THUNDERBOLTS OF THE GODS
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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 = (-720×36526x3600/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”= (-720×36526x3600/88) 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