I have done a combined=C2=A0art and science project that y=
ou can find at http:///itampe.com and the=
happy ending is at the end of this mail (if you are into science) I will t=
ake responsibility for the content but allow anyone copy=C2=A0and add to th=
e content and improve, but the core need to be unchanged if you choose to d=
o so. The basic idea behind the project is that in my mind we are destroyin=
g 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 clai=
m that my ideas are unique. The only way forward is to act as a team ... Pl=
ease copy to anyone that you think=C2=A0should read this.

Copyright

July, 2023 Stefan Israelsson Tampe

Oh, the happy=C2=A0ending, here is the explanation of how nature wor=
ks ;-)

\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 \dra= w[-triangle 90] (0,0) -- +(.1,0);}

\title{On Charge}

\author{Stef= an Israelsson Tampe}

\date{\today}

\begin{document}

\maket= itle

\begin{abstract}

=C2=A0 We will in this document assume that= a charged particle (electron) is built up by (similar to super string theo= ry in a sense) of constellation of loops that has a very peculiar form of i= nteraction 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 ou= r 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 dif= fer slightly between positive and negative charge meaning in the end a diff= erence between particle mass and anti particle mass. Especially we reproduc= e the result that the electron and possitron differes and the resulting mas= s of the positron is corect within measurement errors. We will also show th= at a stable system consists of two almost similar loops or helical paths th= at have opposite sign. We will show that the positive and negative charge i= s constant and the same. We will show how how mass can be calculated and ho= w we can calculate angular momentum which makes it possible to deduce infor= mation 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{a= bstract}

\section{The main model assumptions}

We will base our an= alysis of a basic object that is a stream of charge that has no mass and mo= ve at the speed of light. It will also have the property that it only inter= acts if two infinitesimal line segments are parallel and directed in the sa= me 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 impli= citly assume that each particle is composed of objects that is a basic obje= ct 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 ass= ume that each loop has a fixed amount of charge so that as we enlarge the l= oop, 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 ref= erence frame. The energy is,

$$

q^2 /r

$$

As the speed of refer= ence frame (defined through a limiting procedure) is the same independent o= f if we move the system or not defining the interaction in this frame is he= nce Lorentz invariant.

The observant reader would realize that there= are some issues with these objects. We will consider the streams as a limi= t of a sequence of objects of the kind,

$$

\lim_{n\to \infty, v\to c}= \sum_{i=3D0}^n e_a\frac{\delta_{r_i^a(t) =3D (i/n+vt)\hat x}}{\sqrt{n}}

$$

Where we have $n$ equally spaced normal mass less electron's eve= nly distributed on [0,1] at $t=3D0$ and they move with the speed of light a= long the x-axis. We will assume here that nearby electrons are not interact= ion, but in stead if we take a parallel stream,

$$

\lim_{n\to \infty,= v\to c} \sum_{i=3D0}^n e_a\frac{\delta_{r_i^b(t) =3D (i/n+vt)\hat x + h \h= at 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 th= at 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 larg= e 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 s= treams 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 th= at they do not interact). So will assume that only $r_i^a,r_i^b$ are intera= cting 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 i= t 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 para= llel 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 i= nteraction is done in the rest reference system, one can always consider on= ly the electrostatic interaction when exploring a certain geometrical setup= , which is similar to quantum electrodynamics that does also not have inter= nal 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.

\p= agebreak

\section{The Loop}

\def\dh {2}

\def\h =C2=A0{0.25}

\begin{figure}[h]

=C2=A0 \begin{center}

=C2=A0 =C2=A0 \begin{tikzpic= ture}[scale=3D0.7, transform shape]

=C2=A0 =C2=A0 =C2=A0 \begin{scope}[e= very node/.style=3D{sloped,allow upside down}] =C2=A0 =C2=A0

=C2=A0 =C2= =A0 =C2=A0 =C2=A0 \draw [black] (0, 0) to (\dh*2.5, 0);

=C2=A0 =C2=A0 = =C2=A0 =C2=A0 \draw (2.5*\dh, 0) node {\midarrow} (2.5*\dh,0);

=C2=A0 = =C2=A0 =C2=A0

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [black] (0, 0) to (\dh*= 2.5, 0);

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw (1.5*\dh, 0) node {\midarrow}= (1.5*\dh,0);

=C2=A0 =C2=A0 =C2=A0

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw= (0, 0) circle [radius=3D1.5*\dh];

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw (0,= 0) circle [radius=3D2.5*\dh];

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [domain= =3D10:80] plot ({3*\dh*cos(\x)}, {3*\dh*sin(\x)});

=C2=A0 =C2=A0 =C2=A0 = =C2=A0 \draw [domain=3D15:85] plot ({2*\dh*cos(\x)}, {2*\dh*sin(\x)});

<= br>=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [black] (5,0) to ({3*\dh*cos(10)}, {3*= \dh*sin(10)});

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [black] (3,0) to ({2*\d= h*cos(15)}, {2*\dh*sin(15)});

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \node at (= 2,0.3) {$vr$};

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \node at (4,0.3) {$r$};

= =C2=A0 =C2=A0 =C2=A0 =C2=A0 \node at (5.7,0.4) {$hr$};

=C2=A0 =C2=A0 =C2= =A0 \end{scope}

=C2=A0 =C2=A0 \end{tikzpicture}

=C2=A0 \end{center}**\end{figure}**

Consider 2 concentric loops stacked above each other = forming a double cylindrical entity. the charge in the inner loop is positi= ve and the charge in the outer loop is negative (we will consider a reverse= d version later). We will attache a constant charge density of the streams = will be constant, $e_a$ for negative and $a_b$ =C2=A0for 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 attra= ction if not centered so we expect this selection of geometry to be stable.= let $r_b=3Dvr$ be the inner radius and $r_a=3Dr$ 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 t= he charge density at the outer radius and $e_b$ the inner radius charge den= sity, hence the constraint is,

\begin{figure}[h]

=C2=A0 \begin{ce= nter}

=C2=A0 =C2=A0 \begin{tikzpicture}[scale=3D0.7, transform shape]

=C2=A0 =C2=A0 =C2=A0 \draw [domain=3D80:100] plot ({3*\dh*cos(\x)}, {3*\dh= *sin(\x)});

=C2=A0 =C2=A0 =C2=A0 \draw [domain=3D80:100] plot ({2*\dh*co= s(\x)}, {2*\dh*sin(\x)});

=C2=A0 =C2=A0 =C2=A0 \draw [black] (0,0) t= o ({3*\dh*cos(80)}, {3*\dh*sin(80)});

=C2=A0 =C2=A0 =C2=A0 \draw [black]= (0,0) to ({3*\dh*cos(100)}, {3*\dh*sin(100)});

=C2=A0 =C2=A0 =C2=A0= \node at (0, 4.2) =C2=A0 {$dx_b$};

=C2=A0 =C2=A0 =C2=A0 \node at (2, 4.= 2) {$dx_b =3D r_b\,d\theta$};

=C2=A0 =C2=A0 =C2=A0 \node at (0, 6.2) =C2= =A0 {$dx_a$};

=C2=A0 =C2=A0 =C2=A0 \node at (2, 6.2) {$dx_a=3D r_b\,d\th= eta$};

=C2=A0 =C2=A0 \end{tikzpicture}

=C2=A0 \end{center}

\end{fi= gure}

$$

er_a d\theta =3D (e_ar_a - e_br_b)d\theta =3D (e_a - e_b= v)r_ad\theta =3D (e_a - ve_b)rd\theta.

$$

Or,

$$

e =3D e_a - ve= _b

$$

We shall consider scaling properties and hence it is natural to= consider $e=3Due_a$. The dual setup is considering,

$$

-e =3D be_a -= e_b.

$$

For this case we will considering $e=3Due_b$ as a scaling. W= hen we multiply two of these streams we will consider the ``square root'= ;' of a delta measure that is made stringent by the limiting argument a= bove. 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

\en= d{equation}

For the energy relation below. Also when we want to study th= e ``energy density'' we will mean then that we need to study this e= ffect on the paired =C2=A0$r_i^a,r_i^b$. Then summing the effect on the uni= t length will lead to taking these values.

\begin{equation}

=C2=A0 \l= abel{charge2}

=C2=A0 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. thi= s makes the integrated $e$ vary when you vary the radius, and we will see t= hat 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 signifi= cantly if they is located in 2 parallel planes. And there they interact onl= y on a circle of the radius, say R and along this axis we will scale the ch= arge when we consider the torus as a surface else we will consider it as st= acked 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 =3D k\frac{e_ae_br_ar_b}{r_a-r_b}d\theta =3D 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 $h' =3D hr$ we will see that the forces on o= ne segments in one direction is

$$

F =3D k\frac{e_*^2r_*}{(hr)^2}(1+2= ^{-2}+3^{-2} + \ldots) =3D \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 sid= es we get the energy by integrating $hr$ to,

$$

V_{h,*}d\theta =3D 2\= zeta_h\frac{ke_*^2r_*^2}{hr}d\theta.

$$

So the total energy for one l= oop is,

$$

E =3D (V_{h,1} + V_{h,2} + 2 V_l)2\pi =3D 2\pi k r \Big(\f= rac{2\zeta_h}{h}(e_a^2 + e_b^2v^2) - 2 \frac{e_ae_bv}{|1-v|} \Big).

$$**Using $e =3D e_a-ve_b$ in this expression for the energy, we note that we=
can search to find the stationary point varying $x=3Dve_b$ and keeping the=
rest constant, =C2=A0while also introducing the obvious $A,B$ to this lead=
s to,**

$$

A(2e+2x+2x) - B(e + 2x) =3D 0.

$$

Or,

$$

(2A-B)(= 2x + e) =3D 0.

$$

So,

$$

x =3D ve_b =3D -e/2

$$

This mean= s that $ve_v$ tend to go to zero unless,

\begin{equation}

=C2=A0 \lab= el{cond1}

=C2=A0 2A-B =3D 0 \Leftrightarrow \frac{2\zeta_h}{h} =3D \frac= {1}{|1-v|}.

\end{equation}

For which energy wise it can vary freely! = To simplify the expression for the energy, let first $e_b =3D ue_a$. Let $w= =3Duv$ and note $e =3D e_a(1-w)$. Then,

$$

E =3D 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 =3D 2\pi kre_a^2\frac{2\zeta_h}{h}(1 + w^2 - 2w).

$$=

Complete the square and we get,

\begin{equation}

=C2=A0 \label{en= ergy}

=C2=A0 E =3D 2\pi kr \frac{2\zeta(2)}{h}(e_a(1-w))^2 =3D 4 \pi k r= \frac{\zeta(2)}{h}e^2.

\end{equation}

Note that this condition is inv= ariant of how we combine the charges. To evaluate the energy density and ap= ply limits on them as the system want to scale down in order to minimize en= ergy. Assume the condition \ref{charge2} for evaluating this limit.The char= ge densities at the loop $a$ is,

$$

\rho_a =3D ke_a^2\Big( \frac{2\ze= ta_h}{h} - \frac{u}{1-v} \Big)\frac{1}{r}.

$$

Or using \ref{cond1}, <= br>\begin{equation}

=C2=A0 \label{da}

=C2=A0 \rho_a =3D ke_a^2\frac{2= \zeta_h}{h}(1-u)\frac{1}{r}.

\end{equation}

The density at loop $b$ i= s,

$$

\rho_b =3D ke_a^2\Big( \frac{2\zeta_h}{h}u^2 - \frac{u}{1-v} \B= ig)\frac{1}{r}.

$$

Or again using \ref{cond1},

\begin{equation}

=C2=A0 \label{db}

=C2=A0 \rho_a =3D ke_a^2\frac{2\zeta_h}{h}(u^2-u)\fra= c{1}{r}.

\end{equation}

Hence if these two densities are at a positiv= e and negative limit, we need to have (using \ref{da} and \ref{db})

$$**\rho_a =3D c_a,\quad \rho_b =3D -c_b**

$$

To simplify the analysis of= this, use \ref{cond1} and take,

$$

C_* =3D c_**C =3D c_*\frac{h}{2\z= eta_h k e_a^2} =3D c_* \frac{|1-v|}{k e_a^2}.

$$

Then,

\begin{alig= n}

=C2=A0 \frac{|1-u|}{r} =C2=A0 =C2=A0&=3D C_a =3D c_aC, \label{sys= 1}\\

=C2=A0 u\frac{|1-u|}{r} =C2=A0 &=3D C_b =3D c_bC. \label{sys2}<= br>\end{align}

Note that this result is independent how we combined the = charges to $e$. Hence dividing \ref{sys2} with \ref{sys1},

\begin{equati= on}

=C2=A0 \label{uu}

=C2=A0 e_b/e_a =3D u =3D C_b/C_a =3D c_b/c_a.**\end{equation}**

The constraint \ref{sys1} is,

$$

\frac{|1-u|}{r} = =3D c_a C =3D c_a\frac{|1-v|}{k e_a^2}.

$$

Or,

$$

\frac{e_a^2(|= 1-u|)}{r} =3D c_a\frac{|1-v| }{k}.

$$

Hence,

\begin{equation}

= =C2=A0 \label{vee}

=C2=A0 |1-v| =3D \frac{k e_a^2|1-u|}{r c_a}.

\end{= equation}

in the dual situation we get $u'=3Dc_a/c_b$ and $e_a \to e= _b$ and for this case,

$$

|1-v'|=3D \frac{k e_b^2|1-u'|}{r= 9; c_b} =3D \frac{k e_be_a|1-u|)}{r' c_b} =3D\frac{k e_ae_a|1-u|}{r'= ; c_a} =3D |1-v|\frac{r}{r'}

$$

We can also reformulate the condi= tion for $e$, using \ref{vee} as,

\begin{equation}

=C2=A0 \label{fund= }

=C2=A0 e =3D e_a(1-uv) =3D e_a(1-u) + e_au(1-v) =3D e_a - e_b + \frac{= kue_a^2}{rc_a}(e_a-e_b) =3D \Delta \Big(1 + \frac{ke_be_a}{rc_a}\Big)

\= end{equation}

where we used $\Delta =3D e_a-e_b$.

The dual expression= is then,

\begin{equation}

=C2=A0 \label{fund_d}

=C2=A0 e' =3D= - \Delta \Big(1 + \frac{ke_be_a}{r'c_b}\Big).

\end{equation}

So= in order for $e=3D-e'$ we need,

\begin{equation}

=C2=A0 \label{r= adii}

=C2=A0 r' =3D r/u.

\end{equation}

Hence

$$

(1-v= 9;) =3D -(1-v)u

$$

And also $h' =3D h u$. We can solve for $r$ in= \ref{fund},

\begin{equation}

=C2=A0 \label{radii0}

=C2=A0 r=3D \f= rac{ke_be_a}{c_a\Big(\frac{e}{\Delta}-1 \Big)}.

\end{equation}

But no= t only this, we also note that starting with,

$$

e =3D e_a|1-u|+e_au|= 1-v|

$$

And using \ref{sys1},

$$

e =3D D\frac{rc_a}{e_a} + e_au= |1-v|,

$$

=C2=A0with,

$$

D =3D \frac{h}{2\zeta_h k}.

$$

A= ssuming $h,v,u$ constant we can minimize the energy by minimizing $e$ to ge= t,

\begin{equation}

=C2=A0 \label{e_a}

=C2=A0 e_a =3D \sqrt{\frac{= D r c_a}{u|1-v|}} =3D \sqrt{\frac{\frac{h}{2\zeta_h} r c_a}{k u|1-v|}} =3D = \sqrt{\frac{r c_a}{k u}}.

\end{equation}

Hence from \ref{sys1},

$$=

\frac{r c_a}{k u}|1-u| =3D r c_a \frac{|1-v|}{k}.

$$

Or,

\begi= n{equation}

=C2=A0 \label{uv}

=C2=A0 |1-u| =3D u|1-v|

\end{equatio= n}

Hence,

\begin{equation}

=C2=A0 \label{ee}

=C2=A0 e =3D 2 e_a= |1-u| =3D 2\sqrt{\frac{r c_a}{k u}}|1-u|.

\end{equation}

Also,

\be= gin{equation}

=C2=A0 \label{ee2}

=C2=A0 e =3D 2 \frac{e_a}{c_a}|c_a-c= _b|.

\end{equation}

The constraint \ref{cond1} implies,

\begin{equ= ation}

=C2=A0 \label{h_v}

=C2=A0 h =3D |1-v|2\zeta_h =3D \frac{|1-u|2= \zeta_h}{u}

\end{equation}

And $h'=3D hu$. Now the actual pitch i= s $hr$ is then invariant. So the argument for equal charge would that the m= ost energetically favorable action when a negative and positive charge form= is an alignment and hence equal pitch, hence the negative and positive cha= rge 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},

\b= egin{equation}

=C2=A0 \label{ee2}

=C2=A0 e^2 =3D 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 =3D m c^2 =3D 4 \pi kr\fr= ac{\zeta_h}{h}e^2 =3D 2 \pi \frac{kr}{|1-v|}e^2

$$

Using \ref{uv}, w= ith this, we get,

$$

mc^2 =3D \frac{2 \pi k r u e^2}{|1-u|}.

$$

So,

\begin{equation}

=C2=A0 \label{masss}

=C2=A0 m =3D \eta \frac= {2 \pi k r u e^2}{|1-u|c^2}

\end{equation}

(we will discuss $\eta$ so= on). And hence the dual relationship,

\begin{equation}

=C2=A0 \label{= mass}

=C2=A0 m' =3D \frac{m}{u^2}.

\end{equation}

Note that th= e 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,

$$

\et= a =3D 1\quad [m].

$$

We will need that to not confuse the astute read= er that checks the calculation by examining the units. Hence $m$ will have = the unit $[\text{kg}]$.

\subsection{Stacking into a torus}

Previo= usly we was working with a system where we stacked loops on top of each oth= er to form a cylindrical structure. Now instead we connect all loops so the= y 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 fo= rm,

$$

h' =3D 2\pi R h.

$$

From this we get the dual condit= ion $R'=3DR$. But the stacking of the loop is although possible mathema= tically, hard to motivate for a stable structure. However if we transform t= he 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. I= n 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 repul= sively 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 =3D \frac{R_0}{\gamma}

$$

We will evaluate the interactio= n in the rest frame.=C2=A0 So we stack $n'$ of them and therefore,

$= $

n' h' r =3D 2\pi R.

$$

Or,

$$

hr =3D \frac{1}{n= 9;}

$$

As the distance between the paths are different we realize tha= t this can't be exactly try as we have a contraction in the closest $R-= r$ distance, hence we actually have,

$$

n' h' r =3D 2\pi (R-r= ) =3D \frac{2 \pi R}{f},

$$

with,

$$

f =3D \frac{1}{1-r/R}.

= $$

Hence,

\begin{equation}

hrf =3D \frac{1}{n'}.

\end{equat= ion}

The attractive energy will be as before as that is independent of a= ny 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 to= rus is fixed, e.g. $n'$. Then we know that only the loops will need to = scale. hence we will get =C2=A0from \ref{vee}, \ref{radii0} and \ref{energy= },

\begin{align}

=C2=A0 e =C2=A0&\to xe =C2=A0 =C2=A0 =C2=A0\\

=C2=A0 r =C2=A0&\to r/x^2 =C2=A0 \\

=C2=A0 v =C2=A0&\to v, =C2= =A0 =C2=A0 =C2=A0\\

=C2=A0 u =C2=A0&\to u, =C2=A0 =C2=A0 =C2=A0\\

=C2=A0 h =C2=A0&\to hx^2, =C2=A0 \\

=C2=A0 rh &\to rh, =C2=A0 = =C2=A0 \\

=C2=A0 E =C2=A0&\to x^4E, =C2=A0 =C2=A0 \\

=C2=A0 m =C2= =A0&\to x^4m.

\end{align}

This means that in order to maintain th= e same scaling we must have,

\begin{align}

=C2=A0 R =C2=A0 =C2=A0 &am= p;\to R/x^2, \\

=C2=A0 R_0 =C2=A0 &\to R_0/x^2, =C2=A0 \\

=C2=A0 = r/R =C2=A0 &\to R/r, =C2=A0 \\

=C2=A0 f =C2=A0 =C2=A0 &\to f, = =C2=A0 =C2=A0 \\

=C2=A0 rhf =C2=A0 &\to rhf.

\end{align}

Now a= s the helix will stretch with the $R$ we see that,

\begin{align}

=C2= =A0 E_{tot} &\to E_{tot}, \\

=C2=A0 m_{tot} &\to m_{tot}, \\

= =C2=A0 e_{tot} &\to e_{tot}.

\end{align}

\subsection{Angular = momentum}

The per loop angular momentum is,

$$

l =3D m\gamma(v_h)v= _hR =3D 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 sc= ales as $R$. But as the number so turns along the helix is invariant, we fi= nd that the pitch distance also scales as $R$ which leaves the =C2=A0veloci= ty invariant. Hence $v_h$ is invariant of the scaling and hence the total a= ngular moment which is $n'$ such copies is invariant of the scale.

\= begin{align}

=C2=A0 v_h =C2=A0 =C2=A0 &\to v_h, =C2=A0 \\

=C2=A0 = l =C2=A0 =C2=A0 =C2=A0 &\to l, =C2=A0 =C2=A0 \\

=C2=A0 l_{tot} =C2= =A0&\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} =3D \fr= ac{2\pi R}{L} =3D \frac{2\pi R_0/\gamma}{L_0\gamma} =3D \frac{2\pi R_0}{L_0= }.

$$

Anyhow if we factor in the need to remove from the outer loop t= he same quantity from the inner loop we get,

$$

l =3D m v_h R_0 |1-v|=

$$

$$

l_{tot} =3D n'l =3D n' m v_h R_0 |1-v| =3D n&#= 39; \eta \frac{2 \pi k r u e^2}{|1-u|c^2}\frac{c2\pi R_0}{L_0}R_0|1-v| =3D = \eta \frac{A_0R_0}{L_0}\frac{ k u e^2}{|1-u|c}|1-v|,

$$

with $A_0$ be= ing the torus area e.g,

$$

A_0=3D 2\pi r 2\pi R_0.

$$

Using \re= f{uv} we find,

$$

l_{tot} =3D \eta \frac{A_0R_0}{L_0} \frac{k e^2}{c}= .

$$

In order for the charge to be properly (hopefully) managed we ne= ed,

$$

\frac{A_0 2\pi R_0}{L_0} =3D n'

$$

E.g. we need to s= cale down the area in order to compensate for the extra space the helical p= ath takes. As we have $n'=3D 1/hr$ (forgetting about $f$) identical pit= ches and hence we conclude that,

\begin{equation}

=C2=A0 \label{hbard= ef}

=C2=A0 \hbar =3D \int l_{tot}\,d\theta =3D \eta \frac{A_0 2\pi R_0}{= L_0}\frac{ k u e^2}{|1-u|c}|1-v| =3D \eta n' \frac{k r e^2}{c} =3D \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 mom= entum.

$$

l_{tot} =3D n\hbar

$$

We can solve for $hr$

$$

hr =3D \eta \frac{k e^2}{\hbar c} =3D \alpha \approx \frac{1}{137} =3D \fr= ac{1}{n'}

$$

Which indicate why Wolfgang Pauli's quest to sea= rch for why $1/\alpha$ was almost a natural number (137) may have a partial= answer.

\subsection{Defining the zeta factor}

Consider $N$ charg= es evenly distributed on a unit circle. Let's study the forces on one s= ingle charge. then they are locates on $e^{2\pi k/N},k=3D0,\ldots N-1$. The= force at $k=3D0$ is then. Now we would not like to cancel any of their con= tributions to action at hence we get

$$

V(N) =3D \sum_{k=3D1}^N \frac= {h'r}{R}\frac{1}{|e^{2\pi i k/N}-1|}.

$$

Now,

$$

|e^{2\pi i= k/N}-1|^2 =3D (e^{2\pi i k/N}-1)(e^{-2\pi i k/N}-1) =3D 2 - 2\cos(2\pi k/N= ).

$$

Hence we are left with,

$$

V(N) =3D \frac{1}{\sqrt{2}}\su= m_{k=3D1}^N \frac{h'r}{R}\frac{1}{\sqrt{1-\cos(2\pi k/N)}}.

$$

Us= ing the trigonometric identity for the double cosine's,

$$

1-\cos= (2\pi k/N) =3D 1 - \cos^2(\pi k/N) + \sin^2(\pi k / N) =3D 2\sin^2(\pi k / = N).

$$

Hence

$$

V(N) =3D \frac{1}{2} \sum_{k=3D1}^N \frac{h'= ;r}{R}\frac{1}{\sin(\pi k/N)}.

$$

So we will have,

\begin{equation= }

=C2=A0 \label{zeta}

=C2=A0 \zeta_h(N) =3D N \sum_{k=3D1}^{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}

=C2=A0 =C2=A0\zeta_h(N) > N\sum_{k=3D1}^{N} \fra= c{\pi \alpha}{\pi k/N} > N^2\alpha(\ln(N))

\end{equation}

hence

\begin{equation}

=C2=A0 \label{estimate0}

=C2=A0 \zeta_h(N) > N^2= \alpha\ln(N)

\end{equation}

A direct calculation with $N=3D137$ gives= ,

\begin{equation}

=C2=A0 \label{estimate}

=C2=A0 \zeta_h(N) \appr= ox N 691\alpha.

\end{equation}

\subsection{Numerology}

The followi= ng expression is a quite good equation for the fine structure constant,

= $$

\frac{\alpha}{1+\frac{\alpha}{1 - (2\pi-1)^2}} =3D \frac{1}{137}

$= $

We can explore this further and find another expression,

$$

\fra= c{\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}} =3D \frac{= 1}{137}

$$

We could postulate from this,

$$

\frac{\alpha}{1-\fr= ac{\alpha}{x^2-1}} =3D \frac{1}{137}.

$$

With $x$ satisfying,

$$**x =3D 1 - \frac{2\pi}{1-\frac{4\pi\alpha}{1+\frac{2\pi}{1-\frac{4\pi\alph=
a}{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 concl= uded that

$$

hrf =3D \alpha f =3D \frac{1}{137}.

$$

So,

$$**\frac{\alpha}{1-\frac{r}{R}} =3D \frac{1}{137}.**

$$

Using the found = expression we could match this with the found expresison,

\begin{equatio= n}

=C2=A0 \label{approx}

=C2=A0 \frac{r}{R} \approx \frac{\alpha}{(2\= pi-1)^2-1}.

\end{equation}

But this is the same as,

$$

\frac{r}= {R} =3D \frac{rh}{hR} =3D \frac{\alpha}{hR_0}.

$$

Hence we can identi= fy,

$$

x^2 =3D hR + 1.

$$

Now $x^2 =3D28.7778$ means

$$

h= R_0 =3D 27.7778

$$

Or

$$

\alpha R =3D 27.7778 r.

$$

Or**$$**

R =3D 3807 r.

$$

Now for the velocities we have,

$$

c^2= =3D v_h^2 + v_r^2

$$

But,

$$

v_r/c =3D \frac{2\pi r}{L} =3D \f= rac{2\pi r}{\sqrt{(2\pi r)^2 + (r h 2\pi R)^2}}.

$$

Rearenging we fin= d

$$

v_r/c =3D \frac{1}{\sqrt{1 + (hR)^2}}.

$$

Solving for $hR$= we get,

$$

hR_0/\gamma =3D hR =3D \sqrt{\gamma^2 - 1}.

$$

From= this we can identify

\begin{equation}

=C2=A0 \label{gamma}

=C2=A0= \gamma \approx 5.3

\end{equation}

Finally ass $\alpha$ is invariant = of the duality, we conclude that

$$

R' =3D =C2=A0R/u.

$$

Bu= t $R_0$ is invariant. Hence $\gamma' =3D \gamma u$ whcih means

\begi= n{equation}

=C2=A0 \label{medual}

=C2=A0 \gamma' m' =3D \frac= {\gamma m}{u}.

\end{equation}

\subsection{On $n'$}

For= m \ref{cond1} we see that $h'$ satisfies,

$$

h' =3D hr 2\pi R= =3D \alpha 2\pi R =3D =C2=A0\frac{|1-u|2\zeta_h}{u} \approx \frac{|1-u|}{u= } n'\alpha\ln(n').

$$

Or

$$

R =3D C n'\ln(n').<= br>$$

Which means,

$$

r =3D \frac{r}{R} C n'\ln(n').

$$=

But we also have from \ref{e_a},

$$

r =3D Ce_a^2/n'^2

$$**Equating to,**

$$

e_a^2 =3D C' \frac{r}{R}n'\ln(n') =3D n= '\ln(n')

$$

But as we have an approximate postulated expressi= on 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 $n= 9; =3D 137$.

\subsection{On mass}

If we consider the total mass s= cale invariant we get (using n' copies),

\begin{equation}

=C2=A0 = \label{mdef}

=C2=A0 m_e =3D \gamma n' m =3D \eta n' \gamma =C2= =A0\frac{2 \pi k r u e^2}{|1-u|c^2} =3D \eta n' \gamma \frac{\alpha 2 \= pi k u e^2}{h |1-u|c^2}.

\end{equation}h

e electron mass equation \re= f{mdef} and the condition for $h$, \ref{h_v}, and scaling down to with $\al= pha$,

\begin{equation}

=C2=A0 \label{mz}

=C2=A0 m_e =3D \eta n'= ; \gamma \alpha \frac{2 \pi k u e^2}{\frac{|1-u|2\zeta_h}{u} |1-u|c^2} =3D = \eta n' \alpha \gamma\frac{\pi k e^2}{h\zeta_h c^2} \Big( \frac{u}{|1-u= |}\Big )^2=3D \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 \hba= r}{c} \Big( \frac{u}{|1-u|}\Big )^2.

$$

Taking (\ref{gamma}), $\gamma= =3D 5.3$ we find,

$$

|1-u| \approx \epsilon =3D 8.24\cdot 10^{-9}

$$

Also note that we know that $u =3D 1\pm\epsilon$. Hence from \ref{me= dual},

$$

m_e =3D m/u =3D m/(1\pm\epsilon) \approx m(1\pm \epsilon)**$$**

Taking the lower value of this we get the positron mass,

$$

m= _{positron} =3D 0.510\,998\,946\,2 \quad [Mev/c^2].

$$

Meassured is,<= br>$$

m^*_{positron} =3D 0.510\,998\,946\,1(13)\quad [Mev/c^2].

$$

\subsection{An addition theorem of charge streams and a fundamental sc= aling property}

On the other hand if we overlay many of these geometrica= l structure and span a spherical symmetric object, the only interacting wil= l be done with parallel torus structures if they are separated (far away) a= nd there will be one such pair for every direction. And hence the symmetric= usual Coulomb law naturally applies. Also as the hole constructions is def= ined as a limit between of proper EM theoretical objects, we will understan= d that the magnetic field will properly appear when we change reference fra= me. So in all we have managed to reproduced our macroscopic understanding f= rom these small building blocks.

Consider what will happen when we o= verlay 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 nee= d $rc_a$,$rc_b$ 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 invar= iant as $R$ and the pitch is the same. Also,

$$

E \to Ex, \quad \tex= t{or},\quad m \to mx

$$

Hence in the end, $l \to lx$ for the individu= al systems. This means that we can average naturally the loops in the spher= e and if add only loops pointing towards the upper half uniformly we realiz= e that by vector addition, the overall angular momentum becomes the famous,=

$$

l_z =3D \hbar/2.

$$

\end{document}

\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 \dra= w[-triangle 90] (0,0) -- +(.1,0);}

\title{On Charge}

\author{Stef= an Israelsson Tampe}

\date{\today}

\begin{document}

\maket= itle

\begin{abstract}

=C2=A0 We will in this document assume that= a charged particle (electron) is built up by (similar to super string theo= ry in a sense) of constellation of loops that has a very peculiar form of i= nteraction 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 ou= r 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 dif= fer slightly between positive and negative charge meaning in the end a diff= erence between particle mass and anti particle mass. Especially we reproduc= e the result that the electron and possitron differes and the resulting mas= s of the positron is corect within measurement errors. We will also show th= at a stable system consists of two almost similar loops or helical paths th= at have opposite sign. We will show that the positive and negative charge i= s constant and the same. We will show how how mass can be calculated and ho= w we can calculate angular momentum which makes it possible to deduce infor= mation 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{a= bstract}

\section{The main model assumptions}

We will base our an= alysis of a basic object that is a stream of charge that has no mass and mo= ve at the speed of light. It will also have the property that it only inter= acts if two infinitesimal line segments are parallel and directed in the sa= me 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 impli= citly assume that each particle is composed of objects that is a basic obje= ct 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 ass= ume that each loop has a fixed amount of charge so that as we enlarge the l= oop, 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 ref= erence frame. The energy is,

$$

q^2 /r

$$

As the speed of refer= ence frame (defined through a limiting procedure) is the same independent o= f if we move the system or not defining the interaction in this frame is he= nce Lorentz invariant.

The observant reader would realize that there= are some issues with these objects. We will consider the streams as a limi= t of a sequence of objects of the kind,

$$

\lim_{n\to \infty, v\to c}= \sum_{i=3D0}^n e_a\frac{\delta_{r_i^a(t) =3D (i/n+vt)\hat x}}{\sqrt{n}}

$$

Where we have $n$ equally spaced normal mass less electron's eve= nly distributed on [0,1] at $t=3D0$ and they move with the speed of light a= long the x-axis. We will assume here that nearby electrons are not interact= ion, but in stead if we take a parallel stream,

$$

\lim_{n\to \infty,= v\to c} \sum_{i=3D0}^n e_a\frac{\delta_{r_i^b(t) =3D (i/n+vt)\hat x + h \h= at 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 th= at 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 larg= e 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 s= treams 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 th= at they do not interact). So will assume that only $r_i^a,r_i^b$ are intera= cting 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 i= t 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 para= llel 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 i= nteraction is done in the rest reference system, one can always consider on= ly the electrostatic interaction when exploring a certain geometrical setup= , which is similar to quantum electrodynamics that does also not have inter= nal 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.

\p= agebreak

\section{The Loop}

\def\dh {2}

\def\h =C2=A0{0.25}

\begin{figure}[h]

=C2=A0 \begin{center}

=C2=A0 =C2=A0 \begin{tikzpic= ture}[scale=3D0.7, transform shape]

=C2=A0 =C2=A0 =C2=A0 \begin{scope}[e= very node/.style=3D{sloped,allow upside down}] =C2=A0 =C2=A0

=C2=A0 =C2= =A0 =C2=A0 =C2=A0 \draw [black] (0, 0) to (\dh*2.5, 0);

=C2=A0 =C2=A0 = =C2=A0 =C2=A0 \draw (2.5*\dh, 0) node {\midarrow} (2.5*\dh,0);

=C2=A0 = =C2=A0 =C2=A0

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [black] (0, 0) to (\dh*= 2.5, 0);

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw (1.5*\dh, 0) node {\midarrow}= (1.5*\dh,0);

=C2=A0 =C2=A0 =C2=A0

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw= (0, 0) circle [radius=3D1.5*\dh];

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw (0,= 0) circle [radius=3D2.5*\dh];

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [domain= =3D10:80] plot ({3*\dh*cos(\x)}, {3*\dh*sin(\x)});

=C2=A0 =C2=A0 =C2=A0 = =C2=A0 \draw [domain=3D15:85] plot ({2*\dh*cos(\x)}, {2*\dh*sin(\x)});

<= br>=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [black] (5,0) to ({3*\dh*cos(10)}, {3*= \dh*sin(10)});

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \draw [black] (3,0) to ({2*\d= h*cos(15)}, {2*\dh*sin(15)});

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \node at (= 2,0.3) {$vr$};

=C2=A0 =C2=A0 =C2=A0 =C2=A0 \node at (4,0.3) {$r$};

= =C2=A0 =C2=A0 =C2=A0 =C2=A0 \node at (5.7,0.4) {$hr$};

=C2=A0 =C2=A0 =C2= =A0 \end{scope}

=C2=A0 =C2=A0 \end{tikzpicture}

=C2=A0 \end{center}

Consider 2 concentric loops stacked above each other = forming a double cylindrical entity. the charge in the inner loop is positi= ve and the charge in the outer loop is negative (we will consider a reverse= d version later). We will attache a constant charge density of the streams = will be constant, $e_a$ for negative and $a_b$ =C2=A0for 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 attra= ction if not centered so we expect this selection of geometry to be stable.= let $r_b=3Dvr$ be the inner radius and $r_a=3Dr$ 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 t= he charge density at the outer radius and $e_b$ the inner radius charge den= sity, hence the constraint is,

\begin{figure}[h]

=C2=A0 \begin{ce= nter}

=C2=A0 =C2=A0 \begin{tikzpicture}[scale=3D0.7, transform shape]

=C2=A0 =C2=A0 =C2=A0 \draw [domain=3D80:100] plot ({3*\dh*cos(\x)}, {3*\dh= *sin(\x)});

=C2=A0 =C2=A0 =C2=A0 \draw [domain=3D80:100] plot ({2*\dh*co= s(\x)}, {2*\dh*sin(\x)});

=C2=A0 =C2=A0 =C2=A0 \draw [black] (0,0) t= o ({3*\dh*cos(80)}, {3*\dh*sin(80)});

=C2=A0 =C2=A0 =C2=A0 \draw [black]= (0,0) to ({3*\dh*cos(100)}, {3*\dh*sin(100)});

=C2=A0 =C2=A0 =C2=A0= \node at (0, 4.2) =C2=A0 {$dx_b$};

=C2=A0 =C2=A0 =C2=A0 \node at (2, 4.= 2) {$dx_b =3D r_b\,d\theta$};

=C2=A0 =C2=A0 =C2=A0 \node at (0, 6.2) =C2= =A0 {$dx_a$};

=C2=A0 =C2=A0 =C2=A0 \node at (2, 6.2) {$dx_a=3D r_b\,d\th= eta$};

=C2=A0 =C2=A0 \end{tikzpicture}

=C2=A0 \end{center}

\end{fi= gure}

$$

er_a d\theta =3D (e_ar_a - e_br_b)d\theta =3D (e_a - e_b= v)r_ad\theta =3D (e_a - ve_b)rd\theta.

$$

Or,

$$

e =3D e_a - ve= _b

$$

We shall consider scaling properties and hence it is natural to= consider $e=3Due_a$. The dual setup is considering,

$$

-e =3D be_a -= e_b.

$$

For this case we will considering $e=3Due_b$ as a scaling. W= hen we multiply two of these streams we will consider the ``square root'= ;' of a delta measure that is made stringent by the limiting argument a= bove. 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

\en= d{equation}

For the energy relation below. Also when we want to study th= e ``energy density'' we will mean then that we need to study this e= ffect on the paired =C2=A0$r_i^a,r_i^b$. Then summing the effect on the uni= t length will lead to taking these values.

\begin{equation}

=C2=A0 \l= abel{charge2}

=C2=A0 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. thi= s makes the integrated $e$ vary when you vary the radius, and we will see t= hat 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 signifi= cantly if they is located in 2 parallel planes. And there they interact onl= y on a circle of the radius, say R and along this axis we will scale the ch= arge when we consider the torus as a surface else we will consider it as st= acked 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 =3D k\frac{e_ae_br_ar_b}{r_a-r_b}d\theta =3D 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 $h' =3D hr$ we will see that the forces on o= ne segments in one direction is

$$

F =3D k\frac{e_*^2r_*}{(hr)^2}(1+2= ^{-2}+3^{-2} + \ldots) =3D \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 sid= es we get the energy by integrating $hr$ to,

$$

V_{h,*}d\theta =3D 2\= zeta_h\frac{ke_*^2r_*^2}{hr}d\theta.

$$

So the total energy for one l= oop is,

$$

E =3D (V_{h,1} + V_{h,2} + 2 V_l)2\pi =3D 2\pi k r \Big(\f= rac{2\zeta_h}{h}(e_a^2 + e_b^2v^2) - 2 \frac{e_ae_bv}{|1-v|} \Big).

$$

$$

A(2e+2x+2x) - B(e + 2x) =3D 0.

$$

Or,

$$

(2A-B)(= 2x + e) =3D 0.

$$

So,

$$

x =3D ve_b =3D -e/2

$$

This mean= s that $ve_v$ tend to go to zero unless,

\begin{equation}

=C2=A0 \lab= el{cond1}

=C2=A0 2A-B =3D 0 \Leftrightarrow \frac{2\zeta_h}{h} =3D \frac= {1}{|1-v|}.

\end{equation}

For which energy wise it can vary freely! = To simplify the expression for the energy, let first $e_b =3D ue_a$. Let $w= =3Duv$ and note $e =3D e_a(1-w)$. Then,

$$

E =3D 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 =3D 2\pi kre_a^2\frac{2\zeta_h}{h}(1 + w^2 - 2w).

$$=

Complete the square and we get,

\begin{equation}

=C2=A0 \label{en= ergy}

=C2=A0 E =3D 2\pi kr \frac{2\zeta(2)}{h}(e_a(1-w))^2 =3D 4 \pi k r= \frac{\zeta(2)}{h}e^2.

\end{equation}

Note that this condition is inv= ariant of how we combine the charges. To evaluate the energy density and ap= ply limits on them as the system want to scale down in order to minimize en= ergy. Assume the condition \ref{charge2} for evaluating this limit.The char= ge densities at the loop $a$ is,

$$

\rho_a =3D ke_a^2\Big( \frac{2\ze= ta_h}{h} - \frac{u}{1-v} \Big)\frac{1}{r}.

$$

Or using \ref{cond1}, <= br>\begin{equation}

=C2=A0 \label{da}

=C2=A0 \rho_a =3D ke_a^2\frac{2= \zeta_h}{h}(1-u)\frac{1}{r}.

\end{equation}

The density at loop $b$ i= s,

$$

\rho_b =3D ke_a^2\Big( \frac{2\zeta_h}{h}u^2 - \frac{u}{1-v} \B= ig)\frac{1}{r}.

$$

Or again using \ref{cond1},

\begin{equation}

=C2=A0 \label{db}

=C2=A0 \rho_a =3D ke_a^2\frac{2\zeta_h}{h}(u^2-u)\fra= c{1}{r}.

\end{equation}

Hence if these two densities are at a positiv= e and negative limit, we need to have (using \ref{da} and \ref{db})

$$

$$

To simplify the analysis of= this, use \ref{cond1} and take,

$$

C_* =3D c_**C =3D c_*\frac{h}{2\z= eta_h k e_a^2} =3D c_* \frac{|1-v|}{k e_a^2}.

$$

Then,

\begin{alig= n}

=C2=A0 \frac{|1-u|}{r} =C2=A0 =C2=A0&=3D C_a =3D c_aC, \label{sys= 1}\\

=C2=A0 u\frac{|1-u|}{r} =C2=A0 &=3D C_b =3D c_bC. \label{sys2}<= br>\end{align}

Note that this result is independent how we combined the = charges to $e$. Hence dividing \ref{sys2} with \ref{sys1},

\begin{equati= on}

=C2=A0 \label{uu}

=C2=A0 e_b/e_a =3D u =3D C_b/C_a =3D c_b/c_a.

The constraint \ref{sys1} is,

$$

\frac{|1-u|}{r} = =3D c_a C =3D c_a\frac{|1-v|}{k e_a^2}.

$$

Or,

$$

\frac{e_a^2(|= 1-u|)}{r} =3D c_a\frac{|1-v| }{k}.

$$

Hence,

\begin{equation}

= =C2=A0 \label{vee}

=C2=A0 |1-v| =3D \frac{k e_a^2|1-u|}{r c_a}.

\end{= equation}

in the dual situation we get $u'=3Dc_a/c_b$ and $e_a \to e= _b$ and for this case,

$$

|1-v'|=3D \frac{k e_b^2|1-u'|}{r= 9; c_b} =3D \frac{k e_be_a|1-u|)}{r' c_b} =3D\frac{k e_ae_a|1-u|}{r'= ; c_a} =3D |1-v|\frac{r}{r'}

$$

We can also reformulate the condi= tion for $e$, using \ref{vee} as,

\begin{equation}

=C2=A0 \label{fund= }

=C2=A0 e =3D e_a(1-uv) =3D e_a(1-u) + e_au(1-v) =3D e_a - e_b + \frac{= kue_a^2}{rc_a}(e_a-e_b) =3D \Delta \Big(1 + \frac{ke_be_a}{rc_a}\Big)

\= end{equation}

where we used $\Delta =3D e_a-e_b$.

The dual expression= is then,

\begin{equation}

=C2=A0 \label{fund_d}

=C2=A0 e' =3D= - \Delta \Big(1 + \frac{ke_be_a}{r'c_b}\Big).

\end{equation}

So= in order for $e=3D-e'$ we need,

\begin{equation}

=C2=A0 \label{r= adii}

=C2=A0 r' =3D r/u.

\end{equation}

Hence

$$

(1-v= 9;) =3D -(1-v)u

$$

And also $h' =3D h u$. We can solve for $r$ in= \ref{fund},

\begin{equation}

=C2=A0 \label{radii0}

=C2=A0 r=3D \f= rac{ke_be_a}{c_a\Big(\frac{e}{\Delta}-1 \Big)}.

\end{equation}

But no= t only this, we also note that starting with,

$$

e =3D e_a|1-u|+e_au|= 1-v|

$$

And using \ref{sys1},

$$

e =3D D\frac{rc_a}{e_a} + e_au= |1-v|,

$$

=C2=A0with,

$$

D =3D \frac{h}{2\zeta_h k}.

$$

A= ssuming $h,v,u$ constant we can minimize the energy by minimizing $e$ to ge= t,

\begin{equation}

=C2=A0 \label{e_a}

=C2=A0 e_a =3D \sqrt{\frac{= D r c_a}{u|1-v|}} =3D \sqrt{\frac{\frac{h}{2\zeta_h} r c_a}{k u|1-v|}} =3D = \sqrt{\frac{r c_a}{k u}}.

\end{equation}

Hence from \ref{sys1},

$$=

\frac{r c_a}{k u}|1-u| =3D r c_a \frac{|1-v|}{k}.

$$

Or,

\begi= n{equation}

=C2=A0 \label{uv}

=C2=A0 |1-u| =3D u|1-v|

\end{equatio= n}

Hence,

\begin{equation}

=C2=A0 \label{ee}

=C2=A0 e =3D 2 e_a= |1-u| =3D 2\sqrt{\frac{r c_a}{k u}}|1-u|.

\end{equation}

Also,

\be= gin{equation}

=C2=A0 \label{ee2}

=C2=A0 e =3D 2 \frac{e_a}{c_a}|c_a-c= _b|.

\end{equation}

The constraint \ref{cond1} implies,

\begin{equ= ation}

=C2=A0 \label{h_v}

=C2=A0 h =3D |1-v|2\zeta_h =3D \frac{|1-u|2= \zeta_h}{u}

\end{equation}

And $h'=3D hu$. Now the actual pitch i= s $hr$ is then invariant. So the argument for equal charge would that the m= ost energetically favorable action when a negative and positive charge form= is an alignment and hence equal pitch, hence the negative and positive cha= rge 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},

\b= egin{equation}

=C2=A0 \label{ee2}

=C2=A0 e^2 =3D 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 =3D m c^2 =3D 4 \pi kr\fr= ac{\zeta_h}{h}e^2 =3D 2 \pi \frac{kr}{|1-v|}e^2

$$

Using \ref{uv}, w= ith this, we get,

$$

mc^2 =3D \frac{2 \pi k r u e^2}{|1-u|}.

$$

So,

\begin{equation}

=C2=A0 \label{masss}

=C2=A0 m =3D \eta \frac= {2 \pi k r u e^2}{|1-u|c^2}

\end{equation}

(we will discuss $\eta$ so= on). And hence the dual relationship,

\begin{equation}

=C2=A0 \label{= mass}

=C2=A0 m' =3D \frac{m}{u^2}.

\end{equation}

Note that th= e 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,

$$

\et= a =3D 1\quad [m].

$$

We will need that to not confuse the astute read= er that checks the calculation by examining the units. Hence $m$ will have = the unit $[\text{kg}]$.

\subsection{Stacking into a torus}

Previo= usly we was working with a system where we stacked loops on top of each oth= er to form a cylindrical structure. Now instead we connect all loops so the= y 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 fo= rm,

$$

h' =3D 2\pi R h.

$$

From this we get the dual condit= ion $R'=3DR$. But the stacking of the loop is although possible mathema= tically, hard to motivate for a stable structure. However if we transform t= he 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. I= n 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 repul= sively 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 =3D \frac{R_0}{\gamma}

$$

We will evaluate the interactio= n in the rest frame.=C2=A0 So we stack $n'$ of them and therefore,

$= $

n' h' r =3D 2\pi R.

$$

Or,

$$

hr =3D \frac{1}{n= 9;}

$$

As the distance between the paths are different we realize tha= t this can't be exactly try as we have a contraction in the closest $R-= r$ distance, hence we actually have,

$$

n' h' r =3D 2\pi (R-r= ) =3D \frac{2 \pi R}{f},

$$

with,

$$

f =3D \frac{1}{1-r/R}.

= $$

Hence,

\begin{equation}

hrf =3D \frac{1}{n'}.

\end{equat= ion}

The attractive energy will be as before as that is independent of a= ny 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 to= rus is fixed, e.g. $n'$. Then we know that only the loops will need to = scale. hence we will get =C2=A0from \ref{vee}, \ref{radii0} and \ref{energy= },

\begin{align}

=C2=A0 e =C2=A0&\to xe =C2=A0 =C2=A0 =C2=A0\\

=C2=A0 r =C2=A0&\to r/x^2 =C2=A0 \\

=C2=A0 v =C2=A0&\to v, =C2= =A0 =C2=A0 =C2=A0\\

=C2=A0 u =C2=A0&\to u, =C2=A0 =C2=A0 =C2=A0\\

=C2=A0 h =C2=A0&\to hx^2, =C2=A0 \\

=C2=A0 rh &\to rh, =C2=A0 = =C2=A0 \\

=C2=A0 E =C2=A0&\to x^4E, =C2=A0 =C2=A0 \\

=C2=A0 m =C2= =A0&\to x^4m.

\end{align}

This means that in order to maintain th= e same scaling we must have,

\begin{align}

=C2=A0 R =C2=A0 =C2=A0 &am= p;\to R/x^2, \\

=C2=A0 R_0 =C2=A0 &\to R_0/x^2, =C2=A0 \\

=C2=A0 = r/R =C2=A0 &\to R/r, =C2=A0 \\

=C2=A0 f =C2=A0 =C2=A0 &\to f, = =C2=A0 =C2=A0 \\

=C2=A0 rhf =C2=A0 &\to rhf.

\end{align}

Now a= s the helix will stretch with the $R$ we see that,

\begin{align}

=C2= =A0 E_{tot} &\to E_{tot}, \\

=C2=A0 m_{tot} &\to m_{tot}, \\

= =C2=A0 e_{tot} &\to e_{tot}.

\end{align}

\subsection{Angular = momentum}

The per loop angular momentum is,

$$

l =3D m\gamma(v_h)v= _hR =3D 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 sc= ales as $R$. But as the number so turns along the helix is invariant, we fi= nd that the pitch distance also scales as $R$ which leaves the =C2=A0veloci= ty invariant. Hence $v_h$ is invariant of the scaling and hence the total a= ngular moment which is $n'$ such copies is invariant of the scale.

\= begin{align}

=C2=A0 v_h =C2=A0 =C2=A0 &\to v_h, =C2=A0 \\

=C2=A0 = l =C2=A0 =C2=A0 =C2=A0 &\to l, =C2=A0 =C2=A0 \\

=C2=A0 l_{tot} =C2= =A0&\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} =3D \fr= ac{2\pi R}{L} =3D \frac{2\pi R_0/\gamma}{L_0\gamma} =3D \frac{2\pi R_0}{L_0= }.

$$

Anyhow if we factor in the need to remove from the outer loop t= he same quantity from the inner loop we get,

$$

l =3D m v_h R_0 |1-v|=

$$

$$

l_{tot} =3D n'l =3D n' m v_h R_0 |1-v| =3D n&#= 39; \eta \frac{2 \pi k r u e^2}{|1-u|c^2}\frac{c2\pi R_0}{L_0}R_0|1-v| =3D = \eta \frac{A_0R_0}{L_0}\frac{ k u e^2}{|1-u|c}|1-v|,

$$

with $A_0$ be= ing the torus area e.g,

$$

A_0=3D 2\pi r 2\pi R_0.

$$

Using \re= f{uv} we find,

$$

l_{tot} =3D \eta \frac{A_0R_0}{L_0} \frac{k e^2}{c}= .

$$

In order for the charge to be properly (hopefully) managed we ne= ed,

$$

\frac{A_0 2\pi R_0}{L_0} =3D n'

$$

E.g. we need to s= cale down the area in order to compensate for the extra space the helical p= ath takes. As we have $n'=3D 1/hr$ (forgetting about $f$) identical pit= ches and hence we conclude that,

\begin{equation}

=C2=A0 \label{hbard= ef}

=C2=A0 \hbar =3D \int l_{tot}\,d\theta =3D \eta \frac{A_0 2\pi R_0}{= L_0}\frac{ k u e^2}{|1-u|c}|1-v| =3D \eta n' \frac{k r e^2}{c} =3D \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 mom= entum.

$$

l_{tot} =3D n\hbar

$$

We can solve for $hr$

$$

hr =3D \eta \frac{k e^2}{\hbar c} =3D \alpha \approx \frac{1}{137} =3D \fr= ac{1}{n'}

$$

Which indicate why Wolfgang Pauli's quest to sea= rch for why $1/\alpha$ was almost a natural number (137) may have a partial= answer.

\subsection{Defining the zeta factor}

Consider $N$ charg= es evenly distributed on a unit circle. Let's study the forces on one s= ingle charge. then they are locates on $e^{2\pi k/N},k=3D0,\ldots N-1$. The= force at $k=3D0$ is then. Now we would not like to cancel any of their con= tributions to action at hence we get

$$

V(N) =3D \sum_{k=3D1}^N \frac= {h'r}{R}\frac{1}{|e^{2\pi i k/N}-1|}.

$$

Now,

$$

|e^{2\pi i= k/N}-1|^2 =3D (e^{2\pi i k/N}-1)(e^{-2\pi i k/N}-1) =3D 2 - 2\cos(2\pi k/N= ).

$$

Hence we are left with,

$$

V(N) =3D \frac{1}{\sqrt{2}}\su= m_{k=3D1}^N \frac{h'r}{R}\frac{1}{\sqrt{1-\cos(2\pi k/N)}}.

$$

Us= ing the trigonometric identity for the double cosine's,

$$

1-\cos= (2\pi k/N) =3D 1 - \cos^2(\pi k/N) + \sin^2(\pi k / N) =3D 2\sin^2(\pi k / = N).

$$

Hence

$$

V(N) =3D \frac{1}{2} \sum_{k=3D1}^N \frac{h'= ;r}{R}\frac{1}{\sin(\pi k/N)}.

$$

So we will have,

\begin{equation= }

=C2=A0 \label{zeta}

=C2=A0 \zeta_h(N) =3D N \sum_{k=3D1}^{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}

=C2=A0 =C2=A0\zeta_h(N) > N\sum_{k=3D1}^{N} \fra= c{\pi \alpha}{\pi k/N} > N^2\alpha(\ln(N))

\end{equation}

hence

\begin{equation}

=C2=A0 \label{estimate0}

=C2=A0 \zeta_h(N) > N^2= \alpha\ln(N)

\end{equation}

A direct calculation with $N=3D137$ gives= ,

\begin{equation}

=C2=A0 \label{estimate}

=C2=A0 \zeta_h(N) \appr= ox N 691\alpha.

\end{equation}

\subsection{Numerology}

The followi= ng expression is a quite good equation for the fine structure constant,

= $$

\frac{\alpha}{1+\frac{\alpha}{1 - (2\pi-1)^2}} =3D \frac{1}{137}

$= $

We can explore this further and find another expression,

$$

\fra= c{\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}} =3D \frac{= 1}{137}

$$

We could postulate from this,

$$

\frac{\alpha}{1-\fr= ac{\alpha}{x^2-1}} =3D \frac{1}{137}.

$$

With $x$ satisfying,

$$

$$

Of cause this is very numerological and are simply fined = tuned with the help of trial and error. Can we motivate this? well we concl= uded that

$$

hrf =3D \alpha f =3D \frac{1}{137}.

$$

So,

$$

$$

Using the found = expression we could match this with the found expresison,

\begin{equatio= n}

=C2=A0 \label{approx}

=C2=A0 \frac{r}{R} \approx \frac{\alpha}{(2\= pi-1)^2-1}.

\end{equation}

But this is the same as,

$$

\frac{r}= {R} =3D \frac{rh}{hR} =3D \frac{\alpha}{hR_0}.

$$

Hence we can identi= fy,

$$

x^2 =3D hR + 1.

$$

Now $x^2 =3D28.7778$ means

$$

h= R_0 =3D 27.7778

$$

Or

$$

\alpha R =3D 27.7778 r.

$$

Or

R =3D 3807 r.

$$

Now for the velocities we have,

$$

c^2= =3D v_h^2 + v_r^2

$$

But,

$$

v_r/c =3D \frac{2\pi r}{L} =3D \f= rac{2\pi r}{\sqrt{(2\pi r)^2 + (r h 2\pi R)^2}}.

$$

Rearenging we fin= d

$$

v_r/c =3D \frac{1}{\sqrt{1 + (hR)^2}}.

$$

Solving for $hR$= we get,

$$

hR_0/\gamma =3D hR =3D \sqrt{\gamma^2 - 1}.

$$

From= this we can identify

\begin{equation}

=C2=A0 \label{gamma}

=C2=A0= \gamma \approx 5.3

\end{equation}

Finally ass $\alpha$ is invariant = of the duality, we conclude that

$$

R' =3D =C2=A0R/u.

$$

Bu= t $R_0$ is invariant. Hence $\gamma' =3D \gamma u$ whcih means

\begi= n{equation}

=C2=A0 \label{medual}

=C2=A0 \gamma' m' =3D \frac= {\gamma m}{u}.

\end{equation}

\subsection{On $n'$}

For= m \ref{cond1} we see that $h'$ satisfies,

$$

h' =3D hr 2\pi R= =3D \alpha 2\pi R =3D =C2=A0\frac{|1-u|2\zeta_h}{u} \approx \frac{|1-u|}{u= } n'\alpha\ln(n').

$$

Or

$$

R =3D C n'\ln(n').<= br>$$

Which means,

$$

r =3D \frac{r}{R} C n'\ln(n').

$$=

But we also have from \ref{e_a},

$$

r =3D Ce_a^2/n'^2

$$

$$

e_a^2 =3D C' \frac{r}{R}n'\ln(n') =3D n= '\ln(n')

$$

But as we have an approximate postulated expressi= on 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 $n= 9; =3D 137$.

\subsection{On mass}

If we consider the total mass s= cale invariant we get (using n' copies),

\begin{equation}

=C2=A0 = \label{mdef}

=C2=A0 m_e =3D \gamma n' m =3D \eta n' \gamma =C2= =A0\frac{2 \pi k r u e^2}{|1-u|c^2} =3D \eta n' \gamma \frac{\alpha 2 \= pi k u e^2}{h |1-u|c^2}.

\end{equation}h

e electron mass equation \re= f{mdef} and the condition for $h$, \ref{h_v}, and scaling down to with $\al= pha$,

\begin{equation}

=C2=A0 \label{mz}

=C2=A0 m_e =3D \eta n'= ; \gamma \alpha \frac{2 \pi k u e^2}{\frac{|1-u|2\zeta_h}{u} |1-u|c^2} =3D = \eta n' \alpha \gamma\frac{\pi k e^2}{h\zeta_h c^2} \Big( \frac{u}{|1-u= |}\Big )^2=3D \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 \hba= r}{c} \Big( \frac{u}{|1-u|}\Big )^2.

$$

Taking (\ref{gamma}), $\gamma= =3D 5.3$ we find,

$$

|1-u| \approx \epsilon =3D 8.24\cdot 10^{-9}

$$

Also note that we know that $u =3D 1\pm\epsilon$. Hence from \ref{me= dual},

$$

m_e =3D m/u =3D m/(1\pm\epsilon) \approx m(1\pm \epsilon)

Taking the lower value of this we get the positron mass,

$$

m= _{positron} =3D 0.510\,998\,946\,2 \quad [Mev/c^2].

$$

Meassured is,<= br>$$

m^*_{positron} =3D 0.510\,998\,946\,1(13)\quad [Mev/c^2].

$$

\subsection{An addition theorem of charge streams and a fundamental sc= aling property}

On the other hand if we overlay many of these geometrica= l structure and span a spherical symmetric object, the only interacting wil= l be done with parallel torus structures if they are separated (far away) a= nd there will be one such pair for every direction. And hence the symmetric= usual Coulomb law naturally applies. Also as the hole constructions is def= ined as a limit between of proper EM theoretical objects, we will understan= d that the magnetic field will properly appear when we change reference fra= me. So in all we have managed to reproduced our macroscopic understanding f= rom these small building blocks.

Consider what will happen when we o= verlay 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 nee= d $rc_a$,$rc_b$ 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 invar= iant as $R$ and the pitch is the same. Also,

$$

E \to Ex, \quad \tex= t{or},\quad m \to mx

$$

Hence in the end, $l \to lx$ for the individu= al systems. This means that we can average naturally the loops in the spher= e and if add only loops pointing towards the upper half uniformly we realiz= e that by vector addition, the overall angular momentum becomes the famous,=

$$

l_z =3D \hbar/2.

$$

\end{document}

It's part of the blog, computers tab, I see guile sch=
eme as an art and have a few suggestions of great improvements for it. But =
you have to demand those features if you like it. And if you do not see we =
have some serious issues in the world, just ignore it as much as you like. =
Enjoy your sleep...=C2=A0

On Sun, 2 Jul 2023, 10:24 Maxime Devos <maximedevos@telenet.be> wrote:

I don't see the relation with Gui= le.=C2=A0 What's guile-devel@gnu.org doing

in the To:?

Best regards,

Maxime Devos

I think the =
guile mailing list has come to suffer seriously from a policing of what is =
appropriate for discussion=2E When I first started getting into guile at th=
e beginning of 2021, the list was more open, active and enjoyable=2E Since =
then it has become a ghost town where tumbleweeds blow by in the time that =
remains between intermittent duals=2E But strangely there are still lots of=
people living in the town (ie guile users), they just seldom visit main st=
reet these days=2E

I'm personally intereste=
d to hear about what Stefan and other Guilers are up to beyond Guile, insof=
ar as nobody is directing hate or bigotry=2E But more importantly, I wish t=
hat Guile could again become a space that welcomes discourse insofar as tha=
t discourse isn't perceived by some as causing harm or stirring irrelevant =
controversy=2E

On Jul 5, 2023, at 17:22, =
Maxime Devos <maximedevos@telenet=2Ebe> wrote:

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(Exim 4.90_1) (envelope-from Op 02-07-2023 om 16:43 sch= reef Stefan Israelsson Tampe:It's part of the blog, computers tab, I see guile scheme as an art = and

have a few suggestions of great improvements for it=2E But you hav= e to

demand those features if you like it=2E

The a= rticle and blog post you sent does not contain anything about Guile

Sch= eme -- there are blog posts that do contain stuff about Guile, but

this= wasn't one of them=2EAnd if you do not see we have some

serious issues in the world, j= ust ignore it as much as you like=2E Enjoy

your sleep=2E=2E=2E

This is non-sequitur, because of the following reasons:

1=2E Seeing or not seeing serious issues in the world does not have

= anything to do with the decision to ignore or not ignore the paper,because it is a physics paper, not a serious issues paper=2E

2=2E I never said I ignored it -- in fact, I skimmed it to see if it

contains anything about Guile or serious issues in the world, and

it turned out that the article only contains physics, which while

interesting, is not something I'm _personally_ interested in=2E

<= br> 3=2E I also did not ignore the top part of the e-mail

(=E2= =80=98=2E=2E=2E The basis idea [=2E=2E=2E] and want us all to start doingsomething=2E Please copy to anyone that you think should read

= this=2E=E2=80=99)=2E It is one of the reasons I sent my question on wh= y

should guile-devel@gnu=2Eorg read this article unrelated to Guil= e=2E

4=2E I asked why guile-devel@gnu=2Eorg is in the To, not whe= ther it

should be read or ignored=2E

More to the point, my = issue with this is that apparently it is

completely off-topic for Guile= , because it is not about Guile at all,

and in particular not about Gui= le development=2E There exist many fora on

which you could talk about = such things (*), without resorting to chain

mails=2E

(*) To list= an example:

<https://forums=2Esufficientve= locity=2Ecom/threads/welcome-to-science-technology=2E41297/>=2E

<= br>Best regards,

Maxime Devos