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MTW: Exer. 8.11, pg. 215

  1. Sep 6, 2010 #1
    It is the last part that I am having trouble understanding. It says that if u and w are tangent vectors then,

    [tex]\nabla_{\bold{u}}\bold{w} - \nabla_{\bold{w}}\bold{u} = [\bold{u},\bold{w}][/tex].​


    [tex][\bold{u},\bold{w}] = \partial_{\bold{u}}\partial_{\bold{w}} - \partial_{\bold{w}}\partial_{\bold{u}} = (u^\beta\,v^{\alpha}_{,\beta} - v^\beta\,u^{\alpha}_{,\beta})\bold{e}_\alpha[/tex].


    [tex]\nabla_{\bold{u}}\bold{w} - \nabla_{\bold{w}}\bold{u} = (u^\beta\,w^{\alpha}_{,\beta} - w^\beta\,u^{\alpha}_{,\beta})\bold{e}_\alpha + (u^\beta\,w^{\alpha} - w^\beta\,u^{\alpha})\bold{e}_{\alpha}_{,\beta}[/tex].

    So how do I reason that the second term disappears? Is the derivative of the vector basis zero b/c this is in a local Lorentz frame? In other words, there are no correction terms??

    Thanks in advance for any help you may give me.

  2. jcsd
  3. Sep 6, 2010 #2
    Work in coordinate basis [tex]e_\alpha=\partial_\alpha[/tex]. Use the fact that the Levi-Civita connection is symmetric and
    [tex]\nabla_\alpha e_\beta = ....[/tex]
  4. Sep 6, 2010 #3

    [1] By "Levi-Civita connection" do you mean:
    a) the "connection coefficients" aka
    b) the affine connection aka
    c) the Christoffel symbols??

    [2] In a coordinate basis, the commutator of the partials is zero. The second term is not a commutator of the partials - but of the components of the respective vectors. So it can't be canceled out.

    [3] Finally, [tex]\nabla_{\alpha}\,\bold{e}_{\beta}[/tex] is the connection coefficient, [tex]\Gamma^{\mu}_{\alpha\beta}\bold{e}_{\mu}[/tex]. So then the other term would be: [tex]\nabla_{\beta}\,\bold{e}_{\alpha} = \Gamma^{\mu}_{\beta\alpha}\bold{e}_{\mu}[/tex] ... which when subtracted - and since the Christoffel symbols are symmetric in their lower indices - would then make the second term cancel out!

    So I was too quick to match dummy indices with these two terms! Yes?? (If so thanks dude!!)

  5. Sep 7, 2010 #4
    "would then make the second term cancel out"
    Just remeber: if [tex]A^{\alpha\beta}[/tex] is antisymmetric and [tex]B_{\alpha\beta}[/tex] is symmetric, then [tex]A^{\alpha\beta}B_{\alpha\beta}=0[/tex].
  6. Sep 7, 2010 #5
    Yes! the (uw-wu) is anti-symmetric in a,b and the e_a,b is symmetric in a,b. That's a good one to remember.

    Thanks for all your kind help and patience with me. May God richly bless you, in Jesus' name, amen.

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