Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question

closed as off-topic by tpg2114, centralcharge, 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:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, centralcharge, Emilio Pisanty, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.