If two opposite charges are moving at a velocity of $0.9c$ relative to a laboratory, but are at rest relative to each other (say the charges are moving in parallel), will there be any relativistic corrections to electrostatic attractive force (given by Coloumb's law) exprienced by the charges? In a frame in which the particles are at rest, Coloumb's law would be valid, but in the laboratory frame, the charges, being opposite, would act as antiparallel currents which repel each other. This would lead, I believe, to a reduction in electrostatic attraction on the order of $\gamma^{2}$, where $\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ (cf. https://www.phys.unsw.edu.au/einsteinlight/jw/module2_FEB.htm this works out the equivalent problem but for like, i.e. same sign, charges), but only as viewed from the laboratory frame. Is there any relativistic correction to the frame in which the charges are stationary? I would expect not, but I was just wondering if there is something that I am missing given the relativistic speed of the two particles?

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    $\begingroup$ Of course, in the particles' frame the force is given by the ordinary Coulomb force. In the lab frame the force is a combination of the magnetic force and a modified Coulomb force, e.g. see here. $\endgroup$
    – knzhou
    Aug 1, 2023 at 1:13
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    $\begingroup$ Haha There's some error in the school page, the net force is reduced by gamma, not gamma squared. (Please remove the extra character at the end of the link) $\endgroup$
    – stuffu
    Aug 1, 2023 at 2:51
  • $\begingroup$ If you want to work this problem out for yourself, consider two infinite linear charge densities separated by a distance, at rest. And then reconsider in a moving frame with length contraction and current, etc. By making them infinite, you negate the effects of relativistic squishing of the fields, making it a slightly easier problem to start with, $\endgroup$
    – JEB
    Aug 1, 2023 at 4:34
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    $\begingroup$ @Frobenius If the point charges are side by side its gamma. If they are not side by side, then we need more info to calculate. If charges are placed densely on long lines, that do not contract, only then its gamma squared. $\endgroup$
    – stuffu
    Aug 1, 2023 at 12:25
  • $\begingroup$ I believe it is correct that in this situation $\frac{F_{B}}{F_{E}}$ (ratio of magnetic to electric force) should be equal to $\frac{v^{2}}{c^{2}}$, cf. academic.mu.edu/phys/matthysd/web004/l0220.htm), so the force should be reduced by $\gamma^{2}$. What is the specific error in the school page? $\endgroup$
    – Robert
    Aug 1, 2023 at 13:46

1 Answer 1


$ \newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\tl}[1]{\tag{#1}\label{#1}} $

The relativistic equations for the electromagnetic field of a uniformly moving electric charge $\:q\:$ (see Figure-01) are :

\begin{align} \mb E\plr{\mb x,t} & \e \dfrac{q}{4\pi\epsilon_0\vp}\dfrac{\plr{1\!\m\!\beta^2}}{\plr{1\!\m\!\beta^2\sin^2\!\phi}^{3/2}\vp}\dfrac{\mb r}{\:\:\Vlr{\mb r}^3},\quad \beta\e\dfrac{\upsilon}{c} \tl{01a}\\ \mb B\plr{\mb x,t} & \e \dfrac{\mu_0 q}{\hp{\epsilon} 4\pi\hp{_0}\vp}\dfrac{\plr{1\!\m\!\beta^2}}{\plr{1\!\m\!\beta^2\sin^2\!\phi}^{3/2}\vp}\dfrac{\bl\upsilon\x\mb r}{\:\:\Vlr{\mb r}^3},\quad \mb B\e\dfrac{1}{c^2}\plr{\bl\upsilon\x\mb E}\vphantom{\dfrac{a}{\dfrac{}{}b}} \tl{01b} \end{align}

The "Correction Coefficient" of the electric field (modified Coulomb field) is \begin{equation} \mr{CC} \e \dfrac{\plr{1\!\m\!\beta^2}}{\plr{1\!\m\!\beta^2\sin^2\!\phi}^{3/2}\vp} \tl{02} \end{equation}

So, \begin{equation} \mr{CC} \e \left. \begin{cases} \gamma^{\m 2}\!\!\!\!\!\!& \les 1 \quad \texttt{in Case 1 : } \mb r\,\bl \| \,\bl\upsilon \bl\implies \phi\e 0\\ \:\:\:\gamma & \gr 1 \quad \texttt{in Case 2 : } \mb r\bl \bot \bl\upsilon \bl\implies \phi\e \pi/2\\ \end{cases} \right\} \tl{03} \end{equation} as shown in Figure-02.

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  • $\begingroup$ Thank you! I was mainly considering case 2. This means that laboratory observer will determine a LARGER force than an observer in the particle's frame. How does this square away with time dilation, whereby the laboratory observer should determine that it takes a LONGER time for the two particles to meet? Is this due to relativistic correction to F = ma? $\endgroup$
    – Robert
    Aug 28, 2023 at 20:57

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