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Riemann Tensor Derivation

  1. Aug 8, 2013 #1
    I was working on the derivation of the riemann tensor and got this

    (1) ##\Gamma^{\lambda}_{\ \alpha\mu} \partial_\beta A_\lambda##

    and this

    (2) ##\Gamma^{\lambda}_{\ \beta\mu} \partial_\alpha A_\lambda##

    How do I see that they cancel (1 - 2)?

    ##\Gamma^{\lambda}_{\ \alpha\mu} \partial_\beta A_\lambda - \Gamma^{\lambda}_{\ \beta\mu} \partial_\alpha A_\lambda = 0##

    The only difference is ##\alpha \leftrightarrow \beta##

    First step was ##\left[ D_\alpha, D_\beta \right] A_\mu = D_\alpha (D_\beta A_\mu) - D_\beta (D_\alpha A_\mu)##

    then

    ##D_\beta A_\mu = \partial_\beta A_\mu - \Gamma^{\lambda}_{\mu\beta} A_\lambda = A_{\mu ;\beta} => V_{\mu\beta}##

    then another covariant derivative

    ##D_\alpha V_{\mu\beta} = \partial_\alpha V_{\mu\beta} - \Gamma^{\lambda}_{\ \alpha\mu} V_{\lambda\beta} - \Gamma^{\lambda}_{\ \alpha\beta} V_{\mu\lambda}##

    then plug in

    ## D_\alpha (D_\beta A_\mu) = \partial_\alpha (\partial_\beta A_\mu - \Gamma^{\sigma}_{\ \mu\beta} A_\sigma)
    - \Gamma^{\lambda}_{\ \alpha \mu} (\partial_\beta A_\lambda - \Gamma^{\sigma}_{\ \lambda \beta} A_{\sigma})
    - \Gamma^{\lambda}_{\ \alpha \beta} (\partial_\lambda A_\mu - \Gamma^{\sigma}_{\ \mu\lambda} A_\sigma)##

    And later

    ##-\Gamma^{\lambda}_{\ \alpha \mu} (\partial_\beta A_\lambda - \Gamma^{\sigma}_{\ \lambda \beta} A_{\sigma})##

    which is

    ##-\Gamma^{\lambda}_{\ \alpha \mu} \partial_\beta A_\lambda + \Gamma^{\lambda}_{\ \alpha \mu} \Gamma^{\sigma}_{\ \lambda \beta} A_{\sigma}##

    the 2nd term cancels later, but the 1st one does not (see above)

    Fleisch (Students Guide to Vectors and Tensors) also does this derivation, but he never had two terms like this.
     
    Last edited: Aug 8, 2013
  2. jcsd
  3. Aug 9, 2013 #2
    updated first post a few hours ago.

    Why can't I edit it now?

    ----------------------------------------

    the 2nd calculation (##D_\beta D_\alpha##) should be the same, except that ##\alpha \leftrightarrow \beta##

    could it be that I forgot the product rule for the 2nd term in ##( ... )##?

    I am so stupid :(

    ##\partial_\alpha (\partial_\beta A_\mu - \Gamma^{\sigma}_{\ \mu\beta} A_\sigma)##

    ##= \partial_\alpha \partial_\beta A_\mu - \partial_\alpha (\Gamma^{\sigma}_{\ \mu\beta} A_\sigma)##

    using the product rule on the 2nd term

    ##= \partial_\alpha \Gamma^{\sigma}_{\ \mu\beta} A_\sigma - \Gamma^{\sigma}_{\ \mu\beta} \partial_\alpha A_\sigma##

    doing ##\alpha \leftrightarrow \beta## for the 2nd commutator term

    ##= \partial_\beta \Gamma^{\sigma}_{\ \mu\alpha} A_\sigma - \Gamma^{\sigma}_{\ \mu\alpha} \partial_\beta A_\sigma##

    which just produces the terms I need to cancel the ones from post #1 :)

    -------------------------------------------------------

    Thanks for not posting the answer.

    Sometimes it is hard to see the obvious...

    -------------------------------------------------------

    The full derivation now is

    ##A_{\mu ;\beta \alpha} = \partial_\alpha \partial_\beta A_\mu - \partial_\alpha \Gamma^{\sigma}_{\ \mu\beta} A_\sigma - \Gamma^{\sigma}_{\ \mu\beta} \partial_\alpha A_\sigma - \Gamma^{\lambda}_{\ \alpha\mu} \partial_\beta A_\lambda + \Gamma^{\lambda}_{\alpha\mu} \Gamma^{\sigma}_{\ \lambda\beta} A_\sigma - \Gamma^{\lambda}_{\ \alpha\beta} \partial_\lambda A_\mu + \Gamma^{\lambda}_{\ \alpha\beta} \Gamma^{\sigma}_{\ \mu\lambda} A_\sigma##

    and

    ##A_{\mu ;\alpha \beta} = \partial_\beta \partial_\alpha A_\mu - \partial_\beta \Gamma^{\sigma}_{\ \mu\alpha} A_\sigma - \Gamma^{\sigma}_{\ \mu\alpha} \partial_\beta A_\sigma - \Gamma^{\lambda}_{\ \beta\mu} \partial_\alpha A_\lambda + \Gamma^{\lambda}_{\beta\mu} \Gamma^{\sigma}_{\ \lambda\alpha} A_\sigma - \Gamma^{\lambda}_{\ \beta\alpha} \partial_\lambda A_\mu + \Gamma^{\lambda}_{\ \beta\alpha} \Gamma^{\sigma}_{\ \mu\lambda} A_\sigma##

    subtracting both

    ##A_{\mu ;\beta \alpha} - A_{\mu ;\alpha \beta}##

    using symmetry of the christoffel symbols and ##\partial_\alpha \partial_\beta = \partial_\beta \partial_\alpha## and moving the minus sign out of ##( ... )## we get

    ##\left[ D_\alpha, D_\beta \right] A_\mu = A_{\mu ;\beta \alpha} - A_{\mu ;\alpha \beta} = - R^{\sigma}_{\ \mu\alpha\beta} A_\sigma##

    where

    ##R^{\sigma}_{\ \mu\alpha\beta} = \partial_\alpha \Gamma^{\sigma}_{\ \mu\beta} - \partial_\beta \Gamma^{\sigma}_{\ \mu\alpha} + \Gamma^{\lambda}_{\ \beta\mu} \Gamma^{\sigma}_{\ \lambda\alpha} - \Gamma^{\lambda}_{\ \alpha\mu} \Gamma^{\sigma}_{\ \lambda\beta}##
     
    Last edited: Aug 9, 2013
  4. Aug 9, 2013 #3
    You may also see the derivation in Dirac's book : General theory of Relativity under equation 11.1 if I'm not wrong
     
  5. Aug 9, 2013 #4
    Everything is solved now.

    After some messy messing around and remembering the product rule ( LOL , Thanks Newton ;) ) I got it.

    Can be closed.
     
    Last edited: Aug 9, 2013
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