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:
\{F,G\} \equiv \int_\Sigma d^3x \left(\frac{\delta F}{\delta P^{ab}(\vec{x})}\frac{\delta G}{\delta q_{ab}(\vec{x})} - <br /> \frac{\delta G}{\delta P^{ab}(\vec{x})}<br /> \frac{\delta P}{\delta q_{ab}(\vec{x})}\right)
where \Sigma is some three-dimensional hypersurface in the spacetime,
q - metric induced on \Sigma and P - canonically conjugated momentum.
I would like to compute the following Poisson bracket:
<br /> \{ H_a(f^a),H_b(h^b) \}<br />
where:
<br /> H_a(f^a) := \int_{\Sigma} d^3 x [-2 q_{ac}(D_b P^{bc}) f^a] =<br /> \int_{\Sigma} d^3 x P^{ab} [q_{ac}D_b f^c + q_{cb}D_a f^c]<br />
where D is covariant derivative associated with q_{ab}
and covariant derivative of tensorial density P^{ab} is by definition:
D_a P^{bc} := D_a (P^{bc}/\sqrt{\textrm{det} q })\sqrt{\textrm{det}q}
(since P^{bc}/\sqrt{\textrm{det} q } 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):
<br /> \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<br />
(for partial derivative there is no \sqrt{\textrm{det} q }).
Using equations for H_a(f^a) one can easily find functinal derivatis of it
with respect to generalised positions and momenta and compute mentioned Poisson
bracket:
<br /> \{ H_a(f^a),H_b(h^b) \} = \int_{\Sigma} d^3 x (<br /> [q_{ac}D_b f^c + q_{cb}D_a f^c] [-2 (D_e P^{eb}) h^a] - <br /> [q_{ac}D_b h^c + q_{cb}D_a h^c] [-2 (D_e P^{eb}) f^a])<br />
Because the connection D is torsion-free <br /> [\vec{f},\vec{h}]^a = f^b D_b h^a - h^b D_b f^a and we get:
<br /> \{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a) -2 \int_{\Sigma} d^3 x <br /> (D_e P^{eb}) [q_{ac} h^a D_b f^c - q_{ac} f^a D_b h^c]<br />
The answer should be just:
<br /> \{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a)<br />
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:
\{F,G\} \equiv \int_\Sigma d^3x \left(\frac{\delta F}{\delta P^{ab}(\vec{x})}\frac{\delta G}{\delta q_{ab}(\vec{x})} - <br /> \frac{\delta G}{\delta P^{ab}(\vec{x})}<br /> \frac{\delta P}{\delta q_{ab}(\vec{x})}\right)
where \Sigma is some three-dimensional hypersurface in the spacetime,
q - metric induced on \Sigma and P - canonically conjugated momentum.
I would like to compute the following Poisson bracket:
<br /> \{ H_a(f^a),H_b(h^b) \}<br />
where:
<br /> H_a(f^a) := \int_{\Sigma} d^3 x [-2 q_{ac}(D_b P^{bc}) f^a] =<br /> \int_{\Sigma} d^3 x P^{ab} [q_{ac}D_b f^c + q_{cb}D_a f^c]<br />
where D is covariant derivative associated with q_{ab}
and covariant derivative of tensorial density P^{ab} is by definition:
D_a P^{bc} := D_a (P^{bc}/\sqrt{\textrm{det} q })\sqrt{\textrm{det}q}
(since P^{bc}/\sqrt{\textrm{det} q } 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):
<br /> \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<br />
(for partial derivative there is no \sqrt{\textrm{det} q }).
Using equations for H_a(f^a) one can easily find functinal derivatis of it
with respect to generalised positions and momenta and compute mentioned Poisson
bracket:
<br /> \{ H_a(f^a),H_b(h^b) \} = \int_{\Sigma} d^3 x (<br /> [q_{ac}D_b f^c + q_{cb}D_a f^c] [-2 (D_e P^{eb}) h^a] - <br /> [q_{ac}D_b h^c + q_{cb}D_a h^c] [-2 (D_e P^{eb}) f^a])<br />
Because the connection D is torsion-free <br /> [\vec{f},\vec{h}]^a = f^b D_b h^a - h^b D_b f^a and we get:
<br /> \{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a) -2 \int_{\Sigma} d^3 x <br /> (D_e P^{eb}) [q_{ac} h^a D_b f^c - q_{ac} f^a D_b h^c]<br />
The answer should be just:
<br /> \{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a)<br />
I wonder where I did a mistake or maybe the last integral in my formula
vanishes for some reasons. Can anyone help me?
Thanks.