# Electron (mathematical)

The mathematical electron model

In the mathematical electron model [1], the electron is a mathematical (geometrical) formula fe that embeds (dimensioned) geometrical (Planck unit) objects, yet itself is dimensionless, units = 1 (there is no 'physical' electron), and so is applicable to simulation hypothesis modelling. The dimensioned parameters associated with the electron (mass, wavelength, frequency, charge ...) derive from those embedded Planck units (for mass, length, time, charge ...) the dimension-less mathematical function fe dictating the frequencies of the Planck units, ... where fe is the geometry of 2 dimensionless physical constants, the (inverse) fine structure constant α = 137.035 999 139 (CODATA 2014) and Omega Ω = 2.007 134 9496

${\displaystyle f_{e}=4\pi ^{2}(2^{6}3\pi ^{2}\alpha \Omega ^{5})^{3}=.2389...\;x10^{23}}$, units = 1

### Derivation via Planck units

The Planck units are assigned geometrical objects, and so we can build the electron function fe also as a geometrical object, the construct of these Planck objects. It is these embedded Planck objects that give the electron its physical parameters, furthermore the formula fe also dictates their frequency, and so the geometry of fe has the information of the electron.

Base (Planck) units for mass ${\displaystyle M}$, length ${\displaystyle L}$, time ${\displaystyle T}$, and ampere ${\displaystyle A}$ are constructed as geometrical objects in terms of the fine structure constant α and Omega Ω. A mathematical unit relationship between the objects is dictated by ${\displaystyle n}$ (the unit number);

Geometrical units
Attribute Geometrical object Unit number n
mass ${\displaystyle M=(1)}$ ${\displaystyle 15}$
time ${\displaystyle T=(\pi )}$ ${\displaystyle -30}$
sqrt(momentum) ${\displaystyle P=(\Omega )}$ ${\displaystyle 16}$
velocity ${\displaystyle V=(2\pi \Omega ^{2})}$ ${\displaystyle 17}$
length ${\displaystyle L=(2\pi ^{2}\Omega ^{2})}$ ${\displaystyle -13}$
ampere ${\displaystyle A=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}$ ${\displaystyle 3}$

These objects are not independent of each other but instead are inter-related according to this unit number relationship. This also means that they may overlap and cancel;

${\displaystyle {\frac {(A^{3})(L^{3})}{(T)}}}$, unit number n = 3*3 -13*3 +30 = 0

### Mathematical electron

The electron function (the mathematical formula for the electron) fe incorporates these geometrical base units yet itself is unit-less; units = 1. For example, fe can be defined in terms of σe, where AL as an ampere-meter (ampere-length = e*c) are the units for a magnetic monopole.

${\displaystyle T=\pi ,\;n=-30}$
${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}},\;n=-10}$
${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{2T}}={\frac {(2^{7}3\pi ^{3}\alpha \Omega ^{5})^{3}}{2\pi }},\;n=-10*3+30=0}$
${\displaystyle f_{e}=4\pi ^{2}(2^{6}3\pi ^{2}\alpha \Omega ^{5})^{3}=.23895453...x10^{23},\;units=1}$

##### Electron parameters

Associated with the electron are dimensioned parameters, these parameters however are a function of the base MLTA units, the formula fe dictating the frequency of these units. By setting MLTA to their SI Planck unit equivalents;

electron mass ${\displaystyle m_{e}={\frac {M}{f_{e}}}}$ (M = Planck mass)

electron wavelength ${\displaystyle \lambda _{e}=2\pi Lf_{e}}$ (L = Planck length)

elementary charge ${\displaystyle e=A.T}$ (T = Planck time)

We may interpret this formula for fe whereby for the duration of the electron frequency = 0.2389 x 1023 units of (Planck) time, the electron is represented by AL magnetic monopoles, these then intersect with time T, the units then collapse (units (A*L)3/T = 1), exposing a unit of M (Planck mass) for 1 unit of (Planck) time, which we could define as the mass point-state. Wave-particle duality can then be represented at the Planck level as an oscillation between an electric (magnetic monopole) wave-state (the duration dictated by fe measured in Planck time units), to this M = 1 mass point-state.

By this artifice, although the 'physical' universe is constructed from particles, particles themselves are not physical, they are mathematical. Consequently this approach is applicable to deep universe simulation hypothesis modeling at the Planck scale.

##### Electron Mass

If the particle point-state is a unit of Planck mass, then we have a model for a black-hole electron, the electron function fe centered around this unit of Planck mass. When the wave-state (A*L)3/T units collapse, this black-hole center is exposed for 1 unit of (Planck) time. The electron is 'now' (a unit of Planck) mass M.

Mass in this consideration is not a constant property of the particle, rather the measured particle mass m would refer to the average mass, the average occurrence of the Planck mass point-state as measured over time. As for each wave-state then is a corresponding point-state, and as hv is a measure of the frequency of the wave-state, E = hv = mc2. Notably however the c term is a constant unlike the v term, and so the m term refers to average mass.

If the scaffolding of the universe includes units of (Planck) mass M, then it is not necessary for the particle to have a mass M, instead the point state could be the absence of particle.

##### Quarks

The AL magnetic monopoles may be analogous to quarks, but due to the symmetry and so stability of the geometrical fe there is no fracture point by which an electron could decay, and so resemblance is conjecture. Nevertheless this could explain why the magnitude of electron charge appears to be exactly equivalent to the proton charge (and the sum universe electrically neutral) and so a quark model is worth speculation.

For example, a single AL monopole D = ${\displaystyle \sigma _{e}}$ could equate to a quark with an electric charge of -1/3 e. 3 D quarks would constitute the electron as DDD = (AL)3. There is a candidate for a quark with an charge of 2/3 e, denoted here U = ${\displaystyle \sigma _{t}}$ and is centered on Planck temperature Tp. The symbols U and D are used to illustrate the charge. Note that the units for the U (magnetic monopole) quark are not the same as the units for the D (Planck temperature) quark. Calculating the units directly as units = un gives a basic quark model;

${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}=({2^{7}3\pi ^{3}\alpha \Omega ^{5}}),\;units=u^{-10}}$
${\displaystyle T_{p}={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }},\;units=u^{20},\;...\;\sigma _{t}={\frac {3\alpha ^{2}T_{p}}{2\pi }}=({2^{6}3\pi ^{2}\alpha \Omega ^{5}}),\;units=u^{20}}$

DDD:   ${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{2T}}={\frac {(2^{7}3\pi ^{3}\alpha \Omega ^{5})^{3}}{2\pi }},\;units={\frac {(u^{-10})^{3}}{u^{-30}}}=1,\;charge={\frac {-1}{3}}.{\frac {-1}{3}}.{\frac {-1}{3}}=-1e}$

DDU:   ${\displaystyle \sigma _{e}^{2}\sigma _{t}={(2^{7}3\pi ^{3}\alpha \Omega ^{5})}^{2}{(2^{6}3\pi ^{2}\alpha \Omega ^{5})},\;units=(u^{-10})^{2}(u^{20})=1,\;charge={\frac {-1}{3}}.{\frac {-1}{3}}.{\frac {2}{3}}=0}$

UUU:   ${\displaystyle (2T)^{2}\sigma _{t}^{3}={(2\pi )}^{2}{(2^{6}3\pi ^{2}\alpha \Omega ^{5})}^{3},\;units=(u^{-30})^{2}(u^{20})^{3}=1,\;charge={\frac {2}{3}}.{\frac {2}{3}}.{\frac {2}{3}}=2e}$

UUD:   ${\displaystyle (2T)\sigma _{t}^{2}\sigma _{e}={(2\pi )}{(2^{6}3\pi ^{2}\alpha \Omega ^{5})}^{2}{(2^{7}3\pi ^{3}\alpha \Omega ^{5})},\;units=(u^{-30})(u^{20})^{2}(u^{-10})=1,\;charge={\frac {2}{3}}.{\frac {2}{3}}.{\frac {-1}{3}}=1e}$

All the above return the electron formula fe but in different charge configurations. The proton and neutron quarks would then be constructed upon these configurations, the electron as the fundamental particle.

### Derivation via SI units

#### Magnetic monopole

A magnetic monopole is a hypothesized particle that is a magnet with only 1 pole. The unit for the magnetic monopole is the ampere-meter, the SI unit for pole strength (the product of charge and velocity) in a magnet (A m = e c). A proposed formula for a magnetic monopole σe [2];

${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}ec}{2\pi ^{2}}}=0.13708563....\;x10^{-6},\;units={\frac {Cm}{s}}}$

The following gives a formula for an electron in terms of magnetic monopoles and Planck time (${\displaystyle t_{p}}$).

${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{2t_{p}}}=0.2389....\;x10^{23},\;units={\frac {C^{3}m^{3}}{s^{4}}}}$

However, although this gives us the correct numerical value, this gives us incorrect units when solving the electron mass (${\displaystyle m_{P}}$ = Planck mass).

${\displaystyle m_{e}={\frac {m_{P}}{f_{e}}}=0.910938....\;x10^{30},\;units=kg{\frac {C^{3}m^{3}}{s^{4}}}}$

To resolve this we can consider the possibility that the units for A, m, s are related whereby units ${\displaystyle {\frac {C^{3}m^{3}}{s^{4}}}=1}$

#### Sqrt Planck momentum

The sqrt of Planck momentum is not a recognized constant (it has no SI designation) and so here is denoted as Q with units q whereby

Planck momentum = 2 π Q2, unit = kg.m/s = q2.

${\displaystyle Q=1.019\;113\;411...\;unit=q\;}$

Replacing m with q;

Planck length ${\displaystyle l_{p},\;unit=m=\color {red}{\frac {q^{2}s}{kg}}\color {black}}$

Speed of light ${\displaystyle c,\;unit={\frac {m}{s}}={\frac {q^{2}}{kg}}}$

elementary charge ${\displaystyle e={\frac {16l_{p}c^{2}}{\alpha Q^{3}}},\;units=C={\frac {q^{3}s}{kg^{3}}}}$

The excess electron mass units become;

${\displaystyle {\frac {C^{3}m^{3}}{s^{4}}}={\frac {q^{15}s^{2}}{kg^{12}}}}$

The Rydberg constant R.

Vacuum permeability ${\displaystyle \mu _{0}={\frac {\pi ^{2}\alpha Q^{8}}{32l_{p}c^{5}}}={\frac {4\pi }{10^{7}}},\;units={\frac {kg\;m}{s^{2}A^{2}}}={\frac {kg^{6}}{q^{4}s}}}$

${\displaystyle R_{\infty }={\frac {m_{e}e^{4}\mu _{0}^{2}c^{3}}{8h^{3}}}={\frac {2^{5}c^{5}\mu _{0}^{3}}{3^{3}\pi \alpha ^{8}Q^{15}}},\;units={\frac {1}{m}}=\color {red}{\frac {kg^{13}}{q^{17}s^{3}}}\color {black}}$

This however now gives us 2 solutions for length m, if we conjecture that they are both valid, then there must be a ratio whereby the units q, s, kg overlap and cancel;

${\displaystyle m={\frac {q^{2}s}{kg}}.{\frac {q^{15}s^{2}}{kg^{12}}}={\frac {q^{17}s^{3}}{kg^{13}}};\;thus\;\color {red}{\frac {q^{15}s^{2}}{kg^{12}}}\color {black}=1}$

Which in terms of kg, m, s becomes

${\displaystyle q^{2}={\frac {kgm}{s}};\;q^{30}=({\frac {kgm}{s}})^{15}}$
${\displaystyle ({\frac {q^{15}s^{2}}{kg^{12}}})^{2}={\frac {kg^{9}s^{11}}{m^{15}}}=1}$