# A Particle in the point gravity and the Lorentz field

1. Mar 11, 2017

### zwierz

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. Mar 12, 2017

### zwierz

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