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In Special Relativity, for a perfect fluid (i.e. without heat transference or viscosity) we have a stress tensor $T_{\mu \nu}$

$$ T_{\mu \nu} = -p\eta_{\mu \nu} + (\rho + p)u_\mu u_\nu $$

It is said to be Lorentz covariant, i.e., in other inertial reference frame it is written as

$$ T'_{\mu \nu} = -p'\eta_{\mu \nu} + (\rho' + p')u'_\mu u'_\nu = -p\eta_{\mu \nu} + (\rho + p)u'_\mu u'_\nu $$

So my question is, how can you ensure this if nor pressure $p$ or density $\rho$ are scalars?

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$p$ and $\rho$ are Lorentz scalars, but they only have a simple physical interpretation in some frames. $\rho$ can be interpreted as the density in the instantaneous rest frame of any given fluid element.

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  • $\begingroup$ But if $\rho$ is the density in a instantaneous comoving frame and you make a Lornetz tranformation of it, the result is not again the same density so it cannot be a scalar $\endgroup$ – Vicky Oct 5 '19 at 17:44
  • $\begingroup$ @Vicky $\rho$ is not the density in some other frame. It is the proper density. $\endgroup$ – Andrew Steane Oct 5 '19 at 20:35
  • $\begingroup$ Then, that $\rho$ is always the proper density, then changing the reference frame doesn't change it because in the stress tensor it has to be the proper density always, in all reference frames, by definition of the tensor. And what about pressure, is it 'proper pressure'? After defining 'proper pressure' as $p=dK^i/dS$ with proper force $K^i = ma^i$ being $a^i$ the spatial part of the proper acceleration $a^\mu$ and $m$ the mass of the fluid particle. $dS$ is the transverse to the movement (perpendicular to axis $x^i$ so it doesn't change among reference frames) surface where $dK$ is applied. $\endgroup$ – Vicky Oct 6 '19 at 4:22
  • $\begingroup$ @Vicky Yes that's basically right though I haven't checked in detail. $\endgroup$ – Andrew Steane Oct 7 '19 at 19:12

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