Electron (mathematical)

We can solve the electron parameters; electron mass, wavelength, frequency, charge ... as the frequency of the Planck units themselves, and this frequency is ψ.

${\displaystyle v=11843707.905...,\;units={\frac {m}{s}}}$

${\displaystyle r=0.712562514304...,\;units=({\frac {kg.m}{s}})^{1/4}}$

electron wavelength λ_e = 2.4263102367e-12m (CODATA 2014)

${\displaystyle \lambda _{e}^{*}=2\pi L\psi }$ = 2.4263102386e-12m (L ⇔ Planck length)

electron mass m_e = 9.10938356e-31kg (CODATA 2014)

${\displaystyle m_{e}^{*}={\frac {M}{\psi }}}$ = 9.1093823211e-31kg (M ⇔ Planck mass)

elementary charge e = 1.6021766208e-19C (CODATA 2014)

${\displaystyle e^{*}=A\;T}$ = 1.6021765130e-19 (T ⇔ Planck time)

Rydberg constant R = 10973731.568508/m (CODATA 2014)

${\displaystyle R^{*}=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}}\;u^{13}}$ = 10973731.568508

From the above formulas, we see that wavelength is ψ units of Planck length, frequency is ψ units of Planck time ... however the electron mass is only 1 unit of Planck mass.

Particle mass is a unit of Planck mass that occurs only once per ψ units of Planck time, the other parameters are continuums of the Planck units.

units ${\displaystyle \psi ={\frac {(AL)^{3}}{T}}}$ = 1

This may be interpreted as; for ψ units of Planck time the electron has wavelength L, charge A ... and then the AL combine with time T (A³L³/T) and the units (and scalars) cancel. The electron is now mass (for 1 unit of Planck time). In this consideration, the electron is an event that oscillates over time between an electric wave state (duration ψ units of Planck time) to a unit of Planck mass point state (1 unit of Planck time). The electron is a quantum scale event, it does not exist at the discrete Planck scale (and so therefore neither does the quantum scale).

As electron mass is the frequency of the geometrical Planck mass M = 1, which is a point (and so with point co-ordinates), then we have a model for a black-hole electron, the electron function ψ 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 there 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 the frequency term, it is referring to an average mass (mass which is measured over time) rather than a constant mass (mass as a constant property of the particle at unit Planck time). Thus as noted, when we refer to mass as a constant property, we are referring to mass at the quantum scale, and the electron as a quantum-state particle.

If the scaffolding of the universe includes units of Planck mass M, then it is not necessary for a particle itself to have mass, what we define as electron mass would be the absence of electron.

The charge on the electron derives from the embedded ampere A and length L, the electron formula ψ itself is dimensionless. These AL magnetic monopoles would seem to be analogous to quarks (there are 3 monopoles per electron), but due to the symmetry and so stability of the geometrical ψ 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 embedded within the electron suffice.

Electron formula

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

Time

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

AL magnetic monopole

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

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

If ${\displaystyle \sigma _{e}}$ could equate to a quark with an electric charge of -1/3e, then it would be an analogue of the D quark. 3 of these D quarks would constitute the electron as DDD = (AL)*(AL)*(AL).

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 (A; θ=3, L; θ=-13) units = -10, and if we look at the table of constants, there is no 'units = +10' combination that can include A. We cannot make an inverse electron. However we can make a Planck temperature T_p AV monopole (ampere-velocity).

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

${\displaystyle \sigma _{t}={\frac {3\alpha ^{2}T_{p}}{2\pi }}={\frac {3\alpha ^{2}AV}{2\pi ^{2}}}=({2^{6}3\pi ^{2}\alpha \Omega ^{5}}),\;u^{20},\;scalars={\frac {v^{4}}{r^{6}}}}$

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

The units for ${\displaystyle \sigma _{t}}$ = +20, and so if units = -10 equates to -1/3e, then we may conjecture that units = +20 equates to 2/3e, 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 ψ), 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.

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

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

Numerically:

Adding a proton and electron gives (proton) UUD & DDD (electron) = 2(UDD) = 20 -10 -10 = 0 (zero charge), scalars = 0.

Converting between U and D via U & DDD (electron) = 20 -10 -10 -10 = -10 (D), scalars = ${\displaystyle {\frac {r^{3}}{v^{2}}}}$

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

In this model alpha appears as an orbital constant for gravitational and atomic orbital radius, combining a fixed alpha term with an orbital wavelength term;

${\displaystyle r_{orbital}=2\alpha (\lambda _{orbital})}$

If we replace ${\displaystyle \lambda _{orbital}}$ with the geometrical Planck length L, and include momentum P and velocity V (the 2 components from which the ampere A is derived), then we may consider if the internal structure of the electron involves rotation of this monopole AL super-structure, and this has relevance to electron spin;

${\displaystyle 2\alpha L{\frac {V^{3}}{P^{3}}}=2^{5}\pi ^{5}\alpha \Omega ^{5},\;units=u^{-10},\;scalars={\frac {r^{3}}{v^{2}}}}$

Relativity at the Planck scale can be described by a translation between 2 co-ordinate systems; an expanding (in Planck steps at the speed of light) 4-axis hyper-sphere projecting onto a 3-D space (+ time). In this scenario, particles (with mass) are pulled along by the expansion of the hyper-sphere, this then requires particles to have an axis; generically labeled N-S, with the N denoting the direction of particle travel within the hyper-sphere. Changing the direction of travel involves changing the orientation of the particle N-S axis. The particle may rotate around this N-S axis, resulting in a L-spin or a R-spin.

For simplicity, we can depict the electron as a classically spinning disk, this generates a current which then produces a magnetic dipole, so that the electron behaves like a tiny bar magnet (magnetic fields are produced by moving electric charges). For the classical disk we can use the charge (q), area of the disk (a) and rotation speed (ω) in our calculations.

A thought experiment; if the 3 magnetic monopole quarks are rotating around the electron center (that N-S axis), then they are generating the current (q). As a monopole has the units ampere-meter AL (L a length term), then we can also conjecture something that resembles area (a), and the speed of rotation will give us (ω), and so we can use classical physics to solve our bar magnet electron. The electron is symmetric and so the 3 monopoles are equidistant from each other.

We place our electron in a magnetic field, the electron then starts to orbit this field. If a satellite orbits the earth at radius r, then the distance it travels around the earth = 2πr. However the earth is orbiting the sun, and so the actual distance approximates 2πr + the distance the earth travels. If our electron orbits around a central point (that electric field), then the monopoles will, like the satellite, travel further per orbit (compared to a free electron in space), thus changing the effective area (a), and so our calculations.

The g-factor characterizes the magnetic moment and angular momentum of the electron, it is the ratio of the magnetic moment of the electron to that expected of a classical electron. Could an electron monopole substructure offer a geometrical explanation for this g-factor.