Object at r ——-Light sensing of moving objects ———– (seen as) S
r —— Cosine (wt) + i sine (wt) ——– S = r [cosine (wt) + i sine (wt)]
Particle ————————- Light —————————— Wave
Newton ——– Kepler’s Time dependent ——– Newton’s Time dependent

A-Special theory of relativity

1-Lenght contraction

Line of Sight: r cosine wt: light aberrations
A moving object with velocity v will have when visualized through light sensing a light aberration angle (wt); w = constant and t= time

Also, sine wt = v/c; cosine wt = √ [1-sine² (wt)] = √ [1-(v/c) ²]
Where v = velocity; c = light velocity
A visual object moving with velocity v will be seen as S
S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential

S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y

S x = Visual location along the line of sight = r [√ [1-(v/c) ²]

This Equation is special relativity Length Contraction formula and it is just the visual effects and caused by light aberrations of a moving object along the line of sight.

In a right angled velocity triangle A B C: Angle A = wt
Angle B = 90°; Angle C = 90° -wt
AB = hypotenuse = c; BC = opposite = v; CA= adjacent = c √ [1-(v/c) ²]

2- Time dilatations
Along the line of sight; S x = r cosine wt
Hypotenuse = S x = [c t x] = c t √ [1-(v/c) ²];
Where t = self time; t x = time by others

t x = t √ [1-(v/c) ²]; and
t = {1/√ [1-(v/c) ²]} t x

These are time dilatation equations given by Einstein’s special relativity theory.

3- Δ E= mc²

S = r Exp (ỉ ω t); sin ω t =v/c; v = c sin ω t; r = -(c/ω) cosine ω t;
And r. v = (-c²/ω) sin ω t cosine ω t

P = d S/d t = (v + ỉ ω r) Exp (ỉ ω t); v² = c² sin² ω t; ω² r² = c² cosine² ω t
P² = (v + ỉ ω r). (v + ỉ ω r) Exp [2(ỉ ω t)] = [v² -ω² r² +2ỉ ω (r. v)] Exp [2(ỉ ω t)]

P² = [c² sin² ω t - c² cosine² ω t - 2c²ỉ sin ω t cosine ω t] Exp [2(ỉ ω t)]
P² = – c² [cosine² ω t - sin² ω t + ỉ sin 2ω t] Exp [2(ỉ ω t)]

P² = -c² Exp [4(ỉ ω t)]
E = mP²/2 = – mc²/2 [cosine² 2ω t - sin² 2ω t + 2ỉ sin 2ω t cosine 2ω t]

E = (-mc²/2) {1-2sin² 2ω t + 2ỉ [1- 2sin² ω t] 2[sin ω t cosine ω t]}
E = (-mc²/2) {1- 2(v/c) ² + 4ỉ [1- 2(v/c) ²] (v/c) √ [1- (v/c) ²]}

If v = 0 then E (1) = (-mc²/2); and
If v = c then E (2) = (mc²/2) then

Δ E = E (2) – E (1) = (mc²/2) – (-mc²/2)
Δ E = mc²

B- General Theory of relativity

What is the visual effect for angular velocity along the line of sight? At Perihelion It is called the Advance of perihelion. Let us derive that

Areal velocity is constant: r² θ’ =h Kepler’s Law

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
S = r Exp (ỉ wt); r² θ’= h = S² w’

h = S²w’= [r² Exp (2iwt)] w’=r²θ’; w’ = (θ’) exp [-2(ỉ wt)]
And w’= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]

With w’ = w’ (x) + ỉ w’(y); w’(x) = (h/r²) [1- 2sine² (wt)]
Δ w’= w’(x) – (h/r²) = – 2(h/r²) sine² (wt) = – 2(h/r²) (v/c) ² v/c=sine wt

Angular velocity (h/ r²) (Perihelion/Periastron) = [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²

Δ w’ = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
[180/π; degrees][100years=36526days; century] x [3600; seconds in degree]

Δ w” = (-720×36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century

This equation gives the rate of advance of perihelion of Mercury with better results than all of Albert Einstein’s publications and better than all of published physics.

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<

1- Advance of Perihelion of mercury.

G=6.673x10^-11; M=2x10^30kg; m=.32x10^24kg
ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec
Calculations yields:
v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552
Δ w”= (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century

2- DI Herculis Apsidal motion solution: derived from S= r exp [ỉ ω t]
(See other articles by Joe Nahhas)

W° (ob) = (-720x36526/T) x {[√ (1-ε²)]/ (1-ε) ²} [(v*/c) + (v°/c)] ² degrees/ century

Where v* = v (center of mass) = 106.38km/sec; v° (spin difference) = 0
T = orbital period; ε = eccentricity; c =light speed

Application 3: Gravitational red shift: Pound Rebka Experiment

S = r Exp [î ω t]

1/S = 1/r Exp [-ỉ ω t]
And λ (S) = λ (r) Exp [-ỉ ω t]; λ = wavelength

Then υ(s) = υ(r) Exp [ỉ ω t]; υ = frequency
And υ(S) = υ (r, t) = υ(r, 0) υ (0, t) = υ(r) υ (0, t)
With sin ω(r) t = v/c; cosine ω(r) t = √ [1-(v/c) ²]

Then υ (r, t) = υ(r, 0) {√ [1-(v/c) ²] + ỉ (v/c)} = Real {υ(r, t)} + Imaginary {υ(r, t)}
Real {υ (r, t)} = υ (r, 0) √ [1-(v/c) ²] ≈ υ (r, 0) [1 - 1/2(v/c) ²]

Δ υ (r, t) = real {υ (r, t)} - υ (0, t)
Δ υ (r, t) = -υ (r, 0)/2 [(v/c) ²]
Δ υ(r, t)/υ(r, 0) = -1/2(v/c)²[up]-{1/2(v/c)²[down]} = - (v/c) ²
v² = 2gh; g = 9.81km/s² gravitational acceleration; h = height

Δ υ/υ [Total] =-[2gh/c²]
4- Light bending: Lord Edenton experiment
S = r Exp [ỉ ω t]; From Kepler's Equation: r² θ' = h = 2A/T
h = S²(r, t) θ'(r, t) = r² (θ, t) θ' (θ, t) = r² (θ, 0) Exp [2ỉ ω t] θ' (θ, t) = 2A/t
And θ' (θ, t) = θ' (θ, 0) θ'(0, t) = [h/ r² (θ, 0)] Exp [-2ỉ ω(r) t]
Then θ '(θ, t) = [2A/t r² (θ, 0)] {1 - 2sin²ω(r) t - 2ỉ sin ω(r) t cosine ω(r) t}
Now [t θ'(θ, t)] = [2A/r² (θ' 0)] [1 - 2sin²ω(r) t] -2ỉ [2A/r² (θ, 0)] [sin ω(r) t cosine ω(r) t]
= Δ x + i Δ y
Δ θ = Δ x - [A/r² (θ, 0)] = - [A/r² (θ, 0)][4sin²ω(r)t]; sin ω(r)t = v/c
Δ θ = - [A/r² (θ, 0)](v/c) ²
(v/c) ² ≈ 1.75"; v² = GM/R; G = Gravitational constant; M = Sun mass; R = sun radius
Δ θ = [A/r² (θ, 0)] [1.75"]; A = area
The values depend on near by stars and the measured values fit this equation.
Russians in 1936; Δ θ = 2.74
[A/r² (θ, 0)] = π/2
Δ θ = π/2(1.75") = 2.74"

Application 5: Shapiro time delay (Vikings 6, 7; 1977)
Mars --------------------------- Middle---- Sun ------------- Earth
The center of mass is the sun. The sun produces a velocity field given by
v = √ [GM/a (1- ε²/4)]
From above t =2 arc length/c=2d Δ w/c = (8π r/c) (v/c) ²; Δ w=4π (v/c) ²; r = 2a=d
t = 16πGM/c³ (1-ε²/4); ε = [a (1) -a (2)]/ [a (1) + a (2)] = .2075
t = (8πd/c) (v/c) ²= 8π (377,536,987.5/299792.458) (26.6575872/299792.458)²=250μs
If d = 2a (1-ε²/4), then t = 247.597μs value theorized actual measured value is 250μs
All this is not due to space-time but due to light aberration caused by moving planets.
θ'(0,0) = h(0,0)/r²(0,0) = 2π/T
θ' (0,t) = θ'(0,0)Exp(-2ỉwt)={2π/T} Exp (-2iwt)
θ'(0,t) = θ'(0,0) [cosine 2(wt) - ỉ sine 2(wt)] = θ'(0,0) [1- 2sine² (wt) - ỉ sin 2(wt)]
θ'(0,t) = θ'(0,t)(x) + θ'(0,t)(y); θ'(0,t)(x) = θ'(0,0)[ 1- 2sine² (wt)]
θ'(0,t)(x) – θ'(0,0) = - 2θ'(0,0)sine²(wt) = - 2θ'(0,0)(v/c)² v/c=sine wt; c=light speed
T [θ'(0, t) - θ'(0, 0)] = -4π (v/c) ²
Δ θ = -4π (v/c) ² Earth-Mars
Sun-Photon:
The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²---) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
v=√ [Gm M/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m< ΔΓ = 2 arc length/c = 2[Δ θ] 2d/c = 2[- 4π (v/c) ²] 2d/c; ΔΓ = -8πd/c (v/c) ²;
ΔΓ = 8πd/c³ [GM/a (1-ε²/4)] =16πGM/c³ (1-ε²/4) = Γ0 (1 - ε²/4)
ε = [a (planet 1) - a (planet 2)]/ [a (planet 1) + a (planet 2)] =0.2075 Mars-Earth
Γ0 = 16 πGM/c³= 247.5974607μs=universal constant; ΔΓ = 250μs Mars-Earth.
Joe nahhas1958@yahoo.com All right reserved

Conrad

Conrad J Countess

It seems like you feel that PaleBlueDot was trying to discourage you, when I get the impression he was trying to show you how you could get more people to accept your idea.

For example, I’m not a mathematician. I’ve seen mathematical ‘proofs’ showing that 1=2. They all contained some flaw or false assertion that was not obvious. By the same token, I’ve seen proofs that 0.999…=1 – again with mathematical proofs that looked similar to the ones for 1=2.

Although the idea of 0.999… equaling 1 made sense to me, and I didn’t see any flaws in those proofs, what actually convinced me was the amount of independent verification available for 0.999… being the same value as 1.

By the same token, I don’t have the background to prove or disprove your idea. Having objective opinions and verification of your idea would lend it credibility and authenticity.

Sorry to everyone else to adding ANOTHER comment onto this haha

Heres mathematical proof or did you already see this: http://docs.google.com/Doc?docid=dsn5q6f_101hgtjv9fb&hl=en

Who are you really? I got a feeling that you are not just a meteorology student, because you are pleading the physicist case with to much involvement and sympathy. I know that there are theory assassins out there who are sent out to discredited theories that are threatening to certain parties theory of economic interest. Are you one of them?
I do not wish to hurt anyones pride or livelihood by figuring out something that thousands of educated people who studied for years could not, but I did. And whats more important, what is right or who is right? And whats more important still, bringing the truth about the interconnectedness of the Universe to humanity, so we can be more at home here, or protecting someones hurt feelings because they did not discover this first?
I look at physic as a noble subject studied by noble people. But you are disillusioning me by telling me that physicist are sensitive to some kind of foolish intellectual pride. I would have thought that they were beyond that and would be happy to get at the truth no mater where it came from.

I’ll tell you how I really feel
I just revealed to the world, that the most famous equation in the world “Einstein’s E=mc^2”, when looked at geometrically shows that (E or energy) = (m or rest mass) x (c^2, which is the speed of light in circular rotation), thus revealing the origin of rest mass. This is also the unification of (General Relativity) and (Quantum Theory), another landmark goal in physics, And the hits just keep on coming.
If you think that I should shut up or that you can discourage me, all you are really doing is letting me know that someone is nervous. And that I am at least on the right track. Well I think I am there and that anyone who looks at this objectively will see that I am correct and that you are hiding something. This is too important keep quite about, and what I should really do is “Go Tell it on the Mountaintop” or better still “Go Tell it on the Internet”

The speed of light squared is the speed of light circled and the pivotal point around which a new revolution in physics is taking place. Any interested parties, do not let anyone do your thinking for you, look at the evidence objectively, It speaks for itself.

Short version:http://docs.google.com/Doc?id=dsn5q6f_101hgtjv9fb
Long Version:: http://docs.google.com/Doc?id=dsn5q6f_102hm2wwrhf
Related evidence:http://docs.google.com/Doc?id=dsn5q6f_107jx6qsgdg

Conrad J Countess

Most scientists work for there ENTIRE LIVES to generate the evidence and observations relevant to their great ideas, and there are more than a few that have passed away before they’ve seen their ideas tested, but their process has involved a lot of energy, time, and sacrifice on their part, just to see one experiment happen, and through all that time there is a strong possiblity that the answer they get will totally prove them wrong, but they still do it. Of course, you have to realize that if you come out here with guns blazing with your idea, with no background in the field, telling people that have been wrapped in thought about this stuff for half a CENTURY that they are flat out wrong and you’re right, how does that make them more likely to listen to you?

Personally I’m skeptical of your ideas, which is not a bad thing, skepticism is a healthy thing in science. Unfortunately for you, the burden of proof is on YOUR shoulders. It’s your idea, your passion, prove it to us! I don’t want to see a page with wavy lines and talking smiley faces, show us some more tantalizing evidence! If you’re right, then you get the glory, the right to say I told you so, and the satisfaction that you changed the way the world thinks (not to mention the technological possiblities your idea would enable). If you’re wrong, oh well, there’ve been lots of wrong ideas in the history of science, and nobody will ridicule you for at least giving it an honest shot. I will await your results Conrad!