# How close together can an antiparallel magnetic moments free electron pair get?

See illustration below: Is there a calculation-prediction of the minimum distance separation of an antiparallel electron pair due its like charge repulsion?

There must be an equilibrium between the magnetic attraction and the charge repulsion? (the two electrons in the illustration, antiparallel magnetic moments N-S and S-N will attract the two electrons together forming a stable pair but at the same time the Coulomb charge repulsion force will keep them separated at a distance d).

Any analytical solution to this problem?

I believe I heard or read somewhere once I cannot recall, that this distance d described in the question is equal to the Reduced Compton Wavelength of the electron ƛ=ħ/mc =3.861 592 6796(12) x 10E-13 m = 386 fm, which is also the uncertainty Δχ in this case but I cannot find any analytical solution?

Specifically, the minimum Δχ uncertainty in the position of a free electron at rest is Δχ= 0.5 (ħ/mc) = 0.5ƛ = 193 fm thus half of its reduced Compton wavelength. Notice here that the Reduced Compton Wavelength for the electron also written as ƛe=λe/2π where λe is the Compton wavelength of the electron, represents the radius r of the electron's rest mass field shown in the above illustration (blue sphere).

Therefore, any analytical prediction presented as an answer in this question page must have a minimum uncertainty in the separation distance d for an antiparallel pair of electrons set at d(uncertainty)=2Δχ=ƛe=386 fm.

Thus any value prediction must be written as: d=................ (193 fm), where the value in the parenthesis indicates the ± , uncertainty in the prediction.

Note: Definition of the term rest mass field of the electron used in this question, is (or equals) a sphere volume of radius the total minimum uncertainty in position Δχ of an electron at rest, ±193fm = 386fm. Of course the electron particle by known theory is a massive dimensionless point particle.

• I believe I heard somewhere once I cannot recall that this distance d described in the question is equal to the Reduced Compton Wavelength of the electron ƛ=ħ/mc =3.861 592 6796(12) x 10E-13 m , which is also the uncertainty Δχ in this case but I cannot find any analytical solution? Sep 19 at 13:04
• Very reasonable question. Why the downvote? Sep 21 at 19:40
• This may be obvious, but: The coulomb force of one electron on the other the electron is much greater everywhere than the magnetic force of one electron on the other. Despite the name a "free electron pair" is part of an atom or molecule, not an isolated pair of free particles. This makes this a problem you'll have to solve with quantum mechanics, not classical electricity and magnetism. The electrostatics approach is informative, though - the relevant coulomb force is the net force of all the charged particles in the atom, not just the coulomb force of the electron.
– g s
Sep 25 at 17:05
• Thank you for your very important clarifications and insight. My ansatz is the case of an isolated antiparallel electron pair. Sep 25 at 17:18

According to Alexander A. Mikhailichenko  (eq. 8), reference suggested to me by Prof. Behnam Farid (https://www.researchgate.net/profile/Behnam-Farid-2) this equilibrium linear distance d where the two forces equalize is:

d=√6ƛe~2.5ƛe

where ƛe is the Reduced Compton Wavelength of the electron thus ƛe=386 fm =0.386 picometers.

Therefore resulting in a final value of linear separation between the two electron's magnetic moments in an electron pair at rest of d~2.5x386=965 fm.

That is x1.25 the diameter cross-section of the single electron's rest mass field (blue sphere in illustration of anti parallel electron pair). Meaning in the separated antiparallel electron pair a whole third electron can fit in between!