# In what manner does momentum of a particle with mass decrease due to spatial expansion?

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.

Both the proper and comoving velocity go to zero, but at different rates. The way to see how this works is to start with the Friedmann–Lemaître–Robertson–Walker metric: $$[g_{\mu\nu}] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -a^2(t) & 0 & 0\\ 0 & 0 & -a^2(t) & 0 \\ 0 & 0 & 0 & -a^2(t) \end{array}\right].$$ This metric feeds into the least action principle for a classical massive particle: \begin{align} S &\propto \int \operatorname{d}\tau \sqrt{g_{\mu\nu} \frac{\partial x^\mu}{\partial \tau} \frac{\partial x^\nu}{\partial \tau} } \\ &= \int \operatorname{d}t\, \left(-mc \sqrt{c^2 - a^2 \frac{\operatorname{d} x_i}{\operatorname{d}t} \frac{\operatorname{d} x_i}{\operatorname{d}t} }\right). \end{align} where $x^\mu$ is the comoving (coordinate) position. The integrand in the second line is the Lagrangian ($L$), so the canonical momentum is given by: \begin{align} p_i & \equiv \frac{\partial L}{\partial x_j} \\ & = mc \frac{a^2 \frac{\operatorname{d} x_j}{\operatorname{d} t}}{\sqrt{c^2 - a^2 \frac{\operatorname{d} x_i}{\operatorname{d}t} \frac{\operatorname{d} x_i}{\operatorname{d}t} }}. \end{align} Inverting this relation gives: $$\frac{\operatorname{d} x_j}{\operatorname{d} t} = c \frac{p_j}{a \sqrt{p_i p_i + (mc)^2 a^2}},$$ leading to the Hamiltonian: \begin{align} H &\equiv p_i \frac{\operatorname{d} x_i}{\operatorname{d} t} - L \\ & = c \frac{p_j p_j}{a \sqrt{p_i p_i + (mc)^2 a^2}} + \frac{a m^2 c^3}{\sqrt{ p_i p_i + (mc)^2 a^2}} \\ & = c \sqrt{ \frac{p_i p_i}{a^2} + (mc)^2 }. \end{align}

Notice that in this case the canonical momentum is the quantity conserved because the Lagrangian is invariant under shifts in the comoving coordinate. Thus, for relativistic particles the comomving velocity is $\propto a^{-1}$ for relativistic particles ($p \gg mc a$), and $a^{-2}$ for non-relativistic ones ($p \ll mca$). Yes, even light has a declining comoving velocity in a FLRW coordinate system.

To translate back into physical/proper velocity, recall that $\vec{v}_p = a \frac{\operatorname{d} \vec{x}_{\mathrm{coord}}}{\operatorname{d} t}$, which means that for non-relativistic particles the physical velocity drops like $a^{-1}$, and for ultrarelativistic particles the velocity is given by $\approx c \hat{p} \left(1 - \left[\frac{mca}{p}\right]^2 + \ldots\right)$. Thus, the physical velocity of light, with $m=0$, is fixed, while for massive particles it drops.

Also, if you define the physical momentum in the usual way for special relativity, $$\vec{p}_p \equiv \frac{m \vec{v}_p}{ \sqrt{1 - \left(\frac{v_p}{c}\right)^2}},$$ then the physical momentum ends up being related to the canonical coordinate momentum as: $$\vec{p}_p = \frac{\vec{p}}{a},$$ which is the inverse of the relationship between coordinate and physical velocity, but maintains the falling physical momentum expected from the falling physical velocity.

Also, notice that angular momentum is conserved, as expected by the rotational invariance of the Lagrangian. Note how: $$\vec{L} = \vec{r}_p \times \vec{p}_p = \vec{x} \times \vec{p},$$ the scale factors in the physical momentum and position cancel, giving the same relationship in the coordinate position and canonical coordinate momentum.

You could reach the same conclusions from quantum field theory if you pay careful attention to where all of the scale factors are in the Lagrangian. For example, start from the action: $$S = \int \sqrt{-g}\operatorname{d}^4 x \left[\frac{g^{\mu\nu}}{2} \partial_\mu\phi \partial_\nu \phi - m^2 \phi^2\right],$$ with the metric signature $(+,-,-,-)$, and examine the dispersion relations for the waves, including how physical vs coordinate wave number evolves, etc. You'll find it agrees with the classical massive picture above, as it must by the correspondence principle.

I agree with your conclusion. An example will help understand this.

A rifle, anchored to earth, and pointed at you is fired. The bullet with mass m moves towards you at velocity v, with a momentum of p = mv. In this case it is clear that v is with respect to you because you and the rifle are stationary.

In the next scenario, the rifle moves back with the full speed of the bullet of the prior case (v), so the velocity of the bullet would now be 0, so the momentum of the bullet with respect to you is also 0.

This makes it clear that the momentum p is inversely proportional to the velocity with which the rifle "recedes" away from the bullet and you.

For your particular case, the "rifle" is the universe, and the momentum of any particle would be inversely proportional the scale factor because the particle's velocity is inversely proportional to the scale factor, with respect to the universe.