We have two beams side-by-side: an electron beam and a proton beam. I understand that there will be a force of attraction between them (opposite charges).

But can this force turn repulsive if:

  1. the beams are traveling in different speeds (in the same direction)
  2. the beams are traveling in opposite direction

I saw some similar questions asked already but couldn't find the speed connection anywhere.

  • 1
    $\begingroup$ Can you propose a mechanism that would change the attraction based on speed? $\endgroup$
    – Jon Custer
    Mar 6, 2023 at 17:28
  • 2
    $\begingroup$ I'd rather say that they would repel each other less in the laboratory frame. They never attract each other. $\endgroup$
    – Jon Custer
    Mar 6, 2023 at 17:47
  • 2
    $\begingroup$ I found that two electron beams could attract each other. The accepted answer in the question you linked to in your comment clearly says “The total force between the electrons is always repulsive.” $\endgroup$
    – Ghoster
    Mar 6, 2023 at 18:20
  • 2
    $\begingroup$ I am not sure if we can take the relative velocity to be greater than 'c' w.r.to each other. No, the relative velocity of the beams is always less than $c$. $\endgroup$
    – Ghoster
    Mar 6, 2023 at 18:46
  • 2
    $\begingroup$ @Janaaaa Velocity addition in relativity $\endgroup$
    – Triatticus
    Mar 6, 2023 at 20:51

2 Answers 2


If you are talking about beams, then you can assimilate them as uniformly charged lines moving at their respective velocities. For simplicity, I’ll assume the beams to be parallel. You won’t need relativity, but only basic electromagnetism.

Say the proton (resp. electron) beam has linear charge density $\lambda_+>0$ (resp. $\lambda_-<0$) moving at velocity $v_+$ (resp. $v_-$). Note that the velocity is signed so $v_+v_->0$ when they go in the same direction. The beams therefore carry currents $I_\pm=\lambda_\pm v_\pm$. All the defined quantities are measured with respect to the lab.

Each beam thus generates the EM fields: $$ E=\frac{\lambda_\pm}{2\pi \epsilon_0 r} \\ B=\frac{\mu_0 I_\pm}{2\pi r} \\ $$ So let $d$ be the distance between the two beams you get the electric and magnetic forces (per unit length) to be: $$ F_E=\frac{\lambda_+\lambda_-}{2\pi \epsilon_0 d} \\ F_B=\frac{\mu_0 I_+I_-}{2\pi d} $$ As expected, the electric force is always attractive since they are oppositely charged. However, the magnetic force is attractive when they move in opposite directions and repulsive when they don’t.

You can calculate the ratio of the forces to be: $$ \frac{F_B}{F_E} =\frac{v_+v_-}{c^2} $$ In the vacuum, speeds are smaller than the speed of light, the magnetic force will never compensate the electric force. However in refractive media, the speed of light can be effectively reduced, and it is therefore physically possible for the magnetic force to overcome the electric force.

Hope this helps.

  • $\begingroup$ Hi @lpz: Isn't electric force attractive for opposite charges? $\endgroup$
    – Janaaaa
    Mar 6, 2023 at 20:24
  • 1
    $\begingroup$ Sorry got mixed up. Should be correct now. $\endgroup$
    – LPZ
    Mar 6, 2023 at 20:54

Simply put: no.

There is a sense, which I think you've found already, in which magnetic forces arise from changes to the charge density produced by Lorentz transformations between the frames of different moving charges. (If you haven't read about it, or you're not fully confident about how and why it's a thing, I recommend the relevant chapter in Purcell's EM textbook.)

And, since magnetic forces can be repulsive, it is indeed a natural question to ask whether they can become strong enough to overcome the electrostatic repulsion between two charges.

... and the answer to that is: no.

The reason is that, if you have a beam of protons moving relative to a beam of electrons, you can always place your analysis in the rest frame of the proton beam (or the electron beam; you'll reach the same conclusions). There, the protons are at rest, and the electrons are moving with some velocity. The frame transformation will induce changes to the density of charge in the electron beam, mostly due to length contraction (or the elimination of it, for co-moving beams). But if the two beams consist of charge of only one sign, then the total charge density can never change sign, no matter how creative you get with relativistic length contraction.

And, moreover, magnetic effects don't matter on this new frame. Sure, the electrons are moving, and as such they produce a magnetic field, but the protons are stationary and as such they experience no magnetic Lorentz force at all.

So the interaction will always be between protons and electrons, and it will always be attractive.

Having said this, the analysis above should raise an obvious question: how come magnetic forces between current-carrying wires can be repulsive, then? The main reason is that most current-carrying conductors consist of charges of both signs, moving at different velocities. (Normally, a stationary lattice of ions combined with a slowly moving sea of electrons, but all sorts of combinations are possible or at least conceivable.) And if you have charges of both signs in the wire, then it becomes possible to get creative with Lorentz length contraction to make the total charge on the wire positive, or negative, or zero, because the charges of different signs are moving differently, and as such you can affect them differently with the length contraction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.