The question is from Griffith's Introduction to Electrodynamics, the chapter on Special Relativity:

Picture of the setup, as observed from co-moving frame

A stationary magnetic dipole, $\vec m = m\hat z$, is situated above an infinite uniform surface current, $\vec K = K\hat x$. Suppose that the surface current consists of a uniform surface charge σ, moving at velocity $\vec v = v \hat x$, so that $\vec K = \sigma\vec v$, and the magnetic dipole consists of a uniform line charge λ, circulating at speed v (same v) around a square loop of side l, so that $m = λvl^2$. Examine the same configuration from the point of view of the system $\bar S$, moving in the x direction at speed v. In $\bar S$, the surface charge is at rest, so it generates no magnetic field. Show that in this frame the current loop carries an electric dipole moment.

I read the solution manual for the problem and am stuck at the point where the charge density is transformed. As the velocity of the wire itself as seen from this frame is v, shouldn't the Lorentz factor be $1/\sqrt{1-\frac{v^2}{c^2}}$? However, in the book, the velocity of line charge relative to $\bar S$ is found, and for the line facing towards us, it is 0, and for the line charges in the backside wire, it will be $\frac{2v}{1+\frac{v^2}{c^2}}$, and the velocity used for this backward line charge is this. Why are we using this velocity of the line charges for finding $\gamma$? The length corresponds to the length of the wire, so shouldn't we have the line charge density for the wire on the backside as $\frac{\lambda}{\sqrt{1-\frac{v^2}{c^2}}}$?

  • $\begingroup$ I think the velocity for the line charges in the backside wire would be $\frac{2v}{1+v^2/c^2}$ rather than $\frac{2v}{1-v^2/c^2}$, and for the line facing towards us, it is just zero rather than $v$. $\endgroup$ Commented Jul 30, 2023 at 12:25
  • $\begingroup$ Yeah, I just edited.. but then, why will the Lorentz factor in the case of backside wire be determined by $\frac{2v}{1+\frac{v^2}{c^2}}$ rather than $v$? $\endgroup$
    – V Govind
    Commented Jul 30, 2023 at 13:28

1 Answer 1


For calculating the Lorentz contraction of the distances between two successive electric charges, we first need to calculate the relative velocity of the moving charges WRT the observer at rest in $\bar S$. The relativistic velocity addition implies:

$$u_x=\frac{u'_x+v}{1+u'_xv/c^2} \tag1 $$

Using $u'_x=-v$ for the front wire, we get:

$$u_x=\frac{-v+v}{1-v^2/c^2}=0 \tag2 $$

Recall that the correct direction of the current seems to be opposite of the shown direction for the front wire. Using $u'_x=+v$ for the back wire, we get:

$$u_x=\frac{v+v}{1+v^2/c^2}=\frac{2v}{1+v^2/c^2}\tag3 $$

Finally, for the left- and right-hand side wires, the net velocity of the charges is calculated to be:

$$w_{xy}=\sqrt{\alpha_v^2 v^2+v^2}=v\sqrt{2-v^2/c^2} \tag4$$

  • $\begingroup$ Thanks for the reply! So does this mean that while finding the Lorentz factor for the charge density, what is important is the velocity of the charges and not the velocity of the wire in which it is moving? $\endgroup$
    – V Govind
    Commented Jul 31, 2023 at 12:37
  • $\begingroup$ @VGovind Yes, you're right. The velocity of the wire is not important. $\endgroup$ Commented Jul 31, 2023 at 14:33

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