Gravitation Lagrangian in classical form

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Discussion Overview

The discussion revolves around expressing the classical gravitational Einstein-Hilbert Lagrangian in a more manageable form. Participants explore the mathematical structure of the Lagrangian, particularly focusing on the tensor components involved and their symmetries. The scope includes theoretical considerations and mathematical reasoning related to gravitation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the Einstein-Hilbert action and derives a specific form of the Lagrangian, expressing concerns about the complexity of the tensor involved.
  • Another participant notes that the derived tensor resembles an expression found in Padmanabhan's book, but does not elaborate further.
  • A later reply requests more details about Padmanabhan's version of the expression.
  • One participant finds a version of the gravitational Lagrangian in Padmanabhan's book that matches their own, noting a sign difference due to metric conventions.
  • Another participant proposes a potential redefinition of the tensor to reveal symmetries, suggesting it could be expressed as a combination of simpler tensors.
  • One participant draws a parallel to the Maxwell field Lagrangian, introducing an antisymmetric tensor and discussing its properties, suggesting a natural way to express the gravitational Lagrangian using this tensor.
  • Another participant corrects a previous expression for the tensor and introduces a new symmetric tensor, indicating that this could simplify the overall expression.
  • One participant expresses frustration with the complexity of the derived expressions and considers abandoning the approach.

Areas of Agreement / Disagreement

Participants express varying opinions on the best way to define the tensor involved in the gravitational Lagrangian. There is no consensus on a single "natural" form, and the discussion remains unresolved regarding the simplification and symmetry of the expressions.

Contextual Notes

Participants acknowledge the complexity and potential for simplification in the tensor expressions, but do not resolve the mathematical intricacies or assumptions underlying their formulations.

Barnak
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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) :

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

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

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

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

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

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

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).

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

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

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

\mathscr{L} = \frac{1}{2} g^{\mu \nu} (\, \partial_{\mu} \, \phi \,)(\, \partial_{\nu}\, \phi \,)
 
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Barnak said:
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)

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.
 
George Jones said:
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.

Hmmm, that sounds very interesting. Could you say more about this ?

Can you write down exactly his version ?
 
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 M instead of my H, 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) :

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} ).

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

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 ?
 
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I think I've found a solution to my "problem".

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

\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}).

So I define the following anti-symetric tensor :

M^{\mu \nu \lambda \kappa} = g^{\mu \lambda} g^{\nu \kappa} - g^{\mu \kappa} g^{\nu \lambda},

which has the following properties :

M^{\nu \mu \lambda \kappa} = -\, M^{\mu \nu \lambda \kappa},
M^{\mu \nu \kappa \lambda} = -\, M^{\mu \nu \lambda \kappa},
M^{\lambda \kappa \mu \nu} = M^{\mu \nu \lambda \kappa},
M^{\mu \nu \lambda \kappa} + M^{\mu \lambda \kappa \nu} + M^{\mu \kappa \nu \lambda} = 0.

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

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

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

\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}),

where

\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})

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

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

\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}).

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

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

\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}.

Hmmm, I must admit this is a mess !

I'm throwing all that stuff through the window :mad:
 
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