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A moving charge paradox?

  1. Nov 19, 2008 #1
    Hi there, here I have this paradox and hope you can explain.
    There are 2 charged particles, one is moving at speed v, one is stationary. We know the moving charge gives off a magnetic field which can not exert any force onto the other charge because it does not move.
    Now assume we stay in a frame which moves at a speed of v/2 (the same direction with the moving charge). We can see the first charge moves at v/2 and the second also moves as v/2 (opposite direction) and so there should be some force exerting on both particles?

    Thanks for reading and forgive me if my english is not clear enough.
     
  2. jcsd
  3. Nov 19, 2008 #2
    Well, I'm not sure I understand your question, but it seems reasonable that any charged particle speeding past close enough to another will have some effect.
     
  4. Nov 19, 2008 #3
    I asked myself the same one time ago, and I think in this case you have to transform your reference systems in terms of S/R. It can't be a coincidence that the LORENTZ-Force is subject to this "paradox", after all, heh?

    Honestly, I've no clue.
     
  5. Nov 19, 2008 #4

    jtbell

    User Avatar

    Staff: Mentor

    To resolve this "paradox" you need to use relativity theory. When you switch between reference frames, electric and magnetic fields are "mixed up" by the Lorentz transformation. The electric and magnetic fields individually are different in the two frames, but their combined effect on a charged test particle is the same in both frames, after accounting for length contraction and time dilation where necessary.
     
  6. Nov 19, 2008 #5

    Dale

    Staff: Mentor

    In this frame, the magnetic field of the moving charge does not exert any force on the stationary charge, but the electric field does.
    In this frame there are forces due to both the magnetic field and the electric field. It turns out that, as jtbell mentioned, their sum is equal to the force in the first frame.
     
  7. Nov 20, 2008 #6
    Thanks jtbell and dalespam. The explanation is very concise.
     
  8. Nov 20, 2008 #7
    So then...if a charge was at rest in a uniform magnetic field, then there would be no magnetic force...but if the field was set into motion (while the charge was still), would there be a force on the charge?
     
  9. Nov 20, 2008 #8

    Dale

    Staff: Mentor

    You cannot determine the Lorentz force simply from the magnetic field, you need to specify the electric field also. When you do so and transform the electromagnetic fields to another frame you always obtain the same Lorentz force in each frame.
     
  10. Nov 20, 2008 #9
    So lets say that a charged metal ball was placed at rest on a super-cooled surface. The ball is between 2 walls that act as magnetic poles, thus, it is placed within a relatively uniform magnetic field running from one wall to the other. The electric force on the ball is negligible where it is. If the walls started moving, then what would be the case with the ball?
     
  11. Nov 20, 2008 #10
    Does it (the charged ball) curve in trajectory?
     
  12. Nov 20, 2008 #11
    the ball starts off at rest...the walls then start moving, making it so that the magnetic field moves...what would happen to the ball in this case?
     
  13. Nov 20, 2008 #12

    Dale

    Staff: Mentor

    In terms of the Electromagnetic Tensor:
    [tex]F_{\alpha \beta} = \left( \begin{matrix}
    0 & \frac{-E_x}{c} & \frac{-E_y}{c} & \frac{-E_z}{c} \\
    \frac{E_x}{c} & 0 & B_z & -B_y \\
    \frac{E_y}{c} & -B_z & 0 & B_x \\
    \frac{E_z}{c} & B_y & -B_x & 0
    \end{matrix} \right)[/tex]

    We have the Lorentz Force:
    [tex]f_{\alpha} = \frac{d p_{\alpha}}{d \tau} \, = q \, F_{\alpha \beta} \, u^\beta [/tex]

    So suppose in the wall's frame (ball moving with velocity -v in the x direction) we have a uniform magnetic field of strength B in the z direction with no electric field then:
    [tex]f_{\alpha} = q \, \left( \begin{matrix}
    0 & 0 & 0 & 0 \\
    0 & 0 & B & 0 \\
    0 & -B & 0 & 0 \\
    0 & 0 & 0 & 0
    \end{matrix} \right) \, \left( \begin{matrix}
    c \gamma \\
    -v \gamma \\
    0\\
    0
    \end{matrix} \right) = \left( \begin{matrix}
    0 \\
    0\\
    B q v \gamma \\
    0
    \end{matrix} \right)[/tex]

    Boosting that to the ball's frame (the wall is moving with velocity v in the x direction) we have:
    [tex]f'_{\alpha} = q \, \left( \begin{matrix}
    0 & 0 & -\frac{B v \gamma }{c} & 0 \\
    0 & 0 & B \gamma & 0 \\
    \frac{B v \gamma }{c} & -B \gamma & 0 & 0 \\
    0 & 0 & 0 & 0
    \end{matrix} \right) \, \left( \begin{matrix}
    c \\
    0 \\
    0\\
    0
    \end{matrix} \right) = \left( \begin{matrix}
    0 \\
    0\\
    B q v \gamma \\
    0
    \end{matrix} \right)[/tex]

    Again the electromagnetic fields Lorentz transform such that the Lorentz force on any test charge is the same in all frames. The charged ball experiences a force wether it is the wall or the ball that is moving. In the frame where the wall is at rest we attribute it to the magnetic field and in the frame where the ball is at rest we attribute it to the electric field.
     
    Last edited: Nov 20, 2008
  14. Nov 20, 2008 #13
    I see...thanks
     
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