- #1
zwierz
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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
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
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