As was described in, for example, this post, one can formulate Noether's Theorem also in Hamiltonian Mechanics. Symmetries are then represented by vector fields generated by observables whose Poisson brackets vanish with the Hamiltonian of the system.

My question is: How do you describe transformations of time in this formalism? Flows in the phase space only represent active transformations of the phase space, but they don't change anything about the time variable. It is not possible to derive, for example, conservation of energy as the consequence of a symmetry regarding the system's time-shifts.


2 Answers 2


Yes, it is possible, same as for Lagrangian action. Taking the Hamiltonian action as $$ S_H\{q, p\} = \int_{t_i}^{t_f}{dt\; \left[p\dot{q} - H(q, p, t) \right]} $$ consider an infinitesimal time shift $t' = t +\delta t$, with $q'(t') = q(t) +\delta q$, $p'(t') = p(t) +\delta p$, and assume invariance under time translations: $$ S_H\{q, p\} = \int_{t_i}^{t_f}{dt\; \left[p\dot{q} - H(q, p, t) \right]} = \int_{t'_i}^{t'_f}{dt\; \left[p'\dot{q'} - H(q', p', t) \right]} $$ The 2nd equality above gives $$ \int_{t_i}^{t_f}{dt\; \left[p\dot{q} - H(q, p, t) \right]} = \int_{t_i + \delta t_i}^{t_f + \delta t_f}{dt\; \left[p\dot{q} - H(q, p, t) + p\delta \dot{q} + \dot{q}\delta p - \frac{\partial H}{\partial q}(q, p, t)\delta q - \frac{\partial H}{\partial p}(q, p, t)\delta p \right]} $$ or after slight rearrangement, $$ \left[p\dot{q} - H \right]\Big|_{t_f}\delta t_f - \left[p\dot{q} - H \right]\Big|_{t_i}\delta t_i + \int_{t_i + \delta t_i}^{t_f + \delta t_f}{dt\;\left[p\delta \dot{q} + \dot{q}\delta p + \dot{p}\delta q - \dot{q}\delta p \right]} = \\ = \int_{t_i}^{t_f}{dt\; \frac{d}{dt}\left[ p\dot{q}\delta t - H(q,p,t)\delta t + p\delta q\right]} = \int_{t_i}^{t_f}{dt\; \frac{d}{dt}\left[ 2 p\delta q - H(q,p,t)\delta t \right]} = 0 $$ where use is made of $\dot{q}\delta t = \delta q$ and the Hamiltonian EOMs, $ \dot{q}=\frac{\partial H}{\partial p}$, $\dot{p}=-\frac{\partial H}{\partial q}$. For a time translation that leaves end points invariant such that $\delta q\Big|_{t_i} = \delta q\Big|_{t_f} = 0$, the last equality above reduces to $$ \frac{dH}{dt}(q, p, t) = 0 $$ or $$ H(q, p, t) = const. $$

  • $\begingroup$ Is there any way to describe this time-shift, or mixed transformations in space and time, in the same way it is done in the linked article? That means, using oberservables that create a flow in some manyfold, which represents the transformation that leave the action invariant? $\endgroup$ Dec 27, 2015 at 13:43
  • $\begingroup$ Re: the original article, Qmechanic answered your question separately already. I would add perhaps that if you consider the field theory version with invariance under space-time translations, then you'd end up with conservation of the stress-energy tensor, $\partial^\mu T_\mu^\nu =0$, subsuming energy-momentum conservation. This is probably the "flow-charge" form you are looking for. $\endgroup$
    – udrv
    Dec 28, 2015 at 3:40
  • $\begingroup$ I am sorry, but what you describe as flow here is some combined 4-vector-field (or lorentztensor), whose time and space components satisfy this continuity equation. What I mean when I say flow is some vectorfield (or a flow that is the intral-curves of this vectorfield) that describes the symmetry-transformations in noethers theorem, as it is done in the article I linked. The article does this only for "vertical" transformations, as I am looking for some kind of flow for "horizontal transformations", if I want to use the language of Qmechanic. $\endgroup$ Dec 28, 2015 at 13:58
  • $\begingroup$ What I describe is the conservation law for the "current density" corresponding to / expressing Noether's theorem for space-time translations, only in this case the "current" is tensorial, not vectorial. It is far from "some (arbitrary) combined 4-vector field". Please see at least the basics here: en.wikipedia.org/wiki/Noether%27s_theorem#Field_theory_version. Check paragraph above eq. for $T_\mu^\nu$. $\endgroup$
    – udrv
    Dec 28, 2015 at 20:40
  • $\begingroup$ No offense, I know what this current density is and that it is the expression of conservation in field theories, and that it originates from the space/time-translation invariance of the lagrangian. It is just not what I ment when I posted my question. That's what I was trying to say when I said for now, it is just a lorentztensor whose divergence vanishes, as opposed to a Vektorfield that generates transformations, as in the linked article. $\endgroup$ Dec 29, 2015 at 8:56

Comments to the question (v2):

  1. It should first of all be mentioned that the Hamiltonian generates time-evolution, although I'm sure OP already knows this.

  2. The Hamiltonian Lagrangian $$ \tag{1} L_H(q,\dot{q},p,t) ~:=~\sum_{i=1}^n p_i \dot{q}^i - H(q,p,t)$$
    may be viewed as a first-order Lagrangian system $L_H(z,\dot{z},t)$ in twice as many variable $$\tag{2} (z^1,\ldots,z^{2n}) ~=~ (q^1, \ldots, q^n;p_1,\ldots, p_n).$$

  3. How to apply Noether's theorem for energy conservation in a Lagrangian system is e.g. discussed in my Phys.SE answer here. None of the suggested infinitesimal transformations in my Phys.SE answer are of the form of a vertical vector field $$\tag{3} \delta z^I~=~\varepsilon V^I, \qquad \delta t~=~0, $$
    in such a way that the vector field components $V^I$ do not depend on $\dot{z}$, which seems to be at the core of OP's question.

  4. The corresponding Lagrangian energy function $$\tag{4} h_H(z,\dot{z},t)~:=~ \sum_{I=1}^{2n}\dot{z}^I\frac{\partial L_H(z,\dot{z},t)}{\partial \dot{z}^I} - L_H(z,\dot{z},t)~=~H(z,t) $$ is unsurprisingly the Hamiltonian itself. It will play the role of Noether charge for time translations.

  5. Energy is conserved on-shell if the Hamiltonian Lagrangian $L_H$ (or equivalently, the Hamiltonian $H$), has no explicit time dependence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.