# 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.
Philip Koeck
I've come across a puzzling thought experiment.

Consider two black bodies surrounded by vacuum.
The surrounding temperature is exactly 0 K.

By some ingenious optical device all the radiation from body 1 is focused onto body 2 and vice versa.

If left alone sufficiently long the two bodies will reach equilibrium.
This means that the heat current emitted by body 1 is equal to the heat current absorbed by body 1. The same is also true for body 2.

According to the assumption made the heat current emitted by body 1 is absorbed by body 2 and vice versa.
This also means that the heat currents emitted by the two bodies are equal.

If we also assume that the surfaces of the two bodies are not the same I would conclude that the temperatures of the two bodies have to be different since the emitted heat is proportional to the surface area and to T4.

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aimLow and Demystifier
Small black holes radiate much faster than large black holes. So the equilibrium state will be reached when the smaller BH evaporates.

.Scott said:
Small black holes radiate much faster than large black holes. So the equilibrium state will be reached when the smaller BH evaporates.
Black bodies, not holes. You can imagine a cavity that's perfectly black inside and radiation is allowed to exit and enter through an opening.

Philip Koeck said:
By some ingenious optical device all the radiation from body 1 is focused onto body 2 and vice versa.
I’m on shaky ground but I’m sure that someone else will be able to correct (or expand upon) this if needed...

I think the system you describe is infeasible because it is not possible to have an (optical) system such that “all the radiation from body 1 is focused onto body 2 and vice versa”.

This would violate conservation of etendue. ‘Etendue’ is a measure of how much light (or radiation in general) ‘spreads-out’. Conservation of etendue can be derived from the second law of thermodynamics.

For example, it tells you that you can’t focus sunlight to such an extent that the energy-density corresponds to a temperature higher than that of the sun’s surface.

Take a look at https://en.wikipedia.org/wiki/Etendue and/or do a search.

hutchphd, Demystifier, Vanadium 50 and 3 others
Steve4Physics said:
I’m on shaky ground but I’m sure that someone else will be able to correct (or expand upon) this if needed...

I think the system you describe is infeasible because it is not possible to have an (optical) system such that “all the radiation from body 1 is focused onto body 2 and vice versa”.

This would violate conservation of etendue. ‘Etendue’ is a measure of how much light (or radiation in general) ‘spreads-out’. Conservation of etendue can be derived from the second law of thermodynamics.

For example, it tells you that you can’t focus sunlight to such an extent that the energy-density corresponds to a temperature higher than that of the sun’s surface.

Take a look at https://en.wikipedia.org/wiki/Etendue and/or do a search.
I was thinking the problem must lie in the assumption about the optical device.

I don't think that I run into problems with the second law however, at least not directly.
When the two objects are in equilibrium there is no net heat transfer between them.

It does seem to violate the zeroth law, though, or at least the usual conclusion that objects in equilibrium have to have the same temperature.

aimLow
Philip Koeck said:
I don't think that I run into problems with the second law however, at least not directly.
When the two objects are in equilibrium there is no net heat transfer between them.
One way of stating the second law of thermodynamics is that the (natural) direction of heat flow is from a hotter object to a colder one. (A consequence of this is that when two objects are at the same temperature, the net heat flow between them is zero.)

I guess the link between the second law and etendue is that they both relate to the flow of radiation (in appropriate contexts).

Philip Koeck said:
It does seem to violate the zeroth law, though, or at least the usual conclusion that objects in equilibrium have to have the same temperature.
The zero-th law (as far as I understand it) is about something slightly different. It says if two bodies are both in equilibrium with a third body, the two bodies are in equilibrium with each other. (This allows us to establish the concept of temperature.)

Referring back to original problem, for a system of two black bodies, there should be no doubt that their final temperatures will be equal. Any argument that says otherwise must have one or more mistakes.

Philip Koeck
Steve4Physics said:
One way of stating the second law of thermodynamics is that the (natural) direction of heat flow is from a hotter object to a colder one. (A consequence of this is that when two objects are at the same temperature, the net heat flow between them is zero.)
The usual formulation is that no heat can flow from cold to hot without any other change somewhere.
Some other change could be potential energy being converted into work, for example.
So if no heat flows in effect, I'm not sure if the 2nd law is actually violated.

Another question is of course the process leading to this presumed equilibrium at different temperatures.
That does involve heat flow from cold to hot.

Steve4Physics said:
The zero-th law (as far as I understand it) is about something slightly different. It says if two bodies are both in equilibrium with a third body, the two bodies are in equilibrium with each other. (This allows us to establish the concept of temperature.)
I agree. Often textbooks add that one of the bodies could be a thermometer and therefore the 3 bodies have to have the same temperature.
That's what I meant by "conclusion from the 0th law".

Steve4Physics said:
Referring back to original problem, for a system of two black bodies, there should be no doubt that their final temperatures will be equal. Any argument that says otherwise must have one or more mistakes.
Yes, probably. The mistake ought to be in the assumption about the optical system.

So, in a way, thermodynamics determines what is technically possible in optics.

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Steve4Physics
Steve4Physics said:
I’m on shaky ground but I’m sure that someone else will be able to correct (or expand upon) this if needed...

I think the system you describe is infeasible because it is not possible to have an (optical) system such that “all the radiation from body 1 is focused onto body 2 and vice versa”.

This would violate conservation of etendue. ‘Etendue’ is a measure of how much light (or radiation in general) ‘spreads-out’. Conservation of etendue can be derived from the second law of thermodynamics.

For example, it tells you that you can’t focus sunlight to such an extent that the energy-density corresponds to a temperature higher than that of the sun’s surface.

Take a look at https://en.wikipedia.org/wiki/Etendue and/or do a search.
Not sure if conservation of etendue is the solution to the puzzle.

In the wikipedia page shared above they quote the Brightness theorem, which states that power per unit solid angle and per unit area cannot increase, but that doesn't mean that power per unit area cannot increase.

sophiecentaur, PeroK, Lord Jestocost and 5 others
Does this help: Perfect thermal conductors. Make one body a sphere and the second body a larger concentric spherical shell. The central sphere is kept at temperature T0. In steady state the Touter shell<T0
But if you instead hold outer shell temperatere to Touter shell the inner T0 will never be larger than Touter shell
I think the generalization is obvious.
Does that help?

/

Demystifier
Philip Koeck said:
I've come across a puzzling thought experiment.

Consider two black bodies surrounded by vacuum.
The surrounding temperature is exactly 0 K.

By some ingenious optical device all the radiation from body 1 is focused onto body 2 and vice versa.
Could we eliminate the optical device by assuming one body to be a hollow sphere that contains the other one suspended inside?

We could then assume emisivity or out-radiation of the exterior side of the wall of the hollow sphere to be zero, so all the thermal energy remains inside the two-body system, flowing or not.

Philip Koeck
It depends upon what you are trying to show. I

Lnewqban said:
Could we eliminate the optical device by assuming one body to be a hollow sphere that contains the other one suspended inside?

We could then assume emisivity or out-radiation of the exterior side of the wall of the hollow sphere to be zero, so all the thermal energy remains inside the two-body system, flowing or not.
I considered that too, but the problem is that a lot of the radiation from the surface of the hollow sphere will go straight to the hollow sphere again and not to the solid sphere in the center.
So it's actually more complicated.
I liked the situation where all the radiation from one body is absorbed by the other and vice versa.

The ellipsoid paradox quoted in post 9 is about exactly such a situation.

sophiecentaur and hutchphd
Philip Koeck said:
It does seem to violate the zeroth law, though, or at least the usual conclusion that objects in equilibrium have to have the same temperature.
Are you trying to evade/avoid the forum policy regarding PPMs?

Bystander said:
Are you trying to evade/avoid the forum policy regarding PPMs?
Can't really say. What's a PPM?

Philip Koeck said:
Can't really say. What's a PPM?
A misspelled PMM

(perpetual motion machine)

DrClaude said:
A misspelled PMM

(perpetual motion machine)
Okay, I see.

I did say it's a puzzling thought experiment and I was trying to locate the wrong assumption.

The paper on the ellipsoid paradox you quoted is really to the point and it shows that this paradox has a long history and isn't trivial at all.

Demystifier
Philip Koeck said:
By some ingenious optical device all the radiation from body 1 is focused onto body 2 and vice versa.
It's not much different from the idea of perpetuum mobile; by some ingenious mechanical device all the heat is transferred to work ... The point is that there is no such device, due to the 2nd law. Light naturally tends to spread (rather than focus), which is a version of the law that entropy naturally tends to increase (rather than decrease). A spread light occupies a larger portion of phase space and hence has larger entropy.

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vanhees71, Steve4Physics, hutchphd and 1 other person
Lnewqban said:
Could we eliminate the optical device by assuming one body to be a hollow sphere that contains the other one suspended inside?

We could then assume emisivity or out-radiation of the exterior side of the wall of the hollow sphere to be zero, so all the thermal energy remains inside the two-body system, flowing or not.
I'm back in an old thread I'm afraid.

What about the situation with a solid sphere inside a spherical cavity with the addition that the gap between the sphere and the cavity is filled by some sort of fiber optics that makes sure that essentially all radiation from the sphere reaches the wall of the cavity and vice versa.

Apart from the fact that this is not allowed to happen, what would stand in the way of thermal equilibrium at unequal temperatures in this case?

In connection with this discussion I've started wondering about something I've always taken for granted.

In Young and Freedman's textbook it's stated that the net heat transfer by radiation between an object (o) and its surroundings (s) is given by H = A σ e ( To4 - Ts4).

Does this mean that the heat current absorbed by the object is given by H = A σ e Ts4 ?

Philip Koeck said:
Does this mean that the heat current absorbed by the object is given by H = A σ e Ts4 ?
Yes.

hutchphd said:
Does this help: Perfect thermal conductors. Make one body a sphere and the second body a larger concentric spherical shell. The central sphere is kept at temperature T0. In steady state the Touter shell<T0
But if you instead hold outer shell temperatere to Touter shell the inner T0 will never be larger than Touter shell
I think the generalization is obvious.
Does that help?

/
I dismissed the idea of a sphere in a spherical cavity too quickly (in post 13).

If we add the (idealized) fiber fiber optics I describe in post 19 I'll reconsider.
Alternatively one could work out how much of the radiation from the inner wall of the cavity hits the sphere in the middle and how much goes to some other place on the cavity wall. Bit of geometry.

I don't completely understand your post:
Does the spherical shell give off heat to the outside too?
Are you discussing a steady state situation with two different temperatures or static equilibrium?

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Lnewqban
Apologies if some of this is already covered:

Philip Koeck said:
By some ingenious optical device all the radiation from body 1 is focused onto body 2 and vice versa.
Are you aware of the Etendue Theorem? I think this is more difficult than you imagine
Philip Koeck said:
I don't completely understand your post:
Does the spherical shell give off heat to the outside too?
Are you discussing a steady state situation with two different temperatures or static equilibrium?

An object in any cavity within a black body with a uniform (necessarilly steady I suppose) Temperature will equilibrate to that Temperature. The black body can be radiating externally or not so long as the temperature is steady and uniform.
Does that help?

Lord Jestocost, DrClaude and Philip Koeck
hutchphd said:
Are you aware of the Etendue Theorem? I think this is more difficult than you imagine
I know it in the context of lenses and (focusing) mirrors and the paper quoted in post 9 is also about a system of focusing mirrors, but maybe there's a generalization I haven't picked up.

I'm not sure whether this can be directly translated to a fiber optical system since there's no real focusing going on, or?

If I understand correctly conservation of etendue just means that the more you focus a beam the larger the convergence angle becomes. So the quantity known as "brightness" in microscopy is at best constant in a lens-optical system (for example), but it can decrease due to aberrations etc.

I'll just add a definition here since it seems that terminology is very specific for different fields.
In electron optics the brightness of an electron beam is given by electric current per area of focal spot and solid convergence angle.

The theorem is, to my understanding, a specific statement of energy conservation and has no particular limited applicability

hutchphd said:
The theorem is, to my understanding, a specific statement of energy conservation and has no particular limited applicability
If you think of a beam photons you could in principle focus it into smaller and smaller focal spots without changing the number of photons in the beam and without increasing the convergence angle, so energy is conserved even if you are doing something that's really impossible in optics.

Just a quick remark: It seems that brightness is proportional to the reciprocal of etendue.

hutchphd
Spot size yes but spot size and angle also no. You need to understand the theorem, and it is far from trivial..
As an aside I must confess to practicing optics without ever having had a dedicated optics course (other than general physics). I spent a lot of effort trying to intensify LED light using "light pipes" before finding and fully appreciating this theorm. All is well that ends well.

Gleb1964, DrClaude and Philip Koeck
hutchphd said:
Spot size yes but spot size and angle also no. You need to understand the theorem, and it is far from trivial..
Perfectly agree, but I don't think it's connected to energy conservation.
It must be connected to the second law.

What I'm really looking for is a technical reason (that doesn't rely on the 2nd law)
that the idea with the sphere in the cavity cannot lead to an equilibrium at different temperatures.

Maybe one should do the calculation without the fiber optics to start with.

hutchphd said:
I spent a lot of effort trying to intensify LED light using "light pipes" before finding and fully appreciating this theorm. All is well that ends well.
I would love some electron pipes!

hutchphd
Philip Koeck said:
What I'm really looking for is a technical reason (that doesn't rely on the 2nd law)
that the idea with the sphere in the cavity cannot lead to an equilibrium at different temperatures.

Maybe one should do the calculation without the fiber optics to start with.

I've tried the following now.
It's all in a Word-file, but I'll summarize here.

Fig. 1: A black body sphere (radius r, temperature TS) floats in the center of a spherical black body cavity (radius R, temperature TC).

The fraction f of radiation from the cavity C that hits the sphere S is given by the ratio of the solid angle corresponding to the angle a shown in figure 1 and that corresponding to π/2: f = 1- cos a

I'll leave it to whoever is interested to check my results, but I can send my own derivations on request.

I'm not taking diffraction into account. The sphere could be huge compared to most of the wave lengths involved.

Now the central assumption where the problem might hide:

In equilibrium the radiation emitted by the sphere must be equal to the fraction of the radiation emitted by the cavity which reaches the sphere.
This leads to a ratio (plotted in figure 2)
TS4/TC4 = (1 - cos a)/sin2a

Fig. 2: Ratio between the fourth power of the sphere’s and the cavity’s temperature as a function of the cone’s half angle a.

Assuming I didn't make a mistake in the maths, am I maybe misusing the radiation formula H = A e σ T4?

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Lnewqban
It seems to me that an attempt to eschew Thermodynamics in this description is misguided because of complexity. For instance there is no requirement that the emission from the surface is isotropic. It is more likely to be Lambertian, but such details cannot matter in the bigger picture. So better to think of the black body cavity as a "gas of photons" at a temperature T corresponding to that of the "container". In fact any chunk of ordinary matter is more or less such a partition, and hence the ubiquity of tha Stefan-Boltzmann Law.
In the concentric spherical case by symmetry one can deal easily with the problem posed in the original post. The equilibrium state is T=0 for everything. If instead you supply power P, then the steady state depends upon $$P=4\pi R^2\sigma (T^4-0)$$ for the outer shell. If you supply the power only to the inner sphere it will be warmer than the outer shell according to its area .

hutchphd said:
It seems to me that an attempt to eschew Thermodynamics in this description is misguided because of complexity. For instance there is no requirement that the emission from the surface is isotropic. It is more likely to be Lambertian, but such details cannot matter in the bigger picture. So better to think of the black body cavity as a "gas of photons" at a temperature T corresponding to that of the "container". In fact any chunk of ordinary matter is more or less such a partition, and hence the ubiquity of tha Stefan-Boltzmann Law.
In the concentric spherical case by symmetry one can deal easily with the problem posed in the original post. The equilibrium state is T=0 for everything. If instead you supply power P, then the steady state depends upon $$P=4\pi R^2\sigma (T^4-0)$$ for the outer shell. If you supply the power only to the inner sphere it will be warmer than the outer shell according to its area .
The whole construction is isolated from the rest of the universe so no radiation or heat can escape by any means.
I'm not supplying any power anywhere so the only thing the sphere and the inner wall of the cavity can do is settle into a static equilibrium.

I agree that I maybe shouldn't assume isotropic emission from the cavity wall, but it's the simplest model and I wonder if that detail could make such a big difference.
Or maybe that's exactly what's wrong.

hutchphd said:
For instance there is no requirement that the emission from the surface is isotropic. It is more likely to be Lambertian, ....
According to Wikipedia it's supposed to be isotropic: https://en.wikipedia.org/wiki/Black_body

That might not be trustworthy, of course.

Hi. This is interesting topic.
Even without some special lenses or fiber optics the dilemma is there.

I also made my own calculations and can confirm yours.

For example, if r=1 and R=2 then,
while for temperatures to remain same it needs to receive 25%.
So, big sphere's temperature will settle higher than small sphere's.

this means a surface point equivalently radiates to a lesser solid angle than 2*pi.
Let's check correction factor to 13.4% and to other radii:

for r=1 and R=2 it is 25/13.4= 1.8656

r=1 and R=1.001 gives ratio of 1.045
r=1 and R=1.01 gives ratio of 1.14
r=1 and R=1.1 gives ratio of 1.4166
r=1 and R=2 gives ratio of 1.8656
r=1 and R=3 gives ratio of 1.9425
r=1 and R=4 gives ratio of 1.9685

Seems ratio goes from 1(when radii are almost equal), to 2(when radii ratio is very big).
Which is expected because in Lambertian, a point radiates much more to up than sides.
And when small sphere occupies small solid angle "above" a point on big sphere,
the difference is biggest.

Maybe Lambertian is the solution to the dilemma, or maybe not.

These are calculations, hope there are no mistakes.
I follow this topic to see how this dilemma will be solved.

Lnewqban and Philip Koeck
jonhswon said:
I follow this topic to see how this dilemma will be solved.
Me too.

Maybe the whole approach of comparing outgoing and incoming radiation is wrong, but why?

hutchphd

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