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Defining GR with Poisson Bracket

  1. Nov 11, 2006 #1
    If you have the metric [tex] g_{ab} [/tex] , [tex] \pi _{ab} [/tex] as the metric and "generalized momenta", my question is if you can define GR using Poisson Bracet:

    [tex] \dot g_{ab} =[g_{ab},H] [/tex]

    [tex] \dot \pi _{ab}=[\pi _{ab},H] [/tex]

    and hence use these equations to obtain and solve the metric.:shy:
     
  2. jcsd
  3. Nov 11, 2006 #2
    Karlisbad,

    I don't understand precisely your question.
    Specially, what are the [tex] \pi _{ab} [/tex] ?

    However, since GR derives from a least action principle, I guess that it can be written in an Hamiltonian way. (is it true that "least action" => "hamilton equations" ?)

    If this is right, could it be done along the usual lines?

    Michel
     
  4. Nov 11, 2006 #3

    hellfire

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    There exists a Hamiltonian formulation of general relativity. The "q", from which the "p" is defined, is the spatial metric [itex]h_{ab}[/itex], but not the space-time metric [itex]g_{ab}[/itex]. For the Hamiltonian formalism it is necessary to split space-time into space and time.
     
  5. Nov 11, 2006 #4
    Somebody else asked this question recently in the Diff. Geom. forum. My answer can be found here. Note, however, that some of my factors of [itex]\sqrt{g}[/itex] in the integrals are in fact incorrect. For example, if you take the Hamiltonian density [itex]\mathcal{H}[/itex] to be of weight one, then the Hamiltonian is

    [tex]H = \int_\Sigma d^3x \mathcal{H}[/itex]

    Then the standard equations of motion for the three-metric and momenta are given by

    [tex]g_{ij}(\vec{x}) = \{g_{ij}(\vec{x}),H\} = \int_\Sigma d^3x' \{g_{ij}(\vec{x}),\mathcal{H}(\vec{x}')\},[/tex]
    [tex]\pi^{ij}(\vec{x}) = \{\pi^{ij}(\vec{x}),H\} = \int_\Sigma d^3x' \{\pi^{ij}(\vec{x}),\mathcal{H}(\vec{x}')\}.[/tex]
     
    Last edited: Nov 11, 2006
  6. Nov 19, 2006 #5

    Chris Hillman

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    Wald describes the desired formulation

    Hi, Karlisbad,

    Just wanted to point out that the Hamiltonian formulation you seek is described (correctly) in the textbook by Wald, General Relativity.

    Chris Hillman
     
  7. Nov 19, 2006 #6
    I don't think that's entirely true. IMO, Wald makes a hash of describing the Hamiltonian approach. Of particular relevance to the OP, Wald also doesn't discuss the evolution equations in terms of the Poisson brackets of the metric and its conjugate momentum, and makes only a half-hearted attempt at explaining how Poisson brackets are of relevance to the constraints.

    I hate that appendix in Wald. The fact that he uses an incorrect action for GR is, frankly, embarrassing.
     
  8. Nov 19, 2006 #7

    robphy

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    Please enlighten us.
     
  9. Nov 19, 2006 #8
    Well, Wald is correct to state that the basic action from which the field equations can be derived is the Einstein-Hilbert action:

    [tex]S_{\textrm{E-H}}[g] = \frac{1}{2\kappa}\int_\mathcal{M} d^4x\,\sqrt{-g}R + S_M,[/tex]

    where [itex]S_M[/itex] is a (possibly derivatively coupled) matter action. This is all fine if [itex]\mathcal{M}[/itex] has no boundary. However, if [itex]\partial\mathcal{M}\ne\emptyset[/itex] then in order for the variational principle to be well posed one needs to add the Gibbons-Hawking-York boundary term [itex]S_{\partial\mathcal{M}}[g][/itex]. Then we have

    [tex]S[g] = \frac{1}{2\kappa}\int_\mathcal{M} d^4x\sqrt{-g}R + \frac{1}{\kappa}\int_{\partial\mathcal{M}}d^3y \sqrt{|h|}\textrm{tr}K + S_M,
    [/tex]

    where [itex]h_{ij}[/itex] is a three-metric on [itex]\partial\mathcal{M}[/itex] and [itex]\textrm{tr}K=h^{ij}K_{ij}[/itex] is the trace of the extrinsic curvature of [itex]\partial\mathcal{M}[/itex].

    In fairness, Wald does stress the importance of this boundary contribution to the action, but he concludes that the action above is sufficient to derive sensible field equations. This is untrue. If you evaluate the gravitational action for, say, flat spacetime, then [itex]S_{\textrm{E-H}}[g]=0[/itex]. However, for flat spacetime [itex]S_{\partial\mathcal{M}}[g][/itex] is divergent, making the action effectively infinite. Thus, the action that Wald uses is actually ill defined except when [itex]\mathcal{M}[/itex] is compact. In order to overcome this, one needs to introduce a further correction to the action, meaning that the true action for general relativity is

    [tex] S = S_{\textrm{E-H}}[g] + S_{\partial\mathcal{M}}[g] + S_M - \frac{1}{\kappa}\int_{\partial\mathcal{M}} d^3y\sqrt{|h|}K_0[/tex]

    where [itex]K_0[/itex] is the extrinsic curvature of [itex]\partial\mathcal{M}[/itex] embedded in Minkowski space.
     
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