Deriving the Particles

Here I derive the particle content of the framework developed here. And it will be seen to be very similar to the particle content of the Standard Model. I will not attempt to get the quantitative mass or charge or spin values of the particles. I only attempt to give the number of particles and the comparative levels of charge for the electron and positron, the Weak force bosons, the quarks and the gluons. I also give some indication where the photons and neutrinos come from. I also give some explanation for the three mass generations.

As I show elsewhere, the transition amplitude of a
particle can be obtained from first principles. The complex conjugate of this is
interpreted as the antiparticle partner. Since there are only two amplitudes
that can be obtained from a complex function and its conjugate, this matches the
expectation for the
electron and positron of the Electromagnetic force. Armed with this perspective
a 2^{nd} iteration of the formulism will be shown to duplicate the
properties for the W and Z bosons. The 3^{rd} iteration will
seem to replicate the number of particles and charge levels of the quarks.
Iterating again gives us what appear to be gluons.

If the formulism developed here is correct, then this may possibly provide means of specifying wave functions for W and Z bosons and the quarks, etc. This may help explain why the Weak bosons are so short lived and why the quarks are confined. It may also give means of calculating the relative masses of the particles. The exact nature of interactions and decay might be discerned. And it may also give means of calculating the coupling constants from first principles.

**INTRODUCTION**

In summary, the previous article
starts with the assumption that reality consists of a collection of things all
coexisting in conjunction with each other. Space itself consists of a
conjunction of points, where each point is described with a proposition stating
its coordinates. This results in Equation [3], shown below, where each **q**_{i}

* q_{1}$\wedge $q_{2}$\wedge $q_{3}*$\wedge $...

However, a conjunction between two points results in implications between those points, as shown in Equation [4] below.

**q**_{1}**$\wedge $q**_{2}**→*** *(

So a conjunction implies an implication both ways. And this means that the conjunction of all the points in space implies an implication between any and every two points of space, as expressed in Equation [5] below.

**q**_{i}**→****q**_{i} **→** **
q**_{j})
[5]

This makes it necessary to study what it means to have an implication within a conjunction of points. And it was discovered that such an implication equates to a disjunction of a conjunction of implications as shown in Equation [6] below.

**q**_{0}
**→** **q**_{n}) * $=$*
${\vee}_{j=0}^{n}$
(

This is a recursive relation that can be iterated any number of times. And when iterated an infinite number of times, Equation [7] is derived as shown below.

**q**_{0}
**→** **q**_{n}) ** $=$**
${\vee}_{{i}_{1},{i}_{2},{i}_{3},{i}_{4},\mathrm{...},{i}_{m}=0}^{n}$
(

This is a disjunction of many terms. Each term is a series of implications that
represent a path of steps from start to finish. Since each of the *i's* in the
disjunction runs from 0 to *n*, Equation [7] represents every possible path from **q**_{0}**q**_{n}

However, implication can be expressed in set
theory by the relation of subset. If B is a subset of A, then if A exists, B
also exists, but not the other way around. And a numeric value can be assigned
to the relation of subset using the Dirac measure,
**
$\mathbf{\delta}$**_{x}(A), which is equal to 1 if x is included in A and is 0
if x is not in A. This allows us to map a true proposition a value of 1 and a
false proposition a value of 0. If the set A of the Dirac measure is reduced to
a single element, then the Dirac measure reduces to the Kronecker delta,
_{ij}. This allows us to map disjunction to
addition and conjunction to multiplication. In the continuous case, the
Kronecker delta becomes the Dirac delta function,

This equation can also be iterated an infinite number of times, as shown in Equation [22] below.

$\delta (x-{x}_{0})={\displaystyle {\int}_{R}{\displaystyle {\int}_{R}\cdot \cdot \cdot {\displaystyle {\int}_{R}\delta (x-{x}_{n})\delta ({x}_{n}-{x}_{n-1})\cdot \cdot \cdot \delta ({x}_{1}-{x}_{0})d{x}_{n}d{x}_{n-1}\cdot \cdot \cdot d{x}_{1}}}}$

The complex, Gaussian exponential version of the Dirac delta function is used, as shown below.

[27]

$$\delta ({x}_{q}-{x}_{r})=\underset{{t}_{q}\to {t}_{r}}{\mathrm{lim}}{\left(\frac{m}{2\pi i\hslash ({t}_{q}-{t}_{r})}\right)}^{1/2}\mathrm{exp}\left(\frac{im{({x}_{q}-{x}_{r})}^{2}}{2\hslash ({t}_{q}-{t}_{r})}\right)$$

And this can be shown to solve the recursive property of Equation [20]. When the equation above is substituted into Equation [22], we get,

[28]

${\int}_{-\infty}^{+\infty}{\displaystyle {\int}_{-\infty}^{+\infty}\cdot \cdot \cdot {\displaystyle {\int}_{-\infty}^{+\infty}{(\frac{m}{2\pi i\hslash \Delta {t}_{n}})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{\frac{im{(x-{x}_{n})}^{2}}{2\hslash \Delta {t}_{n}}}\cdot {(\frac{m}{2\pi i\hslash \Delta {t}_{n-1}})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{\frac{im{({x}_{n}-{x}_{n-1})}^{2}}{2\hslash \Delta {t}_{n-1}}}\cdot \mathrm{...}\cdot {(\frac{m}{2\pi i\hslash \Delta {t}_{1}})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{\frac{im{({x}_{1}-{x}_{0})}^{2}}{2\hslash \Delta {t}_{n}}}d{x}_{n}d{x}_{n-1}\mathrm{...}d{x}_{1}}}}$

The exponents of this equation can be manipulated as follows,

$\frac{im{\left(x-{x}_{j}\right)}^{2}}{2\hslash \Delta {t}_{j}}=\frac{im}{2\hslash}{\left(\frac{\Delta {x}_{j}}{\Delta {t}_{j}}\right)}^{2}\Delta {t}_{j}=\frac{im}{2\hslash}{\left(\dot{x}({t}_{j})\right)}^{2}\Delta {t}_{j}$

And when this is substituted into the last integral above, we can collect the exponential terms to get a summation. And we can combine the square-root factors to get Equation [29] below.

[29]

${\int}_{-\infty}^{+\infty}{\displaystyle {\int}_{-\infty}^{+\infty}\cdot \cdot \cdot {\displaystyle {\int}_{-\infty}^{+\infty}{(\frac{m}{2\pi i\hslash \Delta t})}^{n/2}{e}^{\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\scriptscriptstyle \frac{i}{\hslash}}{\displaystyle {\sum}_{j=0}^{\infty}\frac{m}{2}{\left(\dot{x}\left({t}_{j}\right)\right)}^{2}\Delta t}}}}}d{x}_{1}d{x}_{2}\cdot \cdot \cdot d{x}_{n$

Now in the case of a continuum, as $n\to \infty $ and $\Delta t\to 0$, the summation above becomes an integral and we get Equation [30] below,

${\int}_{-\infty}^{+\infty}{\displaystyle {\int}_{-\infty}^{+\infty}\cdot \cdot \cdot {\displaystyle {\int}_{-\infty}^{+\infty}{(\frac{m}{2\pi i\hslash \Delta t})}^{n/2}{e}^{\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\scriptscriptstyle \frac{i}{\hslash}}{\displaystyle {\int}_{0}^{t}\frac{m}{2}{(\dot{x})}^{2}dt}}}}}d{x}_{1}d{x}_{2}\cdot \cdot \cdot d{x}_{n$.

This is Richard Feynman's Path Integral of quantum mechanics for the trajectory of a particle free of any forces acting on it. Again, the details of this derivation are covered in the previous article.

The path integral is another way of obtaining the wave function. It is also called the transition amplitude. Obtaining the wave function is this manner may seem like a trick of mathematical manipulation. But if the particles of the Standard Model emerge from this framework, then perhaps this deserves more attention.

**FINDING PARTICLES**

The last section gives us a starting vocabulary and basic framework. The goal now is to recognize the particle content of this formulism. Shouldn't we expect that the underlying principles that give us quantum mechanics would also give us the particles to which it applies?

The Standard Model of particle physics describes about 17 fundamental particles. There are bosons and fermions, quarks and leptons; the fermions have 3 generations of mass. There is the Higgs boson, and the charged particles have their antiparticle. The table below to the left shows how the particles are arranged. The diagram to the right shows how they interact with each other. These particles make up all the atoms and molecules we see. And the trick will be to find them hiding in the logic of this framework. The attempt will be to show that the particles of the Standard Model can be matched to characteristics of this formulism.

Previously, I suggested that the implication between points could be expressed with complex numbers which was responsible for giving us the particles of the electromagnetic force. And when iterated, the implication between implications would give us the particles that participate in the Weak nuclear force. And when iterated again, the implication between the implication of implications would give us the particles of the Strong nuclear force. This all seems a bit abstract, and so now I'd like to flesh this out.

Because a conjunction implies an implication both ways, **
p$\wedge $q ****→*** *(

And the reverse implication, *q***→** *
p*),

But Equation [33] is also the expression for the transition amplitude for a particle,

and the transition amplitude for the antiparticle is expressed by the complex conjugate,

where
$\Delta t=\left({t}_{q}-{t}_{p}\right)$.

This is possible because an antiparticle can be
thought of as a particle traveling backwards through time. It seems to be a time
reversed version of a particle. But time reversing the
transition amplitude is the same as taking the complex conjugate without time
reversing. Time reversing means taking the negative of
$\Delta t$. If
$\Delta t=\left({t}_{q}-{t}_{p}\right)$ represents a particle moving forward through time, then
$-\Delta t=({t}_{p}-{t}_{q})$
represents an antiparticle moving backwards in time. Yet, since
$\Delta t$
is a cofactor with the *i* in the exponent
and in the square-root of the transition amplitudes, we can transfer the
negative sign on
$\Delta t$
and put it in front of the *i* to get the complex conjugate. Then the
complex conjugate of the transition amplitude represents an antiparticle
traveling between the same two points in the same period of time, where both
particle and antiparticle are now moving in the forward direction of time.

To be concise, the conjunction of all points in
space results in an implication between every two points. And there is nothing
to distinguish one point from another, except arbitrarily assigned coordinates.
So there should be nothing to distinguish one implication from another. One
implication is just as valuable as another. But an implication one way is not
equal to the implication the other way, **p****→*** q*)
$\ne $
(

Now the electromagnetic force acts on two charged particles, the electron and the positron. And since my math representation of implication and its reverse implication is equivalent to the transition amplitude of a particle and antiparticle, this suggests that implication and its reverse, represents an electron and positron, since one is the complex conjugate of the other.

So I will identify the charges of particles with
the unique different ways to configure the various implications within an iteration. The 1^{st} iteration has 2
configurations, an implication and its reverse, and is interpreted as the positive and
negative *electric* charge of the *positron* and *electron*. The 2^{nd}
iteration is characterized by the implication between implications, and there
are 4 different ways to configure the implication between implications and their
reverse, 3 of which are unique. These are interpreted as the *W*^{ +},
*W*^{ -}, and *Z*^{ 0}
particles of the *Weak nuclear* force. The 3^{rd}
iteration is characterized by the implication between the implication of
implications, and its reverse, 5 of which are unique. These will be interpreted
as the *up*, *anti-up*, *down*, *anti-down flavor* property of the first generation
*quarks*. The second and third generation particles will be just a little
more difficult to explain. There is also a fifth *quark* with zero charge found
here; this is not found in the Standard Model. But some think it may explain
dark matter. And finally iterating again, there seems to be 8 unique
configuration which can be interpreted as the 8 different *gluons*. *Photons*
and *neutrinos* are interpreted here as special ways of combining previous
particles.

As you may have guessed, I'm using *italic* words
to indicate my interpretation of my formalism since they seem to match the
properties of the Standard Model.

**THE WEAK ITERATIONS**

In the previous section, the forward implication is represented with the complex, exponential gaussian function.

**p****→** *
q*) $\Rightarrow $

And the reverse implication is represented by its complex conjugate.

**q****→**
*
p*) $\Rightarrow $

We can abbreviate this to

**p****→** *
q*) $\Rightarrow $
${{\left({i}_{1}{A}_{qp}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{+{i}_{1}{E}_{qp}}}}$

and

**q****→**
*
p*) $\Rightarrow $
${{\left(-{i}_{1}{A}_{qp}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{-{i}_{1}{E}_{qp}}}}$ ,

if we let ${E}_{qp}=m{({x}_{q}-{x}_{p})}^{2}/2\hslash ({t}_{q}-{t}_{p})$ and ${A}_{qp}=-m/2\pi \hslash ({t}_{s}-{t}_{r})$ .

A graphical representation of the transition amplitude and its conjugate are shown below in Fig 1A and Fig 1B.

The *p* of Fig 1A and the *e* of Fig 1B are continuous
functions of space and time. The arrow in Fig 1A indicates that the
*p* transition is
from *x _{p}* to

We then get two mathematical entities in
the 1^{st} iteration of the framework. Because this seems to equate to the
transition amplitudes of a particle and antiparticle, I somewhat arbitrarily identify
the forward implication
of the logic with a *positron* and the reverse implication with an *electron*.
To avoid making controversial claims, I will use *italic words*
to label my interpretation of my framework and non-italic words to refer to the
particles of the Standard Model. What I hope to show is that the number of
particles with the various kinds of charge can be accounted for.

Now implication is a part of the 1^{st}
iteration of this framework. But if implication and its reverse can be identified
as a *positron (p)* and *electron* (*e*), then the question becomes do the iterations of this formulism
correspond to particles of the other forces of nature. Let's do the iterations
and count the particles.

In the 1^{st} iteration, a conjunction of two
points gives us an implication both ways, **p$\wedge $q**** →*** *(

In the 1^{st} iteration, implication was represented by the Dirac delta function, which is manipulated into the transition amplitude of Equation [34],

$<{x}_{q}|U(t)|{x}_{p}>\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}{(\frac{m}{2\pi \hslash i({t}_{q}-{t}_{p})})}^{1/2}{e}^{im{({x}_{q}-{x}_{p})}^{2}/2\hslash ({t}_{q}-{t}_{p})}$.

The reverse amplitude is accomplished by interchanging *x _{p}* and

The 2^{nd} iteration is accomplished in the logic by replacing the
propositions in the 1^{st} iteration with implications. In the math this means
replacing the coordinates *x _{p}* and

If the transition amplitudes were for *electrons*,
we would substitute,

Or we could substitute any combination of *electrons* or *positrons*. Substituting
two *positrons*, we would get the amplitude,

Now things are getting more complicated. Instead of the single
exponent we had in the 1^{st} iteration, we now have exponents inside an exponent
in the 2^{nd} iteration. This gives us amplitudes inside
amplitudes. So I will use colors to distinguish the different iterations; black
is for the 1^{st} iteration,
magenta is for
the 2^{nd} iteration, and aqua
is for the 3^{rd} iteration. These colors have nothing to do with
the color charge of the Strong force.

Notice in Equation [40] that there is an imaginary number, *i _{1}*, in the
1

Also notice that I've made a distinction between *m _{e}* and

If we use definition of *E _{qp}* and

${{\left(\frac{{m}_{z}}{2\pi {i}_{2}\hslash ({t}_{\nu}-{t}_{\mu})}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{+{i}_{2}{m}_{z}{\left(\phantom{\rule{8px}{0ex}}{{\left({i}_{1}{A}_{sr}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{+{i}_{1}{E}_{sr}}\phantom{\rule{8px}{0ex}}{-}\phantom{\rule{8px}{0ex}}{\left({i}_{1}{A}_{qp}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{+{i}_{1}{E}_{qp}}}\phantom{\rule{8px}{0ex}}\right)}^{2}/2\hslash ({t}_{\nu}-{t}_{\mu})}}}$.

With similar definitions for ${W}_{\nu \mu}$ and ${B}_{\nu \mu}$ , this can be further abbreviated to,

We can make a graphical representation of Equation [40] in space and time as shown in Fig 2.

The black arrows represent the 1^{st} iteration math written in black. The
magenta arrow represents the 2^{nd}
iteration math written in magenta. Here we
see two positrons, traveling forwards in time, connected to each other. Keep in
mind that everything pictured here is at a small, differential scale. The black
math of Equation [40] came from the Dirac delta function in Equation [27], which
is integrated in Equation [22]. So the math in black is a differential.
Consequently, the
magenta math in Equation [40] is a
differential that must also be integrated to find a wave function. And since
everything is differentially small, it can be represented as a *point particle*,
and we can imagine space potentially being filled with a field of these real or
virtual *particles*.

But it will be easier to simplify Fig 2 to the graph shown in Fig 3.

Then we can talk about the number possible combinations of *e*'s and *p*'s
and the forward and reversed versions of them.

For example, if the *particle* in Fig 3 represents the forward
implication, **r****→***
s*)

If Equation [40] represents the forward amplitude, then the reverse amplitude is obtained by reversing the order of ${t}_{\nu}$ and ${t}_{\mu}$ and reversing the order of subtraction in the exponent. This give us a reverse amplitude,

But since ${\left(a-b\right)}^{2}={\left(b-a\right)}^{2}$ , we can keep the original order of subtraction in the exponent. And we can also keep the original order of ${t}_{\nu}$ and ${t}_{\mu}$ if we take the complex conjugate of ${i}_{2}$ . So in terms of the abbreviated version, if the forward amplitude is,

then the reverse amplitude is,

Technically, the reverse amplitude is not mathematically equal to the forward
amplitude. So the asymmetry between forward and reverse implication maps to an
asymmetry between the forward and reverse amplitude as required. But notice
that
$-1={e}^{i\pi}$.
So we can make
${-{i}_{2}{W}_{\nu \mu}={i}_{2}{W}_{\nu \mu}}}{e}^{{i}_{1}\pi$ . Then we can distribute the
${i}_{1}\pi $ between the two
1^{st} iteration exponents to get,

In other words, the minus sign in the
2^{nd}
iteration exponent only adds a phase angle to the 1^{st}
iteration amplitudes. And both terms inside the square are shifted by the same
phase. However, according to
this article, a phase
factor gets cancelled out in the calculation of any expectation value. So a
phase factor does not have any physical meaning. And both terms inside the
square still essentially represent a positron as before.

Similarly,
^{nd}
iteration exponent to get an amplitude of,

Again, this phase factor is physically meaningless. The resulting particle is
essentially the same as with no minus sign. So what is shown is that the forward
and reverse implication in the
2^{nd}
iteration is actually the same particle. This result applies to every
combination of electrons and positrons substituted in Equation [40].

Also, notice that if the forward implication of *e _{1}*

then the forward implication of *p _{1}*

Although these two amplitudes are not mathematically equal, they are essentially
the same particle, since either way you look at it, it is still connecting a
*positron* with an *electron*.

So out of the many possibilities that we started with in this Section, *e _{1}*

For *e _{1}*

This *particle* has a *negative electric charge*, so I tentatively label it
as a
${W}^{-}$.

For
*p _{1}*

This *particle* has a *positive electric charge*, so I tentatively label it
as a
${W}^{+}$.

For
*e _{1}*

This *particle* has *no electric charge*, so I tentatively label it
as a
${Z}^{0}$.

Thus, the
2^{nd} iteration seems to account for the W and Z bosons of the
Weak nuclear force. We have the correct number of particles and the same
positive, negative, and zero levels of charge. The only difference is that the
Standard model has a -1*e* charge for the W^{-
} and a +1*e* charge for the W^{+}, whereas I get ±2*e*.
If I were to divide by 2, I would get the same electric charge as the Standard
Model for the W and Z bosons. So notice that if I were to ignore the plus or minus
sign of the charge I get and only count the different absolute magnitudes,
|0*e|* and |±2*e|*,
there would be two different levels. If I divide by the number of different absolute values,
there are 2, I
would get the same *electric charge* for my *weak bosons* as the Standard Model.
This trick also works in the 3^{rd} iteration
described below. I don't know why this trick would work. Perhaps there is a
group theoretic reason, or perhaps it occurs by taking into account
superpositions of states. Other reasons that the charge would be ±1*e*
instead of
±2*e *might be perhaps the connection in the
2^{nd} iteration acts like a positive charge reducing the overall
negative charge in the
${W}^{-}$,
and similarly for the
${W}^{+}$.
Or perhaps there is some unusual screening of the electric charge due to the
2^{nd} iteration connection. Or perhaps the electrons and
positrons in these particles are closer to being virtual and don't acquire the
full charge of an electron or positron.

It's curious, though, the Standard Model considers the
W and Z bosons to be elementary particles, not consisting of constituent
sub-particles. But these Standard Model particles can decay into electrons,
positrons and neutrinos with a probability determined by a coupling constant
derived from experimental data. However, the
$W$and
$Z$
*bosons* derived here are
made of constituent *sub-particles*, the *electron* and *positron*.
So here it is easier to see how such decay modes are even possible. The primary
means of propagation comes from the *electrons* and *positrons* since
their transition amplitudes are functions of time and space. And the propagation
of these constituent *electrons* and *positrons* are constrained in
order to maintain the connection of the
2^{nd} iteration that defines those *particles*. The
likelihood of the *W* and *Z* *bosons *
to decay might then depend on how far the
$W$
and *Z* *particles*
can travel without losing their particle-like propagation characteristics.

Consider the
2^{nd} iteration forward amplitude for a
${W}^{+}$ *particle* shown again below (Equation [40]).

The math in black is an amplitude for the propagation of a particle. When integrated with the path integral, this results in a wave function for a particle that propagates through time and space indefinitely with a given velocity. But the whole of Equation [40] will also need to be integrated with the path integral to result in the wave function for a ${W}^{+}$. The integration should be doable, since the differential of what's inside the square of the exponent is include in the integrand. This allows the integral to be treated as if it were the integral of a gaussian. So we should get a similar form as Equation [40] for the wave function for a ${W}^{+}$, with exponents inside exponents.

The question is how does a wave function propagate that has an exponent inside an exponent? I suppose that there is an approximation that can be made near the start of the propagation that appears to look like the normal wave function to which we are accustomed. But farther away, with increasing time, this approximation is no longer valid and second order affects make the wave function diverges, possibly into other particles. Decay rates might be discernable in this equation.

Furthermore,
it's obvious that these
$W$ and
$Z$
*bosons *are more massive than the *electron* or *positron* simply
because each one is constructed of 2 *electrons* and/or *positrons*,
and a heavier mass for the W and Z bosons is what we find in the
Standard Model as well. I've written a different *m _{z}* than for

Now that we appear to have derived the W and Z bosons, we can move on to consider the quarks. All the techniques developed in this section will be applicable to quarks as well.

**THE STRONG ITERATIONS**

The 1^{st} iteration of this framework gives us what appears to be *electrons*
and *positrons*. The 2^{nd} iteration gives us what appears to be
$W$ and
$Z$
*bosons.* We might expect the 3* ^{rd}*
iteration to give us what appears to be

In the 1^{st} iteration the
conjunction of any two points in space results in an implication between those
points and the reverse: **g$\wedge $h**** →*** *(

The 2^{nd} iteration is obtained
by replacing the proposition, **g****h****m****→*** n*)

The 3* ^{rd}* iteration is obtained
by replacing the propositions,

{[(

In the math, the 1^{st} iteration
maps implication to a transition amplitude,

In the 2^{nd} iteration math, these amplitudes
themselves are reinserted the amplitude, one for each of the $x$ variables.
This gives us:

The math in black is the 1^{st}
iteration; the math in magenta
is the 2^{nd} iteration.

Likewise, in the 3* ^{rd}* iteration
math, we reinsert the last equation for each of the $x$
variables in an amplitude. This gives us:

And this is even more complicated. Now we have amplitudes inside
amplitudes inside amplitudes. But with the appropriate definitions for *E* and *A* and *W* and *B* and *Q* and *C*, this simplifies to:

the aqua is for the 3* ^{rd}*
iteration math.

And the space-time graphical representation of Equation [42] is shown in Fig 8.

The *p* stands for *positrons*, and the
${W}^{+}$
stands for the *Weak bosons*. The
black arrows are for 1* ^{st}* iteration

Fig 8 can be simplified by drawing it in 1* ^{st}* iteration
particle space as shown in Fig 9 below.

And this can be simplified even further when drawn in
2* ^{nd}* iteration particle space as shown in Fig 10 below.

In the previous section, the
2* ^{nd}* iteration
$W$ and
$Z$

Using the methods of the previous section, let
${W}^{+}$
and
${W}^{-}$ have an electric charge of +1 and $-$1, respectively. And the
${Z}^{0}$ has an electric charge of 0. Then Fig 10 above has the highest possible
charge. So I tentatively label it an *up quark*,
$u$
. It seems to have a charge
of +2.

Next, consider connecting a ${W}^{-}$ to a ${W}^{-}$ as shown in Fig 11 below.

This *particle* has the lowest possible charge of $-$2. So I
tentatively label it an *anti-up quark*,
$\overline{u}$
.

And there are other possible levels of charge. Consider connecting a ${W}^{-}$ to a ${Z}^{0}$ as shown in Fig 12 below.

This particle has a charge of $-$1. So I tentatively label it an *down quark*,
$d$
.
Its antiparticle is shown below in Fig 13.

This *particle* has a *charge* of +1. So I tentatively label it
an *anti-down quark*,
$\overline{d}$
.

However, there are also combinations of
$W$ and
$Z$
*bosons* that have zero *charge* as shown in Fig 14.

These *particles* are *electrically* neutral with zero charge. I
label it a
$\zeta$, which is the Greek letter sigma used at the end of words. I only use
this symbol because it somewhat looks like a fancy *Z* for zero charge. The two
*particles* of Fig 14 are indistinguishable and represent one *particle*. Since
it has zero charge, it is its own antiparticle.

So the list of *quarks* obtained from this framework are the
$u$,
$\overline{u}$,
$d$,
$\overline{d}$, and the
$\zeta$, with charges +2,
$-$
2, +1,
$-$1, and 0, respectively. These are supposed to represent the property of
flavor for the first generation quarks. If I use the trick from the previous section, I
should divide these charges by the number of different absolute values. There
are 3 of them, |0|, |±1|, |±2|. So if I divide these charges by 3, I get +2/3,
$-$2/3, +1/3,
$-$1/3, and 0. And this matches the electric charge for the quarks of the
Standard Model. Except there is no quark in the Standard Model that has 0 electric
charge. This is beyond the Standard Model. However, some have suggested that perhaps dark matter could
consist of a quark with 0 electric charge,
here. And
there is talk about such a singlet quarks being needed for grand unification
scenarios, here.

But quarks also have what is called a color property of red, green, blue, or anti-red, anti-green, or anti-blue. The color property does not have anything to do with frequency of visible light. These are just archaic names given to this property of quarks. The discipline of Quantum Chromo Dynamics (QCD) studies this property of color; this is what is responsible for the Strong nuclear force that binds quarks, protons, and neutrons together in an atom.

So the question is can we find 3 configurations and their
reverse hiding in each *quark* of this framework. It turns out that we can.
For there are different ways to connect the *electrons* and *positrons* that make up
a *quark* of a given *charge*. For example, consider the ways we can
make *anti-up quark*,
$\overline{u}$. The first way is shown in Fig 15A.

Then second way is shown in Fig 15B. And the third is shown in
Fig 15C. All these configurations have the same *electric charge*, and they have
the same *Weak isospin charge*. So all these configurations should interact with
the other *particles* in the same way. But they are mathematically
different since in the
1* ^{st}* iteration each is connecting different space time points in Equation [42].
And the same can be said for the
$u$ quarks, since we only change

Similarly, the *electrons* and *positrons* of all the other *quarks*
of a given *electric charge*
can be connected in 3 different ways. And since there is no way to measure which
configuration a *quark* is in, all 3 configurations exist in superposition for
each *quark*. Which configuration a *quark* is connected really does not
matter except when you are connecting one *quark* with another as is done
in the 4* ^{th}* iterations with

So perhaps the 3 different ways of connecting the underlying
*electrons* and *positrons* gives us the 3 different colors of the Strong nuclear
force. Then the anti-colors could be obtained by reversing the aqua
arrow in a *quark*. If so, the question becomes why does this create such a strong
force between the *quarks?* I don't
know at this point.

But it does seems intuitive that the quarks should be heavier than
the
$W$ and
$Z$
*bosons* since they are constructed of more *electrons* and *positrons*.
And it seems obvious that the *quarks* should operate on a smaller scale than the
*Weak bosons* since the math of Equation [42] has an extra exponent to make
the equation diverge more quickly from an approximation.

Section
5

Gluons

If we iterate again, we can use the same methods. We simply
count the number of ways that the previous iteration *particles* can be connected
with unique *electric charge*. In the 4* ^{th}* iteration we find the number of
way to uniquely connect 3

In Section 4 we found the *quarks* with the various *flavor* and *color charge*.
These are:

${W}^{-}$**→**
${W}^{-}$ $\overline{u}$,*
anti-up*,
$-$2/3*e*

${W}^{-}$**→**
${Z}^{0}$ $d$,* down*,
$-$1/3*e*

${W}^{-}$**→**
${W}^{+}$ =
${Z}^{0}$**→**
${Z}^{0}$ $\zeta$,*
zero*,
0*e*

${W}^{+}$**→**
${Z}^{0}$ $\overline{d}$,*
anti-down*, +1/3*e*

${W}^{+}$**→**
${W}^{+}$ $u$,* up*,
+2/3*e*

Now let's list every combination of *quarks* with unique *charge*:

$\overline{u}$** →**
$\overline{u}$
,
$-$4/3*e*

$\overline{u}$** →** $d$
,
$-$1*e*

$\overline{u}$** →**
$\zeta$
= $d$** →**
$d$
,
$-$2/3*e*

$\overline{u}$** →** $\overline{d}$
= $d$** →** $\zeta$
,
$-$1/3*e*

$\overline{u}$** →** $u$
= $d$** →** $\overline{d}$
= $\zeta$** →**
$\zeta$
, 0*e*

$d$** →** $u$
= $\zeta$** →**
$\overline{d}$
, +1/3*e*

$\zeta$** →** $u$
= $\overline{d}$** →** $\overline{d}$
, +2/3*e*

$\overline{d}$** →** $u$
, +1*e*

$u$** →**
$u$
, +4/3*e*

Here we seem to have nine *gluons*. The Standard
Model also has nine gluons, two of which combine into one state, giving eight
distinct gluons. Fig 16 below shows the eight gluons as black dots. The colored
triangles labeled with a *q* represent quarks; the upward pointing triangle are
quarks with red, greed, and blue color charge, the downward pointing triangles
are antiquarks with anti-red, anti-green, and anti-blue color charge. Here is a
short video about
quarks and gluons that might be helpful.

Fig 16: Types of gluons and quarks

However, the Standard Model gluons have no electric charge
whereas those developed here do. However, if I divide by the number of distinct
absolute values as I did in the previous sections, there are 5 of them, |0|, |±1/3|, |±2/3|, |±1|, and |±4/3|,
then I divide by 5 and get a much smaller *charge*. Perhaps this is small
enough not to have been detected so far.

The Standard Model gluons also have no mass whereas those developed here
would seem to have a huge mass, more than the *quarks*. Yet, there might be a
means of propagation where the mass converts to a frequency of propagation at
the speed of light.

Or, a 4* ^{th}* iteration may not be the way to get

If *quarks* can be connected, one possibility is shown in Fig 17.

But there are other ways to connect *quarks*, depending on their *color*
configuration. Another way is shown in Fig 18 and Fig 19, where we have changed
the *color* configuration of only the left *quark*.

And we can also change the *color*
configuration of the *quark* on the right. In fact, there are 3 *color*
configurations of the *quark* on the left for each of 3 *color*
configurations of the *quark* on the right. Again, this is a total of nine,
3 of which leave both sides unchanged if that type of connection should occur
again. Perhaps these represent the 3 color-anticolor gluons of the Standard
Model that leave the quark color unchanged if exchanged.

All these possible ways to connect *quarks* exist
in superposition with each other. And the complex phase of things that exist in
superposition can interfere constructively or destructively. It's possible that
they can add to each other or cancel each other out. But in the context of
connecting *quarks*, every possible connection is of
the same length scale. So the phase angles would have more of a tendency to add
than to cancel out. So all these ways of connecting *quarks*
would seem to only strengthen the connection between them. Perhaps this is the
reason that the color force is so strong between quarks.

But how can mere connections from one *quark* to another be considered a *gluon
particle*? In Quantum Field Theory, particles propagation can be described
with Feynman's Path Integrals. But Feynman's integral has a perturbation
expansion which takes into account every possible intermediate exchange with
virtual particles. So if this connection between *quarks*
can be mediated with every possible virtual process, then it is equivalent to the
propagation of a particle. So I need to explain virtual particles in the context
of this framework. Then I can show how the connections explained here are
actually particles in the quantum mechanical sense. This is the subject of the
next section.

However, first notice that there are other ways of connecting the *electrons*
in Fig 17 (repeated below).

Notice here that *e _{2}* is close to

Section
6

Virtual Processes

Many of the processes in quantum theory can be explained in terms of virtual particles. Virtual particles appear out of empty space, they appear for a very brief moment, and then they disappear back into empty space. They appear in pairs from a single point, as particle and antiparticle; they each travel an arbitrary distance to the same place where particle and antiparticle meet up again and cancel each other out. And this process of cancelling each other out results in their having no permanent effects. The virtual particles do not continuously propagate like normal particles. So in that sense they are not considered real. But they can interact with each other and with real particles, so they can change the trajectory of real particle. Yet where do the virtual particles come from?

Virtual particles appear naturally in this framework. In a
previous article, I describe how *virtual particles*
are the result of all the points of space all coexisting in conjunction with
each other. This conjunction of all points means that there is a conjunction of
any two points of space. So recall that
a conjunction of any two points results in an implication between those points
along with its
reverse implication,
**
p$\wedge $q ****→*** *(

The point is that there are virtual particles everywhere in space at all
times. Virtual particles can travel from any point to any other point and
interact with real or virtual particles in the process. The conjunction of all
points mean there is an implication and its reverse everywhere in space. These
are interpreted as virtual *electrons* and *positrons* that exist everywhere.
And if the conjunction of all points necessitates implications everywhere, then
we have a conjunction of implications which necessitates implications of
implications everywhere in space as well. These implications of implications are
interpreted as virtual
*bosons* popping in and out of existence everywhere all the time too.
Similarly, the conjunction of every possible implication between implications
necessitates every possible implication of implications between implications
which is interpreted as virtual *quarks* existing everywhere. And similarly for
the *gluons*. So the existence of space alone necessitates the existence of
all the fields of all the virtual *particles* we know. And these virtual
*particles* are freely available to assist in the creation, propagation, and decay of real
*particles*.

For example, consider how an *electron* might propagate from an initial
position, *p _{i}*, to a final position,

The *electron* initially at *p _{i}* could interact with a virtual pair going from

*Particles* seem to travel through a series of interactions with virtual *particle/anti-particle* pairs. This works for the *electrons* and *positrons*, but it should also work for
*Weak bosons*, *quarks* and *gluons* since they also have *anti-particles* that form virtual *particle/anti-particle* pairs. The *particles* that do not have an
*anti-particle* are electrically neutral; they are their own *anti-particle* and so do not need to be paired with a virtual partner.

But virtual *particles* aid in more than the interaction which define the
propagation of a *particle*. Virtual *particles* also aid in the interaction of
various kinds of *particles* with other kinds of *particles*. Remember
how *quarks* were connected by a *gluon* as shown again in Fig 20.
Here the *quarks*, *u̅ _{1}* and

But there are other ways one *quark* can interact
with another. For instance, a
virtual *quark/anti-quark* pair could emerge for a moment out of the vacuum
as shown in Fig 21.

Here, *u̅ _{v}* and

Section
7

The other particles

The *particles* explained so far are the most general
interpretation of the implications. The end points of the connections have the
most freedom; each is different from the others. However, other kinds of
*particles*
can be constructed if some of the end points are shared in common. For instance,
Fig 22 shows the spacetime configuration for a ${W}^{-}$
*particle*.

Notice that each of the spacetime points, *x _{1}*,

The exponential part of the amplitude for Fig 22 is shown in Equation [43].

But if *x _{2}* and

Here *t _{1}* <

But could *e _{1}* of Fig 23 be changed into
a

Here, both the *electron* and *positron*
are meeting together at *x _{2}*. I suppose
they would immediately annihilate each other before they could be connected into
a single particle. So I don't think this particle can even exist.

Consider, however, the spacetime configuration of an *Up quark* as shown in Fig 25.

If *x _{4}* =

The *positron*, *p _{1}*, is created at

If we do not connect *x _{3}* to

There
are two spatial jumps on each side of the aqua
arrow in both Fig 25 and Fig 26. Each jump generally occurs at different times
in Fig 25. But there are sequences of jumps on each
side in Fig 26.
So it represents changes in direction and more energy associated with those
changes in direction. Fig 26 still
represents a *quark* since the strong arrow is
the only arrow left exposed in the middle and available for connection by a
gluon. It has the same charge as an *Up quark*, but now has more *energy/mass*. So I call
it a *Charm quark*, * c*. And it has its
anti-matter partner, the

We can also start with a *down quark* as shown in Fig 28.

If we let *x _{4}* =

Here again we have a series of spacetime jumps on each side of the strong
arrow which represent more energy associated with changing direction. So it has
the same *electric* charge as the *down quark*, but has more energy. I call it a
*Strange quark*, * s*. And it has its
anti-particle partner, the

Or start again with an *Up quark*, shown again in Fig 30.

If we connect
*x _{2}* to

With these added *positrons*, we now have a series of 3 spacetime jumps on each
side of the strong
arrow. And so this particle has more energy than the previous **
c***quark*. I would call it a *Top quark*, *
t*. Although the
magenta arrows intersect the newly added

And if we start with a *down quark* as shown again
in Fig 32,

we can connect *x _{4}* to

And we get a particle with the same charge as the *down quark* but with more
energy than the *strange quark*. So I call it a Bottom quark, *
b*. Its antiparticle partner is
the

Just out of curiosity, I wonder if there is a particle with the configuration of Fig 34.

This particle is achieved by using the *tau* particle in Fig 26 and connecting *x _{1}* to

But what might *neutrinos* be? Neutrinos have almost no mass, they rarely interact
with other particles, and they only interact through the weak
force. This means that neutrinos are the result of the decay of
$W$ and
$Z$
bosons. For example, a ${W}^{-}$
particle will decay into an electron and a neutrino. Let's draw these
configurations to see how a *neutrino* might be configured. See Fig 35.

Here, a ${W}^{-}$
particle on the left decays into an *electron*, ** e_{1}**, and a

If we can connect the head of one arrow to the tail of another to create a *muon*
as was done in Fig 23, then it should be just as easy to connect the head
of one arrow to its own tail as shown in Fig 36. And since this circle has
just as much of the line going forward as backwards in time, it has a net zero
*electric* charge. Now we have a charge of
$-1$ on both sides of Fig 35, and we have an *electrically neutral neutrino*
just as the Standard Model does. Also, since the black arrow ends at the same
place it started, there is no information nor energy propagated to a different
point in space. But perhaps there is energy transmitted through the
magenta arrow.

However, we still have the tail of the
magenta *weak* arrow floating without a
connection in Fig 36. We can't have this because any arrow here represents a
subtraction of two quantities. A floating tail represents a subtraction of
nothing when both sides of the minus side are required. Since the quantities on
both sides of the subtraction can be commuted to form the anti-particle, both
have equal status, and you cannot go without one or the other.

If the black, *electric* arrow can end at the same point where it began, then the
magenta *weak* arrow can end on the
same black arrow where it began. See Fig 37.

Since space is 3D, having the magenta arrow outside the black arrow is equivalent, topologically, to having the magenta arrow inside the black arrow, as shown in Fig 38.

Now it can be seen that the magenta arrow is pulling together the black arrow so it ends where it began.

But the question is how would such a particle propagate through the spacetime points. If the black arrow can close in on itself, then it can open up to start on a different point than where it began. Then it can close on this different point. It would open and close on different points along a path and in this way propagate from point to point. Perhaps a shorter magenta arrow represents less energy, and perhaps the length of this arrow oscillates, causing the black arrow to seek different starting or ending points.

We also have that the muon decays into a ${W}^{-}$ particle and a neutrino. If we draw this interaction in terms of the framework here, we get Fig 39.

It seems a *neutrino* spontaneously
forms at *x _{2}*

And we might wonder if a *neutrino* could break any
bond of *electrons* or *positrons*? For example, could a *neutrino*
break apart or pull together the *electrons* on only one side of the
*strong* arrow in an *anti-charm* or *anti-top quark*? If it did,
then the *quark* that would result would be unbalanced and would probably not hold together.
There would be more energy on one side of the
*strong* arrow than on the other, and
would fly apart. Or perhaps such particles do exist for the briefest of moments
such that we haven't seen them yet. If we have a wave function for composite
particles such as these, the math would probably show which combinations are
stable, why, and how fast unstable versions would last.

And finally, there is the photon. Photons have no rest mass or charge and always travel at the speed of light. They are created by particle/anti-particle collisions, they annihilate by creating particle/anti-particle pair production, and they interact with electrically charged particles. So photons interact with electrons, muons, and tau particles and their anti-particles, photons interact with quarks and anti-quarks, and photons interact with the W bosons. But photons do not interact with neutrinos, the Z bosons, or gluons, since these don't have a net electric charge.

The question then, is how are photons represented in terms of this framework. Are
*photons* an elementary particle with no constituents, or are they a composite
particle like many described so far? If a photon is created out of the debris of
particle/anti-particle collisions, or if a particle/anti-particles can combine
to form a photon, this suggest that a photon is a composite of at least an *
electron* and a *positron*. But how could the combination of two particles with
mass form a particle with no mass?

If we have
both an *
electron* and a *positron* traveling together from *x _{1}* to

${\left(\frac{m}{2\pi \hslash i({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}+{\left(\frac{m}{2\pi \hslash (-i)({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{-im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}$ .

And the second term can be manipulated to get,

${\left(\frac{m}{2\pi \hslash i({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}+i{\left(\frac{m}{2\pi \hslash i({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{-im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}$.

Which is equal to,

${\left(\frac{m}{2\pi \hslash i({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left({e}^{im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}+i{e}^{-im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}\right)$ .

And since $i={e}^{i\pi /2}$, the above equates to,${\left(\frac{m}{2\pi \hslash i({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left({e}^{im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}+{e}^{i\pi /2}{e}^{-im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}\right)$ .

But when multiplying exponentials of the same number, *e*, we can add the
exponents to get,

${\left(\frac{m}{2\pi \hslash i({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left({e}^{im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}+{e}^{-im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})+i\pi /2}\right)$ .

Then we can take the imaginary number in the denominator of the square-root and modified it to $1/{i}^{1/2}={\left({e}^{-i\pi /2}\right)}^{1/2}={e}^{-i\pi /4}$ , and we get,

${\left(\frac{m}{2\pi \hslash ({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{e}^{-i\pi /4}\left({e}^{im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})}+{e}^{-im{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})+i\pi /2}\right)$ .

The $-i\pi /4$ can be added to each exponent to get,

${\left(\frac{m}{2\pi \hslash ({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left({e}^{i\left(m{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})-\pi /4\right)}+{e}^{-i\left(m{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})-\pi /4\right)}\right)$ .

Now the sum of exponentials is of the form,

${e}^{i\theta}+{e}^{-i\theta}=2\mathrm{cos}(\theta )$,

with $\theta =\left(m{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})-\pi /4\right)$. So finally this gives us a photon amplitude of,

${\left(\frac{m}{2\pi \hslash ({t}_{2}-{t}_{1})}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ $2\mathrm{cos}\left(m{({x}_{2}-{x}_{1})}^{2}/2\hslash ({t}_{2}-{t}_{1})-\pi /4\right)$.

We can recognize that,

$m{\left({x}_{2}-{x}_{1}\right)}^{2}/2\hslash \left({t}_{2}-{t}_{1}\right)=\left(m/2\hslash \right){\left(\left({x}_{2}-{x}_{1}\right)/\left({t}_{2}-{t}_{1}\right)\right)}^{2}\left({t}_{2}-{t}_{1}\right)$ ,

which is equal to, $\left(m{v}^{2}/2\hslash \right)({t}_{2}-{t}_{1})=E({t}_{2}-{t}_{1})/\hslash $.

Notice that *E* has the units of Joules, [J],
$({t}_{2}-{t}_{1})$ has the units of seconds, [s], and
$\hslash $
has the units of Joule-seconds, [Js]. So
$E({t}_{2}-{t}_{1})/\hslash $
has the units of
[J][s]/[Js]=1, which mean that expression is dimensionless, it has no units. And
this is what is require for it to be the argument of the cosine function. The
argument of a sine or cosine function is an angle which has no units of measure.

Yet the cosine function is typical of a wave travelling through space and time.
We have the time portion,
$({t}_{2}-{t}_{1})$, so can we express the energy, *E*, in terms of a variable more
related to a wave, such as frequency? If we let
$E=\hslash \omega $,
where $\omega $ is the angular frequency, then
again the argument of the cosine is dimensionless, as required. The equation,
$E=\hslash \omega $, is known as the Planck-Einstein relation for the energy of a photon. And so our cosine function starts at
*x _{1}* at

Section
8

Interactions

So far the framework of quantum theory has been derived here from first principles. And the implications and higher implications have been correlated to particles. All this seems to match the Standard Model. But these considerations will be more credible if the various interactions of the Standard Model can be explained in terms of this framework. How exactly do the particles here interact?

I've shown how the ${W}^{-}$
and the *muon* decay resulting in other particles and *neutrinos*. But there are quite a
few more interactions to explain. Fig 40 shows what types of particles interact
with which other types of particles. For example, the gluon only interact with
themselves and the quarks. The neutrino interactions always involve a ${W}^{-}$
or ${W}^{+}$
boson. The leptons do not directly interact with the quarks. And photons
do not interact with the neutrinos, the gluon, the ${Z}^{0}$
boson, nor other photons. All these interactions and lack of interactions need
to be explained with this framework.

Individual interactions are usually depicted with Feynman diagrams, where a
particle is depicted as a straight, wavy, or curly line, depending on the type
of particle. In a Feynman diagram a particle or particles are shown to come in
on the left, meet in the middle at a vertex (a dot), and other combinations of
particles are shown to leave at the right. For example, the Feynman diagram of
the $W}^{-$
decaying to an *electron* and a *neutrino* is
shown in Fig 41
below.

Fig 41: $W}^{-$**→** **e**^{−} +
**${\overline{\nu}}_{e}$**

The more complete list of Feynman diagrams with one vertex is shown here. Keep in mind that you can rotate Fig 41 so that any others of those particles are coming in. So, for example, we can rotate Fig 41 so that the electron and neutrino are coming in and the Weak boson is going out as shown in Fig 42.

Fig 42:
**e**^{+} +
**${\nu}_{e}$**** → $W}^{+$**

But now, with time increasing from left to right on the page, the $W}^{-$, electron, and neutrino are going in the reverse direction in time. This makes the Weak boson a $W}^{+$ particle, the electron converts to a positron, and the neutrino converts to its anti-particle. Yet this is still a valid interaction.

In the Standard Model there is no explanation given for why these interactions are permissible. But if the framework here has merit, then it may give a reason for these particular interactions; it may explain what's happening inside the vertices.

The first interaction to consider is the *photon* with the *electron*.
Fig 43 shows a *photon* consisting of *e _{1}* and

The *positron* part of the *photon* establishes a connection with *e _{2}*
to become a different photon than before. Since

conservation of charge connections

Interactions, decays, superposition of all possible interaction to create the field version of the path integral, and how it relates to Feynman diagrams.

coupling constants

Section
9

Conclusion

In the Standard Model, there is assigned a separate mathematical field for each type of particle, a field for the electrons and positrons, a field for the quarks and anti-quarks, a field for the photons, a field for gluons, and a field for the neutrinos, etc. Each type of particle is considered elemental; there is no common substructure to explain them. When particles are capable of influencing each other by interacting, the strength of the interaction is determined by a coupling constant. All the various coupling constants are determined by experimental data. The existence of all the particles and how they interact is determined from direct and indirect observational data alone. There is no fundamental principles to derive quantum theory or the particles to which it applies. The math used is only used because it fits the data from experiment.

Here, however, is described a common substructure from which all the various particles are produced. The reason that this substructure in not discernible by modern methods is that each type of particle is constructed of differentials in the path integral. So each type of particle is again its own differential in a path integral. And as such, differential quantities in an integral are not directly observable, only their accumulative effects can be calculated and observed.