## Divergence of product of killing vector and energy momentum tensor vanishes. Why?

Hi,

in my book, it says:
-----------------------
Beacause of $$T^{\mu\nu}{}{}_{;\nu} = 0$$ and the symmetry of $$T^{\mu\nu}$$, it holds that

$$\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0$$
-----------------------

(here, $$T^{\mu\nu}$$ ist the energy momentum tensor and $$\xi_\mu$$ a killing vector. The semicolon indicates the covariant derivative, i.e. $$()_{;}$$ is the generalized divergence)

I don't understand, why from

" $$T^{\mu\nu}{}{}_{;\nu} = 0$$ and the symmetry of $$T^{\mu\nu}$$ "

it follows, that

$$\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0$$

must hold.

---
derivator

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 Mentor What results when $$\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0$$ is expanded?
 $$\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = \left(T^{\mu\nu})_{;\nu}\xi_\mu\right + T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = 0 + T^{\mu\nu} \left(\xi_\mu\right)_{;\nu}$$ So $$T^{\mu\nu} \left(\xi_\mu\right)_{;\nu}$$ should be equal to 0. But why?

Recognitions:
Homework Help

## Divergence of product of killing vector and energy momentum tensor vanishes. Why?

Use the fact that T^{\mu\nu} is symmetric.

 Lets write $$T^{\mu\nu} \left(\xi_\mu\right)_{;\nu}$$ without Einstein summation convention: $$\sum_\mu\sum_\nu T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = \sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\mu\right)_{;\nu} = - \sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\nu\right)_{;\mu}$$ I see no chance to get it =0 :-(
 Mentor Is $$\sum_\mu\sum_\nu T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = \sum_\alpha\sum_\beta T^{\alpha\beta} \left(\xi_\alpha\right)_{;\beta}$$ correct?
 why shouldn't it be correct? You only changed the names of the indices?!

Mentor
 Quote by Derivator why shouldn't it be correct? You only changed the names of the indices?!

$$\sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\nu\right)_{;\mu} = \sum_\beta\sum_\alpha T^{\alpha\beta} \left(\xi_\alpha\right)_{;\beta}$$

 Recognitions: Science Advisor $$T^{\mu\nu}(\xi_\mu)_{;\nu} = (1/2)\left( T^{\mu\nu}\xi_{\mu;\nu} + T^{\nu\mu}\xi_{\nu;\mu} \right) = (1/2)T^{\mu\nu}\left(\xi_{\mu;\nu} + \xi_{\nu;\mu} \right) = 0$$

 Quote by George Jones What about $$\sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\nu\right)_{;\mu} = \sum_\beta\sum_\alpha T^{\alpha\beta} \left(\xi_\alpha\right)_{;\beta}$$

Yes, it's also correct. But I don't see your point.

Mentor
 Quote by Derivator Yes, it's also correct. But I don't see your point.
You wrote
 Quote by Derivator $$\sum_\mu\sum_\nu T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = \sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\mu\right)_{;\nu} = - \sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\nu\right)_{;\mu}$$
Substitute the relabeled expressions into the left and right sides of the above.

 Oh I see! Thanks! --------- @samalkhaiat how do you justify this step: $$T^{\mu\nu}(\xi_\mu)_{;\nu} = (1/2)\left( T^{\mu\nu}\xi_{\mu;\nu} + T^{\nu\mu}\xi_{\nu;\mu} \right)$$ ? is it true, that you only relabled the summation indices in $$T^{\nu\mu}\xi_{\nu;\mu} \right)$$
 Recognitions: Homework Help Science Advisor That is correct.