Lorentz Transformations For Particle In Uniform Electromagnetic Field

[yusohard]

Homework Statement

A charge q is released from rest at the origin, in the presence of a uniform electric
field and a uniform magnetic field [itex] \underline{E} = E_0 \hat{z} [/itex] and [itex] \underline{B} = B_0 \hat{x} [/itex] 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 [itex] R=m\gamma^2 v / (q B_0) [/itex] What are the equations in [itex] x', y', z, t' [/itex] 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:
[itex] \gamma^2 (y-vt)^2 + (z-R)^2 = R^2 [/itex]

Homework Equations

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

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

Lorentz Force:

[itex]\underline{F} = m \gamma \frac{d v}{d x} = m \gamma \frac{v^2}{R} = q \underline{B} \times \underline{v} [/itex]

The Attempt at a Solution

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

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

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

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

Using the Lorentz force equation above:

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

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

I think it should be:

[itex] (y'^2 + z'^2) = R^2 [/itex]

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

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

[itex] \gamma^2 (y - vt)^2 + z^2 = R^2 [/itex]

Missing the [itex] (z-R)^2 [/itex] term.

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

Any help appreciated. Been driving me crazy.

Thanks in advance!

 

[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 [itex] x', y', z' ,t' [/itex] are:

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

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

[itex] t'=\gamma (t - \frac{v}{c^2} y) [/itex]
[itex] y=\gamma (y' + v t') [/itex]

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

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

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

[itex] \gamma^2(y-vt)^2 + (z-R)^2 = R^2 [/itex]

As required. I'll sleep easy tonight eh.