At what distance does magnetic attraction between two properly aligned electrons overcome electric repulsion?

Question

Electrons have both magnetic dipole moments and charge. Two electrons separated by a distance would repel electrically but it stands to reason they would rotate their spins so their magnetic poles were compatible. This would create magnetic attraction. At what distance would this attraction begin to overpower the repulsion?

Answer 1

As far as I understand, at no distance does this occur.

For macroscopic permanent dipoles, the magnetic dipole-dipole inteaction falls with distance cubed, so for charged permanent magnetic dipoles, if they are repelling each other at some distance, your only hope is to try to bring them closer. You'll probably get some induced electric polarization effects that might lead to some additional attraction, at some length.

But electrons are not like macroscopic permanent dipoles. From a classical perspective, once electrons stop moving relative to one another, they experience just repulsive force due to Coulomb's law; the spin-related effects aren't strong enough to produce a bound state for isolated electrons. In general, other matter must work to help localize electrons near each other (e.g. as in an atom) or mediate an effective bound state (e.g. phonon-mediated Cooper pairs in superconductors).

However, there is an interesting popular science article and associated journal article by M.N. Chernodub in which it is speculated one can "zap" space with a (extremely, extremely) powerful magnetic field to form Cooper pairs in vacuum. It might be fun to read this, even though it is far out of reach of today's technology.

Answer 2

Let's model an electron as a classical ideal point multipole with electric monopole moment $q$ and magnetic dipole moment $p$ (and no higher multipole moments) in CGS units. Then we can tell by dimensional analysis that the distance at which the magnetostatic attraction and the electrostatic repulsion cancel must be on the order of the ratio $p/q$. The exact distance depends on the geometry of the dipole moments' orientations relative to each other and to the spatial vector that connects them.

If I did the algebra right, then if the electrons are aligned like $\rightarrow \rightarrow$, then the attractive force between them is $6 p^2/r^4$, so the balancing distance is $\sqrt{6}\, p/q$. If the electrons are aligned like $\uparrow \downarrow$, then the attractive force is $3 p^2 / r^4$, so the balancing distance is $\sqrt{3}\, p/q$.

We can now go to a semiclassical model in which we plug in $p = \frac{1}{2} g_s \mu_B$, where $g_s$ is the Landé $g$-factor and $\mu_B$ is the Bohr magneton $e \hbar/(2 m_e c)$ (and ignore all other quantum effects). If we ignore corrections from QED (which is definitely the correct thing to do at this level of accuracy), then $g \approx 2$ and $p \approx \mu_B$, so $p / q = \mu_B / e = \hbar/(2 m_e c) = \alpha a / 2 = 1.9 \times 10^{-13}$ meters = 190 femtometers (where $\alpha$ is the fine-structure constant and $a$ is the Bohr radius). This is way smaller than the length scale at which quantum effects become important, which shows that this semiclassical approximation is not self-consistent, and therefore doesn't have much significance in the real world where quantum effects matter.

Answer 3

I don't think there's certain distance at which one 'cancels' out the other and if anything, the energy level would have to be influenced by a outside-acting factor...wouldn't it? Otherwise you have am anti-quantum mechanical phenomenon