I Equilibrium between objects at different temperatures?

[Philip Koeck]

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.

 

[sophiecentaur]

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]

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.

 

[Philip Koeck]

... 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.

 

[sophiecentaur]

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.

 

[Philip Koeck]

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.

 

[sophiecentaur]

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.

 

[Philip Koeck]

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.

 

[sophiecentaur]

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?

 

[Philip Koeck]

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.

 

[sophiecentaur]

 

[sophiecentaur]

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.