We have the Einstein equations $$G_{\alpha\beta}=\frac{8\pi G}{c^4}T_{\alpha\beta}\\R_{\alpha\beta}-\frac12 g_{\alpha\beta}R=\frac{8\pi G}{c^4}T_{\alpha\beta}$$

I have been asked to show that for an electromagnetic field these can be written as $$R_{\alpha\beta}=\kappa (F_{\alpha\gamma}F_\beta^\gamma-\frac14 g_{\alpha\beta}F^{\gamma\delta}F_{\gamma\delta})$$

I am not fully sure what $\kappa$ is, but I suppose it is just some constant. In this case, I'd believe this problem reduces to showing $R=0$, but I don't see how that is true in general. So I am not sure how to show this. Could I get a hint? Preferably not a full solution.

An idea I had was to multiply the above equation by $g^{\alpha\beta}$, which yields $$\frac12R=\frac{6\pi G}{c^4}F_{\beta\gamma}F^{\beta\gamma}$$ where $$T_{\alpha\beta}=F_\alpha^\gamma F_{\beta\gamma}-\frac14 F_{\gamma\delta}F^{\gamma\delta}g_{\alpha\beta}$$ is the energy-momentum tensor. But this does not do anything to help me (I think). I have included it since it may turn out to be helpful.

  • $\begingroup$ @AccidentalFourierTransform Ahh I see - it does mention EM waves, but it said in in brackets after the question, so I assumed that was unimportant. Apologies, I have edited it in. However, I don't see how $T$ is traceless for EM waves, due to the factor of $\frac14$ in the expression for $T$.. could you perhaps explain? $\endgroup$ – John Doe Apr 24 '18 at 21:59
  • $\begingroup$ Hint: $g^{ab}g_{ab}=4$ and $T^a{}_a=0$ . Your approach was on the right track but you made some algebraic mistake somewhere. $\endgroup$ – AccidentalFourierTransform Apr 24 '18 at 22:11
  • $\begingroup$ @AccidentalFourierTransform I thought $g^{ab}$ was defined to be the inverse of $g_{ab}$?? This is why I just cancelled them out rather than putting in a $4$. Is this just because we are in 4D? $\endgroup$ – John Doe Apr 24 '18 at 22:17
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    $\begingroup$ In general, $g^{ab}g_{ab}=\delta^a_a=d$. $\endgroup$ – AccidentalFourierTransform Apr 24 '18 at 22:19
  • $\begingroup$ The title is completely misleading/meaningless. I am changing it $\endgroup$ – magma Apr 26 '18 at 8:02

Let Einstein Field Equations be given

$$R_{\alpha\beta}-\dfrac{1}{2}g_{\alpha\beta}R=k T_{\alpha\beta}.$$

Contract with $g^{\alpha\beta}$ and define $T = g^{\alpha\beta}T_{\alpha\beta}$, so that we have upon using $g^{\alpha\beta}R_{\alpha\beta}=R$


but $g^{\alpha\beta}g_{\alpha\beta}=n$ the dimension of spacetime. Thus


In the case of $n = 4$ we get $kT = -R$. Thus we can substitute $R = -kT$ on Einstein's field equations


Or also


So you have Ricci's tensor written fully in terms of the energy-momentum tensor and its trace. This is just Einstein's equations rewritten. This is the form which you should work with here.

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EDIT: The orijinal question was completely different when I first answered, so I changed the answer for the new version.

First of all, $\kappa \equiv\frac{8 \pi G}{c^4}$ and $G_{\mu\nu} \equiv R_{\mu\nu} - \frac12 g_{\mu\nu} R$ are the definitions.

You can first express $R_{\mu\nu}$ in terms of $T_{\mu\nu}$ by taking the trace of the equation and substitute it back in it. Then, you need to define your stress-energy energy tensor for the EM field.

Since the source-free Maxwell Lagrangian in vacuum is $$ \mathcal{L} = -\frac14 F_{\mu\nu} F^{\mu\nu} $$ you can calculate the stress-energy tensor by its variation with respect to the metric.

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  • $\begingroup$ Hmm, I don't think I have made much progress... taking the trace yields:$$\frac12 R=\frac34\kappa F_{\beta\gamma}F^{\beta\gamma}$$Then this gives$$R_{\alpha\beta}=\kappa T_{\alpha\beta}+\frac34g_{\alpha\beta}F_{\beta\gamma}F^{\beta\gamma}\\=\kappa T_{\alpha\beta}+\kappa g_{\alpha\beta}T_{\mu\nu}g^{\mu\nu}$$What can I do with this? I haven't seen the source-free Maxwell Lagrangian before and am unsure what you mean by its variation wrt the metric. Edit: I think AccidentalFourierTransform's comment on the answer the issues I just listed. Thanks! $\endgroup$ – John Doe Apr 24 '18 at 22:15
  • $\begingroup$ Yeah, I indeed guessed you were just circling around the conclusion. $\endgroup$ – Oktay Doğangün Apr 24 '18 at 22:51

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