1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz Transformations For Particle In Uniform Electromagnetic Field

  1. Jan 4, 2012 #1
    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 [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]


    2. Relevant 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]

    3. 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!
     
  2. jcsd
  3. Jan 6, 2012 #2
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook