I'm going through some notes on how to apply the Hamiltonian formalism to systems with gauge invariance and I found a derivation of Noether's theorem I had never seen before. The idea is roughly that if you have a symmetry transformation that changes the action only by a boundary term:

$$ \delta_{\epsilon} I = \int dt \; \epsilon \frac{dF(t)}{dt}$$

where $\epsilon$ is the infinitesimal parameter of the transformation (constant), then, by promoting $\epsilon$ to a function of the parameter in the worldline $\epsilon (t)$ (satisfying that it vanishes on the boundary of the integral), you should find:

$$ \delta_{\epsilon} I = \int dt \left[ \epsilon(t) \frac{dF(t)}{dt} + \dot{\epsilon}(t) Q(t) \right] $$

This is reasonable since for $\epsilon (t)$ constant you must recover the previous result, and any terms containing higher derivatives of $\epsilon$ can be integrated by parts. Integrating by parts the second term, using that $\epsilon (t)$ is arbitrary and that the variation of the action is zero on shell, you obtain that $F(t)-Q(t)$ is constant along the worldline.

So far so good. My problems start right after the derivation, where a plain sentence asserts that once we have the conserved charge we can recover the symmetry transformation using:

$$ \delta_{\epsilon}f = \epsilon \, \{ f, Q \}_{PB} $$

for any function $f$ on phase space. After a couple of hours struggling with this equation, I can't find why this should be a symmetry transformation. Any ideas? I would be specially interested in arguments not assuming a particular form of the action, since I am dealing with systems with constraints where the action has strange terms included to ensure that these constraints are satisfied. As an example, my action for a relativistic point particle is:

$$ I = \int dt \, \left[ \dot{x}^m p_m - \frac{1}{2}e(p^2 + m^2) \right] $$

with $e(t)$ a Lagrange multiplier.

  • $\begingroup$ Comments to the post (v1): 1. The trick with making $\epsilon$ a function is discussed in e.g. this Phys.SE post. 2. The fact that for a conserved charge $Q$, the Hamiltonian vector field $X_Q$ is a quasi-symmetry of the action is proven in this and this Phys.SE posts. $\endgroup$
    – Qmechanic
    Commented Feb 2, 2016 at 18:40
  • $\begingroup$ Yes, I've seen those posts but the proof always involves assuming the form for the Lagrangian $\dot{q}^I p_I - H(q,p)$, and it is relatively easy to show that the first terms only contrubute a boundary term when transformed as $\delta_{\epsilon}f = \epsilon \{ f,Q\}_{PB}$ and that $H$ doesn't change at all since the time derivative of $Q$ is zero. But in my example, how do you deal with the change in the constraint (i.e., in the term $p^2+m^2$)? $\endgroup$
    – Alex V.
    Commented Feb 2, 2016 at 22:06
  • $\begingroup$ Then it is recommended to edit the question formulation (v1) accordingly, removing irrelevant parts. $\endgroup$
    – Qmechanic
    Commented Feb 3, 2016 at 0:39


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