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

Question

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 Energy-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'Alembert 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?

Answer 1

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.

Answer 2

\begin{align*} \partial_\mu T^{\mu\nu}_{\text{EM}} &= \partial_\mu \left(F^{\mu\lambda} F^\nu_{\ \lambda}\right) - \frac{1}{4} \eta^{\mu\nu} \partial_\mu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) \\ &= F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} + F^\nu_{\ \lambda} \partial_\mu F^{\mu\lambda} - \frac{1}{4} \eta^{\mu\nu} \partial_\mu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) \\ &= F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} + F^\nu_{\ \lambda} \partial_\mu F^{\mu\lambda} - \frac{1}{4} \partial^\nu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) \end{align*}

From Maxwell's Equations:

$$ \partial_\mu F^{\mu\nu} = -J^\nu $$ We have: $$ \partial_\mu T^{\mu\nu}_{\text{EM}} = -F^\nu_{\ \lambda} J^\lambda + F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} - \frac{1}{4} \partial^\nu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) $$ Also we have: \begin{align*} \partial^\nu \left(F^{\lambda\sigma} F_{\lambda\sigma}\right) &= F^{\lambda\sigma} \partial^\nu F_{\lambda\sigma} + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= \eta^{\lambda\tau} \eta^{\rho\sigma} F_{\tau\rho} \partial^\nu F_{\lambda\sigma} + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= F_{\lambda\sigma} \partial^\nu \left(\eta^{\lambda\tau} \eta^{\rho\sigma} F_{\tau\rho}\right) + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} + F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= 2F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \end{align*} and the same: \begin{align} F^{\mu\lambda} \partial_\mu F^\nu_{\ \lambda} &= \left(\eta^{\mu\alpha} \eta^{\beta\lambda} F_{\alpha \beta}\right) \left(\eta_{\mu\gamma} \partial^{\gamma}\right) \left(\eta_{\sigma\lambda} F^{\nu\sigma}\right) \\ &= \eta^{\mu\alpha} \eta^{\beta\lambda} \eta_{\mu\gamma} \eta_{\sigma\lambda} \left(F_{\alpha \beta} \partial^{\gamma} F^{\nu\sigma}\right) \\ &= \delta^\alpha_\gamma \delta^\beta_\sigma \left(F_{\alpha \beta} \partial^{\gamma} F^{\nu\sigma}\right) \\ &= F_{\gamma\sigma} \partial^{\gamma} F^{\nu\sigma} \\ &= F_{\mu\lambda} \partial^\mu F^{\nu\lambda} \end{align} so that: \begin{align} \partial_\mu T^{\mu\nu}_{\text{EM}} &= -F^\nu_{\ \lambda} J^\lambda + F_{\mu\lambda} \partial^\mu F^{\nu\lambda} - \frac{1}{2} F_{\lambda\sigma} \partial^\nu F^{\lambda\sigma} \\ &= -F^\nu_{\ \lambda} J^\lambda + F_{\mu\lambda} \partial^\mu F^{\nu\lambda} - \frac{1}{2} F_{\lambda\mu} \partial^\nu F^{\lambda\mu} \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} F_{\mu\lambda} \partial^\mu F^{\nu\lambda} + \frac{1}{2} F_{\mu\lambda} \partial^\mu F^{\nu\lambda} - \frac{1}{2} F_{\lambda\mu} \partial^\nu F^{\lambda\mu} \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} \left(F_{\mu\lambda} \partial^\mu F^{\nu\lambda} + F_{\lambda\mu} \partial^\lambda F^{\nu\mu} - F_{\lambda\mu} \partial^\nu F^{\lambda\mu}\right) \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} \left(F_{\mu\lambda} \partial^\mu F^{\nu\lambda} + F_{\mu\lambda} \partial^\lambda F^{\mu\nu} + F_{\mu\lambda} \partial^\nu F^{\lambda\mu}\right) \\ &= -F^\nu_{\ \lambda} J^\lambda + \frac{1}{2} F_{\mu\lambda} \left(\partial^\mu F^{\nu\lambda} + \partial^\lambda F^{\mu\nu} + \partial^\nu F^{\lambda\mu}\right) \end{align} Lastly we use Bianchi Identities: $$ \partial^\mu F^{\nu\lambda} + \partial^\lambda F^{\mu\nu} + \partial^\nu F^{\lambda\mu} = 0 \Leftrightarrow \partial^{[\mu} F^{\nu\lambda]} = 0 $$ we get: $$ \partial_\mu T^{\mu\nu}_{\text{EM}} = -F^\nu_{\ \lambda} J^\lambda $$

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I don't know if we can change the $\lambda$ to $\mu$ in the last equation, maybe we can because it is a dummy index.