According to Feynman (Lectures on Physics, Vol. II, 21–6), the potential of a moving point charge is:

$$\phi = \gamma \cdot \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ \gamma^2 \left( x-vt \right)^2 + y^2 + z^2 }}$$

I tried the following relativistic argument to derive it:

Call $S'$ the frame of reference of the particle, and call $S$ our frame. In $S'$, the particle is at rest, and in $S$, it moves toward positive $x$ with velocity $v$.

Now, in $S'$ I have the static potential: $$\begin{align}\phi' & = \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ {x'}^2 + {y'}^2 + {z'}^2 }} \\ \mathbf {A'} & = \mathbf 0 \end{align}$$ Then I make the Lorentz transformations in $\left(t;x,y,z\right)$ and in $\left(\phi ; \mathbf A\right)$ to get them at $S$: $$\begin{align} x & =\gamma\left(x'+vt\right) \\ t & =\gamma\left(v/c^2\cdot x'+t'\right) \\ \phi & = \gamma\phi' \\ \mathbf A & = \gamma \mathbf v \phi' \end{align}$$

Finally, substituting the potential $\phi'$: $$\begin{align}\phi & = \gamma \cdot \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ {x'}^2 + {y'}^2 + {z'}^2 }} \\ & = \gamma \cdot \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ \gamma^2 \left( x-vt \right)^2 + y^2 + z^2 }} \end{align}$$

According to the formula cited above, I arrived at the correct result. But I noticed that I didn't include the transformation of the $\left(\rho;\mathbf j\right)$ four-vector (that is, $\left(\mbox{charge density};\mbox{current density}\right)$). If I included it, I would make this extra substitution in the formula for the potential: $$q' = q/\gamma$$ where $q'$ is the particle's proper charge. Then I would find an extra $\gamma$ in the formula for the field of a moving particle. Since there is no such extra $\gamma$ factor (according to Feynman's formula), there must be something wrong in my $q = \gamma q'$ argument. Where did I misunderstand the Lorentz transformation of four-vectors?

  • 3
    $\begingroup$ Charge is a Lorentz invariant quantity, i.e. $q' = q$, not what you have written. Of course the charge density changes, but that is because the volume factor is modified (precisely by a factor of $\gamma$). $\endgroup$
    – Prahar
    Jan 15, 2016 at 23:17

1 Answer 1


As explained by Prahar, the reason that charge density changes according to the frame of reference is because of the change in the volume factor, and not because of a change in the charge itself:

$$\begin{align} \rho' &= q\cdot \delta\left(x'\right)\delta\left(y'\right)\delta\left(z'\right) \\ \rho &= \gamma \rho'\\ &= \gamma q\cdot\delta\left( \gamma\left(x-vt\right) \right)\delta\left(y\right)\delta\left(z\right) \\ &= \gamma q\cdot \left|\gamma\right|^{-1} \delta\left(x-vt\right)\delta\left(y\right)\delta\left(z\right) \\ &= q\cdot \delta\left(x-vt\right)\delta\left(y\right)\delta\left(z\right) \end{align}$$

In both frames of reference, the charge is: $$\iiint\rho\,dx\,dy\,dz=\iiint\rho'\,dx\,dy\,dz=q$$

I was led to believe that changing the frame of reference changed the charge because in an infinite line of current with current $I$, the charge indeed does change (a neutral wire with a current looks charged when we move past it). Looking at an arbitrary segment of the wire, it wasn't charged, and now it is charged because we're looking at it with from a different frame of reference. Actually, the total charge of the wire becomes infinite! This must stem from the "infinite" part of the "infinite wire" exemple. If I look at a loop of current, the opposite side of the loop has the opposite charge, and then the total charge does not change.


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