25 Aug '09 18:56>2 edits
r ——————>>Exp (ì w t) ———->> S=r Exp (ì wt) Nahhas’ Equation
Orbit——–>> Orbit light sensing——>> Visual Orbit; Exp = Exponential
Particle —->> light sensing of moving objects———— >> Wave
Newton———>>light sensing———->> Quantum
Quantum = Newton x Visual Effects
Quantum – Newton = Relativistic = Optical Illusions
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 = wr
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
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
Orbit——–>> Orbit light sensing——>> Visual Orbit; Exp = Exponential
Particle —->> light sensing of moving objects———— >> Wave
Newton———>>light sensing———->> Quantum
Quantum = Newton x Visual Effects
Quantum – Newton = Relativistic = Optical Illusions
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 = wr
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
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