# 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.