Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  2. jcsd
  3. Nov 20, 2012 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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 :


    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],


    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook