I'll start with the common setup where a free charge is placed near a wire through which current does not flow. The wire is neutral so the free charge does not feel any force upon it. When the a current is generated through the wire (meaning that electrons start moving and the atoms remain in their places, ignoring small oscillations due to thermal velocity), the explanation is that if the free charge is standing still an apparent electric force does not appear since the wire is still neutral. Only when the free charge moves with a velocity along the wire it has a force that acts on it. If we reverse the sign of the free charge's velocity, then the force changes its sign also.

Based on my understanding this picture should be asymmetrical. If the free charge starts moving with the same velocity as the electrons in the wire, then, due to length contraction/dilation, the spacing between the atoms shrink and the the one between the electrons becomes larger. This generates a charge imbalance and thus a force acts upon the free charge. However, if the free charge moves with the same velocity as the electrons but in a direction opposite to the current, then the relative velocity between the electrons and atoms in the wire won't create the same imbalance, unless their velocities are composed classically. So in this case, although the free charge has the same magnitude of its velocity, shouldn't its different direction make the spacing between the charges in the wire slightly different, thus having a different force if the free charge moves in the direction of the current or opposite to it?

  • $\begingroup$ The directions of the 'magnetic' forces are the opposite in the two cases. Maybe you forgot that, or maybe not? $\endgroup$
    – stuffu
    Apr 27, 2023 at 7:38
  • $\begingroup$ Do you know that when you are observing a moving rod and a still standing rod and you start moving yourself, then the moving rod's length changes more that the formerly still standing rod's length? $\endgroup$
    – stuffu
    Apr 27, 2023 at 7:50
  • $\begingroup$ There is indeed an intrinsic asymmetry. In one case you move in the direction of the current and in the other case in the opposite direction. $\endgroup$
    – facenian
    Apr 27, 2023 at 11:49
  • $\begingroup$ The change in direction of the magnetic force is all good. My issue was that changing the velocity of the free charge (same magnitude, different sign) would lead to different velocities of the positive and negative charges in the wire, and since those velocities do not add up as in classical relativity, their perceived values should be slightly different in magnitude. This would imply that the magnetic force would be slightly different in terms of magnitude, not sign. This was the problem. $\endgroup$ Apr 28, 2023 at 8:12

2 Answers 2


Let's consider explicitly two scenarios:

  1. In the lab frame, a positive linear charge density, $\lambda_+$, is moving to the right at speed $v$ while an equal and opposite linear charge density, $\lambda_- = -\lambda_+$, stays fixed in place. The test charge is moving to the right at speed $V$.

  2. Same as #1, but the test charge is moving to the left at speed $V$.

In both cases the current in the lab frame is $\lambda v$.

In scenario 1, to get to the frame of the test charge, we have to boost by $V$ to the right. This results in an apparent contraction of the negative charge carriers, increasing the magnitude of their charge density:

$$ \lambda_-' = - \frac{\lambda_+}{\sqrt{1 - V^2}}$$

(I use units where $c = 1$.)

In the frame of the test charge, the velocity of the positive charge carriers is given by the velocity addition formula between $+v$ and $-V$,

$$ v_r = \frac{v - V}{1 - vV} $$

To get the density of the positive charge carriers when they have velocity $v_r$, we consider a boost from their lab frame (where they have velocity $v$) to their rest frame and then to the frame where they have velocity $v_r$. The result is

$$ \lambda_+' = \sqrt{\frac{1 - v^2}{1 - v_r^2}} \lambda_+ = \frac{1-vV}{\sqrt{1-V^2}} \lambda_+ $$

where we've done some algebra on the right hand side.

The overall charge density is then given by

$$ \frac{\lambda'}{\lambda_+} = \frac{1-vV}{\sqrt{1-V^2}} - \frac{1}{\sqrt{1 - V^2}} = \frac{-vV}{\sqrt{1-V^2}} $$

When we repeat the analysis in scenario 2, we obtain

$$ \frac{\lambda'}{\lambda_+} = \frac{1+vV}{\sqrt{1-V^2}} - \frac{1}{\sqrt{1 - V^2}} = \frac{+vV}{\sqrt{1-V^2}} $$

So when we compare the positive charge densities observed in the rest frame of the test charge in scenarios 1 and 2, they're related in some complicated way, which you've called "asymmetrical". That's true. But when we subtract off the negative charge densities, we get equal and opposite results. These total charge densities support the conclusion that the magnetic forces in the two scenarios are equal and opposite.

Although this exercise is a good "sanity check" on the consistency of electromagnetism with special relativity, you probably shouldn't use such thought experiments as a method of reasoning about relative magnitudes of magnetic effects in general: it simply creates unnecessary work. Instead, you should just use the laws of magnetism directly (i.e. the Lorentz force law plus either Ampère's law or the Biot–Savart law).

  • $\begingroup$ This looks like the thing that I was looking for, and yes, I was interested in this way of looking at it as a sanity check, meaning that qualitative results from Lorentz force/Biot-Savart/Ampere's Law should derive from this approach. $\endgroup$ Apr 28, 2023 at 14:17

It's much more complicated. All fields and their derivitives, the charge and current densities, are transformed according to

ct= Cosh[u] ct' + Sinh[u] z'
z = Cosh[u] z' + Sinh[u] ct'

and, with a differential d in front, their differentials too. The D and H fields are surface 2-forms describing charge and current density. With D=0, the surface H vector is parallel to the current direction with magnitude H=I/ (2pi R), as if the current is concentrated in the surface.

As a surface density it has the form

H =  I/(2 pi R)  (r dphi) ^ dz 

and its dual, the force field

B  =  I/(2 pi R) dct ^ dr

As a map in the directions t,r its a force infuencing the radius by

dr = B dt (v' = v x B )

in any radial direction while leaving the coordinates z, phi moving freely .

The complete solution in the outer space of the cylinder of Radius R is

 H =  I/(2 pi) If[r<R,0,R/r]  (r dphi) ^ dz 

with differential (observe cancellation of field 1/r and angualar scale r)

dH = j = (I/(2 pi R)  If[r==R,1,0]  * (r (dphi/(2pi)) ^ dz 

Lorentz boost applied to th system of reference of the observer moving in r=const direction, that is

j' = (I/(2 pi R) If[r==R,1,0]    (r (dphi/(2pi))  ^ (chu dz' + shu dct')

that is, the so-called Lorenz contraction is a Lorentz dilation of dz-intervals and the additional creation of an electric current density in the angular direction.

This all rests on the assumption, that the conductor is grounded, i.e. in its rest system has zero electric field. If somebody has experience with surface charges of some 1000 V, its of course evident how important zero electric fields were the detection of the laws of magnetism.

So, by these considerations, a current of opposite charges in a grounded conductor with electrical outer field zero must have different densities of charge, if the currents are equal. At equal densities the moving negative cloud would produce am electric field ~ sin(u)= v/c/Sqrt(1-v^2/c^2).

A moving observer just has different coordinates. Nothing real changes by coordinate change. As is mostly not mentioned in this not so easy experiment context, are the undelying assumptions: that all physical systems, including the light as the synchonizing signal for distant clocks, relatively at rest, and the ticking mechanism in the spin-spin system of the atomic clocks, all are representations in different spaces of the same universal abstract Poincare group with defining representation in R(-1,1,1,1).

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    Apr 27, 2023 at 11:59

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