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Variation of the Einstein-Hilbert action in noncoordinate basis

  1. Jul 10, 2014 #1
    The variation of the Einstein Hilbert action is usually done in coordinate basis where there is a crucial divergence term one can neglect which arise in the variation of the Ricci tensor, and is given by ##g^{ab}\delta R_{ab} = \nabla_c w^c## where
    $$w^c = g^{ab}(g^{db} \delta \Gamma^{c}_{db} - g^{cb} \delta\Gamma^c_{db})$$.
    However, when one varies in a noncoordinate basis, one supposedly (see link below) get's an extra term and arrive at
    $$g^{ab}\delta R_{ab} = \nabla_c w^c - C^c_{cd} \delta \Gamma^d_{ab}g^{ab}.$$

    How is this result derived?

    My calculations so far is the following: From the Riemann tensor in a noncoordinate basis ##\{e_a \}## with structure constants ##[e_b, e_c] = C^a_{bc}e_a## given by
    $$R^a_{bcd} = e_c \Gamma^a_{db} - e_d\Gamma^a_{cb} + \Gamma^f_{db} \Gamma^a_{cf} - \Gamma^f_{cb} \Gamma^a_{df} - C^f_{cd} \Gamma^a_{fb}$$
    the variation yields
    $$\delta R^a_{bcd} = e_c \delta \Gamma^a_{db} - e_d\delta \Gamma^a_{cb} + \delta\Gamma^f_{db} \Gamma^a_{cf} + \Gamma^f_{db} \delta\Gamma^a_{cf} - \delta\Gamma^f_{cb} \Gamma^a_{df} - \Gamma^f_{cb} \delta \Gamma^a_{df} - C^f_{cd} \delta \Gamma^a_{fb}- \delta C^f_{cd} \Gamma^a_{fb}$$
    From here we can extract three terms of the respective covariant derivatives ##\nabla_d \delta \Gamma^a_{cb}## and ##\nabla_c \delta \Gamma^a_{db}##; however the terms ##\Gamma^f_{cd}\delta \Gamma^a_{fb}## and ##\Gamma^f_{dc} \delta \Gamma^a_{fb}## are not present. Taking these into account, I get
    $$\delta R^a_{bcd} = \nabla_c \delta \Gamma^a_{db} - \nabla_d \delta \Gamma^a_{cb} + (\Gamma^f_{cd} - \Gamma^f_{dc}) \delta \Gamma^a_{fb} - C_{cd}^f\delta\Gamma^a_{fb} - \delta C_{cd}^f\Gamma^a_{fb}= \nabla_c \delta \Gamma^a_{db} - \nabla_d \delta \Gamma^a_{cb}- \delta C_{cd}^f\Gamma^a_{fb}.$$
    If there are no error in my calculations, where do I go from here? Assuming I have done nothing wrong so far, it seems like one must achieve the equality ##\delta C_{cd}^f\Gamma^c_{fb} g^{db} = C^c_{cd} \delta \Gamma^d_{ab}g^{ab}##.. Any ideas on how this equality might be obtained?


    Equation (2.9) in: http://scitation.aip.org/content/aip/journal/jmp/15/6/10.1063/1.1666735
     
  2. jcsd
  3. Jul 10, 2014 #2
    As the subject of the thread states, I wonder about noncoordinate bases. I.e. bases for which ##e_a \neq \partial_a##.
     
  4. Jul 10, 2014 #3

    Mentz114

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    Gold Member

    Tangent space basis vectors like ##e_a = f \partial_a## commute, so the problem goes away for them.
    If ##{C^f}_{cd}## is constant, then is ##\delta {C^f}_{cd} = 0## ?

    In which case you need ## {C^c}_{cd} = 0##, which could be the case.

    [Edit]
    I guess you saw my deleted post.
     
  5. Jul 12, 2014 #4
    When one varies the components of the
    metric (expressed in a basis), one also varies the associated basis, and since the structure constants are spacetime functions associated to a particular basis, these varies too. Thus, it seems like the variation of the structure constant does not disappear.
     
  6. Jul 15, 2014 #5
    To correct my previous answer to you, I do agree that the structure constants are constant, since the variation is performed for a fixed basis.

    Thus, the mystery where the additional term involving ##C_{cd}^{\ \ c}## comes from is even bigger.
     
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