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Magnetic Moment and Spin of a Neutron |
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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. 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.
The magnetic moment of a neutron, measured in nuclear magneton units, is −1.9130. The nuclear magneton is defined as
where e is the unit of electrical charge, h is the reduced
Planck's constant, m_{p} 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
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
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
If the particle is a spherical shell with outer radius R_{o} and inner radius R_{i} and then the magnetic moment would be
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.
In another study it was found that the relativistic angular momentum of a spherical particle of radius R and mass m_{0} spining at ω radians per second is given by
where β_{m} is average tangential velocity on the sphere.
The solution can be found in terms of λ=β_{m}^{2/3} where λ is the solution to the equation
where σ=(m_{0}cR/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
The solution for λ is approximately λ= 0.3097 and thus β_{m}=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
Therefore for the neutron
This means a neutron is rotating at a rate of
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., 4.74x10^{21} rotations per second. See Nuclear Rotation. < !--
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
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 q_{N}=−2/3, R_{No}=1.1133 fermi, and R_{Ni}=0.3 fermi. For the neuton's inner positive shell q_{P}=+2/3, R_{Po}=0.3 fermi, and R_{Pi}=0. fermi. Thus
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.
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
The angular momentum L of a neutron is then
The ratio of this to the reduced Planck's constant is
If this is equal to the Bohr-Mottleson term (I(I+1))^{½}
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
If the neutron is considered to consist of two shells of q=±4/3 instead of q=±2/3 then the product would be
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 = 1.1659x10^{23} radians/sec = 1.8605x10^{22} rotations per second
The tangential velocity v at the equator of a rotating sphere of radius R is ωR. In the case of a neutron it is
Relative to the speed of light this is
So the computed rate of rotation of the neutron is compatible with the Special Theory of Relativity.
(To be continued.) -->
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 1.847x10^{22} times per second.
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