# Electrodynamics in Robertson Walker spacetime

1. Jun 20, 2015

### vanhees71

I've a (perhaps somewhat stupid) question. Is there a good source, where one finds the plane-wave solutions (or what comes closest to it) for electrodynamics for the closed and open (non-flat) Friedmann-Lemaitre-Robertson-Walker metric. I try this myself for a while, because to my surprise I couldn't find this anywhere in the literature. The background is that I don't like the usual "derivation" for the redshift formula in terms of photons, because I don't think that photons are easy to describe for a general Robertson-Walker space-time. The best I could come up with, and which is what I try to explain my students in the next recitation session for the cosmology lecture of my boss, is the standard treatment of the redshift in terms of the leading-order eikonal solution for the free Maxwell equations in covariant Lorenz gauge, which leads to the same formula as the naive argument with the "photon", which is clear, because the eikonal approach is the way to go from wave to ray optics and thus a naive photon picture.

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 spacial 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.

2. Jun 20, 2015

### atyy

3. Jun 20, 2015

### vanhees71

Thanks for the quick reply. I know the second book, but this deals with exact solutions for Einstein's field equations of gravitation, what I look for is an exact solution of the Maxwell equations in a FLRW-background space time. I'll have a look whether I can find something about this in these books.

4. Jun 20, 2015

### WannabeNewton

5. Jun 20, 2015

### vanhees71

Thanks a lot. I'll have a careful look at these papers.

6. Jun 20, 2015

### George Jones

Staff Emeritus
I don't have any answers, but I do have some questions.

I don't quite understand the above.

Is this actually a solution to Einstein's equation, or do we assume that the plane wave rides on top of Minkowski spacetime "for free", i.e., that we don't take account of the electromagnetic wave's energy-momentum tensor in Einstein's equation?

Similarly, are you searching for:

1) "plane waves" (for comoving observers) that don't contribute to the stress-energy tensor;

2) an electrovac solution to Einstein's equation that has the same symmetries as an FLRW spacetime (same Killing vectors), and that probably doesn't exist;

3) an electrovac solution to Einstein's equation that is a "slight" perturbation, due to the electromagnetic wave's energy-momentum tensor, and "looks like" plane waves for comoving observers?

I think that you want 1), but I am not quite sure what this means. Do you want a plane wave in coordinates more tied to what comoving observers see, like Riemann- or Fermi normal coordinates?

The literature probably concentrates on 3, e.g., paper that WannabeNewton referenced writes:

7. Jun 20, 2015

### vanhees71

I think, I didn't make my problem clear enough. It's much simpler than what you suggest: I simply look for the solution of the Maxwell equations in a given (fixed) FLRW spacetime. I think, I've found another paper by Googling which seems to provide right this, making use of the Debye potentials:

J. M. Cohen, L .S. Kegeles, Electromagnetic fields in curved spaces: A constructive procedure, PRD 10, 1070 (1974)
http://dx.doi.org/10.1103/PhysRevD.10.1070

8. Jun 20, 2015

### bcrowell

Staff Emeritus
Could you explain more what you think the issue is? It seems like a non-issue to me.

9. Jun 20, 2015

### vanhees71

Well, quantum field theory in curved space time is a pretty delicate subject, and I think it's overkill, because the redshift formula $\omega a=\text{const}$ is easily derived through the eikonal approximation of the em. field in curved spacetime,
$$g^{\mu \nu} \partial_{\mu} \psi \partial_{\nu} \psi=0.$$
For the FLRW metric with the conformal time for radial rays this reads
$$(\partial_{\tau} \psi)^2-(\partial_{\chi} \psi)^2=0.$$
For a ray moving from a distant source radially to the origin, where the observer is located, this means
$$\partial_{\tau} \psi=-\partial_{\chi} \psi.$$
Since in the eikonal equation, the overall scale factor of the metric $a(\tau)$ can be divided out, leaving an equation which is independent of $\tau$ and thus the canonical momentum $\omega=\partial_{\tau} \psi=\omega(\chi)$. This implies that
$$\psi=\omega(\chi) \tau+\psi_2(\chi),$$
The eikonal equation thus gives
$$\partial_{\chi} \psi=\omega'(\chi) \tau+\psi_2'(\chi)=-\partial_{\tau} \chi=\omega(\chi).$$
Since the right-hand side is independent of $\tau$ we find $\omega'=0$ and thus $\omega=\text{const}$. This gives finally
$$\psi=\omega(\tau-\chi)+\text{const}.$$
So in leading-order eikonal approximation you get
$$A_{\mu}(t,\chi)=a_{\mu} \exp[-\mathrm{i} \omega (\tau-\chi)].$$
On the other hand
$$\mathrm{d} \tau=\frac{1}{a(t)} \mathrm{d} t.$$
Thus in comoving coordinates
$$\tilde{\omega}=\frac{\partial \psi}{\partial t} = \frac{\partial \psi}{\partial \tau} \frac{\mathrm{d} \tau}{\mathrm{d} t}=\frac{\omega}{a(t)}$$
or
$$\tilde{\omega} a(t)=\text{const}.$$
Thus if light is emitted at $\chi=\chi_e$ at time $t_e$ with frequency $\tilde{\omega}_e$ the observer at $\chi=0$ observes the light at a later time $t_o$ with another frequency $\omega_o$. According to the above derivation we have
$$\tilde{\omega}_e a(t_e)=\tilde{\omega}_o a(t_0) \; \Rightarrow \; \tilde{\omega}_o=\tilde{\omega}_e \frac{a(t_e)}{a(t_o)}.$$
Since the universe is expanding this makes $\omega_o<\omega_e$, i.e., it implies a redshift of spectral lines of the light emitted from a distant objects, the Hubble redshift.

The above derivation shows that the same holds true for the wave number $k=2 \pi/\lambda=\omega$ and thus the wave number is also red-shifted in comoving coordinates, i.e., a fundamental observer finds
$$\lambda_o=\lambda_e \frac{a(t_o)}{a(t_e)}=:(1+z) \lambda_e.$$

For a given $\chi$ the times $t_o$ and $t_e$ can be calculated from radial null-geodesics, which describes the propagation of the surfaces of constant phase of the light wave. Since for the co-moving coordinates the geodesics are simply given by $\mathrm{d} s^2=0$ (as can be derived from the geodesic equation, using the FLRW metric in the form with the time coordinate $t$), i.e.,
$$\mathrm{d} t=-a(t) \mathrm{d} \chi \; \Rightarrow \chi=\int_{t_e}^{t_o} \mathrm{d} t \frac{1}{a(t)}.$$
Now all this uses the eikonal approximation for the electromagnetic waves. So I've asked myself, in how far one can derive deviations from this behavior for the redshift in the case of $K \neq 0$. For our universe these seems not to be very relevant since due to the standard model of cosmology the universe (large-scale coarse-grained) seems to be described by a flat FLRW metric with $K=0$, where the electromagnetic field obeys the flat-space Maxwell equations in conformal coordinates, and thus plane waves $A_{\mu} \propto \exp(-\mathrm{i} \omega (\tau-\chi))$ are indeed exact solutions and thus the above considerations are valid for the strict Maxwell equations.

Last edited: Jun 20, 2015
10. Jun 20, 2015

### bcrowell

Staff Emeritus
Is this in reply to my #8? You don't need to use quantum mechanics. You can just describe a classical wave packet.

11. Jun 20, 2015

### WannabeNewton

As far as the redshift goes I don't quite understand the issue either. The Eikonal approximation is valid when $K\neq 0$ as well, so long as you consider modes whose wavelength is much smaller than the radius of curvature of the space-time, a condition which must be met in the $K = 0$ case as well.

The redshift formula $\omega a = \text{const.}$ is valid in both the $K = 0$ and $K \neq 0$ cases, a fact which is extremely easy to derive if one uses Killing vectors. C.f. Wald pp. 103-104.

12. Jun 21, 2015

### vanhees71

Sure, in the eikonal approximation it's very straight forward, as shown in my previous posting. I just thought, it would be as easy by constructing an exact solution of the Maxwell equations in a FLRW spacetime which corresponds to a plain wave in flat space time. For $K=0$ it's simple, because there the electrodynamics is as in flat Minkowski space due to the conformal symmetry of the free Maxwell equations and with the conformal time, in this case the FLRW metric is conformally flat as also shown above. For the non-flat cases, it's not so easy.

Yesterday night I read the above quoted article
J. M. Cohen, L .S. Kegeles, Electromagnetic fields in curved spaces: A constructive procedure, PRD 10, 1070 (1974)
http://dx.doi.org/10.1103/PhysRevD.10.1070

The trick is to work with the gauge invariant fields and to use Debye potentials. For the FLRW metric they obey simple separable wave equations, whith the conformal time entering in the form $\partial_{\tau}^2 \chi$, and thus you have in all three cases of the curvature multipole expansions for the free electromagnetic field $\propto \exp(-\mathrm{i} \omega \tau)$. This is exact, and thus the red-shift formula established. Of course, I've not fully understood the details yet, because unfortunately my versatility with the Cartan calculus is a bit rusty, but that seems to be the most elegant way to understand what's going on.

On Friday, I've figured out another way, based on the very nice treatment of electrodynamics in curved space-time in Landau-Lifshitz vol. 2. There, everything is formulated in the 1+3-dimensional formalism, making the entire set of equations looking like the Maxwell equations in three-dimensional form. There you establish also quite easily that for the FLRW metric using the conformal time as a coordinate, the electric and magnetic field components all obey wave equations with $\partial_{\tau}^2 \vec{E}$ + purely spacial derivatives. From this it's also immideately clear that the usual redshift formula is exact. Perhaps one can work out the complete solution using the Debye potentials for $\vec{B}$ in a much more naive way than in the above cited paper, analogous to this technique in Minkowski space.

The derivation in Wald also uses the eikonal approximation. The trick with the Killing vectors is of course a very appealing approach.

Last edited: Jun 21, 2015
13. Jun 21, 2015

### ChrisVer

I wonder... can't you derive for a general metric (so for an FRW metric too) the Maxwell equations for the EM wave?

14. Jun 21, 2015

### hunt_mat

Here may be a naive suggestion (since my GR is very rusty), you know the condition to derive the FRW metric right? So why not just apply those conditions with the electromagnetism tensor as part of Einstein's equations?

15. Jun 23, 2015

### vanhees71

Sure. That's no big problem. You just use the generally covariant action (for the free Maxwell field,
$$A=-\frac{1}{4} \int \mathrm{d}^4 q \sqrt{-g} F_{\mu \nu} F^{\mu \nu}, \quad F_{\mu \nu}=\nabla_{\mu} A_{\nu}-\nabla_{\nu} A_{\mu}.$$
$$\nabla_{\mu} F^{\mu \nu}=0, \quad \epsilon^{\mu \nu \rho \sigma} \nabla_{\nu} F_{\rho \sigma}=0,$$
where the latter (the inhomogeneous Maxwell equations) is identically fulfilled through the ansatz via the vector potential $A_{\mu}$.

The trouble is that even in a so symmetric case as the FLRW metric, it's not trivial to find an ansatz for the potentials, where the covariant D'Alembertian applied to the vector field yields separate equations as in flat space. The trick is to introduce an appropriate generalization of the Debye potentials as shown in the cited paper.

I'm not yet very far with understanding all the details, but the key issue seems to be that there exists a generalized form of the angular-momentum operator (in flat 3-space $\vec{r} \times \vec{\nabla}$), which should exist, because the FLRW metric is isotropic and thus admits a rotation group. Since this is due to the symmetries, this must be related to the Killing vectors of the space-time. So there must be a simpler way to see this for the FLRW (and perhaps also the Schwarzschild) metric, but in the paper they show that it even works for the Kerr metric, which is less symmetric (?).

16. Jun 28, 2015

### TrickyDicky

As already pointed out by several posters there is no problem for the local approximation for any case, now you seem to be asking for a global plane wave solution and that is not possible for any of those spacetimes including FRW with K=0 and Minkowski spacetime, unlike the case in Euclidean space(where it is not really useful physically anyway since being infinite spatially and for all t they preclude propagation and are not found in nature, only superposition approximations to it are observed).

17. Jun 28, 2015

### vanhees71

Is the following somehow wrong, because there I indeed find a kind of plane-wave solution valid for all FLRW metrics!

The covariant Maxwell equations look as follows
$$F_{\mu \nu} = \nabla_{\mu} A_{\nu}-\nabla_{\nu} A_{\mu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
and (in Heaviside-Lorentz units)
$$\nabla_{\mu} F^{\mu \nu}=j^{\nu}.$$
This can be rewritten, using the commutator of the covariant derivative
$$\nabla_{\mu} \nabla_{\nu} A^{\mu}=\nabla_{\nu} \nabla_{\mu} A^{\mu} -R_{\mu \nu} A^{\nu},$$
using the sign conventions (from [Adler])
$$\nabla_{\mu} \nabla_{\nu} A_{\rho} =-R_{\rho \sigma \mu \nu} A^{\nu}, \quad R_{\sigma \nu}={R^{\rho}}_{\sigma \rho \nu}.$$
Thus we have in Lorenz gauge
$$\nabla_{\mu} A^{\mu}=\frac{1}{\sqrt{-g}} \partial_{\mu} (\sqrt{-g} A^{\mu})=0$$
the generalized wave equation
$$\nabla_{\mu} \nabla^{\mu} A^{\nu}+{R^{\nu}}_{\mu}A^{\mu}=j^{\nu}.$$
The FLRW metric in conformal coordinates reads
$$\mathrm{d} s^2=a^2(\tau) [\mathrm{d} \tau^2 - \mathrm{d} \chi^2 - S_K^2(\chi)(\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi)].$$
You have
$$S_K(\chi)=\begin{cases} \sinh \chi &\text{for} \quad K=-1,\\ \chi & \text{for} \quad K=0,\\ \sin \chi & \text{for} \quad K=+1. \end{cases}$$
For the case $K=0$ the square bracket is the Minkowski metric. This means that, because of the conformal invariance of Maxwell equations all solutions for $A_{\mu}$ can be taken from Minkowski space, among them the plane-wave solution
$$A_{\mu}=a_{\mu} \exp(-\mathrm{i} [\tilde{\omega} (\tau-\chi)], \quad \text{with} \quad a_{\mu}=(0,0,0,A_0), \quad A_0=\text{const}, \quad \tilde{\omega}=\text{const}.$$
The Mathematica Notebook attached to this posting shows explicitly that this is indeed a solution of the Maxwell equations in all FLRW metrics (I don't know, how to upload a mathematica notebook; so I uploaded a pdf-printout).

Of course, in the usual fundamental coordinates you have the time coordinate $t$ instead of the "conformal time" $\tau$, which are related by
$$\mathrm{d} \tau a(\tau)=\mathrm{d} t \; \Leftrightarrow \; \mathrm{d} \tau=\frac{\mathrm{d} t}{a(t)},$$
where $a(t)$ is the usual scale factor in fundamental coordinates in cosmology. This implies the Hubble-redshift formula. In conformal coordinates (in the following written with a twidle over the vector/tensor symbols) you have
$$\tilde{k}_{\mu}=\tilde{\omega}(1,1,0,0) \; \Rightarrow \; k_{\mu} = \frac{\partial \tilde{q}^{\nu}}{\partial q^{\mu}} \tilde{k}_{\nu}=(\tilde{\omega}/a(t),\omega,0,0).$$
For a fundamental observer (at rest wrt. the fundamental reference frame) you have the four-velocity $u^{\mu}=(1,0,0,0)$ and thus he measures the frequency
$$\omega=\tilde{\omega}/a(t_o)$$
Thus if the light is emitted from a light source at rest wrt. the fundamental coordinates from a radially symmetric surface at $\chi=\chi_s$ you get
$$\omega_e=\frac{\tilde{\omega}}{a(t_e)}$$
and for an observer located at $\chi_o$
$$\omega_o=\frac{\tilde{\omega}}{a(t_o)}.$$
So you get for the red shift $z$
$$1+z=\frac{\omega_e}{\omega_o}=\frac{a(t_o)}{a(t_e)}.$$
The time of observation and emission are related by the coordinate distance $\chi_o-\chi_e \simeq \chi_o$ of source and observer by the fact that the surfaces of constant phase move along the light-like radial geodesics, given in conformal coordinates by $\chi=\tau+\text{const}$. In fundamental coordinates you simply get this by solving
$$\mathrm{d} s^2=\mathrm{d} t^2-a^2(t) \mathrm{d} \chi^2 \; \Rightarrow \; \chi_o-\chi_e=\int_{t_e}^{t_o} \mathrm{d} t \frac{1}{a(t)}.$$

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18. Jun 29, 2015

### TrickyDicky

If the solution is valid for Minkowski spacetime and FLRW spacetimes are locally minkowskian, it appears to follow that at least locally the solution is valid in FLRW(except obviously at the BB singularity that is anyway generally considered to be outside the manifold) regardless of K, no?

But my point was that the solution in Minkowski spacetime is valid only for spacelike hypersurface slices on the past light cone(for retarded solutions), being singular at the observer's present t=0 hypersurface(no absolute simultaneity), unlike the Euclidean plane waves where the solution is valid from t=-infinity to t=+infinity (absolute simultaneity). It seems odd to call wave to something that finds such mathematical obstacles to propagate from past to future. At least in the FLRW cosmologies the preferred slicing makes thing easier wrt this issue.

19. Jun 30, 2015

### vanhees71

Sure, it's of course not a plane but a kind of spherical wave.

Of course this most simple exact solution is not what you want to describe the radiation from a distant "pointlike" source. This becomes clear that in Minkowski space this solution, written in terms of the usual normalized spherical coordinates is
$$\vec{E}=\frac{E_0}{4 \pi r \sin \vartheta} \vec{e}_{\varphi} \cos[\omega(t-r)].$$
It's singular not only at the origin $r=0$ but along the entire $z$ axis $\vartheta=0,\pi$.

$$\vec{E} \simeq \frac{E_0}{4 \pi r} \vec{e}_{\vartheta} \cos[\omega(t-r)].$$
Of course the fields are singular for $r \rightarrow 0$, but there's of course the source, and the free-field solutions are not valid inside the source. The radiation comes from a surface at $r>0$. The eikonal solution as the far-field approximation of an exact solution.