# Motion of a particle in an EM field

1. Jul 7, 2015

### fluidistic

1. The problem statement, all variables and given/known data
There's a uniform EM field given by $\vec E=E\hat y$, $\vec B=B\hat z$
with respect to an inertial reference frame K.
A charged particle with rest mass m and charge q>0 moves in the field with an initial velocity orthogonal to $\vec B$.
1)Write down the equation of motion of the particle.
2)Show that if $E<cB$ then there exist an inertial frame of reference K' that moves with constant velocity with respect to K such that the electric field vanishes.
3)Solve the equation of motion in K' and show that the trajectory is circular in a plane orthogonal to $\vec B$, with a constant frequency of oscillation.
2. Relevant equations
Lorentz force.

3. The attempt at a solution
1)$\frac{d\vec p}{dt}=q(E\hat y + \vec v \times B \hat k)$. I can choose $\vec v(t=0)$ to be worth $v_0 \hat x$ if I have to solve that equation of motion. I am not sure whether my answer is ok or if they want it more simplified.
2)Not sure how to tackle this one. Should I solve the equation in 1) (which I'm not even sure how to do due to the term involving the velocity) and realize that if I move with constant velocity with respect to K then $\vec E$ vanishes? Or should I start with E<cB and simplify some things out?

Thank you.

Last edited: Jul 7, 2015
2. Jul 8, 2015

### Orodruin

Staff Emeritus
You need to look at the transformation properties of the electromagnetic field under Lorentz transformations. Once you do that, things should become clear.

3. Jul 8, 2015

### fluidistic

Ok thanks Orodruin.
So I've solved part 2), I get that $\vec v=-\frac{E}{B} \hat x$.
I'm unable to solve part 3) though.
My attempt is: $\frac{d\vec p}{dt}=q(\vec E' + \vec v \times \vec B')=q \vec v \times \vec B'$. Using Lorentz transformation for fields I got that $\vec B'=\gamma \left ( B- \frac{vE}{c^2} \right ) \hat z$.
And so, skipping the algebra steps, $\frac{d\vec p}{dt}=\frac{qE}{\gamma}\hat y$. I don't recognize any circular motion there.

4. Jul 8, 2015

### Orodruin

Staff Emeritus
You have inserted the velocity of the frame K' as the velocity of the motion you want to solve for. This is not the case. You need to consider a general velocity in K', where there is no electric field.

5. Jul 8, 2015

### fluidistic

I see. Before I replaced v, I had reached that $\frac{d\vec p}{dt}=qv\gamma \left ( B- \frac{vE}{c^2} \right ) \hat y$. And I didn't and still don't see any circular motion.

6. Jul 8, 2015

### Orodruin

Staff Emeritus
No, you need to be much more careful. You are still mixing the relative velocity between the frames, which appears in the transformation, and the velocity of the particle, which appears in the force law. These are not the same velocities!

7. Jul 8, 2015

### Orodruin

Staff Emeritus
No, you need to be much more careful. You are still mixing the relative velocity between the frames, which appears in the transformation, and the velocity of the particle, which appears in the force law. These are not the same velocities!

8. Jul 8, 2015

### fluidistic

Ok I think I see my mistake. So $\vec B' = \gamma \left ( B-\frac{vE}{c^2} \right ) \hat z$ where v=E/B, so $\vec B'$ is a constant vector whose magnitude I'll call B'.
Now $\frac{d\vec p}{dt}=qB'(\vec u \times \hat z)$ where $\vec u$ is the velocity of the particle which is orthogonal to $\vec B$ (and $\vec B'$) initially.
I don't see how to deal with the cross product nor how to continue from there.

9. Jul 8, 2015

### Orodruin

Staff Emeritus
Try writing it down in component form. You should get two coupled differential equations.

10. Jul 9, 2015

### fluidistic

Ok thanks a lot, I think I solved the problem albeit I'm not sure whether I picked a quick way to do so.
The coupled system of the 2 ODE's I got is $m \dot u_x=qu_yB'$, $m\dot u_y =-qu_xB'$. I transformed it into a single 2nd order DE: $\ddot u_x + \underbrace{\frac{q^2B'^2}{m^2}}_{k^2}u_x=0$ with initial conditions $u_x(0)=u_{x_0}$ and $\dot u_x(0)=\dot u_{x_0}$.
I first solved the DE and expressed the solution in terms of complex exponentials but I didn't the insight of the path of the motion so I went back and expressed it in terms of a cosine, namely $u_x=A\cos (kt+\phi )$. This yielded $u_y=-\frac{mkA}{qB'} \sin (kt+\phi )$.
I then integrated both expressions with respect to time to get x(t) and y(t), that are a sine and a cosine (plus a constant which is the position at time t=0) with the same amplitude and frequency, therefore the path is a circle in a plane orthogonal to $\vec B$ and $\vec B'$.
The frequency is qB'/m. While the radius of the circle is "mA/(qB')" in my case, that I must determine by imposing the initial conditions. If I didn't make any mistake, if I choose phi to be 0, then $A=u_{x_0}$, the speed of the particle in the x direction at time t=0.