# 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.

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.