Electron (mathematical)

The mathematical electron model

In the mathematical electron model, the electron is a geometrical formula (fe). Although dimensionless, this formula embeds the Planck units and also dictates the frequency of these Planck units, such that it is from this formula fe and these Planck units that the electron parameters (mass, wavelength, frequency, charge) are derived. As such this electron formula incorporates the complete electron.

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


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

Planck objects
The Planck units are assigned geometrical objects MLTVA

These MLTVA objects were chosen so that they can be linked via a unit number relationship, and so can be defined in terms of only 2 dimensioned quantities (scalars), here are used r and v.

Mathematical electron
The electron function fe incorporates dimensioned units yet itself is dimension-less; units = 1, and thus is a mathematical constant.

Here fe is defined in terms of σe, where AL is an ampere-meter (ampere-length = e*c are the units for a magnetic monopole).


 * $$T = \pi,\; unit = u^{-30},\;scalars = \frac{r^9}{v^6}$$


 * $$\sigma_{e} = \frac{3 \alpha^2 A L}{2\pi^2} = {2^7 3 \pi^3 \alpha \Omega^5},\; unit = u^{(3 \;-13 \;= \;-10)},\; scalars = \frac{r^3}{v^2}$$


 * $$f_e = \frac{\sigma_{e}^3}{2 T} = \frac{(2^7 3 \pi^3 \alpha \Omega^5)^3}{2\pi},\; unit = \frac{(u^{-10})^3}{u^{-30}} = 1, scalars = (\frac{r^3}{v^2})^3 \frac{v^6}{r^9} = 1$$


 * $$f_e = 4\pi^2(2^6 3 \pi^2 \alpha \Omega^5)^3 = .23895453...x10^{23},\;unit = 1$$ (unit-less)

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

electron mass    $$m_e^* = \frac{M}{f_e}$$ (M =  Planck mass) = 0.910 938 232 11 e-30

electron wavelength $$\lambda_e^* = 2\pi L f_e$$ (L = Planck length) = 0.242 631 023 86 e-11

elementary charge   $$e^* = A\;T$$ (T =  Planck time) = 0.160 217 651 30 e-18

Rydberg constant $$R^* = (\frac{m_e}{4 \pi L \alpha^2 M}) = \frac{1}{2^{23} 3^3 \pi^{11} \alpha^5 \Omega^{17}}\frac{v^5}{r^9}$$ = 10 973 731.568 508

We may note that electron wavelength is the sum of fe (0.2389 x 1023) units of length object L (electron frequency then fe units of time) whereas the mass object M (Planck mass) occurs only once per fe (once per 1023 units of time).

This suggests that for the duration of the electron frequency fe = 0.2389 x 1023 units of (Planck) time, the electron is represented by AL magnetic monopoles, the electron is in the electric-state, these monopoles then intersect with a unit of (Planck) time T, the units then collapse (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 (M = 1 is a point).

Wave-particle duality might 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 entities that exist at unit (Planck) time, but rather are oscillating events that occur over time (oscillating from electric wave to mass point).

Electron Mass
If the particle mass is a unit of Planck mass per oscillation, and as M=1 is a point (with point co-ordinates), 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 (point) is exposed for 1 unit of (Planck) time. The electron is 'now' (a unit of Planck) mass.

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 over time. The formula E = hf is a measure of the frequency f 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 = hf = mc2. Notably however the c term is a fixed constant unlike the f 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, only that in this 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.

Electron formula
 * $$f_e = 2^{20} \pi^8 3^3 \alpha^3 \Omega^{15},\; unit = 1, scalars = 1$$

Time
 * $$T = \pi \frac{r^9}{v^6},\; u^{-30}$$

AL magnetic monopole
 * $$\sigma_{e} = \frac{3 \alpha^2 A L}{2\pi^2} = {2^7 3 \pi^3 \alpha \Omega^5},\; u^{-10}, \;scalars = \frac{r^3}{v^2}$$


 * $$f_e = \frac{\sigma_{e}^3}{2 T} = \frac{(2^7 3 \pi^3 \alpha \Omega^5)^3}{2\pi} = 2^{20} 3^3 \pi^8 \alpha^3 \Omega^{15},\; unit = \frac{(u^{-10})^3}{u^{-30}} = 1, scalars = (\frac{r^3}{v^2})^3 \frac{v^6}{r^9} = 1$$

If $$\sigma_{e}$$ could equate to a quark with an electric charge of $-1⁄3$e, then it would be an analogue of the D quark. 3 D quarks would constitute the electron as DDD = (AL)3 (we are adding the exponents).

We would assume that the charge on the positron (anti-matter electron) is just the inverse of the above, however there is 1 problem, the AL units = -10, if we look at the table of constants, there is no units = +10 combination that can include A (units=3). However we can make a Planck temperature Tp monopole.


 * $$T_p = \frac{2^7 \pi^3 \Omega^5}{\alpha},\; u^{20}, \;scalars = \frac{r^9}{v^6}$$


 * $$\sigma_{t} = \frac{3 \alpha^2 T_p}{2\pi} = \frac{3 \alpha^2 A V}{2\pi^2} = ({2^6 3 \pi^2 \alpha \Omega^5}),\; u^{20},\;scalars = \frac{v^4}{r^6}$$


 * $$f_e = (2T) \sigma_{t}^2 \sigma_{e} = 2^{20} 3^3 \pi^8 \alpha^3 \Omega^{15},\; unit = (u^{-30}) (u^{20})^2 (u^{-10}) = 1, scalars = (\frac{r^9}{v^6}) (\frac{v^4}{r^6})^2  \frac{r^3}{v^2} = 1$$

The units for $$\sigma_{t}$$ = +20, and so if units = -10 equates to $-1⁄3$e, then we may conjecture that units = +20 equates to $2⁄3$e, which would be the analogue of the U quark. Our plus charge now becomes DUU, and so although the positron has the same wavelength, frequency, mass and charge magnitude as the electron (both solve to fe), internally its charge structure resembles that of the proton, the positron is not simply an inverse of the electron. This could have implications for the missing anti-matter, and for why the charge magnitude of the proton is exactly the charge magnitude of the electron.


 * $$D = \sigma_{e},\; unit = u^{-10},\; charge = \frac{-1e}{3}, \;scalars = \frac{r^3}{v^2}$$


 * $$U = \sigma_{t},\; unit = u^{20},\; charge = \frac{2e}{3}, \;scalars = \frac{v^4}{r^6}$$

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;


 * $$\sigma_{e} = \frac{3 \alpha^2 e c}{2 \pi^2} = 0.13708563.... \;x 10^{-6},\; units = \frac{C m}{s}

$$

The following gives a formula for an electron in terms of magnetic monopoles and Planck time ($$t_p$$).


 * $$f_e = \frac{\sigma_e^3}{2 t_p} = 0.2389.... \;x 10^{23},\; units = \frac{C^3 m^3}{s^4}$$

However, although this gives us the correct numerical value, this gives us incorrect units when solving the electron mass ($$m_P$$ = Planck mass).


 * $$m_e = \frac{m_P}{f_e} = 0.910938.... \;x 10^{30},\; units = kg \frac{C^3 m^3}{s^4}$$

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

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.


 * $$Q = 1.019\; 113\; 411...\; unit = q \;$$

Replacing m with q;

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

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

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

The excess electron mass units become;


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

The Rydberg constant R$∞$.

Vacuum permeability $$\mu_0 = \frac{\pi^2 \alpha Q^8}{32 l_p c^5} = \frac{4 \pi}{10^7},\; units = \frac{kg \;m}{s^2 A^2} = \frac{kg^6}{q^4 s}$$


 * $$R_\infty = \frac{m_e e^4 \mu_0^2 c^3}{8 h^3} = \frac{2^5 c^5 \mu_0^3}{3^3 \pi \alpha^8 Q^{15}},\; units = \frac{1}{m} = \color{red}\frac{kg^{13}}{q^{17} s^3}\color{black} $$

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;


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

Which in terms of kg, m, s becomes


 * $$q^2 = \frac{kg m}{s};\; q^{30} = (\frac{kg m}{s})^{15}$$


 * $$(\frac{q^{15} s^2}{kg^{12}})^2 = \frac{kg^{9} s^{11}}{m^{15}} = 1$$