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Killing vector in Stephani 
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#1
Jan1711, 10:41 PM

PF Gold
P: 7,120

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]? 


#2
Jan1711, 11:28 PM

Emeritus
Sci Advisor
PF Gold
P: 5,197

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? 


#3
Jan1911, 02:33 AM

PF Gold
P: 7,120

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.



#4
Jan1911, 06:03 PM

Sci Advisor
HW Helper
P: 11,894

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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