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I'm trying to write down the stress-energy tensor for a single photon in GR, but I'm running into trouble with its transformation properties. I'll demonstrate what I do quickly and then illustrate the problem. Given a photon with wavevector p, we write

[tex]{\bf T} = \int \frac{\mathrm{d}^3 p}{p^0} f {\bf p \otimes p}[/tex]

where

[tex]f = \int \mathrm{ d\tau} \delta^4({\bf x} - {\bf x'}(t))\delta^3({\bf p} - {\bf p'}) = \delta^3({\bf x} - {\bf x'}(\tau))\delta^3({\bf p} - {\bf p'})[/tex]

which I think should be invariant even considering the nontrivial transformation properties of the Dirac distribution in curved space, although I actually have not been able to find a good reference for this (if anyone has one I would really appreciate it). So then we proceed immediately to

[tex]{T^{\mu \nu}} = \frac{p^{\mu}p^{\nu}}{p^0}\delta^3({\bf x} - {\bf x'}),[/tex]

and then averaging over a spacetime volume,

[tex]{T^{\mu \nu}} = \frac{1}{\sqrt{-g}\Delta^3 x\Delta x^0}\int_{\Delta^3 x \Delta x^0} \sqrt{-g}{\mathrm d}^3 x{\mathrm d} x^0 \frac{p^{\mu}p^{\nu}}{p^0}\delta^3({\bf x} - {\bf x'}(\tau)),[/tex]

[tex]\begin{equation}{T^{\mu \nu}} = \frac{1}{\Delta^3 x}\frac{p^{\mu}p^{\nu}}{p^0}.\end{equation} ~~~~~~~~~~~~~(1)[/tex]

So one can find this last expression all over the place, but I'm running into trouble with it even under Lorentz transformations, although it indeed looks covariant (d^3 x p^0 is invariant, and p \otimes p is a tensor). Here's a reduced example: suppose I'm in flat spacetime and I have two frames, A and B. In frame A I have a wavevector p_A = {p_A^0, p_A^1, 0, 0}. I construct T_A from p_A using equation (1):

[tex]{T^{\mu \nu}_A} = \frac{1}{\Delta^3 x}\left[ \begin{array}{ccc}

p^0_A & p^1_A & 0 & 0 \\

p^1_A & \frac{(p_A^1)^2}{p_A^0} & 0 & 0 \\

0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 \end{array} \right][/tex]

Now I boost p_A -> p_B and T_A -> T_B as shown:

[tex]p_B^{\mu} = \Lambda^{\mu}_{~~\nu} p_A^{\nu} = ( \gamma p_A^0 - \beta \gamma p_A^1, -\beta \gamma p_A^0 + \gamma p_A^1,0,0 )[/tex]

[tex]T^{\mu \nu}_B = \Lambda^{\mu}_{~~\mu'} \Lambda^{\nu}_{~~\nu'} T^{\mu' \nu'}_A = \frac{1}{\Delta^3 x}\left[ \begin{array}{ccc}

\gamma^2\frac{(p_A^0-\beta p_A^1)^2}{p_A^0} & \gamma^2\frac{(\beta p_A^0 - p_A^1)(\beta p_A^1-p_A^0)}{p_A^0} & 0 & 0 \\

\gamma^2\frac{(\beta p_A^0 - p_A^1)(\beta p_A^1-p_A^0)}{p_A^0} & \gamma^2 \frac{(-\beta p_A^0 +p_A^1)^2}{p_A^0} & 0 & 0 \\

0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 \end{array} \right][/tex]

Whereas constructing T_B with p_B via equation 1 we get a slightly different matrix T'_B, which I won't reproduce here because I'm typing too much. At any rate, for each non-zero element we have

[tex]T'_B/T_B = \frac{p_A^0}{(p_A^0-\beta \gamma p_A^1)}.[/tex]

This seems to imply that the wavevector should be picking up a scale factor beyond what it gets through the Lorentz transformation, but this makes no sense to me. The factor appears to be the difference between p_A^0 and p_B^0, but always evaluating the p^0 in the "original" frame doesn't seem very covariant to me.

Anyhoo, if anybody sees a flaw in what I've done so far or has related insight I'd be very glad to hear it. I also tried setting up T for a single (non-photon) particle using T = rho u \otimes u, with a Dirac distribution definition of rho, but that ultimately brought me back to equation 1 (which doesn't seem consistent with this: https://www.physicsforums.com/showthread.php?t=178692 ...)

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# Stress-energy tensor for a single photon

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