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
  • 117,350
emacs drives me nuts
  • 960

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
  • 117,350