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Killing vector in Stephani

  1. Pengwuino

    Pengwuino 7,118
    Gold Member

    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. jcsd
  3. cepheid

    cepheid 5,194
    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 7,118
    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 12,310
    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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