﻿ The Relativistic Angular Momentum, Magnetic Moment and Spin of an Electron
San José State University

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The Relativistic Angular Momentum,
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).

For an electron

#### 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.

## Relativistic Angular Momentum

In another study it was found that the relativistic angular momentum of a spherical particle of radius R and mass m0 spining at ω radians per second is given by

#### L = (m0cR)βm/(1 − βm²)3/2

where βm is average tangential velocity on the sphere.

The solution can be found in terms of λ=βm2/3 where λ is the solution to the equation

#### (1 − λ³) = σλ

where σ=(m0cR/L)2/3.

The first step toward a solution for an electron is the evaluation of the parameter σ. The immediate problem 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.

Thus

### σ = [(9.109x10−31)(3x108)(0.84x10−15)/(9.274)(0.527x10−34)]2/3 = [0.4697x10−3]2/3 = 0.6042x10−2 = 6.042x10−3

To get the solution for λ by iteration let λn be the n-th iterate. Then

#### λn+1 = [1 − σλn]1/3

The iteration can be started at λ0=1.

The iteration quickly converges to λ=0.997986 and thus βm=0.996980526.

This is the mean relative tangential velocity. The relationship between the mean and maximum tangentential veocities for a spherical ball at velocities far below the speed of light is

#### βm = (2/5) βmax

But at velocities close to the speed of light

#### βmax = βm

For the electron vmax is very close to the speed of light.

## Rotation Rate

This means an electron is rotating at a rate of

### ω = βmaxc/R = (0.996980526)(2.9979x108/(0.84x10−15) = 3.5582x1023 radians per second = 5.663x1022 times 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 comparable to the high rates found for nuclei in general; i.e., 4.74x1021 rotations per second. See Nuclear Rotation. .

The relativistic method of computing the rotation rate handled the problem that it might imply the matter of the material of the electron could be traveling faster than the speed of light. The general problem of the determination of rotation rates taking into account Special Relativity is dealt with in Relativistic Angular Momentum.

## Conclusion

Taking into account the relativistic nature of angular momentum the measured magnetic moment of a electron is consistent with it deriving from it being a rotating spherical electrostatic charge. Its computed rate of rotation is about 5.663x1022 times per second.