I've been reading about the origin of the magnetic force and learnt that the magnetic and electric forces are actually the same forces viewed from different frames of reference. One common explanation is the one below where the moving charge "experiences" a net repulsive force due to the length contraction of the "positive charges" from the particle's frame of reference. We see that in this case the force is perpendicular to the velocity.

enter image description here

But what happens if the particle is moving perpendicular to the current? (as in the picture below). Then there will be no length contraction from the particles frame of reference and thus no net force. But from what I know, there should be a force perpendicular to the velocity according to the cross product/right hand rule. So what is the origin of this force?

enter image description here

  • $\begingroup$ It's not just the length contraction of the positive charges, there is also a dilation of the negative charges (a la Bell's Spaceship Paradox); moreover, the plane of simultaneity matters, too. While the +/- charges are aligned in your diagram, they are not so in a moving reference frame. $\endgroup$
    – JEB
    Dec 6, 2022 at 15:26
  • $\begingroup$ Related : Why don't stationary charge feel force from a current carrying wire?. $\endgroup$
    – Frobenius
    Dec 6, 2022 at 19:45
  • $\begingroup$ Relative velocity of the +/- charges wrt the outside test charge is not zero in the second diagram (can you say why?). With this fact, can you calculate the length contraction for the +/- current? $\endgroup$
    – KP99
    Dec 7, 2022 at 6:42
  • $\begingroup$ @KP99 Because they move diagonally downwards? I just have a hard time figuring out how this would be analogous to the first picture where we have length contraction in the horizontal direction? $\endgroup$ Dec 7, 2022 at 10:38
  • $\begingroup$ @JEB Yes, I think I got it, thanks! $\endgroup$ Dec 7, 2022 at 10:39

1 Answer 1


Not all problems work out nicely in 3 + 1 dimensions. Consider two 4-currents in the (unprimed) wire rest frame:

$$ j^+_{\mu} = (\rho^+, 0, 0, 0) $$

$$ j^-_{\mu} = (\rho^-, v_d\rho^-, 0, 0) $$

With the charge densities satisfying $\rho^+ = -\rho^-$ (the wire is neutral) and the drift velocity is leftward, $|v_d| = -v_d$, the total four current density is:

$$ j_{\mu} = (0, v_d\rho, 0, 0) $$

(note: many complain the drift velocity is too small to invoke $\gamma$, which is true, but if you do the Lorentz transform parallel to the current you will see it is the relativity of simultaneity that dominates the change in charge density).

The moral of that parenthetical story is: when in doubt, Lorentz transform. So that's what we need to do here. With ($c=1$) and the test charge's (primed) frame moving along $+y$:

$$ u_{\mu} = \gamma(1, 0, v, 0) $$

you get:

$$ \Lambda_{\mu\nu} = \left [\begin{array}{cccc} \gamma & 0 &-v\gamma & 0 \\ 0 & 1 & 0 & 0 \\ -v\gamma & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \end{array}\right ]$$

$$ j'_{\mu}= \Lambda_{\mu\nu}j_{\nu} $$ $$ j'_{\mu}= (0, v_d\rho, 0, 0) = j_{\mu} $$

So the force in the primed frame is caused by a magnetic field, no 3 + 1 magic here.

Of course, this is wholly unsatisfying compared with the original problem, which works out in spite on being not at all manifestly covariant.

It also raises the question of "where is the moving charge?". If you're moving above the wire, then the transformation of a magnetic field is:

$$ B'_{\parallel} = B'_{\parallel}$$

That is fine, but if you're moving towards the wire (in the plane of the current and charge):

$$ B'_{\perp} = \gamma\big(B_{\perp} - \vec v \times \vec E\big) =\gamma B_{\perp}$$

as $\vec E = 0$.

So now we've picked up a $\gamma$: the primed field is stronger (and not circular symmetric about the wire). That is not evident from the simple transformation of $j_{\mu}$.

I think the resolution here is to go to Jefimenko's Eq for a magnetic field generated by currents and charges on the past light cone:

$$\vec B'(\vec r', t') = -\frac{\mu_0}{4\pi}\int{\Big[ \frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^3} \times \vec J'(\vec r'', t'_r) + \frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^2} \times \frac 1 {c} \frac{\vec \partial J'(\vec r'', t_r)}{\partial t'} \Big]dV''}$$

Since the current is constant:

$$\vec B'(\vec r', t') = -\frac{\mu_0}{4\pi}\int{\Big[ \frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^3} \times \vec J'(\vec r'', t'_r)\big]dV''}$$

(Note the double primed coordinate here is the integration variable, not any reference frame).

Computing this a bit of a project. In the primed frame you have a current density looking like:

$$\vec j'(x', y', z', t') = v_d Q\delta(y'-vt')\delta(z')\hat x' $$ where Q is the limit of $\rho$ in an infinitely thin wire. So this is a line of moving current.

Then fix ${\vec r'}$ to be the position of the charge, in the primed frame:

$$ \vec r' = (x', y', z') = (0, 0, 0) $$

At this point, it is best fix $t' \ne 0$ so you have some distance from the wire.

Then compute the retard time $t'_r$. The $\delta(y'-vt')$ will filter over the past light cone along the moving wire, giving some kind of conic section, but it should all work out, as it always does. Have at it. $\gamma$ is in there, somewhere.


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