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I Lorentz force on a charge due to a moving magnet

  1. Feb 4, 2017 #1
    Let's say I have a charge q which is viewed from its rest frame. So it's velocity v is 0. So the so-called magnetic component of its Lorentz force, which is q v x B, is 0. But I can have a magnet moving in this frame of reference.

    Let's say the velocity of this magnet according to the charge's rest frame is u. The magnet possesses a magnetic vector potential at each point in space, including the point where charge q is located. So as the magnet moves through space, the magnetic vector potential on the charge changes with time.

    So even in the rest frame of the charge, there should be Lorentz force on it, but in this particular frame of reference, it is due to the electric field E produced by the changing magnetic vector potential. This electric field equals -∂A/∂t. But there should be another way to express this. I could express this in terms of u.

    Considering only the translational motion of the magnet (i.e. ignoring rotations), could we represent the induced electric field as (u) A? But this is a gauge-dependent field, and the electric field should not be gauge-dependent. I notice that (u) A = A(uA) - u x ∇ x A (per Feynman subscript notation), and I cannot help but wonder if the term A(uA) was somehow neglected in the calculation. Suppose I say that the electric scalar potential was affected by the motion of the magnet, and let's hypothesize a correction to the electric scalar potential's contribution to the electric field of -A(uA). As a result, the effective scalar potential would be altered by the quantity uA. The result would be that the induced electric field instead becomes - u x ∇ x A.

    In the rest frame of the magnet, the charge would have a velocity equal and opposite to u. Let's call this velocity v which is a different v than above, mind you. So v = -u, where v is the velocity of the charge in the rest frame of the magnet, and u is the velocity of the magnet in the rest frame of the charge. So our force on the charge becomes q v x B, (where B = ∇ x A) which is exactly what the Lorentz force should give us.

    I have two questions:

    • Is the electric scalar potential at charge q in its rest frame affected by the motion of a magnet, providing an electric field of -A(uA) in addition to (u) A, yielding an electric field of - u x ∇ x A = - u x B = v x B?
    • If not, then would its failure to do so indicate that the Lorentz force on a charge moving in the presence of a stationary magnet is missing an additional term, which would be -A(vA)?

    Sincerely,

    Kevin M.
     
    Last edited: Feb 4, 2017
  2. jcsd
  3. Feb 4, 2017 #2
    Regarding the "moving magnet and conductor problem" which is highlighted in the following article:

    https://en.wikipedia.org/wiki/Moving_magnet_and_conductor_problem#Conductor_frame

    It speaks of a solution solving for E' in the conductor frame that results from the magnet moving in this frame.

    The problem with solution is that it is not unique. E' could differ from v x B by the gradient of a scalar field and still be a solution for the Maxwell-Faraday equation. This was not addressed by the Wikipedia article.
     
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