# Trace of the stress-energy tensor

1. Sep 27, 2013

### fluidistic

0. The problem statement, all variables and given/known data
Hi guys,
I must show that the trace of the stress energy tensor is zero.
The definition of it is $T^{\mu \nu }=\frac{1}{4\pi} \left ( F^{\mu \sigma } F^{\nu \rho} \eta _{\sigma \rho}-\frac{1}{4} \eta ^{\mu \nu } F^{\sigma \rho} F_{\sigma \rho} \right )$.

1. The attempt at a solution
I know it's pure algebraic manipulations but for some reason I get stuck. Trace is $T^\mu _\mu =\eta_{\mu\nu}T^{\mu\nu}=\frac{1}{4\pi}\left ( \eta_{\mu\nu}F^{\mu\rho}F^{\nu\rho}\eta _{\sigma\rho} - \frac{1}{4} \eta_{\mu\nu}\eta^{\mu\nu}F^{\sigma\rho}F_{\sigma \rho}\right )$
$=\frac{1}{4\pi}\left ( F^\sigma _\nu F^\nu_\sigma-\frac{1}{4} \delta ^\mu _\mu F^{\sigma \rho}F_{\sigma \rho} \right ) = \frac{1}{4\pi} \left ( F^\sigma _\nu F^\nu _\sigma - F^{\sigma\rho}F_{\sigma \rho} \right )$.
This is where I'm stuck. I don't know how to show that the terms in the parenthesis are equal; if they are. I'd appreciate any comment. Thanks.

2. Sep 28, 2013

### vanhees71

First of all it is of utmost importance to thoroughly keep track about the indices. Thus you should write
$$F^{\mu \nu} \eta_{\nu \rho}={F^{\mu}}_{\rho},$$
because the Faraday tensor is antisymmetric.

Given that caveat you are already done with your prove. You just have to use that
$${F^{\sigma}}_{\nu} {F_{\sigma}}^{\nu}=F^{\sigma \nu} F_{\sigma \nu}.$$
Just prove this identity!

3. Sep 28, 2013

### WannabeNewton

Just to add on a bit, you can raise and lower indices under contraction at whim in this context. So you can just say $F^{\sigma}{}{}_{\nu}F_{\sigma}{}{}^{\nu} = F_{\sigma\nu}F^{\sigma\nu}$. All this is just from implicitly using the metric tensor to raise and lower indices twice (recall what I told you about musical isomorphisms). You can do this with as much fluidity as you want, without having to show that you're actually using the metric tensor each time; after a while the operation of raising and lower indices will become natural and you'll be doing it almost subconsciously in your calculations :)

4. Sep 28, 2013

### fluidistic

Thanks guys but I still don't understand anything nor why my attempt fails:
$F^\sigma _\nu F^\nu _\sigma =F^{\sigma \rho}\eta _{\rho \nu}\eta ^{\nu \rho}F _{\sigma \rho}=F^{\sigma \rho} \delta ^\nu _\rho F_{\sigma \rho}=4F^{\sigma \rho}F_{\sigma \rho}$.

5. Sep 28, 2013

### TSny

On the right, you've used the notation $\rho$ for the dummy summation index for the first two factors and then you used the same notation $\rho$ for the dummy summation index for the last two factors. But the dummy summation index for the last two factors is completely independent of the dummy summation index for the first two factors. So, you can use $\rho$ for the summation index for the first two factors, but you should use some other notation for the summation index for the last two factors.

A general rule is that if you have any term made up of any number of factors, a dummy summation index in that term should appear once and only once as an upper index and once and only once as a lower index.

6. Sep 28, 2013

### vela

Staff Emeritus
By the way, you should take a look at our posts to see how to get the indices spaced correctly in $\LaTeX$. It's not clear whether $F^\mu_\nu$ means $F^\mu{}_\nu$ or $F_\nu{}^\mu$.

7. Sep 28, 2013

### fluidistic

Ah I see, thanks... I hope I got it right in my next attempt.
Oh I see, I never paid any attention to this, didn't know there was a difference between $F^\mu{}_\nu$ and $F_\nu{}^\mu$.
New attempt but still stuck: $F^\sigma{}_\nu F_\sigma{}^\nu =F^{\sigma \chi}\eta _{\chi \nu}\eta _{\sigma\alpha}F^{\alpha \nu}$. I don't see how I could simplify this to $F^{\sigma\nu}F_{\sigma\nu}$.

8. Sep 28, 2013

### Dick

To get closer to you what you want you should raise the $\nu$ in the first factor of $F^\sigma{}_\nu F_\sigma{}^\nu$ and lower it in the second.

9. Sep 28, 2013

### WannabeNewton

Just use your two instances of the metric tensor to lower both indices on $F^{\alpha\nu}$.

And yes $F^{\mu}{}{}_{\nu}$ is definitely not the same as $F_{\nu}{}{}^{\mu}$ unless the tensor in question is symmetric.

10. Sep 28, 2013

### fluidistic

Thanks guys, I finally got the tooth out of the mouth. It was worst than a tooth extraction for me.
Here is my attempt, this time I think I got it: $F^\sigma{}_\nu F_\sigma{}^\nu =F^{\sigma \chi}\eta _{\chi \nu}\eta _{\sigma\alpha}F^{\alpha \nu}=F^{\sigma\chi} \eta_{\chi\nu}F_\sigma{}^\nu=F^{\sigma\chi}\eta_{\chi\nu}(-F^\nu{}_\sigma)=-F^{\sigma\chi}F_{\chi\sigma}=F^{\chi\sigma}F_{\chi\sigma}$.

11. Sep 28, 2013

### Dick

That's fine. But you really didn't need to use antisymmetry. $\eta_{\chi\nu}F_\sigma{} ^\nu=F_{\sigma\chi}$. Alternatively, $F^\sigma{}_\nu F_\sigma{}^\nu = F^{\sigma \alpha} \eta_{\alpha \nu} \eta^{\beta \nu} F_{\sigma \beta}$. The product of the two etas in the middle is $\delta^\beta_\alpha$.

Last edited: Sep 28, 2013
12. Sep 28, 2013

### fluidistic

Thanks, I see. And to finish this off, the delta is worth the identity for alpha=beta and worth the zero matrix otherwise right? So that one falls over the solution.

13. Sep 29, 2013

### Dick

It's worth 1 if alpha=beta and 0 otherwise (not matrices). So it gives you $F^{\sigma \alpha}F_{\sigma \alpha}$ for example, which is equivalent to what you want. I'm not sure what 'falls over the solution means'.