﻿ !DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0//EN"> The Relativistic Angular Momentum, Magnetic Moment and Spin of a Neutron
San José State University

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The Relativistic Angular Momentum,
Magnetic Moment
and Spin of a Neutron

## Background

In 1922 the physicists Otto Stern and Walther Gerlach ejected a beam of silver ions 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 charged ions 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. However here 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 a neutron, measured in nuclear magneton units, is −1.9130. The nuclear magneton is defined as

#### ½he/mp in the SI system and ½he/(mpc) in the cgs system

where e is the unit of electrical charge, h is the reduced Planck's constant, mp is the rest mass of a proton 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.

The magnetic moment of a proton is 2.79285. The ratio of these two magnetic moments is −0.685, intriguingly close to −2/3. There is only a 2.7 percent difference. This suggests that the ratio of the intrinsic magnetic moments of the neutron and proton might be precisely −2/3.

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

#### qekR²ω = μ[½he/mp] but the constant e cancels out, so qkR²ω = μ(½h)/mpand hence ω = μ(½h)/(qkmpR²)

If the particle is a spherical shell with outer radius Ro and inner radius Ri and then the magnetic moment would be

#### μ = qek(Ro²−Ri²)ω

For a neutron there are two spherical shells, an outer negative charge and an inner positive charge.

This makes the relativistic analysis of the spin of a neutron considerably more complicated than that of a proton. It requires the computation for a negative shell extending from the outer radius down to R=0 then the computation for a negative charge extending from the inner radius down to R=0. The angular momentum for the shell of negative is the difference of those two figures. The angular momentum for the inner shell of positive charge can be based on only the inner shell radius.

Rather than attempt this complicated procedure what will be done below is compute the rotation rate assuming the outer negative charge extends from the outer radius of the neutron to its centeR.

## 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 a neutron is the evaluation of the parameter σ. The radius of a neutron; i.e. 1.1133 fermi.

Thus

### σ = [(1.6749x10−27)(2.9979x108)(1.1133x10−15) /(1.913)(0.527x10−34)]2/3 = [5.5449]2/3 = 3.1328

The solution for λ is approximately λ= 0.3097 and thus βm=0.1724

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) βmaxhence βmax = (5/2)βm

Therefore for the neutron

## Rotation Rate

This means a neutron is rotating at a rate of

### ω = βmaxc/R = (0.431)(2.9979x108/(1.1133x10−15) = 1.1606x1023 radians per second = 1.847x1022 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. < !--

## Rotation Rate

Let the characteristics of these outer and inner shells be denoted by subscripts of N and P for negative and positive charges. The above equation for the spin rate then becomes

## Charge Structure of the Neutron

According to the quark model of a neutron it consists of two down quarks each with a charge of −1/3 electrostatic units and one up quark with a charge of 2/3 of an electrostatic unit. The two down quarks are equivalent to one spherical shell with a charge of −2/3 electrostatic unit. Thus, for the neuton's outer negative shell qN=−2/3, RNo=1.1133 fermi, and RNi=0.3 fermi. For the neuton's inner positive shell qP=+2/3, RPo=0.3 fermi, and RPi=0. fermi. Thus

### ω = −1.9130(0.527x10−34)/[(2/5)(1.6749x10 −27) ((−2/3)(1.1133x10−15)²−(0.3x10−15)²) +(2/3)(0.3x10−15)²)] = −1.0082x10−34/ [(0.669x10 −27)(−0.6463x10−30)] = −1.0082x10−34/(0.4324x10 −57) = 2.3318x1023 radians/sec = 3.711x1022 rotations per second

This is an increditably high rate but it is what it has to be to generate its measured magnetic moment. It is similar to the increditably high rates found for nuclei in general. See Nuclear Rotation.

## Angular Momentum

Note that the moment of inertia J of the particle is equal to kmR² where m is the mass of the particle.

The mass of a neutron is 1.674929x10 −27 kg. With a radius of 1.1133x10 −15 m its moment of inertia J is

### J = 1.674929x10 −27(2/5)(1.1133x10 −15)² = 1.674929x10 −27(0.4958x10 −30) = 8.304x10 −58 kg m²

The angular momentum L of a neutron is then

### L = Jω = (8.304x10 −58)(2.3318x1023) = 1.93632x10−34 kg m²/sec

The ratio of this to the reduced Planck's constant is

#### L/(h) = 1.93632x10−34/(1.054x10−34) = 1.83721

If this is equal to the Bohr-Mottleson term (I(I+1))½

#### I(I+1) = (1.83721)² = 3.3753

No integer generates this product. If the neutron is considered to consist of two shells of q=±1 instead of q=±2/3 then the product would be

#### I(I+1) = (1.2248)² = 1.5001

If the neutron is considered to consist of two shells of q=±4/3 instead of q=±2/3 then the product would be

#### I(I+1) = (1.4142)² = 2.0

So the positive and negative charges are twice what the quark model proposes. This may be another instance of the g-ratio phenomena. In any case it means the rate of rotation of a neutron is one half of what was computed above; i.e., ω = 1.1659x1023 radians/sec = 1.8605x1022 rotations per second

## Compatibility with Special Relativity

The tangential velocity v at the equator of a rotating sphere of radius R is ωR. In the case of a neutron it is

### v = ωR = (1.1659x1023)(1.1133x10−15) = 1.298x108 m/s.

Relative to the speed of light this is

#### v/c=1.298x108/3x108 = 0.433 so the relativisticcorrection factor is γ = 1/(1 −(v/c)²)½ = 0.9016

So the computed rate of rotation of the neutron is compatible with the Special Theory of Relativity.

(To be continued.) -->

## Conclusion

The measured magnetic moment of a neutron is consistent relativistically with it deriving from it being a rotating spherical electrostatic charge. Its rate of rotation is about 1.847x1022 times per second.