Question 12.69: A charge q is released from rest at the origin, in the prese...

\text { Rewrite Eq. } 12.109 \text { with } x \rightarrow y, y \rightarrow z, z \rightarrow x:

\begin{aligned}&\bar{E}_{x}=E_{x}, \quad \bar{E}_{y}=\gamma\left(E_{y}-v B_{z}\right), \quad \bar{E}_{z}=\gamma\left(E_{z}+v B_{y}\right) \\&\bar{B}_{x}=B_{x}, \quad \bar{B}_{y}=\gamma\left(B_{y}+\frac{v}{c^{2}} E_{z}\right), \quad \bar{B}_{z}=\gamma\left(B_{z}-\frac{v}{c^{2}} E_{y}\right)\end{aligned}                               (12.109)

This gives the fields in system \overline{ S } moving in the y direction at speed v.

\text { Now } E =\left(0,0, E_{0}\right) ; B =\left(B_{0}, 0,0\right), \text { so } \bar{E}_{y}=0, \bar{E}_{z}=\gamma\left(E_{0}-v B_{0}\right), \bar{E}_{x}=0.

(The condition E_{0} / B_{0}<c guarantees that there is no problem getting to such a system.)

With this, \bar{B}_{y}=0, \bar{B}_{z}=0, \bar{B}_{x}=\gamma\left(B_{0}-\frac{v}{c^{2}} E_{0}\right)=\gamma B_{0}\left(1-\frac{v^{2}}{c^{2}}\right)=\gamma B_{0} \frac{1}{\gamma^{2}}=\frac{1}{\gamma} B_{0} ; \quad \overline{ B }=\frac{1}{\gamma} B_{0} \hat{ x }.

The trajectory in \overline{ S }: Since the particle started out at rest at the origin in S, it started out with velocity -v \hat{ y } \text { in } \overline{ S }. According to Eq. 12.71 it will move in a circle of radius R, given by

p = Q B R                               (12.71)

p=q B R, \text { or } \gamma m v=q\left(\frac{1}{\gamma} B_{0}\right) R \Rightarrow R=\frac{m \gamma^{2} v}{q B_{0}}.

The actual trajectory is given by \bar{x}=0 ; \bar{y}=-R \sin \omega \bar{t} ; \bar{z}=R(1-\cos \omega \bar{t}) ; \text { where } \quad \omega=\frac{v}{R}.

The trajectory in S: The Lorentz transformations Eqs. 12.18 and 12.19, for the case of relative motion in the y-direction, read:

\begin{matrix} \text { (i) } \bar{x}=\gamma(x-v t) \text {, } \\ \text { (ii) } \bar{y}=y \text {, } \\ \text { (iii) } \bar{z}=z \text {, } \\ \text { (iv) } \bar{t}=\gamma\left(t-\frac{v}{c^{2}} x\right) \text {. } \end{matrix}                           (12.18)

\left. \begin{matrix} \text { (i′) } x=\gamma(\bar{x}+v \bar{t}), \\ \text { (ii′) } y=\bar{y} \text {, } \\ \text { (iii′) } z=\bar{z} \text {, } \\ \text { (iv′) } t=\gamma\left(\bar{t}+\frac{v}{c^{2}} \bar{x}\right). \end{matrix} \right\}                                     (12.19)

So the trajectory in S is given by:

x=0 ; y=\gamma(-R \sin \omega \bar{t}+v \bar{t})=\gamma\left\{-R \sin \left[\omega \gamma\left(t-\frac{v}{c^{2}} y\right)\right]+v \gamma\left(t-\frac{v}{c^{2}} y\right)\right\} , or

\underbrace{y\left(1+\gamma^{2} \frac{v^{2}}{c^{2}}\right)}_{\gamma^{2} y\left(1-\frac{v^{2}}{c^{2}}+\frac{v^{2}}{c^{2}}\right)=\gamma^{2} y}=\gamma^{2} v t-\gamma R \sin \left[\omega \gamma\left(t-\frac{v}{c^{2}} y\right)\right]\}(y-v t) \gamma=-R \sin \left[\omega \gamma\left(t-\frac{v}{c^{2}} y\right)\right];

z=R\left(1-\cos ^{2} \omega \bar{t}\right)=R\left[1-\cos \omega \gamma\left(t-\frac{v}{c^{2}} y\right)\right].

We can get rid of the trigonometric terms by the usual trick:

\left.\begin{array}{l}\gamma(y-v t)=-R \sin \left[\omega \gamma\left(t-\frac{v}{c^{2}} y\right)\right] \\z-R=-R \cos\left[\omega \gamma\left(t-\frac{v}{c^{2}} y\right)\right]\end{array}\right\} \Rightarrow \gamma^{2}(y-v t)^{2}+(z-R)^{2}=R^{2}.

Absent the \gamma^{2}, this would be the cycloid we found back in Ch. 5 (Eq. 5.9). The \gamma^{2} makes it, as it were, an elliptical cycloid — same picture as Fig. 5.7, but with the horizontal axis stretched out.

(y-R \omega t)^{2}+(z-R)^{2}=R^{2}                            (5.9)