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

Derivator

Hi,

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

[tex]\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0[/tex]
-----------------------

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


I don't understand, why from

" [tex]T^{\mu\nu}{}{}_{;\nu} = 0[/tex] and the symmetry of [tex]T^{\mu\nu}[/tex] "

it follows, that

[tex]\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0[/tex]

must hold.


---
derivator

George Jones

What results when

[tex]\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0[/tex]

is expanded?

Derivator

[tex]
\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}
[/tex]

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

nicksauce

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

Derivator

Lets write [tex]

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

[/tex] without Einstein summation convention:

[tex]

\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}

[/tex]

I see no chance to get it =0
:-(

George Jones

Is

[tex]\sum_\mu\sum_\nu T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = \sum_\alpha\sum_\beta T^{\alpha\beta} \left(\xi_\alpha\right)_{;\beta}[/tex]

correct?

Derivator

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

George Jones

What about

[tex]\sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\nu\right)_{;\mu} = \sum_\beta\sum_\alpha T^{\alpha\beta} \left(\xi_\alpha\right)_{;\beta}[/tex]

samalkhaiat

[tex]
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
[/tex]

Derivator

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

George Jones

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

Derivator

Oh I see! Thanks!

---------

@samalkhaiat

how do you justify this step:
[tex]

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

[/tex]

?

is it true, that you only relabled the summation indices in
[tex]

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

[/tex]

nicksauce

That is correct.