Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In a theoretical physics homework problem, I have to show the following: $$\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$$

Where $T$ is the Energie-Momentum-Tensor, $j$ the generalized current and $F$ the Field-Tensor. We use the $g$ for the metric tensor, I think in English the $\eta$ is more common.

I know the following relationships:

  • Current and magnetic potential with Lorenz gauge condition: $$\mathop\Box A^\mu = \mu_0 j^\mu$$

  • Energy-Momentum-Tensor: $$T^{\mu\nu} = \frac1{\mu_0} g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu} + \frac1{4\mu_0} g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$$

  • Field-Tensor: $$F^{\mu\nu} = 2 \partial^{[\mu} A^{\nu]} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$

  • d'Alambert operator: $$\mathop\Box = \partial_\mu \partial^\mu$$

  • Bianchi identity: $$\partial^{[\mu} F^{\nu\alpha]} = 0$$

So far I have set all the definitions into the formula I have to show, but I only end up a lot of terms from antisymmetrisation and product rule. I also drew all what I have in Penrose graphical notation, but I still cannot see how to tackle this problem.

Could somebody please give me a hint into the right direction?

share|improve this question
    
Look at $F_{\alpha \beta}$ in $T_{\mu\nu}$ I think that $\beta$ is not right because the free indices are $\mu \nu$ and you have an extra free index $\beta$ –  Jorge Jan 21 '13 at 13:59
1  
The first term in the expression for $T^{\mu\nu}$ should be something like $F^{\mu\alpha}F^{\nu}_{\alpha}$ –  twistor59 Jan 21 '13 at 14:20
    
Indeed, I fixed it. I just typed it wrong here, that was not source of my confusion so far. –  queueoverflow Jan 21 '13 at 15:21
2  
I think you're missing the most important equation of all: that $\partial_\mu F^{\mu \nu} = \mu_0 j^\nu$. –  Muphrid Jan 21 '13 at 15:43
2  
@queueoverflow By the way, in English $g$ is used for any general metric, while $\eta$ is reserved for the Minkowski metric. –  Chris White Jan 21 '13 at 19:53
show 3 more comments

1 Answer 1

up vote 1 down vote accepted

Let's look at different terms from differentiating $T^{\mu\nu} $.

The first from differentiating $ g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}$ is $$\partial_\nu g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}= g^{\mu\alpha} F_{\alpha\beta} (\partial_\nu F^{\beta\nu}) +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}= - \mu_0 F_{\alpha\beta} j^\beta +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$

The first term is exactly what you want, the second cancels against the stuff you get from differentiating $g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$:

$$\partial^\mu F_{\kappa\lambda} F^{\kappa\lambda}=2 F_{\kappa\lambda} (\partial^\mu F^{\kappa\lambda})=-2 F_{\kappa\lambda} (\partial^\kappa F^{\lambda\mu}+\partial^\lambda F^{\mu\kappa}) =-4 (\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$ where in the second equality sign we have used Bianchi identity and in the last equality we have used $$ F_{\kappa\lambda} \partial^\kappa F^{\lambda\mu} \underset{\text{relabel indecies}}= F_{\nu\beta}\partial^\nu F^{\beta \mu} \underset{\text{antisym. of $F$}}= F_{\beta\nu}\partial^\nu F^{\mu\beta} $$ This exactly cancels the second term in the first equation.

share|improve this answer
    
With Murphrid's comment in mind, I am able to follow your answer, except for the very last equality sign. I renamed the indices in my notes to match yours, but I do not see why $\partial^\beta F^{\nu\mu} = \partial^\nu F^{\mu\beta}$ holds. –  queueoverflow Jan 21 '13 at 19:33
1  
It doesn't. You should split the two terms (including the $F_{\kappa \lambda}$) and relabel the dummy indices on the second term. –  Vibert Jan 21 '13 at 20:20
    
Now I see it. Thanks! –  queueoverflow Jan 22 '13 at 15:38
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.