# Quantum gravity (Planck)

Relativistic Planck gravity for Simulation Hypothesis modeling

A gravitational force between macro-objects is simulated by a ( fine structure constant based) pixel lattice geometry supporting rotating particle-particle orbiting pairs. As this approach uses digital time, it is applicable for programming Planck-unit Simulation Hypothesis models [1]. Orbits between macro-objects are the averaged sum of the underlying particle-particle orbitals. Although dimension-less (using rotating circles to simulate gravity), these orbitals can be defined by units of dimensioned ${\displaystyle \hbar c}$.

3 body (3 mass points, 3 orbitals) orbit, each point begins with equal (anti-clockwise) momentum

### Planck unit gravity

In Planck level simulations, the simulation clock-rate is measured in discrete units of incrementing Planck time (digital time frames) forming a universe time-line against which the frequencies of Planck events can be mapped. The mathematical electron model is applicable to digital Planck unit gravity simulations as it assigns to particles an oscillation between an electric wave-state (particle frequency) to a discrete unit of Planck-mass (at unit Planck-time) mass point-state. Mass can then be treated as a digital event rather than a constant property of the particle such that for any chosen unit of Planck time, all those particles that are simultaneously in the mass point-state can form individual rotating orbital pairs with each other. Consequently information regarding orbiting macro-objects is not required as gravitational orbits naturally emerge from the averaged sum of the underlying rotating particle-particle orbital pairs.

4 body orbit; (4 mass points, 6 orbitals) with central 3 sub orbitals followed by 10 sub orbitals

For objects whose mass is less than Planck mass, there will be units of Planck time when the object has no particles in the point-state and so no gravitational interactions. As such gravity, as a function of mass, is also a discrete event, the magnitude of the gravitational interaction approximating the magnitude of the strong force, the gravitational coupling constant representing a measure of the frequency of these interactions and not the magnitude of the gravitational force itself. Gravity and mass are therefore interchangeable terms describing different aspects of the same process.

All particles simultaneously in the mass point-state at any unit of Planck time are connected to each other by individual rotating gravitational orbitals (which can be represented as units of ${\displaystyle \hbar c}$). The velocity of the gravitational orbit is summed from these individual particle-particle velocities. Gravitational potential and kinetic energies are a measure of the alignment of the underlying orbitals. The orbital angular momentum of the planetary orbits derives from the sum of the planet-sun particle-particle orbital angular momentum irrespective of the angular momentum of the sun itself and the rotational angular momentum of a planet includes particle-particle rotational angular momentum.

8-body (8 mass points, 27 orbitals) orbit

### Gravitational coupling constant

The Gravitational coupling constant αG characterizes the gravitational attraction between a given pair of elementary particles in terms of the electron mass to Planck mass ratio;

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

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 t, satellite (B) will have ${\displaystyle 1kg/m_{P}=45.9x10^{6}}$ particles in the point-state. The earth (A) will have ${\displaystyle 5.97\;x10^{24}kg/m_{P}=0.274\;x10^{33}}$ particles in the point-state. For any given unit of Planck time the number of links between the earth and the satellite will sum to;

${\displaystyle N_{links}={\frac {m_{A}m_{B}}{m_{P}^{2}}}=0.126\;x10^{41}}$

The gravitational orbit can be characterized by :${\displaystyle N_{links}\hbar c}$

If A and B are respectively Planck mass particles then N = 1. If A and B are respectively electrons then the probability that any 2 electrons are simultaneously in the mass point-state for any chosen unit of Planck time becomes N = αG and so a gravitational interaction between these 2 electrons will occur only once every 1045 units of Planck time.

### Planck unit formulas

dimensionless alpha pixel

(inverse) fine structure constant α = 137.03599...,

np = pixel number

λobject = Schwarzschild radius

mP = Planck mass

${\displaystyle N}$ = number of Planck mass point-states per unit of Planck time

#### Dimension-less structures

Between 2 rotating (Planck mass) points;

${\displaystyle n_{g}={\frac {n_{p}}{\sqrt {N}}}}$ (pixel to mass ratio)
${\displaystyle d={\sqrt {2\alpha }}{\frac {n_{p}}{\sqrt {N}}}=({\sqrt {2\alpha }})n_{g}}$ (pixel aggregate)
${\displaystyle r=2\alpha n_{p}^{2}=d^{2}N}$ (orbital length)
B's orbit relative to A's time-line axis

While B (satellite) has a circular orbit around A (planet) in 3-D space co-ordinates, when mapped in an expanding relativistic hyper-sphere, it follows a cylindrical orbit (from B1 to B11) around the A time-line axis in hyper-sphere co-ordinates. If A is moving with the universe expansion (albeit stationary in 3-D space) then B is orbiting A at the speed of light, td (orbital period allowing for relativistic hyper-sphere motion) emerges along the A time-line axis;

We can simulate this 3d cylinder by projecting it onto a 2d plane as the difference between 2 orbits;

${\displaystyle t_{small}=2\pi {\frac {dN}{2}}}$
${\displaystyle t_{large}=2\pi d^{3}N}$
${\displaystyle t=t_{large}-t_{small}=2\pi (d^{3}-{\frac {d}{2}})N}$
${\displaystyle v={\frac {2\pi r}{t}}={\frac {d}{d^{2}-{\frac {1}{2}}}}}$

Orbital momentum is simulated by the incrementing angle as a function of t. A 'Planck' orbital as minimum distance between 2 point mass (i.e.: with no wave-state) would have an orbital period;

${\displaystyle t_{p}=2\pi (d^{3}-{\frac {d}{2}})=28456.6547...;\;n_{g}=1}$
${\displaystyle angle={\frac {1}{(d^{3}-{\frac {d}{2}})}}}$

In orbits with multiple points, N>1. Example: a 4-body (4-point) orbit comprising 1-point (start 13623, 0) orbiting a 3-point center (see 4-body orbit diagram). The 3 center points (centered 0, 0) are then repeated to simulate increasing mass (λorbit = 2N, kr = 28456.6547....)

Each orbiting pair is calculated independently. All orbits are then summed and averaged before the next unit of Planck time is incremented. Thus the universe simulation can be updated in real-time on a serial processor.

Orbital parameters
total points j orbit period k ng N (mass) barycenter r/j = r-(2αnp2);
j1 = 1+3 k1 = 5306671 n = 5 N = 1.5 3434, 144
j2 = 1+6 k2 = 3032251 (k1/k2 = 7/4) n = 5/2 N = 6.83 1953, 176
j3 = 1+9 k3 = 2122395 (k1/k3 =10/4) n = 5/3 N = 16.13 1366, 228
j4 = 1+12 k4 = 1632445 (k1/k4 = 13/4) n = 5/4 N = 29.388 1064, 214
j5 = 1+15 k5 = 1326037 (k1/k5 = 16/4) n = 5/5 N = k5/kr = 46.598 865, 214

#### SI units

The above can be measured in Planck units. As ${\displaystyle n_{g}}$ can therefore be used as a measure of length to mass, when ${\displaystyle n_{g}}$ =1, N = ${\displaystyle n_{p}^{2}}$ and so between objects A and B (where mass MA >> mB), 2Nlp = λA. Converting to SI units gives (simplified for non-relativistic orbits);

${\displaystyle N_{points}={\frac {M_{A}}{m_{P}}}}$ (number of particles in the Planck mass point-state per unit of Planck time)
${\displaystyle r_{g}=d^{2}N_{points}l_{p}=\alpha n_{g}^{2}\lambda _{A}}$ (converting to Schwarschild radius)
${\displaystyle v_{g}={\frac {c}{d}}}$ (gravitational orbit velocity)
${\displaystyle T_{g}={\frac {2\pi r_{g}}{v_{g}}}}$ (standard gravitational orbit period)
${\displaystyle a_{g}={\frac {v_{g}^{2}}{r_{g}}}}$ (gravitational acceleration)
${\displaystyle L_{oam}=N_{links}n_{g}{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}}$ (orbital angular momentum)
${\displaystyle L_{ram}=({\frac {2}{5}})N_{links}\;n_{rot}{\frac {h}{2\pi }},\;{\frac {kgm^{2}}{s}}}$ (rotational angular momentum)

Example: Earth orbits

${\displaystyle N_{points}={\frac {M_{earth}}{m_{P}}}}$

Earth surface orbit

rg = 6371.0 km
ag = 9.820m/s^2
Tg = 5060.837s
vg = 7909.792m/s


Geosynchronous orbit

rg = 42164.0km
ag = 0.2242m/s^2
Tg = 86163.6s
vg = 3074.666m/s


Moon orbit (d = 84600s)

rg = 384400km
ag = .0026976m/s^2
Tg = 27.4519d
vg = 1.0183km/s


Example: Planetary orbits

${\displaystyle N_{points}={\frac {M_{sun}}{m_{P}}}}$
mercury: rg = 57909000km, Tg = 87.969d, vg = 47.872km/s
venus: rg = 108208000km, Tg = 224.698d, vg = 35.020km/s
earth: rg = 149600000km, Tg = 365.26d, vg = 29.784km/s
mars: rg = 227939200km, Tg = 686.97d, vg = 24.129km/s
jupiter: rg = 778.57e9m, Tg = 4336.7d, vg = 13.056km/s
pluto: rg = 5.90638e12m, Tg = 90613.4d, vg = 4.740km/s


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).

${\displaystyle E_{orbital}={\frac {hc}{2\pi r_{6371}}}-{\frac {hc}{2\pi r_{42164}}}=0.412x10^{-32}J}$
${\displaystyle N_{links}={\frac {M_{earth}m_{satellite}}{m_{P}^{2}}}=0.126x10^{41}}$
${\displaystyle E_{total}=E_{orbital}N_{links}=53MJ/kg}$

### Angular momentum

The orbital angular momentum of the planets derives from the angular momentum of the orbital pairs (and so is independent of the orbital angular momentum of the sun).

${\displaystyle N_{sun}={\frac {M_{sun}}{m_{P}}}}$
${\displaystyle N_{planet}={\frac {M_{planet}}{m_{P}}}}$
${\displaystyle N_{links}=N_{sun}N_{planet}}$

#### orbital angular momentum

${\displaystyle L_{oam}=2\pi {\frac {Mr^{2}}{T}}=N_{sun}N_{planet}{\frac {h}{2\pi }}{\sqrt {\frac {2\alpha }{N_{sun}}}}n_{p}=N_{links}n_{g}{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}}$
${\displaystyle n_{g}={\frac {n_{p}^{*}}{\sqrt {N_{sun}}}}}$
${\displaystyle L_{oam}=N_{links}{\frac {hd}{2\pi }},\;{\frac {kgm^{2}}{s}}}$

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 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.

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

#### rotational angular momentum

${\displaystyle N_{links}=(N_{planet})^{2}}$

The rotational angular momentum ${\displaystyle L_{ram}}$ contribution to planet rotation.

${\displaystyle v_{rot}={\sqrt {N_{points}}}{\frac {c}{2\alpha n_{p}}}={\frac {c}{2\alpha n_{rot}}}}$
${\displaystyle T_{rot}={\frac {2\pi r}{v_{rot}}}}$
${\displaystyle L_{ram}={\frac {2}{5}}{\frac {2\pi Mr^{2}}{T}}=({\frac {2}{5}})N_{links}\;n_{rot}{\frac {h}{2\pi }},\;{\frac {kgm^{2}}{s}}}$

Earth:

Trot = 83847.7s (86400)

vrot = 477.8m/s (463.3)

${\displaystyle L_{ram}=.727\;x10^{34}\;{\frac {kgm^{2}}{s}}}$ (.705)

Mars:

Trot = 99208s (88643)

vrot = 214.7m/s (240.29)

${\displaystyle L_{ram}=.187\;x10^{33}{\frac {kgm^{2}}{s}}\;}$(.209)

Rotational angular momentum combined with vrot

${\displaystyle L_{ram}v_{rot}=({\frac {2}{5}})N_{links}{\frac {hc}{2\pi 2\alpha }},\;{\frac {kgm^{3}}{s^{2}}}}$

### Orbital plane rotation

8-body (8 mass points, 27 orbitals) orbit

#### relativistic orbits

${\displaystyle t_{o}={\frac {2\pi r_{g}}{c}}}$
${\displaystyle t_{d}=T_{g}{\sqrt {1-{\frac {v_{g}^{2}}{c^{2}}}}}={\sqrt {T_{g}^{2}-t_{o}^{2}}}=t_{o}{\sqrt {{2\alpha n_{g}^{2}}-1}}=t_{o}{\sqrt {d^{2}-1}}}$

If we consider the smaller orbit as a rotation of the orbital plane itself (for both orbits, orbital velocity = c) then we can re-write the above as;

${\displaystyle t_{small}=2\pi {\frac {dN}{2}}{\frac {l_{p}}{c}}}$
${\displaystyle t_{large}=2\pi d^{3}N{\frac {l_{p}}{c}}}$
${\displaystyle t_{d}=t_{large}-t_{small}=2\pi (d^{3}-{\frac {d}{2}})N{\frac {l_{p}}{c}}}$
${\displaystyle v_{plane}={\frac {c}{2d^{3}}}}$
${\displaystyle t_{plane}=4\pi d^{5}N{\frac {l_{p}}{c}}}$
${\displaystyle t_{d}={\frac {2\pi r_{g}}{(v_{g}+v_{plane})}}}$

#### precession

The ellipticity of the B orbit around A.

semi-minor axis: ${\displaystyle b=\alpha l^{2}\lambda _{A}}$

semi-major axis: ${\displaystyle a=\alpha n^{2}\lambda _{A}}$

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

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

arc secs per 100 years:

${\displaystyle 1296000{\frac {3n^{2}}{2\alpha l^{4}}}{\frac {100T_{earth}}{T_{planet}}}}$
Mercury = 42.98
Venus = 8.62
Earth = 3.84
Mars = 1.35
Jupiter = 0.06


The orbital plane becomes

${\displaystyle t_{plane}=2\pi ({\sqrt {2\alpha }})^{5}nl^{4}N_{sun}{\frac {l_{p}}{3c}}}$

Mercury

${\displaystyle t_{plane}{\frac {42.98}{1296000}}=100*365.245*24*3600}$s

#### wave-state

In order to have a point-state for each unit of Planck time, an object must have a minimum mass = Planck mass. For a simple H orbital, there are only the electron and proton, the point-state will seldom occur and so we must include the orbital wavelength (in Ta) as an additional term (the atomic orbital shape and so the orbit will reflect the geometry of the proton and electron and so tplane). The atomic Bohr model;

${\displaystyle v_{a}={\frac {c}{2\alpha n}}}$
${\displaystyle r_{a}=\alpha n^{2}(\lambda _{orbital})}$

For example, the following best fit geometries for tplane compare with H (1s-2s) transition = 2466061413 MHz [2]. The wavelengths of the electron and proton are included.

${\displaystyle T_{a}=2\pi {\frac {\alpha (\lambda _{e}+\lambda _{p})}{v_{a}}}}$
${\displaystyle t_{plane}={\frac {2{\sqrt {6}}d}{\alpha ^{2}}}={\frac {4{\sqrt {3}}}{{\sqrt {\alpha }}^{3}}}}$
${\displaystyle t_{d}=T_{a}{\sqrt {1-t_{plane}^{2}}}}$
${\displaystyle {\frac {3}{4t_{d}}}=2466061423\;MHz}$

and Positronium (1s-2s)= 1233607216 MHz [3]

${\displaystyle T_{a}=2\pi {\frac {\alpha (2\lambda _{e})}{v_{a}}}}$
${\displaystyle t_{plane}={\frac {28{\sqrt {2}}d}{3\alpha ^{2}}}={\frac {56}{3{\sqrt {\alpha }}^{3}}}}$
${\displaystyle t_{d}=T_{a}{\sqrt {1+t_{plane}^{2}}}}$
${\displaystyle {\frac {3}{4t_{d}}}=1233607221\;MHz}$

### Planck force

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

a) ${\displaystyle M_{a}=m_{b}}$

${\displaystyle F_{g}={\frac {F_{p}}{{(4\alpha n^{2})}^{2}}}}$

b) ${\displaystyle M_{a}>>m_{b}}$

${\displaystyle F_{g}={\frac {\lambda _{b}F_{p}}{{(2\alpha n^{2})}^{2}\lambda _{a}}}={\frac {m_{b}c^{2}}{2\alpha ^{2}n^{4}\lambda _{a}}}=m_{b}a_{g}}$

### Orbital transition

Atomic electron transition is the change of an electron from one energy level to another. The following redefines the Rydberg formula in terms of physical' orbitals, where transition is an orbital replacement, the electron plays no pre-dominant role.

Consider the Hydrogen Rydberg formula for transition between and initial ${\displaystyle i}$ and a final ${\displaystyle f}$ orbit. The incoming photon ${\displaystyle \lambda _{R}}$ causes the electron to jump' from the ${\displaystyle n=i}$ to ${\displaystyle n=f}$ orbit.

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

The above could be interpreted as referring to 2 photons;

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

Let us suppose a region of space between a free proton ${\displaystyle p^{+}}$ and a free electron ${\displaystyle e^{-}}$ which we may define as zero. This region then divides into 2 waves of inverse phase which we may designate as photon (${\displaystyle +\lambda }$) and anti-photon (${\displaystyle -\lambda }$) whereby

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

The photon (${\displaystyle +\lambda }$) leaves (at the speed of light), the anti-photon (${\displaystyle -\lambda }$) however is trapped between the electron and proton and forms a standing wave orbital. Due to the loss of the photon, the energy of (${\displaystyle p^{+}+e^{-}+-\lambda )<(p^{+}+e^{-}>+0}$) and so stable.

Let us define an (${\displaystyle n=i}$) orbital as (${\displaystyle -\lambda _{i}}$). The incoming Rydberg photon ${\displaystyle {\lambda _{R}}=(+\lambda _{i})-(+\lambda _{f})}$ arrives in a 2-step process. First the ${\displaystyle (+\lambda _{i})}$ adds to the existing (${\displaystyle -\lambda _{i}}$) orbital.

${\displaystyle (-\lambda _{i})+(+\lambda _{i})=zero}$

The (${\displaystyle -\lambda _{i}}$) orbital is canceled and we revert to the free electron and free proton; ${\displaystyle p^{+}+e^{-}+0}$ (ionization). However we still have the remaining ${\displaystyle -(+\lambda _{f})}$ from the Rydberg formula.

${\displaystyle 0-(+\lambda _{f})=(-\lambda _{f})}$

From this wave addition followed by subtraction we have replaced the ${\displaystyle n=i}$ orbital with an ${\displaystyle n=f}$ orbital. The electron has not moved (there was no transition from an ${\displaystyle n_{i}}$ to ${\displaystyle n_{f}}$ orbital), however the electron region (boundary) is now determined by the new ${\displaystyle n=f}$ orbital ${\displaystyle (-\lambda _{f})}$.