- #1

gionole

- 281

- 24

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.