I have done a combined art and science project that you can find at http:///
itampe.com and the happy ending is at the end of this mail (if you are into
science) I will take responsibility for the content but allow anyone
copy and add to the content and improve, but the core need to be unchanged
if you choose to do so. The basic idea behind the project is that in my
mind we are destroying our world completely mindlessly and want us all to
start doing something. I will surely act as much as I can to help this
movement and I do not claim that my ideas are unique. The only way forward
is to act as a team ... Please copy to anyone that you think should read
this.
Copyright
July, 2023 Stefan Israelsson Tampe
Oh, the happy ending, here is the explanation of how nature works ;-)
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{pictex}
\usepackage[all]{xy}
\usepackage{tikz}
\usepackage[a4paper]{geometry}
\usetikzlibrary{arrows}
\newcommand{\midarrow}{\tikz \draw[-triangle 90] (0,0) -- +(.1,0);}
\title{On Charge}
\author{Stefan Israelsson Tampe}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
We will in this document assume that a charged particle (electron) is
built up by (similar to super string theory in a sense) of constellation of
loops that has a very peculiar form of interaction that is as simple as one
can possible think of. That this model has a chance of explaining the
normal analytical treatment of charges in our macroscopic world is a bit if
a challenge to explain. We will assume that there is a limit for how much
energy density we can have and they will differ slightly between positive
and negative charge meaning in the end a difference between particle mass
and anti particle mass. Especially we reproduce the result that the
electron and possitron differes and the resulting mass of the positron is
corect within measurement errors. We will also show that a stable system
consists of two almost similar loops or helical paths that have opposite
sign. We will show that the positive and negative charge is constant and
the same. We will show how how mass can be calculated and how we can
calculate angular momentum which makes it possible to deduce information on
this model. We will also be able to conclude why $\alpha \approx 1/137$ and
why this is so and why not exactly $1/137$ and why the specific value is
137. We will show why $\hbar$ is a fundamental constant.
\end{abstract}
\section{The main model assumptions}
We will base our analysis of a basic object that is a stream of charge that
has no mass and move at the speed of light. It will also have the property
that it only interacts if two infinitesimal line segments are parallel and
directed in the same direction and if we draw the tangent lines these
elements are located at the smallest distance to each other. We will assume
for the specific case the basic Coulomb's law apply for these special
segments. We will implicitly assume that each particle is composed of
objects that is a basic object that is overlaid in all possible directions
and that for two particles, there is always a a matched pair that can
express the normal electrostatic law so that we can reproduce the usual
macroscopic interaction. We will assume that each loop has a fixed amount
of charge so that as we enlarge the loop, it will be less dense. In a sense
it is a closed system That can scale. We will assume that there is a energy
density limit, one for each sign of the charge that is almost the same.
\section{Lorentz invariance}
for two segments to be interacting they most be parallel. And in their
reference frame. The energy is,
$$
q^2 /r
$$
As the speed of reference frame (defined through a limiting procedure) is
the same independent of if we move the system or not defining the
interaction in this frame is hence Lorentz invariant.
The observant reader would realize that there are some issues with these
objects. We will consider the streams as a limit of a sequence of objects
of the kind,
$$
\lim_{n\to \infty, v\to c} \sum_{i=0}^n e_a\frac{\delta_{r_i^a(t) =
(i/n+vt)\hat x}}{\sqrt{n}}
$$
Where we have $n$ equally spaced normal mass less electron's evenly
distributed on [0,1] at $t=0$ and they move with the speed of light along
the x-axis. We will assume here that nearby electrons are not interaction,
but in stead if we take a parallel stream,
$$
\lim_{n\to \infty, v\to c} \sum_{i=0}^n e_a\frac{\delta_{r_i^b(t) =
(i/n+vt)\hat x + h \hat y}}{\sqrt{n}}
$$
As we will close them int0 loops, We will assume that each current loop has
the same number of charges and we will assume that this is an invariant of
the world. Note that as we increase the velocity they will contract and in
order to get a nonzero charge density we need to spread the charges out
more and more in their reference frame. Hence when we define the
interaction in that frame, any two parallel segments need to be perfectly
matched. Also in that frame the next charge is infinitely large distance
away so there is no self interaction and if they are offset and not at the
closest distance they will be infinitely far away. Also if the streams
interact the probability of two steams hitting each other is zero in a
sense so we could hand wave away that part as well (e.g. they sync so that
they do not interact). So will assume that only $r_i^a,r_i^b$ are
interacting as normal electric charges and the rest does not. And we will
demand the streams to be parallel and likewise directed and also located so
that it is in a sense closes as possible if we consider the tangent lines
of the streams. Hence in any geometrical constellation one need to search
for parallel tangents that are not offsets e.g. if you draw the tangents
they need to be parallel and the pairing need to be at the closest
distance. As the interaction is done in the rest reference system, one can
always consider only the electrostatic interaction when exploring a certain
geometrical setup, which is similar to quantum electrodynamics that does
also not have internal magnetic terms. As we defined the energy in the rest
frame we are free to also put a limit there of the energy density. One for
each charge.
\pagebreak
\section{The Loop}
\def\dh {2}
\def\h {0.25}
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=0.7, transform shape]
\begin{scope}[every node/.style={sloped,allow upside down}]
\draw [black] (0, 0) to (\dh*2.5, 0);
\draw (2.5*\dh, 0) node {\midarrow} (2.5*\dh,0);
\draw [black] (0, 0) to (\dh*2.5, 0);
\draw (1.5*\dh, 0) node {\midarrow} (1.5*\dh,0);
\draw (0, 0) circle [radius=1.5*\dh];
\draw (0, 0) circle [radius=2.5*\dh];
\draw [domain=10:80] plot ({3*\dh*cos(\x)}, {3*\dh*sin(\x)});
\draw [domain=15:85] plot ({2*\dh*cos(\x)}, {2*\dh*sin(\x)});
\draw [black] (5,0) to ({3*\dh*cos(10)}, {3*\dh*sin(10)});
\draw [black] (3,0) to ({2*\dh*cos(15)}, {2*\dh*sin(15)});
\node at (2,0.3) {$vr$};
\node at (4,0.3) {$r$};
\node at (5.7,0.4) {$hr$};
\end{scope}
\end{tikzpicture}
\end{center}
\end{figure}
Consider 2 concentric loops stacked above each other forming a double
cylindrical entity. the charge in the inner loop is positive and the charge
in the outer loop is negative (we will consider a reversed version later).
We will attache a constant charge density of the streams will be constant,
$e_a$ for negative and $a_b$ for positive. We will be a bit sloppy in the
mathematical rigor and consider that all interaction terms is a limit in
$L_2$ of their combination.
Loops of different charge sign will attract if they are concentric and have
very little attraction if not centered so we expect this selection of
geometry to be stable. let $r_b=vr$ be the inner radius and $r_a=r$ be the
outer radius, then we will assume a scaling so that the effective charge
density in the outer cylinder to have the radian contribution constraint
assuming $e_a$ to be the charge density at the outer radius and $e_b$ the
inner radius charge density, hence the constraint is,
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=0.7, transform shape]
\draw [domain=80:100] plot ({3*\dh*cos(\x)}, {3*\dh*sin(\x)});
\draw [domain=80:100] plot ({2*\dh*cos(\x)}, {2*\dh*sin(\x)});
\draw [black] (0,0) to ({3*\dh*cos(80)}, {3*\dh*sin(80)});
\draw [black] (0,0) to ({3*\dh*cos(100)}, {3*\dh*sin(100)});
\node at (0, 4.2) {$dx_b$};
\node at (2, 4.2) {$dx_b = r_b\,d\theta$};
\node at (0, 6.2) {$dx_a$};
\node at (2, 6.2) {$dx_a= r_b\,d\theta$};
\end{tikzpicture}
\end{center}
\end{figure}
$$
er_a d\theta = (e_ar_a - e_br_b)d\theta = (e_a - e_bv)r_ad\theta = (e_a -
ve_b)rd\theta.
$$
Or,
$$
e = e_a - ve_b
$$
We shall consider scaling properties and hence it is natural to consider
$e=ue_a$. The dual setup is considering,
$$
-e = be_a - e_b.
$$
For this case we will considering $e=ue_b$ as a scaling. When we multiply
two of these streams we will consider the ``square root'' of a delta
measure that is made stringent by the limiting argument above. and hence
will we use
\begin{equation}
\label{charge1}
e_ae_br_ar_b d \theta, \quad e_a^2r_a^2d \theta, \quad e_b^2r_b^2d \theta
\end{equation}
For the energy relation below. Also when we want to study the ``energy
density'' we will mean then that we need to study this effect on the paired
$r_i^a,r_i^b$. Then summing the effect on the unit length will lead to
taking these values.
\begin{equation}
\label{charge2}
e_ae_a, \quad e_a^2, \quad e_b^2.
\end{equation}
To see that we will assume a normalized condition on the energy density at
the singly scaled pairing
You may say, this does not cut it. this makes the integrated $e$ vary when
you vary the radius, and we will see that this changes. Now this is a
correct observation and in a sense it works out that way. But we will stack
multiple such loops and form a torus with radius $R$. If you examine two of
these torus-es they only interact significantly if they is located in 2
parallel planes. And there they interact only on a circle of the radius,
say R and along this axis we will scale the charge when we consider the
torus as a surface else we will consider it as stacked circles the
effective charge on all those concentric loops will be $e$. This is a bit
tricky to understand but it is as it is in this model and the task is not
to dismiss it, but see if there is any explainable power in this model. Not
to make a water tight theory as we first need to pass the first floor of
the theoretical castle.
Anyway, the charge condition is a scale invariant condition in order for
the final charge to be correct as argues. The attractive energy per radian
of the loops are (where we use the special Coulomb's law and the
observation \ref{charge1})
$$
V_ld\theta = k\frac{e_ae_br_ar_b}{r_a-r_b}d\theta =
k\frac{e_ae_bvr}{1-v}d\theta
$$
Similarly as we stack loops (or helix)) right on top of each other with a
pitch $h' = hr$ we will see that the forces on one segments in one
direction is
$$
F = k\frac{e_*^2r_*}{(hr)^2}(1+2^{-2}+3^{-2} + \ldots) =
\zeta(2)\frac{ke_*^2}{(hr)^2}
$$
Now this is a simplification e.g. if we connect it and turn it into a torus
or helix, so we will just assume that this part transforms as $\zeta_h/h$,
where we punt for now what $\zeta_h$ is. Hence if we consider the force on
both sides we get the energy by integrating $hr$ to,
$$
V_{h,*}d\theta = 2\zeta_h\frac{ke_*^2r_*^2}{hr}d\theta.
$$
So the total energy for one loop is,
$$
E = (V_{h,1} + V_{h,2} + 2 V_l)2\pi = 2\pi k r
\Big(\frac{2\zeta_h}{h}(e_a^2 + e_b^2v^2) - 2 \frac{e_ae_bv}{|1-v|} \Big).
$$
Using $e = e_a-ve_b$ in this expression for the energy, we note that we can
search to find the stationary point varying $x=ve_b$ and keeping the rest
constant, while also introducing the obvious $A,B$ to this leads to,
$$
A(2e+2x+2x) - B(e + 2x) = 0.
$$
Or,
$$
(2A-B)(2x + e) = 0.
$$
So,
$$
x = ve_b = -e/2
$$
This means that $ve_v$ tend to go to zero unless,
\begin{equation}
\label{cond1}
2A-B = 0 \Leftrightarrow \frac{2\zeta_h}{h} = \frac{1}{|1-v|}.
\end{equation}
For which energy wise it can vary freely! To simplify the expression for
the energy, let first $e_b = ue_a$. Let $w=uv$ and note $e = e_a(1-w)$.
Then,
$$
E = 2\pi kre_a^2\Big( \frac{2\zeta_h}{h}(1 + w^2) - 2\frac{w}{|1-v|} \Big).
$$
Or using \ref{cond1},
$$
E = 2\pi kre_a^2\frac{2\zeta_h}{h}(1 + w^2 - 2w).
$$
Complete the square and we get,
\begin{equation}
\label{energy}
E = 2\pi kr \frac{2\zeta(2)}{h}(e_a(1-w))^2 = 4 \pi k
r\frac{\zeta(2)}{h}e^2.
\end{equation}
Note that this condition is invariant of how we combine the charges. To
evaluate the energy density and apply limits on them as the system want to
scale down in order to minimize energy. Assume the condition \ref{charge2}
for evaluating this limit.The charge densities at the loop $a$ is,
$$
\rho_a = ke_a^2\Big( \frac{2\zeta_h}{h} - \frac{u}{1-v} \Big)\frac{1}{r}.
$$
Or using \ref{cond1},
\begin{equation}
\label{da}
\rho_a = ke_a^2\frac{2\zeta_h}{h}(1-u)\frac{1}{r}.
\end{equation}
The density at loop $b$ is,
$$
\rho_b = ke_a^2\Big( \frac{2\zeta_h}{h}u^2 - \frac{u}{1-v} \Big)\frac{1}{r}.
$$
Or again using \ref{cond1},
\begin{equation}
\label{db}
\rho_a = ke_a^2\frac{2\zeta_h}{h}(u^2-u)\frac{1}{r}.
\end{equation}
Hence if these two densities are at a positive and negative limit, we need
to have (using \ref{da} and \ref{db})
$$
\rho_a = c_a,\quad \rho_b = -c_b
$$
To simplify the analysis of this, use \ref{cond1} and take,
$$
C_* = c_**C = c_*\frac{h}{2\zeta_h k e_a^2} = c_* \frac{|1-v|}{k e_a^2}.
$$
Then,
\begin{align}
\frac{|1-u|}{r} &= C_a = c_aC, \label{sys1}\\
u\frac{|1-u|}{r} &= C_b = c_bC. \label{sys2}
\end{align}
Note that this result is independent how we combined the charges to $e$.
Hence dividing \ref{sys2} with \ref{sys1},
\begin{equation}
\label{uu}
e_b/e_a = u = C_b/C_a = c_b/c_a.
\end{equation}
The constraint \ref{sys1} is,
$$
\frac{|1-u|}{r} = c_a C = c_a\frac{|1-v|}{k e_a^2}.
$$
Or,
$$
\frac{e_a^2(|1-u|)}{r} = c_a\frac{|1-v| }{k}.
$$
Hence,
\begin{equation}
\label{vee}
|1-v| = \frac{k e_a^2|1-u|}{r c_a}.
\end{equation}
in the dual situation we get $u'=c_a/c_b$ and $e_a \to e_b$ and for this
case,
$$
|1-v'|= \frac{k e_b^2|1-u'|}{r' c_b} = \frac{k e_be_a|1-u|)}{r' c_b}
=\frac{k e_ae_a|1-u|}{r' c_a} = |1-v|\frac{r}{r'}
$$
We can also reformulate the condition for $e$, using \ref{vee} as,
\begin{equation}
\label{fund}
e = e_a(1-uv) = e_a(1-u) + e_au(1-v) = e_a - e_b +
\frac{kue_a^2}{rc_a}(e_a-e_b) = \Delta \Big(1 + \frac{ke_be_a}{rc_a}\Big)
\end{equation}
where we used $\Delta = e_a-e_b$.
The dual expression is then,
\begin{equation}
\label{fund_d}
e' = - \Delta \Big(1 + \frac{ke_be_a}{r'c_b}\Big).
\end{equation}
So in order for $e=-e'$ we need,
\begin{equation}
\label{radii}
r' = r/u.
\end{equation}
Hence
$$
(1-v') = -(1-v)u
$$
And also $h' = h u$. We can solve for $r$ in \ref{fund},
\begin{equation}
\label{radii0}
r= \frac{ke_be_a}{c_a\Big(\frac{e}{\Delta}-1 \Big)}.
\end{equation}
But not only this, we also note that starting with,
$$
e = e_a|1-u|+e_au|1-v|
$$
And using \ref{sys1},
$$
e = D\frac{rc_a}{e_a} + e_au|1-v|,
$$
with,
$$
D = \frac{h}{2\zeta_h k}.
$$
Assuming $h,v,u$ constant we can minimize the energy by minimizing $e$ to
get,
\begin{equation}
\label{e_a}
e_a = \sqrt{\frac{D r c_a}{u|1-v|}} = \sqrt{\frac{\frac{h}{2\zeta_h} r
c_a}{k u|1-v|}} = \sqrt{\frac{r c_a}{k u}}.
\end{equation}
Hence from \ref{sys1},
$$
\frac{r c_a}{k u}|1-u| = r c_a \frac{|1-v|}{k}.
$$
Or,
\begin{equation}
\label{uv}
|1-u| = u|1-v|
\end{equation}
Hence,
\begin{equation}
\label{ee}
e = 2 e_a|1-u| = 2\sqrt{\frac{r c_a}{k u}}|1-u|.
\end{equation}
Also,
\begin{equation}
\label{ee2}
e = 2 \frac{e_a}{c_a}|c_a-c_b|.
\end{equation}
The constraint \ref{cond1} implies,
\begin{equation}
\label{h_v}
h = |1-v|2\zeta_h = \frac{|1-u|2\zeta_h}{u}
\end{equation}
And $h'= hu$. Now the actual pitch is $hr$ is then invariant. So the
argument for equal charge would that the most energetically favorable
action when a negative and positive charge form is an alignment and hence
equal pitch, hence the negative and positive charge is constrained to be
the same and as we see below this also imply that the $\hbar$ must be the
same. Anyway \ref{h_v} and squaring \ref{ee},
\begin{equation}
\label{ee2}
e^2 = 4\frac{r c_a}{k u}|1-u|^2.
\end{equation}
Special relativity means that we can deduce the masses per loop from
\ref{energy} as,
$$
E = m c^2 = 4 \pi kr\frac{\zeta_h}{h}e^2 = 2 \pi \frac{kr}{|1-v|}e^2
$$
Using \ref{uv}, with this, we get,
$$
mc^2 = \frac{2 \pi k r u e^2}{|1-u|}.
$$
So,
\begin{equation}
\label{masss}
m = \eta \frac{2 \pi k r u e^2}{|1-u|c^2}
\end{equation}
(we will discuss $\eta$ soon). And hence the dual relationship,
\begin{equation}
\label{mass}
m' = \frac{m}{u^2}.
\end{equation}
Note that the unit is $kg/m$ with $\eta$ currently an unknown unit. However
the loop is like a delta measure and you can see it as the result of taking
the limit with a scaled mass and thinner small cylindrical shell. Hence,
$$
\eta = 1\quad [m].
$$
We will need that to not confuse the astute reader that checks the
calculation by examining the units. Hence $m$ will have the unit
$[\text{kg}]$.
\subsection{Stacking into a torus}
Previously we was working with a system where we stacked loops on top of
each other to form a cylindrical structure. Now instead we connect all
loops so they form a torus. When we do this we will consider the pitch
defined by,
$$
hr2\pi R.
$$
E.g. the old $h$ is now $h'$ which is in the form,
$$
h' = 2\pi R h.
$$
>From this we get the dual condition $R'=R$. But the stacking of the loop is
although possible mathematically, hard to motivate for a stable structure.
However if we transform the loops to helical path's along the helix with a
velocity $v$ we have indeed produced a system that stabilizes as each path
is non interacting. In the reference frame of the system, where we move
with the particles along the big circle we will still make a loop and the
helix will interact repulsively with a similar part one pitch away. As the
number of pitches is the same, e.g. $hr$ we realizes that we have two
radius's of the torus. One in the system of the lab $R$ and one in the rest
frame $R_0$. and we have,
$$
R = \frac{R_0}{\gamma}
$$
We will evaluate the interaction in the rest frame. So we stack $n'$ of
them and therefore,
$$
n' h' r = 2\pi R.
$$
Or,
$$
hr = \frac{1}{n'}
$$
As the distance between the paths are different we realize that this can't
be exactly try as we have a contraction in the closest $R-r$ distance,
hence we actually have,
$$
n' h' r = 2\pi (R-r) = \frac{2 \pi R}{f},
$$
with,
$$
f = \frac{1}{1-r/R}.
$$
Hence,
\begin{equation}
hrf = \frac{1}{n'}.
\end{equation}
The attractive energy will be as before as that is independent of any
movements of the loops orientation. The repulsion will however be active on
only on two parts of the loops where they interact. The energy will be the
mean which is the same as using the center ($R$) distance. However, the
energy density that we use need to be analyzed at the $R-r$ distance where
it is the most extreme. we will do so by doing the transformation $c_* \to
c_*/f$. In this new parameterisation. The unit of $h$ is here $[1/m]$.
\subsection{Scaling}
Consider scaling. As the number of loops per torus is fixed, e.g. $n'$.
Then we know that only the loops will need to scale. hence we will get
from \ref{vee}, \ref{radii0} and \ref{energy},
\begin{align}
e &\to xe \\
r &\to r/x^2 \\
v &\to v, \\
u &\to u, \\
h &\to hx^2, \\
rh &\to rh, \\
E &\to x^4E, \\
m &\to x^4m.
\end{align}
This means that in order to maintain the same scaling we must have,
\begin{align}
R &\to R/x^2, \\
R_0 &\to R_0/x^2, \\
r/R &\to R/r, \\
f &\to f, \\
rhf &\to rhf.
\end{align}
Now as the helix will stretch with the $R$ we see that,
\begin{align}
E_{tot} &\to E_{tot}, \\
m_{tot} &\to m_{tot}, \\
e_{tot} &\to e_{tot}.
\end{align}
\subsection{Angular momentum}
The per loop angular momentum is,
$$
l = m\gamma(v_h)v_hR = m v_h R_0.
$$
The question is how $v_h$ scales. If the length of the helix scales as $R$
and hence the time it takes to move one turn scales as $R$. But as the
number so turns along the helix is invariant, we find that the pitch
distance also scales as $R$ which leaves the velocity invariant. Hence
$v_h$ is invariant of the scaling and hence the total angular moment which
is $n'$ such copies is invariant of the scale.
\begin{align}
v_h &\to v_h, \\
l &\to l, \\
l_{tot} &\to l_{tot}.
\end{align}
If we let the length of the helix as $L$ then $v_h$ satisfy (in the rest
frame),
$$
\frac{v_h}{c} = \frac{2\pi R}{L} = \frac{2\pi R_0/\gamma}{L_0\gamma} =
\frac{2\pi R_0}{L_0}.
$$
Anyhow if we factor in the need to remove from the outer loop the same
quantity from the inner loop we get,
$$
l = m v_h R_0 |1-v|
$$
$$
l_{tot} = n'l = n' m v_h R_0 |1-v| = n' \eta \frac{2 \pi k r u
e^2}{|1-u|c^2}\frac{c2\pi R_0}{L_0}R_0|1-v| = \eta \frac{A_0R_0}{L_0}\frac{
k u e^2}{|1-u|c}|1-v|,
$$
with $A_0$ being the torus area e.g,
$$
A_0= 2\pi r 2\pi R_0.
$$
Using \ref{uv} we find,
$$
l_{tot} = \eta \frac{A_0R_0}{L_0} \frac{k e^2}{c}.
$$
In order for the charge to be properly (hopefully) managed we need,
$$
\frac{A_0 2\pi R_0}{L_0} = n'
$$
E.g. we need to scale down the area in order to compensate for the extra
space the helical path takes. As we have $n'= 1/hr$ (forgetting about $f$)
identical pitches and hence we conclude that,
\begin{equation}
\label{hbardef}
\hbar = \int l_{tot}\,d\theta = \eta \frac{A_0 2\pi R_0}{L_0}\frac{ k u
e^2}{|1-u|c}|1-v| = \eta n' \frac{k r e^2}{c} = \eta \frac{k r e^2}{hr c}.
\end{equation}
Note that we here consider one helix turn per pitch, but, as discussed
above, this can also be any integer number of pitches hence we actually
have the Bohr condition of angular momentum.
$$
l_{tot} = n\hbar
$$
We can solve for $hr$
$$
hr = \eta \frac{k e^2}{\hbar c} = \alpha \approx \frac{1}{137} =
\frac{1}{n'}
$$
Which indicate why Wolfgang Pauli's quest to search for why $1/\alpha$ was
almost a natural number (137) may have a partial answer.
\subsection{Defining the zeta factor}
Consider $N$ charges evenly distributed on a unit circle. Let's study the
forces on one single charge. then they are locates on $e^{2\pi
k/N},k=0,\ldots N-1$. The force at $k=0$ is then. Now we would not like to
cancel any of their contributions to action at hence we get
$$
V(N) = \sum_{k=1}^N \frac{h'r}{R}\frac{1}{|e^{2\pi i k/N}-1|}.
$$
Now,
$$
|e^{2\pi i k/N}-1|^2 = (e^{2\pi i k/N}-1)(e^{-2\pi i k/N}-1) = 2 -
2\cos(2\pi k/N).
$$
Hence we are left with,
$$
V(N) = \frac{1}{\sqrt{2}}\sum_{k=1}^N
\frac{h'r}{R}\frac{1}{\sqrt{1-\cos(2\pi k/N)}}.
$$
Using the trigonometric identity for the double cosine's,
$$
1-\cos(2\pi k/N) = 1 - \cos^2(\pi k/N) + \sin^2(\pi k / N) = 2\sin^2(\pi k
/ N).
$$
Hence
$$
V(N) = \frac{1}{2} \sum_{k=1}^N \frac{h'r}{R}\frac{1}{\sin(\pi k/N)}.
$$
So we will have,
\begin{equation}
\label{zeta}
\zeta_h(N) = N \sum_{k=1}^{N} \frac{\pi \alpha}{\sin(\pi k/N)}
\end{equation}
Now as
$$
\sin(\pi k / N) < \pi k / N
$$
then, including that we have N charges, we get
\begin{equation}
\zeta_h(N) > N\sum_{k=1}^{N} \frac{\pi \alpha}{\pi k/N} >
N^2\alpha(\ln(N))
\end{equation}
hence
\begin{equation}
\label{estimate0}
\zeta_h(N) > N^2\alpha\ln(N)
\end{equation}
A direct calculation with $N=137$ gives,
\begin{equation}
\label{estimate}
\zeta_h(N) \approx N 691\alpha.
\end{equation}
\subsection{Numerology}
The following expression is a quite good equation for the fine structure
constant,
$$
\frac{\alpha}{1+\frac{\alpha}{1 - (2\pi-1)^2}} = \frac{1}{137}
$$
We can explore this further and find another expression,
$$
\frac{\alpha}{1+\frac{\alpha}{1 - \Big(1-\frac{2\pi}{1 -
\frac{4\pi\alpha}{1 +
\frac{2\pi}{1-\frac{4\pi\alpha}{1-2\pi/(1+2\alpha)}}}}\Big )^2}} =
\frac{1}{137}
$$
We could postulate from this,
$$
\frac{\alpha}{1-\frac{\alpha}{x^2-1}} = \frac{1}{137}.
$$
With $x$ satisfying,
$$
x = 1 -
\frac{2\pi}{1-\frac{4\pi\alpha}{1+\frac{2\pi}{1-\frac{4\pi\alpha}{x}}}}
$$
Of cause this is very numerological and are simply fined tuned with the
help of trial and error. Can we motivate this? well we concluded that
$$
hrf = \alpha f = \frac{1}{137}.
$$
So,
$$
\frac{\alpha}{1-\frac{r}{R}} = \frac{1}{137}.
$$
Using the found expression we could match this with the found expresison,
\begin{equation}
\label{approx}
\frac{r}{R} \approx \frac{\alpha}{(2\pi-1)^2-1}.
\end{equation}
But this is the same as,
$$
\frac{r}{R} = \frac{rh}{hR} = \frac{\alpha}{hR_0}.
$$
Hence we can identify,
$$
x^2 = hR + 1.
$$
Now $x^2 =28.7778$ means
$$
hR_0 = 27.7778
$$
Or
$$
\alpha R = 27.7778 r.
$$
Or
$$
R = 3807 r.
$$
Now for the velocities we have,
$$
c^2 = v_h^2 + v_r^2
$$
But,
$$
v_r/c = \frac{2\pi r}{L} = \frac{2\pi r}{\sqrt{(2\pi r)^2 + (r h 2\pi
R)^2}}.
$$
Rearenging we find
$$
v_r/c = \frac{1}{\sqrt{1 + (hR)^2}}.
$$
Solving for $hR$ we get,
$$
hR_0/\gamma = hR = \sqrt{\gamma^2 - 1}.
$$
>From this we can identify
\begin{equation}
\label{gamma}
\gamma \approx 5.3
\end{equation}
Finally ass $\alpha$ is invariant of the duality, we conclude that
$$
R' = R/u.
$$
But $R_0$ is invariant. Hence $\gamma' = \gamma u$ whcih means
\begin{equation}
\label{medual}
\gamma' m' = \frac{\gamma m}{u}.
\end{equation}
\subsection{On $n'$}
Form \ref{cond1} we see that $h'$ satisfies,
$$
h' = hr 2\pi R = \alpha 2\pi R = \frac{|1-u|2\zeta_h}{u} \approx
\frac{|1-u|}{u} n'\alpha\ln(n').
$$
Or
$$
R = C n'\ln(n').
$$
Which means,
$$
r = \frac{r}{R} C n'\ln(n').
$$
But we also have from \ref{e_a},
$$
r = Ce_a^2/n'^2
$$
Equating to,
$$
e_a^2 = C' \frac{r}{R}n'\ln(n') = n'\ln(n')
$$
But as we have an approximate postulated expression from \ref{approx},
$$
\frac{r}{R} \approx \frac{\alpha}{1-(2\pi-1)^2}.
$$
So,
$$
e_a^2 \approx C'' \alpha n'\ln(n') \approx C'' \ln(n')
$$
This indicate why we have $n' = 137$.
\subsection{On mass}
If we consider the total mass scale invariant we get (using n' copies),
\begin{equation}
\label{mdef}
m_e = \gamma n' m = \eta n' \gamma \frac{2 \pi k r u e^2}{|1-u|c^2} =
\eta n' \gamma \frac{\alpha 2 \pi k u e^2}{h |1-u|c^2}.
\end{equation}h
e electron mass equation \ref{mdef} and the condition for $h$, \ref{h_v},
and scaling down to with $\alpha$,
\begin{equation}
\label{mz}
m_e = \eta n' \gamma \alpha \frac{2 \pi k u e^2}{\frac{|1-u|2\zeta_h}{u}
|1-u|c^2} = \eta n' \alpha \gamma\frac{\pi k e^2}{h\zeta_h c^2} \Big(
\frac{u}{|1-u|}\Big )^2= \eta n' \gamma \alpha^2 \frac{\pi \hbar}{\zeta_h
c} \Big( \frac{u}{|1-u|}\Big )^2,
\end{equation}
plugging in \ref{estimate}, we find,
$$
m_e \approx \eta \gamma \frac{\alpha}{691} \frac{\pi \hbar}{c} \Big(
\frac{u}{|1-u|}\Big )^2.
$$
Taking (\ref{gamma}), $\gamma = 5.3$ we find,
$$
|1-u| \approx \epsilon = 8.24\cdot 10^{-9}
$$
Also note that we know that $u = 1\pm\epsilon$. Hence from \ref{medual},
$$
m_e = m/u = m/(1\pm\epsilon) \approx m(1\pm \epsilon)
$$
Taking the lower value of this we get the positron mass,
$$
m_{positron} = 0.510\,998\,946\,2 \quad [Mev/c^2].
$$
Meassured is,
$$
m^*_{positron} = 0.510\,998\,946\,1(13)\quad [Mev/c^2].
$$
\subsection{An addition theorem of charge streams and a fundamental scaling
property}
On the other hand if we overlay many of these geometrical structure and
span a spherical symmetric object, the only interacting will be done with
parallel torus structures if they are separated (far away) and there will
be one such pair for every direction. And hence the symmetric usual Coulomb
law naturally applies. Also as the hole constructions is defined as a limit
between of proper EM theoretical objects, we will understand that the
magnetic field will properly appear when we change reference frame. So in
all we have managed to reproduced our macroscopic understanding from these
small building blocks.
Consider what will happen when we overlay two loops at a certain point. To
maintain the overall limit balance we need $c_a\to xc_a$, $c_b\to xc_b$. To
leave charge invariant we then need $rc_a$,$rc_b$ to be invariant as seen
by \ref{sys1} and \ref{sys2} to be constant. Thus mean $r\to r/x$. This
imply $h\to hx$ and $v\to v$ are invariant as $R$ and the pitch is the
same. Also,
$$
E \to Ex, \quad \text{or},\quad m \to mx
$$
Hence in the end, $l \to lx$ for the individual systems. This means that we
can average naturally the loops in the sphere and if add only loops
pointing towards the upper half uniformly we realize that by vector
addition, the overall angular momentum becomes the famous,
$$
l_z = \hbar/2.
$$
\end{document}
