Momentum of a moving object in FRW metric according to an observer comoving with cosmic expansion

I would like to show that in an FRW metric the momentum of a freely falling object decays as the inverse of the scale factor. I know there are many proofs and arguments for this but I am trying to get this starting from geodesics and having some trouble.

General logic

My general approach is the following.

The object is moving along some path. I want to find the momentum of the object that would be measured by a coincident, locally inertial observer, i.e. a detector that is freely falling and locally cannot tell that spacetime is curved.

In the locally inertial frame where the observer is momentarily at rest the 4-velocity of the observer has components $$u^\mu\equiv d\xi^\mu/d\sigma= (1,0,0,0)$$, where $$\xi^\mu$$ are the coordinates of this local inertial (cartesian) frame and $$\sigma$$ is the observer's proper time (in this frame $$d\sigma=d\xi^0$$). In this same frame the object has 4-momentum $$p^\mu \equiv md\xi^\mu/d\tau$$, where $$\tau$$ is the object's proper time, and $$m$$ the rest mass.

Since special relativity holds in this frame the components of $$p^\mu$$ have the values $$p^\mu=(E_\mathrm{obs},\mathbf{p}_\mathrm{obs})$$, the energy and momentum that would be measured by the observer at rest in this frame. They obey $$E_\mathrm{obs}^2-\mathbf{p}_\mathrm{obs}^2=m^2$$. We can retrieve the energy $$E$$ by contracting $$u^\mu$$ with $$p^\mu$$:

$$E_\mathrm{obs}=g_{\mu\nu}u^\mu p^\nu,$$ where $$g_{\mu\nu}=\text{diag}(1,-1,-1,-1)$$ are the components of the metric in this locally inertial coordinate system.

The magnitude of the 3-momentum in this system is just

$$|\mathbf{p}_\mathrm{obs}| = \sqrt{(g_{\mu\nu}u^\mu p^\nu)^2 - m^2}.$$

The right hand side is invariant under general coordinate transformations. Therefore, I think we can calculate the RHS in any coordinate system we want and the value will be the magnitude of the 3-momentum as measured by the observer corresponding to the path $$u^\mu$$ in a locally inertial frame.

Is this ok so far?

FRW metric

$$x^\mu = (t, \mathbf{x})$$ are the coordinates of the FRW metric, defined by:

$$d\tau^2 = dt^2 - a(t)^2 \left(d\mathbf{x}^2 + K \frac{(\mathbf{x}\cdot d\mathbf{x})^2}{1-K\mathbf{x}^2} \right),$$ where $$K =0$$, $$+1$$, or $$−1$$.

The geodesic paths can be found using the Euler-Lagrange equations (see this post). The resulting equations are

\begin{align} 0 &= \frac{d^2 t}{d\tau^2} + a\dot{a} \left(\mathbf{x}'^2 +\frac{K(\mathbf{x}\cdot \mathbf{x}')^2}{1-K \mathbf{x}^2}\right)\\ \mathbf{0} &= \frac{d}{d\tau} \left[ a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right) \right] - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} a^2 \left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right), \end{align} where a prime means $$d/d\tau$$ and $$\dot{a}=da/dt$$ is a function of $$t$$.

The simple solutions to these equations are $$t=\tau, \mathbf{x}=\text{const}$$. These solutions correspond to the "comoving" observers that move along with the cosmic expansion. The world lines of these observers all have 4-velocity $$u^\mu=(1,0,0,0)$$ in this coordinate system. When the object is at some location $$\mathbf{x}$$ at time $$t$$ I want to get the momentum as measured in a locally inertial frame in which the comoving observer that sits at position $$\mathbf{x}$$ is momentarily at rest.

To get the equations of motion for an arbitrary freely falling object you have to integrate the above equations. I was able to do this only for the second equation but I think that's enough.

Define $$\mathbf{f}$$ by $$\mathbf{f} \equiv a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right),$$ to write the second equation as $$\mathbf{0} = \frac{d \mathbf{f}}{d\tau} - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} \mathbf{f}.$$ Multiply through by the integrating factor $$\sqrt{1-K\mathbf{x}^2}$$ to get

\begin{align} \mathbf{0} &= \sqrt{1-K\mathbf{x}^2}\frac{d \mathbf{f}}{d\tau} - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{\sqrt{1-K\mathbf{x}^2}} \mathbf{f} \\ &= \sqrt{1-K\mathbf{x}^2}\frac{d \mathbf{f}}{d\tau} + \mathbf{f} \frac{d}{d\tau}\sqrt{1-K\mathbf{x}^2} \\ &= \frac{d}{d\tau}\left(\mathbf{f} \sqrt{1-K\mathbf{x}^2}\right). \end{align}

The solution is \begin{align} \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} = a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right), \end{align} for some constant 3-vector $$\mathbf{c}$$.

The term $$(\mathbf{x}\cdot\mathbf{x}')$$ can be isolated by dotting this equation with $$\mathbf{x}$$:

\begin{align} \frac{\mathbf{c}\cdot\mathbf{x}}{\sqrt{1-K\mathbf{x}^2}} &= a^2\left((\mathbf{x} \cdot \mathbf{x}') + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}^2}{1-K\mathbf{x}^2}\right)\\ &= a^2 (\mathbf{x} \cdot \mathbf{x}') \left(1+ \frac{K\mathbf{x}^2}{1-K\mathbf{x}^2}\right)\\ &= a^2 \frac{\mathbf{x} \cdot \mathbf{x}'}{1-K\mathbf{x}^2}. \end{align}

Now $$\mathbf{x}'$$ can be written in terms of $$\mathbf{x}$$: \begin{align} a^2 \mathbf{x}' &= \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} - a^2 \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\\ &= \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} - \frac{K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{\sqrt{1-K\mathbf{x}^2}}\\ &= \frac{\mathbf{c} - K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{\sqrt{1-K\mathbf{x}^2}}. \end{align}

Finally the spatial part of the object's 4-momentum is \begin{align} p^i = m\frac{d\mathbf{x}}{d\tau} = m \frac{\mathbf{c} - K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{a^2\sqrt{1-K\mathbf{x}^2}}. \end{align}

Now to try to use the $$|\mathbf{p}_\mathrm{obs}| = \sqrt{(g_{\mu\nu}u^\mu p^\nu)^2 - m^2}$$ formula I had earlier.

Since $$u^\mu=(1,0,0,0)$$, $$g_{00}=1$$, and $$g_{0i}=0$$ at $$\mathbf{x}$$, $$|\mathbf{p}_\mathrm{obs}| = \sqrt{(p^0)^2 - m^2}$$.

Writing out $$g_{\mu\nu}p^\mu p^\nu = m^2$$ gives \begin{align} m^2 &= (p^0)^2 - m^2 a^2\left(\mathbf{x}'^2 + K \frac{(\mathbf{x}\cdot \mathbf{x}')^2}{1-K\mathbf{x}^2} \right) \\ &= (p^0)^2 - m^2 (\mathbf{f} \cdot \mathbf{x}') \\ &= (p^0)^2 - m^2 \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} \cdot \frac{\mathbf{c} - K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{a^2\sqrt{1-K\mathbf{x}^2}} \\ & = (p^0)^2 - m^2 \frac{\mathbf{c}^2 - K(\mathbf{c}\cdot\mathbf{x})^2}{a^2 (1-K\mathbf{x}^2)}, \end{align}

and I get

$$|\mathbf{p}_\mathrm{obs}| = \sqrt{(p^0)^2 - m^2} = \frac{m}{a}\sqrt{\frac{\mathbf{c}^2 - K(\mathbf{c}\cdot\mathbf{x})^2}{1-K\mathbf{x}^2}}.$$

But I don't think this is proportional to $$1/a$$ in general.

If the object passes through the origin then everything works as I expect. In this case $$\mathbf{c}$$ is parallel to $$\mathbf{x}'$$ when $$\mathbf{x}=0$$ and the equation of motion keeps it moving along a straight line parallel to this initial $$\mathbf{x}'$$ so $$\mathbf{x}$$ remains parallel to $$\mathbf{c}$$ and $$(\mathbf{c} \cdot \mathbf{x})^2 = \mathbf{c}^2\mathbf{x}^2$$ and the term in the square root is just a constant $$|\mathbf{c}|$$.

But for an object flying somewhere else, not passing through the origin I think the angle between $$\mathbf{c}$$ and $$\mathbf{x}$$ will change as it goes (and in particular it will not always be $$0$$) and so the momentum observed by a locally inertial observer will not simply scale as $$1/a$$.

I guess maybe the angle between $$\mathbf{c}$$ and $$\mathbf{x}$$ changes as $$\mathbf{x}^2$$ changes in just such a way as to keep the term in the square root constant but I don't see how to show that.

Or do I have the entire logic of the exercise wrong?

It turns out that last factor in the square root is actually a constant -- so everything works nicely and momentum decays as $$1/a$$ for any object moving on a geodesic as it's supposed to.
I took the derivative of the term in the square root: $$\frac{\mathbf{c}^2-K(\mathbf{c}\cdot \mathbf{x})^2}{1-K\mathbf{x}^2} \equiv A.$$
This introduces the terms $$\mathbf{c}\cdot \mathbf{x}'$$ and $$\mathbf{x}\cdot \mathbf{x}'$$, which we can get rid of using some of the formulas in the original post:
\begin{align} \mathbf{c}\cdot \mathbf{x}' &= \frac{\mathbf{c}^2-K(\mathbf{c}\cdot \mathbf{x})^2}{a^2\sqrt{1-K\mathbf{x}^2}} = \frac{\sqrt{1-K\mathbf{x}^2}}{a^2}A \\ \mathbf{x}\cdot \mathbf{x}' &= \frac{1}{a^2}\sqrt{1-K\mathbf{x}^2}(\mathbf{c}\cdot \mathbf{x}) \end{align}
\begin{align} \frac{dA}{d\tau} &=\frac{-2K (\mathbf{c}\cdot \mathbf{x})(\mathbf{c}\cdot \mathbf{x}')}{1-K\mathbf{x}^2} + \frac{\left(\mathbf{c}^2-K(\mathbf{c}\cdot \mathbf{x})^2\right)2K(\mathbf{x}\cdot \mathbf{x}')}{(1-K\mathbf{x}^2)^2}\\ &= \frac{-2K (\mathbf{c}\cdot \mathbf{x})}{1-K\mathbf{x}^2}\frac{\sqrt{1-K\mathbf{x}^2}}{a^2}A + \frac{A}{1-K\mathbf{x}^2} \frac{2K}{a^2}\sqrt{1-K\mathbf{x}^2}(\mathbf{c}\cdot \mathbf{x})\\ &= 0. \end{align}
So $$|\mathbf{p}_\mathrm{obs}| = \frac{m}{a}\sqrt{A}= \frac{\text{const}}{a}.$$