Stanford Perimeter institute + All space-time Physics

+ NASA = 0 Physics

Visual Effects and the confusions of “Modern” physics

r ——————Light sensing of moving objects ——- S

Actual object————— Light —————— Visual object

r – ——-cosine (wt) + i sine (wt) – S = r [cosine (wt) + i sine (wt)]

Newton– Kepler’s time visual effects — Time dependent Newton Wave Equation

Line of Sight: r cosine wt

r ——————- r cosine (wt) line of sight light aberrations

A moving object with velocity v will be visualized by

light sensing through an angle (wt);w = constant and t= time

Also, sine wt = v/c; cosine wt = √ [1-sine² (wt)] = √ [1-(v/c) ²]

A visual object moving with velocity v will be seen as S

S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential

S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y

S x = Visual along the line of sight = r [√ [1-(v/c) ²]

This Equation is special relativity length contraction formula and it is just the visual effects caused by light aberrations of a moving object along the line of sight.

In a right angled velocity triangle A B C: Angle A = wt; angle B = 90°; Angle C = 90° -wt

AB = hypotenuse = c; BC = opposite = v; CA= adjacent = c √ [1-(v/c) ²]

And what is the visual effect for angular velocity of Perihelion of Mercury along the line of sight?

Perihelion of Mercury

Kepler (demolish) Vs Einstein’s Space-jail of time

Ending Einstein’s space jail of time in 2009 that led to fraud Symbol E=mc²

Areal velocity is constant: r² θ’ =h Kepler’s Law

h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity

r² θ’= h = S² w’

S = r exp (ỉ wt) ——————————————————————————–>

h = [r² Exp (2iwt)] w’=r²θ’

w’ = (θ’) exp [-2(i wt)]

w’= (h/r²) [cosine 2(wt) – ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) – ỉ sin 2(wt)]

w’ = w'(x) + ỉ w'(y) ; w'(x) = (h/r²) [ 1- 2sine² (wt)]

Δ w’= w'(x) – (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’ = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second

{x [180/π;degrees]x[100years=36526days;century]x[3600;seconds in degree]

Δ w” = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century

This Kepler’s Equation solves all the problems Einstein and all physicists could not solve

DI Her Binary starts systems

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

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: The 43″ seconds of arc of advance of perihelion of Planet Mercury (General relativity) is given by Kepler’s equation better than all of Published papers of Einstein. Kepler’s Equation can solve Einstein’s nemesis DI Her Binary stars motion and all the other dozens of stars motions posted for past 40 years on NASA website SAO/NASA as unsolved by any physics.

Anyone dare to prove me wrong? I just showed that:

Physicists and astrophysicists still can not read a telescope after 400 years!

]]>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

http://www.chem.leeds.ac.uk/People/CMR/images/scco24.jpg

Compare the former idea of “string net liquid”, forming the vacuum:

http://www.newscientist.com/article.ns?id=mg19325954.200

The blog of yours is excellent for me. It covers exactly the areas, into which I’m interested too.

]]>