# A Spherical Symmetry of the Schwarzschild Metric

1. Mar 8, 2017

### mjordan2nd

In one of the lectures I was watching it was stated without proof that the Schwarzschild metric is spherically symmetric. I thought it would be a good exercise in getting acquainted with the machinery of GR to show this for at least one of the vector fields in the algebra. The Schwarzschild metric is given as follows

$$g_{00} = 1-\frac{2gm}{r}$$
$$g_{11} = - \left( 1 - \frac{2gm}{r} \right)^{-1}$$
$$g_{22} = -r^2$$
$$g_{33} = -r^2 sin \theta$$

and all nondiagonal components are zero. Right now I'm just trying to show that the Lie derivative vanishes with respect to the following vector field:

$$X=sin \phi \frac{\partial}{\partial \theta} + \cot \theta \cos \phi \frac{\partial}{\partial \phi}$$.

By the way, if my chart map is $Q$ then the convention I'm using is $t = Q^0$, $r=Q^1$, $\theta = Q^2$ and $\phi = Q^3$.

In components the Lie derivative is given by

$$(\ell_X g)_{ij} = X^m \frac{\partial}{\partial x^m}(g_{ij}) - \frac{\partial X^s}{\partial x^i} g_{sj} - \frac{\partial X^s}{\partial x^j}g_{is}.$$

I've been able to show that all components of the Lie derivative are zero except for the (33) component. For the (33) component of the Lie derivative the only term that shows up from the first term is when m=2, and it is $-2r^2 \sin \theta \cos \theta \sin \phi$. For the next two terms, the only contribution will be when s=3, and the two terms will be equal so that term can be written as

$$-2 \frac{\partial X^3}{\partial \phi} g_{33}$$
$$2 \cot \theta \sin \phi g_{33}$$
$$-2 r^2 \cot \theta \sin \phi sin^2 \theta$$
$$-2r^2 \cos \theta \sin \theta \sin \phi$$

So overall I'm getting

$$(\ell_X g)_{ij} = -4r^2 \sin \theta \cos \theta \sin \phi.$$ I feel like maybe I'm just dropping a negative somewhere, but I'm not seeing it. Any help figuring this out would be appreciated. Thanks.

2. Mar 8, 2017

### mjordan2nd

Never mind. I now see that my formula for the Lie derivative is wrong. The minuses should be pluses.

3. Mar 9, 2017

### Orodruin

Staff Emeritus
Even if you solved it already, I would just point out that it should be quite clear from the beginning, as it is an assumption Schwarzschild used to derive it. The angular part of the metric is just the regular metric on a sphere - which displays rotational symmetry. It is the only part of the metric that is affected by rotations.

4. Mar 9, 2017

### mjordan2nd

Thank you for pointing this out. I have a tendency to miss these kind of connections and get lost in the heap of algebra, especially when learning something new.