text:
\section*{Introduction}

After 30 years of effort the so-called ``Standard Model"
of the elementary particles has not come up with
a precise theoretical determination of the masses
of either the mesons and baryons or of the leptons,
which means that neither the mass of the fundamental
electron nor the mass of the fundamental proton
have been explained. The quarks, which have been
introduced by Gell-Mann [1] forty years ago, are said to
explain the mesons and baryons. But the standard model
does not explain neither the mass, nor the charge, nor the
spin of the baryons and leptons, the three
fundamental time-independent properties of the particles.
The fractional electric charges imposed on the quarks do not
explain the charge e$^\pm$, neither does the spin imposed
on the quarks explain the spin. The measured values of the
properties of the particles are given in the Review of Particle
Physics [2]. There are many other attempts to explain the
elementary particles or only one of the particles, too many
to list them here. For example Skyrme [3] has proposed a
unified theory of the mesons and baryons, and El Naschie has,
 in the last years, proposed a topological theory for high
energy particles and the spectrum of the quarks [4-7].

   The need for the present investigation has been expressed by
Feynman [8] as follows: ``There remains one especially unsatisfactory
feature: the observed masses of the particles, m. There is no theory
that adequately explains these numbers. We use the numbers in all
our theories, but we do not understand them - what they are, or where
they come from. I believe that from a fundamental point of view, this
is a very interesting and serious problem". Today, twenty years
later, we still rely on the standard model for an explanation of the
masses of the elementary particles but have not succeeded. It is
time to try something different.

\section {The spectrum of the masses of the particles}

As we have done before [9] we will focus attention on the so-called
``stable" mesons and baryons
whose masses are reproduced with other data in Tables 1 and 2.
It is obvious that any attempt to explain the masses of the mesons and
baryons should begin with the particles that are affected by the fewest
parameters. These are certainly the particles without isospin (I = 0) and
without spin (J = 0), but also with strangeness S = 0, and charm C = 0.
Looking at the particles with I,J,S,C = 0 it is startling to find that
their masses are quite close to integer multiples of the mass of the
$\pi^0$\,meson. It is m($\eta) = (1.0140 \pm 0.0003)\,\cdot$\,4m($\pi^0$),
and the mass of the resonance $\eta^\prime$ is m($\eta^\prime$) = (1.0137
$\pm$ 0.00015)\,$\cdot$\,7m($\pi^0$). Three particles seem hardly to be
sufficient to establish a rule. However, if we look a little further we
find that m($\Lambda$) = 1.0332\,$\cdot$\,8m($\pi^0$) or m($\Lambda$) =
1.0190\,$\cdot$\,2m($\eta$). We note that the $\Lambda$ baryon has spin
1/2, not spin 0 as the $\pi^0$,\,$\eta$ mesons. Nevertheless, the mass of
$\Lambda$ is close to 8m($\pi^0$). Furthermore we have m($\Sigma^0$) =
0.9817\,$\cdot\,$9m($\pi^0$), m($\Xi^0) = 0.9742\,\cdot\,$10m($\pi^0$),
m$(\Omega^-)$ =
1.0325\,$\cdot\,$12m($\pi^0)$ = 1.0183\,$\cdot$\,3m($\eta$), ($\Omega^-$
is charged and has spin 3/2). Finally the masses of the charmed baryons
are m($\Lambda_c^+$) = 0.99645\,$\cdot$\,17m($\pi^0$) =
1.024\,$\cdot$\,2m($\Lambda$), m($\Sigma_c^0$) =
1.00995\,$\cdot$\,18m($\pi^0$),
m($\Xi_c^0$) = 1.0170\,$\cdot$\,18m($\pi^0$), and m($\Omega_c^0$) =
0.99925\,$\cdot$\,20m($\pi^0$).

	\begin{table}\caption{The ratios m/m($\pi^0$) of the $\pi^0$ and
$\eta$ mesons
\newline
\hspace*{1.5cm}
 and of the baryons of the $\gamma$-branch.}
	\begin{tabular}{lllllcl}\\
\hline\hline\\
 & m/m($\pi^0$) & multiples & decays & fraction & spin &
mode
 \\
 & & & & (\%) & &\\
[0.5ex]\hline
\\
$\pi^0$ & 1.0000 & 1.0000\,\,$\cdot$\,\,$\pi^0$ & $\gamma\gamma$ & 98.798
& 0 & (1.)\\
 & & & $e^+e^-\gamma$ & \,\,\,1.198 & &\\
\\
$\eta$ & 4.0563 & 1.0141\,$\cdot$\,\,4$\pi^0$ & $\gamma\gamma$ & 39.43 & 0
&
(2.)\\
 & & & 3$\pi^0$ & 32.51 & &\\
 & & & $\pi^+\pi^-\pi^0$ & 22.6 & &\\
 & & & $\pi^+\pi^-\gamma$ & \,\,\,4.68 & &\\
\\
$\Lambda$ & 8.26575 & 1.0332\,$\cdot$\,\,8$\pi^0$ & p$\pi^-$ & 63.9 &
$\frac{1}{2}$
& 2$\ast$(2.)\\
 & & 1.0190\,$\cdot$\,\,2$\eta$ & n$\pi^0$ & 35.8 & &\\
\\
$\Sigma^0$ & 8.8359 & 0.9817\,$\cdot$\,\,9$\pi^0$ & $\Lambda \gamma$ & 100
&
$\frac{1}{2}$ & 2$\ast (2.) + (1.)$\\
\\
$\Xi^0$ & 9.7412 & 0.9741\,$\cdot$\,10$\pi^0$ & $\Lambda\pi^0$ & 99.52 &
$\frac{1}{2}$ &
2$\ast(2.) + 2(1.)$\\
\\
$\Omega^-$ & 12.390 & 1.0326\,$\cdot$\,12$\pi^0$ & $\Lambda$K$^-$ & 67.8
&
$\frac{3}{2}$ & 3$\ast(2.)$\\
 & & 1.0183\,$\cdot$\,\,3$\eta$ & $\Xi^0\pi^-$ & 23.6 & &\\
 & & &  $\Xi^-\pi^0$ & \,\,\,8.6 & &\\
\\
$\Lambda_c^+$ & 16.939 & 0.99645\,$\cdot$\,17$\pi^0$ & many & &
$\frac{1}{2}$ &
2$\ast(2.) + (3.)$\\
 & & 0.9630\,$\cdot$\,17$\pi^\pm$\\
\\
$\Sigma_c^0$ & 18.179 & 1.0099\,$\cdot$\,18$\pi^0$ & $\Lambda_c^+\pi^-$ &
$\approx$100 & $\frac{1}{2}$ & $\Lambda_c^+ + \pi^-$\\
\\
$\Xi_c^0$ & 18.307 & 1.0170\,$\cdot$\,18$\pi^0$ & eleven & (seen) &
$\frac{1}{2}$   &
2$\ast(3.)$\\
\\
$\Omega_c^0$ & 19.985 & 0.99925\,$\cdot$\,20$\pi^0$ & seven & (seen) &
$\frac{1}{2}$ &
2$\ast(3.) + 2(1.)$\\
& & 0.9854\,$\cdot$\,\,5$\eta$ & & &\\
[0.2cm]\hline\hline
\vspace{0.1cm}
\end{tabular}
\footnotemark{\footnotesize The modes apply to neutral particles only. The
$\ast$
 marks coupled modes.}

	\end{table}

   Now we have enough material to formulate the
\emph{integer multiple rule} of the particle masses,
according to which the masses of the
$\eta$,\,$\Lambda$,\,$\Sigma^0$,\,$\Xi^0$,\,
$\Omega^-$,\,$\Lambda_c^+$,\,$\Sigma_c^0$,\,
$\Xi_c^0$, and $\Omega_c^0$
particles are, in a first approximation, integer multiples of the mass
of the $\pi^0$\,meson, although some of the particles have spin,
and may also have charge as well as strangeness and charm. A
consequence of the integer multiple rule must be that the ratio of the
mass of any meson or baryon listed above divided by the mass of another
meson or baryon listed above is equal to the ratio of two integer numbers.
And indeed, for example m($\eta$)/m($\pi^0$) is practically two times
(exactly 0.9950$\,\cdot$\,2) the ratio m($\Lambda$)/m($\eta$). There is
also the ratio m($\Omega^-$)/m($\Lambda$) = 0.9993\,$\cdot$\,3/2.
We have furthermore e.g. the ratios m($\Lambda$)/m($\eta$) =
1.019\,$\cdot$\,2, m($\Omega^-$)/m($\eta$) = 1.018\,$\cdot$\,3,
m($\Lambda_c^+$)/m($\Lambda$)
= 1.0247\,$\cdot$\,2, m($\Sigma_c^0$)/m($\Sigma^0$) =
1.0287\,$\cdot$\,2,  m($\Omega_c^0$)/m($\Xi^0$) = 1.0258\,$\cdot$\,2,
and m($\Omega_c^0$)/m($\eta$) = 0.9854\,$\cdot$\,5.

   We will call, for reasons to be explained soon, the particles
discussed above, which follow in a first approximation the integer
multiple rule, the \emph{$\gamma$-branch} of the particle spectrum.
The mass
ratios of these particles are in Table 1. The deviation of the mass ratios
from exact integer multiples of m($\pi^0$) is at most 3.3\%, the average
of the factors before the integer multiples of m($\pi^0$) of the nine
$\gamma$-branch particles in Table 1 is 1.0066 $\pm$ 0.0184. From a least
square analysis follows that the masses of the ten particles on Table 1
lie on a straight line given by the formula

\begin{equation} \mathrm{m}(N)/\mathrm{m}(\pi^0) = 1.0065\,N - 0.0043
\qquad  N\,>\,1,
\end{equation}
where N is the integer number nearest to the actual ratio of the particle
mass divided by m($\pi^0$). The correlation coefficient in Eq.(1)
has the nearly perfect value R$^2$ = 0.999.

   The integer multiple rule applies to more than just the stable mesons
and baryons. The integer multiple rule applies also to the $\gamma$-branch
baryon resonances which have spin J = 1/2 and the meson resonances with
I,J  $\leq$ 1, listed in [2] or in Table 3 of Appendix A. The
$\Omega^-$\,baryon will not be considered because it has
spin 3/2 but would not change the following equation significantly.
 If we consider all mesons and baryons
of the $\gamma$-branch in Table 3, ``stable" or unstable,
then we obtain from a least square analysis the formula

\begin{equation}    \mathrm{m}(N)/\mathrm{m}(\pi^0) = 0.999\,N +
0.0867\qquad    N\,>\, 1,
\end{equation}
with the correlation coefficient  0.9999. The line through the points is
shown in Fig.\,1 which tells that 41 particles of the $\gamma$-branch
of different spin and isospin, strangeness and charm; five I,J = 0,0
$\eta$\,\,mesons, fifteen J = 1/2 baryons, ten I = 0, J = 0,1
c$\bar{\mathrm{c}}$ mesons, ten I = 0, J = 0,1 b$\bar{\mathrm{b}}$
mesons and the
 $\pi^0$\,meson with I,J = 1,0, lie on a straight line with slope
0.999. In other words they approximate the integer multiple rule
very well. Spin 1/2 and spin 1 does not seem to affect the integer
 multiple rule, i.e. the ratios of the particle masses, neither
does strangeness S $\neq$ 0 and charm C $\neq$ 0.

\begin{figure}

	\includegraphics{Fig1.eps}

	\begin{quote}

Fig.\,1: The mass of the mesons and baryons of the $\gamma$-branch,
  stable or unstable, with I\,$\leq$ 1,\,J\,$\leq$ 1 in units of
m($\pi^0$)
 as a function of the integer  N, demonstrating the integer multiple rule.
	\end{quote}
\end{figure}

   Searching for what else the
$\pi^0$,\,$\eta$,\,$\Lambda$,\,$\Sigma^0$,\,$\Xi^0$,\,$\Omega^-$
particles have in common, we find that the principal decays (decays
with a fraction\,$>$\,1\%) of these particles,
 as listed in Table 1, involve primarily $\gamma$-rays, the
 characteristic case is $\pi^0 \rightarrow \gamma\gamma$  (98.8\%). We
will later on discuss a possible explanation for the 1.198\% of the decays
of $\pi^0$ which do not follow the $\gamma\gamma$ route but decay via
$\pi^0 \rightarrow$ e$^+$ + e$^-$ + $\gamma$. After the
$\gamma$-rays the next most
frequent decay product of the heavier particles of the $\gamma$-branch are
$\pi^0$\,mesons which again decay into $\gamma\gamma$. To describe the
decays in another way, the principal decays of the particles listed above
take place $\emph{always without the emission of neutrinos}$\,; see Table
1. There the decays and the fractions of the principal decay modes are
given,
taken from the Review of Particle Physics. We cannot consider decays
with fractions $<$\,1\%. We will refer to the particles whose masses are
approximately integer multiples of the mass of the $\pi^0$\,meson, and
which decay without the emission of neutrinos, as the
$\gamma$-$\emph{branch}$ of the particle spectrum.

   To summarize the facts concerning the $\gamma$-branch. Within
0.66\% on the average the masses of the stable particles of the
$\gamma$-branch  in Table\,1 are integer multiples
(namely 4,\,8,\,9,\,10,\,12, and even 17,\,18,\,20) of the mass of the
$\pi^0$\,meson.
It is improbable that nine particles have masses so close to integer
multiples of m($\pi^0$) if there is no correlation between them and the
$\pi^0$\,meson. It has, on the other hand, been argued that the integer
multiple rule is a numerical coincidence. But the probability that the
mass ratios of nine particles of the $\gamma$-branch fall by coincidence
on integer numbers between 1 and 20 instead on all possible numbers
between 1 and 20 with two decimals
after the period is smaller than 10$^{-20}$, i.e.\,nonexistent. The
integer multiple rule is not affected by more than 3\% by the spin, the
isospin, the strangeness, and by charm. The integer multiple rule seems
even to apply to the $\Omega^-$ and $\Lambda_c^+$ particles, although
they are charged. In order for the integer multiple rule to be valid the
deviation of the ratio m/m($\pi^0$) from an integer number must be smaller
than 1/2N, where N is the integer number closest to the actual ratio
m/m($\pi^0$). That means that the permissible deviation decreases rapidly
with increased N. All particles of the $\gamma$-branch have deviations
smaller than 1/2N.

   The remainder of the stable mesons and baryons are the
$\pi^\pm$, K$^{\pm,0}$,\,p,\,\,n,\quad D$^{\pm,0}$, and D$_s^\pm$
particles which make up the neutrino-branch ({$\nu$-branch}) of the
particle spectrum. The ratios of their masses are given in Table 2.

	\begin{table}\caption{The $\nu$-branch of the particle
spectrum}
	\begin{tabular}{llllrcl}\\

\hline\hline\\
 & m/m($\pi^\pm$) & multiples & decays
 & fraction & spin & mode
 \\
 & & & & (\%) & &\\
[0.5ex]\hline
\\
$\pi^\pm$ & 1.0000 & 1.0000\,\,$\cdot$\,\,$\pi^\pm$ & $\mu^+\nu_\mu$ &
\,\,99.9877 & 0 & (1.)\\
\\
K$^{\pm,0}$ & 3.53712 & 0.8843\,$\cdot$\,\,4$\pi^\pm$ & $\mu^+\nu_\mu$ &
\,\,63.43 & 0 &
(2.) + $\pi^0$\\
 & & & $\pi^\pm\pi^0$ & \,\,21.13 && \,\,\,\,(K$^\pm$)\\
 & & & $\pi^+\pi^-\pi^+$ & \,\,\,\,\,5.58 && (2.) + $\pi^\mp$\\
 & & & $\pi^0 e^+ \nu_e$ (K$_{e3}^+$)& \,\,\,\,\,4.87 &
&\,\,\,\,(K$^0$,$\overline{{\mathrm{K}}^0}$)\\
 & & & $\pi^0\mu^+\nu_\mu$ (K$_{\mu3}^+$) & \,\,\,\,\,3.27 & &\\
\\
n & 6.73185 & 0.8415\,$\cdot$\,\,8$\pi^\pm$ & $p\,e^-\overline{\nu}_e$ &
100.\,\,\,\,\,\,\,\,  & $\frac{1}{2}$ & 2$\ast$(2.)\\
 & & 0.9516\,$\cdot$\,\,2K$^\pm$  & & & & \,\, + $2\pi^\pm$\\
 & &  0.9439\,$\cdot$\,\,(K$^0$ + $\overline{\mathrm{K}}^0$) & & & &\\
\\
D$^{\pm,0}$ & 13.393 & 0.8370\,$\cdot$\,16$\pi^\pm$ & $e^+$ anything &
\,\,17.2 &
0 & 2(2$\ast$(2.)\\
 & & 0.9466\,$\cdot$\,\,4K$^\pm$ & K$^-$ anything & \,\,24.2 & & \,\, +
$2\pi^\pm$)\\
 & & 0.9954\,$\cdot$\,(p + $\bar{\mathrm{n}}$) & $\overline{\mathrm{K}}^0$
anything\\
 & &                       & \,\,+\,K$^0$ anything & \,\,59\\
 & &                       & $\eta$ anything & $<$\,13\\
&&                          &K$^+$ anything &5.8\\
\\
D$^\pm_s$ & 14.102 & 0.8295\,$\cdot$\,17$\pi^\pm$ & K$^-$ anything &
\,\,13 & 0 &
body\\
 & & 0.9967\,$\cdot$\,\,4K$^\pm$ & $\overline{\mathrm{K}}^0$
 anything & & &  centered\\
 & &                            & \,\,+\,K$^0$ anything & \,\,39 & &
cubic\\
 & &                            & K$^+$ anything & \,\,20\\
 & &                            & $e^+$ anything & 8\\
[0.2cm]\hline\hline

\vspace{0.1cm}
	\end{tabular}

	\footnotemark{\footnotesize The particles with negative
	charges have conjugate charges of the listed  decays. Only the
	decays of K$^\pm$ and D$^\pm$ are listed. The oscillation modes carry
            one electric charge. The $\ast$ marks coupled modes.}

	\end{table}

These particles are in general charged, exempting the K$^0$ and
D$^0$  mesons and the neutron n, in contrast to the particles of the
$\gamma$-branch, which are in general neutral. It does not make a
significant difference whether one considers the mass of a particular
charged or neutral particle. After the $\pi$\,mesons, the largest mass
difference between charged and neutral particles is that of the K\,mesons
(0.81\%), and thereafter all mass differences between charged and neutral
particles are $<0.5\%$. The integer multiple rule does not immediately
apply to the masses of the $\nu$-branch particles if m($\pi^\pm$) (or
m($\pi^0$)) is used as reference, because m(K$^\pm$) =
0.8843\,$\cdot$\,4m($\pi^\pm$). 0.8843\,$\cdot$\,4 = 3.537 is far from
integer. Since the masses
 of the $\pi^0$\,meson and the $\pi^\pm$\,mesons differ by only
3.4\% it has been argued that the $\pi^\pm$\,mesons are, but for the
isospin, the same particles as the $\pi^0$\,meson, and that therefore
the $\pi^\pm$\,mesons cannot start another particle branch. However,
this argument is not supported
by the completely different decays of the $\pi^0$\,mesons
and the $\pi^\pm$\,mesons. The $\pi^0$\,meson decays almost
exclusively into $\gamma\gamma$ (98.8\%),  whereas the
$\pi^\pm$\,mesons decay practically exclusively into $\mu$\,mesons
and neutrinos, as in $\pi^+$ $\rightarrow$
$\mu^+$  + $\nu_\mu$  (99.9877\%). Furthermore, the lifetimes of the
$\pi^0$  and the $\pi^\pm$ mesons differ by nine orders of magnitude,
being $\tau$($\pi^0$) = 8.4\,$\cdot$\,10$^{-17}$ sec versus
$\tau$($\pi^\pm$)
= 2.6\,$\cdot$\,10$^{-8}$ sec.

   If we make the $\pi^\pm$\,mesons the reference particles of the
$\nu$-branch, then we must multiply the mass ratios m/m($\pi^\pm$) of the
above listed particles with an average factor 0.848 $\pm$ 0.025, as
follows from the mass ratios on Table 2.
The integer multiple rule may, however, apply
directly if one makes m(K$^\pm$) the reference for masses larger than
m(K$^\pm$). The mass of the neutron is 0.9516\,$\cdot$\,2m(K$^\pm$),
 which is only a fair approximation to an integer multiple. There are, on
the other
hand, outright integer multiples in m(D$^\pm$) = 0.9954\,$\cdot$\,(m(p) +
m($\bar\mathrm{n}$)), and in m(D$_s^\pm$) = 0.9968\,$\cdot$\,4m(K$^\pm$).
A least square
analysis of the masses of the $\nu$-branch in Table 2 yields the formula

\begin{equation} \mathrm{m}(N)/0.853\mathrm{m}(\pi^\pm) = 1.000\,N +
0.00575\qquad  N\,>\, 1 ,
\end{equation}
\noindent
with R$^2$ = 0.998. This means that the particles of the $\nu$-branch are
integer multiples of m($\pi^\pm$) times the factor 0.853. One must,
however, consider that the $\pi^\pm$\,mesons are not necessarily the
perfect reference for all $\nu$-branch particles, because $\pi^\pm$ has
I = 1, whereas for example K$^\pm$ has I = 1/2 and S = $\pm$1 and
the neutron has also I = 1/2. Actually the
factor 0.853 in Eq.(3) is only an average. The mass ratios indicate that
this factor  decreases slowly with increased m(N).
 The existence of the factor and its decrease will be explained later.

\vspace{1cm}
   Contrary to the particles of the $\gamma$-branch, the $\nu$-branch
particles decay preferentially with the emission of neutrinos, the
foremost example is $\pi^\pm \rightarrow \mu^\pm$ +
$\nu_\mu(\bar{\nu}_\mu)$ with a
fraction of 99.9877\%. Neutrinos characterize the weak interaction. We
will refer to the particles in Table 2 as the $\emph{neutrino branch}$
($\nu$-branch) of the particle spectrum. We emphasize that a weak decay of
the particles of the $\nu$-branch is by no means guaranteed. Although the
neutron decays via n $\rightarrow$ p + e$^-$ + $\bar{\nu}_e$ in 887 sec
(100\%), the proton is stable. There are, on the other hand, weak decays
such as e.g. K$^+
\rightarrow \pi^+\pi^-\pi^+$ (5.59\%), but the subsequent decays of the
$\pi^\pm$\,mesons lead to neutrinos and e$^\pm$.

   To summarize the facts concerning the $\nu$-branch of the mesons
and bary-ons. The masses of these particles seem to follow the integer
multiple rule if one uses the $\pi^\pm$\,mesons as reference, however
the mass ratios share a common factor 0.85 $\pm$ 0.025.

   To summarize what we have learned about the \emph{integer
multiple rule}: In spite of differences in charge, spin, strangeness, and
charm the masses of the ``stable" mesons and baryons of the
$\gamma$-branch  are integer multiples of the mass
 of the $\pi^0$\,meson within at most 3.3\% and on the average
within 0.66\%. Correspondingly, the masses of the ``stable"
particles of the $\nu$-branch are, after multiplication with a factor 0.85
$\pm$ 0.025, integer multiples of the mass of the $\pi^\pm$\,mesons.
The integer multiple rule has been anticipated
much earlier by Nambu [10], who wrote in 1952 that ``some regularity
[in the masses of the particles] might be found if the masses were
measured in a unit of the order of the $\pi$-meson mass". A similar
suggestion has been made by Fr\"ohlich [11]. The integer multiple rule
suggests that the particles are the result of superpositions of modes
and higher modes of a wave equation.

\section {Standing waves in a cubic lattice}

 We will now study, as we
have done in [12], whether the so-called ``stable" particles of the
$\gamma$-branch cannot be described by the frequency spectrum of
standing waves in a cubic lattice, which can accommodate automatically
the Fourier frequency spectrum of an extreme short-time collision by
which the particles are created.
The investigation of the consequences of lattices for particle theory
was initiated by Wilson [13] who studied a cubic fermion lattice. His
study has developed over time into lattice QCD.

   It will be necessary for the following to outline the most elementary
aspects of the theory of lattice oscillations. The classic paper describing
 lattice oscillations is from Born
 and v.\,Karman [14], henceforth referred to as B\&K. They looked at first
at the oscillations of a one-dimensional chain of points with mass m,
separated by a constant distance $\emph{a}$. This is the
$\emph{monatomic}$ case, all lattice points have the same mass. B\&K
assume
that the forces exerted on each point of the chain originate only from the
two neighboring points. These forces are opposed to and proportional to
the displacements, as with elastic springs (Hooke's law). The equation of
motion is in this case

\begin{equation} \mathrm{m}\ddot{u}_{n} = \alpha(u_{n+1} - u_n)
 - \alpha(u_n - u_{n-1})\,. \end{equation}
\noindent
The $u_n$ are the displacements of the mass points from their
equilibrium position
which are apart by the distance \emph{a}. The dots signify, as usual,
differentiation with respect
to time, $\alpha$ is a constant characterizing the force between the
lattice points, and n is an integer number. For \emph{a} $\rightarrow$ 0
Eq.(4) becomes the wave equation c$^2\partial^2$u/$\partial$x$^2$ =
$\partial^2$u/$\partial$t$^2$ (B\&K).

   In order to solve Eq.(4) B\&K set

\begin{equation} u_n = Ae^{i(\omega\,t\, +\, n\phi)}\,,
\end{equation}

\noindent
which is obviously a temporally and spatially periodic solution or
describes \emph{standing waves}. n is an
integer, with n  $<$ N, where N is the number of points in the chain.
$\phi$ = 0 is the monochromatic case. We also consider higher modes,
 by replacing n$\phi$ in Eq.(5) by n$^\prime\phi$, where n$^\prime$
is n times \,an integer number $>$\,1. The wavelengths are then shorter
by one over the integer number n$^\prime$/n. At n$\phi$ = $\pi$/2 are
nodes, where for all times $\emph{t}$ the displacements are zero, as
with standing waves f(x,t) = A\,cos($\omega$t)\,cos(n$\phi$) =
A\,cos($\omega$t)\,cos(kx).
If a displacement is repeated after n points we have n$\emph{a}$ =
$\lambda$, where $\lambda$ is the wavelength, $\emph{a}$ the lattice
constant, and it must be n$\phi$ = 2$\pi$ according to (5). It follows that

\begin{equation} \lambda = 2\pi\emph{a}/\phi\,.
\end{equation}

Inserting (5) into (4) one obtains a continuous frequency spectrum of
the standing waves as given by Eq.(5) of B\&K

\begin{equation}  \omega = \pm\,2\sqrt{\alpha/\mathrm{m}}
\,\mathrm{sin}(\phi/2)\,. \end{equation}
\noindent
 B\&K point out that there is not only a continuum
of frequencies, but also a  \emph{maximal frequency} which
is reached when $\phi$ = $\pi$, or at the
minimum of the possible wavelengths $\lambda$ = 2$\emph{a}$. The
boundary conditions are periodic, that means that $u_n$ = $u_{n + N}$,
where N is the number of points in the chain. Born referred to the
periodic boundary condition as a ``mathematical convenience". The number
of normal modes must be equal to the number of particles in the lattice.

    Born's model of the crystals has been verified in great detail by X-ray
 scattering and even in much more complicated cases by neutron scattering.
 The theory of lattice oscillations has been pursued in particular by
Blackman [15], a summary of his and other studies is in [16].
Comprehensive reviews of the results of linear studies of lattice dynamics
have been written by Born and Huang [17], by Maradudin et al. [18], and by
Ghatak and Kothari [19].

\section {The masses of the $\gamma$-branch particles}

   We will now assume, as seems to be quite natural, that the particles
 \emph{consist of the same particles into which they
decay}, directly or ultimately. We know this from atoms, which
consist of nuclei and electrons, and from nuclei, which consist of
protons and neutrons. Quarks
have never been observed among the decay products of
elementary particles. For the $\gamma$-branch particles
our assumption means that they consist of photons. Photons
and $\pi^0$\,mesons are the principal decay products of the
$\gamma$-branch particles, the characteristic example
 is $\pi^0$ $\rightarrow$ $\gamma\gamma$  (98.8\%). Table 1
shows that there are decays of the $\gamma$-branch particles which
lead to particles of the $\nu$-branch, in particular to pairs of $\pi^+$
and $\pi^-$ mesons.  It appears that this has to do with pair
production in the $\gamma$-branch particles. Pair production
is evident in the decay $\pi^0 \rightarrow$ e$^+$ + e$^- +
\gamma$ (1.198\%) or in the $\pi^0$\,meson's third most frequent
decay $\pi^0 \rightarrow$ e$^+$e$^-$e$^+$e$^-$
(3.14$\cdot10^{-3}$\%). Pair production requires the
presence of electromagnetic waves of high energy. Anyway,
the explanation of the $\gamma$-branch particles must begin
with the explanation of the most simple example of its kind, the
$\pi^0$\,meson, which by all means seems to consist of photons.
The composition of the particles of the $\gamma$-branch suggested here
offers a direct route from the formation of a $\gamma$-branch particle,
through its lifetime, to its decay products. Particles that are made of
photons are necessarily neutral, as the majority of the particles of the
$\gamma$-branch are.

   We also base our assumption that the particles of the $\gamma$-branch
are made of photons on the circumstances of the formation of the
$\gamma$-branch particles. The most simple and straightforward creation of
a $\gamma$-branch particle are the reactions $\gamma$ + p $\rightarrow$
$\pi^0$ + p, or in the case that the spins of $\gamma$ and p are parallel
\,\,$\gamma$ + p $\rightarrow$
$\pi^0$ + p + $\gamma^\prime$. A photon impinges on a proton and creates
a $\pi^0$\,meson. The considerations which follow apply as well for other
 photoproductions such as $\gamma$ + p $\rightarrow \eta$ + p or $\gamma$
+ d  $\rightarrow \pi^0$ + d and to the photoproduction of $\Lambda$  in
$\gamma$ + p $\rightarrow \Lambda$ + K$^+$, but also for the
electroproductions e$^-$ + p $\rightarrow
\pi^0$ + e$^-$ + p or e$^-$ + d $\rightarrow$   $\pi^0$  + e$^-$ + d,
see Rekalo et al.\,\,[20]. The most simple example of the creation of a
$\gamma$-branch particle by a strong interaction is the reaction
p + p $\rightarrow$ p + p + $\pi^0$. The electromagnetic energy
accumulated in a proton during its acceleration reappears as the
$\pi^0$\,meson.

   In $\gamma$ + p $\rightarrow \pi^0$ + p   the pulse of the
incoming electromagnetic wave is in 10$^{-23}$\,sec  converted
into a continuum of electromagnetic waves with frequencies ranging
from 10$^{23}$ sec$^{-1}$ to $\nu$ $\rightarrow$ $\infty$ according
to Fourier analysis. There must be a cutoff frequency,
 otherwise the energy in the sum of the frequencies would exceed the
 energy of the incoming electromagnetic wave. The wave packet so
created decays, according to experience, after
8.4\,$\cdot$\,10$^{-17}$\,sec into two
electromagnetic waves or $\gamma$-rays. It seems to be very
unlikely that Fourier analysis does not hold for the case of an
electromagnetic wave impinging on a proton. The question then arises of
what happens to the electromagnetic waves in the timespan of 10$^{-16}$
seconds between the creation of the wave packet and its decay into two
$\gamma$-rays\,? We will show that the electromagnetic waves
can continue to exist for the 10$^{-16}$ seconds until the wave packet
decays.

   If the wave packet created by the collision of a $\gamma$-ray
with a proton consists of electromagnetic
waves, then the waves cannot be progressive because the
wave packet must have a \emph{rest mass}. The rest mass is
the mass of a particle whose center of mass does not move. However
\emph{standing electromagnetic waves}  have a rest mass. Standing
electromagnetic waves are equivalent to a lattice, because in standing
waves the waves travel back and forth between the nodes, just as
lattice points oscillate between the nodes of the lattice oscillations.
The oscillations in the lattice take care of the continuum of
 frequencies of the Fourier spectrum of the collision. So we assume that
the very many photons in the wave packet are held together in a cubic
 lattice. It is not unprecedented that photons have been
considered to be building blocks of the elementary particles. Schwinger
[21] has once studied an exact one-dimensional quantum electrodynamical
model in which the photon acquired a mass $\sim$ $e^2$.

   We will now investigate the standing waves in a cubic photon
lattice. We assume that the lattice is held together by a weak force
acting from one
lattice point to its nearest neighbors. We assume that the range
of this force is 10$^{-16}$\,cm, because the range of the weak nuclear
force is on the order of 10$^{-16}$\,cm, as stated e.g.\,\,on p.25 of
 Perkins [22].  We set the lattice constant at
\begin{equation}
\emph{a} = 1\cdot 10^{-16}\,\mathrm{cm}\,, \end{equation}
as we have done originally in [23]. The lattice constant
of a cubic lattice can be derived from lattice theory,
see Appendix B.  For the sake of simplicity we set the
sidelength of the lattice at 10$^{-13}$\,cm, there are then 10$^9$ lattice
points. The exact size of the nucleon is given in [26] and will be used
later. As we will see the ratios of the masses of the particles are
independent of the sidelength of the lattice. Because it is the
most simple case, we assume that a
central force acts between the lattice points. We cannot consider
spin, isospin, strangeness or charm of the particles. The frequency
equation for the waves in an isotropic monatomic cubic lattice with
central forces is, in the one-dimensional case, given by Eq.(7). The
direction of the waves is determined by the direction of the
incoming  $\gamma$-ray.

   According to Eq.(13) of B\&K the force constant $\alpha$ is

\begin{equation}   \alpha = \emph{a}\,(c_{11} - c_{12} - c_{44})\,,
\end{equation}

\noindent
where c$_{11}$, c$_{12}$ and c$_{44}$ are the elastic constants in
continuum mechanics which applies in the limit $\emph{a}$ $\rightarrow$
0. If we consider central forces then c$_{12}$ = c$_{44}$ which is the
classical Cauchy relation. Isotropy requires that c$_{44}$ =
(c$_{11}\,-\,$  c$_{12}$)/2.
The waves are longitudinal. Transverse waves in a cubic lattice
with central forces are not possible according to [19]. All frequencies
that solve Eq.(7) come with either a plus or a minus sign which is, as we
will see, important. The reference frequency in Eq.(7) is

\begin{equation}  \nu_0 = \sqrt{\alpha/4\pi^2\mathrm{m}}
= \mathrm{c}/2\pi \emph{a}\,,
\end{equation}
\noindent
as we will see, using Eq.(12). c is the velocity of light.

   The \emph{limitation of the group velocity} has
now to be considered. The group velocity is given by

\begin{equation}  \mathrm{c}_g = \frac{d\omega}{dk} =
\emph{a}\sqrt{\frac{\alpha}{\mathrm{m}}}\cdot\frac{df(\phi)}{d\phi}\,.
\end{equation}
\noindent
The group velocity of the photons has to be equal to the
velocity of  light  c throughout the entire frequency spectrum, because
photons move with the velocity of light. In order to learn how this
requirement affects the frequency distribution we have to know the value
of $\sqrt{\alpha/\mathrm{m}}$ in a photon lattice. But we do not have
information
about what either $\alpha$ or m might be in this case. In the following
we set $\emph{a}\sqrt{\alpha/\mathrm{m}}$ = c, which means,
since $\emph{a} = 10^{-16}$\,cm, that $\sqrt{\alpha/\mathrm{m}}$ =
3\,$\cdot$\,10$^{26}$ sec$^{-1}$, or that the corresponding period is
$\tau$ = 1/3\,$\cdot$\,10$^{-26}$ sec, which is the time it takes for a
wave to travel with the velocity of light over one lattice distance. With

\begin{equation} \mathrm{c} = \emph{a}\sqrt{\alpha/\mathrm{m}}\,
\end{equation}

\noindent
the equation for the group velocity is

\begin{equation} \mathrm{c}_g = \mathrm{c}\cdot df/d\phi\,.
\end{equation}

  For photons that means, since c$_g$
must then always be equal to the velocity of light c,
that df/d$\phi$ = 1. This requirement determines the form
of the frequency distribution regardless of the
order of the mode of oscillation or it means that instead of the sine
function in Eq.(7) the frequency is given by

\begin{equation}\nu = \pm\,\nu_0(\,\phi + \phi_0\,)\,. \end{equation}

   For the time being we will disregard $\phi_0$ in Eq.(14) because
$\phi_0$ = 0 when the boundary condition is periodic. The frequencies
 of the corrected spectrum in Eq.(14)
must increase from $\nu$ = 0 at the origin $\phi$ = 0 with slope
1 (in units of $\nu_0$) until the maximum is reached at $\phi = \pi$.
The energy  contained in the oscillations  must be proportional to
 the sum of all frequencies (Eq.15). The \emph{second mode}  of
the lattice oscillations contains 4 times as much
energy as  the basic mode, because the frequencies are twice the
frequencies of the basic mode, and there are twice as many oscillations.
 Adding, by superposition, to the second mode different
numbers of basic modes or of second modes will give exact
integer multiples of the
energy of the basic mode. Now we understand the integer multiple
rule of the particles of the $\gamma$-branch. There is, in the framework
of this theory, on account of Eq.(14), no alternative but $\emph{integer
multiples}$ of the basic mode for the energy contained in the frequencies
of the different modes or for superpositions of different modes. In other
words, the masses of the different particles are integer multiples of the
mass of the $\pi^0$\,meson, if there is no spin, isospin,
strangeness or charm.

   We remember that the measured masses in Table 1, which
incorporate different spins, isospins, strangeness and charm, spell
out the integer multiple rule within on the average 0.65\% accuracy. It is
worth noting that $\emph{there is no free parameter}$ if one takes
 the ratio of the energies contained in the frequency distributions of
the different modes, because the factor $\sqrt{\alpha/\mathrm{m}}$
in Eq.(7) or $\nu_0$ in Eq.(14) cancels. This means,
in particular, that the ratios of the frequency distributions, or the mass
ratios, are independent of the mass of the photons at the lattice points,
as well as of the magnitude of the force between the lattice points.

   It is obvious that the integer multiples of the sum of the frequencies
in the particles are only a
first approximation of the theory of lattice oscillations and of the mass
ratios of the particles. The equation of motion in the lattice Eq.(4) does
not apply in the eight corners of the cube, nor does it
apply to the twelve edges nor, in particular, to the six sides of the
cube. A cube with 10$^9$ lattice points is not correctly described by
the periodic boundary condition we have used to derive Eq.(7), but
is what is referred to as a microcrystal.
 A phenomenological theory of the frequency distributions in
microcrystals, considering in particular the surface energy, can be
found  in Chapter 6 of Ghatak and Kothari [19]. The surface energy
may account for the small deviations of the mass ratios of the mesons
and baryons from the integer multiple rule of the oscillations in a cube.
However, it seems to be futile to pursue
a more accurate determination of the oscillation frequencies before
we know what the interaction of the electron with mass is. The mass
of the electron is 0.378\% of the mass of the $\pi^0$\,meson and
hence is a substantial part of the deviation of the mass ratios from
the integer multiple rule.

   Let us summarize our findings concerning the $\gamma$-branch.
The particles of the $\gamma$-branch consist of standing
electromagnetic waves. The $\pi^0$\,meson is the basic mode.
The $\eta$ meson corresponds to the
second mode, as is suggested by m($\eta$) $\approx$
4m($\pi^0$). The $\Lambda$ baryon corresponds to the superposition
 of two second modes, as is suggested by m($\Lambda) \approx$
2m($\eta$). This superposition apparently results in the creation of
spin 1/2. The two modes would then have to be coupled. The
 $\Sigma^0$ and $\Xi^0$ baryons are superpositions
 of one or two basic modes on the $\Lambda$  baryon. The
$\Omega^-$ particle corresponds to the superposition of three coupled
second modes  as is suggested by m($\Omega^-$) $\approx$
3m($\eta$). This procedure apparently causes spin 3/2. The charmed
$\Lambda_c^+$ baryon seems to be the first particle incorporating a
third mode. $\Sigma_c^0$ is apparently the superposition of
a negatively charged basic mode on  $\Lambda_c^+$, as is
suggested by the decay of $\Sigma_c^0$. The easiest explanation
of $\Xi_c^0$ is that it is the superposition of two coupled
third modes. The superposition of two modes of the same type is,
as in the case of $\Lambda$, accompanied by spin 1/2. The $\Omega_c^0$
baryon is apparently the superposition of two basic modes on the $\Xi_c^0$
particle. All neutral particles of the $\gamma$-branch are thus accounted
for. The explanation of the charged $\gamma$-branch particles
$\Sigma^\pm$  and $\Xi^-$ has been described in [67].
The modes of the particles are listed in Table 1.

   We have also found the $\gamma$-branch $\emph{antiparticles}$.
The rest mass of the antiparticles follows from the sum of the energies
h$\nu$ in the negative frequencies which solve Eq.(7) or Eq.(14). As
the particles, the antiparticles of the $\gamma$-branch
consist of standing electromagnetic waves. They have the same rest
mass as the particles. Antiparticles have always been associated with
 negative energies. Following Dirac's argument for electrons and positrons,
 we associate the masses with the negative frequency distributions with
antiparticles. We emphasize that the existence of antiparticles is an
automatic consequence of our theory.

   All particles of the $\gamma$-branch are unstable with lifetimes on the
order of 10$^{-10}$ sec or shorter. Born [24] has shown that the
oscillations in cubic lattices held together by central forces are
unstable. It seems, however, to be possible that the particles can be
unstable for reasons other than the instability of the lattice which
apparently causes the most frequent (electromagnetic) decay of the
$\pi^0$\,meson $\pi^0 \rightarrow \gamma\gamma$ (98.798\%),
or the most frequent (electromagnetic) decay of the
$\eta$ meson $\eta \rightarrow \gamma\gamma$ (39.43\%).
 Pair production seems to make it possible to understand the decay
of the $\pi^0$\,meson $\pi^0 \rightarrow$ e$^-$ + e$^+ + \gamma$
(1.198\%).  Since in our model the $\pi^0$\,meson consists of a
multitude of electromagnetic waves it seems that pair
production takes place within the $\pi^0$\,meson, and even more so
in the higher modes of the $\gamma$-branch where the electrons and
positrons created by pair production tend to settle on mesons, as e.g.
in $\eta \rightarrow \pi^+ +  \pi^- + \pi^0$ (22.6\%) or in the decay
$\eta \rightarrow \pi^+ + \pi^- + \gamma$ (4.68\%), where the origin
of the pair of charges is more apparent. Pair production is also evident
in the decays $\eta \rightarrow$ e$^+$e$^-\gamma$ (0.6\%)  or
$\eta \rightarrow$ e$^+$e$^-$e$^+$e$^-$ (6.9$\cdot10^{-3}$\%).

   Finally we must explain the reason for which the photon lattice or the
$\gamma$-branch particles are limited in size to a particular value of
about $10^{-13}$ cm, as
the experiments show. Conventional lattice theory using the periodic
boundary condition does not limit the size of a crystal, and in fact very
large crystals exist. If, however, the lattice consists of standing
electromagnetic waves the size of the lattice is limited by the radiation
pressure. The lattice will necessarily break up at the latest when the
outward directed radiation pressure is equal to the inward directed
elastic force which holds the lattice together. For details we refer to
[25].

\section {The rest mass of the $\pi^0$\,meson}

   So far we have studied the ratios of the masses of the particles. We
will now determine the mass of the $\pi^0$\,meson in order to validate
that the mass ratios link with the actual masses of the particles. The
energy of the $\pi^0$\,meson is \vspace{0.5cm}

\centerline{E(m($\pi^0$)) = 134.9766\,MeV =
 2.16258\,$\cdot\,10^{-4}$\,erg.}
\vspace{0.5cm}
\noindent
The sum of the energies  E = h$\nu$ of the frequencies of the
one-dimensional waves in $\pi^0$ seems to be given by the equation

\begin{equation}\mathrm{E}_\nu =
\frac{\mathrm{Nh}\nu_0}{2\pi}\,\,\int\limits_{-\pi}^{\pi}\,f(\phi)d\phi\,.
\end{equation}
\noindent
 N is the number of all lattice
points. The total energy of the frequencies in a cubic lattice is equal
to the number N of the oscillations times the average of the energy of
the individual frequencies. In order to arrive at an exact value of N in
Eq.(15) we have to use the correct value of the radius of the proton,
for which we use, as in [9],
\begin{equation}
r_p = (0.880\pm0.015)\cdot10^{-13}\,\mathrm{cm}\,,
\end{equation}
\noindent
according to [26], or it is
r$_p$ = (0.883 $\pm$  0.014)\,$\cdot$\,10$^{-13}$\,cm according to [27].
The Review of Particle Physics gives for the charge radius of the proton
the value r$_p$ = (0.875 $\pm$ 0.007)$\,\cdot\,10^{-13}$ cm. With
\emph{a} = 10$^{-16}$\,cm it follows from Eq.(16) that the number
of all lattice points in the cubic lattice is

\begin {equation} \mathrm{N} = 2.854\cdot10^9 \cong 1\,418^{\,3}\,.
 \end{equation}
\noindent
When the number of the grid points of a cubic lattice is derived from the
volume of a sphere one cannot arrive at an integer number to the third.
The radius of the $\pi^\pm$\,mesons has also been measured [28] and after
further analysis [29] was found to be 0.83\,$\cdot$\,10$^{-13}$\,cm. A much
earlier measurement [30] found r$_\pi$ at (0.86 $\pm$ 0.14)$\,\cdot\,10^{-13}$
cm. The Review of Particle Physics does not give a value for r$_\pi$.
Within the uncertainty of the radii we have r$_p$ = r$_\pi$.
And according to [31] the charge radius of $\Sigma^-$ is (0.78 $\pm$
0.10)\,$\cdot$\,$10^{-13}$\,cm.

  If the oscillations are parallel to an axis, the limitation
of the group velocity is taken into account, that means if
Eq.(14) applies and the absolute values
of the frequencies are taken, then the value of
the integral in Eq.(15) is $\pi^2$. With N = 2.854\,$\cdot$\,10$^9$
and $\nu_0$ = c/2$\pi\emph{a}$  follows from Eq.(15) that the sum
of the energy of the frequencies of the basic
mode corrected for the group velocity limitation is
E$_{corr}$ = 1.418\,$\cdot$\,10$^9$\,erg. That means
that the energy is 6.56\,$\cdot\,10^{12}$ times larger than E(m($\pi^0$)).
This discrepancy is inevitable, because the basic frequency of the
Fourier spectrum after a collision on the order of
10$^{-23}$ sec duration is $\nu$ = 10$^{23}$ sec$^{-1}$, which means,
when E = h$\nu$, that one basic frequency alone contains an energy of
about 9\,m($\pi^0$)c$^2$.

   To eliminate this discrepancy we use, instead of the simple
form E = h$\nu$, the complete quantum mechanical energy of a linear
oscillator as given by Planck

\begin{equation} E = \frac{h\nu}{e^{h\nu/kT} -\,1}\,\,.
\end{equation}
\\
\noindent
This equation was already used by B\&K for the determination of the
specific heat of cubic crystals or solids. Equation (18) calls into
question the value of the temperature T in the interior of a particle. We
determine T empirically with the formula for the internal energy of solids

\begin{equation} u = \frac{R\Theta}{e^{\Theta/T} -1}\,\,,
\end{equation}
\noindent
which is from Sommerfeld [32]. In this equation R = N\,$\cdot$\,k =
2.854\,$\cdot\,10^9$\,k,
where k is Boltzmann's constant, and $\Theta$ is the
characteristic temperature introduced by Debye [33] for the explanation of
the specific heat of solids. It is $\Theta = h\nu_m$/k, where $\nu_m$ is a
maximal frequency. In the case of the oscillations making up the
$\pi^0$\,meson the maximal frequency is $\nu_m = \pi\nu_0$,  therefore
$\nu_m = 1.5\cdot10^{26}$ sec$^{-1}$, and we find that $\Theta =
7.2\cdot10^{15}$\,K.

  In order to determine T we set the internal energy u equal to
m$(\pi^0)$c$^2$. It then follows from Eq.(19) that $\Theta$/T =
30.20, or T = 2.38\,$\cdot$\,10$^{14}$\,K. That means that Planck's
formula (18) changes Eq.(15) into
\begin{equation}\mathrm{E}_\nu(\pi^\pm) =
\frac{\mathrm{Nh}\nu_0}{2\pi(\mathrm{e^{h\nu/kT}}\,\mathrm{-}\,1)}
\,\int\limits_{-\pi}^{\pi}\,\phi\,d\phi\,, \end{equation}
\noindent
This introduces a factor
\begin{equation}
1/(\mathrm{e}^{\Theta/T} - 1\,) \cong 1/\mathrm{e}^{30.2} =
1/(1.305\cdot10^{13}) \,,  \end{equation}
into Eq.(15). In other words, if we determine the temperature T of the
particle through Eq.(19), and correct Eq.(15) accordingly then we arrive
at a sum of the oscillation energies in the $\pi^0$\,meson which is
\begin{equation} \sum_1^N\mathrm{E}_\nu = 1.0866\cdot10^{-4}\,
\mathrm{erg} = 67.82\,\mathrm{MeV}\,,\end{equation}
whereas m($\pi^0)$c$^2$/2 = 67.488\,MeV.
 The sum of the energies of N one-dimensio-nal photon oscillations
in the $\pi^0$\,meson lattice  is 0.502\,E(m($\pi^0$)). We
have to double this amount because standing waves consist  of
two waves traveling in opposite direction with
the same absolute value of the frequency. The sum of the
energy of the oscillations in the $\pi^0$\,meson is therefore
\begin{equation} \mathrm{E}_\nu(\pi^0)(theor) =
2.1732\cdot10^{-4}\,\mathrm{erg} = 135.64\,\mathrm{MeV} =
1.005\,\mathrm{E(m}(\pi^0))(exp)\,,\end{equation}
if the oscillations are parallel to the $\phi$ axis. The energy in the
measured mass of the $\pi^0$\,meson and the energy
in the sum of the oscillations agree fairly well, considering the
uncertainties of the parameters involved. The theoretical mass of the
$\eta$ meson is then m($\eta$)(\emph{theor}) = 4\,$\cdot$\,m($\pi^0$) =
542.56\,MeV = 0.991\,m($\eta$)(\emph{exp}), and the theoretical mass
of the $\Lambda$\,baryon, the superposition of two $\eta$ mesons, is
then m($\Lambda$)(\emph{theor}) = 8\,$\cdot$\,m($\pi^0$)
 = 1085.1\,MeV = 0.9726\,m($\Lambda$)(\emph{exp}).

   To sum up: The $\pi^0$\,meson is formed when a $\gamma$-ray
collides with a proton, $\gamma$ + p $\rightarrow \pi^0$ + p. By the
collision the incoming $\gamma$-ray is converted into a packet of standing
electromagnetic waves, the $\pi^0$\,meson. After $10^{-16}$ seconds
the wave packet decays into two electromagnetic waves,
$\pi^0 \rightarrow \gamma\gamma$. Electromagnetic waves prevail
throughout the entire process. The energy in the rest mass
of the $\pi^0$\,meson
 and the other particles of the $\gamma$-branch is correctly given
by the sum of the energy of  standing electromagnetic waves in a cube,
if the energy of the oscillations is determined by Planck's formula for
the energy of a linear oscillator. \emph{The $\pi^0$\,meson is like an
adiabatic, cubic black body filled with standing electromagnetic waves}.
A black body of a given size and temperature can certainly contain the
energy in the rest mass of the $\pi^0$\,meson, which is O($10^{-4})$
erg, if only the frequencies are sufficiently high. We know from
Bose's work [34] that Planck's formula applies to a photon gas as well.
For all $\gamma$-branch particles we have found a simple mode of
standing electromagnetic waves. Since the equation determining
 the frequency of the standing waves is quadratic it follows
\emph{automatically} that for each positive frequency there
is also a negative
frequency of the same absolute value, that means that for each particle
there exists also an antiparticle. For the explanation of the stable
mesons and baryons of the $\gamma$-branch $\emph{we use
only photons, nothing else}$. This is a rather conservative
 explanation of the $\pi^0$\,meson and the $\gamma$-branch
particles.  \emph{We do not use hypothetical particles}.

   From the frequency distributions of the standing waves follow
the ratios of the masses of the particles which obey the integer
multiple rule. It is important to note that in this theory the ratios of
the masses of the $\gamma$-branch particles to the mass of the
$\pi^0$\,meson $\emph{do not depend}$ on the sidelength of
the lattice, and the distance between
the lattice points, neither do they depend on the strength
of the force between the lattice points nor on the mass of the
lattice points. The mass ratios are determined only by the spectra
of the frequencies of the standing electromagnetic waves.

\section {The neutrino branch particles}

The masses of the neutrino branch, the $\pi^\pm$, K$^{\pm,0}$, n,
D$^{\pm,0}$ and D$^\pm_s$ particles, are integer multiples of the
mass of the $\pi^\pm$\,mesons times a factor $0.85\,\pm\,0.02$ as
we stated before. We assume,
as appears to be quite natural, that the $\pi^\pm$\,mesons and the
 other particles of the neutrino branch \emph{consist of the
same particles into which they decay}, that means in the case of the
$\pi^\pm$\,mesons of muon neutrinos $\nu_\mu$, anti-muon neutrinos
$\bar{\nu}_\mu$, electron neutrinos $\nu_e$, anti-electron neutrinos
$\bar{\nu}_e$ and of an electron or positron,  as exemplified by the
decay sequence
$\pi^\pm$  $\rightarrow$  $\mu^\pm$ + $\nu_\mu$($\bar{\nu}_\mu$),
\,$\mu^\pm$ $\rightarrow$ e$^\pm$ + $\bar{\nu}_\mu$($\nu_\mu$)
+ $\nu_e$($\bar{\nu}_e$). The absence of an electron neutrino $\nu_e$
in the decay branches of $\pi^-$ or of an anti-electron neutrino
$\bar{\nu}_e$ in the decay
branches of $\pi^+$ can be explained with the composition of
the electron or positron, which will be discussed in Section 11. The
existence of neutrinos and antineutrinos is unquestionable.
 Since the particles of the $\nu$-branch decay through weak
decays, we assume, as appears likewise to be natural, that
\emph{the weak nuclear force holds the particles of the $\nu$-branch
together}. This assumption has far reaching consequences, it is not
only fundamental for the explanation of the $\pi^\pm$\,mesons, but
 leads also to the explanation of the $\mu^\pm$\,mesons and
ultimately to the explanation of the mass of the electron.
The existence of the weak nuclear force is unquestionable. Since
the range of the weak interaction, which is about 10$^{-16}$ cm [22],
 is only about a thousandth of the diameter of the particles, which is
about 10$^{-13}$ cm, the weak force can hold particles together only
if the particles have a lattice structure, just as macroscopic crystals
are  held together by microscopic forces between atoms. In the
absence of a central force which originates in the center of the particle
and extends throughout the entire particle, as the Coulomb force,
the configuration of the particle is not spherical but cubic, reflecting
the very short range of the weak nuclear force.  We will show
that the energy in the rest mass of the $\nu$-branch particles is
the energy in the oscillations of a cubic lattice consisting of electron
neutrinos and muon neutrinos and their antiparticles, plus the energy
in the rest masses of the neutrinos, plus a small part with the energy
in the electric charges the particle carries.

   First it will be necessary to outline the basic aspects of diatomic
lattice
oscillations. In $\emph{diatomic}$ lattices the lattice points have
alternately the masses m and M, as with the masses of the electron
 neutrinos m($\nu_e$) and muon neutrinos m($\nu_\mu$).
 The classic example of a diatomic lattice is the salt
crystal with the masses of the Na and Cl atoms in the lattice points.
The theory of diatomic harmonic lattice oscillations was started by
Born and v.\,Karman [14]. They first discussed a diatomic chain. The
equation of motions in the chain are according to Eq.(22) of B\&K

\begin{equation} \mathrm{m}\ddot{u}_{2n} = \alpha(u_{2n+1} +
 u_{2n-1} - 2u_{2n})\, ,\end{equation}

\begin{equation} \mathrm{M}\ddot{u}_{2n+1} = \alpha(u_{2n+2} +
 u_{2n} - 2u_{2n+1}) \,, \end{equation}
\noindent
where the u$_n$ are the displacements, n an integer number and $\alpha$ a
constant characterizing the force between the particles. Eqs.(24,25) are
solved with

\begin{equation}u_{2n} = Ae^{i(\omega\,t\,+\,2n\phi)} ,\end{equation}

\begin{equation}u_{2n+1} =
Be^{i(\omega\,t\,+\,(2n+1)\phi)}\,,\end{equation}

\noindent
where A and B are constants and $\phi$ is given by $\phi =
2\pi\emph{a}/\lambda$ as in Eq.(6). $\emph{a}$ is the lattice constant as
before and $\lambda$ the wavelength, $\lambda$ = n$\emph{a}$. The
solutions of Eqs.(26,27) are obviously periodic in time and space and
describe again standing waves. Using (26,27) to solve (24,25) leads to a
secular equation from which according to Eq.(24) of B\&K the frequencies
of the oscillations of the chain follow from

\begin{equation}4\pi^2\nu^2_\pm  = \alpha/\mathrm{Mm}\cdot((\mathrm{M+m})
\pm\sqrt{(\mathrm{M-m})^2 +
4\mathrm{mM}\mathrm{cos}^2\phi}\,)\,.\end{equation}

Longitudinal and transverse waves are distinguished by the minus or plus
sign in front of the square root in (28).

\section{The masses of the $\nu$-branch particles}

The characteristic case of the neutrino branch particles are the
 $\pi^\pm$\,mesons which can be created in the process $\gamma$ + p
$\rightarrow \pi^-$ +  $\pi^+$  + p. A photon impinges on a proton and is
converted in $10^{-23}$ sec into a pair of particles of opposite charge.
A simple example of the creation of a $\nu$-branch particle by strong
interaction is the case p + p $\rightarrow$  p + p + $\pi^-$  + $\pi^+$.
Fourier
analysis dictates that a continuum of frequencies must be in the collision
products. The waves must be standing waves in order to be part of the
rest mass of a particle. The $\pi^\pm$\,mesons decay via
 $\pi^\pm$ $\rightarrow$  $\mu^\pm$ + $\nu_\mu(\bar{\nu}_\mu)$
(99.98770\%) followed by
$\mu^\pm \rightarrow$ e$^\pm$ + $\bar{\nu}_\mu(\nu_\mu)$
+ $\nu_e(\bar{\nu}_e)$ ($\approx$ 100\%).
Only $\mu^\pm$\,mesons,
 which decay into e$^\pm$ and neutrinos, and neutrinos
result from the decay of the $\pi^\pm$\,mesons.  If the
particles consist of the particles into which they decay, then
 the $\pi^\pm$\,mesons  are made of neutrinos, antineutrinos
 and e$^\pm$. Since neutrinos interact through the weak
force which has a range of O(10$^{-16}$)\,cm according to p.25 of [22],
and since the size of the $\pi^\pm$\,mesons [30] is on the order of
10$^{-13}$\,cm, \emph{the $\nu$-branch particles must have a lattice
 with  N neutrinos}, N being the same as in Eq.(17). It is not known
with certainty that neutrinos actually have a
rest mass as was originally suggested by Bethe [35] and Bahcall [36]
and what the values of m($\nu_e$) and m($\nu_\mu$) are. However, the
results of the Super-Kamiokande [37] and the Sudbury [38] experiments
indicate that the neutrinos have rest masses. Different rest masses of the
electron neutrino, muon neutrino and tau neutrino guarantee that the
three neutrino types are different. And, as the experiments show,
they are different. Otherwise the neutrinos of the three types do not differ,
they do not have charge and have the same spin.

The neutrino lattice must be
diatomic, meaning that the lattice points have alternately larger
(m($\nu_\mu$)) and smaller (m($\nu_e$)) masses. We will retain the
traditional term diatomic. \emph{The term neutrino lattice will refer to a
lattice consisting of neutrinos and antineutrinos}. The lattice we
consider
is shown in Fig.\,2. Since the neutrinos have spin 1/2 this is a
four-Fermion lattice which is required for the explanation of the weak
decays. The first investigation of cubic Fermion lattices in context with
the elementary particles was made by Wilson [13]. A neutrino lattice is
electrically neutral. Since we do not know the interaction of the electron
with a neutrino lattice we cannot consider lattices with a charge.

	\begin{figure}[h]

	\hspace{3cm}
	\includegraphics{crys12ed.eps}
	\vspace{-0.2cm}
	\begin{quote}
Fig.\,2: A cell in the neutral neutrino lattice of the $\pi^\pm$\,mesons.
Bold lines mark the forces between neutrinos
and antineutrinos. Thin lines mark the forces between
either neutrinos only, or antineutrinos only.
	\end{quote}
	\end{figure}

The neutrino lattice oscillations take care of the continuum of
frequencies which must, according
to Fourier analysis, be present after the high energy collision
which created the particle. We will, for the sake of simplicity, first set
the sidelength of the lattice at 10$^{-13}$\,cm that means approximately
equal to the size of the nucleon. The lattice then contains about 10$^9$
lattice points. The sidelength of the lattice does not enter Eq.(28)
for the frequencies of diatomic oscillations. The calculation of the
ratios of the masses  is consequently
 independent of the size of the lattice, as was the case with the
$\gamma$-branch. The size of the lattice can be explained with
the pressure which the lattice oscillations exert on a crossection of the
lattice. The pressure cannot exceed Young's modulus of the lattice.
We require that the lattice is isotropic.

	\begin{figure}[h]
	\hspace{1.5cm}
	\includegraphics{Fig6.eps}
	\vspace{-.5cm}
	\begin{quote}
Fig.\,3: The frequency distribution $\nu_-/\nu_0$ of the basic diatomic
mode
according to Eq.(24) with M/m = 100. The dashed line shows the
distribution of the frequencies corrected for the group velocity
limitation.
	\end{quote}
	\end{figure}

   From the frequency distribution of the axial diatomic oscillations
(Eq.28), shown in Fig.\,3, follows the group velocity
   $\mathrm{d}\omega/\mathrm{dk}
= 2\pi\emph{a}\,\,d\nu/d\phi$\, at each point $\phi$. With $\nu =
\nu_0f(\phi)$ and $\nu_0$ = $\sqrt{\alpha/4\pi^2\mathrm{M}}$ =
c/2$\pi\emph{a}$
 as in Eq.(10) we find

\begin{equation} \mathrm{c}_g = d\omega/dk =
\emph{a}\sqrt{\alpha/\mathrm{M}}\cdot df(\phi)/d\phi\,.
\end{equation}
\noindent
In order to determine the value of d$\omega/dk$ we have to know the value
of $\sqrt{\alpha/\mathrm{M}}$. From Eq.(9) for $\alpha$ follows that
$\alpha =
\emph{a}\,c_{44}$ in the isotropic case with central forces. The group
velocity is therefore

\begin{equation} \mathrm{c}_g = \sqrt{a^3c_{44}/\mathrm{M}}\cdot df/d\phi\,.
\end{equation}

\noindent
In Eq.(29) we now set $\emph{a}\sqrt{\alpha/\mathrm{M}}$ = c,
as in Eq.(11), where c  is the velocity of light. It follows that

\begin{equation} \mathrm{c}_g = \mathrm{c}\cdot df/d\phi\,,
\end{equation}

\noindent
as it was with the $\gamma$-branch, only that now on account of
the rest masses of the neutrinos the group velocity must be smaller
than c, so the value of df/d$\phi$ is limited to $<$\,1, but c$_g
\cong$\, c,
which is a necessity because the neutrinos in the lattice soon approach
the velocity of light as we will see. Equation (31) applies regardless
whether we consider $\nu_+$ or $\nu_-$ in
Eq.(28). That means that there are no separate transverse
oscillations with their theoretically higher frequencies.

  The rest mass M of the heavy neutrino can be determined with lattice
theory from Eq.(30) as we have shown in [12]. This involves the
 inaccurately known compression modulus of the proton. We will,
therefore, rather determine the rest mass
of the muon neutrino with Eq.(33), which leads to m($\nu_\mu$)
$\approx$  50\,milli-eV/c$^2$. It can be verified easily that
 m($\nu_\mu$) = 50\,milli-eV/c$^2$ makes sense. The energy
of the rest mass of the $\pi^\pm$ mesons is 139\,MeV, and we
 have N/4 = 0.7135$\cdot10^9$
muon neutrinos and the same number of anti-muon neutrinos with
an energy of about 50\,milli-eV. It follows that the energy in the
rest masses of all muon and anti-muon neutrinos in $\pi^\pm$  is
71.35\,MeV, that is 51\% of the energy of the rest mass of the
$\pi^\pm$ mesons, m($\pi^\pm$)c$^2$ = 139.57\,MeV. A very
small part of m($\pi^\pm$)c$^2$ goes, as we will see, into the
electron neutrino masses, the rest of the energy in $\pi^\pm$ is
in the lattice oscillations.

   The energy in the rest mass of the $\pi^\pm$\,mesons is the sum
 of the oscillation energies plus the sum of the energy in the rest
masses of the neutrinos, plus the energy in e$^\pm$.\\
\centerline{ \emph{The $\pi^\pm$\,mesons are like cubic black
bodies filled with oscillating neutrinos}.}
 For the sum of the energies of the frequencies
we use Eq.(20), with the same N and $\nu_0$ we used
for the $\gamma$-branch. For the  integral in Eq.(20) of the axial
diatomic frequencies corrected for the group velocity limitation we find
the value $\pi^2$/2 as can be easily derived from the plot of the
corrected frequencies in Fig.\,3. The value of the integral in
Eq.(20) for the axial diatomic frequencies $\nu = \nu_0\phi$ is 1/2 of
the value $\pi^2$ of the same integral in the case of axial monatomic
frequencies, because in the latter case the increase of the corrected
frequencies continues to $\phi$ = $\pi$, whereas in the diatomic case the
increase of the corrected frequencies ends at $\pi$/2, see Fig.\,3.  We
consider c$_g$ to be so close to c that it does not
change the value of the integral in Eq.(20) significantly. It can be
calculated that the time average of
the velocity of the electron neutrinos in the $\pi^\pm$\,mesons is
$\bar{v}$ = 0.99989\,c if m($\nu_e$) = 0.365\,milli-eV/c$^2$
as will be shown in Eq.(67). Consequently we find that the sum of the
 energies of the corrected diatomic neutrino
frequencies is 0.5433$\cdot$10$^{-4}$\,erg = 33.91\,MeV. We double
 this amount because we deal with standing waves or the superposition
of two waves of the  same energy and find with Eq.(20) that the
energy of the neutrino oscillations in $\pi^\pm$ is

\begin{equation} \mathrm{E}_\nu(\pi^\pm)  \cong
1/2\cdot\mathrm{E}_\nu(\pi^0)  = 67.82\,\mathrm{MeV} =
0.486\,\mathrm{m}(\pi^\pm)\mathrm{c}^2\,,
\end{equation}
\noindent
or that \emph{ $\approx$ 1/2 of the energy of $\pi^\pm$  is in the
oscillation energy E$_\nu(\pi^\pm)$}.

   In order to determine the sum of the rest masses of the neutrinos we
make use of E$_\nu(\pi^\pm)$ and obtain an approximate value of the
sum of the  rest masses of the neutrinos in $\pi^\pm$ from
\begin{equation} \mathrm{m}(\pi^\pm)\mathrm{c}^2
- \mathrm{E}_\nu(\pi^\pm) = \sum\,[m(\nu_\mu) +
 m(\bar{\nu}_\mu) + m(\nu_e) + m(\bar{\nu}_e)]\mathrm{c}^2 =
71.75\,\mathrm{MeV}\,,
\end{equation}
\noindent
and find that \emph{$\approx$ 1/2 of the energy of $\pi^\pm$  is in
the neutrino rest masses}.\\ Since nothing else but the electric charge
contributes to the rest mass of $\pi^\pm$ it appears that in a good
approximation the oscillation energy in $\pi^\pm$ is equal to the
energy in the sum of the neutrino rest masses in $\pi^\pm$, i.e.
\begin{equation} \mathrm{E}_\nu(\pi^\pm) \cong
\Sigma\,\mathrm{m(neutrinos)(\pi^\pm)c^2} =
\mathrm{N}/2\cdot(\mathrm{m}(\nu_\mu) +
\mathrm{m}(\nu_e))\mathrm{c}^2
\cong 1/2\cdot\mathrm{m(\pi^\pm)c^2}.  \end{equation}
This applies only to the neutral neutrino lattice of the pion, the
consequences of the charge of $\pi^\pm$ have not been considered.

   If m($\nu_e$) $\ll$ m($\nu_\mu$) and m($\nu_\mu$)
 = m($\bar{\nu}_\mu$), as we will justify later, we arrive
with Eq.(33) and N/2 = 1.427$\cdot10^9$ at an approximate value
for the rest mass of the muon neutrino\\
\smallskip
\hspace{4.5cm} m($\nu_\mu$) $\approx$ 50\,milli-eV/$\mathrm{c}^2$.\\
\smallskip
An accurate value of m($\nu_\mu$) will be given later, Eq.(70).

   The sum of the energy of the rest masses of all neutrinos in $\pi^\pm$, Eq.(33),
plus the oscillation energy, Eq.(32), neglecting the energy in e$^\pm$, gives
 the theoretical rest mass of the $\pi^\pm$\,mesons which is, since we
used m($\pi^\pm)$  in the determination of the neutrino rest masses with
Eq.(33), equal to the experimental rest mass of 139.57\,MeV/c$^2$.

   A cubic lattice and conservation of neutrino numbers during the reaction
 $\gamma$  + p $\rightarrow \pi^+ + \pi^-$  + p  \emph{necessitates}
that the $\pi^+$ and $\pi^-$  lattices contain just as many  electron
 neutrinos as anti-electron neutrinos. If the lattice is cubic it
 must have a center neutrino (Fig.\,4). Conservation of neutrino numbers
 requires furthermore that the center neutrino of  $\pi^+$ is matched
by an antineutrino in $\pi^-$. In the decay sequence of (say)
 the $\pi^-$\,meson $\pi^- \rightarrow  \mu^- + \,\bar{\nu}_\mu$ and
$\mu^- \rightarrow$ e$^-$ + $\nu_\mu\,+\,\bar{\nu}_e $ an electron
 neutrino $\nu_e$ does not appear. But since (N\,-\,1)/4 electron
neutrinos  $\nu_e$ must be in the $\pi^-$ lattice it follows that
(N\,-\,1)/4 electron neutrinos must go with the electron emitted
in the $\mu^-$ decay.

\begin{figure}[h]
	\hspace{2.2cm}
	\includegraphics{NaCl.eps}
	\begin{quote}
Fig.\,4: The center of a NaCl lattice. (After Born and Huang).
	\end{quote}
\end{figure}

    We must now be more specific about N,
 which is an odd number, because a cubic lattice (Fig.\,4)
 has a center particle, just as the NaCl lattice.
In the $\pi^\pm$\,mesons there are then (N\,-\,1)/4 muon
neutrinos $\nu_\mu$ and the same number of
anti-muon neutrinos $\bar{\nu}_\mu$, as well as
 (N\,-\,1)/4 electron neutrinos $\nu_e$ and the same number
of anti-electron neutrinos $\bar{\nu}_e$, \emph{plus} a center
neutrino or antineutrino. We replace
 N\,-\,1 by N$^\prime$. Since N$^\prime$ differs from N by
 only one in $10^9$ we have N$^\prime$ $\cong$ N. Although the
numerical difference between N and N$^\prime$ is negligible we
cannot consider any integer number N because that would mean
that there could be fractions  of a neutrino. N$^\prime$ is an even
number, because each cell (Fig.\,2) of the lattice (Fig.\,4) consists
 of four pairs of neutrinos.

   The \emph{antiparticle} of the $\pi^+$\,meson is the particle in
which all frequencies of the
neutrino lattice oscillations have been replaced by frequencies with
the opposite sign, all neutrinos replaced by their antiparticles and the
positive charge replaced by the negative charge. If, as we will show, the
antineutrinos have the same rest mass as the neutrinos it follows that the
antiparticle of the $\pi^+$\,meson has the same mass as $\pi^+$ but
opposite charge, i.e. is the $\pi^-$\,meson. As we will see, the
explanation of the mass of the $\pi^\pm$\,mesons opens the door to
the explanation of the mass of the muon and of the electron.

   Now we turn to the K\,mesons whose mass is m(K$^\pm$) =
0.8843\,$\cdot$\,4m($\pi^\pm$). The primary decay of the K$^\pm$\,mesons
K$^\pm \rightarrow \mu^\pm + \nu_\mu(\bar{\nu}_\mu)$ (63.5\%)
 leads to the same end products as the $\pi^\pm$\,meson decay
$\pi^\pm \rightarrow \mu^\pm + \nu_\mu(\bar{\nu}_\mu)$ (99.98\%).
From this and the decay of the $\mu^\pm$\,mesons we learn
that the K\,mesons must, at least partially, be made of the same four
neutrino types as in the $\pi^\pm$\,mesons namely of muon neutrinos,
anti-muon neutrinos, electron neutrinos and anti-electron neutrinos and
their oscillation energies. However the K$^\pm$\,mesons cannot be
solely the second mode of the lattice oscillations of the
$\pi^\pm$\,mesons, because the second mode of the
$\pi^\pm$\,mesons has an energy of
\begin{eqnarray}
{ \mathrm{E}((2.)\pi^\pm) = 4\mathrm{E}_{\nu}(\pi^\pm) +
\mathrm{N}/2\cdot(\mathrm{m}(\nu_\mu) +
\mathrm{m(\nu_e))\,c^2}\nonumber }\\
\cong 2\mathrm{m(\pi^\pm)c^2} + 1/2\cdot
\mathrm{m(\pi^\pm)c^2} = 348.92\,\mathrm{MeV}\,,\end{eqnarray}
with 2E$_\nu(\pi^\pm$) $\cong$ m($\pi^\pm$)c$^2$ and
N/2\,$\cdot$\,(m($\nu_\mu$) + m($\nu_e$)) $\cong$ m($\pi^\pm$)/2 from
Eqs.\, (33,34). The 348.9\,MeV characterize
the second or (2.)\,\,mode of the $\pi^\pm$ mesons, which fails
m(K$^\pm$)c$^2$ = 493.7\,MeV by a wide margin.

   The concept that the K$^\pm$\,mesons are alone a higher mode of
the $\pi^\pm$\,mesons also contradicts our point that the particles
consist of
the particles into which they decay. The decays K$^\pm$ $\rightarrow$
$\pi^\pm \,+ \,\pi^0$ (21.13\%), as well as K$^+$ $\rightarrow$ $\pi^0$ +
e$^+ + \nu_e$ (4.87\%), called K$^+_{e3}$, and
K$^+$ $\rightarrow$ $\pi^0 + \mu^+ + \nu_\mu$ (3.27\%), called
K$^+_{\mu3}$, make up 29.27\% of the K$^\pm$\,meson
 decays. A $\pi^0$\,meson figures
in each of these decays. If we add the energy in the rest mass of a
$\pi^0$\,meson m($\pi^0$)c$^2$ = 134.97\,MeV to the
348.9\,MeV in the second mode of the $\pi^\pm$\,mesons then
we arrive at an energy of 483.9\,MeV, which is 98.0\% of
m(K$^\pm$)c$^2$. Therefore we conclude that the
K$^\pm$\,mesons consist of the second
 mode of the $\pi^\pm$\,mesons $\emph{plus}$ a $\pi^0$\,meson
or are the state (2.)$\pi^\pm$ + $\pi^0$. Then
it is natural that  $\pi^0$\,mesons from the $\pi^0$ component
in the K$^\pm$\,mesons are among the decay products of the
K$^\pm$\,mesons.

   The average factor 0.85 $\pm$ 0.025 which appears in Eq.(3)
for the ratios of the masses
of the particles of the $\nu$-branch to the mass of the
$\pi^\pm$\,mesons is a consequence of the neutrino rest masses.
They make it impossible that the ratios of the particle masses are
outright integer multiples because the particles consist of the
energy in the neutrino oscillations and in the neutrino
rest masses which are independent of the order of the lattice
oscillations. Since the contribution in percent of the neutrino rest
masses to the $\nu$-branch particle masses decreases with
increased particle mass
 the factor in front of the mass ratios of the $\nu$-branch particles
must decrease with increased particle mass.

The K$^0$,$\overline{{\mathrm{K}}^0}$\,mesons have a rest
mass m(K$^0$,$\overline{{\mathrm{K}}^0}$) = 1.00809\,m(K$^\pm$),
or it is m(K$^0$,$\overline{{\mathrm{K}}^0}$) =
0.99984\,(m(K$^\pm$) + $\alpha_f\cdot4$m$(\pi^\mp$)).
 We obtain the K$^0$\,meson if we superpose onto the
second mode of the $\pi^\pm$\,mesons instead of
a $\pi^0$\,meson a basic mode of the $\pi^\pm$\,mesons
with a charge opposite
to the charge of the second mode of the $\pi^\pm$\,meson.
The K$^0$ and $\overline{{\mathrm{K}}^0}$\,mesons,
or the state (2.)$\pi^\pm$ + $\pi^\mp$, is made
of neutrinos and antineutrinos only, without a photon component,
 because the second mode of $\pi^\pm$ as well as the basic mode
$\pi^\mp$ consist of neutrinos and antineutrinos only.  The
K$^0$\,meson has a measured mean square charge radius
 $\langle$r$^2$$\rangle$ = $-$\,0.077 $\pm$ 0.010\,fm$^2$ according
 to [39], which can only be if there are two charges of opposite sign
within K$^0$, as this model implies. Since the mass of a
$\pi^\pm$\,meson is by 4.59\,MeV/c$^2$ larger than
the mass of a $\pi^0$\,meson the mass of K$^0$ should be larger
than m(K$^\pm$), and indeed m(K$^0$)\,$-$\,m(K$^\pm$) =
3.972\,MeV/c$^2$ according to [2]. Similar differences occur
with m(D$^\pm$)\,$-$\,m(D$^0$) and
m($\Xi_c^0$)\,$-$\,m($\Xi_c^+$). The decay
K$^0_S \rightarrow \pi^++ \pi^-$ (68.6\%) creates directly the
$\pi^+$ and $\pi^-$ mesons which are part of the (2.)$\pi^\pm$ +
$\pi^\mp$ structure of K$^0$ we have suggested.
 The decay K$^0_S \rightarrow \pi^0 + \pi^0$ (31.4\%) apparently
originates from the 2$\gamma$ branch of electron positron annihilation.
Both decays account for 100\% of the decays of K$^0_S$. The
decay K$^0_L \rightarrow 3\pi^0$ (21.1\%) apparently comes from the
3$\gamma$ branch of electron positron annihilation. The two decays of
K$^0_ L$ called K$^0_{\mu3}$ into $\pi^\pm\,\mu^\mp\,\nu_\mu$
(27.18\%) and K$^0_{e3}$ into $\pi^\pm$\,e$^\mp\,\nu_e$
(38.79\%) which together make up 65.95\% of
the K$^0_{L}$ decays apparently originate from the decay of
the second mode of the $\pi^\pm$\,mesons in the
K$^0$ structure, either into $\mu^\mp$ + $\nu_\mu$
 or into e$^\mp$ + $\nu_e$. The same types of decay,
 apparently tied to the (2.)$\pi^\pm$ mode, accompany also the
K$^\pm$  decays K$^\pm$ $\rightarrow \pi^\pm\pi^0$ (20.92\%)
in which, however, a $\pi^0$\,meson replaces the
$\pi^\pm$\,mesons in  the K$^0_L$ decay products. Our rule
that the particles consist of the particles into which they decay also
holds for the K$^0$ and $\overline{{\mathrm{K}}^0}$\,mesons. The
explanation of the K$^0$,$\overline{{\mathrm{K}}^0}$\,mesons with the
state (2.)$\pi^\pm$ + $\pi^\mp$ confirms that the state (2.)$\pi^\pm$ +
$\pi^0$ was the correct choice for the explanation of the K$^\pm$\,mesons.
The state (2.)$\pi^\pm$ + $\pi^\mp$ is also crucial for the explanation of
the absence of spin of the K$^0$,$\overline{{\mathrm{K}}^0}$\,mesons,
as we will see later.

   The neutron whose mass is m(n) = 0.95156\,$\cdot$\,2m(K$^\pm$)
is either the superposition of a K$^+$ and a K$^-$\,meson or of
a K$^0$\,meson and a $\overline{{\mathrm{K}}^0}$\,meson. As
has been shown in [67], the spin rules out
  a neutron consisting of a K$^+$ and a K$^-$\,meson. On the
other hand, the neutron can be the superposition of a K$^0$
and a $\overline{{\mathrm{K}}^0}$\,meson which guarantees that
the neutron consists of neutrinos without a photon component.
 In this case the neutron lattice contains at each lattice point a
$\nu_\mu,\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$ neutrino quadrupole
because in each K$^0$  and $\overline{{\mathrm{K}}^0}$\,meson
are neutrino pairs at the lattice points, and there is a single quadrupole
of positive and negative electric charges because each K$^0$  and
$\overline{{\mathrm{K}}^0}$\,meson carries a pair of opposite
elementary electric charges. There must be opposite charges in
the neutron because it has a mean square charge radius
$\langle$r$^2\rangle$ = $-$ 0.1161\,fm$^2$ [2].
The lattice oscillations in the neutron must be a coupled  pair in
order for the neutron to have spin 1/2, just as the $\Lambda$
baryon with spin 1/2 is a superposition of two $\eta$\,mesons. With
m(K$^0$)(\emph{theor}) = m(K$^\pm$) + 4\,MeV/c$^2$ =
 487.9\,MeV/c$^2$
 from above it follows that m(n)(\emph{theor}) $\approx$
2m(K$^0)$(\emph{theor}) $\approx$ 975.8\,MeV/c$^2$
= 1.04\,m(n)(\emph{exp}).

   The proton, whose mass is m(p) = 0.99862\,m(n), does not decay
and does not tell which particles it is made of.
 However, we learn about the structure of the proton through the
decay of the neutron n $\rightarrow$ p + e$^- +  \,\bar{\nu}_e$ (100\%).
An electron and one single anti-electron neutrino is emitted when the
neutron decays and 1.29333 MeV are released. But there is no place
for a permanent vacancy of a single missing neutrino and for a small
amount of permanently missing oscillation energy in a nuclear lattice. As
it appears all anti-electron neutrinos are removed from the structure of
the neutron in the neutron decay and converted into
 the kinetic energy of the decay products. This type of
process will be explained in Section 9. On the other hand,
 it is certain that the proton consists
 of a neutrino lattice because the neutron has a neutrino lattice. The
proton carries a net positive elementary electric charge because
the neutron carries an e$^+$e$^-$e$^+$e$^-$ quadrupole, of
which one e$^-$ is lost  in the $\beta$-decay.
The concept that the proton carries just one elementary electric charge
has been abandoned a long time ago when it was said that the
proton consists of three quarks carrying fractional electric charges.
Each elementary charge in the proton has a magnetic moment,
 all of them point in the same direction because the spin
 of the one e$^-$ must be opposite to the spin of the two e$^+$.
Each magnetic moment of the elementary charges has a g-factor $\cong$\,\,2.
All three electric charges in the proton must then have a g-factor
$\approx$\,\,6, whereas the measured g-factor of the magnetic
moment of the proton is g(p) = 5.585 = 0.93\,$\cdot$\,6.

  The D$^\pm$\,mesons with m(D$^\pm$) = 0.9954\,(m(p) +
m($\bar{\mathrm{n}}$))
are the superposition of a proton and an antineutron of opposite spin or
of their antiparticles, whereas the superposition
of a proton and a neutron with the same spin creates the deuteron
 with spin 1 and a mass m(d) = 0.9988\,(m(p) + m(n)). In this
case the proton and neutron interact with the strong force, nevertheless
the deuteron consists of a neutrino lattice with standing waves. The
D$_s^\pm$\,mesons seem to be made of a body centered cubic
lattice (Fig.\,5) as described in [40].

\begin{figure}[h]
	\hspace{4cm}
	\includegraphics{Cube.eps}
	\begin{quote}
Fig.\,5: A body-centered cell. (After Born and Huang).
\indent\qquad\qquad\,\,\,\,In the center of a D$^\pm_s$ cell is a
$\tau$\,neutrino,
 in the corners are\\
\indent $\nu_{\mu}$,$\bar{\nu}_\mu$,$\nu_e$,$\bar{\nu}_e$
neutrinos, as in Fig.\,2.\\
	\end{quote}
\end{figure}

   Summing up: The particles of the $\nu$-branch consist of oscillating
neutrinos and one or more positive and/or negative elementary electric
charges. The characteristic feature of the $\nu$-branch particles is
the cubic lattice consisting of $\nu_\mu,\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$
neutrinos. The rest mass of the $\nu$-branch particles is the sum of
the rest masses of the neutrinos and antineutrinos in the lattice plus the
mass in the energy of the  lattice oscillations plus the mass in the electric
charges. The existence of the neutrino
 lattice is a necessity if one wants to explain the spin, or the absence
 of spin, of the $\nu$-branch particles. \emph{We do not use hypothetical
particles} for the explanation of the $\nu$-branch particles, just as we
did not use hypothetical particles for the explanation of the
$\gamma$-branch particles.

\begin{center}
\section{The weak force in the interior of the particles}
\end{center}

\noindent
After we have explained the masses of the stable mesons and baryons
with cubic lattices consisting of either photons or of  neutrinos,
we can now determine the strength of the weak and the strength
of the strong nuclear forces. Both are 70 years old puzzles.
 We will use lattice theory to determine the strength of the weak
force which holds the lattices of the elementary particles together.
We will then show that the strong force between two elementary
particles is nothing but the sum of the unsaturated weak forces
emanating from the lattice points at the surface of the lattice.

In order to determine the force in the interior of the cubic lattices
with which we have explained the particle masses we will,
as we have done before in [23], use a classical paper by Born and
Land\'{e}  [41], (B\&L), dealing with the potential and compressibility
of regular ionic crystals. It is essential to realize that,
\bigskip

\begin{itemize}
\item
 \emph{for the existence of a cubic lattice it is necessary that the force
between the lattice points has an attractive part and a repulsive part}.
\end{itemize}
\begin{itemize}
\item
{Otherwise the lattice would not be stable and collapse}.
\end{itemize}
For the ionic crystals considered by B\&L the Coulomb force between the ions
 is the attractive force, whereas the repulsive force originates from the electron
clouds surrounding the ions. When the electron clouds of the ions approach
each other during the lattice oscillations they repel each other. The magnitude
of the repulsive force is not known per se and has to be determined from the
properties of the crystal.

We will follow exactly the procedure in B\&L in order to see whether their
theory is also applicable to a cubic lattice made of neutrinos. In this case the
Coulomb force is, of course, irrelevant. As B\&L do, we say that the potential
 of a cell of the neutrino lattice has an attractive part $-$ a/$\delta$ and a
repulsive part + b/$\delta^\mathrm{n}$ with the unknown exponent n. $\delta$
is the distance in the direction between two neutrinos of the same type, either
muon neutrinos and anti-muon neutrinos or electron neutrinos and
anti-electron neutrinos.

   The potential of a cell in an ionic  cubic  lattice is of the form
\begin{equation}
\phi = -\,\mathrm{a}/\delta + \mathrm{b}/\delta^\mathrm{n}\,, \end{equation}
Eq.(1) of B\&L. The constant b is eliminated with the equilibrium condition
d$\phi$/d$\delta$ = 0. Consequently
\begin{equation} \mathrm{b} = \frac{\mathrm{a}}{\mathrm{n}}
\,\delta_0^{\mathrm{n} -1}\,, \end{equation}
where $\delta_0$ is the lattice constant.
The unknown exponent n of $\delta$ in Eq.(36) was determined by B\&L with
the help of the compression modulus $\kappa$ which is known for ionic
crystals. $\kappa$ is defined by
\begin{equation} \kappa = -\,1/V\cdot dV/dp\,. \end{equation}

   The compression modulus  of an ionic lattice is given by
\begin{equation} \kappa = 9\,\delta_0^4/\mathrm{a}\,(\mathrm{n} -1)\,,
\end{equation}
Eq.(4) in B\&L.
The interaction constant a of the Coulomb force in cubic ionic
crystals resulting from the contributions of all ions on a single ion is given
by Eq.(5) of B\&L, it is
\begin{equation}\mathrm{a} = 13.94\,\mathrm{e}^2 = 3.2161\cdot10^{-18}\,\mathrm{erg\cdot cm}\,,\end{equation}
where e is the elementary electric charge. This equation is fundamental for
the theory of ionic lattices and is based on an earlier
paper by Madelung [42]. Consequently we find that
\begin{equation} (\mathrm{n} -1) = 10.33\,\mathrm{r}_0^4/\mathrm{e}^2\kappa\,,
\end{equation}
where r$_0$ = $\delta_0$/2 is the distance between a pair of neighboring
Na and Cl ions. For the alkali-halogenids, such as NaCl or KCl, B\&L found
that n $\approx$ 9. If n = 9 is used in Eq.(39) to determine theoretically the
compression modulus $\kappa$, then the theoretical values of $\kappa$
agree, in a first approximation, with the experimental values of $\kappa$,
thus confirming the validity of the ansatz for the potential in Eq.(36).

   We now apply Eq.(41) to the neutrino lattice of the elementary particles in
order to determine the potential in the interior of the particles. We must
use for r$_0$  the distance between two neighboring neutrinos in the lattice,
which is equal to the range of the weak nuclear force. The range of the weak
force is 1$\cdot10^{-16}$ cm, as in Eq.(8), and so we have
\begin{equation} \mathrm{r}_0 = 1\cdot 10^{-16}\,\mathrm{cm}\,. \end{equation}
We have used this value of r$_0$ throughout our
explanation of the masses of the mesons and baryons, though r$_0$
was previously designated by the symbol \emph{a}. We must, furthermore,
replace e$^2$ by the interaction constant g$_w^2$ of the weak force which
holds the nuclear lattice together. According to Perkins [22]
\begin{equation} \mathrm{g}_w^2 = 4\pi\hbar\,\mathrm{c}\cdot 1.05\cdot10^{-2}\,
(\mathrm{M}_W/\mathrm{M}_p)^2\,,  \end{equation}
where M$_W$ is the mass of the W boson, M$_W$ = 80.403\,GeV/c$^2$,
and M$_p$ is the mass of the proton, M$_p$ = 0.938\,272\,GeV/c$^2$.
That means that
\begin{equation} \mathrm{g}_w^2 = 2.9976\cdot 10^{-17}
\,\mathrm{erg\cdot cm}. \end{equation}

   We must also use the compression modulus $\kappa$ of the nucleon.
The value of the compression modulus of the nucleon  has been determined
theoretically by Bhaduri et al. [43], following earlier theoretical and experimental
investigations of the compression moduli of nuclei. Bhaduri et al. found that the
compression modulus K$_\mathrm{A}$(1) of the nucleon ranges from 900 to
1200\,MeV, or is 940\,MeV  or  900\,MeV for particular sets of parameters.
We determine $\kappa$ of the nucleon with
\begin{equation} \kappa = 9/\rho_\#\mathrm{K}_\mathrm{nm}\,,
\end{equation}
from Eq.(1) in [43]. Bhaduri et al. write that ``the compression modulus
K$_\mathrm{nm}$ of nuclear matter is calculated by considering the
nucleons as point particles", which they are not. Other assumptions are also
sometimes made such as infinite nuclear matter, periodic boundary conditions,
etc. Recent theoretical studies of the compressibility of ``nuclear matter" [44,45,46]
place the compressibility K$_\mathrm{{nm}}$ at values from between
250 to 270 MeV, not much different from what is was twenty years earlier
in [43]. Considering the uncertainty of K$_\mathrm{nm}$ it seems to be
justified to set, in the case of the nucleon, K$_\mathrm{nm}$ =
K$_\mathrm{A}$, where K$_\mathrm{A}$ is defined as the compression
modulus for a finite system with A nucleons. It then follows from Eq.(45)
with the number density $\rho_\#$ being the number density per fm$^3$,
and with the radius of the nucleon R$_0$ = 0.88\,$\cdot$\,10$^{-13}$\,cm,
and with 1\,MeV = 1.6022\,$\cdot$\,10$^{-6}$\,erg that the compression
modulus of the nucleon is
\begin{equation} \kappa_n = 1.603\cdot10^{-35}\,\,\mathrm{cm^2/dyn}\,,
\end{equation}
if we use for K$_\mathrm{A}$(1) the value 1000\,MeV. We will
keep in mind
that $\kappa_n$ is not very accurate because K$_\mathrm{A}$(1) is not very accurate.

If we insert (42), (44), and (46) into n $-$ 1 = 10.33\,r$_0^4$/e$^2\kappa$
(Eq.41) we find an equation for the exponent  n of the term
b/$\delta^\mathrm{n}$  in the repulsive part of the potential in a nuclear lattice,

\begin{equation} \mathrm{n} = 1 + 2.164\cdot10^{-12} = 1 + \epsilon\,,
\end{equation}
with r$_0^4$ = O(10$^{-64}$), g$_w^2$ = O(10$^{-17}$) and $\kappa$ =
O(10$^{-35}$).

\bigskip

That means that

\bigskip
\emph{the potential $\phi$ in the interior of an elementary
particle is given by}
\begin{equation} \phi = - \,\frac{\mathrm{a}}{\delta} + \frac{\mathrm{b}}
{\delta^{1 + \epsilon}} = \frac{\mathrm{a}}{\delta}\,[\frac{(\delta_0/\delta)\,^\epsilon}{\mathrm{n}} - 1\,]\,,
\end{equation}
which we can reduce with n $-$ 1 = $\epsilon$, neglecting a term multiplied
by $\epsilon\,^2$ = O(10$^{-24}$), using also a = 13.94\,e$^2$ (Eq.40)
and 1/n $\cong$ (1$-$ $\epsilon$), to
\begin{equation} \phi \cong -\,\frac{\mathrm{a}\,\epsilon}{\delta}\,
[1 - \mathrm{ln}(\delta_0/\delta)] \cong -\,\frac{13.94\,\mathrm{g}_w^2\,\epsilon}
{\delta} \, [1 - \mathrm{ln}(\delta_0/\delta)\,]\,. \end{equation}
In equilibrium the value of $\phi$ in the nuclear lattice is about
g$_w^2\cdot\epsilon$/e$^2$ $\approx$ 2.7$\cdot$10$^{-10}$ times
smaller than the corresponding electrostatic potential in an ionic lattice. A
graph of the potential in Eq.(49) versus $\delta$ is shown in Fig.\,6..

\begin{figure}[h]
	\vspace{0.5cm}
	\hspace{2.5cm}
	\includegraphics{forces.eps}
	\vspace{-0.2cm}
	\begin{quote}

Fig.6: The potential $\phi$ of the weak force as a function of $\delta$.\\
\indent\hspace{1.1cm} After [23].

	\end{quote}
\end{figure}

\bigskip

 The minimum of the curve marks the equilibrium. From Eq.(49) follows with
F$_\mathrm{w}$ = d$\phi$/d$\delta$  and $\delta$ = 2r that

\bigskip

\hspace{1cm}\emph{the weak force in the interior of the nuclear lattice is}

\begin{equation} \mathrm{F}_\mathrm{w} \cong \,\frac {3.485\cdot \mathrm{g}_w^2\,\epsilon}
{\mathrm{r}^2}\cdot \mathrm{ln}(\frac{\delta}{\delta_0})\,.
\qquad(\mathrm{dyn})\end{equation}
For all distances $\delta$ $>$ $\delta_0$ the force F$_\mathrm{w}$ is attractive,
for all distances $\delta$ $<$ $\delta_0$ the force is repulsive. The small value of
$\epsilon$ $\approx$ 10$^{-12}$ means that small displacements
$\delta/\delta_0$ \,$<$\,1 of the neutrinos from their equilibrium position,
which carry the neutrinos into the domain of their neighboring neutrino,
cause a very strong repulsive force between both neutrinos.

   We have thus determined the potential of the weak force in the interior of
the lattice of the elementary particles with lattice theory. Let us consider how
this was done.

\bigskip
\noindent
\emph{Following exactly the procedure used by B\&L to determine
the potential in the interior of an ionic crystal},

\bigskip
\noindent
\emph{we have determined the
potential  in the interior of the lattice of an elementary particle}

\medskip
\noindent
by using the parameters of the nuclear lattice. As in an
ionic lattice the potential in a nuclear lattice has an attractive and a repulsive
part, as is necessary for the stability of the lattice. We have found that in
equilibrium in a nuclear lattice the absolute values of the attractive and
repulsive terms of the potential are very nearly the same, because
$\epsilon$ is on the order of 10$^{-12}$. One wonders, in view of
the extraordinarily
small value of $\epsilon$, whether the potential in Eq.(48) is
not an academic result. Therefore it has to be shown that Eq.(48) is
indeed relevant for elementary particles. We can show that
the mass of the muon neutrino m($\nu_\mu$) depends on n $-$ 1 =
$\epsilon$ = 2.164\,$\cdot\,10^{-12}$ and yields a correct mass
m($\nu_\mu$) only if n $-$ 1 is on the order of 10$^{-12}$, for details
see [9], Eq.(33) therein.

We have thus confirmed the validity of the potential $\phi$  =
$-$\,\,a/$\delta$ + b/$\delta^\mathrm{n}$ = $-$ a/$\delta$ +
b/$\delta^{1 + \epsilon}$.
That means \emph{we have found the potential of
the weak force which holds the nuclear lattice together}.

\begin{center}
\section{The strong force between two
elementary particles}
\end{center}

   Crucial for the understanding of the existence of a strong force between
the sides of two cubic lattices is the observation that

\bigskip
\begin{itemize}
\item
 \emph{the sides of  two halves of a cubic lattice cleaved in vacuum
exert a strong, attractive force on each other}.
\end{itemize}
 It is an automatic consequence of lattice theory that
\begin{itemize}
\item
\emph{the weak force, which holds the lattice together,
is accompanied by a strong force which emanates from
the sides of the lattice}.
\end{itemize}
 This seems to contradict the simple
observation that two salt crystals stacked upon each other can be
separated without any effort. This is only so because the surface of
a cubic crystal cleaved in air oxidizes immediately, and then the sides
do not attract each other any longer. The origin of the force emanating
from the sides of a cubic ionic lattice is the Coulomb force between
ions of opposite polarity, i.e.\,\,the force which
 holds the lattice together. The attractive force emanating from
the side of a crystal cleaved in vacuum has a macroscopic value. The
existence of this force has tangible consequences in space technology.
The force between the sides of two cubic lattices was first studied
by Born and Stern [47] (B\&S).

   If U$_{12}$ is the potential between two halves of a crystal with the
equal surfaces A, or the work that is necessary to separate the two halves
of a cleaved crystal, then the capillary constant $\sigma$ is given by
Eq.(2) of B\&S
\begin{equation} \sigma = -\,\mathrm{U}_{12}/\mathrm{A}\,. \end{equation}
The capillary constant is, in the following, taken at zero degree absolute
and against vacuum for the square outside area A of a cubic crystal. $\sigma$
has been explained by B\&S, but their formula cannot be
used here because they use the value  n = 9 of the alkali-halegonids.
We will instead use Eq.(463) from Born [48] for the capillary constant
$\sigma_{100}$ of the (100) front surface of an ionic cubic crystal
\begin{equation} \sigma_{100} = -\,\frac{\mathrm{e}^2}{\mathrm{r}_0^3}\,
\frac{s_0(1)}{2}\cdot[\,1 - \frac{1}{\mathrm{n}}\,\frac{s_0(\mathrm{n})}
{s_0(1)}\cdot\frac{S_0(1)}{S_0(\mathrm{n})}\,]\,. \end{equation}
The sums s$_0$(n) and S$_0$(n) originate from the contributions of the
different lattice points to the repulsive part of the potential. The sign of the
second term on the right hand side in Eq.(52) comes from the repulsive
part of the potential in Eq.(36). For the capillary constant in a nuclear lattice
we set e$^2$ =  g$_w^2$, n = 1 + $\epsilon$, (Eq.47), and
s$_0$(1) = $-$\,0.0650 from [48] p.743. We find that
s$_0$(n) $\cong$ s$_0$(1) since n = 1 + $\epsilon$ and
$\epsilon \cong$ 10$^{-12}$.  Similarly we have
S$_0$(n) $\cong$ S$_0$(1). Then we have
\begin{equation} \sigma_{100} \cong \frac{0.065}{2}\,\frac{\mathrm{g}_w^2}
{\mathrm{r}_0^3}\, \epsilon\,. \qquad(\mathrm{dyn/cm}) \end{equation}

The work required to separate one half of a nuclear lattice from the other
half is according to Eq.(51) given by
\begin{equation} \mathrm{U}_{12} \cong -\,\frac{0.065}{2}\,\,
\frac{\mathrm{g}_w^2\,\epsilon}{\mathrm{r}_0^3}\cdot\mathrm{A}\,.
\end{equation}
\noindent
We determine the area A with the number of the lattice points in the
cubic nuclear lattice

\begin{equation} \mathrm{N}  = 2.854\cdot10^9\,, \end{equation}
\noindent
from Eq.(17). It follows that A = ($\sqrt[3]{\mathrm{N}}\,\cdot\,$r$_0)^2$.
And it follows that  the strong attractive force between the sides
of two nuclear lattices  is

\medskip

\begin{equation} \mathrm{F_s} = \frac{\mathrm{dU}_{12}}{\mathrm{dr}} =
-\,\frac{\mathrm{d}\sigma}{\mathrm{dr}}\cdot\mathrm{A} =
\frac{3\cdot0.065\,\mathrm{g}_w^2\,\epsilon}
{2\,\mathrm{r}^4}\cdot\mathrm{A}\,. \end{equation}

\medskip
\emph{The force emanating from the front surface of a cubic nuclear lattice,
\indent
the strong force, is}

\begin{equation} \mathrm{F_s} = \frac{0.0975\,\,\mathrm{g}_w^2\,\,\epsilon}
{\mathrm{r}^4}\cdot(\sqrt[3]{\mathrm{N}}\cdot\mathrm{r}_0)^2\,.
 \qquad (\mathrm{dyn}) \end{equation}

\medskip
\noindent
The strong force depends first of all on the \emph{weak interaction constant}
g$_w^2$, and secondly on the number of lattice points on the side of
the lattice, ($\sqrt[3]{\mathrm{N}})^2\,$ = 2.012\,$\cdot\,10^6$.
 The strong force is also inversely proportional to the \emph{fourth} power
of the distance between the particles. In our model the strong force depends,
other than on constants and r$^{-4}$, on the number N$^{2/3}$ of the lattice
points at the side of the lattice. That means that the force which
emanates from the sides of the $\pi$\,mesons is the same as the force
which emanantes from the sides of the proton, because both have, in this
model, the same number of lattice points. We also note that the strong force,
(Eq.57), is independent of a charge. The entire force which goes out from
the surface of the lattice is six times as much as given by Eq.(57).

   The ratio of the strong force 6\,F$_\mathrm{s}$ emanating from the entire
surface of a cubic nuclear lattice to the weak force F$_\mathrm{w}$, Eq.(50),
between the neutrinos in the same lattice is
\begin{equation}6\,\mathrm{F_s} =
\frac{0.338\cdot10^6}{\mathrm{(r/r_0)^2\,ln}(\mathrm{r/r_0})}
\cdot \mathrm{F}_w\,.
\end{equation}
The prime factor in the ratio of the strong and weak forces is the number of lattice
points at a side of the lattice ($\sqrt[3]{\mathrm{N}})^2$ = 2.012$\cdot10^6$.
The ratio F$_\mathrm{s}$/F$_\mathrm{w}$ is very sensitive to the ratio r/r$_0$.
For r $>$ r$_0$ the strong force decreases with increasing r, for r $\rightarrow$
r$_0$ we have F$_s$ $\rightarrow \infty$, and for r $<$ r$_0$ the strong force is
 repulsive, as it must be when one lattice enters another lattice.

\medskip

   To summarize: According to lattice theory
\begin{itemize}
\item
\emph{the existence of the
strong nuclear force between two elementary particles is an automatic
consequence of our explanation of the masses of the elementary
particles with cubic nuclear lattices,}
\end{itemize}
held together by the weak nuclear force.  The lattices we have used for
the explanation of
the masses of the particles consist of photons or neutrinos. That means:
\begin{center}
\emph{We do not use hypothetical particles.}
\end{center}

We have found long sought after answers to the questions what is the weak
nuclear force and the strong nuclear force, and why is the strong force so much
stronger than the weak force\,?  The strong force is nothing but the sum
of the large number of unsaturated weak forces at the surface of the nuclear
 lattice.

   In order to understand the origin of the strong nuclear force
one has to understand the structure of the elementary particles, which
we have explained with nuclear lattices.  We have now also understood
the strength of the weak force which holds the nuclear lattice together and the
cause of the strong force between two nuclear lattices.

\section{The rest mass of the muon}

Surprisingly one can also explain the mass of the $\mu^\pm$\,mesons
with our explanation of the $\pi^\pm$\,mesons. The existence of the
muons has been a puzzle since their discovery 70 years ago.
The muons belong to the lepton
 family which is distinguished from the mesons and baryons by the
absence of strong interaction with the mesons and baryons. The charged
leptons make up 1/2 of the number of stable elementary particles. The
standard model of the particles does not deal with the lepton masses.
Barut [49] has given a simple and quite
accurate empirical formula relating the masses of the electron,
muon and $\tau$\,lepton, which formula has been
extended by Gsponer and Hurni [50] to the quark masses.

    The origin of most of the $\mu^\pm$\,mesons is the decay
 of the $\pi^\pm$\,mesons, $\pi^\pm$ $\rightarrow \mu^\pm
+ \nu_\mu (\bar{\nu}_\mu$), or the decay of the K$^\pm$\,mesons,
K$^\pm \rightarrow \mu^\pm + \nu_\mu(\bar{\nu}_\mu)$.
The mass of the muons is m($\mu^\pm$) = 105.658\,369 $\pm$
9$\cdot$$10^{-6}$\,MeV/c$^2$, according to the Review of
Particle Physics [2]. The mass of the muons is usually compared to
the mass of the electron and is very often said to be
\begin{equation}
\mathrm{m}(\mu^\pm) = \mathrm{m(e}^\pm)\cdot(1 + 3/2\alpha_f) =
 206.554\,\mathrm{m(e}^\pm) =  0.99896\,\mathrm{m}(\mu^\pm)(exp)\,,
\end{equation}
(with the fine structure constant $\alpha_f$ = 1/137.036).
The experimental value of m($\mu^\pm$) is 206.768\,m(e$^\pm$).
This formula for m($\mu^\pm$) was given by Barut [51] following
an earlier suggestion by Nambu [10] that the mass of the $\pi$\,meson
is $\approx 2/\alpha_f\,\cdot\,$m(e$^\pm$) and that m($\mu^\pm$)
$\approx$  3/2$\alpha_f$\,$\cdot$\,m(e$^\pm$). The muons are
``stable", their lifetime
$\tau(\mu^\pm) = 2.19703\cdot10^{-6} \pm 4\cdot10^{-11}$\,sec is
about a hundred times longer than the lifetime of the $\pi^\pm$\,mesons,
that means longer than the lifetime of any other elementary particle, but
for the electrons, protons and neutrons.

   Comparing the mass of the $\mu^\pm$\,mesons to the mass of the
$\pi^\pm$\,mesons  from which the $\mu^\pm$\,mesons emerge
we find, with m($\pi^\pm$) = 139.570\,18\,MeV/c$^2$, that
\begin{eqnarray}
\mathrm{m}(\mu^\pm)/\mathrm{m}(\pi^\pm) = 0.757027 = 1.00937\cdot3/4\\
\cong 3/4 + \alpha_f = 0.757297 \nonumber\\
= 1.00036\cdot\mathrm{m}(\mu^\pm)/\mathrm{m}(\pi^\pm)(\emph{exp})
\nonumber\,.
\end{eqnarray}

 \noindent
The term  +\,$\alpha_f$ in the preceding equation will be explained
later, Eq.(100).
The mass of the $\mu^\pm$ mesons is in a good approximation 3/4
of the mass of the $\pi^\pm$\,mesons. We have also m($\pi^\pm)\,-$
m($\mu^\pm$) = 33.9118\,MeV/c$^2$ = 0.24297m($\pi^\pm$)
or approximately 1/4\,$\cdot$\,m$(\pi^\pm$). The
 mass of the electron is approximately 1/207 of the mass of the muon,
 the contribution of m(e$^\pm$) to m($\mu^\pm$) will therefore
be neglected in the following. We assume, as we have done before
and as appears to be natural, that the particles, including the muons,
$\emph{consist of the particles into}$ \emph{which they decay}. The
$\mu^\pm$\,mesons decay via
$\mu^\pm \rightarrow$ e$^\pm$ + $\bar{\nu}_\mu(\nu_\mu)$
+ $\nu_e(\bar{\nu}_e)$ ($\approx$ 100\%). The muons are
apparently composed of an elementary electric charge e$^\pm$
and some of the neutrinos,
antineutrinos and their oscillations which are, according to
Section 6,  present in the cubic neutrino lattice of
the $\pi^\pm$\,mesons from which the $\mu^\pm$\,mesons
come. The $\mu^\pm$\,mesons with a mass
m($\mu^\pm$) $\cong$ 3/4\,$\cdot$\,m($\pi^\pm$) seem to
be related to the $\pi^\pm$\,mesons
rather than to the electron with which the $\mu^\pm$\,mesons have
been compared traditionally, although m($\mu^\pm$) is separated
from m(e$^\pm$) by a factor $\cong$ 207.

   From Eq.(33) followed that the rest mass of a muon neutrino
should be about 50\,milli-eV/c$^2$. Provided that
the mass of an electron neutrino m($\nu_e)$ is small as compared
to m($\nu_\mu$), as will be shown by Eq.(72), we find,
 with N = 2.854$\cdot10^9$, that:
\bigskip

(a) The difference of the rest masses of the $\mu^\pm$ and $\pi^\pm$
mesons is nearly equal to the sum of the rest masses of all muon
neutrinos, respectively anti-muon neutrinos, in the $\pi^\pm$\,mesons.
\bigskip

\noindent
    m($\pi^\pm)\,-\,$m$(\mu^\pm$) = 33.912\,MeV/$\mathrm{c}^2$
\quad versus \quad $\mathrm{N^\prime}/4\cdot\mathrm{m}(\nu_\mu)$
$\approx$ 35.68\,$\mathrm{MeV}/\mathrm{c}^2$\,.

\bigskip

(b) The energy in the oscillations of all
$\nu_\mu,\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$
 neutrinos in the $\pi^\pm$\,mesons
 is nearly the same as the energy in the oscillations of
all $\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$, respectively
$\nu_\mu,\bar{\nu}_e,\nu_e$, neutrinos in the $\mu^\pm$\,mesons.
The oscillation energy is the rest mass of a particle minus the sum of
the rest masses of all neutrinos in the particle as in Eq.(33). With
m($\nu_\mu$) = m($\bar{\nu}_\mu$) and m($\nu_e$) =
m($\bar{\nu}_e)$ from Eqs.(68,71) we have
\begin{equation} \mathrm{E}_{\nu}(\pi^\pm) =
 \mathrm{m}(\pi^\pm)\mathrm{c}^2 -
\mathrm{N^\prime}/2\cdot[\mathrm{m}(\nu_\mu) +
\mathrm{m}(\nu_e)]\mathrm{c}^2 = 68.22\,\mathrm{MeV}
 \end{equation}
\quad versus
\begin{equation} \mathrm{E}_{\nu}(\mu^\pm) =
 \mathrm{m}(\mu^\pm)\mathrm{c}^2 -
\mathrm{N^\prime/4}\cdot \mathrm{m}(\nu_\mu)\mathrm{c}^2 -
\mathrm{N^\prime}/2\cdot\mathrm{m}(\nu_e)\mathrm{c}^2 =
69.98\,\mathrm{MeV}\,.\end{equation}
\noindent
Equation (62) means that either all N$^\prime$/4 muon neutrinos or
all N$^\prime$/4 anti-muon neutrinos have been removed from the
$\pi^\pm$  lattice in its decay.  If, e.g.,  any $\nu_\mu$ neutrinos
were to remain in the $\mu^+$\,meson after the decay of the
$\pi^+$\,meson they ought to appear in the decay of  $\mu^+$,
but they do not.

   We attribute the  1.768\,MeV difference
 between the left and right side of (a) to the second order effects
which cause the deviations of the masses of the particles from
the integer multiple rule. There is also the difference that the left
side of (a) deals with two charged particles, whereas the right side
deals with neutral particles. (b) seems to say that the
oscillation energy of all neutrinos in the $\pi^\pm$ lattice is conserved
in the $\pi^\pm$ decay, which seems to be necessary because the
oscillation frequencies in $\pi^\pm$ and $\mu^\pm$ must follow Eq.(14)
as dictated by the group velocity limitation. If indeed
\begin{equation} \mathrm{E}_\nu(\pi^\pm) = \mathrm{E}_\nu(\mu^\pm)
\end{equation}
then it follows from the difference of Eqs.(61) and (62) that
\begin{equation}
 \mathrm{m}(\pi^\pm)\,-\,\mathrm{m}(\mu^\pm) =
\mathrm{N}^\prime/4\cdot\mathrm{m}(\nu_\mu) =
\mathrm{N}^\prime/4\cdot \mathrm{m}(\bar{\nu}_\mu)\,. \end{equation}

The N$^\prime$/4 electron neutrinos, respectively anti-electron neutrinos,
which come, as we will see in Section 11, into the
$\mu^\pm$\,mesons with the elementary electric charge e$^\pm$, are
the recipients of 1/4 of the oscillation energy E$_\nu(\pi^\pm$) of the
$\pi^\pm$\,mesons which becomes available when the muon neutrinos or
anti-muon neutrinos leave the $\pi^\pm$ lattice in the $\pi^\pm$ decay.
The neutrinos coming with e$^\pm$ into $\mu^\pm$ make it possible that
$\mathrm{E}_\nu(\pi^\pm) = \mathrm{E}_\nu(\mu^\pm)$. After the $\pi^\pm$
decay the muon neutrinos in $\mu^\pm$ retain their original oscillation
energy E$_\nu(\pi^\pm$)/4, the electron neutrinos and anti-electron
neutrinos
in $\mu^\pm$ retain their original oscillation energies E$_\nu(\pi^\pm$)/4
as well. The remaining oscillation energy E$_\nu(\pi^\pm$)/4 of the
$\pi^\pm$
lattice so far not accounted for is picked up by the electron neutrinos,
 respectively anti-electron neutrinos, brought into $\mu^\pm$ by the
elementary electric charge. Without a recipient for this oscillation energy
E$_\nu(\pi^\pm$)/4 a stable new particle can apparently not be formed
in the $\pi^\pm$ decay, that means there is no $\mu^0$\,meson.

   We should note that in the $\pi^\pm$ decays only \emph{one single}
 muon neutrino or single anti-muon neutrino is emitted, not N$^\prime$/4
of them, but that in the
 $\pi^\pm$ decay 33.912\,MeV are released. Since according to (b) the
oscillation energy of the neutrinos in the $\pi^\pm$ mesons is conserved
in their decay the 33.912\,MeV released in the $\pi^\pm$ decay can come
from \emph{no other source} then from the rest masses of the muon
neutrinos or the anti-muon neutrinos in the $\pi^\pm$\,mesons. The
average kinetic energy  of the
neutrinos in the $\pi^\pm$ lattice is about 50\,milli-eV, so it is not
possible that a single neutrino in $\pi^\pm$ possesses an energy
 of 33.9\,MeV. The 33.9\,MeV can come only from the sum of the
muon neutrino or anti-muon neutrino rest masses in the
$\pi^\pm$\,mesons. However, what happens then to the neutrino
numbers\,?

   Either conservation of neutrino numbers is violated or the
decay energy comes from equal numbers  of muon neutrinos and
anti-muon neutrinos. Equal numbers N$^\prime$/8  muon neutrinos
and  N$^\prime$/8 anti-muon neutrinos would then be in
the $\mu^\pm$\,mesons. This would not make a difference in either the
oscillation energy or in the sum of the rest masses of the neutrinos or in
the spin of the $\mu^\pm$\,mesons. A situation similar to the $\pi^\pm$
decay occurs in the $\mu^\pm$ decay. The 105.147\,MeV released in
the $\mu^\pm$ decay comes, in our model, mainly from the energy in
the rest masses of either N$^\prime$/4$\cdot$$\nu_\mu$  or
N$^\prime$/4$\cdot$$\bar{\nu}_\mu$
neutrinos and their oscillations, because the rest masses of the
 $\nu_e,\bar{\nu}_e$ neutrinos, which are also in $\mu^\pm$, are so
small. Conservation of neutrino numbers in the $\mu^\pm$ decay
requires that N$^\prime$/8 muon neutrinos and N$^\prime$/8
anti-muon neutrinos are in the
$\mu^\pm$ lattice. Since m($\nu_\mu$) = m($\bar{\nu}_\mu$)
we will, however, for the sake of simplicity, use the term
N$^\prime$/4\,$\cdot$\,m($\nu_\mu$)
or N$^\prime$/4\,$\cdot$\,m($\bar{\nu}_\mu$) for
N$^\prime$/8\,$\cdot[$m($\nu_\mu$) + m($\bar{\nu}_\mu$)].

   Inserting m($\pi^\pm$)\,$\mathrm{-}$\,m($\mu^\pm$) =
N$^\prime$/4\,$\cdot$\,m($\nu_\mu$)
from Eq.(64) into Eq.(62) we arrive at an equation for \emph{the
theoretical value of the mass of the $\mu^\pm$\,mesons}. It is
\begin{equation} \mathrm{m}(\mu^\pm)\mathrm{c}^2 =
1/2\cdot[\,\mathrm{E}_\nu(\pi^\pm) +
\mathrm{m}(\pi^\pm)\mathrm{c}^2 +
\mathrm{N^\prime\,m}(\nu_e)\mathrm{c}^2/2\,] =
103.95\,\mathrm{MeV}\,\,,\end{equation}
which is 0.9838\,m($\mu^\pm$)c$^2$(\emph{exp}) and  expresses
 m($\mu^\pm$) through the well-known mass of $\pi^\pm$,
the calculated oscillation energy of $\pi^\pm$, and a small contribution
(0.4\%) of the electron neutrino and anti-electron neutrino masses.
Eq.(65) shows that
our explanation of the mass of the $\mu^\pm$\,mesons comes close to
the experimental value m($\mu^\pm$) = 105.658\,MeV/c$^2$. With
E$_\nu(\pi^\pm$) = E$_\nu(\mu^\pm$) and with m($\pi^\pm)$ from
Eq.(34) we find a different form of  Eq.(65) which is, with m($\nu_e$)
=  m($\bar{\nu}_e$), in the case of  the $\mu^+$\,meson,
\begin{equation}\mathrm{m}(\mu^+) =
 \mathrm{E}_\nu(\mu^\pm)/\mathrm{c}^2 +
\mathrm{N^\prime m}(\bar{\nu}_\mu)/4 +
\mathrm{N^\prime}\mathrm{m}(\nu_e)/2  \,\,. \end{equation}
As Eq.(66) tells, the rest mass of the muons is the sum of the rest masses
of the muon neutrinos, respectively anti-muon neutrinos, and of the masses
of the electron neutrinos and anti-electron neutrinos which are in the muon
lattice, plus
the oscillation energy of these neutrinos, neglecting the mass of e$^\pm$.
 The ratio  m($\mu^\pm$)/m($\pi^\pm$) is 3/4, as it must be, if we
divide Eq.(66) by m($\pi^\pm$) which follows from Eq.(34) and if we
neglect the small masses of the electron neutrinos  and anti-electron neutrinos.

   \emph{The  $\mu^\pm$\,mesons cannot be point particles} because
they have a neutrino lattice.  The commonly held belief that the
$\mu^\pm$\,mesons are point particles  is based on the results of scattering
 experiments. But at a true point the density of a ``point particle" would
be infinite, which poses a problem. It is odd that the $\mu^\pm$\,mesons,
which emerge from the $\pi^\pm$\,mesons and have a mass which is
3/4\,$\cdot$\,m($\pi^\pm$), should have a mass which is concentrated in a point,
whereas it is accepted and measured that the $\pi^\pm$\,mesons have
a body of finite size. Since, on the other hand, neutrinos do not interact, in
a very good approximation, with charge or mass it will not be possible
to determine the size of the $\mu^\pm$\,meson lattice through conventional
scattering experiments. The $\mu^\pm$\,mesons  \emph{appear} to be
point particles because only the electric charge of the muons
participates in the scattering process and the elementary electric
charge scatters like a point particle. \emph{99.5\% of the muon mass
consists of non-interacting neutrinos}. There is therefore currently no
measurable difference in e$^-$-p and $\mu^-$-p scattering.

   Finally we must address the question for what reason do the
muons or leptons not interact strongly with the mesons
and baryons\,? We have shown in  Section 8 that a strong force emanates
from the sides of a cubic lattice caused by the unsaturated weak
forces of about $10^6$  lattice points at the surface of the lattice of
the mesons and baryons. This follows from the study of Born and Stern [47]
which dealt with the forces between two parts of a cubic lattice cleaved
in vacuum. The strong force between two particles is an automatic
consequence of the weak internal force which holds the particles together.
If the muons have a lattice consisting of N/8 muon neutrinos and N/8
anti-muon neutrinos and of N/4 electron neutrinos and N/4 anti-electron
neutrinos their lattice surface is not the same as the surface of the
cubic  $\nu_\mu,\,\bar{\nu}_\mu,\,\nu_e,\,\bar{\nu}_e$ lattice of the
mesons and baryons. Therefore  the muon lattice does
not bind with the cubic lattice of the mesons and baryons.

   To summarize what we have learned about the $\mu^\pm$\,mesons.
Eq.(66) says that the energy in m($\mu^\pm$)c$^2$ is the sum of
the oscillation energies plus the sum of the energy of the rest masses of
the neutrinos and antineutrinos in m($\mu^\pm$), neglecting the energy in
e$^\pm$. The three neutrino types in the lattice of the $\mu^\pm$\,mesons
shown in Fig.\,7 are the remains of the cubic neutrino lattice in the
$\pi^\pm$\,mesons. Since  N/8\,$\cdot$\,$\nu_\mu$ and
N/8\,$\cdot$\,$\bar{\nu}_\mu$ neutrinos have been
 removed from the $\pi^\pm$ lattice in the $\pi^\pm$ decay
the rest mass of the $\mu^\pm$\,mesons must be
$\cong$ 3/4\,$\cdot$\,m($\pi^\pm$), as the experiments find.
The $\mu^\pm$\,mesons are not point particles.

\begin{figure}[h]
	\vspace{0.5cm}
	\hspace{0.2cm}
	\includegraphics{mulated.eps}
	\vspace{-0.3cm}
	\begin{quote}
Fig.\,7: A section through the central part of the neutrino\\
\indent \hspace{1.1cm} lattice of the $\mu^+$ meson without its charge.
	\end{quote}
\end{figure}

   The mass of the $\tau^\pm$ leptons, whose mass is m($\tau^\pm$) =
1776.99\,MeV/c$^2$  = 16.8182\,m($\mu^\pm$) or
1.8939\,m(p), follows from the most frequent
leptonic decay of the D$^\pm_s$\,mesons, D$^\pm_s
\rightarrow \tau^\pm + \nu_\tau(\bar{\nu}_\tau)$ (6.4\%). It can be shown
 readily that the oscillation
energies of the lattices in D$^\pm_s$ and in $\tau^\pm$ are the same.
From that follows that the energy in the rest mass
 of the $\tau^\pm$ leptons is the sum of the oscillation energy in the
$\tau$  lattice plus the sum of the energy of the rest masses of all
neutrinos and antineutrinos in the $\tau$ lattice, just as with the
$\mu^\pm$\,mesons. We will skip the details. The tau leptons are
not point particles either.

\section{The neutrino masses}

   Now we come to the neutrino masses. As mentioned before, the
experiments show that the electron neutrinos, muon neutrinos and tau
neutrinos are different. It is undisputed that all of them have no charge
and the same spin. However, if the three neutrino types have different
rest masses then the electron neutrinos, muon neutrinos and tau
neutrinos are different.  There is no certain knowledge
what the neutrino masses are. Numerous values for m($\nu_e$) and
m($\nu_\mu$) have been proposed and upper limits for them have
been established experimentally which have, with time, decreased
steadily. The Review of Particle Physics [2] gives
for the mass of the electron neutrino the value $<$\,2\,eV/c$^2$. Neither
the Superkamiokande [37] nor the Sudbury [38] experiments determine
a neutrino mass, however, both experiments make it very likely that the
neutrinos have rest masses. We will now determine the neutrino masses
from the composition of the $\pi^\pm$\,mesons and from the
$\beta$-decay of the neutron.

   If the same principle that applies to the decay of the
 $\pi^\pm$\,mesons, namely that in the decay the oscillation energy
 of the decaying particle is conserved and that an entire neutrino type
supplies the energy released in the decay, as expressed by
Eqs.(61,62), also applies to the decay of
the neutron n $\rightarrow$ p + e$^-\,+\,\bar{\nu}_e$, then the mass of
the anti-electron neutrino can be determined from the known difference
$\Delta$ =  m(n)\,$-$\,m(p) = 1.293\,332\,MeV/c$^2$ $\cong
 5/2\,\cdot$\,m(e)/c$^2$ [2]. Nearly one half of $\Delta$ comes
from the energy lost by the emission of the electron, whose mass is
0.510\,9989\,MeV/$\mathrm{c}^2$. N anti-electron neutrinos are in the neutrino
 quadrupoles in the neutron, one-fourth of them is carried away by the
emitted electron. We have seen in the paragraph above Fig.\,4 that
the decay sequence of the $\pi^\pm$\,mesons requires
that the electron carries with it N$^\prime$/4 electron neutrinos, if the
 $\pi^\pm$\,mesons consist of a lattice with a center neutrino or
antineutrino
and equal numbers of $\nu_e,\bar{\nu}_e,\nu_\mu,\bar{\nu}_\mu$
neutrinos as required by conservation of neutrino numbers during
the creation of $\pi^\pm$. The composition of the electron will be
discussed in detail in Section\,11. The electron can carry
 N$^\prime$/4 anti-electron neutrinos as well as N$^\prime$/4 electron
neutrinos, see Appendix C. Since, as we will see shortly, m$(\nu_e$) =
 m($\bar{\nu}_e$) this does not make a difference energetically but is
relevant for the orientation of the spin vector of the emitted electron.
After the neutron
has lost N$^\prime$/4 anti-electron neutrinos to the electron
emitted in the $\beta$-decay the other 3/4$\cdot$N$^\prime$
 anti-electron neutrinos in the neutron provide
 the energy ($\Delta$ $-$ m(e$^-$))c$^2$ = 0.782\,333\,MeV
$\cong$ 3/2$\cdot$m(e)c$^2$
 released in the decay of the neutron. From this follows, after division by
 3/4$\cdot$N$^\prime$, that the rest mass of the anti-electron neutrino is
\begin {equation} \mathrm{m}(\bar{\nu}_e) =
0.365\,\mbox{milli-eV}/\mathrm{c}^2\,.\end{equation}

\noindent
 Since theoretically the anti-neutron decays as
$\bar{\mathrm{n}} \rightarrow\bar{\mathrm{p}}$ + e$^+$ + $\nu_e$ it
follows from the same considerations as with the decay of the neutron that
\begin{equation} \mathrm{m}(\nu_e) = \mathrm{m}(\bar{\nu}_e)\,.
\end{equation}
We note that
\begin{equation}\mathrm{N^\prime}/4\cdot \mathrm{m}(\nu_e) =
 \mathrm{N^\prime}/4\cdot \mathrm{m}(\bar{\nu}_e) =
 0.51\,\mathrm{m(e}^\pm)\,. \end{equation}
This equation is, as we will see, fundamental for the explanation of
the mass of the electron.

   Inserting Eq.(67) into Eq.(33) for the sum of the masses of the
neutrinos in $\pi^\pm$  we find that
\begin{equation} \mathrm{m}(\nu_\mu) =
49.91\,\mbox{milli-eV}/\mathrm{c}^2\,.
\end{equation}
Since the same considerations apply for either the $\pi^+$ or
the $\pi^-$ meson it follows that
\begin{equation}\mathrm{m}(\nu_\mu) = \mathrm{m}(\bar{\nu}_\mu)\,.
\end{equation}
Experimental values for the rest masses of the different neutrino types are
not available. However, it appears that for the $\nu_\mu \leftrightarrow
\nu_e$
oscillation the value for $\Delta$m$^2$ = m$^2_2$ $\mathrm{-}$ m$^2_1$  =
3.2$\times10^{-3}$\,eV$^2$ given on p.1565 of [37] can be used to
 determine m$_2$ = m($\nu_\mu$) if m$_1$ = m$(\nu_e$) is much smaller than
 m$_2$. We have then m($\nu_\mu$) $\approx$ 56.56\,milli-eV/c$^2$,
which is compatible with the value of m($\nu_\mu$) given in Eq.(70).

From Eqs.(67,70) follows that
\begin{equation}\mathrm{m}(\nu_e) = 1/136.74\cdot \mathrm{m}(\nu_\mu)
 \cong \alpha_f\mathrm{m}(\nu_\mu)\,. \end{equation}
\noindent
1/136.74 is 1.0021 times the fine structure constant $\alpha_f$ =
e$^2/\hbar$c = 1/137.036. It does not seem likely that Eq.(72) is just
a coincidence. The probability for this being a coincidence is zero
considering the infinite pool of numbers on which the
ratio m($\nu_e$)/m($\nu_\mu)$ could settle.

   The mass of the $\tau$\,neutrino $\nu_\tau$ can be determined from
the decay D$_s^\pm$  $\rightarrow$
$\tau^\pm$ + $\nu_\tau\,(\bar{\nu}_\tau)$,
 and  the subsequent decay $\tau^\pm \rightarrow \pi^\pm + \bar{\nu}_\tau
(\nu_\tau)$, stated in [2]. The appearance of $\nu_\tau$
in the decay of D$^\pm_s$
and the presence of $\nu_\mu,\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$ neutrinos
in the $\pi^\pm$ decay product of the $\tau^\pm$\,leptons means that
 $\nu_\tau,\bar{\nu}_\tau,\nu_\mu,\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$
neutrinos must be  in the D$^\pm_s$ lattice.
The additional $\nu_\tau$ and $\bar{\nu}_\tau$ neutrinos can be
accomodated in D$^\pm_s$ by a body-centered cubic
lattice, in which there is in the center of each cubic cell one particle
different
from the particles in the eight cell corners (Fig.\,5). In a body-centered
cubic lattice are N$^\prime$/4  cell centers, if N$^\prime$ is the number
of lattice points without the cell centers. If the particles in the cell
centers are tau neutrinos then N$^\prime$/8 \,$\nu_\tau$ neutrinos and
N$^\prime$/8 anti-tau neutrinos \,$\bar{\nu}_\tau$  must be present,
because of conservation of neutrino numbers. From m(D$^\pm_s$) =
1968.2\,MeV/c$^2$ and m($\tau^\pm$) =
1777\,MeV/c$^2$  follows that

\begin{equation} \mathrm{m}(\mathrm{D}^\pm_s) - \mathrm{m}(\tau^\pm) =
191.2\,\mathrm{MeV/c^2} = \mathrm{N^\prime}/8
\cdot\mathrm{m}(\nu_\tau)\,.\end{equation}
\noindent
The rest mass of the $\tau$\,neutrinos is therefore
\begin{equation} \mathrm{m}(\nu_\tau) = \mathrm{m}(\bar{\nu}_\tau) =
0.536\,\mathrm{eV/c}^2\,.
\end{equation}
\noindent
From the neutrino masses given by Eq.(74) and Eq.(70) follows that
\begin{equation}
\mathrm{m}(\nu_\tau) = 10.76\,\mathrm{m}(\nu_\mu) =
1.048\,(\alpha_w/\alpha_f)\mathrm{m}(\nu_\mu)\,,
\end{equation}
 where $\alpha_w$  is the weak coupling constant $\alpha_w$  =
$g_w^2/4\pi\hbar$c, (Eq.43),
and $\alpha_f$  is  the fine  structure constant. We keep in mind that
g$_w^2$ in $\alpha_w$ is not nearly as accurately known as
e$^2$  in $\alpha_f$. With Eq.(72) we find that
\begin{equation}
\mathrm{m}(\nu_\tau) = 1.048\cdot(\alpha_w/\alpha_f^2)\cdot
\mathrm{m}(\nu_e) = 1474\,\mathrm{m}(\nu_e)\,. \end{equation}

   According to Eqs.(67,70,74) the sum of the masses of the electron
neutrino, muon neutrino and tau neutrino is 0.586\,eV/c$^2$, primarily
because of the mass of the tau neutrino. According to the Review of
Particle Physics (2004,\,p.439) it follows from astrophysical data that the
sum of the neutrino masses $\sum_{i}$\,m${(\nu{_i})}$ $\leq$
0.7\,eV/c$^2$. We arrive at essentially the same result.

   To summarize what we have learned about the masses of the leptons:
We have found an explanation for the mass of the $\mu^\pm$\,mesons
and $\tau^\pm$ leptons. We have also determined the rest masses of
the \emph{e},\,$\mu,\tau$ neutrinos and antineutrinos. In other words,
 we have found the masses of all leptons, exempting the electron,
which will be dealt with in the next Section.

\section{The electron}

The electron differs from the other particles we have considered in so far
as it appears that its electric charge cannot be separated from its mass,
whereas in the other charged particles the mass of the electric charge is,
 in a first
approximation, unimportant for the mass of the particles. Even in the
muons the mass of the electron contributes only five thousandth of the
muon mass. On the other hand, the electron is fundamental for the
stability of the charged
particles whose lifetime is sometimes orders of magnitude larger than the
lifetime of their neutral counterparts. For example the lifetime of the
$\pi^\pm$\,mesons is eight orders of magnitude larger than the lifetime of
$\pi^0$, the lifetime of the proton is infinite, whereas the neutron
decays
in about 900 seconds and, as a startling example, the lifetime of
$\Sigma^\pm$ is O($10^{-10}$) seconds, whereas the lifetime of
$\Sigma^0$ is O($10^{-20}$) seconds. There is something particular
to the interaction of the elementary electric charge with the particle masses.

   After J.J. Thomson [52] discovered the small corpuscle which soon
 became known as the electron an enormous amount of theoretical work
 has been done to explain the existence of the electron. Some of the most
 distinguished physicists have participated in this effort. Lorentz [53],
Poincar\'{e} [54], Ehrenfest [55], Einstein [56], Pauli [57], and others
showed that it is fairly certain that the electron cannot be explained as
a purely electromagnetic particle. In particular it was not clear
 how the elementary electric charge could be held together in its
small volume because the internal parts of the charge repel each other.
Poincar\'{e} [58] did not leave it at showing that such an electron
 could not be stable, but suggested a solution for the problem by
introducing what has become known as the Poincar\'{e} stresses whose origin
however remained unexplained. These studies were concerned with the
 static properties of the electron, its mass m(e) and its electric
charge e; the positron, the spin and the neutrinos were not known at that
time. In order to explain the electron with its existing mass and
charge it appears to be necessary to add to Maxwell's equations a
non-electromagnetic mass and a non-electromagnetic force which could
hold the electric charge together. We shall see what this mass and force is.

   The discovery of the spin of the electron by Uhlenbeck and
Goudsmit [59] increased the difficulties of the problem in so far as it now
 had also to be explained how the angular momentum $\hbar$/2 and the
magnetic moment $\mu_e$  come about.
The spin of a point-like electron seemed to be explained by Dirac's [60]
equation, however it turned out later [61] that Dirac type equations can
be constructed for any value of the spin. Afterwards Schr\"{o}dinger [62]
tried to explain the spin and the magnetic moment of the electron with
 the so-called Zitterbewegung. Dirac [63] suggested a model of an electron
without spin, consisting of a charged, hollow sphere held together by surface
tension. The first higher mode of oscillation appeared to be the muon, to
quote ``one can look upon the muon as an electron excited by radial
oscillations". Many other models
of the electron were proposed. On p.74 of his book ``The Enigmatic
Electron" Mac\,Gregor [64] lists more than thirty such models.
At the end none of these models has been successful
because the problem developed a seemingly insurmountable difficulty
 when it was shown through electron scattering experiments
that the charge radius of the electron must be smaller
than $10^{-16}$\,cm [65], in other words that the electron appears
to be a point particle, at least by three orders of magnitude smaller
than the classical electron radius r$_e$  =  e$^2$/m$_e$c$^2$  =
2.8179$\cdot10^{-13}$\,cm. This, of
course, makes it very difficult to explain how a particle can have
a finite angular momentum when the radius goes to zero, and how an
electric charge can be confined in an infinitesimally small volume. If the
elementary electric
charge would be in a volume with a radius of O($10^{-16}$)\,cm the
 Coulomb self-energy would be orders of magnitude larger than the
rest mass of the electron, which is not realistic. The choice is between
a massless point charge and a finite size particle with a non-interacting
mass to which an elementary electric charge is attached. It seems fair
to say that at present, more than 100 years after the discovery of the
electron,
we do not have an accepted theoretical explanation of the electron.

   We propose in the following, as in [66], that the non-electromagnetic
mass which seems  to be necessary in order to explain the electron
consists of neutrinos. This is actually a necessary consequence of our
 model  of the mass of the $\pi^\pm$\,mesons and of the decay
sequence of $\pi^\pm$. And we propose that the non-electromagnetic
force required to hold the  electric charge and the neutrinos in the electron
together is the weak nuclear force which, as we
 have suggested, holds together the masses of the mesons and
baryons and also the mass of the muons. Since the range of the weak
nuclear force is on the order of $10^{-16}$\,cm the neutrinos must be
arranged in a lattice with the weak force extending from each lattice
point only to the nearest neighbors.  The O($10^{-13}$)\,cm size
of the neutrino lattice in the electron does not at all contradict the results
of the scattering experiments, just as the  explanation of the mass of
the muons  with our model does not contradict the
apparent point particle characteristics of the muon, because neutrinos
are in a very good approximation non-interacting and therefore are
not noticed in scattering experiments with electrons.

   The rest mass of the electron is m(e) = 0.510\,998\,92 $\pm$
4$\cdot10^{-8}$\,MeV/c$^2$ and the electrostatic charge
 of the electron is e = 4.803\,2044\,$\cdot\,10^{-10}$\,esu, as stated
in the Review  of Particle Physics [2].  \emph{ The objective of a
theory  of the electron must, first of all, be the explanation of
m(e$^\pm$) and e$^\pm$}, but also of s(e$^\pm$) and of the
magnetic moment $\mu_e$. We will first
explain the rest mass of the electron, making use of what we
 have learned  about the explanation of the mass of the
$\mu^\pm$\,mesons in Section 9. The muons are leptons,
 just as the electrons, that means
that they interact  with other particles exclusively through
the electric force. The muons have a mass which is 206.768
times larger than the mass of the electron, but they have the same
elementary electric charge as the electron or positron and the same
spin. Scattering experiments tell that the $\mu^\pm$\,mesons are point
particles  with a size $<$\,$10^{-16}$\,cm, just as the electron. In other
words, the muons have the same characteristics as the electrons and
 positrons but for a mass which is about 200 times larger.
Consequently the muon is often referred to as a ``heavy"  electron.
If a non-electromagnetic  mass is required to explain
 the mass of the electron then a non-electromagnetic mass 200 times
 as large as in the electron is required to explain the mass of the muons.
These non-electromagnetic masses must be \emph{non-interacting},
otherwise scattering experiments could not find the size of either the
electron or the muon at 10$^{-16}$\,cm.

   We have explained the mass of the muons
in Section 9. According to our model the muons consist of an
elementary electric charge and an oscillating lattice of neutrinos.
Neutrinos are the only non-interacting matter we know  of.
In the muon lattice  are, according to our model,
 (N\,-\,1)/4 = N$^\prime$/4 muon neutrinos $\nu_\mu$
(respectively anti-muon  neutrinos $\bar{\nu}_\mu$),
N$^\prime$/4 electron neutrinos $\nu_e$ and the same
 number of anti-electron neutrinos $\bar{\nu}_e$,  one
elementary electric charge and the energy of the lattice oscillations.
The letter N stands for the number of all neutrinos and antineutrinos in
 the cubic lattice of the $\pi^\pm$ mesons, N = 2.854$\cdot10^9$,
Eq.(15). It is, as explained in Section 9, a necessary consequence
of the composition of $\pi^\pm$ and the decay sequence $\pi^-$
$\rightarrow$  $\mu^-$ + $\bar{\nu}_\mu$ and
$\mu^-$  $\rightarrow$ e$^-$ + $\bar{\nu}_e$ + $\nu_\mu$,
that \emph{there must be N$^\prime$/4 electron neutrinos $\nu_e$
in the emitted electron}.  The explanation of the $\pi^\pm$\,mesons
led to the  explanation of the $\mu^\pm$\,mesons and now leads
to the explanation of the mass of e$^\pm$.
For the mass of the electron neutrinos and anti-electron neutrinos
we found in Eq.(67)  that m($\nu_e$) = m($\bar{\nu}_e$) =
0.365\,milli-eV/c$^2$. The sum of
the energies in the rest masses of the N$^\prime$/4 neutrinos
or antineutrinos in the lattice of the electron or positron is then
\begin{equation} \sum{\,\mathrm{m(\nu_e)c^2}} =
 \mathrm{N}^\prime/4\cdot\mathrm{m}(\nu_e)\mathrm{c}^2
= 0.260\,43\,\mathrm{MeV}
 = 0.5096\,\mathrm{m(e^\pm)}\mathrm{c}^2\,. \end{equation}
\noindent
In other words, \emph{1/2 of the rest mass of the
electron is approximately equal to the sum of the rest masses of
the neutrinos in the electron}. In modern parlance this is the ``bare"
part of the electron. The bare part is not observable.
The other half of the rest mass of the
electron must originate from the energy in the electric charge
carried by the electron.

   From pair production
$\gamma$ + M   $\rightarrow$  e$^-$ + e$^+$ + M, (M being any
 nucleus), and from conservation of neutrino numbers follows
necessarily that there must also be a neutrino lattice
 composed of N$^\prime$/4 anti-electron neutrinos, which make
up the lattice of the positrons, which lattice has, since m($\nu_e$) =
m($\bar{\nu}_e$), the same rest mass as the neutrino lattice of the
electron, as it must be for the antiparticle of the electron.
Conservation of charge requires conservation of neutrino numbers.

   Fourier analysis dictates that a continuum of high frequencies must be
in the electrons or positrons created by pair production in a timespan of
$10^{-23}$ seconds. We will now determine the energy
E$_\nu$(e$^\pm$) in the oscillations in the  interior of the electron.
Since we want to explain the \emph{rest mass} of the electron we can
only consider the frequencies of non-progressive waves, either standing
waves or circular waves. The sum of the energies of the lattice
oscillations is, in the case of the $\pi^\pm$\,mesons,  given by

\begin{equation} \mathrm{E}_\nu(\pi^\pm) =
\frac{\mathrm{Nh}\nu_0}{2\pi(\mathrm{e^{h\nu/kT}}\,\mathrm{-}\,1)}
\,\int\limits_{-\pi}^{\pi}\,\phi\,d\phi\,. \end{equation}
\noindent
This is Eq.(20) which was used to determine the oscillation energy in
the $\pi^0$ and $\pi^\pm$ mesons. This type of equation
was introduced by Born and v.\,Karman [14] in order to explain the
internal energy of cubic crystals.  If we apply Eq.(78) to the
 oscillations  in the electron which has N$^\prime$/4 electron neutrinos
 $\nu_e$ we arrive at  E$_\nu$(e$^\pm)$  = 1/4$\cdot$E$_\nu(\pi^\pm$),
which is mistaken because E$_\nu(\pi^\pm$) $\approx$
m($\pi^\pm$)c$^2$/2 and m($\pi^\pm$) = 273\,m(e$^\pm$).
Eq.(78) must be modified in order to be suitable for the oscillations in
the electron. It turns out that we must use
\begin{equation} \mathrm{E}_\nu(\mathrm{e}^\pm) =
\frac{\mathrm{Nh}\nu_0\cdot\alpha_f}{2\pi(\mathrm{e^{h\nu/kT}}\,
\mathrm{-}\,1)}\,\int\limits_{-\pi}^{\pi}\,\phi\,d\phi\,,
\end{equation}
\noindent
where $\alpha_f$ is the fine structure constant. As is well-known the
fine structure constant $\alpha_f$ characterizes the strength of the
electromagnetic forces. The appearance of $\alpha_f$ in Eq.(79)
means that the nature of the oscillations in the electron is different
from the oscillations in the $\pi^0$ or $\pi^\pm$ lattices. With
$\alpha_f$  = e$^2/\hbar$c
and $\nu_0$ = c/2$\pi$\emph{a} we have
\begin{equation} \mathrm{h}\nu_0\alpha_f = \mathrm{e}^2/\emph{a}\,,
\end{equation}
 which shows that the oscillations in the electron are \emph{electric
oscillations}. The appearance of e$^2$  in Eq.(80) guarantees that
the oscillation energy of the electron and positron are the same, as it
must be.

   There must be N$^\prime$/2 oscillations of the elements of the electric
charge  in e$^\pm$, because we deal with non-progressive waves, which
are the superposition of two waves. That means that  2$\times$N$^\prime$/4
$\cong$ N/2 oscillations are in Eq.(79). As we will see later the spin requires
that the oscillations are circular. From Eqs.(78,79) then follows that
\begin{equation} \mathrm{E_\nu(e}^\pm) =
 \alpha_f/2\cdot\mathrm{E}_\nu(\pi^\pm)\,. \end{equation}
\noindent
E$_\nu(\pi^\pm)$  is the oscillation energy in the $\pi^\pm$\,mesons
which  can be calculated with Eq.(78). According to Eq.(32) it is
\begin{equation} \mathrm{E}_\nu(\pi^\pm) = 67.82 \,\mathrm{MeV} =
0.486\,\mathrm{m}(\pi^\pm)\mathrm{c}^2 \approx
\mathrm{m}(\pi^\pm)\mathrm{c}^2/2\,.
 \end{equation}
 With E$_\nu(\pi^\pm$) $\approx$ m($\pi^\pm$)c$^2$/2 =
139.57/2\,MeV and $\alpha_f$  = 1/137.036  follows from Eq.(81)
that the oscillation energy of the electron or positron is
\begin{equation} \mathrm{E_\nu(e}^\pm) = \frac{\alpha_f}{2}\cdot
\frac{\mathrm{m}(\pi^\pm)\mathrm{c}^2}{2} =
0.254\,623\,\mathrm{MeV} =
0.996\,570\,\mathrm{m}(\mathrm{e}^\pm)\mathrm{c}^2/2\,.
\end{equation}
\noindent
If we replace in Eq.(83) the experimental value for m($\pi^\pm$) by
the good empirical approximation
m($\pi^\pm$) $\cong$ m(e$^\pm$)(2/$\alpha_f$), Eq.(102), then it
follows that
\begin{equation}\mathrm{E}_\nu(\mathrm{e}^\pm) \cong
1/2\cdot \mathrm{m(e^\pm)c}^2 \,.\end{equation}
In other words,\,\,\emph{ 1/2 of the energy in the rest mass of the
electron is made up by the oscillation energy in} e$^\pm$. This equation
corresponds to Eq.(34) for the oscillation energy in the
$\pi^\pm$\,mesons.

   In Eq.(83) we have determined the value of the oscillation energy in
e$^\pm$  from the product of the very accurately known fine structure
constant and the very accurately
known rest mass of the $\pi^\pm$\,mesons which establishes
a firm value of the oscillation energy of e$^\pm$.  The other half of
the energy  in m(e$^\pm$) is in the rest masses of the neutrinos in
the electron, Eq.(77).  We can confirm Eq.(83) without
 using E$_\nu(\pi^\pm)$ with the formula for the oscillation energy
 in the form of Eq.(88) with N/2 = 1.427$\cdot10^9$,
e = 4.803$\cdot10^{-10}$\,esu,
 $\emph{a}$ = 1$\cdot10^{-16}$\,cm, f(T) = 1/1.305$\cdot10^{13}$ and,
with the integral being $\pi^2$, we obtain E$_\nu$(e$^\pm$) =
0.968\,m(e$^\pm$)c$^2$/2. This calculation involves more
parameters than in Eq.(83) and is consequently less accurate than Eq.(83).

   In a good approximation the oscillation energy of e$^\pm$ in Eq.(83)
is equal to the sum of the energies in the rest masses
 of the electron neutrinos in the e$^\pm$ lattice in Eq.(77), or
E$_\nu$(e$^\pm$) $\cong$ N$^\prime$/4\,$\cdot$\,m($\nu_e$)c$^2$ =
N$^\prime$/4\,$\cdot$\,m($\bar{\nu}_e$)c$^2$. Since
 \begin{equation} \mathrm{m(e}^\pm)\mathrm{c}^2
 =  \mathrm{E}_\nu(\mathrm{e}^\pm) +
\sum{\,\mathrm{m}(\nu_e)\mathrm{c}^2}
= \mathrm{E_\nu(e^\pm)} + \mathrm{N^\prime/4\cdot m(\nu_e)c^2}\,,
\end{equation}
\noindent
it follows from Eqs.(83) and (77) that
\begin{equation} \mathrm{m(e^\pm)c^2}(theor) =
0.5151\,\mathrm{MeV} =
1.0079\,\mathrm{m(e^\pm)c^2}(exp)\,. \end{equation}
The theoretical rest mass of the electron or positron agrees within the
accuracy of the parameters N and m($\nu_e)$ = m($\bar{\nu}_e$)
with the measured rest mass.

   From Eq.(81) follows with E$_\nu(\pi^\pm)$ $\cong$
m($\pi^\pm$)c$^2$/2  from Eq.(82) that

\vspace{0.5cm}
\centerline{2E$_\nu$(e$^\pm) = \alpha_f$E$_\nu(\pi^\pm)$ $\cong$
$\alpha_f$m$(\pi^\pm)$c$^2$/2  $\cong$
m(e$^\pm$)c$^2$\,,}

\vspace{0.5cm}
\noindent
or that
\begin{equation} \mathrm{m(e^\pm)}\cdot2/\alpha_f  =
 274.072\,\mathrm{m(e^\pm)}
= 1.0034\,\mathrm{m(\pi^\pm)}\,, \end{equation}
whereas the actual ratio of the mass of the $\pi^\pm$\,mesons  to
the mass of the electron is m($\pi^\pm$)/m(e$^\pm$) =
273.132 or 0.9966\,$\cdot$\,2/$\alpha_f$. The ratio of
m($\pi^\pm$)/m(e$^\pm$) which follows from Eq.(87) is similar to
what we will find in Eq.(102). This is a necessary condition for the
validity of our model of the electron.

   We have thus shown that the \emph{rest mass of the electron or
positron can be explained} by the sum  of the rest masses of the
electron neutrinos or anti-electron neutrinos in a cubic  lattice with
N$^\prime$/4 electron neutrinos or N$^\prime$/4 anti-electron
neutrinos, plus the mass in the sum of the energy of
N/2 standing electric oscillations in the lattice, Eq.(83). The one
oscillation added to the even numbered N$^\prime$/4 oscillations
is the oscillation at the center of the lattice, Fig.\,8. From this model
follows, since it deals with a cubic neutrino lattice, that \emph{the
electron is not a point particle}. However, since neutrinos are
non-interacting their presence will not be detected in
electron scattering experiments.

\begin{figure}[h]
\vspace{0.5cm}
\hspace{2.2cm}
\includegraphics{elat.eps}
\vspace{-0.2cm}
\begin{quote}
Fig.\,8: Horizontal or vertical section through the central part \\
\indent\hspace{1.1cm} of the electron lattice.
\end{quote}
\end{figure}

   The charge radius of the electron determined by electron scattering
experiments is $<$\,$10^{-16}$\,cm [65] and seems to contradict the
model of the electron proposed here, whose size is on the order of
10$^{-13}$\,cm as shown in Appendix D. However, the
experimental size does not apply to the circumstances considered
here. One has to find the scattering formula for finite size charged cubic
lattices  and analyze the experimental data with such a scattering formula
in order to see whether our model is in contradiction to the experiments.
If the electron lattice behaves like a rigid body then electron
scattering should take place at the center of the lattice, which is a point [64].

   Let us compare the rest mass of the electron to the \emph{rest mass of
the muon}, which was explained in Section 9 with an oscillating lattice of
muon neutrinos, electron neutrinos and anti-electron neutrinos and an
elementary electric charge.  The electron
has the most simple neutrino lattice, consisting of only one neutrino type,
either electron neutrinos or anti-electron neutrinos, and it has the smallest
sum of the rest masses of the neutrinos in a particle. The heavy weight
of the  muon m($\mu^\pm$) = 206.768\,m(e$^\pm$) is a
consequence of the heavy weight of the
 N$^\prime$/4 muon neutrinos or N$^\prime$/4 anti-muon neutrinos in
the muon lattice. The mass of either a muon neutrino or an anti-muon
neutrino is 137 times the mass of an electron or anti-electron neutrino,
Eq.(72). That makes the electron neutrinos and anti-electron neutrinos in
$\mu^\pm$ as well as the mass of the
electric charge in a first approximation negligible. It then follows from
 Eq.(98) that m($\mu^\pm$)(\emph{theor}) $\cong$
3/2\,$\cdot$\,(m($\nu_\mu$)/m($\nu_e$))\,$\cdot$\,m(e$^\pm$) =
3/2$\alpha_f\,\cdot$\,m(e$^\pm$) = 205.554\,m(e$^\pm$) = 0.99413\,m($\mu^\pm$)(\emph{exp}),
which proves that the heavy mass of the muons is caused by the
heavy $\nu_\mu,\bar{\nu}_\mu$ neutrinos.

   In order to confirm the \emph{validity} of our preceding explanation
of the mass
of the electron we must show that the sum of the charges in the electric
oscillations in the interior of the electron is equal to the elementary
electric charge of the electron.
We recall that Fourier analysis requires that, after pair production,
there must be a continuum of frequencies in the electron and positron.
With h$\nu_0\alpha_f$ = e$^2$/\emph{a} from Eq.(80) follows
from Eq.(79) that the oscillation energy in e$^\pm$ is the sum of
2$\times$(N$^\prime$/4 + 1) $\cong$ N/2  electric oscillations

\begin{equation} \mathrm{E}_\nu(\mathrm{e}^\pm) =
\frac{\mathrm{N}}{2} \cdot \frac{\mathrm{e^2}}{\emph{a}}\cdot
\frac{f(T)}{2\pi}\,
\int\limits_{-\pi}^{\pi}\,\phi\,\mathrm{d}\phi\,, \end{equation}
with f(T) = 1/(e$^{h\nu/kT} \mathrm{-}$ 1) = 1/1.305$\cdot10^{13}$
from Eq.(21). Inserting the values for N, f(T) and \emph{a} we find
that E$_\nu$(e$^\pm$) = 0.968\,m(e$^\pm$)c$^2$/2 $\cong$
m(e$^\pm$)c$^2$/2, similar to Eq.(83). The discrepancy
between m(e$^\pm$)c$^2$/2 and E$_\nu$(e$^\pm$) so calculated
must originate from the uncertainty of the parameters N, f(T) and \emph{a}
in Eq.(88).

   In order to determine the charge e in the electric oscillations we
replace the integral divided by 2$\pi$ in Eq.(88), which has the value
$\pi$/2, by the sum $\Sigma\,\phi_k\Delta\,\phi$, where k is an integer
number with the maximal value k$_m$ = (N/4)$^{1/3}$. $\phi_k$ is
equal to k$\pi$/k$_m$  and we have

\begin{displaymath}  \Sigma\,\phi_k\,\Delta\phi =
\sum_{k=1}^{k_m}\,\frac{k\pi}{k_m}\cdot\frac{1}{k_m}
= \frac{ k_m(k_m + 1)\pi}{2\,k_m^2} \cong \frac{\pi}{2}\,,
\end{displaymath}
\noindent as it must be. The energy in the individual electric oscillation
with index k is then
\begin{equation} \Delta\mathrm{E}_\nu(k) = \phi_k\,\Delta\phi =
k\pi/k_m^2\,.
\end{equation}

      Suppose that the energy of the electric oscillations is correctly
described by the self-energy of an electric charge Q
\begin{equation}\mathrm{U} = 1/2\,\cdot\,\mathrm{Q}^2/\mathrm{r}\,.
\end{equation}
The self-energy of the elementary electric charge is normally used to
determine the mass of the electron from its charge, here we use Eq.(90)
the other way around, we determine the charge from the energy
in the oscillations.

   The charge of the electron is contained in the electric oscillations.
That means that the electric charge of e$^\pm$ is not concentrated in a
point, but is distributed over N/4 = O($10^9)$ charge elements
 Q$_k$. \emph{The charge elements are distributed in a cubic lattice}
and the resulting electric field is cubic, not spherical. In the absence of a
central force which originates at the center of the particle and affects all
parts of the particle the configuration of the particle is not spherical but
cubic, just as it was with
the shape of the $\pi^\pm$\,mesons. For distances large as compared
to the sidelength of the cube, (which is O($10^{-13}$)\,cm), say at the
first Bohr radius which is on the order of $10^{-8}$\,cm, the deviation
of the cubic field from the spherical field will be reduced by about
$10^{-10}$. The charge in all electric oscillations in the electron is
\begin{equation} \mathrm{Q} = \sum_{k}\,\mathrm{Q}_\mathrm{k}\,.
\end{equation}

   Setting the radius r in the formula for the self-energy  equal to
2\,\emph{a}, one charge element is separated from the nearest other
by 2\emph{a},  we find, with Eqs.(88,89,90), that the charge in the
individual electric oscillations is

\begin{equation} \mathrm{Q_k} =
\pm\,\sqrt{2\pi\,N\,e^2f(T)/k_m^2}\,\cdot\,\sqrt{k}\,.
\end{equation}

\noindent
and with k$_m$ = 1/2\,$\cdot$\,(N/4)$^{1/3}$ = 447 and
\begin{equation}  \sum_{k=1}^{k_m}\,\sqrt{k} = 6310.8\, \end{equation}

\noindent follows, after we have doubled the sum over $\sqrt{k}$,
because for each index k there is a second oscillation on the negative
axis of $\phi$, that

\begin{equation} \mathrm{Q} = \Sigma\,\mathrm{Q_k} =
\pm\,5.027\cdot10^{-10}\,\,\mathrm{esu}\,,\end{equation}

\noindent whereas the elementary electric charge is e$^\pm$ =
$\pm$\,4.803\,$\cdot\,10^{-10}$\,esu.  That means that our theoretical
charge of the electron or positron is 1.047 times the elementary electric
 charge. Within the uncertainty of the parameters the theoretical charge
of the electron agrees with
the experimental charge e. We have confirmed that it follows from our
explanation of the mass of the electron that the electron has, within a
5\% error, the correct electric charge.

   Each element of the charge distribution is surrounded in the horizontal
plane by four electron neutrinos as in Fig.\,8, and in vertical direction by
an electron neutrino above and also below the element. The electron
neutrinos hold the charge elements in place.  We must assume that
the charge elements are bound to the neutrinos by the weak nuclear
force. The weak nuclear force plays here a role similar
to its role in holding, for example,  the $\pi^\pm$ or $\mu^\pm$ lattices
together. It is not possible, in the absence of a definitive explanation
of the neutrinos, to give an explanation for the electro-weak
interaction between the electric oscillations and the neutrinos.
However, the presence of the range
 \emph{a} of the weak nuclear force in e$^2$/\emph{a} is a sign
that the weak force is involved in the electric oscillations. The
attraction of the charge elements by the neutrinos
overcomes the Coulomb repulsion of the charge elements.
 The weak nuclear force is the missing non-electromagnetic force
or the Poincar\'{e} stress which holds the elementary electric charge
together. The same considerations apply
for the positive electric charge of the positron, only that then the
electric oscillations are all of the positive sign and that they are
bound to  anti-electron neutrinos.

   Finally we learn that  Eq.(88) precludes the possibility that the
charge of the electron sits only on its surface. The number N in
Eq.(88) would then be on the order
of $10^6$, whereas N must be on the order of $10^9$ so that
E$_\nu$(e$^\pm$) can be m(e$^\pm$)c$^2$/2 as is necessary.
In other words, the charge of the electron must be distributed throughout
the interior of the electron, as we postulated.

   Summing up: The rest mass of the electron or positron originates from
the sum of the rest masses of N$^\prime$/4 electron neutrinos or
anti-electron neutrinos in cubic lattices plus the mass in the energy of
the electric oscillations in their neutrino lattices. The neutrinos, as
well as the electric oscillations, make
 up 1/2 of the rest mass of e$^\pm$ each.
 The  electric oscillations are bound to the
neutrinos by the weak nuclear force. The sum of the charge elements
 of the electric oscillations accounts for the elementary charge of the
electron, respectively positron. The electron or the positron are
not point particles.

   One hundred years of sophisticated theoretical work have made it
abundantly clear that the electron is not a purely electromagnetic particle.
There must be something else in the electron but electric charge,
otherwise the electron could not be stable. It is equally clear from the
most advanced scattering experiments that the  ``something else"  in the
electron must be non-interacting, otherwise it could not  be that we find that
the charge radius of the electron must be smaller than $10^{-16}$\,cm.
The only non-interacting matter we know of with certainty are the neutrinos.
So it  seems to be natural to ask whether neutrinos are not part of the
electron. Actually we did
not introduce the neutrinos in an axiomatic manner but rather as a
consequence of our explanation of the stable mesons, baryons
and muons. It follows necessarily from this model that after the
decay of, say, the $\mu^-$\,meson there must be electron neutrinos in the
emitted electron, and that they make up one half of the rest mass of the
electron. The other half of the energy in the electron originates from the
energy of the electric oscillations. We have thus explained the rest
masses of the electron or positron which agree, within 1\% accuracy,
with the experimental
 value of m(e$^\pm$). We have learned that the charge of the electron
 is not concentrated in a single point, but rather is distributed over
O(10$^9$) elements which are held together with the neutrinos by the
weak nuclear force. The sum of the charges in the electric oscillations
is, within the accuracy of
the parameters, equal to the elementary charge of the electron. From the
explanation of the mass and charge of the electron follows, as we will
show, the correct spin and magnetic moment of the electron, the other
two fundamental properties of the
electron. With a cubic lattice of anti-electron neutrinos we also arrive
with the same considerations as above at the correct
mass, charge, spin and magnetic moment of the positron.

\section{The ratios m($\mu^\pm$)/m(e$^\pm$), m($\pi^\pm$)/m(e$^\pm$) \\
and  m(p)/m(e)}

   In order to check on the \emph{validity of our explanation of m($\pi^\pm$),
m($\mu^\pm$) and also of m(e$^\pm$)} we will now look at the ratios
m($\mu^\pm$)/m(e$^\pm$), m($\pi^\pm$)/m(e$^\pm$) and
 m(p)/m(e). In order to determine m($\mu^\pm$)/m(e$^\pm$) we first
modify Eq.(69) by setting  N$^\prime$/4\,$\cdot$\,m($\nu_e$)
= 0.5\,m(e$^\pm$), not at 0.51\,m(e$^\pm$). In other words we say that
1/2 of the mass of the electron is made of neutrinos. If the other half of
the mass of the electron originates from the electric charge of the
electron, as we have
shown in the preceding Section,  then the mass of the electron
 is twice the mass of the sum of the rest masses of the neutrinos
 in the electron (Eq.69) and we have

\begin{equation}  \mathrm{m(e}^\pm) = \mathrm{N}^\prime/2\cdot
\mathrm{m}(\nu_e) \quad\mathrm{or}\quad
\mathrm{N}^\prime/2\cdot\mathrm{m}(\bar{\nu}_e)\,.
\end{equation}
 We also set E$_\nu(\pi^\pm$) = 0.5\,m($\pi^\pm$)c$^2$, not
at 0.486\,m($\pi^\pm$)c$^2$
as in Eq.(32). With E$_\nu(\pi^\pm$) = E$_\nu(\mu^\pm$) from Eq.(63)
follows with Eq.(61) and m($\pi^\pm$)c$^2$ = 2\,E$_\nu(\pi^\pm$)
that  \begin{equation} \mathrm{E}_\nu(\mu^\pm) =
\mathrm{N}^\prime/2\cdot[\mathrm{m}(\nu_\mu) +
\mathrm{m(\nu_e)]c}^2\,. \end{equation}
From Eq.(66) and with Eqs.(68,71) then follows that
\begin{equation} \mathrm{m}(\mu^\pm) = 3/4\cdot\mathrm{N^\prime}
\mathrm{m}(\nu_\mu) + \mathrm{N^\prime}\mathrm{m}(\nu_e)\,,
\end{equation}
considering only the neutrino masses, not the charge in $\mu^\pm$.
With m(e$^\pm$) = N$^\prime$/2\,$\cdot$\,m($\nu_e$) from
Eq.(95) we have
 \begin{equation} \frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)}(theor) =
\frac{3}{2}\cdot\frac{\mathrm{m}(\nu_\mu)}{\mathrm{m}(\nu_e)} + 2\,,
\end{equation}
or with m($\nu_\mu$)/m($\nu_e$) $\cong$ 1/$\alpha_f$ from Eq.(72)
it turns out that

\begin{equation} \frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)}(theor)
\cong \frac{3}{2}\cdot \frac{1}{\alpha_f} + 2 = 207.55
= 1.0038\cdot\frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)}(exp)\,,
 \end{equation}
\noindent
with m($\mu^\pm$)/m(e$^\pm$)(\emph{exp}) =  206.768. In order
to arrive at the proper ratio of m($\mu^\pm$)/m(e$^\pm$)
\begin{itemize}
\item
\emph{it must be that m($\nu_e$) = $\alpha_f$\,$\cdot$\,m($\nu_\mu$)},
\end{itemize}
as in Eq.(72).

   The mass of the muon is, according to Eq.(98), much larger than
the mass of the electron because the mass
 of the muon neutrino is much larger than the mass of the electron
 neutrino, m($\nu_\mu)$ $\cong$ 137\,m($\nu_e$). The ratio of the
mass of the muon to the mass of the electron
is independent of the number of the neutrinos in both lattices.

   Equation (99) is nearly the same as Barut's [51] empirical formula
(Eq.59) according to which the muon mass ratio is
\begin{displaymath}\mathrm{m}(\mu^\pm)/\mathrm{m(e}^\pm)(\emph{emp}) =
3/2\alpha_f + 1 = 206.55 = 0.9989\,\mathrm{m}(\mu^\pm)/\mathrm{m(e}^\pm)
(\emph{exp})\,. \end{displaymath}
\noindent
A much better approximation to the experimental mass ratio is obtained
when the +\,1 in Barut's formula is replaced by +\,1.25. The thus calculated
m($\mu^\pm$)/m(e$^\pm$) = 206.804
differs then from the measured m($\mu^\pm$)/m(e$^\pm$) =
206.7683 by the factor 1.00017.

   A better relation between the mass of the muon and the mass of the
pion can now be established. According to Eq.(60) it is, empirically,
\begin{equation} \mathrm{m}(\mu^\pm)/\mathrm{m}(\pi^\pm) \cong
3/4 + \alpha_f\,. \end{equation}
With $\alpha_f$E$_\nu(\pi^\pm$) = 2E$_\nu$(e$^\pm$), (Eq.81),
and m($\pi^\pm)$c$^2$ = 2E$_\nu(\pi^\pm$), as well as
m(e$^\pm$)c$^2$ = 2E$_\nu$(e$^\pm$), follows that
\begin{equation}  \mathrm{m}(\mu^\pm) \cong 3/4\cdot\mathrm{m}(\pi^\pm) +
2\,\mathrm{m}(\mathrm{e}^\pm) = 1.00039\,\mathrm{m}(\mu^\pm)(exp)\,.
 \end{equation}
 The reason for the additional term 2m(e$^\pm$) on the right hand side
of Eq.(101) has still to be found.

\medskip
   Similarly we obtain for the $\pi^\pm$\,mesons the ratio
\begin{equation} \frac{\mathrm{m}(\pi^\pm)}{\mathrm{m(e^\pm)}}(theor)
= 2\,[\frac{\mathrm{m}(\nu_\mu)}{\mathrm{m}(\nu_e)} +1]
 \cong \frac{2}{\alpha_f} + 2 = 276.07 =
1.0108\,\frac{\mathrm{m}(\pi^\pm)}{\mathrm{m(e^\pm)}}(exp)\,,
\end{equation}
with m($\pi^\pm$)/m(e$^\pm$)(\emph{exp}) =  273.1321. We have,
 however, only considered the ratio of the rest masses of the neutrinos in
$\pi^\pm$  and e$^\pm$, not the consequences of either the charge
or the spin in $\pi^\pm$. Nambu's (improved) empirical
formula for the mass ratio m($\pi^\pm$)/m(e$^\pm$) is
\begin{equation} \frac{\mathrm{m}(\pi^\pm)}{\mathrm{m}(\mathrm{e}^\pm)}(emp)
= \frac{2}{\alpha_f}\,-\,1 = 273.07 =
0.99978\,\frac{\mathrm{m}(\pi^\pm)}{\mathrm{m}(\mathrm{e}^\pm)}(exp)\,.
 \end{equation}

  The rest mass of the $\pi^\pm$\,mesons is
273 $\cong$ $2\times1/\alpha_f$ times larger than the
rest mass of the electron because both, the muon neutrino as well
 as the  anti-muon neutrino masses in $\pi^\pm$, are $\cong\,$137
times larger than the masses of the electron or anti-electron
neutrinos in the electron or positron. If we divide Barut's Eq.(59)
for m($\mu^\pm$)/m(e$^\pm$) by Eq.(103) for the ratio
m($\pi^\pm$)/m(e$^\pm$), replacing +\,1 in (59) by +\,1.25, we arrive,
neglecting a term with $\alpha_f^2$, at m($\mu^\pm$)/m($\pi^\pm$) =
3/4 + $\alpha_f$,  a very good approximation of the experimental mass
ratio m($\mu^\pm$)/m($\pi^\pm$) in Eq.(60), as it must be.

\medskip
   For the $\pi^0$\,meson we find that
\begin{equation} \mathrm{m}(\pi^0)/\mathrm{m(e}^\pm)(emp) =
1.000\,268\,(2/\alpha\, - \,10) = 264.1426\,, \end{equation}
as it must be,  because
m($\pi^\pm$)\,$-$\,m($\pi^0$) = 0.9988\,$\cdot$\,9m(e$^\pm$),
and because of Eq.(103).
The masses of the $\gamma$-branch particles are, according to Eq.(1), in a
good approximation integer multiples of the mass of the $\pi^0$\,meson.
Their masses are consequently multiples of the right hand side of
Eq.(104). For example we find that m($\eta$)/m(e$^\pm$)(\emph{exp})
= 1.0144\,$\cdot$\,4\,$\cdot$\,(2/$\alpha$\,$-$\,10) and we have
m($\Lambda$)/m(e$^\pm$)(\emph{exp}) =
1.0335\,$\cdot$\,8\,$\cdot$\,(2/$\alpha$\,$-$\,10).

\medskip
   In order to determine m(n)/m(e$^\pm$) we start with K$^0$ =
(2.)$\pi^\pm$ + $\pi^\mp$ and E((2.)$\pi^\pm$) = 4E$_\nu(\pi^\pm$) +
N$^\prime$/2\,$\cdot$\,[m($\nu_\mu)$ + m$(\nu_e)$]c$^2$, Eq.(35),
and with m($\pi^\pm$) = N\,[m($\nu_\mu$) + m$(\nu_e$)], Eq.(34).
Then m(K$^0$) =  7N$^\prime$/2\,$\cdot$\,[m($\nu_\mu)$ + m($\nu_e$)],
 and with m(n) $\cong$ m(K$^0$ +  $\overline{{\mathrm{K}}^0}$) =
 2m(K$^0$)  follows that
\begin{equation} \frac{\mathrm{m(n)}}{\mathrm{m(e^\pm)}}(theor) =
14\,[\frac{\mathrm{m}(\nu_\mu)}{\mathrm{m}(\nu_e)} + 1] =
14/\alpha_f + 14 = 1932.5 =
1.051\frac{\mathrm{m(n})}{\mathrm{m(e^\pm)}}(exp)\,,
\end{equation}
with m(n)/m(e)(\emph{exp}) = 1838.68. But we have only
considered the mass of the neutrino lattice in the neutron, not the
consequences of the two positive and two negative elementary
charges in the neutron. The empirical value of m(n)/m(e$^\pm$)
is 14/$\alpha_f$ $-$ 0.9977\,$\cdot$\,80.

The ratio of the mass of the proton to the mass of the electron, for
which an explanation has been looked for since about a hundred years,
is m(p)/m(e)(\emph{exp}) = 1836.15. From our theoretical explanation
of the neutron follows that
\begin{equation} \frac{\mathrm{m(p)}}{\mathrm{m(e)}}(theor) \cong
 14\,[\frac{\mathrm{m}(\nu_\mu)}{\mathrm{m}(\nu_e)} + 1] - 5/2
= \frac{14}{\alpha_f} + 23/2 = 1930.00\,, \end{equation}
because the energy lost in the $\beta$-decay of the neutron is
1.29333\,MeV or 1.01239\,$\cdot$\,5/2\,$\cdot$\,m(e$^\pm)\,$c$^2$.
1930.0  is 1.051 times the experimental mass ratio 1836.15. We
have, again, considered only the mass of the neutrino lattice in the
proton, not the consequences of the three elementary electrical
charges in the proton.  The empirical formula for m(p)/m(e) is
 \begin{equation}
\mathrm{m(p)}/\mathrm{m(e)}(emp) = 14\,[1/\alpha_f - 6]  =
14/\alpha_f - 84 =  0.9901\,\mathrm{m(p)}/\mathrm{m(e)}(exp)\,.
\end{equation}

   The persistent appearance of the fine structure constant $\alpha_f$
in the leading term of the ratio of the masses of the particles to the
mass of the electron is a consequence of the preponderance of the
mass of the muon neutrinos in the lattices of the particles, following
the relation m($\nu_e$) = $\alpha_f\,\cdot$\,m($\nu_\mu$).
\medskip

   We arrive at a better agreement between the theoretical values
of m(e$^\pm$), m($\mu^\pm$)/m(e$^\pm$) and
m($\pi^\pm$)/m(e$^\pm$) and their actual values if
we introduce a modified oscillation energy of the electron
E$^\prime_\nu$(e$^\pm$) given by
\begin{equation} \mathrm{E}^\prime_\nu(\mathrm{e}^\pm) =
\mathrm{E}_\nu(\mathrm{e}^\pm)(1 + \alpha_f/2)\,.
\end{equation}
 With Eq.(83) we then have
\begin{equation} \mathrm{E}^\prime_\nu(\mathrm{e}^\pm) =
1.000\,206\,\mathrm{m(e^\pm)c}^2/2\,. \end{equation}
The agreement of E$^\prime_\nu$(e$^\pm$) given by Eq.(109) with
the actual m(e$^\pm$)c$^2$/2 is an order of magnitude better
than it was with Eq.(83).

   The modification of the oscillation energy applies only to the electron,
not to either $\mu^\pm$ or $\pi^\pm$, because according to Eq.(83)
E$_\nu$(e$^\pm$) is proportional to m($\pi^\pm$) and if m($\pi^\pm$)
were also modified by (1 + $\alpha_f$/2) then E$^\prime_\nu$($\pi^\pm$)
would be proportional to (1 + $\alpha_f$/2)$^2$ and would be a
worse approximation then Eq.(32).

   With m(e$^\pm$)$^\prime$ standing for the mass of the electron
calculated with the modified oscillation energy E$_\nu^\prime$(e$^\pm$),
and assuming that m(e$^\pm)^\prime$c$^2$ =
2E$_\nu^\prime$(e$^\pm$), as seems to be the case according to
Eq.(109), we obtain

\begin{equation} \frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)^\prime} =
\frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)(1 + \alpha_f/2)} \cong
\frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)}(1
\mathrm\,{-}\,\alpha_f/2)\,,
\end{equation}
and with Eq.(99) we have
\begin{equation} \frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)^\prime} =
( \frac{3}{2\alpha_f} + 2)(1\mathrm\,{-}\,\alpha_f/2) =
\frac{3}{2\alpha_f} + 1.25 \,\mathrm\,{-}\,\alpha_f\,,\end{equation}
which agrees remarkably well with the slightly modified form of Barut's
Eq.(59) in which the +\,1 is replaced by +\,1.25. So we have
\begin{equation}
\frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)^\prime}(theor) =
206.7967 =
1.000\,137\,\frac{\mathrm{m}(\mu^\pm)}{\mathrm{m(e}^\pm)}(exp)\,.
\end{equation}

   Similarly we obtain for m($\pi^\pm$)/m(e$^\pm$)$^\prime$ with Eq.(81)
\begin{equation} \frac{\mathrm{m}(\pi^\pm)}{\mathrm{m(e}^\pm)^\prime} =
\frac{2\mathrm{E}_\nu(\pi^\pm)}{2\mathrm{E}_\nu^\prime(\mathrm{e}^\pm)}
\cong \frac{2\mathrm{E}_\nu(\pi^\pm)(1\,\mathrm{-}\, \alpha_f/2)}
{\alpha_f/2\cdot2\mathrm{E}_\nu(\pi^\pm)}\,,
\end{equation}
or \begin{equation}
\frac{\mathrm{m}(\pi^\pm)}{\mathrm{m(e}^\pm)^\prime}(theor)
= \frac{2}{\alpha_f}(1\,\mathrm{-}\,\alpha_f/2) =
\frac{2}{\alpha_f}\,\mathrm{-}\,1
= 273.072\,,\end{equation}
whereas the experimental m($\pi^\pm$)/m(e$^\pm$) is 273.132 =
1.000\,22\,(2/$\alpha_f$\,$\mathrm{-}$\,1).

\medskip
   Our theoretical calculations of  m($\pi^\pm$)/m(e$^\pm$) and of
m($\mu^\pm$)/m(e$^\pm$) agree within the percent range with their
experimental values. This can only be if our explanation of m($\pi^\pm$),
m($\mu^\pm$)
and m(e$^\pm$) are correct in the same approximation, and if the ratio
m($\nu_e$) = $\alpha_f$m($\nu_\mu$) (Eq.72) is valid. In other words,
our theoretical values of
m($\pi^\pm$)/m(e$^\pm$) and of m($\mu^\pm$)/m(e$^\pm$)
\emph{confirm the \emph{validity} of our explanation of m($\pi^\pm$),
m($\mu^\pm$) and of m(e$^\pm$), as well as the validity of the relation
m($\nu_e$) = $\alpha_f$m($\nu_\mu$)}.

\section{The spin of the $\gamma$-branch particles}

  It appears to be crucial for the validity of a model of the elementary
particles that the model can also explain the spin of the particles
without additional assumptions. The spin or the intrinsic angular
momentum is, after the mass and charge, the third most important
property of the elementary particles. The standard model does
not explain the spin, the spin is imposed on the quarks.
As is well-known the spin of the electron
was discovered by Uhlenbeck and Goudsmit [59] more than 80
years ago. Later on it was established that the baryons have spin as
well, but not the mesons. We have proposed an explanation of the spin
of the particles in [67].  For current efforts to
understand the spin of the nucleon see Jaffe [68] and of the
spin structure of the $\Lambda$ baryon see G\"ockeler et al.\,[69].
Rivas has described his own model of the spin and other spin models
in his book [70]. The explanation of the spin requires an unambiguous
answer, the spin must be 0 or 1/2 or integer multiples thereof, nothing
else. At present we do not have an accepted explanation of the spin.

    For the explanation of the spin of the particles it seems
to be necessary to have
an explanation of the structure of the particles. The spin of
a particle is, of course, the sum of the angular momentum
 vectors of the waves in the particle, plus the sum of the spin
vectors of its  neutrinos and antineutrinos, plus the spin of the
electric charges which the particle carries. It is
 striking that the particles which consist of
a single mode do not have spin, as the $\pi^0,
\pi^\pm$ and $\eta$ mesons do, see Tables 1 and 2.
 It is also striking that particles whose mass is approximately
twice the mass of a smaller particle have spin 1/2 as is the
case with the $\Lambda$ baryon,
m($\Lambda$) $\approx$ 2m($\eta$),  and with the nucleon
 m(n) $\approx$ 2m(K$^\pm$) $\approx$ 2m(K$^0$). The $\Xi^0_c$
baryon which is a doublet of one mode has also spin 1/2. Composite
particles which consist of a doublet of one mode plus one or two other
single modes have spin 1/2, as the  $\Sigma^0$, $\Xi^0$\,
and $\Lambda_c^+$, $\Sigma_c^0$, $\Omega_c^0$ baryons do.
 The only particle which seems to be the triplet of a single mode,
 the $\Omega^-$ baryon with m($\Omega^-$) $\approx$ 3m($\eta$),
 has spin 3/2. It appears that the relation between
the spin and the modes of the particles is straightforward.

  In our model of the $\gamma$-branch particles the $\pi^0$
and $\eta$ mesons consist of N = 2.85$\cdot10^9$ standing
electromagnetic waves, each with its  own frequency.
Each of the electromagnetic waves in the $\pi^0$ and
$\eta$  mesons may have spin s = 1,
 because circularly polarized electromagnetic waves have an angular
momentum as was first suggested by Poynting [71] and verified by,
among others, Allen [72]. The creation of the $\pi^0$\,meson in the
reaction
$\gamma$ \,+\, p $\rightarrow \pi^0$ + p and conservation of angular
momentum dictates that the sum of the angular momentum
vectors of the N electromagnetic waves in the $\pi^0$\,meson must
be zero, $\sum_ij(s_i)$ = 0.  Either the sum of the spin vectors of
the electromagnetic waves in the $\pi^0$\,meson is zero, or each
electromagnetic wave in the $\pi^0$\,meson has zero spin which
would mean that they are linearly polarized. As is well-known, circularly
polarized electromagnetic waves have spin s = 1, or an
angular momentum. Superposing on
such a wave an electromagnetic wave with the same frequency
and same amplitude traveling in opposite direction, as is the case
for a standing electromagnetic wave, introduces an angular momentum
of the opposite direction. In a standing electromagnetic
 wave the angular momentum of both waves cancel.
Linearly polarized electromagnetic waves are not expected to
have angular momentum. That this is actually so was proven by
Allen [72].  Since the electromagnetic waves
 in the $\pi^0$ and $\eta$ mesons do not have an angular
momentum or since the sum of the spin vectors $s_i$ of the
circular electromagnetic  waves is zero, the intrinsic angular
momentum of the $\pi^0$ and $\eta$ mesons is zero, or
\begin{equation} j(\pi^0,\eta) =  \sum_{i} \,j(s_i) =
0 \quad (0\,\le\,i\le \mathrm{N})\,.\end{equation}
In this model the $\pi^0$ and $\eta$ mesons
do not have an intrinsic angular momentum or spin, as it must be.

  We now consider particles such as the $\Lambda$ baryon whose
 mass is m($\Lambda$) = 1.0190\,$\cdot$\,2m($\eta$). The $\Lambda$
baryon seems to consist of the superposition of two waves of equal
 amplitudes and of frequencies $\omega$ and $\mathrm{-}\,\omega$,
 $|\mathrm{-}\,\omega |$ = $\omega$, at each of the N points of the
lattice. The waves in such particles must be coupled what we have
marked in Tables 1,2 by the $\ast$ sign. The particles contain then
N circular waves, each with its own frequency and each
having an angular momentum of $\hbar$/2 as we will see.

  The superposition of two perpendicular linearly polarized traveling
waves of equal amplitudes and frequencies shifted in phase by
$\pi$/2 leads to a circular wave with the constant angular momentum
j = $\hbar$. As is well-known, the total energy of a traveling wave is
the sum of the potential and the kinetic energy. In a traveling wave the
kinetic energy is always equal to the potential energy. From

\begin {equation} \mathrm{E}_{pot} + \mathrm{E}_{kin} =
\mathrm{E}_{tot} = \hbar\omega\,,\end {equation}
\noindent
follows \begin{equation} \mathrm{E}_{tot} = 2\mathrm{E}_{kin} =
2\frac{\Theta\omega^2}{2}\, = \hbar\omega
,\end{equation}
\noindent
with the moment of inertia $\Theta$. It follows that the angular momentum
 is

\begin{equation} j = \Theta\omega = \hbar\,.\end{equation}
\noindent
   This applies to a traveling wave  and corresponds to spin s = 1, or
 to a circularly polarized photon.

  We now add to one monochromatic circular wave
with frequency $\omega$ a  second circular wave with
$\mathrm{-}\,\omega$  of the same absolute value as $\omega$ but
 shifted in phase by $\pi$,  having the same amplitude, as we have
done in [67]. Negative frequencies are permitted solutions of the
equations for the lattice oscillations, Eq.(7). In other words we consider
the circular waves

\begin{equation} x(t) = exp[i\omega t] + exp[-\,i(\omega t +
\pi)]\,,\end{equation}

\begin{equation} y(t) = exp[i(\omega t + \pi/2)] + exp[-\,i(\omega t +
  3\pi/2)]\,\,.\end{equation}
\noindent
This can also be written as
\begin{equation} x(t) = exp[i\omega t] - exp[-i\omega t]\,,\end{equation}

\begin{equation} y(t) = i\cdot(exp[i\omega t] + exp[-i\omega t]
)\,.\end{equation}
If we replace \emph{i} in the Eqs. above by $-$\,\emph{i} we have
a circular wave turning in opposite direction.The energy of the
superposition of the two waves is the sum of the
energies of both individual waves, and since in circular
waves E$_{kin}$ = E$_{pot}$  we have with Eq.(117)

\begin{equation} 4\mathrm{E}_{kin} = 4\Theta\omega^2/2 =
\mathrm{E}_{tot} = \hbar\omega\,,\end{equation}
\noindent
from which follows that the circular wave has an angular momentum

\begin{equation} j = \Theta\omega = \hbar/2\,. \end{equation}
\noindent
The superposition of two circular monochromatic  waves of equal
 amplitudes and frequencies $\omega$ and $\mathrm{-}\,\omega$
satisfies the necessary condition for spin \\s = 1/2 that the angular
momentum is  j = $\hbar$/2.

    Our model of the particles treats the $\Lambda$
baryon, which has spin s = 1/2 and a mass m($\Lambda$) =
1.0190\,$\cdot$\,2m($\eta$), as the superposition of two
particles  of the same type with N standing electromagnetic
waves. The waves are circular because they are the
superposition of two waves with the same absolute value
of the frequency and the same amplitude.  The angular
momentum  vectors  of all circular waves in the lattice cancel,
except for \emph{the wave at the center of the crystal}. Each
oscillation with frequency $\omega$ at $\phi$\,$>$\,0 has at its mirror
position $\phi$\,$<$\,0 a wave with the frequency $-\,\omega$,
which has a negative angular momentum, since j =
mr$^2\omega$  and $\omega = \omega_0\phi$.
Consequently the angular momentum vectors of both waves cancel.
The center of the lattice oscillates, as all other lattice points do, but
with the frequency $\nu$(0) which is determined by the longest
possible wavelength, which is twice the sidelength d of the lattice,
so $\nu$(0) = c/2d. As the other circular waves in the lattice the
circular wave at the center has the angular momentum $\hbar$/2
according to Eq.(124). The angular momentum of the center wave
is the only angular momentum which is not canceled by a
wave of opposite circulation. There are, as it must be, three
possible orientations of the axis of the circular waves.

    The net angular momentum of the N circular waves in the
lattice  reduces to the angular momentum of the center wave and
is $\hbar$/2.  Since the circular waves in the $\Lambda$ baryon
are the only possible contribution to an angular momentum the
intrinsic angular momentum of the $\Lambda$ baryon is $\hbar$/2 or
\begin{equation} j(\Lambda) = \sum_i\,j(\omega_i) = j(\omega_0) =
\hbar/2\,.\end{equation}
 We have thus explained that the $\Lambda$ and likewise
 the $\Xi_c^0$ baryon satisfy the necessary condition
that j = $\hbar$/2 for s = 1/2. The intrinsic angular momentum of the
$\Lambda$ baryon is the consequence of the superposition of two
circular waves of the same amplitude and the same absolute value
of the frequency. Spin 1/2 is caused by the composition of the
 particles, it is not a contributor to the mass of a particle.

   The other particles of the $\gamma$-branch, the
$\Sigma^0$, $\Xi^0$, $\Lambda^+_c$, $\Sigma^0_c$ and
$\Omega^0_c$ baryons are composites of a baryon with spin 1/2
plus one or two $\pi$\,mesons
 which do not have spin. Consequently the spin of these particles is
1/2. The spin of all particles of the $\gamma$-branch, exempting the
spin of the $\Omega^-$\,baryon, has thus been explained.
 For an explanation of  s($\Sigma^{\pm,0}$)  = 1/2, of s($\Xi^{-,0}$)
 = 1/2 and of s($\Xi_c^{0,+})$  = 1/2, regardless
 whether the particles  are charged or neutral, we refer to [67]. In these
cases the elementary electric charge does not seem to be added as
an electron or positron to the neutral baryons, but rather through either
$\pi^-$ or $\pi^+$\,mesons, which do not not have spin. The presence
of the $\pi^\pm$\,mesons in the charged versions of $\Sigma^0$, $\Xi^0$
and $\Xi^0_c$ is documented by the appearance of  $\pi^\pm$\,mesons
in the decay products of $\Sigma^{\pm}$, $\Xi^{-}$  and $\Xi_c^{+}$,
whereas in the decays of $\Sigma^0$ and $\Xi^0$
charged mesons do not appear.

\section{The spin of the particles of the $\nu$-branch}

The characteristic particles of the neutrino-branch are the
$\pi^\pm$\,mesons which have zero spin. At first glance it seems
to be odd that the $\pi^\pm$\,mesons do not have spin, because
 it seems that the $\pi^\pm$\,mesons should have spin 1/2 from
the spin of the charges e$^\pm$ in $\pi^\pm$. What happens to
 the spin of e$^\pm$ in $\pi^\pm$\,? The solution of
 this puzzle is in the composition of the $\pi^\pm$\,mesons which
are, in our model of the particles, made of a lattice of
neutrinos and antineutrinos (Fig.\,2) each having spin 1/2,
the lattice oscillations, and an elementary electric charge.

   There is a fundamental difference between
the $\pi^0$ and $\pi^\pm$\,mesons. All lattice points
 in $\pi^\pm$ have spin per se, because neutrinos have spin,
whereas in the $\pi^0$\,meson the lattice points do not have
spin because they are standing electromagnetic waves.
 Longitudinal oscillations in the neutrino lattice of the
$\pi^\pm$\,mesons do not cause an angular momentum,
 $\sum_i\,j(\nu_i)$ = 0, because for longitudinal oscillations
$\vec{r}\,\times\,\vec{p}$  =  0. In the
cubic lattice of N = O($10^9$) neutrinos and
antineutrinos of the $\pi^\pm$\,mesons the spin of nearly
 all neutrinos and antineutrinos must cancel because conservation of
angular momentum during the creation of the $\pi^\pm$\,mesons
requires that the total angular momentum around a central axis is $\hbar$/2.
In fact the spin vectors of all but the neutrino or antineutrino in the
center of the lattice cancel. In order for this to be so
 the spin vector of any particular neutrino in the lattice
has to be opposite to the spin vector of the neutrino at its
mirror position. As is well-known only left-handed neutrinos and
right-handed antineutrinos exist. From $\nu$ = $\nu_0\phi$ (Eq.14) follows
that the direction of motion of the neutrinos in e.g.\,\,the upper right
quadrant ($\phi$\,$>$\,0) is opposite to the direction of motion in the
lower left quadrant ($\phi$\,$<$\,0). Consequently the spin vectors of all
neutrinos or antineutrinos in opposite quadrants are opposite and cancel.
 The only angular
momentum remaining from the spin of the neutrinos of the lattice is the
angular momentum of the neutrino or antineutrino at the \emph{center of
the lattice} which does not have a mirror particle.
Consequently the electrically neutral neutrino lattice consisting of
N$^\prime$/2 neutrinos and N$^\prime$/2 antineutrinos and the center
 particle, each with spin j($n_i$) = 1/2, has an intrinsic
angular momentum j = $\sum_i\,j(n_i)$ =
\emph{j(n$_0$}) = $\hbar$/2.

   But electrons or positrons added to the neutral neutrino
lattice have likewise spin 1/2. If the spin of the electron or positron
added to the neutrino lattice is opposite to the spin of the neutrino
or antineutrino in the center of the lattice then the net spin of the
 $\pi^+$ or $\pi^-$ mesons is zero, or
 \begin{equation} j(\pi^\pm) = \sum_i\,j(\nu_i) +\sum_i\,j(n_i) +
 j(\mathrm{e}^\pm) = j(n_0) + j(\mathrm{e}^\pm) = 0
\,\quad(0\leq i \leq \mathrm{N})\,.\end{equation}
It is important for the understanding of the structure of
the $\pi^\pm$\,mesons to realize that s($\pi^\pm$) = 0 can only be
explained if the $\pi^\pm$\,mesons consist of a \emph{neutrino lattice}
to which an electron or positron is added whose spin is opposite to the
net spin of the neutrino lattice. \emph{Spin 1/2 of the elementary
electric  charge can only be canceled by something that has also spin
1/2, and in $\pi^\pm$ the only conventional choice for that is a single neutrino}.

   \emph{The spin, the mass and the decay of $\pi^\pm$ require that
the $\pi^\pm$\,mesons are made of a cubic neutrino lattice and an
elementary electric charge} e$^\pm$.

   The spin of the K$^{\pm}$\,mesons is zero. With the spin of the
 K$^\pm$\,mesons we
encounter the same oddity we have just observed with the spin of the
$\pi^\pm$\,mesons, namely we have a particle which carries an
elementary electric charge with spin
1/2, and nevertheless the particle does not have spin. The explanation
of s(K$^\pm$) = 0 follows the same lines as the explanation of
the spin of the $\pi^\pm$\,mesons. In our model the
 K$^\pm$\,mesons are described by the state (2.)$\pi^\pm$ + $\pi^0$,
 that means by the second mode of the lattice oscillations
of the $\pi^\pm$\,mesons plus a $\pi^0$\,meson. The second mode
 of the longitudinal oscillations of a neutral neutrino lattice does
not have a net intrinsic angular momentum $\sum_i\,j(\nu_i)$ = 0. But
 the spin of the neutrinos contributes an angular momentum $\hbar$/2,
which originates from the neutrino or antineutrino in the center of the
lattice, just as it is with the neutrino lattice in the
$\pi^\pm$\,mesons, so $\sum_i\,{ j(n_i)} = j(n_0) = \hbar/2$. Adding
an elementary electric charge with a spin opposite to the net intrinsic
angular momentum of the neutrino lattice creates the charged
(2.)$\pi^\pm$ mode which has zero spin
\begin{equation}j((2.)\pi^\pm) = \sum_i\,{j(n_i)} + j(\mathrm{e}^\pm)
= j(n_0) + j(\mathrm{e}^\pm) = 0\,.\end{equation}
 As discussed in Section 6 it is
necessary to add a $\pi^0$\,meson to the second mode of the
$\pi^\pm$\,mesons in
order to obtain the correct mass and the correct decays of the
K$^\pm$\,mesons. Since the $\pi^0$\,meson does not
have spin the addition of the $\pi^0$\,meson does not add to the
intrinsic angular momentum of the K$^\pm$\,mesons. So
 s(K$^\pm$) = 0 as it must be.

  The explanation of s = 0 of the K$^0$ and
$\overline{{\mathrm{K}}^0}$\,mesons described by the
state (2.)$\pi^\pm$ + $\pi^\mp$  is different because there is
now no electric charge whose spin could cancel the spin of the
neutrino lattice.  The longitudinal oscillations of the second mode
of the neutrino oscillations of $\pi^\pm$  in K$^0$ as well
 as of the basic $\pi^\mp$ mode do not create an angular momentum,
$\sum_i\,j(\nu_i)$ = 0. The sum of the spin vectors of the neutrinos in
K$^0$ and $\overline{{\mathrm{K}}^0}$ is determined by the
neutrinos in the second mode of the $\pi^\pm$\,mesons,
or the (2.)$\pi^\pm$ state, and the basic  $\pi^\mp$ mode,
each have N$^\prime$/2 neutrinos and
N$^\prime$/2 antineutrinos plus a center neutrino or antineutrino, so
the number of all neutrinos and antineutrinos in the sum of both states,
the K$^0$,$\overline{{\mathrm{K}}^0}$\,mesons, is 2N. Since the
size of the lattice of the K$^\pm$\,mesons and the
 K$^0$\,mesons is the same it follows that two neutrinos are at each
lattice point of the K$^0$ or $\overline{{\mathrm{K}}^0}$\,mesons.
We assume that Pauli's exclusion principle applies to neutrinos
as well. Consequently each neutrino at each lattice point must
share its location with an antineutrino. That means that the contribution
of the spin of all neutrinos and antineutrinos to the intrinsic angular
momentum of the K$^0$\,meson is zero or $\sum_i\,j(2n_i)$ = 0.
The sum of the spin vectors of the two opposite charges in
either the K$^0$  or the  $\overline{{\mathrm{K}}^0}$\,mesons,
or in the (2.)$\pi^\pm$ + $\pi^\mp$ state, is also
zero. Since neither the lattice oscillations nor the spin of the
neutrinos and antineutrinos
nor the  electric charges contribute an angular momentum
\begin{equation} j(\mathrm{K}^0) = \sum_i\,j(\nu_i) +
\sum_i\,j(2n_i) + j(\mathrm{e}^+ + \mathrm{e}^-) = 0\,. \end{equation}
 The intrinsic angular momentum
of the K$^0$ and $\overline{{\mathrm{K}}^0}$\,mesons
 is zero, or s(K$^0$,$\overline{{\mathrm{K}}^0}$) = 0, as it
must be. In simple terms since, e.g., the structure of K$^0$ is
(2.)$\pi^+$ + $\pi^-$, the spin of K$^0$ is the sum of the spin of
(2.)$\pi^+$ and of $\pi^-$, both of which do not have spin.
It does not seem possible to arrive at
 s(K$^0$,$\overline{{\mathrm{K}}^0}$) = 0  if both particles
do not contain the N pairs of neutrinos and antineutrinos required by the
 (2.)$\pi^\pm$ + $\pi^\mp$ state which we have suggested in Section 6.

   In the case of the neutron one
must wonder how it comes about that a particle which seems to be the
superposition of two particles without spin ends up with spin 1/2. The
neutron,
 which has  a mass $\approx$ 2m(K$^\pm$) or 2m(K$^0$), is either
the superposition of a K$^+$ and a K$^-$ meson or of a K$^0$ and a
$\overline{\mathrm{K}^0}$ meson. The intrinsic angular
momentum of the superposition of K$^+$ and K$^-$ is either 0 or $\hbar$,
which means that the neutron cannot be the superposition of K$^+$
and K$^-$. For a proof of this statement we refer to [67].

   On the other hand the neutron can be the superposition of a
K$^0$  and a $\overline{\mathrm{K}^0}$\,meson. A significant change
in the lattice occurs when a K$^0$ and a  $\overline{\mathrm{K}^0}$ meson
 are superposed. Since each K$^0$\,meson contains N neutrinos
and N antineutrinos, as we discussed
in context with the spin of K$^0$, the number of all
neutrinos and antineutrinos in superposed K$^0$  and
$\overline{\mathrm{K}^0}$ lattices is 4N. Since the size of the
lattice of the proton as well of the neutron is the same as the
size of K$^0$ each of the N lattice
points of the neutron now contains four neutrinos, a muon neutrino
and an anti-muon neutrino as well as an
electron neutrino and an anti-electron neutrino. The
$\nu_\mu,\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$ quadrupoles
oscillate just like individual neutrinos do because we learned from
Eq.(7) that the ratios of the oscillation frequencies are independent
of the mass as well as of the interaction constant between the lattice
points. In the neutrino quadrupoles the spin of the neutrinos and
antineutrinos cancels, $\sum_i\,j(4n_i)$ = 0. The superposition of
two circular neutrino lattice oscillations,  that means the circular
oscillations of frequency $\omega_i$,  contribute as before
the angular momentum of the center circular oscillation, so
$\sum_i\,j(\omega_i) = j(\omega_0) = \hbar/2$. The spin
 and charge of the four electrical charges e$^+$e$^-$e$^+$e$^-$
 hidden in the sum of the K$^0$ and $\overline{\mathrm{K}^0}$
mesons cancel, j(4e$^\pm$) = 0. It follows
 that the intrinsic angular momentum of a neutron created by the
superposition of  a K$^0$  and a $\overline{\mathrm{K}^0}$ meson
comes from the circular neutrino lattice oscillations only and is
\begin{equation} j(\mathrm{n}) = \sum_i\,j(\omega_i) + \sum_i\,j(4n_i) +
 j(4\mathrm{e}^\pm) =  \sum_i\,j(\omega_i) = j(\omega_0) = \hbar/2\,,
\end{equation}
as it must be. In simple terms, the spin of the neutron originates from
the superposition of two circular neutrino lattice
oscillations  with the frequencies $\omega$ and $\mathrm{-}\,\omega$
 shifted in phase by $\pi$ at all lattice points. From those only the angular
momentum $\hbar$/2 of the oscillation at the center of the lattice remains.

   The spin of the proton is 1/2 and is unambiguously defined by the decay
of the neutron n $\rightarrow$ p + e$^-$ + $\bar{\nu}_e$. We have
suggested in Section 10 that 3/4$\cdot$N$^\prime$ anti-electron neutrinos
of the neutrino lattice of the neutron are removed in
 the $\beta$-decay of the neutron and that the other N$^\prime$/4
 anti-electron neutrinos leave with the emitted electron. The intrinsic
angular momentum of the proton originates then from the spin of the
central $\nu_\mu\bar{\nu}_\mu\nu_e$ triplet, from the spin of
 the e$^+$e$^-$e$^+$ triplet which is part of the remains of the neutron,
 and from the angular momentum of the center of the lattice oscillations
with the superposition of two circular oscillations. The spin of the
central $\nu_\mu\bar{\nu}_\mu\nu_e$ triplet
is canceled by the spin of the e$^+$e$^-$e$^+$ triplet. The intrinsic
angular momentum of the proton is

\begin{equation}
 j(\mathrm{p}) = j(\nu_\mu\bar{\nu}_\mu\nu_e)_0\,  +
j{(\mathrm{e}^+\mathrm{e}^-\mathrm{e}^+})
+ j(\omega_0) = j(\omega_0) = \hbar/2\,,  \end{equation}
\noindent as it must be.

   The other mesons of the neutrino branch, the D$^{\pm,0}$ and
D$_s^\pm$ mesons, both having zero spin, are
superpositions of a proton and an antineutron of opposite spin, or of
their antiparticles, or of a neutron and an antineutron of opposite spin
in D$^0$.  The spin of D$^\pm$ and D$^0$ does therefore not
pose a new problem.

   For an explanation of the spin of $\mu^\pm$ we refer to [73].
Since  all muon or anti-muon
neutrinos have been removed from the $\pi^\pm$ lattice in the
$\pi^\pm$ decay it follows that a neutrino vacancy is at the center of
the $\mu^\pm$ lattice (Fig.\,7). Without a neutrino in the center of
 the lattice the sum of the spin vectors of all neutrinos in the
$\mu^\pm$ lattice is zero. However the $\mu^\pm$\,mesons consist of
the neutrino lattice plus an electric charge e$^\pm$ whose spin is 1/2.
The spin of  the $\mu^\pm$\,mesons originates from the spin of the
elementary electric charge carried
by the $\mu^\pm$\,mesons and is consequently s($\mu^\pm$) = 1/2.
The same considerations apply for the spin of $\tau^\pm$,
 s($\tau^\pm$) = 1/2.

   An explanation of the spin of the mesons and baryons can only be
valid if the same explanation also applies to the antiparticles of these
particles whose spin is the $\emph{same}$ as that of the ordinary
particles. The antiparticles of the $\gamma$-branch consist of
electromagnetic waves whose frequencies differ from the frequencies
of the ordinary particles only by their sign.
The angular momentum of the superposition of two circular oscillations
with $\mathrm{-}\,\omega$ and $\omega$  has the same angular
momentum as the superposition of two circular oscillations
 with frequencies of opposite  sign, as in $\Lambda$. Consequently
the spin of the antiparticles of the $\gamma$-branch is the same as
the spin of the ordinary
particles of the $\gamma$-branch. The same considerations apply
to the circular neutrino lattice oscillations which cause the spin of the
neutron, the only particle
of the $\nu$-branch which has spin. In our model of the particles
the spin of the neutron and the anti-neutron is the same.

   Let us summarize: The spin of the particles of the $\nu$-branch
originates from the angular
momentum vectors of the lattice oscillations and the spin vectors
of the neutrinos in the particles and the spin vector
of the elementary electric charge or charges a particle carries. The
contribution of all or all but one of the O($10^9$) oscillations and
O($10^9)$  neutrinos to the intrinsic
angular momentum of the particles must cancel, otherwise the
spin cannot be either 0 or 1/2. It requires the symmetry of a cubic
lattice for this to happen. The center of the lattices alone determines
the intrinsic angular momentum of the oscillations and neutrinos in the
lattice.
Adding to that the spin vector of one (or more) elementary electric
charges with spin 1/2 and we arrive at the total intrinsic angular
momentum of a particle. The most illuminating case are the
$\pi^\pm$\,mesons which do not have spin although they carry
an elementary electric charge. Actually the neutrino lattice
of the $\pi^\pm$\,mesons has spin 1/2 from its central neutrino,
but this spin vector is canceled by the spin of the electric charge
e$^\pm$, so s($\pi^\pm$) = 0.

   From the foregoing we arrive also at an understanding of the reason for
the astonishing fact that the intrinsic angular momentum or spin of the
particles is independent of the mass of the particles,
as exemplified by the spin $\hbar$/2 of the electron being the
same as the spin $\hbar/2$ of the proton, notwithstanding the fact that
 the mass of the proton is 1836 times larger than the mass of the
electron. However,
in our model, the spin of the particles including the electron
is determined solely by the angular momentum $\hbar$/2 at the
 center of the lattice, the other angular momentum vectors in the
particles cancel. The spin is independent of the
number of the lattice points in a cubic lattice. Hence the mass
of the particles  in the other
$10^9$  lattice points is inconsequential for the intrinsic angular
 momentum of the particles.  In this model of the particles the spin
is independent of the mass of the particles, as it must be.

\section{The spin and magnetic moment \\ of the electron}

    The model of the electron we have proposed in Section 11
 has, in order to be valid, to pass a crucial test; the model has
to explain satisfactorily the spin and the magnetic moment of the
 electron. When Uhlenbeck and Goudsmit [59] (U\&G) discovered
 the existence of the spin of the electron they also proposed that the
electron has a magnetic moment with a value equal to Bohr's
magnetic moment $\mu_B$ = e$\hbar$/2m(e$^\pm$)c. Bohr's
magnetic moment results from the motion of an electron on a circular
orbit around a proton. The magnetic moment of the electron postulated
by U\&G has been confirmed experimentally, but has been corrected
by  about  0.11\% for the anomalous magnetic moment.
If one tries to explain the magnetic moment of the electron
  with an electric charge moving on a circular orbit around the
particle center, analogous to the magnetic moment of hydrogen, one ends
up with velocities larger than the velocity of light, which cannot be, as
already noted by U\&G. It remains to be explained how the magnetic
 moment of the electron comes about.

   We will have to explain the spin of the electron first. The spin, or
the intrinsic angular momentum of a particle is, of course, the sum of the
angular momentum vectors of all components of the particle. In the
 electron these are the neutrinos and the electric oscillations. Each
neutrino has spin 1/2 and in order for the electron to have
s = 1/2 all, or all but one, of the spin vectors of the neutrinos in their
lattice must cancel.
If the neutrinos are in a simple cubic lattice as in Fig.\,8 and the
 center particle of the lattice is not  a neutrino, as in Fig.\,8, the
spin vectors of all neutrinos cancel, $\Sigma\,j(n_i)$ = 0, provided
that the spin vectors of the
 neutrinos of the lattice point in opposite direction at their mirror
 points in the lattice. Otherwise the spin vectors of the neutrinos
would add up and make a very large angular momentum. We
follow here the procedure we used in [73] to explain the spin
of the muons. The spin vectors of all electron neutrinos in the electron
cancel, just as the spin vectors of all muon and electron neutrinos
 in the muons cancel, because there is
 a neutrino vacancy at the center of their lattices, Figs.\,7,8.

   We will now see whether the electric oscillations in the electron
contribute to its angular
 momentum. As we said in context with Eq.(81) there must be two
times as many electric oscillations in the electron lattice than there are
neutrinos. The oscillations can either be standing waves consisting
of two linear oscillations or two circular oscillations  with the
frequencies $\omega$ and $-\,\omega$.  Both the standing
 waves and the circular oscillations must be non-progressive in order to
be part of the \emph{rest mass} of a particle. We will now assume that
the electric oscillation are circular.  Circular oscillations have an angular
momentum $\vec{j} = m\,\vec{r}\times\vec{v}$. And, as in the case
of the spin vectors of the neutrinos, all or all but one of the O$(10^9)$
angular momentum vectors of the electric oscillations must cancel in
order for the electron to have spin 1/2. We describe the
superposition of the two oscillations by

\begin{equation} x(t) = exp[i\omega t] + exp[-\,i(\omega t +
\pi)]\,,\end{equation}

\begin{equation} y(t) = exp[i(\omega t + \pi/2)] + exp[-\,i(\omega t +
                 3\pi/2)]\,\,,
\end{equation}

\noindent
that means by the superposition of an oscillation with the
frequency $\omega$ and a second oscillation with the frequency
 $\mathrm{-}\,\omega$ as in Eqs.(119,120). The oscillation with
$\mathrm{-}\,\omega$ is shifted in phase by $\pi$.
 Negative frequencies are permitted solutions of the equations
of motion in a cubic lattice, Eqs.(7,14). As is well-known
 oscillating electric charges should emit radiation. However,
 this rule does already not hold in the hydrogen atom, so we
will assume that the rule does not hold within the electron either.

   In circular oscillations the kinetic energy is always equal to the
potential energy  and the sum of both is the total energy. From
\begin{equation} \mathrm{E}_{pot} + \mathrm{E}_{kin} =
 2\,\mathrm{E}_{kin} = \mathrm{E}_{tot} \end{equation}
follows, as discussed in Eqs.(123,124), that the angular
 momentum of the superposition of the two circular oscillations is
\begin{equation} j = \Theta\,\omega = \hbar/2\,. \end{equation}
That means that each of the O$(10^9$) superposed circular
electric oscillations has an angular momentum  $\hbar$/2.

   The circulation of the circular oscillations in Eqs.(131,132) is
opposite
for all $\phi$ of opposite sign. It follows from the equation for
the displacements u$_n$  of the lattice points Eq.(5)
\begin{equation} u_n = Ae^{i(\omega\,t\, +\, n\phi)}\,,
\end{equation}
\noindent
 that the velocities of the lattice points are given by
\begin{equation} v_n = \dot{u}_n = i\,\omega_n\,u_n\,. \end{equation}
The sign of $\omega_n$ changes with the sign of $\phi$ because
the frequencies are given by Eq.(14), that means by
\begin{equation} \omega_n = \pm\,\omega_0(\,\phi_n + \phi_0\,)\,.
\end{equation}

Consequently the circulation of the electric oscillations is opposite
to the circulation at the mirror
 points in the lattice and the angular momentum vectors cancel,
but for the angular momentum vector of the electric oscillation at
the\emph{ center of the lattice}. The center circular oscillation has,
as all other electric oscillations, the angular momentum  $\hbar$/2
as Eq.(134) says. Since the spin vectors n$_i$  of the neutrinos
in the electron cancel the angular momentum of the entire electron
 lattice is
\begin{equation} j(\mathrm{e}^\pm) = \sum_i\,j(n_i) + \sum_i\,j(el_i) =
 j(el_0) =\hbar/2\,,
\end{equation}
as it must be for spin s = 1/2. The explanation of the spin of the
electron
given here follows the explanation of the spin of the baryons, as well
as the explanation of the absence of spin in the mesons. A valid
explanation of the spin must be applicable to all particles, in particular
to the electron, the prototype of a particle with spin.

   We will now turn to the magnetic moment of the electron which is
 known with extraordinary accuracy, $\mu_e$  =
1.001\,159\,652\,186\,$\mu_B$, according to the Review of Particle
Physics [2], $\mu_B$ being Bohr's magneton. The decimals after
1.00\,$\mu_B$  are caused by the anomalous magnetic moment which
we will not consider. As is well-known the magnetic dipole moment of
a particle with spin is, in Gaussian units, given by
\begin{equation} \vec{\mu} = g\,\frac{e\hbar}{2mc}\,\vec{s}\,,
 \end{equation}
where g is the dimensionless Land\'{e} factor, m the rest mass of
the particle that carries the charge e and $\vec{s}$  the spin vector.
The g-factor has been introduced in order to bring the magnetic moment
 of the electron into agreement with the experimental facts. As U\&G
 postulated and as has been confirmed experimentally the
g-factor of the electron is 2. With the spin s = 1/2  and g = 2
the magnetic dipole moment of the electron is then
\begin{equation} \mu_e = e\hbar/2m(e^\pm)c\,,
 \end{equation}
or one Bohr magneton $\mu_B$  in agreement with the experiments,
neglecting the anomalous moment. For a structureless point particle
Dirac [60] has explained why g = 2 for the electron. However
we consider here an electron with a finite size and which is at rest.
 When it  is at rest the electron has still its magnetic moment. Dirac's
theory does therefore not apply here. In order to arrive at an
explanation of the magnetic moment of the electron it will be necessary
to understand the structure of the electron and the origin of the spin.

  The only part of Eq.(140) that can be changed in order to explain the
g-factor of an electron with structure is the ratio e/m which deals with
the spatial distribution of charge and mass. If part of the mass of the
electron is non-electromagnetic and the non-electromagnetic part of
the mass does
not contribute to the magnetic moment of the electron, which to all
that we know is true for neutrinos, then the ratio e/m in Eq.(139) is not
 e/m(e$^\pm$)  in the case of the electron. The elementary charge e
certainly remains unchanged, but  e/m depends on what fraction of the
mass of the electron is of electromagnetic origin and what fraction of the
mass is  non-electromagnetic.  Only the current, not the mass of a
current loop, determines the magnetic moment of a loop. From the very
accurately known values of $\alpha_f$, m($\pi^\pm$)c$^2$
and m(e$^\pm)$c$^2$ and from Eq.(83)
for the energy in the electric oscillations in the electron
$\mathrm{E}_\nu$(e$^\pm$) =
$\alpha_f$/2\,$\cdot$\,m($\pi^\pm$)c$^2$/2 =
0.996570\,m(e$^\pm$)c$^2$/2 follows
that  very nearly one half of the mass of the electron is of electric
origin, the other half of m(e$^\pm$)  is made of neutrinos, Eq.(77),
and neutrinos do not contribute to the magnetic moment. That means
that in the electron the mass that carries the charge e  is approximately
m(e$^\pm$)/2. The magnetic moment of the electron is then
\begin{equation}
\vec{\mu}_e = g \frac{e\hbar}{2m(e^\pm)
/2\cdot c}\vec{s}\,, \end{equation}
and with s = 1/2 we have $\mu_e$ = g\,$e\hbar$/$2m(e^\pm)c$.
Because of Eq.(140) the g-factor must be equal to one and is
 unnecessary. What we have found is akin to what Perkins [23, p.320]
states as follows: ``The magnetic moment of a charged particle depends
on the ratio e/m and thus, classically, for a rotating structure, on the spatial
distributions of charge and mass. \emph{ If the two distributions are the
same, a value g = 1 is obtained on classical argu-
ments}."(emphasis added).

   In other words, if exactly one half of the mass of the
electron consists of neutrinos, then it follows automatically that
the electron has the correct magnetic moment
$\mu_e$ = $e\hbar/2m(e^\pm)c$ without the artificial
g-factor. For the explanation of the magnetic moment of the
muon see Appendix E.

\section*{Conclusions}

   We conclude that this model of the particles solves a number of
problems for which an answer heretofore has been hard to come by.
From the creation of the $\pi^0$\,meson and its decay into
$\gamma\gamma$  after 10$^{-16}$ sec follows that the $\pi^0$\,meson
and the other particles of the $\gamma$-branch must consist of standing
electromagnetic waves. The rest masses of the particles of the
$\gamma$-branch obey the integer multiple rule.
From the creation of the $\pi^\pm$\,mesons and their decay
follows that the $\pi^\pm$\,mesons and the other particles of
the $\nu$-branch must consist of a lattice of
$\nu_\mu,\bar{\nu}_\mu,\nu_e,\bar{\nu}_e$ neutrinos, their
oscillation energies and one or more elementary electric charges.
 From the explanation of the $\pi^\pm$\,mesons follows
that the rest mass of the $\mu^\pm$\,mesons is $\cong$ 3/4
of the rest mass of the $\pi^\pm$\,mesons and it also follows
that 1/2 of the rest mass of the electron or
positron must consist of electron neutrinos or anti-electron neutrinos.
The other half of the rest mass of the electron is in the energy of
electric oscillations. The elementary electric charge of the electron is
the sum of the charges carried by the individual electric oscillations.
The rest mass of the $\mu^\pm$\,mesons is so much larger than
the rest mass of the electron or positron because the rest mass of
the muon neutrinos is so much larger than the rest mass of the
electron or anti-electron neutrinos. We can determine the strength of
the weak force which holds the nuclear lattice together, and explain
the strong force between two elementary particles.
We have verified the validity of
our explanation of m(e$^\pm$), m($\mu^\pm$) and m($\pi^\pm$)
and of the relation m($\nu_e$) = $\alpha_f$\,$\cdot$\,m($\nu_\mu$)
by showing that the calculated ratios m($\mu^\pm$)/m(e$^\pm$)
and m($\pi^\pm$)/m(e$^\pm$) agree within 1\% with the
experimental values of these mass ratios.
Only photons, neutrinos, charge and the weak nuclear force are
needed to explain the  masses of the electron, the muon and of the
stable mesons and baryons and their antiparticles. We can explain
the  spin of the baryons and the absence of spin in the mesons, and
the spin of the electron and muon  as well; without making
 additional assumptions. We have also determined the rest masses
of the electron neutrino, the muon neutrino and the tau neutrino and
found that the mass of the electron neutrino is equal to the mass of
the muon neutrino times the fine structure constant. Our model is based
on the pioneering studies of lattice theory by Born and his students,
Born and v.\,Karman, Born and Land\'{e} and Born and Stern.

\vspace{1cm}

   {\bfseries Acknowledgments}. I gratefully acknowledge the contributions

of
Dr. T. Koschmieder to this study. I thank Professor A. Gsponer \\
\indent for information about the history of the integer
multiple rule.

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[54] Poincar\'{e},\,H. 1905. Compt.Rend. {\bfseries 140},1504. \\
\indent
Translated in:\\
\indent \,\,Logunov,\,A.A. 2001. \emph{On The Articles by Henri Poincare\\
\indent``On The Dynamics of the Electron"}. Dubna JINR.

\smallskip
\noindent
[55] Ehrenfest,\,P. 1907. Ann.Phys. {\bfseries 23},204.

\smallskip
\noindent
[56] Einstein,\,A. 1919. Sitzungsber.Preuss.Akad.Wiss. {\bfseries20},349.

\smallskip
\noindent
[57] Pauli,\,W. 1921. \emph{Relativit\"{a}tstheorie}. B.G. Teubner.\\
\indent
Translated in: \,\,1958. Theory of Relativity. Pergamon Press.

\smallskip
\noindent
[58] Poincar\'{e},\,H. 1906. Rend.Circ.Mat.Palermo {\bfseries21},129.

\smallskip
\noindent
[59] Uhlenbeck,\,G.E. and Goudsmit,\,S. 1925. Naturwiss. {\bfseries13},953 .

\smallskip
\noindent
[60] Dirac,\,P.A.M. 1928. Proc.Roy.Soc.London A{\bfseries117},610.

\smallskip
\noindent
[61] Gottfried,\,K. and Weisskopf,\,V.F. 1984. \emph{Concepts of Particle\\
\indent
Physics}. \,\,Vol.1,\,p.38. Oxford University Press.

\smallskip
\noindent
[62] Schr\"{o}dinger,\,E. 1930. Sitzungsber.Preuss.Akad.Wiss.
{\bfseries24},418.

\smallskip
\noindent
[63] Dirac,\,P.A.M. 1962. Proc.Roy.Soc. {\bfseries268},57

\smallskip
\noindent
[64] Mac\,Gregor,\,\,M.H. 1992. \emph{The Enigmatic Electron}. Kluwer.

\smallskip
\noindent
[65] Bender,\,D. et al. 1984. Phys.Rev. D {\bfseries30},515.

\smallskip
\noindent
[66] Koschmieder,\,E.L. 2005. http://arXiv.org/physics/0503206.

\smallskip
\noindent
[67] Koschmieder,\,E.L. 2003. Hadr.J. {\bfseries26},471,\\
\indent\,\,2003. http://arXiv.org/physics/0301060.

\smallskip
\noindent
[68] Jaffe,\,R.L. 2001. Phil.Trans.Roy.Soc.Lond.A {\bfseries359},391.

\smallskip
\noindent
[69] G\"ockeler,\,M. et al. 2002. Phys.Lett.B {\bfseries545},112.

\smallskip
\noindent
[70] Rivas,\,M. 2001. \emph{Kinematic Theory of Spinning Particles}.
Kluwer.

\smallskip
\noindent
[71] Poynting,\,J.H. 1909. Proc.Roy.Soc.A {\bfseries82},560.

\smallskip
\noindent
[72] Allen,\,J.P. 1966. Am.J.Phys. {\bfseries34},1185.

\smallskip
\noindent
[73] Koschmieder,\,E.L. 2003. http://arXiv.org/physics/0308069,\\
\indent\,\,2005. \emph{Muons: New Research},\,1. Nova.

\smallskip
\noindent
[74] Palazzi,\,P. $<$http://particlez.org/p3a/abstract/2004-001.html$>$.

\smallskip
\noindent
[75] Ablikim,\,M. et al. 2007. http://arXiv.org/abs/0705.4500.

\section*{Appendix A}

\bigskip

\begin{center}
\large
{The integer multiple rule}
\normalsize
\end{center}

\medskip
\noindent
When the integer multiple rule was first proposed [12]
we relied on the data available at that time, Barnett et al. Rev.Mod.Phys.
 {\bfseries68},611,1996. Since then the data on the particle masses have
multiplied and become more precise. It seems now that particles with
masses even 70\,$\times$\,m($\pi^0$) follow the integer multiple rule.
A list of the particles which obey the integer multiple rule
 is given in the following Table 3.

\begin{table}\caption{The particles following the integer multiple rule}
\begin{tabular}{lcll|lcll}\\
\hline\hline\\
&J&m/m($\pi^0$) & multiples & &J&m/m($\pi^0$) & multiples\\
[0.5ex]\hline\\

$\pi^0$ & 0&1.0000 & 1.0000\,$\cdot$\,1$\pi^0$ &  $\eta_c$(1S) & 0 &
22.0809 & 1.0037\,$\cdot$\,22$\pi^0$\\

$\eta$ & 0 & 4.0563 & 1.0141\,$\cdot$\,4$\pi^0$ & J/$\psi$ & 1 & 22.9441 &
0.9976\,$\cdot$23$\pi^0$\\

$\eta^\prime$(958) & 0 &  7.0959 &1.0137\,$\cdot$\,7$\pi^0$ &
$\chi_{c0}$(1P) & 0 & 25.2989 & 1.0119\,$\cdot$\,25$\pi^0$\\

$\eta$(1295) & 0 & 9.5868 & 0.9587\,$\cdot$\,10$\pi^0$ & $\chi_{c1}$(1P) &
1 & 26.0094 & 1.0004\,$\cdot$\,26$\pi^0$\\

$\eta$(1405) & 0 & 10.445 & 1.0445\,$\cdot$\,10$\pi^0$ & $\eta_c$(2S) & 0
& 26.9528 & 0.9983\,$\cdot$\,27$\pi^0$\\

$\eta$(1475) & 0 & 10.9352 & 0.9941\,$\cdot$\,11$\pi^0$ & $\psi$(2S) & 1 &
27.3091 & 1.0115\,$\cdot$\,27$\pi^0$\\

& & & &  $\psi$(3770) & 1 & 27.9389 & 0.9978\,$\cdot$\,28$\pi^0$\\

$\Lambda$ & 1/2 & 8.2658 & 1.0332\,$\cdot$\,8$\pi^0$ & $\psi$(4040) & 1 &
29.9237 & 0.9975\,$\cdot$\,30$\pi^0$\\

$\Lambda$(1405) & 1/2 & 10.4166 & 1.0417\,$\cdot$\,10$\pi^0$ &
$\psi$(4191) & 1 & 31.0543 & 1.00175\,$\cdot$\,31$\pi^0$\\

$\Lambda$(1670) & 1/2 & 12.3725 & 1.0310\,$\cdot$\,12$\pi^0$ &
$\psi$(4415) & 1 & 32.7538 & 0.9925\,$\cdot$\,33$\pi^0$\\

$\Lambda$(1800) & 1/2 & 13.335 & 1.0258\,$\cdot$\,13$\pi^0$ & & & &\\

$\Sigma^0$ & 1/2 & 8.8359 & 0.9818\,$\cdot$\,9$\pi^0$ & $\Upsilon$(1S) & 1
& 70.0884 & 1.0013\,$\cdot$\,70$\pi^0$\\

$\Sigma$(1660) & 1/2 & 12.2984 & 1.0249\,$\cdot$\,12$\pi^0$ &
$\chi_{b0}$(1P) & 0 & 73.0455 & 1.0006\,$\cdot$\,73$\pi^0$\\

$\Sigma$(1750) & 1/2 & 12.9652 & 0.9973\,$\cdot$\,13$\pi^0$ &
$\chi_{b1}$(1P) & 1 & 73.2926 & 1.0040\,$\cdot$\,73$\pi^0$\\

$\Xi^0$ & 1/2 & 9.7412 & 0.9741\,$\cdot$\,10$\pi^0$ & $\Upsilon$(2S) & 1 &
75.8102 &
0.9975\,$\cdot$\,76$\pi^0$\\

$\Omega^-$ & 3/2 & 12.3907 & 1.0326\,$\cdot$\,12$\pi^0$ & $\chi_{b0}$(2P)
& 0 & 75.8094 & 0.9975\,$\cdot$\,76$\pi^0$\\

$\Lambda_c^+$ & 1/2 & 16.9397 & 0.9965\,$\cdot$\,17$\pi^0$ &
$\chi_{b1}$(2P) &1 & 75.9795 & 0.9997\,$\cdot$\,76$\pi^0$\\

$\Lambda_c$(2593)$^+$ & 1/2 & 19.2285 & 1.0120\,$\cdot$\,19$\pi^0$ &
$\Upsilon$(3S) & 1 & 76.7185 & 0.9963\,$\cdot$\,77$\pi^0$\\

$\Sigma_c$(2455)$^0$ & 1/2 & 18.1792 & 1.00995\,$\cdot$\,18$\pi^0$ &
$\Upsilon$(4S)
& 1 & 78.3795 & 1.0049\,$\cdot$\,78$\pi^0$\\

$\Xi_c^0$ & 1/2 & 18.3069 & 1.01705\,$\cdot$\,18$\pi^0$ &
$\Upsilon$(10860) & 1 &
80.4954 & 1.0062\,$\cdot$\,80$\pi^0$\\

$\Xi^{\prime0}_c$ & 1/2 & 19.0996 & 1.0052\,$\cdot$19$\pi^0$ &
$\Upsilon$(11020)
& 1 & 81.6364 & 0.9956\,$\cdot$\,82$\pi^0$\\

$\Xi_c$(2790) & 1/2 & 20.6843 & 0.9850\,$\cdot$\,21$\pi^0$\\

$\Omega_c^0$ & 1/2 & 19.9849 & 0.9992\,$\cdot$\,20$\pi^0$\\
[0.3cm]\hline\hline
\end{tabular}
\end{table}

  The masses are taken from the Review of Particle
Physics [2]. Only particles, stable or unstable, with J\,$\le$\,1
are listed. To be
on the safe side, we use only particles which the Particle Data
Group considers to be ``established". The $\Omega^-$ baryon
with J = 3/2 is also given for comparison, but is not included
in the least square analysis. In all there are 41 particles which
follow the integer multiple rule. The data of Table 3 are plotted in
Fig.\,1, the line on Fig.\,1 is determined by Eq.(2).

   Palazzi [73] has studied the correlation of the masses of the
different $\eta$ mesons using a
mass unit of 33.86\,MeV/c$^2$, which is 1.0034\,m($\pi^0$)/4.
In his Fig.\,2a he shows that the masses of the $\eta$\,\,mesons
are integer multiples of his mass unit, having a nearly perfect
correlation coefficient. This agrees with our integer multiple
rule. For the mass of $\psi$(4160) in Table 3 we have used a
recent measurement of Ablikim et al.\,\,[74] which places
m($\psi$(4160)) at 4191.6\,MeV/c$^2$,
which is 1.00175\,$\cdot$\,31m($\pi^0$).

\section*{Appendix B}

\bigskip

\begin{center}

\large
{The lattice constant}

\normalsize
\end{center}

\noindent
An equation for the lattice constant in cubic lattices has been
given by Born and Land\'{e} [41], Eq.(6) therein. Suppose
there are N lattice points in a diatomic cubic crystal. There are then
 N/2 masses m$_1$ and N/2 masses m$_2$  and the mass in
the crystal is N/2$\cdot$(m$_1$ + m$_2$). In each cell are eight
neighboring particles, i.e.\,\,there are N/8 such cells. The volume of
each cubic cell is \emph{a}$^3$ and the total volume of the crystal
is N\emph{a}$^3$/8. The density $\rho$ is then
$\rho$ = N/2$\cdot$(m$_1$ + m$_2$)/(N\emph{a}$^3$/8),
from which follows that the lattice constant is given by
\begin{equation} \emph{a}^3 = 4\,(\mathrm{m}_1 + \mathrm{m}_2)/\rho\,.
\end{equation}

   We determine the lattice constant of the neutrino lattice with
m($\nu_e$) = 0.365\,milli-eV/c$^2$ and m($\nu_\mu)$ =
49.91\,milli-eV/c$^2$, from Eqs.(67,70), and with the density of
the $\pi^\pm$\,mesons $\rho(\pi^\pm)$ =
 m($\pi^\pm$)/Vol($\pi^\pm$) or with $\rho$ =
139.57\,MeV/c$^2$Vol($\pi^\pm)$. The volume of the
$\pi^\pm$\,mesons can be determined from the
measured radius of the $\pi^\pm$\,mesons,
r$_\pi$ = 0.880$\cdot$10$^{-13}$\,cm = r$_p$,
from Eq.(16). The third power of the lattice constant of
the neutrino lattice of the $\pi^\pm$\,mesons is then
\begin{equation} \emph{a}^3 = \frac{4\,(49.91 +
0.365)\,\mbox{milli-eV}}{139.57\,\mathrm{MeV/Vol}(\pi^\pm)} =
\frac{201.1\, \mbox{milli-eV}}{139.57\,\mathrm{MeV}}\times
\mathrm{Vol}(\pi^\pm) \end{equation}
or \begin{equation} \emph{a}^3 = 1.44085\cdot 10^{-9}\cdot
\mathrm{Vol}(\pi^\pm)
\end{equation}
The volume of the $\pi^\pm$\,mesons is
 4\,$\pi$/3\,$\cdot$\,r$_\pi^3$  with r$_\pi$ =
0.88\,$\cdot$\,10$^{-13}$\,cm, Eq.(16). The measured
r$_\pi$  is, however, not equal to the sidelength d of the cubic
lattice, which is an integer multiple of the lattice distance \emph{a}.
We must therefore replace the measured r$_\pi$ by
d/$\sqrt[3]{4\pi/3}$. The volume of the cubic $\pi^\pm$ meson is
then equal to (r$_\pi$(exp))$^3$ and it follows that

\begin{equation} \emph{a} = 0.9939\cdot10^{-16}\,\mathrm{cm}\,.
\end{equation}

This agrees qualitatively with the neutrino lattice constant we use
\emph{a} = 1\,$\cdot$\,10$^{-16}$\,cm, which we postulated in Eq.(8),
the difference with \emph{a} = 1$\cdot$10$^{-16}$\,cm is well within
the uncertainty of r$_\pi$. This
agreement is, of course, a consequence of our determination of the
 neutrino masses m($\nu_e$)  and m($\nu_\mu$) with the help of our
postulated \emph{a}. It is useful to know that lattice theory, expressed by
Eq.(6) of B\&L or by Eq.(142), leads to the neutrino lattice constant. If the
neutrino masses could be determined independently from our calculations,
the lattice constant of a neutrino lattice could be calculated from Eq.(142)
without making an assumption about \emph{a}.

\section*{Appendix C}

\bigskip

\begin{center}
\large{Four types of electrons}
\normalsize

\medskip
\emph{How come that e$^-$ can carry N\,$^\prime$/4 electron neutrinos\\
{\bfseries as well as} N\,$^\prime$/4 anti-electron neutrinos}\,?
\end{center}

\smallskip
\noindent
Since the rest mass and the spin of $\nu_e$ is the same as that of
$\bar{\nu}_e$\\
the most obvious difference between $\nu_e$ and $\bar{\nu}_e$ is the
\begin{center}
\emph{direction of the spin vector.}
\end{center}

  In $\gamma \rightarrow$ e$^-$e$^+$  the spin of e$^-$ and e$^+$ must be
the same. Conservation of angular momentum (s($\gamma$) = 1) requires that
the spin vectors of e$^-$ and e$^+$ point in the same direction. From
conservation of neutrino numbers follows that in e$^-$  the same number
 N$^\prime$/4 of electron neutrinos must be as anti-electron neutrinos are
in e$^+$. From the explanation of the electron in Section 11, in
particular from Fig.\,8, we learned that the electron neutrinos or
anti-electron neutrinos in e$^\pm$ do not contribute to the spin of
e$^\pm$. That means that the direction of the spin is determined by the
direction of the electric field ef$_\pm$ in the center of the lattice.
The direction of the electric field in the center of the lattice has to
have the same direction as the other field elements of the lattice, so
that the sum of the electric elements is equal to the elementary electric
charge. On the other hand, the sum of the spins of all charge elements
of the lattice must cancel, but for the spin of the center element. If no
neutrino is in the center of the lattice, the spin of  all neutrinos in the lattice
cancels.

   So the electron with an upward directed spin s(e$^-)\uparrow$ consists of
the upward directed electric field ef$_-\uparrow$ and N$^\prime\nu_e$/4
neutrinos.

\bigskip

Or \quad s(e$^-)\uparrow$ = ef$_-\uparrow$ + N$^\prime\nu_e$/4
\quad and $\quad$ s(e$^-)\downarrow$ = ef$_-\downarrow$ +
N$^\prime\bar{\nu}_e$/4.

\bigskip

\noindent
The \emph{antiparticles} of e$^-\uparrow$ and e$^-\downarrow$ are

\bigskip

\qquad\,\, s(e$^+)\uparrow$ = ef$_+\uparrow$  + N$^\prime\bar{\nu}_e$/4
\quad and \quad s(e$^+)\downarrow$ = ef$_+\downarrow$ + N$^\prime\nu_e$/4.

\bigskip
In the case $\gamma \rightarrow$ e$^-$e$^+$, when e$^-$ and e$^+$ have the
same spin\,$\uparrow$,\\
 that would mean that e$^-$ has to be ef$_-\uparrow$ + N$^\prime\nu_e$/4,\\
and would mean that e$^+\uparrow$ has to be ef$_+\uparrow$ +
N$^\prime\bar{\nu}_e$/4.

\bigskip
\noindent
In the case that e$^-$  and e$^+$ have spin $\downarrow$
then e$^-$ has to be ef$_-\downarrow$ + N$^\prime\bar{\nu}_e$/4,\\
and e$^+$ has to be ef$_+\downarrow$ + N$^\prime\nu_e$/4.

\bigskip
That means that \emph{the neutrino lattice of the electron can consist of
 either N$\,^\prime\nu_e$/4 neutrinos or N$\,^\prime\bar{\nu}_e$/4 antineutrinos}.

\section*{Appendix D}

\bigskip

\begin{center}

\large{The electron radius}

\normalsize

\end{center}

\noindent
As is well-known [2] the classical electron radius is given by
\begin{equation} \mathrm{r(e)}_{c\,l} =
\mathrm{e}^2/\mathrm{m(e^\pm)}\,\mathrm{c}^2
= 2.817940\cdot 10^{-13}\,\mathrm{cm}\,. \end{equation}
This equation is based on the premise that the electron has a symmetric
spherical charge distribution and that the entire mass of m(e)  is of
electric origin. As mentioned in Section 11 the electron scattering
experiments do not confirm that the electron radius has this value, rather
the charge radius of the electron has been found to be on the order of
10$^{-16}$\,cm instead of 10$^{-13}$\,cm.

   In Section 11 we have explained the mass of the electron with a cubic
 lattice consisting to one half of electric oscillations and to the other half of
electron neutrinos, Eqs.(77) and (83).
The position of the charge elements of our model of the electron  is shown
in Fig.\,8, they are separated by the distance 2\emph{a} = 2\,$\cdot\,10^{-16}$\,cm.
With N/4 charge elements the volume of the cubic charge distribution
is N/4\,$\cdot$\,(2\emph{a})$^3$ = 5.708\,$\cdot$\,10$^{-39}$\,cm$^3$,
from which follows that the sidelength of the cubic charge distribution
of the free electron is
\begin{equation} \mathrm{d(e)}_{cu} = 1.787\cdot10^{-13}\,\mathrm{cm}\,.
\end{equation}
The volume of the cubic charge
distribution of a free electron corresponds to a sphere with the radius
\begin{equation} \mathrm{r(e)}_{cu} = 1.10866\cdot10^{-13}\,\mathrm{cm}\,.
\end{equation}
For a comparison, it follows from the measured charge radius of the
$\pi^\pm$\,mesons (Eq.16) that\\
\centerline{ r$(\pi^\pm)$ = 0.88$\cdot$\,10$^{-13}$\,cm,}
and that the sidelength of the cubic lattice of the $\pi^\pm$\,mesons is\\
\centerline{d$(\pi^\pm)$ = 1.4186\,$\cdot$\,10$^{-13}$\,cm.}

\medskip
   The sidelength of a free cubic elementary charge distribution,
Eq.(147), is by 25\% larger than the sidelength d($\pi^\pm$) of the
$\pi^\pm$\,mesons. The dimension of the electric charge e$^\pm$
\emph{in}  the $\pi^\pm$\,mesons must be equal to the measured
charge radius of $\pi^\pm$, which is based solely on the
interaction of electrons with $\pi^\pm$\,mesons, the non-interacting
neutrinos in $\pi^\pm$  do not contribute to the scattering. The effective
radius of a free cubic  electron, Eq.(148), is larger than the effective
radius r($\pi^\pm$) of the same charge in the neutrino lattice of a
$\pi^\pm$\,meson.
The charge elements in the interior of the $\pi^\pm$\,mesons
are closer together than in the free electron. When the charge e$^\pm$
is introduced into the neutral neutrino lattice of the $\pi^\pm$\,mesons
the electric charge is compressed. That means that a binding
energy must be involved when an electron is added to the neutral
lattice of a $\pi^\pm$\,meson. We will pursue this topic in a later paper.

Comparing the classical electron radius Eq.(146) to the effective radius
of a free cubic electron, Eq.(148),  we find that
\begin{equation} \mathrm{r(e)}_{c\,l} = 2.5417\,\mathrm{r(e})_{cu} \,.
 \end{equation}
The apparent contradiction between our theoretical effective charge
radius of the cubic charge distribution of the electron in Eq.(148) and the
experimentally measured charge radius of the electron, which is on
the order of $10^{-16}$\,cm [65], is a consequence of scattering
formulas which do not deal with a rigid cubic charge distribution of a
finite size.

\section*{Appendix E}

\bigskip

\begin{center}

\large{The magnetic moment of the muon}

\normalsize

\end{center}

\noindent
   The explanation of the magnetic moment of the electron in Section\,15
has to pass a critical test, namely it has to be shown that the same
 considerations lead to a correct explanation of the magnetic moment of
the muon $\mu_\mu$ = e$\hbar$/2m($\mu^\pm$)c,  which is
about 1/200th of the magnetic  moment of the electron but is known
with nearly the same accuracy as $\mu_e$.  Both magnetic moments
are related through the equation
\begin{equation} \frac{\mu_\mu}{\mu_e} =
\frac{\mathrm{m(e}^\pm)}{\mathrm{m}(\mu^\pm)}
= \frac{1}{206.768}\,, \end{equation}
as follows from Eq.(139) applied to the electron and muon.
This equation agrees with the experimental results to the sixth decimal.
The muon has, as the electron, an anomalous magnetic moment which
is too small to be considered here.

   As shown in Section 9 the muons consist of a lattice of
N$^\prime$/4 muon neutrinos $\nu_\mu$, respectively anti-muon
neutrinos $\bar{\nu}_\mu$, of N$^\prime$/4 electron neutrinos and
the same number of anti-electron neutrinos plus an elementary electric
charge. For the explanation of the magnetic moment of the muon we
follow the same reasoning we have used for the explanation of the
magnetic moment of the electron. We say that m($\mu^\pm)$  consists
of two parts, one part which causes the magnetic moment and another
part which does not contribute to the magnetic moment.  The part
of m($\mu^\pm$) which causes the magnetic moment must contain the
electric charge and circular electric oscillations without which there would
be no magnetic moment. It becomes immediately clear from the small
mass of the electron neutrinos and from Eq.(79) for the energy of the
 electric oscillations in the electron that $\Sigma$\,m($\nu_e$) and
 E$_\nu$(e$^\pm$) are too small, as compared to the energy in the
rest  masses of all neutrinos in the muons, to make up m($\mu^\pm)$/2.
If, however, the electric charge elements  in the muon lattice bind to the
muon neutrinos instead to the electron neutrinos, as in the case of the electron,
then one obtains m($\mu^\pm$)/2 from the sum of the oscillation energy
of the muon neutrinos, (E$_\nu(\mu^\pm$)/4), plus the sum of the
energy in the rest masses of the muon neutrinos plus the energy
E$_\nu$(e$^\pm$) in the electric oscillations
\vspace{0.2cm}
\begin{eqnarray}
\lefteqn {1/4\cdot\mathrm{E}_\nu(\mu^\pm) \,+\,
\sum\,{\mathrm{m(\nu_\mu)c}^2} +
\mathrm{E}_\nu(\mathrm{e}^\pm)  }\nonumber \\
&  =  & 1/4\cdot\mathrm{E}_\nu(\mu^\pm) +
 \mathrm{N}/4\cdot \mathrm{m}(\nu_\mu)\mathrm{c}^2 +
\mathrm{m(e^\pm})\mathrm{c}^2/2 \nonumber\\
  & = & 53.3125\,\mathrm{MeV}
  =  0.50457\,\mathrm{m}(\mu^\pm)\mathrm{c}^2\,,
\end{eqnarray}

\noindent
where we have used E$_\nu(\pi^\pm)$ = E$_\nu(\mu^\pm)$ according
to Eq.(63), with E$_\nu(\pi^\pm$) = m($\pi^\pm)$c$^2$/2 =
69.7851\,MeV  and also E$_\nu$(e$^\pm$) = m(e$^\pm$)c$^2$/2 from
Eq.(84), as well as  m($\nu_\mu$)c$^2$ = 49.91\,milli-eV, (Eq.70),
m($\mu^\pm$)c$^2$ = 105.6583\,MeV and N$^\prime$ $\cong$ N.
Eq.(151) says that the part of m($\mu^\pm$) which carries the
electric charge and causes the magnetic moment is $\approx$
m($\mu^\pm$)/2, provided that the charge elements bind to the muon
neutrinos instead of the electron neutrinos. The remaining part of
m($\mu^\pm$) which does not carry charge and does not contribute
to the magnetic moment is given by
\begin{eqnarray}
\lefteqn{ 3/4\cdot\mathrm{E}_\nu(\mu^\pm) \,+\,
\sum\,(\mathrm{m}(\nu_e) + \mathrm{m}(\bar{\nu}_e))\mathrm{c}^2 + \mathrm{m(e^\pm)}\mathrm{c}^2/2}\nonumber \\
&  =  & 3/4\cdot\mathrm{E}_\nu(\mu^\pm)
+ 3/4\cdot\mathrm{N}\mathrm{m}(\nu_e)\mathrm{c}^2 \nonumber \\
 & = & 53.1201\,\mathrm{MeV}
  =  0.50275\,\mathrm{m}(\mu^\pm)\mathrm{c}^2\,.
\end{eqnarray}
The additional m($\mathrm{e}^\pm)$c$^2$/2 on the top line of Eq.(152)
originates from the  energy in the mass of the N/4 electron neutrinos which
make up the neutral part of the electron. For m($\nu_e)$  = m$(\bar{\nu}_e)$
we use the value 0.365\,milli-eV/c$^2$ as in Eq.(67). The sum of the
mass of the charged part of m($\mu^\pm$) plus the neutral part of
m($\mu^\pm$) is 1.0073\,m($\mu^\pm$). It is important to
note that Eqs.(151,152) depend critically on the validity of
E$_\nu(\mu^\pm$) = E$_\nu(\pi^\pm$), from Eq.(63).

   If the electric charge elements bind to the muon neutrinos in the muon
lattice and if the charged part of the muon makes up 1/2 of the mass of
the muon as in Eq.(151) then it follows from Eq.(139) that the magnetic
moment of the muon is given by
\begin{equation} \vec{\mu}_\mu = \frac{\mathrm{e}\hbar}
{2\mathrm{m}(\mu^\pm)/2\cdot\mathrm{c}}\cdot\vec{s}\,.  \end{equation}
With s = 1/2 we have
\begin{equation}\mu_\mu = \mathrm{e}\hbar/2\mathrm{m}(\mu^\pm)\mathrm{c}\, \end{equation}
 as it must be, without the artificial g-factor.

We have thus shown that we can explain the magnetic moment of the muon
with the same concept that we have applied to the explanation of the
magnetic moment of the electron, namely that 1/2 of the mass of the
electron does not contribute to the magnetic moment because
this half of the mass does not carry charge. In the case of the muon the
same is true, provided that the charge elements bind to the muon neutrinos
in the muon lattice.

\end{document}

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