Question about momentum in the FLRW metric I'm reading through Modern Cosmology by Dodelson and Schmidt 2nd edition, and I am wondering if anyone can comment more on the following part.
In Section 2.2, we define the Euclidean FLRW metric by $g_{\mu\nu} = \text{diag}(-1, a^{2}, a^{2}, a^{2})$ where  $a=a(t)$ is the scale factor. At the end of the subsection, the book has the following passage.

One final comment about the relation between energy and momentum for a massless
particle as expressed in Eq. (2.29). If we define
$$ p^{i} = aP^{i}, $$
we have $E^{2} = δ_{ij}p^{i}p^{j}$, so that we can identify $p$ with the physical momentum (while $P^{i}$ is the momentum defined with respect to the comoving grid). In terms of the physical momentum, the well-known relation
$$ E = p \quad\text{ where }\quad p ≡ |p| $$
continues to hold, which is why we will mostly use this version of particle momentum.

Some notes:

*

*The book is talking about massless particles here.

*The book doesn't give a definition for $P^{\alpha}$ apart from saying that it is the 4-momentum of a particle.

I see that the energy-momentum relationships of special relativity hold if we define "physical momentum" as $p^{i} = aP^{i}$, but these energy-momentum relationships that are mentioned should hold for SR, not necessarily for GR. Is there a stronger argument for why this should be considered "physical momentum?"
 A: You can define the word "momentum" to be anything you want. The question is what useful property motivates your definition in a particular context.
In general in GR, for a massless particle, one defines the momentum via
\begin{equation}
P^\mu = \frac{dx^\mu}{d\lambda}
\end{equation}
where $\lambda$ is an affine parameter of the null geodesic that the photon follows. We identify $P^0$ as the energy and $P^i$ as the 3-momentum. Because massless particles follow null geodesics, the momentum obeys the relationship
\begin{equation}
g_{\mu\nu} P^\mu P^\nu=0
\end{equation}
We can now see what this formula implies in cosmology. Assuming an FLRW metric, one finds that for a massless particle,
\begin{equation}
P^0 = E = a |\vec{P}|
\end{equation}
Now, since we build our intuition from flat space, it is useful to define $\vec{p} = a \vec{P}$. If we make this definition, we find that
\begin{equation}
E = |\vec{p}|
\end{equation}
and therefore we can apply our intuition from flat space to the variables $E$ and $\vec{p}$. If we ever need to return to the general GR definition of momentum, we can always rescale $\vec{p}$ by $1/a$ to get back to $\vec{P}$.
There's no "deeper" explanation than this. It is simply a useful choice of variables, motivated by the fact that it lets us use our flat space intuition. You can describe physics in any variables you want to so you are completely free to make any definitions you want so long as you use them consistently; however a good choice of variables can let you make connections with other areas or make the math easier or otherwise make what is going on more transparent.
