Скачать или смотреть Planck Area gravity equals the Landauer Limit using Hawking Temperature

This is a video tying in GR and quantum mechanics.
Quantum Entanglement as Source of Gravity

I have found a way to describe gravity from a Landauer Limit perspective which describes the minimum amount of energy to erase one quantum bit and how that value correlates to General Relativity's gravitational energy per Planck Area. To do this I calculated the energy of a given 1KM radius Schwarzschild black hole and used General Relativity to calculate mass and then converted to total energy and then divide by the number of Planck Areas of that event horizon to find a gravitational energy per Planck Area. Next I compare the GR Joules per Planck Area to the Landauer Principle using Hawking temperature as an input to see what relationship exists between General Relativity and Quantum Entanglement.

Steps:
Calculate mass of 1KM radius black hole Schwarzschild Metric
Calculate energy of Black Hole E = MC2
Calculate number of Planck Areas of given black hole = NP
Divide total energy by the number of Planck Areas
Record the answer
Calculate black hole temperature using Hawking temperature
Use Hawking temperature to plug into Landauer Limit
Record results and compare with GR
Results were GR= 1.261 x 10-30J and using Landauer Limit = 1.74 x 10-30J


Calculating Black Holes Gravitational Energy
Find the mass, M, of black hole using Schwarzschild metric-

(1000m) x (299,792,458 m/s)2 M= 1000m x 8.98755179 x 1016 m2/s2 M = 6.735 x 1029 kg
2 x 6.67430 x 10-11 m3 kg-1 s-2 1.33486 x 10-10 m3 kg-1 s-2

Calculate Gravitational Energy E=MC2
E = (6.735 x 1029kg) x (299,792,458 m/s)2 = E = 6.0736 x 10 Joules

Calculating Energy Per Planck Area
6.0736 x 1046J E per Planck Area, 1.261 x 10-31J
4.813 x 1076

Calculate Hawking Temperature
TH = ħC3
8Pi GmkB
Constants:

Reduced Planck constant ħ = h/2Pi = 1.054571817 X 10-34 Joules
Planck Constant h = 6.62607015 x 10-34 Joules

Calculate TH: TH =
(1.0547187 x 10-34 Js) x (288,792,458 m/s)3
8Pi x 6.67430 x 10-11 m3 kg-1 s-2 x 6.735 x 1029kg x 1.380649 x 10-23 J/K

TH = N/DTH = 2.8477 x 10-9 J m3 / s2
1.5616 x 10-2 J m3 / s2 per K

Hawking Temperature: TH = 1.823 x 10-7K

Calculate Landauer Energy at TH
E Landauer = kBTH ln 2 E Landauer =
(1.380649 x 10-23 J/K) x (1.823 x 10-7 K) x 0.6931 E Landauer = 1.74 x 10-30Joules




Comparison Between Landauer and GR

Value from General Relativity via E=MC2 = 1.261 x 10-30
Value from Landauer using Hawking Temperature = 1.74 x 10-30

Difference 0.7247
Possible Implication The Landauer minimum is a static value and should be a bit higher statically than dynamically. A possible reason for this difference is a dynamic gravity emission duty cycle near RMS.

Other implications:
It is apparent from these results that free flying photons do not affect the local ST geometry. Gravity comes not from kinetic energy but from the energy of constrained degrees of freedom. This suggests photons only contribute to gravity when they interact with matter or form pair production.
An analogy can be drawn from Maxwell's Demon requiring a bit of energy in order to be able to obtain knowledge Its as if event horizons have a Maxwell's Demon measuring at high rates at every Planck Area of the horizon.

At a mass of a black hole at 100KM, the two orders of magnitude larger express themselves as two orders of magnitude less in terms of energy.

100KM black hole GR = 1.261 x 10-32 And the Landauer minimum = 1.74 x 10-32

Thank you for considering this proposal to discover how gravity emerges.

Bill Manuel
16800 Bell Ave
Eastpointe MI 48021
(586) 306-5354