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Divergence of the energy momentum tensor

  • #1
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I need to prove that in a vacuum, the energy-momentum tensor is divergenceless, i.e.
$$ \partial_{\mu} T^{\mu \nu} = 0$$
where
$$ T^{\mu \nu} = \frac{1}{\mu_{0}}\Big[F^{\alpha \mu} F^{\nu}_{\alpha} - \frac{1}{4}\eta^{\mu \nu}F^{\alpha \beta}F_{\alpha \beta}\Big]$$ Here ##F_{\alpha \beta} ## is the electromagnetic field tensor and ##\eta^{\mu \nu}## is the metric.
I've worked through all the steps. I need to know if my answer is correct.
Now we have
$$ \partial_{\mu} \Big(F^{\alpha \mu } F^{\nu}_{\alpha}\Big) = F^{\alpha \mu} \Big(\partial_{\mu}F^{\nu}_{\alpha}\Big) + F^{\nu}_{\alpha}\Big( \partial_{\mu} F^{\alpha \mu}\Big)$$
The second term vanishes because of Maxwell's equation with ##J^{\alpha} =0 ##.
$$F^{\alpha \mu}\Big( \partial_{\mu} F^{\nu}_{\alpha} \Big) =\frac{F^{\alpha \mu} (\partial_{\mu} F^{\nu}_{\alpha}) + F^{\mu \alpha} (\partial_{\alpha} F^{\nu}_{\mu})}{2}$$ In the second term, we let ##\mu \rightarrow \alpha## and ##\alpha \rightarrow \mu##, since they're both dummy variables. But since ##F^{\mu \alpha} = -F^{\alpha \mu} ##, the expression becomes
$$ \partial_{\mu}\Big(F^{\alpha \mu } F^{\nu}_{\alpha}\Big) = \frac{F^{\alpha \mu} \Big(\partial_{\mu}F^{\nu}_{\alpha} - \partial_{\alpha}F^{\nu}_{\mu}\Big)}{2}$$ Now, $$
\Big(\partial_{\mu}F^{\alpha \beta}\Big)F_{\alpha \beta} = \eta^{\gamma \alpha}\eta^{\delta \beta} \eta_{\theta \alpha} \eta_{\phi \beta} F^{\theta \phi} \Big( \partial_{\mu} F_{\gamma \delta}\Big) = \delta^{\gamma}_{\theta} \delta^{\delta}_{\phi}F^{\theta \phi} \Big( \partial_{\mu} F_{\gamma \delta}\Big) = F^{\theta \phi}\Big( \partial_{\mu} F_{\theta \phi}\Big) = F^{\alpha \beta}\Big( \partial_{\mu} F_{\alpha \beta}\Big)
$$
So differentiating the second term of the energy momentum tensor gives us,
$$\partial_{\mu} \Big(\eta^{\mu \nu} F^{\alpha \beta} F_{\alpha \beta}\Big) = -2 F^{\alpha \mu} \partial^{\nu} F_{\mu \alpha}$$
Where I replaced ##\beta## by ##\mu## and flipped the indices.
Combining the previous two expressions, we have
$$ \partial_{\mu} T^{\mu \nu} = \frac{1}{2 \mu_{0}} F^{\alpha \mu} \Big[\partial_{\mu} F^{\nu}_{\alpha} - \partial_{\alpha} F^{\nu}_{\mu} + \partial^{\nu} F_{\mu \alpha} \Big]$$
Now consider the term in the square brackets:
$$\partial_{\mu} F^{\nu}_{\alpha} - \partial_{\alpha} F^{\nu}_{\mu} + \partial^{\nu} F_{\mu \alpha} = \eta^{\nu \theta}\partial_{\mu} F_{\alpha \theta} - \eta^{\nu \theta}\partial_{\alpha} F_{\mu \theta} + \eta^{\nu \theta} \partial_{\theta} F_{\mu \alpha} = \eta^{\nu \theta}\Big( \partial_{\mu} F_{\alpha \theta} + \partial_{\alpha} F_{\theta \mu} + \partial_{\theta} F_{\mu \alpha}\Big)$$
Where I flipped the indices in the second term.The term in brackets is zero because of Maxwell's other equation. Therefore,
$$ \partial_{\mu} T^{\mu \nu} = 0$$

Are all the steps correct? Thank you for your help!
 

Answers and Replies

  • #2
TSny
Homework Helper
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2,841
Looks good
 

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