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A What is right expression for momentum constraint in GR?

  1. Jul 30, 2016 #1
    Momentum constraint in GR in ADM formalism is written in the form

    $$\mathcal M_i=\gamma_{ij}D_k\pi^{kj},~~~~~~~~~~(1a)$$ or equivalently

    $$\mathcal M_i=D_k\pi^{k}_i,~~~~~~~~~~(1b)$$ where
    ##\pi^{ij}=-\gamma^{1/2}\left(K^{ij}-\gamma^{ij}K\right)~##, ##K=\gamma^{ij}K_{ij}~##, ##\gamma=\det \gamma_{ij}~## and ##D_i~## is covariant derivative. This is from DeWitt1967 parer and original ADM parer.

    However, those who deal with numerical relativity uses $$\mathcal
    M_i=D_jK^j_i-D_iK.~~~~~~~~~~~~~~~(2)$$

    What formula is right? (they coincides only if ##\gamma## does not depend on spatial coordinates, which is evidently not the case.
     
  2. jcsd
  3. Jul 30, 2016 #2
    I think the equations 1a, 1b and 2 are all same (besides a factor of ##-\sqrt{\gamma}## in equation 2). To establish the equality you need to use the fact that the intrinsic covariant derivative is (pullback) metric compatible. Also ##D_i\sqrt{\gamma}=\frac{1}{2}\sqrt{\gamma}\gamma^{ab}D_i\gamma_{ab}=0##
     
  4. Jul 30, 2016 #3
    Let me ask, why we could not write ##D_i\sqrt {\gamma}=\partial_i \sqrt {\gamma}\sim\gamma^{ab}\partial_i\gamma_{ab}\ne0##? I ask this because it is well known that ##d\gamma\sim\gamma^{ab}d\gamma_{ab}##, where ##d## is usual differencial. From the other hand it seems that ##D_i\gamma=\partial_i \gamma##.
     
    Last edited: Jul 30, 2016
  5. Jul 30, 2016 #4
    Last edited: Jul 30, 2016
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