Muhammad Idrees Ahmad

3 Responses to “How Muslims are Treated”

1. Alex said

   March 30, 2008 at 5:07 pm

   I know that kind of abuse and willful neglect happens but it’s equally shocking and appalling every time.
2. ExEx Blogger said

   March 31, 2008 at 12:19 am

   I am also moved by this video. I’m glad that those Americans stood up against such racism and religious profiling.
3. joe nahhas said

   January 29, 2009 at 8:09 am

   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

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