperature of the other object is the same or different? It can’t. Therefore overall emission doesn’t decrease as temperatures equalize.

57. I, like many, continue to be puzzled why so many people seemingly just cannot grasp the concept behind the GHE and/or see it as being in violation of the 2nd Law. The GHE is fundamentally a radiative resistance effect. Much of the upwelling radiation from the surface and lower atmosphere is absorbed by the atmosphere and subsequently re-radiated back downward toward the surface (direct downward LW at the surface or so-called ‘back radiation’ has little to do with it and is major misnomer). Now much of what is absorbed is re-radiated back up in the same direction it was going pre-absorption (something climate science seems to have overlooked when quantifying GHG ‘forcing’, but that’s another issue), but the key point is some is radiated back down in like post albedo solar power is radiating down in. In order to ‘push through’ the same amount of power as is arriving from the Sun (about 240 W/m^2), the surface and lower atmosphere has to emit a higher rate because not all of what is emitted up acting to cool can pass through the remaining atmosphere into space.

As Frank has pointed out, on an individual particle or molecular level, there is no requirement that a photon’s energy has to travel soley from warm to cold or will only be absorbed if it originated from a warmer source.

• RW: We can be somewhat more precise about “re-radiation”. When a GHG in the atmosphere absorbs an upward infrared photon, “that energy” has a 50-50 chance of being emitted as an upward or downward photon. However, the same is true for downward infrared photons (emitted by GHGs or clouds in the atmosphere) and for photons from the sun that don’t reach the surface and aren’t reflected back into space. I put quotes around “that energy” because we can’t actually account for the fate of the individual packets of energy delivered by photons. The energy lost by radiation is emitted in all directions. When net energy transfer by convection and conduction is negligible and temperature isn’t changing, then the energy delivered and lost by radiation must be equal and exactly half will be “re-directed” in a direction opposite from which it came.

You also say: “In order to ‘push through’ the same amount of power as is arriving from the Sun (about 240 W/m^2), the surface and lower atmosphere has to emit a higher rate because not all of what is emitted up acting to cool can pass through the remaining atmosphere into space.” Calculations suggest that the temperature of the surface of the earth would need to rise to nearly 350 degK to “push through” the power arriving from the sun and DLR. This makes the greenhouse effect nearly 100 degK, not 33 degK. Fortunately, an unstable lapse rate is produced long before the surface temperature can rise this high, allowing convection to remove the energy that can’t escape through our fairly opaque lower troposphere to space. About 160 W/m2 is delivered to the surface by the sun; 100 W/m2 leaves via convection and 60 W/m2 via net IR radiation (390 up minus 330 down). IMO, skeptics should stop trying to poke holes in radiative forcing from increasing GHG’s and focus their attention on how convection will change with increasing GHG’s.

Increasing GHG’s will increase DLR; the average DLR photon arriving at the surface have been emitted from a lower altitude where it is warmer. However, this “surface radiative forcing” amounts to only 1 W/m2 for 2X CO2. That forcing alone would produce a surface temperature rise of only 0.2 W/m2 even if none of the extra energy were removed by increased convection. 3.7 W/m2 of radiative forcing at the tropopause requires a 1 degK temperature at the altitude from which the average photon escaping to space and this figure is doubled if you assume that relative humidity remains constant. How can this trivial increase in surface temperature (0.2 degL) be compatible with a large increase in the upper atmosphere (2 degK)? This can only happen if less energy – not more energy – is temporarily convected upward. And the 2 degK warming in the upper troposphere needed to restore radiative equilibrium can only be produced by an increase in convection. I can’t integrating radiative forcing at the surface and tropopause. Can anyone else?

• Frank,

The defining situation for forcing is the one, where the only thing that has changed in troposphere is the CO2 concentration, moisture levels and temperatures are unchanged. The radiative balance at TOA is changed by the amount of forcing. All that change goes to warming of troposphere and surface, by far most to the warming of surface. Thus the overall balance at surface is not very different from that at tropopause.

The changes in radiative heat exchange would warm the tropopause much faster. The troposphere warms therefore slightly more than the surface and that slows down the convection. The convection settles automatically to the level that’s required by the energy balance of the troposphere.

In the hypothetical no-feedback case the temperature differential between the surface and any fixed altitude of the troposphere changes very little, but the tropopause is pushed a little higher allowing for a larger temperature difference between the surface and the new tropopause. As the no-feedback case is a theoretical counter factual construct, different people may have slightly different views on its properties trough differing ways of defining the concept.

As the altitude of tropopause is raised the convection extends higher up. Thus we do certainly have some altitudes where convection is increased, while it’s forced to be weaker near surface by the energy balance argument.

• Pekka: Let’s imagine a step function doubling of CO2. About 1 W/m2 of DLR arrives at the surface and about 3.7 W/m2 less escape into space, but are still trapped inside the atmosphere. The retained energy needs to reach higher altitudes where radiation can escape more easily to space. Now what happens?

An unstable lapse rate is necessary for convection, but it doesn’t directly determine how much energy (power) flows upwards. Flow continues as long as the gradient is unstable – once the upper atmosphere has warmed enough, flow must stop. However, there is a way out of this limitation. Convection reaches much higher altitudes in the tropics than the temperature zone, transferring energy to altitudes where radiative cooling is more efficient – and more efficient cooling allows an unstable lapse rate to persist for longer. The higher the flow; the longer the flow and the cooler the earth’s surface. Assuming the “hot spot” in the upper tropical troposphere is really missing, climate models appear to have problems modeling the mutually-reinforcing relationship between higher convection, increased radiative cooling, and more persistent convection. More convection will strengthen the trade winds, which will increase surface evaporation. And more moisture decreases the stability of any lapse rate. Another mutually reinforcing element in the system. Does anyone ever calculate the height of convective columns from fundamental principles?

• Frank,

What you describe is workings of the present atmosphere. More CO2 adds a little to that but is not expected to lead to any new phenomena.

I’m not an expert at all on this, but I’m sure that meteorological models can describe reasonably well the convective uplift in the tropical atmosphere, and that these models are based on physical equations.

• Pekka wrote: I’m not an expert at all on this, but I’m sure that meteorological models can describe reasonably well the convective uplift in the tropical atmosphere, and that these models are based on physical equations.

Frank replies: The climate models (I don’t know about “meteorological models”) used by the IPCC perform calculation on cells that are much too big to represent the features in convective towers. This is why “cloud resolving models” were created, but they can’t be used on a planetary scale. Isaac Held showed some results from CRMs on his blog.

Furthermore, all GCMs make arbitrary choices of parameters that control clouds and precipitations. AR4 referred to the set of climate models they use as “an ensemble of opportunity” – not a group of models intended to explore the full range of possible futures compatible with known physics.
This Stainforth paper shows how dramatically and chaotically results change as parameters are modified within the known range of uncertainty. The process by which parameters in climate models have been “optimized” one by one in the past appears meaningless and potentially subject to bias. Note that this paper covered only 6 of 21 adjustable parameters and used a slab ocean, avoiding additional serious uncertainties about transport of heat into the deep ocean.

D. A. Stainforth et al, Nature 433, 403-406 (27 January 2005) | doi:10.1038/nature03301;
Uncertainty in predictions of the climate response to rising levels of greenhouse gases

Abstract: The range of possibilities for future climate evolution needs to be taken into account when planning climate change mitigation and adaptation strategies. This requires ensembles of multi-decadal simulations to assess both chaotic climate variability and model response uncertainty. Statistical estimates of model response uncertainty, based on observations of recent climate change admit climate sensitivities—defined as the equilibrium response of global mean temperature to doubling levels of atmospheric carbon dioxide—substantially greater than 5 K. But such strong responses are not used in ranges for future climate change because they have not been seen in general circulation models. Here we present results from the ‘climateprediction.net’ experiment, the first multi-thousand-member grand ensemble of simulations using a general circulation model and thereby explicitly resolving regional details. We find model versions as realistic as other state-of-the-art climate models but with climate sensitivities ranging from less than 2 K to more than 11 K. Models with such extreme sensitivities are critical for the study of the full range of possible responses of the climate system to rising greenhouse gas levels, and for assessing the risks associated with specific targets for stabilizing these levels.

First Paragraph: As a first step towards a probabilistic climate prediction system we have carried out a grand ensemble (an ensemble of ensembles) exploring uncertainty in a state-of-the-art model. Uncertainty in model response is investigated using a perturbed physics ensemble in which model parameters are set to alternative values considered plausible by experts in the relevant parameterization schemes. Two or three values are taken for each parameter (see Methods); simulations may have several parameters perturbed from their standard model values simultaneously. For each combination of parameter values (referred to here as a ‘model version’) an initial- condition ensemble is used, creating an ensemble of ensembles. Each individual member of this grand ensemble (referred to here as a ‘simulation’) explores the response to changing boundary conditions by including a period with doubled CO2 concentrations.

Methods: Perturbations are made to six parameters, chosen to affect the representation of clouds and precipitation: the threshold of relative humidity for cloud formation, the cloud- to-rain conversion threshold, the cloud-to-rain conversion rate, the ice fall speed, the cloud fraction at saturation and the convection entrainment rate coefficient. This is a subset of those explored by ref. 9. In each model version each parameter takes one of three values (the same values as those used by ref. 9); for cloud fraction at saturation only the standard and intermediate values are used. As climateprediction.net continues, the experiment is exploring 21 parameters covering a wider range of processes and values.

Key paragraph: Can we coherently predict the model’s response to multiple parameter perturbations from a small number of simulations each of which perturbs only a single parameter? The question is import- ant because it bears on the applicability of linear optimization methods in the design and analysis of smaller ensembles. Figure 2c shows that assuming that changes in the climate feedback parameter l combine linearly provides some insight, but fails in two important respects. First, combining uncertainties gives large fractional uncertainties for small predicted and hence large uncertainties for high sensitivities. This effect becomes more pronounced the greater the number of parameters perturbed. Second, this method systematically underestimates the simulated sensitivity, as shown in Fig. 2c, and consequently artificially reduces the implied likelihood of a high response. Furthermore, more than 20% of the linear predictions are more than two standard errors from the simulated sensitivities. Thus, comprehensive multiple-perturbed-parameter ensembles appear to be necessary for robust probabilistic analyses.

• Frank,

What you write about the climate models is the reason for my referral to meteorological models. I do think that they can answer the question you posed, i.e. tell about the nature of the tropical uplift.

If we have such better models then why we have all the uncertainties of the climate models?

The reason is in the fact that meteorological models are initialized to observations and can make forecasts for limited periods only. Their use does not require that the models are well behaved over longer periods. They might, e.g., violate severely some essential conservation laws when run over too long periods. Many of the mechanisms and relationships that are essential for projecting climate are of little importance in weather forecasting.

There are, however, so many similarities that some models are used both in weather forecasting and in climate analysis, but even these models must be run in different domains, and some parts of the models must be quite different for the different applications.

58. on May 6, 2013 at 3:45 am | Reply John Millett

Frank,

“On a microscopic level, radiation’s interactions with matter are determined by the Schwartzschild equation:
dI/ds = emission – absorption = no*B(T) – no*I_0

and

“In other words, after light travels through a homogeneous material for long enough, the material emits blackbody radiation”.

Then, from the equation, the product of molecular number density and emission/absorption cross section must become unity. What causes that evolution?

• John,
The equation is written for the intensity of radiation of a specific direction and how that intensity changes when we move in the same direction. It does inot tell anything about anything else.

• Frank wrote: “On a microscopic level, radiation’s interactions with matter are determined by the Schwarzschild equation:

dI/ds = emission – absorption = no*B(T) – no*I_0

“In other words, after light travels through a homogeneous material for long enough, the material emits blackbody radiation”.

John replied: Then, from the equation, the product of molecular number density and emission/absorption cross section must become unity. What causes that evolution?

Frank expands: After light travels through a homogeneous material for long enough, emission and absorption eventually will approach equilibrium. a) If a relatively intense beam of light at a given wavelength enters and passes through a relatively cool medium, the intensity of the light a certain distance into the medium will be effected more by losses due to absorption instead of gains from emission. A laboratory spectrophotometer has a lamp with a source at several 1000 degK. When I put my sample (295 degK) in the beam and I record an absorption spectrum, I ignore the relatively few photons emitted by the sample.) However, if my sample absorbs 99.9999999999…% of the incident photons, emission eventually will become significant. b) If I point an IR thermometer at empty sky at night, essentially all of the photons (except the background from the Big Bang and a few distant stars) reaching the IR detector will have been emitted by the atmosphere (which is much warmer than space). In this case, emission dominates in the upper atmosphere, but absorption begins to become important once the intensity has grown. c) The earth’s surface and atmosphere are similar in temperature, so the radiation leaving the earth for space depends on emission by both the surface and the atmosphere, as well as absorption by the atmosphere.

Eventually emission and absorption will approach equilibrium, meaning that dI/ds will approach 0 and the intensity of the radiation won’t change as the radiation travels further through the medium. (As Pekka kindly pointed out, I is the intensity of radiation traveling in a particular direction and s is the distance traveled in that direction. I_0 is the intensity where s=0, and s is often chosen to be zero where light enters the medium.) When dI/ds = 0, mathematically either 1) I = B(T) for that wavelength – the intensity expected for blackbody radiation, or 2) n = 0 or o = 0 – there are no molecules present or the molecules present don’t absorb or emit that wavelength).

If you look down on the earth from space, you see some photons coming from the earth’s surface through the “atmospheric window” with an intensity appropriate for 280-300 degK. If you look at wavelengths strongly absorbed by water vapor, few of the photons with those wavelengths make it from the surface to space. However, the amount of water vapor diminishes rapidly with altitude and those wavelengths become fairly transparent in the upper troposphere. So the intensity you see at those wavelengths is appropriate for a blackbody at the temperature of the mid-troposphere, perhaps 25-40 degK cooler than the surface. The atmosphere doesn’t become transparent at the wavelengths most strongly absorbed by carbon dioxide until well into the stratosphere, so the edges of that absorption has the intensity expected for the temperature at the tropopause (215 degK) and the very center of the band comes from the warmer stratosphere. You’ve probably seen figures such as the one below many places (including SOD). To be honest, most of these figures are calculated from radiative transfer theory and I’m not sure where to find real observations from space. (Some calculations are done looking down from 20 km, the top of the troposphere, and then you don’t see the spikes from CO2 emission from the warmer stratosphere.)

I’m probably getting myself onto very shaky ground scientifically, but the Schwarzschild equation appears to apply to radiation traveling through liquids and even solids, as well as gases. In liquids and solids, the individual lines from rotational energy levels are smeared out by perturbations from neighboring molecules, in much the same way that they are smeared out in gases at high pressure. Since solids and liquids are dense and often have large absorption/emission cross-sections, the radiation traveling though them becomes blackbody radiation at all wavelengths after a very short distance. Therefore the radiation that emerges from most surfaces has the intensity expected for blackbodies, except that some of the radiation is reflected/scattered at the surface. Scattering/reflection occurs at the interface both on the way out and equally on the way in – so the emissivity and absorptivity of materials are equal. So the Planck function (for blackbody radiation from solid surfaces) appears in the Schwarzschild equation for good reasons. For energy to be conserved on a microscopic level (Schwarzschild eqn) or a macroscopic level (Kirckhoff’s law), absorption needs to be equal to emission when a radiation field is in equilibrium with the surrounding matter. So the absorption and emission coefficients must be equal at the molecular level and the absorptivity and emissivity must be equal at the macroscopic level.

59. on May 6, 2013 at 6:13 am | Reply John Millett

Pekka,

“The quantum mechanical formula for absorption of a single photon is identical to the formula for the emission of an identical photon, when the material system does the reverse transition between two states”.

Where does Einstein’s Coefficients A&B, which I understand to be related but not equal, come into this?

Reverting to my question about the function performed by equal and opposite radiation fluxes between two bodies at the same temperature:

A cavity at steady state at temperature T, its constituent energy waves darting around every which way. The system, lacking any structure or order, its entropy is at a maximum. Introduce two bodies each at temperature T into the cavity. Local directional radiation fluxes emerge between the two bodies (the work of Prevost’s ghost?). That is, order has evolved from disorder, a prohibited reduction in entropy?

• John,

Einstein’s A could be considered the absorptivity of a single molecule, B is similarly the emissivity of a single molecule multiplied by the Planck function. Thus B takes into account the probability that the molecule is in the particular excited state. (There are actually two B’s, the other is for stimulated emission and important only in strong electromagnetic field like that inside a laser.)

The value of A is essentially the absolute square of the quantum mechanical transition amplitude.

I formulated my description in more general terms to make it true for solid and liquid bodies as well, not only for molecules in a gas.

I don’t understand your last point. You are telling about a system that’s changed by external action (introducing two bodies). Thus 2LoT has nothing to say about the change.

60. on May 6, 2013 at 7:50 am | Reply John Millett

Frank,

“On the earth and in the atmosphere, molecules exist in an intense, continuous shower of photons – about a thousand photons per second per molecule”.

Isn’t this sufficient reason to think that radiation exchange occurs one- way directly between a body and its radiative surroundings rather than 2-way between bodies?

• John: Why would an intense shower of photons suggest one-way transfer. The intense shower is traveling in all directions.

• on May 9, 2013 at 8:37 am John Millett

Frank, the question should have read: Isn’t this sufficient reason to think that exchange occurs directly between a body and its radiative surroundings rather than between two bodies? Whether the exchange is one or two way is a separate issue.

• John,

The intensity of the photon shower alone does not make it useful to use the concept of radiative surroundings. It’s not a useful concept, when the radiation comes from a single source or a few sources.

It may be useful, when the radiation originates from many sources and in particular when it originates from optically dense gas (or plasma) that surrounds the body. In other cases the properties of the photons depend on the source in such a way that it’s best to take both the source and the target into directly into account.

• John: A molecule at the surface of the earth or in the lower troposphere may see a photon passing nearby every few microseconds, but those photons are traveling 300 m/microsecond and often travel farther than this before being absorbed. The radiative “surroundings” are pretty big on the earth, so I agree with Pekka that it isn’t a very useful concept.

You can get into difficulties if you think that the “surroundings” are filled with the electric and magnetic fields (rather than photons). Fields can cancel and used to calculate energy flux. Some the “Sky Dragons appear to use canceling fields to “prove” one-way (rather than two-way) flux of energy between the surface and the atmosphere. The wave theory of electromagnetic radiation probably died when Mitchelson and Morley found that “ether” did not exist, but it is still useful today for calculating how a large number of photons will behave.

• Frank,

Wave theory of electromagnetic radiations keeps on going as strong as ever, but it determines the outcome only as far as there’s coherence. The radiation is normally not coherent, in case of IR, lasers produce coherent radiation, not molecules of atmosphere or surfaces of warm bodies. A warm body cannot stop incoming photons because it cannot create coherent EM fields to stop it.

• on May 10, 2013 at 1:34 pm DeWitt Payne

Pekka,

I haven’t seen the original papers in Phil. Mag., but according to Vaughan Pratt, Wood failed to distinguish between coherent and incoherent superposition of light when he tried to prove that his poorly documented greenhouse experiment produced the correct results. He was wrong on both counts.

• DeWitt,

I write the following as it may be illuminating to some.

I don’t think that the original simple experiment of Wood was outright wrong, but his extension to the atmosphere certainly was.

Whether trapping of IR by the window glazing is important or not depends on the prevailing conditions and on the type of glazing. A single glass window has essentially the same temperature on both surfaces. Under appropriate conditions it gets as hot as the bottom surface heated by the sun. In that case it radiates as much IR as it stops radiation from the surface. In that case it does not affect the temperature through its influence on IR. This was probably the explanation for Wood’s result.

A very different situation would be a greenhouse with double glass windows in cold climate. In that case the outer glass is likely to be much colder than the interior of the greenhouse and radiate much less. To have the effect we must have a sizable temperature difference between the bottom surface and the outer surface of the window, exactly as the GHE is dependent on the sizable temperature difference between Earth surface and upper tropopause.

Concerning the coherence discussed in the second note of Wood and the response to that, the issue is not the same that I discussed in my previous comment. My comment was about the lack of coherence between to photons, while Wood and others discuss coherence that is true even for single photons. After all single photons do have reflection and diffraction, but they do not interact to a significant degree with other photons – except in lasers and some similar special cases.

• Another example about the influence of glazing. Our house has triple glass windows. I have pointed my IR thermometer to the window when the outside temperature is -20C, it’s reading has been +17C or higher. The room temperature was about +20 C. If the window would not cause “a GHE” I would feel very uncomfortable near the window in winter as I would radiate much more IR towards the window that receive from that direction. The radiation from -20C is only 55% of that from +20C, while that from the window at +17C is 96%.

I feel almost as warm close to the window as I feel elsewhere in the room only because glass stops IR and my window creates a radiative GHE.

• Pekka and DeWitt: The glass in a car is single pane. Isn’t the heat that builds up in a car on a sunny day the best example of the “greenhouse effect” that is produced by glass?

It would be interesting to point an IR thermometer at a variety of glass surfaces under a variety of conditions. Glass with and without a low emissivity coating at the same temperature (both completely inside a room) should appear to have different temperatures according to an IR thermometer.

• Frank,

I cannot be sure, but I wouldn’t expect that replacing the glass by another window material transparent to IR would make a big difference for a car. in sunshine.

Reading the first Wood note we can see that he understood that there’s a radiative effect as well, the question is, how strong it is relative to the other effects in each specific case. The most important factor that enters in determination of the strength of the radiative effect is the temperature difference between the internal surfaces like the ground or the top of the dashboard and the outer surface of the window. The top of the dashboard is certainly hotter. Thus there’s an effect, but how large that is in comparison with the influence of other surfaces of the car is another matter.

In an otherwise well insulated home in cold climate replacing glass windows with IR transparent crystals would be a significant effect like 150 W/m^2 of window area.

• Pekka wrote: “Wave theory of electromagnetic radiations keeps on going as strong as ever, but it determines the outcome only as far as there’s coherence. The radiation is normally not coherent, in case of IR, lasers produce coherent radiation, not molecules of atmosphere or surfaces of warm bodies. A warm body cannot stop incoming photons because it cannot create coherent EM fields to stop it.”

Frank replies: One can observe many light phenomena typical of waves even when radiation is not coherent or even monochromatic. For example, I can hold two fingers slightly apart in front of a light source and see one or more bands where the light waves “cancel”.

I was mistaken to suggest that the absence of an “ether” dealt a fatal blow wave theory. Transverse “electromagnetic waves” of perpendicular electric and magnetic fields moving at the speed of light are solutions for Maxwell’s equations in empty space. In “A First Course on Atmospheric Radiation”, Grant Petty spends 10 pages on classical wave theory and applies it to numerous situations (especially phenomena like scattering involving objects of similar size to wavelength), and then admits on the next few pages that sometimes you need to use QM to get the right answer. Things become even more confusing when one adds the idea magnetic fields are [merely] a consequence of electric fields and special relativity. In his book “QED”, Feynman spends several pages several pages emphasizing that light is not a wave and then illustrates how some wave phenomena are produced by particle [photons] following the (simplified) principles of quantum mechanics that he introduces. The description of how light takes all possible paths from one point to another while reflecting off a mirror is particularly illuminating when he describes scraping away some of the mirror and creating a diffraction grating.

It appears as if QM explains all of the behavior of light, and electromagnetic waves only explains part of the behavior of light. Why aren’t electromagnetic waves a “dead” theory? Well, Newton’s Laws of motion need to be corrected for relativity, but they are still great for calculating how things behave under many circumstances. Even so, these laws are still “fundamentally incorrect”. Likewise EM waves are great for calculating certain aspects of how light behaves, but waves appears to me to be as “fundamentally incorrect”. Unfortunately, most of the world acts as it there are two equally valid explanations for light, photons and EM waves, which means that I’m probably confused.

• Frank,

I largely agree on what you write. My following comments go a little deeper in physical theories

Quantum theory differs is in many ways from the customary way of looking at physics. The full theory cannot be solved exactly even when we leave out all strong interactions. The very successful theory of interaction of electromagnetic radiation with matter (Quantum Electrodynamics, QED) is based on a specific approach for doing approximate calculations known as perturbation theory. That works because photons interact so weakly with other photons and weakly enough also with matter.

Order is kept in the calculation of the perturbation theory with the help of Feynman’s diagrams. Each of the diagrams represents a term in the approximate calculation. That’s helpful, because the contribution to the results goes rapidly down, when the diagrams get more complex.

The simplest full process is formed by a diagram that describes an emission of a photon and the subsequent absorption of it at another location. Every full process described by such a diagram is independent and thus incoherent with the other processes. That is the same thing as saying that photons are emitted and later absorbed in a way that does not significantly depend on any other photons. (Coherence is a very essential factor in QM.)

There are more complex diagrams that include more than one line that corresponds to a photon. Such diagrams are needed for the most accurate calculations, and they are also needed if this approach is used to analyze lasers, where the more complex ones are not that small due to stronger electromagnetic fields. In the case of such strong fields other approaches may, however, be useful, and we me be back using Maxwell’s equations. Quantum theory is important also in lasers, but it can be introduced in a different way.

It’s typical for quantum theory that the same physics can be looked at in many different ways. All of these ways corresponds to a particular approximation of the full equations. For some questions on approach is best, for other questions some other. It’s often possible to analyze the same problem in different ways. When that can be done, the different approaches agree on the final outcome within the range of expected accuracy of the methods.

Wave-particle duality was perhaps the first important example of looking at the same physical system in different ways. For electromagnetic radiation the duality means that the question is not only about waves, but in many ways photons act as massless particles. For electrons and also heavier particles it means that they have also wavelike properties.

Maxwell’s theory of electromagnetism has not ceased to be valid, but it’s not a full theory, but then no existing theory is. In many application QED developed by Feynman and and others is the best extension, but in some cases other ways of extending it to include quantum effects work better.

Physicists are still struggling to understand fundamentals of QED in spite of the success of the theory to produce incredibly accurate results that agree with experiments. Putting together all good partial theories to form a single consistent theory has not succeeded. It’s possible and even likely that QED alone is fundamentally faulty even when it’s extremely successful in practice, and that those faults can be corrected only in a wider theory that covers “everything”.

Diffraction and reflection of light is based on the coherence between the alternative routes one single photon may take between the point of emission and point of absorption. That’s the “wave” in wave-particle duality. The coherence that I mention in the 4th paragraph is coherence between two photons. That’s not needed for diffraction or reflection and does actually play almost no role in that. (I include the word ‘almost’, because it does have an important role in reflections inside a laser, and a really small effect elsewhere.)

61. on May 7, 2013 at 8:32 am | Reply John Millett

DeWitt Payne

“And finally, how does an excited state on the surface of an object or an excited gas molecule know whether the temperature of the other object is the same or different? It can’t. Therefore overall emission doesn’t decrease as temperatures equalize”.

Popular though the practice is, IMO, anthropomorphising inanimate matter doesn’t lead to better understanding – a surface or molecule can’t know its own temperature any less than it can’t know that of the surrounding medium. How could it know that its temperature is above absolute zero and it should start to radiate; or that it needs to radiate at an intensity befitting its temperature? A more fruitful approach, hopefully, would recognise that entropy law requires an atom in an excited state, at any temperature, to revert to its ground state. Emission and temperature fall would be the results of that internal dynamic and they would cease with achievement of the ground state. Conversely, absorption and temperature rise result from interaction with the surrounding radiation which is governed by the wavelength-dependent energy of the visiting photon and the energy state of the host atom. A ground-state atom would be more likely than an excited one to absorb the photon. A cold, long wavelength, low energy photon would less likely be accommodated by the host atom than would a high energy photon. When the photon lacks the power to energise the atom from its ground state absorption and temperature rise would cease. As the temperature difference between the matter and the surrounding radiation approaches zero, so too do the alternating emissions and absorptions. Well, that’s a novice’s take on the issue.

• John,

entropy law requires an atom in an excited state, at any temperature, to revert to its ground state.

Entropy is not a concept that can be applied to individual atoms or molecules. It’s true that an individual molecule in an excited state will usually revert to ground state trough the emission of a photon. (In some cases like that of a nitrogen molecule, that’s, however, not possible due to the requirement of conservation of angular momentum or a similar reason.) The average lifetime of a molecular vibrational state may be as long as one second.

Emission and temperature fall would be the results of that internal dynamic and they would cease with achievement of the ground state.

Your reference to ground state seems to refer to a system with zero absolute temperature.

Conversely, absorption and temperature rise result from interaction with the surrounding radiation

Perhaps so, but depends on, how we interpret the sentence

.. which is governed by the wavelength-dependent energy of the visiting photon and the energy state of the host atom. A ground-state atom would be more likely than an excited one to absorb the photon.

There are no general rules of that type. Each state of a gas molecule can absorb at some specific wavelengths. The ground state absorbs at some wavelengths, the excited states at some other wavelengths. There’s no reason to think that the excited states would be less likely to absorb than the ground state. There are, however, more molecules in the ground state than in any specific excited state.

A cold, long wavelength, low energy photon would less likely be accommodated by the host atom than would a high energy photon.

Again no general rule of that type. The answer depends on the the set excited states that can be formed. As an example CO2 can absorb very low energy very long wavelength IR photons that correspond to rotational states of the molecule as well as much higher energy and shorter wavelength photons that lead to the vibrational states. There’s a wide gap between these energy levels, and CO2 cannot absorb to a significant degree photons with energies in this range.

When the photon lacks the power to energise the atom from its ground state absorption and temperature rise would cease.

Right.

As the temperature difference between the matter and the surrounding radiation approaches zero, so too do the alternating emissions and absorptions. Well, that’s a novice’s take on the issue.

No. When the temperature difference approaches zero, the alternating emissions and absorptions remain as frequent as before. What approaches zero is the difference between the rate of emission and the rate of absorption.

• on May 7, 2013 at 8:55 pm | Reply DeWitt Payne

John Millet,

I specified an excited state. That means the temperature of the system must be above absolute zero.

In a system, the site or molecule itself has no temperature. Temperature is a collective property of the system. It only has energy. If its energy is equal to or above the energy level of the next state, it will be in an excited state. The distribution of energy in the system between all the possible energy levels of all the sites or molecules depends on the temperature of the system and the magnitude of the energy levels, if it’s in local thermodynamic equilibrium. The emission rate of the system for a particular wavelength depends only on the number of excited atoms, molecules or whatever at the particular transition energy level and the decay coefficient for that energy level, the Einstein A21 coefficient for a molecule in the first excited state, for example. There is no dependence on any other part of the system, such as the temperature of another object in a closed system.

For CO2 molecules at 300K, about 7% are in the first vibrational level for the symmetric vibration of the carbon atom with respect to the oxygen atoms. The state is degenerate because you can have two orthogonal vibrations. A much smaller fraction is in the second excited state. The rotational energy levels are much smaller and many are populated. The pure rotational spectrum is in the microwave region as I remember. But since the rotational quantum number can change by ± 1 when the vibrational state decays to the ground state, you get two branches at shorter and longer wavelength as well as a central peak.

62. on May 12, 2013 at 11:24 am | Reply John Millett

On Wood’s experiment:

A contemporary, C.G. Abbot, agreed with Wood that ”convection is the main means of transport of heat away from the surface; but that it would be a small factor if the air was transparent to longwave”. Abbot moves from there directly to his conclusion “that the atmosphere is important as a trapping agent to increase surface temperature”. Abbot seems to say that the longwave-trapping atmosphere prevents the surface from cooling by radiation direct to space and forces it to do so by convection instead. But isn’t the convection-driving property of the atmosphere, its negative vertical pressure gradient coupled with differential surface heating by the sun, independent of its longwave-trapping property?

Frank,

Two cars in the sun, windows up in one, down in the other. Inside the latter is cooler because the cooler denser outside air displaces the inside air allowing heat to escape. Lack of convection rather than IR trapping by glass is the key factor.

Pekka,

At -20, I expect there is a furnace somewhere in the system? Air is a poor conductor and the window incorporates two isolated air cells. Wouldn’t the triple glazing serve a predominately insulating function, allowing the desired room temperature to be maintained with a lower furnace setting and fuel consumption? Were radiation the main factor, the inside panel would be at 17 both sides. It would heat the middle panel which would heat the outside one. All three panels would be at about the same temperature. The cooling rate of the system at 17 facing -20 outside air would, adversely, require the furnace setting to be raised in order to maintain desired room temperature.

• John,

Concerning the Abbot comment. It atmosphere where transparent to longwave the surface would be so cold that little convection would ensue.

In my example about the window. To keep the house warm without excessive heating power, the window must prevent both strong convection and radiative energy losses. Multiple glazing is important for both.

• on May 13, 2013 at 1:43 pm | Reply DeWitt Payne

John Millet,

Prevention of convective energy transfer is critical to maintaining anything at a temperature different from ambient, refrigerators, ovens, houses, etc., not just greenhouses. If only convection were important, however, then why are modern double or triple glazed windows constructed with low emissivity glass? Low emissivity means high reflectivity. So the glazing minimizes both radiative heating in the summer and radiative cooling in the winter. The better the insulation and the greater the difference in temperature between inside and outside, the more important is the control of radiative loss or gain. Imagine your home oven with an IR transparent window, for example. A car with closed IR transparent windows would be cooler than a car with closed normal glass windows. How much cooler would depend on the the thermal conductivity of the rest of the car.

63. on May 12, 2013 at 12:42 pm | Reply John Millett

Pekka Pirila

May 6: “I don’t understand your last point. You are telling about a system that’s changed by external action (introducing two bodies). Thus 2LoT has nothing to say about the change”.

What I’m getting at is that body-body radiative exchange at equilibrium performs no physical function – it isn’t transporting energy – but, against nature’s way, it does sustain in the radiation field a local ordered region between the bodies in contrast to the disorder prevailing elsewhere.

May 9: “It (surrounding radiation) may be useful, when the radiation originates from many sources and in particular when it originates from optically dense gas (or plasma) that surrounds the body. In other cases the properties of the photons depend on the source in such a way that it’s best to take both the source and the target into directly into account”.

In the surface-atmosphere case, what is the source? The boundary condition for DLR at TOA is I = 0 (ignoring solar LW). DLR is produced at the altitude at which OLR escapes to space and is approximately equal to it (earth’s sphericity requires that a slightly larger proportion of emission at any altitude is directed surface-ward than space-ward). DLR proceeds
surface-ward, iteratively, moving always to denser air, implying a high probability of zero transmission. The source of DLR would be the surface layer of the atmosphere. Emissions being isotropic and 4/6 of them laterally directed, photons in this layer could originate from nearby elevated surfaces.

• John,

In a case of two bodies in empty space (very far from all other bodies) the radiation is ordered in the sense that all radiation comes from either one of the two bodies. (In free space we have the 3 K background microwave radiation, everything else comes from specific sources.

In the surface-atmosphere case, what is the source?

Surface and atmosphere.

The boundary condition for DLR at TOA is I = 0 (ignoring solar LW). DLR is produced at the altitude at which OLR escapes to space and is approximately equal to it (earth’s sphericity requires that a slightly larger proportion of emission at any altitude is directed surface-ward than space-ward).

At any single point the emission is totally isotropic.

DLR proceeds surface-ward, iteratively, moving always to denser air, implying a high probability of zero transmission. The source of DLR would be the surface layer of the atmosphere.

DLR originates at all levels in the atmosphere. At wavelengths of high emissivity the radiation that reaches the surface originates very close and has the intensity that corresponds to the temperature of near surface atmosphere. At other wavelengths the point of origin is typically higher up and the intensity lower.

Emissions being isotropic and 4/6 of them laterally directed, photons in this layer could originate from nearby elevated surfaces.

That 4/6 is not true, not even near. I have no idea, what made you propose such a wrong value. Only a small fraction of radiation that hits the surface originates from surface elsewhere. In dense forest the radiation to the surface comes from the canopy, but then we can consider the canopy as proxy for the surface.

• on May 14, 2013 at 8:18 am John Millett

Pekka,

The 4/6 is pure geometry. Emission everywhere is isotropic, or in all directions. Resolve along 3 orthogonal axes each with 2 directions – one up, one down, 4 sideways. The plane parallel assumption dispenses with all the sideways – the heuristic models of the greenhouse atmosphere have equal fluxes up and down only. Resolved into up and down directions only, the globe’s sphericity means a slightly higher proportion of emission up or space-ward directed (I got it back to front originally).

“At other wavelengths the point of origin is typically higher up and the intensity lower”.

Given that DLR, as postulated, consists of GHGs emitting to GHGs at higher mass density, layer by layer, it is not obvious which wavelengths, if any, would be transmitted from upper layers to the surface.

Abundant particulate matter in the atmospheric surface layer could scatter sideways-directed photons towards the surface, rendering the source uncertain..

• John,

You must include a lot that’s not nearly horizontal to get the value 4/6. Counting axes is far too simplistic, so badly that I couldn’t imagine that anybody would do something like that.

Exactly horizontal has zero share as has any exact direction. On the average the share of radiation originating at the surface is perhaps a few percent of radiation that hits the surface, and even that cancels actually out in a more comprehensive analysis. Thus it can excluded with full justification in the calculation of the overall surface energy balance, but must be considered carefully in measurements of downwelling IR.

• on May 14, 2013 at 7:50 pm DeWitt Payne

John Millet,

Emission everywhere is isotropic, or in all directions.

It’s not isotropic as viewed from an area on a plane parallel to the surface at altitude z measured from the surface. If we take φ as the angle from the zenith and θ as the bearing parallel to the surface, then the incidence for any given φ for all θ is the direct normal incidence in W/m²/steradian multiplied by cosine(φ). Irradiance is defined as the energy passing through (or emitted by) that plane per unit area or the integral of I(θ,φ)cos(φ)dφdθ for θ from 0 to 2π and φ from 0 to π (or -π/2 to π/2) That’s fairly simple for emitted energy as it’s assumed that there is no angular dependence on emission for a flat gray body. But it’s more complicated for downwelling atmospheric radiation as the direct normal incidence varies with φ. Counting axes is not going to cut it.

64. Pekka: Thanks for the long answer to my question. Maxwell’s equations describe many phenomena correctly: the energy lost to space as electromagnetic radiation when electrons flow through a wire. One the other hand, they don’t describe the behavior of electrons moving/orbiting in the vicinity of a nucleus. Maxwell’s equations say those electrons should rapidly lose energy to space as electromagnetic radiation and spiral into the nucleus. In this case, Maxwell’s theory of electromagnetism has ceased to be valid. I suspect the theory is always wrong under circumstances where the Uncertainty Principle will be violated.

• Frank,

You are certainly right that Quantum Theory is needed whenever uncertainty principle affects results, or whenever Plank’s constant is needed. Planck presented his radiation formula before Quantum Mechanics was developed, but his formula could not be explained without QM, For many years it was just a formula based on an ad hoc postulate that was the first step towards QM.

• on May 15, 2013 at 1:53 pm | Reply DeWitt Payne

Frank,

One the other hand, they don’t describe the behavior of electrons moving/orbiting in the vicinity of a nucleus. Maxwell’s equations say those electrons should rapidly lose energy to space as electromagnetic radiation and spiral into the nucleus.

Point charges in orbit around the nucleus was always an overly simplistic description of the structure of an atom. It was proposed before we knew much about the strong and weak forces that apply at short ranges. An electron can only be captured by an unstable nucleus with a superabundance of protons. It doesn’t need to ‘orbit’. You still need QM to correctly describe atomic structures, of course.

• Dewitt: Pekka and I were discussing why we still view light as the electromagnetic radiation described by Maxwell’s equations, when we know that light is composed of massless particles (photons) that obey the laws of quantum mechanics. We know the laws of quantum mechanics produce phenomena characteristic of waves without a need for Maxwell’s equations. So, are Maxwell’s equations obsolete or, even worse, wrong? Maxwell’s equations are great for all type of macroscopic phenomena, like the electromagnetic radiation emitted when electrons flow through an antenna. However, they are wrong for an electron (or electrons) near a proton (or nucleus). For an electron to remain near a proton, but not smash into it; classical mechanics says it must be in some kind of orbit and Maxwell’s equations are supposed to require the electron emit electromagnetic radiation (just like they do in an antenna).

Above Pekka suggested that Maxwell’s equations are still valid, despite the fact that light is composed of particles that follow the laws of QM. I think we both now agree that they aren’t valid when their use will lead to a violation of the uncertainty principle.

The parallel below suggests that we should be able to derive Maxwell’s equations from QED:

Molecular Behavior Macroscopic Phenomena
Kinetic Theory Gases -> Statistical Mechanics -> Classical Thermodynamics
Photons/QED -> ???? -> Maxwell’s Equations

• Frank,

A little more on the validity of Maxwell’s equations. They are equations for the electromagnetic field in presence of source terms. One of the source terms represents charge and the other current, i.e. moving charge. In order to get correct results the charged matter must be handled according to QM, but Maxwell’s equations remain unchanged.

It may appear surprising that it’s not necessary to introduce photons to get correct results, but that’s the case for most of the situations, where photons are routinely discussed. A quantum mechanical description of a system that includes both the charged particles that act as the source of radiation and charged particles that act as the sink has solutions that have all the properties of a process that involves photons even when the photons are not introduced at all. That follows from the quantization of the possible transitions in the system of particles in combination with conservation of energy and momentum and the Maxwell’s equations.

Maxwell’s equations are a valid part of the equations of QED, using Feynman’s diagrams and the related concepts of photons is a practical way of solving the set of equations very precisely. In quantum field theories the separation between particles and fields becomes complex. That’s related to the unresolved philosophical problems of observation and measurement in all QM. In these theories everything is described by wave functions or fields that extend everywhere but never specify completely where a particle is or is not. They tell only, how likely it is that we observe a particle in a small volume if we choose to check whether it is there.

In ordinary QM the total number of particles is fixed, only their locations are not known precisely. In quantum filed theories even the total number of particles is variable as particles and particle-antiparticle pairs are created and destroyed all the time. QED is a limited version of quantum field theory where the set of particles considered as fields is limited mainly to photons and electrons. Other massive particles than electrons play a simpler role as they are so much heavier that they act more like fixed charges. (When their creation gets relevant, other interactions get important as well and QED starts to fail.)

• on May 16, 2013 at 2:43 pm DeWitt Payne

Frank,

You specifically stated that orbiting electrons prove that Maxwell’s equations are not valid for an atom. I pointed out that your example is flawed as the electrons do not, in fact, orbit the nucleus.

• on May 16, 2013 at 2:59 pm DeWitt Payne

Frank,

Another interesting point about Maxwell’s equations is that they prefigure special relativity. Electricity and magnetism behave the way they do so that the universe looks the same to observers moving at different velocities. So don’t let anyone tell you that special relativity is only important at velocities approaching the speed of light. GPS requires special and general relativistic correction to the orbiting clocks.

• DeWitt and Pekka: I fully understand that electrons don’t orbit the nucleus. However, we do know that matter consists mostly of empty space with a positively charge nucleus surrounded by electrons. Most or all of these electrons remain attached to one atom. Given classical mechanics and Maxwell’s equations, is there any other solution besides orbiting that permits an electron to remain near a nucleus indefinitely? Isn’t an orbiting electron required to give off electromagnetic radiation and spiral into the nucleus?

Since classical theory and observation disagree, something is wrong with the classical picture. I now see that DeWitt is correct when insisting that the failure could (and does) point to a problem with classical mechanics rather than with Maxwell’s equations. We can’t tell if Maxwell’s equations are right or wrong on the atomic scale, if we can’t precisely specify where the electron is and how fast it is moving. In fact, we probably don’t know whether electrons move at all inside atoms and molecules. (Moving electrons are sometimes invoked to explain Van der Waals force, but that is fairly indirect evidence of any motion.)

After some reading, I vaguely understand that in quantum field theory, Maxwell’s equations describe the electric and magnetic fields associated with photons, the particles that carry forces between charged particles. The solutions to Maxwell’s equations in a vacuum apparently demand motion at the speed of light and permit sinusoidal solutions with a frequency/wavelength.

Can E = hv can’t be derived from Maxwell’s equations?

• Frank,

This discussion has gone into the question of minimal assumptions needed to derive certain results. Answering reliably such questions is difficult, because few textbooks, if any, discuss these issues carefully. It’s much more common to introduce a whole standard set of postulates in combination with the most common way of doing the calculations. That has often led to the erroneous idea that this is the only correct approach. With quantum theory that is not a valid conclusion. Everything can be calculated in several ways, which may appear very different. A particular approach may be clearly the best for a specific problem, but probably never the only one.

After this introduction I tell that my educated guess is that the formula E = hv can be derived from the combination of Maxwell’s equations and quantum mechanical description of matter. QM is certainly needed, but probably only for matter. I’m not certain of this, but I think that this is the case. In such a description the formula is not about photons as photons are not part of the theory, but the matter still behaves in the same way as in a theory that includes the concept of photon. it’s easier to use the concept of photons, but perhaps not absolutely required.

This answer should be qualified also by the comment that certain essential inconsistencies of the theory have not been resolved without full QED, and not totally satisfactorily even with it. (Renormalization works very well in practice and a lot is understood about it, but open questions do remain at the level of fundamentals.)

65. on May 16, 2013 at 2:38 am | Reply John Millett

DeWitt Payne,

“It’s not isotropic as viewed from an area on a plane parallel to the surface at altitude z measured from the surface”.

Nor if viewed from an area on a plane perpendicular to the surface, of which there are 4 compared with 2 parallel ones.

The trigonometric integrations which give irradiance presumably apply to all 6 “surfaces”.

How do you get from a “point” to a “surface area”? A volume of gas incorporates neither. But the radiatively-active constituent, the molecule able to stretch, bend and rotate and absorb and emit photons, comes closer to a point than an area. The trick must be that isotropic emissions from the point source that pass through a nearby plane are deemed to be sourced from the plane to which “unit area” can apply? What you write follows from there?

• John,

The concept of a surface is an idealization that works very well for many cases but that cannot be extended at all to considerations of the kind you present. For such considerations the better approach is to take the radiation from every small volume as fully isotropic and then analyze further, how it gets transmitted or absorbed.

It’s typical that every point at surface radiates isotropically to all open directions. The cosine law that DeWitt presented is a geometric factor that’s related to the fact that any planar surface area looks the narrower the closer we get to the plane of the surface.

• on May 17, 2013 at 7:37 am John Millett

Pekka,

It’s not the cosine law that puzzles me, it’s the Wm-2 in DeWitt’s formula – Watts per square metre of what?

• By my comment I wanted to tell that sentences like

The trigonometric integrations which give irradiance presumably apply to all 6 “surfaces”.

do not make sense at all except for a real cube opaque to the radiation.

The plane parallel approximation can be formulated precisely and is a very useful approach, but I don’t think that anything like that can be done with “all 6 surfaces”.

• on May 17, 2013 at 1:56 pm DeWitt Payne

John Millet,

W/m² of electromagnetic radiation from (or to) a planar surface with an area of one square meter integrated over a wavelength range and averaged over all angles, like the Stefan-Boltzmann equation is the integration of the Planck equation. See http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html and scroll down to the Planck Radiation Formula box.

66. on May 18, 2013 at 1:08 pm | Reply John Millett

DeWitt Payne,

Thanks for the link which included:

To find the radiated power per unit area from a surface at this temperature, multiply the energy density by c/4.

The question remains: where in a gas body is the radiating surface? The only tangible matter in a gas body are molecules, a small proportion of which are radiators.

• John Millet,

The surface for gases is arbitrary, you pick it to define the thickness of the layer of gas of interest. You need a sufficient number of molecules in the layer so that temperature is defined and the concept of local thermodynamic equilibrium is applicable. It’s the radiation through the plane that’s important. For calculating atmospheric absorption/emission for any given total path length, you use enough layers that increasing the number of layers doesn’t change the result significantly. You need to do this in the atmosphere because it isn’t isotropic and isothermal. For monochromatic radiation in an isotropic, isothermal system, a single calculation is all that’s necessary

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