1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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

    [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]


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