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When we observe two point charges moving parallel to each other we can see two forces acting on each of the charges:

  • the Coulomb force
  • the magnetic force ($\mathbf{F}=q\mathbf{v}×\mathbf{B}$) (similar to the force between two parallel wires with the same current)

However, if we change to the charges' frame of reference, there will only be one force - the Coulomb one, and moreover, the amplitude of this force should be the same as the previous one's (the distance between the charges is constant and the force is independent of velocity). The resulting force should be independent of the frame of reference, yet it isn't and I can't find the missing piece. I know that a magnetic field is just an electric field viewed from a different frame of reference, but it doesn't help in this specific case.

I found this question to be similar: Two electron beams exert different forces on each other depending on frame of reference?, but in one charge scenario the charge density has no meaning.

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neverneve
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    The Lorentz Force, as I've seen it defined, takes into account both Electric and Magnetic contributions $\mathbf{F}_{\text{Lorentz}} = q(\mathbf{E}+\mathbf{v} \times \mathbf{B})$. I know there are other definitions, so clarify which you're using. See http://en.wikipedia.org/wiki/Lorentz_force – eepperly16 Jun 08 '15 at 18:07
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    The missing piece is that you have to correctly transform the actual field of a moving charge, which is not the same as the field of a stationary current. – CuriousOne Jun 08 '15 at 18:14
  • @eepperly16 Ah, that's right. What I meant by Lorentz force was only the magnetic contribution, $\mathbf{F}=q\mathbf{v}×\mathbf{B}$. I'll correct the question. – neverneve Jun 08 '15 at 18:15
  • Not a full answer, but see this related video on relativity and magnetism https://www.youtube.com/watch?v=1TKSfAkWWN0 – eepperly16 Jun 08 '15 at 18:19
  • That's exactly the video from which I learned about electric and magnetic fields being the same thing viewed from different frames. I understand the situation in the video, but there's no way to explain my scenario with length contraction (maybe time dilation then?). – neverneve Jun 08 '15 at 18:23
  • Here was a question about violation about Netwons 3d law and the answer see http://physics.stackexchange.com/questions/114466/apparent-violation-of-newtons-3rd-law-and-the-conservation-of-angular-momentum/185629#185629. It is related to your question. – HolgerFiedler Jun 08 '15 at 20:52
  • Some explanation why we get a magnetic field see here – HolgerFiedler Jun 08 '15 at 20:57
  • Just to be clear, the magnitude of the force in the lab frame should not be equal to the magnitude of the force in the charges' frame; $\vec{F} = d\vec{p}/dt$, and while $d\vec{p}$ will be the same in this case (since it's perpendicular to the relative motion of the frames), $dt$ will not be due to relativistic effects. – Michael Seifert Jun 25 '15 at 18:26

2 Answers2

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I'm only beginning to study electromagnetism, so please jump on any mistakes I may have made.

Coulomb's law only holds for electrostatic situations, so applying Coulomb's law in the observer's frame of reference is invalid, since the charges are moving from the observer's perspective. The electric and magnetic field of dynamic systems is instead given by the four Maxwell Equations. Even for a simple scenario for this one, the math can get very complicated very quickly, but I'll try to explain the relevant physics.

As the particle moves through space, the electric and magnetic fields in space change as a function of time. Maxwell's equations tell us that changing magnetic fields contribute to the electric field and changing electric fields contribute to the magnetic field. This link between the two fields makes sense from the sense of special relativity since the electric and magnetic fields are relativistically the same field.

After doing all the math (which involves a couple of second order partial differential equations), you will be able to calculate the electric and magnetic forces on each electron and sum them to find that the force is indeed equivalent to the simple Coulombic force from the electron's frame of reference.

eepperly16
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If the charges are moving parallel to each other, in the reference frame of of the one charge, the charges aren't moving. So the magnetic field is zero and the force is $\bar F=q \bar E $. For the charge to be still, the force must be zero, and thus the electric field also. But if the electric field can't be zero, then there is another force that holds the charges in their positions. In the general reference frame we have that(for the one charge): $F=0 --> \bar E = -\bar u \times \bar B $.

Hope this helps.

Constantine Black
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