# Equilibrium between objects at different temperatures?

• I
• Philip Koeck
In summary, the two black bodies in a vacuum will reach equilibrium and have the same temperature. However, this is not possible because an "infallible... optical device" is needed to achieve this.
vanhees71 said:
What's observed is the energy density or in this case the energy-current density (Poynting vector). Given the Poynting vector ##\vec{S}## the energy flowing through an arbitrarily oriented surface per unit time (power) is given by
$$P(t)=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{S}(t,\vec{x}).$$
I completely agree with that, but I can't see how to apply that to emission from a black body. The surface of a black body gives off radiation from its surface. No energy is flowing through the surface as far as I can see.

I'd really like to focus the discussion on thermal emission. So no scattering or reflection.
Unfortunately the Wiki shared by Sophie seems to jump between these topics and many sources only discuss scattering, as pointed out by Sophie.

What it is clear, I think is the following:

A black body is a Lambertian emitter and it always looks flat no matter what shape it is.
For example a sphere looks like a disk. (See several statements in the Wiki and other sources.)

This means (to me) that the current per solid angle and projected surface area is independent of θ.

The above means that the current per solid angle and actual surface area is proportional to cos(θ).

I believe the above statement is Lambert's law for emission, but I might be wrong.

Based on the above statement I argued in post 63 that Lambert's law is not due to the projection being smaller than the actual surface by a factor cos(θ), but that it has to come from the properties of the "elementary emitters", whatever they might be.

I'd really like to know what people think of this argument

An afterthought:
If I don't assume that the "elementary sources" sit on the surface, but rather in a small volume just below the surface, then I would get the cos-dependence even for isotropic emission (on average) from the individual sources.
Maybe that's a better model.

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Philip Koeck said:
it has to come from the properties of the "elementary emitters", whatever they might be.
I see you've built in an 'unknown' into that statement and I don't think that's the best way to try for an explanation. (valiant attempt tho')

Personally, I don't have a problem with using the experience of the measurer to define a phenomenon. Whatever you measure radiation with will have an aperture and the number of photons admitted through this aperture from the emitting area will follow a cosine law.

This will apply to any size of emitting surface and the same argument also applies to the cos law 'spreading' effect when a tilted surface is illuminated by a plane wave front (from infinity). There's a distinct difference between receiving illumination from a single distant source and emitting radiation in all directions from a diffuse source. That difference provides a bit of a cognitive dissonance, particularly when we think too hard about the situation.

sophiecentaur said:
... the number of photons admitted through this aperture from the emitting area will follow a cosine law.
I don't think that's consistent with what I read in various sources.

According to the Wiki and a book I have (Max Garbunny) a sphere (for example) that is a Lambertian emitter will look flat, like a disk, to an observer. To me that means that the photon current passing through an aperture and reaching the detector is independent of the orientation of the surface.
That's just an experimental finding!

Philip Koeck said:
I completely agree with that, but I can't see how to apply that to emission from a black body. The surface of a black body gives off radiation from its surface. No energy is flowing through the surface as far as I can see.

I'd really like to focus the discussion on thermal emission. So no scattering or reflection.
Unfortunately the Wiki shared by Sophie seems to jump between these topics and many sources only discuss scattering, as pointed out by Sophie.

What it is clear, I think is the following:

A black body is a Lambertian emitter and it always looks flat no matter what shape it is.
For example a sphere looks like a disk. (See several statements in the Wiki and other sources.)

This means (to me) that the current per solid angle and projected surface area is independent of θ.

The above means that the current per solid angle and actual surface area is proportional to cos(θ).

I believe the above statement is Lambert's law for emission, but I might be wrong.

Based on the above statement I argued in post 63 that Lambert's law is not due to the projection being smaller than the actual surface by a factor cos(θ), but that it has to come from the properties of the "elementary emitters", whatever they might be.

I'd really like to know what people think of this argument

An afterthought:
If I don't assume that the "elementary sources" sit on the surface, but rather in a small volume just below the surface, then I would get the cos-dependence even for isotropic emission (on average) from the individual sources.
Maybe that's a better model.
It's thermal radiation. So from any surface element (in its rest frame) there's simply a radially symmetric emission from each surface element of the emitting surface.

vanhees71 said:
It's thermal radiation. So from any surface element (in its rest frame) there's simply a radially symmetric emission from each surface element of the emitting surface.
In that case, if the "emitting units" sit on the surface and emit isotropically, I refer back to my post 63.

Philip Koeck said:
That's just an experimental finding!
All theories stand by being confirmed by measurement(?).
Philip Koeck said:
In that case, if the "emitting units" sit on the surface and emit isotropically, I refer back to my post 63.
It may help is you assume that there must be something wrong with your argument and perception. That is often the biggest hurdle. I think there is a lot of sloppy use of terms about Mr Lambert and the cosine law is often taken (wrongly) out of context. It's only since reading around with an open mind that I am happy with it.

The cosine law applies to incident radiation; it works at the Earth's poles and at the equator. PV cells are mounted with a tilt because of the cos law.
When you observe the Moon, it appears of uniform brightness despite the fact that it's a sphere because you can see a bigger area of poorly illuminated (but isotropic) moon surface at the edges. That all makes sense and agrees with our intuition.

For an emitter you assume a black body, which can be treated as a hole in a large sphere with radiation coming out of it and NOT a set of isotropic emitters on the surface. Wiki goes on to say:

"The observer at angle θ to the normal will be seeing the scene through the same aperture of area dA0 (still corresponding to a dΩ wedge) and from this oblique vantage the area element dA is foreshortened and will subtend a (solid) angle of dΩ0 cos(θ). "

So then you have the same situation as that of an illuminated patch and the angle relative to the normal introduces "foreshortening". I just realised where the apparent paradox lies. Whaddya think?

Philip Koeck
sophiecentaur said:
For an emitter you assume a black body, which can be treated as a hole in a large sphere with radiation coming out of it and NOT a set of isotropic emitters on the surface.
Here I completely agree with you. If we think of a black body as a hole in a cavity and assume that the radiation inside the cavity is isotropic and homogeneous then the radiation coming out from the hole will follow the cos-law simply because the effective area of the hole will be proportional to cos(θ).
That's completely in line with what vanhees71 wrote too.
Here we actually have radiation that comes from inside the cavity and traverses a hole that is tilted by θ with respect to the radiation.

What I was thinking of was an actual surface made of "black body material", not a hole.
In that case I don't see that geometry alone will lead to the cosine law if the elementary emitters are in the surface. If the emitters were in a thicker layer below the surface the situation would be different. In fact it would be a bit like the cavity with a hole in it.

Philip Koeck said:
What I was thinking of was an actual surface made of "black body material", not a hole.
I appreciate that but the evidence is that the cosine law has been found, by experiment, to apply. That implies that a lambertian emitting surface behaves as if it were a hole in a surface with a large (enough) volume beneath it with all the walls at the temperature of the surface.
1. We can't argue against the evidence of cosine law behaviour
2. This behaviour for emission is the reciprocal of an absorbing surface and there is a very plausible argument for cosine behaviour in emission.
3. I have a feeling that having different laws for absorption and emission would involve some basic failure of thermodynamics laws - and would require a Maxwell Demon to explain it.

It may be annoying but is there anywhere else to go?

DrClaude and Philip Koeck
sophiecentaur said:
I appreciate that but the evidence is that the cosine law has been found, by experiment, to apply. That implies that a lambertian emitting surface behaves as if it were a hole in a surface with a large (enough) volume beneath it with all the walls at the temperature of the surface.
1. We can't argue against the evidence of cosine law behaviour
2. This behaviour for emission is the reciprocal of an absorbing surface and there is a very plausible argument for cosine behaviour in emission.
3. I have a feeling that having different laws for absorption and emission would involve some basic failure of thermodynamics laws - and would require a Maxwell Demon to explain it.

That's sort of what I've found so far.
Experimental evidence shows that the cosine law applies and the cosine-law is necessary for the 2nd law.
Just as you say, if absorption is proportional to cos(θ) then emission must be too. Otherwise you can't get equilibrium at equal temperatures.

That's similar to what I found in my calculation for a specific case (sphere in a spherical cavity) discussed in an earlier post (55?). The sphere and cavity can only be in equilibrium at equal temperatures if the cosine law for emission holds.

So, in summary, what we have is that the cosine law has to hold for emission in order for the 2nd law to hold.

sophiecentaur said:
It may be annoying but is there anywhere else to go?

I think there might be.

The second law doesn't really explain why something happens, at least not in the sense that it provides a mechanism.
In a way the second law is a result but not a cause.

In the case of the cavity with hot inner surfaces and a hole in the wall where radiation escapes the cosine law for emission (and absorption) follows from pure geometry. That's a very simple "mechanistic" explanation.

For a solid surface I don't see that, at least not if the radiation sources are isotropic and only located on the surface. (see earlier post 63).

Maybe we can conclude that even in the case of a solid surface thermal radiation doesn't come from the surface but from deeper inside the solid.
Then the cosine-law would make perfect sense even if the "radiation inside the solid" (whatever that might be) is isotropic, or, if you prefer, each elementary source is isotropic.

Ok but we have yet to find an example of the second law failing. There would be incredible knock-ins from that. So your hoped for model would need to survive a very stringent test.

I do agree that I get brain pain when I try to imagine what must happen to EM waves to make them appear to emerge from a virtual image at infinity (or even less).

Philip Koeck said:
In a way the second law is a result but not a cause.
Are we basically down to the nature of a lambertian surface? How's this idea?

It's definitely not flat because it needs to scatter equally in all directions. That would affect the ray paths from all points on the surface. More and more surface peaks would obscure the view from a low angle. Because of the equilibrium condition all over the surface there would be no change in temperature where the scattered rays hit the obscuring peaks so the net emission in a given direction would depend only on the rays that get through without being absorbed by peaks. (It's the thermal equilibrium over the surface that's the clever bit.)

So you 'make' a lambertian surface and test it by its variation of scattering performance and then you heat it up. It's then the geometry of the rough surface that makes it lambertian for emission too.

AND . . . . as the surface material increases in thermal resistivity, the surface will approach isotropic behaviour. (That bit just came to me)

Philip Koeck
I still do not understand, where your problem is. The energy flow of electromagnetic waves is described by the Poynting vector. That's it!

vanhees71 said:
I still do not understand, where your problem is. The energy flow of electromagnetic waves is described by the Poynting vector. That's it!

vanhees71 said:
I still do not understand, where your problem is. The energy flow of electromagnetic waves is described by the Poynting vector. That's it!
I think it is meant for this thread.

I assume you are referring to your statement from post 70, which I answered in post 71.

I can understand that reasoning if radiation hits a surface or if radiation escapes through a hole, but I can't understand it if radiation actually originates in a surface.
In post 71 I explain in more detail how I think.

If radiation doesn't originate in the surface but in some finite volume below the surface the situation is just like that for a cavity with a hole in it. In that case I completely buy this reasoning.

Philip Koeck said:
I think it is meant for this thread.
The Poynting vector can't help in a situation where many different random sources of EM are littered over a three dimensional ( or even two dimensional) region. Poynting needs E and H vectors (other vectors are available) before it can be defined. It isn't even a diffraction problem. And how does the cosine law emerge from all this? I think we'd need a few more details and a path through solving that maths.

Philip Koeck said:
An afterthought:
If I don't assume that the "elementary sources" sit on the surface, but rather in a small volume just below the surface,
Bearing in mind that a lambertian surface has to be 'rough', with unlimited depth, in order to scatter with a cosine law then the emitting surface would have to be the same. If we accept that it's to do with geometry then all that's necessary is to appreciate what happens when radiation from one part of the surface is obscured by another part. That fraction of the 'current' is re-admitted into the surface; the current emitted from the obscuring part of the surface can't be added to by the radiation it obscures because of the temperature equilibrium over the surface.

That doesn't explain the specific cosine nature of the law but it will, 'just by arm waving', have a zero parallel to the surface and have a generically cosine shape. I wouldn't know where to start but I'd bet there's a man, on PF who could.

Philip Koeck
sophiecentaur said:
Bearing in mind that a lambertian surface has to be 'rough', with unlimited depth, in order to scatter with a cosine law then the emitting surface would have to be the same. If we accept that it's to do with geometry then all that's necessary is to appreciate what happens when radiation from one part of the surface is obscured by another part. That fraction of the 'current' is re-admitted into the surface; the current emitted from the obscuring part of the surface can't be added to by the radiation it obscures because of the temperature equilibrium over the surface.

That doesn't explain the specific cosine nature of the law but it will, 'just by arm waving', have a zero parallel to the surface and have a generically cosine shape. I wouldn't know where to start but I'd bet there's a man, on PF who could.
That's definitely an interesting idea!

Philip Koeck said:
That's definitely an interesting idea!
It can only ever be an idea imo. No one makes lambertian scatterers but near enough is good enough Even the Moon seems to do a fairly good job and that is multicoloured with rocks and all.

sophiecentaur said:
It can only ever be an idea imo. No one makes lambertian scatterers but near enough is good enough Even the Moon seems to do a fairly good job and that is multicoloured with rocks and all.
Maybe one shouldn't expect the same mechanism for Lambertian scatterers and emitters.
A black body is supposed to be a Lambertian emitter, but a black body doesn't scatter at all so it can't be a Lambertian scatterer.

For scattering the surface structure ought to be important for the cosine law.

For emission I could imagine that the cosine law emerges even for atomically flat surfaces.
Maybe the reason for the cosine law is simply that radiation originates below the surface rather than on the surface, so the situation is like that of a cavity with a hole in its wall.

Philip Koeck said:
For emission I could imagine that the cosine law emerges even for atomically flat surfaces.
'Imagining' is not really a basis for a theory. I often fall into this trap too.

A thermodynamic argument demands that there must be some reciprocity between scattering and emission. If not then there could be a structure in which the laws are violated. But the lambertian surface and the black body are not real things; they are just models to work with. Trying to explain 'how' either of them could be manufactured is pretty much pointless.

It's interest to note that the nearest thing to a total absorber of light which can be used in TV studios (where black level was important) was a hole, cut in a white board with a void behind which had black velvet lining. The void was a few cm deep and larger than the hole. It was a good enough black level to use in a brightly lit studio from a range of angles. Afaik, no one tried heating the internal surface to measure the emitted thermal spectrum but you could expect the absorbance of the hole would obviously follow a cosine law. (Just throwing that one in of fun.)

Philip Koeck
sophiecentaur said:
A thermodynamic argument demands that there must be some reciprocity between scattering and emission. If not then there could be a structure in which the laws are violated.
Do you have a reference or some web resource that discusses this?
Sounds interesting.

Philip Koeck said:
Do you have a reference or some web resource that discusses this?
Sounds interesting.
The Maxwell Demon was a thought experiment to demonstrate how heat could be made to flow 'uphill' but only by violating the second law of thermodynamics. That basic principle must hold in the scenario of the OP. Try a search on violating the second law.

Trying to find a 'loophole' in this by involving lambertian surfaces is hardly worth doing because the lambertian cosine law is only a concept and the second law doesn't hang on the characteristic of a particular surface. We're sort of chasing our tails here, I think.

As Wiki says "The second law of thermodynamics is a physical law based on universal experience " so isn't that enough? (Wiki is not 100% reliable but I couldn't find counter evidence against the second law and it really would be BIG NEWS .)

Philip Koeck, vanhees71 and Lord Jestocost
sophiecentaur said:
The Maxwell Demon was a thought experiment to demonstrate how heat could be made to flow 'uphill' but only by violating the second law of thermodynamics. That basic principle must hold in the scenario of the OP. Try a search on violating the second law.
Trying to find a 'loophole' in this by involving lambertian surfaces is hardly worth doing because the lambertian cosine law is only a concept and the second law doesn't hang on the characteristic of a particular surface. We're sort of chasing our tails here, I think.
I'm not looking for any loophole, quite on the contrary.
I'm looking for a confirmation of the 2nd law from outside thermodynamics.

I'll summarize what I've learnt so far:

It seems that two black bodies can only be in equilibrium with each other at equal temperatures if emission from the surface of a black body is lambertian.

I'll post a little thought experiment that shows that in a very direct way I believe.
And yes, I know that a black body is an idealization.

The only thing I'm still wondering about is whether one can come up with the cosine law for emission from some other part of physics such as solid state physics or electromagnetism.

I can see that the cosine law arises from pure geometry in the case where a surface or a hole is hit by incoming radiation, but I don't see how that idea can be applied to radiation originating from a surface.

sophiecentaur said:
... the lambertian cosine law is only a concept and the second law doesn't hang on the characteristic of a particular surface. We're sort of chasing our tails here, I think.
Here's the thought experiment:

The two black curves are spherical surfaces (not drawn so well) made of (idealized) "black body material".
The blue lines represent two cones made of "cat's eye material" (also idealized).
The purpose of the cones is to reflect back all radiation that doesn't go through the pin hole in the middle of the construction.
The two red lines are examples of rays that do pass through the pin hole.
Only radiation that goes through the hole needs to be considered when studying the thermal equilibrium between surfaces A and B.

Notice that surface A is tilted out of the symmetric orientation by some angle θ.

The whole contraption is thermally isolated to the outside.

It's easy to see that the surfaces A and B can only be in radiative equilibrium with each other if the current going through the pin hole from A to B is equal to the current going from B to A.

A is larger than B by a factor 1/cosθ.
Therefore the current emitted per area has to be proportional to cosθ in order for the surfaces to be in equilibrium at equal temperatures.

From this I would conclude that the cosine law is essential for the 2nd law to hold in the case of black bodies in radiative equilibrium with each other.

Philip Koeck said:
A is larger than B by a factor 1/cosθ.
Can you give me a derivation of that? It isn't obviously the case so it needs to be justified, I think.

sophiecentaur said:
Can you give me a derivation of that? It isn't obviously the case so it needs to be justified, I think.
I realize that I was sloppy, but it should hold approximately for narrow cones.

The horizontal distance between the pin hole and surface A should be kept the same as that between the pin hole and B to make things easier.

Surface B can be split into many small patches (square or hexagonal for example).
Each of these patches is projected onto surface A by rays going through the pin hole.
We can also define a central ray for each patch, which is the one coming from the center of a patch on B.

Now each of the projected patches on A is tilted with respect to the central ray (defined above) by an angle θ and is therefore larger by a factor 1/cosθ than the corresponding patch on B.

For narrow cones the angle θ is almost the same for every patch.

There is also a small effect due to the fact that half of surface A moves closer to the pin hole and half moves further from the pin hole as you increase θ, but those two effects partly cancel and they are small for narrow cones.

So all in all I would say that A is larger than B by a factor of roughly 1/cosθ for narrow cones.

Alternatively you can also look at my calculation with the sphere inside a sphere.

I believe that is exact and it also shows that the cosine law for emission is necessary for radiative equilibrium at equal temperatures.

PS: It might turn out that if surface A is constructed properly (so that each patch on A is tilted from the central ray by exactly θ) it would be larger than B by exactly 1/cosθ even if the cone is not narrow, but surface A would not be spherical then.
This sounds like quite a difficult geometry problem.

Philip Koeck said:
I realize that I was sloppy, but it should hold approximately for narrow cones.
Wouldn't you expect a "squared" in there, as the areas are involved? I really don't follow your argument and where's it going? You invent a complicated, approximate model and neither you nor I have the Maths to solve it.
I notice that the rest of PF has avoided this thread; probably with good reason.

sophiecentaur said:
Wouldn't you expect a "squared" in there, as the areas are involved? I really don't follow your argument and where's it going? You invent a complicated, approximate model and neither you nor I have the Maths to solve it.
Why should there be a "squared"? The tilting only introduces a change normal to the tilt axis so the ratio between the tilted area and the corresponding untilted area becomes 1/cosθ.

Which part of the argument don't you follow?

I would say the model is simple, but I agree it's approximate. The previous one with concentric spheres is exact, but a bit more involved.

Philip Koeck said:
exact, but a bit more involved.
The result of both models must be that the second law works but the spheres must be less involved than the reflecting cones. Also, you have to remember that equilibrium will occur with or without lambertian surfaces or you could have a Maxwell Demon by just choosing appropriate surfaces.
I'm beginning to lose interest in this thread because nothing new or helpful has turned up for quite a while. Shall we call it a day?

sophiecentaur said:
The result of both models must be that the second law works but the spheres must be less involved than the reflecting cones. Also, you have to remember that equilibrium will occur with or without lambertian surfaces or you could have a Maxwell Demon by just choosing appropriate surfaces.
I'm beginning to lose interest in this thread because nothing new or helpful has turned up for quite a while. Shall we call it a day?
I'm quite sure (though not 100 %) that for black bodies to be in thermal equilibrium at equal temperatures they have to be lambertian emitters.
For me that's an interesting finding and I would have liked to see some confirmation from outside thermodynamics that black bodies are in deed lambertian emitters.

For other surfaces the analysis is far more complicated since we have to take into account scattering and reflection as well. So I would not conclude that you can get equilibrium at non-equal temperatures by considering non-lambertian and/or non-black surfaces.
Don't worry. I am completely convinced that the 2nd law holds for all kinds of objects.

Fine if we stop the thread. Thanks for all the input. I definitely learnt a lot about Lambert's law.

Philip Koeck said:
So I would not conclude that you can get equilibrium at non-equal temperatures by considering non-lambertian and/or non-black surfaces.
Lol. Too many negatives there but no worries. I don't think Lambert is needed because no one could make one.

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