Killing vecotrs of Schwarzschild metric

[Vrbic]

Hello,
I have a question, whether is possible to looking for Killing Vectors (KV) in this way (I know about general solution):
From Schwarzschild metric I can see two KV \frac{\partial}{\partial t} and \frac{\partial}{\partial\phi}. Then I see that other trivial KV arent there. Metric in dt and dr is independant on \theta, \phi so I suppose I can "split" metric and looking for KV just in spherical part d\theta^2+\sin^2{\theta}d\phi^2.
Can I suppose transformation this metric to the form: d\alpha^2+d\beta^2 and claim the \frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta} are KV?

 

[WannabeNewton]

No. You won't be able to find such a transformation.

Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of $so(3)$, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.

 

[Vrbic]

No. You won't be able to find such a transformation.

Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of $so(3)$, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.

Do I need know all transformation? I thought that differencial form of it could be enough. If I suppose \alpha=\alpha(\theta,\phi), \beta=\beta(\theta,\phi), than I can write d\alpha=\alpha_{\theta}+\alpha_{\phi}, d\beta=\beta_{\theta}+\beta_{\phi}, where subscript means differentation. When I put it back to the ds^2=d\alpha^2+d\beta^2 I get 3 eq.:
1=\alpha_{\theta}^2+\beta_{\theta}^2
0=\alpha_{\theta}\alpha_{\phi}+\beta_{\theta}\beta_{\phi}
\sin^2{\theta}=\alpha_{\phi}^2+\beta_{\phi}^2.
From them (if from first guess \alpha_{\theta}=\cos{\phi}, \beta_{\theta}=\sin{\phi}) I get d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi, d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi.
Multiplying by g^{\mu\nu} I hope can get \frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta}. What is wrong?

 

[Bill_K]

Can I suppose transformation this metric to the form: d\alpha^2+d\beta^2 and claim the \frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta} are KV?

If you could put the metric in that form, it would be FLAT. It's not flat, it's a sphere.

 

[WannabeNewton]

From them (if from first guess \alpha_{\theta}=\cos{\phi}, \beta_{\theta}=\sin{\phi}) I get d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi, d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi.

This clearly doesn't reduce to the metric on the 2-sphere: simple substitution yields $d\alpha^2 + d\beta^2 = d\theta ^2 + \sin^2 \theta d\phi^2 + (\sin \theta \cos\phi - \sin 2\phi )\sin\theta d\theta d\phi$.

You simply won't be able to put the 2-sphere metric in the form $ds^2 = d\alpha^2 + d\beta^2$ and Bill has elucidated above why this is impossible.