1
$\begingroup$

According to standard cosmology theory the physical momentum $p$ of both massive and massless particles decay like: $$p \propto \frac{1}{a(t)}$$ where $a(t)$ is the scale factor as function of cosmological time $t$ (for a derivation see page 12 in these cosmology lecture notes).

Does this imply that a gas inside a rigid box cools over cosmological timescales?

In other words does the space enclosed by the box expand through its walls even though they are rigid?

We know that the walls of the box constrain the gas particles to stay inside the box; but surely they do not constrain the expanding space itself though which the particles move?

$\endgroup$
  • 2
    $\begingroup$ I would think "rigid" normally implies (among other things) that the box is not subject to cosmological expansion. Did you mean a box that grows along with $a(t)$ but has no other dynamical behavior? $\endgroup$ – David Z Oct 24 '15 at 12:22
  • $\begingroup$ Could the space enclosed by the box expand through the walls even though they are rigid? $\endgroup$ – John Eastmond Oct 24 '15 at 12:25
  • $\begingroup$ I don't know, it's your question... but I would imagine not. Otherwise, why are there walls at all? $\endgroup$ – David Z Oct 24 '15 at 13:51
2
$\begingroup$

The binding of matter with the four forces, gravitational, weak, electromagnetic, strong is much stronger than the expansion of the universe, the famous raisin bread analogy.

raisin bread

Animation of an expanding raisin bread model. As the bread doubles in width (depth and length), the distances between raisins also double.

The loaf (space) expands as a whole, but the raisins (gravitationally bound objects) do not expand; they merely grow farther away from each other.

Your box is one of the raisins. It is held together by strong electromagnetic forces Its dimensions are held constant by these forces whether there is gas inside or not. In addition , the walls are a boundary for the gas starting with its initial temperature and pressure again through electromagnetic forces. ( the gas will cool due to black body radiation but that is another story).

The whole raisin is held together by the standard forces, only the space between raisins is affected by the expansion. It might help you to think quantum mechanically: Quantum mechanically there is no empty space where a gas exists with temperature and pressure. There are a great number of fields of operators of elementary particles, with creation and annihilation happening continuously to keep the gas at that temperature and pressure.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But in principle one could imagine a rigid metal box that is many light-years in width. Surely a photon that has reflected off one wall will be redshifted by the time it reaches the opposite wall? I guess this process will repeat until half the wavelength of the photon is the width of the box. At that point the photon energy will remain constant. $\endgroup$ – John Eastmond Oct 24 '15 at 13:48
  • $\begingroup$ Photons are quantum mechanical entities.Gravitation has not been consistently quantized to really answer such detailed question. From quantum mechanics there cannot exist a rigid box light years in dimension because all the forces are transmitted through electromagnetic interactions, and the rigidity depends on the possibility of exchanging photons to keep the dimensions of "box". One would have to solve the specific problem, as at such distances the definitions are given by the gravitational field. $\endgroup$ – anna v Oct 24 '15 at 14:19
  • $\begingroup$ @JohnEastmond: The size of the box doesn't matter for your argument. What does matter is that the photon gas is in thermal equilibrium with gravity. As the universe expands the temperature of the thermal gravitational background will go towards zero and since the photon gas is coupling to the gravity, so will the temperature of the photon gas, even in a perfectly closed box. There are two complications with this picture: for one the temperature of the black hole of the mass of the box is finite and the thermalization timescale is probably longer than the black hole decay time scale. $\endgroup$ – CuriousOne Oct 24 '15 at 18:53
  • $\begingroup$ @CuriousOne: So you agree that a rigid box of gas does lose energy due to the effect of cosmological expansion on the moving gas molecules in addition to the much larger effect of black body radiation from the box itself? $\endgroup$ – John Eastmond Oct 24 '15 at 20:47
  • $\begingroup$ @JohnEastmond: If cosmological expansion actually lowers the temperature of the gravitational background, then, yes. That, of course, is not a proven fact and I remember having seen a paper which made some claims that accelerated expansion is actually heating the universe. I think it would also be next to impossible to perform this experiment in reality because the coupling timescales are probably much, much larger than the timescales of the universe and we are limited from below by finite temperatures due to the boundary conditions, anyway. $\endgroup$ – CuriousOne Oct 24 '15 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.