Poisson brackets in general relativity

[shoehorn]

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

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

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

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

 

[samalkhaiat]

My question is this: what exactly is

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

$$\frac{1}{2}(\delta^k_i\delta^l_j + \delta^l_i\delta^k_j)$$


sam

 

[Entropia]

The Poisson bracket is a fundamental concept in classical mechanics, but it can also be extended to field theories, including general relativity. In general relativity, the phase space variables are the metric tensor g_{ij} and its conjugate momentum \pi^{ij}. The Poisson brackets for these variables can be defined in a similar way as in the discrete theory, but there are some additional considerations due to the symmetry of the metric tensor.

To understand the Poisson brackets in general relativity, we first need to understand the functional derivatives involved. In the discrete theory, the functional derivatives are simply partial derivatives with respect to the phase space variables q^i and p_i. However, in the field theory, these derivatives become functional derivatives, which are defined as follows:

\frac{\delta F}{\delta q^i} = \lim_{\epsilon \to 0} \frac{F(q^i + \epsilon \delta^i, p_i) - F(q^i, p_i)}{\epsilon}

where \delta^i is a small variation in the variable q^i. In general relativity, this variation is not just a small change in the metric tensor, but it must also preserve the symmetry of the tensor. This means that the variation \delta^i must satisfy the following condition:

\delta^i_{\phantom{i}j} = \delta^j_{\phantom{j}i}

This condition ensures that the variation preserves the symmetry of the metric tensor. With this in mind, we can now define the Poisson brackets for general relativity 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)

where the functional derivatives are now defined with the appropriate variation that preserves the symmetry of the metric tensor. This ensures that the Poisson brackets satisfy the correct symmetry properties.

To answer your question about \frac{\delta g_{ij}}{\delta g_{kl}}, it is not a simple expression like \delta^i_k\delta^j_l. Instead, it is a more complicated expression that involves the variation of the metric tensor and its derivatives. It can be derived using the definition of the functional