- #1
paweld
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In hamiltonian formulation of GR there appears some constraints (it may be found
e.g. in "Modern canonical quantum GR" by Theimann, ch. 1.2).
I would like to find a Dirac algebra of the constraints (i.e. compute
Poisson bracket between constraints), but my results are not consistent
with formulas given by Thiemann (eq. 1.2.15).
Below I present my calculations:
Let's assume that the Poisson bracket is defined as follows:
[tex]\{F,G\} \equiv \int_\Sigma d^3x \left(\frac{\delta F}{\delta P^{ab}(\vec{x})}\frac{\delta G}{\delta q_{ab}(\vec{x})} -
\frac{\delta G}{\delta P^{ab}(\vec{x})}
\frac{\delta P}{\delta q_{ab}(\vec{x})}\right)[/tex]
where [tex]\Sigma[/tex] is some three-dimensional hypersurface in the spacetime,
q - metric induced on [tex]\Sigma[/tex] and P - canonically conjugated momentum.
I would like to compute the following Poisson bracket:
[tex]
\{ H_a(f^a),H_b(h^b) \}
[/tex]
where:
[tex]
H_a(f^a) := \int_{\Sigma} d^3 x [-2 q_{ac}(D_b P^{bc}) f^a] =
\int_{\Sigma} d^3 x P^{ab} [q_{ac}D_b f^c + q_{cb}D_a f^c]
[/tex]
where [tex] D[/tex] is covariant derivative associated with [tex] q_{ab}[/tex]
and covariant derivative of tensorial density [tex]P^{ab} [/tex] is by definition:
[tex] D_a P^{bc} := D_a (P^{bc}/\sqrt{\textrm{det} q })\sqrt{\textrm{det}q}[/tex]
(since [tex] P^{bc}/\sqrt{\textrm{det} q } [/tex] is ordinary tensor). The second equality
in above equation holds because the following way of doing integration by parts works
for any tensors F, G (I omit indicies, and assume that boundary term vanishes):
[tex]
\int_{\Sigma} d^3 x \sqrt{\textrm{det} q } F D_a G = -\int_{\Sigma} d^3 x \sqrt{\textrm{det} q } G D_a F
[/tex]
(for partial derivative there is no [tex]\sqrt{\textrm{det} q } [/tex]).
Using equations for [tex] H_a(f^a) [/tex] one can easily find functinal derivatis of it
with respect to generalised positions and momenta and compute mentioned Poisson
bracket:
[tex]
\{ H_a(f^a),H_b(h^b) \} = \int_{\Sigma} d^3 x (
[q_{ac}D_b f^c + q_{cb}D_a f^c] [-2 (D_e P^{eb}) h^a] -
[q_{ac}D_b h^c + q_{cb}D_a h^c] [-2 (D_e P^{eb}) f^a])
[/tex]
Because the connection D is torsion-free [tex]
[\vec{f},\vec{h}]^a = f^b D_b h^a - h^b D_b f^a [/tex] and we get:
[tex]
\{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a) -2 \int_{\Sigma} d^3 x
(D_e P^{eb}) [q_{ac} h^a D_b f^c - q_{ac} f^a D_b h^c]
[/tex]
The answer should be just:
[tex]
\{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a)
[/tex]
I wonder where I did a mistake or maybe the last integral in my formula
vanishes for some reasons. Can anyone help me?
Thanks.
e.g. in "Modern canonical quantum GR" by Theimann, ch. 1.2).
I would like to find a Dirac algebra of the constraints (i.e. compute
Poisson bracket between constraints), but my results are not consistent
with formulas given by Thiemann (eq. 1.2.15).
Below I present my calculations:
Let's assume that the Poisson bracket is defined as follows:
[tex]\{F,G\} \equiv \int_\Sigma d^3x \left(\frac{\delta F}{\delta P^{ab}(\vec{x})}\frac{\delta G}{\delta q_{ab}(\vec{x})} -
\frac{\delta G}{\delta P^{ab}(\vec{x})}
\frac{\delta P}{\delta q_{ab}(\vec{x})}\right)[/tex]
where [tex]\Sigma[/tex] is some three-dimensional hypersurface in the spacetime,
q - metric induced on [tex]\Sigma[/tex] and P - canonically conjugated momentum.
I would like to compute the following Poisson bracket:
[tex]
\{ H_a(f^a),H_b(h^b) \}
[/tex]
where:
[tex]
H_a(f^a) := \int_{\Sigma} d^3 x [-2 q_{ac}(D_b P^{bc}) f^a] =
\int_{\Sigma} d^3 x P^{ab} [q_{ac}D_b f^c + q_{cb}D_a f^c]
[/tex]
where [tex] D[/tex] is covariant derivative associated with [tex] q_{ab}[/tex]
and covariant derivative of tensorial density [tex]P^{ab} [/tex] is by definition:
[tex] D_a P^{bc} := D_a (P^{bc}/\sqrt{\textrm{det} q })\sqrt{\textrm{det}q}[/tex]
(since [tex] P^{bc}/\sqrt{\textrm{det} q } [/tex] is ordinary tensor). The second equality
in above equation holds because the following way of doing integration by parts works
for any tensors F, G (I omit indicies, and assume that boundary term vanishes):
[tex]
\int_{\Sigma} d^3 x \sqrt{\textrm{det} q } F D_a G = -\int_{\Sigma} d^3 x \sqrt{\textrm{det} q } G D_a F
[/tex]
(for partial derivative there is no [tex]\sqrt{\textrm{det} q } [/tex]).
Using equations for [tex] H_a(f^a) [/tex] one can easily find functinal derivatis of it
with respect to generalised positions and momenta and compute mentioned Poisson
bracket:
[tex]
\{ H_a(f^a),H_b(h^b) \} = \int_{\Sigma} d^3 x (
[q_{ac}D_b f^c + q_{cb}D_a f^c] [-2 (D_e P^{eb}) h^a] -
[q_{ac}D_b h^c + q_{cb}D_a h^c] [-2 (D_e P^{eb}) f^a])
[/tex]
Because the connection D is torsion-free [tex]
[\vec{f},\vec{h}]^a = f^b D_b h^a - h^b D_b f^a [/tex] and we get:
[tex]
\{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a) -2 \int_{\Sigma} d^3 x
(D_e P^{eb}) [q_{ac} h^a D_b f^c - q_{ac} f^a D_b h^c]
[/tex]
The answer should be just:
[tex]
\{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a)
[/tex]
I wonder where I did a mistake or maybe the last integral in my formula
vanishes for some reasons. Can anyone help me?
Thanks.