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I was asked in a recent homework to "show" that the for the stress-energy tensor $$T^{\mu\nu}=[(\rho +p)]U^{\mu}U^{\nu}+P\eta^{\mu\nu}$$
For a perfect fluid that the conservation law holds, that is $T^{\mu\nu},_{\nu}=0$ (all in Minkowski space-time). I was quite surprised and confused about this question, as it seemed that this, while not strictly postulated, came from the field equations as the Einstein tensor does banish. And while I understand some physical arguments in term of local conservation of energy and momentum can be given, I was told that I could obtain the law by tensor algebra. What some peers did was to consider the four-velocity on MCRF such that $U^{T}=(c^2,0,0,0)$ and that here the derivative must banish while also saying that the density and pressure must be constant for it to be a perfect fluid.

I think this is incorrect as the MCRF is only valid at a particular point, because if not that would mean you could find a global frame in which the velocity field is always 0 for the 3-velocity component, but I can't seem to find any viable construction for such a frame. Also, $\frac{\partial U^{\beta}}{\partial {x^{\alpha}}}$ must require to consider changes in a neighborhood of that point, so while at the that particular point the four velocity is 0, I don't think you can conclude the derivative must banish. Also, I haven't found a definition of a perfect fluid that imposes such strong hypothesis for the pressure and density, and I've seen examples were they aren't constant.

I suspect the question is ill-posed, or either just a trivial consequence of it being a stress energy tensor (we saw in class the Einstein tensor banishes so we can use this fact). So I would like to ask if it makes sense as it is formulated?

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It makes sense to prove this using incopressibility and the equations of motion for the fluid. Start with:

$\nabla _{\mu } T^{\mu\nu}=(\rho +p)\nabla _{\mu }(U^{\mu}U^{\nu})+p\nabla _{\mu }g^{\mu\nu} $

The second term vanishes for the Levi-Civita connection (metric compatibility). The first term is :

$\nabla _{\mu }(U^{\mu}U^{\nu})=U^{\nu}\nabla_{\mu}U^{\mu} +U^{\mu}\nabla_{\mu}U^{\nu}$

If the fluid is incompressible: $\nabla_{\mu}U^{\mu}=0$

If the fluid flows on geodesics: $U^{\mu}\nabla_{\mu}U^{\nu}=(\nabla _{U}U)^{\nu}=0$

So those 2 conditions imply conservation of energy-momenum.

Edit: As pointed out by OP, this treatment is partial and assumes both incompressibility and constant pressure. In the full treatment, one take the equation $\nabla _{\mu } T^{\mu\nu}=0$ , substitutes T for a perfect fluid as above, and projects the equation on different directions to get the 2 equations:

$U^\mu\partial _\mu \rho =-(\rho +p)\nabla_{\mu}U^{\mu} $
$(\rho +p)U^{\mu}\nabla_{\mu}U_{\nu}=-\nabla_{\nu}p-U_{\nu }U^{\mu }\nabla_{\mu}p $

The first equation is a conservation equation $\rho$, and the second is a conservation equation for the momentum.

Reference: Relativistic Fluid Dynamics notes by Jason Olsthoorn Sections 3.1-3.3 http://mathreview.uwaterloo.ca/archive/voli/2/olsthoorn.pdf

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  • $\begingroup$ Thanks for the answer! But I can't really find any source that says a perfect fluid must be incompressible. Also shouldn't you also consider the leibniz rule on the pressure and density terms? $\endgroup$ – Fedro May 25 '18 at 16:31
  • $\begingroup$ You are right, I should take derivatives of p and $\rho$ as well. If I do so, I expect I will get Navier-Stokes equations with the right $\nabla p$ term. I will edit my answer and edit the pressure dependence. $\endgroup$ – tsufli May 26 '18 at 6:50

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