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I have to derive the electromagnetic energy-momentum tensor from Noether's theorem and translation invariance. Due to translation invariance and gauge transformation: $$\delta A_\mu= a^\nu F_{\mu\nu}$$ (with $a^\nu=a^\nu(x)$)and then simply by reading off from $$\delta S[A]=-\int d^4x\>\partial_\mu a_\nu\,T^{\mu\nu}$$ I should get $T^{\mu\nu}$.

My attempt: $$\delta S[A]=\int \delta \mathcal{L}\,d^4x$$ So I need to find $\delta \mathcal{L} = -\frac{1}{4} \delta(F^2)$. With $$\delta(F^2)=2F^{\mu\nu} \delta F_{\mu\nu}=2F^{\mu\nu} a^\rho (\partial_\mu F_{\nu\rho} -\partial_\nu F_{\mu\rho})=2F^{\mu\nu} a^\rho (\partial_\mu F_{\nu\rho} +\partial_\nu F_{\rho\mu}) $$ Now using the Bianchi identity $\partial_\mu F_{\nu\rho}+\partial_\nu F_{\rho\mu}+\partial_\rho F_{\mu\nu}=0$. $$\delta(F^2)=-2F^{\mu\nu}a^\rho\,\partial_\rho F_{\mu\nu}$$ I can rewrite this but I am not sure if this is beneficial: $$-2F^{\mu\nu}a^\rho\,\partial_\rho F_{\mu\nu}= (\partial_\rho a^\rho)F^2+\partial_\rho(-a^\rho F^2)$$
But then I would have $$\delta \mathcal{L} = -\frac{1}{4} ((\partial_\rho a^\rho)F^2+\partial_\rho(-a^\rho F^2))$$
Now I am not sure, what to do with the second term. Would it vanish because it is a total derivative? But then there would be just one term in $T^{\mu\nu}$.
What I am doing wrong?

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