Killing vector in Stephani

  1. Pengwuino

    Pengwuino 6,942
    Gold Member

    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]?
  2. jcsd
  3. cepheid

    cepheid 5,189
    Staff Emeritus
    Science Advisor
    Gold Member

    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?
  4. Pengwuino

    Pengwuino 6,942
    Gold Member

    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.
  5. dextercioby

    dextercioby 12,327
    Science Advisor
    Homework Helper

    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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