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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?

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2  
Guessing: Use the equations of motion, they're more or less equivalent to energy-conservation. Also, use the symmetry of F and of Gamma to twiddle indices. That should do the trick. –  WIMP Apr 6 '13 at 11:39
1  
Ehm, $\nabla_a F_{bc} \neq \nabla_a F^{bc}$. –  Vibert Apr 6 '13 at 14:07
    
Thanks @Vibert. Corrected it. –  Aftnix Apr 6 '13 at 14:12
    
You know the covariant divergence of the Faraday tensor is zero; so too is the covariant curl $\nabla_{[a} F_{bc]}$ equal to zero. Try permuting the indices on what you have to construct that quantity and see what you can conclude about the individual terms. –  Muphrid Apr 6 '13 at 16:26

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