Quantum Newtonian “free fall”?
3 selective thought experiments in favor to final rational thought:
1. A helium balloon trapped inside a speeding vehicle
2. A man falling free inside an elevator
3. Odd 5 quantum math: say vacuum gas has “free fall choice” inversely related to its density and, say you have odd 5 denser balls. Which way they “fall”?
By subtracting the “momentum follower definition” from first two thought experiments I have concluded third key question.
To achieve quantum gravity (linear efficiency) gravitational force must be applied in the first place to a speeding photon inside a vacuum. A vacuum gas qualifies then best to preserve such “free light fall event” in all directions of spacetime in relation to photon energy. Light speed inside a vacuum becomes then the preserved “free fall event” itself. Gravitational force is transferred to a lower gas density property (only momentum follower). The curvature of spacetime and its associate particles measures inefficient (extra) energy and gets erased. Therefore quantum gravity diverge into zero spacetime curvature plus mass-gravity equivalence (cm^2). Force radiates inverse lower density (Higgs-vacuum) gas resuming mass constantly converts into gravity and vice versa at 99,Planck% the speed of light inside a vacuum according to Newton’s inverse square law. A divergent black hole event horizon converts energy into mass and not mass into energy (99,9% square newtonian force inside a vacuum). A black hole should only radiate Planck energy and amplify almost twice Higgs energy inside a vacuum. The mass-gravitation equation for black hole density emission inside a vacuum becomes cm^2 (99,9% valid). A black hole should radiate like gravity (Planck radiation), a black hole’s volume should weigh 99,9% square Higgs density and 0,1% gravitational wavelength, and a black hole should be invisible to any physical observation like gravity. And that’s my idea on quantum gravity. And apparently you have missed the backdoor variable square mass equation.