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Further via an argument using the equivalence principle it is also pretty straight forward to derive the relation between distance and apparent magnitude of far-distant objects, which is important for the distance-redshift measurement (the Hubble Law) and its relation to the history of the universe through the Friedmann equations.

What I'd like to understand for myself is, whether there is a simple solution close to a plane wave in Minkowski space for the general Robertson-Walker spacetime. So far I couldn't find one (neither in the literature nor by myself).

My status of understanding is as follows. You start from the Lagrangian for the electromagnetic field,

$$\mathcal{L}=-\frac{1}{4} \int \mathrm{d}^4 q \sqrt{-g} F_{\mu \nu} F^{\mu \nu},$$

where

$$F_{\mu \nu}=\mathrm{D}_{\mu} A_{\nu} - \mathrm{D}_{\nu} A_{\mu}=\partial_{\mu} A_{\nu} - \partial_{\mu} A_{\nu}.$$

The first thing one observes is that this action is conformal invariant, i.e., under the transformation

$$q^{\mu} \rightarrow q^{\mu}, \quad g_{\mu \nu} \rightarrow \lambda g_{\mu \nu}, \quad g^{\mu \nu} \rightarrow \frac{1}{\lambda} g^{\mu \nu},$$

where ##\lambda=\lambda(q)## is an arbitrary scalar field.

Now using the FLRW metric with conformal time one has

$$\mathrm{d} s^2=a^2(\tau) [\mathrm{d} \tau^2 -\mathrm{d} \chi^2 -S_K^2(\chi) (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2)]$$

with

$$S_K=\begin{cases}

\sin \chi & \text{for} \quad K=+1, \\

\chi & \text{for} \quad K=0,\\

\sinh \chi & \text{for} \quad K=-1.

\end{cases}$$

Of course you can as well use any of the other forms for the spatial coordinates, which doesn't change much with my problem, as far as I can see at the moment.

Now it's clear that the electrodynamics in these coordinates is independent on the scale parameter ##a(\tau)##, and for this purpose one can work with a fictituous spacetime with a static metric

$$ \mathrm{d} \tilde{s}^2= \mathrm{d} \tau^2 -\mathrm{d} \chi^2 -S_K^2(\chi) (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$

This implies that in these coordinates for the case of a flat universe, ##K=0##, it's all very simple, because then this is simply the Minkowski metric with ##(\chi,\vartheta,\varphi)## standard spherical coordinates for the spatial part. You can as well work with Cartesian coordinates and immediately you have the plane-wave solutions

$$A_{\mu}(x)=a_{\mu} \exp[-\mathrm{i} \omega (\tau-z)], \quad (a_{\mu}=(0,a_x,a_y,0).$$

Then, going to comoving coordinates, i.e., to the time coordinate, for which the FLRW metric becomes

$$\mathrm{d} s^2=\mathrm{d} t^2 -a^2(t) [\mathrm{d} \chi^2 + S_K^2(\chi) \mathrm{d} \vec{\Omega}^2].$$

Then one has

$$\mathrm{d} \tau=\mathrm{d} t a(t)$$

and thus from the phase for the flat-space plane-wave solution one immiately finds the standard formula for the redshift

$$\omega_e a(t_e)=\omega_o a(t_o),$$

where the subscripts stand for emission and observation, leading to the redshift

$$1+z=\frac{\omega_o}{\omega_e}=\frac{a(t_o)}{a(t_e)}$$

one also gets from the eikonal approximation also for the non-flat cases, ##K=\pm 1##.

If one tries to find an exact solution for the free Maxwell equations in these cases, it becomes amazingly complicated, even taking into account the fact that of course you can also work with the fictitious static metric with the conformal time. I thought, maybe one can use the alternative spatial coordinate ##\rho##, for which the FLRW metric reads

$$\mathrm{d} s^2 = a^2(\tau) [\mathrm{d} \tau^2 - \frac{1}{(1+K \rho^2/4)^2}(\mathrm{d} \rho^2 + \rho^2 \mathrm{d} \vec{\Omega}^2) ],$$

which you can obviously rewrite into a form which comes very close to Cartesian coordinates

$$\mathrm{d} s^2=a^2(\tau) [\mathrm{d} \tau^2 - \frac{1}{(1+K (x^2+y^2+z^2)/4)^2} \mathrm{d} \vec{x}^2].$$

Of course, for ##K=0## this gives the Minkowski metric for the static space-time useful for the free Maxwell equations, but for ##K \in \{\pm 1 \}## I don't see a simple plane-wave like exact solution for the Maxwell equation.

So, if somebody knows a paper or book, dealing with such questions, I'd be very interested.