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Stress-energy tensor for a single photon

  1. Jan 26, 2014 #1
    Hi,

    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 ...)
     
    Last edited: Jan 26, 2014
  2. jcsd
  3. Jan 26, 2014 #2

    pervect

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    Being a classical theory, I'd rather talk about the stress energy tensor of a null dust than a single photon.
    I did that already in https://www.physicsforums.com/showthread.php?t=681172#post4323336

    I would expect that E=pc, so I think you might have the same result I did - perhaps you make to make this substitution to make the results come out the same.

    You should find that everything scales as the doppler factor^2, which should be (1+v/c)/(1-v/c).
     
  4. Jan 26, 2014 #3

    bcrowell

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    I think pervect has it right. Some material in section 10.6 of my SR book http://www.lightandmatter.com/sr/ may also be helpful. (There's an example where I do the stress-energy tensor of a plane wave.)
     
  5. Jan 26, 2014 #4
    Thanks pervect and bcrowell for your thoughts; it does seem sensible to treat the photon as a wave packet and then transform the Poynting flux; that may be what I do. However, I'm still not sure that I think the situation for null dust is fully consistent -- I can set p_A^0 = p_A^1 = E and recover pervect's result (without loss of generality, in fact), but I still encounter the issue where if I transform the individual wave vector and then rebuild the stress tensor in the boosted frame (B) I get a different answer than I do by transforming the tensor from the unboosted frame (A). Without associating the energy of the null dust with a lengthscale/timescale via e.g. a frequency in which case we don't need an additional factor when transforming the four-momentum, I still think we should be able to reconcile the two transformation methods... but I really have no idea how.
     
  6. Jan 27, 2014 #5
    Actually, I think I've figured it out (this may be what you were suggesting this whole time):

    Basically, the 1/d^3 x transforms as a density of particles rather than as a coordinate volume. This is required by the invariance of p^0 d^3 x, and can be shown instructively by observing the change of the separation of two particles under a Lorentz transformation. However, for two frames in which the spacetime volume elements are fixed, this implies a scale factor properly associated with each four-momentum:

    [tex]p_B^{\mu} = \frac{p_A^0}{p_B^0} \frac{\mathrm{d}x_B^{\mu}}{\mathrm{d}x_A^{\nu}}p_A^{\nu}[/tex]

    So the formula of the stress tensor remains the same as (1), with the p^{\mu} no longer representing simply the Lorentz boosts of the four-momenta. This neatly recovers the anomalous factor I had found before.
     
    Last edited: Jan 27, 2014
  7. Jan 27, 2014 #6

    bcrowell

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    In #1, what quantity is f supposed to represent? Is it supposed to be something classical? Something quantum-mechanical? In quantum-mechanics, you don't express a wavefuction as a function of *both* position and momentum; the two representations are complementary but separate (they're Fourier transforms of each other). There is also the issue that the photon doesn't actually have a wavefunction in the usual technical sense.

    The other thing to keep in mind is that a propagating pointlike wave packet is not a solution of Maxwell's equations, or of any other wave equation. (It would violate the diffraction limit.) If you write down a wave that is not a solution of Maxwell's equations, then the stress-energy tensor is in general not going to be divergenceless, which violates conservation of energy-momentum.
     
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