My original question was far too long, so I am editing it down to this much smaller version:

The Lorentz force for an electron traveling through a uniform magnetic field with no electric field is $F=q*v*B$.

Is $v$ just the velocity of the electron, or is it the velocity of the electron relative to the magnetic field. To me it seems obvious that it can not be "just the velocity" because this could change for different inertial frames. That would mean different observers would see different accelerations on the electron. Seemingly leading to a contradiction that you could make the electron both hit a wall and not hit a wall depending on which inertial frame you are in.

Consider the following experiment:

We have a space ship with an electron hovering between 2 bar magnets and the whole system is drifting through space at a constant velocity. Will there be a force on the electron, or will there be no force because there is no difference in velocity between the magnets and the electron?


You're hitting on some of the considerations that led Einstein to his 1905 "Elektrodynamik bewegter Körper" paper.

In form of the Lorentz force law you state, the force is the electron's velocity relative to your present frame. The magnetic field $B$ is also as measured in your present frame. You are right to be worried that the whole lot might not be consistent between frames, and this concern is the reason we must think not of magnetic and electric fields alone, but the full Faraday tensor.

If you want to work things out from a relatively moving frame, then you need to write the Lorentz force in its Lorentz covariant form, that is:

$$F^\mu = q \,\mathscr{F}^\mu{}_\nu\,v^\nu$$

where $F$ is the four-force, $\mathscr{F}$ the Faraday tensor and $v$ the four velocity. $\mathscr{F}$ transforms as a rank 2 mixed tensor, mixing electric and magnetic fields, as one shifts between frames. $v$ naturally transforms as a four vector, as does the four-force $F$. These transformations act together to make the electron's behavior consistent to all observers.

So, in your spaceship, the electron stationary wrt to the ship's frame feels no force from the magnets which are stationary wrt to the same frame.

Now let's be an observer watching the ship coasting by. The electron is moving and it sees a magnetic field from the magnets. But now $\mathscr{F}$ transforms so that the effect of the magnetic field is exactly offset by an electric field component of the Faraday tensor. So we still conclude that the electron coasts by showing the same inertial motion as the spaceship.


The reason this interests me is that no mater how fast the magnet is spinning about its axis there is no difference in the strength or direction of the magnetic field at all points in space

Exactly. Your claim about the velocity of the electron being relative to the velocity of the magnet is correct. However, a spinning magnet is just that, a spinning magnet. There is symmetry and no part of the magnetic field changes, therefore no matter how fast you spin the magnet, electron would react all the same. Take a look at the magnetic dipole for instance. There is no angular velocity of the dipole rotation in the equations, because it is symmetric.

However, if you take a magnet that is deformed, this will actually cause the strength of the magnet to be different at various points in space, simply because the magnetic field lines would be time dependant. Hypothetically, you can say that a domain orientations in a bar magnet are not perfectly symmetric, and that a real bar magnet does not have such properties. In that case, yes, the electron would react differently, but with an unnoticeable amount, really.

  • $\begingroup$ I was afraid of an answer like this. I understand your reasoning, but then in my mind claim 2 must be false. I have added another experiment which I should have originally posted instead, I posted the spinning magnet because I felt it would be more easy to test in practice. Would you mind giving a response to my "question 3"? $\endgroup$ – Andrew Jul 13 '16 at 17:48
  • $\begingroup$ Why would be the claim 2 false? You always have to look at the velocities with regards of the frame you're looking from. For the 3rd question, just ask what would you see if you would be able to sit on the electron. From that perspective, you see only a uniform, time-invariant magnetic field. An observer from a stationary place would see that you and the magnets (infinitely long, for the record) are moving and compute the Lorentz equation accordingly. $\endgroup$ – bluecore Jul 13 '16 at 18:06
  • $\begingroup$ Edit: Reading it again I see where you're heading. But really, is there a difference between uniform magnetic field and two bar magnets moving the same speed as the electron, thus creating a uniform magnetic field? Such field would be produced only by sufficiently long magnets. Whether they stay in place or move is irrelevant, as long as the output is uniform, time invariant magnetic field. $\endgroup$ – bluecore Jul 13 '16 at 18:08
  • $\begingroup$ but that is the reason I think it contradicts claim 2. From the electrons point of view (or any observer traveling with the same velocity) the electron is stationary and there is no velocity and no Lorentz force. From our point of view the electron has a velocity so the Lorentz force is not zero. $\endgroup$ – Andrew Jul 13 '16 at 18:11
  • $\begingroup$ The magnets don't need to be long. Normal (short) bar magnets also produce a field between them . $\endgroup$ – Andrew Jul 13 '16 at 18:12

The field is just assumed to be fixed over a certain region of space. Then, $v$ indicates the velocity of the charged particle moving uniform w.r.t the inertial frame of magnet. Then the force is given by


This is valid for a point charge. Now, suppose a person $A$ is moving on a rail cart. He has a magnet in his hand. The cart is moving at a constant speed (The speed is assumed to be ordinary, i.e., $v<<c$). According to him, he see the magnet at rest w.r.t him(or the cart). But, when it comes to a person standing on the ground, he see the magnet is in motion. So, it is necessary to spot the reference frame you select in order to explain the magnetic force.

Now, consider a charged particle fitted on a cart. The cart is moving at a constant velocity $v$, with $v<<c$. A person stationary on the cart sees the charge at rest. So, he calculate the electric field due to it at some distance given by Coulomb's law. From that he could calculate the force it could exert on some other charge at this distance. But, for a person stationary w.r.t the ground, he sees the charge is moving with a velocity $v$. A moving charge creates a magnetic field. He see a current (obviously, for a fraction of a second) there and what he identifies is the magnetic force, not an electric one. Magnetism is just a relativistic phenomenon. Another point is that as per the symmetry of relativity, both see the same value of force eventhough each perceived it as a different manifestation of a single force.

What I meant is that your thinking is right. To state such an equation of magnetic force, the charge should be in constant velocity w.r.t the magnetic field.


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