8/3/2019 CERN GPS frame of reference shift error of 20 meters faster than the speed of light

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Nahhas' "Faster than the speed of light" Puzzle solution I 

Joe Nahhas Alfred Nobel Wrong

Arab American real time first physicist Joe Nahhas explains CERN Geneva - San

Grasso Opera experiment Dr Dario Autiero 2011 error of "faster than the speed of light"

Abstract: frame of reference shift from San Grasso to GPS satellite amounts to

an error of 20 meters and is not counted for in Opera experiment and Dr Dario

Autiero 20 meters "faster than the speed of light" calculations are all wrong

Proof:

1 = 1 is self evident; 2 = 2 is self evident

And r0= r 0 is self evident; add and subtract r

Then r 0 = r + (r 0 - r); divide by r

Then (r 0 /r) = 1 + (r 0 - r)/ r; multiply by f 

Then (r 0 /r) f = f  + [(r - r 0)/ r] f --------------------Eq-1 

Also, f 0= f 0; add and subtract f 0

Then f 0 = f + (f 0- f) --------------------------------Eq-2 

Comparing Eq-1to Eq-2

Then f 0 = (r0 /r) f; or f 0 /f = r0 /r; subtract 1 from both sides

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8/3/2019 CERN GPS frame of reference shift error of 20 meters faster than the speed of light

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Then (f 0 /f) - 1= (r0 /r) -; or, (f 0 - f) /f = (r 0 - r)/ r

Or, - d f/f = - d r /r; and d f/f = d r /r -------------------Eq-3 

Solving Eq-3 using separation of variables 

Then d f/f = d r /r = λ+ í ω = translation + rotation 

Or, f = f 0 e (λ+ í ω) t and r = r0 e (λ+ í ω) t 

With r = r x + í r y = r0 e λ t (cosine ω t + í sine ω t)

Then r x = r0 e λ t cosine ω t; and r y = r0 e λ t sine ω t 

And (r y / r x) = tan ω t; and ω t = tan-1 (r y / r x)

With r x = r0 e λ t cosine ω t = r0 e λ t cosine tan-1 (r y / r x)

Then (r x / r0) = e λ t cosine tan-1 (r y / r x); e λ t = (r x / r0)/cosine tan-1 (r y / r x)

And λ t = L n [(r x / r0)/cosine tan-1 (r y / r x)]

= L n [cosine tan-1 (r y / r x)/cosine tan-1 (r y / r x)] = L n 1 = 0

With r y = r0 e λ t sine ω t = r0 e λ t sine tan-1 (r y / r x)

Then (r y / r0) = e λ t sine tan-1 (r y / r x)

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r0 r y

r x

8/3/2019 CERN GPS frame of reference shift error of 20 meters faster than the speed of light

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And λ t = L n [(r y / r0)/sine tan-1 (r y / r x)]

= L n [sine tan-1 (r y / r x)/sine tan-1 (r y / r x)] = L n 1 = 0

This means decay or birth factor λ = 0 or the decay or birth factor e λ t = 1

And r = r0 e í ω t = r x + í r y = r0 (cosine ω t + í sine ω t)

What this math means?

If a car is at a distance r0; its visual is r = r0 e (λ+ í ω) t 

If r0 is the same not shopped or welded e λ t = 1; then r0 visual is r = r0 e í ω t 

With r = r0 e í ω t; divide by time t

Then velocity v = v 0 e í ω t 

And (r. v) = (r0. v 0) e2 í ω t 

Then (r. v) = (r0. v 0) (cosine 2ω t + í sine 2ω t) 

Or (r. v) = (r. v) x + í (r. v) y = (r0. v 0) (cosine 2ω t + í sine 2ω t) 

Along the line of sight (r. v) x= (r0. v 0) cosine 2ω t 

For an elliptical motion (r0. v 0) = 2 π a b/T

With a = semi major axis a and semi minor axis b and orbital period T

And r0 = a (1 - ε2)/ (1 + ε cosine θ); b = a √ (1 - ε2); ε = eccentricity 

Then (r0. v 0) = 2 π a b/T = 2 π a2 [√ (1 - ε2)]/T

At perihelion θ = 0 and r0 (perihelion) = a (1 - ε2)/ (1 + ε cosine 0°) 

And r0 (perihelion) = a (1 - ε

2

)/ (1 + ε) = a (1 - ε) 

Angular velocity at perihelion is θ' (perihelion) = (r. v) x / [r0 (perihelion)] 2 

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8/3/2019 CERN GPS frame of reference shift error of 20 meters faster than the speed of light

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And θ' (perihelion) = (r0. v 0) cosine 2ω t/ [r0 (perihelion)] 2 

= {2 π a2 [√ (1 - ε2)]/ a2 (1 - ε) 2 T} cosine 2ω t

= {2 π [√ (1 - ε2)]/ (1 - ε) 2 T} cosine 2ω t

= {2 π [√ (1 - ε2)]/ (1 - ε) 2 T} (1 - 2 sine2 ω t)

= {2 π [√ (1 - ε2)]/ (1 - ε) 2 T} (1 - 2 sine2 ω t)

And θ' (perihelion) - {2 π [√ (1 - ε2)]/ (1 - ε) 2 T}

= - 4 π {[√ (1 - ε2)]/ (1 - ε) 2 T} sine2 ω t in radians per period - seconds

Angle per second multiply by T

Angle (θ perihelion) = - 4 π {[√ (1 - ε2)]/ (1 - ε) 2} sine2 ω t radians 

In degree per second multiply by 180/ π 

= - 720 {[√ (1 - ε2)]/ (1 - ε) 2} sine2 ω t 

In arc second multiply by 3600

Angle (θ perihelion) = (- 720 x 3600) {[√ (1 - ε2)]/ (1 - ε) 2} sine2 ω t 

Per century multiply by (t0 /T) = century/period; θ perihelion = θ p 

Angle (θ perihelion) = (- 720 x 3600) (t/T) {[√ (1 - ε2)]/ (1 - ε) 2} sine2 ω t 

With t = 36526; T = 88; ε = 0.206; ω t = tan-1 (v/c) = aberration angle

And v = orbital velocity - spin velocity = 47.9

And c = 300,000; {[√ (1 - ε2)]/ (1 - ε) 2} = 1.552

Finally θ p = (- 720 x 3600) (t0 /T) {[√ (1 - ε2)]/ (1 - ε) 2} sine2 ω t 

= (- 720 x 3600) (36526/88) (1.552) sine

2

[tan

-1

(47.9/300,000)]= 43 arc second per century Google 50 solutions of Mercury's Perihelion 

Same as measured and same as Einstein validating this formula validity

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8/3/2019 CERN GPS frame of reference shift error of 20 meters faster than the speed of light

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GPS satellite has a velocity of 14000 km/hr and a period T = ½ day

and eccentricity of ε = 0 

The velocity of the GPS satellite is 14,000,000/0.5 x 24 x 3600 = 3.8888889

Earth Spin = 0.4651 and ω t = tan-1

[(3.89 - 0.4651)/300,000]

With θ p = (- 720 x 3600) (t0 /T) {[√ (1 - ε2)]/ (1 - ε) 2} sine2 ω t 

With ε = 0; θ p = (- 720 x 3600) (t0 /T) sine2 ω t 

With t0 = 1 day and T = 0.5 days

With 24 hours = 360°; or 1 second = 15 arc second

Then T d = [(- 720 x 3600)/15] (t0 /T) sine2 ω t seconds

Then T= [(- 720 x 3600)/15] (1/0.5) sine2 tan -1 [(3.89 - 0.4651)/300,000]

seconds = 0.000045043 seconds per day or per Earth revolution

Same as measured to justify this method

In radial distance: 0.000045043 x 300,000/2 π = 13.51289/2 π = 2.15 km

With San Grasso moving with respect to the satellite the orbit aberration is

With θ p = (- 720 x 3600) (t0 /T) {[√ (1 - ε2)]/ (1 - ε) 2} sine2 ω t 

With ε = 0; θ p = (- 720 x 3600) (t0 /T) sine2 ω t 

With 24 hours = 360°; or 1 second = 15 arc second

Then T error = [(- 720 x 3600)/15] (t0 /T) sine2 ω t seconds

With ω t = tan -1 [(3.89 - 0.4651)/300,000]

And (t0 /T) = (2.15/730)

T error = [(- 720 x 3600)/15] (2.15/730) sine2 tan -1 [(3.89 - 0.4651)/300,000] 

= 0.00000006635 seconds and 0.00000006635 x 300,000,000 = 19.9 meters

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