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I would like to know how to derive the Electromagnetic Stress-Energy Tensor in curved spacetime.

I would like to arrive at the result

$$T^{\mu\nu} = \frac{1}{\mu_0} \left[ F^{\mu \alpha}F^\nu{}_{\alpha} - \frac{1}{4} \eta^{\mu\nu}F_{\alpha\beta} F^{\alpha\beta}\right] \,.$$

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Start with Hamiltonian density, the quantity in the integrand from the definition of the Hamiltonian: $$H = \int d^3x \left( \psi_{,0} \frac{\partial \mathcal{L}}{\partial \psi_{,0}} - \mathcal{L} \right) \equiv \int d^3x \,\mathcal{H}$$ $\mathcal{L}$, of course, denotes the Lagrangian density. Since $\mathcal{H}$ corresponds to Hamiltonian density, it should be the $(00)$ component of the energy-momentum tensor, i.e. $T_{0}^{0}. $Instead of some generic field $\psi$, plug in the photon field $A_\mu$ and upgrade the equation to the full covariant form: $$T_{\mu}^{\nu} = A_{\sigma ,\mu} \frac{\partial \mathcal{L}}{\partial A_{\sigma ,\nu}} - \delta^{\nu}_{\mu} \mathcal{L}_{ED}$$ Since $$\mathcal{L}_{ED} = -\frac{1}{4}F^{\mu \nu} F_{\mu \nu}$$ $$F^{\mu \nu} = A^{\nu,\mu} - A^{\mu,\nu}$$ a straightforward calculation gives $$T_{\mu}^{\, \, \nu} = \frac{1}{4} \left( -A^{\sigma}_{,\mu}F^{\nu}_{\,\, \sigma} + \frac{1}{4} \delta^{\nu}_{\mu} F^{\sigma \rho} F_{\sigma \rho} \right).$$ This is the canonical energy-momentum tensor, which is generally not symmetric nor gauge invariant. To fix that, you simply make the tensor symmetric by adding a suitable (basically irrelevant) term $S_{\sigma \mu \nu}$ such that: $$S_{\sigma \mu \nu} = - S_{\mu \sigma \nu}$$ $$\bar{T}_{\mu}^{\, \, \nu} = \bar{T}_{\nu}^{\, \, \mu} = T_{\mu}^{\, \, \nu} + \partial^\sigma S_{\sigma \mu \nu}$$ which should give you the right expression. Incidentally, the term you should get is $S_{\mu \nu \sigma} = A_\sigma F_{\mu\nu}$. You can simply plug it in as you would do with any other ansantz and see what it does.

I haven't done the complete calculation myself for the purposes of writing this answer, so I might be off by a minus sign, multiplicative constant or up to a permutation/relabeling of indices. Tell me if you see something wrong (or edit the answer yourself). But this should be enough to give you an idea how to derive it.

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    $\begingroup$ This makes a lot of sense! Thanks. As an aside, I was also able to derive this by varying the action from the relativistic dust, the gravitational field etc. $\endgroup$ – Anshuman Chhabra Oct 3 '16 at 12:55
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This is my derivation \begin{equation} T_{\mu\nu} = \frac{-2 c}{\sqrt{-g}} \frac{\delta S_{M}}{\delta g^{\mu\nu}} \; . \end{equation}

\begin{equation} S_{EM}[g^{\mu\nu},A^\mu] = \frac{-1}{4 \mu_0}\int d^4x \sqrt{-g} F_{\alpha\beta} F^{\alpha\beta} \; , \end{equation} \begin{eqnarray} \delta_g S_{EM} &=& \frac{-1}{4 \mu_0}\int d^4x \bigg[ \delta_g(\sqrt{-g})F_{\alpha\beta} F^{\alpha\beta} + \sqrt{-g} \delta_g (F_{\alpha\beta} F^{\alpha\beta} ) \bigg] \; ,\\ &=&\frac{-1}{4 \mu_0}\int d^4x \bigg[ - \frac 1 2 d^4x \sqrt{-g} g_{\mu\nu} \delta g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} + \sqrt{-g} \delta_g (F_{\alpha\beta} F^{\alpha\beta} ) \bigg]\; .\\ \frac{\delta S_{EM}}{\delta g^{\mu\nu}} &=& \frac{-1}{4 \mu_0} \bigg[ - \frac 1 2 d^4x \sqrt{-g} g_{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} + \sqrt{-g} \frac{\delta}{\delta g^{\mu\nu}} (F_{\alpha\beta} F^{\alpha\beta} ) \bigg]\; . (1) \end{eqnarray} Consider the last term in the vielbein form \begin{equation} F_{\alpha\beta} F^{\alpha\beta} = e^I_\alpha e^J_\beta F_{IJ} e^\alpha_K e^\beta_L F^{KL}\; .(2) \end{equation} We have done here for isolated the flat structure ($g_{\mu\nu}$-independent) from curved structure. Next, we will use the chain rule \begin{equation} \frac{\delta\,\,}{\delta g^{\mu\nu}} = \frac{\delta\,\,}{\delta e^\lambda_P} \; \frac{\delta e^\lambda_P\,}{\delta g^{\mu\nu}}\;. (3) \end{equation} From $g^{\mu\nu} = \eta^{MP}e^\mu_M e^\nu_P \;$ we have \begin{equation} \delta g^{\mu\nu} = 2 \eta^{MP}e^\mu_M \delta^\nu_\lambda \, \delta e^\lambda_P\; .(4) \end{equation} By using (2), (3) and (4) we can calculating the last term of (1)as \begin{eqnarray} \frac{\delta (F_{\alpha\beta} F^{\alpha\beta} )}{\delta g^{\mu\nu}} &=& \frac{\delta}{\delta e^\lambda_P} (e^I_\alpha e^J_\beta F_{IJ} e^\alpha_K e^\beta_L F^{KL} ) \; \frac{\delta e^\lambda_P\,}{\delta g^{\mu\nu}}\;\\ &=& 4 e^I_\alpha e^J_\beta e^\alpha_K \frac{\delta e^\beta_L}{\delta e^\lambda_P} F_{IJ} F^{KL} \;\frac{\delta e^\lambda_P\,}{\delta g^{\mu\nu}}\;\\ &=& ( 4 e^I_\alpha e^J_\beta e^\alpha_K \delta^\beta_\lambda \delta^P_L F_{IJ} F^{KL} )(\frac 1 2 \eta_{MP} e^M_\mu \delta^\lambda_\nu )\;,\\ &=& 2 e^I_\alpha e^J_\beta e^\alpha_K \delta^\beta_\lambda \delta^P_L \delta^\lambda_\nu \, e^M_\mu \eta_{MP} \, F_{IJ}F^{KL}\;,\\ &=& 2 e^I_\alpha e^J_\nu e^\alpha_K e_{L \mu } F_{IJ} F^{KL}\;,\\ &=& 2 F_{\alpha \nu} F^\alpha {}_\mu = 2 g^{\alpha\beta} F_{\alpha \mu} F_{\beta \nu} \; . \end{eqnarray} Then we obtain \begin{equation} \frac{\delta S_{EM}}{\delta g^{\mu\nu}} = \frac{1}{8 \mu_0} d^4x \sqrt{-g} g_{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} - \frac{1}{4 \mu_0} d^4x \sqrt{-g} (2 g^{\alpha\beta} F_{\alpha\mu} F_{\beta\nu})\;, \end{equation} and the energy-momentum tensor of the electtromagnetic field reads \begin{equation} T_{\mu\nu} = \frac{-2 c}{\sqrt{-g}} \frac{\delta S_{EM}}{\delta g^{\mu\nu}} = \frac c {\mu_0}g^{\alpha\beta} F_{\alpha\mu} F_{\beta\nu} -\frac {c} {4 \mu_0} g_{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} \; . \label{TEM} \end{equation}

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