Media has no bounds when it comes to Muslim Bashing

…No. The print media is faltering because the writing in many newspapers over the past decade have turned to the far, ugly right. Conservative neo-fascism disguised as free speech used to bully and intimidate anyone who has a viewpoint that they dislike. And it hasn’t been just one newspaper. It has been most of the major newspapers. The writing has been filled with hate-mongering, and vicious slander of a religion and a people.  >>>>>

This entry was posted on Sunday, August 17th, 2008 at 9:21 pm and is filed under Palestine. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Media has no bounds when it comes to Muslim Bashing”

  1. Joe Nahhas Says:
    January 29, 2009 at 4:08 am | Reply

    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

  2. Alexander Says:
    February 11, 2009 at 1:09 pm | Reply

    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 << 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) = (-720×36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years

    W” (ob) = (-720×26526x3600/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<<M; Solar system
    v (M) = √ [Gm² / (m + M)a(1-ε²/4)] ≈ 0; m<<M

    Application 1: Advance of Perihelion of mercury.

    G=6.673×10^-11; M=2×10^30kg; m=.32×10^24kg; ε = 0.206; T=88days
    c = 299792.458 km/sec; a = 58.2km/sec; 1-ε²/4 = 0.989391
    ρ (m) = 0.696×10^9m; ρ(m)=2.44×10^6m; 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) = (-720×36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ²
    W” (ob) = (-720×36526x3600/88) x (1.552) (48.14/299792)² = 43.0”/century

    V1143Cgyni Apsidal Motion Solution

    W° (ob) = (-720×36526/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

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