Stress-energy tensor for a single photon

In summary, the problem pervect and bcrowell are discussing is that the stress-energy tensor of a single photon does not seem to be invariant under Lorentz transformations, even though it should be.
  • #1
salvador_dali
5
0
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 ...)
 
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  • #2
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).
 
  • #3
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.)
 
  • #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.
 
  • #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.
 
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  • #6
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.
 

1. What is the stress-energy tensor for a single photon?

The stress-energy tensor for a single photon is a mathematical object that describes the energy and momentum of a single photon. It is a 4x4 matrix that represents the distribution of energy and momentum in space and time.

2. How is the stress-energy tensor for a single photon calculated?

The stress-energy tensor for a single photon can be calculated using the Maxwell's equations of electromagnetism, which describe the behavior of electromagnetic fields. It is also possible to derive the tensor from the quantum mechanical wave function of the photon.

3. What is the physical significance of the stress-energy tensor for a single photon?

The stress-energy tensor for a single photon is important in understanding the behavior of electromagnetic fields and how they interact with matter. It also plays a crucial role in the theory of general relativity, as it is used to describe the curvature of spacetime due to the presence of energy and momentum.

4. How does the stress-energy tensor for a single photon differ from that of a classical particle?

The stress-energy tensor for a single photon is different from that of a classical particle because photons are massless particles and therefore have no rest mass. This means that the energy and momentum of a photon are solely determined by its frequency and polarization, while for a classical particle, the mass also plays a role in determining its energy and momentum.

5. Can the stress-energy tensor for a single photon be used to describe the energy and momentum of a group of photons?

Yes, the stress-energy tensor for a single photon can be extended to describe the energy and momentum of a group of photons. This is done by summing up the individual tensors for each photon in the group. However, in this case, the tensor will not represent the properties of a single photon, but rather the collective properties of the group.

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