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Consider a system of stationary 3-D ideal gas in the rest-frame $S$. This system is described by $PV=Nk_BT.$ Also, from the principle of equipartition, $E=\frac{3}{2}Nk_BT.$

Now we introduce a boost (and a co-moving system $S'$). I am assuming $N$ and $k_B$ are Lorentz-invariant.

we get a Lorentz-Fitzgerald contraction along the axis of movement, so $V$ is downscaled by a factor of $\gamma$: $ \ \ V'=\frac{V}{\gamma}.$

As the gas is stationary in S, its momentum in S is 0 so the energy transformation gives $E'=\gamma E$ so from equipartition we get $T'=\gamma T.$

If we assume $P$ is Lorentz invariant (as I have always thought was the case) we get a contradiction! From the ideal gas law we get $T'=\frac{T}{\gamma},$ while equipartiton gives us $T'=\gamma T.$ This is obviously not the case!

We must therefore assume that pressure is non-Lorentz-invariant, then we get $$P'V'=Nk_BT' \implies \frac{P'V}{\gamma}=\gamma Nk_BT \implies P'=\frac{P}{\gamma^2}.\ $$

Why isn't pressure Lorentz invariant? Is it derivable from it being the trace of a stress-energy tensor, or from being force per unit area using the force transformation? Where does the factor $\frac{1}{\gamma^2}$ come from? Were I critically wrong along the way?

In short: How do thermodynamic sizes transform under a Lorentz boost?

EDIT: Could it be that the actual answer is that the ideal gas law as we know it (namely, $PV=Nk_BT$) is only correct for the rest frame, and it's true form for a general frame contains some factor $f(\gamma)$ such that $f(1)=1?$

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Pressure is part of the energy-momentum tensor. For an ideal fluid, and with the convention $g_{\mu\nu}= {\rm diag}(1,-1,-1,-1)$, this tensor can be written as
$$ T_{\mu\nu}= (\varepsilon +p) u_\mu u_\nu -p g_{\mu\nu}. $$ Here $u^\mu$ is the four velocity $$ u^\mu=\gamma(1,{\bf v}) $$ of the fluid. In the local rest frame of the fluid $u_\mu= (1,0,0,0)$, so this becomes $$ T_{\mu\nu}=\left[\matrix{\varepsilon & 0& 0& 0\cr 0& p &0 & 0\cr 0&0& p & 0\cr 0&0&0&p}\right]_{\mu\nu}. $$ So, although the energy-momentum tensor looks quite different in different frames, once you diagonalize you will get the same $p$ and $\varepsilon$. In this sense $p$ is an invariant, because it is an eigenvalue of a matrix.

Further to answer you other point. Yes ideal gas kinetic theory only holds in the rest frame Actually for non-deal systems the notion of "rest frame" is not well defined. Even for a single component gas you can have a frame in which there is no energy flux (Landau frame), a frame in which there is no baryon-number flux (Eckart frame), a frame with no entropy flux, and frame the total momentum density is zero. In general all are different.

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  • $\begingroup$ So if pressure is Lorentz-invariant, where were I wrong to get $P'=\frac{P}{\gamma}?$ $\endgroup$ – A. Ok Jun 25 '17 at 10:21
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    $\begingroup$ @A. OK. Because to get to the moving frame you need to transform $T_{mu\nu}$ as a tensor. There really is no other easy way. In particular temperature is only defined in the local rest frame. In any other frame there is an energy flux --- so the fluid is not in equilibrium. $\endgroup$ – mike stone Jun 25 '17 at 10:42
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    $\begingroup$ I don't think there is agreement that "pressure" is invariant. nature.com/articles/s41598-017-17526-4 $\endgroup$ – Rob Jeffries Nov 2 at 19:30

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