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Physics
Classical Physics
Electromagnetism
Hamiltonian for a charged particle in a magnetic field
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[QUOTE="etotheipi, post: 6488337"] Looks absolutely fine! The first Hamilton equation gives ##\dot{x}_i = \frac{1}{m} \left( p_i - \frac{e}{c}A_i \right)## and the second Hamilton equation gives\begin{align*}\dot{p}_i = m\ddot{x}_i + \frac{e}{c} \sum_j \left[ \frac{\partial A_i}{\partial x_j} \dot{x}_j \right] &=\frac{e}{mc} \sum_j \left[ \left( p_j - \frac{e}{c}A_j \right) \frac{\partial A_j}{\partial x_i} \right]\\ &= \frac{e}{c} \sum_j \left[ \frac{\partial A_j}{\partial x_i} \dot{x}_j \right]\end{align*}so you end up with\begin{align*}m\ddot{x}_i = \frac{e}{c} \sum_j \left[ \left( \frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j} \right) \dot{x}_j \right] = \frac{e}{c} \sum_j \left[ F_{ij} \dot{x}_j \right] = \frac{e}{c} \left(\mathbf{v} \times \mathbf{H} \right)_i \end{align*}Note that in this last equation the indices of the Maxwell tensor ##F_{ij}## run over the set ##\{ 1,2,3 \}##, so here we are only dealing with magnetic components. To obtain the entire Lorentz force, you need to modify the original Hamiltonian to include an additional term ##e\phi##. [/QUOTE]
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Physics
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Electromagnetism
Hamiltonian for a charged particle in a magnetic field
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