I am trying to solve Hughston Tod question (2.15):

Find the general solution of the Lorentz equation $$m\dot{V}_i=q(E_i+\epsilon_{ijk}V_jB_k)$$for constant $E_i$ and $B_i$, given the initial position and velocity of the particle.

I found following solution in chapter 2.2 of Motion of Charged Particles in Fields

I went about it in the following way: $$\vec{v}=\vec{v}_\parallel+\vec{v}_\perp $$ Similarly $$\vec{E}=\vec{E}_\parallel+\vec{E}_\perp $$

Where the parallel and orthogonal directions are wrt to the magnetic field. Using this the equation becomes:


Clearly it should be possible to isolate the equations: $$m\dot{\vec{v}}_\parallel=q\vec{E}_\parallel$$ $$m\dot{\vec{v}}_\perp=q(\vec{E}_\perp+\vec{v}_\perp\times\vec{B})$$

The first of the two equations is easily solved. But I didn't understand in the reference how he solved for the second part of the equation.

Help is appreciated.

I actually am practicing solving equations with index notation so here is what i have got now:

multiplying the Lorentz equation by $\epsilon_{lim}B_m$ yields:


This simplifies to: $$\frac{d}{dt}(m\underbrace{\epsilon_{lim}V_iB_m}_{=\frac{m}{q}\dot{V}_l-E_l})=q(\epsilon_{lim}E_iB_m+V_mB_mB_l-B_mB_mV_l)$$ However since $E_l$ is constant the derivative of it vanishes and what remains is: $$\left(\frac{m}{q}\right)^2\ddot{V}_l=\epsilon_{lim}E_iB_m+V_mB_mB_l-B_mB_mV_l$$


I can see that the drift acceleration results from the first term whereas the circular motion from the last term. But I don't know how to interpret the 2nd term and how come there is no drift due to the electric field?


1 Answer 1


You have to decompose the vector $$\vec{v}_\perp=\vec i v_{x}+\vec j v_{y}$$ and the second equation with its cross product into x- and y-components. Then you have three equations (with $\vec B$ in z-direction): $$\begin{align}m\dot{{v}}_z &=qE_z\\ m\dot{{v}}_x &=q(E_x+v_yB_z)\\ m\dot{{v}}_y &=q(E_y-v_xB_z)\end{align}$$

  • $\begingroup$ You don't need to thank @freecharly; we are a community and as a fellow comrade, I would contribute where possible as an editor. $\endgroup$
    – user36790
    Dec 15, 2016 at 4:09

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