5

Suppose we have an indestructible box that doesn't let through any matter or radiation or whatever. In the box, there is matter which is evenly distributed (state A). The energy content of the box is thus $$E=mc^2$$ where $m$ is the rest mass of the matter$^1$.

Now suppose the box contains 50% matter and 50% antimatter that reacts in such a way that the outcome of the annihilation is 100% radiation (photons, state B). As energy is conserved, the energy of state B should be $$E=pc$$ where $p$ is the sum over the absolute values of all photon momenta.

Question: Could an external observer distinguish between state A and state B?

According to $E^2=p^2c^2+m^2c^4 = \text{const}$, state B should behave just like a box with (evenly distributed) mass $m=p/c$, i.e. exactly like state A?

What confuses me: Energy from mass (density) does not look the same in the stress-energy-tensor of the Einstein field equations. Even if the momentum vectors are cancelling each other out (which they should), leaving only the time component (energy), what happens with the immense radiation pressure of state B which does not cancel out? Does this lead to observable$^2$ differences between A and B? Where is pressure and stress being accounted for in $E^2=p^2c^2+m^2c^4$? Or is that equation just an approximation that assumes no pressure?

$^1$Assuming thermal energy, pressure and stress etc. can be neglected.

$^2$ Observable from ouside the box.

Dale
  • 99,825
emacs drives me nuts
  • 952
  • @Dvij D.C.: I didn't write state B is massless, just that it has no rest mass (photons clearly do not have rest mass). My question is whether B behaves as if it had mass $m=p/c$. – emacs drives me nuts Sep 11 '21 at 15:21
  • I removed my old comment because I just noticed that you are defining $p$ to be the sum of norms of three-momenta of all the photons and then the assertion is correct. But again, your comment now is incorrect. State B does have rest mass, it's $m$, namely, the exact same value as for the rest mass of the old system. –  Sep 11 '21 at 15:24
  • 1
    This is basically a duplicate of How is an ideal mirrored box of photons distinguishable from massive particles? but you raise an interesting question about the radiation pressure that makes me reluctant to close this. – John Rennie Sep 11 '21 at 15:28
  • 2
    To address your massless photon point, one photon doesn't have a rest mass, but two photons do (unless they are moving parallel), and a hot bowl of photon soup definitely has a rest mass. –  Sep 11 '21 at 15:29
  • @Dvij D.C.: Thanks for that clarification. – emacs drives me nuts Sep 11 '21 at 15:34

1 Answers1

3

You did not explicitly state it, but I assume that you also intend the material of the box to be very rigid such that when the pressure increases inside the resulting increased stress does not produce any measurable increased strain. I.e. the size of the box is unchanged before and after.

This scenario is well studied in the literature for a spherical box. When the pressure inside the box increases there is a stress produced in the walls. That stress is a tension in the walls. The increased tension in the shell effectively cancels out the increased pressure inside the box. The result is that there is no change outside the box. This is what we would expect due to Birkhoff’s theorem.

I am not aware of any investigation with non-spherical geometries. It certainly is possible that there would be an external change of some sort due to dipole or quadrupole effects, but certainly it would be small since there would be no monopole change.

Dale
  • 99,825
  • Thanks, sounds reasonable. Some answer to a similar question clains there is a difference ouside the box (using the argument that the momenum tensor differs in either case -- which is a flawed argument, IMO). So that other answer is wrong? My thinko was that the box is ideal and could thus be beglected; which is not the case because it might carry non-negligible stress, and will do so in case B. Presumably, that stress is what "fixes" the different momentum tensors? – emacs drives me nuts Sep 11 '21 at 15:57
  • 1
    Yes, the shape / volume of the box shall remain the same. Otherwise, something would escape the original dimensions. – emacs drives me nuts Sep 11 '21 at 16:04
  • I wouldn’t say that the other answer was wrong as much as that the other question is a mess. This one is a much better question. Having a massless box is a problem because it will violate energy conditions, particularly when you try to put it in tension. So I think that the other question is not very answerable. That said, the answer to that question would not be a good answer for your question. – Dale Sep 11 '21 at 17:48
  • I think both questions (and their answers) show clearly that there are limits to idealization: here it does not make physical sense to ignore the nature of the box - enclosing and isolating a physical system is not a trivial task! – Stéphane Rollandin Sep 12 '21 at 07:51