Calculating Killing vectors of Schwarzschild metric

  • #1
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I am trying to understand the solution to exercise 7.10(e) on pages 175-176 of Robert Scott's student's manual to Schutz's textbook.

He writes the following:
Table 7.1 rows four, five and six, lists the three Killing vector fields associated with invariance for rotations about the three spatial Cartesian axes.
Because Schwarzschild also has spherical symmetry it enjoys the same Killing vector fields.
We can transform these into the spherical coordinates of (ii) using relations in Appendix B giving:
$$\vec{Q}=\vec{e_t}$$
$$\vec{R}=\vec{e_\phi}$$
$$\vec{S}=\bigg(\frac{\partial \theta}{\partial x} S^x +\frac{\partial \theta}{\partial z} S^z\bigg)\vec{e_\theta}+\bigg(\frac{\partial \phi}{\partial x} S^x+\frac{\partial \phi}{\partial z} S^z\bigg)\vec{e_\phi}=\cos \phi \vec{e_\theta}-\cot \theta \sin \phi \vec{e_\phi}$$
$$\vec{T}=\bigg(\frac{\partial \theta}{\partial x} T^x +\frac{\partial \theta}{\partial z} T^z\bigg)\vec{e_\theta}+\bigg(\frac{\partial \phi}{\partial x} T^x+\frac{\partial \phi}{\partial z} T^z\bigg)\vec{e_\phi}=\sin \phi \vec{e_\theta}-\cot \theta \cos \phi \vec{e_\phi}$$

I don't understand how to find ##S^x, S^z## or ##T^x,T^z## from the metric or from the cartesian representation of the rotation vectors?
The derivatives are calculated with spherical coordinates which I understand how to achieve them.

Any help?
 
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Answers and Replies

  • #2
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Ok, I believe I found my answer: ##S^x=z , S^z=-x## and ##T^x=0, T^z=-y##.
 
  • #3
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There's typo in the text, it should be ##T^y=z , T^z=-y##.
 

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