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\begin{document}
\vspace*{2in}
\begin{center}
{\bf Production of the Standard Model
From a Kaluza-Klein Theory}\\[.1in]
Peter Gillan\\
gillan@alpha.fdu.edu\\
c/o Peter Schaeffer, Becton Hall\\
Fairleigh Dickinson University, 1000 River Road,
Teaneck, New Jersey  07666, USA\\
Received 31 October 2001\\[.1in]
{\bf Abstract}
\end{center}

The Standard Model plus gravitation (SM) is
produced from the Kaluza-Klein theory of pure gravity
in six dimensions with fields depending on $\theta$ and $\phi$,
two compactified coordinates. The theory is called
KK($\theta,\phi$). When KK($\theta,\phi$) is quantized,
the Planck masses of the massive
Fourier modes, representing the dependence upon $\theta$ and $\phi$,
are quantum corrected downward to elementary particle masses,
rendering them observable. KK($\theta,\phi$) explains the origins of the
elementary particles, Maxwell's equations, the Dirac equation,
the weak interactions and the strong interactions.
Quark confinement, asymptotic freedom and chiral
symmetry breaking are achieved. Ten parameters unexplained in the SM
are produced from KK($\theta,\phi$).
Superweak vectors with precisely determined masses are predicted.
Testable predictions that new leptons exist are made.
In KK($\theta,\phi$), the cosmological constant vanishes naturally.
KK($\theta,\phi$) is trivially anomaly-free.
The relationship between quantum mechanics and general relativity
is demonstrated.

\begin{center}
PACS numbers: 04.50.+h, 12.10.$-$g, 12.20.Ds, 11.27.+d\\[.1in]
Keywords: Kaluza-Klein, Standard Model, Theory of everything.\\
\clearpage
{\bf Part I. The Classical Theory}\\[.1in]
{\bf 1. Introduction}
\end{center}

In special relativity, Einstein put the dimension of time on the same
footing as the dimensions of space.  In effect, he added one dimension
of time to the three dimensions of space.
Likewise, we add two dimensions of time to the existing four
dimensions.  This brings the number of dimensions of time up to three.
There are three dimensions of time because there are three dimensions
of space.
This finishes the job of making time symmetrical to space.

In general relativity, Einstein took as the Lagrangian for his theory
of gravitation the curvature scalar $R$.
Likewise, we take as the Lagrangian for our theory the 
curvature scalar $\w R$ in six dimensions.  Thus, our theory is
Kaluza-Klein [1,2] in six dimensions.

This paper produces the Standard Model [3-5] plus gravitation (SM)
from the Kaluza-Klein theory of pure gravity in six
dimensions (6D) with fields allowed to depend on the two
compactified coordinates.
I call this theory KK($\theta,\phi$).
The compactified coordinates $\theta$ and $\phi$
are the usual spherical coordinates.

The original Kaluza-Klein theory
is not realistic.  For example, Kaluza's original theory produced
only two terms in its Lagrangian --- those for the gravitational and
electromagnetic fields.  All other fields and terms that we now know
exist do not appear in the Lagrangian.  Kaluza assumed fields were
independent of the fifth coordinate. When this restriction is lifted and
with natural amplifications, the balance of the SM is produced.

Klein compactified the fifth
dimension, resulting in an infinite tower of Fourier modes.  The lowest
order mode was massless.  The higher order particles, which
represented the dependence on the fifth coordinate, had Planck masses.
This dependence produces the fields and terms that
we now know exist.  The Planck masses of the particles representing
compactified coordinate dependence are quantum corrected downward
to elementary particle masses when KK($\theta,\phi$)
is quantized.

Table~1. Tower of Fourier modes in KK($\theta,\phi$).
Quantum numbers for the vectors and leptons are $|lm\rangle$.
\begin{center}
\begin{tabular}{ccccccccccccccc} \hline\hline
          $l$&$\backslash m$&$-3$&$-\tfrac52$&$-2$&$-\tfrac32$&
          $-1$&$-\tfrac12$&$0$&$\ \tfrac12$&$\ 1$&
          $\ \tfrac32$&$\ 2$&$\ \tfrac52$&$\ 3$\\[.03in] \hline
          $3$& &$A_\mu^{3,-3}$& &$A_\mu^{3,-2}$& 
                &$A_\mu^{3,-1}$& &$A_\mu^{30}$& &\ $A_\mu^{31}$& 
                &\ $A_\mu^{32}$& &\ $A_\mu^{33}$\\
           & & & & & & & & & & & & & & \\
          $\tfrac52$& & &$\nu_{\tau1}$& &$\nu_{\tau2}$&
                                 &$\tau_1^-$& &\ $\nu_{\tau3}$&
                                 &\ $\tau_2^+$& &\ $\nu_{\tau4}$& \\
           & & & & & & & & & & & & & & \\
          $2$& & & &$A_\mu^{2,-2}$& &$A_\mu^{2,-1}$& &$A_\mu^{20}$& 
                            &\ $A_\mu^{21}$& &\ $A_\mu^{22}$& & \\
           & & & & & & & & & & & & & & \\ 
          $\tfrac32$& & & & &$\nu_{\mu1}$& &$\mu_1^-$& &\ $\nu_{\mu2}$& 
           &\ $\mu_2^+$& & & \\
           & & & & & & & & & & & & & & \\
          $1$& & & & & &$W^-$& &$Z^0$& &\ $W^+$& & & & \\
           & & & & & & & & & & & & & & \\
          $\tfrac12$& & & & & & &$e^-$& &\ $\nu_e$& & & & & \\
           & & & & & & & & & & & & & & \\
         $0$& & & & & & & &$\gamma$& & & & & & \\ \hline\hline
\end{tabular}
\end{center}

One of the new features of this work is a new way of doing quantum
field theory. This could be called finite QED. It is QED formulated for
extended particles. Extended particles are not point particles, but rather
have radii of the order of the Planck length. In the classical theory, therefore, 
they have Planck masses. The quantum mass correction equation in this
version of QED converts these Planck masses to elementary particle masses.
This renders the particles representing the
compactified coordinate dependence
observable. I will show these are the elementary particles found in nature.
See Table~1.

In addition, allowing derivatives with respect to the two compactified
coordinates produces many terms in the Lagrangian $\w R$.
The nonvanishing terms that appear automatically within $\w R$
are precisely those found in nature.  This is the realization of
Einstein's dream [6--8] of accounting for all physical phenomena
from the ``pure marble" of geometry, without the ``base wood"
of additional matter fields.

We will produce the Lagrangian for the observable features of the SM from the
6D curvature scalar, the Lagrangian for KK($\theta,\phi$).
Each term in the SM is a separate assumption.
These terms may be produced by just one assumption, the curvature
scalar for KK($\theta,\phi$).  

Ten unexplained masses and coupling constants are produced in Part~II.
Neither the symmetry group
$SU(3)\times SU(2)\times U(1)$ nor the Higgs mechanism will be derived.
These are shown to be superfluous in Part~II.
Theories of interactions in the Standard Model are based on quantum field theory.
They are renormalizable. This is a requirement for point particles.
But if particles are extended, as claimed by Superstrings, renormalizability
and the gauge symmetry that preserves it become unnecessary.
Likewise, the Higgs mechanism, which is assumed to generate mass,
becomes superfluous.

This is unlike today's Kaluza-Klein theories, where observed elementary
particles lie in the zero-mass sector of the Kaluza-Klein theory and receive
a mass from the Higgs mechanism. Here a determination of the masses of the
elementary particles is not possible.  They are almost completely unexplained.
In KK($\theta,\phi$), one can convert the Planck masses
obtained from the classical
theory to elementary particle masses from quantum corrections,
thereby rendering a complete explanation for them.

I will now present a comparison of KK($\theta,\phi$) with other Kaluza-Klein
theories.  This will follow closely the review by Overduin and Wesson [9].
DeWitt [10] was the first to suggest incorporating the
non-Abelian symmetry $SU(2)$ of the SM into a Kaluza-Klein theory.
Others [11--13] took up the challenge, ending with Cho and Freund [14,15].
The main difference between these Kaluza-Klein theories
and KK($\theta,\phi$) is the origin of vectors in the theories.
In the former, each dimension produces one vector in the usual way.
In the latter, one dimension produces all vectors needed for the SM.
This procedure is superior because it accounts for the
$SU(2)\times U(1)$ structure of electroweak symmetry.
Here the photon is associated with the lowest order $l=m=0$
spherical harmonic and the $W^\pm$ and $Z^0$ are associated
with the next highest order $l=1$, $m=\pm1$ and $m=0$ harmonics.
See Table~1.
Thus, KK($\theta,\phi$) explains why there are three weak vectors,
while there is one electromagnetic vector. In addition, it explains why
the photon is massless, while the W and Z have mass.
Also explained is why the $W^\pm$ has charge, while the photon
and $Z^0$ are neutral.

A disadvantage of the non-Abelian Kaluza-Klein theories is that they
naively require one dimension for each vector produced. Thus,
many extra dimensions are required to produce the group
$SU(3)\times SU(2)\times U(1)$ of the SM. In addition, there is no
reason why some of the higher dimensions should be different than others.
This would be required to explain the group
$SU(3)\times SU(2)\times U(1)$. In particular, the question
of why the strong interactions are different than the electroweak
is left unanswered.

These two types of Kaluza-Klein theories, however, are similar in the
following respects: All matter fields are contained within the
higher-dimensional curvature scalar. There are no external, additional
matter fields. This embodies Einstein's vision of
nature being the result of pure geometry. In addition, each of these
Kaluza-Klein theories is pure gravitation with indices allowed to
run to values greater than four.

References [16--23] give the physical implications of compactified
coordinate dependence for
general relativity in the macroscopic realm --- such as its effect on the
advance of the perihelion of Mercury. KK($\theta,\phi$) gives their implications
for microscopic physics  --- elementary particles.

All unified field theories that include gravity (theories of everything)
suffer from the same problem --- the inability to make sufficient
contact with low-energy phenomenology.  These include
the original Kaluza-Klein theory in five dimensions, the extensions
of this theory with more than one higher dimension (intended to account
for non-Abelian symmetry), Supergravity, Superstrings,
and the present-day theory of everything, M-theory.

For example, none of these theories produces the right particle spectrum.
Kaluza-Klein theory without supersymmetry has only bosons.
The incorporation of fermions in such a theory is apparently without explanation.
Supersymmetry ameliorates the situation somewhat. However in this case,
the most abundant type of particles is scalars.  d=11, N=8 Supergravity
has 70 of them.  This is in contradiction to observation, which indicates
that there are no scalars.  The presently-accepted way out is to assume
that the scalars are too massive to be detected.  This means up to 70
additional assumptions.

The solution to this problem is to incorporate the branch of mathematics
known as spinor theory into Kaluza-Klein theory.  Spinor theory dictates
that a null vector (the photon) may be equated to a pair of spinors
(fermions).  When this is combined with 6D Kaluza-Klein with
compactified coordinate dependence, all of the observed fermions
are produced. And because there is no relation between fermions and
scalars, this produces no scalars. Thus, one has agreement
with observation.

With only two higher dimensions that form an ordinary 2-sphere,
compactification in KK($\theta,\phi$) is trivial, just as it is for the original
Kaluza-Klein theory with one higher dimension. We do not have to worry about
compactification as in, for example, the theories of Cho and Freund
[24--28], in which the higher dimensions have general curvature.
In this case, one runs into difficulties with the consistency
of the field equations when the dependence upon the compactified coordinates is
eliminated (Duff [29--32]). KK($\theta,\phi$) does not have this
problem, but if it did we would solve it by allowing fields to depend on the
compactified coordinates.  We would agree with
Cho's call [33--36] for abandonment of the ``zero modes approximation."

In KK($\theta,\phi$),
one does not need any of the various compactification mechanisms
such as altering Einstein's equations by incorporating torsion [37--40],
adding higher-derivative terms such as $R^2$ to the Lagrangian [41],
or by adding matter fields.  This last method, with the right matter
terms, is known as spontaneous compactification.  The first example of 
this is that of Cremmer and Scherk [42,43].

The current method of choice for the incorporation of fermions into
Kaluza-Klein theory is supersymmetry.  This is more natural than their 
incorporation by hand, but there are drawbacks. For example, the
fermionic superpartners of Kaluza-Klein's bosons have not been 
detected.  Similarly, the bosonic superpartners of the observed fermions
are not observed.  There is little evidence from experiment
for supersymmetry.  By comparison, the evidence for
KK($\theta,\phi$) is all fermionic physics in the SM as it reproduces
the observable features of the SM.  Spinor theory
dictates that the photon may be equated to all possible pairs of
fermions in the Lagrangian.  Spinor theory in KK($\theta,\phi$) is proven
every day in particle accelerators. It accounts for pair production 
by producing spinor-vector interaction terms in the Lagrangian
(the assumption of minimal coupling in the SM.)  With two spinors and one vector
as factors, this term represents a Feynman diagram with two
external fermionic lines and one external bosonic line.

In addition, one of the new terms to appear as a result
of compactified coordinate dependence is the term for the Dirac equation.
This equation and all it implies must be assumed in d=11 Supergravity.

One advantage of Supergravity is that it provides a number for the
dimensionality of spacetime.  Nahm [44] showed 11 is the 
maximum number of dimensions of spacetime for supersymmetric gravity.
In 11 dimensions, nature is maximally supersymmetric.
Witten [45] proved 11 is the minimum number of dimensions
required for the incorporation of the group $SU(3)\times SU(2)\times U(1)$
of the SM.  However, this assumes the one-dimension-for-one-vector
aspect, along with its disadvantages, of the non-Abelian theories mentioned above.
Cremmer, Julia and Scherk [46] showed that in 11 dimensions
there is only one choice for the additional matter fields.
Freund and Rubin [47] showed that d=11 Supergravity naturally compactifies
to four macroscopic and seven microscopic dimensions.

KK($\theta,\phi$) also provides a number for the dimensionality of spacetime.
Instead of a symmetry between bosons and fermions, we postulate a
symmetry between space and time. This implies there are three dimensions of time.
There are three dimensions
of time because there are three dimensions of space.  Spacetime has
six dimensions.  In six dimensions the entire Standard Model, in all its
complexity and with all its idiosyncracies, can be explained.

The theory of d=11 Supergravity lost its status as the theory of everything
in the mid 1980's.
There were several reasons for this.  First, the compactified spacetime
did not contain quarks or leptons, nor the gauge group for the SM.
Second, one cannot build chirality, necessary for an accurate description
of fermions in the Standard Model, into a theory with an odd number of
dimensions.  Third, it had a large cosmological constant, contradicting
observation. Fourth, it had anomalies.
KK($\theta,\phi$) naturally accounts for what appears to be the gauge group
for the SM with its expansions in terms of spherical harmonics
for electroweak theory
and the higher-dimensional part of the 6D Lorentz group for
the strong interactions.  In KK($\theta,\phi$), these expansions, together
with spinor theory, do indeed produce quarks and leptons.
In six dimensions, chirality is possible.  The large cosmological
constant appearing in the 4D version of Kaluza-Klein theory
is naturally eliminated in KK($\theta,\phi$).

Superstrings [48] is supposed to be able to predict the parameters of the
Standard Model.  It does not do this, at least not at the present time.
I will show KK($\theta,\phi$) does this, at least better than Superstrings.
There are about 20 unexplained parameters in the SM. KK($\theta,\phi$)
explains ten of these at the present time.  This is ten more than Superstrings
explains.

String theory has three separate uniqueness problems.
First, the number of supersymmetric generators can have any value
from N=1 to N=8. Second, there five different classes of string
theory. Finally, within each class there are thousands of different
string theories, each corresponding to a different way one can
compactify seven higher dimensions. In contrast,
there is only one KK($\theta,\phi$).
KK($\theta,\phi$) does not have supersymmetry,
with its multiple values for N.  Particles in KK($\theta,\phi$) are not strings,
which are classified into five types. Finally, there is only
one way to compactify the two higher dimensions.  It is a trivial compactification
to the 2-sphere, similar to the compactification of the one HD of original
Kaluza-Klein to the 1-sphere or circle.  Thus, KK($\theta,\phi$) does not suffer
from a uniqueness problem.

Superstrings has matter terms
known as Chapline-Manton terms [49] that must be added to the
Lagrangian.  In six dimensions, KK($\theta,\phi$) has no matter terms.

Green and Schwarz [50] and Gross et al. [51] showed all anomalies
vanish for SO(32) and $E_8\times E_8$ string theory, respectively.
String theory may provide an anomaly-free path to quantum gravity [52].
However,
I claim quantum gravity is not the unification of gravity and quantum theory.
I can produce the entire Standard Model plus gravity without it.
How can a theory whose effects are entirely negligible be important?
The relationship between gravity and quantum theory is given
below. But if you must have a quantum theory of gravity,
my new way of doing quantum field theory should help.

This being said,
I will now show why KK($\theta,\phi$) is anomaly-free.
First, as I will show below, the 6D metric tensor
in KK($\theta,\phi$) contains all types of 4D matter fields.
The 6D curvature scalar in KK($\theta,\phi$) contains all types of 4D
matter terms.  This is accomplished by expanding the vector $\w g_{5\mu}$ in
KK($\theta,\phi$) in terms of spherical harmonics, spinors and 6D
Lorentz generators. Now, the Lagrangian for KK($\theta,\phi$)
is the 6D curvature scalar.
There are no other terms, which would be called 6D matter terms.
In other words, in six dimensions, spacetime is empty. There are no
currents. Thus, there are no conservation laws.
Now, anomalies are quantum mechanical violations
of conservation laws.
Without conservation laws there are no anomalies.
Therefore, KK($\theta,\phi$) is anomaly-free.

Quantum mechanics is not philosophical in origin,
but results from the properties of waves. For example, it is
the fact that one cannot simultaneously determine the position and
momentum of a wave that leads to the uncertainty principle.
Quantum mechanics is also known as wave mechanics.
The waves follow from wave equations, which are field equations.
The field equations originate from the Lagrangian.
The Lagrangian is the 6D curvature scalar.
The curvature scalar is the Lagrangian for general relativity.
Therefore, quantum mechanics is derived from general relativity.
This is the relationship between quantum mechanics and general relativity.

The worst disadvantage of string theory is that it does not make
predictions. They reside at the Planck mass.
M-theory is worse than string theory because it is nonperturbative.
The electroweak sector, which contains the testable predictions,
of KK($\theta,\phi$) is perturbative.
KK($\theta,\phi$) has a mechanism
for converting Planck masses to elementary particle masses.
Thus, it makes clear-cut physical predictions.  KK($\theta,\phi$) predicts the
Standard Model as I will show.

The purpose of this paper is to establish enough evidence for
KK($\theta,\phi$) to warrant further investigation.
This is suggested by the demonstration that KK($\theta,\phi$)
produces the observable features of the SM.  The standard model
is quite an achievement, but KK($\theta,\phi$) is far superior.
One must allow for the possibility that such a simple and effective
theory exists.

The organization of Part I is as follows:   
Section~2 breaks down the 6D metric tensor into
4D plus higher-dimensional quantities. Section~3 specifies the Lagrangian.
Four-dimensional gravitation is derived from KK($\theta,\phi$)
in Sec.~4.  The cosmological constant is eliminated in this section.
In Sec.~5, Maxwell and mass terms
for the photon, W and Z are obtained. Methods for dealing with
the extra coordinates in each term in the Lagrangian
are given in Sec.~6. Section~7 derives the
Dirac equation from the 6D curvature scalar.
Interaction terms for leptons and the W and Z are deduced
in Sec.~8. Section~9 derives the strong interactions.
Quarks, confinement, asymptotic freedom and chiral symmetry
breaking are produced in Sec.~10.


\begin{center}
{\bf  2. The 6D metric tensor}
\end{center}

In this section, we break down the 6D metric tensor into 4D and
higher-dimensional
quantities in preparation for its substitution into the 6D
curvature scalar.  The postulate of KK($\theta,\phi$) implies a
6D spacetime.  Included in this postulate of six dimensions
is the size, shape, spacelike or timelike nature and
connection to 4D spacetime of the higher dimensions.  Therefore, 
we will further postulate that the two higher dimensions form a sphere
of constant Planck length radius $r_0$.
The reason the two higher dimensions are compactified is cosmological
in origin. Einstein did not present cosmology in his papers
on special and general relativity. Likewise, we will not
deal with cosmology in this paper. This will come later.

In addition to being compactified, we postulate the higher dimensions
to be timelike.
It is widely believed that the higher dimensions in a Kaluza-Klein theory
must be spacelike. This is wrong. The truth is, the 
signature of the higher dimensions must be positive $(+ +)$. However,
a positive signature can represent either a timelike dimension
or a spacelike dimension. When one opens the front cover
of a copy of Misner, Thorne and Wheeler [53], one finds 20
authors that represent timelike dimensions with a plus
sign in the
signature. That is, they all assume the signature $(- - - +)$.
The signature for our metric tensor is $(- - - + + +)$.

The differential line element for two timelike dimensions that 
form a sphere of radius $r_0$ is
\begin{equation}
ds^2=\w g_{ij}d\w x^{\,i}d\w x^{\,j}=
     r_0^{\ 2}\,d\theta^2+r_0^{\ 2}\sin^2\!\!\theta\,d\phi^2,\label{eq:2.2}
\end{equation}
where $i,j$ take the value five or six and $\theta$ and $\phi$ are
the usual spherical coordinates. A hat over a symbol denotes a
6D quantity.

Instead of choosing $\w x^{\,5}=\theta$ and $\w x^{\,6}=\phi$ with
$\w g_{55}=r_0^{\ 2}$ and $\w g_{66}=r_0^{\ 2}\sin^2\!\!\theta$,
as is usually done, it is equivalent to set
\begin{eqnarray}
\widehat x^{\,5}&=&r_0\,\theta,\label{eq:2.2a}\\
\widehat x^{\,6}&=&r_0(\sin\!\theta)\,\phi,\label{eq:2.2b}\\
\widehat g_{55}&=&\widehat g_{66}=1,\label{eq:2.2c}\\
\widehat g_{56}&=&\widehat g_{65}=0.\label{eq:2.2d}
\end{eqnarray}

The incorporation of the factor of
$r_0$ into $\w x^{\,5}$ and $\w x^{\,6}$
makes the compactified coordinates more like the
4D ones, which have the units of distance. This suggests
we incorporate the factor of $\sin\!\theta$ into $\w x^{\,6}$ instead
of $\w g_{66}$. This eliminates the cosmological constant
while keeping the higher dimensions compactified. This is described in Sec.~4.
The value 1 for $\widehat g_{66}$ mimics
Ricci-flatness of the higher dimensions.
This means [30] the cosmological constant is zero and there is no
Higgs mechanism. Both these features are not typical of today's
Kaluza-Klein theories. 

For use in calculating the Lagrangian in Sec.~3, I tabulate here the
differentials and derivatives of the two compactified
coordinates $\w x^{\,5}$ and $\w x^{\,6}$ according to
Eqs.~(\ref{eq:2.2})--(\ref{eq:2.2b}) 
\begin{eqnarray}
d\w x^{\,5}&=&r_0\,d\theta,\label{eq:2.3a}\\ 
\frac{\partial}{\partial\w x^{\,5}}&=&
\frac{1}{r_0}\frac{\partial}{\partial\theta},\label{eq:2.3b}\\
d\w x^{\,6}&=&r_0\sin\!\theta\,d\phi,\label{eq:2.3c}\\
\frac{\partial}{\partial\widehat x^{\,6}}
&=&\frac{1}{r_0\sin\!\theta}\frac{\partial}{\partial\phi}.\label{eq:2.3d}
\end{eqnarray}

The 6D metric tensor is denoted by
$\w g_{\alpha\beta}$,
where indices in the
beginning of the Greek alphabet such as $\alpha,\ \beta,\ \gamma,$
and $\delta$ run from one
to six.  The indices $\mu,\ \nu,\ \rho,$ 
and $\sigma$ from the middle of the Greek alphabet range from
one to four. The first four coordinates
of the 6D spacetime are those of the ordinary 4D
spacetime of experience.

I postulate the connection between the compactified sphere
and 4D spacetime is as follows:
The 4D worldline of the particle under consideration
is in the $z$-direction of the embedding space
of the compactified sphere.
Therefore, the circular sixth dimension,
parameterized by the coordinate $\phi$,
is perpendicular to the $z$-axis, a line in 4D spacetime.
Taking the line to be $x^1, x^2, x^3$, or $x^4$, we have $\w g_{6\mu}=0$.
This is because the coordinate $\phi$, along with a line of 4D
spacetime, forms a cylinder and the metric tensor for these coordinates
of a cylinder is zero.  This is due to the simple reason that the
coordinates are perpendicular.  It is true of any right-circular cylinder.
It is not realized that this most common type of compactified geometry
(sometimes called cylindrical) causes the vector in original Kaluza-Klein
to be zero.

The fifth dimension is parameterized by the coordinate $\theta$.
It is measured from the positive  $z$-axis. It is not,
in general, perpendicular to the $z$-direction consisting of 4D spacetime.
Therefore, the vector $\w g_{5\mu}$ in KK($\theta,\phi$) is not zero.

The relation 
$g_{\mu\nu}dx^\mu{d}x^\nu 
=\widehat g_{\alpha\beta}d\widehat x^{\,\alpha}{d}\widehat x^{\,\beta}$
equates a distance in 4D 
spacetime, where $g_{\mu\nu}$ is the metric tensor,
to the same distance considered as part
of a 6D spacetime.  From this equation we may derive
a relation between $\widehat g_{\mu\nu}$ and $g_{\mu\nu}$.
The result is
\begin{equation}
\widehat g_{\mu\nu}=g_{\mu\nu}+A_{\mu}A_\nu,\label{eq:2.4}
\end{equation}
where
\begin{equation}
A_\mu=-\frac{\partial \widehat x^{\,5}}{\partial x^\mu}\,.\label{eq:2.4a}
\end{equation}
Here we have used the fact derived from above that
\begin{equation}
g_{\mu\nu}=\w g_{\alpha\beta}
\frac{\p \w x^{\,\alpha}}{\p x^\mu}\,
\frac{\p \w x^{\,\beta}}{\p x^\nu}\label{eq:2.5}
\end{equation}
and the definition
\begin{equation}
A_\mu\equiv\w g_{\mu5}=\w g_{5\mu}.\label{eq:2.6}
\end{equation}

To summarize the results for the 6D metric tensor
in terms of 4D quantities, we have
\begin{equation}
\left(
\begin{array}{ccc}
\w g_{\mu\nu}&\w g_{\mu5}&\w g_{\mu6}\\
\w g_{5\nu}&\w g_{55}&\w g_{56}\\
\w g_{6\nu}&\w g_{65}&\w g_{66}
\end{array}
\right)=\left(
\begin{array}{ccc}
g_{\mu\nu}+A_\mu A_\nu&A_\mu&0\\
A_\nu&1&0\\
0&0&1
\end{array}
\right).\label{eq:2.8a}
\end{equation}
The contravariant metric tensor, introduced via the relations
$\w g^{\,\alpha\beta}\w g_{\beta\gamma}=
\w \delta_{\,\ \gamma}^{\,\alpha},$
is
\begin{equation}
\left(
\begin{array}{ccc}
\w g^{\,\mu\nu}&\w g^{\,\mu5}&\w g^{\,\mu6}\\
\w g^{\,5\nu}&\w g^{\,55}&\w g^{\,56}\\
\w g^{\,6\nu}&\w g^{\,65}&\w g^{\,66}
\end{array}
\right)=\left(
\begin{array}{ccc}
g^{\mu\nu}&-A^\mu&0\\
-A^\nu&
1+A^\mu A_\mu&0\\
0&0&1
\end{array}
\right).\label{eq:2.8b}
\end{equation}

In general, any arbitrary field should depend
upon all six coordinates, including the two extra ones.  
This dependence on the extra coordinates
is represented by a Fourier expansion in terms of
spherical harmonics.  We have
\begin{equation}
g_{\mu\nu}(\w x^{\,\alpha})=
\sum_{l=0}^\infty
\sum_{m=-l}^l
g_{\mu\nu}^{lm}(x^\rho)
Y_{lm}(\theta,\phi),\label{eq:2.9a}
\end{equation}
\begin{equation}
A_\mu(\w x^{\,\alpha})=
\sum_{l=0}^\infty
\sum_{m=-l}^l
A_\mu^{lm}(x^\nu)
Y_{lm}(\theta,\phi),\label{eq:2.9b}
\end{equation}
where the $Y_{lm}(\theta,\phi)$ are the spherical harmonics.


\begin{center}
{\bf 3. The Lagrangian}
\end{center}

The Lagrangian density for KK($\theta,\phi$) is 
\begin{equation}
{\cal L}=k\w R,\label{eq:3.1}
\end{equation}
where $\w R$
is the 6D curvature scalar and 
$k=c^3/16\pi G$, where $c$ is the speed of light and $G$ is the
constant of gravitation.
We have
\begin{equation}
\w R=\w g^{\,\a\bb}\w R_{\a\bb},\label{eq:3.1a}
\end{equation}
where $\w R_{\a\bb}$ is defined from the commutator of covariant derivatives
\begin{equation}
\w R_{\a\bb}=
\p_\g\w\G^{\,\g}_{\,\ \a\bb}
-\w\G^{\,\dd}_{\ \,\bb\g}\w\G^{\,\g}_{\ \,\a\dd}
-\p_\bb\w\G^{\,\g}_{\,\ \a\g}
+\w\G^{\,\dd}_{\,\ \a\bb}\w\G^{\,\g}_{\ \,\dd\g},\label{eq:3.1b}
\end{equation}
where 
\begin{equation}
\w\G^{\,\g}_{\,\ \a\bb}=
\w g^{\,\g\dd}\w\G_{\dd\a\bb},\label{eq:3.1c}
\end{equation}
where
\begin{equation}
\w\G_{\dd\a\bb}=
\tfrac{1}{2}\left(
-\p_\dd\w g_{\a\bb}
+\p_\a\w g_{\bb\dd}
+\p_\bb\w g_{\dd\a}\right).
\label{eq:3.1d}
\end{equation}

The action integral for KK($\theta,\phi$) is
\begin{equation}
I=k\int\w R\sqrt{-\w g}\,\,d^{\,6}\w x,\label{eq:3.2}
\end{equation}
where $\w g$ is the determinant of the 6D
metric tensor.  This determinant may be expressed in terms of
the 4D metric tensor $g_{\m\n}$.
If we multiply the middle column of Eq.~(\ref{eq:2.8a})
by $A_\n$ and subtract the resulting column from the first, we obtain
\begin{equation}
\sqrt{-\w g}=
\left(-\left|
\begin{array}{ccc}
g_{\m\n}&A_\m&0\\
0&1&0\\
0&0&1
\end{array}\right|\right)^{1/2}=
\sqrt{-g}\,.
\label{eq:3.2a}
\end{equation}
Using Eqs.~(\ref{eq:3.2a}),~(\ref{eq:2.3a}) and~(\ref{eq:2.3c}), 
the 6D volume element becomes
$\sqrt{-g}\,r_0^{\ 2}\sin\!\theta\,d\theta\,d\phi\,d^4x$.
The constant $r_0^{\ 2}$ may be ignored
as the entire Lagrangian is multiplied by it and the field
equations remain the same after its removal.

The dependence upon the two extra coordinates has a profound effect
on 6D Kaluza-Klein.  Instead of obtaining only the terms for
four-dimensional gravitation and electromagnetism
(the Kaluza-Klein miracle) in the Lagrangian,
one now obtains hundreds of new, odd-looking terms.
This is because terms with $\partial_5$ or $\partial_6$ are not zero now.
This embarrassment of riches may be why Kaluza did not
allow dependence upon the extra coordinate.

To analyze the large number of terms, we determine the
free-field terms for each field of interest.
These are obtained by setting all other fields to zero
in each term in the Lagrangian.
Thus, the free-field terms have only
the field under consideration in them as nontrivial factors.
Interaction terms have more than one type of field in them.

There are three classes of terms in the Lagrangian.
First are the terms for the
free-$g_{\m\n}$ field.  These contain only $g_{\m\n}$ or
$g^{\,\m\n}$; they are obtained by setting $A_\m=0$.  These
terms are important macroscopically, where the sources of
$g_{\m\n}$ add and those for the $A_\m$ field cancel.
Second are the terms for the free-$A_\m$ field.  These are
obtained by setting $g_{\m\n}=\eta_{\m\n}$, the Minkowski metric
tensor.  Note one cannot set $g_{\m\n}=0$; however it can be
frozen out by setting it equal to a constant.
This causes all derivatives of $g_{\m\n}$
and $g^{\,\m\n}$ to vanish.  These terms are important 
microscopically, where the sources of the $A_\m$ field do
not cancel and those for the $g_{\m\n}$ field are very
small.  Third are the terms for the interaction of the
$g_{\m\n}$ and $A_\m$ fields.  These contain at least one
factor of $A$ and at least one derivative of $g_{\m\n}$
or $g^{\,\m\n}$.  These terms vanish both macroscopically
and microscopically, where $A_\m$ vanishes and $g_{\m\n}$
is constant, respectively.  We will not consider this
class of terms further.


\begin{center}
{\bf 4. Four-dimensional gravitation}
\end{center}

There are three types of free-field  terms in the Lagrangian for
$g_{\mu\nu}$: the 4D curvature scalar $R$, terms
with two factors of $\gh$ and terms with four factors of
$\gh$. The first two of these types are 
\begin{eqnarray}
&k&R\label{eq:4.1}\\
-&k&\GG\p_5^{\ 2}\gh\label{eq:4.2}\\
-&k&\GG\p_6^{\ 2}\gh\label{eq:4.3}\\
-\tfrac{1}{2}&k&\p_5\GG\p_5\gh\label{eq:4.4}\\
-\tfrac{1}{2}&k&\p_6\GG\p_6\gh\label{eq:4.5}
\end{eqnarray}

Because the size of the compactified sphere is of the order
of the Planck length, one cannot observe the compactified
coordinate dependence of $\gh$.  To eliminate it, one must
integrate over the compactified coordinates, thereby
averaging over them and obtaining a 4D description of the
6D curvature scalar.

Term~(\ref{eq:4.1}) contains the set of terms for
4D gravitation.  These are obtained by setting
$\gh=g_{\m\n}^{00}|00\rangle$ and
$\GG=\overline g^{\,\m\n}_{00}\langle00|$.  One may group all the kets
together and all the bras together to yield a single bra
$\langle00|$ and ket $|00\rangle$, which when multiplied
together yield one.  The result is the curvature scalar
$R(g_{\mu\nu}^{00})$, which contains $g_{\m\n}^{00}$ and
$g_{00}^{\,\m\n}$ only.  We interpret $g_{\m\n}^{00}$
as the graviton field.
Therefore, KK($\theta,\phi$) contains 4D gravitation.

The procedures for taking the derivatives with respect to
the compactified coordinates
and integrating over $\theta$ and $\phi$ for 
Terms~(\ref{eq:4.2})--(\ref{eq:4.5})
are the same as those for Terms~(\ref{eq:5.2})--(\ref{eq:5.4a})
described in Sec.~6.  Considering only $g_{\mu\nu}^{lm}$ with
$l=1$ or 0, we arrive at the following result for these terms
\begin{eqnarray}
\frac{3\sqrt3\,-9}{12r_0^{\ 2}}&k&
\overline g_{00}^{\,\m\r}\overline g_{00}^{\,\n\s}
g_{\r\s}^{1,-1}g_{\m\n}^{11}\label{eq:4.6a}\\
+\ \ \frac{3\sqrt3\,-9}{12r_0^{\ 2}}&k&
\overline g_{00}^{\,\m\r}\overline g_{00}^{\,\n\s}
g_{\r\s}^{11}g_{\m\n}^{1,-1}\label{eq:4.6b}\\
-\ \ \frac{2\sqrt3\,+\sqrt{30}\,}{12r_0^{\ 2}}&k&
\overline g_{00}^{\,\m\r}\overline g_{00}^{\,\n\s}
g_{\r\s}^{10}g_{\m\n}^{10},\label{eq:4.6c}
\end{eqnarray}
where we have considered only the $g_{00}^{\,\m\n}$ term in
$\GG$ and only the $g_{\m\n}^{00}$ field in the factor of
$\sqrt{-g}\,$ in the differential volume element.  

As $g^{\,\m\n}_{00}$ is the graviton field, 
Terms~(\ref{eq:4.6a})--(\ref{eq:4.6c}) are mass
terms for the $g_{\m\n}^{1m}$ fields.  The field $g_{\m\n}^{00}$
does not have a mass term because its associated spherical harmonic
$Y_{00}$ does not depend on the compactified coordinates.
Therefore, $g_{\m\n}^{00}$ is massless, which is consistent
with our interpretation of it as the graviton.  The factor
$k$ in the coefficient of these terms is not absorbed into the
definition of the fields as it will be for vectors and spinors
so the calculation of masses from these coefficients is
slightly different from the usual procedure.  A reasonable
method has $km^2/2$ as the coefficient of each term.
The factor of $r_0^{\ 2}$ in the denominator of
the coefficients of these terms implies that the $g_{\m\n}^{1m}$
have Planck masses.  Possibly,
quantum corrections [54] reduce
these masses to the usual elementary particle mass scale.
Because the $l>0$ fields have masses, they are weaker than the massless
graviton field in small-scale spacetime.  This is for the same reason
the weak interaction is weaker than the electromagnetic.
Thus, they, like the graviton, may be neglected in this arena.
In large-scale spacetime, 
they play no role because their masses limit the range of their
interaction.  This is where they differ from the graviton and are
eliminated from physical importance.

Therefore, one may approximate $\gh$ with $g_{\m\n}^{00}$,
which becomes the 4D metric tensor.
Thus, $g_{00}^{\,\m\n}$ and $g_{\m\n}^{00}$ raise and lower
4D indices, respectively. In addition, only
$g_{\m\n}^{00}$ appears in the factor of $\sqrt{-g}\,$
in the volume element. The tensor $g_{\mu\nu}$ contains
only one known elementary particle, the graviton. As I will imply
later, all other elementary particles, including the W and Z, leptons,
quarks and gluons are contained within $A_\mu$. This is accomplished
by expanding $A_\mu$ in terms of spherical harmonics, spinors and/or
6D Lorentz group generators.
The various components of $A_\mu$ thus produced
will be identified with the elementary particles. 

There arise in the Lagrangian $k\widehat R$ the terms    
\begin{equation}
-k\w g^{\,55}\w g^{\,66}\p_5^{\ 2}\w g_{66}
-\tfrac12\,k\w g^{\,55}\p_5\w g^{\,66}\p_5\w g_{66}.
\label{eq:4.8}
\end{equation}
With our choice of $\w x^{\,5}$ and $\w x^{\,6}$, the field 
$\w g_{66}=\w g^{\,66}=1$. Therefore, these terms are zero. However,
had we made the usual choice for the compactified coordinates
$\w x^{\,5}=\theta$ and $\w x^{\,6}=\phi$, then
$\w g_{66}=r_0^{\ 2}\sin^2\!\!\theta$,
$\w g^{\,66}=1/(r_0^{\ 2}\sin^2\!\!\theta)$ and
$\w g^{\,55}=1/r_0^{\ 2}+A^\mu A_\mu$.
Terms~(\ref{eq:4.8}) would then contain the term $2k/r_0^{\ 2}$,
which is a cosmological constant roughly the size of the Planck mass.
Incorporating $\sin\!\theta$ into the coordinate $\w x^{\,6}$
instead of the scalar $\w g_{66}$ eliminates this large
cosmological constant.


\begin{center}
{\bf 5. The electroweak vectors}
\end{center}

In this section I derive Maxwell and mass terms for the photon,
W and Z.
The free-$A_\m$-field terms have anywhere from one to six factors
of $A$.  Of these, we consider here only terms with one or two of
these factors.  These terms are more important and are
easier to obtain.  They are
\begin{eqnarray}
&k&\e^{\m\n}
\e^{\,\r\s}\left(\tfrac{1}{2}\,
\p_\r A_\m\p_\n A_\s
-\tfrac{1}{2}\,
\p_\r A_\m\p_\s A_\n\right)\label{eq:5.1}\\
+&k&\e^{\m\n}\left(2\p_\m\p_5A_\n\right)\label{eq:5.2}\\
+&k&\e^{\m\n}\left(-\p_5^{\ 2}A_\m A_\n-A_\m\p_5^{\ 2}A_\n\right)
\label{eq:5.3}\\
+&k&\e^{\m\n}\left(-2\p_5A_\m\p_5A_\n\right)\label{eq:5.4}\\
+&k&\e^{\m\n}\left(-\tfrac{1}{2}\,\p_6A_\m\p_6A_\n\right).
\label{eq:5.4a}
\end{eqnarray}

Like many calculations in general relativity,
arriving at these terms is a straightforward, but lengthy calculation.
One should limit oneself to terms with two or less factors of A
as soon as possible in the calculation. Terms~(\ref{eq:5.1})
are part of the Kaluza-Klein miracle --- they lead to Maxwell's
equations for the photon when the compactified coordinate
dependence of $A_\mu$ is neglected and $\partial_5=\partial_6=0$.
This implies Terms~(\ref{eq:5.2})--(\ref{eq:5.4a}) are zero.
In addition, Terms~(\ref{eq:5.3})--(\ref{eq:5.4a})  are similar
to Terms~(\ref{eq:4.2})--(\ref{eq:4.5}), which are much easier to
obtain. Terms like~(\ref{eq:5.2}),
with only one factor of A, appear only twice
in the Lagrangian before they are added together.
Because these terms have one second derivative, one may neglect the
terms within the products $\Gamma\Gamma$,
which have two first derivatives,
in the curvature scalar, when deriving them.

Terms~(\ref{eq:5.1}) will lead to Maxwell's equations for the photon,
W and Z and predicted superweak vectors.   Term~(\ref{eq:5.2}) will
lead to the Dirac equation for fermions.
Terms~(\ref{eq:5.3})--(\ref{eq:5.4a}) will lead to mass terms
for vectors and interaction terms for spinors with vectors.
Terms with three to six factors of A are
self-interaction terms for vectors.

The procedures for evaluating these terms
by differentiating and integrating
over the spherical harmonics in them are described in the next section.
Because $g_{00}^{\,\m\n}=\e^{\m\n}$, then
$\e^{\m\n}$ acts as the contravariant 4D metric
tensor and is associated with the bra $\langle00|$ in these
terms. Because the vector $A^\m$ is really two separate fields,
$\e^{\m\n}$ and $A_\n$, it should be written as $\e^{\m\n}A_\n$
when integrating over $\theta$ and $\phi$; otherwise the result of
integration will be incorrect. This is explained in the next section.
The expansion for $A_\m$ is carried to $l=1$.  

Terms~(\ref{eq:5.1}) are the only terms
with two factors of $A$ and two 4D derivatives.
The result for these terms is 
\q
-\tfrac{k}{4}\,
F^{\m\n}_{00}F_{\m\n}^{00}
-\tfrac{k}{4\sqrt3}\,
F^{\m\n}_{1,-1}F_{\m\n}^{11}
+\tfrac{k}{4\sqrt3}\,
F^{\m\n}_{10}F_{\m\n}^{10}
-\tfrac{k}{4\sqrt3}\,
F^{\m\n}_{11}F_{\m\n}^{1,-1},\label{eq:5.5}
\qq
where $F_{\m\n}^{lm}=\p_\m A_\n^{lm}-\p_\n A_\m^{\m\n}$.
We will identify $A_\m^{00}$ as the photon field shortly.
It was thought that the Maxwell term for the photon
had to be of negative energy for timelike higher dimensions.
This is true when the timelike components
of the Minkowski metric tensor are negative.
However, it is the spacelike components
of the metric tensor that must be negative because
they represent the physical degrees of freedom.
The timelike components must be positive.
Therefore, the higher dimensions must be timelike if the Maxwell
term for the photon is to be of positive energy.
Despite this, the sign of the Maxwell term for $A_\m^{10}$
indicates negative energy.
This will also be the case for its mass term.
This is acceptable as long as the ground state
is defined to be the state closest to zero energy instead
of lowest energy.
Positive energy states are stable because the negative
energy states are no closer to zero energy than the
positive states.

The fields are now redefined to yield the conventional Maxwell term
for the photon
\begin{equation}
A_\m^{\pp lm}=\left(\frac{c^4}{16\pi G}
\right)^{1/2}A_\m^{lm},\label{eq:5.6}
\end{equation}
where we have included the factor of $c$ from $dx^4$ in the differential
volume element in the redefinition.
Terms~(\ref{eq:5.5}) now become
\q
-\tfrac{1}{4}\,
F^{\,\pp\m\n}_{00}F_{\m\n}^{\,\pp00}
-\tfrac{1}{4\sqrt3}\,F^{\,\pp\m\n}_{1,-1}
F_{\m\n}^{\,\pp11}
+\tfrac{1}{4\sqrt3}\,
F^{\,\pp\m\n}_{10}F_{\m\n}^{\,\pp10}
-\tfrac{1}{4\sqrt3}\,
F^{\,\pp\m\n}_{11}
F_{\m\n}^{\,\pp1,-1},\label{eq:5.5'}
\qq
where $F_{\m\n}^{\pp lm}=
\p_\m A_\n^{\pp lm}
-\p_\n A_\m^{\pp lm}$.
Henceforth, we will drop the prime for $A_\m^{\pp lm}$.
These terms appear in the Lagrangian for the SM.
They are the free-field (Maxwell) terms for the photon,
W and Z.  The photon is identified with $A_\mu^{00}$;
the $W_\mu^-$ is $A_\mu^{1,-1}$;
the $W_\mu^+$ is $A_\mu^{11}$,
and the $Z_\mu^0$ is $A_\mu^{10}$.

Thus,
\begin{eqnarray}
g_{\mu5}&=&A_\mu\label{1}\\
        &=&\sum_{l=0}^{\infty}\sum_{m=-l}^{l}A_\mu^{lm}Y_{lm}(\theta,\phi)\\
        &=&A_\mu^{00}Y_{00}+A_\mu^{1,-1}Y_{1,-1}+A_\mu^{10}Y_{10}
         +A_\mu^{11}Y_{11}+\ldots\\
        &=&A_\mu^{00}Y_{00}+W_\mu^{-}Y_{1,-1}+Z_\mu^{0}Y_{10}
         +W_\mu^{+}Y_{11}+\ldots
\end{eqnarray}
The photon is denoted by $A_\mu^{00}$, so as not to be confused with
$A_\mu$, which is given in Eq.~(\ref{1}).  I use only the physical fields;
there is no auxiliary vector ${\bf B_\mu}$ nor consequent mixing.

Note that we have given the W and Z the same normalization as the
photon. This is the simplest thing to do and will be absolutely necessary
in order to obtain the masses of the W and Z from the coefficients
of their mass terms. This normalization for the W and Z leaves a factor
of $1/\sqrt3$ in their Maxwell terms. This does not
contradict observation as it would for the photon because one does
not detect the W and Z directly; one can only observe their decay
products.

The result for Term~(\ref{eq:5.2}) is
$\tfrac{\sqrt 3}{2}\,\pi r_0^{\ -1}k\e^{\m\n}\p_\m A_\n^{10}$.
This term is a divergence, which can be transformed into a
surface integral at infinity, where the fields vanish. Therefore,
Term~(\ref{eq:5.2}) is zero for the $A_\m^{lm}$.

Terms~(\ref{eq:5.3})--(\ref{eq:5.4a}) are the only ones 
with two factors of A and two higher-dimensional derivatives.  
The result for these terms is
\begin{eqnarray}
+\frac{9+4\sqrt3\,}{12r_0^{\ 2}}\,
&\e^{\m\n}&A_\m^{1,-1}
A_\n^{11}\label{eq:5.7a}\\
+\frac{9+4\sqrt3\,}{12r_0^{\ 2}}\,
&\e^{\m\n}&A_\m^{11}
A_\n^{1,-1}\label{eq:5.7b}\\
-\frac{\sqrt{30}\,}{3r_0^{\ 2}}\,
&\e^{\m\n}&A_\m^{10}
A_\n^{10}.\label{eq:5.7c}
\end{eqnarray}
These are mass terms for the W and Z.  Like
$g_{\m\n}^{00}$, the field $A_\m^{00}$ does not have a
mass term because its associated spherical harmonic $Y_{00}$ does
not depend on the compactified coordinates. Therefore
$A_\m^{00}$ is massless, which is consistent with our
identification of it as the photon.
Thus, KK($\theta,\phi$) gives a reason why the photon is massless
while the W and Z have masses. Because $r_0$ is the Planck
length, the W and Z have Planck masses.  These are
converted to their actual masses in Part~II.

Thus, we have derived the term that leads to Maxwell's equations
using the Kaluza-Klein miracle. This refers to the calculation
of the 5D curvature scalar, which miraculously leads to the
4D curvature scalar plus the $-\frac14FF$ term when it is
assumed that fields do not depend on the extra coordinates.
This is the origin of Maxwell's equations.

Thus, KK($\theta,\phi$)
leads to the existence of the W and Z along with the terms that
describe them. This is due to an expansion of $A_\mu$ in terms of
spherical harmonics, which represents the compactified coordinate
dependence of fields in KK($\theta,\phi$). Orthogonality of the
spherical harmonics produces one Maxwell and one mass term for each
of the W's and Z when the Lagrangian density is integrated over
$\theta$ and $\phi$. This derivation of the existence of the photon,
W and Z explains the origin of what appears to be the symmetry group
$SU(2)\times U(1)$.


\begin{center}
{\bf 6. How to differentiate and integrate
over the $Y_{lm}(\theta, \phi)$}
\end{center}

This section may be omitted in a first reading.  It describes how to
go from `raw' terms in the Lagrangian such as 
Terms~(\ref{eq:5.1})--(\ref{eq:5.4a}) to their more final forms such as
Terms~(\ref{eq:5.7a})--(\ref{eq:5.7c}). 
Without an organized procedure for this, one soon runs into
expressions that cannot be evaluated because of infinities.
In addition, there will be more than one way to do certain calculations,
each with a different result. The right way must be specified.

If one attempts to multiply three or more expansions in terms of
spherical harmonics found in terms with three or more factors of $A$
in the Lagrangian, one soon runs into a mess consisting of too many
factors of $\sqrt\pi$ as well as a plethora of other square roots.
These do not all disappear after the term is integrated over
$\theta$ and $\phi$ as they do in terms with only two expansions.
One gets the sense that something is wrong here.
There must be a better way.  Indeed there is:  One must convert
the spherical harmonics to kets.
Then products of two kets may be written as single
kets using Clebsch-Gordan coefficients.  This process is continued
until there is just one ket for each of the contravariant and
covariant groups of factors. Then the ket for the contravariant group of
factors is written as a bra and the final bra and ket are multiplied
together representing the integral over $\theta$ and $\phi$.
Or, if one chooses, the final bra and ket may be reconverted
to spherical harmonics.  Because we are integrating here over just two
spherical harmonics, the excess factors of $\sqrt\pi$ and
other square roots appearing in each spherical harmonic will not appear
in the answer.  Thus, we will have converted integrals over three 
or more spherical harmonics into those over just two.  Integrals were
meant to be taken over two spherical harmonics.  

Before the procedures for taking the higher-dimensional derivatives
and integrating over the spherical harmonics are discussed, I will
show how to deal with the bras and kets in each term. 
As described in Sec.~4, $\gggm$ is the 4D
contravariant metric tensor and
raises 4D indices. Therefore, we have equations like
\begin{equation}
A_{lm}^{\m}=g^{\,\m\n}_{00}A_{\n}^{lm}.
\label{eq:A1}
\end{equation}
This shows that $g_{00}^{\,\m\n}$ is the contravariant
part of $A_{lm}^{\m}$. Since every seemingly contravariant
tensor may be written in this
fashion, then $\gggm$ is the only true contravariant tensor.
It matters whether $A_{lm}^{\m}$ or $\gggm$ is the
contravariant tensor, which will later be associated with a bra,
because the results of integrating over $\theta$ and $\phi$
depend upon which is chosen. For example, if $A_{lm}^{\m}$ 
is the contravariant tensor, we would have
\begin{equation}
\overline A_{11}^{\m}\langle11|A_{\m}^{11}|11\rangle
=\overline A_{11}^{\m}A_{\m}^{11},\label{eq:A2}
\end{equation}
while if $\gggm$ is the contravariant tensor, then the same expression
could be written as 
\begin{equation}
g_{00}^{\,\m\n}\langle00|A_{\m}^{1,-1}
|1,-1\rangle A_{\n}^{11}|11\rangle
=\tfrac{1}{\sqrt3}\,\overline A_{11}^{\m}A_{\m}^{11},\label{eq:A3}
\end{equation}
where the kets $|1,-1\rangle$ and $|11\rangle$ are combined first.
There is a factor of $\sqrt3$ difference between these expressions.
Equation~(\ref{eq:A3}) is correct because $\gggm$ is the only
true contravariant tensor.

Because factors are expanded in terms
of spherical harmonics, which are then converted into noncommuting
kets, factors in a term must be ordered properly.  This order
is determined by the above definition of the curvature scalar
in terms of the metric tensor. We will be dealing with terms
with no more than two nontrivial factors. In this case the order
of the two factors does not matter because integration over
the only bra $<00|$ we will be using and two kets
acts like a dot product between the two kets.

As the kets are nonassociative, the order of the multiplication
of factors in a term must be specified.  This order is largely
implied by the structure of the equations that define the
curvature scalar.  First, the factors of $A$ in each of
Eqs.~(\ref{eq:2.8a}) and~(\ref{eq:2.8b}) are combined.  
Then the factors within
$\w\G^{\,\dd}_{\ \a\bb}$, $\w R_{\a\bb}$ and
$\w R$ are combined in that order.
In practice, these rules are not necessary because we will be
dealing with terms with no more than two nontrivial factors.

Note that the order of factors and the order of
multiplications in a term pertain only to the covariant factor
[for example, $A_{\n}^{lm}$ in Eq.~(\ref{eq:A1})] within the
seemingly contravariant factor [$A_{lm}^{\m}$ in Eq.~(\ref{eq:A1})].
The contravariant factors $\gggm$ are not tied down to a particular
position in the term nor in the order of multiplications like the
covariant factors, because they are associated with the ket
$|00\rangle$.  
All of the 4D contravariant metric tensors in a
term are grouped together and placed to the left
of the group of covariant factors. The adjoint is taken of the 
group of $g_{00}^{\,\m\n}$.  The reason the $\gggm$ are grouped 
together is that we will want to combine the kets of the  
contravariant factors and (separately) those of the covariant
factors and write the adjoint of the resultant ket for the
contravariant factors as a bra so that there will be one bra and
one ket, representing the contravariant and covariant factors,
respectively.  The final bra and ket are then combined, representing
one integral over $\theta$ and $\phi$.  If the 
contravariant factors were not grouped together, we would have
more than one integral over $\theta$ and $\phi$ per term.
The contravariant group of factors is placed to the left
of the covariant group because bras must be placed to the
left of kets if their products are to represent integrals.  
Because the only contravariant tensors that will ever have
to be considered are the $\gggm$, the only bra that will ever
have to be considered is $\langle00|$.

Terms~(\ref{eq:5.3}) have the factor $\partial_5^{\ 2}$, which equals
$(1/r_0^{\ 2})\partial^{\ 2}_\theta$, according to Sec.~2.
Here $\partial_\theta=\partial/\partial\theta$.
To see how these terms are integrated over $\theta$ and $\phi$,
we examine the case where $l=0$ or 1 in the expansion for
$A_{\m}$.  The derivatives with respect to $\theta$
are taken: $\partial^{\ 2}_\theta Y_{00}=0$ and $\partial^{\ 2}_\theta
Y_{1m}=-Y_{1m}$.
If $l>1$, this simplification for $\partial^{\ 2}_\theta Y_{1m}$ is
not possible and the expression for $\partial_\theta$ described in the
procedure for Term~(\ref{eq:5.4}) must be used.

The spherical harmonics of the contravariant and covariant factors
are now written as kets $Y_{lm}\to|lm\rangle$ in preparation for
combining them. Next, the two covariant expansions are 
multiplied and the products of kets are written in terms of
single kets using Clebsch-Gordan coefficients.  For example,
\begin{equation}
|11\rangle|1,-1\rangle=\tfrac{1}{\sqrt6}\,|20\rangle
+\tfrac{1}{\sqrt2}\,|10\rangle+\tfrac{1}{\sqrt3}\,|00\rangle.\label{eq:A4}
\end{equation}
As explained in Sec.~5, the special relativity Minkowski metric
applies to the free-$A_\mu$-field terms, including Terms~(\ref{eq:5.3}).
Therefore, we set $g^{\,\mu\nu}_{00}=\eta^{\,\mu\nu}$,
where $\eta^{\,\mu\nu}$ now has the ket $|00\rangle$.
The adjoint of the
contravariant factor is taken and its ket is written as a bra.
The contravariant and covariant expressions are now multiplied
together.  The integrals over $\theta$ and $\phi$ may be 
expressed in terms of the orthonormality of the kets:
$\langle00|lm\rangle=\delta_{0l}\delta_{0m}$. 

If $l>1$, factors of $e^{\pm2i\phi}$ must be combined with the kets
of the differentiated factor before the kets of the two covariant
factors are combined. For example, $e^{2i\phi}|2,-1\rangle=-|21\rangle$.
This must be right because it leads to the equation
$\partial_\theta^{\ 2}Y_{1m}
=-Y_{1m}$ for the case $l=1$. (Note, however, that $e^{2i\phi}|2,-2
\rangle$ cannot be combined preliminarily because it does not equal
another ket.)

Term~(\ref{eq:5.4}) contains two factors of $\partial_5$. It
is dealt with in the same way as Terms~(\ref{eq:5.3}),
except that $\partial_\theta$ is rewritten
\begin{equation}
\partial_\theta=\frac{1}{2\hbar}\left(e^{-i\phi}L_+-e^{i\phi}L_-\right),
\label{eq:A5}
\end{equation}
where $L_+$ and $L_-$ are the raising and lowering operators,
respectively, for the spherical harmonics.  We have
\begin{equation}
L_\pm Y_{lm}=\left[l(l+1)-m(m\pm1)\right]^{1/2}\hbar Y_{l,m\pm1}.
\label{eq:A6}
\end{equation}
This definition for $\partial_\theta$ follows from combining
$L_+=L_x+iL_y$ and $L_-=L_x-iL_y$, where
\begin{eqnarray}
         L_x&=&i\hbar(\sin\!\phi\,\partial_\theta
             +\cot\!\theta\cos\!\phi\,\partial_\phi),\label{eq:A7a}\\
         L_y&=&i\hbar(-\cos\!\phi\,\partial_\theta
             +\cot\!\theta\sin\!\phi\,\partial_\phi),\label{eq:A7b}
\end{eqnarray}
which are the $x$ and $y$ components of the angular momentum vector
${\bf L}$ written in spherical coordinates.

Rewriting the operator $\partial_\theta$ is necessary to write
$\partial_\theta Y_{lm}$ properly in terms of spherical harmonics,
which can then be rewritten as kets.
One must write products of kets in terms of single kets because
integrals over spherical harmonics were meant to take place over
two spherical harmonics.

When one uses this expression for $\partial_\theta$, one winds up
with factors of $e^{i\phi}$ in the term. In order to evaluate
these expressions, these terms have their final bra and ket
rewritten as spherical harmonics according to the transformations
$\langle00|\to Y_{00}$ and $|lm\rangle\to Y_{lm}$ and the terms
are explicitly integrated over $\theta$ and $\phi$ in the usual
way. Note that the factors of $e^{i\phi}$ must cancel if the integral
over $\phi$ and the term is to be nonzero.
For terms without functions of $\phi$, use may be made of
the orthonormality of the kets.

Term~(\ref{eq:5.4a}) contains two factors of $\partial_6$, where
$\partial_6=\partial/(r_0\sin\!\theta\partial\phi)$ from Sec.~2.
This term is dealt with in the same way as Terms~(\ref{eq:5.3}).
First, the derivatives with respect to $\phi$ of the $Y_{lm}$ are taken
$\partial_\phi Y_{lm}=imY_{lm}$. In order to evaluate terms with the two
factors of $1/\sin\!\theta$ from the factors of $\partial_6$,
the terms have the spherical harmonics of both covariant
factors written in terms of spherical 
harmonics with their $l$ and $|m|$ values decreased by one. (The
quantum number $m$ is increased by one if it is negative.)
Some examples of this are
\begin{eqnarray}
Y_{1,\pm1}&=&\mp\sqrt{\tfrac32}\,Y_{00}\sin\!\theta e^{\pm i\phi},
\label{eq:A8a}\\
Y_{2,\pm2}&=&\mp\sqrt{\tfrac54}\,Y_{1,\pm1}\sin\!\theta e^{\pm i\phi},
\label{eq:A8b}\\
Y_{3,\pm1}&=&\mp\sqrt{\tfrac{7}{12}}\left(\sqrt5\,Y_{20}+Y_{00}\right)
\sin\!\theta e^{\pm i\phi}.\label{eq:A8c}
\end{eqnarray}
The factors of $\sin\!\theta$ in the $\partial_6$'s cancel with those
extracted from the covariant factors as in
Eqs.~(\ref{eq:A8a})--(\ref{eq:A8c}).
This eliminates the factor of $1/\sin^2\theta$,
which would otherwise result in infinity when integrating over $\theta$.
The spherical harmonics are now written as kets and combined.

\begin{center}
{\bf 7. Leptons}
\end{center}

This section derives the Dirac equation for leptons.
It is possible to introduce fermions into a purely bosonic Kaluza-Klein theory
by expanding the vectors in KK($\theta,\phi$) in terms of 
spinors [53]. For any vector $W_\m$,
\begin{equation}
W_\m=-\h\pa w_{B \dot V},\label{eq:6.1}
\end{equation}
where $w_{B\dot V}$ is a second rank spinor.  A sum is implied over the
indices $B$ and $\dot V$.  These indices take the values
1, 2 and $\dot 1$, $\dot 2$, respectively.  The 
$\oo\s^{B\dot V}$ are the Pauli spin matrices and
$\s_0^{B\dot V}$ is the unit matrix.

If the vector field is null as for the case of the photon, the
second rank spinor may be equated to the product of
two 2-component spinors.
For example, for the photon we have
\q
\am|00\rangle=-\h\pa\xi_B\eta_{\dot V}
\tfrac{1}{\sqrt2}\k\kk
+\h\pa\tau_B\omega_\wdv\tfrac{1}{\sqrt2}\kk\k,\label{eq:6.2}
\qq
where
\begin{equation}
|00\rangle=\tfrac{1}{\sqrt2\,}\k\kk
            -\tfrac{1}{\sqrt2\,}\kk\k.\label{eq:6.2a}
\end{equation}
The spinor $\xi_B$ or $\omega_\wdv$ has the ket $\k$, while $\eta_\wdv$ or
$\tau_B$ is in the state $\kk$. The first ket in each pair in
Eq.~(\ref{eq:6.2}) is identified with the spinor with an undotted index,
while the second is matched with the dotted-index spinor. We have
\begin{equation}
\am=-\h\pa\xi_B\eta_\wdv\label{eq:6.2b}
\end{equation}
for the first pair of kets and
\begin{equation}
\am=-\h\pa\tau_B\omega_\wdv\label{eq:6.2c}
\end{equation}
for the second.

The spinors in this expansion may recombine to form $A_\m^{10}$ (the Z).
We have
\q
A_\m^{10}|10\rangle=-\h\pa\xi_B\eta_{\dot V}
\tfrac{1}{\sqrt2}\k\kk
-\h\pa\tau_B\omega_\wdv\tfrac{1}{\sqrt2}\kk\k,\label{eq:6.3}
\qq
where
\begin{equation}
|10\rangle=\tfrac{1}{\sqrt2\,}\k\kk
            +\tfrac{1}{\sqrt2\,}\kk\k.\label{eq:6.3a}
\end{equation}

The derivative $\partial_5$ in Term~(\ref{eq:5.2}), which will lead to the
Dirac equation, eliminates $A_\mu^{00}|00\rangle$ from the expansion of
$A_\mu$, leaving only $A_\mu^{10}|10\rangle$.
Substituting this into Term~(\ref{eq:5.2}) and reconverting the spinor
kets to $|10\rangle$ and $|00\rangle$ (for example, 
$\kk\k=\tfrac{1}{\sqrt2}|10\rangle-\tfrac{1}{\sqrt2}|00\rangle$), we obtain
\begin{equation}
\frac{\sqrt{3}\,\pi k}{8r_0}\mm\pa(\xi_B\partial_\n\eta_\wdv
+\partial_\n\xi_B\eta_\wdv),\label{eq:6.5}
\end{equation}
plus a similar set of terms involving the spinors $\tau_B$ and
$\omega_\wdv$.  The spinors $\eta_{\dot V}$ and $\xi_B$ will be
identified with the electron field and its complex conjugate,
while $\omega_\wdv$ and $\tau_B$ will be the neutrino and its
conjugate. These identifications are suggested by the
associations these fields have with their kets above, according to
their quantum numbers $l$ and $m$ given in Table~1.

We will ignore the neutrino terms for now and concentrate
on deriving the Dirac equation for the electron.
The fields are redefined
\begin{equation}
\xi^\prime_B=\left(\frac{\sqrt3\,c^3}{128\hbar Gr_0}
\right)^{1/2}\xi_B,\label{eq:6.6}
\end{equation}
with an identical redefinition for $\eta_\wdv$. We have included the factor
of $c$ from $dx^4$ in the differential volume element in the redefinition.
The result for Term~(\ref{eq:5.2}) is now
\begin{eqnarray}
&i\hbar c&\mm\pa\xi^\prime_B\partial_\n
\eta^\prime_\wdv\label{eq:6.7a}\\
+&i\hbar c&\mm\pa\partial_\n\xi^\prime_B\eta^\prime_\wdv.\label{eq:6.7b}
\end{eqnarray}
I have multiplied these terms by $i$ to render them real
when complex spinor fields are used. This factor does not
follow from the 6D curvature scalar; multiplying by it is
allowed because the terms are the only free-field terms for the spinors.
The field equations remain the same after its introduction.

Spinor mass terms do not follow directly from the Lagrangian.
That is, they are not part of the 6D curvature scalar.
This is because one must introduce spinors into the Lagrangian
with the above-mentioned substitution for vectors, which does not
allow terms of the form $-m\xi^\prime_B\xi^{\prime B}$.
However, leptons acquire masses from
their charges via the classical radius
equation to be considered in Part~II. Therefore, we introduce the mass terms
\begin{equation}
-m\xi^\prime_B\xi^{\prime B}-m\eta^{\prime\wdv}
\eta^\prime_\wdv.\label{eq:6.8}
\end{equation}

Each of Terms~(\ref{eq:6.7a}) or~(\ref{eq:6.7b}) will lead to the
Dirac equation. We will
consider only Term~(\ref{eq:6.7a}) in what follows.

Lagrange's equations for $\xi'_B$ and $\eta'_\wdv$ are
\begin{eqnarray}
&i\hbar c&\mm\pa\partial_\n\xi'_B=m\eta^{\prime\wdv},\label{eq:6.9a}\\
&i\hbar c&\mm\pa\partial_\n\eta'_\wdv=m\xi^{\prime B}.\label{eq:6.9b}
\end{eqnarray}
Using the equations [53] $\zeta^B\xi'_B=-\zeta_B\xi^{\prime B}$
and $\eta^{\prime\wdv}=\varepsilon^{\dot V\dot W}\eta'_{\dot W}$,
where the only nonzero components of $\varepsilon^{\dot V\dot W}$
are $\varepsilon^{\dot1\dot2}=-\varepsilon^{\dot2\dot1}=1$,
Eq.~(\ref{eq:6.9a}) becomes
\begin{equation}
-i\hbar c\s^\m_{B\dot V}\partial_\m\xi^{\prime B}
=m\eta'_\wdv,\label{eq:6.9a'}
\end{equation}
where the $\s^\m_{B\dot V}$ are the associated basic spin matrices.
They are computed from the $\pa$ via the relations
\begin{equation}
\s^\m_{B\dot V}\s_\n^{B\dot V}=-2\delta^\m_{\ \n}.\label{eq:6.10}
\end{equation}
The $\s^\m_{B\dot V}$ are equal to minus the $\pa$, except for
$\s^2_{B\dot V}$, which equals $\s_2^{B\dot V}$.

Equations~(\ref{eq:6.9a'}) and~(\ref{eq:6.9b}) may be put in matrix form
\begin{eqnarray}
\left(i\partial_0+i\oo\sigma\cdot\oo\nabla\right)\xi&=&m\eta,
     \label{eq:6.11z}\\
\left(i\partial_0-i\oo\sigma\cdot\oo\nabla\right)\eta&=&m\xi,\label{eq:6.11}
\end{eqnarray}
where the $\oo\s$ are the Pauli spin matrices and
\begin{equation}
\xi=\left(\begin{array}{c}
\xi^{\prime1}\\ 
\xi^{\prime2}\end{array}\right),
\qquad\eta=\left(
\begin{array}{c}\eta^\prime_{\dot1_{}}\\ 
\eta^\prime_{\dot2}
\end{array}\right).\label{eq:6.11a}
\end{equation}
We have set $\hbar=c=1$, which applies for the remainder of this
section. Adding and subtracting Eqs.~(\ref{eq:6.11z}) and~(\ref{eq:6.11}),
we obtain
\begin{eqnarray}
         i\partial_0\varphi+i\oo\s\cdot\oo\nabla\chi
         &=&m\varphi,\label{eq:6.111'}\\
        -i\partial_0\chi-i\oo\s\cdot\oo\nabla\varphi
         &=&m\chi,
\label{eq:6.11'}
\end{eqnarray}
where
$\varphi=\frac{1}{\sqrt2\,}(\xi+\eta)$ and
$\chi=\frac{1}{\sqrt2\,}(\xi-\eta)$. 

Equations~(\ref{eq:6.111'}) and~(\ref{eq:6.11'}) may be recast as
\begin{equation}
i\gamma^\m\partial_\m\psi=m\psi,\label{eq:6.11''}
\end{equation}
where
\begin{equation}
\psi=\left(\begin{array}{c}\varphi\\ \chi\end{array}
\right),\label{eq:6.11''a}
\end{equation}
\begin{equation}
\gamma^0=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right),\qquad
\oo\gamma=\left(\begin{array}{cc}0&\oo\s\\-\oo\s&0\end{array}\right),
\label{eq:6.11''b}
\end{equation}
where each element of the matrices in Eqs.~(\ref{eq:6.11''b}) is itself a
$2\times2$ matrix.
Eq.~(\ref{eq:6.11''}) with~(\ref{eq:6.11''a})
and~(\ref{eq:6.11''b}) constitute the Dirac
equation for the electron in the Dirac representation for $\psi$.

A similar derivation of the Dirac equation for the neutrino
involving the spinors $\tau_B$ and $\omega_\wdv$ may be made.
The photon may be expanded in terms of spinors with $l>\h$.
The muon and its neutrino are associated with the kets
$\llle\frac32,-\h\rr$ and $\llle\frac32\,\h\rr$, respectively.
The tau and its neutrino are in the states
$\llle\frac52,-\h\rr$ and $\llle\frac52\,\h\rr$, respectively.
Continuing in this manner, KK($\theta,\phi$) predicts an infinite
number of fermions, each corresponding to a state $\llle lm\rr$,
where $l$ and $m$ are half-odd integral. Thus, the assumption of
two extra coordinates, along with the resulting expansions in
terms of spherical harmonics and spinors, produces multiple
generations of leptons. Part~II explains why there appears to
be a small number of generations. The SM contains
three generations.

I claim that the fact that the positron appears to have negative energy
is not merely appearance but is deep-rooted and cannot be
transformed away.  The negative energy of the positron is dealt with
in the same way as that for the $Z^0$: The ground state is defined
to be that closest to zero energy instead of lowest energy.

In order to reproduce the SM fully, one must produce the chirality
of the weak interactions.  Chirality is not merely allowed in
KK($\theta,\phi$) because it has an even number of dimensions,
but is demanded by it. This follows from the fact that a massless
particle must have its right-handed field equal to zero.
The nearly zero mass of the neutrino is derived in Part~II.
Therefore, $\nu_R=0$ and $\nu=\nu_L$ because $\psi=\psi_L+\psi_R$.
This prevents an interaction term with the W, right-handed neutrinos
and electrons.

It is noteworthy that the Dirac equation is derived from the 
6D curvature scalar of KK($\theta,\phi$).
This occurs because the divergence
$\partial_\mu A^\mu_{10}$ splits into terms of the form of the Dirac
term $\overline\psi\,\gamma^\mu\partial_\mu\psi$ when the spinor form
$\overline\psi\gamma^\mu\psi$ of the vector $A^\mu_{10}$ is used.

One may wonder why the Maxwell term does not apply to spinors.
If a vector is equal to its spinor equivalent, one could substitute its
spinor form into the Maxwell Terms~(\ref{eq:5.1}). However, examination of
Terms~(\ref{eq:5.1}) and~(\ref{eq:5.2}) reveals that the Dirac term,
which contains $\partial_5$, is larger by a factor $1/r_0$.
Therefore, the spinor form of the Maxwell term is negligible
compared to the Dirac term.


\begin{center}
{\bf 8. Interaction terms}
\end{center}

Interaction terms for intermediate vector bosons
and leptons may be derived from the
vectors' mass Terms~(\ref{eq:5.7a})--(\ref{eq:5.7c}).
This is accomplished by expanding
one of the vectors in terms of spinors. 
For example, Term~(\ref{eq:5.7c}) without the factor of $k$
absorbed by the two factors of $A_\mu^{10}$ starts off as
\begin{equation}
-\ \ \frac{\sqrt{30}}{3r_0^{\ 2}}\,k\eta^{\mu\nu}A_\mu^{10}A_\nu^{10}.
\label{eq:7.1}
\end{equation}
This term ends up as
\begin{equation}
\frac83\,\left(\frac{10}{\pi}\right)^{1/2}\sqrt{\hbar c}\,Z_\mu^0\left(
  \overline e\,\gamma^\mu e+{\overline\nu}_e\gamma^\mu\nu_e\right),
  \label{eq:7.1'}
\end{equation}
Thus, the bare weak charge
of the electron or neutrino interacting with the $Z^0$ from the 
6D curvature scalar is
\begin{equation}
g_{\widehat R}=\frac83\left(\frac{10}{\pi}\right)^{1/2}\sqrt{\hbar c}\,
              =4.75766\sqrt{\hbar c}\,.\label{eq:7.3}
\end{equation}
To see where this coefficient comes from, I write each of the factors
that go into its formation. The coefficient is equal to
\begin{equation}
\left(-\frac{\sqrt{30}}{3r_0^{\ 2}}\,\right)_1
\left(\frac{c^3}{16\pi G}\right)_2
\left(c\right)_3
\left(\frac{16\pi G}{c^4}\right)_4^{\ 1/2}
\left(\frac{128\hbar Gr_0}{\sqrt3\,c^3}\right)_5
\left(-\frac12\right)_6
\left(\frac12\right)_7,\label{eq:7.3'}
\end{equation}
where I have numbered the factors for future reference.

The 1st factor is the coefficient of the Z mass term from Sec.~5.
We note that $r_0=L=(\hbar G/c^3)^{1/2}$ is the
Planck length. Factor 2 is the constant $k$ from Sec.~3, by which
the entire Lagrangian is multiplied. This gives the Lagrangian the
units of energy. The 3rd factor comes from the relativistic
time coordinate $dx^4=cdt$ in the differential volume element of the
action integral of the term.
Factor 4 results from the redefinition of one vector field in
the term as described in Sec.~5. Factor 5 is from the redefinition
of two spinors in the Dirac term 
described in Sec.~7.
Factor 6 is from Eq.~(\ref{eq:6.1}), which equates vectors
to spinors. The 7th factor arrises when $|10\rangle$ is
written in terms of $l=1/2$ kets, each pair of which are
multiplied by the Clebsch-Gordan coefficient $1/\sqrt2$. This
Clebsch-Gordan coefficient is squared when the spinors' $l=1/2$ kets
are reconverted to the kets $|10\rangle$ and $|00\rangle$ in order
to arrive at the interaction or Dirac term. Another way
of looking at factor 7 is that the  $Z^0$ is composed of
one-half an electron-positron pair
and one-half a neutrino-antineutrino pair.

We now consider the case of the muon $\mu_1^-$.
The Clebsch-Gordan coefficient that relates $|10\rangle$ to the
muon-antimuon's $l=3/2$ pair of spinor kets is $1/\sqrt{20}$.
This means $A_\mu^{10}$ (the Z) may also be considered to be 1/20th a
muon-antimuon pair. This produces a factor of 1/20 in factor 7 in
Eq.~(\ref{eq:7.3'}) instead of the factor of 1/2 for the electron.
Thus, the coefficient of the weak interaction term from $\widehat R$
for the muon would appear to be 1/10th that of the electron.
Because the above-mentioned Clebsch-Gordan coefficient for the
muon is 1/10th that of the electron in the Dirac term as well,
the resulting Dirac equation
for the muon would have a factor of 1/10 as coefficient for
the Dirac term as well as the weak-interaction term. Since, at
this point, these are the only two terms in the Dirac equation
for the muon, one may multiply the entire equation by 10 and
recover the Dirac term with coefficient one as well as a weak
charge from $\widehat R$ for the muon that is the same as that for
the electron.

The weak charge of the tau from the 6D curvature scalar
remains the same as that for the electron by the same reasoning
used for the muon. Note that one does not expand $A_\mu^{30}$
in terms of the $l=3/2$ muon, for example, but rather it is $A_\mu^{10}$
that must appear in the weak interaction term and what one
does for the interaction term one must do for the Dirac term.
Thus, the compactified coordinate dependence of these terms
remains the same as for the case of the electron.

Spinors may combine to form the $W^+$ and $W^-$
\begin{eqnarray}
A_\mu^{11}|11\rangle&=&-\tfrac12\sigma_\mu^{B\dot V}
         \xi_B\omega_{\dot V}\k\k,\label{eq:7.4a}\\
         A_\mu^{1,-1}|1,-1\rangle&=&-\tfrac12\sigma_\mu^{B\dot V}
         \tau_B\eta_{\dot V}\kk\kk,\label{eq:7.4b}
\end{eqnarray}
which, when substituted into Terms~(\ref{eq:5.7a})
and~(\ref{eq:5.7b}), yield
\begin{eqnarray}
-\ \frac{16+12\sqrt3}{3\sqrt\pi}\,\sqrt{\hbar c}\,& W_\mu^+&
         {\overline\nu}_e\gamma^\mu e\label{eq:7.5a}\\
        -\ \frac{16+12\sqrt3}{3\sqrt\pi}\,\sqrt{\hbar c}\,& W_\mu^-&
         \overline e\,\gamma^\mu\nu_e.\label{eq:7.5b}
\end{eqnarray}

As it stands, there is no interaction between
photons and electrons.
This is obtained from a photon mass term in Part~II.

Interaction terms are an important part of the SM, which
produces them through a covariant derivative used in the Dirac term.
This is done to maintain the symmetry $SU(2)\times U(1)$. Here, however,
we have an entirely different but equally effective method for generating
interaction terms. When the spinor form $\overline\psi\,\gamma^\mu\psi$
of the vector $A^\mu$ is substituted
into a mass term of the form $A^\mu A_\mu$ a term of the
form $\overline\psi\,\gamma^\mu\psi A_\mu$ is created.

One may substitute two spinors for both vectors in terms like (\ref{eq:7.1}).
Four-spinor interaction terms result.  These are negligible compared to
vector-spinor interaction terms because they contain an extra factor of
$r_0$ from the additional spinor redefinition.  [See factor 5 of
Eq.~(\ref{eq:7.3'}).]  This explains why there are no four-point
Fermi interactions and why the weak interactions are carried by
intermediate vector bosons.


\begin{center}
{\bf 9. The strong interactions}
\end{center}

General relativity is the result of the Lorentz group in four
dimensions.  But what about the Lorentz group in six dimensions?
When one makes the transition from four to six dimensions, nine
additional generators are needed to describe a Lorentz transformation.
Each of these additional generators corresponds to a rotation or boost
in a coordinate surface with at least one compactified coordinate.
What force is produced by these nine additional types of
Lorentz transformation?  It is the strong interaction.

The total number of generators for the Lorentz group in six dimensions
is 15.  I will denote the 15 generators corresponding to an
infinitesimal 6D Lorentz
transformation by $\lambda_a$, where $a=1,\ldots,15$.
The six 4D generators are white and produce gravitation.
The nine higher-dimensional generators are colored and produce
the strong interaction.

Following the example set by the electroweak interaction above,
where we expanded $A_\mu$ in terms spherical harmonics,
we expand $g_{\mu\nu}^{lm}$ and $A_\mu^{lm}$
in terms of the 15
generators $\lambda_{a}$ of the 6D Lorentz group
\begin{eqnarray}
g_{\m\n}^{\,lm}&=&\sum_{a=1}^{15}g_{\m\n}^{\,lma}\la_{a},\label{eq:8.1}\\
A_\m^{lm}&=&\sum_{a=1}^{15}A_\m^{lma}\la_{a}.\label{eq:8.2}
\end{eqnarray}
The Latin indices $a,b,c$ range from one to 15.
Henceforth we will observe the convention of summing over
these indices when they appear twice in a term.  These expansions
in terms of 6D Lorentz generators are analogous to those in terms of
spherical harmonics. The notation for fields previously
defined is now changed---$A_\m^{00}$ and $g_{\m\n}^{00}$ are no
longer the photon and graviton.  These are now denoted by
$A_\m^{00a'}$ and $g_{\m\n}^{00a'}$, respectively; the 
$A_\m^{1ma'}$ are now the $W$'s and $Z$. Here the index $a'$
takes the values corresponding to the six white generators
of the 6D Lorentz group.

In the 4D differential volume element, the
$\sqrt{-g}$ refers to $\eta_{\m\n}$ in microscopic spacetime.
The tensor $\eta_{\m\n}$ is not expanded in terms of 6D Lorentz group
generators because it is constant and after a 6D Lorentz transformation,
the field's values are oriented the same way
in spacetime.  Therefore, it has no gravitational or strong charge
with respect to a 6D Lorentz transformation.  Similarly,
in microscopic spacetime, the $A^\m_{lma}$ are written in terms
of the $A_\m^{lma}$ using the 4D contravariant
metric tensor $\kw$ according to the formula
\begin{equation}
A_{lma}^\m=\kw A_\n^{lma}.\label{eq:8.3}
\end{equation}
Thus, the only contravariant fields we will have to
consider are the $\kw$ and the only generator we will have to
consider for the contravariant fields is $\la_{a'}$.

Before we proceed to consider terms with factors expanded in terms
of 6D Lorentz generators, we describe how to deal with these
generators in each term.

To eliminate the 6D Lorentz generators from the terms in the Lagrangian
and make the terms scalars, we determine one generator for each of
the covariant and contravariant groups of factors in a term and
multiply the two generators together in a scalar or dot product. This
product is defined such that
\begin{equation}
\lambda_{a}\cdot\lambda_{b}=\delta_{ab},\label{eq:B1}
\end{equation}
where $\delta_{ab}$ is the Kronecker delta. This is similar to
integrating over the orthogonal kets representing the covariant 
and contravariant groups of factors.

Each factor in a term is expanded in terms of the 15
generators $\lambda_a$, nine of which are colored and six
are white. It remains to determine which type of products
exist between factors with these generators within each of the
covariant and contravariant groups.

There are three possible products for the $\lambda_a$ vectors
of each of the covariant and contravariant factors:
scalar (dot), vector (cross), or tensor. In order to produce one
$\lambda_a$ for the covariant or contravariant factors and yet
involve all of them, a vector product seems most appropriate
\begin{equation}
\lambda_a\times\lambda_b=[\lambda_a,\lambda_b]=
\sum_{c} c_{abc}\lambda_{c},\label{eq:B3}
\end{equation}
where the $c_{abc}$ are the structure constants of the 6D Lorentz group.
This is similar to the expression of products of kets in terms
of single kets using Clebsch-Gordan coefficients. 

Because the generators are not associative under vector
multiplication, parentheses must be used to specify the
order of multiplications.  This would be the same as that
given in Sec.~6.

According to the vector identity ${\bf A\cdot(B\times C)
=(A\times B)\cdot C}$, the scalar product
may be interchanged with the last vector 
product in the term without affecting the value of the term.
This would be how to obtain the familiar form of mass and
Maxwell terms with two strongly interacting fields $A_\mu$
--- with the scalar or dot product between these factors as opposed
to between the covariant factors of $A_\mu$
and contravariant $\eta^{\mu\nu}$.

Expanding Terms~(\ref{eq:5.1}) first in
terms of spherical harmonics and then in terms of 6D Lorentz generators,
we arrive at the terms
\q
-\tfrac14\,
F^{\,\m\n}_{00a}F_{\m\n}^{00a}
-\tfrac{1}{4\sqrt3}\,F^{\,\m\n}_{1,-1a}F_{\m\n}^{11a}
+\tfrac{1}{4\sqrt3}\,
F^{\,\m\n}_{10a}F_{\m\n}^{10a}
-\tfrac{1}{4\sqrt3}\,
F^{\,\m\n}_{11a}F_{\m\n}^{1,-1a},\label{eq:8.4}
\qq
where $F_{\m\n}^{lma}=\partial_\m A_\n^{lma}
                         -\partial_\n A_\m^{lma}$.

The term with one factor of $A$, with only this factor expanded
in terms of colored Lorentz generators, is zero due to the orthogonality
of the generators.  (With only one factor of colored $\lambda_a$, there
is no way to make the term a scalar.)

Doubly expanding the
factors of $A$ in Terms~(\ref{eq:5.3})--(\ref{eq:5.4a}), we obtain
\begin{eqnarray}
+\ \frac{9+4\sqrt3\,}{12r_0^{\ 2}}\,
&\kw &A_\m^{\,1,-1a}A_\n^{11a}\label{eq:8.5a}\\
        +\ \frac{9+4\sqrt3\,}{12r_0^{\ 2}}\,
&\kw &A_\m^{\,11a}A_\n^{1,-1a}\label{eq:8.5b}\\
        -\ \frac{\sqrt{30}\,}{3r_0^{\ 2}}\,
&\kw &A_\m^{\,10a}A_\n^{10a},\label{eq:8.5c}
\end{eqnarray}  
which are  mass terms for the $A_\m^{1ma}$.

The Lagrangian for KK should
contain terms with anywhere from two to six strongly interacting
vectors or gluons.  However, strong interactions with diagrams with
five or six external gluon lines, corresponding to terms with five or six
gluons, are not observed. The explanation
for this is as follows. If one examines the definition of the
curvature scalar
described in Sec.~3, one finds that the only way to
obtain terms with five or six factors of $A_\mu$ is through the
presence of $\widehat g^{\,55}=1+g^{\mu\nu}A_\mu A_\nu$. Now
since the strong interaction cross product between the two factors
of $A$ here is purely antisymmetric, while the multiplication of these two
factors by $g^{\mu\nu}$ symmetrizes them, the resulting terms
with five or six gluons are zero.


\begin{center}
{\bf 10. Quarks, confinement, asymptotic freedom and chiral symmetry breaking}
\end{center}


The main aspects of strong interaction phenomenology are quarks,
quark confinement, asymptotic freedom and chiral symmetry breaking.
These are not hard to achieve in KK($\theta,\phi$).
The existence of quarks can be explained by expanding the photon
in terms of 6D Lorentz generators and spinors. In final form
we have
\q
A_\m^{00}|00\rangle=-\tfrac12\s_\m^{B\dot V}
\varpi_{\dot V}^p\chi_B^q\lambda_{pq}\tfrac{1}{\sqrt2}\k\kk
+\tfrac12\s_\m^{B\dot V}\iota_{\dot V}^p\kappa_B^q
\lambda_{pq}\tfrac{1}{\sqrt2}\kk\k\label{eq:9.1}
\qq
We have neglected to consider the terms corresponding to the
six white generators in the expansion for
$A_\m^{00}$.  Each of these is the photon. 
The index for the nine colored generators of the 6D Lorentz group
has been rewritten as $pq$ where $p$ and $q$ take the values
r,g,b, which stand for red, green and blue.  A sum over $p$ and
$q$ is implied.  The spinors $\chi_B^q$ and $\iota_{\dot V}^p$
are associated with the ket $\k$, while $\varpi_{\dot V}^p$ and
$\kappa_B^q$ have $\kk$ as their higher-dimensional angular
momentum representation. Transforming the spinors according to 
Sec.~7, we find the up quark $u^p$ corresponds to $\iota_{\dot V}^p$
while the down quark $d^p$ corresponds to $\varpi_{\dot V}^p$.

By expanding $A_\m^{00}$ in terms of Lorentz generators and spinors
with $l>\frac12$, quarks in other families may be produced. The
strange and charmed quarks are associated with the states
$\left |\tfrac32,-\tfrac12\right >$ and
$\left|\frac32\frac12\right>$, respectively.  The top (bottom)
quark is in the state $\left|\frac52\frac12\right>$
$\left(\left|\frac52,-\frac12\right>\right)$. According to
KK($\theta,\phi$), there are an infinite number of quarks, each 
corresponding to a state $|lm\rangle$, where $l$ and $m$ are
half-odd integral. Thus, expanding $A_\mu$ in terms of spherical
harmonics, Lorentz generators and spinors produces multiple
generations of quarks. It is not unlikely that the number of
generations appears to be limited by the same reason given in Part~II
for leptons.

Terms for the interactions of quarks with
gluons may be derived from the gluons' mass terms.
Schematically, we have
\begin{eqnarray}
         c_{1}\overline u_p\ga u_qA_\m^{00pq}
       &+&c_{2}\overline d_p\ga d_qA_\m^{00pq}\label{eq:9.2a}\\
        +c_{3}\overline u_p\ga u_qA_\m^{10pq}
       &+&c_{4}\overline d_p\ga d_qA_\m^{10pq}\label{eq:9.2b}\\
        +c_{5}\overline u_p\ga d_qA_\m^{11pq}
       &+&c_{6}\overline d_p\ga u_qA_\m^{1-1pq},\label{eq:9.2c}
\end{eqnarray}
 where $c_{1},\ldots,c_{6}$ are constants.

It is highly unlikely that quarks have charges exactly one-third
that of the electron because charge
will be shown in Part~II to be the result of quantum corrections
and these are notoriously uneven in value.
There does not appear to be any way of achieving one-third integral
charges in Part~II.
Thus, we will subscribe to the original theory of the strong interactions,
the Han-Nambu model [55,56],
in which quarks have integral charges. This is possible because our theory
does not have an exact strong interaction symmetry of the Lagrangian.
Quarks of different colors have different charges. The symmetry is broken.

Instead of quarks having color, they are non-singlet $SU(3)'$ states in
the Han-Nambu model. I will term this property $L(6)$ color, named
for the group for KK($\theta,\phi$)'s theory of the strong interactions,
the higher-dimensional part of the Lorentz group in six dimensions.
The term `$L(6)$ color' replaces retroactively the term `color' used above.

Terms describing free quarks (the Dirac term), quark masses and the
electromagnetic interactions of single quarks are zero.  This is because the
$L(6)$ colored generators of the quarks are orthogonal to the white generators
of $\eta^{\m\nu}_{a'}$ or $A_\m^{00a'}$ (the only other fields in
the terms) and their dot or scalar product is zero.  With only one
$L(6)$ colored generator there is no way to make the term a scalar.

The SM assumes the Dirac or free-quark term
in its Lagrangian, thereby postulating free quarks.
It then goes to great lengths to try to eliminate
free quarks. Why not just leave out the free-quark term?
In KK($\theta,\phi$),  the free-quark term is zero.  This implies there are no
free quarks.  This is the simplest method for achieving
quark confinement:  To eliminate free quarks, simply leave out the
free-quark term in the Lagrangian.

Although there are no mass terms for quarks in $\widehat R$,
masses for quarks can be generated by their $L(6)$ color, in much the same
way that the classical radius equation can be used to generate a
mass for the electron from its charge. Thus, we would have only strong
interaction and mass terms for quarks. New terms for quarks, however, can
be generated by expanding the white vectors $A_\mu^{lma'}|lm\rangle$
in terms of spinors that are white combinations of quarks.
In this way, we arrive at free-field and photon interaction terms
for white combinations of quarks or hadrons. Because the
free-quark terms vanish but the free-hadron terms do not, quarks
can only be found in white combinations or hadrons.

Confinement in KK($\theta,\phi$) may be tested by detecting
free gluons.  These are predicted because the
free-field (Maxwell) terms for gluons do not vanish.
Free gluons are forbidden by confinement in the SM.

The chiral symmetry breaking of the strong interactions
is caused by a nonzero pion mass.
According to the classical radius equation presented in Part~II,
a particle that has charge must have mass.  The pion has a mass
because the quarks and gluons of which it is composed have
masses due to their strong interaction charges or $L(6)$ color.
Therefore, KK($\theta,\phi$) explains chiral symmetry breaking.

We now describe the potential for confinement in KK($\theta,\phi$).
Each of the gluons with which the quarks interact
must have a mass because it has $L(6)$ color,
which would, like electric charge,
yield a mass by the classical radius equation to be discussed in Part~II.
Because free quarks are forbidden by the vanishing of the
free-quark term they must always be interacting.  Now, in order
for two quarks to remain interacting, they must be within range
of the strong interaction.  Therefore, they cannot be separated by
a distance greater than the Compton wavelength
$\lambda=h/mc$ of a gluon.
The mathematical expression for this is
an infinite square-well potential with radius equal to the 
Compton wavelength of a gluon, surrounding each quark.
Inside this infinite square well, the quarks are free to
move about because the gluons that influence
the quarks have masses due to their $L(6)$ color, rendering the
interaction weak.  Thus, aside from the infinite square-well
confining potential, the strong interaction is really weak.
This explains quark confinement and asymptotic freedom. 


\clearpage
\begin{center}
{\bf Part II. The Quantum Theory}
\end{center}


\begin{center}
{\bf 11. Quantum field theory in KK($\theta,\phi$)}
\end{center}


This part derives the values of ten
unexplained parameters in the Standard Model plus gravity (SM)
from the Kaluza-Klein theory of pure gravity in six dimensions with
fields allowed to depend on the extra coordinates [KK($\theta,\phi$)].
The parameters are from the electroweak sector.

The derivations are made possible by converting Planck
masses to elementary particle masses.
This conversion process is the quantum mass correction in QED.
This correction contains a cutoff $\Lambda$, which is set to the
Planck mass instead of infinity.  This results from extended particles,
which are not points, but have sizes of the order of the Planck length.
Particles were thought to have to be pointlike in order to avoid
violations of either special relativity or causality, but Superstrings
has shown that extended particles are, in fact, viable.

In QED the zero size and masslessness of the photon
cause infinities, which have given rise
to special procedures developed to handle them.
Regularization introduces a cutoff, which allows
one to see how a certain parameter goes to infinity.
Renormalization is the process of canceling infinities for a
parameter.  Gauge dependence allows one to obtain the 
zero-mass limit of the photon without infinities.
The Higgs mechanism preserves the symmetry
$SU(2)\times U(1)$, necessary for renormalizability.
In KK($\theta,\phi$) the photon is extended and has a mass.
Therefore, the infinities associated with regularization,
renormalization and the zero-mass limit for the photon
never develop.  One does not have to
regularize, renormalize, assume gauge dependence,
a symmetry group, nor the Higgs mechanism.
These become superfluous.

I now describe how 
regularization becomes superfluous in KK($\theta,\phi$).
Quantum corrections in QED consist of
vacuum polarization, the electron self-energy,
and the vertex correction. 
The one-loop corrections entail integrals over
momentum.
The integral for vacuum
polarization is over the momentum of the electron.
The remaining two integrals are over the
momentum of the photon. In QED each of the three
integrals is ultraviolet divergent. The divergence
arises from infinite momentum.
We consider the integrals over the momentum of the photon.
In KK($\theta,\phi$) the momentum $\hbar k$
of the photon never reaches infinity.
It is cut off at $\Lambda c$,
where   $\Lambda=\alpha^2\hbar/(2cr_\gamma)
             =2.37\times10^{-10}$g
is calculated in Secs.~12 and~13.
Here $r_\gamma$ is the radius of the photon.
The mass $\Lambda$ in QED
is the large mass of a ficticious massive photon.  In KK($\theta,\phi$),
$\Lambda$ is the classical mass of the usual photon.
That is, $\Lambda$
is the photon's mass before it is
quantum corrected downward to its actual value
$\mu=5.31\times10^{-348}\mbox{g}.$
This is described in Sec.~15.
Thus, the integral over the momentum of the photon
can be calculated directly, with $\Lambda c$
as the upper limit.  The lower limit is $\mu c$.
The integral is over momentum $\hbar k=m_\gamma c$,
where $m_\gamma$ is the (variable) rest mass of the photon.
Thus, the integral is really over mass.

Section~14 implies $\Lambda$ is also the mass
of the electron before it is quantum corrected
downward to its actual mass in Sec.~15.  Therefore,
the upper limit of the integral over the momentum
of the electron is also $\Lambda c$.
The lower limit is the
quantum corrected (actual) mass
of the electron times the speed of light.
The upper limits of the integrals
over momentum eliminate the need for regularization.

An object of mass $m$ at rest contains an
amount of energy given by $E=mc^2$.
The object may be converted to energy by combining
it with antimatter.
Then $E=pc$ for the resulting photons.
We set $mc^2=pc$.  Therefore,
the momentum $p$ contained in a material object
at rest with mass $m$ is given by
\q
p=mc.
\qq
This equation is similar to $E=mc^2$.
This completes the analogy with relativity.

Gauge dependence in QED is introduced to obtain
the zero-mass limit of the photon without infinities. 
The infinities occur when division by the photon mass
$\mu$ takes place.
KK($\theta,\phi$) solves this problem by using the classical
photon mass $\Lambda$ instead of the actual, quantum
corrected mass $\mu$.
An example of this is in the classical propagator.
The classical photon mass must be used because we are dealing
with the classical propagator.
Because this mass is nearly the size of the Planck mass,
one does not have to worry about division by zero
or some other very small number.
Therefore, gauge dependence is not needed.
This avoids the difficult problem of eliminating gauge
dependence of physical quantities. One does not have
to resort to any of the various devices proposed to
do this, such as the Gupta-Bleuler
indefinite metric quantization.

One of the assumptions of QED
is the gauge fixing term $-\frac12\lambda
(\partial\cdot A)^2$, where $\lambda$ is the
gauge parameter. When one tries to derive this
term from the 6D curvature scalar of KK($\theta,\phi$),
one finds that all terms of this type cancel.
In other words the coefficient is zero.
Therefore,
\q
\lambda=0.
\qq
This eliminates the gauge-fixing term from the free-$A$
field terms and leaves us with a Maxwell and
mass term for the photon.  Taking the divergence of the
equations for $A$ that result from these terms, we find
\q
\partial\cdot A=0.
\qq
This eliminates the gauge degree
of freedom for $A$. Thus, one cannot
incorporate various types of vectors
into the phase factor for $A$ in KK($\theta,\phi$).
There is no covariant derivative.
There is no
unitary symmetry for the Lagrangian.
The Higgs mechanism is not needed
to maintain unitary symmetry.

The quantization of the electroweak sector of KK($\theta,\phi$)
presented below uses the equations from standard QED.
The equations are used differently.
The one-loop quantum corrections calculated from these equations
should be correct because they
will lead to the correct masses and
coupling constants for the elementary particles.
It would be more proper to derive the equations according to the new
procedures just described above. This should yield the same results.
I will not show this and use the equations from standard QED.

Renormalization is the method for handling infinities
when particles are pointlike, but if particles are extended,
the infinities never develop.
Instead, there are cutoffs.
The cutoffs cause quantum corrections to be finite,
which allows finite bare quantities.
This eliminates the need for renormalization
and allows the calculation of the relevant parameter.
For example, consider the equation [54]
\begin{equation}
m=m_0+\delta m
\end{equation}
from QED.  Here $m$ is the mass of the electron,
$m_0$ its bare mass and $\delta m$ the quantum mass correction.
In a renormalizable model, both $m_0$ and $\delta m$ are infinite
and cancel and $m$ must be inserted by hand.  Thus, $m$ here is
completely unexplained.  However in KK($\theta,\phi$) both $m_0$
and $\delta m$ are finite, allowing a determination of $m$ to be made.
In this way, one can calculate masses.
Similarly, the equation in QED for the quantum charge correction
allows one to calculate coupling constants.

I will now show how to determine $m$.
The equation for the one-loop electron
quantum mass correction [54] for electromagnetism is
\begin{equation}
m=m_0+\delta m,\label{eq:13.1}
\end{equation}
where
\begin{equation}
\delta m=m\frac{3\alpha}{4\pi}\left(\ln\frac{\Lambda^2}{m^2}
+\frac12\right).\label{eq:13.2}
\end{equation}
Here $\alpha$ is the fine structure constant. 
The mass $m_0$ for the electron is its classical free mass.
This is the mass when the charge of the
electron is zero and without a quantum mass correction.
The mass $\Lambda$ is the classical
interacting mass for the electron.  
This is the mass of the electron when it has charge, but before the
quantum mass correction is made.
The actual electron mass $m$ is the quantum corrected interacting mass.
This is the mass of the electron with charge after the quantum mass
correction has been taken.

In KK($\theta,\phi$) the mass $m_0$ is zero [from Part~I].  This is
because, unlike vectors and tensors, there are no mass terms in 
the 6D curvature scalar 
for spinors.  In addition, because $m_0$  is calculated with
its charge equal to zero, the electron does
not acquire a mass from its charge.  Producing
mass from charge is described in the next section.
We have
\begin{equation}
m_0=0.\label{eq:13.3}
\end{equation}
One now combines Eqs.~(\ref{eq:13.1}), (\ref{eq:13.2}) and~(\ref{eq:13.3})
and solves for $m$. The result is the mass formula
\begin{equation}
m=\exp\left(\frac14-\frac{2\pi}{3\alpha}
    \right)\Lambda.\label{eq:13.4}
\end{equation}

This is the most important equation in this paper. It converts Planck
masses $\Lambda$ into elementary particle masses $m$.
This renders observable compactified coordinate dependence in a Kaluza-Klein
theory by converting the masses of the massive Fourier modes from
Planck masses to elementary particle masses.

The problem of deriving the mass of the electron on purely
theoretical grounds now translates into finding the proper values
for $\Lambda$ and $\alpha$. If one substitutes the Planck mass for
$\Lambda$ and sets $\alpha$ to 1/137, the mass $m$ obtained is
much smaller than the electron mass.  The question arrises,
``What will $\alpha$ have to be to convert a Planck mass to an
ordinary mass?"  If Eq.~(\ref{eq:13.4}) is solved for $\alpha$,
with $\Lambda$ the Planck mass and $m$ the mass of the electron,
one finds $\alpha$ in Eq.~(\ref{eq:13.4}) must be larger than
1/137.  Perhaps $\alpha$ in Eq.~(\ref{eq:13.4}) is the bare fine
structure constant. The bare fine structure constant should be
larger than 1/137.

As will be shown, the fine structure constant $\alpha$ starts
out at 1, is then reduced to 1/19 by charge screening due to
vacuum polarization and finally reduced to 1/137 by the
vertex correction. Because charge screening due to vacuum
polarization is allowed by the Feynman diagram for the mass of
the electron, while the vertex correction, which has an external
photon line, is not, the value for $\alpha$ in Eq.~(\ref{eq:13.4})
will be 1/19.

Thus, the ordinary mass scale
is actually the quantum-corrected Planck mass.
Equation~(\ref{eq:13.4}) allows predictions at the Planck mass
to be tested at low energies.  By this I mean that predicted particles
will have elementary particle masses instead of Planck masses.
Equation~(\ref{eq:13.4})
solves the mystery of why there are two widely separated levels
of mass in physics, the Planck mass and the elementary
particle mass scale.  This is referred to as the hierarchy problem.

Section~12 derives an expression for $\Lambda$.
This expression contains the photon's radius, which is
calculated in Sec.~13.
Section~14 determines the value 1/19 for $\alpha$ in
the quantum mass correction equation.
Section~15 derives the mass of the electron
and the value 1/137 for the fine structure constant. Masses for the
muon, tau and the neutrinos are deduced in Sec.~16.
The reason why only the observed elementary particle interactions
take place is explained in Sec.~17.
Section~18 explains the masses of the $W^\pm$ and $Z^0$ from scratch
(instead of the weak interaction coupling constant $G_W$).
New superweak vectors with precisely calculated masses are predicted
in Sec.~19.  Section~20 makes the testable prediction that a new lepton
degenerate with the muon exists.


\begin{center}
{\bf 12. The classical photon mass $\Lambda$}
\end{center}

The electron does not have a mass term in the 
6D curvature scalar of KK($\theta,\phi$).  However, the electron
acquires a mass through its electric field.
An electric field contains energy and since the electron
always carries this field with it, this energy may be
considered to be a rest mass.  Equivalently, the source of the
electron's mass may be ascribed to its charge.  It requires energy
to keep together an assemblage of like electric charges, which tend to
repel each other. This is represented by the equation for the
electron's classical radius
\begin{equation}
mc^2=\frac{e^2}{r_c},\label{eq:14.1}
\end{equation}
where $e$ is the electron's charge and $r_c$ the size of its
charge distribution.  Equation~(\ref{eq:14.1}) shows why
a particle that has charge must have mass.
This source of mass is one
of the two sources of mass in KK($\theta,\phi$).  The other is explicit mass
terms in the 6D curvature scalar.
One reason the classical radius source of
mass is not more accepted is that the classical radius equation
concept as it stands is flawed; the situation is fixed
in Sec.\ 13.

I contend that the classical radius equation applies to the photon
as well as the electron.  If this is so, then the
photon must have some type of charge.  This appears to be the case
because anything that interacts with the photon can be said to have 
charge and one photon interacts with another during
photon-photon scattering [54]. The coupling constant for this
interaction is $\alpha^2$.  As there are two photons involved,
this type of charge for the photon is $e_\gamma=e^2$.
Note that this type of charge is really charge squared
and does not have the units of ordinary electric charge. In fact,
because the fine structure constant $\alpha=e^2/\hbar c=1/137$ is
dimensionless, $e^2$ has the same units as $\hbar c$. Thus, 
it is possible to consider this type of charge to be
something mechanical, but I will
keep it electromagnetic in what follows.

The remainder of this section is essentially a derivation of the
classical radius equation for the case of the photon.  However,
since this equation is the same regardless of the
internal distribution of charge for the particle,
it may be omitted in a first reading. It also demonstrates how
to construct an elementary particle.

In analogy with the electron, the classical mass for the photon will 
be equal to its electrostatic energy $H$. Equation~(\ref{eq:14.1})
for the classical radius of the electron is derived from classical
electrodynamics but is valid for quantum electrodynamics as well.
Therefore, we will derive the classical mass of the photon from
classical electrodynamics. We have
\begin{equation}
\Lambda c^2=H,\label{eq:14.2}
\end{equation}
where the electrostatic energy is derived from the electric field
${\bf E}$
\begin{equation}
H=\frac{1}{8\pi}\int {\bf E}^2\,d^3x.\label{eq:14.2a}
\end{equation}
The electric field is obtained from the potential $\Phi({\bf x})$
\begin{equation}
{\bf E}=-\overrightarrow\nabla\Phi({\bf x}).\label{eq:14.2b}
\end{equation}
Finally, the potential is derived from the charge density $\rho({\bf x})$
\begin{equation}
\Phi({\bf x})=\int\frac{\rho({\bf x'})}{|{\bf x}-{\bf x}'|}
\,d^3x'.\label{eq:14.2c}
\end{equation}
Because ${\bf x'}$ is the variable of integration and ${\bf x}$
is the observation point, it is useful to separate the variables in
$1/|{\bf x-x'}|$
\begin{equation}
\frac{1}{|{\bf x-x'}|}=4\pi\sum^\infty_{l=0}\sum^l_{m=-l}
\frac{1}{(2l+1)}\,\frac{r^{\prime l}}{r^{l+1}}Y^*_{lm}(\theta',\phi')
Y_{lm}(\theta,\phi),\label{eq:14.2d}
\end{equation}
where $r=|{\bf x}|$ and the $Y_{lm}(\theta,\phi)$ are the spherical
harmonics. The next step is to determine the charge density
$\rho({\bf x})$.

The charge is located, of course, where the
particle is located. The particle is located in ordinary 
three-dimensional space (3DS). However, one cannot localize the
photon field to a distance smaller than $2r_\gamma$, where $r_\gamma$
is the radius of the photon, along its direction of motion 
because this would mean that it would be localized
on the compactified sphere. This localization
is impossible because the photon's higher-dimensional wave function, which is
the spherical harmonic $Y_{00}(\theta,\phi)=1/\sqrt{4\pi}$,
must be spread out over the entire compactified sphere.
The coordinate $\theta$ of the sphere projects
onto the photon's 4D world line, which is in the $z$
direction of the embedding space of the compactified sphere.
(The coordinate $\theta$ is the ordinary spherical 
coordinate measured from the positive $z$-axis.)
The photon's quantum
mechanical spin causes the distance $2r_\gamma$ to form a
sphere of radius $r_\gamma$ in 3DS. The photon is therefore
a line segment of a certain type of charge
of length $2r_\gamma$ spinning about its
midpoint.

Because the charge is located where the particle is and the probability
of finding the photon at a given point is equal to the magnitude
squared of its wave function, the charge density will be proportional to
the magnitude squared of its wave function
\q
\rho({\bf x})=e_\gamma|\psi({\bf x})|^2.\label{eq:14.3a}
\qq
The wave function
$\psi({\bf x})$ of a photon as localized as possible in 3DS is equal 
to its radial wave function $R(r)$ times its spin wave function
$S(\theta,\phi)$
\begin{equation}
\psi({\bf x})=R(r)S(\theta,\phi).\label{eq:14.3}
\end{equation}
The radial wave function is given by the projection of the higher-dimensional
wave function onto the 4D world line parameterized by the coordinate
$z$. This projection will be given by the higher-dimensional wave function 
multiplied by a weight factor, which is the differential area
$dA$ of the sphere per length $dz$. The differential area of the
compactified sphere is $2\pi r_0 \sin\theta$,
which is the circumference of a circle centered about the $z$-axis,
multiplied by the arc length $r_0 d\theta$.
Thus,
\begin{equation}
\frac{dA}{dz}=\frac{2\pi r_0 (\sin\theta) r_0 d\theta}{dz}.\label{eq:14.4}
\end{equation}
From the spherical coordinate relation $z=r_0\cos\theta$, we have
$dz=-r_0\sin\theta d\theta$. Therefore,
\begin{equation}
\frac{dA}{dz}=-2\pi r_0=c,\label{eq:14.4'}
\end{equation}
where $c$ is a constant.
The projection of the photon's higher-dimensional wave function
$Y_{00}$ onto the $z$-axis is
\begin{equation}
R(r)=Y_{00}\frac{dA}{dz}=cY_{00}.\label{eq:14.5}
\end{equation}
Therefore, the radial wave function $R(r)$ in 3DS is proportional to the
higher-dimensional wave function $Y_{00}$.
Of course, this procedure could have been applied to any $Y_{lm}$,
with the result that the projection of $Y_{lm}$ onto the $z$-axis
is proportional to $Y_{lm}$. 

After normalization, the radial function is given by a new constant
\begin{eqnarray}
R(r)&=&c',\qquad 0\le\ r \le r_\gamma\label{eq:14.5a'}\\       
R(r)&=&0, \qquad r_\gamma<\ r<\infty.\label{eq:14.5b'}
\end{eqnarray}
The spin wave function is given by
\begin{equation}
\frac{1}{\sqrt3}\left(
\begin{tabular}{c}
1\\
0\\
1
\end{tabular}
\right)
\to \tfrac{1}{\sqrt3}Y_{11}(\theta,\phi)
+\tfrac{1}{\sqrt3}Y_{1,-1}(\theta,\phi)
=S(\theta,\phi),
\label{eq:14.6}
\end{equation}
where we have omitted the longitudinal polarization
state $Y_{10}(\theta,\phi)$ of the photon.
The constant $c'$ is determined by the condition
\begin{equation}
\int \rho({\bf x})\,d^3 x=e_\gamma.\label{eq:14.7}
\end{equation}
We find
\begin{eqnarray}
\rho({\bf x})&=&\frac{9e_\gamma}{8\pi r_\gamma^{\ 3}}\,
\sin^2\!\!\theta,\qquad 0\le\ r\le r_\gamma\label{eq:14.8a}\\
\rho({\bf x})&=&0,\qquad\qquad\qquad r_\gamma<\ r<\infty.\label{eq:14.8b}
\end{eqnarray}
Substituting this into Eq.~(\ref{eq:14.2c}),
we find from Eqs.~(\ref{eq:14.2a})--(\ref{eq:14.2d}) that
\begin{equation}
H=\frac{e_\gamma^{\ 2}}{2r_\gamma}.\label{eq:14.9}
\end{equation}
This is the monopole contribution in Eq.~(\ref{eq:14.2c}). The higher order
terms are negligible. Therefore, the dependence of $H$ upon the
angular and radial dependence of $\rho({\bf x})$ is negligible.
The energy behaves as if the charge were uniformly distributed
throughout a sphere of radius $r_\gamma$.

Equations~(\ref{eq:14.9}) and~(\ref{eq:14.2}) yield
for the classical mass of the photon
\begin{equation}
\Lambda=\frac{e_\gamma^{\ 2}}{2c^2 r_\gamma}
         =\frac{\alpha_\gamma\hbar}{2c\,r_\gamma},\label{eq:14.10}
\end{equation}
where $\alpha_\gamma=e_\gamma^{\ 2}/\hbar c$.
Equation (\ref{eq:14.10}) is
the classical radius equation.
The distance appearing in the classical radius equation is
that between the two most widely separated points of charge
in the distribution.  For a spherical distribution it is therefore
the diameter.  This is why a factor of two multiplying the radius
appears in our equation.  This factor does not appear in the usual
definition of the classical radius equation, which is only an
order-of-magnitude estimate.
Using the relation
$\alpha_\gamma=\alpha^2$, we have
\begin{equation}
\Lambda=\frac{\alpha^2\hbar}{2c\,r_\gamma}.\label{eq:14.10'}
\end{equation}
It remains to determine the radius $r_\gamma$ of the photon.


\begin{center}
{\bf 13. The radius of the photon}
\end{center}

In order to eliminate the infrared divergences in QED the photon must
have a small mass.
Therefore, I treat it in what follows as a massive particle, like
the electron. Among other things, this means the photon may be
brought to rest. If a particle is at rest and located at the
origin of 3DS then its 
4D world line is in the ordinary time direction.
However, particles are never really at rest
because they are spinning. Setting the photon's spin angular
momentum equal to $\Lambda v \langle r \rangle$, we obtain
\begin{equation}
\left[s(s+1)\right]^{1/2}\hbar=\Lambda v \langle r \rangle.\label{eq:15.1}
\end{equation}
A simple way of determining the radius at which the particle
rotates is to place its entire mass at a particular average radius
$\langle r \rangle$. 

The higher-dimensional wave function $Y_{00}$ of the photon
projected onto its 4D world line, which is in the $z$
direction of the embedding space of the compactified sphere,
is distributed uniformly over the length of the world line.  Thus,
$\langle r\rangle=\frac 12\,r_\gamma$.
Setting $s=1$ in Eq.~(\ref{eq:15.1}), we obtain
\begin{equation}
\Lambda v \tfrac 12\,r_\gamma=\sqrt 2\,\hbar.\label{eq:15.1'}
\end{equation}
Substituting Eq.~(\ref{eq:14.10}) into Eq.~(\ref{eq:15.1'}), we have
\begin{equation}
v=\frac{4\sqrt2\, c}{\alpha_\gamma}.\label{eq:15.2}
\end{equation}

Note that because $\alpha_\gamma=\alpha^2=(1/137)^2$, the velocity
of spin for the photon is much greater than the speed of light.
This is one objection to the concept of the classical radius
equation.  There is a way around this problem, however.  The trick
is to use the bare fine structure constant
$\alpha_{\gamma0}=\alpha_0^{\ 2}$, which will later be determined
to be unity, in the denominator of Eq.~(\ref{eq:15.2}) instead of the
actual one.
As I will show, this larger denominator will bring the
velocity of spin for the photon to something below the speed of
light.  The justification for substituting the bare fine structure
constant for the actual one is that only the bare photon is
spinning.  Its surrounding outer shell of polarized vacuum
and quantum `charge' correction does not spin along with it.

I will define the bare fine structure constant $\alpha_{\gamma0}$
for the photon to be
\begin{equation}
\alpha_{\gamma0}=\frac{e_{\gamma0}^{\ \ 2}}{4\pi\hbar c},\label{eq:15.3}
\end{equation}
while the actual fine structure constant $\alpha_\gamma$ is
$\alpha_\gamma=e_\gamma^{\ 2}/(\hbar c)$.
These definitions introduce a factor of $4\pi$ into the
denominator of Eq.~(\ref{eq:15.2}) when the actual fine structure constant
$e_\gamma^{\ 2}/(\hbar c)=\alpha_\gamma=\alpha^2$
is converted to the bare one
$e_{\gamma0}^{\ \ 2}/(\hbar c)=4\pi\alpha_{\gamma0}=4\pi\alpha_0^{\ 2}$.
Thus,
\begin{equation}
v=\frac{\sqrt2\,c}{\pi\alpha_0^{\ 2}}.\label{eq:15.2'}
\end{equation}

I now show why $\alpha_0=1$.
We start with the photon mass term
\q
+\ \frac{\mu^2}{2}A_{00}^\mu A_\mu^{00}.
\qq
We now convert one of the vectors in this term to spinors
using the prescription from Part~I. The result is
\q
e_0 \overline\psi\gamma^\mu A_\mu^{00}\psi,
\qq
where $e_0$ is the charge of the electron before quantum corrections.
Here the electron field $\psi$ is derived from the photon
$A_\mu^{00}$ instead of the $Z_\mu^0=A_\mu^{10}$
as in the Dirac term. Therefore we may choose a different 
normalization for $\psi$ here than in the Dirac term.
We choose the normalization such that
\q
e_0=\sqrt{4\pi\hbar c}.
\qq
Substituting this into $\alpha_0=e_0^{\ 2}/(4\pi\hbar c)$,
we find
\q
\alpha_0=1.
\qq

Substituting this into Eq.~(\ref{eq:15.2'}), we have
\begin{equation}
v=\frac{\sqrt2}{\pi}\,c.\label{eq:15.2''}
\end{equation}
Because $\sqrt2$ is less than $\pi$, the velocity of spin is less
than the speed of light.  This is the velocity of spin of a point
midway out from the center of a spinning photon; the velocity of a
point on the surface of it is twice as great but is still less
than the speed of light because $2\sqrt2$ is still less than $\pi$.

One would expect that the size of the charge distribution in 3DS
to be given by its projection from the 4D world line onto a spatial
axis in 3DS. Then $r_\gamma=r_0 \sin\psi$, where $\psi$
is the angle between the 4D world line and the ordinary $t$-axis.
Thus it would seem that the radius for the photon is less than $r_0$.
However, this is not the case.   The compactified sphere forms a
boundary in 3DS because the higher dimensions overlap 3DS.
The surface charge density is specified on this boundary.
The distribution of charge that is equivalent to a spherical
surface charge density of radius $r_\gamma$ inside the compactified
sphere plus a boundary
at the compactified sphere is a spherical
surface charge density outside the compactified sphere.  This follows from
the method of images. Therefore, we have
\q
r_\gamma^{\ \prime}=\frac{r_0^{\ 2}}{r_\gamma},
\qq
where the radius of the photon is effectively $r_\gamma^\prime$.
Substituting $r_\gamma=r_0\sin\psi$ into this equation,
we find
\q
r_\gamma^{\ \prime}=\frac{r_0}{\sin\psi}.
\qq
Henceforth we will drop the prime from $r_\gamma^{\ \prime}$
and refer to the radius for the photon as
\q
r_\gamma=r_0/\sin\psi.\label{eq:15.4}
\qq

From the photon's velocity of spin determined above, one may calculate
the angle $\psi$ from the following equation
\begin{equation}
\tan\!\psi=\frac{\Delta x}{c\Delta t}=\frac{v}{c}=
             \frac{\sqrt2}{\pi},\label{eq:15.5}
\end{equation}
where $x$ is the spatial direction in which the photon moves
and the time axis is labeled $ct$.
Therefore,
\begin{equation}
r_\gamma=2.4361r_0.\label{eq:15.4'}
\end{equation}

The distance $r_0$ remains an unexplained parameter.  Presumably the
compactified sphere has been shrinking, while ordinary 4D spacetime
has been expanding.  Thus, cosmological considerations
would determine the radius $r_0$ of the compactified sphere.  In the absence of
such considerations, it is easy to guess that the most important
distance in KK($\theta,\phi$), namely $r_0$, is equal to the most important
distance in physics, namely the Planck length $L$.  We have
\begin{equation}
r_0=L=1.62\times 10^{-33}\ \mbox{cm}.\label{eq:15.6}
\end{equation}
I claim the Planck length is important because it is the radius of the
two higher dimensions.

Combining Eqs.~(\ref{eq:15.6}), (\ref{eq:15.4'}) and (\ref{eq:14.10'})
we obtain the value for the classical mass $\Lambda$ of the photon 
\begin{equation}
\Lambda=2.37\times10^{-10}\ \mbox{g}.\label{eq:15.12}
\end{equation}


\begin{center}
{\bf 14. The screened fine structure constant}
\end{center}

We now consider the effects of charge screening due to
vacuum polarization on $\alpha_0$. The one-loop expression for the
fine structure constant $\alpha_\gamma$ after charge screening via
the photon is [54]
\begin{equation}
\alpha_\gamma=\left(1+\frac{\alpha_0}{3\pi}
  \,\ln\frac{\Lambda^{\prime2}}{m^2}\right)^{-1}\alpha_0.\label{eq:16.1}
\end{equation}
The screened fine structure constant represented here by $\alpha_\gamma$
bears no relation to the photon's fine structure constant
$\alpha_\gamma=\alpha^2$ considered above.
The cutoff $\Lambda'$ is defined according to [54]
\begin{equation}
\sum_{s=1}^S C_s\ln\lambda_s^2=-\ln\frac{\Lambda^{\prime2}}
                                   {m^2},\label{eq:16.2}
\end{equation}
where the $\lambda_s m$ are the large masses of the $S$ spinors.
The quantum mass correction formula~(\ref{eq:13.4}) suggests that
the large mass of each spinor is $\Lambda$. Therefore,
\begin{equation}
\lambda_sm=\Lambda.\label{eq:16.3}
\end{equation}
Combining Eqs.~(\ref{eq:16.2}) and (\ref{eq:16.3}), we have
\begin{equation}
\sum_{s=1}^SC_s\ln\frac{\Lambda^2}{m^2}
  =-\ln\frac{\Lambda^{\prime2}}{m^2}.\label{eq:16.2'}
\end{equation}
According to Ref.~52,
\begin{equation}
\sum_{s=1}^SC_s=-1.\label{eq:16.4}
\end{equation}
Substituting this into Eq.~(\ref{eq:16.2'}), we find
\begin{equation}
\Lambda'=\Lambda.\label{eq:16.5}
\end{equation}

In addition to photons producing an electron-positron pair, one might
have vacuum polarization involving the $Z^0$. To calculate the effect of
this process on $\alpha_\gamma$, we use the equation
\q
\alpha_{\gamma Z}=\left(1+\frac{\alpha_\gamma}{3\pi}
\ln\frac{\Lambda_Z^{\ \,2}}{m^2}\right)^{-1}\alpha_\gamma,
        \label{eq:16.6}
\qq
which is similar to Eq.~(\ref{eq:16.1}). Here $\alpha_{\gamma Z}$
is the fine structure constant after screening via the photon and $Z^0$ and
$\Lambda_Z$ is the classical mass of the $Z^0$ due to its weak charge. 
The classical mass $\Lambda_Z$
is determined by an equation of the same form as that which
determines the classical mass of the photon
\q
\Lambda_Z=\frac{g^2}{2c^2r_Z},\label{eq:16.7}
\qq
where the weak charge $g$ of the $Z^0$ is given by the definition
of the weak interaction coupling constant $G_W$
\q
g^2=2^{5/2}\frac{G_W}{(\hbar c)^3}M_W^{\ \ 2}\hbar c,\label{eq:16.7a}
\qq
where $M_W$ is the mass of the $W$.

We obtain the radius $r_Z$ of the $Z^0$ in a manner similar to that
of the photon. The radial function for the $Z^0$'s weak charge
distribution is
\q
R(r)=\left(\frac{3}{r_Z}\right)^{1/2}\frac{r}{r_Z}.\label{eq:16.8}
\qq
This is obtained by noting the higher-dimensional wave function
for the $Z^0$
is $Y_{10}$, which is proportional to $\cos\theta$.  This, in turn,
is equal to $z/r_Z$ according to the definition of spherical coordinates.
The line segment along the coordinate $z$ 
rotates in 3DS to become a radial coordinate.
The average value for $r$ is given by
\q
\left<r\right>=\int_0^{r_Z} R^*(r)\,rR(r)\,dr=\tfrac34\,r_Z.\label{eq:16.9}
\qq

In the velocity calculation, $\alpha_0$ appears instead of $\alpha_0^{\ 2}$
because self interactions with two vertices are allowed for the $Z^0$
as opposed to the case of the photon, where only photon-photon 
scattering, which is a self interaction with four vertices,
is permitted. In addition,
the value of the bare fine structure constant $\alpha_0$ is given by
\q
\alpha_0=l(l+1),\label{eq:16.10}
\qq
where, from its above-mentioned spherical harmonic, $l=1$ for the $Z^0$. 
The above considerations lead to a radius for the $Z^0$
\q
r_Z=6.7389r_0.\label{eq:16.11}
\qq
Equations (\ref{eq:16.7}), (\ref{eq:16.7a}), (\ref{eq:16.11}),
and (\ref{eq:15.6}) yield 
for the classical mass of the $Z^0$ due to its weak charge
\q
\Lambda_Z=6.87\times10^{-7}\ \mbox{g}.\label{eq:16.12}
\qq

Substituting the actual mass of the electron for $m$
in Eqs.~(\ref{eq:16.1}) and~(\ref{eq:16.6}) and 
$\alpha_0=1$ into Eq.~(\ref{eq:16.1}) and using Eqs.~(\ref{eq:16.5}),
(\ref{eq:15.12}), (\ref{eq:16.6}) and~(\ref{eq:16.12}),
we obtain a value for the screened fine structure constant
\q
\alpha_{\gamma Z}=\frac{1}{19.71}.\label{eq:16.13}
\qq


\begin{center}
{\bf 15. The electron}
\end{center}

We have yet to consider the effect of the quantum charge vertex
correction on $\alpha_{\gamma Z}$, however this is
unnecessary for quantum mass corrections because it requires one
external photon line, which does not appear in the quantum mass
correction diagram. Thus, the fine structure constant used
in the quantum mass correction formula (\ref{eq:13.4})
is $\alpha_{\gamma Z}$.  To summarize, the equations that determine
the electron's mass are
\begin{eqnarray}
         \alpha_\gamma&=&\left(1+\frac{\alpha_0}{3\pi}
         \ln\frac{\Lambda^2}{m^2}\right)^{-1}\alpha_0,\label{eq:17.1a}\\ 
         \alpha_{\gamma Z}&=&\left(1+\frac{\alpha_\gamma}{3\pi}
         \ln\frac{\Lambda_Z^{\ \,2}}{m^2}\right)^{-1}
         \alpha_\gamma,\label{eq:17.1b}\\
         m&=&\exp\!\left(\frac14-\frac{2\pi}{3\alpha_{\gamma Z}}
         \right)\Lambda,\label{eq:17.1c}
\end{eqnarray}
where $\alpha_0=1$ and $\Lambda$ and $\Lambda_Z$ are given
by Eqs.~(\ref{eq:15.12}) and~(\ref{eq:16.12}), respectively.
Solving the three Eqs.~(\ref{eq:17.1a})--(\ref{eq:17.1c})
for the three unknowns
$\alpha_\gamma$, $\alpha_{\gamma Z}$ and $m$, we obtain
\q
m=\exp\!\left(\frac94-\frac{6\pi}{\alpha_0}\right)
    \Lambda^5\Lambda_Z^{\ \,-4}.\label{eq:17.2}
\qq

The value for $m$ obtained from Eq.~(\ref{eq:17.2}) is
$1.15\times10^{-4}$
MeV, which is much less than the actual electron
mass of 0.511 MeV. This is because we have calculated the
mass of the electron due to its electric charge. To this 
we must add the contribution due to its weak charges.

The electron has two weak charges:  that from weak isospin
and from an explicit weak interaction term in the 6D curvature
scalar $\widehat R$ of KK($\theta,\phi$).  The electromagnetic $\alpha_0$
was also determined by two charges:  the $z$-component of
weak isospin and the electromagnetic
charge, which was zero, from $\widehat R$.

The mass of the electron due to its weak charges is obtained from
Eq.~(\ref{eq:17.2}), but with $\alpha_0$
determined by the electron's weak charge from its
weak isospin $T$ or its weak charge from the 6D curvature
scalar $\widehat R$. The weak charge $g_{\widehat R}$ of the electron
from $\widehat R$ in Part~I
is $4.758\sqrt{\hbar c}$. Thus,
\q
\alpha_0=\frac{g_{\widehat R}^{\ \ 2}}{4\pi\hbar c}
          =1.8013.\label{eq:17.3}
\qq
Substituting this into Eq.~(\ref{eq:17.2}), we obtain for the weak mass
of the electron 0.510 MeV, which has an error of three tenths
of one percent. 
The remaining weak charge $g_T$ of the electron due to its weak isospin
$T=L=\left[l(l+1)\right]^{1/2}$ is 
$\left[l(l+1)4\pi\hbar c\right]^{1/2}$, where from Table~1,
$l=\frac12$ for the electron. Exchanging $g_T$ for $g_{\widehat R}$
in Eq.~(\ref{eq:17.3}) we find the mass of the electron due to
its weak isospin is negligible.
Thus, the prediction for the electron's mass stands at an accurate
0.510 MeV. We have determined
our first unexplained parameter --- the mass of the electron.

The observed value 1/137 of the fine structure constant $\alpha$ 
is determined from
$\alpha_{\gamma Z}$ by the final quantum charge, or vertex, correction.
The one-loop expression for the fine structure constant $\alpha_V$ after
the vertex quantum correction is [54]
\q
\alpha_V=\left(1+\frac{\alpha_{\gamma Z}}{3\pi}
  \ln\frac{m_e}{\mu}\right)^{-2}\alpha_{\gamma Z},\label{eq:17.4}
\qq
where $\alpha_{\gamma Z}$ is not infinite but given by Eq.~(\ref{eq:16.13}),
$m_e$ is the mass of the electron,
and $\mu$ is the actual mass of the photon. To calculate $\mu$, we note
that, as explained above, the photon has effective `charge' $e^2$ and
fine structure constant $\alpha^2$, both of which are the square of the
respective quantities for the electron.
Considering only charge screening on the bare fine structure
constant for the photon, we have 
\q
\alpha_{\gamma Z}=\left(\frac{1}{19.27}\right)^2.\label{eq:17.5}
\qq
This value for $\alpha_{\gamma Z}$ contains 1/19.27
instead of 1/19.71 because we use the electromagnetic $\alpha_0=1$
for the photon instead of the weak $\alpha_0=1.8013$ for the electron.

Substituting Eq.~(\ref{eq:17.5}) into Eq.~(\ref{eq:17.1c}), we obtain
\q
\mu=5.31\times10^{-348}\ \mbox{g}.\label{eq:17.6}
\qq
This mass for the photon is far below the experimental limit of
$4\times10^{-48}\ \mbox{g}$.  Although the mass of the photon is
not generally considered to be a parameter (it is thought to be zero),
it must be small but nonzero in order to eliminate infrared divergences
in QED. The mass of the photon is the second unexplained parameter
of the SM to be explained by KK($\theta,\phi$).

Because $\mu\ne0$ the quantum vertex
correction in Eq.~(\ref{eq:17.4}) is not infinite. Equation (\ref{eq:17.4})
then yields
\q
\alpha_V=\frac{1}{486},\label{eq:17.7}
\qq
which is multiplied by a factor of $\sqrt{4\pi}$ to obtain $\alpha$.
To see where the $\sqrt{4\pi}$ comes from, we note that
because there is a factor of $4\pi$ in the denominator for the
bare fine structure constant $\alpha_0=e_0^{\ 2}/(4\pi\hbar c)$
while the actual fine structure constant $\alpha=e^2 /(\hbar c)$
does not have this factor, at some point in the calculation
for $\alpha$, we will have one equation for the fine structure
constant with a factor of $4\pi$ in the denominator and the next
without it.  This introduces a factor of $4\pi$.
Apparently, the fine structure constant for the electron must be
derived from that of the photon. The photon is a much more important
particle than the electron in KK($\theta,\phi$). For example, it is the
classical mass $\Lambda$ of the photon that is the electromagnetic cutoff.
Therefore, we have
$\alpha_V^{\ 2}=e_\gamma^{\ 2}/(4\pi\hbar c)$, while
$\alpha^2=e_\gamma^{\ 2}/(\hbar c)$.
These equations produce a factor of $\sqrt{4\pi}$
\q
\alpha=\sqrt{4\pi\alpha_V^{\ 2}}\,=\sqrt{4\pi}\,\alpha_V.\label{eq:17.8}
\qq
Thus, our derived value for $\alpha$ is 1/137.1, which differs from
the actual value of 1/137.036 by less than one tenth of one percent. 
This is the third unexplained parameter in the SM 
explained by KK($\theta,\phi$).


\begin{center}
{\bf 16. The muon, tau and neutrinos}
\end{center}

The masses of the muon and tau are calculated in a manner similar
to that of the electron. Their electromagnetic masses remain the 
same because $\alpha_0$ stays at unity. Their weak masses, however,
increase drastically because their weak charges $g_T$ due to 
isospin increase from 
$\left[\frac12\left(\frac12+1\right)4\pi\hbar c\right]^{1/2}$
for the electron to $\left[l(l+1)4\pi\hbar c\right]^{1/2}$, where
from Table~1, $l=\frac32$ for the muon and $l=\frac52$ for the
tau and the mass depends on the weak charge exponentially. 
Thus, it is the exponential dependence of the mass upon the
weak charge that causes the muon to be so much more massive
than the electron.

As with the electron, the weak charge for the muon 
is that due to isospin or from $\widehat R$.
The former is 6.865 $\sqrt{\hbar c}$.
This is calculated in a manner similar to the electron's
$g_T$. From Eq.~(\ref{eq:17.3}) we have $\alpha_0=3.75$.
Equation (\ref{eq:17.2}) yields
$m=117.3$ MeV, which is 11\% higher than the actual
muon mass of 105.7 MeV.
The weak charge for the muon $g_{\widehat R}$
from the 6D curvature
scalar of KK($\theta,\phi$) is the same as that for the
electron [Part~I]. The mass contribution from this weak charge for the muon
is the same as that for the electron
(one electron mass) and is therefore negligible.
The mass of the muon is the fourth unexplained parameter
of the SM to be explained by KK($\theta,\phi$).

Like the muon, the mass of the tau due to its weak charge from
$\widehat R$ is one electron mass and is negligible.
Therefore, its weak $\alpha_0$ is given by $l(l+1)$,
where $l=\frac52$. Its calculated mass is 2067 MeV, which is 16\%
higher than its actual mass of 1784 MeV.
This is the fifth unexplained parameter explained by KK($\theta,\phi$).

I now show how to derive the Gell-Mann relation
for the weak interactions. We have
\q
L_z\psi=m\hbar\psi,
\qq
where $L_z$ is the $z$-component of angular momentum
due to motion of any particle on the compactified sphere. 
This angular momentum is commonly known as isospin.
Here $m$
does not represent a mass but rather the quantum number
for the z-component of isospin.
The electron's charge quantum number,
which is integral and $m$ are related by elementary
quantum mechanics according to the relation [57]
\begin{equation}
m=c+\mbox{int.},\label{eq:15.7}
\end{equation}
where $c$ is a constant for the multiplet and int.\ is an integer.
According to Ref.~57,
$c=\frac 12$ for spinors. 
Equation (\ref{eq:15.7}) is the Gell-Mann relation
$Q=t_3+\frac12 y$, with the following identifications:  charge
$Q=\mbox{int.}$, the z-component of weak isospin $t_3=m$ and one-half
the weak hypercharge $\frac12 y=-c$. Thus, the origin of the Gell-Mann
relation is explained by KK($\theta,\phi$).

We will now derive the masses of the neutrinos.
Solving Eq.~(\ref{eq:15.7}) for
the integer and substituting $c=\frac12$ and $m=\frac12$ for the
neutrino, we find the charge on the neutrino is zero. This means
neutrinos do not couple to the photon. Therefore, their electromagnetic
masses are zero. Neutrinos have the same weak charges 
$g_T$ and $g_{\widehat R}$ as their charged leptonic counterparts.
However, since the photon is absent, their weak $\alpha_0$ 
is screened only via the $Z^0$. Thus, there will be only one
vacuum polarization equation.  In addition, the electromagnetic
cutoff $\Lambda$ will be absent from the quantum mass correction
equation. It is replaced by a weak cutoff described below.

The equations that determine the masses of neutrinos are
\begin{eqnarray}
  \alpha_Z&=&\alpha_0\left(1+\frac{\alpha_0}{3\pi}
             \ln\frac{\Lambda_Z^{\ \,2}}{m_\nu^{\ \,2}}\right)^{-1},
                      \label{eq:18.1a}\\
         1&=&\frac{3\alpha_Z}{4\pi}\left(\ln
             \frac{\Lambda_Z\Lambda_Z'}{m_\nu^{\ 2}}+\frac12\right),
                      \label{eq:18.1c}
\end{eqnarray}
where Eq.~(\ref{eq:18.1c}) is the quantum mass correction equation
before it is solved for $m_\nu$ as described in Sec.~12.
One of the $\Lambda$'s in this equation must be
$\Lambda_Z'=\Lambda_Zm_\nu^{\ 2}/M_W^{\ \ 2}$. In other words,
$\Lambda_Z'$ is simply $\Lambda_Z$ with the $W$ mass squared replaced
by the neutrino mass squared. This is necessary in order to allow for
the weakness of the weak interaction where the neutrino emits a $Z^0$ at the
first vertex of the quantum mass correction diagram.
This process is a weak one due to the large mass of the $Z^0$.
However, all subsequent processes described by
the vacuum polarization and quantum mass correction diagrams should
not be `tagged' as weak because the $Z^0$ has already been produced
and its energy and momentum must be maintained if it is to be paid
back according to the uncertainty principle.
This means that all $\Lambda$'s in the 
vacuum polarization and mass correction equations besides
the $\Lambda_Z'$ mentioned above must be $\Lambda_Z$.

In the electron's quantum mass correction
equation, the classical mass of the electron was given by the sum of its
classical electromagnetic and weak masses, but the classical weak mass
$\Lambda_Z'=\Lambda_Zm_e^{\ 2}/M_W^{\ \ 2}=2.63\times10^{-17}$ g
was negligible compared to the electromagnetic
$\Lambda= 2.37\times10^{-10}$ g.
Now that the photon's effects are absent from the quantum mass
equation, $\Lambda$ is replaced by $\Lambda_Z'$.

Note that the neutrino mass $m_\nu$ cancels in
Eq.~(\ref{eq:18.1c}). The relative smallness of $\Lambda_Z'$ is the
reason why neutrinos have such small masses.

The solution of Eqs.~(\ref{eq:18.1a}) and~(\ref{eq:18.1c}) is
\q
   m_\nu=\Lambda_Z^{-1/8}\left(\frac{\Lambda_Z'}{m_\nu^{\ 2}}\right)^{-9/8}
   \exp\!\left(\frac{3\pi}{2\alpha_0}-\frac{9}{16}\right).\label{eq:18.2}
\qq
Because the fraction involving $\alpha_0$ in the exponential in
Eq.~(\ref{eq:18.2}) does not have a minus sign in front of it as
for charged leptons, the masses of neutrinos will increase
for decreasing $\alpha_0$, as an examination of this equation
will show. Thus, the smaller of $g_T$ or $g_{\widehat R}$
will yield the masses of the neutrinos instead of the larger.
The larger of these two weak charges
will yield a mass that is negligible because the mass depends
on the weak charge exponentially.

Each neutrino will have the same weak charges $g_T$ and $g_{\widehat R}$
as its charged leptonic counterpart.  For the electron neutrino
$g_T=3.07\sqrt{\hbar c}$ (leading to $\alpha_0=3/4$) and
$g_{\widehat R}=4.758\sqrt{\hbar c}$ (leading to $\alpha_0=1.8013$).
Substituting these separately
into Eq.~(\ref{eq:18.2}), we find that the mass of the electron neutrino is
$6.16\times10^{-7}$ eV, which is
well below the experimental limit of 20 eV. The masses of the muon 
and tau neutrinos are determined from $g_{\widehat R}=4.758\sqrt{\hbar c}$.
They are the same at $1.57\times10^{-8}$ eV,
far lower than their experimental limits of 0.2 MeV and 35 MeV,
respectively.
The masses of the neutrinos are the sixth, seventh and eighth unexplained
parameters of the SM to be derived from KK($\theta,\phi$).


\begin{center}
{\bf 17. Elementary particle interactions}
\end{center}

Just as a given particle may decay only into certain decay products,
a given ket may only have certain other kets as its factors.
In fact, the relationship is one-to-one, with the ket for each
particle determining which decays can take place.
For example, one may ask why the $W^+$ decays into the particles
it does, instead of the decay products for the $W^-$
or the $Z^0$.  To be sure, these decays can be ruled out by
the law of conservation of charge, but then one must ask why
the law of conservation of charge holds.  I claim this law follows
from the rules for combining kets given by
Clebsch-Gordan coefficients.

One may know which
elementary particle interactions can and cannot take place by
determining whether the relevant Clebsch-Gordan coefficient is
zero or not. An elementary particle interaction can take place if the
coefficient is not zero. This is the case if
and only if $m=m_1+m_2$ and $|l_1-l_2|\le l\le|l_1+l_2|$. 
In terms of the bare charge quantum number $Q$ for vectors,
we have $Q=m$, for spinors
$Q=m-\tfrac12$ and for antispinors $Q=m+\tfrac12$.
I claim the relation $m=m_1+m_2$ is
the law of conservation of charge.

A short cut to obtaining the spinor expansion of a vector is to
write the vector's ket in terms of half-odd integral $l$ kets
using Clebsch-Gordan coefficients. One then writes the particle
symbols of the spinors with the quantum numbers of the spinor
kets below the spinor kets. The quantum
numbers of spinors are given in Table~1. Finally, note that the
first spinor ket in each pair is associated with an antiparticle
by convention.

An example of elementary particle interactions being dictated by
nonzero Clebsch-Gordan coefficients is as follows:  The ket 
$|11\rangle$ may be transformed into the product of kets
$\left|\tfrac12\,\tfrac12\right>\left|\tfrac12\,\tfrac12\right>$
because the relevant Clebsch-Gordan coefficient is not zero but one.
We have
\begin{equation}
|11\rangle=
\left|\tfrac12\,\tfrac12\right>\left|\tfrac12\,\tfrac12\right>.
\label{eq:21.2}
\end{equation}

After identifying $|11\rangle$ above with the $W^+$, the first
$\left|\tfrac12\,\tfrac12\right>$ with ${\overline e}^{\,+}$
(a positron)
and the second $\left|\tfrac12\,\tfrac12\right>$ with $\nu_e^0$
(a neutrino) we have the decay
$W^+\to {\overline e}^{\,+}+\nu_e^0$.
From the Lagrangian standpoint, this interaction is justified by
substituting a spinor expansion
for one factor of $W_\mu^{+}$ in its mass term,
thereby generating an interaction term as described in Part~I.

Note, for example, the $W^+$ (ket $|11\rangle$)
cannot decay into two neutrinos (kets $\kk\k$) because
the Clebsch-Gordan coefficient for this process is zero.
Similarly, the $W^+$ cannot decay into two electrons, two positrons,
a positron and an electron, or an electron and a neutrino because
the Clebsch-Gordan coefficients for all these processes are zero.
These violate $m=m_1+m_2$, which is the law of conservation of charge.

There is one addition to the above-mentioned rules for obtaining
nonzero Clebsch-Gordan coefficients involving leptons.
There is nothing to prevent $l$ from being
expanded in terms of kets with different values for half-odd
integral $l_1$ and $l_2$.
For example, if $l=1$, $l_1=\tfrac12$ and $l_2=\tfrac32$, then it would
appear one could have
$|11\rangle=
\sqrt{\tfrac{1}{4}}
\left|\tfrac12\,\tfrac12\right>\left|\tfrac32\,\tfrac12\right>
-\sqrt{\tfrac{3}{4}}
\left|\tfrac12\,,-\tfrac12\right>\left|\tfrac32\,\tfrac32\right>$,
with the corresponding interactions $W^+\to{\overline e}^{\,+}
+\nu_{\mu1}^0$
or $W^+\to{\overline\nu}_e^{\,0}+\mu^+_2$. Both of these decays violate
electron and muon number conservation. One needs a reason why this
cannot occur because it contradicts observation.
Here it is: All spinors in KK($\theta,\phi$) are ultimately
derived from the photon because this is the only vector that is null.
This is a requirement for writing a vector in terms of two spinors.
Now one cannot expand the ket for the photon $|00\rangle$ in terms
of kets with different values for $l_1$ and $l_2$ because then the
minimum value for $l$ will be greater than zero, while the photon
must have $l$ equal to zero. The minimum value for $l$ in the above
example was $|\tfrac12\,-\tfrac32|=1$. Remember the
minimum value for $l$ is given by $|l_1-l_2|$ from the relation
$|l_1-l_2|\le l\le|l_1+l_2|$. This means that we must have
$l_1=l_2$ and $|00\rangle$ can only be expanded in terms of kets
from one family of leptons at a time. For example, one could expand
$|00\rangle$ in terms of $l_1=l_2=\tfrac12$ kets or $l_1=l_2=\tfrac32$
kets but not both in the same expansion. In effect, only one family of
leptons may exist at one time.  This adds the requirement
$l_1=l_2$ to the above rules for obtaining nonzero Clebsch-Gordan
coefficients involving leptons and ensures that electron and muon
number is conserved.

The elementary particles
are described by the Lagrangian for KK($\theta,\phi$).
The Lagrangian for KK($\theta,\phi$) is derived from 6D spacetime.
I claim that conjugation of charge, parity and time reversal
is an invariance
because reversal of all six dimensions leads to an identical spacetime.


\begin{center}
{\bf 18. The W and Z}
\end{center}

One source of mass in KK($\theta,\phi$) is the natural appearance of explicit
mass terms in the 6D curvature scalar.
For spinors, the graviton and the photon this mass is zero, while for
the $Z^0$ and $W^\pm$ these masses [from Part~I] are 1.9109\,$\hbar/cr_0$
and 1.6293\,$\hbar/cr_0$, respectively. (Note that $r_0$ is the Planck
length and $\hbar/cr_0$ is the Planck mass.) To these masses we should
add the mass due to the particle's charges. One obtains this type of
mass by substituting the particle's charge into the classical
radius equation.
The predicted masses of the $W$ and $Z$ without these masses will
turn out to be close enough to the
actual values to allow us not to bother with this calculation.

Now that we have the above-mentioned Planck masses of the
$W^\pm$ and $Z^0$, the
question remains how to reduce these masses to their
observed values. We noted above that a vector can be
considered to be a combination of two spinors. Thus, we can
use the quantum mass correction developed for spinors for vectors.
Both the wave functions for the $Z^0$ and the $W^\pm$ are made
up of those for the electron and electron neutrino.  From the first
of these spinor expansions one can say that the $Z^0$ is one half an
electron-positron pair and one half a neutrino-antineutrino pair.
The factors of one half arise from the Clebsch-Gordan coefficients
of $1/\sqrt2$ multiplying each of the two spinor pairs of kets.
The mass of the neutrino-antineutrino pair is negligible compared
to that of the electron-positron pair. Thus, the $Z^0$'s mass
contributions are from the electron-positron pair and are
one-half of twice the electron's mass or simply one electron mass.
The $W^+$ is a positron-neutrino pair and the $W^-$ is an
electron-antineutrino pair. Because the neutrino masses are
negligible, each $W$ mass is based on one electron mass.

Because the masses of the $Z^0$ and $W^\pm$ are based on one electron
mass, their quantum mass corrections will be that of the electron.
Therefore,
\q
M_W=\exp\!\left(\frac14-\frac{2\pi}{3\alpha_{\gamma Z}}\right)
  \Lambda(W_\mu^\pm),\label{eq:20.3} 
\qq
where we have denoted the above-mentioned classical mass of the
$W$ by $\Lambda(W_\mu^\pm)$. In addition,
$\alpha_{\gamma Z}=1/19.71$, which is the value that yields the
mass of the electron in Eq.~(\ref{eq:17.1c}).
Dividing Eq.~(\ref{eq:20.3}) by Eq.~(\ref{eq:17.1c})
when  $\alpha_{\gamma Z}$ equals $1/19.71$, we obtain
\q
\frac{M_W}{m_e}=\frac{\Lambda(W_\mu^\pm)}{\Lambda}.\label{eq:20.4}
\qq
Using Eq.~(\ref{eq:20.4}) to solve for the mass of the $W^\pm$, we find
$M_W=76.18$ GeV. Similarly,
\q
\frac{M_Z}{m_e}=\frac{\Lambda(Z_\mu^0)}{\Lambda},\label{eq:20.5}
\qq
where $\Lambda(Z_\mu^0)$ is the above-mentioned classical Planck mass of the
$Z$. This leads to a $Z^0$ mass of 89.34 GeV. 

These masses are in fairly good 
agreement with experiment [58], where $M_W=80.40\pm0.2$ GeV and
$M_Z=91.1867\pm0.0021$ GeV. The errors are presumably due to
lack of consideration of the mass contributions due to their charges.
The SM does not explain the masses of the W and Z
because it derives them from the weak interaction
coupling constant $G_W$, which comes from experiment.
KK($\theta,\phi$) explains the masses of
these two particles from scratch, allowing, instead, one to derive $G_W$
from them.  These are parameters nine and ten of the SM.
I will not derive any more parameters from the SM.
Most of these fall within the realm of the strong interaction sector
of quantum KK($\theta,\phi$) (quark masses, for example).  
The calculation of masses and coupling constants for
this sector may be inferred from that for the electroweak sector.

QED is considered correct because it predicts the right numbers for
certain properties of the electron.  This is despite the appearance of 
divergences.  Likewise, KK($\theta,\phi$) should be considered correct 
because it predicts the right numbers for the masses and coupling constants
for the elementary particles.
And KK($\theta,\phi$) eliminates the appearance of
divergences in QED.


\begin{center}
{\bf 19. Superweak interactions}
\end{center}

The vector $A_\mu$ in KK($\theta,\phi$)
is expanded in terms of spherical harmonics $Y_{lm}$
\q
A_\mu=\sum_{l=0}^\infty \sum_{m=-l}^l A_\mu^{lm}Y_{lm}.\label{eq:21.1}
\qq
We neglect the strong interactions for now.
The $l=0$ coefficient $A_\mu^{00}$ in this expansion is the photon.
The $l=1$ coefficients $A_\mu^{1m}$ are the $W$ and $Z$.
The $l=2$ and $l=3$ coefficients $A_\mu^{2m}$ and $A_\mu^{3m}$
are vectors for two new superweak interactions.

The procedures used to obtain the masses of the
$A_\mu^{2m}$ from the spinors of which they are composed are
the same as those described for the $A_\mu^{1m}$ in Sec.~18.
Remember that in order to obtain nonzero Clebsch-Gordan coefficients
for leptons, $l_1=l_2$, which means that one cannot expand
$l=2$ kets in terms of $l_1=\frac32$ and $l_2=\frac12$ kets
as one might expect, but rather kets with $l_1=l_2=\frac32$.
The wave function of the vector $A_\mu^{2m}$ is composed of those of
the muon and the muon neutrino.  The mass contributions of these particles
add. The neutrino contribution is negligible.  An exact prediction
for the mass of $A_\mu^{2m}$ can be made if we multiply the classical
mass ratio $\Lambda(A_\mu^{2m})/\Lambda$ by the mass of the muon
[similar to Eq.~(\ref{eq:20.4})].

From $\Lambda(A_\mu^{20})=3.403\,\hbar/cr_0$ (where $\hbar/cr_0$
is the Planck mass), we obtain for $A_\mu^{20}$ a mass of 33.08 TeV.
Similarly, $\Lambda(A_\mu^{2,\pm1})=3.216\,\hbar/cr_0$, which leads to a
mass of 31.24 TeV for $A_\mu^{2,\pm1}$. The classical mass
$\Lambda(A_\mu^{2,\pm2})=0.4699\,\hbar/cr_0$ yields the actual mass
4.56 TeV for the $l=2$, $m=\pm2$ particle.  

The vectors $A_\mu^{3m}$ are composed of a muon and a muon neutrino.
Their masses are obtained by multiplying
their classical mass ratios by the mass of the muon. We have
$\Lambda(A_\mu^{30})=3.996\,\hbar/cr_0$,
$\Lambda(A_\mu^{3,\pm1})=3.032\,\hbar/cr_0$,
$\Lambda(A_\mu^{3,\pm2})=2.818\,\hbar/cr_0$ and
$\Lambda(A_\mu^{3,\pm3})=0.7538\,\hbar/cr_0$.
The masses of these particles are 38.62 TeV, 29.31 TeV, 27.24 TeV and
7.29 TeV, respectively.

The charges of the $A_\mu^{2m}$ are not, in general, given by
their quantum number $m$.  This is because $m$ is the vector's
bare charge quantum number, which is not necessarily its
actual charge quantum number.  The vector's actual charge
quantum number is given by
the sum of the actual charges of the fermions in its spinor expansion.
This is because the vector may be thought to come about from the
combination of the spinors from which it is composed.
Another way of looking at this is that a vector and its spinor expansion
must be mathematically and physically identical.  For example, consider
the $Z^0$.  From its spinor expansion, we find the $Z^0$ is composed of
one-half an electron-positron pair and one-half a neutrino-antineutrino
pair.  Thus, the charge of the $Z^0$ is given by 
$\frac12(-1+1)+\frac12(0+0)$, which equals zero.
Had one of the numbers in the parentheses been different the $Z^0$
would have wound up with a different charge ---
and a fractional one at that. One of the numbers will
be different for the case of the spinor with bare charge quantum number $-2$
and actual charge quantum number 0.  This spinor is in the
muon's multiplet and is described in Sec.~20 (it has $Q=-2$). Therefore,
instead of having a $-2$ here we will have a 0.  This is why the charge of
$A_\mu^{lm}$ is not, in general, given by $m$ and why it may have a
fractional charge.

The charges of the next two generations of intermediate vector bosons ---
the $A_\mu^{lm}$ where $l=2$ or 3 are as follows.  
The charge of the superweak vector $A_\mu^{2,\pm2}$
is $\pm1$.  The vector $A_\mu^{20}$
is neutral, while $A_\mu^{2,\pm1}$ has a charge of $\pm\frac34$.
The charge of the superweak vector $A_\mu^{3,\pm3}$
is $\pm1$.  The vector $A_\mu^{30}$
is neutral, while $A_\mu^{3,\pm1}$ has a charge of $\pm\frac35$.
The charge of $A_\mu^{3,\pm2}$ is $\pm1$.

KK($\theta,\phi$) predicts no Higgs particle and no supersymmetric
particles. Instead, it predicts new superweak vectors.
The least massive of these are $A_\mu^{2\pm2}$ with a mass of
4.56 TeV and $A_\mu^{3\pm3}$ at 7.29 TeV. These should be
detected by the Large Hadron Collider.


\begin{center}
{\bf 20. Predicted leptons}
\end{center}


The particle with $l=\tfrac32$ and $m=-\tfrac32$ in Table~1 has
$Q=-2$ according to the Gell-Mann Relation (\ref{eq:15.7}) and
therefore, $\alpha_0=Q^2=4$. This would put its electromagnetic mass
at about the muon's weak mass as the muon's weak $\alpha_0=3.75$.
Because it has $l=\frac32$, which is the same as that of the muon, its
weak mass would be that of the muon. Its total mass would therefore, be
about double that of the muon (260 MeV to be exact).
However, a lepton with this mass is not observed.
This must be due to the fact that the 
electromagnetic $\alpha_0$ is too large for the higher-loop
series [54] in terms of
$\alpha_0$ to converge. This electromagnetic series for the
multiplicative quantum charge correction due to vacuum polarization
becomes infinite,
resulting in an infinite denominator for $\alpha_\gamma$. The parameter
$\alpha_\gamma$ is therefore zero. This means the charge is 
completely screened by the photon. In effect, the particle has no charge
and does not couple to the photon. Therefore, it is a neutrino.
Because it is in the muon's multiplet, it is a muon neutrino. It would have
the same mass as, but be different than, the ordinary muon neutrino,
which has $m=\frac12$.

The new leptons in the tau's multiplet have $l=\frac52$ and
$m=\pm\frac52,\pm\frac32$, according to Table~1.
These $m$ quantum numbers distinguish the new particles from
the tau and its neutrino, which have $m=\pm\frac12$.
The bare charge quantum numbers of these new particles are
$Q=-3,-2,1,2$. By the same reasoning used below for the muon's
multiplet, the particle with $Q=1$ has properties similar to
the antitau, while the others are new tau neutrinos.

Fermions in the next spinor multiplet beyond the tau's have $l=\frac72$
and $-\frac72\le m \le\frac72$. Because none of these particles
are observed, the weak $\alpha_0$ must be too large
for the higher-loop series
in terms of this $\alpha_0$ to converge.
Because this series in terms of the weak $\alpha_0$ does not
converge, the weak charges of these $l=\frac72$ particles are zero,
by the same argument used above for the vanishing of the electric charge.
Therefore, they do not couple to the $Z^0$.  This would be the reason
why there appears to be only three families of leptons.

The particles with $Q=0$ or $|Q|\ge2$ also have no electric charge
and therefore, have no electromagnetic or weak masses because
$\alpha_{\gamma Z}=\alpha_{Z}=0$. These particles would acquire a
small mass through their superweak couplings to $A_\mu^{lm}$,
$l\ge 2$. These superweak masses should be smaller than
ordinary neutrino masses, which are small due to the weakness of
the weak interaction.

The particle with $Q=1$ has electric charge but no weak charge and
is screened by the photon only. The equations that determine the mass
of this particle are
\begin{eqnarray}
  \alpha_\gamma&=&\left(1+\frac{\alpha_0}{3\pi}
  \ln\frac{\Lambda^2}{m^2}\right)^{-1}\alpha_0,\label{eq:19.1a}\\
         m&=&\exp\!\left(\frac14-\frac{2\pi}{3\alpha_\gamma}\right)
           \Lambda.\label{eq:19.1b}
\end{eqnarray}
The solution of Eqs.~(\ref{eq:19.1a}) and~(\ref{eq:19.1b}) is
\q
m=\exp\!\left(\frac{9}{20}-\frac{6\pi}{5\alpha_0}
    \right)\Lambda.\label{eq:19.2}
\qq
Substituting $\alpha_0=1$ into Eq.~(\ref{eq:19.2}),
we find $m=4.81\times10^{9}$ TeV.

Because their electric and weak charges are the same, spinors with
$l>\frac72$ will have the same masses as those with $l=\frac72$.
That is, particles with $Q=0$ or $|Q|\ge2$ will be nearly massless
and those with $|Q|=1$ will have the mass $4.81\times10^{9}$ TeV.

It may be possible to test KK($\theta,\phi$) before 2007,
when the Large Hadron Collider begins taking data. KK($\theta,\phi$)
makes the testable prediction of the existence of a new lepton,
which I call $\mu_2^+$ and is degenerate with the muon. Table~1
gives its quantum numbers as $l=\frac32$ and $m=\frac32$.
According to the Gell-Mann relation (\ref{eq:15.7}), its bare
charge quantum number is $Q=1$.  $\mu_2^+$
has the same electromagnetic and weak
bare fine structure constants and therefore, mass, as the usual
muon $\mu_1^-$.
Because its charge is reversed it is similar to
the antimuon ${\overline\mu}_1^+$.
However, it is not the same as the antimuon because
its quantum number $m=\frac32$ is different than
the antimuon's $m=\frac12$.  Because this particle has the same
mass as the usual muon $\mu_1^-$,
there is no reason why it should not have been
detected already. I believe we have been detecting it all along
but have been mistaking it for the antimuon.

If one writes the $l=1$ ket $|11\rangle$ in terms of pairs of 
$l_1=l_2=\tfrac32$ kets using Clebsch-Gordan coefficients
(as we did above for the $l=1$ ket $|11\rangle$
in terms of one pair of $l_1=l_2=\tfrac12$ kets) one obtains
\begin{equation}
|11\rangle=
\sqrt{\tfrac{3}{10}}
\left|\tfrac32\,,-\tfrac12\right>\left|\tfrac32\,\tfrac32\right>
+\sqrt{\tfrac{3}{10}}
\left|\tfrac32\,\tfrac32\right>\left|\tfrac32\,,-\tfrac12\right>
-\sqrt{\tfrac{2}{5}}
\left|\tfrac32\,\tfrac12\right>\left|\tfrac32\,\tfrac12\right>.
\label{eq:21.3}
\end{equation}
Next, the symbol for each particle is written below the ket
corresponding to it in Table~1. One then obtains three separate decays
of the $W^+$.  This is because when this expansion is substituted
for one $W^+$ in its mass term in the Lagrangian, three separate
interaction terms result. The first pair of kets leads to the
decay of the $W^+$ in terms of the new type of muon,
$W^+\to\mu^+_2+{\overline\nu}_{\mu1}^{\,0}$. 
The second pair of kets
does not lead to anything interesting. The third and last pair
of kets leads to the familiar decay of the $W^+$ into the
antimuon and the muon neutrino,
 $W^+\to{\overline \mu}^{\,+}_1+\nu_{\mu1}^0$.

The prediction of KK($\theta,\phi$)
is that the $W^+$ decays not only into the antimuon and the usual
muon neutrino, as one would expect, but also into the new particle,
which has been mistaken for the antimuon,
and the muon antineutrino. The new decay products contain
the antiparticle of the neutrino one would expect.
The case of the $W^-$ produces a similar situation. In addition
to the familiar decay of this particle into the muon and the muon
antineutrino, one would also have the $W^-$ decaying into the
antiparticle  of the new lepton, which has been mistaken for
the muon and the regular muon neutrino. Again, the new lepton
is produced with the antiparticle of the neutrino one would
expect, resulting in another prediction.

In contrast to the $W^\pm$, decays of the $Z^0$ do not produce
anything interesting because the new and old decays are indistinguishable.
Here, the new particle is not produced with the antiparticle
of the neutrino one would expect, but rather with its own antiparticle.
Thus, the new decay products are the new particle \ $\mu^+$ and its
antiparticle ${\overline \mu}^{\,-}$, which are indistinguishable
from the old decay products of the muon \ $\mu^-$ and its antiparticle
${\overline\mu}^{\,+}$. 

I claim the existence of a new lepton $\mu_2^+$ does
not spoil the agreement between the calculation of the decay width of
the $Z^0$ in the Standard Model and that derived from experiment,
which suggests that only the known leptons exist.
The effects of the new particle are already included in the
established decay width.
This is implied by the expansion of $Z_\mu^{0}\left|10\right>$
in terms of the electron and its neutrino and its expansion 
in terms of the muon and
its neutrino. We have
\begin{equation}
|10\rangle=
\sqrt{\tfrac{1}{2}}
\left|\tfrac12\,\tfrac12\right>\left|\tfrac12\,,-\tfrac12\right>
-\sqrt{\tfrac{1}{2}}
\left|\tfrac12\,,-\tfrac12\right>\left|\tfrac12\,\tfrac12\right>.
\label{eq:21.4}
\end{equation}
Identifying each ket with its particle from Table~1, we obtain 
\q
Z_0^\mu=\sqrt{\tfrac{1}{2}}{\overline e}^+ \gamma^\mu e^-
      +\sqrt{\tfrac{1}{2}}{\overline\nu}^0_e \gamma^\mu \nu^0_e.
\label{eq:21.5}
\qq
In addition,
\begin{eqnarray}
|10\rangle= 
&\sqrt{\tfrac{9}{20}}&
\left|\tfrac32\,\tfrac32\right>\left|\tfrac32\,,-\tfrac32\right>
+\sqrt{\tfrac{9}{20}}
\left|\tfrac32\,,-\tfrac32\right>\left|\tfrac32\,\tfrac32\right>\\
-&\sqrt{\tfrac{1}{20}}&
\left|\tfrac32\,\tfrac12\right>\left|\tfrac32\,,-\tfrac12\right>
-\sqrt{\tfrac{1}{20}}
\left|\tfrac32\,,-\tfrac12\right>\left|\tfrac32\,\tfrac12\right>.
\label{eq:21.6}
\end{eqnarray}
Identifying the kets with their particles from Table~1, we have
\q
Z_0^\mu=\sqrt{\tfrac{9}{20}}{\overline\nu}_{\mu2}^0 
                                                \gamma^\mu \nu_{\mu2}^0
     +\sqrt{\tfrac{9}{20}}{\overline\mu}_2^-  \gamma^\mu \mu_2^+
     -\sqrt{\tfrac{1}{20}}{\overline\mu}_1^+  \gamma^\mu \mu_1^-
     -\sqrt{\tfrac{1}{20}}{\overline\nu}_{\mu1}^0  \gamma^\mu
                                       \nu_{\mu1}^0.
\label{eq:21.7}
\qq
Note that the sum of the squares of the Clebsch-Gordan coefficients
add up to one for each of the expansions.  Further, note that the
sum of the squares of the coefficients for the usual muon $\tfrac1{20}$,
and the new type of muon $\tfrac9{20}$ make the same contribution
of $\tfrac12$ to the expansion of the $Z^\mu_0$ as the electron.
This is significant
because the effects of a possible new particle are absent in the electron's
multiplet because there are no new particles there.

When the expansion for the Z in terms of the new and old types of muon
is substituted for one factor of Z in its mass term in the Lagrangian,
the coefficient $g$, which is roughly the weak interaction coupling constant,
of the new terms is split into pieces for the new
and old types of muon. Now the partial decay width of the Z 
for what was called the muon in Ref.~58 is proportional to $g^2$.
The partial decay width
of the Z is the same when calculated in the new way, 
with the existence of a new type of muon and the old way,
according to Ref.~58,
because $g^2$ is the same in both cases. In the former, $g^2$ is
proportional to the sum of the squares of the contributions
from $\mu_2^+$ and $\mu_1^-$, namely
$\left(\tfrac{9}{20}+\tfrac1{20}\right)$
and in the latter, it is proportional to the $\tfrac12$
of what was called the muon in Ref.~58.
Remember, the new type of muon is so much like the antimuon,
we have been detecting it and its contribution to $g^2$ all along.

This procedure results in different values for the coefficient $g$
for the lepton-Z interaction terms for $\mu_2^+$ and $\mu_1^-$.
This would seem to contradict observation, which indicates that
each lepton has the same weak interaction coupling constant $g$.
Apparently, after the decay width of the Z is calculated,
the Dirac term comes into play and all leptons wind up with
the same value for $g$. See Sec.~8 of Part~I.
 
The existence of a new lepton does not spoil the agreement
between the Standard Model and experiment for the ratio of cross sections
for $e^+e^-$ annihilation into hadrons and muons for the same reason
it does not do this for the agreement between the Standard Model
and experiment for the $Z^0$ decay width. Both the cross sections
and the decay width depend [58] on $g^2$.
These arguments would apply to the new neutrino in the muon's
multiplet and the new particles in the tau's multiplet as well.

The new type of muon $\mu_2^+$ should be detectable in pion decay.
The pion may produce a virtual W as follows --- $\pi^\pm\to W^\pm$,
with the decay continuing as above.  It appears to me, the only way
to distinguish the new type of muon from the antimuon is the type of
neutrino with which it is produced.  This is because fermions are produced
in particle-antiparticle pairs.  Therefore, $\mu_2^+$, being a particle,
is produced with an antiparticle, ${\overline\nu}_{\mu1}^0$
the usual muon antineutrino. ${\overline\mu}_1^+$, being
an antiparticle, is produced with the particle $\nu_{\mu1}^0$,
the usual muon neutrino. Thus, the new particle is produced with
an antineutrino instead of a neutrino.

In elementary particle physics experiments, 
neutrinos are not normally detected.
For example, in large particle accelerators,
neutrinos pass through the detectors
completely unnoticed. One has to design a neutrino detection experiment.
However, it seems that the only neutrino detection experiments that
have been performed to date are not sufficient to prove the existence of
the new particle. Nor can they rule it out. For example,
in the first high-energy neutrino experiment [59], which distinguished
between the electron neutrino and the muon neutrino, both the decays
\q
\pi^+\to{\mu}_2^+ +{\overline\nu}_{\mu1}^0\label{500}
\qq
and
\q
\pi^-\to\mu_1^-+{\overline\nu}_{\mu1}^0\label{501}
\qq
were produced, which would mask the intended effect. Detection of muon
antineutrinos could mean either the new particle or the usual muon was produced.
One would need an experiment where, for example, only positive pions decay
and not negative pions. Then, when a muon antineutrino is detected one would
know that it came from the $\pi^+$ and not a $\pi^-$.  This would imply the decay
in Eq.~(\ref{500}) and rule out that in Eq.~(\ref{501}).
The existence of a new lepton would be established.
KK($\theta,\phi$) would be confirmed.

It is not difficult to obtain a source of pure
positive pions.  In any material, negative pions that are slowed down and
come to rest are attracted to the nuclei and captured by them.
This was the case in the first cosmic ray experiment
[60] that detected the pion. In this experiment the emitted neutrinos were not detected,
disallowing the possibility of detecting a muon antineutrino coming from a positive pion.

KK($\theta,\phi$) predicts no Higgs particle and no supersymmetric
particles. Instead, it makes the telltale testable
prediction that a new lepton degenerate with the muon exists.


\begin{center}
{\bf ACKNOWLEDGMENTS}
\end{center}
I would like to thank N.P. Chang of City College of City University
of New York for helpful suggestions.\newline
Thanks be to God.


\begin{center}
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\end{center}
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\end{flushleft}



\end{document}
