1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A problem I'm having with Noether's Theorem

  1. Jul 1, 2012 #1
    The biggest problem I'm having with Noether's theorem is that I can't seem to find it stated precisely enough anywhere. The standard statement seems to be just that 'for any continuous symmetry of a system there is a corresponding conserved quantity'. I think I understand this fine when the symmetry is a symmetry of the Lagrangian itself, since then the theorem is relatively easy to prove. Where I have difficulty is when the Lagrangian changes under a transformation but the equations of motion do not, in which case I think Noether's theorem is still supposed to apply but it doesn't seem to in this example:

    If you take a 1D harmonic oscillator, the Lagrangian is: [itex]L = (1/2)m\dot{q}^{2} - (1/2)kq^{2}[/itex]

    Now consider the transformation of multiplying q by a constant. If you think about SHM motion it's clear that this is a symmetry, if I'm interpreting what a symmetry means correctly, because it's a transformation that maps solutions of the equation of motion to other solutions of the equation of motion. However, the Lagrangian is not invariant under this transformation.
    If you apply an infinitesimal version of this transformation to a given solution (q' = (1 + ε)q), the new Lagrangian along the path as a function of time will be related to the old by:

    L' = L + 2εL

    If Noether's theorem applied, and I am applying it correctly, the conserved quantity associated with this symmetry would then be:

    [itex]q(\partial L/\partial\dot{q}) - 2I[/itex]

    Where dI/dt = L.

    The trouble is L is not the total time derivative of any function, if it were then the action along any path would depend only on the end points. So there doesn't seem to be any conserved quantity associated with this symmetry.

    I'd appreciate it if anyone could point out what I'm doing wrong here. Maybe under the rigorous statement of Noether's theorem it isn't supposed to apply in this case, maybe I'm misunderstanding what a symmetry is, and maybe it does apply but I'm applying it wrongly. I'd like to get this resolved anyway. Thanks in advance.
     
    Last edited: Jul 1, 2012
  2. jcsd
  3. Jul 2, 2012 #2
    Are you allowed to bump threads?
     
  4. Jul 2, 2012 #3
    I hope so because I'm curious to hear the answer to this question myself. I've only admired Noether's theorem from afar so any definite questions always interest me...
     
  5. Jul 3, 2012 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    A symmetry in the sense of Noether's theorem is defined by the demand that the variation of the action is invariant under the corresponding transformation. That's not the case for your harmonic-oscillator action and dilatation symmetry. It cannot be since the action contains dimensionful quantities like the mass and k.
     
  6. Jul 3, 2012 #5
    But neither the mass nor k have dimensions of length, which is the only dimension being affected by the transformation I am using.

    What do you mean by the statement that the variation of the action must be invariant? In this example any path that minimises the action between two endpoints prior to the transformation will still minimise the action between its new endpoints once the transformation has been applied to it, is that not sufficient to qualify as a symmetry then?
     
    Last edited: Jul 3, 2012
  7. Jul 14, 2012 #6
    Symmetry as mentioned in Noether's Theorem refers to symmetry of the Lagrangian which would mean L' = L.

    I think you're trying to see if symmetry in the equations of motion will lead to a conserved quantity which as you can see from your example is not quite true. The equations of motion are derived from the Lagrangian so symmetry in the Lagrangian leads to symmetry in the equations of motion, however as you can see from your counter-example, symmetry in the equations of motion doesn't necessarily lead to symmetry in the Lagrangian.

    So, just because there's a symmetry in the equations of motion, doesn't mean there's a conserved quantity. It's symmetry of the system i.e. the Lagrangian that you're looking for :)
     
  8. Jul 15, 2012 #7
    Hmm but this seems to contradict the book I have. It says that the Lagrangian can change as long as it changes by the total time derivative of some function, in which case a conserved quantity can still be found. I can also see that if the Lagrangian changes in such a way then the equations of motion will be invariant even though the Lagrangian is not, and the transformation can be called a symmetry. What I can't make sense of is the idea that the converse is true, that if the equations of motion are invariant then the Lagrangian must change by the total time derivative of a function and a conserved quantity exists. The example in my original post seems to contradict this. From the responses I've had including yours I guess that the converse just isn't true, and symmetries of the equations of motion don't necessarily lead to conserved quantities, so thanks for your help.

    Applications of Noether's theorem where the Lagrangian changes do seem to crop up though. For example in the book I have on Lagrangian mechanics it shows how a conserved quantity (relating to centre of mass) can be derived from a change of velocity transformation. This happens despite the fact that the Lagrangian is not invariant under such a transformation (it depends explicitly on a particle's speed). It is only the equations of motion in this case that are invariant. I've also started a book on Quantum Field Theory (it was the use of Noether's Theorem here which got me on the line of thought that lead to this thread). In field theory, Noether's theorem still applies, but the Lagrangian is no longer invariant under displacement, only the equations of motion are, so to derive conservation of energy and momentum you need to handle cases where the Lagrangian changes.
     
  9. Jul 15, 2012 #8

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    As I've already written earlier in the thread. The weakest assumption (at least as far as I know) for a symmetry is that the first variation (functional derivative) of the action is invariant under the corresponding operation on the dynamical quantities of the system.
     
  10. Jul 15, 2012 #9
    Ah, you're book is probably referring to how the Lagrangian of a system is not unique. What you're actually interested in is the action i.e. the integral of the Lagrangian with respect to time. The aim is then to find an extremum of this action and this extremum for the action of L will be the same as the extremum for the action of L' = aL (for any real number a) and also for L' = L + df/dt (for any arbitrary function f).

    Once you pick a Lagrangian you want to use, then that's the thing you look for a symmetry in. Some Lagrangians are easier to spot symmetries in than others (usually the obvious one will do!)
     
  11. Jul 16, 2012 #10
    Ok thank you both for your help, so a symmetry in the equations of motion is not sufficient for Noether's theorem to apply then. I'm happy to accept what both of you are saying on this, since you seem to agree and it explains the example in my original post. It does seem to contradict my book though, which is 'Lagrangian and Hamiltonian mechanics' by M.G. Calkin. I'll post the quotes which seem to contradict what you're saying and you can see what you think:

    With this definition of invariance transformations Noether's theorem is then quoted as:

    And the derivation given makes use of the assumption that if the equations of motion are invariant then the Lagrangian must change by a total time derivative.

    So this does seem to be saying that a symmetry in the equations of motion is sufficient for Noether's theorem to apply, because in this case the Lagrangian must apparently change by a total time derivative. But I don't think it does necessarily have to change by a total time derivative, the example in my original post seems to contradict this, and this is why I was getting confused.

    It does the same thing in a later derivation of the same result using the action and active rather than passive transformations:

    And from there it goes on to rederive Noether's theorem. Again I have a problem with that very last sentence, that the change in the action has to depend only on the endpoints. Noether's theorem seems only to apply if that is true, but that seems to be a stronger condition than just 'actual paths are taken into other possible actual paths'. I came up with the example in my post by trying to find a case where the difference was not a function only of the end points, but depended on the path taken and was just also minimised for the same paths the action was. An obvious choice is making the Lagrangian change by a multiple of itself, and that seems to be what the book would call an 'invariance transformation' but there is no conserved quantity.

    So do you guys agree that the book is wrong? Or am I still misunderstanding something about what you and/or the book is saying?
     
  12. Jul 17, 2012 #11
    I'm gonna take a bold leap here and say I disagree with the book. I learned this from my lecture notes and whilst the "invariance transformation" approach might be equivalent, it's not the same as what I'm used to. Then again this just might be a convention thing and I'm unfamiliar with it.

    Feel free to look at this, page 24 onwards might be the most relevant:
    http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf
     
  13. Jul 17, 2012 #12

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    As far as I can see, there's nothing wrong with the derivation given on pages 23ff. However, it's not the most general case of a symmetry considered. This approach will fail for Galileo boosts, which in fact are a symmetry of Newtonian space time and lead to the conservation of the center-mass velocity. For this, you need to consider the more general case that the Lagrangian is allowed to change by the total time derivative of a function of the generalized coordinates and time, and you still have a symmetry.
     
  14. Jul 18, 2012 #13
    Yes those notes seem to refer to cases where the Lagrangian is invariant under some transformation, but I think Noether's theorem can be used more generally than that.

    Vanhees, do you think that the quotes I posted from the book are wrong?
     
  15. Jul 18, 2012 #14

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    What you quoted from the book seems to be perfectly fine, because it says that the old and the new Lagrangian are allowed to deviate by a total dervative of a function of the generalized coordinates and time. At least in this form, all fundamental conservation laws can be derived, as far as I know.
     
  16. Jul 19, 2012 #15
    Are you sure that's what it's saying? I read it as saying that whenever the equations of motion are unchanged by a transformation (what the book defines as an invariance transformation) then the Lagrangian must change by a total time derivative and a conserved quantity therefore exists.

    This seems to contradict the example in my original post where the transformation doesn't change the form of the equations of motion (therefore satisfying the book's definition of an invariance transformation) but the Lagrangian doesn't change by a total time derivative so a conserved quantity does not exist.

    I am basically disputing the claim the book seems to make that a transformation which does not change the form of the equations of motion must change the Lagrangian by a total time derivative.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A problem I'm having with Noether's Theorem
  1. Noether's Theorem (Replies: 36)

  2. Noether's Theorem (Replies: 3)

Loading...