Mini Review Volume 1 Issue 4
Park Ave, Saint John, Canada
Correspondence: Paul TE Cusack, BScE, Dule 23 Park Ave,Saint John, NB E2J 1R2, Canada
Received: May 08, 2018 | Published: July 17, 2018
Citation: Cusack PTE. The speed of light, spin, superforce threshold energy and astro-theology. Open Acc J Math Theor Phy. 2018;1(4):139-141. DOI: 10.15406/oajmtp.2018.01.00022
In this paper, we examine some relationships between the speed of light and factors contained in the cosmology paper Astrotheology by the same author. We consider Newton’s Gravitational Equation; Angular momentum at the atomic level; the Threshold Energy; the RLC Universal Circuit; and finally, the Domino comparison modelling the speed of light. New equations are developed form these comparisons.
Keywords: astro-theology, speed of light, threshold energy, spin, superforce
In this brief and final paper, I provide some calculations involving Astrothoelogy parameters previously published, and the speed of light (or the speed of Electromagnetism). We show that the Superforce (S.F.) is stored in Mass and Spin where the Threshold Energy is reached. Everything in the universe functions on the S.F. We show, through calculations, that the speed of light is a reaction with the ether it travels through.1 Corrections are made for the Michaelson Morley experiments and the drift of the ether found by Einstein. We begin with the Newtonian gravitation Equation.
Newton’s equation
F=GM1M2/R2
Superforce=S.F.=F=8/3=2.667
G=2/3=6.67
M=Mass Gap=1.5
F=G M1M2/8/3=2/3 (1.5)
R2=1/F=3/8
R=√3/√8
=√3/√2√4
=sin 60°/√2
=F s
=Moment
Recall
det|A|=4=Moment
det|B|=| cos θ −sin θ , sin θ cos θ|
=(1/2 × 1/2) − (−√3√/2 −√3√/2)
=1/4 - (3/2)
=0.25-1/5
=1/25 = Emin
=1/8
det|A| ∗ det|B|=Det P
4 ∗ 1/8=1/2 = det P=cos 60°=t/dM/dt
Mom. ∗ dM/dt= t/ det |B|
Mom. ∗ Emin=1/2 =tmin
Mom. R2= Mom.3
Emin= t/ (dM/dt)* 1/Mom.
=t/(Fd) 1/ dM/dt
=t/R ∗1/(dM/dt)
1/8=Emin.=1./R (1/2)
R=4
F=GM1M2/R2
4(8/3)=2/3(M1M2)
64=M1M2
M1=2=64/ M2
M2=32~31.8=Frequency of the Human mind.
Emin=1/8=1..25
1/Emin=tmin=8
8/3=S.F.=tmin/3=t/(d/t)
=t2/d2
=12/12=1
=t2(Ln t)=1
F=x Ln x
(Ln x)2=1/t
y2=y,
y=√y,
y=E=1=√1=±1
F=πLn π
=359
≈1/K.E.
K.E.=1/F=1/2 Mv2
2=FMv2
3/8 (2)=Mv2
6/8=3/4=Mv2
=1/s
3/4=(1.5)v2
v=3/4 (2/3)
=0.25∼251=T
The reason light can travel for billions of years without any loss of velocity is that, as it travels, it converts mass of the ether into Energy (Figure 1) (Figure 2). Once the light beings to propagate, it is a reaction started that never ends until it encounters a surface that it can’t break down (which is also a wave). I’ve developed an equation derived from AT Math that explains why light (EM) travels at 2.9979.
In the Michelson- Morley Experiment, no “either drift” was detected. Why? Because light t is dependent on the either. What would be experienced is a bend in the light beam over great distances. That is what Einstein found that he thought proved his incorrect theorem. He though gravity bent light around the Sun. What he really was seeing was ether drift.
Light is like a domino path. They can only drop at one speed where they are going forward or backwards. The energy in one is transmitted like a fire as they being to fall one after the other.2 The only thing that stops the transition is a domino that won’t fall which is mass.
One can also think of a series of matches in a sting. If one match is lite, the next will be lit until the fire lights us a surface of mass.
c2=(1/ ρ)∫π1Ln t dt
freq.=1/T=1/251~4=Moment=Fd
1/F=R=√3/√8
=6123=d
Speed of Light = ℂ=csc θ
csc 60°=2/√3=c=v=d/t
d=√3/√8
=√3/√8 t
3=2√8 t
3/2=√8 t
1./5=√8 t
t=530
E=hf
=6.626 (1/251)
=1/38
E=1/38=√3 × M
=1.519
=1/G
E=hf=6.626(1/251)=26.314
E=hf=6.616(31.8)=2.109
2.109/2.6314=1/8=Emin
1/(4(2))=1/8=hπ/ (1/T)= vdiffusion
c=(1−sin1) h/π
PV=nRT
=19905 /A=6.023(8.314)T
T=8/3L/6.023(8314)
L=s/3=|E||t|sinθ
s=3 sin 60°
3(√3/2)
=1.5√3
=0.5
L3=1.25=Emin
T=88.7K
This is the temperature at which there should be zero resistance to electricity (Figure 3).
c2=100%/[h sin 1][dM/dt]
c2=100%/hVL
Emin=hVL=1/c2
E=Mc2
E/c=1/c2=M
M=1/c2=sin θ
csc θ=c2
c=√csc θ
M=1.074
E=Mc2
dE/dt=dM/.dt∗s
dE/dt=s=|E||t|sin θ
1/s=1/sin θ=c=3
s=1/3
s=(E)(t)sin θ
1/3=(1)(√3)sin θ
sin θ=1/√3=cot 60°
E=Mc2
dE/dt=dM/dt∗2c=(1.98)(2(2.9979))=118.7
This is the Mass for the last element of the Periodic Table (Figure 4) (Figure 5).
We see that Astrotheology mathematics explains a lot of observations in the cosmos.
This is my final paper. I owe a debt of gratitude to many scientists and mathematicians. I would like to single out Dr. Owen Dunn, Chemistry teacher at the illustrious St. Malachy’s High School. In addition, Mr. Paul Assaff taught us to think like mathematicians. Mr. Terry Comeau taught us more that just English; He also taught work ethic. Mr. Bill Coffee, History teacher taught us perspective and the necessity of God. High School teachers are the most influential. Despite the fact I received no scholarship from them, I owe them a debt of gratitude for the sake of the future of human knowledge. Thanks to deceased Prof Eric Garland, UNB Fredericton, who taught us communications, who had confidence instilled in us. Thanks to Prof Byron Walton,
UNB Saint John who supervised my thesis and wrote umpteen recommendations for me. Education should be broad and encompass at least two specialities and thanks to the Canadian people who provided me with an income while studying, although I could have used a job. Thanks to the province of New Brunswick who supported me while worked …..on God’s masterpiece.
The author declares that there is no conflict of interest.
©2018 Cusack. This is an open access article distributed under the terms of the, which permits unrestricted use, distribution, and build upon your work non-commercially.