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]

<br /> \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} <br />

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

 

[nicksauce]

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

 

[Derivator]

Lets write <br /> <br /> T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} <br /> <br /> without Einstein summation convention:

<br /> <br /> \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}<br /> <br />

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]

<br /> 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<br />

 

[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:
<br /> <br /> T^{\mu\nu}(\xi_\mu)_{;\nu} = (1/2)\left( T^{\mu\nu}\xi_{\mu;\nu} + T^{\nu\mu}\xi_{\nu;\mu} \right)<br /> <br />

?

is it true, that you only relabled the summation indices in
<br /> <br /> T^{\nu\mu}\xi_{\nu;\mu} \right)<br /> <br />

 

[nicksauce]

That is correct.