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Killing vector in Stephani

by Pengwuino
Tags: killing, stephani, vector
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Pengwuino
#1
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

[tex]\xi_{a,b}+\xi_{b,a}=0[/tex]

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

[tex]\xi_{a,bc}+\xi_{b,ac}=0[/tex]
[tex]\xi_{b,ca}+\xi_{c,ba}=0[/tex]
[tex]\xi_{c,ab}+\xi_{a,cb}=0[/tex]

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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cepheid
#2
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?
Pengwuino
#3
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.

dextercioby
#4
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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