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Killing vector in Stephani

by Pengwuino
Tags: killing, stephani, vector
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Jan17-11, 10:41 PM
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In Stephani's "Relativity", section 33.3, equation (33.9), he has the Killing equations for cartesian coordinates as


From there he says upon differentiation, you can get the following three equations


Now, I'm not use to the ,; notation, but doesn't the first equation mean

[tex]\partial_b \xi_a + \partial_a \xi_b=0[/tex]?

If so, I don't understand the other 3 equations then. If for example, the first one is suppose to be subsequent differentiation by [tex]\partial_c[/tex], then wouldn't it be[tex]\xi_{a,b,c}+\xi_{b,a,c}=0[/tex]?
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Jan17-11, 11:28 PM
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I think that it is supposed to be a second derivative, and the second comma is omitted. So:

[tex] \xi_{b,ca} = \partial_a(\partial_c\xi_b)[/tex]

EDIT: If you assume that, then does it work?
Jan19-11, 02:33 AM
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As far as i can tell, no. He seems to be permuting the indices but I don't know what about the killing vector allows one to do that.

Jan19-11, 06:03 PM
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Killing vector in Stephani

If you're working in a flat space without a torsion, then the partial derivatives commute when applied to any covector, be it Killing or not.

So from the Killing equation [itex] \xi_{(a,b)} = 0 [/itex], differentiating it by [itex] x^c [/itex], one obtains succesively

[tex] \xi_{(a,b)c} = \xi_{(a,bc)} = 0 {}[/tex] ,

thing which allows you, Stephani and everybody else to permute the indices in every of the 6 possible cases, without changing anything.

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