# Imagining the Tenth Dimension

Ouch, my brain hurts.

Thanks a lot for showing me this Devin. Now my head is going to explode.

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### One response to “Imagining the Tenth Dimension”

1. Joe Nahhas

Do not do that because it does not exist
Einstein’s Nemesis: DI Herculis Eclipsing Binary Stars Solution
The problem that the 100,000 PHD Physicists could not solve

This is the solution to DI Her “Quarter of a century” Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney
Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics
For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton’s equation and time dependent Kepler’s equation that accounts for Quantum – relativistic effects and it explains these effects as visual effects. Here it is

Universal- Mechanics

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

= m γ + 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)

F = [d²(m r)/dt² – (m r)θ’²]r(1) + (1/mr)[d(m²r²θ’)/d t]θ(1) = [-GmM/r²]r(1)

d² (m r)/dt² – (m r) θ’² = -GmM/r²; d (m²r²θ’)/d t = 0

Let m =constant: M=constant

d²r/dt² – r θ’²=-GM/r² —— I

d(r²θ’)/d t = 0 —————–II
r²θ’=h = constant ————– II
r = 1/u; r’ = -u’/u² = – r²u’ = – r²θ'(d u/d θ) = -h (d u/d θ)
d (r²θ’)/d t = 2rr’θ’ + r²θ” = 0 r” = – h d/d t (du/d θ) = – h θ'(d²u/d θ²) = – (h²/r²)(d²u/dθ²)
[- (h²/r²) (d²u/dθ²)] – r [(h/r²)²] = -GM/r²
2(r’/r) = – (θ”/θ’) = 2[λ + ỉ ω (t)] – h²u² (d²u/dθ²) – h²u³ = -GMu²
d²u/dθ² + u = GM/h²
r(θ, t) = r (θ, 0) Exp [λ + ỉ ω (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)
r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]
r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

r(0,t)= Exp[λ(r) + ỉ ω (r)]t; Exp = Exponential

r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì ω(r)]t} Nahhas’ Solution

If λ(r) ≈ 0; then:

r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[ỉ ω(r)t]

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

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
h = 2πa²√ (1-ε²); r (0, 0) = a (1-ε)

θ’ (0,0) = h/r²(0,0) = 2π[√(1-ε²)]/T(1-ε)²
θ’ (0,t) = θ'(0,0)Exp(-2ỉwt)={2π[√(1-ε²)]/T(1-ε)²} 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

Δ θ’ = [θ'(0, t) – θ'(0, 0)] = -4π {[√ (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second
{(180/π=degrees) x (36526=century)

Δ θ’ = [-720×36526/ T (days)] {[√ (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

This is the T-Rex equation that is going to demolished Einstein’s space-jail of time

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 = v (center of mass); v is the sum of orbital/rotational velocities = v(cm) for DI Her
Let m = mass of primary; M = mass of secondary

v (m) = primary speed; v(M) = secondary speed = √[Gm²/(m+M)a(1-ε²/4)]