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

[Derivator]

Hi,

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

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

 

[George Jones]

What results when

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

is expanded?

 

[Derivator]

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

So $$T^{\mu\nu} \left(\xi_\mu\right)_{;\nu}$$ should be equal to 0. But why?

 

[nicksauce]

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

 

[Derivator]

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
:-(

 

[George Jones]

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?

 

[Derivator]

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

 

[George Jones]

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

What about

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

 

[samalkhaiat]

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

 

[Derivator]

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.

 

[George Jones]

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

You wrote

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

Substitute the relabeled expressions into the left and right sides of the above.

 

[Derivator]

Oh I see! Thanks!

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@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
$$T^{\nu\mu}\xi_{\nu;\mu} \right)$$

 

[nicksauce]

That is correct.