- #1
mjordan2nd
- 173
- 1
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.
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.