Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Killing Vector and Ricci curvature scalar

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm currently self-studying Carroll's GR book and get stuck by proving
    the following identity:

    [tex] K^\lambda \nabla _\lambda R = 0 [/tex]

    where K is Killing vector and R is the Ricci Scalar


    2. Relevant equations

    Mr.Carroll said that it is suffice to show this by knowing:

    [tex] \nabla _\mu \nabla _\sigma K^\mu = R_{\sigma \nu}K^\nu [/tex]

    Bianchi identity [tex] \nabla ^ \mu R_{\rho \mu} = \frac{1}{2} \nabla _\rho R [/tex]

    and Killing equation [tex] \nabla _\mu K_\nu + \nabla _\nu K_\mu = 0 [/tex]



    3. The attempt at a solution

    The work I done so far :

    [tex] K^\lambda \nabla _\lambda R = 2 K^\lambda \nabla ^\mu R_{\mu \lambda} = 2 \left( \nabla ^\mu R_{\mu \lambda} K^\lambda -R_{\mu \lambda} \nabla ^\mu K^\lambda \right) = 2 \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma [/tex]

    Note that [tex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/tex], since

    [tex]R_{\mu \lambda} \nabla ^\mu K^\lambda = - R_{\mu \lambda} \nabla^\lambda K^\mu = -R_{\lambda\mu} \nabla^\lambda K^\mu = -R_{\mu \lambda} \nabla ^\mu K^\lambda[/tex]

    And I can't get any further :cry:

    Could someone help?? Thanks in advace
     
  2. jcsd
  3. Nov 30, 2008 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [itex]\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma = 0[/itex] in the same way that [itex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/itex].
     
  4. Nov 30, 2008 #3
    Hi~

    [itex]R_{\mu \lambda} \nabla ^\mu K^\lambda =0 [/itex]
    is due to the Killing Equation and the symmetry of Ricci tensor [tex] R_{\mu \lambda} = R_{\lambda \mu}[/tex]

    But it seems like
    [tex]\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma \neq \nabla _\sigma \nabla ^\mu \nabla _\mu K^\sigma [/tex] ??

    Thanks for Reply!
     
    Last edited: Nov 30, 2008
  5. Nov 30, 2008 #4
    I think I figure it out! :redface:

    [tex] \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma [/tex]

    [tex] = (g^{\mu \sigma} \nabla_\sigma) (g_{\sigma \mu}\nabla^\mu)(\nabla_\mu K^\sigma)[/tex]

    [tex] = g^{\mu \sigma}g_{\sigma \mu} (\nabla _\sigma \nabla^\mu \nabla_\mu K^\sigma) + g^{\mu \sigma}(\nabla^\mu \nabla_\mu K^\sigma)(\nabla_\sigma g_{\sigma \mu})[/tex]

    [tex] = \nabla_\sigma \nabla^\mu \nabla_\mu K^\sigma [/tex]

    the second term of the third line vanishes due to metric compatibility.

    Is the above correct??:uhh:
     
    Last edited: Nov 30, 2008
  6. Nov 30, 2008 #5

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, metric compatibility is used, but be careful with the indices. You have used [itex]\mu[/itex] and [itex]\sigma[/itex] four times each in the product [itex](g^{\mu \sigma} \nabla_\sigma) (g_{\sigma \mu}\nabla^\mu)(\nabla_\mu K^\sigma)[/itex].
     
  7. Nov 30, 2008 #6
    I noticed that. But I can't derive it out if I use different indices...

    I don't know how to turn

    [tex]g^{\mu \sigma}g_{\lambda \rho}(\nabla_\sigma \nabla^\rho \nabla _\mu K^\lambda)[/tex]

    into

    [tex] \nabla _\lambda \nabla^\mu \nabla_\mu K^\lambda [/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook