I'm trying to prove Noether's theorem in the context of (point-particle) classical mechanics, however, I'm a bit unsure on a few things.

To keep things as simple as possible I'm only considering the one-dimensional case. As such I'm starting with the full variation of the path $$\begin{align}q(t)\rightarrow q'(t')&=q'(t)+\dot{q}'(t)\delta t=q(t)+\delta q(t)+\dot{q}(t)\delta t\\ \\ \dot{q}(t)\rightarrow \dot{q}'(t')&=\dot{q}'(t)+\ddot{q}'(t)\delta t=\dot{q}(t)+\delta\dot{q}(t)+\ddot{q}(t)\delta t,\end{align}\tag{0.1}$$ to first-order in $\delta t$. This leads to the following variations (to first-order) $$\begin{align}\delta_{_{T}}q&=q'(t')-q(t)=\delta q(t) +\dot{q}(t)\delta t\\ \\ \delta_{_{T}}\dot{q}&=\dot{q}'(t')-\dot{q}(t)=\delta\dot{q}(t)+\ddot{q}(t)\delta t,\end{align}\tag{0.2}$$ where the subscript $T$ is to remind us that we are deforming the time, $t\rightarrow t+\delta t$ as well as the path (a so-called "total" or "full" variation). Now, assuming that this is a symmetry of the classical action $$S[q(t)]=\int dt\,L\left(q(t),\dot{q}(t),t\right)\tag{1}$$ we have that $(1)$ changes by at most a surface term, i.e. $$\delta_{_{T}}S=\int dt\,\frac{d}{dt}G\left(q(t),t\right)\tag{2}$$ Now, the left-hand side of $(2)$ is given by $$\begin{align}\delta_{_{T}}S&=\int\delta_{_{T}}(dt)\,L+\int dt\,\delta_{_{T}}(L)=\int dt\frac{d(\delta t)}{dt}\,L+\int dt\,\left[\frac{\partial L}{\partial q}\delta_{_{T}}q +\frac{\partial L}{\partial\dot{q}}\delta_{_{T}}\dot{q}+\frac{\partial L}{\partial t}\delta t\right]\\&=\int dt\frac{d(\delta t)}{dt}\,L+\int dt\,\left[\frac{\partial L}{\partial q}\Big(\delta q(t) +\dot{q}(t)\delta t\Big) +\frac{\partial L}{\partial\dot{q}}\Big(\delta\dot{q}(t) +\ddot{q}(t)\delta t\Big)+\frac{\partial L}{\partial t}\delta t\right]\\&=\int dt\,\left[\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\delta q +\left(\frac{\partial L}{\partial q}\dot{q}+\frac{\partial L}{\partial\dot{q}}\ddot{q}+\frac{\partial L}{\partial t}\right)\delta t+\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\delta q\right)+L\frac{d(\delta t)}{dt}\right]\\&=\int dt\,\left[\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\delta q +\frac{dL}{dt}\delta t+\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\delta q\right)+L\frac{d(\delta t)}{dt}\right]\\&=\int dt\,\left[\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\delta q +\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\delta q +L\delta t\right)\right]\end{align}\tag{3}$$ Assuming that $q(t)$ satisfies the Euler-Lagrange equation then we have that $$\delta_{_{T}}S=\int dt\,\left[\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\delta q +L\delta t\right)\right]=\int dt\,\frac{d}{dt}G\left(q(t),t\right)\tag{4}$$ which implies that $$\int dt\,\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\delta q +L\delta t-G\right)=0\qquad\Rightarrow\qquad\frac{\partial L}{\partial\dot{q}}\delta q +L\delta t-G=\text{constant}\tag{5}$$ That is, the quantity: $$\Lambda\big(q(t),\dot{q}(t),t\big)=\frac{\partial L}{\partial\dot{q}}\delta q +L\delta t-G,\tag{6}$$ is a constant of motion.

However, I have some doubts about what I've done so far, and this has lead me to ask the following questions:

$1.$ Have I used the correct total variation of the path in the first place?

$2.$ If I have used the correct variation are the steps in the proof correct?

$3.$ My original motivation to try and replicate a proof of Noether's theorem was to prove energy conservation as a consequence of time translation. It seems that in this case one assumes that the total variation vanishes, i.e. $$\delta_{_{T}}q=q'(t')-q(t)=0\tag{7}$$ (c.f. QMechanic's answer here). Why is this the case? What is the justification?

I realise that this type of question has been asked several times before, but having read the posts that I could find I haven't found that any of them have completely answered my questions. Any help will be much appreciated.

  • $\begingroup$ I've edited my question to reflect Qmechanic's suggestions and have attempted a more general proof of Noether's theorem. $\endgroup$
    – user35305
    Aug 17, 2017 at 11:06

1 Answer 1


OP's post seems to have a dual purpose:

  1. Prove Noether's theorem (NT) in general: Then OP's assumptions that

    • (i) $L$ has no explicit time dependence and
    • (ii) $\delta t$ does not depend on $t$

    should be removed. For an example that violates assumption (ii), see e.g. this Phys.SE post.

    Also note the subtlety that $\delta_T$ and $\frac{d}{dt}$ do not commute, which OP seems to assume in eq. (3).

  2. Prove via NT that a Lagrangian without explicit time dependence leads to energy conservation: In this case it is natural to try the transformation $$q^{\prime}(t^{\prime})~=~q(t)$$ in order to secure a symmetry of the action $$S[t\mapsto q(t)]~=~ S[t^{\prime}\mapsto q^{\prime}(t^{\prime})],$$ which may then be used in NT, cf. this Phys.SE post.

  • $\begingroup$ Thanks for your response. Ideally I would like to prove it in general, but I wasn't quite sure how to do it, so for now I've restricted to no explicit time dependence in the Lagrangian and $\delta t$ time-independent. Also, why is it natural to try $q'(t')=q(t)$ in the case of time translation? Is it because one defines that the coordinates change as $q'(t)=q(t+\delta t)$ under time translation, and then one can simply make a change of variables, $t\rightarrow t-\delta t$ such that $q'(t')=q(t)$? $\endgroup$
    – user35305
    Aug 16, 2017 at 20:57
  • $\begingroup$ I've updated my OP to reflect your suggested edits. $\endgroup$
    – user35305
    Aug 17, 2017 at 11:00

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