# Electron (mathematical)

The mathematical electron model

In the mathematical electron model [1], the electron is a mathematical formula that describes a geometrical object (fe). Embedded within this formula are geometrical objects that embody the functions (attributes) of the Planck units. It is these objects as Planck units that give the measured electron parameters of mass, wavelength, charge ... the formula (fe) itself is dimensionless, units = 1, there is no 'physical' electron, and so this approach is applicable to simulation hypothesis modelling.

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

### Base (Planck) units

As the electron formula embeds the geometrical base structures (objects) for mass M, length L, time T and charge A, these MLTA objects are also the geometry of the dimensionless constants alpha and Omega. As they combine within the electron formula Lego-style, we can assign a mathematical (unit number θ) to define the relationship between them, this unit number has equivalence to the SI units (kg ⇔ 15, m ⇔ -13, s ⇔ -30, A ⇔ 3). For example, if we combine amperes object A with length object L (the ampere-meter is the unit for the magnetic monopole), and then divide by object time T, the unit numbers cancel.

${\displaystyle {\frac {(A^{3})(L^{3})}{(T)}}}$ , θ = 3*3 -13*3 +30 = 0
Geometrical units
Attribute Geometrical object Unit number θ
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 ${\displaystyle A=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}$  ${\displaystyle 3}$

As the MLTA objects have numerical solutions, we may convert them to SI equivalents (or any system of units) using dimensioned scalars. For example, V = 2πΩ2 = 25.3123819;

scalar v = 11843707.905m/s gives c = V*v = 299792458m/s (SI units)

As the scalars, being dimensioned, also follow the unit number relationship, we find that we need only 2 scalars to define the rest (if we know the numerical values for 2 constants such as c and Planck mass, then we can solve their scalars (as alpha and Omega have fixed values), and from these scalars we can solve the dimensioned physical constants (G, h, e, me, kB ...).

### Mathematical electron

The electron function fe incorporates these geometrical base units yet itself is unit-less; units = 1 (θ = 0), and the scalars cancel (scalars = 1), and thus is of itself a mathematical constant (as the scalars cancel, it will have the same numerical value regardless of which system of (Planck) units (for charge, length, time, mass) are used).

Here fe is defined in terms of σe, where AL as an ampere-meter (ampere-length = e*c) are the units for a magnetic monopole. The scalars (v, r) are used where for v; θ = 17 and for r = 0.712 562 514 304 ...; θ = 8.

${\displaystyle T=\pi ,\;unit^{-30},\;scalars={\frac {r^{9}}{v^{6}}}}$  (θ = -30)
${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}},\;unit^{-10},\;scalars={\frac {r^{3}}{v^{2}}}}$  (θ = 3 -13 = -10)
${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{2T}}={\frac {(2^{7}3\pi ^{3}\alpha \Omega ^{5})^{3}}{2\pi }},\;unit={\frac {(unit^{-10})^{3}}{unit^{-30}}}=1,scalars=({\frac {r^{3}}{v^{2}}})^{3}{\frac {v^{6}}{r^{9}}}=1}$
${\displaystyle f_{e}=4\pi ^{2}(2^{6}3\pi ^{2}\alpha \Omega ^{5})^{3}=.23895453...x10^{23},\;unit=1}$  (unit-less)

##### 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 (θ = 3*3 -13*3 +30 = 0), 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 discrete 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 discrete Planck mass point-state as measured over time. The formula E = hv is a measure of the frequency v of occurrence of Planck's constant h and applies to the electric wave-state. As for each wave-state then is a corresponding mass point-state, then for a particle E = hv = mc2. Notably however the c term is a fixed constant unlike the v term, and so the m term is referring to average mass rather than constant mass.

If the scaffolding of the universe includes units of (Planck) mass M, then it is not necessary for the particle itself to have mass. The electron can then be defined as an event that oscillates between an electric wave-state according to the formula fe, and a mass point-state (in which state the electron has no dimensioned ALT parameters).

##### Quarks

The charge on the electron derives from the embedded ampere A and length L, the electron formula fe itself is dimensionless. These AL magnetic monopoles would seem to be analogous to quarks, but due to the symmetry and so stability of the geometrical fe there is no clear fracture point by which an electron could decay and so this would be difficult to test. We can however conjecture on what a quark solution might look like, the advantage with this approach being that we do not need to introduce new 'entities' for our quarks, the Planck units suffice.

${\displaystyle f_{e}=2^{20}\pi ^{8}3^{3}\alpha ^{3}\Omega ^{15},\;unit=1,scalars=1}$

The AL magnetic monopole 'quark'

${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}},\;unit^{-10},\;scalars={\frac {r^{3}}{v^{2}}}}$

An AV Planck temperature 'quark'

${\displaystyle T_{p}={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }},\;unit^{20},\;scalars={\frac {r^{9}}{v^{6}}}}$
${\displaystyle \sigma _{t}={\frac {3\alpha ^{2}T_{p}}{2\pi }}={\frac {3\alpha ^{2}AV}{2\pi ^{2}}}=({2^{6}3\pi ^{2}\alpha \Omega ^{5}}),\;unit^{20},\;scalars={\frac {v^{4}}{r^{6}}}}$
${\displaystyle (2T)^{2}\sigma _{t}^{3}=4\pi ^{2}(2^{6}3\pi ^{2}\alpha \Omega ^{5})^{3},\;unit=(unit^{-30})^{2}(unit^{20})^{3}=1,scalars=({\frac {r^{9}}{v^{6}}})^{2}({\frac {v^{4}}{r^{6}}})^{3}=1}$

For example, at the base level 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 unit for the D (magnetic monopole) quark is not the same as the unit for the U (Planck temperature) quark. For example;

${\displaystyle D=\sigma _{e},\;unit^{-10},\;scalars={\frac {r^{3}}{v^{2}}},\;charge={\frac {-1}{3}}}$
${\displaystyle U=\sigma _{t},\;unit^{30},\;scalars={\frac {v^{4}}{r^{6}}},\;charge={\frac {2}{3}}}$

DDD:   ${\displaystyle \sigma _{e}\sigma _{e}\sigma _{e}=2^{21}\pi ^{9}3^{3}\alpha ^{3}\Omega ^{15},\;(unit^{-10})(unit^{-10})(unit^{-10})=unit^{-30},\;scalars={\frac {r^{9}}{v^{6}}},\;charge={\frac {-1}{3}}.{\frac {-1}{3}}.{\frac {-1}{3}}=-1e}$

DDU:   ${\displaystyle \sigma _{e}\sigma _{e}\sigma _{t}=2^{20}\pi ^{8}3^{3}\alpha ^{3}\Omega ^{15},\;(unit^{-10})(unit^{-10})(unit^{20})=1,\;scalars=1,\;charge={\frac {-1}{3}}.{\frac {-1}{3}}.{\frac {2}{3}}=0}$

DUU:   ${\displaystyle \sigma _{e}\sigma _{t}\sigma _{t}=2^{19}\pi ^{7}3^{3}\alpha ^{3}\Omega ^{15},\;(unit^{-10})(unit^{20})(unit^{20})=unit^{30},\;scalars={\frac {v^{6}}{r^{9}}},\;charge={\frac {-1}{3}}.{\frac {2}{3}}.{\frac {2}{3}}=1e}$

We see that the DDU formula (neutron) has no units or scalars. It corresponds to the formula fe. The DDD formula (electron) and DUU formula (proton) are separated from the DDU formula by a unit of T unit-30 and 1/T unit30 respectively. It may therefore be possible for the neutron to split into an electron and proton without breaking any mathematical rules.

The above formulas refer only to the 'quark' component. According to the mass of the electron, there is no binding energy required for the AL quarks, however the mass of the proton and neutron imply that a configuration that includes a U quark and a D quark requires significant binding energy, this has not been included in the DDU and DUU formulas.

### 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

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