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We have an action of the form:

$$S=\int d^4x\sqrt{-g}\left(\frac{R}{2\kappa^2}+\frac14F_{\mu\nu}F^{\mu\nu}+\frac12m^2A_{\mu}A^{\mu}\right).$$

Here $R$ is the curvature scalar, $A_{\mu}$ is a vector field, $F^{\mu\nu}$ is the Faraday tensor and ($\kappa,m$) are constants. Using the variational method I varied the metric in order to obtain the field equations and got this:

$$\frac{1}{4\kappa^2}g_{\mu\nu}R+\frac{1}{8}g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}+\frac{1}{4}m^2g_{\mu\nu}A_{\sigma}A^{\sigma}=\frac{1}{2\kappa^2}R_{\mu\nu}$$

but I did this by only varying the terms $\sqrt{-g}$, $R_{\mu\nu}$ and $g^{\mu\nu}$. Do I have to write $F_{\alpha\beta}F^{\alpha\beta}=g^{\alpha\lambda}g^{\beta\rho}F_{\alpha\beta}F_{\lambda\rho}$ and apply the variation to those two metrics too? By the way my attempt, so far, is correct isn´t it?

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  • $\begingroup$ Please note that the last term of the left hand side is wrong. Metric's missing $\endgroup$
    – Noone
    Commented Oct 17, 2020 at 17:18
  • $\begingroup$ also, I think you dropped a factor of $\frac{1}{2k^2}$ on the right hand side. $\endgroup$ Commented Oct 17, 2020 at 19:00
  • $\begingroup$ Thank you for your feedback! I think its correct now (Still not counting the two metrics in the field strength tensor and the other in the vector field) $\endgroup$ Commented Oct 17, 2020 at 20:22
  • $\begingroup$ I can understand your notation with the index μ being used three times but it is preferred to change the μ that denotes summation to some other letter. It's abuse of notation. $\endgroup$
    – Noone
    Commented Oct 18, 2020 at 11:07

3 Answers 3

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Yes, your action is of the form \begin{equation} S=\int\text{d}^{4}x\sqrt{-g}(\mathcal{L}_{\text{EH}}+\mathcal{L}_{\text{M}}), \end{equation} where \begin{equation} \mathcal{L}_{\text{EH}}=\frac{R}{2k^{2}} \end{equation} is the part whose variation with respect to $g_{\mu\nu}$ gives you the Einstein tensor in the equations of motion, and \begin{equation} \mathcal{L}_{\text{M}}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^{2}A_{\mu}A^{\mu} \end{equation} is the part corresponding to matter, whose variation with respect to $g_{\mu\nu}$ gives you the energy-momentum tensor. In order to find the correct equations, you have to write $\mathcal{L}_{\text{M}}$ as \begin{equation} \mathcal{L}_{\text{M}}=\frac{1}{4}g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}F_{\mu\nu}+\frac{1}{2}m^{2}g^{\mu\alpha}A_{\alpha}A_{\mu} \end{equation} and apply the variation to the three metric tensors that appear there.

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Yes. You have to vary $g^{ab}$ everywhere it appears.

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Yes. The field strength tensor is $F_{\mu \nu}$. The only two fundamental fields in your action are $g_{\mu \nu}$ and $A_\mu$ (unless you use Palatini variation and treat the connection as independent of the metric). So before preforming the variation, you should write the action in terms of these fundamental fields, and vary w.r.t each of them. The only other option would be to consider $A_\mu, A^\mu, F_{\mu \nu}, F^{\mu \nu}$ as independent fields, which would lead to problems like the lack of kinetic terms for all of them.

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  • $\begingroup$ Thats the only thing left for me to do right? I dont have to write $A^{\mu}=A_{\nu}g^{\mu\nu}$ and vary that one too or do I? $\endgroup$ Commented Oct 16, 2020 at 23:06
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    $\begingroup$ You do. As I said, your only fundamental fields are $g_{\mu nu}$ and $A_\mu$. not $A^\mu$ $\endgroup$
    – Rd Basha
    Commented Oct 17, 2020 at 8:05

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