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Conservation law for any potential field?

  1. Dec 29, 2015 #1
    Consider a free particle moving in a general time-dependent scalar potential. Energy & momentum are not conserved. However, there is a symmetry in the lagrangian: the velocity appears only as its square, so we can rotate it without affecting the value of L. What conservation law results from this symmetry?
  2. jcsd
  3. Dec 29, 2015 #2


    Staff: Mentor

    Conservation of angular momentum.
  4. Dec 29, 2015 #3
    No, that's in the case of a central potential, where the symmetry is a rotation of both the position & velocity. I am asking about a completely arbitrary potential, and noting that rotation of the velocity alone should still be a symmetry.
  5. Dec 29, 2015 #4


    Staff: Mentor

    Hmm, that is a good point that I missed. You would still calculate the conserved quantity using Noether's theorem, but I don't know what it would be offhand.
    Last edited: Dec 29, 2015
  6. Dec 31, 2015 #5
    One bump...
  7. Dec 31, 2015 #6
    Isn't this just time reversal symmetry?
  8. Dec 31, 2015 #7
    No, I'm talking about rotating the velocity by a general angle in any direction. This should be a continuous symmetry.
  9. Dec 31, 2015 #8


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    You cannot just rotate the velocity without rotating the coordinate system itself. The transformations covered by Noether's theorem are of the form ##t \to t +ks## and ##\vec x \to \vec X(t,s,\vec x)##, not transformations of the velocities.

    You can do more general canonical transformations in Hamiltonian mechanics, but based on the symmetries of the Lagrangian this is not the case. In order to have a symmetry of the Lagrangian you therefore need to have rotational symmetry of the potential as well, resulting in conservation of angular momentum.
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