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What is Laplace-Runge-Lenz vector

  1. Jul 23, 2014 #1

    This vector is a constant of the motion for an inverse-square force law (as are the angular momentum vector and the energy).

    It points to the periapsis of the orbit (the position of closest approach).

    It is proportional to the eccentricity of the orbit (for that reason, a scaled version of it is sometimes called the eccentricity vector).

    Carried over into quantum mechanics, it makes possible an elegant derivation of the energy spectrum of a hydrogenic atom.


    We start with the equations of motion:
    [itex]\frac{d{\mathbf r}}{dt} = \frac{\mathbf p}{m} ,\ \frac{d{\mathbf p}}{dt} = - \frac{K \mathbf r}{r^3}[/itex]
    for position r, momentum p, mass m, and force constant K.

    The energy is
    [itex]E = \frac{p^2}{2m} - \frac{K}{r}[/itex]

    The angular momentum is
    [itex]{\mathbf L} = {\mathbf r} \times {\mathbf p}[/itex]

    The LRL vector is
    [itex]{\mathbf A} = {\mathbf p} \times {\mathbf L} - \frac{mK \mathbf r}{r}[/itex]

    Its magnitude is
    [itex]A^2 = 2mEL^2 + (mK)^2[/itex]

    and it has direction constraint
    [itex]{\mathbf A}\cdot{\mathbf L} = 0[/itex]

    The eccentricity vector is
    [itex]{\mathbf e} = \frac{1}{mK}{\mathbf A}[/itex]

    Extended explanation

    Derivation of conic-section orbit shape from the LRL / eccentricity vector.

    First, show that the orbit is in a plane that goes through the origin, using conservation of angular momentum around the origin.
    [itex]{\mathbf L}\cdot{\mathbf r} = 0[/itex]

    Multiply the LRL vector by r:
    [itex]{\mathbf A}\cdot{\mathbf r} = L^2 - mKr[/itex]

    Rearrange and square:
    [itex] (mK)^2 r^2 = (L^2 - {\mathbf A}\cdot{\mathbf r})^2 [/itex]

    This equation is manifestly a quadratic equation in r. With planarity, this shows that the orbit is a two-dimensional conic section: a line, circle, ellipse, parabola, or hyperbola. Here is a derivation using an elliptical orbit; the other cases have very similar derivations, and can be derived from the elliptical case. Let the orbit be
    [itex]{\mathbf r} = a \{\cos u - f, \sqrt{1-e^2} \sin u, 0 \}[/itex]

    with a the semimajor axis, e the eccentricity, f an unknown factor, and u the "eccentric anomaly", the circle angle for the ellipse as a squashed circle. We first find that (vector L) is {0,0,L} and (vector A) is {A,0,0}, and that
    [itex]r = \frac{1}{mK}(L^2 - {\mathbf A}\cdot{\mathbf r}) = \frac{1}{mK}((L^2 + A a f) - A a \cos u)[/itex]

    This constrains |f| = |e|, and we can take f = e without loss of generality, making the distance [itex]r = a (1 - e \cos u)[/itex]. With one conic-section focus being at the origin, we thus complete the derivation of Kepler's first law.

    This also yields not only the directions, but also the magnitudes of the conserved quantities as functions the orbit's size and shape: [itex]L = \sqrt{mKa(1-e^2)}[/itex], [itex]A = mKe[/itex], and [itex]E = - \frac{K}{2a}[/itex]

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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