In some inertial reference frame $S$, there is a uniform electric field $\mathbf{E}$ directed along the positive $y$-direction and a uniform magnetic field $\mathbf{B}$ directed along the positive $z$ direction. The magnitudes of the fields are $E$ and $B$, respectively, with $E<c B$. Show that it is possible to find a reference frame in which the electric field vanishes, and determine the relative speed $\beta c$ of the two frames and the magnitude of the magnetic field in the new frame.

[Hint: You may assume that under a standard Lorentz boost with speed $v=\beta$ c along the $x$-direction, the electric and magnetic field components transform as

$$\left(\begin{array}{c} E_{x}^{\prime} \\ E_{y}^{\prime} \\ E_{z}^{\prime} \end{array}\right)=\left(\begin{array}{c} E_{x} \\ \gamma(\beta)\left(E_{y}-v B_{z}\right) \\ \gamma(\beta)\left(E_{z}+v B_{y}\right) \end{array}\right) \quad a n d \quad\left(\begin{array}{c} B_{x}^{\prime} \\ B_{y}^{\prime} \\ B_{z}^{\prime} \end{array}\right)=\left(\begin{array}{c} B_{x} \\ \gamma(\beta)\left(B_{y}+v E_{z} / c^{2}\right) \\ \gamma(\beta)\left(B_{z}-v E_{y} / c^{2}\right) \end{array}\right),$$

where the Lorentz factor $\gamma(\beta)=\left(1-\beta^{2}\right)^{-1 / 2}$.]

A point particle of mass $m$ and charge $q$ moves relativistically under the influence of the fields $\mathbf{E}$ and $\mathbf{B}$. The motion is in the plane $z=0$. By considering the motion in the reference frame in which the electric field vanishes, or otherwise, show that, with a suitable choice of origin, the worldline of the particle has components in the frame $S$ of the form

$$\begin{aligned} &c t(\tau)=\gamma(u / c) \gamma(\beta)\left[c \tau+\frac{\beta u}{\omega} \sin \omega \tau\right] \\ &x(\tau)=\gamma(u / c) \gamma(\beta)\left[\beta c \tau+\frac{u}{\omega} \sin \omega \tau\right] \\ &y(\tau)=\frac{u \gamma(u / c)}{\omega} \cos \omega \tau \end{aligned}$$

Here, $u$ is a constant speed with Lorentz factor $\gamma(u / c), \tau$ is the particle's proper time, and $\omega$ is a frequency that you should determine.

Using dimensionless coordinates,

$$\tilde{x}=\frac{\omega}{u \gamma(u / c)} x \quad \text { and } \quad \tilde{y}=\frac{\omega}{u \gamma(u / c)} y$$

sketch the trajectory of the particle in the $(\tilde{x}, \tilde{y})$-plane in the limiting cases $2 \pi \beta \ll u / c$ and $2 \pi \beta \gg u / c$.