Types of symmetries in Lagrangian mechanics

[gionole]

Bear with me on this and please read carefully, flaws of my logic might be hidden in there.

We can have 2 types of symmetry: Lagrangian symmetry and action symmetry.

Lagrangian symmetry is when we do transformation and $L$ doesn't change even by one bit, i.e L' = L (exactly equal, exactly the same thing) i.e $dL = 0$. If this situation occurs, then we definitely have conserved quantity: $\frac{\partial L}{\partial \dot q} K$ (K could be dependent on all of the coordinates or none of the coordinates and could be a number).

Action symmetry is when we do transformation, but now, $L$ changes(i.e $dL \neq 0$) (note that action symmetry is still present when $dL = 0$, but I'm talking about the fact that we might still have action symmetry even if $dL \neq 0$).
then if $dL \neq 0$, in order to have action symmetry, the new Lagrangian($L'$) must be different from old lagrangian($L$) by total time derivative of some function(i.e $L' = L + \frac{d}{dt}f(q,t)$)

Question 1: Is there other cases when we still have action symmetry/invariance other than the cases of
1) $dL =0$ and 2) $dL \neq 0$, && $L' = L + \frac{d}{dt}f(q,t)$)

Question 2: If $L = \frac{1}{2}m\dot x^2$, and we do transformation $x(t) -> x(t) -> \epsilon(t)$ (note that I do time-dependent transformation), then new $L'$ becomes $\frac{1}{2}m\dot x^2 + m\dot x \dot \epsilon + \frac{1}{2}m \dot \epsilon^2$ and we have: $L' = L + m\dot x \dot \epsilon$(ignoring $O(\dot \epsilon^2)$). Now, we know momentum is conserved for freely moving particle, but looking at: $L' = L + m\dot x \dot \epsilon$, neither $dL = 0$ nor $m\dot x \dot \epsilon$ can be represented by total time derivative, which means we don't have lagrangian symmetry and we don't have action symmetry, but we still know momentum is conserved. I thought, that in order to have momentum conserved, either we should have action symmetry or lagrangian symmetry, but we got none. I know if I had used $\epsilon$, instead of $\epsilon(t)$, we would have lagrangian symmetry, but I wanted to go with $\epsilon(t)$. Doesn't noether consider time dependent transformations at all ? What's going on ? also, I know that even if we do $\epsilon(t)$ transforrmation, we can prove by least action principle that momentum is conserved, but my question is only related with the fact that we must get conservation only if lagrangian OR action is symmetric.

 

[vanhees71]

Bear with me on this and please read carefully, flaws of my logic might be hidden in there.

We can have 2 types of symmetry: Lagrangian symmetry and action symmetry.

Lagrangian symmetry is when we do transformation and $L$ doesn't change even by one bit, i.e L' = L (exactly equal, exactly the same thing) i.e $dL = 0$. If this situation occurs, then we definitely have conserved quantity: $\frac{\partial L}{\partial \dot q} K$ (K could be dependent on all of the coordinates or none of the coordinates and could be a number).

Action symmetry is when we do transformation, but now, $L$ changes(i.e $dL \neq 0$) (note that action symmetry is still present when $dL = 0$, but I'm talking about the fact that we might still have action symmetry even if $dL \neq 0$).
then if $dL \neq 0$, in order to have action symmetry, the new Lagrangian($L'$) must be different from old lagrangian($L$) by total time derivative of some function(i.e $L' = L + \frac{d}{dt}f(q,t)$)

That's rather the criterion for the even weaker demand that a symmetry shouldn't change the variation of the action functional, i.e., that the two Lagrangians are equivalent leading to the same equations of motion.

Question 1: Is there other cases when we still have action symmetry/invariance other than the cases of
1) $dL =0$ and 2) $dL \neq 0$, && $L' = L + \frac{d}{dt}f(q,t)$)

If the time is also transformed the most general case is that
$$\frac{\mathrm{d} t'}{\mathrm{d} t} L'[q',\dot{q}',t']=L[q,\dot{q},t] + \frac{\mathrm{d}}{\mathrm{d} t} f(q,t).$$
This case is, e.g., needed if you want to describe symmetry under Lorentz transformations in special relativity, where the time is also transformed.

Question 2: If $L = \frac{1}{2}m\dot x^2$, and we do transformation $x(t) -> x(t) -> \epsilon(t)$ (note that I do time-dependent transformation), then new $L'$ becomes $\frac{1}{2}m\dot x^2 + m\dot x \dot \epsilon + \frac{1}{2}m \dot \epsilon^2$ and we have: $L' = L + m\dot x \dot \epsilon$(ignoring $O(\dot \epsilon^2)$). Now, we know momentum is conserved for freely moving particle, but looking at: $L' = L + m\dot x \dot \epsilon$, neither $dL = 0$ nor $m\dot x \dot \epsilon$ can be represented by total time derivative, which means we don't have lagrangian symmetry and we don't have action symmetry, but we still know momentum is conserved. I thought, that in order to have momentum conserved, either we should have action symmetry or lagrangian symmetry, but we got none. I know if I had used $\epsilon$, instead of $\epsilon(t)$, we would have lagrangian symmetry, but I wanted to go with $\epsilon(t)$. Doesn't noether consider time dependent transformations at all ? What's going on ? also, I know that even if we do $\epsilon(t)$ transforrmation, we can prove by least action principle that momentum is conserved, but my question is only related with the fact that we must get conservation only if lagrangian OR action is symmetric.

That's not a symmetry a la Noether. This is understandable since this transformation means to go to a general accelerated frame of reference, and indeed Newton's equations are not form-invariant under such transformations. It's only invariant under the Galilei symmetry (or, for special cases, some larger symmetry like, e.g., for the Kepler problem, where you have an O(4) symmetry, and the Runge-Lenz vector as extra conserved quantities).

For the general case, I have only my German manuscript on mechanics (Sect. 3.5):

https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf

 

[gionole]

Thanks for the great answer. Couple of points that I wanna make sure about.

Note that I ask the questions below in such a way that I answer them by myself, so all you gotta do is agree or not.

For simplicity and the reason I haven't learned special relativity yet, we can forget about special case you mentioned.

Question 1: So then, we only got 2 cases($dL =0$ - lagrangian symmetry and $L' = L + \frac{d}{dt}f(q,t)$ - Action symmetry). Is this correct what I call lagrangian symmetry and what I call Action symmetry ?

Question 2: If I understood correctly, using $\epsilon(t)$ is not such a transformation that Noether has her theorem based on. She only discusses $\epsilon K(q)$ transformations where $\epsilon$ is constant. Correct ?

Question 3: If Question 2 is correct, then what I do is check if either $\epsilon K(q)$ transformation causes either $dL = 0$ or $L' = L + \frac{d}{dt}f(q,t)$ and if either of this happens, then I definitely can say there is some quantity that is conserved and if none of it occurs, then I don't have conserved quantity. Correct ?

Sorry, don't know German.

 

[vanhees71]

Thanks for the great answer. Couple of points that I wanna make sure about.

Note that I ask the questions below in such a way that I answer them by myself, so all you gotta do is agree or not.

For simplicity and the reason I haven't learned special relativity yet, we can forget about special case you mentioned.

Question 1: So then, we only got 2 cases($dL =0$ - lagrangian symmetry and $L' = L + \frac{d}{dt}f(q,t)$ - Action symmetry). Is this correct what I call lagrangian symmetry and what I call Action symmetry ?

I've never read these terms in the literature. Noether's original paper was about field theories, and it was very general and complete.

When used in point-particle mechanics in the Lagrangian version of the action principle you can allow for transformations of the most general type,
$$t'=t+\epsilon \Theta(q,\dot{q},t), \quad q'=q+\epsilon Q(q,\dot{q},t).$$
That's called a symmetry if to order $\epsilon$ the variation of the action is unchanged. That implies that there must exist a function $\tilde{\Omega}$ such that
$$\frac{\mathrm{d} t'}{\mathrm{d} t} L(q',\dot{q}',t')=L(q,\dot{q},t) + \epsilon \frac{\mathrm{d}}{\mathrm{d} t} \Omega(q,t).$$
Working out the expansion to order $\epsilon$ of the left-hand side leads to the "symmetry condition" (Einstein summation convention applies):
$$Q_k \frac{\partial L}{\partial q_k} + \left (\frac{\mathrm{d} Q_k}{\mathrm{d} t}-\dot{q}_k \frac{\mathrm{d}\Theta}{\mathrm{d} t} \right) p_k + L \frac{\mathrm{d}\Theta}{\mathrm{d} t} + \Theta \frac{\partial L}{\partial t} + \frac{\mathrm{d} \tilde{\tilde{\Omega}}}{\mathrm{d} t}=0.$$
Working out all the total time derivatives and using the Euler-Lagrange equations then leads to the conservation law
$$\frac{\mathrm{d}}{\mathrm{d} t} (Q_k p_k-\Theta H + \tilde{\Omega})=0.$$

Question 2: If I understood correctly, using $\epsilon(t)$ is not such a transformation that Noether has her theorem based on. She only discusses $\epsilon K(q)$ transformations where $\epsilon$ is constant. Correct ?

Question 3: If Question 2 is correct, then what I do is check if either $\epsilon K(q)$ transformation causes either $dL = 0$ or $L' = L + \frac{d}{dt}f(q,t)$ and if either of this happens, then I definitely can say there is some quantity that is conserved and if none of it occurs, then I don't have conserved quantity. Correct ?

Sorry, don't know German.