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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, 2017 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, 2017 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 $\endgroup$
    – ProfRob
    Nov 2, 2019 at 19:30
  • $\begingroup$ Can we say that the radiation pressure is also invariant? Is it important in which way the pressure is produced in order to determine whether or not it is an invariant, or your calculation is applicable to all kinds of pressures? $\endgroup$ Jun 6, 2020 at 8:08
  • $\begingroup$ The pressure is usually defined inthe rest frame of the radiation. i.e the frame in which the temperature looks the same in all directions. For eaxmple cosmic background radiation does not have the same apparent temperture in all directions because the solar system is moving with respect to the CMB Once that motion is taken into accound we see the same temperature (that is always done in CMB plots) and that is the frame where the pressure is defined. $\endgroup$
    – mike stone
    Jun 6, 2020 at 12:36

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