Focussing on classical mechanics of a point particle, WLOG since it captures the relevant information for field theory and generalises to the quantum case, how do we show -- in general -- that conserved charges generate the associated transformations of the variables?

Start with infinitesimal transformations $x(t) \rightarrow x(t) + \epsilon \xi(x)$ which, at least for a potential $V(x)$ would seem to imply $p(t) \rightarrow p + \epsilon m \dot{\xi}(x) = p + \epsilon p \cdot \nabla \xi(x) + \epsilon m \partial_{t} \xi$. Then in one-dimension (for simplicity) the conserved charge associated with this symmetry would be $$ \epsilon Q = \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x - \epsilon f $$ where $f$ is the function with which $\mathcal{L} \rightarrow \mathcal{L} + \epsilon\frac{d}{dt}f$ changes by a total derivative.

Question 1: f will in general depend upon $x$ and $\dot{x}$ but am I supposed to rewrite this in terms of $x$ and $p$?

Continuing, let's write $ Q= p\xi(x) - f$ and consider whether this generates the transformation through Poisson brackets. We'd like $$\delta x = \epsilon \{x, Q\} ~~~\textrm{and}~~~ \delta p = \epsilon \{p, Q\} $$ Using the form of Q we get $$\{x, Q\} = \xi - \frac{\partial f}{\partial p}~~~\textrm{and}~~~ \{p, Q\} = -p \frac{\partial \xi}{\partial x} + \frac{\partial f}{\partial x}.$$ Question 2: To reproduce the transformations of the variables this would seem to require two conditions $$\frac{\partial f}{\partial p} = 0 ~~~\textrm{and}~~~ \frac{\partial f}{\partial x} = m\frac{\partial \xi}{\partial t} + 2p \frac{\partial \xi}{\partial x}.$$ Are these conditions automatically satisfied for some reason? Must they be imposed for consistency? What happens if they're not?

  • $\begingroup$ Linked. $\endgroup$ – Cosmas Zachos Sep 14 '19 at 20:53
  • $\begingroup$ Thank you, Cosmas. I wonder if you have any thoughts on the current question? I mean I know the statement that charges generate transformations is true but how can it be proved in this way? $\endgroup$ – lux Sep 15 '19 at 16:01
  • $\begingroup$ No thoughts beyond linked and linked therein. Your time-dependent surmise appears obscure and not canonical. Perhaps the answer to this question might be helpful; perhaps not. $\endgroup$ – Cosmas Zachos Sep 15 '19 at 16:39
  • $\begingroup$ More linked. $\endgroup$ – Cosmas Zachos Sep 15 '19 at 19:31

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