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A Particle in the point gravity and the Lorentz field

  1. Mar 11, 2017 #1
    Hi I will write physics terms but actually this is a math problem.

    Consider a particle which moves in accordance with the following equation

    $$m\boldsymbol{\ddot r}=\boldsymbol B\times\dot{\boldsymbol r}-\frac{\gamma}{r^3}\boldsymbol r,\quad r=|\boldsymbol r|\qquad (*)$$
    and ##\gamma=const>0,\quad \boldsymbol B=\boldsymbol{const}##
    It is a Hamiltonian system with three degrees of freedom.
    What about integration of this problem ? it looks like a classical one. It would be great to have three involutive integrals but I know only pair: the energy integral and the following one
    $$F(\boldsymbol r,\dot{\boldsymbol r})=(\boldsymbol{ K},\boldsymbol B)-\frac{1}{2}\Big(B^2r^2-(\boldsymbol B,\boldsymbol r)^2\Big),\quad \boldsymbol K=m\boldsymbol r\times\dot{\boldsymbol r},\quad B=|\boldsymbol B|$$
    The phase flow to the system with Hamiltonian ##F## (after we introduce impulses in accordance with system (*)) generates a group of symmetry to system (*) and if one could integrate the the system with Hamiltonian ##F## explicitly then one can reduce the initial system to the Hamiltonian system with two degrees of freedom. But that is not enough for explicit integration of (*). Perhaps the reduced system admits some separation of variables. That's all I can say
     
    Last edited: Mar 11, 2017
  2. jcsd
  3. Mar 12, 2017 #2
    The reduction of order.


    Let ##Oxyz## be an inertial frame, ##\vec{Om}=\boldsymbol r##. Introduce a rotating reference frame ##O\xi\eta\zeta##. This frame has angular velocity $$\boldsymbol\omega=\frac{1}{2m}\boldsymbol B.$$

    Relative this new frame of reference the equation (*) takes the form

    $$m\boldsymbol a_r=\frac{1}{4m}\boldsymbol B\times(\boldsymbol B\times r)-\gamma\frac{\boldsymbol r}{r^3},\qquad (**)$$

    here ##\boldsymbol a_r## is the acceleration of the particle relative to the frame ##O\xi\eta\zeta##.


    Let ##\boldsymbol v_r## be the relative velocity and let ## \boldsymbol K_r=m\boldsymbol r\times \boldsymbol v_r## stand for the relative angular momentum.
    It is easy to see that the quantity ##(\boldsymbol K_r, \boldsymbol B)## is the first integral to (**);
    Thus one can reduce system (**) to system with two degrees of freedom by introducing for example cylindrical coordinates about an axis ##\boldsymbol B##.
    The Lagrangian has the form
    $$L=\frac{m}{2}|\boldsymbol v_r|^2-V,\quad V=\frac{1}{8m}|\boldsymbol B\times \boldsymbol r|^2-\frac{\gamma}{r}.$$

    The problem is still open...
     
    Last edited: Mar 12, 2017
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