In "The Feynman Lectures on Physics" book, Vol 2, Chapter 13, there is a section titled "The relativity of the magnetic and electric fields". There Dr. Feynman describes magnetic force between the parallel conducting wires as a relativistic effect.
As the electrons move in one wire, relative to the positively charged nuclei, there is a length contraction effect "observed" by the electrons moving in the other wire. So the linear density of the positive nuclei of the first wire appear to increase, as for the electrons of the other wire. Therefore, the first wire is not electrically neutral anymore, and it attracts the electrons of the second wire by merely Coulomb force. Therefore the magnetic force doesn't really exist. https://www.feynmanlectures.caltech.edu/

I am having trouble understanding this effect when a free charge is moving perpendicular to a conducting wire. According to classical electromagnetism a conducting wire creates a magnetic field around it. If a free charge moves towards the wire - the Lorentz force must act on it, which is perpendicular to the motion of the charge and the magnetic field. The result is a force parallel to the wire that acts on the free charge.
How can you explain this situation considering magnetism as merely relativistic effect?

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    $\begingroup$ Where does Feynman say that the magnetic force doesn’t exist? (Perhaps it doesn’t exist in a particular Lorentz frame, but that’s not the same as not existng at all.) Or that magnetism is “merely a relativistic effect”? If you have a magnetic-only field in one Lorentz frame, there is no other Lorentz frame in which it goes away. I think you’ve drawn some incorrect conclusions from the chapter, which is really about how fields transform. $\endgroup$
    – Ghoster
    Commented May 9 at 5:36
  • $\begingroup$ Imagine you have a free charge at rest near the conducting wire. Because the electrons are moving in the wire, the free charge at rest "sees" them with length contraction. I conclude that the charge at rest must experience Coulomb force as well, does it really? $\endgroup$ Commented May 9 at 5:49
  • $\begingroup$ Feynman effectively argues that E and B transform into each other. In this particular example there are even frames in which either E or B vanishes. $\endgroup$
    – my2cts
    Commented May 9 at 8:08
  • $\begingroup$ Veritasium has a video on this. How Special Relativity Makes Magnets Work. It was confusing, so dXoverdteqprogress came up with Veritasium's 'How Special Relativity Makes Magnets Work' - EXPLAINED (better) $\endgroup$
    – mmesser314
    Commented May 9 at 14:39
  • $\begingroup$ @mmesser314 Not only Veritasium. I would say that every youtube channel on physics made a video on this topic, including Dr. Don Lincoln form Fermilab. But the very origin of that idea is the Feynman's book. Yet nobody would give a satisfactory answer on what happens when a charge moves perpendicularly towards a conducting wire. $\endgroup$ Commented May 10 at 1:42

3 Answers 3


As Ric alluded to, the reason Feynman's example works is in some ways an accident of its unusual geometry. Assuming familiarity with the Lorentz force in the context of magnetostatics, it is reasonable to consider Lorentz boosts parallel to the symmetry axis of the current carrying wire, because these always result in static fields in the primed coordinate system (the system is presumably in a steady state, and the geometry is constant.) Things are more complicated when the boost is oblique to the symmetry axis or orthogonal, because in that case the wire is moving in the primed coordinates, and hence its magnetic field is dynamic rather than static. A slightly different approach is needed to explain the electric field experienced by the particle.

The somewhat unsatisfying (but possibly instructive) explanation involves thinking about how Feynman's explanation could be generalized. For example, for what sorts of questions would it be just as useful to consider a hypothetical current carrying wire in place of a 'real' one (understanding that the infinite wire envisioned by Feynman is itself an idealization?) In fact, if you think about Feynman's example carefully, you will realize that what is important about it isn't that the wire is perfectly straight and infinitely long, but that the magnetic field it generates can be approximated by one generated by such a hypothetical ideal wire. In fact, we don't even need a global approximation: a local approximation near the test particle is enough. You then realize that any magnetic field can be approximated this way, at least locally, and away from singularities that are never quite as ideal as the theorists imagine them to be.

So in answer to your question, given a charged test particle moving towards the wire, you can determine the electric field experienced by the particle at each moment in time in its own co-moving frame by carrying out Feynman's calculation for any hypothetical current-carrying wire parallel to the direction of motion of the test charge (read: orthogonal to the other, real/existant wire) that generates a field consistent with the field experienced by the particle (incidentally, this also explains why the magnitude of the magnetic force is the same in each case.)

In general, electric and magnetic fields transform under Lorentz boosts, and Feynman's argument lets you sort of preview the relationship between electric and magnetic fields without having to use the machinery of covariant tensors. If $\vec v\propto \hat x$, and $\beta = v / c$, $\gamma = 1/\sqrt{1 -\beta^2}$,

$E' = (E_x, \gamma E_y - \beta B_z, \gamma E_z + \beta B_y)$

$B' = (B_x, \gamma B_y + \beta E_z, \gamma B_z -\beta E_y)$

(or if the initial electric field is zero)

$E' = (0, -\beta B_z, \beta B_y)$,

$B' = (B_x, \gamma B_y, \gamma B_z)$.

  • $\begingroup$ Thank you of course for you detailed answer. I had a discussion about this question with a YouTube user, who is apparently a physicist. I feel that the answers I get are not unanimous, different people tell me different things. $\endgroup$ Commented May 23 at 15:40
  • $\begingroup$ The YouTube user called DrDeuteron told me this: every thing is time dependent in the cat frame. In the cat frame the (scalar) charge density of the protons is: rho(x, y, z, t) = (rho_0)delta(y-b)delta(z-vt) for impact parameter b, moving at v=v_z. It also has vector current density that isn't static, even though it is zero in the wire rest frame. Then the current flowing electrons have both an x and z component in the cat's rest fame...so..go ahead and compute the fields analytically...and get back to me. Maybe next month? It's not easy. $\endgroup$ Commented May 23 at 15:42
  • $\begingroup$ The only way that his argument could be right is that after all that calculation (which he admits not being easy) you get that there is a charge density imbalance in the wire, between right and left sides of the test charge moving perpendicularly to the wire, so that's why you get a Coulomb force sideways. I told him that, no further comments from him. $\endgroup$ Commented May 23 at 15:50
  • $\begingroup$ To be honest I like the Feynman's idea, the idea of magnetic force without magnetic field. It resembles the 19th century Weber's and Ampere's formulas of magnetic force calculation without magnetic field, using only two moving charges or two current elements. But problems arise when try to calculate magnetic force of a moving test charge towards a conducting wire using Faynman's idea. $\endgroup$ Commented May 23 at 16:00
  • $\begingroup$ DrDeuteron is correct, except that the calculation isn't really that difficult if you know how electromagnetic fields transform under Lorentz boosts. Have you learned about magnetic induction yet? (i.e. that a time-dependent magnetic flux generates an EMF?) I ask only because you have asked some very smart questions that you probably wouldn't have thought to pose if you had already rushed through an AP or first year college physics curriculum. $\endgroup$
    – TLDR
    Commented May 24 at 23:38

To maybe get at what may have been the point, consider a very simple example of a single moving charge, say one electron, there's both an electric field and a magnetic field in this observer's frame of reference.

If you perform a Lorentz boost to a different frame of reference, so that the charge is stationary, there's no longer a magnetic field, just a static electric field.

Not all configurations containing a magnetic field can be Lorentz transformed to remove the magnetic field from all spacetime, like magnetism as you point out, or EM waves is another example.

  • 1
    $\begingroup$ Watch this video to get my point. I am talking about length contraction of moving electrons due their velocity. The very origin of the idea shown in the video is the Feynman's book. Now I would appreciate if someone tells me what happens when a charge moves perpendicularly towards a conducting wire. $\endgroup$ Commented May 10 at 2:00
  • $\begingroup$ Thanks for reference. Re a charge moving perpendicularly to a conducting wire, the equation for force on charge is $$\vec F=q\vec v \times \vec B$$ In cylindrical coordinates, wire current in $+z$ direction at $r=0$, thus $\vec B$ is only in $+\theta$ depending only on $r$. Say the charge is initially at $z=0$, $r=r_0$, $\theta=0$, moving in the $-r$ direction straight towards the wire with some velocity $\vec v_0$, the initial force on the charge would therefore be in the $-z$ direction. The $-z$ velocity component grows, and the force is in the $+r$ direction, so end up with circular path. $\endgroup$
    – Ric
    Commented May 10 at 3:10
  • $\begingroup$ That's correct. But can you get the same result in the same fashion as Veritasium did in his video, considering only length contraction of of moving electrons, without considering the existence of the actual magnetic field B? $\endgroup$ Commented May 10 at 3:31
  • $\begingroup$ Hmm, on second thought, not circular, since $\vec B$ not uniform. I think we can agree that you get the same result in any inertial frame of reference. The initial force in the cat's frame of reference has $\vec v=0$, so only force from "contracted length Electric field" exists and magnetic field does not initially contribute a force, which must work out to same result in guy's frame of reference where cat is moving and there's no such Electric field. As I understand it, Maxwell's equations tells us how solenoids work, led to Lorentz developing the Lorentz transform, which led Einstein to SR. $\endgroup$
    – Ric
    Commented May 10 at 4:03

As the electrons move in one wire, relative to the positively charged nuclei, there is a length contraction effect "observed" by the electrons moving in the other wire. So the linear density of the positive nuclei of the first wire appear to increase, as for the electrons of the other wire. Therefore, the first wire is not electrically neutral anymore, and it attracts the electrons of the second wire by merely Coulomb force.

Note: We think there is no change in neither force or charge density. So Feynman is confused. "We" are observers in the lab frame.

As a proton moves transversely between wires, relative to the moving electrons, there is a relativity of simultaneity effect "observed" by the proton. So the positions, where the points forming a moving field line of a moving electron, are at the same instant, appear to be such that the proton feels a force longitudinally along the wires.

Note: We think there is change in force direction, but not in field line direction. "We" are observers in the lab frame.


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