Stress-energy tensor for radiation from a star I have been puzzling over an exercise in Schutz's A First Course in General Relativity:

Show that, in the rest frame $\mathcal{O}$ of a star of constant
  luminosity $L$ (total radiated energy per second), the stress-energy
  tensor of the radiation from the star at the event $(t,x,0,0)$ has
  components $T^{00}=T^{0x}=T^{x0}=T^{xx}=L/(4\pi x^2)$.  The star sits
  at the origin.

Thus far in the text, however, we have only discussed up to perfect fluids.  Without a fluid in space, what is the physical meaning of say, the momentum $T^{0x}$ and pressure $T^{xx}$?
Secondly, what is the basic approach to deriving these facts about $T$?  I can see why these quantities would fall off at $1/x^2$ heuristically, but I do not see how this follows from the first principles of the stress-energy tensor.
 A: Maybe I found your question a bit late, but hope it helps:
First, we can think of radiation basically as a fluid. Then pressure and density follow from our perspective. 
Secondly,  the star radiates spherically. $L$ is the total energy per second, so if I divide by the total area I'll get the energy flux at any moment 
$$ \frac{L}{4\pi r^2}$$ 
In our particular point $(t, x, 0, 0)$, $r = x$. And the value above is equal to $T^{0x}$ (radius $x$) since at this point of the sphere I can place also a surface of constant $x$. By symmetry $T^{x0}$ has the same value. 
As far as I know, radiation from stars is mostly photons (at least in this problem) and photons move radially from this source. In our particular point, photons must move exclusively in the $+x$ direction and will have only spatial momentum $p^x$ and the other components will be zero. Then any component like $T^{yz}$ will be zero, since no movement occurs in the $y,z$ directions. However, it is also true that a photon's momentum equals its energy (since they're massless and using natural units). Then the $x$ momentum flux in this point will be equal to the energy flux. As before, this point on the sphere is also one where I can place a surface of constant $x$. Then $T^{xx} = \frac{L}{4\pi x^2}$.
The components like $T^{xy}$ are null since photons move parallel to surfaces of constant $y,z$ and cannot cross them (no flux). Similarly for components like $T^{y0}$ (at this point we cannot place surface of constant $y,z$).
For fixed $t$, $L$ is simply the total energy. Dividing by the area gives us the energy density or $T^{00}$. This is analogous to the reason why, in the same chapter, we can define the vector $\vec{N}$ as having both density and flux components so easily, thanks to natural units. 
