David Kirkpatrick

May 15, 2008

Nanowire solar cells and black holes

From KurzweilAI.net, nanotech that may boost solar efficiency and black holes may have an escape hatch of sorts

Nanowires may boost solar cell efficiency, engineers say
PhysOrg.com, May 14, 2008

University of California, San Diego electrical engineers have created experimental solar cells spiked with nanowires that could lead to highly efficient thin-film solar cells of the future.

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Physicists Demonstrate How Information Can Escape From Black Holes
PhysOrg.com, May 14, 2008

Physicists at Penn State and the Raman Research Institute in India have discovered such a mechanism by which information can be recovered from black holes.

They suggest that singularities do not exist in the real world. “Information only appears to be lost because we have been looking at a restricted part of the true quantum-mechanical space-time,” said Madhavan Varadarajan, a professor at the Raman Research Institute. “Once you consider quantum gravity, then space-time becomes much larger and there is room for information to reappear in the distant future on the other side of what was first thought to be the end of space-time.”

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1 Comment »

  1. Kepler (demolish)Vs Einstein’s space jail of time
    r ————– Exp (i wt) ———–S= r Exp (ì wt) Nahhas’ Equation
    Orbit location———–Orbit light sensing ————– Visual orbit location
    Particle/Newton —————- –Visual ————————– Wave/Quantum
    Quantum – Newton=visual effects=relativistic effects=space-time confusions
    S= visual distance; r = actual distance; v = speed and c = light speed
    S = r Exp (i wt) = r [cosine (wt) + î sine (wt)]
    P =d S/d t = v Exp (ì w t) + ì r w Exp (ì w t); v=d r/d t; v=w r
    = v (1+ ì) [Exp (ì wt)] = visual velocity
    E (definition) = m/2(m v + m’ r) ²; E = mc²/2 If v = 0; m’ r=mc
    E (visual) = mp²/2 = mv²/2(1+ì) ² Exp 2(ì w t)
    E (visual) = mv²/2(1 + 2ì -1) [cos2wt + ì sin2wt]
    E (visual) = ì (mv²) [1-2sin²wt + 2i [sin (wt)] [cosine (wt)]
    If wt = (2n+1) π/4
    E (visual) = ì (mv²) [1-1 ± ỉ] = ± (mc²); v = c
    2-Central force law Areal velocity is constant: r² (d θ/d t) =h Kepler’s Law
    h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
    r² (d θ/d t) = h = S² (d w/d t)
    Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] (d w/d t)
    (d w/d t) = (h/r²) exp [-2(i wt)]
    d w/d t= (h/r²) [cosine 2(wt) – ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) – ỉ sin 2(wt)]
    d w/d t = d w(x)/d t + d w(y)/d t; d w(x)/d t = (h/r²) [ 1- 2sine² (wt)]
    d w(x)/d t – (h/r²) = – 2(h/r²)sine²(wt) = – 2(h/r²)(v/c)² v/c=sine wt
    (h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²
    Δ w/d t = (d w/d t – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
    Δ w/d t = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians
    Δ w°/d t = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π
    Δ w°/d t = (-720×36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years
    Δ w”/d t = (-720×3600/T) {[√ (1-ε²)]/(1-ε) ²} (v/c) ² seconds of arc multiplication by 3600
    Δ w/d t = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century
    The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system
    Application 3: 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
    Calculations yields:
    v =48.14km/sec
    [√ (1- ε²)] (1-ε) ² = 1.552
    Δ w”= (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century
    E ≠ mc² (special-relativity) and the 43" seconds of arc of advance of perihelion of Planet mercury (general-relativity) is are caused by deformed space-time physicists "thought" and not deformed space (x, y, z).

    Anyone dare to prove me wrong?
    E (Energy by definition) = mv²/2 = mc²/2; if v = c
    m = mass; v= speed; c= light speed; w= angular velocity; t= time
    S = r Exp (ì w t) = r [cos (wt) + ì sin (wt)] Visual effects
    P = visual velocity = change of visual location
    P = d S/d t = v Exp (ì w t) + ì w r Exp (ì w t)
    = (v + ì w r) Exp (ì w t) = v (1 + ì) Exp (ì w t) = visual speed; v = w r
    E (visual energy= what you see in lab) = m p²/2; replace v by p in E = mv²/2
    = m p²/2 = m v²/2 (1 + ì) ² Exp (2ì wt)
    = mv²/2 (2ì) [cosine (2wt) + ì sine (2wt)]
    =ì mv² [1 – 2 sine² (wt) + 2 ì sine (wt) cosine (wt)];v = speed; c = light speed
    wt = π/2
    E (visual) = ìmv² (1 – 2 + 0)
    E (visual) = -ì mc² ≡ mc² (absolute value;-ì = negative complex unit) If v = c
    w t = π/4
    E (visual) = imv² [1-1 +ỉ] =-mc²; v = c
    wt =-π/4+ỉln2/2; 2ỉ wt=-ỉπ/2 – ln2
    Exp (2i wt) = Exp [-ỉπ/2] Exp [ln(1/2)]=[-ỉ (1/2)]
    E (visual) = imv² (-ỉ/2) =1/2mc² v = c
    Conclusion: E = mc² is the visual Illusion of E = mc²/2 joenahhas1958@yahoo.com. All rights reserved.
    PS: In case of E=mc² claims to be rest energy claims then
    E=1/2m (m v + m’ r) ² = (1/2m) (m’ r) ²; v = 0
    E = (1/2m) (mc) ²; m’ r =mc

    Comment by Joe Nahhas — January 26, 2009 @ 5:20 am

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