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Physics
Classical Physics
Electromagnetism
Hamiltonian for a charged particle in a magnetic field
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[QUOTE="zhangnous, post: 6488393, member: 690274"] So the first Hamilton equation gives ##\dot{x}_i = \frac{1}{m} \left( p_i - \frac{e}{c}A_i \right)##, which is just the canonical momentum ##p_{i}=m\dot x_{i}+\frac{e}{c}A_{i}## in this special situation, and can also be used into the second Hamilton equation. It seems strange for me and looks like these two equation play with themselves. But I remember the situation of one free particle. Hamiltonian is ##H = \frac{p^2}{2m} + V(x)##. First gives##\dot x = \frac{\partial H}{\partial p}=\frac{p}{m}## and ##\dot p = m\ddot x## then second gives ##\dot p = -\frac{\partial H}{\partial x}=-\frac{\partial V(x)}{\partial x}=F##. Finally have Newton's equation ##F=m\ddot x##. Which is also the two equation play with each other. This seems I just playing a joke with my own mind. Thanks for you comprehensive explanation [/QUOTE]
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Physics
Classical Physics
Electromagnetism
Hamiltonian for a charged particle in a magnetic field
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