User:Platos Cave (physics)/sandbox

Simulating gravitational and atomic orbits via rotating particle-particle orbital pairs at the Planck scale

An orbital simulation program is described that uses a geometrical approach to emulate gravitational and atomic orbits at the Planck scale.

For simulating gravity, orbiting objects A, B, C... are sub-divided into discrete points, each point representing 1 unit of Planck mass mP (for example, a 1kg satellite would be divided into 1kg/mP = 45940509 points). Each point in object A forms an orbital pair with every point in objects B, C..., resulting in a universe-wide, n-body network of rotating point-to-point orbital pairs. Each orbital pair rotates 1 unit of Planck length lp per unit of Planck time tp at velocity c (c = lp/tp) in hypersphere space co-ordinates, when mapped over time, gravitational orbits emerge between the objects A, B, C... The basic simulation uses only the start position of each point, as it maps only rotations of the points within their respective orbital pairs, information regarding the macro objects A, B, C...; momentum, center of mass, barycenter etc ... is not required.

Each point represents 1 unit of Planck mass, however particles have a mass much less than Planck mass, and so a point could consist of 1020 or more particles. The point itself can then be sub-divided into particle-particle orbital pairs, the classic example is the H atom (with a single proton-electron orbital pair), the H atom electron transition frequencies used as reference for studying particle-particle orbital pairs.

The simulation treats particles, not as distinct entities but as oscillations between an electric wave-state (duration particle frequency) to mass point-state (duration 1tp). As the mass point-state occurs seldom relative to the electric wave-state (duration Planck mass/electron mass = 1023 units of Planck time), the atomic orbital is predominately a wave-state rotation effect. Therefore, to ensure for an object that at every unit of Planck time there is a particle in the mass point-state (i.e.: a gravity effect will occur at every unit of Planck time), and if for example we have only electrons, then we would require 1023 electrons, such that on average 1 electron will be in the mass point state at that unit time.

Gravity is therefore not weaker than the electric force, rather it is stronger at the Planck scale (point-point orbitals rotate faster than wave-wave), its apparent weakness is simply because point-point rotations when mapped over time seldom occur relative to wave-wave in orbitals (the probability of occurrence as the inverse of the gravitational coupling constant). This also means that gravitational orbits as we observe them are time emergent properties of rotating orbitals, at the Planck scale there is no gravity or electric force.

For mapping atomic orbitals, the simulation includes an additional alpha term to account for the slower wave-wave rotation, other than this, the principal difference between gravitational and atomic orbitals is one of scale. There are not 2 separate forces used by the simulation, instead particles are treated as oscillations between the 2 states (electric and mass), the orbitals themselves however are essentially the same and so the same program can be used for both states.



N-body orbitals


The simulation itself is dimensionless, to translate to familiar dimensioned gravitational orbitals, we can use the Planck units whereby points (of mass equivalent to Planck mass) are assigned co-ordinates on a 2-D (x-y) plane (representing 3-D space). Every point is then connected to every other point to form a circular orbital pair resulting in a universe-wide n-body network of point-point orbital pairs.

The simulation increments in discrete steps (each step translates to a unit of Planck time), during each step (each unit of Planck time), the orbitals rotate 1 unit of Planck length at velocity c = lp/tp in 4-axis hyper-sphere coordinates.

These rotations are then summed and averaged giving new point co-ordinates. As this occurs for every point before the next increment to the simulation clock (the next unit of Planck time), the orbits can be updated in 'real time' (simulation time) on a serial processor. As the orbitals are circular, the barycenter for each orbital is its center, the points located at each orbital 'pole'.

Although orbital and so point rotation occurs at c, the hyper-sphere expansion is equidistant and so `invisible' to the observer. Instead observers (being constrained to 3D space) will register these 4-axis orbits (in hyper-sphere co-ordinates) as a circular motion on a 2-D plane (in 3-D space). An apparent time dilation effect emerges as a consequence.



Simulation
Each point in the simulation is assigned initial (x, y) 2-D point co-ordinates (representing 3-D space), forming orbital pairs that rotate around each other on a 2-D plane according to an angle β as defined by the orbital pair radius (the atomic orbital β has an additional alpha term).


 * $$\beta = \frac{1}{r_{orbital} \sqrt{r_{orbital}}}$$

The total distance travelled, 1 unit of (Planck) length per increment to the simulation clock (1 (Planck) time unit) is given in (x, y, z) co-ordinates, where the (z) axis represents the hypersphere expansion axis.

As the simulation treats each (point-point) orbital independently (independent of all other orbitals), no information regarding the points (other than their initial start coordinates) is required by the simulation.

For n-body orbits, to reduce computation we can use the relative mass, in the earth moon example the earth is 81x more massive than the moon, and so we need only represent the center (earth) with 81 points and the periphery (moon) with 1 point. In the following orbits, 1 point is assigned as the orbiting point, the remaining points forming the 'central' mass. The only distinction being that the central mass points are assigned (x, y) co-ordinates relatively close to each other, and the orbiting point is assigned (x, y) co-ordinates distant from the central points (this becomes the orbital radius). The simulation however treats all points equally, the center points also orbiting each other according to their orbital radius.

After every orbital has rotated 1 (Planck) length unit anti-clockwise, the new co-ordinates for each rotation per point are then averaged and summed, the process then repeats. After 1 complete orbit (return to the start position by the orbiting point), the period tsim (as the number of increments to the simulation clock) and the (x, y) plane orbit length lsim (distance as measured on the 2-D plane) are noted.

Key:

1. i; number of 'physical' center points in the orbit (the center mass).

2. j = i*x + 1; number of virtual center points (to reduce computation time, i*x virtual points are added to increase center mass up to j = jmax.

3. jmax; maximum number of mass points per orbital radius.

4. x, y; start co-ordinates for each point (2-D plane).

5. rα; a radius constant, here rα = sqrt(2α) = 16.55512; where alpha = inverse fine structure constant = 137.035 999 084 (CODATA 2018).


 * $$r_{orbital} = {r_{\alpha}}^2 \;*\; r_{wavelength} $$

Fixed radius, variable mass (2D plane)
The radius is fixed according to i and $$j_{max}$$, we continuously add mass j (there is only 1 orbiting point) to the center until we reach j = $$j_{max}$$ and that radius is now 'saturated'. To add further mass requires increasing radius.

Equations for the 2-D plane:


 * $$n_g = \frac{j_{max}}{j}$$ ratio of mass to maximum mass per orbital radius


 * $$r_{outer} = {r_{\alpha}}^2 \;*\;2 (\frac{ j_{max}}{i})^2$$, orbital radius


 * $$r_{barycenter} = \frac{r_{outer}}{j}$$, barycenter


 * $$v_{outer} = \frac{j}{r_{\alpha} j_{max}}$$, orbiting point velocity


 * $$v_{inner} = \frac{1}{r_{\alpha} j_{max}}$$, orbited point(s) velocity


 * $$t_{outer} = \frac{2 \pi r_{outer}}{v_{outer}} = \frac{4 \pi {j_{max}}^3}{i^2 j} {r_{\alpha}}^3$$, orbiting point period


 * $$l_{outer} = 2 \pi (r_{outer} - r_{barycenter})$$, distance travelled

Simulation data:


 * length $$l_{sim} = \frac{t_{sim}(j-1)}{j_{max}r_{\alpha}}$$


 * radius $$r_{sim} = (\frac{j}{j-1}) \frac{l_{sim}}{2 \pi} = \frac{t_{sim}}{2\pi n_g r_{\alpha}}$$


 * velocity $$v_{sim} = (\frac{j}{j-1}) \frac{l_{sim}}{t_{sim}} = \frac{1}{n_g r_{\alpha}}$$

Example:

i = 81, j = jmax = 32*81+1 = 2593 (3321 orbitals)

tsim = 58430803.84

lsim = 3528109.12

Fixed mass, variable radius (2D plane)
The mass is fixed and radius is variable, for example, to model a 1kg satellite to earth orbit will require earth mass/Planck mass = 0.2744 x1033 points and 1kg/Planck mass = 45940510 points. We can reduce calculation by using only relative mass and then use the dimensionless ng to assign the start parameters. For example, from the standard gravitational parameters, the earth to moon mass ratio approximates 81:1 and so we can reduce to 1 point orbiting a center of mass comprising 81 points.


 * $$j = \frac{3.986004418\;x10^{14}}{4.9048695\;x10^{12}} = 81.2663$$

There is 1 orbiting point (distant point) and 81 central points (points in close vicinity)


 * $$i = j - 1$$

To calculate ng


 * $$r_{earth-moon}$$ = 384400km


 * $$\lambda_{Earth}$$ = 0.00887m (Schwarzschild radius)


 * $$n_g = \sqrt{ \frac{2 r_{earth-moon}} {\lambda_{Earth}} } \frac{1}{r_{\alpha}}$$ = 17783.25

This gives


 * $$j_{max} = n_g j$$ = 1445178.5

Converting from dimensionless numbers to SI Planck units using lp and c;


 * $$t_{outer} = 4 \pi (\frac{ {j_{max}}^3}{i^2 j}) {r_{\alpha}}^3 (\frac{l_p}{c}) = 0.1772\; 10^{-25}$$s


 * $$r_{outer} = 2 (\frac{ j_{max}}{i})^2 {r_{\alpha}}^2 (l_p) = 0.2872\; 10^{-23}$$m


 * $$v_{Moon} = (c) \frac{j}{j_{max}{r_{\alpha}}} = 1018.3m/s$$


 * $$v_{Earth} = (c) \frac{1}{j_{max} r_{\alpha}} = 12.53m/s$$


 * $$barycenter = \frac{r_{earth-moon}}{j} = 4730km$$

We can use the actual radius and period to translate between values.


 * $$t_{earth-moon}$$ = 27.322 days


 * $$\lambda_0 = \frac{2 j^2}{i^2}$$


 * $$\frac{r_{earth-moon}}{r_{outer}} \lambda_0 m_p = 0.59738\; 10^{25}$$kg


 * $$\frac{t_{outer} 0.59738\; 10^{25} kg} {\lambda_0 m_P } = 2371851 = 27.452$$ days

The above assumes a circular orbit, to form an elliptical orbital we can use unaligned orbitals.

Gravitational coupling constant
In the above, the points were assigned a mass as a theoretical unit of Planck mass. Conventionally, the Gravitational coupling constant αG characterizes the gravitational attraction between a given pair of elementary particles in terms of a particle (i.e.: electron) mass to Planck mass ratio;


 * $$\alpha_G = \frac{G m_e^2}{\hbar c} = (\frac{m_e}{m_P})(\frac{m_e}{m_P}) = 1.75... x10^{-45}$$

For the purposes of this simulation, particles are treated as an oscillation between an electric wave-state (duration particle frequency) and a mass point-state (duration 1 unit of Planck time). This inverse αG then represents the probability that any 2 electrons will be in the mass point-state at any unit of Planck time (wave-mass oscillation at the Planck scale ).


 * $${\alpha_G}^{-1} = \frac{m_P^2}{m_e^2} = 0.57... x10^{45}$$

As mass is not treated as a constant property of the particle, measured particle mass becomes the averaged frequency of discrete point mass at the Planck level. If 2 dice are thrown simultaneously and a win is 2 'sixes', then approximately every (1/6)x(1/6) = (1/36) = 36 throws (frequency) of the dice will result in a win. Likewise, the inverse of αG is the frequency of occurrence of the mass point-state between the 2 electrons. As 1 second requires 1042 units of Planck time ($$t_p = 10^{-42}s$$), this occurs about once every 3 minutes.


 * $$\frac{{\alpha_G}^{-1}}{t_p}$$

Gravity now has a similar magnitude to the strong force (at this, the Planck level), albeit this interaction occurs seldom (only once every 3 minutes between 2 electrons), and so when averaged over time (the macro level), gravity appears weak.

If particles oscillate between an electric wave state to Planck mass (for 1 unit of Planck-time) point-state, then at any discrete unit of Planck time, a number of particles will simultaneously be in the mass point-state. If an assigned point contains only electrons, and as the frequency of the electron = fe, then the point will require 1023 electrons so that, on average for each unit of Planck time there will be 1 electron in the mass point state, and so the point will have a mass equal to Planck mass (i.e.: experience continuous gravity at every unit of Planck time).


 * $$f_e = \frac{m_P}{m_e} = 10^{23}$$

For example a 1kg satellite orbits the earth, for any given unit of Planck time, satellite (B) will have $$1kg/m_P = 45940509$$ particles in the point-state. The earth (A) will have $$5.9738 \;x10^{24} kg/m_P = 0.274 \;x10^{33}$$ particles in the point-state, and so the earth-satellite coupling constant becomes the number of rotating orbital pairs (at unit of Planck time) between earth and the satellite;


 * $$N_{orbitals} = (\frac{m_A}{m_P})(\frac{m_B}{m_P})  = 0.1261\; x10^{41}$$

Examples:

1. 1kg satellite at a synchronous orbit radius


 * $$j = N_{orbitals} = 0.1261 \;x10^{41}$$


 * $$i = 5.9738 \;x10^{24}kg/m_P = 0.27444 \;x10^{33}$$ (earth as the center mass)


 * $$2 i l_p = \lambda_{earth} = 0.00887$$ (Schwarzschild radius)


 * $$r_{o} = 42164.17 km$$ (synchronous orbit)


 * $$n_g = \sqrt{\frac{r_{o}}{i l_p}} = 5889.674$$


 * $$j_{max} = n_g j$$


 * $$t_{outer} = 4 \pi (\frac{ {j_{max}}^3}{i^2 j}) {r_{\alpha}}^3 (\frac{l_p}{c}) = 0.1325265 \;x10^{-11} s

$$


 * $$r_{outer} = 2 (\frac{ j_{max}}{i})^2 {r_{\alpha}}^2 (l_p) = 0.648515 \;x10^{-9} m $$


 * $$v_{outer} = \frac{c j}{j_{max}r_{\alpha}} = 3074.66 m/s$$


 * $$\lambda_0 = 2 \frac{j^2}{i^2}$$


 * $$\frac{t_{outer} i}{\lambda_0} = \frac{\pi n_g^3 r_{\alpha}^3 \lambda_{earth}}{c} = 86164.09165 s$$


 * $$\frac{r_{outer} i}{\lambda_0} = \frac{n_g^2 r_{\alpha}^2 \lambda_{earth}}{2} = 42164.17 km$$

2. The energy required to lift a 1 kg satellite into geosynchronous orbit is the difference between the energy of each of the 2 orbits (geosynchronous and earth).


 * $$E_{orbital} = \frac{h c}{2 \pi r_{6371}} - \frac{h c}{2 \pi r_{42164}} = 0.412 x10^{-32}J$$ (energy per orbital)


 * $$N_{orbitals} = \frac{M_{earth}m_{satellite}}{m_P^2} = 0.126 x10^{41}$$ (number of orbitals)


 * $$E_{total} = E_{orbital} N_{orbitals} = 53 MJ/kg$$

3. The orbital angular momentum of the planets derived from the angular momentum of the respective orbital pairs.


 * $$N_{sun} = \frac{M_{sun}}{m_P} $$


 * $$N_{planet} = \frac{M_{planet}}{m_P} $$


 * $$N_{orbitals} = N_{sun}N_{planet} $$


 * $$n_g = \sqrt{\frac{R_{radius} m_P}{2 \alpha l_p M_{sun}}} $$


 * $$L_{oam} = 2\pi \frac{M r^2}{T} = N_{orbitals} n_g\frac{h}{2\pi} \sqrt{2 \alpha},\;\frac{kg m^2}{s} $$

The orbital angular momentum of the planets;

mercury = .9153 x1039 venus   = .1844 x1041 earth   = .2662 x1041 mars    = .3530 x1040 jupiter  = .1929 x1044 pluto  = .365 x1039

Orbital angular momentum combined with orbit velocity cancels ng giving an orbit constant. Adding momentum to an orbit will therefore result in a greater distance of separation and a corresponding reduction in orbit velocity accordingly.


 * $$L_{oam}v_g = N_{orbitals} \frac{h c}{2\pi},\;\frac{kg m^3}{s^2} $$



Freely moving points
The simulation calculates each point as if freely moving in space, and so is useful with 'dust' clouds where the freedom of movement is not restricted (i.e.: in the above example, the earth particles do not follow gravitational orbits around each other). When measuring the orbit of a single point around a larger mass, after each complete orbit we can note that the orbit period and radius reduces (as a function of center mass and start radius distance).

In this animation, 32 mass points begin with random co-ordinates (the only input parameter here are the start (x, y) coordinates of each point; i, j, r ... are not pre-set). We then fast-forward 232 steps to see that the points have now clumped to form 1 larger mass and 2 orbiting masses. The larger center mass is then zoomed in on to show the component points are still orbiting each other, there are still 32 freely orbiting points, only the proximity between them has changed.



Orbital trajectory (circular vs. straight)
Orbital trajectory is a measure of alignment of the orbitals. In the above examples, all orbitals rotate in the same direction = aligned. If all orbitals are unaligned the object will appear to 'fall' = straight line orbit.

In this example, for comparison, onto an 8-body orbit (blue circle orbiting the center mass green circle), is imposed a single point (yellow dot) with a ratio of 1 orbital (anti-clockwise around the center mass) to 2 orbitals (clockwise around the center mass) giving an elliptical orbit.

The change in orbit velocity (acceleration towards the center and deceleration from the center) derives automatically from the change in the orbital radius (there is no barycenter).

The orbital drift (as determined where the blue and yellow meet) is due to these orbiting points rotating around each other.

Precession
semi-minor axis: $$b = \alpha l^2 \lambda_A$$

semi-major axis: $$a = \alpha n^2 \lambda_A$$

radius of curvature :$$L = \frac{b^2}{a} = \frac{a l^4 \lambda_A}{n^2}$$


 * $$\frac{3 \lambda_A}{2 L} = \frac{3 n^2}{2 \alpha l^4}$$

arc secs per 100 years (drift):


 * $$T_{earth}$$ = 365.25 days

drift = $$\frac{3 n^2}{2 \alpha l^4} 1296000 \frac{100 T_{earth}}{T_{planet}}$$

Mercury (eccentricity = 0.205630) T = 87.9691 days a = 57909050 km (n = 378.2734) b = 56671523 km (l = 374.2096) drift = 42.98

Venus (eccentricity = 0.006772) T = 224.701 days a = 108208000 km (n = 517.085) b = 108205519 km (l = 517.079) drift = 8.6247

Earth (eccentricity = 0.0167) T = 365.25 days a = 149598000 km (n = 607.989) b = 149577138 km (l = 607.946) drift = 3.8388

Mars (eccentricity = 0.0934) T = 686.980 days a = 227939366 km (n = 750.485) b = 226942967 km (l = 748.843) drift = 1.351



Hyper-sphere orbit
An expanding hyper-sphere forms the scaffolding of the `universe'. The hyper-sphere expands in uniform incremental steps (the simulation clock-rate) as the origin of the speed of light, and so (hyper-sphere) time and velocity are constants. Particles are pulled along by this expansion, the expansion as the origin of motion, and so all objects, including orbiting objects, travel at, and only at, the speed of light in these hyper-sphere co-ordinates. Time becomes time-line.

While B (satellite) has a circular orbit period on a 2-axis plane (the horizontal axis representing 3-D space) around A (planet), it also follows a cylindrical orbit (from B1 to B11) around the A time-line (vertical expansion) axis (td) in hyper-sphere co-ordinates. A is moving with the universe expansion (along the time-line axis) at (v = c), but is stationary in 3-D space (v = 0). B is orbiting A at (v = c), but the time-line axis motion is equivalent (and so `invisible') to both A and B, as a result the orbital period and velocity measures will be defined in terms of 3-D space co-ordinates by observers on A and B. In dimensionless terms;


 * $$d = r_{\alpha} n_g$$


 * $$t_0 = 2 \pi r = 2 \pi \frac{t}{2 \pi d}$$


 * $$v_{outer} = \frac{1}{d}$$

For object B


 * $$t_d = \sqrt{t^2 - {t_0}^2} = t \sqrt{1 - v_{outer}^2}$$

For object A


 * $$t_d = t \sqrt{1 - v_{inner}^2}$$

Planck force

 * $$F_p = \frac{m_P c^2}{l_p}$$


 * $$M_a = \frac{m_P \lambda_a}{2 l_p} ,\;m_b = \frac{m_P \lambda_b}{2 l_p}$$


 * $$F_g = \frac{M_a m_b G}{R^2} = \frac{\lambda_a \lambda_b F_p}{4 R_g^2} = \frac{\lambda_a \lambda_b F_p}{4 \alpha^2 n^4 (\lambda_a + \lambda_b)^2} $$

a) $$M_a = m_b$$


 * $$F_g = \frac{F_p}{{(4 \alpha n^2)}^2} $$

b) $$M_a >> m_b$$


 * $$F_g = \frac{\lambda_b F_p}{{(2 \alpha n^2)}^2 \lambda_a} = \frac{m_b c^2}{2 \alpha^2 n^4 \lambda_a} = m_b a_g$$

Atomic orbitals
Particles are treated as an electric wave to (Planck) mass point oscillation, the wave-state as the duration of particle frequency in Planck time units, the point-state duration (this state can be assigned mapping coordinates) as 1 unit of Planck time, the particle itself is an oscillation between these 2 states and not a fixed entity. An electron has a frequency = 1023 units of Planck time and so an object would require 1023 electrons in order that, on average, 1 electron will be in the mass-state at any given unit of Planck time. If we require 1020 assorted particles to make 1 Planck mass, then each gravity point has 1020 particles. This point would then have an average mass = Planck mass and so gravity (orbital pair rotation) can occur at each unit of Planck time.

We can then zoom in and create an orbital which has 2 particles instead of 2 mass points. This orbital is predominately electric-wave to electric-wave, the simplest example is the H atom (an electron to proton orbital pair), and as transition frequencies have been precisely measured, we can use them as reference to analysis individual orbitals. The gravitational orbital can then be considered as the scaling up of the underlying atomic orbitals, the gravitational orbit as the time averaging of gravitational orbitals.

Features;

1. The electron and proton wavelengths are combined $$\lambda_H = \lambda_e + \lambda_p$$ and then divided into alpha units to form the orbital (Bohr) radius.

2. Orbital rotations occur in $$\lambda_H$$ steps. After each step the electron has point (point-state) co-ordinates and so can be mapped.

3. During transition, the orbital radius absorbs (or ejects) the photon in alpha unit steps, the velocity component can then be calculated for each step giving a precise average velocity. The orbital radius is treated as physically analogous to the photon albeit of inverse or reverse phase, the electron itself has a passive role in transition.

4. An orbital radius variable δ is used to correlate radius with observed transition frequencies.


 * $$\lambda_{orbital} = (2\alpha + \delta) \lambda_H$$

In atomic orbitals fig. 1 δ is mapped against transition energies in eV, in atomic orbitals fig.2 δ is mapped against the principal quantum number n2 (Lyman series). Crossover (δ = 0) occurs close to n2 = 2 at half the ionization energy = 6.7992 eV.

Simulation


The orbital radius is divided into sub-segments (alpha units) joined together in series. The wavelength of the orbital radius is the sum of these segments. To reduce computation, the $$\lambda_{e+p}$$ (wavelength) component is added later.


 * $$\alpha_{unit} = \frac{1}{2 \pi 2\alpha} = 0.000581$$

The angle of rotation is β (see also gravitational orbitals), the β of atomic orbitals include an extra sqrt(2α) term.


 * $$\beta = \frac{1}{r_{orbital} \sqrt{r_{orbital}}\sqrt{2\alpha}}$$

The simulation assigns co-ordinates to the electron point-state. After each wave to point oscillation cycle, the electron point jumps (the actual motion of the electron occurs during the electron wave-state) 1 step according to β (as with gravitational orbitals), thus plotting over time an orbit around a center.

The radius of a basic orbital (δ = 0).


 * $$r _{orbital} = 2 \alpha n^2$$

On the 2-D plane:


 * $$r _{orbital} = 2 \alpha n^2$$


 * $$v_{orbital} = \frac{1}{2 \alpha n}$$


 * $$t_{orbit} = \frac{2\pi r}{v} = 2\pi 4 {\alpha}^2 n^3$$

And so for the basic orbital, where the orbital number n=1 and δ = 0, to complete 1 rotation of the alpha orbital by the electron would require a period approximately 471964 steps, where each step is the duration of 1 wave to point oscillation (measured in Planck time units). However for the purpose of the simulation (to reduce computation), only the alpha units are calculated. The (inverse) alpha = 137.035999177 is used as this is the more familiar. The Bohr radius then becomes a physical construct of these 471964 alpha units added together in series.


 * $$t _{ref} = 2 \pi 4 \alpha^2 \sim 471964$$



Transition (theory)
The orbital radius is treated as physically analogous to the photon albeit of inverse or reverse phase, and as such it is the orbital radius that absorbs or ejects the photon during transition, in the process the orbital radius is extended (until the photon is completely absorbed). Conversely the orbital radius may eject a photon, the above in reverse. This process occurs in steps, at each step the orbital radius continues to rotate, the electron, being pulled along by this rotation according to angle β, thereby describes a spiral path as the orbital radius length changes (the electron has a passive role in the transition phase).

The incoming (or ejected photon) is also a construct of alpha units, denoted $$r_{incr}$$ to distinguish from $$\alpha_{unit}$$. Note that the minus sign indicates that the $$r_{incr}$$ unit is of opposite phase to the alpha unit. The wave-state component is added later.


 * $$r_{incr} = -\frac{1}{2 \pi 2\alpha} = -0.000581 $$


 * $$\alpha_{unit} + r_{incr} = zero$$

During the transition phase, for each transition step an $$r_{incr}$$ unit is exchanged (transferred) between the orbital radius and the photon.


 * $$r_{orbital} = r_{orbital} + r_{incr}$$ (per transition step)

If the wavelength of $$\lambda_{photon}$$ = the wavelength of the orbital radius $$\lambda_{orbital}$$, and as these waves are of inverse phase, the orbital radius will be deleted. This is defined as ionization.


 * $$\lambda_{orbital} + \lambda_{photon} = zero$$

However an incoming photon is actually 2 photons as per the Rydberg formula.


 * $$\lambda_{photon} = R.(\frac{1}{n_i^2}-\frac{1}{n_f^2}) = \frac{R}{n_i^2}-\frac{R}{n_f^2}$$


 * $$\lambda_{photon} = (+\lambda_i) - (+\lambda_f) = (+\lambda_i) + (-\lambda_f)$$

The (+$$\lambda_i$$) will subtract from the orbital radius as described above, however the (-$$\lambda_f$$), because of the Rydberg minus term, will conversely increase the orbital radius. And so for the duration of the (+$$\lambda_i$$) photon wavelength, the orbital radius does not change as the 2 photons cancel;


 * $$r_{orbital} = r_{orbital} + (\lambda_i) - (\lambda_f) = r_{orbital} + r_{incr} - r_{incr}$$

The ($$\lambda_f$$) has the longer wavelength, and so after the ($$\lambda_i$$) photon has been absorbed, and for the remaining duration of the ($$\lambda_f$$) wavelength, the orbital radius will be extended in $$-r_{incr}$$ steps ($$-r_{incr}$$ = +0.000581).


 * $$r_{orbital} = r_{orbital} + 0.000581$$ (per transition step)

For an $$n_i$$=1 ($$\lambda_i = 1t_{ref}$$) to $$n_f$$=2 ($$\lambda_f = 4t_{ref}$$) orbital transition, the $$\lambda_i$$ photon absorption by the $$n_i$$=1 orbital requires 1$$t _{ref}$$ steps, the remaining $$\lambda_f$$ still has 3$$t _{ref}$$ segments (of $$r_{incr}$$) left, and so transition continues for another 3$$t _{ref}$$ steps. A $$n_i$$=2 to $$n_f$$=3 transition would require $$t = 4t _{ref} + (9-4)t _{ref}$$ steps.



Transition (method)

 * $$r_{orbital} = (2\alpha + \delta)$$

Period of orbit; $$l_{step}$$ is the distance traveled along the x-y plane giving $$t_{orbital}$$ as the period measured along the orbital timeline (z-axis) in hypersphere co-ordinates (see Hypersphere orbits).


 * $$t_{orbital} = n_i 2 \pi r_{orbital} \frac{\sqrt{1-l_{step}^2}}{l_{step}}$$

Electron transition, raising the electron to a higher energy level, can occur when a photon strikes. The incoming photon separates, $$t_{orbital}$$ is the period of the $$\lambda_i$$.


 * $$\lambda_{photon} = (+\lambda_i) + (-\lambda_f)$$

Once $$\lambda_i$$ has been absorbed, $$\lambda_f$$ continues added to the orbital radius until it is also absorbed, giving $$t_{transition}$$. For a transition from an (n = i) initial orbital to (n = f) final orbital, $$l_{step}$$ is a function of angle β and thus of radius (β, as a function of radius, reduces as the radius extends), and so a $$l_{step}$$ value is calculated for each step, $$t_{transition}$$ then becoming the total period summed (from each $$l_{step}$$) over the transition. Periods $$t_{orbital}$$ and $$t_{transition}$$ are then added.


 * $$H_{n_i-n_f} = (n_f - n_i)\;\frac{2 c}{(\lambda_e + \lambda_p)}\;\frac{1}{(t_{orbital} + t_{transition})}$$

To calibrate the average radius (2α + δ) over the transition process we can use the following frequencies.

H(1s-2s) = 2466 061 413 187.035 kHz

H(1s-3s) = 2922 743 278 665.79 kHz

H(1s-4s) = 3082 581 563 822.63 kHz

δ values for each n


 * $$r_{averaged} = 2\alpha + \delta_{ns}$$


 * $$\delta_{2s} = -0.000954660863,\;4 r_{1s}-r_{2s} = 0.00124$$


 * $$\delta_{3s} = -0.001483071027,\;9 r_{1s}-r_{3s} = 0.00547$$


 * $$\delta_{4s} = -0.001672037031,\;16 r_{1s}-r_{2s} = 0.0123$$

Extrapolating:


 * $$n^2 = 2, \;\delta = -0.000044356388$$


 * $$\delta = 0,\; n^2 = 1.951048816$$


 * $$eV = 6.79922, \;\delta = -0.00004437128$$ (ionisation energy/2)


 * $$\delta = 0,\; eV = 6.62856911$$


 * $$n = 2^{24}, \;\delta = -0.0019181476$$


 * $$eV = 13.59843797, \;\delta = -0.0019181479$$ (ionisation energy)

Measuring transition energy in eV, whereby 0eV would be a state of no-transition (the electron remaining in the n = 1s orbital). See atomic orbitals fig. 1, 2.


 * $$eV = 0, \;\delta = 0.00163335174$$


 * $$\delta = 0.00163335174, \;n = 1.00000723202$$

For clarity, if we shift our graph by $$\delta = -0.00004437128$$ then we see an apparent correlation between δ=0 and n=sqrt(2) giving a classical Bohr radius $$\lambda_{orbital} = 2\alpha (\lambda_e + \lambda_p)n^2$$.


 * $$eV = 6.79922,\; \delta = 0,\; n^2 = 2.000016823$$ (ionisation energy/2)

The Positronium $$1s-2s$$ transition


 * $$P(f_{1s-2s}) = 1233 607 216.4$$MHz


 * $$r_{orbital} = 2\alpha + 0.01311882852$$



Helium
The above considered a charge equivalence, 1 electron to 1 proton. If we expose the electron to more charge, then we may anticipate further changes to the orbital radius. To illustrate, in this example the orbital radius is divided into 4 parts each of $$t_{ref}/4$$ = 117986 and each part is equivalent to 13.59844 eV, the ionization energy of H. This means that if an electron orbits at a radius where $$t_{orbit}$$ = 117986, then it will require 3*13.59844eV to reach a base H orbital ($$t_{ref}$$ = 117986 + 3*117986) and then a further 13.59844eV to ionize from there. Total ionization energy = 4*13.59844eV = 54.4eV.

In this He animation, both orbiting electrons initially occupy the same orbital radius ($$t_{orbit}$$ = 247310). As the first He electron (#1 red) is being ionized (absorbing momentum), the remaining He electron (#2 blue) simultaneously drops to a lower orbital ($$t_{orbit}$$ = 117986), transferring momentum to electron #1 in the process and thus subsidizing the ionization of electron #1. After dissociation of the red electron, another photon strikes and the blue electron transitions from its now n= 1 to an n= 2 orbital.



Diatomic H
Diatomic Hydrogen radius = 37pm. The H Bohr radius was set above at 2α * (λe + λp) = 105.89pm. To simulate as a 'gravitational' orbit using only an anti-clockwise rotation with no allowance for charge, we set;

electrons; mass = 1 point, start co-ordinates (-99, 0) and (99, 0)

protons; mass = 1836 points, (0, 37) and (0, -37)

orbit center = (0, 0)

number of point to point orbitals = 6747301

... thereby setting the distance from each electron to each proton = 105.89 respectively and electron to electron at 2*99.46. The H2 ionization energy (15.426eV) is 1.1344x greater than for the H atom (13.59844eV). Likewise combining the 2 electron-electron radius (105.89/99.46 = 1.06727) gives 1.13454.

If we reduce proton-proton separation, the protons act as a single center mass and the electrons follow a circular orbit. By increasing the proton-proton separation, the electron orbits increase proportionately. This separation distance (74pm) gives a symmetrical orbit.