In studying analytical mechanics (or it's quantum analog), one will come across statements such as:
$$f(x^{i}+\delta x^{i})=f(x^{i})+\delta f(x^{i})=f(x^{i})+\frac{\partial f(x^{i})}{\delta x^{i}}\delta x^{i}=f(x^{i})+\{f,g^{i}\}\delta x^{i}$$
where $g$ is the infinitesimal generator of translation and summation over all coordinates (i) is implied. Of course in the quantum case, we (roughly) have a switch to commutators.
But there is already a method to translate a function with more precision, the translation operator. Switching to vector notation:
$$f(\overrightarrow{x}+\overrightarrow{a})=e^{\overrightarrow{a}\cdot\nabla}f(\overrightarrow{x}).$$
Is the Poisson bracket just a first order approximation of this? (Note that for a general space or spacetime, $\nabla$ will be the covariant derivative) letting $\overrightarrow{a}\longmapsto\overrightarrow{\delta x}$ one obtains:
$$f(\overrightarrow{x}+\overrightarrow{\delta x})=e^{\overrightarrow{\delta x}\cdot\nabla}f(\overrightarrow{x}).$$
Since we're interested in finding $\delta f(\overrightarrow{x})$,
$$\delta f(\overrightarrow{x})=f(\overrightarrow{x}+\overrightarrow{\delta x})-f(\overrightarrow{x})=(e^{\overrightarrow{\delta x}\cdot\nabla}-1)f(\overrightarrow{x}).$$
One can parameterize a general path through spacetime with the parameter $\tau$, such that: $x^{i}=\phi^{i}(\tau)$. One then has $$\delta x^{i}=\delta\tau\frac{\partial\phi^{i}}{\partial\tau}=\delta\tau\overset{\centerdot}{\phi^{i}}$$ (note that for the 4-path of a massive particle $\tau$ corresponds to the proper time). One then obtains:
$$\delta f(\overrightarrow{x})=(e^{\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i}}-1)f(\overrightarrow{x})=\{\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i}+(\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i})^{2}+\cdots\}f(\overrightarrow{x})$$
which, to first order, has some noticable similarities to the quantum lagrangian:
$$\delta f(\overrightarrow{x})\approx\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i}f(\overrightarrow{x}).$$
As a final consideration, if $f(\overrightarrow{x})$ is a small three-volume element $\delta V (\overrightarrow{x})$, then the variation in the volume element as it propagates is:
$$\delta(\delta V(\overrightarrow{x}))=\{\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i}+(\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i})^{2}+\cdots\}\delta V(\overrightarrow{x}).$$
The latter term in brackets can be rewritten:
$$(\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i})^{2}\delta V(\overrightarrow{x})=(\delta\tau\frac{\partial x^{i}}{\partial\tau}\frac{\partial}{\partial x^{i}})^{2}\delta V(\overrightarrow{x})=(\delta\tau\frac{D}{d\tau})^{2}\delta V(\overrightarrow{x})$$
Which is related to the Ricci tensor (equation 7).
$$=-(\delta\tau\delta V)R_{\mu\nu}\overset{\cdot}{\phi^{\mu}}\overset{\cdot}{\phi^{\nu}}+(\delta\tau)^{2}\frac{\partial^{2}}{\partial\tau^{2}}\delta V.$$
The first term can be rewritten:
$$-(\delta\tau\delta V)R_{\mu\nu}\overset{\cdot}{\phi^{\mu}}\overset{\cdot}{\phi^{\nu}}=-\frac{1}{2}(\delta\tau\delta V)R_{\mu\nu}g^{\mu\nu}(\overset{\cdot}{\phi_{\nu}}\overset{\cdot}{\phi^{\nu}}).$$
If our volume element is encompassing a mass such that $\tau$ is proper time, then $\overset{\cdot}{\phi^{i}}=U^{i}$ a component of the four velocity. Rewriting in terms of the four-velocity (and making use of the (- +++) metric):
$$=-\frac{1}{2}(\delta\tau\delta V)R_{\mu\nu}g^{\mu\nu}(U_{\nu}U^{\nu})=-\frac{1}{2}(\delta\tau\delta V)R_{\mu\nu}g^{\mu\nu}(-c^{2})$$
$$=\frac{1}{2}(\delta\tau^{2}\delta V)Rc^{2}.$$
Inserting this result back, we obtain:
$$\delta(\delta V(\overrightarrow{x}))=\{\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i}+(\delta\tau)^{2}\frac{c^{2}}{2}R+\frac{(\delta\tau)^{2}}{2}\frac{\partial^{2}}{\partial\tau^{2}}\}\delta V(\overrightarrow{x})$$
but $\frac{\partial^{2}}{\partial\tau^{2}}=\frac{force}{mass}$ and that term composed with the volume element becomes something akin to a potential energy V such that:$$\delta(\delta V(\overrightarrow{x}))=\{\delta\tau\overset{\cdot}{\phi^{i}}\nabla_{i}+(\delta\tau)^{2}\frac{c^{2}}{2}R-(\delta\tau)^{2}V\}\delta V(\overrightarrow{x}).$$
Now, the last term, I didn't follow through precisely, but nevertheless an equation of similar form of the last one here should result. This looks a LOT like the lagrangian. Are poisson brackets and commutators just a first order approximation to applying the translation operator and what exactly is their relationship to the lagrangian?
