- #1
shoehorn
- 420
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Hi. I've been wondering about the following and haven't made much progress on it. 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)<br />
Then, for example, we have the following fundamental Poisson brackets
<br /> \{q^i,p_j\} = \delta^i_{\phantom{i}j},<br /> \{q^i,q^j\} = \{p_i,p_j\} = 0.<br />
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}'),<br /> \{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
<br /> \{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).<br />
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}}?
\{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)<br />
Then, for example, we have the following fundamental Poisson brackets
<br /> \{q^i,p_j\} = \delta^i_{\phantom{i}j},<br /> \{q^i,q^j\} = \{p_i,p_j\} = 0.<br />
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}'),<br /> \{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
<br /> \{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).<br />
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}}?
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