# Generalized Poisson brackets

1. Oct 2, 2006

### shoehorn

Hi. I've been wondering about the following and haven't made much progress on it. (Note that I've also posted this in the relativity section since the ultimate aim of this is to apply it to canonical relativity but since this is essentially a question about tensors I thought I'd put a copy here too. Sorry if this is a hassle.)

To set the scene, consider the following. Suppose that we have some sort of discrete theory in which the phase space variables are $$q^i$$ and $$p_i$$. If we have some functions $$F(q,p)$$ and $$G(q,p)$$ we can define their Poisson bracket as

$$\{F,G\} = \sum_i\left( \frac{\partial F}{\partial q^i}\frac{\partial G}{\partial p_i} - \frac{\partial F}{\partial p_i}\frac{\partial G}{\partial q^i}\right)$$

Then, for example, we have the following fundamental Poisson brackets

$$\{q^i,p_j\} = \delta^i_{\phantom{i}j}, \{q^i,q^j\} = \{p_i,p_j\} = 0.$$

Now suppose that we go to a field theory in which the fundamental `variables' are the position-dependent $$q^i(\vec{x})$$ and $$\pi_j(\vec{x})$$. The Poisson brackets of two functionals $$F(q,\pi)$$ and $$G(q,\pi)$$ are then

$$\{F,G\} \equiv \sum_i \int_\mathcal{M} dv \left( \frac{\delta F}{\delta q^i}\frac{\delta G}{\delta \pi_i} - \frac{\delta F}{\delta \pi_i}\frac{\delta G}{\delta q^i}\right).$$

With this definition we obtain

$$\{q^i(\vec{x}),\pi_j(\vec{x}')\} = \delta^i_{\phantom{i}j}\delta(\vec{x}-\vec{x}'), \{q^i(\vec{x}),q^j(\vec{x}')\} = \{\pi_i(\vec{x}),\pi_j(\vec{x}')\} = 0.$$

That's all fine and good. But, in the Hamiltonian version of general relativity the fundamental variables are actually the three-dimensional metric $$g_{ij}(\vec{x})$$ and $$\pi^{ij}(\vec{x})$$, both of which are symmetric tensors. I can define the Poisson brackets for these variables easily as

$$\{F,G\} \equiv \int_\mathcal{M}\left( \frac{\delta F}{\delta g_{ij}}\frac{\delta G}{\delta \pi^{ij}} - \frac{\delta F}{\delta\pi^{ij}}\frac{\delta G}{\delta g_{ij}}\right).$$

My question is this: what exactly is

$$\frac{\delta g_{ij}}{\delta g_{kl}}?$$

I assume that it's not something simple like $$\delta^i_k\delta^j_l$$ and that it has to reflect the symmetry of the metric tensor, but I can't seem to figure it out. Since I can't figure it out, neither can I work out explicitly what the fundamental Poisson brackets for GR should be.

Can anyone shed some light on this? An immediate extension that springs to mind is the general case of the above, i.e., if $$T^{a_1\ldots a_r}$$ is some arbitrary (r,0) tensor then what is

$$\frac{\delta T^{a_1\ldots a_r}}{\delta T^{b_1\ldots b_r}}$$?

2. Oct 10, 2006

### mtiano

I'm not sure if this will help but I'll give it a try. It may depend on the configuration space you are using for relativity. For example one common configuration space is to take the space of all riemannian metrics on your manifold and mod out by the diffeomorphism group. Now if done this way Marsden and Fischer found that whenever you try to write the hamiltonian down you always end up with zero.

3. Oct 12, 2006

### coalquay404

I don't see how this is related to the OP's question. For example, the fact that the Hamiltonian for general relativity is zero (on-shell) has nothing to do with Fischer and Marsden - it's a consequence of the reparametrization invariance of general relativity.

Going back to the OP's question, I'll point out that what you're asking does not in fact depend on the configuration space at all - the key lies in some pretty straightforward functional derivation. I'm not going to go through rigorous proofs of anything, but I will show you some shorthand ways of remembering things. You know that if you have a (0,2) tensor $$A_{ij}$$ you can write it as

$$A_{ij} = A_{(ij)} + A_{[ij]}$$

where the brackets denote symmetrization and antisymmetrization, respectively. I define symmetrization and antisymmetrization of a (0,2) tensor by

$$A_{(ij)} = \frac{1}{2}\bar{\delta}^{kl}_{ij}A_{kl}$$
$$A_{[ij]} = \frac{1}{2}\delta^{kl}_{ij}A_{kl}$$

where

$$\bar{\delta}^{kl}_{ij} = \delta^k_i\delta^l_j + \delta^k_j\delta^l_i$$
$$\delta^{kl}_{ij} = \delta^k_i\delta^l_j - \delta^k_j\delta^l_i$$

So, since the three-metric $$g_{ij}$$ is symmetric we can write $$g_{ij} = \frac{1}{2}\bar{\delta}^{kl}_{ij}g_{kl}$$, which immediately impies

$$\delta g_{ij} = \frac{1}{2}\bar{\delta}^{kl}_{ij}\delta g_{kl}$$

and similarly for the momenta. Now regard the phase space for general relativity as being composed of the pairs $$(g_{ij},\pi^{kl})$$ (beware: this isn't really a manifold at all - it's a stratified manifold and if you want to prove anything rigorous regarding it you need to be *extremely* careful). Then, for phase space functionals F and G, define the Poisson brackets as

$$\{F,G\} \equiv \int_\Sigma d^3x \left(\frac{\delta F}{\delta g_{mn}(\vec{x})}\frac{\delta G}{\delta \pi^{mn}(\vec{x})} - \frac{\delta G}{\delta g_{mn}(\vec{x})}\frac{\delta F}{\delta \pi^{mn}(\vec{x})}\right)$$

where $$\Sigma$$ is some three-dimensional hypersurface in the spacetime. If you use the above result for the variation of $$g_{ij}$$ and note that

$$\frac{\delta g_{ij}(\vec{x})}{\delta g_{kl}(\vec{y})} = \frac{1}{2}\bar{\delta}^{kl}_{ij}\delta^{(3)}(\vec{x}-\vec{y})$$

where $$\delta^{(3)}(\vec{x}-\vec{y})$$ is the three-dimensional Dirac delta function, it's then straightforward to work out explicitly the fundamental Poisson brackets for general relativity. If I remember correctly, you should get

$$\{g_{ij}(\vec{x}),\pi^{kl}(\vec{x}')\} = \frac{1}{2}\bar{\delta}^{kl}_{ij}\delta^{(3)}(\vec{x}-\vec{x}')$$
$$\{g_{ij}(\vec{x}),g_{kl}(\vec{x}')\} = \{\pi^{ij}(\vec{x}),\pi^{kl}(\vec{x}')\} = 0$$

You can then go through the usual steps of the Hamiltonian method. In particular, since the Hamiltonian for general relativity vanishes on-shell, i.e., on the constraint surface, we can define a Hamiltonian that's good over the entire phase space simply as a linear combination of the constraints:

$$H = \int_\Sigma d^3x \sqrt{g}\mathcal{H} = \int_\Sigma d^3x\sqrt{g}(N\mathcal{C} + N_i \mathcal{C}^i)$$

where ($$N,N^i$$) are the lapse and shift and ($$\mathcal{C},\mathcal{C}^i$$) are the Hamiltonian and momentum constraints, respectively (see chapter 21 of MTW for a definition of these constraints - they call them the superHamiltonian and the supermomentum). Some tedious (and quite hard) algebra then gives you the evolution equations for the gravitational field:

$$\dot{g}_{ij}(\vec{x}) = \int_\Sigma d^3x'\sqrt{g}\{g_{ij}(\vec{x}),\mathcal{H}(\vec{x}')\}$$
$$\dot{\pi}^{ij}(\vec{x}) = \int_\Sigma d^3x'\sqrt{g}\{\pi^{ij}(\vec{x}),\mathcal{H}(\vec{x}')\}$$

(I may have gotten some of the factors of $$\sqrt{g}$$ wrong here - check them for yourself.) The final step in the Hamiltonian method is then to look at the Poisson brackets of the constraints with each other. You take the set of constraints and define the Poisson brackets above in order to give you the Dirac algebra (this is simply an algebra with the constraints as elements and the Poisson bracket as the algebraic operation). You can prove that the constraints are satisfied as the theory evolves forward in time by calculating out all of the Poisson brackets to prove that the algebra closes.

Oh, and by the way, your question about extending the variation to tensors of arbitrary rank can be answered by the method above. Just express the tensor as a sum of symmetric and antisymmetric parts, carry out the variation, and then do a functional derivative. There are some rigorous properties that the resultant functional derivative must have but if you follow the simple rules you don't need to worry about these.

Last edited: Oct 13, 2006
4. Nov 23, 2010

### paweld

I decided not to start new topic because my question concerns the matter
discussed above. Let's assume that the Poisson bracket is defined as follows:
$$\{F,G\} \equiv \int_\Sigma d^3x \left(\frac{\delta F}{\delta P^{ab}(\vec{x})}\frac{\delta G}{\delta q_{ab}(\vec{x})} - \frac{\delta G}{\delta P^{ab}(\vec{x})} \frac{\delta P}{\delta q_{ab}(\vec{x})}\right)$$
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:
$$H_a(f^a) := \int_{\Sigma} d^3 x [-2 q_{ac}(D_b P^{bc}) f^a] = \int_{\Sigma} d^3 x P^{ab} [q_{ac}D_b f^c + q_{cb}D_a f^c]$$
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:
$$\{ H_a(f^a),H_b(h^b) \} = \int_{\Sigma} d^3 x ( [q_{ac}D_b f^c + q_{cb}D_a f^c] [-2 (D_e P^{eb}) h^a] - [q_{ac}D_b h^c + q_{cb}D_a h^c] [-2 (D_e P^{eb}) f^a])$$
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:
$$\{ H_a(f^a),H_b(h^b) \} = -H_a([\vec{f},\vec{h}]^a) -2 \int_{\Sigma} d^3 x (D_e P^{eb}) [q_{ac} h^a D_b f^c - q_{ac} f^a D_b h^c]$$
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.