# Show that $\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$

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?

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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 $\beta$ – Nivalth Jan 21 at 13:59

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The first term in the expression for $T^{\mu\nu}$ should be something like $F^{\mu\alpha}F^{\nu}_{\alpha}$ – twistor59 Jan 21 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 at 15:21

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I think you're missing the most important equation of all: that $\partial_\mu F^{\mu \nu} = \mu_0 j^\nu$. – Muphrid Jan 21 at 15:43

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@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 at 19:53
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## 1 Answer

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.

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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 at 19:33

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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 at 20:20
Now I see it. Thanks! – queueoverflow Jan 22 at 15:38
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