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.

A Perturbation Theory Proof

To prove the validity of the perturbation expansion, let's start with a simple and obvious statement:

It seems intuitive that a variable x can be the sum of two terms containing two new variables. In fact, the same can be said of variable a1:

And, a2 can be the sum of two terms containing two new variables. We can repeat this exercise as many times as we like until we reach a(sub-n):

Using equations 2 through 5 we make a series of substitutions which transform equation 1 into the familiar perturbation expansion:

However, equation 6 is an exact solution which can't be had unless we know the value of a(sub-n). Let's assume we don't. Let's eliminate a(sub-n). If we do, we get an approximate solution:

Now, how do we apply this approximate series known as a perturbation series? Consider a function of x:

Let's assume this function of x is too hard to solve. So we re-write f(x) above to show the easy, solvable part and the seemingly unsolvable part:

Let's suppose the hard part is g(x). Wouldn't it be great if we could just get rid of it? Not permanently. Just for the time being. We do away with g(x) by multiplying it by epsilon, and epsilon has a zero limit.

We end up with equation 12 which is easy to solve. The solution can be found at 13. However, we didn't solve variable x--we solved what is called an unperturbed x or x0. We can now plug that solution into equation 7 and equation 8. Because x is approximately equal to x0 plus the rest of the series, we can make a substitution at equation 11 to get the following:

At 15 we set the equation to zero. The idea here is to solve the xi coefficients. This is often done by grouping terms on the basis of their epsilon powers. The equation is broken up into smaller, simpler equations that are easy to solve. At 18 below, the solved coefficients are plugged into the series and epsilon is set back to 1 (giving us back the equivalent to the missing hard part of the original equation). At 19 and 20 we simply add up the coefficients to get a close approximation of the true value of x.

Since the intent of this post was to do a general proof of perturbation theory, no specific problem-solving examples were provided. Click here to see an example of a specific problem solved using perturbation theory.

Why the Graviton Can't Be Found

Why hasn't the graviton been discovered yet? A thought experiment could shed some light on this question. Imagine a universe with only a Higgs field and nothing else. No strong, weak or electromagnetic interactions, no spacetime as we understand it. The basis of this universe is just the Higgs, so the only boson available is the Higgs boson. It's true the Higgs can decay into other particles, but let's focus on it while it is a Higgs.

Now, in such a universe, there should be no gravity, since there are no gravitons, right? (We performed a similar thought experiment in a previous post involving photons. Click here to read all about it.) Let's lay out the mathematics and see. First, we define the variables:

If we find gravity in our Higgs-only universe, that would explain why the graviton hasn't been found--it isn't necessary--so let's begin with the Higgs Lagrangian (L) at equation 1 below. At 2 we convert the Lagrangian to the Hamiltonian (H). To make the math less cumbersome we set the kinetic term equal to chi at 3.

We make a substitution at 4. Equation 5 is a Hamiltonian (H') with the same energy as H, but a different mass and kinetic energy. Equations 4 and 5 represent two adjacent fields whose centers of mass are r distance apart. At 6 we show the equality or conserved energy of the two fields. At 7 and 8 we equate the kinetic and potential energy differences.

Here is an overly simplified, crude diagram for illustrative purposes only:

As you can see the two adjacent fields are outlined with imaginary boxes and labeled blue (high kinetic energy/low mass) and red (low kinetic energy/high mass). The white dots represent the masses. Now, equation 8 fails to take into account distance r, so let's convert mass m as follows:

At equation 11 we have distance r where we want it. Equation 11 is the value of the kinetic-energy difference between the two fields. Classical kinetic energy is a function of velocity squared. What we want to know is the value of the velocity squared:

Now that we know the value of velocity squared, we can do one more step and determine the value of the gravitational constant for this Higgs universe (Gh):

We made a substitution at 14 above and end up with Newtonian gravity! And no gravitons! Equation 14 reveals that gravity is the net velocity squared of kinetic energy differences. If we divide both sides by another r, we get gravitational acceleration. Given these results, one could postulate that gravity is the net motion resulting from motion differences. And motion differences are caused by mass differences. Einstein suggested that matter curves spacetime. However, that assertion is very specific to our universe. A more general assertion is mass disturbs the status quo, whatever that may be, and causes kinetic energy variations. At the quantum scale, gravity does not seem to need its own boson. Information is passed using whatever boson is available. In this case, it's the Higgs.

Now, for extra credit, let's derive Einstein's field equations from equation 14:

If you are feeling ambitious, you can work backwards and derive the Higgs Langrangian from Einstein's field equations.

Update: Here is a couple of videos that falsify the graviton:

Resolving the Cosmological Constant Problem

Imagine you're a quantum physicist and you just got through calculating the energy density of the vacuum. To your horror you find that your calculation is around a hundred orders of magnitude greater than vacuum energy density measurements! This, in a nutshell, is the cosmological constant problem. According to your numbers, the cosmological constant should be huge. According to WMAP measuremments, the cosmological constant is too small to tweet about. In this post we shed some light on this problem. First, here are the variables we will be using:

Below we derive the cosmological constant value based on the WMAP vacuum energy density measurement of approximately 10^-10 Joules per cubic meter (see equations 3 and 4):

The next order of business is to consider the limitations caused by the Compton wavelength formula and the Heisenberg uncertainty principle--so we derive equation 8:

There are two ways we can interpret equation 8: 1. If delta-x grows smaller, energy (delta-E) grows bigger. This makes perfect sense when you consider that smaller spaces contain shorter wavelengths, and shorter wavelengths correspond to higher energies. 2. If delta-x grows smaller, delta-E becomes more uncertain. In any case, we use equation 8 as our means to make sense of the cosmological constant problem.

Equation 9 below is equation 8 modified. On the left side we have the vacuum energy measurement (E) and one meter (x). Of course the product of these values is far greater than Planck's constant, so the right side has the scale factor n:

OK, here's what we need to do: we need to reduce the vacuum's volume while maintaining a constant energy density. This can be done, but only up to a point. The question becomes, up to what point? Or, if you prefer, down to what point? How small can we make the energy and space before things get weird? Here's the answer:

Equation 10 above is the formula for finding the cutoff, or, the Heisenberg uncertainty limit. The product of the energy and space can't go below h-bar/2 without violating the uncertainty principle.

At 11 and 12 below, we plug-n-chug:

At the cutoff, the distance is around 10^-4 meters. The volume is 10^-12 cubic meters. The energy is approximately 10^-22 Joules. If the space is reduced further, the energy value will explode! With an upper limit of infinity! Obviously, if the energy skyrockets, or becomes more uncertain, the energy density won't be the same--and any calculation done at that level will yield ridiculous results.

The diagram below shows vacuum energy density matches observations at or above the cutoff. Any meaningful measurement is taken above that point. But if you prefer infinities and uncertainties, feel free to go smaller than the Heisenberg limit.

Thus, the cosmological constant problem appears to be a natural consequence of doing calculations and measurements below the cutoff limit.

An Incredibly Simple Formula For Removing Unwanted Infinities

Reality says, "It's finite," but your math says,"It's infinite." Why? And, is there a simple way to solve the problem? Today we present an incredibly simple formula for removing unwanted infinities. First, let's define the variables we will use:

Now that we got that out of the way, let's examine why infinities occur. One way to get infinity is to add up an infinite number of numbers as the following integral-summation shows:

You may have noticed something missing in equation 1 above. The thing that's missing makes the infinity finite:

The missing ingredient is none other than dx -- a really tiny number. When the infinity is multiplied by dx it magically becomes a finite number.

So one way to rid ourselves of an unwanted infinity is to multiply it by dx or something equivalent. Equation 5 below is a general formula that always provides a finite value by canceling infinity should one arise.

The first factor is equivalent to dx. The second factor could be equivalent to an infinite sum. Let's apply equation 5 to a classical example. Suppose we have a lattice with N cells. Each lattice cell contains some fraction (epsilon) of the total energy (E):

We could just add up all the parts to get the total energy (E):

We could calculate the average energy per cell (bar-epsilon*E) and multiply that value by the number of cells (N) to get the total energy:

But let's see what happens when we apply our new formula:

At equations 13 and 14, the dx equivalent is simply 1 which is often true in a classical case. It's a bit redundant but yields the right answer as do the more familiar methods above. In this example, it's like counting bricks to get the total. Very simple but effective. It's effective because the average energy per cell is simply the total energy divided by N--the number of cells. Thus, the bar-epsilon in our formula is the reciprocal of N--and the two variables cancel, leaving the total energy E. But look what happens when bar-epsilon doesn't equal the reciprocal of N:

In the above example we have infinite or infinitely uncertain energy in each cell, but the bar-epsilons and N's cancel, the uncertainty and/or infinity is nullified and we end up with the finite energy E.

The following example shows how infinity can rear its ugly head in a seemingly classical lattice.

Here's a data table and chart for different values of n:

Notice how the density increases as n decreases. As n increases the density levels off. The same is true for a 3D lattice:

When n is increased to infinity, the density is virtually constant and behaves like bricks. Increasing and decreasing the number of bricks does not change their mass density.

When n is reduced to zero, the density blows up to infinity! No more brick-density behavior:

We can translate the lattice equation into something resembling perturbation theory:

At equation 29 above we take epsilon to infinity (which is equivalent to taking n to zero) and the answer is 1 + infinity. We can use a famous ad hoc method to rid ourselves of the infinity: simply subtract it or ignore it. By pretending it isn't there, we get a legitimate finite energy value. On the other hand, it would make more mathematical sense to reduce epsilon to zero:

Let's apply our new formula to this problem. We begin with the original lattice equation and reduce n to zero to get the infinite result. At equation 33 we multiply the infinite density by the lattice volume to get the total energy--which is also infinite.

At 33b we restate the formula for convenience. At 34 we plug in the values we get from equation 33. At 35 we see the total energy is no-longer infinite, but the correct finite value.

Now, let's consider an even more classical setup. Here's the lattice diagram:

Surely the density should always behave like bricks, since we have one dot per cell. Sixteen dots per sixteen cells has the same density as one dot per one cell:

But even bricks have a limit. If you divide a brick up into smaller and smaller pieces, eventually you end up with something that no-longer resembles a brick. We can simulate this by reducing n to 1, and k to 0 (see 39 and 40). Once again we get the dreaded infinity--but we can fix that using the new formula (see 41 through 44).

How about a real-world example? At 45 below we set the total energy of two electrons equal to their Coulomb energy at a fixed radius (ro). You will note the total energy (2Ee) is quantized at a fixed integer value (no).

If the radius were to shrink to zero, one would think the energy would explode to infinity. To solve this problem we do some algebra to derive equation 52 below:

At 51 and 52 we see the radius cancelling itself. The total energy of the electrons is determined by the quantum number n. Apparently it is possible to have zero to infinite Coulomb energy (or force) without impacting the overall energy? The crude diagrams below sheds some light on this:

The diagrams show two Coulomb energies: E1 and E2 with radii of r1 and r2, respectively. There is a unit area A. Over a large surface area there are more units of A and vice versa. There are two circles; each representing two different surface areas. Using this information we derive equation 58:

Equation 58 tells us that a large Coulomb energy over a small surface area is equal to a small Coulomb energy over a large surface area, and the total energy is the Coulomb energy over unit area A times the surface area. The total energy is, of course, the total energy of the two electrons--and that energy is finite and conserved! This result is consistent with our formula:

If we plug in the following values and do the math, we get the finite energy of the two electrons:

Thus, we now have an incredibly simple and redundant formula that shows the connection between the parts and the whole at the classical and quantum levels.

Update: Below is a proof of the formula: