|San José State University|
& Tornado Alley
to the Spins of its Component Quarks
There are many physicists who assert that subatomic particles do not really revolve around an axis. The term particle spin was coined because such a phenomenon would account for the splitting of a beam of particles upon passage through a strongly spatially varying magnetic field. Spin is said to be metaphorical rather than real.
For a bit of charge of dq that is spinning at a rate of ω (radians per second) or ν (turns per second) the effective current is di=dqν=dqω/(2π). The magnetic moment generated by the motion of this charge in its circular trajectory is the product of the effective current times the area surrounded by the path of the charge. The area of the loop which the current surrounds is πr². Thus
where v is the tangential velocity of the charge. This is analogous to the angular momentum of a particle. The angular momentum involves the mass of the particle rather than the term (dq/2).
The total magnetic moment of a charge Q=∫dq is then
On the other hand, the angular momentum L of a mass M=∫dm is given by
It is worthwhile at this point to establish the relationship between the magnetic moment and angular momentum. Let angular momentum be denoted by L. The if L is divided by the mass of the particle and the result multiplied by the charge divided by 2 the result is the magnetic moment μ.; i.e.,
Angular momentum is quantized so magnetic moment is inversely proportional to mass. Thus the magnetic moment of an electron should be 1836 times the magnetic moment of a proton. The magnetic moment of an electron is 9284.76×10−27 J/sec whereas that of a proton is 14.106×10−27 J/sec. The ratio of these two figures is 658.2135. This is 1836/2.79.
If the mass and charge are distributed uniformly so dq=σdm and σ=Q/M then
Another way of looking at the problem is that the above relation give the quantum of angular momentum; i.e.
For the electron the quantum of angular momentum is
The value the reduced Planck's constant is 1.0545718×10−34 J-sec.
This is amazingly close. However what theory requires is ½
h. The problem is resolved
by introducing a g-ratio equal to 2 in the analysis. One could equally well account for the result by saying the minimum spin
quantum number is 2 instead of 1.
For the proton the quantum of angular momentum is
The ratio of this angular momentum to ½
h is 5.586. If a Bohr-Mottelson √(I(I+1)) rule is involved, that ratio
can be accounted for by I=5 since √(5*6)=5.4772. The whole problem of the spin of a proton is covered by an article in
the July 21, 2014 issue of Scientific American.
In non-relativistic physics angular momentum is Jω, where J is the moment of inertia based upon the distribution of mass. It is presumed here that the mass and electrical charge are located in a solid spherical charge of uniform density. The moment of inertia of mass M distributed over a solid sphere of radius r is (2/5)Mr². The charge radius of a proton is 0.875 fermi=8.75×10−16 meter. Thus
Thus the apparent angular rate of rotation is
This is about 80,000 billion billion rotations per second, or equivalently, 80 billion trillion rotations per second.
The question immediately arises as to how high is the equatorial tangential velocity of the proton relative to the speed of light. The velocity is given by rω. Since the charge radius 0.875 fermi
Since the speed of light is only 3×108 m/sec this figure is incompatible with Special Relativity. To make the analysis compatible with Special Relativity it is necessary to redetermine the angular momentum taking into acount the relativistic adjustment of mass with velocity on a rotating sphere.
The analysis of angular momentum taking into account the conditions of Special Relativity has not been worked out and so some of the following may not be correct but it illustrates how a rate of rotation of a proton may be determined which is compatible with Special Relativity.
First consider a thin ring of radius R and rest mass M0 rotating at an angular rate of ω. Its tangential velocity is then ωR. Its apparent mass is then
where c is the speed of light.
Thus the angular momentum of the ring, instead of being M0R²ω is
For a rotating spherical surface the angular momentum would have to be integrated over the rings of radius R=R0cos(φ), where φ is the latitude angle ranging from −π/2 to +*pi;/2. Since the angular momentum J→∞ as ωRrarr;c for any finite J there is an ω such that ωR<c and the angular momentum of the rotating spherical surface is equal to the given J.
If the radius of the spherical surface is equal to the charge radius of the proton then the the surface area A and volume V of the proton are
The surface density σ of the mass of 1.6727×10−27 kg distributed over this surface area is
The volume density μ of the mass of 1.6727×10−27 kg distributed over the charge volume is
(To be continued.)
HOME PAGE OF Thayer Watkins,