ABSTRACT:
This paper shows a new mathematical algorithm that allows weak-force bosons to have mass and leaves photons massless while giving mass to neutrinos and other leptons.
The right side of equation 1 below is the Higgs vector used to ensure that photons don't have mass and that the weak-force gauge bosons have mass. This same vector also ensures that leptons will have mass except for neutrinos.
Neutrinos, however, are not massless. This fact indicates that the electroweak theory is not complete. To fix the theory we need something like the vector on the right side of equation 2:
This new vector compensates for whatever contributes to neutrino mass. If we use this new vector when determining masses for leptons, and, use the old vector for determining masses for gauge bosons, we end up with the status quo and the extra bonus of slightly massive left-handed neutrinos.
To theoretically justify this new vector we'll examine how the old vector was derived from a Lagrange potential. We will make a minor adjustment to this Lagrange potential without changing its value and its gauge translation invariance. The minor adjustment will allow a derivation of the new vector as well as the old. Here is the Lagrange potential in its original form:
Equations 3 through 6 demonstrate gauge translation invariance and lead to equation 7. Next, we take the derivative with respect to phi to acquire the minimum potentials:
At 11 and 12 above we have the minimum potentials we find in the Higgs vector at equation 1. To derive the new vector at equation 2 we do the following:
Two new terms are added to the potential that cancel each other, so the potential is the same. Once again gauge translation invariance is demonstrated. After all is said and done, we have two useful equations: 18 and 19. If we substitute the zero value at 18 into 19 we have the original Lagrange potential. Once again, we can take the derivative and derive the original Higgs vector. Or, we can take the derivative of equation 19 without the substitution:
At 23 we end up with two non-zero solutions. If phi is small, we have the first approximate solution. If phi is larger we have the second approximate solution. This is consistent with, say, an electron being more massive than its family neutrino. We now have what we need to create the new vector (see equation 2).
The next step is to show how this new vector is applied. When the mathematics is normally done, all terms containing h(x) are discarded at the end, but the solutions don't change if we discard h(x) early or leave it out of the vector. Doing so greatly streamlines the math. Thus the new vector becomes
There are three families of leptons. Since the same mathematics applies to all three, let's just focus on the electron family. The normal interaction Lagrangian for the electron family is
The problem with this Lagrangian is it assumes the left-handed neutrino has no mass, so we need to adjust the Yukawa coupling Ge to G.
Since the right-handed electron doesn't have a corresponding right-handed neutrino, the Greek letter nu with an R subscript has a zero limit at equation 30. After performing the matrix operations we get
Now let's set up some substitutions and define the modified Yukawa coupling G to include neutrino and electron masses:
The final results are below. At 37 we have the left-handed neutrino's mass. At 38 we have the electron's mass (notice how the adjusted Yukawa coupling coupled with the new groundstate equals the standard Yukawa coupling coupled with the original groundstate. Both terms equal the electron's mass.)
The forgoing exercise can be repeated for the other two families of leptons. To account for different masses, simply use different Yukawa couplings.
In conclusion, to give neutrinos mass requires a new Lagrange potential that can yield two field vectors: one for gauge bosons and one for leptons. Also, Yukawa couplings for massive neutrinos need to be added. When these requirements are met, the final solution shows that left-handed neutrinos do indeed have mass.
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