The forces exerted by an electromagnetic field on a moving charge

Question

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?

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.