# Quantum gravity (Planck)

Planck quantum gravity in Simulation Hypothesis modeling

A gravitational force between macro-objects is simulated by a discrete fine structure constant based pixel lattice geometry d supporting discrete rotating particle-particle orbital pairs (as units of $\hbar c$ ) applicable for programming Planck-unit Simulation Hypothesis models . Orbits between macro-objects become the sum of the underlying particle-particle orbitals.

### 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 digital Planck events can be mapped. The mathematical electron model is applicable to Planck 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 where the Planck units are interacting geometrical objects. With this approach mass can 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 simultaneously in the mass-state can form rotating pairs with each other. Consequently information regarding orbiting macro-objects is not required as gravitational orbits derive from the sum of these underlying orbital pairs.

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.

Each particle that is in the mass point-state per unit of Planck time is linked to every other particle simultaneously in the mass point-state by a pixel lattice 'space' defined by the fine structure constant alpha, these links constitute mass, velocity and length with the orbital itself as a unit of $\hbar c$ . The velocity of any 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.

### 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;

$\alpha _{G}={\frac {Gm_{e}^{2}}{\hbar c}}={\frac {m_{e}^{2}}{m_{P}^{2}}}=1.75...x10^{-45}$ 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 $1kg/m_{P}=45.9x10^{6}$ particles in the point-state. The earth (A) will have $5.97\;x10^{24}kg/m_{P}=0.274\;x10^{33}$ 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;

$N_{links}={\frac {m_{A}m_{B}}{m_{P}^{2}}}=0.126\;x10^{41}$ The gravitational orbit can be characterized by :$N_{links}\hbar c$ 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

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

np = pixel number

mP = Planck mass

Between 2 rotating point mass

$r=2\alpha n_{p}^{2}l_{p}$ (orbital length)
$v={\frac {c}{{\sqrt {2\alpha }}n_{p}}}$ (orbital velocity)

Between objects A and B where mass MA >> mB:

np* = average np (average of particle links between A and B)

$N_{points}={\frac {M_{A}}{m_{P}}}$ (number of particles in the Planck mass point-state per unit of Planck time)
$n_{g}={\frac {n_{p}^{*}}{\sqrt {N_{points}}}}$ (analogous to the atomic principal quantum number n)
$d={\sqrt {2\alpha }}{\frac {n_{p}^{*}}{\sqrt {N_{points}}}}=({\sqrt {2\alpha }})n_{g}$ (pixel aggregate)

For any given distance rg, the pixel number np can support multiple units of Planck mass and so ng can be treated as analogous to atomic energy levels. When ng = 1 that np level is full and additional units of Planck mass will go to the next np level.

$r_{g}=d^{2}N_{points}l_{p}=\alpha n_{g}^{2}\lambda _{M}$ (converting to Schwarschild radius)
$v_{g}={\sqrt {N_{points}}}v={\frac {c}{d}}={\frac {c}{{\sqrt {2\alpha }}n_{g}}}$ (gravitational orbit velocity from summed point velocities)
$a_{g}={\frac {v_{g}^{2}}{r_{g}}}$ (gravitational acceleration)
$T_{g}={\frac {2\pi r_{g}}{v_{g}}}$ (gravitational orbit period)
$L_{oam}=N_{links}n_{g}{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}$ (orbital angular momentum)
$L_{ram}=({\frac {2}{5}})N_{links}\;n_{rot}{\frac {h}{2\pi }},\;{\frac {kgm^{2}}{s}}$ (rotational angular momentum)

Example: Earth orbits

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

$N_{points}={\frac {M_{sun}}{m_{P}}}$ 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).

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

$N_{sun}={\frac {M_{sun}}{m_{P}}}$ $N_{planet}={\frac {M_{planet}}{m_{P}}}$ $N_{links}=N_{sun}N_{planet}$ #### orbital angular momentum

$L_{oam}=2\pi {\frac {Mr^{2}}{T}}=N_{sun}N_{planet}{\frac {h}{2\pi }}{\sqrt {\frac {2\alpha }{N_{sun}}}}n_{p}=N_{links}n_{g}{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}$ $n_{g}={\frac {n_{p}^{*}}{\sqrt {N_{sun}}}}$ $L_{oam}=N_{sun}N_{planet}{\frac {hd}{2\pi }},\;{\frac {kgm^{2}}{s}}$ The orbital angular momentum of the planets is independent of the orbital angular momentum of the sun;

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.

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

$N_{links}=(N_{planet})^{2}$ The rotational angular momentum $L_{ram}$ contribution to planet rotation.

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

Trot = 83847.7s (86400)

vrot = 477.8m/s (463.3)

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

Mars:

Trot = 99208s (88643)

vrot = 214.7m/s (240.29)

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

Rotational angular momentum combined with vrot

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

#### relativistic orbits

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;

$t_{o}={\frac {2\pi r_{g}}{c}}$ $t_{d}=T_{g}{\sqrt {1-{\frac {v_{g}^{2}}{c^{2}}}}}={\sqrt {T_{g}^{2}-t_{o}^{2}}}=t_{o}{\sqrt {{2\alpha n_{g}^{2}}-1}}=t_{o}{\sqrt {d^{2}-1}}$ We can simulate this cylinder as a 2d orbital plane by assigning orbital velocity = c and orbital radius

$r=d^{3}Nl_{p}$ $t_{d}=2\pi (d^{3}N-{\frac {dN}{2}}){\frac {l_{p}}{c}}$ If we suppose a rotation of the A-B orbital plane itself to account for that 2nd term

$v_{plane}={\frac {c}{2d^{3}}}$ $t_{plane}=4\pi d^{5}N{\frac {l_{p}}{c}}$ $t_{d}={\frac {2\pi r_{g}}{(v_{g}+v_{plane})}}$ The earth surface orbit example; rg = 6371.0km

$\delta _{t}=T_{g}-t_{d}={\frac {\pi dNl_{p}}{c}}=.17615\;x10^{-5}s$ $s=\delta _{t}{\frac {1296000}{T_{g}}}=.4511\;x10^{-3}$ (arc seconds)
$t_{plane}=.1454\;x10^{14}s$ (1296 000 arc seconds)
$t_{plane}=T_{g}$ (.4511 x10-3 arc seconds)

#### precession

The ellipticity of the B orbit around A.

semi-minor axis: $b=\alpha l^{2}\lambda _{A}$ semi-major axis: $a=\alpha n^{2}\lambda _{A}$ radius of curvature :$L={\frac {b^{2}}{a}}={\frac {al^{4}\lambda _{A}}{n^{2}}}$ ${\frac {3\lambda _{A}}{2L}}={\frac {3n^{2}}{2\alpha l^{4}}}$ Combining L with vplane gives the precession

$T_{precession}={\frac {4\pi {\sqrt {2\alpha }}\alpha ^{2}nl^{4}\lambda _{sun}}{3c}}$ arc secs per 100 years:

Mercury = 42.98
Venus = 8.62
Earth = 3.84
Mars = 1.35
Jupiter = 0.06


note: 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, the proton and electron are simultaneously in the point-state (Nlinks) only once every 1/30th sec and so the atomic orbital physics dominates. Although the atomic orbital shape will depend on the geometry of the proton and electron, we may use the same orbital link for the atomic and the gravitational orbitals. The atomic Bohr model

$v_{a}={\frac {c}{2\alpha n}}$ $r_{a}=\alpha n^{2}(\lambda _{e}+\lambda _{p})$ For example, a best fit tplane compares with H (1s-2s) transition = 2466061413187035 Hz .

$t_{plane}={\frac {16\pi {\sqrt {3\alpha }}(\lambda _{e}+\lambda _{p})}{c}}$ $t_{d}=({\frac {3}{4}}){\sqrt {T_{a}^{2}-t_{plane}^{2}}}=2466061423MHz$ ### 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) $M_{a}>>m_{b}$ $F_{g}={\frac {\lambda _{b}F_{p}}{{(2\alpha n^{2})}^{2}\lambda _{a}}}={\frac {m_{b}c^{2}}{2\alpha ^{2}n^{4}\lambda _{a}}}=m_{b}a_{g}$ ### 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 role.

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

$\lambda _{R}=R.({\frac {1}{n_{i}^{2}}}-{\frac {1}{n_{f}^{2}}})={\frac {R}{n_{i}^{2}}}-{\frac {R}{n_{f}^{2}}}$ The above could be interpreted as referring to 2 photons;

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

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

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

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

$0-(+\lambda _{f})=(-\lambda _{f})$ From this wave addition followed by subtraction we have replaced the $n=i$ orbital with an $n=f$ orbital. The electron has not moved (there was no transition from an $n_{i}$ to $n_{f}$ orbital), however the electron region (boundary) is now determined by the new $n=f$ orbital $(-\lambda _{f})$ .