Quantum gravity (Planck)

Method for programming gravitational orbitals (gravitons) as units of ${\displaystyle \hbar c}$ for mathematical universe simulation hypothesis models

The following describes a method in which gravitational orbitals (gravitons) are defined as discrete units of ${\displaystyle \hbar c}$ that are used to link orbiting objects[1]. In an article on the mathematical electron particles are simplified to oscillations between an electric wave-state (over time) to a discrete Planck-mass at unit Planck-time ( Planck black-hole) point-state. For orbiting objects a graviton is assigned as a link between each individual particle in that object that is in the mass point-state per unit of Planck time. The gravitational interaction between particles derives from these (physical) graviton (${\displaystyle \hbar c}$) links between particle units of Planck mass (the particle point-states) and can be defined in terms of a gravitational equivalent to the principal quantum number ${\displaystyle n}$ and the number of orbital links (a function of object mass). Gravitational orbits are then the sum of these underlying orbitals; the orbital angular momentum of the planetary orbits derives from the sum of the planet-sun particle-particle ${\displaystyle \hbar }$ orbital angular momentum irrespective of the angular momentum of the sun itself and the rotational angular momentum of a planet approximated by the sum of its particle-particle rotational angular momentum. All particle point-states share a common time frame as measured in discrete units of Planck time (digital time), thus this approach is suitable for Simulation Hypothesis models.

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 we replace wave-particle duality with an electric wave-state to Planck-mass (for 1 unit of Planck-time) point-state oscillation then at any discrete unit of Planck time t a certain number of particles will simultaneously be in the mass point-state. For example a 1kg satellite orbits the earth, for any t, satellite (A) will have ${\displaystyle 1kg/m_{P}=45.9x10^{6}}$ particles in the point-state. The earth (B) will have ${\displaystyle 5.97\;x10^{24}kg/m_{P}=0.274\;x10^{33}}$ particles in the point-state. If we assign a graviton (gravitational orbital) to link each respective Planck-mass point-state then for any given unit of Planck time the number of gravitons linking the earth to the satellite;

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

The observed satellite orbit around the earth derives from the sum of these ${\displaystyle 0.126\;x10^{41}}$ graviton angular momentum. If A and B are respectively Planck mass particles then ${\displaystyle N_{gravitons}}$ = 1. If A and B are respectively electrons then

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

The frequency of an electron oscillation cycle = ${\displaystyle (m_{P}/m_{e})t_{p}}$ and so the probability that any 2 electrons are simultaneously in the mass point-state for any chosen unit of Planck time t equals αG, and so the average frequency of occurrence = ${\displaystyle 1/\alpha _{G}(t_{p})}$. ${\displaystyle N_{gravitons}}$ is the sum of all the respective particle αG's between the orbiting objects at any unit of Planck time, as a consequence for objects whose mass is less than Planck mass there will be units of time when there are no graviton links and wave-state interactions will predominate. Gravity becomes the sum of discrete (graviton) interactions between (particle-particle) units of Planck mass.

Quantum (Bohr) gravity

Although the atom has a complex geometry, gravitational orbits are an average of all the underlying gravitons (gravitational orbitals) and so more closely approximate a classical geometry, it is therefore not necessary to know precisely the structure of an individual graviton. Consequently we can adapt the atomic Bohr model to gravitational orbits albeit the gravitational quantum number n, being an average of all the individual graviton n's, is not an integer, and N as the average number of particle Planck-mass point-states per unit of Planck-time where;

${\displaystyle N_{point-states}={\frac {M}{m_{P}}}}$

As we are calculating gravity only between point-states (units of Planck mass per unit Planck time), then for any 2 points (i,j)

${\displaystyle r=\alpha n_{i,j}^{2}2l_{p}}$

If object B with mass = ${\displaystyle m_{P}}$ orbits (planet) A (${\displaystyle m_{A}>>m_{B}}$) then we may calculate the average distance between point B and each individual point in A giving

${\displaystyle r_{average}=\alpha n_{a,b}^{2}2l_{p}}$

However this average ${\displaystyle n_{a,b}}$ would apply to the distance between the mass center point of A to point B (i.e.: ${\displaystyle N_{point-states}=1}$), in order to include the mass of A we assign;

${\displaystyle n^{2}={\frac {n_{a,b}^{2}}{N_{A}}}}$

This divides ${\displaystyle r_{average}}$ into discrete segments of length ${\displaystyle \alpha n^{2}2l_{p}}$. The (inverse) fine structure constant α = 137.03599...

${\displaystyle r_{g}=N_{point\;states}(\alpha n^{2}2l_{p})}$

${\displaystyle v_{g}={\frac {c}{{\sqrt {2\alpha }}n}}}$

${\displaystyle a_{g}={\frac {1}{N_{point-states}}}{\frac {c^{2}}{2\alpha ^{2}n^{4}2l_{p}}}}$

${\displaystyle T_{g}={\frac {r_{g}}{v_{g}}}=N_{point-states}({\frac {2\pi {\sqrt {2\alpha }}\alpha n^{3}2l_{p}}{c}})}$

Example: Earth radius = 6371 km

${\displaystyle \mu _{earth}=3.986004418(9)x10^{14}}$ standard gravitational parameter

${\displaystyle N_{earth}={\frac {M_{earth}}{m_{P}}}=.2744385886...x10^{33}}$

rg = 6371.0 km (n = 2289.408...)
ag = 9.820m/s^2
Tg = 5060.837s
vg = 7909.792m/s


Geosynchronous orbit

rg = 42164.0km  (n = 5889.66...)
ag = 0.2242m/s^2
Tg = 86163.6s
vg = 3074.666m/s


Moon orbit (d = 84600s)

rg = 384400km  (n = 17783.25...)
ag = .0026976m/s^2
Tg = 27.4519d
vg = 1.0183km/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).

${\displaystyle E_{graviton}={\frac {hc}{2\pi r_{6371}}}-{\frac {hc}{2\pi r_{42164}}}=0.412x10^{-32}J}$

${\displaystyle N_{gravitons}={\frac {M_{earth}m_{satellite}}{m_{P}^{2}}}=0.126x10^{41}}$

${\displaystyle E_{total}=E_{graviton}\;.\;N_{gravitons}=53MJ/kg}$

Example: Planetary orbits

${\displaystyle M_{sun}={\frac {1.32712440018x10^{20}}{G}}=.19889x10^{31}}$ standard gravitational parameter

${\displaystyle N_{sun}={\frac {M_{sun}}{m_{P}}}=.9137...x10^{38}}$

${\displaystyle n={\sqrt {\frac {r_{(sun-planet)}}{N_{sun}\alpha 2l_{p}}}}}$

${\displaystyle T=N_{sun}({\frac {2\pi {\sqrt {2\alpha }}\alpha n^{3}2l_{p}}{c}})}$

mercury: r = 57909000km, T = 87.969d
venus: r = 108208000km, T = 224.698d
earth: r = 149600000km, T = 365.26d
mars: r = 227939200km, T = 686.97d
jupiter: r = 778.57e9m, T = 4336.7d
pluto: r = 5.90638e12m, T = 90613.4d


Angular momentum

Orbital angular momentum

By linking all respective mass-states between 2 orbiting objects with units of momentum ${\displaystyle \hbar }$ the total orbit momentum ${\displaystyle L_{oam}}$ becomes the sum of the underlying momentum.

${\displaystyle N_{gravitons}=({\frac {M_{planet}M_{sun}}{m_{P}^{2}}})}$

${\displaystyle L_{oam}=2\pi {\frac {Mr^{2}}{T}}=N_{gravitons}\;n{\frac {h}{2\pi }}{\sqrt {2\alpha }},\;{\frac {kgm^{2}}{s}}}$

The angular momentum of a single point-point orbital (${\displaystyle N_{gravitons}=1}$);

${\displaystyle L_{oam}=({\frac {m_{P}^{2}}{m_{P}^{2}}})\;n{\frac {h}{2\pi }}{\sqrt {2\alpha }}=n{\frac {h}{2\pi }}{\sqrt {2\alpha }}}$

Orbital angular momentum ${\displaystyle L_{oam}}$ of the planets;

mercury = .9153 x1039 (n = 378.2733)
venus    = .1844 x1041 (n = 517.0853)
earth    = .2662 x1041 (n = 607.9927)
mars     = .3530 x1040 (n = 750.4850)
jupiter   = .1929 x1044 (n = 1387.0157)
pluto   = .365 x1039 (n = 3820.2628)


Mean orbital velocity ${\displaystyle v_{g}}$

mercury = 47.87km/s  (47.87km/s)
venus    = 35.02km/s  (35.02km/s)
earth    = 29.78km/s (29.78km/s)
mars     = 24.13km/s (24.13km/s)
jupiter   = 13.06km/s  (13.07km/s)
pluto   = 4.74km/s (4.72km/s)


We note that the angular momentum term ${\displaystyle L_{oam}}$ depends on ${\displaystyle n}$ leading to the dilemma whereby infinite distance results in infinite angular momentum. We also note that the orbital velocity ${\displaystyle v_{g}}$ decreases proportionately suggesting the graviton combines angular momentum with velocity. From;

${\displaystyle L_{oam}v_{g}=(m_{P}v_{g}r_{g})(v_{g})=(m_{P}{\frac {c}{{\sqrt {2\alpha }}n}}\alpha n^{2}2l_{p})({\frac {c}{{\sqrt {2\alpha }}n}}){\frac {1}{2\pi }}={\frac {hc}{2\pi }}}$

Rotational angular momentum

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

${\displaystyle T_{rot}={\frac {2\pi r_{orbital}}{v_{rot}}}=2\pi \alpha n^{2}\lambda _{orbital}{\frac {2\alpha n}{c}}={\frac {4\pi \alpha ^{2}n^{3}\lambda _{orbital}}{c}}}$

${\displaystyle v_{rot}={\frac {c}{2\alpha n}}}$

${\displaystyle r_{orbital}=\alpha n^{2}\lambda _{orbital}}$

${\displaystyle L_{ram}={\frac {2}{5}}{\frac {2\pi Mr^{2}}{T}}=({\frac {2}{5}})N_{orbitals}\;n{\frac {h}{2\pi }},\;{\frac {kgm^{2}}{s}}}$

Giving:

nearth = 2289.4 (6371km)

Trot = 83847.7s (86400)

vrot = 477.8m/s (463.3)

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

nmars = 5094.7 (3390km)

Trot = 99208s (88643)

vrot = 214.7m/s (240.29)

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

Combining rotational angular momentum with velocity;

${\displaystyle L_{ram}v_{rot}=(m_{P}{\frac {c}{2\alpha n}}\alpha n^{2}2l_{p})({\frac {c}{2\alpha n}}){\frac {1}{2\pi }}={\frac {hc}{2\pi 2\alpha }}}$

The energy of a photon in an atomic orbital transition (Rydberg formula)

${\displaystyle E={\frac {L_{ram}v_{rot}}{r_{a}}}={\frac {hc}{2\pi 2\alpha }}{\frac {1}{\alpha \lambda _{orbital}}}({\frac {1}{n_{i}^{2}}}-{\frac {1}{n_{f}^{2}}})}$

re-written in terms of amperes ${\displaystyle A^{2}}$ where j is a dimensionless constant (see mathematical electron model)

${\displaystyle {\frac {hc}{2\pi \alpha ^{2}}}={\frac {j^{2}A^{2}}{2^{8}2\pi t_{p}}}}$

Time dilation

Velocity

Velocity: In the article 'Programming Relativity in a Planck unit Universe'[2], 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 ${\displaystyle T_{g}c}$ at radius ${\displaystyle r_{g}}$ and orbital velocity ${\displaystyle v_{g}}$. If A is moving with the universe expansion (albeit stationary in 3-D space) then the orbital time ${\displaystyle t_{g}}$ alongside the A time-line axis becomes;

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

Gravitational

${\displaystyle v_{s}=v_{escape}={\sqrt {2}}.v_{g}}$

${\displaystyle {\sqrt {1-{\frac {2GM}{r_{g}c^{2}}}}}={\sqrt {1-{\frac {v_{s}^{2}}{c^{2}}}}}}$

Nuclear binding energy

Binding energy in the nucleus can be simplified using the same approach.

${\displaystyle m_{nuc}=m_{p}+m_{n}}$

${\displaystyle \lambda _{s}={\frac {l_{p}m_{P}}{m_{nuc}}}}$

${\displaystyle r_{0}={\sqrt {\alpha }}\lambda _{s}}$

${\displaystyle R_{s}=\alpha \lambda _{s}}$

${\displaystyle v_{s}^{2}={\frac {c^{2}}{\alpha }}}$

The gravitational binding energy (μ) is the energy required to pull apart an object consisting of loose material and held together only by gravity.

${\displaystyle \mu _{G}={\frac {3Gm_{nuc}^{2}}{5R_{s}}}={\frac {3m_{nuc}c^{2}}{5\alpha }}={\frac {3m_{nuc}v_{s}^{2}}{5}}}$

${\displaystyle a_{c}={\frac {3e^{2}}{20\pi \epsilon r_{0}}}}$

${\displaystyle E={\sqrt {(}}\alpha )a_{c}={\frac {3m_{nuc}c^{2}}{5\alpha }}={\frac {3m_{nuc}v_{s}^{2}}{5}}=\mu _{G}}$

Average binding energy in the nucleus = μG = 8.22MeV/nucleon.

Anomalous precession

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

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

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

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

${\displaystyle precession={\frac {3n^{2}}{2\alpha l^{4}}}.1296000.(100T_{earth}/T_{planet})}$

Mercury = 42.9814
Venus = 8.6248
Earth = 3.8388
Mars = 1.3510
Jupiter = 0.0623


Planck force

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

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

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

a) If ${\displaystyle M_{a}=m_{b}}$, the object mass is not required

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

b) If ${\displaystyle M_{a}>>m_{b},\;(\lambda _{a}+\lambda _{b}=\lambda _{a})}$, then relative mass is used and ${\displaystyle F_{g}=m_{b}a_{g}}$

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

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

Orbital transition

Atomic electron transition is defined as a change of an electron from one energy level to another, theoretically this should be a discontinuous electron jump from one energy level to another although the mechanism for this is not clear. The following uses the wavelengths (frequency in units of Planck time) of the orbitals and the Rydberg formula as a means to time' the transition period.

Let us 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 ${\displaystyle (-\lambda _{f})}$.