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Given, \begin{equation} T^{\mu\nu} = F^{\mu\lambda} F^\nu{}_{\lambda} - \frac{1}{4} \eta^{\mu\nu} F^{\lambda\sigma} F_{\lambda\sigma}. \end{equation}

Here $(T^{\mu\nu})$ is the energy-momentum tensor of electromagnetic fields.

Staying in four-dimensional tensor notation, demonstrate that the tensor obeys the energy-momentum conservation equation ("the tensor is conserved"):

\begin{equation} \partial_\mu T^{\mu\nu} = 0 \end{equation}

as long as Maxwell's equations hold:

\begin{align} &\partial_\mu F^{\mu\nu} = 0, \\ &\partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} + \partial_\lambda F_{\mu\nu} = 0. \end{align}

My Solution:

Let's begin by taking 4-derivative of $T^{\mu\nu}$

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

The second term goes to $0$ since $\partial_{\mu}\eta^{\mu \nu} = 0$ and similarly $\partial_{\mu}F^{\lambda\sigma} F_{\lambda\sigma} = 0$ since $F^{\lambda\sigma} F_{\lambda\sigma}$ is a scalar (indices contraction).

Thus, $$\partial_{\mu}T^{\mu\nu} = \partial_{\mu}\left( F^{\mu\lambda} F^\nu{}_{\lambda}\right)$$

Now we apply product rule,

\begin{align} \partial_{\mu}T^{\mu\nu} &= \partial_{\mu}\left( F^{\mu\lambda} F^\nu{}_{\lambda}\right)\\ &= F^\nu{}_{\lambda}{\partial_{\mu}\left( F^{\mu\lambda} \right)} + F^{\mu\lambda} \partial_{\mu}\left(F^\nu{}_{\lambda}\right) \end{align}

The first term goes to $0$ because of Maxwell's law.

Now we want to show that $$F^{\mu\lambda} \partial_{\mu}\left(F^\nu{}_{\lambda}\right) = 0$$ Observe that we can write $$F^\nu{}_{\lambda} = F^{\nu\alpha}F_{\alpha\lambda}$$ where $\alpha$ is a dummy index.

$\color{red}{\text{Not sure if above step is correct or not. It feels right since dummy variable $\alpha$ would contract.}}$$\color{red}{\text{If this is wrong then my next thought is to use $\eta$ to raise and lower the indices.}}$

Keeping that in consideration, we get,

\begin{align} F^{\mu\lambda} \partial_{\mu}\left(F^\nu{}_{\lambda}\right) &= F^{\mu\lambda} \partial_{\mu}\left(F^{\nu\alpha}F_{\alpha\lambda}\right) \end{align} applying product rule, \begin{align} F^{\mu\lambda} \partial_{\mu}\left(F^\nu{}_{\lambda}\right) &= F^{\mu\lambda} F_{\alpha\lambda}\partial_{\mu}(F^{\nu\alpha}) + F^{\mu\lambda} F^{\nu\alpha}\partial_{\mu}(F_{\alpha\lambda}) \end{align}

I am kinda stuck here. Few thoughts,

  1. Can I cancel the first term $F^{\mu\lambda} F_{\alpha\lambda}\partial_{\mu}(F^{\nu\alpha})$ using Maxwell's law? Not sure since the lower index of $\partial$ is not same as first index of F. On the other hand, can I use antisymmetry of F to write $F^{\nu \alpha} = -F^{\alpha \nu}$ and then relabel $\alpha$ as $\mu$? That doesn't feel right since $\mu$ is not a free index here.
  2. Also to show the second term $F^{\mu\lambda} F^{\nu\alpha}\partial_{\mu}(F_{\alpha\lambda})$ is $0$, how can I use Maxwell's law? Do I antisymmetrize it?

Regards,

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    $\begingroup$ we can write $F^\nu{}_{\lambda} = F^{\nu\alpha}F_{\alpha\lambda}$ No. That doesn’t make sense on dimensional grounds. $\endgroup$
    – Ghoster
    Commented Feb 19 at 5:50
  • $\begingroup$ ‘The second term goes to 0’. No. $\endgroup$
    – my2cts
    Commented Feb 19 at 7:08
  • $\begingroup$ ‘Observe that we can write’. No. $\endgroup$
    – my2cts
    Commented Feb 19 at 7:09
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    $\begingroup$ ‘Maxwell's equations’. You mean, source-free Maxwell's equations. Also, you don’t need the second equation. $\endgroup$
    – my2cts
    Commented Feb 19 at 7:12
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    $\begingroup$ You really should call it a divergence, not a derivative, as it's a sum of $n$ derivatives if $n$ is the spacetime dimension. $\endgroup$
    – J.G.
    Commented Feb 19 at 8:35

1 Answer 1

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Using the product rule, Maxwell's equations (with $j^\mu =0$), the antisymmetry of the field strength tensor and some index gymnastics yields $$ \begin{align} \partial_\mu (F^{\mu \lambda}F^\nu_{\; \lambda}) &=\underbrace{(\partial_\mu F^{\mu \lambda})}_{=0}\,F^\nu_{\; \lambda}+F^{\mu \lambda}\,\partial_\mu F^\nu_{\;\lambda}\\[2pt] &=F^{\mu \lambda} \, \partial_\mu F^\nu_{\; \lambda}\\[3pt] &=F_{\mu \lambda}\, \partial^\mu F^{\nu \lambda}\\[3pt] &= F_{\sigma \lambda}\,\partial^\sigma F^{\nu \lambda} \\&=-F_{\lambda \sigma} \, \partial^\sigma F^{\nu \lambda} \end{align}\tag{1} \label{1}$$ and $$\begin{align} \partial_\mu (\eta^{\mu \nu}F^{\lambda \sigma}F_{\lambda \sigma})&=\partial^\nu (F^{\lambda \sigma}F_{\lambda \sigma})\\[3pt] &=(\partial^\nu F^{\lambda \sigma})F_{\lambda \sigma}+F^{\lambda \sigma}\partial^\nu F_{\lambda \sigma}\\[3pt] &=2 F^{\lambda \sigma}\partial^\nu F_{\lambda \sigma}\\[3pt] &=2F_{\lambda \sigma} \, \partial^\nu F^{\lambda \sigma}\\[3pt] &= -2F_{\lambda \sigma}(\partial^\lambda F^{\sigma \nu}+\partial^\sigma F^{\nu \lambda})\\[3pt] &=-4F_{\lambda \sigma}\,\partial^\sigma F^{\nu \lambda}. \end{align}\tag{2} \label{2} $$ Combining \eqref{1} and \eqref{2}, one finds the desired result: $$\begin{align} \partial_\mu T^{\mu \nu}&= \partial_\mu \left(F^{\mu \lambda}F^\nu_{\;\lambda}- \frac{1}{4}\eta^{\mu \nu}F^{ \lambda \sigma} F_{ \lambda \sigma}\right) \\[3pt]&=0 . \tag{3}\end{align}$$

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  • $\begingroup$ How did you go from equation 1 to the next line? Did you use Minkowski metric to raise the index for F^{\nu}{}_{\lambda}? And then use it again to lower the indices for F^{\mu \ lambda}? Can you elaborate on that step? $\endgroup$ Commented Feb 19 at 16:21
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    $\begingroup$ @TheWanderer If two tensors $A^{\mu \ldots}$, $B_\mu^{\; \;\ldots}$ are contracted, you can always use $A^{\mu \ldots} B_\mu^{\; \; \ldots}= A_\mu^{\; \; \ldots} B^{\mu \ldots}$. You only have to watch out that one of the two summation (dummy) indices is upstairs and the other one is downstairs. On the other hand, a free index always keeps its original position. $\endgroup$
    – Hyperon
    Commented Feb 19 at 16:42

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