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Poisson brackets for Hamiltonian descriptions

  1. Feb 5, 2009 #1

    haushofer

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    Hi, I have a (maybe rather technical) question about the Hamiltonian formulation of gauge theories, which I don't get.

    With an infinitesimal symmetry on your space-time M one can look at the corresponding transformation of the canonical variables in phase-space PS. This can be done by a phase space function H[]:M --> PS. The generators of these symmetries give you global charges.

    If I generate my symmetry with a vector field [itex]\xi[/itex] or [itex]\alpha [/itex], the statement which is often made is that

    [tex]
    \{H[\xi],H[\alpha] \} = H [[\xi,\alpha]]
    [/tex]

    In words: the Poisson algebra {} of the charges (LHS) is isomorphic (=) to the Lie algebra of the corresponding symmetries on your space-time (RHS).

    My question is: why is this so natural to assume? Henneaux and Brown wrote an article about the details and adjustments of this assumption (central charges in the canonical realization...), but I don't see why this should be "natural" in the first place. I'm missing something here.

    Thanks in forward :)
     
  2. jcsd
  3. Feb 7, 2009 #2

    samalkhaiat

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    I hope, I have uderstood your question correctly!

    In the real life, we use Noether procedure to write [itex]H_{\alpha}[/itex] in terms of the dynamical variables on PS:

    [tex]H_{\alpha} = \int dx \pi \mathcal{L}_{\alpha} \phi \ \ \ (R)[/tex]

    where [itex]\mathcal{L}_{\alpha}[/itex] is Lie derivative along the vector field [itex]\alpha[/itex].
    It follows immediately from eq(R) that

    1) canonical transformations generated by the charge/constraint correspond precisely to the diffeomorphisms generated by the vector field [itex]\alpha[/itex] ,i.e.,

    [tex]
    \{ H_{\alpha} , \phi \} = \mathcal{L}_{\alpha} \phi, \ \ \mbox{etc.} \ (1)
    [/tex]

    2) charges/constraints satisfy the algebra

    [tex]\{ H_{\alpha}, H_{\xi} \} = H_{[\alpha , \xi ]} \ \ \ (2)[/tex]

    To confirm the consistency between eq{1) and eq(2), we use the Jacobi identity

    [tex]\{ \{ H_{\alpha} , H_{\xi} \} , \phi \} = \{\{H_{\alpha}, \phi\} , H_{\xi} \} - \{\{H_{\xi} , \phi \} , H_{\alpha} \}[/tex]

    This gives

    [tex]\{\{H_{\alpha} , H_{\xi} \} , \phi \} = \mathcal{L}_{\alpha} \mathcal{L}_{\xi} \phi - \mathcal{L}_{\xi}\mathcal{L}_{\alpha} \phi = \mathcal{L}_{[\alpha , \xi ]} \phi[/tex]

    However, the algebra of constraints is not, in general, a proper Lie algebra: the RHS of eq(2) may contain structure functions rather than structure constants. This means, in the BRST terminology, that we may have an open algebra. Indeed, it has been shown [Bargmann and Komar] that the algebra of constraints is not isomorphic to the Lie algebra of the "gauge group" of tetrad-gravity.

    *****

    Ok, here is a math question for you:

    Given two function [itex] f,g : PS \rightarrow \mathbb{R}[/itex], their Poisson bracket is defined by

    [tex]\{ f , g \} = \mathcal{L}_{X_{f}}g = - \mathcal{L}_{X_{g}}f[/tex]

    this means that PB turns the vector space of functions on PS into Lie algebra. Use this Lie-bracket to show that the linear map [itex] f \rightarrow X_{f}[/itex] (which associates to f its (Hamiltonian) vector field) takes Poisson brackets of functions into commutators of vector fields:

    [tex][X_{f} , X_{g}] = X_{\{f,g\}} \ \ (3)[/tex]

    compair this with eq(2).

    regards

    sam
     
  4. Feb 9, 2009 #3

    haushofer

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    A, nice! I will look at this tomorrow! Sam to the rescue, again :D
     
  5. Feb 11, 2009 #4

    samalkhaiat

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    It is a pleasure, very time.
     
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