How Muslims are Treated
March 29, 2008
Normal people being heroes and villains – silence is complicity.
Thanks to Sabbah’s Blog – “Listen carefully to the results at the end of the clip. They are scary!” (15% approving of the racism, 54% silent, 32% challenging the racism).
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I know that kind of abuse and willful neglect happens but it’s equally shocking and appalling every time.
I am also moved by this video. I’m glad that those Americans stood up against such racism and religious profiling.
Einstein’s Nemesis: DI Her Eclipsing Binary Stars Solution
The problem that the 100,000 PHD Physicists could not solve
This is the solution to the “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’ = [-720x36526/ 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)]
v (cm) = [m v(m) + M v(M)]/(m + M) All rights reserved. joenahhas1958@yahoo.com