Let's say I have two infinitesimal boxes filled with gases in curved spacetime. These points are time-like separated. They will obey different thermal distributions based on the metric. What does the zeroth law of thermodynamics

If a body C is in thermal equilibrium with two other bodies, A and B, then A and B are in thermal equilibrium with one another.

mean over here? If two boxes have temperatures $T_1$ and $T_2$ respectively and $T_1 = T_2$, are they in thermal equilibrium?

Or, if I have boxes A and B, can I define a box C that is in thermal equilibrium with A and B (even though they are separated and obey different distributions)?

  • $\begingroup$ A box containing gas does not occupy a point in space time, but follows a world line. For them to be in thermal equilibrium they will need to have constant thermal contact, at least for some finite part of that path. That means each must, at a minimum, have part of their world line in the other's forward light cone. And box C would need to have thermal contact with at least one of A and B. $\endgroup$
    – Dan
    Jan 19, 2022 at 14:33
  • $\begingroup$ Sorry I had meant a moving point. Yes you can assume those conditions. I think the question holds though? As the change in distribution of temperature affects things? $\endgroup$ Jan 19, 2022 at 14:36
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    $\begingroup$ First you have to ask yourself how the The zeroth law of thermodynamics in special relativity (2020) works: there are a lot of misconceptions and this paper is the most clear and modern point of view. $\endgroup$
    – Quillo
    Nov 29, 2023 at 9:41

1 Answer 1


We can define thermodynamic equilibrium and formulate zeroth law of thermodynamics for static spacetimes, when the metric could be written as: $$ ds^2=-g_{00}dt^2+g_{ij}dx^idx^j. $$ At equilibrium the matter also would be static, but unlike the flat space the temperature (as measured by thermometer at given point) would gain spatial dependence, $T(x)$, of the form: $$ T(x)=\frac{T_0}{\sqrt{|g_{00}|}}, $$ where $T_0$, the red-shifted temperature, has the meaning of a temperature at a point where gravitational potential is zero (and so the $|g_{00}|=1$). It is this red-shifted temperature that is constant across the region of thermal equilibrium. This spatial dependence of temperature is known as Tolman temperature gradient (after R.C. Tolman who discovered this effect in 1930).

For a modern discussion of the effect see:

What does the zeroth law of thermodynamics <…> mean over here?

It means that thermodynamic equilibrium introduces an equivalence relation between thermodynamic (sub)systems. The precise nature of thermal contact between systems A and B does not matter: it could be photon radiation within the reservoir enveloping both A and B, or vibrational degrees of freedom of some solid conductive material stretching between A and B, or gas of massive particles, or a sequence of different subsystems etc., as long as we know the temperature at one point, Tolman relation gives us equilibrium temperature everywhere else.

More details, discussions and related results could be found in PhD thesis of J. Santiago:

  • Santiago, J. (2019). On the Connections between Thermodynamics and General Relativity, PhD thesis, Victoria University of Wellington, arXiv:1912.04470.
  • $\begingroup$ Thanks for this answer, I gave you +1. However, I wonder why the question had a score of 0, if it was worth you spending all this time on it? It has a score of +1 now because I upvoted it, but I was curious why it had a score of 0 at the time. $\endgroup$ Jan 26, 2022 at 5:05
  • $\begingroup$ @user1271772: The question has a downvote. My guess is that someone didn't like initial formulation of the question that was mixing spacetime points with worldlines. $\endgroup$
    – A.V.S.
    Jan 26, 2022 at 7:25
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    $\begingroup$ Thanks for pointing me again to the early version. I'm still sometimes baffled by the voting practices on this site. An HNQ can get a hundred upvotes in one day, but an equally good question (or better quality question) can go several months without any votes at all. $\endgroup$ Jan 26, 2022 at 7:45
  • $\begingroup$ @avs the zeroth law of thermodynamics holds for black holes as well no? But that is of different form than the metric u mention. Does this mean the analysis is incomplete? $\endgroup$ Jun 1, 2022 at 12:24
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    $\begingroup$ @MoreAnonymous: Schwarzschild and Reissner-Nordstöm black holes are static. Kerr and Kerr-Newman are stationary. Extension of the analysis on the stationary solutions is discussed in Santiago's thesis (and the original paper). Also keep in mind, that when people talk about zeroth law in the context of black hole they mean related but different thing. The bridge between two meanings is in the QFT in curved spacetime where quantum fields provide a natural medium for thermodynamics. $\endgroup$
    – A.V.S.
    Jun 2, 2022 at 7:01

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