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*Einstein and Eddington*. Am i writing about biography of those two great scientist? Actually, ‘*Einstein and Eddington* is a movie title from the BBC, (see here for the official site of the movie).

The story revolves around the two scientists and a solar eclipse, and the world in the middle of war (around World War I). Now that is intriguing. Why? Because during the tension between the two waring nations, and at the same war affect most of the scientist in the countries. Despite the danger of being labelled traitors, the two men began a unique correspondence. An eclipse in Africa provided an opportunity to prove Einstein’s theories to the world. Eddington, an unlikely hero, set out on a journey that would change people’s perceptions of the universe forever. Hey, did i see Smeagol Gollum? Well, i am not going to review the movie though, but it is a nice movie anyway.

But what i have thought are:

*First*, an eclipse, is a natural phenomenon, and in particular an astronomical phenomenon. We can see the phenomenon either as an inspiring or a scientific phenomenon, or both. Either way, we can enjoy that phenomenon in anyway we like. So whatever the situation, politic, social, or whatever it is, the phenomenon is the same to all, and belong to all.

*The second*, how those scientist can step aside the differences of the political situation, standing still on the scientific ground, and finally able to speak on the same language of understanding in the scientific language of our nature.

*The third*, the scientific understanding can overcome all of the differences, if not, there will be no language of nature for us all.

But the history sometimes can repeat itself, after 90 years, the eclipse will return. Eclipse will always the same, it is between the Sun, the Moon and the Earth. But in the same time, the tensions between countries also repeat. Why? Why should be war happen in our life? Well, i don’t have proper answer. For me, i just see something from the other side.

Next week there will be an eclipse, and some other eclipses also will occur, and one total solar eclipse should make a big hit in this year. And at the same time, this is a long year of astronomy, with the spirit ‘yours to discover’. If only this ‘yours’ is shared by all of us, maybe we can learn from Einstein and Eddington. Well, we don’t have to be great scientists to enjoy the phenomenon, but we have all the right to enjoy the awe inspiring phenomenon from the sky. And from the beneath of mounting tasks that i should do, not to forget the preparation for upcoming eclipse, i just spare my time to ponder, if only i could share the two minutes of the eclipse to the world, will that make a difference to the humanity? Nahh .. i don’t know, maybe i ponder too much because of too much coffee, i guess.

Science writer working on several projects in astrophysics, history and cultural of astronomy.

September 28, 2013

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V1143Cgyni Binary Stars Apsidal motion Puzzle solution

The motion puzzle that Einstein MIT Harvard Cal-tech NASA and all others could not solve.

Introduction: For 350 years Physicists Astronomers and Mathematicians missed Kepler’s time dependent equation that changed Newton’s equation into a time dependent Newton’s equation and together these two equations combine classical mechanics and quantum mechanics into one mechanics explains “relativistic” effects as the difference between time dependent measurements and time independent measurements of moving objects and solve all motion in all of Mechanics posted on Smithsonian NASA website SAO/NASA that Einstein and all 100,000 space-time “physicists” could not solve by space-time physics or any published physics.

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location

r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location:

P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

= change of location + change of mass

= m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

F = d P/d t = d²S/dt² = Total force

= m(d²r/dt²) +2(dm/dt)(d r/d t) + (d²m/dt²)r

= mγ + 2m’v +m”r; γ = acceleration; m” = mass acceleration rate

In polar coordinates system

r = r r(1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)

Proof:

r = r [cosθ î + sinθĴ] = r r (1); r (1) = cosθ î + sinθ Ĵ

v = d r/d t = r’ r (1) + r d[r (1)]/d t = r’ r (1) + r θ'[- sinθ î + cos θĴ] = r’ r (1) + r θ’ θ (1)

θ (1) = -sinθ î +cosθ Ĵ; r(1) = cosθî + sinθĴ

d [θ (1)]/d t= θ’ [- cosθî – sinθĴ= – θ’ r (1)

d [r (1)]/d t = θ’ [ -sinθ’î + cosθ]Ĵ = θ’ θ(1)

γ = d [r’r(1) + r θ’ θ (1)] /d t = r” r(1) + r’ d[r(1)]/d t + r’ θ’ r(1) + r θ” r(1) +r θ’ d[θ(1)]/d t

γ = (r” – rθ’²) r(1) + (2r’θ’ + r θ”) θ(1)

F = m[(r”-rθ’²)r(1) + (2r’θ’ + rθ”)θ(1)] + 2m'[r’r(1) + rθ’θ(1)] + (m”r) r(1)

= [d²(mr)/dt² – (mr)θ’²]r(1) + (1/mr)[d(m²r²θ’)/dt]θ(1) = [-GmM/r²]r(1)

d²(mr)/dt² – (mr)θ’² = -GmM/r² Newton’s Gravitational Equation (1)

d(m²r²θ’)/dt = 0 Central force law (2)

(2) : d(m²r²θ’)/d t = 0 m²r²θ’ = [m²(θ,0)φ²(0,t)][ r²(θ,0)ψ²(0,t)][θ'(θ, t)]

= [m²(θ,t)][r²(θ,t)][θ'(θ,t)]

= [m²(θ,0)][r²(θ,0)][θ'(θ,0)]

= [m²(θ,0)]h(θ,0);h(θ,0)=[r²(θ,0)][θ'(θ,0)]

= H (0, 0) = m² (0, 0) h (0, 0)

= m² (0, 0) r² (0, 0) θ'(0, 0)

m = m (θ, 0) φ (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential

φ (0, t) = Exp [ λ (m) + ỉ ω (m)]t

r = r(θ,0) ψ(0, t) = r(θ,0) Exp [λ(r) + ì ω(r)]t

ψ(0, t) = Exp [λ(r) + ỉ ω (r)]t

θ'(θ, t) = {H(0, 0)/[m²(θ,0) r(θ,0)]}Exp{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}} ——I

Kepler’s time dependent equation that Physicists Astrophysicists and Mathematicians missed for 350 years that is going to demolish Einstein’s space-jail of time

θ'(0,t) = θ'(0,0) Exp{-2{[λ(m) + λ(r)]t + ỉ[ω(m) + ω(r)]t}}

(1): d² (m r)/dt² – (m r) θ’² = -GmM/r² = -Gm³M/m²r²

d² (m r)/dt² – (m r) θ’² = -Gm³ (θ, 0) φ³ (0, t) M/ (m²r²)

Let m r =1/u

d (m r)/d t = -u’/u² = -(1/u²)(θ’)d u/d θ = (- θ’/u²)d u/d θ = -H d u/d θ

d²(m r)/dt² = -Hθ’d²u/dθ² = – Hu²[d²u/dθ²]

-Hu² [d²u/dθ²] -(1/u)(Hu²)² = -Gm³(θ,0)φ³(0,t)Mu²

[d²u/ dθ²] + u = Gm³(θ,0)φ³(0,t)M/H²

t = 0; φ³ (0, 0) = 1

u = Gm³(θ,0)M/H² + Acosθ =Gm(θ,0)M(θ,0)/h²(θ,0)

mr = 1/u = 1/[Gm(θ,0)M(θ,0)/h(θ,0) + Acosθ]

= [h²/Gm(θ,0)M(θ,0)]/{1 + [Ah²/Gm(θ,0)M(θ,0)][cosθ]}

= [h²/Gm(θ,0)M(θ,0)]/(1 + εcosθ)

mr = [a(1-ε²)/(1+εcosθ)]m(θ,0)

r(θ,0) = [a(1-ε²)/(1+εcosθ)] m r = m(θ, t) r(θ, t)

= m(θ,0)φ(0,t)r(θ,0)ψ(0,t)

r(θ,t) = [a(1-ε²)/(1+εcosθ)]{Exp[λ(r)+ω(r)]t} Newton’s time dependent Equation ——–II

If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then

θ'(0,t) = θ'(0,0) Exp{-2ì[ω(m) + ω(r)]t}

r(θ, t) = r(θ,0) r(0,t) = [a(1-ε²)/(1+εcosθ)] Exp[i ω (r)t]

m = m(θ,0) Exp[i ω(m)t] = m(0,0) Exp [ỉ ω(m) t] ; m(0,0)

θ'(0,t) = θ'(0, 0) Exp {-2ì[ω(m) + ω(r)]t}

θ'(0,0)=h(0,0)/r²(0,0)=2πab/Ta²(1-ε)²

= 2πa² [√ (1-ε²)]/T a² (1-ε) ²; θ'(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²

θ'(0,t) = {2π[√(1-ε²)]/T(1-ε)²}Exp{-2[ω(m) + ω(r)]t

θ'(0,t) = {2π[√(1-ε²)]/(1-ε)²}{cos 2[ω(m) + ω(r)]t – ỉ sin 2[ω(m) + ω(r)]t}

θ'(0,t) = θ'(0,0) {1- 2sin² [ω(m) + ω(r)]t – ỉ 2isin [ω(m) + ω(r)]t cos [ω(m) + ω(r)]t}

θ'(0,t) = θ'(0,0){1 – 2[sin ω(m)t cos ω(r)t + cos ω(m) sin ω(r) t]²}

– 2ỉ θ'(0, 0) sin [ω (m) + ω(r)] t cos [ω (m) + ω(r)] t

Δ θ (0, t) = Real Δ θ (0, t) + Imaginary Δ θ (0.t)

Real Δ θ (0, t) = θ'(0, 0) {1 – 2[sin ω (m) t cos ω(r) t + cos ω (m)t sin ω(r)t]²}

W(ob) = Real Δ θ (0, t) – θ'(0, 0) = – 2 θ'(0, 0){(v°/c)√ [1-(v*/c) ²] + (v*/c)√ [1- (v°/c) ²]}²

v ° = spin velocity; v* = orbital velocity; v°/c = sin ω (m)t; v*/c = cos ω (r) t

v°/c << 1; (v°/c)² ≈ 0; v*/c << 1; (v*/c)² ≈ 0

W (ob) = – 2[2π √ (1-ε²)/T (1-ε) ²] [(v° + v*)/c] ²

W (ob) = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² radians

W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² degrees; Multiplication by 180/π

W° (ob) = (-720×36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years

W” (ob) = (-720x26526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² seconds /100 years

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)

v (m) = √ [GM²/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system

v (M) = √ [Gm² / (m + M)a(1-ε²/4)] ≈ 0; m<<M

Application 1: Advance of Perihelion of mercury.

G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg; ε = 0.206; T=88days

c = 299792.458 km/sec; a = 58.2km/sec; 1-ε²/4 = 0.989391

ρ (m) = 0.696×10^9m; ρ(m)=2.44×10^6m; T(sun) = 25days

v° (M) = 2km/sec ; v° = 2meters/sec

v *= v(m) = √ [GM/a (1-ε²/4)]; v(M) = √[Gm²/(m + M)a(1-ε²)] ≈ 0

v°(m) = 2m/sec (Mercury) v°(M)= 2km/sec(sun)

Calculations yields: v = v* + v° =48.14km/sec (mercury); [√ (1- ε²)] (1-ε) ² = 1.552

W” (ob) = (-720x36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ²

W” (ob) = (-720x36526x3600/88) x (1.552) (48.14/299792)² = 43.0”/century

V1143Cgyni Apsidal Motion Solution

W° (ob) = (-720×36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years

v° = -v°(m) + v°(M)

v* = 2v(cm) + σ

v°(m) = spin velocity of primary

v°(M) = spin velocity of secondary

v(cm) = [m v(m) + M v(M)]/(m + M) center of mass velocity

σ = √ {{[v(m) – v(cm)]² + [v(M) – v(cm)]²}/2} = standard deviation

W° = 3.36°/century as reported in many articles