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Killing vectors in KS coordinates

  1. Jun 21, 2011 #1


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    I want to prove (or disprove) that the vector with components
    [tex]\xi^u = \frac{v}{2 r_s}\hspace{5 mm} \xi^v = \frac{u}{2 r_s}[/tex]

    is a Killing vector of the KS space-time with line element

    [tex]\frac{4 r_s^3}{r} e^{-\frac{r}{r_s}} \left( du^2 -dv^2\right) + r^2 \left( d\theta^2 + sin^2 \theta d\phi^2\right) [/tex]

    Here r is an implicit function of u,v, which is handled by a constraint equation.

    However, I seem to be having a hard time getting GRT to handle the contraints. I try

    [tex]\left( \frac{r}{r_s} - 1 \right) e^{\frac{r}{r_s}} = u^2 - v^2[/tex] directly, but it doesn't simplify.

    I try to feed it the following

    {\frac {\partial }{\partial u}}r \left( u,v \right) =2\,{{\it r\_s}}^{
    2}u \left( r \left( u,v \right) \right) ^{-1} \left( {e^{{\frac {r
    \left( u,v \right) }{{\it r\_s}}}}} \right) ^{-1}


    but it complains about "illegal use of object as a name", I can't see what it's objecting to.

    So a) -does this look like the right expression for the Killing Vector? And b) - how does one successfully get the constraints into GrTensor?
    Last edited: Jun 21, 2011
  2. jcsd
  3. Jun 21, 2011 #2


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    Don't need any complicated algebra, it's true almost by inspection. Just show directly that the Lie derivative of gμν with respect to ξμ is zero. The coordinate transformation is u → u + εv, v → v + εu. Then u2 - v2 → u2 - v2 so it obeys the constraint, and du2 - dv2 → du2 - dv2 so it's an isometry, and that's really all there is to it!
  4. Jun 21, 2011 #3


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    I set r_s to 1, and renamed r to rr,and finally got GRT to crunch through it,but your way is both much easier and more insightful. Thanks!
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