﻿ The Magnetic Moment and the Spin of a Proton
San José State University

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The Magnetic Moment
and the Spin of a Proton

## 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. 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 a proton, measured in nuclear magneton units, is +2.79285. 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 neutron is −1.9130. 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

## Angular Momentum

Note that the moment of inertia J of the 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 = μ[(½h)(m/mp)/q or, equivalently L/(½h) = (μ/q)(m/mp)

For a proton q=1 and m=mp thus

#### L/(½h) = 2.79285

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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 = 2.79285((½h) and therefore ( I(I+1))½ = ½(2.79285) = 1.396425 which means (I(I+1)) = 1.95 ≅ 2

Thus, clearly the degree of freedom of the charged spherical proton is 1. It has to be that or magnetic moments would not be measurable.

## Rotation Rate

From a previous equation

### ω = μ(½h)/(qkmpR²) = 2.79285(0.527x10−34)/((2/5)(1.6726x10−27)(0.84x10−15)²) = 1.47183195x10−34/(0.472075x10−57 = 3.117795x1023 radians/second = 4.962x1022 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; i.e., 4.74x1021 rotations per second for a deuteron. See Nuclear Rotation.

## Compatibility with Special Relativity

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

#### v = ωR = (3.117795x1023)(0.84x10−15) = 2.6189x108 m/s.

Relative to the speed of light this is

#### v/c=2.6189x108/3x108 = 0.873 so the relativisticcorrection factor is γ = 1/(1 −(v/c)²)½ = 0.4877

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

## Prediction

With a knowledge that a spherical charge has only one degree of freedom the coefficient of the Bohr-Mottelson term would be √2. That would give a magnetic moment for a proton of 2√2=2.8284 magnetons, a deviation from experimentally measured value of only 1.27 percent.

## Conclusion

The measured magnetic moment of a proton is consistent with it deriving from it being a rotating spherical electrostatic charge. Its computed rate of rotation is about 5x1022 times per second.