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I have the following problem to solve:

A particle of mass $m$ and charge $e$ moves in the laboratory in crossed, static, uniform, electric and magnetic fields. $\mathbf{E}$ is parallel to the $x$-axis; $\mathbf{B}$ is parallel to the $y$-axis. Find the EOM for $|\mathbf{E}|<|\mathbf{B}|$ and $|\mathbf{B}|<|\mathbf{E}|$.

I was planning on using the following: $$\vec{E}'=\gamma(\vec{E}+\vec{\beta}\times\vec{B})-\frac{\gamma^2}{\gamma +1}\vec{\beta}(\vec{\beta}\cdot \vec{E})$$ $$\vec{B}'=\gamma(\vec{B}-\vec{\beta}\times\vec{E})-\frac{\gamma^2}{\gamma +1}\vec{\beta}(\vec{\beta}\cdot \vec{B})$$ along with $$t'=\gamma (t-\vec{\beta}\cdot\vec{x})$$ $$ \vec{x}'=\vec{x}+\frac{(\gamma -1)}{\beta^2}(\vec{\beta}\cdot \vec{x})\vec{\beta}-\gamma \vec{\beta}t$$ with the addition constraints $$\frac{d\vec{p}}{dt}=e\left[ \vec{E}+\frac{\vec{u}}{c}\times \vec{B}\right]$$ and $$\frac{dU}{dt}=e\vec{u}\cdot \vec{E}$$ To solve this I am going to switch to a frame with $$\vec{\beta}=\frac{E}{B}\hat{z}$$for the first case. With this case $dU/dt=0$ and i can solve the equations of motion to find $\vec{x}(t)$ directly, and then boost back to get the trajectories in the original frame. However for the the second case I was wondering if my procedure is correct. I am going to switch to a frame with $\vec{\beta}=(B/E) \hat{z}$ to remove the magnetic field. Now here it seems that $\vec{u}_0$ is perdepdicular to $\vec{E}$ to start in the new frame, but that it will be accelerated in the $x$ direction and hence $dU/dt\neq 0$ and I can't just straightforwardly solve the EOMs. How would I proceed from here?


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closed as off-topic by tpg2114, Dimensio1n0, Emilio Pisanty, Qmechanic Nov 10 '13 at 16:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

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You are asked to give the equations of motion. Do you also need to solve them? Are you to assume the motion is nonrelativistic? –  Ron Maimon Apr 30 '12 at 5:02
Thanks for your response Ron. After sleeping I gained more stamina and took another shot at problem solving and came out on top. I solved $du^{\alpha}/d\tau = F^{\alpha \beta}u_{\beta}$ directly for components of the velocity perpendicular and parallel to $\vec{E}$. I got the equations of motion then. –  kηives May 1 '12 at 2:08
Then you can post your solution as an answer, and the question will be done with. –  Ron Maimon May 1 '12 at 3:41
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1 Answer 1

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After switching to the frame mentioned above, we are left with only a static electric field, perpendicular to the initial velocity of the particle. Now we consider $$\frac{d u^\alpha}{d\tau}=\frac{e}{mc}F^{\alpha\beta}u_\beta$$ This decomposes as $$\frac{du^0}{d\tau}=\frac{e}{mc}F^{0\beta}u_\beta=\frac{e}{mc}F^{0i}u_i=\frac{e\gamma}{mc}\vec{E}\cdot\vec{v}$$ and $$\frac{du^i}{d\tau}=\frac{e}{mc}F^{i\beta}u_\beta=\frac{e}{mc}(F^{i0}u_0 -F^{ij}u_j)=\frac{e}{mc}(\gamma c)\vec{E}$$ since $\vec{B}$ is zero in this frame. Now I write all the velocities as parallel and perpendicular to the electric field and define $\omega_E=\frac{eE}{mc}$ $$\frac{d(\gamma c)}{d\tau}=\omega_E (\gamma v_{||})$$ $$\frac{d(\gamma v_{||})}{d\tau}=\omega_E (\gamma c)$$ $$\frac{d(\gamma v_{\perp})}{d\tau}=0$$ Then differentiating the second equation by $d/d\tau$ we get $$\frac{d^2(\gamma v_{||})}{d\tau^2}=\omega_E \frac{d(\gamma c)}{d\tau}=\omega_{E}^2(\gamma v_{||})$$ solutions to this are $(\gamma v_{||})=A\sinh(\omega_E \tau)+B\cosh(\omega_E \tau)$ which implies $(\gamma c)=A\cosh(\omega_E \tau)+B\sinh(\omega_E \tau)$ and $\gamma v_{\perp}=\text{const.}$. Now at $\tau=0$ we know that $v_{\perp}=v_0$ and $v_{||}=0$. Let $$\gamma_0=\frac{1}{\sqrt{1-\frac{v_{0}^{2}}{c^2}}}$$ then the initial conditions demand that $B=0$, $A=\gamma_0 c$, and $\text{const.}=\gamma_0 v_0$. Then we have $$(\gamma c)=\gamma_0 c \cosh(\omega_E \tau)\implies \gamma=\gamma_0 \cosh(\omega_E \tau)$$ $$(\gamma v_{||})=\gamma_0 \cosh(\omega_E \tau)v_{||}=(\gamma_0 c)\sinh(\omega_E \tau)\implies v_{||}=c\tanh(\omega_E \tau)$$ $$\gamma v_{\perp}=\gamma_0 \cosh(\omega_E \tau)v_{\perp}=\gamma_0 v_0\implies v_{\perp}=\frac{v_0}{\cosh(\omega_E \tau)}$$ Now $dt/d\tau=\gamma$ so that $$t=\int_{0}^{\tau}\gamma d\tau'=\int_{0}^{\tau}\gamma_0 \cosh(\omega_E \tau')d\tau'$$ then $$\frac{t \omega_E}{\gamma_0}=\sinh(\omega_E \tau)$$ The important one though is $$\cosh(x)=\sqrt{1+\sinh^2(x)}\implies \cosh(\omega_E \tau)=\sqrt{1+\frac{t^2 \omega_{E}^{2}}{\gamma_{0}^{2}}}$$ after plugging these into the above equations for $v_{\perp}$ and $v_{||}$ we can solve for $x_{||}$ and $x_{\perp}$ from $$dx=v\, dt\implies x_{||}=\int_{0}^{t}v_{||}(t')dt'\quad \text{and}\quad x_{\perp}=\int_{0}^{t}v_{\perp}(t')dt'$$ These are what I was looking for.

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