# Quantum gravity (Planck)

Quantum gravity formulas in Planck units

Gravitational formulas listed in terms of discrete Planck-units, $\hbar c$ and the fine structure constant for application in Simulation Hypothesis modelling.

### Discussion

The following uses the mathematical electron model to simulate gravitational interactions between particles at the Planck level in a digital time frame where time is measured in discrete units of Planck time (the simulation clock-rate). In this model particles are treated as oscillations between an electric wave-state (over time) to a discrete unit of Planck-mass (at unit Planck-time) mass point-state. Mass is therefore not considered a constant property of the particle, consequently 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. Gravity, as with mass, is also not treated as a constant property but rather as a discrete event, the magnitude of the gravitational interaction per unit time approximates the magnitude of the strong force, the gravitational coupling constant represents 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 unit of Planck mass mP, velocity v2 and distance r whereby $m_{P}v^{2}r=\hbar c$ (defined as a gravitational orbital). The velocity of a gravitational orbit is summed from these individual particle-particle v2. 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 t 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 (A) will have $1kg/m_{P}=45.9x10^{6}$ particles in the point-state. The earth (B) 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 gravitational coupling links between the earth and the satellite will sum to;

$N_{links}={\frac {m_{A}m_{B}}{m_{P}^{2}}}=0.126\;x10^{41}$ 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 t, N = αG and so a gravitational interaction between these 2 electrons will occur only once every 1045 units of Planck time.

### Planck unit gravitational formulas

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

np = number of Planck units

$r=\alpha n_{p}2l_{p}$ (distance from a point mass)
$v={\frac {c}{\sqrt {2\alpha n_{p}}}}$ (orbital velocity associated with a point mass)
$T={\frac {2\pi r}{v}}$ (orbital period)
$N_{points}={\frac {M}{m_{P}}}$ (number of particles in the Planck mass point-state per unit of Planck time per object mass M)
$n={\sqrt {\frac {n_{p}}{N_{points}}}}$ (gravitational analogue to the principal quantum number)
$r_{g}=\alpha n_{p}2l_{p}=\alpha n^{2}N_{points}2l_{p}=\alpha n^{2}\lambda _{M}$ (distance from a center of mass)
$v_{g}={\sqrt {N_{points}}}v={\frac {c}{{\sqrt {2\alpha }}n}}$ (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)
$L_{oam}=N_{links}n{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}$ (orbital angular momentum)
$L_{ram}=({\frac {2}{5}})N_{links}\;n{\frac {h}{2\pi }},\;{\frac {kgm^{2}}{s}}$ (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

$L_{oam}=2\pi {\frac {Mr^{2}}{T}}=N_{sun}N_{planet}{\frac {h}{2\pi }}{\sqrt {\frac {2\alpha n_{p}}{N_{sun}}}}=N_{links}n{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}$ 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 reduces to a unit of $\hbar c$ regardless of orbit distance.

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

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

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

Npoints = Nearth

Trot = 83847.7s (86400)

vrot = 477.8m/s (463.3)

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

Mars:

Npoints = Nmoon

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

$L_{ram}v_{rot}=({\frac {2}{5}})N_{links}{\frac {hc}{2\pi 2\alpha }},\;{\frac {kgm^{3}}{s^{2}}}$ Comparison with (Planck) Bohr atomic orbitals

$r=\alpha n_{p}2l_{p}$ (wavelength of atomic orbital)
$v={\frac {c}{2\alpha {\sqrt {n_{p}}}}}$ (orbital velocity associated with atomic orbital)
$n={\sqrt {\frac {n_{p}}{f_{o}}}}$ (fo; dimensionless orbital function)
$n_{p}=n^{2}f_{o}$ $r_{a}=\alpha n_{p}2l_{p}=\alpha n^{2}f_{o}2l_{p}=\alpha n^{2}\lambda _{o}$ $v_{a}={\sqrt {f_{o}}}v={\frac {c}{2\alpha n}}$ $m_{P}{v_{a}}^{2}{r_{a}}={\frac {m_{P}l_{p}c^{2}}{2\alpha }}={\frac {hc}{2\pi 2\alpha }}$ ### Time dilation

#### Velocity

Velocity: In the article 'Programming Relativity in a Planck unit Universe', a model of a virtual hyper-sphere universe expanding in Planck steps was proposed. In that model objects are pulled along by the expansion of the hyper-sphere irrespective of any motion in 3-D space. As such, while B (satellite) has a circular orbit in 3-D space co-ordinates it has a cylindrical orbit around the A (planet) time-line axis in the hyper-sphere co-ordinates with orbital period $T_{g}c$ at radius $r_{g}$ and orbital velocity $v_{g}$ . If A is moving with the universe expansion (albeit stationary in 3-D space) then the orbital time $t_{g}$ alongside the A time-line axis becomes;

$t_{g}={\sqrt {(T_{g}c)^{2}-(2\pi r_{g})^{2}}}=(T_{g}c){\sqrt {1-{\frac {v_{g}^{2}}{c^{2}}}}}$ #### Gravitational

$v_{s}=v_{escape}={\sqrt {2}}.v_{g}$ ${\sqrt {1-{\frac {2GM}{r_{g}c^{2}}}}}={\sqrt {1-{\frac {v_{s}^{2}}{c^{2}}}}}$ ### Anomalous precession

semi-minor axis: $b=\alpha l^{2}\lambda _{sun}$ semi-major axis: $a=\alpha n^{2}\lambda _{sun}$ $L={\frac {b^{2}}{a}}={\frac {al^{4}\lambda _{sun}}{n^{2}}}$ ${\frac {3\lambda _{sun}}{2L}}={\frac {3n^{2}}{2\alpha l^{4}}}$ $precession={\frac {3n^{2}}{2\alpha l^{4}}}.1296000.(100T_{earth}/T_{planet})$ Mercury = 42.9814
Venus = 8.6248
Earth = 3.8388
Mars = 1.3510
Jupiter = 0.0623


### 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 $(-\lambda _{f})$ .