Noether's theorem with non-finite transformations

  • #1
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Homework Statement:

Use Noether's theorem to find the constants of motion of the transformation:
## q_1' = e^{\epsilon} q_2 ##
## q_2' = e^{-\epsilon} q_1 ##

Relevant Equations:

Lagrangian:
## L = \dot q_1 \dot q_2 - \omega q_1 q_2 ##
Transformation:
## q_1' = e^{\epsilon} q_2 ##
## q_2' = e^{-\epsilon} q_1 ##
Noether's theorem:
## K = \frac {\partial L} {\partial \dot q_i} \delta q_i + (L - \frac {\partial L} {\partial \dot q_i} \dot q_1) \delta t ##
Hi!

I am given the lagrangian:

## L = \dot q_1 \dot q_2 - \omega q_1 q_2 ##

(Which corresponds to a 2D harmonic oscillator) And I am given two transformations and I am asked to say if there is a constant of motion associated to each transformation and to find it (if that's the case).

I am not sure which is the literature my course is following, but I understand I have to:
  1. Check that the transformation is a symmetry of the action. That is, if:
    ## S[q_1, q_2] = \int L(q_1, q_2, t) dt ##
    ## S'[q_1', q_2'] = \int L'(q_1'(t'), q_2'(t'), t') dt' ##
    Then when ## S = S' ##, it means there is a constant of motion associated to that transformation.
  2. Use Noether's theorem to calculate the constant of motion ##K##:
    ## K = \frac {\partial L} {\partial \dot q_i} \delta q_i + (L - \frac {\partial L} {\partial \dot q_i} \dot q_1) \delta t ##
Case 1

So, first transformation is:

## q_1' = e^{\epsilon} q_1 ' ##
## q_2' = e^{-\epsilon} q_2 ' ##

Which is easy to verify as a symmetry of the action and, as ##\epsilon## is really small, I can do ##e^{\epsilon} \approx 1 + \epsilon ## to identify ## \delta q_1 = \epsilon## and ## \delta q_2 = -\epsilon## to plug them into Noether's theoren and calculate the constant of motion.

Case 2
Second transformation is where I am stuck:

## q_1' = e^{\epsilon} q_2 ##
## q_2' = e^{-\epsilon} q_1 ##

If I plug that transformation into ##S'[q_1', q_2']##, it is easy to check it is a symmetry of the action which suggests there exists a constant of motion for that transformation, but the derivation of Noether's theorem we did in my course (and the one I have always found on the internet) is for transformations of the shape:

## q_1' = q_1 + \delta q_1 ##

Which my professor labeled as ##finite## in the sense that if I make the variation ##\delta q_1## really small, ## q_1'## always maps back to ##q_1##. But notice that my transformation, after doing ##e^{\epsilon} = 1 + \epsilon ## is:

## q_1' = q_2 + \epsilon q_2 = q_2 + \delta q_1 ##

Where ## \delta q_1 = \epsilon q_2 ##. My transformation is mapping ## q_1'## to ##q_2## and viceversa, which sugggests I can't use Noether's theorem as is. At least, not as I wrote it above.

How should I proceed here?

I was thinking of doing Noether's theorem derivation again but for transformations of the kind ## q_i' = q_j + \delta q_i ## to get the constant of motion ## K ## associated to that kind of transformations, but I am not sure if this would be a good approach or if there is something easier I could do or where I can read about these type of transformations.
 
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Answers and Replies

  • #2
Delta2
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I am not very experienced with applying Noether's theorem but here is my thought

IF the lagrangian is ##L=\dot q_1 \dot q_2 -\omega q_1 q_2## (though in the OP you claim it to be ##L=\dot q_1 \dot q_2 -\omega q_1 q_1## i guess the second ##q_1## at the end is a typo) then it is symmetric w.r.t to ##q_1## and ##q_2##, that is if we put ##q_1## in place of ##q_2## and vice versa the lagrangian remains unchanged. Therefore any transform that maps ##q_1## to ##q_2## and ##q_2## to ##q_1## is a symmetry of the action.
 
  • #3
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I am not very experienced with applying Noether's theorem but here is my thought

IF the lagrangian is ##L=\dot q_1 \dot q_2 -\omega q_1 q_2## (though in the OP you claim it to be ##L=\dot q_1 \dot q_2 -\omega q_1 q_1## i guess the second ##q_1## at the end is a typo) then it is symmetric w.r.t to ##q_1## and ##q_2##, that is if we put ##q_1## in place of ##q_2## and vice versa the lagrangian remains unchanged. Therefore any transform that maps ##q_1## to ##q_2## and ##q_2## to ##q_1## is a symmetry of the action.
Indeed it was a typo, but I have fixed it. Thank you!

And I have no trouble recognizing that the transformation I am given (or any transformation that maps ##q_1## to ##q_2##) is going to leave the action unchanged, but I am asked to calculate the constant of motion associated to that transformation and the problem is I am not sure that the following equation (which we derived from Noether's theorem) will still be valid in this case:

## K = \frac {\partial L} {\partial \dot q_i} \delta q_i + (L - \frac {\partial L} {\partial \dot q_i} \dot q_1) \delta t ##
 
  • #4
Delta2
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I think ##K## is the conserved quantity for ANY transform that is a symmetry of the action, regardless if you have only prove it for transforms of the type ##q_i'=q_i+\delta q_i##. There must be a generalized proof.
 
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  • #5
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I think ##K## is the conserved quantity for ANY transform that is a symmetry of the action, regardless if you have only prove it for transforms of the type ##q_i'=q_i+\delta q_i##. There must be a generalized proof.
I don't agree completely on that as you can have two different transformations which are symmetries of the action but yield different conserved quantities, but I do think that the two transformations I was given are going to have the same constants of motion associated. But I was hoping someone could have seen that "generalized proof" that I could check and use as a mathematical argument.
 
  • #6
Delta2
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I ll just mention @vanhees71 @PeroK here, maybe they are kind enough and can enlighten us.
 
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  • #7
vanhees71
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I don't understand what's said in #4. Either ##K## is a conserved quantity or not. If it is a conserved quantity it's the generator of a symmetry transformation. A symmetry transformation is a transformation which leaves the variation of the action invariant.

The most general formulation of this Noether theorem (i.e., that any conservation law implies a one-parameter Lie symmetry and vice versa) is in the Hamilton formulation of the action principle, where a symmetry is represented by any canonical transformation, that leaves the variation of the action invariant.
 
  • #8
Delta2
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@vanhees71 , what is the formula for the conserved quantities in the most general formulation of Noether's theorem, and what is the type of the transformations that the theorem uses?

Because the OP thinks that the formula he gives for K holds only if the transforms are of the type ##q'_i=q_i+\delta q_i##.
 
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  • #9
vanhees71
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This is the special case of transformations in configuration space. It's of course sufficient for classical mechanics, because there everything can be formulated in terms of the Lagrange formalism, where you work with the action as a functional of trajectories in configuration space. As I said, I have this only in a German manuscript:

https://itp.uni-frankfurt.de/~hees/publ/theo1-l3.pdf
Noether's theorem in the Lagrange formulation is at p. 88ff. I think the OP's ##K## is essentially right for the special case that the action itself is invariant. The more general case is that only the variation of the action is invariant, and this leads to the addition of an arbitrary function ##\Omega(q,t)## in Eq. (3.5.4) of my manuscript.
 
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