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Gravitation Lagrangian in classical form

  1. Nov 20, 2012 #1
    I'm trying to express the classical gravitation Einstein-Hilbert lagrangian into some nice way, and I'm having a problem.

    It is well known that the Einstein-Hilbert action is the following (I don't write the constant in front of the integral, to simplify things) :

    [itex]S_{EH} = \int R \, \sqrt{-g} \; d^4 x[/itex].

    After removing the hypersurface term, the lagrangian is this (this is a well known result) :

    [itex]\mathscr{L}_{EH} = g^{\mu \nu} \, (\, \Gamma_{\lambda \kappa}^{\lambda} \; \Gamma_{\mu \nu}^{\kappa} - \Gamma_{\mu \kappa}^{\lambda} \; \Gamma_{\lambda \nu}^{\kappa})[/itex].

    When I substitute the definition of the Christoffel symbols and simplify things, I get this :

    [itex]\mathscr{L}_{EH} = \frac{1}{4} H^{\mu \nu \lambda \kappa \rho \sigma} (\, \partial_{\mu} \, g_{\lambda \kappa})(\, \partial_{\nu} \, g_{\rho \sigma})[/itex],

    where I defined this horrible thing (this is the source of my problem) :

    [itex]H^{\mu \nu \lambda \kappa \rho \sigma} = \left( \, g^{\mu \nu} (\, g^{\lambda \rho} \, g^{\kappa \sigma} - g^{\lambda \kappa} \, g^{\rho \sigma}) + 2 \, g^{\nu \kappa} (\, g^{\mu \lambda} \, g^{\rho \sigma} - g^{\mu \rho} \, g^{\lambda \sigma}) \right)[/itex].

    This expression is ugly : it isn't symetric in [itex]\lambda \kappa[/itex] and [itex]\rho \sigma[/itex]. Of course, I can make it symetric, but then the [itex]\mu \nu[/itex] indices are getting in the way and make things more complicated.

    Is there a "natural" way to define a proper [itex]H^{\mu \nu \lambda \kappa \rho \sigma}[/itex] ? Any thoughts on this ?

    The lagrangian above is nice because it is very similar to the real scalar field lagrangian without mass :

    [itex]\mathscr{L} = \frac{1}{2} g^{\mu \nu} (\, \partial_{\mu} \, \phi \,)(\, \partial_{\nu}\, \phi \,)[/itex]
     
    Last edited: Nov 20, 2012
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  3. Nov 20, 2012 #2

    George Jones

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    This looks like the same expression that Padmanabhan writes down in his book Gravitation: Foundations and Frontiers. It doesn't look like he does anything more with this.
     
  4. Nov 20, 2012 #3
    Hmmm, that sounds very interesting. Could you say more about this ?

    Can you write down exactly his version ?
     
  5. Nov 20, 2012 #4
    I've found a sample of Padmanabhan's book, from this place :

    http://www.filestube.com/g/gravitation+foundations+and+frontiers

    On page 243, he gives an expression for the gravitation lagrangian that is indeed exactly like mine, with the symbol [itex]M[/itex] instead of my [itex]H[/itex], and with a global sign difference (he's using the other sign convention for the metric).

    So this is reassuring. :cool:

    Sadly, his expression isn't made symetric in some of its indices, exactly like mine.

    I'm pretty sure (from intuition) that this ugly tensor could be expressed in some nice way, but I don't know how.
    I feel it's a combination of simpler tensors, that could reveal some symetries. Something like this (but it's yet unsatisfying) :

    [itex]H^{\mu \nu \lambda \kappa \rho \sigma} = g^{\mu \nu} (\, g^{\lambda \rho} \, g^{\kappa \sigma} - g^{\lambda \kappa} \, g^{\rho \sigma} ) + g^{\nu \kappa} (\, g^{\mu \lambda} \, g^{\rho \sigma} - g^{\mu \rho} \, g^{\lambda \sigma} ) + g^{\mu \rho} (\, g^{\nu \sigma} \, g^{\lambda \kappa} - g^{\nu \lambda} \, g^{\kappa \sigma} )[/itex].

    That tensor should be explicitely symetric for the indices [itex]\lambda \kappa[/itex], [itex]\rho \sigma[/itex] and also under the triple exchange [itex]\mu \lambda \kappa \leftrightarrow \nu \rho \sigma[/itex].

    It could also be interpreted as a kind of "super-metric", or a kind of "metric" in the metric-space, but I'm not sure of this yet.

    Any suggestions ?
     
    Last edited: Nov 20, 2012
  6. Nov 21, 2012 #5
    I think I've found a solution to my "problem".

    In the simpler case of the Maxwell field, the lagrangian is this :

    [itex]\mathscr{L}_{\rm Maxwell} = -\, \frac{1}{4} \, F^{\mu \nu} F_{\mu \nu} \equiv \frac{1}{2} (\, g^{\mu \kappa} g^{\nu \lambda} - g^{\mu \nu} g^{\lambda \kappa})(\, \partial_{\mu} \, A_{\lambda})(\, \partial_{\nu} \, A_{\kappa})[/itex].

    So I define the following anti-symetric tensor :

    [itex]M^{\mu \nu \lambda \kappa} = g^{\mu \lambda} g^{\nu \kappa} - g^{\mu \kappa} g^{\nu \lambda}[/itex],

    which has the following properties :

    [itex]M^{\nu \mu \lambda \kappa} = -\, M^{\mu \nu \lambda \kappa}[/itex],
    [itex]M^{\mu \nu \kappa \lambda} = -\, M^{\mu \nu \lambda \kappa}[/itex],
    [itex]M^{\lambda \kappa \mu \nu} = M^{\mu \nu \lambda \kappa}[/itex],
    [itex]M^{\mu \nu \lambda \kappa} + M^{\mu \lambda \kappa \nu} + M^{\mu \kappa \nu \lambda} = 0[/itex].

    That [itex]M[/itex] tensor is also the fundamental representation of the Lorentz generators, which satisfies some commutation relations. The Maxwell lagrangian can then be written like this :

    [itex]\mathscr{L}_{\rm Maxwell} \equiv \frac{1}{2} M^{\mu \lambda \kappa \nu} (\, \partial_{\mu} \, A_{\lambda})(\, \partial_{\nu} \, A_{\kappa})[/itex].

    So it should be "natural" to express the gravitation lagrangian using this [itex]M[/itex] tensor. Once all symmetrizations were made, I've found this :

    [itex]\mathscr{L}_{\rm EH} = \frac{1}{16} \, \mathcal{G}^{\mu \nu \lambda \kappa \rho \sigma} (\, \partial_{\mu} \, g_{\lambda \kappa})(\, \partial_{\nu} \, g_{\rho \sigma})[/itex],

    where

    [itex]\mathcal{G}^{\mu \nu \lambda \kappa \rho \sigma} = g^{\lambda \rho} M^{\mu \kappa \nu \sigma} + g^{\kappa \rho} M^{\mu \lambda \nu \sigma} + g^{\lambda \sigma} M^{\mu \kappa \nu \rho} + g^{\kappa \sigma} M^{\mu \lambda \nu \rho} + g^{\lambda \kappa} (\, M^{\nu \sigma \rho \mu} + M^{\nu \rho \sigma \mu}) + g^{\rho \sigma} (\, M^{\mu \kappa \lambda \nu} + M^{\mu \lambda \kappa \nu})[/itex]

    This expression is fully symetric under the exchanges [itex]\lambda \kappa \leftrightarrow \kappa \lambda[/itex], [itex]\rho \sigma \leftrightarrow \sigma \rho[/itex] and the triple exchange [itex]\mu \lambda \kappa \leftrightarrow \nu \rho \sigma[/itex].

    Now, I suspect that this last expression could be simplified a bit (I don't like the last terms in parenthesis). Any idea ?
     
    Last edited: Nov 21, 2012
  7. Nov 21, 2012 #6
    I made a mistake in the last expression. The correct tensor is this one, which is pretty heavy :grumpy: :

    [itex]\mathcal{G}^{\mu \nu \lambda \kappa \rho \sigma} = 2 \, g^{\mu \nu} (\, M^{\lambda \rho \sigma \kappa} + M^{\lambda \sigma \rho \kappa}) + g^{\mu \rho} (\, M^{\nu \lambda \sigma \kappa} + M^{\nu \kappa \sigma \lambda}) + g^{\mu \sigma} (\, M^{\nu \lambda \rho \kappa} + M^{\nu \kappa \rho \lambda}) + g^{\nu \lambda} (\, M^{\mu \rho \kappa \sigma} + M^{\mu \sigma \kappa \rho}) + g^{\nu \kappa} (\, M^{\mu \rho \lambda \sigma} + M^{\mu \sigma \lambda \rho})[/itex].

    EDIT : If I introduce a symetric tensor instead of [itex]M[/itex], the previous expression could be simplified a bit :

    [itex]Q^{\mu \nu \lambda \kappa} = g^{\mu \lambda} g^{\nu \kappa} + g^{\mu \kappa} g^{\nu \lambda}[/itex], with the properties [itex]Q^{\nu \mu \lambda \kappa} = Q^{\mu \nu \kappa \lambda} = Q^{\mu \nu \lambda \kappa}[/itex], then :

    [itex]\mathcal{G}^{\mu \nu \lambda \kappa \rho \sigma} = 2 \, g^{\mu \nu} Q^{\lambda \kappa \rho \sigma} + 2 \, g^{\lambda \kappa} Q^{\mu \nu \rho \sigma} + 2 \, g^{\rho \sigma} Q^{\mu \nu \lambda \kappa} - g^{\mu \rho} Q^{\nu \sigma \lambda \kappa} - g^{\mu \sigma} Q^{\nu \rho \lambda \kappa} - g^{\nu \lambda} Q^{\mu \kappa \rho \sigma} - g^{\nu \kappa} Q^{\mu \lambda \rho \sigma} - 4 \, g^{\mu \nu} g^{\lambda \kappa} g^{\rho \sigma}[/itex].

    Hmmm, I must admit this is a mess ! :yuck:

    I'm throwing all that stuff through the window :mad:
     
    Last edited: Nov 21, 2012
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