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Newtonian formulation/proof of Noether's theorem

  1. Dec 30, 2015 #1
    Hi.

    I've only ever seen Noether's theorem formulated ond proven in the framework of Lagrangian mechanics. Is it possible to do the same in Newtonian mechanics, essentially only using F=dp/dt ?

    The "symmetries" in the usual formulation of the theorem are symmetries of the action with respect to continuous transformations. How would this translate to Newtonian mechanics?
     
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  3. Dec 30, 2015 #2

    Vanadium 50

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    Newton's Laws are examples of equations of motion, and the equations of motion are the outcome of the Lagrangian approach. I don't see how you could start with the outcome and then recover a theorem about the process - a process that you never undergo, since you started with the outcome.
     
  4. Dec 31, 2015 #3
    Newton's laws use F without describing how to work out what the force is. You need to include a separate verbal description of the forces for all possibe states of the system, and there is no standard format for expressing that- short of rederiving Lagrangian mechanics. Therefore, the tools you would need in order to define symmetries in this context to not exist.
     
  5. Dec 31, 2015 #4
    THere are equation of motions that are not derivable from Lagrangians, so I guess the question might have some relevance.
     
  6. Jan 3, 2016 #5

    samalkhaiat

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    Yes, you can.
    1) If the potential [itex]V[/itex] does not depend explictly on the position, i.e., invariant under translation, then Newton equation gives you the law of momentum conservation: [itex]\frac{dp}{dt} = 0 \ \Rightarrow \ p = \mbox{constant}[/itex].
    2) Multiplying Newton equation by [itex]\dot{x}[/itex], you get
    [tex]\frac{d}{dt} \left( \frac{1}{2} m v^{2} + V \right) = 0 .[/tex]
    Thus, symmetry under time-translation gives you the law of energy conservation.
    3) Take the cross product of Newton equation with [itex]\vec{r}[/itex], you get
    [tex]\vec{r} \times \frac{d\vec{p}}{dt} = \frac{d}{dt} (\vec{r} \times \vec{p}) = - \vec{r} \times \nabla V .[/tex]
    So, if the potential depends only on [itex]|\vec{r}|[/itex], i.e., rotationally invariant, then
    [tex]\frac{d}{dt} (\vec{r} \times \vec{p}) \equiv \frac{d \vec{L}}{dt} = 0,[/tex]
    and this says that angular momentum conservation follows from the invariance under rotation.
    As you see, these are Noether-type statements. However, the beauty of the Noether theorem can only be appriciated in the Lagrangian formalism.
     
  7. Jan 4, 2016 #6
    How can the potential not depend on position? It always depends on it, or it's not a potential. If it's constant then it doesn't affect anything.
     
  8. Jan 4, 2016 #7

    samalkhaiat

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    Yes, when "it doesn't affect anything" it means you have translation invariance.
     
  9. Jan 11, 2016 #8
    I don't quite understand this one. Where exactly did you use symmetry under time-translation? I can only see you starting from Newton's equation and arriving at energy conservation without further assumptions.
     
  10. Jan 12, 2016 #9

    samalkhaiat

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    What is the difference between
    [tex]\frac{dV}{dt} = \frac{\partial V}{\partial t} + \frac{\partial V}{\partial x} \dot{x} ,[/tex]
    and
    [tex]\frac{dV}{dt} = \frac{\partial V}{\partial x} \dot{x} ,[/tex]
    and which one did I use?
     
  11. Jan 12, 2016 #10
    Okay, you assume that the potential has no explicit time dependance.

    So in this Newtonian formulation, all symmetries are in fact symmetries of the potential? In the Lagrangian formalism, the full Lagrangian (which in the simplest case ##L=T-V## also contains kinetic energy) needs to be symmetric. Are these definitions of "symmetry" really equivalent?
     
  12. Jan 13, 2016 #11

    samalkhaiat

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    This happens when the dynamical variables satisfy the equation of motion (Newton's equation in our case), and the symmetry in this case is called on-shell symmetry. In general, i.e., off-shell symmetry of the action integral leads to the so-called Noether identity
    [tex]
    \left( \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) \right) \delta x - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \delta x - L \delta t \right) = 0
    [/tex]
     
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