# Is pressure Lorentz-invariant?

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 $$f(1)=1?$$

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.
• So if pressure is Lorentz-invariant, where were I wrong to get $P'=\frac{P}{\gamma}?$ – A. Ok Jun 25 '17 at 10:21
• @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. – mike stone Jun 25 '17 at 10:42