Yes, it's true, the U.S. Patent Office is going to open its doors to let all of us see what's been filed, and even better yet, make our own comments on the applications.
(All via the miracle of the Internet, of course.)
Kinda makes you wonder how that bad boy of the Patent Office, Albert Einstein, who used to labor there a century ago (the Swiss Patent Office, that is), would think about that idea.
And would it be a mathematical answer? 
Check out this report for more details.
And thanks to Nobelprize.org for the early photograph.
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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
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) = (-720x36526/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 v (M) = √ [Gm² / (m + M)a(1-ε²/4)] ≈ 0; m
Application 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; 1-ε²/4 = 0.989391
ρ (m) = 0.696x10^9m; ρ(m)=2.44x10^6m; 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) = (-720x36526/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