Curing Divergences without Supersymmetry and Renormalization

Abstract:

Supersymmetry or ad hoc methods such as renormalization are often used to tame infinities that result from divergent functions in quantum physics. Although SUSY particles have yet to be discovered and may be too massive to fulfill their purpose, and, renormalization seems to lack mathematical rigor. Here we offer an alternative method that employs the least-action and Heisenberg uncertainty principles.

Imagine a Lagrangian with divergent terms. One strategy is to renormalize it. Simply discard the divergent terms, especially if they are infinite. However, Paul Dirac had this to say about such methods: "I must say that I am very dissatisfied with the situation because this so-called 'good theory' does involve neglecting infinities which appear in its equations, ignoring them in an arbitrary way. This is just not sensible mathematics. Sensible mathematics involves disregarding a quantity when it is small – not neglecting it just because it is infinitely great and you do not want it!"

Another strategy is to add superpartners that each have the same mass as their respective standard-model counterparts but make an opposite contribution to the Lagrangian. As a result, the divergence vanishes. Albeit, there is a slight problem: the symmetry of Super-symmetry is broken--the superpartners are believed to be more massive than their standard-model partners. This deflates the balloon of vanishing divergences. To make matters worse, there is a complete and total lack of empirical evidence supporting these superpartners.

If renormalization seems like bad math and SUSY particles are nowhere to be found, what other options are there? How about the least-action and Heisenberg uncertainty principles? Let's first examine the least-action principle:

A particle typically takes the shortest path possible between two points. For that to happen, delta-s, at equation 1, cannot be a large, divergent quantity. It should be zero units of action or time multiplied by energy. However, the following is true:

Line 3 shows that time multiplied by energy is greater than or equal to h-bar. To get delta-s to equal zero requires steps 4 through 6:

At 7 we set up another substitution. The final equations are 8 and 9 below:

Equations 8 and 9 show why there's a least action principle and why energy is generally conserved. Suppose we have a conserved energy L. The divergent energy, delta-E, can be interpreted as energy borrowed from the vacuum. Because it's borrowed, it must vanish within time delta-t. The larger this energy, the shorter its lifespan. As a result, the energy L that you start with is the energy you end up with. It is conserved. Also, the action is the least action.

At equation 10 we have a Lagrangian where there is no borrowed energy. Because no energy is borrowed, time delta-t is infinite. In other words, this scenario can last indefinitely and create the impression that energy is always conserved.

At equation 11 we have the opposite extreme: a Lagrangian that diverges to infinity. The good news is delta-t is zero, which shows that infinite borrowed energy does not exist. We can also infer that large borrowed energies exist for too short of a time to be meaningfully observed and measured, so the energy we do observe and measure is small by comparison. Thus, renormalization works despite its ad hoc nature because nature wipes out divergences by means of the uncertainty principle and least action. The only time it is appropriate to keep the divergent terms is when divergent energy is added to the system and not borrowed from nothing.

Now, let's suppose L is a Lagrangian for vacuum energy (see equation 12). A Higgs boson (m-sub-H) pops into existence and has a lifespan of t-sub-H. A too-large Higgs mass would have a lifespan too short to provide a meaningful opportunity to observe it, so the mass we are most likely to observe is a smaller mass.

More examples: Equation 13 below takes into account multiple particles. Equation 14 takes into account a Lagrangian or function with multiple terms and parameters.

Since delta-s must be zero to minimize the action, then delta-s along D dimensions must also be zero. Further, both delta-s and s have units of momentum multiplied by position. If we integrate over position and/or momentum space, the following must be true:

The uncertainty of knowing a particle's position is cancelled by knowing its momentum and vice versa. As a result, the particle's action is minimized along with its position path and momentum.

In conclusion, divergences are tamed if the least-action and uncertainty principles are applied. SUSY particles are not needed and ad hoc methods such as renormalization can be set aside.

References:

1. Lincoln, Don. 2013-05-21. What is Supersymmetry? Fermilab.

2. Martin, Stephen P. 1997. A Supersymmetry Primer. Perspectives on Supersymmetry. Advanced Series on Directions in High Energy Physics. Vol. 18.

3. Susskind, Leonard. 2012. Supersymmetry and Grand Unification Lectures. Stanford University

4. McMahon, David. 2008. Quantum Field Theory Demystified. McGraw Hill

5. Baez, John. 11/14/2006. Renormalizability. math.ucr.edu

6. Renormalization. Wikipedia

Giving Neutrinos Mass by Adjusting the Higgs and Electroweak Mathematics

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.