I've read that the momentum of particles declines due to the universe's expansion. In particular, that $p \propto \frac{1}{a}$, where $a$ is the scale factor. For light, this momentum reduction happens via redshift. For particles with mass, I've seen the formula:
$p = \frac{mv}{\sqrt{1-v^2}} \propto \frac{1}{a}$
(This comes from page 12 of these Cambridge University lecture notes, where $v$ is presumably expressed in units where $c=1$.)
I'm interested in whether such a particle slows down in proper peculiar velocity as measured by the galaxies it passes e.g.:
Q: Is a particle released from Earth at 50% $c$ always observed as moving at 50% $c$ (in proper units) by observers in a distant galaxy at the time it passes through their galaxy)?
I've had trouble working this out from the formulae I've seen. It depends on subtleties of what type of units $v$ is expressed in. The Cambridge notes say that the $v$ in the numerator is in comoving coordinates, while the $v$ in the denominator is in proper coordinates. If so, it seems to me that the particle wouldn't slow down in the sense above (its peculiar proper velocity wouldn't decline) since even a constant proper peculiar velocity like the speed of light is proportional to $\frac{1}{a}$ in comoving coordinates.
However the notes go on to say that this shows the particle converges to the Hubble Flow, which seemed at odds with this. There are multiple meanings this could have:
1) The particle's proper peculiar velocity approaches zero.
2) The particle's comoving velocity approaches zero.
Do people know which of these is meant?
I'm aware that in an exponentially expanding universe, (2) is true for all particles, massless or not, accelerating or not. So it seems odd if that is all the notes are saying, but it seems to be the only version implied if the $v$ in the numerator is in comoving coordinates.
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