In summary, the Tolman law describes how the temperature in a fixed gravitational field depends on the position. Here, I provide a simple general derivation of the law. It's generally in the sense that it is applicable to any (thermalized) physical degrees of freedom for which one can define Lagrangian and Hamiltonian. It's simple in the sense that, in the derivation, one does not need to care about any details of those physical degrees of freedom, while general relativity is applied at a rather elementary level. To achieve simplicity, at some points I use heuristic arguments based on physical intuition, thus avoiding mathematical rigor. After the derivation, I briefly discuss a few elementary examples.
  • #36
PeterDonis said:
the observed Unruh temperature will also vary from bottom to top. But this has nothing to do with Tolman's law, because there is no gravitational potential; spacetime is flat. Tolman's law is about the effects of curved spacetime.
That's wrong. The variation of Unruh temperature is exactly a manifestation of Tolman law. It appears because ##g_{00}## is not a constant in an accelerated frame. Tolman law does not require a curvature.
 
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  • #37
I just looked up Einstein's "Über die spezielle und allgemeine Relativitätstheorie". Methinks he uses exactly my line of argument to justify the redshift for a rotating disk from the special relativistic transformation of time and then argues that this argument carries over to general gravitational field. I don't see why this line of argument should not be applicable to temperature.
 
  • #38
The difference is that the frequency of a wave is the time-component of the wave-four vector, ##(\omega/c,\vec{k})##, and temperature is a scalar (field). Also note that relativistic thermodynamics before around the mid 1960ies was a mess with several different conventions! For an early review, see

N. G. van Kampen, Relativistic thermodynamics of moving systems, Phys. Rev. 173, 295 (1968),
https://doi.org/10.1103/PhysRev.173.295
 
  • #39
In Landau/Lifshetz Vol. 5, they derive as condition for thermodynamical equilibrium the condition ##T\sqrt{g_{00}}=const.##. In the following they remark that "this can also be written as ## T=const. \cdot \frac{dx_0}{ds}## which allows to consider the body not only in a system where he is at rest, but also in a system where he is moving (rotates as a whole). To that end, the derivative has to be taken along the world line passing through a given point of the body." So he goes exactly the same way I was proposing to extend the Tolman relation to moving bodies.
 
  • #41
Demystifier said:
The variation of Unruh temperature is exactly a manifestation of Tolman law. It appears because ##g_{00}## is not a constant in an accelerated frame. Tolman law does not require a curvature.
Mathematically, if you assume spacetime is flat and that you have a perfect fluid whose worldlines are the Rindler congruence, yes, you will obtain Tolman's Law.

However, those assumptions are not really consistent, because "perfect fluid" implies nonzero stress-energy, and flat Minkowski spacetime is not a solution of the EFE for nonzero stress-energy. Similar remarks apply to Unruh radiation; giving the radiation a nonzero energy density, which is required for it to have a nonzero temperature, is not really consistent with spacetime being flat. (Calling the energy density "vacuum energy density" doesn't help; that just means you have a nonzero cosmological constant, and flat Minkowski spacetime is not a solution of the EFE for a nonzero cosmological constant either.)

Also, if you have a perfect fluid with accelerated worldlines, something has to be producing that acceleration. A static, spherically symmetric configuration of matter can do that without expending energy because of the particular way it curves spacetime. But in flat spacetime, you can't just magically say the fluid accelerates; something has to be making it accelerate, and that something has to be expending energy (like a rocket engine). And you have to include the energy involved in that in your analysis as well. That means there is even more stress-energy present in your model, which isn't compatible with flat spacetime, and it also means there is the possibility of other heat transfer that your model does not include.

I understand the use of the flat spacetime Rindler case as a heuristic; I just don't think it can do all the work that is being ascribed to it in this thread.
 
  • #42
PeterDonis said:
But in flat spacetime, you can't just magically say the fluid accelerates; something has to be making it accelerate, and that something has to be expending energy (like a rocket engine). And you have to include the energy involved in that in your analysis as well. That means there is even more stress-energy present in your model, which isn't compatible with flat spacetime, and it also means there is the possibility of other heat transfer that your model does not include.
When you really fear that you need a rocket so heavy, that its gravity would disturb the system, which I consider quite unlikely, you could imagine other sources of acceleration like reflection of a laser beam.
 
  • #43
DrDu said:
When you really fear that you need a rocket so heavy, that its gravity would disturb the system, which I consider quite unlikely, you could imagine other sources of acceleration like reflection of a laser beam.
Sure, a rocket of ordinary size, even counting all of its fuel and rocket exhaust, won't be enough stress-energy to curve spacetime--but it also won't provide enough acceleration to serve as any kind of illustration of Tolman's law; the temperature gradient due to the variation in ##g_{00}## with position will be many, many orders of magnitude too small to detect. If you're going to say we could in principle detect such small temperature differences, then we could just as well in principle detect such small amounts of spacetime curvature. The two go together; you can't just arbitrarily take one into account and ignore the other.

Similar remarks would apply to any other source of acceleration, like a laser beam.
 
  • #44
PeterDonis said:
Sure, a rocket of ordinary size, even counting all of its fuel and rocket exhaust, won't be enough stress-energy to curve spacetime--but it also won't provide enough acceleration to serve as any kind of illustration of Tolman's law; the temperature gradient due to the variation in ##g_{00}## with position will be many, many orders of magnitude too small to detect. If you're going to say we could in principle detect such small temperature differences, then we could just as well in principle detect such small amounts of spacetime curvature. The two go together; you can't just arbitrarily take one into account and ignore the other.

Similar remarks would apply to any other source of acceleration, like a laser beam.
For a laser beam, it should be possible to estimate the effect of the energy density of the laser. The accelerated system can only see the gravity of the light already reflected. This intensity is maximal and constant in time (judged from the accelerated system) but proportional to acceleration directly at the reflector. The gravitational potential due to light reflected at earlier times should be integrable due to redshift. Maybe one could estimate whether it increases linearly or sub-linearly with acceleration?
 
  • #45
DrDu said:
For a laser beam, it should be possible to estimate the effect of the energy density of the laser.
You don't have to estimate; the stress-energy tensor of an EM field is well known.

https://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

DrDu said:
The accelerated system can only see the gravity of the light already reflected.
Not at all. The source of the laser, which must contain enough energy to supply it, will already be in the past light cone of the accelerated system, along with its gravity.
 
  • #46
The intensity of the laser will have to increase steadily to maintain constant acceleration. The accelerated observer will always see the reactor only as a point source of gravity (retarded) which is more and more far away. The laser he will only notice when it hits him and probably the gravity of the light reflected.
 
  • #47
DrDu said:
The intensity of the laser will have to increase steadily to maintain constant acceleration.
So what?

DrDu said:
The accelerated observer will always see the reactor only as a point source of gravity (retarded)
If you are using Newtonian gravity, this is not correct; there is no retardation in Newtonian gravity.

If you are using relativity, gravity is not a force, it's spacetime curvature, sourced by stress-energy. All of the stress-energy that is going to be pumped into the laser is already in the past light cone of the accelerated observer before the laser even turns on. So retardation is irrelevant.

DrDu said:
The laser he will only notice when it hits him
He will only notice the acceleration due to the laser when it hits him, yes. But that does not mean he only notices the gravity due to the stress-energy that gets pumped into the laser when it hits him. See above.

DrDu said:
and probably the gravity of the light reflected
As already stated, all of the stress-energy that causes the "gravity" the accelerated observer senses is already in his past light cone before the laser even turns on.

The reflection of the laser light does put some of that stress-energy closer to the accelerated observer than it would otherwise be, which I think (I haven't done the detailed math to check) will increase somewhat the gravity the accelerated observer measures, from what it would have been if the laser had not fired. But that doesn't mean the accelerated observer measures no gravity at all until the laser hits him. See above.
 

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