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The Momenta and Spins of Neutrinos


In the empirical analysis of the beta decay of nuclides it was found that there was missing energy. Physicists were confronted with the possibilty that the hallowed principle of the conservation of energy might not hold in beta decay. Wolfgang Pauli came up with a plausible explanation. He conjectured that the missing energy was being carried off by a neutral particle. That fit with the general intuition of physicists and Pauli's conjecture was accepted. Enrico Fermi gave the neutral particle the name neutrino, the little neutral one.

For a very long time there was no evidence for the existence of neutrinos except the missing energy in beta decays. Finally in a huge tank of water in an underground cavern the interactions of neutrinos with ordinary matter were detected. One bit of evidence for their existence of comes from their high speed. While no material object can travel faster than the speed of light in a vacuum some can travel faster than light travels in the local material such as water. When they do so they emit radiation, called Cerenkov radiation, which has been detected.

Neutrino Flavors

The neutrino was somehow associated with the electron. When the electron-like muon and tauon particles were were found it was speculated that they had their own particular varieties of neutrino-like particles. These became known as flavors.

From the work of P.A.M. Dirac it was believed that each flavor of neutrino had its own anti-particle. So there supposedly are six different types of neutrino.

The Quantum Principle of
a Minimum Momentum

Linear momentum L is a function of mass m and velocity v. The fundamental principle of quantum theory is that a particle has a quantum of momentum for each degree of freedom it has. The quantum of momentum for linear motion is ½h, where h is Planck's Constant divided by 2π.

For rotorary motion the situation is somewhat different. Aage Bohr and Dan Mottelson found that an object with d degrees of freedom has angular momentum of (d(d+1))½h. Therefore for a spherical object with one degree of freedom the minimum angular momentum would be (√2)h. This is taken note of but not utilized in the computation which follows.

Classical and Relativistic
Linear Motion of Particles

Classically the linear momentum p of a partical of mass m0 traveling at a velocity v is

p =m0v

If p is necessarily equal to ½h then classically, without limit,

v = ½h/m0

For a particle of sufficiently small mass this could produce, contrary to the Special Theory of Relativity, a velocity greater than c, the speed of light in a vacuum. This violation of Special Relativity does not mean particles do not have literal spin; it means classical formulas do not apply for situations involving velocities compable to the speed of light. To avoid this violation of Special Relativity a formula for relativistic momentum must be used.

The way to derive the formula for relativitic momentum is to construct the relativistic Lagrangian for the physical system and take the derivative of it with respect to v. That quantity is the relativitic momentum.

The combined rest mass energy and kinetic energy of a particle is given by

E = m0c²/(1−β²)½
β = v/c

The rest-mass energy of the particle is m0c². There is no potential energy function for the free particle. Therefore the relativistic momentum, which is the partial derivative of the Lagrangian with respect to v, is equal to

(∂E/∂v) = (−½)m0c²(1−β²)−3/2(−2v/c²)
which reduces to
(∂E/∂v) = m0v(1−β²)−3/2

Thus relativistic momentum p can be expressed as

p = (m0c)β(1−β²)−3/2

To solve for β as a function of p multiply both sides of the above equation by (1−β²)3/2 and divide by p to obtain

(1−β²)3/2 = ((m0c/p)β

Now raise both sides to the (2/3) power and thus obtain

(1−β²) = ((m0c/p)2/3β2/3

Let β2/3=λ so β23. Furthermore let (m0c/p)2/3 be denoted as σ.

The above equation then takes the form

(1 −λ3) = σλ

This equation is easily solved numerically to any degree of accuracy.

That process starts with computing the value of the parameter σ. For very small values of σ the solution looks like

Obviously the solution for λ and hence for β cannot possibly be greater than unity.

For rotorary motion the parameter σ takes the form

σ = (m0cR/L)2/3

where R is the radius of the spherical particle and L is the quantum of angular momentum ½h.

The Mass of the
Electron's Neutrino

For a long time it was thought that a neutrino was massless. Then some experiments indicated it was not massless but could not give a definite estimate of its mass. Experimenters have been working hard to come up with a definite estimate. A recent estimate puts the value of m0 at the near infinitesimal figure of 0.06 of an electron volt. In kilograms this is

m0 = (0.06)1.7826627x10−36 kg
m0 = 1.07x10−37 kg

The Size of a Neutrino

It is quite plausible that the mass densities of particles are essentually the same. Then the ratio of the volumes of two types of particles would be the same as the ratio of their masses. The volumes are proportional to the cubes of their radii so the ratio of their radii would be equal to the cube root of the ratio of their masses.

For the neutrino and the proton this is

Rν/RP = (1.07x10−37)/(1.672623x10−27))1/3
Rν/RP = (0.0697x10−9)1/3
Rν/RP = (0.3999x10−3) = (3.999x10−4)

Since RP = 0.84x10−15 m

Rν = 3.359x10−19 m.

The Spin of an
Electron's Neutrino

The crucial variable is the tangential velocity relative to the speed of light, β, at the equator of aherical particle. This is given by

β = ωR/c.

As derived above, the equation to be solved is

(1 −λ3) = σλ

where λ=β2/3

The Rotation Rate
of a Neutrino

The parameter σ for a neutrino is given by

σ = (m0cRν/L)2/3
σ = ((1.07x10−37)(2.9979x108)(3.359x10−19)/(5.27x10−35))2/3
σ = (0.20446x10−12))2/3
σ = (0.3471x10−8) = (3.471x10−9)

The solution for λ is essentually 1 and likewise βm and βmax are both essentually 1. That gives the rotation rate of a neutrino as ω=c/R; i.e.,

ω = (2.9979x108)/(3.359x10−19)
ω = (8.925x1026) radians per second
= (1.42x1026) times per second.

This is an order of magnitude higher than the fantastically high rotation rate of an electron.

The Different Flavors
of Neutrinos

There is evidence that the neutrinos of different flavors transform into each other. That implies that they have equal masses and hence the same radii. That implies they have the same rotation rates.

But a muon has a mass 207 times that of an electron, If the neutrino of the muon has the same mass density as the neutrino of the electron then the ratio of their radii is given by

Rμ/Re =` (207)1/3 = 5.9155

On the other hand, the tauon has a mass 3477 times of an electron. thus the ratio of the their radii is

Rτ/Re =` (3477)1/3 = 15.15

These imply

σμ = ( 5.9155)2/3σν = (3.2709)(3.471x10−9) = (1.13531x10−8)
στ = ( 15.15)2/3σν = (6.1227)(3.471x10−9) = (2.1252x10−8)

These would imply slower rates of rotation for their corresponding neutrinos but the βmax for the neutrino of an electron is so close to 1 those ratios make little difference.

Change of Shape
of Particles

Of course βm not is not exactly what is needed. What is needed that a spinning particle satisfies Special Relativity is βmax not exceeding 1. To get this the change in the shape of the particle at higher rates of rotation has to be taken into account. There is a contraction in length a higher speeds. This means the circumference of the sphere contracts more at the equator than those at higher latitudes. Consequently at higher rotation rates the shape of the particle becomes more like a cylinder.

For a cylinder βmaxm.


Taking into account the relativistic nature of angular momentum the spin of a neutrino can be derived from it being a rotating spherical object. Its computed rate of rotation is about 1.42x1026 times per second as compared with that of an electron of 4.195x1025 times per second.

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