My favorite toy model of cosmic microwave background in the expanding universe is a cube with totally reflective walls. If CMB is a black-body radiation, it must be isotropic and homogeneous - the same number of photons go through a side of the cube, so it can be reflective and the effect is the same. This model (remnant of my Olbers paradox period) seem to work also if the cube is expanding - it gives redshifted that is colder CMB.

Recently I imagined free (non-relativistic) particles instead of photons in the cube. This again requests that the universe is homogeneous and motion of particles is isotropic. But now I have a problem - if the walls of the cube are moving (the cube or the universe is expandind), this model suggests that the particles slowly come to rest (they reflect from receding walls).


2 Answers 2


The peculiar velocity (i.e. the velocity measured by an observer fixed to the comoving frame) of a free particle in an expanding cosmology is defined as:

$$\vec{v}=\frac{{\rm d}\vec{r}}{{\rm d}t} - H(t)\vec{r}$$

Here $\vec{r}$ is the proper distance, $t$ is the proper time, and $H(t)$ is the time-evolving Hubble parameter. We can check whether the peculiar velocity changes with time by taking its derivative:

$$\frac{{\rm d}\vec{v}}{{\rm d}t}=\frac{{\rm d}^2\vec{r}}{{\rm d}t^2} - \vec{r}\frac{{\rm d}H}{{\rm d}t} - \frac{{\rm d}\vec{r}}{{\rm d}t}H(t)$$

On inspection it's clear that a particle with non-zero peculiar velocity in a universe with a non-zero Hubble parameter must have $\frac{{\rm d}\vec{v}}{{\rm d}t}\neq 0$, and with a bit of work it can be shown that in a universe like ours they decay as $1/a$, where $a$ is the scale factor.

You might be tempted to pick an observer stuck to the particle ($\vec{r}(t)=0$), in which case you would be tempted to conclude that the peculiar velocity is constant, but this doesn't work since the peculiar velocity is defined for an observer fixed to the comoving frame, i.e. with zero peculiar velocity themselves.

Your model doesn't really work for free particles, because it involves collisions (either with the walls of the box, or other particles), which means the particles aren't really free. Further, the reflections from receding walls are a poor model because in that picture the particles lose energy at discrete times, whereas in a cosmological model the process is continuous. But your conclusion isn't completely wrong: since the peculiar velocity of every particle is decaying, in the absence of gravitational amplification of the velocities, eventually they will all come to equilibrium at rest.

  • $\begingroup$ So if the expanding universe would contain ideal gas the relative velocities of its particles during collisions would gradually decrease (to zero in infinite time)? $\endgroup$
    – Leos Ondra
    Mar 1, 2019 at 8:55
  • $\begingroup$ @LeosOndra yes, provided the expansion continues. And they would have 0 velocity relative to the comoving frame, they would of course continue to expand with the rest of the Universe. $\endgroup$
    – Kyle Oman
    Mar 1, 2019 at 9:11

I don’t see the problem. There is no energy conservation here for the collection of balls/photons, the wall absorbs it. The photons too will lose energy upon bouncing of the moving mirrors. It is only that photons are not allowed to slow down, so they will redshift, asymptotically approaching infinite redshift. The balls’ only way to lose energy is slowing down.


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