Consider a charge $q$ moving in a uniform magnetic field $\vec{B}=B\hat{y}$. The initial velocity of the charge is $\vec {v_{0}}=v\hat{x}$ ($v$ is a constant). Due to the magnetic field, the charge experiences a force $\vec{F}=qvB\hat{z}$ at $t=0$ and proceeds to exhibit cyclotron motion. Now if we were to move with velocity $v\hat{x}$, then the velocity of the charge would be $\vec{0}$ with respect to us at $t=0$. Therefore, the magnetic field w.r.t to us is also $\vec{0}$. Note that as the magnetic field is uniform, $\frac{\partial \vec{B}}{\partial t}=\vec{0}$ w.r.t to the new frame; hence, no electric field will be produced. Then how do we explain the acceleration of the charge w.r.t to the new frame?

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    $\begingroup$ When changing reference frames, the magnetic and electric fields mix up. You'll see an electric force at $t=0$ $\endgroup$ Commented May 29, 2023 at 16:40

1 Answer 1


Let S be the frame of reference in which the charge is moving with $v_0=v\hat x$. Let S' be the frame in which it is at rest. In S, the initial rate of change of momentum (z-component; there is no other) is $$\frac{dp_z}{dt}= qvB_y$$

In S' the charge is at rest, so there is no magnetic Lorentz force, even though there is still a magnetic field. But, as Nickolas Alves points out, there is an electric field, $E'_z$, in the z-direction, so $$\frac{dp'_z}{dt'}= qE'_z.$$ From the standard theory of field transforms, $$E'_z=\gamma (E_z+vB_y)\ \ \ \ \ \ \text{so in this case}\ \ \ \ \ E'_z=\gamma vB_y.$$ Turning our attention to the left hand sides of the equations for rate of change of momentum, suppose that in S, q acquires an increment of transverse momentum $\Delta p_z$ in a time $\Delta t$, whereas in S', q acquires an increment of transverse momentum $\Delta p'_z$ in a time $\Delta t'$.

Transverse momentum is the same in S and S', so $\ \ \ \ \ \Delta p'_z=\Delta p_z.$

A time interval at the position of q in its rest frame, S', is a proper time, whereas in S it is an improper time, so $\Delta t'=\frac 1{\gamma}\Delta t$. Therefore we have $$\frac{dp'_z}{dt'}= \gamma \frac{dp_z}{dt}.$$ Substituting $\gamma \frac{dp_z}{dt}$ for $\frac{dp'_z}{dt'}$ and $\gamma vB_y$ for $E'_z$, in the equation of motion in S' we find it to be identical to the equation of motion in S.

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    $\begingroup$ @KBhatta123. Thank you for accepting my answer. Note that the field transforms between frames MUST be such that the equation of motion in the S' frame implies the equation of motion in the S frame, as no inertial frame is favoured; all are equally valid. We can, in fact turn the argument in my answer round the other way, taking the equivalence of the equations of motion as our premiss and deducing the transform $E'_z=\gamma vB_y$ ! $\endgroup$ Commented May 30, 2023 at 17:31

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