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Τμ. = Σον δ(x – x (1)) = Σ c2P,P δ(x – x(t)) = Τυμ, (7.19) E. η 1Question 7.1* (i) The energy-momentum tensor of a perfect fluid, with 4-velocity field UM, is given in an arbitrary frame by

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Answer #1

Given the energy momentum tensor for the perfect fluid,
T^{\mu\nu}=(\epsilon +P)\frac{U^\mu U^\nu}{c^2}-Pg^{\mu\nu}
(i)
Contracting the above expresion with the metric, we get
  T=g_{\mu\nu}T^{\mu\nu}=(\epsilon +P)\frac{g_{\mu\nu}U^\mu U^\nu}{c^2}-P g_{\mu\nu}g^{\mu\nu}
And by using the fact the 4-velocity by definition has the norm
  g_{\mu\nu}U^\mu U^\nu=c^2
And also in four dimensions
g_{\mu\nu}g^{\mu\nu}=4
So, we get
T=(\epsilon +P)-4P
  \Rightarrow T=\epsilon -3P

(ii)
In the local inertial frame, in which the fluid is at rest, the energy density \epsilon is simply given by
  \epsilon=\rho c^2
And also, in the local frame, the 4-velocity is given by
U^{\mu}\equiv (c,0,0,0)
And locally, we have
  g^{\mu\nu}\equiv \eta^{\mu\nu}=\text{diag}(1,-1,-1,-1)
And so, we get in the local inertial frame,
T^{\mu\nu}=(\rho c^2+P)\delta^{\mu}_{0}\delta^\nu_{0}-P\eta^{\mu\nu}
And so, we get
  T^{\mu\nu}=\text{diag}(\rho c^2,P,P,P)
So, in this frame, the trace of the stress-tensor is given by
T=\eta_{\mu\nu}T^{\mu\nu}=\rho c^2-P-P-P
  \Rightarrow T=\rho c^2-3P
This is not exactly same as the answer (i). But, it is almost same in a sense only the energy density is given by the local energy density \epsilon=\rho c^2 .

(iii)
The stress-energy tensor of a system of particles as given in equation (7.19) in the question
T^{\mu\nu}=\sum_{n}c^2 \frac{p_n^{\mu}p_n^{\nu}}{E_{n}} ~\delta(x-x_{n}(t))
So, the trace in this is case means
T=\eta_{\mu\nu}T^{\mu\nu}=\sum_{n}c^2 \frac{\eta_{\mu\nu} p_n^{\mu}p_n^{\nu}}{E_{n}} ~\delta(x-x_{n}(t))
Now, for photons, the norm of the momentum is zero (i.e., photons are massless), i.e.,
  \eta_{\mu\nu} p_n^{\mu}p_n^{\nu}=0
And so, we get
T=0
Now, we use the result obtained in part (ii) combined with the above traceless condition (T=0) to get for photons as a perfect fluid
  T=\rho c^2-3P=0
\Rightarrow \rho =\frac{3P}{c^2}
This is the equation of state, i.e., the energy density as a function of the pressure.


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