Quantum gravity (Planck)

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

An orbital simulation program is described that emulates gravitational and atomic orbitals as the sum of individual particle-particle orbital pair rotations at the Planck scale.

By 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 used. The 26 points orbit each other resulting in 325 point-point orbitals.

For simulating gravity, orbiting objects A, B, C... are sub-divided into discrete points, each point representing 1 unit of Planck mass m_P (for example, a 1kg satellite would be divided into 1kg/m_P = 45940509 points). Each point in object A then 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 l_p per unit of Planck time t_p at velocity c (c = l_p/t_p) in hypersphere space co-ordinates, when these orbital pair rotations are summed and mapped over time, gravitational orbits emerge between the objects A, B, C...

For simulating electron transition within the atom, the mass points are subdivided into their component atomic orbitals. The quantum energy levels are derived geometrically via a hyperbolic spiral (the electron path) that emerges during transition.

The basic simulation uses only the start position (x, y coordinates) 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 orbital is calculated independently of all other orbitals). Atomic orbitals require the co-ordinates of the nucleus and electron. As the simulation is dimensionless, the only physical constant used is the fine structure constant alpha.

Theory

The simulation itself is a dimensionless geometrical model, but it can be measured in Planck units. In the simulation, particles are treated as an electric wave-state to (Planck) mass point-state oscillation, the wave-state as the duration of particle frequency in Planck time units, the point-state duration as 1 unit of Planck time (as a point, this state can be assigned mapping coordinates), the particle itself is an oscillation between these 2 states (i.e.: the particle is not a fixed entity). For example, an electron has a frequency (wave-state duration) = 10²³ units of Planck time followed by the mass state (1 unit of Planck time). The background to this oscillation is given in the mathematical electron model.

If the electron is mass (1 unit of Planck mass) for 1 unit of Planck time, and then no mass for 10²³ units of Planck time (the wave-state), then in order for a (hypothetical) object composed only of electrons to have 1 unit of Planck mass at every unit of Planck time, the object will require 10²³ electrons. This is because orbital rotation occurs at each unit of Planck time and so the simulation requires this object to have a unit of Planck mass at each unit of Planck time (i.e.: on average there will always be 1 electron in the mass point state). We would then measure the mass of this object as 1 Planck mass (our measured mass of an object reflects the average number of units of Planck mass per unit of Planck time). For the simulation program, this object can now be defined as a point (it will have point co-ordinates at each unit of Planck time and so can be mapped). As the simulation is dividing the mass of objects into Planck mass size points and then rotating these points around each other as point-to-point orbital pairs, then by definition gravity becomes a mass to mass interaction.

Nevertheless, although this is a mass-point to mass-point rotation, and so referred to here as a point-point orbital, it is still a particle to particle orbital, albeit the particles are both in the mass state. We can also map particle to particle orbitals for which both particles are in the wave-state, the H atom is a well-researched particle-to-particle orbital pair (electron orbiting a proton) and so can be used as reference. To map orbital transitions between energy levels, the simulation uses the photon-orbital model, in which the orbital (Bohr) radius is treated as a 'physical wave' akin to the photon albeit of inverse or reverse phase. The photon can be considered as a moving wave, the orbital radius as a standing/rotating wave (trapped between the electron and proton).

It is the rotation of the orbital radius that pulls the electron, resulting in the electron orbit around the nucleus. Furthermore, orbital transition (between orbitals) occurs between the orbital radius and the photon, the electron has a passive role. Transition (the electron path) follows a specific hyperbolic spiral for which the angle component periodically cancels into integers which correspond with the orbital energy levels (at 360° radius =4r, 360+120°=9r, 360+180°=16r, 360+216°=25r ... 720°=∞r). As these spiral angles (360°, 360+120°, 360+180°, 360+216° ...) are linked directly to pi, we may ask if quantization of the atom has a geometrical origin.

In summary, both gravitational and atomic orbitals reflect the same particle-to-particle orbital pairing, the distinction being the state of the particles; mass to mass or wave to wave. There are not 2 separate forces used by the simulation, instead particles are treated as oscillations between the 2 states (electric wave and mass point).

The gravitational orbits that we observe are the time averaging sum of the underlying gravitational orbitals. Gravity is therefore not weaker than the electric force, rather it is stronger at the Planck scale (for point-point orbitals rotate faster than wave-wave), its apparent weakness is simply because point-point rotations seldom occur relative to wave-wave orbitals (the probability of occurrence as the inverse of the gravitational coupling constant which for each electron would be the ratio $1:10^{23}$).

N-body orbitals

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

The simulation universe is a 4-axis hypersphere expanding in increments with 3-axis (the hypersphere surface) projected onto an (x, y) plane with the z axis as the simulation timeline (the expansion axis). Each point is assigned start (x, y, z = 0) co-ordinates and forms pairs with all other points, resulting in a universe-wide n-body network of point-point orbital pairs. The barycenter for each orbital pairing is its center, the points located at each orbital 'pole'.

The simulation itself is dimensionless, simply rotating geometries. To translate to dimensioned gravitational or atomic orbits, we can use the Planck units (mass m_P, length l_p, time t_p), such that the simulation increments in discrete steps (each step assigned as 1 unit of Planck time), during each step (for each unit of Planck time), the orbitals rotate 1 unit of (Planck) length (at velocity c = l_p/t_p). These rotations are then all summed and averaged to give 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.

Orbital pair rotation on the (x, y) plane occurs in discrete steps according to an angle β as defined by the orbital pair radius (the atomic orbital β has an additional alpha term).

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

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.

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.

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

2 body orbits (x, y plane)

For simple 2-body orbits, to reduce computation only 1 point is assigned as the orbiting point and the remaining points are assigned as the central mass. For example the ratio of earth mass to moon mass is 81:1 and so we can simulate this orbit accordingly. However we note that the only actual distinction between a 2-body orbit and a complex orbit 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) ... this is because the simulation treats all points equally, the center points also orbiting each other according to their orbital radius, for the simulation itself there is no difference between simple 2-body and complex n-body orbits.

The Schwarzschild radius formula in Planck units

$r_{s}={\frac {2l_{p}M}{m_{P}}}$

As the simulation itself is dimensionless, we can remove the dimensioned length component $2l_{p}$, and as each point is analogous to 1 unit of Planck mass $m_{P}$, then the Schwarzschild radius for the simulation becomes the number of central mass points. We then assign (x, y) co-ordinates (to the central mass points) within a circle radius $r_{s}$ = number of central points = total points - 1 (the orbiting point).

After every orbital has rotated 1 (Planck) length unit (anti-clockwise in these examples), 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 (as the number of increments to the simulation clock) and the (x, y) plane orbit length l (distance as measured on the 2-D plane) are noted.

Key:

1. $r_{s}$ = i; number of center points in the orbit (the center mass).

2. j = total number of points (for simple 2 body orbits with only 1 orbiting point, j = i + 1 = 82).

3. j_max = radius to mass co-efficient.

4. x, y = start co-ordinates for each point (on a 2-D plane), z = 0.

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). This constant adapts the simulation specifically to gravitational and atomic orbitals.

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

Orbital formulas (2-D plane)

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

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

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

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

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

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

Simulation data:

period $t_{sim}$

length $l_{sim}=t_{sim}{\frac {i}{j_{max}r_{\alpha }}}$

radius $r_{sim}={\frac {l_{sim}}{2\pi }}$

velocity $v_{sim}={\frac {l_{sim}}{t_{sim}}}={\frac {i}{r_{\alpha }j_{max}}}$

barycenter $b_{sim}={\frac {x_{max}+x_{min}}{2}}$

Object leaving a gravitational circular orbit (j_max = j) with constant outward motion follows the same alpha hyperbolic spiral as an ionizing electron

For example; 8 mass points (28 orbitals) divided into j = 8 (total points), i = j - 1 (7 center mass points). After 1 complete orbit, actual period t and distance travelled l are noted and compared with the above formulas.

1) j_max = i+1 = 8

period $t=74465.0516,\;t_{outer}=74471.6125$

length $l=l_{sim}=3935.7664,\;l_{outer}=3936.1032$

radius $r_{sim}=626.3951$

velocity $v_{sim}=1/18.920137$

barycenter $b_{sim}=89.5241,\;r_{barycenter}=89.4929$

2) j_max = 32*i+1 = 225

period $t=1656793370.3483,\;t_{outer}=1656793381.3051$

length $l=l_{sim}=3113519.1259,\;l_{outer}=3113519.1385$

radius $r_{sim}=495531.959$

velocity $v_{sim}=1/532.128856$

barycenter $b_{sim}=70790.283,\;r_{barycenter}=70790.280$

3) Moon orbit.

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 i = 81 points, j = i + 1.

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

$r_{earth-moon}$ = 384400km

$M_{earth}$ = 0.597378 10²⁵kg

Solving $j_{max}$

$r_{outer}={r_{\alpha }}^{2}\;*\;2({\frac {j_{max}}{i}})^{2}={\frac {2r_{earth-moon}m_{P}}{M_{earth}l_{p}}}$

$j_{max}=1440443$

Gives

$t_{outer}=4\pi {({\frac {j_{max}{r_{\alpha }}}{i}})}^{3}({\frac {l_{p}}{c}})=0.8643\;10^{-26}$s

$t_{outer}{\frac {M_{earth}}{m_{P}}}=2371844$s (27.452 days)

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

$v_{Earth}=(c){\frac {1}{j_{max}r_{\alpha }}}=12.57m/s$

$r_{barycenter}={\frac {r_{earth-moon}}{j}}=4688km$

Gravitational coupling constant

In the above, the points were assigned a mass as a theoretical unit of Planck mass. 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;

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

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 α_G 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 α_G 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 = f_e, 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:

$i={\frac {M_{earth}}{m_{P}}}=0.27444\;x10^{33}$ (earth as the center mass)

$i2l_{p}=0.00887$ (earth Schwarzschild radius)

$s={\frac {1kg}{m_{P}}}=45940509$ (1kg orbiting satellite)

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

1) 1kg satellite at earth surface orbit

$r_{o}=6371000km$ (earth surface)

$j_{max}={\frac {j}{r_{a}}}{\sqrt {\frac {r_{o}}{il_{p}}}}=0.288645\;x10^{44}$

$n_{g}={\frac {j_{max}}{j}}=2289.41$

$r=r_{\alpha }^{2}n_{g}^{2}il_{p}=r_{o}$

$v={\frac {c}{n_{g}r_{\alpha }}}=7909.7924$ m/s

$t=2\pi {\frac {r_{outer}}{v_{outer}}}=5060.8374$ s

2) 1kg satellite at a synchronous orbit radius

$r_{o}=42164.17km$

$j_{max}={\frac {j}{r_{a}}}{\sqrt {\frac {r_{o}}{il_{p}}}}=0.74256\;x10^{44}$

$n_{g}={\frac {j_{max}}{j}}=5889.674$

$r=r_{\alpha }^{2}n_{g}^{2}il_{p}=r_{o}$

$v={\frac {c}{n_{g}r_{\alpha }}}=3074.66$ m/s

$t=2\pi {\frac {r_{outer}}{v_{outer}}}=86164.09165$ s

3) 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)

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

4) 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.

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). 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, they have formed planets.

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

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

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

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 resource: Relativity (Planck)

Each point moves 1 unit of (Planck) length per 1 unit of (Planck) time in x, y, z (hyper-sphere) co-ordinates, the simulation 4-axis hyper-sphere universe expanding in uniform (Planck) 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.

$d=r_{\alpha }n_{g}$

$t_{0}=2\pi r=2\pi {\frac {t}{2\pi d}}$

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

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

Bohr radius during ionization, as the H atom electron reaches each n level, it completes 1 orbit (for illustration) then continues outward (actual velocity will become slower as radius increases according to angle β)

In the atom we find individual particle to particle orbitals, and as such the atomic orbital is principally a wave-state orbital (during the orbit the electron is predominately in the electric wave-state). The wave-state is defined by a wave-function, we can however map (assign co-ordinates to) the mass point-states and so follow the electron orbit, for example, in 1 orbit at the lowest energy level in the H atom, the electron will oscillate between wave-state to point-state approximately 471960 times.

During electron transition between orbitals, we find the electron follows a hyperbolic spiral, this is significant because periodically the spiral angle components cancel reducing to integer radius values (360°=4r, 360+120°=9r, 360+180°=16r, 360+216°=25r ... 720°=∞r).

As these spiral angles (360°, 360+120°, 360+180°, 360+216° ...) are linked directly to pi, we may ask if quantization of the atom has a geometrical origin. .

Simulation

The simulation program can be modified for atomic orbitals by including an additional alpha term in the rotation angle β.

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

The following example simulates an electron transition, the electron begins at radius $r=r_{orbital}$ and moves in incremental steps to higher orbitals. The nucleus is a set of 249 mass points arranged in the center, period and length are measured at the specific spiral angles (360°; x=r, y=0 ... 720°; x=4r, y=0 ... 900°; x=-16r, y=0 ... )

Key: nucleus = 249 mass points and the electron = 1 mass point, t_sim = period and l_sim = distance travelled by the electron, the radius coefficient n_f is divided by $r_{orbital}$ for clarity. As the electron travels further from the center, the nucleus expands.

H atom electron transition spiral plotting the nucleus and barycenter as the electron transitions from n=1 to n=8

$j_{atom}=250$ (atomic mass)

$i_{nucleus}=j_{atom}-1=249$ (relative nucleus mass)

$r_{wavelength}=2({\frac {j_{atom}}{i_{nucleus}}})^{2}$

$r_{orbital}=2\alpha \;*\;r_{wavelength}$ (radius)

$t_{calc}={\frac {t_{sim}}{r_{wavelength}(n^{2}-1)}}$

$l_{n}={\frac {l_{sim}}{2\pi r_{orbital}}}$

$r_{b}=r_{sim}-{\frac {r_{sim}}{j_{atom}}}$

$r_{n}={\frac {r_{b}}{r_{orbital}}}$

Experimental values for H(1s-ns) transitions (n the principal quantum number).

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

H(1s-∞s) = 3288 086 857 127.60 kHz (n_f = ∞)

R = 10973731.568157 (Rydberg constant)

α =137.035999177 (inverse fine structure constant

proton/electron mass ratio pe = 1836.152673426

$\lambda _{H}={\frac {2c}{\lambda _{e}+\lambda _{p}}}=8\pi c\alpha ^{2}R{\frac {pe}{pe+1}}$

Dividing wavelength by transition returns a dimensionless number;

$t_{ns}={\frac {\lambda _{H}}{H(1s-ns)}}$

Including the n variable gives

$t_{expt}=t_{ns}(n^{2}-1)$ (the number of transition 'segments')

There is no n=1 to n=1 transition, but the ionization energy will be virtually equivalent to the n=1 orbital, and so for this orbital we can use the n=infinity transition as a close approximation.

$t_{1s}={\frac {\lambda _{H}}{H(1s-\infty s)}}$

Note: For a theoretical Rydberg atom (of point size, infinite mass and disregarding wavelength), each segment is a multiple of $t_{ref}$

$t_{ref}=2\pi 2\alpha 2\alpha =471964.333$

Electron transition (mass = 250 n=1 to n=5)

angle	rn	tref	texpt	tsim	lsim	ln	x, y (electron)	x, y (nucleus)	x, y (barycenter)
360° n=1	0.9999836	471964.333	471959.243*	471957.072	3457.8864	0.99999	550.3341, 0.0036	-2.2125, -0.0075	-0.0023, -0.0075
720° n=2	4.0010691	1887857.333	1887839.826	1888356.002	6917.7127	3.00054	2202.8558, 0.0001	-7.9565, -1.9475	0.8868, -1.9397
840° n=3	8.9997519	4247679.375	4247634.049	4247567.984	13831.460	4.99993	-2472.8741, 4296.3067	13.5590, -10.3184	3.6133, 6.9081
900° n=4	15.999424	7551429.333	7551347.553	7551165.283	20747.092	6.99987	-8814.9421, 13.3054	25.6293, 13.3056	-9.7330, 13.3056
936° n=5	24.99631	11799108.333	11798979.543**	11798590.909	31118.741	8.99931	-11157.39, -8079.27	16.573, 39.086	-28.123, 6.612

* at infinity the second photon energy reduces to irrelevancy.

** Thiele interpolation

Comparative data

H atom transitions spiral data for different nucleus mass; green=64, dark blue=126, orange=250, blue=experimental

If we include simulations for mass = 64 and mass = 126 as reference and calculate all data relative to each respective $t_{n=1}$ (i.e.: divide by n²) then we can compare results and we find the simulation and experimental data graphs agree.

We also find that the differences in the n=2 data (m_sim/m_expt) are proportionate to the nucleus mass and so by varying nucleus shape and mass it may be possible to duplicate the experimental results and this is the next step.

Note: A simulated Rydberg atom gives a straight line (no peaks) - i.e.: the barycenter is always (0, 0) and the nucleus is a point (with mass but no size) and so does not expand.

Comparison

n	m64	m126	m250	mexpt
1	471957.4090	471957.1876	471957.0721	471959.2428
2	472480.7459	472220.6477	472089.0004	471959.9566
3	471949.6292	471951.8661	471951.9982	471959.3387
4	471946.4217	471947.1847	471947.8302	471959.2221
5	471942.8053	471942.7795	471943.6364	471959.1075

Electron at different n level orbitals

Theory

Bohr model

The H atom has 1 proton and 1 electron orbiting the proton, the electron can be found at fixed radius (the Bohr radius) from the proton (nucleus), these radius represent different energy levels (orbitals) at which the electron may be found orbiting the proton and so are described as quantum levels. Electron transition (to higher energy levels) occurs when an incoming photon provides the required energy (momentum). Conversely emission of a photon will result in electron transition to lower energy levels.

The principal quantum number n denotes the energy level for each orbital. As n increases, the electron is at a higher energy and is therefore less tightly bound to the nucleus (as n increases, the electron is further from the nucleus). Each n (electron shell) can accommodate up to n² electrons (1, 4, 9, 16, 25...), and accounting for two states of spin, 2n². As these orbitals are fixed according to integer n, the atom can be said to be quantized.

The basic (alpha) radius for each n level uses the fine structure constant alpha (α = 137.036) whereby;

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

Such that at n = 1, the start radius r = 2α. We can map the electron orbit around the orbital as a series of steps with the duration of each step the frequency of the electron + proton wavelengths ($\lambda _{p}+\lambda _{e}$). The steps are defined according to angle β;

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

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

At specific n levels;

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

This gives a length travelled per step as the inverse of the radius

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

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

The number of steps (orbital period) for 1 orbit of the electron then becomes

$t_{orbital}={\frac {2\pi r_{orbital}}{v_{orbital}}}=2\pi 2\alpha 2\alpha n^{3}$

A base (reference) orbital (n=1)

$t_{ref}=2\pi 4\alpha ^{2}$

Photon orbital model

The electron can jump between n levels via the absorption or emission of a photon. In the photon-orbital model, the orbital (Bohr) radius is treated as a 'physical wave' akin to the photon albeit of inverse or reverse phase such that $orbital\;radius+photon=zero$ (cancel).

The photon can be considered as a moving wave, the orbital radius as a standing/rotating wave (trapped between the electron and proton). It is the rotation of the orbital radius that pulls the electron, resulting in the electron orbit around the nucleus (orbital momentum comes from the orbital radius). Furthermore, orbital transition (between orbitals) occurs between the orbital radius and the photon, the electron has a passive role.

The photon is actually 2 photons as per the Rydberg formula (denoted initial and final).

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

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

The wavelength of the ($\lambda _{i}$) photon corresponds to the wavelength of the orbital radius. The (+$\lambda _{i}$) will then delete the orbital radius as described above (orbital + photon = zero), however the (-$\lambda _{f}$), because of the Rydberg minus term, will have the same phase as the orbital radius and so conversely will 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 each other;

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

However, the ($\lambda _{f}$) has the longer wavelength, and so after the ($\lambda _{i}$) photon has been absorbed, and for the remaining duration of this ($\lambda _{f}$) photon wavelength, at each transition step the orbital radius will be extended until the ($\lambda _{f}$) is also absorbed. At each step, as the orbital radius increases, the orbital rotation angle β will conversely decrease, and as the velocity of orbital rotation depends on β, the velocity will adjust accordingly.

For example, the electron is at the n = 1 orbital. To jump from an initial $n_{i}=1$ orbital to a final $n_{f}=2$ orbital, first the ($\lambda _{i}$) photon is absorbed ($\lambda _{i}+\lambda _{orbital}=zero$ which corresponds to 1 complete n = 1 orbit by the electron, the orbital phase), then the remaining ($\lambda _{f}$) photon continues until it too is absorbed (the transition phase).

$\lambda _{i}=1t_{ref}$

$\lambda _{f}=4t_{ref}$ (n = 2)

After $t_{ref}$ steps, the ($\lambda _{i}$) photon is absorbed, but the ($\lambda _{f}$) photon still has $\lambda _{f}=(4-1)t_{ref}$ steps remaining until it too is absorbed.

orbital transition during orbital rotation

Instead of a discrete jump between energy levels by the electron, absorption/emission takes place in steps, each step corresponds to a unit of $r_{incr}$;

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

As $r_{incr}$ has a minus value, the ($\lambda _{i}$) photon will shrink the orbital radius accordingly, per step

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

Conversely, because of its minus term, the ($\lambda _{i}$) photon will extend the orbital radius accordingly;

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

The transition frequency is a combination of the orbital phase and the transition phase.

${\frac {{n_{f}}^{2}-{n_{i}}^{2}}{t_{orbital}+t_{transition}}}$

The model assumes orbits also follow along a timeline z-axis

$t_{orbital}=t_{ref}{\sqrt {1-{\frac {1}{(v_{orbital})^{2}}}}}$

The orbital phase has a fixed radius, however at the transition phase this needs to be calculated for each discrete step as the orbital velocity depends on the radius;

$t_{transition}=t_{ref}{\sqrt {1-{\frac {1}{(v_{transition})^{2}}}}}$

Alpha spiral

Hyperbolic spiral

A hyperbolic spiral is a type of spiral with a pitch angle that increases with distance from its center. As this curve widens (radius r increases), it approaches an asymptotic line (the y-axis) with the limit set by a scaling factor a (as r approaches infinity, the y axis approaches a).

In its simplest form, a fine structure constant spiral (or alpha spiral) is a specific hyperbolic spiral that appears in electron transitions between atomic orbitals in a Hydrogen atom.

It can be represented in Cartesian coordinates by

$x=a^{2}{\frac {cos(\varphi )}{\varphi ^{2}}},\;y=a^{2}{\frac {sin(\varphi )}{\varphi ^{2}}},\;0<\varphi <4\pi$

This spiral has only 2 revolutions approaching 720° as the radius approaches infinity. If we set start radius r = 1, then at given angles radius r will have integer values (the angle components cancel).

$\varphi =(2)\pi ,\;r=4$ (360°)

$\varphi =(4/3)\pi ,\;r=9$ (240°)

$\varphi =(1)\pi ,\;r=16$ (180°)

$\varphi =(4/5)\pi ,\;r=25$ (144°)

$\varphi =(2/3)\pi ,\;r=36$ (120°)

Starting with $\varphi =0,\;r=2\alpha$ (n=1), for each step during transition;

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

$\varphi =\varphi +\beta$

As $\beta$ is proportional to the radius, as the radius increases the value of $\beta$ will reduce correspondingly (likewise reducing the orbital velocity).

Further information: Fine-structure_constant_(spiral)

Setting t = step number (FOR t = 1 TO ...), we can calculate the radius r and the $n_{f}^{2}$ at each step .

$r=2\alpha +{\frac {t}{2\pi 2\alpha }}$ (number of increments t of $r_{incr}$)

$n_{f}^{2}=1+{\frac {t}{2\pi 4\alpha ^{2}}}$ ($n_{f}^{2}$ as a function of t)

$\varphi =4\pi {\frac {(n_{f}^{2}-n_{f})}{n_{f}^{2}}}$ ($\varphi$ at any $n_{f}^{2}$)

fig 5. H atom orbital transitions from n1-n2, n2-n3, n3-n1 via 2 photon capture, photons expand/contract the orbital radius. The spiral pattern emerges because the electron is continuously pulled in an anti-clockwise direction by the rotating orbital.

H atom

The Bohr radius for an ionizing electron (H atom) follows this hyperbolic spiral. At specific spiral angles, the angle components (for this particular spiral) cancel returning an integer value for the radius (360°=4r, 360+120°=9r, 360+180°=16r, 360+216°=25r ... 720°=∞r).

The wavelength of the electron in terms of the Rydberg constant = 10973731.568539(55) m⁻¹

$\lambda _{e}={\frac {2}{t_{ref}R}}$

In the classical Bohr model, the electron orbits around the barycenter (center of mass) and for this is used the reduced mass (the CODATA proton-electron mass ratio μ = 1836.152673426(32)).

$\mu _{n}={\frac {m_{e}+m_{p}}{m_{p}}}=1+{\frac {1}{\mu }}$ = 1.000544 617 021

However, the $H_{1s-\infty s}$ (ionization) vs. Rydberg constant shows slight divergence

${\frac {R}{H(1s-\infty s)}}$ = 1.000533 776 387

The formula for transition, including the variable $\mu _{n}$ term;

$f{(n_{i}\;to\;n_{f})}=({\frac {t_{ref}Rc}{\mu _{n}}})({\frac {{n_{f}}^{2}-{n_{i}}^{2}}{(t_{orbital}+t_{transition})}})$

We can then determine the precise value for $\mu _{n}$ for each energy level using the literature values as reference:

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

H atom transition (n=1 to n=64); alpha orbital spiral angles (pi) vs transition energies (eV)

Gives for these n levels

n = 2, $\mu _{n}$ = 1.000539 387 875

n = 3, $\mu _{n}$ = 1.000536 337 888

n = 4, $\mu _{n}$ = 1.000535 460 372

orbital transition frequencies (Rydberg - H) in eV vs spiral angle (pi) for transitions from n=1 up to n=64

We can also map the (continuous) transition energy difference (in eV) between a Rydberg atom (nucleus mass >> electron mass) and the lighter proton (y-axis = Rydberg atom eV - H atom eV).

External links

• Fine structure constant hyperbolic spiral
• 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 modelling at the Planck scale using geometrical objects