# Heuristic Evaluation of Stress-Energy Tensor

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1. Feb 4, 2015

### Itserpol

I've been reading through Schutz's A First Course in General Relativity, and my solution to a particular problem has got me wondering if I'm being careful enough in my approach. The problem states:

Show that, in the rest frame $\mathcal{O}$ of a star of constant luminosity $L$ (total energy radiated 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.​

Starting with the $T^{0x}$ term, I know that it is the flux of 0-momentum (energy) across a surface of constant $x$. Since the luminosity is the total power output of the star and the photons all move radially away from the source, I need simply divide $L$ by the surface area of a sphere of radius $x$, and I obtain the desired quantity. And because $T$ is symmetric, I get $T^{x0}$ as well for free. For the $T^{xx}$ term, I know that the momentum of a photon is $h\nu$ (assuming units such that $c=1$) which is the same as the photon's energy, so the momentum flux should be identical to the energy flux. Finally, for the $T^{00}$ term (energy flux across a surface of constant $t$, i.e. energy density), I use the fact that for a photon, $\Delta t=\Delta x$. So rather than dividing the total energy by some volume $\Delta x\Delta y\Delta z$, I can use $\Delta t\Delta y\Delta z$ and I again obtain luminosity over the area of a sphere.

So, although I got the correct result, I wonder if I've been too "hand-wavey" about it. It's been a few years since I finished my undergrad, and I've lost some confidence in my ability to heuristically solve problems. Any comments?

2. Feb 4, 2015

### Staff: Mentor

All correct.

Yes; the more usual term here is "pressure" instead of "momentum flux", but your reasoning is correct.

This sort of works, but I think a less hand-waving argument is simply to observe that, just as momentum flux is equal to energy flux, momentum density is equal to energy density for light. You've already shown that momentum density is equal to energy flux, because $T$ is symmetric ($T^{x0}$ is the momentum flux across a surface of constant $t$, i.e., momentum density), so that shows that energy density is also equal to energy flux.

3. Feb 4, 2015

### Itserpol

You're right, that is a much better solution! Honestly, that was the part I felt the most iffy about. It follows a nice parallel to the others as well. I should try to look a little harder for symmetries like that.

Thanks! :)