# covarient derivative of electromagnetic field tensor

I'm trying to prove the energy momentum tensor in curved spacetime for Electromagnetic field is Divergence-less directly(Without using general lie derivative method which can prove any energy momentum tensor obtained from varying $g^{ab}$ .

From free lagrangian

$$L = -1/4 F_{ab}F_{cd}g^{ac}g^{bd}$$

We get

$$T_{ab} = 1/4(F_{ac}F_{bd}g^{cd} -1/4g_{ab}F_{mk}F_{nl}g^{mn}g^{kl})$$

Now what we have in hand is

$$\nabla_aF^{ab} = 0$$

But To prove $\nabla_aT^{ab} = 0$ from the above expression of $T_{ab}$

We have to compute stuff like $\nabla_aF_{bc}$

But I'm actually getting no where, i have started like following:

$$\nabla_a F_{bc}$$ $$= g_{bn}g_{cn}\nabla_a F^{bc}$$ $$= g_{bn}g_{cn}( \partial_aF^{bc} + \Gamma^b_{af}F^{cf} + \Gamma^c_{af}F^{bf})$$

From the divergence relation I can get

$$\partial_aF^{ab} + \Gamma^a_{af}F^{bf} + \Gamma^b_{af}F^{af} = 0$$

$$\Rightarrow \partial_aF^{ab} + \Gamma^a_{af}F^{bf} = 0$$

What should i do from here?

I can also use the covarient curl relation,

$$\nabla_{[a}F_{bc]} = 0$$

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This is probably off topic for this site. (See the FAQ.) For future reference Math.StackExchange would be a better venue. //// Now to your question: you have to use the curl relation. Using that $\nabla$ commutes with $g$, you have $$\nabla^a (F_{ac}F_b{}^c) = \nabla^aF_{ac}F_{b}{}^c + F_{ac} \nabla^a F_b{}^c$$ The first term evaluates to zero. The curl relation gives $$0 = F^{ac}\nabla_a F_{bc} + F^{ac} \nabla_b F_{ca} + F^{ac} \nabla_c F_{ab}$$ Antisymmetry in $ac$ from $F$ means that the first and third terms are equal. So $$\nabla^a (F_{ac}F_b{}^c) = \frac12 F^{ac} \nabla_b F_{ac}$$ –  Willie Wong Apr 9 '13 at 11:02