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I think I am misunderstanding the concept of symmetry in Lagrangian mechanics or maybe I am misunderstanding the content of Noether's theorem. Let me elaborate:

Suppose $L(q,\dot q,t)$ is the Lagrangian of some physical system, then by using the Euler-Lagrange equation you can determine the equation of motion (one equation in this case) of the coordinate $q$. It turns out that $L^´:=L+\frac{d}{dt}f(q,t)$ leads to the exact same equations of motion (for any function $f(q,t)$), so we can consider $L$ and $L^´$ equivalent Lagrangians. Notice however that this only works when $f$ is not dependent on $\dot q$. If $f$ depends on $\dot q$ the Lagrangians might in fact lead to different equations of motion.
Now here is the concept of symmetry that I thought was correct (however I am not sure if this is really correct):

When we do a little infinitesimal coordinate transformation $q \rightarrow q' = q + \epsilon \chi $ then the variation of the Lagrangian (under that transformation) has to be of the following form: $$\delta L = \epsilon \frac{d}{dt} f(q,t)$$ Since $f$ is only depending on $q$ and $t$, we know that the Lagrangian has essentially not changed (leads to the same equations of motion) and is hence symmetric under the coordinate transformation.
But when I looked at the proof of noethers theorem, I found that it is really not necessary for $f$ to not depend on $\dot q$. The condition$$ \delta L = \epsilon \frac{d}{dt} f(q,\dot q ,t)$$ is sufficient for Noether's theorem (in the sense that you will get a conserved quantity).
So what am I missing here? I thought Noether's theorem said that every symmetry implies a conserved quantity and every conserved quantity implies a symmetry. But looking at this, it seems like there are some conserved quantities that are not related to some symmetry. Or am I getting the concept of symmetry wrong?

Diracology
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Daniel W.
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2 Answers2

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Noether's theorem actually says that continuous symmetries of the action imply in constants of motion. The devil is in details and the details here are the terms symmetries of the action and constants of motion. By symmetries of the action we shall understand transformations that leave the action invariant or quasi-invariant, i.e., that do not change the equations of motion. Constants of motion are quantities which are constant along the time evolution of the system.

When one does an infinitesimal transformation $q\rightarrow q+\epsilon \chi$, the lagrangian change as $$\delta L=\frac{\partial L}{\partial q}\epsilon\chi+\frac{\partial L}{\partial \dot q}\epsilon\dot\chi=\epsilon (\dot p\chi +p\dot\chi)=\epsilon\frac{d}{dt}(p\chi).$$ In order to the action be at most quasi-invariant, $S$ and $S'$ have to differ at most by a term $f(q,t)$, since in that case Hamilton's principle imply in the same equations of motion. Hence we can consider, $$S'-S=\epsilon\int\delta Ldt=\epsilon\int \frac{df(q,t)}{dt}dt, $$ and therefore $$\frac{d}{dt}(p\chi)=\frac{df(q,t)}{dt},$$ or $$C=p\chi-f(q,t),$$ is a constant of motion along the time evolution of the system.

If $$\delta L=\epsilon \frac{df(q,\dot q,t)}{dt},$$ then $$S'-S=\epsilon\int\delta Ldt=\epsilon\int \frac{df(q,\dot q,t)}{dt}dt.$$ However we cannot say $$C'=p\chi-f(q,\dot q,t)$$ is a constant of motion because the equations of motion itself have changed. The system has one dynamic evolution before the transformation and another after. It is meaningless to compare the value of some dynamical variable by using different dynamic evolutions.

Diracology
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OP wrote (v3):

[...] It turns out that $L^´:=L+\frac{d}{dt}f(q,t)$ leads to the exact same equations of motion (for any function $f(q,t)$), so we can consider $L$ and $L^´$ equivalent Lagrangians. Notice however that this only works when $f$ is not dependent on $\dot{q}$. If $f$ depends on $\dot{q}$ the Lagrangians might in fact lead to different equations of motion. [...]

If the variational/functional derivatives exists, then the Euler-Lagrange (EL) equations does not depend on $f$ even if it depends on $\dot{q}$, cf. e.g. this Phys.SE post.

[...] But when I looked at the proof of Noether's theorem, I found that it is really not necessary for $f$ to not depend on $\dot q$. [...]

Correct.

[...] I thought Noether's theorem said that every symmetry implies a conserved quantity and every conserved quantity implies a symmetry. [...]

Only the former is Noether's theorem. The latter would be the inverse Noether's theorem, which only holds under additional assumptions.

Qmechanic
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