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Magnetic field seen by electron in thomas precession

  1. Nov 14, 2012 #1
    According to Larmor's theorem, the magnetic field seen by a rotating frame is changed. In a frame that rotates at the Larmor frequency, the magnetic field appears nonexistent.

    In the treatment of thomas precession, the magnetic field in the instantaneous resting frame (which is rotating to an observer in the laboratory frame) is as if the frame is not rotating. Why?
  2. jcsd
  3. Nov 14, 2012 #2


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    This is a misunderstanding. The E and B fields change under a velocity boost, but are unaffected by a rotation. A rotating frame causes a Coriolis force, and at the Larmor frequency the Coriolis force is just sufficient to balance the (unchanged) B field.
  4. Nov 15, 2012 #3

    Thank you! I understand that the actual B field is not changed during Larmor precession.

    However, the equaiton of motion in the rotating frame should read

    dL(angular momentum)/dt = gamma*L x (B-B') while B' is the fictitious magnetic field.

    In Thomas precession, one writes

    dL/dt (lab frame) = dL/dt (rotating frame) + omega X L


    dL/dt (rotating frame) = gamma*L x (B - v X E/c)

    The fictitious B field is missing in (B-vXE/c). Still don't understand why.
  5. Nov 15, 2012 #4


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    rays, There's nothing mysterious going on. In both cases, (dL/dt)rot = (dL/dt)nonrot - ω x L, where ω is the Larmor precession frequency in one case and the Thomas precession frequency in the other. The only reason they look different is because in the Larmor case they rewrite ω as a fictitious B field, and in the Thomas case they leave it as ω.
  6. Nov 15, 2012 #5
    Thanks again.

    In the rotating frame of thomas precession, the B field from the non rotating frame in its entirety is used to describe the motion in the rotating frame.

    This seems in contradiction to the treatment of other scenarios. For example, in nuclear magnetic resonance, when one decribes the motion of nuclear spins in the Larmor rotating frame the static B field does not enter into the equation of motion.

    Souldn't one use a B field in the rotating frame of the thomas precession that is different from the actual B field (in non rotating frame)? If one does that, then

    dL/dt(rot) = magnetic moment X (B-a*omeag)

    dL/dt(nonrot) = dL/dt(rot) + omeag X L = magnetic moment X B

    The correction term from a*omega disapears from the energy calculation which is performed in the nonrotating frame.

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