Τμ. = Σον δ(x – x (1)) = Σ c2P,"P δ(x – x(t)) = Τυμ, (7.19)...

Given the energy momentum tensor for the perfect fluid,

(i)
Contracting the above expresion with the metric, we get
  
And by using the fact the 4-velocity by definition has the norm
  
And also in four dimensions

So, we get

  

(ii)
In the local inertial frame, in which the fluid is at rest, the energy density is simply given by
  
And also, in the local frame, the 4-velocity is given by

And locally, we have
  
And so, we get in the local inertial frame,

And so, we get
  
So, in this frame, the trace of the stress-tensor is given by

  
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 .

(iii)
The stress-energy tensor of a system of particles as given in equation (7.19) in the question

So, the trace in this is case means

Now, for photons, the norm of the momentum is zero (i.e., photons are massless), i.e.,
  
And so, we get

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
  

This is the equation of state, i.e., the energy density as a function of the pressure.