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

1. Apr 10, 2010

Derivator

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

2. Apr 10, 2010

George Jones

Staff Emeritus
What results when

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

is expanded?

3. Apr 10, 2010

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?

4. Apr 10, 2010

nicksauce

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

5. Apr 10, 2010

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
:-(

6. Apr 10, 2010

George Jones

Staff Emeritus
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?

7. Apr 10, 2010

Derivator

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

8. Apr 10, 2010

George Jones

Staff Emeritus

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

9. Apr 10, 2010

samalkhaiat

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

Last edited: Apr 10, 2010
10. Apr 10, 2010

Derivator

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

11. Apr 10, 2010

George Jones

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

12. Apr 10, 2010

Derivator

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

Last edited: Apr 10, 2010
13. Apr 10, 2010

nicksauce

That is correct.