# Second law of thermodynamics in general relativity

I recently came across some posts in Stack Exchange, which say that simultaneity doesn't have any meaning in General Relativity (see Is simultaneity well-defined in general relativity?).

I came up with a thought experiment which shows that the second law of thermodynamics is compatible with special relativity:

Considering a rod which has two temperatures $$T_1>T_2$$, where $$T_1$$ is the temperature in the left half and $$T_2$$ is the temperature in the right half. If I intentionally hypothesize that the speed of light can be surpassed by heat, then there will exist a point $$A$$ on the right end of the rod which has higher temperature $$T_3>T_2$$.

However since this is outside the light cone emanating from the middle, one can then construct a linear simultaneity hypersurface which goes through the point $$A$$ and say just to the right of the middle point where the temperature is $$T_2$$.

However since the heat is a physical flow its direction must be preserved in different coordinate frames (the heat vector can have different components in different frames, but the physical flow will be in the same direction). In particular the heat flow will be from left to right.

However, that means that heat is flowing from cold to hot and thereby the 2nd law will be violated. Thus, the violation of SR (Special Relativity) has led to a violation of TD (Thermodynamics). Maybe this thought experiment is wrong. Please point out if so.

But if my thought experiment is correct and simultaneity really doesn't have a meaning in GR (General Relativity) and one can really make “almost anything” simultaneous then one can't really formulate the 2nd law properly in GR. Or does this have to do something with the fact that in GR we are not dealing with inertial frames of reference?

• Why would you think that heat flow can exceed the speed of light? Aug 5, 2019 at 20:16
• I don't think it. I intentionally assume that in order to show the compatibility of the 2nd law with special relativity. Aug 5, 2019 at 20:29
• "If I intentionally hypothesize that the speed of light can be surpassed by heat, then there will exist a point A on the right end of the rod which has higher temperature T 3 >T 2" Why should be so? Aug 6, 2019 at 15:48
• @MicheleGrosso should have added that $T_3$ is measured is at a later time, such that it is outside the light cone at the centre of the rod. Aug 7, 2019 at 12:13

## 2 Answers

I came up with a thought experiment which shows that the second law of thermodynamics is compatible with special relativity

You claim to have come up with a thought experiment showing that for a certain type of violation of SR, there will be problems with thermodynamics as well. That isn't the same thing as showing that thermodynamics is compatible with SR.

The title of the question refers to general relativity, but there is nothing in the text of the question that has anything to do with general relativity.

However since the heat is a physical flow its direction must be preserved in different coordinate frames

This is not true even in Galilean relativity. Say a car is driving to the east, and heat is flowing from the engine compartment into the passenger compartment, in the car's frame. In the frame of the earth, heat is flowing to the west.

• I can maybe agree with the first statement that it doesn't show the compatibility of SR and TD. I don't agree with the last two points: Because in context of the thought a notion of simultaneity is required. However in GR according sources in this forum simultaneity is not an absolute notion. I think there was a misunderstanding with the last point your raised. I agree that the components of vectors are different in different frames of reference, however the vector points in the same direction. I.e. in your example the heat is always flowing from the engine compt. to the passenger compt. Aug 7, 2019 at 12:14
• @user139383 Two points: 1) Simultaneity is not an absolute notion already in SR. Infact depends on the reference frame. What is different in GR is that you can not define a global inertial reference frame. 2) A vector is a geometrical object and as such frame-independent. The dot product with another vector is a scalar which is an invariant, so the angle between two vectors is an invariant. Aug 8, 2019 at 12:42

Here a formulation of the second law of thermodynamics in GR.

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. It remains constant in ideal cases where the system is in thermodynamic equilibrium or increases in case of irreversible processes. However the statistical interpretation of entropy, as the number of microstates accessible to a macroscopic state, requires it to be an invariant, that is independent of the reference frame.

To state this in GR, you firstly define the thermodynamic macrovariables transformation law in SR and secondly via the Einstein equivalence principle you extend it to GR.

1)
In SR given the laboratory reference frame $$O$$ and the macroscopic object reference frame $$O'$$ moving with uniform relative velocity $$v$$ vs. the laboratory, you define:
$$dQ = \gamma (v) dQ'$$
where:
$$dQ$$ amount of heat supplied to the macroscopic system in the laboratory frame
$$\gamma (v)$$ Lorentz factor
$$dQ'$$ amount of heat supplied to the macroscopic system in its rest frame
Note: it is the same transformation law of the energy, i.e. in compliance with the interpretation of the amount of heat as a form of energy.

As the entropy $$S$$ is defined as $$dS = dQ / T$$, where $$T$$ is the absolute temperature, to have it invariant you define:
$$T = \gamma (v) T'$$

2)
The Einstein equivalence principle states that locally the physical laws can be formulated according to SR. The extension to GR is realized via the tensorial formalism.

• I don't think this is right. I agree that entropy is Lorentz invariant. But for the rest, you need to introduce the system's momentum vector. Also, temperature is usually rised to a Lorentz invariant, so it shouldn't get any Lorentz factor. This subject is actually subtle and some parts are still controversial, even today.
– Cham
Aug 8, 2019 at 14:14
• @Cham. The thermodynamic temperature is a measure of the average energy of the translational, vibrational and rotational motions of matter's particle constituents; therefore the Lorentz factor is consistent. As for the three-momentum, in the macroscopic system rest frame $O'$ it is zero by definition. Aug 8, 2019 at 15:31
• Temperature is a measure of average energy in the rest frame (like rest mass) and makes sense only in that frame. It's an invariant, like entropy and rest mass. This was a huge source of confusion in old litterature and is well documented today. But it's still a source of confusion.
– Cham
Aug 8, 2019 at 16:37
• As far as I know there is no real consensus on relativistic thermodynamics. Not even in special relativity. One can find papers where authors have argued for different ways Temperature and Entropy transform. Aug 8, 2019 at 16:58
• @Cham and OP. Thanks for emphasizing the matter is today still open. Let us consider my post as a contribution to the discussion. Aug 9, 2019 at 14:46