﻿ The Magnetic Moment and Spin of an Electron
San José State University

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The Magnetic Moment
and Spin of an Electron

## Background

In 1922 the physicists Otto Stern and Walther Gerlach ejected a beam of silver atoms into a sharply varying magnetic field. The beam separated into two parts. In 1926 Samuel A. Goudsmit and George E. Uhlenbeck showed that this separation could be explained by the valence electrons of the silver atoms having a spin that is oriented in either of two directions. It has been long asserted that this so-called spin is not literally particle spin. It is often referred to as intrinsic spin whatever that might mean. However here in the material that follows it is accepted that the magnet moment of any particle is due to its actual spinning and the spin rate can be computed from its measured magnetic moment.

## Magnetic Moments

The magnetic moment of an electron, measured in Bohr magneton units, is −9.2740. The Bohr magneton is defined as

#### ½he/me in the SI system and ½he/(mec) in the cgs system

where e is the unit of electrical charge, h is the reduced Planck's constant, me is the rest mass of an electron and c is the speed of light. Thus the magneton has different dimensions in the different systems of units. In the SI system it has the dimensions of energy per unit time (Joules per second).

The magnetic moment μ of a charge Q revolving about a point R distance away at a rate of ω radians per second is given by

#### μ = QR²ω = qeR²ω

where the constant e is the unit of electrostatic charge and q is the charge in electrostatic units.

When the charge is distributed uniformly over a spherical surface or volume the formula takes the form

#### μ = qekR²ω

where k=2/3 if it is a spherical surface and k=2/5 if it is a spherical volume.

For a particle then in the SI system

## Angular Momentum

Note that the moment of inertia J of a particle is equal to kmR² where m is the mass of the particle. Thus the angular momentum L of the particle is given by

#### L = Jω = (kmR²)(μ/q)(½h)/(kmeR²) and therefore L/(½h) = (μ/q)(m/me)

For an electron q=1 and m=me thus

#### L/(½h) = 9.274

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Aage Bohr and Ben Mottleson found nuclei rotations satisfy the h√I(I+1) rule, where I is an integer representing the number of degrees of freedom of the rotating object. The number of degrees of freedom for a sphere is a bit uncertain. It could be three for rotations about three orthogonal axes. It could be just one for a charged sphere.

The angular momentum according the Bohr-Mottelson Rule is

Thus

#### h(I(I+1))½ = 9.274((½h) and therefore ( I(I+1))½ = ½(9.274) = 4.637 which means (I(I+1)) = 21.5 ≅ 4(5)

Thus, this suggests that the number of degrees of freedom of the charged spherical electron is 4. There is evidence that the number of degrees of freedom for a charged sphere is 1 or 3. Four degrees of freedom would require a torus. One degree of freedom would be rotation as a vortex ring. A second degree of freedom would be rotation as a wheel. Another two degrees of freedom would come as the flipping of a coin about two orthogonal axes.

## Rotation Rate

First the case of a spherical electron charge will be investigated. From a previous equation for a spherical ball

### ω = μ(½h)/(qkmpR²)

The question is what would be the radius of an electron? Some physicists are prone to declaring particles to be spatially points despite the fact a charged point particle would have infinite energy. Thus there is not enough energy in the entire Universe, even if all of its mass were converted to energy, to create a single charged point particle. Any evidence for a point particle is equally well evidence for a charged spherical figure: surface, shell or ball.

It is plausible, perhaps even likely, that the radius of a charged particle is a function only of its charge. This would make the radius of an electron the same as the radius of a proton; i.e. 0.84 fermi.

From the previous equation for the rotation rate of a charged sphere

### ω = 9.274(0.527x10−34)/((2/5)(9.109x10−31)(0.84x10−15)²) = 4.8874x10−34/(2.5709x10−61 = 1.901x1027 radians/second = 3.021x1026 rotations per second

This is an increditably high rate but it is what it would have to be to generate its measured magnetic moment. It is similar to the increditably high rates found for nuclei in general; i.e., 4.74x1021 rotations per second for a deuteron but 5 orders of magnitude higher. See Nuclear Rotation.

## Compatibility with Special Relativity

Classically the tangential velocity v at the equator of a rotating sphere of radius R is ωR. In the case of an electron it would be

#### v = ωR = (1.901x1027)(0.84x10−15) = 1.5968x1012 m/s.

Relative to the speed of light this is

#### v/c=1.5968x1012/3x108 = 5.323x103

so the tangential speed cannot be determined from the classical formula. The above computed rate of rotation of the electron is not compatible with the Special Theory of Relativity. The problem stems from the classical quantum theoretic principle

#### L = Jω = nhand hence ω = nh/J

Therefore, according to this principle, the smaller the moment of inertia J of a particle the faster is its rate of rotation without any limit.

The relativistic method of computing the rotation rate can be handled but not here. It along with the possibility of a torus shaped electron will be dealt with elsewhere. The general problem of the determination of rotation rates taking into account Special Relativity is dealt with in Relativistic Angular Momentum.