Gravitation Lagrangian in classical form

In summary, the Gravitation Lagrangian in classical form is a mathematical expression that describes the dynamics of a system under the influence of gravitational forces. It includes the position and velocity of all particles, as well as potential and kinetic energy. It is used to derive equations of motion and can also be applied to other branches of physics. It differs from the Gravitation Hamiltonian, which is based on the conservation of energy.
  • #1
Barnak
63
0
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]
 
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  • #2
Barnak said:
[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 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.
 
  • #3
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 ?
 
  • #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 ?
 
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  • #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 ?
 
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  • #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:
 
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1. What is the Gravitation Lagrangian in classical form?

The Gravitation Lagrangian in classical form is a mathematical expression that describes the dynamics of a system under the influence of gravitational forces. It is derived from the Lagrangian mechanics, which is a mathematical framework for analyzing the motion of a system.

2. What are the variables included in the Gravitation Lagrangian?

The Gravitation Lagrangian includes the position and velocity of all the particles in the system, as well as the gravitational potential energy and kinetic energy of the system.

3. How is the Gravitation Lagrangian used in classical mechanics?

The Gravitation Lagrangian is used to derive the equations of motion for a system under the influence of gravitational forces. These equations can then be solved to determine the trajectory and behavior of the system over time.

4. What is the difference between the Gravitation Lagrangian and the Gravitation Hamiltonian?

The Gravitation Lagrangian and the Gravitation Hamiltonian are two different mathematical approaches to analyzing the dynamics of a system under the influence of gravitational forces. The Lagrangian approach is based on the principle of least action, while the Hamiltonian approach is based on the conservation of energy.

5. Can the Gravitation Lagrangian be applied to systems beyond classical mechanics?

Yes, the Gravitation Lagrangian can also be used in other branches of physics, such as quantum mechanics and general relativity. In these cases, it is known as the Quantum Gravitation Lagrangian and the General Relativity Lagrangian, respectively.

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