momentum of particles in FLRW universe

The expression for the line element in a FLRW universe is
$$ds^2 = c^2dt^2 - a^2(t)[\frac{dr^2}{(1-\kappa r^2)} + r^2(d\theta^2+\sin^2\theta d\phi^2)]$$

The geodesic equation, used to describe the kinematics of particles or photons in the manifold is
$$\frac{d^2x^{\mu}}{ds^2} + \Gamma^{\mu}_{\alpha\beta} u^{\alpha}u^{\beta} = 0$$

Let’s look at $\frac{dt}{ds}$ specifically,

$$\frac{d^2t}{ds^2} + \Gamma^{t}_{\alpha\beta} u^{\alpha}u^{\beta} = 0$$
say $u^{\theta}$ and $u^{\phi}$ are zero

And if you compute all of the relevant christoffel symbols i.e $\Gamma$s, you will see that $\Gamma^{t}_{rr}$ is the only non-zero component.
$$\frac{d^2t}{ds^2} + \Gamma^{t}_{rr} {u^{r}}^2 = 0$$
and since we know that $u^r = \frac{dr}{ds}$
$$\frac{d^2t}{ds^2} + \Gamma^{t}_{rr} {\frac{dr}{ds}}^2 = 0$$
If we substitute the value of $\Gamma^t_rr$, we get
$$\frac{d^2t}{ds^2} + \frac{a \dot{a}}{(1-\kappa r^2)} \frac{dr}{ds}^2 = 0$$

Now, we also know that $u_{\mu}u^{\mu} = 1$ for particles.

$$u_{t}u^{t} +u_{r}u^{r} +u_{\theta}u^{\theta} +u_{\phi}u^{\phi} = 1$$
$$u_{t}u^{t} +u_{r}u^{r} = 1$$
$$g_{tt}{u^{t}}^2 +g_{rr}{u^{r}}^2 = 1$$
$${u^{t}}^2 -\frac{a^2(t)}{1-\kappa r^2}{u^{r}}^2 = 1$$
$${\frac{dt}{ds}}^2 -\frac{a^2(t)}{1-\kappa r^2}{\frac{dr}{ds}}^2 = 1$$
$$[{{\frac{dt}{ds}}^2 -1}]\frac{1-\kappa r^2}{a^2(t)} = {\frac{dr}{ds}}^2$$
Substituting the above expression in the expression below, we get
$$\frac{d^2t}{ds^2} + \frac{a \dot{a}}{(1-\kappa r^2)} \frac{dr}{ds}^2 = 0$$
$$\frac{d^2t}{ds^2} + \frac{\dot{a}}{a}[{{\frac{dt}{ds}}^2 -1}] = 0$$
say $$V = \frac{dt}{ds}$$
$$\frac{dV}{ds} + \frac{\dot{a}}{a}[{V^2 -1}] = 0$$
$$\frac{dV}{ds} + \frac{1}{a}\frac{da}{dt}\frac{ds}{ds}[{V^2 -1}] = 0$$
$$\frac{dV}{ds} + \frac{1}{a}\frac{da}{ds}\frac{ds}{dt}[{V^2 -1}] = 0$$
$$\frac{dV}{ds} + \frac{1}{a}\frac{da}{ds}\frac{1}{V}[{V^2 -1}] = 0$$
$$\frac{dV}{ds} = - \frac{1}{a}\frac{da}{ds}\frac{1}{V}[{V^2 -1}]$$
$$\frac{VdV}{(V^2 -1)} = - \frac{da}{a}$$
$$\frac{1}{2}ln(V^2 -1) = - ln(a) + c$$

$$(V^2 -1)^{\frac{1}{2}} \propto \frac{1}{a}$$
$$(\frac{dt}{ds}^2 -1) \propto \frac{1}{a^2}$$

we wrote earlier that $u^{\mu}u_{\mu} = 1$
$$u^{t}u_{t} -u^{i}u_{i} = 1$$
where $i$ represents the spatial components.
$$u^{t}u_{t} -1 =u^{i}u_{i}$$
$$\frac{dt}{ds}^2 -1 =u^{i}u_{i}$$

$$u^{i}u_{i} \propto \frac{1}{a^2}$$
$$|p| \propto \frac{1}{a}$$
where $|p|^2 = \sigma_{ij}p^i p^j$.

Note that we are working in units of $c=1$. If I’m missing factors anywhere, I apologize.

We basically get that the spatial momentum of any free particle in an FLRW universe is inversely proportional to the scale factor $a(t)$ and since $a(t)$ is always positive and increasing, the momentum keeps decreasing with time.

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