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GR Lie Derivative of metric vanish <=> metric is independent

  1. Jul 18, 2017 #1
    1. The problem statement, all variables and given/known data

    How to show that lie deriviaitve of metric vanish ##(L_v g)_{uv}=0## <=> metric is independent of this coordinate, for example if ##v=\partial_z## then ##g_{uv} ## is independent of ##z## (and vice versa)

    2. Relevant equation

    I am wanting to show this for the levi-civita symbol as the connection i.e. metric compatability were we have ## \nabla_{a}g^{uv} =0 ## so the first term of ##(L_v g)_{uv}=0## vanishes trivially.

    3. The attempt at a solution

    ##(L_ug)_{uv} = U^{\alpha}\nabla_{\alpha}g_{uv}+g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}## (1)

    If the metric is independent of some coordinate, e.g ##z## then we have ## \partial_z g_{ab} =0## but , assuming metric compatibility ##\nabla_{\alpha}g_{uv}=0## anyway, so there's no need to split it up into the partial and connection term, which seemed to me the most obvious way and only way I can see to substitute the information of metric independence of (say) ##z## into (1)

    So we have ##(L_ug)_{uv} = g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}##, where there are no derivatives on the metric in these two terms, so I'm pretty stuck..

    Many thanks
     
  2. jcsd
  3. Jul 18, 2017 #2
    This is not so in general. You can have metric invariance along a vector which is not in one of the coordinate directions. The metric of an ordinary sphere is a example.
    So, you should try to show that vanishing of the Lie derivative of the metric along a vector ##\Rightarrow## metric is invariant along that vector.
     
  4. Jul 18, 2017 #3

    stevendaryl

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    First of all, the Lie derivative does not involve the connection, so it's actually the case that:

    [itex](L_U g)_{\mu \nu} = U^\alpha \partial_\alpha g_{\mu \nu} + g_{\mu \alpha} \partial_\nu U^\alpha + g_{\alpha \nu} \partial_\mu U^\alpha[/itex]

    The first term doesn't vanish in general (although the corresponding expression involving the covariant derivative does: [itex]U^\alpha \nabla_\alpha g_{\mu \nu} = 0[/itex])

    Now, let's look at the special case where [itex]U[/itex] is a coordinate basis vector, [itex]e_\lambda[/itex], that means that [itex]U^\lambda = 1[/itex] and [itex]U^\alpha = 0[/itex] if [itex]\alpha \neq \lambda[/itex]. So for this particular choice of [itex]U[/itex], [itex]\partial_\mu U^\alpha = \partial_\nu U^\alpha = 0[/itex]. So we have:

    [itex](L_U g)_{\mu \nu} = \partial_\lambda g_{\mu \nu} + 0 + 0[/itex]
     
  5. Jul 20, 2017 #4
    perfect, thank you
     
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