Lorentz Transformations For Particle In Uniform Electromagnetic Field

1. Jan 4, 2012

yusohard

1. The problem statement, all variables and given/known data

A charge q is released from rest at the origin, in the presence of a uniform electric
field and a uniform magnetic field $\underline{E} = E_0 \hat{z}$ and $\underline{B} = B_0 \hat{x}$ in frame S.
In another frame S', moving with velocity along the y-axis with respect to S, the electric field is zero.
What must be the velocity v and the magnetic field in the frame S' ?

Show that the particle moves in S' in a circle of radius $R=m\gamma^2 v / (q B_0)$ What are the equations in $x', y', z, t'$ which describe the trajectory of the particle in the moving frame S' ?

By transforming from frame S’ show that the path of the particle in the original frame S is:
$\gamma^2 (y-vt)^2 + (z-R)^2 = R^2$

2. Relevant equations

Transformations of electric and magnetic fields for boosts in y-direction:

$E'_x = \gamma (E_x + \beta c B_z)$
$E'_y = E_y$
$E'_z = \gamma (E_z - \beta c B_x)$
$B'_x = \gamma (B_x - (\beta / c) E_z )$
$B'_y = B_y$
$B'_z = \gamma (B_z + (\beta / c) E_x )$

Lorentz Force:

$\underline{F} = m \gamma \frac{d v}{d x} = m \gamma \frac{v^2}{R} = q \underline{B} \times \underline{v}$

3. The attempt at a solution

Only the E field in the z-axis exists and, as stated in the problem, is zero:

$E'_z = \gamma (E_0 - \beta c B_0) = 0 \rightarrow v=E_0/B_0$

And similarly only the B field in the x-axis has a solution, and using the equation for v above:

$B'_x = \gamma (B_0 - (\beta / c) E_0 ) = \gamma B_0 (1 - E^2_0/c^2 B^2_0 ) = B_0 / \gamma$

Using the Lorentz force equation above:

$m \gamma \frac{v^2}{R} = q B_0/\gamma \rightarrow R=m\gamma^2 v / (q B_0)$

Now i doubt how to write equations in $x', y', z, t'$ which describe the trajectory of the particle in the moving frame S'.

I think it should be:

$(y'^2 + z'^2) = R^2$

As it is in the yz-plane(right?)

If i transform back with $y' = \gamma (y - vt)$ and $z'=z$ I just get:

$\gamma^2 (y - vt)^2 + z^2 = R^2$

Missing the $(z-R)^2$ term.

Can anyone see where i've gone wrong or what i've missing?

Any help appreciated. Been driving me crazy.

2. Jan 6, 2012

yusohard

Solved it.

For those interested:

It is indeed going in a circle in the zy-plane in S'. Since it starts at the origin in S' and goes clockwise, the equations of motion in terms of $x', y', z' ,t'$ are:

$x'=0$
$y'=-R Sin(\omega t')$
$z'=R(1-Cos(\omega t'))$
Where $\omega = v/R$

Transform via usual lorentz transforms for boosts in y-axis:

$t'=\gamma (t - \frac{v}{c^2} y)$
$y=\gamma (y' + v t')$

Subbing in for the primed parameters in and following through etc,etc,etc,etc,etc,etc:

$y= vt - \frac{R}{\gamma} Sin(\omega \gamma (t - \frac{v}{c^2}y))$
$z= R(1- \gamma R Cos(\omega \gamma (t - \frac{v}{c^2}y))$

Using your favourite trig formula (no prizes for that one), we can see indeed:

$\gamma^2(y-vt)^2 + (z-R)^2 = R^2$

As required. I'll sleep easy tonight eh.