Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.