I'm confused about the Lorentz force in relation to the frame of reference. There are many questions about that here, but I still don't get it. I've tried to break it down to the most basic example.

Lets say we have a single charge moving into the screen with velocity $v$ between two infinitely long magnet bars. The Lorentz force law tells us that the charge will experience a force to the side.

diagram of moving charge

So far so good. But now I change the frame of reference into one where the magnets move at speed $v$ out of the screen and the charged particle remains stationary. The relative motion is the same, so I expect the same force.

Since the charged particle isn't moving, there is no Lorentz force.

Answers to the other questions explain it with the electric field. Where does the electric field come from in the second frame of reference?

  • The magnets have no net charge, so Gauss' Law can't be it, right?

  • The two magnets are now moving along the axis on which they have infinite length, so the motion is not really changing anything observable. They might as well be static. In particular, the magnetic field does not change, so Faraday's law of induction shouldn't create an electric field either.

  • 2
    $\begingroup$ "The two magnets are now moving along the axis on which they have infinite length, so the motion is not really changing anything observable. They might as well be static. " That's untrue. $\endgroup$ Commented Jul 12, 2016 at 14:05
  • $\begingroup$ You are cutting across magnetic lines of force, which are observable: upload.wikimedia.org/wikipedia/commons/5/57/Magnet0873.png. The only reason why you aren't observing them is because you have decided not to put iron shavings into your setup. $\endgroup$
    – CuriousOne
    Commented Jul 12, 2016 at 20:36
  • $\begingroup$ I try to avoid iron shavings. They are a pain to clean up afterwards. Without performing the lorentz transformation, the magnetic field vector at the position of the charge is always downwards and does not change in magnitude either. The magnetic field would indeed be static from the reference frame of the charge. Now, when doing the transformation as one has to, I have no clue how the result looks like. But there's going to be an electric field somehow. Still hoping for a hint on how that might look like. $\endgroup$
    – Stefan
    Commented Jul 12, 2016 at 22:22
  • $\begingroup$ @CuriousOne Magnetic field lines are not observable. They’re a helpful diagrammatic tool to show what a vector field looks like by tracing some selected paths through them. The only reason iron shavings show field lines is because they naturally clump together. $\endgroup$ Commented Dec 1, 2018 at 16:14

2 Answers 2


You can't get away from special relativity, which is what "unifies" electric and magnetic phenomena. The electromagnetic field really is a single field (actually a tensor, but don't worry about that), and its components mix together in the Lorentz transform in a similar way to how the x,y,z components of a usual vector field mix together in a rotation.

There are two ways to find the electric field here: You can perform a kinematic Lorentz transform to obtain the source charges/currents in the moving frame. You can also obtain the electric field component directly from the Lorentz transform of the electromagnetic field.

Lorentz Transform on Sources

An object in motion appears length contracted to observers: $L' = L_0/\gamma$, where $L_0$ is the length of the object its rest frame and $\gamma$ is the Lorentz factor associated with the velocity of the object:

$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$

Consequently, if the object is charged, the charge density increases when in motion:

$$\rho' = \gamma \rho_0$$

The magnetic field in the rest frame arises from the magnetization $\vec M$ (magnetic dipole density) of the magnets, which is pointed up along the N-S axis. Each magnetic dipole is effectively a small current loop with current $\vec J = \rho \vec u$, where $\vec u$ is the tangential velocity around the N-S axis.

enter image description here

Then in the moving frame, the speed $u$ is not uniform around the loop, and charge densities at the points of the loop flowing with and against the direction of motion are different: $$u_\pm = \frac{u \pm v}{1 \pm uv/c^2} $$

$$\rho_\pm = \gamma_\pm \rho_0 = \frac{\rho_0}{\sqrt{1 - u_\pm^2/c^2}}$$

Therefore $u_+ > u_-$, which means $\rho_+ > \rho_-$, so each current loop (magnetic dipole) acquires an electric dipole moment. Each magnetic dipole pointed up, so each electric dipole is pointed to the right. This is the source of the electric field in the moving frame.

Lorentz Transform of E and B fields

If you begin with fields $\vec E$, $\vec B$, and want the fields in an inertial frame moving at relative speed $\vec v$, the new fields are:

\begin{align} \vec E_{\parallel}' &= \vec E_{\parallel} \\ \vec B_{\parallel}' &= \vec B_{\parallel} \\ \vec E_\perp' &= \gamma(\vec E_\perp + \vec v \times \vec B) \\ \vec B_\perp' &= \gamma(\vec B_\perp - \frac{1}{c^2} \vec v \times \vec E) \end{align}

In our case $\vec E = 0$ and $\vec B_\parallel = 0$, so the above simplifies to \begin{align} \vec E' &= \gamma \vec v \times \vec B \\ \vec B' &= \gamma \vec B \end{align}

If the $\vec B$ field is up (from S to N) and $\vec v$ is into the page, this gives an $\vec E$ field pointed to the right.

  • $\begingroup$ This answer isn’t satisfying for me. You still didn’t explain the source of the E field after the transformation. Let’s instead say you started in the second frame of reference where the charge was stationary. In that frame, the charge wouldn’t move and when transforming the fields to the first frame you’d find the charge still doesn’t move. This is inconsistent. $\endgroup$ Commented Dec 1, 2018 at 16:06
  • $\begingroup$ Surely Maxwell’s equations must be satisfied in BOTH frames of reference. However, how are they satisfied in the second frame of reference where you say there is an electric field? According to Maxwell’s equations, an electric field can only exist if there is a charge density or changing B to produce it. There is no changing B or charge density so it seems that Maxwell’s equations predict an E field that is zero everywhere. $\endgroup$ Commented Dec 1, 2018 at 16:09
  • $\begingroup$ Figured it out: Maxwell’s equations don’t guarantee that E is zero. They only guarantee that curl E and div E are zero. This doesn’t uniquely define E without boundary conditions. E could be a uniform field for example. So point is: E can exist without any sources. $\endgroup$ Commented Dec 2, 2018 at 2:18
  • $\begingroup$ @hddh You've got it. The infinite extent of the bars adds some confusion because there is no source charge, or equivalently it is "at infinity". If the bars were finite then they would have a surface charge on either end in the moving frame, sourcing the dipole electric field. However Maxwell's equations are just fine with having an E field without sources in the infinite case - it just has a very restricted form. $\endgroup$
    – adipy
    Commented Dec 7, 2018 at 6:19

If you think of a magnet as being made up of individual dipoles all aligned with one another (this is what happens in Iron or Neodymium bar magnets), then you can ask "how does each dipole transform under Lorentz transformation?" Under Lorentz transformation magnetic dipoles become electric dipoles. The electric field from the electric dipoles causes the Lorentz force in this frame.

  • $\begingroup$ With electromagnetism, whenever I ask "why?" once too often, I get hit by special relativity and quantum :-) I expect you are right, but 16 pages of equations is a bit too much for me to digest right now. Maybe you could answer your own formulation of my question. How would the electric field generated by the electric dipoles look like? $\endgroup$
    – Stefan
    Commented Jul 12, 2016 at 16:34
  • $\begingroup$ If the two reference frames move with v to each other, the electric fields differ by a v x B term (assuming v << c). If we can calculate the magnetic field in one frame, the effect on the electric field in the other is easily calculated. The magnetic field itself is not significantly effected. $\endgroup$
    – Stefan
    Commented Jul 18, 2016 at 15:49

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