Motion of Charged Particles in EM fields [gnuplot]

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Simulation [gnuplot]

Modeling

Lorentz force $$\begin{equation} \label{eq:lorentz} m\boldsymbol{\ddot{x}} = \color{red}{q\boldsymbol{E}} + \color{blue}{q (\boldsymbol{v}\times \boldsymbol{B})} \end{equation}$$ Assume $\boldsymbol{E}=\left(-kx, -ky, 2kz\right)\ \left(k\ge 0\right), \boldsymbol{B}=\left(0, 0, B\right), q=-e$.Then from ($\ref{eq:lorentz}$) $$\begin{eqnarray} \ddot{x} &=& \frac{e}{m}\left(kx-\dot{y}B\right) \label{eq:x} \\ \ddot{y} &=& \frac{e}{m}\left(ky+\dot{x}B\right) \label{eq:y} \\ \ddot{z} &=& -\frac{2ek}{m}z \label{eq:z} \end{eqnarray}$$

em_motion_gif.plt

em_motion.gif

Extra

Analytical solution

Solving ($\ref{eq:z}$), we obtein $$\begin{equation} z=C\cos\left(\sqrt{2}\omega_{0}t+\alpha\right), \end{equation}$$ where $${\omega_{0}}^2=\frac{ek}{m}.$$ Rearranging ($\ref{eq:x}$) and ($\ref{eq:y}$), we have $$\begin{eqnarray} \left\{ \label{eq:xy} \begin{array}{l} \ddot{x}-{\omega_{0}}^2 x+2b\dot y=0\\ \ddot{y}-{\omega_{0}}^2 y+2b\dot x=0, \end{array} \right. \end{eqnarray}$$ where $$b=\frac{eB}{2m}$$ By putting $x=Ae^{i\omega t}$ and $y=Be^{i\omega t}$, substitution of these in ($\ref{eq:xy}$) leads to ($\ref{eq:xy2}$). $$\begin{eqnarray} \left\{ \label{eq:xy2} \begin{array}{l} -\left(\omega^2+{\omega_{0}}^{2}\right)A+2ib\omega B = 0 \\ -2ib\omega A-\left(\omega^2+{\omega_{0}}^{2}\right)B = 0． \end{array} \right. \end{eqnarray}$$ To get the solutions except for $A=B=0$, solve ($\ref{eq:det}$): $$\begin{equation} \left| \begin{array}{cc} -\left(\omega^2+{\omega_{0}}^{2}\right) & 2ib\omega \\ -2ib\omega & -\left(\omega^2+{\omega_{0}}^{2}\right) \end{array} \right| = 0 \label{eq:det} \end{equation}$$ $$\therefore\omega^2+{\omega_0}^2 = \pm 2b\omega \label{eq:det2}$$ When $b$ is greater than $\omega_0$, we obtain $$\begin{eqnarray} \label{eq:omega} \left\{ \begin{array}{l} \pm\omega_1 = b+\sqrt{b^2-{\omega_0}^2} \\ \pm\omega_2 = b-\sqrt{b^2-{\omega_0}^2}. \end{array} \right. \end{eqnarray}$$ Inserting ($\ref{eq:omega}$) into ($\ref{eq:xy2}$), we get $$\begin{eqnarray} \frac{B}{A} = \begin{cases} -i && (+\omega_1, +\omega_2)\\ i && (-\omega_1, -\omega_2). \end{cases} \end{eqnarray}$$ $$\begin{eqnarray} \label{eq:solution} \therefore\left\{ \begin{array}{l} x = A_1 e^{i\omega_1 t}+A_2 e^{-i\omega_1 t}+A_3 e^{i\omega_2 t}+A_4 e^{-i\omega_2 t} \\ y = -i A_1 e^{i\omega_1 t}+i A_2 e^{-i\omega_1 t}-i A_3 e^{i\omega_2 t}+i A_4 e^{-i\omega_2 t} \\ \end{array} \right. \end{eqnarray}$$ We put $$\begin{eqnarray} \label{eq:Ralpha} \begin{cases} R_1 = 2\sqrt{A_1 A_2} \\ R_2 = 2\sqrt{A_3 A_4} \\ \alpha_1 = \tan^{-1}\left[\frac{i\left(A_1-A_2\right)}{A_1+A_2}\right]\\ \alpha_2 = \tan^{-1}\left[\frac{i\left(A_3-A_4\right)}{A_3+A_4}\right]. \end{cases} \end{eqnarray}$$ Combining ($\ref{eq:solution}$) and ($\ref{eq:Ralpha}$), we obtain $$\begin{eqnarray} \left\{ \begin{array}{l} x = R_1 \cos\left(\omega_1 t+\alpha_1\right)+R_2 \cos\left(\omega_2 t+\alpha_2\right) \\ y = R_1 \sin\left(\omega_1 t+\alpha_1\right)+R_2 \sin\left(\omega_2 t+\alpha_2\right). \end{array} \right. \end{eqnarray}$$

EM_field.plt

E_field.png

B_field.png