Confusion about Noether's Theorem

Question

In classical mechanics, a transformation $q \rightarrow q + \delta q$ is a symmetry if the resultant change in the Lagrangian is a total derivative, $$ \delta L = \frac{dF}{dt}.$$ If we derive the change in the Lagrangian corresponding to an arbitrary transformation, we get \begin{align} \delta L &= \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \frac{d\delta q}{dt} \tag{2.1} \end{align} Observe that \begin{equation} \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \delta q \right) = \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \delta q + \frac{\partial L }{\partial \dot{q}} \frac{d \delta q}{dt} \tag{2.2} \end{equation} Assuming we are on a solution trajectory, we may use the Euler-Lagrange equation, which states that \begin{equation} \frac{d}{dt}\frac{\partial L }{\partial \dot{q}} = \frac{\partial L}{\partial q} \tag{2.3} \end{equation} Solving for the second term on the RHS of (2.2) and substituting the result of (2.3), we have \begin{equation} \frac{\partial L }{\partial \dot{q}} \frac{d \delta q}{dt} = \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \delta q \right) - \frac{\partial L}{\partial q} \delta q \tag{2.4} \end{equation} Finally, putting this back into (2.1), we obtain \begin{equation} \delta L = \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \delta q \right) \tag{2.5} \end{equation} Which states that {any} transformation about a solution trajectory is a symmetry because the induced change in the Lagrangian is a total derivative.

My question, now, is very simple: Obviously not all transformations are symmetries unlike my derivation above shows, so what went wrong?

Edit: Traditionally, we would end the argument by equating (2.5) to $\frac{dF}{dt}$ and then by calling $\frac{\partial L}{\partial \dot{q}} \delta q - F$ the conserved quantity. The problem I see with this is: why do we look for another function $F$ such that $\delta L = \frac{dF}{dt}$ is satisfied when $\delta L$ is already a total time derivative.

Answer 1

Let us start from $$\delta L= \dfrac{\partial L}{\partial q}\delta q+\dfrac{\partial L}{\partial \dot q}\delta\dot q.$$

We can use the product rule for derivatives to rewrite

$$\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot q}\delta q\right) = \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot q}\delta q+\dfrac{\partial L}{\partial \dot q}\delta \dot q.$$

This means that $\delta L$ for a generic variation, takes the form $$\delta L = E\delta q+\dfrac{d\Theta}{dt},\quad E\equiv \dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot q},\quad \Theta \equiv \dfrac{\partial L}{\partial \dot q}\delta q.$$

The quantity $E$ determines the equation of motion $E = 0$, while $\Theta$ is an object called pre-symplectic potential. In fact, if we define the canonical momentum $$p \equiv \dfrac{\partial L}{\partial \dot q}$$ then $\Theta = p\delta q$. The observation now is that given the action $S$ we observe that the contribution to $\delta S$ from $\Theta$ is a boundary term. More precisely, if the path is defined on the interval $[a,b]$ the contribution is gives $p(b)\delta q(b)-p(a)\delta q(a)$. If we are considering variations keeping endpoints fixed, $\delta q(a)=\delta q(b)=0$ and therefore this contribution drops out.

As a result, demanding $\delta S=0$ for all variations keeping endpoints fixed yields $E=0$ because the $\Theta$ term doesn't contribute. If you then pick a path that solves the equations of motion, i.e., that has $E=0$, obviously you will get $\delta L = \frac{d\Theta}{dt}$, a total derivative, but that doesn't say the variation is a symmetry, because it relied on using a path that solves the equations of motion.

A (quasi)-symmetry is a variation that applied to any path (solving EOM or not), yields $\delta L = \frac{dK}{dt}$. The reason is that in this case $$\dfrac{dK}{dt} = E\delta q + \dfrac{d\Theta}{dt} \Longrightarrow \dfrac{d(K-\Theta)}{dt} = E\delta q,$$ and when restricted to a classical solution, $E=0$ and therefore $P=K-\Theta$ is conserved.

In fact, a symmetry is the special case $K=0$ where $\delta L=0$ and the Lagrangian is preserved. The whole motivation to define a quasi-symmetry like that is exactly that it is a minor modification to Noether's theorem that allows for a conserved charge also.

By the way, a nice by-product of this kind of analysis is the appearance of $\Theta$. When you study Hamiltonian mechanics you will learn that there is an object called symplectic form, which is the exterior derivative of $\Theta$, namely, $\Omega = \delta p\wedge \delta q$. This is going to be a two-form in phase space that determines the Poisson bracket. What this analysis shows is that the symplectic potential - and hence the symplectic form - can be read off from varying the Lagrangian and writing it in the form as we did here.

Answer 2

It is not enough that the transformation leaves the Lagrangian unchanged for solutions of the equation of motion. To be a symmetry, the Lagrangian should be unchanged (up to a total time derivative) for any trajectory, solution or not.

Notice that when the Euler-Lagrange equations are derived from the principle of stationary action, the requirement of for a solution $q(t)$ is that precisely that it leaves the action unchanged for infinitesimal variations $q \rightarrow q + \delta q$. In other words that $$\delta S = \int d t \delta L(q, \dot{q}, t) = 0,$$

What you've shown is just the same, but in reverse - that trajectories that satisfy the Euler-Lagrange equation are stationary points of the action.