Electron to electron interaction

Question

I am interested to know if it would be possible to calculate the magnetic attraction between two electrons at zero space between their magnetic poles against electrostatic repulsion between them in simple words can magnetic force of spin hold together two electrons nevertheless there is a electric repulsion.Thanks.

Answer 1

As far as we know, electrons are point particles, so you don't want to try to calculate anything when they are at zero separation. You'd just get infinite forces.

Classically, the magnetostatic dipole-dipole interaction dominates over the electrostatic interaction when two electrons are separated by less than about one Compton wavelength. This is because the former varies as the inverse fourth power of the separation, while the latter varies as the inverse square of the separation.

Magnetically, an electron behaves like a point dipole with magnetic moment equal to the Bohr magneton,

$$\mu_B=\frac{e\hbar}{2m},$$

if we neglect tiny QED corrections to the magnetic moment. Here $m$ is the mass of the electron.

The force exerted by one magnetic dipole $\mathbf{m}_1$ on another magnetic dipole $\mathbf{m}_2$ is

$$\mathbf{F}=\frac{3\mu_0}{4\pi r^5}\left[(\mathbf{m}_1\cdot\mathbf{r})\mathbf{m}_2+(\mathbf{m}_2\cdot\mathbf{r})\mathbf{m}_2+(\mathbf{m}_1\cdot\mathbf{m}_2)\mathbf{r}-\frac{5(\mathbf{m}_1\cdot\mathbf{r})(\mathbf{m}_2\cdot\mathbf{r})}{r^2}\mathbf{r}\right]$$

where $\mathbf{r}$ is the vector from $\mathbf{m}_1$ to $\mathbf{m}_2$. (See Wikipedia.) When the dipoles are aligned along the same axis, with separation $d$, this formula gives a magnetostatic attractive force of magnitude

$$F_\text{m}=\frac{3\mu_0 m_1 m_2}{2\pi d^4}.$$

In the case of two electrons, $m_1=m_2=\mu_B$, so

$$F_\text{m}=\frac{3\mu_0e^2\hbar^2}{8\pi m^2d^4}.$$

The electrostatic repulsion of the two electrons at separation $d$ is given by Coulomb's Law,

$$F_\text{e}=\frac{e^2}{4\pi\epsilon_0d^2}.$$

Note that $F_\text{m}\sim 1/d^4$ while $F_\text{e}\sim 1/d^2$. There is a critical distance

$$d_\text{crit}=\sqrt\frac{3\epsilon_0\mu_0\hbar^2}{2m^2}=\sqrt{\frac{3}{2}}\frac{\hbar c}{m}=\sqrt{\frac{3}{2}}\overline\lambda_C,$$

where $\overline\lambda_C$ is the reduced Compton wavelength of the electron, at which the two forces are equal in magnitude. Farther apart, electrostatic repulsion dominates. Closer together, magnetostatic attraction dominates. (Note that the critical separation is thus unstable. The electrons don't want to stay at this separation.)

The total force can be written in the dimensionless form

$$\frac{F}{F_C}=\left(\frac{\overline\lambda_C}{d}\right)^2-\frac{3}{2}\left(\frac{\overline\lambda_C}{d}\right)^4$$

where

$$F_C=\frac{e^2}{4\pi\epsilon_0\overline\lambda_C^2}$$

is the electrostatic force at one reduced Compton wavelength.

A graph of the total force looks like this

where the horizonal axis is $d/\overline\lambda_C$ and the vertical axis is $F/F_C$. The maximum repulsion occurs at $d=\sqrt{3}\,\overline\lambda_C$ and has magnitude $F_C/6$.

Since the Compton wavelength is a standard measure of where quantum effects start to be important, this classical analysis can't be taken too seriously. But it indicates that spin-spin interactions are important at short distances. In quantum field theory, this is apparent because the scattering of two charged spin-1/2 particles is different from the scattering of two charged spin-0 particles.