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Given an action of the form

\begin{equation}S=-\frac{1}{4}\int d^4x\eta^{\mu\nu}\eta^{\lambda\rho}F_{\mu\lambda}F_{\nu\rho}\end{equation}

where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$, $\eta_{\mu\nu}=g_{\mu\nu}/a^2(\eta)$, where $g_{\mu\nu}$ is given by the line element:


I would like to solve for $A_{\mu}$, and standard solution is

\begin{equation}A_{\mu}^{(\alpha)}=e_{\mu}^{(\alpha)}e^{ik_\nu x^\nu}.\end{equation}

I am interested in knowing how to derive this result.

My approach is first write the Lagrangian from action and use EL eq

\begin{equation}\frac{\partial \mathcal{L}}{\partial A_{\mu}}-\frac{d}{d x^{\nu}}\frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A_{\mu})}=0\end{equation}

My main problem is mathematical difficulty in evaluating the EL eq. Can anyone please help me on this?

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Are you working with QFT in curved spacetime? In that case, the volume element is $d^4x\sqrt{-g}$ where $g$ is the determinant of the metric (this is 1 if the metric is just Minkowski). –  Danu Jun 11 '14 at 14:27

1 Answer 1

The action for an electromagnetic field in curved space is given by,

$$S=-\frac{1}{4}\int \mathrm{d}^4 x \, \sqrt{|g|} \, F_{\mu\nu}g^{\mu\lambda}g^{\sigma \nu}F_{\lambda \sigma}$$

for a generic metric, $g_{\mu\nu}$ - notice the correct volume element is with $\sqrt{|g|}$. The equations of motion or equivalently Euler-Lagrange equations are,

$$\partial_\nu \left( \sqrt{|g|}F^{\mu\nu}\right)=0$$

in vacuum, where we have chosen to hide the additional factors of the metric by raising the index of the field-strength tensor. In your question, your solution is a plane wave, for $g_{\mu\nu}=\eta_{\mu\nu}$. If you wanted to work in the spacetime background you provided,

$$\mathrm{d}s^2 = a(t)^2 \left[ \mathrm{d}t^2-\mathrm{d}x^2-\mathrm{d}y^2-\mathrm{d}z^2\right]$$

you must raise tensors with that metric, and include the volume factor. In your case the action becomes,

$$S = -\frac{1}{4}\int \mathrm{d}^4x \, \, a(t)^4 \, F_{\mu\nu}g^{\mu\lambda}g^{\sigma \nu}F_{\lambda \sigma}$$

$$\partial_\nu \left( a(t)^4 F^{\mu\nu}\right)=0 \quad \implies \partial_i F^{\mu i}=-\left(\partial_0+\frac{4\dot{a}(t)}{a(t)} \right)F^{\mu 0}$$

where $F^{\mu\nu}$ is raised with your curved metric.

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But, this does not really solve the problem. And, I am not sure if the EL equation you wrote is correct. –  titanium Jun 12 '14 at 4:33
@titanium: The equations of motion I wrote are correct; the result is trivial and can be found in any textbook on field theory in curved space. Now, regarding the solution you have - it's just a plane wave, it actually solves the equations in flat Minkowski space, so just plug it in and check. –  JamalS Jun 12 '14 at 5:36
@titanium: If you prefer, you can write them as $\nabla_{\mu}F^{\mu\nu}=0$, with a covariant derivative, and supplemented by the Bianchi identiy, but it's essentially the same thing I wrote. –  JamalS Jun 12 '14 at 5:41

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