# The spin

## Spin of the photon

According to mass-energy equivalence and to the Planck

$E=h\nu$

and Einstein

$E=mc^2$

formulae, the photon may be considered as a small mass

$m=\frac{h\nu}{c^2}$

concentrated in a ring of radius r. It moves at the velocity c of light and rotates at the velocity of light. The linear and rotational velocities add according to the Lorentz transformation thus limiting the maximum velocity to c.

$R=\frac{c}{2\pi \nu}$

One deduces the angular momentum of the photon:

$L=mcR=\frac{mc^2}{2\pi\nu}=\frac{h\nu}{2\pi\nu}=\hbar$

The photon, according to Rocard (Rocard, Y, Thermodynamique, Masson, Paris, 1957, p. 250) may thus be modelled as a rotating ring of mass

$h\nu/c^2$

## Spin of the electron

The electron, has a mass at rest and a variable velocity; therefore the calculation is more complicated than for the photon. MacGregor, in his book, The Enigmatic Electron (MacGregor M.H., The Enigmatic Electron, Kluwer, Dordrecht, 1992), shows that one has to take into account the relativistic variation of the mass due to the rotation. He assumes that the velocity at the equator of the spherical electron is equal to c, the light velocity. Therefore, according to relativity, the density varies in the electron from center to equator. Let us compute the mass m of a sphere (ball with no empty interior) with radius R in relativistic rotation at velocity $v=\omega r$ Assuming that the electron is a rigid solid, with infinite mechanical strength, the angular velocity $\omega$ is constant along the radius. The relativistic specific mass is $\rho(r)$ at a distance r from the center of the axis of rotation and $\rho_0$ on the axis. If the radius of the electron is unchanged We have:

$\rho(r) =\frac{\rho_0}{\sqrt{1-\frac{v^2(r)}{c^2}}}$

Let us consider an elementary cylinder of height h. The linear velocity of rotation at a distance r from the axis is

$v=\omega r$

Being c at the equator, the angular velocity is $\omega=c/R$

The mass of the sphere is then:

$m=\int_0^R \rho(r)2\pi(2h)rdr =2\pi\rho_0 R^2 \int_0^R\frac{2\sqrt{R^2-r^2}}{\sqrt{1-\frac{(\omega r)^2}{c^2}}}d\frac{r^2}{2R^2} =2\pi\rho_0 R^3 \int_0^R\frac{\sqrt{1-\frac{r^2}{R^2}}}{\sqrt{1-\frac{c^2 r^2}{R^2 c^2}}}d\frac{r^2}{R^2} =2\pi\rho_0 R^3 \int_0^Rd\frac{r^2}{R^2}=2\pi\rho_0 R^3$

The same method applies to the moment of inertia:

$I=\int_0^R r^2 dm=\int_0^R r^2\rho_0(r)2\pi(2h)rdr =2\pi\rho_0 R^3 \int_0^R\frac{2\sqrt{1-\frac{r^2}{R^2}}}{\sqrt{1-\frac{r^2}{R^2}}}d\frac{r^4}{4R^4}

=\pi\rho_0 R^3 \int_0^R d\frac{r^4}{R^4}=\pi\rho_0 R^5=\frac{1}{2}mR^2$

This value is not very different from the classical value:

$\frac{3}{5}mR^2$

The calculation is intractable in the general case but simplifies crucially in this special case.

The relations of Einstein and Planck give the eigen frequency of the electron:

$\nu=\frac{mc^2}{h}$

The vibration may be assimilated to the angular rotation velocity:

$\omega=2\pi\nu=2\pi\frac{mc^2}{h}$

$R= \frac{c}{\omega}=\frac{\hbar}{mc}=R_C$

MacGregor obtains in this manner the intrinsic angular momentum of the electron:

$L=I\omega=\frac{1}{2}mR_C^2 2\pi\nu=\frac{1}{2}\frac{h\nu}{c^2}\left(\frac{c}{2\pi\nu}\right)^22\pi\nu= \frac{1}{2}\hbar$

in accord with experiment, giving a spin de 1/2 in units of $\hbar$.

This calculation, as the preceding shows that the angular momentum depends on the mass distribution in the particle still unknown from experiment. The same dependence on the charge distribution happens for the magnetic moment.

## Magnetic moment

The spin is related to the magnetic moment of a charged particle such as an electron, a proton or even a neutron. The rotation of the electric charges generates a circular current giving a magnetic moment. A simple assumption is that the electric current is a circle but it may well be distributed in the whole volume of the particle. The corresponding magnetic moment is given by the formula

$\mu=IS=\frac{e\pi R^2}{T}$

where the period is

$T=\frac{2\pi R}{v}$

Finally

$\mu=\frac{e\pi R^2v}{2\pi R}=\frac{evR}{2}$

The angular momentum being

$L=MvR$

the magnetic moment is

$\mu=\frac{eL}{2M}$

According to the Bohr postulate, the angular momentum is quantized and equal to $\hbar$ for a circular trajectory which is the case for the electron on its orbit in the hydrogen atom. But the electron itself has an intrinsic orbital momentum which is half the orbital momentum although the orbital and intrinsic magnetic moments of the electron are equal in the first order. The orbital and intrinsic magnetic moments of the electron are measurable but it does not seem possible to measure their angular momenta.

Then we have for the orbital magnetic moment or Bohr magneton

$\mu_B=\frac{e\hbar}{2M}$

and for the intrinsic electron magnetic moment

$\mu_e=g\frac{e\hbar}{4M}=g\mu_B$

where g=2.00232. g is called g-factor, Landé factor or gyromagnetic ratio. g of the orbital or Bohr magneton is equal to one. When g ≠ 1, g is called anomalous. The anomaly is defined by a = (g - 2)/2. The intrinsic magnetic moment of the electron is 0.1% larger than the orbital magnetic moment. The anomaly of the electron being small one may conclude that the electric charges are situated on the equator. The proton and the neutron have large anomalies g_p = 2.8 atomic units (au) and g_n = -1.9 au. It means that the electric charges and/or the masses are differently distributed.