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.
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.”
David Kirkpatrick 
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
Conclusions:
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=mc²/2
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
E=mc²/2
Comment by Joe Nahhas — January 26, 2009 @ 5:20 am