39 The Laplace-Runge-Lenz Vector - ckw

By Wesley Mcdonald,2014-03-18 09:38
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39 The Laplace-Runge-Lenz Vector - ckwThe,the,CKW,runge

3.9. The Laplace-Runge-Lenz Vector

    Besides the angular momentum vector L, the Kepler problem possesses another conserved vector

    called the Laplace- Runge- Lenz vector. To see how it comes about, consider a central force of

    the form

    r (3.79) pf,,fr;;r

    so that

    d pLpLpL;,;?;;;dt

     [] ,;pLL0

    r ,;;frmrr;;;;r

    fr;;2?? (3.80) ,,?mrrrrr;;?,r


    21d1drrr rrrr,,,;;2dt2dt

    eq(3.80) becomes

    frrrr;;d;?22 pLrr;,?mrrr,?mfrr;;;;;;~(2dtrrr?)

    dr;?2 (3.81) ,?mfrr;;~(dtr?)

    For the Kepler problem,

    k (3.49) fr,?;;2r

    so that

    ddr;? pL;,mk;;~(dtdtr?)

    r (3.82) ApL,;?,mkconstr

    where A is called the Laplace- Runge- Lenz vector. Since

     and LpL,;,0Lrrpr,,;,,0;;;;

    eq(3.82) implies

     (3.83) LA,,0

    which means A must lie in some fixed direction in the plane of the orbit. Denoting the angle

    between A and r by , eq(3.82) gives

     (3.84) Ar,,Arcos,;,?pLrmkr;;


    2 pLrrpL;,,;,,l;;;;

    eq(3.84) becomes

    2 Arlmkrcos,?

    2lr Amkcos

    1mkA;? (3.85) ,?cos1~(2rlmk?)

    which is simply the orbit (3.51) of the Kepler problem with

    ( (perihelion point) 0


    22El (3.86) Amke,?mk12mk

To summarize, we have shown that for the Kelper problem, A, L, and E are constants of the

    motion. Since the problem has 3 degrees of freedom, the motion contains exactly 6

    rindependent constants of motion, e.g., the initial values of r and . Thus, not all of the 7 scalar

    quantities in A, L, and E can be used as independent constants to specify the motion. In fact, A,

    L, and E can only be used to select an (entire) orbit for the motion. Having chosen an orbit, one still need a scalar constant to specify the initial position of the particle on the orbit. Therefore, there can only be 5 independent scalars among A, L, and E. Of the 2 relations connecting the

    latter, one is given by eq(3.83), and the other by eq(3.86), which can be written as

    2222 (3.87) AmkmEl,?2

    Thus, the independent contants may be chosen as E, L, and 1 scalar quantity out of A.

As shown in ?3.6, only the inverse-square and the Hooks law forces give rise to closed orbital

    motion. Now, an open orbit motion is ergodic, i.e., the particle never retraces its own path so that it will eventually pass through every point on the constant energy surface in phase space. In real space, the orbit fills out the annular region between the apsidal distances (radii of turning points) [see Fig.3.7]. The function is therefore infinitely valued. Since E , L and the r;;

    initial position are all single valued quantities, the last constant of motion must convey this

    characteristic and be infinitely valued. Hence, for open orbits, cannot be a simple algebraic

    function of r and p like the Laplace- Runge- Lenz vector. As for the only other closed orbit case,

    ndi.e., the Hooks law force, the conserved quantity turns out to be a 2 rank tensor [see ?9.7].

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