# Do free particles slow down in expanding universe?

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.