- #1

zhangnous

- 2

- 5

the Hamiltonian of a charged particle in a magnetic field(ignore the electric field) is $$H=\sum_{i} \left\{ \frac{1}{2m} \left[ p_{i}-\frac{e}{c}A_{i}(x) \right]\left[ p_{i}-\frac{e}{c}A_{i}(x) \right] \right\},$$and$$p_{i}=m\dot x_{i}+\frac{e}{c}A_{i}(x)$$and Hamilton's equations of motion is $$\dot q_{i} = \frac{\partial H}{\partial p_{i}}$$$$\dot p_{i} = -\frac{\partial H}{\partial q_{i}}$$then I solve these equations in x. $$\dot x = \frac{\partial H}{\partial p_{x}} = \frac{p_{x}-\frac{e}{c}A_{x}}{m}$$and$$\dot p_{x}=m\ddot x + \frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{x}}{\partial y}\dot y+\frac{\partial A_{x}}{\partial z}\dot z \right)$$$$-\frac{\partial H}{\partial x}=-\left[ \frac{p_{x}-\frac{e}{c}A_{x}}{m}(-\frac{e}{c}\frac{\partial A_{x}}{\partial x})+\frac{p_{y}-\frac{e}{c}A_{y}}{m}(-\frac{e}{c}\frac{\partial A_{y}}{\partial x})+\frac{p_{z}-\frac{e}{c}A_{z}}{m}(-\frac{e}{c}\frac{\partial A_{z}}{\partial x}) \right]=\frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{y}}{\partial x}\dot y+\frac{\partial A_{z}}{\partial x}\dot z \right)$$then$$m\ddot x + \frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{x}}{\partial y}\dot y+\frac{\partial A_{x}}{\partial z}\dot z \right)=\frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{y}}{\partial x}\dot y+\frac{\partial A_{z}}{\partial x}\dot z \right)$$and get $$ma_{x}=\frac{e}{c}(B_{z}\dot y-B_{y}\dot z)$$

Am I a right way to work it out? I am not sure if it is right, because it seems like a math structure game.