Unique Double Pulsar Proves Einstein's "Spacetime" Theories

Albert Einstein was famously annoyed about God playing dice with the universe.  Perhaps as an apology, God has set up the perfect demonstration of the theory of general relativity - it seems that if He moves in mysterious ways, those ways are at least consistent with four-dimensional spacetime.

A fundamental concept in relativity is the idea of curved spacetime, the idea that large masses bend surrounding space (and which has led to more people imagining bowling balls on sheets of rubber than could possibly have happened otherwise).  The problem is that even our largest local mass might as well be called "The wee tiny little o'Sun-chan".  It looks pretty big, but that's only because we're really small.

The double pulsar PSR J0737−3039A/B has a number of properties that make it the perfect test for relativity.  For one thing it doesn't need that great big bloody name, because it's the only double pulsar we've ever observed: just say "the double pulsar" to an astrophysicist and they'll know what you mean, so there's no need to copy out that serial number-looking code for your own studies.

The pulsars orbit each other every two and a half hours (rather than the month it takes the Earth and the Moon with, which have half the separation).  It's basically a vast mangler of spacetime, more than twice the mass of the Sun and made of the densest state of matter known to human science.  The only way it could be a more perfect test is if the orbit was edge-on to the Earth, and what do you know, it is.

This means that twice every orbit we get 30 seconds of pure signal from one of the pulsars while the other is eclipsed behind it.  By analyzing this data, scientists from McGill University have conducted the most extreme testing of Einstein's theories possible to date.  The math is pretty complicated, and has a nasty tendency to include terms like "The c-squared-sigma-B over G ratio is point nine four plus or minus point one three", but the conclusion is one that we can all understand:

That Einstein fella?  He was pretty damn right.

Posted by Luke McKinney.

http://www.nsf.gov/news/news_summ.jsp?cntn_id=111831&org=NSF&from=news

Comments

[slider]

Ok, so how do we know any frame dragging is going on? Why does this prove Einstein was correct? Is there a page two to this article?

[Joe Nahhas]

Einstein's bad physics of Space-time reversed Darwin's theory
Man ascended to a great "white" Ape
beleiving in space-time that can be expalined by high school math
For 400 years kepler's time dependent equation and was missed by physicists and I can prove that
any takers?

[joe nahhas]

There is One and Only One Mechanics: Universal Mechanics:
Ending Einstein's space-jail of time in 2009
joenahhas1958@yahoo.com
Introduction to Universal Mechanics: For 350 years Physicists Astronomers and Mathematicians missed Kepler's time dependent equation introduced here and transformed Newton's equation into a time dependent Newton' equation and together these two equations explain Quantum - Relativistic effects; it combines classical mechanics and quantum mechanics into one mechanics and explains "relativistic" effects as the difference between time dependent measurements and time independent measurements of moving objects. Time is not a structure like space to scientifically accept space-to imaginary time - back to space jumping continuum (x, y, z, it) regardless of what all 100,000 living space-time physicists and Astrophysicists have said about it because Physics is a business and not necessarily science or scientific and like every business Physics comes with fraud and fraud is "junk" experimentation and junk based "thought" experimentation that led to fraud physics E=mc².

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)

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) = (-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 3: 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

Application 4: Gravitational red shift: Pound Rebka Experiment

r (θ, t) = r(θ, 0) Exp [î w(r)t] = λ(r) Exp[ỉ ω(r)t]; λ(r) = 0
1/r(θ, t) = 1/r(θ,0) = (1/λ) {Exp [-(î wt)]}

υ (θ, t) = υ (θ,0) υ(0, t) = υ (θ, 0) Exp (-ì wt) = υ (θ, 0) [cos (ω t) + ỉ sin (ω t)]
sin ω(r)t = v/c; cos ω(r)t = √[1-(v/c)²]

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

Δ υ (0, t) = real {υ (0, t)} - υ (0, 0)
Δ υ (0, t) = -υ (0, 0)/2 [(v/c) ²]

Δ υ(0, t)/υ(0, 0) = -1/2(v/c)²[up]-{1/2(v/c)²[down]} = - (v/c) ²
v² = 2gh=2(9.806) (22.6m)

Δ υ/υ [Total]=[2x9.806x22.6/299792.458]=4.93169x10^-15
Observed value: 5.1±5x10^-15

5- Light bending: Lord Eddington experiment

θ' (θ, t) = θ' (θ, 0) θ'(0, t) = [h/r²(θ,0)] Exp { -2ỉ[ ω(m) + ω(r)]t}; ω (m) = 0
θ '(θ, t) = [2A/tr²(θ, 0)]{1 - 2sin²ω(r)t - 2ỉ sin ω(r)t cos ω(r)t}

[t θ'(θ, t)] = [2A/r²(θ' 0)][1 - 2sin²ω(r)t] -2ỉ[2A/r²(θ, 0)][sin ω(r)t cos 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 6: 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.

[Alexander Nahhas]

Einstein's Physics Dollar Store on Campus
MIT Harvard Cal-Tech
Sponsored by NASA
Why Relativity theory is not Physics and why Einstein's "thought" = 0
Walking the walk and talking the talk taking on all space-time confusion of physics by
MIT Harvard and Cal-Tech and all other Physics dollar stores departments
And why LHC burned itself

Visual Effects and the confusions of "Modern" physics

r --------- Light sensing of moving objects ------- S
Actual object----- Light --------- Visual object
r - -------cosine (wt) + i sine (wt) - S = r [cosine (wt) + i sine (wt)]
Newton-- Kepler's time visual effects -- Time dependent Newton
Particle -------------- Visual effects -------------------- Wave

Line of Sight: r cosine wt

r ------------------- r cosine (wt) line of sight light aberrations

A moving object with velocity v will be visualized by
light sensing through an angle (wt);w = constant and t= time
Also, sine wt = v/c; cosine wt = √ [1-sine² (wt) = √ [1-(v/c) ²]

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 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 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]

[joe nahhas]

Relativity theories is not worth the ink

As Camelopardis Binary Stars High Rate Apsidal Motion Puzzle Solution
By Joe Nahhas

Abstract: This is the solution to the 30 years most studied Binary Stars with high rate orbit axial rotations puzzle that made astrophysicists wipe their glasses and wipe their high tech telescopes eyepieces and sent Einstein's space-time physics research papers solutions back to sender and said "NO" to the 100,000 space-time Physicists and Astrophysicists in their hideouts after they could not solve this motion puzzle by any said or published Physics for thirty years including 109 years of Nobel Prize winner Physics and physicists and 400 years of astronomy and is dedicated to the two DRS KH. F. Khailullin and V.S. Kozyreva of Moscow University who posted this motion puzzle in 1983 as not solvable by space-time physics or any said or published Physics and still posted as the motion puzzle to solve on Smithsonian-NASA website SAO/NASA.

Universal Mechanics Solution: For 350 years Physicists Astronomers and Mathematicians missed Kepler's time dependent equation introduced here and transformed Newton's equation into a time dependent Newton' equation and together these two equations explain Quantum - Relativistic effects; it combines classical mechanics and quantum mechanics into one mechanics and explains "relativistic" effects as the difference between time dependent measurements and time independent measurements of moving objects and in practice it amounts to "Visual" effects.

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]

v = r' r (1) + r θ' θ (1)

Then θ (1) = -sine θ î +cosine θ Ĵ; r(1) = cosine θ î + sine θ Ĵ

And, d [θ (1)]/d t= θ' [- cosine θ î - sine θ Ĵ= - θ' r (1)
And, d[r (1)]/d t = θ' [-sine θ' î + cosine θ 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
If v°/c << 1; (v°/c) ² ≈ 0; And If 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) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/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)]
v (M) = √ [Gm² / (m + M)a(1-ε²/4)]

As Camelopardis Apsidal motion solution:
Data T=3.431; r(m) =0.1499 m=3.3M(0) R(m) =2.57R(0) [v°(m),v°(M)]=[40,30]
ε = 0.1695; 1-ε = 00.8305; r(M) =0.1111 M=2.5M(0) R(M) = 2.5R(0) ;m + M=5.8M(0)

G=6.673x10^-11; M (0) = 1.98892x10^30kg; R (0) = 0.696x10^9m
1- ε²/4 = 0.9928; [√ (1-ε²)]/ (1-ε) ² = 1.43
a = [R (m)/r (m)] = (2.57/0.1499) (0.696x10^9) m
With a (1-ε²/4) = (2.57/0.1499) (0.696x10^9) (0.9988) = 11.8470x10^9m
And v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 110.1786325km/sec
And v (M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 145.435795km/sec
And v (cm) = ∑m v/∑m = 125.3756853km/sec
σ =√ {∑ [v-v (cm)] ²/2}
=√ {[(110.1786325-125.3756853)² + (145.435795-125.7356853)²]/2}
σ = 25.1659669
Spin: v° = v° (m) + v° (M) = 40km/s + 30km/s = 70km/sec
And v (m) = 110.1786325km/sec
Also, v (M) = 145.435795km/sec; v (cm)= 125.3756853km/sec
σ = 25.1659669km/sec

1- With v* = v (m) + v (M) +σ=350,7803944km/sec; [√ (1-ε²)]/ (1-ε) ² = 1.43
T = 3.431days
W° (observed) = (-720x36526/T) x {√ [(1-ε²)] (1-ε) ²} {[v* + v°]/c} ²= 15.0°/century
Dr Guinan: W°= 15°/century 1989

2- With v * = 2v (cm) + σ =
And v* = 2v (cm) + σ =
2[m v(m) + M v(M)]/(m + M) + √{{[v(m)-v(cm)]² + [v(M)-v(cm)]²}/2}
= 275.9176729km/sec
Then v* + v° = 275.9176729 + 70 = 345.9176729km/sec
W° = (-720x36526/T) x {[√ (1-ε²)]/ (1-ε) ²} {[v* + v°]/c} ²= 14.6°/100 years
3- Khailullin: 1983 v (p) =110.4; v(s) = 145.8; σ=25.2685
2∑ m v/∑m + σ + 70=346.0185
W°= 14.6 °/century same as reported [same as published]
References:
Apsidal motion of As Cameloparids by Khailullin: 1983
Apsidal motion of As Camelopardis Edward Guinan and Frank Maloney: 1986
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