Dirac algebra of constraints in GR

[paweld]

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:
$$\{ H_a(f^a),H_b(h^b) \}$$
where:
$$H_a(f^a) := \int_{\Sigma} d^3 x [-2 q_{ac}(D_b P^{bc}) f^a] =<br /> \int_{\Sigma} d^3 x P^{ab} [q_{ac}D_b f^c + q_{cb}D_a f^c]$$
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):
$$\int_{\Sigma} d^3 x \sqrt{\textrm{det} q } F D_a G = -\int_{\Sigma} d^3 x \sqrt{\textrm{det} q } G D_a F$$
(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:
$$\{ H_a(f^a),H_b(h^b) \} = \int_{\Sigma} d^3 x (<br /> [q_{ac}D_b f^c + q_{cb}D_a f^c] [-2 (D_e P^{eb}) h^a] - <br /> [q_{ac}D_b h^c + q_{cb}D_a h^c] [-2 (D_e P^{eb}) f^a])$$
Because the connection D is torsion-free $$[\vec{f},\vec{h}]^a = f^b D_b h^a - h^b D_b f^a$$ and we get:
$$\{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a) -2 \int_{\Sigma} d^3 x <br /> (D_e P^{eb}) [q_{ac} h^a D_b f^c - q_{ac} f^a D_b h^c]$$
The answer should be just:
$$\{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a)$$
I wonder where I did a mistake or maybe the last integral in my formula
vanishes for some reasons. Can anyone help me?
Thanks.

 

[Omega137]

I'll work it out in a few days...

 

[paweld]

Any idea what's wrong with my calculation?

 

[marcus]

One idea would be to write email to a PhD student who has probably studied from the same book that you are using. Since you are using Thiemann's book, here is the list of people in his group at Erlangen:
http://theorie3.physik.uni-erlangen.de/people.html

One of the PhD students there, whose papers I have seen and consider excellent, is Christian Boehmer. The list of Thiemann's group gives his email as
christian.boehmer_at_theorie3.physik.uni-erlangen.de
replace _at_ by @

There are also postdocs there who work on LQG, like Enrique Borja.
Maybe if you write to Boehmer and he does not have time to look at it, he could still recommend and give you the email address of one of the others.

Here at PF, the most expert person is f-h.
You can also try writing a private message (PM) upper right corner of the PF screen
addressed simply to f-h and giving the link to your thread, with the equations. He may not have seen your post.
Here is the link to your thread:
https://www.physicsforums.com/showthread.php?p=3006419#post3006419

Sorry I can't give you better help than this.

 

[Careful]

one can easily find functinal derivatis of it
with respect to generalised positions and momenta and compute mentioned Poisson
bracket:
$$\{ H_a(f^a),H_b(h^b) \} = \int_{\Sigma} d^3 x (<br /> [q_{ac}D_b f^c + q_{cb}D_a f^c] [-2 (D_e P^{eb}) h^a] - <br /> [q_{ac}D_b h^c + q_{cb}D_a h^c] [-2 (D_e P^{eb}) f^a])$$

Didn't you make a mistake in this line ? When computing the Poisson brackets, did you take as well the q_{ab} and the Levi-Civita connection into account?

 

[paweld]

I think that this is correct if the second equality below holds:
$$H_a(f^a) := \int_{\Sigma} d^3 x [-2 q_{ac}(D_b P^{bc}) f^a] =<br /> \int_{\Sigma} d^3 x P^{ab} [q_{ac}D_b f^c + q_{cb}D_a f^c]$$
Or maybe my expression for functional derivatives of H are not good:
$$\frac{\delta H_a(f^a)}{\delta q_{ab}} = [-2 (D_e P^{eb}) f^a]$$
$$\frac{\delta H_a(f^a)}{\delta P^{ab}} = [q_{ac}D_b f^c + q_{cb}D_a f^c]$$

 

[Careful]

It appears to me that your first functional derivative might not be right.

 

[paweld]

Yes, you are right Careful.
I didn't take into account the fact that covarinat derivative depends on
metric. The first functional derivative should be:
$$\frac{\delta H_a(f^a)}{\delta q_{ab}} = P^{db} D_d f^a + P^{da} D_d f^b - D_d (P^{ab} f^d)$$
Now I get correct expression for Poisson bracket of constraints.
Thanks for help.