Quantum gravity (Planck)

Emulating gravitational and atomic orbits via n-body discrete rotating particle-particle orbital pairs at the Planck scale

This orbit model uses a geometrical approach instead of the dimensioned physical constants (G, h, c) to emulate gravitational and atomic orbits via an n-body network of rotating particle-to-particle orbital pairs. Each particle in the 'universe' is connected to every other particle by a discrete circular rotating orbital (representing a unit of momentum) with the particles at each orbital pole, thereby forming a universe-wide network of particle-to-particle orbital pairs .

Although dimensionless (the model does not use formulas that require dimensioned constants), orbit period and radius can be measured in Planck units, and so this approach is applicable to modelling gravitational orbitals at the Planck scale. In dimensioned terms, per unit of Planck time t_p each orbital pair rotates by 1 unit of Planck length l_p at velocity c (v = l_p/t_p) in 4-axis hypersphere co-ordinates, all orbitals are then summed and averaged to give the new particle co-ordinates. The process then repeats, gravitational orbits between macro-bodies in 3-D space emerging as the averaged (over time) sum of these individual rotating orbitals. The particles (of the orbiting bodies) are connected with each other directly via these orbitals, information regarding the macro orbiting bodies or associated barycenter(s) is not required.

In a simple orbit where a small mass (such as an electron or satellite) is orbiting a larger mass (nucleus, planet ...), the radius of orbit follows the formula

$r_{orbit}=r_{constant}\;*\;r_{wavelength}$

To emulate gravitational and atomic orbit(al)s, the (inverse) fine-structure constant alpha is used. Although the orbit itself occurs in the hypersphere at the speed of light, this orbital constant returns period and velocity for a 2-D plane (of 3-D space).

$r_{constant}=2\alpha$

Atomic orbitals are a geometrical subset of these gravitational orbitals, and so likewise can be interpreted as physical units of momentum rather than regions of probability where the particle can be found. They are treated as analogous in physical characteristics to the photon, with electron transition between orbitals occurring when the orbital itself absorbs (or ejects) a photon. The electron has no direct role in the transition.

Although simply the sum of rotating circles, the equations that emerge to describe simple orbits resemble the Bohr model.

By simply selecting the start co-ordinates on a 2-D plane for each point (unit of mass) accordingly, we can 'design' the required orbits. No other parameters are required.

8-body (8 mass points, 28 orbitals), the resulting orbit is a function of the start positions of each point

Particle-particle orbitals

Particles are assigned `point' (representing a discrete unit of mass) co-ordinates within a 4-axis hyper-sphere `universe'. Every particle is then connected to every other particle by a circular orbital. The hyper-sphere expands in incremental steps (FOR age = 1 TO ... representing a discrete unit of time), this expansion causes the orbitals to rotate in steps (representing a discrete unit of length) correspondingly.

In a simulation, for each value of incrementing variable age (the simulation clock-rate), all (n-body) orbitals rotate by 1 step (driven by the hypersphere expansion) and the co-ordinates calculated. These are then summed and averaged giving new particle co-ordinates. As this occurs for every particle before the next increment to age, the process can be updated in 'real time' on a serial processor. As the orbitals are circular, the barycenter for each orbital is its center, the particles at each orbital 'pole'.

As these steps are discrete, they may be measured in terms of the Planck units, each point as a unit of Planck mass, with the orbitals rotating 1 unit of Planck length l_p per 1 unit of Planck time t_p (i.e.: 1 increment to age) at velocity c = l_p/t_p in hyper-sphere coordinates.

Although orbital and so particle motion occurs at c, the hyper-sphere expansion is equidistant and so `invisible' to the observer. Instead observers 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.

N-body orbit simulation

A simulation program comprising n-body rotating orbitals is described. Each particle 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.

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

The rationale for angle β is introduced in the section on atomic orbitals (below).

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

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

Fixed radius, variable mass

For the following simple 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 (in the simulation there is no mass greater than 1 unit).

After every orbital has rotated 1 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 t_sim (as the number of increments to the simulation clock age) and the (x, y) plane orbit length l_sim 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 = j_max

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

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

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

Symmetrical 4 body orbit; (3 center mass points, 1 orbiting point, 6 orbital pairs). Note that all points orbit each other.

Example: i = 81, j = j_max = 32*81+1 = 2593 (3321 orbitals)

$r_{orbital}=r_{constant}\;*\;r_{wavelength}={r_{\alpha }}^{2}\;*\;2({\frac {j_{max}}{i}})^{2}$

t_sim = 58430803.8405

l_sim = 3528109.1197

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

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

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

This gives the following equations for the 2-D plane

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

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

Fixed mass, variable radius

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 n_g to assign the start parameters. For example, from the standard gravitational parameters, the earth to moon mass ratio approximates 81:1.

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

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

$i=j-1$

To calculate n_g

$r_{earth-moon}$ = 384400km

$\lambda _{Earth}$ = 0.00887m (Schwarzschild radius)

$n_{g}={\sqrt {\frac {2r_{earth-moon}}{\lambda _{Earth}}}}{\frac {1}{r_{\alpha }}}$ = 17783.25

This gives

$j_{max}=n_{g}j$ = 1445178.5

Converting from dimensionless numbers to SI Planck units using l_p 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

$\lambda _{0}={\frac {2j^{2}}{i^{2}}}$

${\frac {r_{earth-moon}}{r_{outer}}}\lambda _{0}m_{p}=0.59738\;10^{25}$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 (PE vs KE).

Gravitational coupling constant

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

$\alpha _{G}={\frac {Gm_{e}^{2}}{\hbar c}}={\frac {m_{e}^{2}}{m_{P}^{2}}}=1.75...x10^{-45}$

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). The above value α_G then represents the probability that 2 electrons will be in the mass point-state at any unit of Planck time (wave-particle duality at the Planck level represented by an electric-wave to mass-point oscillation ).

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 36 throws (frequency) of the dice will result in a win. The inverse of α_G is the frequency of occurence of the mass point-state between the 2 electrons. As 1 second is 10⁴² units of Planck time, this occurs about once a minute. Gravity now has a similar magnitude to the strong force (at this, the Planck level), albeit this interaction occurs seldom (the Planck level), 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 in the universe will simultaneously be in the mass point-state. For example a 1kg satellite orbits the earth, for any given unit of time, satellite (B) will have $1kg/m_{P}=45.9x10^{6}$ 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 number of orbital links (the gravitational coupling constant) between the earth and the satellite will sum to the number of orbitals;

$N_{orbitals}={\frac {m_{A}m_{B}}{m_{P}^{2}}}=0.1261\;x10^{41}$

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)

$2il_{p}=\lambda _{earth}=0.00887$ (Schwarzschild radius)

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

$n_{g}={\sqrt {\frac {r_{o}}{il_{p}}}}=5889.674$

$j_{max}=n_{g}j$

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

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

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

$\lambda _{0}=2{\frac {j^{2}}{i^{2}}}$

${\frac {t_{outer}i}{\lambda _{0}}}={\frac {\pi n_{g}^{3}r_{\alpha }^{3}\lambda _{earth}}{c}}=86164.09165s$

${\frac {r_{outer}i}{\lambda _{0}}}={\frac {n_{g}^{2}r_{\alpha }^{2}\lambda _{earth}}{2}}=42164.17km$

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).

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

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

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

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

$N_{sun}={\frac {M_{sun}}{m_{P}}}$

$N_{planet}={\frac {M_{planet}}{m_{P}}}$

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

$n_{g}={\sqrt {\frac {R_{radius}m_{P}}{2\alpha l_{p}M_{sun}}}}$

$L_{oam}=2\pi {\frac {Mr^{2}}{T}}=N_{orbitals}n_{g}{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}$

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 n_g 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.

$L_{oam}v_{g}=N_{orbitals}{\frac {hc}{2\pi }},\;{\frac {kgm^{3}}{s^{2}}}$

32 mass points (496 orbitals) begin with random co-ordinates, after 2³² steps they have clumped to form 1 large mass and 2 orbiting masses.

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 preset). 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.

8-body circular orbit plus 1-body with opposing orbitals 1:2

PE vs. KE (opposing orbitals)

Gravitational potential and kinetic energy are measures of alignment of the orbitals. In the above examples, all orbitals rotate in the same direction = kinetic energy. If all orbitals are unaligned the object will appear to 'fall' = potential energy.

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.

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 :$L={\frac {b^{2}}{a}}={\frac {al^{4}\lambda _{A}}{n^{2}}}$

${\frac {3\lambda _{A}}{2L}}={\frac {3n^{2}}{2\alpha l^{4}}}$

arc secs per 100 years (drift):

$T_{earth}$ = 365.25 days

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

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

Illustration of B's cylindrical orbit relative to A's time-line axis

Hyper-sphere orbit

Main article: Relativity (Planck)

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 (t_d) 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

$F_{p}={\frac {m_{P}c^{2}}{l_{p}}}$

$M_{a}={\frac {m_{P}\lambda _{a}}{2l_{p}}},\;m_{b}={\frac {m_{P}\lambda _{b}}{2l_{p}}}$

$F_{g}={\frac {M_{a}m_{b}G}{R^{2}}}={\frac {\lambda _{a}\lambda _{b}F_{p}}{4R_{g}^{2}}}={\frac {\lambda _{a}\lambda _{b}F_{p}}{4\alpha ^{2}n^{4}(\lambda _{a}+\lambda _{b})^{2}}}$

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

The atomic orbital in this model is a specific case of the gravitational orbital. It is also treated as a distinct unit of rotational momentum (rather than a region of probability), and has physical similarities to the photon (during orbital transition, it is the orbital radius which absorbs/ejects the photon thereby lengthening or shortening, the electron itself has a passive role). The orbital dimensions are a function of the fine structure constant alpha and the electron-nucleus wavelength. The gravitational orbital simulation can be applied to atomic orbitals, the difference being the angle of rotation β (into this angle is included an extra alpha term).

electron (blue dot) moves 1 step anti-clockwise along the alpha orbital circumference

Base orbital

The basic alpha orbital is a circular orbital with a radius 2α. In contrast to the gravitational orbital, the atomic orbital has an additional ${\sqrt {2\alpha }}$ component (note: in this atomic orbital model the n variable is analogous to the principal quantum number).

$r_{orbital}=2\alpha n^{2}$

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

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

$\beta ={\frac {1}{r_{orbital}{\sqrt {r_{orbital}}}{\sqrt {2\alpha }}}}={\frac {1}{4{\alpha }^{2}n^{3}}}$

The value for CODATA 2018 (inverse) alpha α = 137.035999084, thus the alpha orbital (where n = 1), will require about t_ref = 2π4α² ~ 471964 steps to make 1 complete rotation. To reduce computational time, the electron-nucleus frequency component f of the orbital radius is added later. For a simple Hydrogen atom, f is taken from the sum of the wavelengths of the electron + proton (λ_H = λ_e + λ_p).

The frequency component f determines the duration of the alpha orbital step, each step corresponding to 1 frequency cycle, if we use this λ_H example, then the frequency is about f_H = 0.11954 x10²³ t_p (units of Planck time), thus for the duration of 1 step;

time required = 0.11954 x10²³ t_p

distance travelled = wavelength/(2α n)

however for computational reasons the simulation calculates only the alpha orbital component.

orbital transition in steps

Orbital transition

The photon is divided into segments $r_{incr}$.

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

Photon absorption:

A photon will be absorbed by the orbital radius such that after each transition step the orbital radius will be reduced by r_incr (as r_incr has a minus value).

$r_{orbital}=r_{orbital}+r_{incr}$

Conversely, by ejecting a photon the orbital radius will increase.

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

$\lambda _{photon}=R.({\frac {1}{n_{i}^{2}}}-{\frac {1}{n_{f}^{2}}})={\frac {R}{n_{i}^{2}}}-{\frac {R}{n_{f}^{2}}}$

$\lambda _{photon}=(+\lambda _{i})-(+\lambda _{f})$

As $r_{incr}$ is minus, the '(+$\lambda _{i}$)' photon will decrease per transition step the orbital radius in $r_{incr}$ increments as 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;

$r_{orbital}=r_{orbital}+(\lambda _{i})-(\lambda _{f})=r_{orbital}+r_{incr}-r_{incr}$

However 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 ($-r_{incr}$ = +0.000581).

$r_{orbital}=r_{orbital}-r_{incr}$

As with gravitational orbitals, the velocity of rotation derives from the radius of the orbital (from $\beta$) and so adjusts as the orbital radius changes. The simulation calculates the velocity per step to determine the final transition velocity component.

We may also note that of the base orbital radius r, although r = 2α, it is a construct of these $r_{incr}$ segments, as we transition from r = 0 to r = 2α in $r_{incr}$ increments, we require $t_{ref}$ steps.

Relativistic orbital

If we include the frequency component f in the orbital radius (which is not practical for a simulation where f ~ 10²³), we find that the electron travels 1 Planck length per unit of Planck time in hypersphere coordinates, as with gravitational orbitals.

H atom

H ($f_{1s-2s}$) = 2466061413.187GHz is the most precisely measured of the spectra and can be used as a reference.

$f_{H}={\frac {2}{\lambda _{e}/(2\pi )+\lambda _{p}/(2\pi )}}$

Including the hypersphere relativistic term, the transition from an n_initial to the higher n_final orbit by photon absorption;

$t_{orbital}=n_{i}t_{ref}{\sqrt {1-{v_{orbital}}^{2}}}$

$t_{spiral}=(n_{f}-n_{i})t_{ref}{\sqrt {1-{v_{average}}^{2}}}$

$H\;f_{n_{i}-n_{f}}={\frac {(n_{f}-n_{s})f_{H}c}{(t_{orbital}+t_{spiral})n_{s}}}$

H atom orbital transitions from n1-n2, n2-n3, n3-n1 via 2 photon capture (photons expand (contract) the orbital radius, the electron continues its anti-clockwise rotation.

In the animation (right), the electron is moving in an anti-clockwise direction, it is the orbital radius that is extending (or contracting) in $r_{incr}$ steps, the electron has a passive role. Note that we are calculating only for a 2-D plane (whereas the electron travels in 3-D space), furthermore there are no assumptions made regarding complex geometries and so we can only suggest models. For example, we can compensate (and so improve accuracy in line with experimental values) by truncating the base orbital period from $t_{ref}$ to $t_{o}$. Here we set $t_{o}$ = 471961.8898 for the H atom n_i=1 to n_f=2 transition using the (CODATA 2018 ) wavelengths for the electron and proton. There are other approachs, the results are similiar.

electron; 2.42631023867e-12m

proton; 1.32140985539e-15m

t_o = 471961.8898 gives H(1s-2s) 2466061413MHz

Optimal values for these transitions;

$t_{o}$ = 471960.1758 gives H(1s-3s) 2922743278MHz

$t_{o}$ = 471959.8404 gives H(1s-4s) 3082581563MHz

We can extrapolate to get approximate values

$t_{o}$ = 471959.715 for H(1s-5s) 3156563436MHz

$t_{o}$ = 471959.626 for H(1s-6s) 3196751153MHz

The theorectical value for ionization of the H electron;

$t_{o}$ = 471959.242135 for 3288086800MHz

One possible explanation could involve the Doppler effect as each ejected photon carries away different 'quantities' of momentum with it. If we set t_ref = 2π2α2α as the reference = 1.0 then the shift per photon would correspond to

H(1s-2s) = 1.002628

H(1s-3s) = 1.002980

H(1s-4s) = 1.003098

Ionization = 1.003297

In summary, photon #1 strikes and deletes a section of the orbital radius whilst photon #2 adds to the orbital radius (t_orbital), as a consequence the orbital radius does not change (the electron orbit continues as per normal). After photon #1 has been absorbed, the radius will begin expanding until photon #2 is also completely absorbed (t_transition). The electron is now in the new orbit(al).

Thus the quantum nature of the orbitals also derives from the photon wavelengths, this suggests that the photon wavelengths are not independent of the electron wavelength.

He dissociation of electron 1 (red) then electron 2 (blue) transition n=1 to n=2

Helium

Higher atoms have more complex geometries but the principal remains the same. In this He animation, both orbiting electrons occupy the same orbital radius. As electron 1 (red) is ionized (absorbing momentum), electron 2 (blue) simultaneously drops to a lower orbital radius (emitting momentum), thus electron 2 subsidizes the ionization of electron 1.

After dissociation of the red electron, the blue electron then transitions from its new n= 1 to an n= 2 orbital. The axes measure only the alpha orbital component.

gravitational orbit with mass and distance ratio equivalent to diatomic H

Diatomic H

Diatomic Hydrogen radius = 37pm. The H Bohr radius was set above at 2α * (λ_e + λ_p) = 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 the optimal solution for a gravitational orbit.

External links

• Physical constant anomalies
• Planck units as geometrical objects
• The mathematical electron
• Programming relativity at the Planck scale
• Programming the cosmic microwave background at the Planck level
• The sqrt of Planck momentum
• The Programmer God
• Simulation hypothesis modeling at the Planck scale using geometrical objects