0
$\begingroup$

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:

$$m(\dot{\vec{v}}_\parallel+\dot{\vec{v}}_\perp)=q(\vec{E}_\parallel+\vec{E}_\perp+\vec{v}_\perp\times\vec{B})$$

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:

$$m\epsilon_{ilm}\dot{V}_iB_m=q(\epsilon_{lim}E_iB_m+\epsilon_{ijk}\epsilon_{lim}V_jB_kB_m)$$

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: $$\left(\frac{m}{q}\right)^2\ddot{V}_l=\epsilon_{lim}E_iB_m+V_mB_mB_l-B_mB_mV_l$$

$$\left(\frac{m}{Bq}\right)^2\ddot{V}_l=\epsilon_{lim}E_iB_m/B^2+V_mB_mB_l/B^2-V_l$$

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?

$\endgroup$
1
$\begingroup$

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): $$\begin{align}m\dot{{v}}_z &=qE_z\\ m\dot{{v}}_x &=q(E_x+v_yB_z)\\ m\dot{{v}}_y &=q(E_y-v_xB_z)\end{align}$$

$\endgroup$
  • $\begingroup$ 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 '16 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.