# Can relativity explain the magnetic attraction between two parallel electrons or electron beams comoving in a vacuum? (No wires)

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.

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, a magnetic field is an electric field in a different reference frame.

• 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? Apr 8 '17 at 0:33
• 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. Apr 8 '17 at 3:58
• @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? Apr 8 '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). Apr 8 '17 at 12:46

In the frame of reference in which the charges are at rest ("the charges' frame") the charges experience a Coulomb's law repulsion force of magnitude qE in which E is the magnitude of the electric field due to one charge in the vicinity of the other. In the "lab frame", 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 is a straightforward transform of transverse force. It can be justified by a very simple argument: the charges repel and acquire transverse momentum, which is a Lorentz invariant, but the time for a given momentum change to occur is dilated by $\gamma$ in the lab frame compared with the charges' frame.]

It's tempting to assume that

magnetic force = repulsive force in lab frame – repulsive force in charges' frame.


Giving: magnetic force = $\frac{qE}{\gamma}-qE$

More detailed analysis shows that it's not as simple as this. Using the transforms for transverse $\vec{E}$ and $\vec{B}$ we find that

In lab frame, electric force on q is $\gamma qE$, magnetic force =$-\gamma q E \frac{v^2}{c^2}$

So the electric repulsive force is actually greater in the lab frame, but the magnetic attraction more than compensates. Do check that the two forces add to give a resultant of $\frac{qE}{\gamma}$, just as we showed at the beginning!