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Symmetry behind Laplace-Runge-Lenz vector conservation?

  1. Jan 13, 2008 #1
    For gravitation and any inverse-square forces, the Laplace-Runge-Lenz vector is conserved. (see http://en.wikipedia.org/wiki/Laplace-Runge-Lenz_vector" [Broken])
    Any conserved quantity is associated with a symmetry of the Hamiltonian with respect to some coordinate, according to the Noether theorem.

    I would like to know what this symmetry coordinate(s) is (are) when the conservation of the LRL vector is involved.

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 13, 2008 #2
    i dont know if i got right your question, but i think that the answer is spherical coordinate, since H=P^2+V(r), and you have anly radial dependence on the V.
    Last edited by a moderator: May 3, 2017
  4. Jan 13, 2008 #3

    The spherical symmetry is not enough to imply the conservation of the LRL vector.
    The conservation of the LRL is a consequence of the 1/r² dependence of the force.
    But what is then the symmetric coordinate(s) behind this conservation law?

    This might also be partly related to the fact that the trajectories are closed which only happen for 1/r² and r² potentials. Similarly, other dependences (1/r4, ...) lead to non-closed trajectories.
  5. Jan 14, 2008 #4
    yes i think you are right, the spherical symmetry gives you angular momentum conservation. But if you're system is kepler's one you get also that Lenz vector is conserved, in fact in the proof you use only this 2 facts...
    and the coordinate you use is the only radial one since the forces are central.
    since we live in 3-dim world the dependence on V is r^-2, but we colud have log(r) depndence in flat world... i think thats why we find This vector.
    a good exercice is to calculate what kind of vector we find in other dimension...

    bye marco
  6. Jan 14, 2008 #5


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  7. Jan 18, 2008 #6

    You are right, it looks like an answer, but I have two problems with it:

    1) I don't understand it
    2) I want to relate the conservation of the LRL vector to a symmetry and the Noether theorem

    I don't see how the scaling symmetry can fit in the Noether theorem and answer my question, since the hamiltonian is changed by this scaling. But since the scaling of the Hamiltonian is known and simple, maybe the answer might reduce to a tiny extension of the Noether theorem or to small re-definition of the Kepler Hamiltonian so as to restore a perfecty symmetry and make the Noether theorem directly applicable (without changing any physics).

    There is probably a simple answer.

    (note: I should check how the scaling affects the motion)
    Last edited: Jan 18, 2008
  8. Jan 31, 2011 #7
    The answer is much simplier than you can think. The Lenz-Vector is defined as follows:

    A=p[tex]\timesL[/tex] - m[tex]\alpha[/tex]e[tex]_{r}[/tex]

    m[tex]\alpha[/tex]e_{r} is a constant. Therefore

    [tex]\frac{d}{dt}[/tex] m[tex]\alpha[/tex]e_{r}=0

    As p[tex]\timesL[/tex] is a constant too,

    [tex]\frac{d}{dt}[/tex]A=p[tex]\timesL[/tex] - m[tex]\alpha[/tex]e[tex]_{r}[/tex]=0

    shows that the Lenz-Vector is preserved.

    There is no symmetry behind it itself, since it is formed from the product of angular and linear momentum and their preservation is derived from their symmetries.

    (sorry for the formatation)
  9. Feb 8, 2011 #8
    Well observed D'Alembert !
    It is clear that any combination of several constants of motion is also a constant of motion.
    Obviously, one could generate many related constant of motion, and each of them do not need to be associated with a symmetry.
    So, the Noether theorem goes from symmetry to a constant, and not the reverse.

    Thanks for the observation!
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