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How can relativity explain the magnetic attraction of two electrons (or two electron beams) comoving in a vacuum at some certain constant velocity?

It is well known (https://acceleratorinstitute.web.cern.ch/acceleratorinstitute/ACINST89/Schindl_Space_Charge.pdf) that two parallel electrons or electron beams at the same velocity will experience an attractive Lorentz force that increases in magnitude and eventually equals the repulsive Coulomb force at the hypothetical velocity of c (if you could get to that velocity that is).

It seems that relativity would suggest that comoving electrons would of course view each other as being at rest and hence their Coulomb fields and forces would not be Lorentz contracted. In addition, they would not view any created magnetic fields and hence no Lorentz forces. This would seem to suggest that the electrons, from the comoving or proper frame of reference, would merely repel each other according to Coulomb's law. But, it is experimentally known that two parallel electrons or electron beams do in fact attract when moving at a given constant velocity.

When there is no wires involved, there seems no way to 'introduce' via length contraction, time dilation, or otherwise, a means via relativity to explain the attractive Lorentz force that negates and or counters the Coulomb force for these comoving electrons.

Please note, I am specifically NOT talking about wires, so I kindly request no references or explanations based on a Lorentz contraction of the positive atomic lattice in said wires as is generally done. By the way, I do not dispute the wire-based answers at all as they are entirely logically consistent.

The only things involved in the question, are the comoving parallel electrons or electrons beams and a vacuum through which they are moving at some constant velocity.

Qmechanic
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user141720
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2 Answers2

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The argument from the electrons' frame of reference is correct. The total force between the electrons is always repulsive. In the lab frame, the repulsive force looks weaker. There are two ways to explain this, both equivalent:

  1. There is a magnetic force between the charges, since moving charges create magnetic fields and parallel currents attract--just like a wire.
  2. The electrons are experiencing time dilation, and so move apart more slowly than they would if they were at rest in the lab frame.

At the speed of light, the electrons (from the lab frame) would not experience the passage of time, so they wouldn't be able to move apart. That's why the magnetic attraction equals the electric repulsion at $c.$

This is similar to explanations of what causes a free moving electron to curve in the presence of a magnetic field.

  1. In the rest frame of the magnet, the electron experience the Lorentz force $F=q\vec{v}\times \vec{B}$.
  2. In the rest frame of the electron, there is no magnetic force since $\vec{v} = 0$. So, the force must be from an electric field. From Maxwell's equations, we can calculate the field from $\partial\vec{B}/\partial{t}.$

In short, what looks like a pure magnetic field or a pure electric field to one observer can look like a combination of the two to another observer in a different state of motion.

Mark H
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  • So, the time dilation also occurs then with parallel electron currents in wires? Does this conflict with the length contraction explanation when wires are involved? – user141720 Apr 08 '17 at 00:33
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    Wires are different due to the neutrality of the wires (positively charged nuclei included). With no electric field in the lab frame, magnetic effects (or, Lorentz-boosted electric effects) dominate. I'm not seeing a contradiction. – Mark H Apr 08 '17 at 03:58
  • @MarkH-In your second second point you state that in the rest frame of the electron there is no magnetic force, and you can calculate $E$ from $\frac{\partial \vec B}{\partial t}$. But isn't this zero because the magnetic force is zero (and stays zero if the charges get accelerated). Can't you calculate the electric field between the electrons in their rest frame (which becomes their CM frame if they get opposite velocities due to their acceleration) simply by applying the Coulomb law between two charges? – Deschele Schilder Apr 08 '17 at 11:56
  • I see now that you are talking in the second part about a single electron moving through the magnetic field of a magnet. In the rest frame of the electron then, there is no Lorenz force though instead of no magnetic force, as you state. The magnetic force in the rest frame of the electron isn't zero, and varies as the electron moves through the $\vec B$-field, which is why you can apply $\frac{\partial \vec B}{\partial t}$ for calculating the electric force (which must be equal to the Lorenz force, not to the magnetic force). – Deschele Schilder Apr 08 '17 at 12:46
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    @AndrewSteane I've edited the summary remark. – Mark H Jun 25 '22 at 13:21
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The answer to the question in the title is "Yes it can, but one needs to proceed carefully; there are some subtleties."

In the frame of reference in which the charges are at rest ("the charges' frame, C") the charges experience a Coulomb's law repulsion force of magnitude $qE_\text C$ in which $E_\text C$ is the magnitude of the electric field due to one charge in the vicinity of the other, in the charges' frame. Thus, $$E_\text C=\frac q{4\pi \epsilon_0 r^2}.$$ In the "lab frame, L", in which the charges are moving, abreast of each other, in parallel paths at speed $v$, the force between them is smaller: it is multiplied by a factor of $\frac{1}{\gamma}$. [This transform of a transverse force can be justified very simply: the charges repel and would acquire, if allowed to do so, transverse momentum, which is a Lorentz invariant. But the time for a given small momentum change to occur is dilated by $\gamma$ in the lab frame compared with the charges' frame.] For these transverse forces, then, $$F_{\text L}=\frac{\Delta p_{\text L}}{\Delta t_\text L}=\frac{\Delta p_{\text C}}{\gamma\Delta t_\text C}=\frac 1 \gamma F_\text C$$

To summarise the story so far:

Total electromagnetic force on a charge in charges' frame = $qE_\text C$.

Total (repulsive) electromagnetic force on a charge in lab frame = $q\frac 1{\gamma}E_\text C$.

We need to partition this lab-frame force into electric and magnetic parts. Fortunately we know that...

In lab frame, (repulsive) electric force on $q$ = $q(\gamma E_\text C)$.

[We can derive this informally by considering one charge to be surrounded, in its rest frame, by a Gaussian sphere, on which surface we shall place the other charge, let us say on the sphere's equator. Consider lines of latitude a distance $\Delta \lambda$ apart, close to and either side of the equator. In the lab frame $\Delta \lambda$ is contracted to $\frac 1\gamma \Delta \lambda$. But the electric flux, $\Delta \phi$, leaving the Gaussian surface between given lines of latitude (like the total flux leaving the whole surface) won't be changed by observing from a different reference frame. So the electric field strength on the equator will be increased from $E_\text C= \frac {\Delta \phi}{2 \pi r \Delta \lambda\epsilon_0}$) in the charges' frame to $\frac {\Delta \phi}{2 \pi r (\Delta \lambda/\gamma)\epsilon_0}=\gamma E_\text C$ in the lab frame.]

The magnetic force between the charges in the lab frame must be the difference between the total electromagnetic force and the electric field force, so...

$$\text{magnetic force}\ =\ q\tfrac 1{\gamma}E_\text C - q\gamma E_\text C\ =-q\gamma E_\text C \tfrac{v^2}{c^2}\ =-\frac{q^2}{4\pi\epsilon_0 r^2} \gamma\tfrac{v^2}{c^2}$$ The minus signifies attraction. Using $c^2=\frac 1 {\mu_0 \epsilon_0}$ we present the force as $$\text{magnetic force} =-q\left(\gamma \frac{\mu_0 q v}{4\pi r^2}\right)v.$$

This is of the form of a magnetic Lorentz force on one of the charges, the content of the big bracket being the magnitude, $B$, of the magnetic flux density due to the other charge. Apart from the factor of $\gamma$, this is the value of $B$ given by the Biot-Savart rule, using $qv$ for the current element $I\delta l$. Note that the Biot-Savart treatment of a non-steady current is only a low speed approximation: the $\gamma$ arising from our relativistic derivation is correct.

Even though, on the face of it, the two point charges moving on parallel paths is a simpler set-up than two parallel current-carrying wires, its treatment using just Coulomb/Gauss and elementary concepts of SR (time dilation, length contraction and charge invariance) has proved to be trickier for the point charges than it is for the wires. In particular we took account of the (repulsive) electric field force between the charges being greater in the lab frame than in the charges' frame.

Philip Wood
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  • In my opinion you must take a look in my answer therein : Charges and relative motion. All are exactly relativistic (not approximate). And more general (angle $\phi$). I must note the two methods to find the fields and forces in the ''lab'' frame. One method is by direct use of the Liénard–Wiechert potentials. The other is by Lorentz transforming the fields under a Lorentz boost. The two results are of course identical. – Frobenius Jun 25 '22 at 21:49
  • @Frobenius Thank you. A lot of work went into your excellent answer. My aim, though, was to deal carefully with the specific case forming the subject of this question, using the most basic of Special Relativity concepts. I believe that my treatment, like yours, is exactly in accordance with classical electromagnetism and SR (not approximate), and that we agree on the result for this specific case. – Philip Wood Jul 23 '22 at 12:15