This is the common problem of a charged particle moving in a static electric and magnetic field. Say $\textbf{E}=(E_x,0,0)$ and $\textbf{B}=(0,0,B_z)$.

In the inertial frame of reference, the equation of motion is (1): \begin{equation} \frac{d \textbf{v} }{dt} = -\frac{q \textbf{B} }{m}\times \textbf{v} + \frac{q}{m}\textbf{E} \end{equation}

We can find equations for $v_x$ an $v_y$ and see that the resulting motion is a circular orbit with a constant drift velocity $v_d=\frac{E_x}{B_z}$.

Surely I should get the same answer if I solve the problem in a rotating frame of reference?

I know that (2): $$ \frac{d \textbf{v} }{dt} \vert_{Inertial} = \frac{d \textbf{v} }{dt} \vert_{Rotational} + \boldsymbol{\omega}\times\textbf{v};$$

If I use Eq. (1) as the LHS of Eq. (2), and choose $ \boldsymbol{\omega}=-\frac{q \textbf{B} }{m}$, then I get (3):

$$ \frac{d \textbf{v} }{dt} \vert_{Rotational} = \frac{q}{m}\textbf{E};$$

How do I obtain a constant drift velocity (as mentioned before) from this? Have I used any formula incorrectly? Does the electric field $\textbf{E}=(E_x,0,0)$ change form in the rotating frame?


1 Answer 1


Sure you have to transform the fields too. Let's say that your system rotates around the z-axis, so B remains unchanged, but E will move around a circle in the rotating system, so its coordinates will be:

$$ E=(E_x*\cos(\omega_{rot} t), E_x*\sin(\omega_{rot} t),0) $$

where I used $E_x$ as an amplitude, and $\omega_{rot}$ as angular frequency of the rotating system.

  • $\begingroup$ Thanks. If I now solve the equations I find that $v_x$ and $v_y$ have cos or sin dependence, so the velocity vector is rotating. But the rotation should already be incorporated in the rotating frame, so I am still missing a horizontal drift velcoty $\endgroup$
    – SuperCiocia
    Mar 30, 2014 at 17:23
  • $\begingroup$ That horizontal drift velocity apprears in a rotating motion in the rotating system, and the rotation if the angular frequencies are equal appears as linear motion. $\endgroup$ Mar 30, 2014 at 18:10
  • $\begingroup$ I get $ \textbf{v} \vert_{Rot} = \frac{E_x}{B_z} (\sin(\omega t)\textbf{i} - \cos(\omega t)\textbf{j})$, How do you transform back to the rotational frame to show that it's just a linear velocity? $\endgroup$
    – SuperCiocia
    Mar 30, 2014 at 21:29

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