Defining GR with Poisson Bracket

Karlisbad
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If you have the metric g_{ab} , \pi _{ab} as the metric and "generalized momenta", my question is if you can define GR using Poisson Bracet:

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

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

and hence use these equations to obtain and solve the metric.:shy:
 
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Karlisbad,

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

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
 
There exists a Hamiltonian formulation of general relativity. The "q", from which the "p" is defined, is the spatial metric h_{ab}, but not the space-time metric g_{ab}. For the Hamiltonian formalism it is necessary to split space-time into space and time.
 
Karlisbad said:
If you have the metric g_{ab} , \pi _{ab} as the metric and "generalized momenta", my question is if you can define GR using Poisson Bracet:

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

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

and hence use these equations to obtain and solve the metric.:shy:

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 \sqrt{g} in the integrals are in fact incorrect. For example, if you take the Hamiltonian density \mathcal{H} to be of weight one, then the Hamiltonian is

H = \int_\Sigma d^3x \mathcal{H}[/itex]<br /> <br /> Then the standard equations of motion for the three-metric and momenta are given by<br /> <br /> g_{ij}(\vec{x}) = \{g_{ij}(\vec{x}),H\} = \int_\Sigma d^3x&amp;#039; \{g_{ij}(\vec{x}),\mathcal{H}(\vec{x}&amp;#039;)\},<br /> \pi^{ij}(\vec{x}) = \{\pi^{ij}(\vec{x}),H\} = \int_\Sigma d^3x&amp;#039; \{\pi^{ij}(\vec{x}),\mathcal{H}(\vec{x}&amp;#039;)\}.
 
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Wald describes the desired formulation

Hi, Karlisbad,

Karlisbad said:
If you have the metric g_{ab} , \pi _{ab} as the metric and "generalized momenta", my question is if you can define GR using Poisson Bracet:

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

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

and hence use these equations to obtain and solve the metric.:shy:

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

Chris Hillman
 
Chris Hillman said:
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

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.
 
coalquay404 said:
I hate that appendix in Wald. The fact that he uses an incorrect action for GR is, frankly, embarrassing.

Please enlighten us.
 
robphy said:
Please enlighten us.

Well, Wald is correct to state that the basic action from which the field equations can be derived is the Einstein-Hilbert action:

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

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

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,<br />

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

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 S_{\textrm{E-H}}[g]=0. However, for flat spacetime S_{\partial\mathcal{M}}[g] is divergent, making the action effectively infinite. Thus, the action that Wald uses is actually ill defined except when \mathcal{M} 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

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

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