Stanford University most stupid professor Leonard Susskind

Arabs real time AstronomyThe 30 solutions of Mercury’s Perihelion

(I/O) θ’0 = θ’0 + [(I – O)/O] θ’0Real time physics = event time physics + time delay physics

Quantum = classical + [email protected]

I am Joe Nahhas greatest and only physicist alive since 1258 AD telling idiots like the two on my left Buzz off: Time travel based “modern Physics” is not physics but Einstein’s and Western “University” academia garbage imposed in classrooms and used wrong in scientific calculations to send 500,000 “physicists” to go work at Macdonald’s without loss of subject. My name is Professor Joe Nahhas founder of Arabs real time physics July 4th 1973.

There is one and only one physics Arabs real time physics. Leonard Susskind is not only an idiot but this is the proof of his and Einstein’s stupidity

This is me in October 2009 flashing my 1979 picture stapled to my thermo book congratulating myself and buzzing off 500 years of “western” stupidity of physics and buzzing off physicists and re - introducing real time physics that Arabs in 10th century started with 1001 new Arabs real time physics formulas to change physics and history

Confronting criminals in Alfred Nobel suit is my promise

mailto:[email protected]

Greetings: My name is Joe Nahhas and I was expelled from University of Michigan in 1991 and PHD wiped for exposing criminals in Alfred Nobel suit Headquartered in USA department of energy

Frederick Reines 1953 US physicists of US DOE and Einstein’s dumbest thought E = mc2

Modern physics which is based on space – to time - back to space travel fiction is not physics but Einstein’s and Alfred Nobel criminals’ stupidity imposed in classrooms and used in scientific

calculation terrorizing students with University “Professors’” stupidity

I Joe Nahhas accuse Nobel Prize winner Dr Stephen Chu of the crime of fraud including scientific fraud money laundering in the amounts of billions of dollars along with other Nobel Prize winners. Dr Stephen Chu is shelling billions in

research money when he knows it is all fraud. Crime started with Lorentz and Einstein and the Alfred Nobel Prize Institution and USA department of energy is where the crimes happened and keep happening. In 1953 Frederick Reines rigged physics while he was working at USA department of energy to create the dark energy multi billion dollars scam and the dark energy scam have/had reached an unprecedented levels that criminals has to be confronted and Dr Stephen Chu is a criminal in Alfred Nobel suit and it is the subject of this article

Western physicists after the 1932 two chemists discovery that the heavy metal of Uranium can split releasing heat they raced for a way to speed up the splitting process to make a heat bomb we call nuclear bomb. At this stage western physicists realized that they can tell the USA congress and get all the money they want and they used visual effects to claim new discovery fooling and blackmailing Congress the American people and humanity and the time has come to expose them and who can expose them? I can!

There is no dark energy because it is based on modern physics and modern physics is made up physics and can be deleted in its entirety without loss of subject. I am that good and can show how modern physics is Garbage

Dark energy Neutrino crime:

Frederick Reines Neutrino crime

The Nobel Prize in Physics 1995 was awarded "for pioneering experimental contributions to lepton physics" jointly with one half to Martin L. Perl "for the discovery of the tau lepton" and with one half to Frederick Reines "for the detection of the neutrino". Frederick Reines and many others used visual effects to claim discovery when there is no discovery and he did it while working for America’s most wanted crime hub called US department of energy

Einstein is fraud and E = mc2/2 and not E = mc2 Terrorist Einstein talk about is dark energy visual Illusion after 106 years till can never be found

I can show you using 5th grade math that he is fraud by showing that 5th grade math can solve the same problems Einstein solved and the problems Einstein cannot solve. I can show you how USA department of energy used visual effect to verify

Einstein’s theory when his theory and the experimental proofs are nothing but fraud and a crime

Modern Physics is Garbage: I can use 5th grade math to beat Alfred Nobel physics and physicists

A = A; add and subtract B to getA = B + (A – B); divide by B to getA/B = 1 + (A – B)/B; multiply by C to get (A/B) C = C + [(A – B)/B] CIf a Car has a velocity C at a distance A and the Car is moving away; then,

the car velocity would be measured as (A/B) C at a distance B and the difference between the two velocities measurements at A and B is (A/B) C – C = [(A – B)/B] C

C = classical physics = event time measurement of C(A/B) C = real time measurements of C = quantum mechanics [(A – B)/B] C = time delay Illusion = relativisticOr Quantum = Classical + relativisticMeaning when light was discovered and used as measuring tool physicists

started seeing Illusions confused for real thing and called it modern physics. Real time illusions only use was speeding up the information age called modern technology. Everything scientist said that it exist they have no proof because it is an Illusion

(A/B) C = C + [(A – B)/B] C

This equation ends 500 years of modern physicsUniversity of Michigan time travel bus drivers and criminals like these idiots

who believe in dark energy when dark energy is all made up fraud

Chairs of Nuclear Engineering Department past 30 years

James J. Duderstadt Glenn Knoll Bill Martin Ron Gilgenbach

Arabs real time particle motion:All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z). The state of any object in the Universe can be expressed as the product S = m r; State = mass x location:

P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rate

In polar coordinates system

r = r r (1) ;v = r' r(1) + r θ' θ(1) ; γ = (r" - rθ'²)r(1) + (2r'θ' + r θ")θ(1)r = location; v = velocity; γ = accelerationF = m γ + 2m'v +m" rF = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + (m" r) r (1) = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/mr) [d (m²r²θ')/d t] θ (1) = 0 Light force lawProof:First r = r [cosine θ î + sine θ Ĵ] = r r (1)Define r (1) = cosine θ î + sine θ Ĵ Define v = d r/d t = r' r (1) + r d[r (1)]/d t = r' r (1) + r θ'[- sine θ î + cosine θĴ] = r' r (1) + r θ' θ (1)

Define θ (1) = -sine θ î +cosine θ Ĵ;And with r (1) = cosine θ î + sine θ ĴThen d [θ (1)]/d t= θ' [- cosine θ î - sine θ Ĵ= - θ' r (1) And d [r (1)]/d t = θ' [-sine θ î + cosine θ Ĵ] = θ' θ (1)

Define γ = d [r' r (1) + r θ' θ (1)] /d t = r" r (1) + r'd [r (1)]/d t + r' θ' r (1) + r θ" r (1) +r θ'd [θ (1)]/d t γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 If light mass m = constant, then With d² r/dt² - r θ'² = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2 With d² r/dt² - r θ'² = 0 E q – 1 And d (r²θ')/d t = 0 Eq-2 From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant

Differentiate with respect to timeThen 2rr'θ' + r²θ" = 0Divide by r²θ' Then 2(r'/r) + θ"/θ' = 0And 2(r'/r) = - θ"/θ' = 2[λ (r) + ỉ ω (r)]Also, r = r0 (θ, 0) e [λ (r) + ỉ ω (r)] t

And θ' = θ' 0 (θ, 0) e -2[λ (r) + ỉ ω (r)] t

For a fixed orbit: λ r = 0Also, r = r 0 (θ, 0) e ỉ ω t

And θ' = θ' 0 (θ, 0) e -2ỉ ω t

With d² r/dt² - r θ'² = 0Let u = 1/r; r = 1/u; r²θ' = h = θ’ /u² And d r/d t = (d r/ d u) (d u /d θ) (d θ/ d t) = (- 1/u ²) (θ’) (d u/ d θ) = - h (d u/ d θ)And d² r/ d t² = - h (θ’) (d² u/ d θ ²) = [- h²/r²] (d² u/ d θ ²) = - h² u² (d² u/ d θ ²)With d² r/dt² - r θ'² = 0 E q – 1 And - h² u² (d² u/ d θ ²) – (1/u) (h u²) ² = 0Then (d² u/ d θ ²) + u = 0Then u = A e - í θ

Then r 0 (θ, 0) = (1/A) e í θ And r 0 (θ, 0) = r 0 (0, 0) e í θ And r = r 0 (0, 0) e í θ e ỉ ω t

Or, r = r 0 e í (θ + ω t) This is real time particle motionLight is emitted and not an emitter and humans have no means of measuring light in motion and at best they can take a snap shot.In mathematical languageThen r = r 0 e í θ at t = 0And r = r 0 e í (θ + ω t) Then ∆ r = r 0 e í θ [1 - e í ω t]Or ∆ r = r [1 - e í ω t] = ∆ r (x) + í ∆ r (y)∆ r (x) = r [1 – cosine ω t] = - 2r sine ² ω t/2∆ r (x) = - 2r sine ² ω t/2 Eq - 3

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Dark energy Neutrino crime:

Frederick Reines Neutrino crime

The Nobel Prize in Physics 1995 was awarded "for pioneering experimental contributions to lepton physics" jointly with one half to Martin L. Perl "for the discovery of the tau lepton" and with one half to Frederick Reines "for the detection of the neutrino".

Anti-neutrino + proton ---------> neutron + positronPositron + electron --------------> Gamma + GammaGamma caught by a sensor and amplified to a signal 1 --> signal 1 on Oscilloscope 1Neutron + Cadmium --> [Cadmium eats neutron] --> coughs GammaGamma caught by a sensor and amplified to a signal 2 --> signal 2 on Oscilloscope 2

Reines and Cowan who claimed that if they install a tank of water close to a nuclear reactor building and shield it from cosmic radiation then the Neutrino can be found from seeing two γ- gamma signals.The first signal γ- gamma signal is an indication of an Antineutrino capture by a neutron and the second γ- gamma signal is an indication of a neutron capture by cadmium and excited cadmium spits another γ- gamma signal

A Water Tank with 200 liters of water and 40 CdCl2 or a mixture of tri ethyl benzene

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* --------------------------------------------------------------------- * signal 1 captured 25 μ seconds later signal 2 capturesSignal 1 and two differ by a small size captured on oscilloscope screens

Water has a refractive index of 4/3 and try ethyl benzene has a refractive index of 1.44

There was a time window of 25 micro seconds shots were taken and when a # 1 signal appeared in this 25 micro-seconds window the second window showed another signal appearing within 0- 5 microseconds

A signal moving inside a medium is expected to get delayed from 0-5 micro

seconds in a 0-25 micro second windowThe diffraction slows the motion inside a medium with refractive index n:

And r = r 0 e í (θ + ω t) With θ = 0

Then r = r 0 e í ω t Dividing by cThen T = T (0) e ỉ ω t

T = T (0) e ỉ ω t; ω t = arc tan (1/ n); n v = c; n = index of refraction; v = velocity of light in medium; c = light speed in vacuumT = T (0) [cosine ω t + ỉ sine ω t] = T (0) [1 - 2sine² (ω t/2) + ỉ sine ω t] = T (x) + ỉ T (y)T (x) = T (0) [1 - 2sine² (ω t/2)] = T (0) {1 - 2sine² {[(arc tan (1/n)]/2}}Δ T = T (x) - T (0) = - 2 T (0) sine² {[arc tan (1 /n)]/2}Δ T = - 2 T (0) sine² {[arc tan (1 /n)]/2}T (x) = T (0) cosine arc tan [1 /n)] T (x) = T (0) cosine arc tan [3/4] = 0.8 T (0)T (x) - T (0) = 0.2 T (0) Δ T = 0.2 T (0) The assumption that the free neutrino is detected at the Savannah River nuclear power plant experiment in 1953 and published in nature 1956 by Reines and Cowan and advertised by United Nations in 1958 is not only wrong but dead wrong. The claim of a second signal detection delayed 5 micro second after the pair annihilation indicating a neutron capture by Cadmium and proving the existence of the neutrino to support Wolfgang Pauli Neutrino theory and relativity theory is not justified. Here is whyProof: The gate length is T (0) = 25 μ s (microseconds)The Container that had Cadmium is basically Tri ethyl benzene mixed with water. The refractive index of water at 25°C is 1.333 ~ 4/3 and the refractive index of tri ethyl benzene is 1.44 with 5 to 1 ratio water to tri ethyl benzene which is a minor distraction to water refractive index.

Δ T = - 2 x 25 sine² {[arc tan (~ 3/4)/2]} = 5 μ s time delaysΔ T = 0.2 T (0) = 0.2 x 25 = 5 μ s time delays The coincidence of Pair annihilation gammas and Neutron capture gammas is not justified

If all the mixture was Tri ethyl benzene because the literature says it is a mixture and it say it is tri ethyl benzeneΔ T = - 2 x 25 {[sine² arc tan (1/1.44)]/2} = 4.46575 μ s time delays

The delayed signal within the 5 μ s intervals is not a sign of Neutrino capture but a flash of light caught traveled and caught delayed. Meaning it is the same gamma and not two different gammas. What Reines and Cowan did is measuring the same signal in a 25 micro second interval at the start they measured the signal and in that 25 micro seconds interval the signal had 5 micro seconds signal time delay that was measured as a second event. Reines and Cowan claimed two signals the first event was the signal of pair annihilation and the second event is the neutron capture and releasing a second gamma ray when in fact looking at the same signal. This is the basis of all dark energy fraud

California institute of technology Ahmed Zewail Femto Chemistry historical mistake is the same repeated stupidity

Γ = t ℮ ỉ θ; θ = arc tan (v/c)

The Nobel Prize in Chemistry 1999"for his studies of the transition states of chemical reactions using femto second spectroscopy"Dr Ahmed Zewail Nobel Lecture

Femtochemistry: Atomic-Scale Dynamics of the Chemical Bond Using Ultrafast LasersOr finding tIn his Nobel work Dr Zewail used Cyanogenic iodide with λ = 306 nanometer; 1 nanometer = 10 - 9 and watched it fluoresces at λ = 388.9; t = 200 femto seconds; 1 femto second = 10 - 15 second He shot a Laser beam to stir a solution that contained crystal of Cyanogenic iodideThen every 10 femto seconds he shot again to see what happens and after the last shot he concluded the stir is dead and Cyanogenic iodide brokeICN ----------------> CN + I

The index of refraction of Cyanogenic iodide is n = 1.64 [US Army articles on the net]With λ = c T: T (1) = 306 x 10 - 9/3 x 10 8 = 102 x 10 - 15

With λ = c T: T (2) = 388.9 x 10 - 9/3 x 10 8 = 129.633 x 10 - 15

Δ Γ = T (1) - T (2) = - 27.6333 x 10 - 15

And Δ Γ = - 2 t sine ² arc tan (v/c)And Δ Γ = - 2 t sine ² arc tan (1/n)

And -2 sine ² arc tan (v/c) = - 0.1455And t = Δ Γ / - 2 t sine ² arc tan (1/n) = - 27.6333 x 10 - 15/ - 0.1462 = 190 x 10 - 15

Or t = 190 x 10 - 15

The last shot is just a wasteDr Zewail number is 200 +/- 30 femotsecondsOther Nobel Prize winner used the same visual trick at Micro scaleThen Nano technology came and used the same trick at nano scale And then Pico technology came and used it at pico scaleI am not saying Dr Ahmed Zewail is wrong what I am saying is Dr Zewail and all others are wrong and Dr Zewail is another Nobel and no Nobility.

But that is not all Einstein used the same crime

Lord Eddington used the same crime

So did Copernicus Galileo, Tyco Brahe, Kepler, Newton, Hooke, Le Verrier, Laplace Von Soldner, ………….

The entire western civilization 500 years of physics

The entire western civilization 500,000 scientists used the same crime

All rights reserved

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Chapter one: What is Arabs real time astronomy?

Car size at distance 4




Observer line: …………………Observer ………………If we stand in the middle of the street and watch a car moving directly away from us we see as if the size of the car shrunk. If we take a picture of the car at different distances then we see that the pictures of the car look like as if the car contracted in size. The further the car away from us the smaller image of the car in the picture. The size of the car did not change but real time light data reflected from the car and ran through air and into our eyes changes the real time image of the car to indicate a different location of the car and not a different size of the car. Our eyes see real time images of a moving car. If the car stops, then the image of the car freezes. A picture or a snap shot of a moving car is a frozen image of the moving car. When a car moves its real time image change size but the car does not change size. The car as a moving object at different distances has a variable real time image that our eyes see in real time and depends on distance and motion as perceived by our eyes. In short statementsOur eyes see objects in real time as different objects at different distances and different speeds. Or, our eyes see real time objects we call images that change when measured from different distances or when moving. Our eyes do not see actual objects but Images of actual objects and our eyes see different images of same actual objects when at different distances of moving In mathematical form:

Page 2

1 = 1 self evident2 = 2 is self evidentImage = image = I is self evident line - 1 Object = object = O = self evident line - 2And, - Object = - objects = - O = self evident line - 3I = I line - 1O = O line - 2And, - O = - O line - 3 Adding: line - 1 and line - 2 and line - 3I + O – O = O + (I – O) line - 4I = O + (I – O); O – O = 0 one left side

(I/O) = (O/O) + (I – O)/O; dividing by on both sides line - 5 Or, (I/O) = 1 + (I – O)/O; O/O = 1 line - 6 (I/O) O = O + [(I – O)/O] O Line - 7 Real time object = event time Object + time delay objectReal time astronomy = event time astronomy + time delay astronomy



R1 R2 O2 Car size at distance 2

D D D R1 O1 d

R1 Car size at distance 1

Observer line: …………………Observer ………………The image of R1 is R2; or R2 in real time is R1

Page 3

Let us do R1 real time analysis R2 = R2

R2 = R1 + (R2 – R1)Dividing by R1

(R2/R1) = (R1/R1) + (R2 – R1)/ R1Visual measurement scale(R2/R1) = 1 + [(R2 – R1)/ R1]

Distance R1 when measured using light source and not a measuring tape is not the same as R1 but is measured as R2. R2 is the visual distance of R1. Using an eye as an instrument

visual distance R2 is smaller than actual distance R1. The percentage of visual distance R2 to actual distance R1 is equal to (R2/R1).

(R2/R1) = 1 + [(R2 – R1)/ R1] = 1 + [(R2/ R1) – 1] Percentage = 1 + [percentage – 1]

Multiplying by R1

(R2/R1) R1 = R1 + [(R2 – R1)/ R1] R1Visual = actual + (Visual – actual)

In clear words when we try to measure the distance R1 using light source reflected from the car we see a percentage of R1 which is (R2/R1) R1 and that percentage times R1 is the

visual we measure R2 Real time distance = event time distance + delay time distance

Visual distance = distance + visual effect Visual measurement scale using distanceReal time astronomy scale using distance

(R2/R1) = 1 + [(R2 – R1)/ R1]Visual measurement scale using VelocityReal time astronomy scale using distance

(V2/V1) = 1 + [(V2 – V1)/ V1]Visual measurement scale using angleReal time astronomy scale using angle

(θ 2/ θ 1) = 1 + [(θ 2 – θ 1)/ θ 1]Visual measurement scale using angular velocityReal time astronomy scale using angular velocity

(θ’ 2/ θ’ 1) = 1 + [(θ’ 2 – θ’ 1)/ θ’ 1]Visual measurement scale using timeReal time astronomy scale using time

(t 2/ t 1) = 1 + [(t 2 – t 1)/ t 1]

This formula is real time astronomical scale(V/A) = 1 + [(V – A)/ A]

(Visual/Actual) = 1 + [(Visual – Actual)/ Actual]

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Arabs real time astronomy application(I/O) θ’0 = θ’0 + [(I – O)/O] θ’0

Real time physics = event time physics + time delay physics

In old termsQuantum = classical + relativistic

(R2/R1) = 1 + [(R2 – R1)/ R1](R2/R1) θ’0 = θ’0 + [(R2 – R1)/ R1] θ’0

This is Arabs real time astronomy distance scale to measure the angular velocity θ’0

(V2/V1) = 1 + [(V2 – V1)/ V1](V2/V1) θ’0 = θ’0 + [(V2 – V1)/ V1] θ’0

This is Arabs real time astronomy Velocity Scale to measure the angular velocity θ’0 Planet Mercury angular velocity accumulation in arc second per hundred years

Or, (V2/V1) δ θ’0 = δ θ’0 + [(V2 – V1)/ V1] δ θ’0Planet Mercury Angular velocity θ’0 = 2 π/Tm = 2(3.1415)/88 x24x3600Tm = Planet Mercury orbital period = 88 days = 88 x 24 x 3600 seconds Planet Mercury angular velocity accumulation in arc second per century is: δ θ’0 = (2 π/ Tm) (1 century/Tm) (180/ π) (3600)

Or, δ θ’0 = (2 π/ Tm) (1 century/Tm) (180/ π) (3600) = (2x3.1415/88x24x3600) (36526days/88days) (180/3.1415) (3600) = 70.75

(R2/R1) θ’0 = θ’0 + [(R2 – R1)/ R1] θ’0(R2/R1) δ θ’0 = δ θ’0 + [(R2 – R1)/ R1] δ θ’0

What we see = what there is + visual illusionModern astronomers see = 70.75 + 43What there is = 70.75The visual Illusion is = 43

[(R2 – R1)/ R1] δ θ’0 = 43Page 5

As an example: when Astronomers look at planetary motion around the sun and they saw the angular velocity in arc second per century:Equal to δ θ’ m = (v m /r m) (180/ π) [36526/T m (days)] (3600) = 70.75 arc second per centuryAstronomers also saw the visual effect Of: [(r m - r0)/r0] δ θ’ m = [(r m - r0)/r0] (v m /r m) (180/ π) [36526/T m (days)] (3600)And it is equal to [(r m - r0)/r0] δ θ’ m = 43Where r m = Mercury – Sun distance = 58.2 X 106kmAnd v m = Mercury around the sun orbital speed Sun distance = 47.9 km/secAnd r 0 = r e = Earth – Sun distance = 149.6 X 106kmAnd T m = Mercury around the sun orbital period

When astronomers measure the angular velocity from two different distances they make the following error: [(r m - r0)/r0] δ θ’ m = 43 When astronomers measure Planet mercury angular speed around the Sun from Earth r0 and not from the sun r m they make the following errorError: [(r m - r0)/r0] δ θ’ m = 43 Arabs real time astronomy is:

Real time astronomy = event time astronomy + time delay astronomyVisual = actual + IllusionsVisual = actual + (visual – actual)Quantum = classical + relativisticReal time classical = classical + relativistic illusions Quantum mechanics and relativity theory is 20th century Garbage for physics.Measurements = actual + errorsActual + errors = actual + errors

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Chapter TwoThe confrontation between Arabs real time Astronomy andWestern Nobel prize winner modern space time astronomy

Arabs astronomers:

0 M Mercury S Sun

Planet Distance rX 106km

PlanetOrbit T

Orbit speed v in km/sec

Less Earth speed

Spin speedkm/sec

Angular velocity; v/r arc sec/ century

Mercury 58.2 88 47.9 18.1 .002 70.75Venus 108.2 224.7 35.05 5.7 6.52 10.86Earth 149.6 365.26 29.8 .46511 4.1Mars 227.936 687 24.14 0.2411Jupiter 778.412 4333 13.06 12.6Saturn 1,426.725 10760 9.65 9.87Uranus 2,870.97 30690 6.80 2.59Neptune

4,498 60180 5.43 2.68

Pluto 5906.4 90730 4.74

That:

1- All object classified as planets Suns Moons has a real time circular motionAs: r = r 0 e í (θ + ω t) And not1 – Kepler - Newton’s ellipse r= [a (1 - ε²)/ (1 + ε cosine θ)] Or Einstein’s rotating ellipse: r= a (1 - ε²)/ [1 + ε cosine (θ - Ψ)] 2- The difference between two real time circular motions is a visual rotating ellipseOr, the difference between r1 = r 10 e í [θ (1) + ω (1) t] and r2 = r 20 e í [θ (2) + ω (2) t] Is Arabs real time ellipse: r= [a (1 - ε²)/ (1 + ε cosine θ)] e í [θ + ω t]

3 - Suns moons planets do not rotate around each other That is: r= [a (1 - ε²)/ (1 + ε cosine θ)] e í [θ + ω t]

Is a visual ellipse 4 – Gravity does not exist between planets moons and sun That Newton’s F = - G m M/r² is wrongThat Einstein’s F = - G m M/r² + k/r4 is wrong5 – Gravity is due to atmosphere density and spin periodThat G = G [Ρ, T°] = G [atmospheric density, spin period]

In short: Arabs real time astronomy says

500 years ago Europeans took 10th century Arabs real time Astronomy and changed it to Space- time astronomy and in the process physics was changed from an Arabic science to European – western stupidity globalized through planet of the Apes Institution and best known as Alfred Nobel Prize winner Institution

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Alfred Nobel Prize winner physics theoryApe Man Modern Physicist Nobel Prize winner physicist Ape Ass.

Alfred Nobel: The Man behind the Nobel Prize; Joe Nahhas: The Man after Alfred Nobel Institution head Since 1901, the Nobel Prize has been honoring men and women from all corners of the globe for outstanding achievements in physics. The foundations for the prize were laid in 1895 when Alfred Nobel wrote his last will, leaving much of his wealth to the establishment of the Nobel Prize. But who was Alfred Nobel? Meet Alfred Nobel - scientist, inventor, entrepreneur, author and pacifist. What is modern physics? Did anyone hear the big bang? Did anyone saw a black hole? Did anyone find Einstein’s gravity waves? Did anyone hook up a hose to the sky and found dark energy to verify E = mc2? Does anyone have a proof of Newton’s action at a distance? Does anyone have a proof of Einstein time travel? Does anyone figure out why humans have to accept Alfred Nobel prize winners claim that the universe is made up of particles that dies and rebirth billion billion times a second? Hey dude I am a redneck one grandfather was a damn farmer and my other grandfather was a damn Carpenter and my father was a Carpenter by inheritance but I never knew what a dumb ass really is till I met my science teacher! What did Alfred Nobel honor? The difference between an Alfred Nobel Prize winner physicists and fortune tellers is a flip of a coin. Space - Time is not physics but Alfred Nobel physicists’ flip of a coin between a crime and stupidity. Alfred Nobel Institution honored criminals in Alfred Nobel suit for all of past century. Modern space – time physics changed physics from science to stupidity and physicists from researchers to criminals. In writing: Yes I am after Alfred Nobel Institution head on my plate and I accept nothing less. Space – time physics is stupidity itself. Ending Nobel Prize for stupidity from classrooms and scientific calculations is the only Nobel thing left to do. Alfred Nobel Prize became 1001 times more worth in cash value. Big Institutions backed wrong physics and physicists for all of past century for a prize and a billion dollars jack pot and garbage for physics. I am Joe Nahhas and I am unknown to the world of physics but I have 1001 new physics formulas to cut Alfred Nobel Institution head and delete Alfred Nobel Prize winner physics and physicists without loss of subject. This book is just the beginning and it is prime time and it is Arabs real time astronomy

Page 8

Chapter three: 10th century Arabs real time astronomyArab Physicists of Baghdad of 950 Ad had said:

I: Measurements are made along the line of sight or line of measurement

Arabs in 950 AD said that light going vertical (blue) to the eye it will go through the eye to be seen. The Arabs added that light coming oblique (pink) to the line of sight (blue line) we would see the part of light (pink) that is equal to the amount of light of horizontal amount (blue) or its vertical projection. What the Arabs concluded is that human eye and machines are unidirectional that can only recognize the amount of light along the line of sight or line of measurements. The Arabs mean that any point on the oblique pink line will be measured as its vertical projection as the blue line. The Arabs added that the horizontal projection of the pink line is the amount that would be measured as Zero or not seen at all. A point location on the pink line would be measured as equal to its vertical projection on the blue line and that part would be seen detected and measured by the eyes and instruments and its horizontal projection would not be detected or measured by the eyes or instruments and would be completely missed.

r0 sine ω t r0

ω t r0 cosine ω t The angle between the pink line and the blue line is called light aberration angle. The aberration angle between the line where an object is (pink) and an object projection is with the eye as center. Light aberrations angle equals to: ω t = arc tan (v/c) where v = relative velocity of the moving object and c is light velocity. The light coming from a distant object is seen along the line of sight and that meant we do not measure actual distance of an object r0 (pink line) but we measure its image or vertical projection which is r0 cosine ω t (blue line) and we escape measuring the horizontal projection r0 sine ω t which is perpendicular to the line of sight (black line).

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In Short, light coming from objects at distance r0 will be seen at r0 cosine ω tUsing real time mathematics to explain what we can see and/or detect or/and measure is what the Arabs have/had said and that would mean that if we try to measure a distance r0 we would not find r0 but we would find: Arabs real time distance is seen in real time along the line of sight.

Or, r = r0 cosine ω t + í r0 cosine ω t = r0 e í ω (r) t Eq - 1 Or r = real (blue = along line of sight) + í imaginary (black = perpendicular to line of sight) Or, r = r0 e í ω (r) t. The absolute value of r = the absolute value of r0

This statement said that light coming from objects when measured it is measured along line of sight and that meant we see light projections of objects (blue line) of objects positioned at positions r0 (pink). We do not see objects at r0 but see images of objects light projections and images of objects light projection along the line of sight is r0 cosine ω t and images of object light projections perpendicular along the line of sight is r0 sine ω t (blue line).

The Arabs said that we do not see or measure light reflected from objects but we measure light speed projections as constant or light speed projections is measured as constant.

c c c

The Arabs claim that we do not see objects position r0 but we see their visual position r. What humans see is r0 measured along the line of sight and miss r0 perpendicular to the line of sight. In other words an object at a distance r0 humans are not aware that they do not see r0 but they see a visual of r0 which is r = r0 e í ω (r) t and projection of r0 along the line of sight that humans eye see and instruments measure is r0 cosine ω t and the projection of r0 perpendicular to the line of sight is r0 cosine ω t. The Arabs added that when accounted for, the difference between an object image or visual position r (quantum mechanics) and object location or position r0 (classical mechanics) is equal to a time delay effect (relativistic mechanics)

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Modern physics is based on 3 different mechanics classical quantum and relativistic mechanics and their combination known as strings

Arabs said that there is one and only one mechanics

Quantum = classical + relativistic

Arabs real time dependent mechanics (quantum mechanics): r = r0 e í ω (r) t

Arabs real time independent mechanics (classical mechanics) is r = r0; t = 0 The difference between r = r0 e í ω (r) t; and r = r0 Or, r0 e í ω (r) t - r0 = Arabs real time mechanics visual effects (relativistic mechanics)

Or r = r = visual distance and r0 = actual distance Or, r = r0 + (r - r0)

Or visual distance = actual distance + (visual distance – actual distance)Dividing by r0And r /r0 = r0/r0 +(r – r0)/r0

And r /r0 = 1+ (r– r0)/r0What are wrong in all of modern physics are all wrong physics measurementsWhat this law said is:If the car width is r0 moves away the visual of the car width is r. The quantity r/r0 is the percentage that humans eye see and instruments measure and it is equal to 1 and the percentage difference between the car width and what the human eye see of the car width or a time delay effect or a visual distance effect of (r– r0)/r0 If someone attempts to measure a quantity θ’0; then multiply by θ’0And (r /r0) θ’0= θ’0 + [(r - r0)/r0] θ’0

What this formula said everything about modern astronomy is all wrongWhat is wrong with all of astronomy? If we to measure the angular velocity of moving planets around a star ,or, a star moving around another star in the skies θ’0, then we do not measure actual θ’0 but we measure θ’0 + [(r - r0)/r0] θ’0Or we have antidote to western space – time stupidity [(r - r0)/r0] θ’0

This is the basis of correct measurements. The entire western civilization 500 years of physics and the world 500,000 “modern” western civilization educated physicists made/make this mistake and globalize their stupidity and to correct this mistake is to delete western civilization physicists without loss of subject.There is nothing in “modern” physics that can be considered as correct because modern physics is based on wrong measurements. Modern physics is “wrong time physics” that can be deleted without loss of subject. Arabs real time physics is the past present and future of all of physics and it is the subject of this book.

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Arabs real time astronomy: The Arabs Astronomers and Physicists said that:We can not see or/and detect or/and measure something in space that did not exist; we can not see or/and detect or/and measure something that had not happened. We can only see or/and detect or/and measure things and events in space that had happened. What we see or/and detect or/and measure is not necessarily what there is and not necessarily what happened. We see or/and detect or/and measure traces and images of objects of past event. We see or/and detect or/and measure in present time of images of object and events that happened in past time. That is we measure images and record data of objects of past events in

present time. An event changes in time and when an event is measured it is measured after it happened.

Present time = present timePresent time = past time + (present time – past time)

Present time = past time + time difference

Present time = real time; past time = event time; (present time – past time) = time delay, or, Real time = event time + time delay

Real time Astronomy = Event time Astronomy + time delays Astronomy

In Modern Physics language the Arabs would have saidQuantum = classical + relativistic What there is = classicalWhat we see and measure and detect and partially record is = quantum What we see and measure and has no detection because it is an Illusion = relativisticTime dependent measurements = time independent measurements +(Time dependent measurements – time independent measurements)

Measured = actual + (measured – actual) We do not see object in real time but we see images of objects in real time In common Terms: Image = object + (image – object)

With Image = I; object = OI = O + (I – O)I/O = O/O + (I – O)/OI/O = 1 + (I – O)/OArabs real time measurements equation:Is: (I/O) q = q + [(I – O)/O] q; q measured quantity ------------- Eq – 1That is a quantity q when measured in real time it would appear as q + [(I – O)/O] q 500 years of modern physics and 500,000 of modern physicists is all wrong because of this principle (I/O) q = q + [(I – O)/O] ----------------------- Eq - 1Or quantum = classical + relativistic If Eq – 1 is applied it would expose modern physics as stupidity itself and it would expose modern astronomy as astrology itself. The thing about this principle given by Eq – 1: (I/O) q = q + [(I – O)/O] q is that it is flawless physics law or truth itself

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There are objects at location r0 (actual or classical distance) and when measured in real time they appear as at a location r (quantum distance) = r 0 e í ω (r) t

That is the distance in the Universe is a classical distance but when a distance measured in

real time it appears as quantum distance and the difference between classical distance and quantum distance is relativistic distance

Quantum (real time distance) = classical (event distance) + [Quantum – classical]Quantum distance = classical distance + relativistic distanceRelativistic distance = quantum distance – classical distance

We see detect and measure quantum distance r = r0 e í ω (r) t of an actual or classical distance r0 and the difference between quantum distance r = r0 e í ω t and actual or classical distance r0 is relativistic distance Take r = r0 e í ω (r) t

Quantum (real time) = classical (event time) + [Quantum – classical]

Quantum time = classical time + relativistic time

Relativistic time = quantum time – classical time

We see detect and measure Arabs real time physics

Or, t = t0 e í ω (t) t of an actual or classical time t0 ------------------ Eq - 3

And the difference between Arabs t = t0 e í ω (t) t and actual or classical time t0 is relativistic time

Quantum = classical + [Quantum – classical]Quantum = classical + relativistic time

Relativistic = quantum – classical

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z). The state of any object in the Universe can be expressed as the product S = m r; State = mass x location is all of astronomy needed Equation -1That is if mass of objects in space is known and if the location of object in space is know then the state S = m r = mass x location of any object in space can be known and that is all needed to be known to describe space motion of any object and that is what is called astronomyThat is mass m location r and time t = Astronomy and from this we can conclude:

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Quantum (mass) = classical (mass) + [Quantum – classical]Quantum mass = classical mass + relativistic mass

Relativistic mass = quantum mass – classical massWe see detect and Arabs real time mass m = m0 e í ω (m) t ------------- Eq- 4 of an actual or classical time m0 and the difference between Arabs real time mass m = m0 e í ω (m) t and actual or classical time m0 is Arabs real time mass delaySuch a claim by the Arabs that Astronomical measurements are made along the line Is saying Quantum mass = classical mass + relativistic mass

And saying (I/O) q = q + [(I – O)/O] q; q measured quantity Eq - 1And real time distance r = r0 e í ω (r) t Eq - 2

And real time t = t0 e í ω (t) t Eq - 3And real time mass m = m0 e í ω (m) t Eq - 4And S = m r, the state equation Eq - 5 That is Arabs see detect and measure quantum distance quantity r = r0 e í ω (r) t

When there is classical distance quantity r0 And the difference explains that the universe is seen in real time and not in silly Western ideas of wrong space time. That is the difference between an object location and an object measurement of its location is “relativistic” and that 500 years of “western” Modern physics can be deleted without loss of subject and 500,000 “Modern” physicists can be told idiots with no fear or regrets.That is Arabs detect and measure quantum time quantity t = t0 e í ω (t) t

When there is classical time quantity t0 That is Arabs see detect and measure quantum mass quantity m = m0 e í ω (m) t

When there is classical mass quantity m0 What is wrong in all of Modern physics is this experimental mistake:[(I – O)/O] q in everything that western modern space – timers do from space science to nuclear structure and this mistake was made by every physicist. This painful mistake is the one mistake when corrected it deletes western civilization physics and physicists explain the universe correctly and matches experimental results with unprecedented accuracy to delete “modern” physics of past 500 years and delete 500,000 “modern” physicists without loss of subject. What I am going to do1- Show how this one mistake was/is made daily by all western civilization educated “modern” physicists2- How this one mistake won the majority of Nobel prizes and changed physics from 1258 Arabic science to 2010 western educated stupidity3 – Show how correcting this mistake will solve any and all unsolved physical motion problems that is not solvable by space – time physics or any read or published physics

Arabs real time physics and astronomy is the past present and future of all of physics

Page 14

Chapter four





Observer line: …………………Observer ………………Humans do not see detect or measure cars but humans see detect and measures image of the car. In other words humans see detect and measure images of objects in real time.

Page 15

The size of the car object is the same but the size of the car image is dependent on car distance and transmitted signal. Meaning we see detect and measure an image of size I in real time (present time) of an object O seen detected and measured in event (past time). That is when images of cars are seen these images of the cars are seen in real time. And when these car objects move the car see itself the same size. That is we measure images of object in real time and not objects. The difference between images and objects are visual illusions Arabs knew about and described in 10th century Ad. Conclusion: Is that humans see detect and measure images of objects in real time

Why all Modern Physics and modern astronomy is all wrong?With Image = I; object = OI = O + (I – O)I/O = O/O + (I – O)/OI/O = 1 + (I – O)/OAnd (I/O) q = q + [(I – O)/O] q; q measured quantity --------------------- Eq - 1

Modern physics is all wrong because Modern physics is wrong time physics based on wrong physics measurements that misses by a factor of [(I – O)/O] q and this quantity [(I – O)/O] q is taken as real experimental result and not as a relativistic illusion correction. In other words:Correct Physics is: (I/O) q = q + [(I – O)/O] q

Modern Physics is wrong because(I/O) q = q + [(I – O)/O] q

What we measure = what there is + visual effectsQuantum = classical + relativisticQuantum = we see in real time = observer dependent = time dependent measurementsClassical = object independent of observer = time independent measurementsRelativity = illusions In mathematical terms:Relativity = time dependent measurements – time independent measurements Real time scale

Real time = event time + (real time – event time)Real time = event time + time delay; time delay = (real time – event time)With real time = Γ; event time = t; time delay = (Γ – t)Γ = t + (Γ – t) = t + ∆ Γ; ∆ Γ = (Γ – t)Γ/t = t/t + (Γ – t)/tΓ/t = 1 + (Γ – t)/t To measure orbital speed using time, then multiply by orbital speed to get the following equation: (Γ/t) θ’0= θ’0 + [(Γ – t)/t] θ’0

And θ’0 = orbital speed The quantity θ’0 measurement is independent of time The quantity (Γ/t) θ’0 measurement is time dependent

The difference = [(Γ – t)/t] θ’0 = relativistic = Illusions

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With (Γ/t) θ’0= θ’0 + [(Γ – t)/t] θ’0

Modern physics is wrong when: In general1 - (Γ/t) θ’0= θ’0 Quantum = classical or when someone looks at the car from different distances and wonder why he/she does not see the same size car.

An observer sees a car passing by at close distance

An observer sees a car passing by at far distance

Each observer will find the car size different

2 – Modern physics is wrong when (Γ/t) θ’0= [(Γ – t)/t] θ’0

When quantum = relativistic

3 – Modern physics is wrong when θ’0= [(Γ – t)/t] θ’0

When classical = relativistic

Let us demonstrate the power of Arabs real time Physics and astronomy and see how Arabs real time astronomy works but first let us do some definitions.

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Real time distance scaleMeasured distance = actual distance + [measured distance – actual distance]Or, r = r0 + (r x – r0)And r /r0 = r0/r0 +(r – r0)/r0

And r /r0 = 1+(r– r0)/r0And (r /r0) θ’0= θ’0 [(r - r0)/r0] θ’0

Real time Velocity Scale vWith r = r0 + (r – r0) Divide by tThen r/t = r0/t + (r/t – r0/t)And v = v0 + (v – v0); dividing by v0And v/ v0 = 1 + (v – v0)/ v0Multiplying by θ’0And (v/ v0) θ’0= θ’0+ [(v – v0)/ v0] θ’0

Real time acceleration scale γWith r = r0 + (r – r0) Divide by t²Then r/ t² = r0/ t² + (r/ t² – r0/ t²)

And γ = γ 0 + [(γ – γ 0)]; dividing by v0

And γ / γ 0 = 1 + [(γ – γ 0)/ γ 0]Multiplying by θ’0

And (γ / γ 0) θ’0= θ’0+ [(γ – γ 0)/ γ 0] θ’0The angular speed is a 5th grade math that is measured wrong by Astronomers expose the stupidity of modern physics and the stupidity of modern astronomy The circumference of a planet circular motion is C = 2 π r m where r m is the radius. In 1969 I knew about circular speed v m = 2 π r m/ T m where Tm is the period of rotation. Also, in 1969 I knew about angular speed θ’ m = 2 π /Tm = (v m/ r m). Or C = 2 π r m; v m = 2 π r m/Tm; θ’ m = 2 π /Tm = (v m/ r m)

θ’ m = 2 π /Tm= (v m/r m)

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Angular velocity is θ’ m = 2 π /Tm= (v m/r m)In arc second per century δ θ’ m = (v m/r m) (180/ π) [36526/Tm (days)] (3600)

Arabs Real time Astronomy distance scale:

Is: (r m /r0) θ’ m= θ’ m [(r m - r0)/r0] θ’ m

Multiplying by δ θ’ m

Then (r m /r0) δ θ’ m= δ θ’ m + [(r m - r0)/r0] δ θ’ m

In arc second per century δ θ’ m = (v m/r m) (180/ π) [36526/Tm (days)] (3600)The distance Error is: [(r m - r0)/r0] δ θ’ m

The velocity Error is: [(v m - v0)/v0] δ θ’ m

The acceleration error is: [(γ m – γ 0)/ γ 0] δ θ’ mThe angular velocity real time distance scale Error:Is: [(r m - r0)/r0] δ θ’ m = [(r m - r0)/r0] (v m/r m) (180/ π) [36526/Tm (days)] (3600)The angular velocity real time velocity scale Error:Is: [(v m - v0)/v0] δ θ’ m = [(v m - v0)/v0] (v m /r m) (180/ π) [36526/T m (days)] (3600)The angular velocity real time acceleration scale Error:

Is: [(γ m – γ 0)/ γ 0] δ θ’ m = [(γ m – γ 0)/ γ 0] δ θ’0 (v m /r m) (180/ π) [36526/T m (days)] (3600)As an example: when Astronomers look at planetary motion around the sun and they saw the angular velocity in arc second per century:Equal to δ θ’ m = (v m /r m) (180/ π) [36526/T m (days)] (3600)

= 70.75 arc second per centuryAstronomers also saw the visual effect Of: [(r m - r0)/r0] δ θ’ m = [(r m - r0)/r0] (v m /r m) (180/ π) [36526/T m (days)] (3600)And it is equal to [(r m - r0)/r0] δ θ’ m = 43Where r m = Mercury – Sun distance = 58.2 X 106kmAnd v m = Mercury around the sun orbital speed Sun distance = 47.9 km/secAnd r 0 = r e = Earth – Sun distance = 149.6 X 106kmAnd T m = Mercury around the sun orbital period

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This quantity of: [(r m - r0)/r0] δ θ’ m = [(r m - r0)/r0] (v m /r m) (180/ π) [36526/T m (days)] (3600)And it is equal to [(r - r0)/r0] δ θ’ m = 43 is taken as the proof of the existence of silly Modern Astronomy notion that the Universe is not made of three dimensional Space (x, y, z) = (length, width, height) but made of a more complex space – to imaginary time – back to space Universe (x, y, z, it) = (length, width, height, imaginary time)

[(v m – v e)/v e] δ θ’0 = [(v m – v e)/v e] (v m /r m) (180/ π) [36526/T m (days)] (3600)And it is equal to [(r - r0)/r0] δ θ’ m = 43[(γ m – γ e)/ γ e] δ θ’ m = [(γ m – γ e)/ γ e] (v m /r m) (180/ π) [36526/T m (days)] (3600)[(γ – γ 0)/ γ 0] δ θ’ m= 43

Space – time is Einstein’s crime against physics


PlanetOrbit T


Less Earth speed

Spin speedkm/sec



4,498 60180 5.43 2.68

Pluto 5906.4 90730 4.74

I am the Greatest Nuclear Physicist of all time And I have 1001 all new physics formulas to prove it

My name is Joe Nahhas and I am the greatest Physicist and Astronomer of past present and the future because as a teenager I stood up to criminals with Nobel prizes and this book is how to expose criminals with Nobel prizes in physics and the fact that most criminals with Nobel prizes are a product of the big western institutions has nothing to do with science but has something to do with western capitalistic greed fraud and arrogance that changed physics from science to stupidity and the time has come to delete stupidity and that is to delete western physics and physicists without exception since the rise of western civilization 500 years ago starting with Copernicus. There is nothing correct in all western civilization physics that is taught in classrooms and used in scientific calculations. Yes, I am saying that all of physics is all wrong and this book is the proof of it and there is nothing better than to start with the grand master of stupidity, German – American moron of the 20th century, the one and only Albert Einstein.

Page 20

Einstein greatest achievement in space said Planet Mercury travel in space at an angular speed of 70.75 arcs second per centuryAnd in time at angular speed of 43 arcs second per centuryThe Advance of Planet Mercury’s Perihelion given by Euclid’s Geometry [2 π/T] [180/π][36526/88][3600] = [2 π /88 x 24x3600] [180/π][36526/88][3600] = 70.75 Einstein’s time travel space – time formula: ψ = - 6 π GM/a c² (1 - ε²) [180/ π] [36526/T][3600] = 43 angular velocity time travel of planet Mercury G = 6.673 x 10-11= gravitational acceleration constantM =2x1030 kilogram = mass of the Sun Eccentricity of Mercury’s orbit = ε = 0.206T= 88 days = period of Planet mercury’s rotation around the SunAnd c = 299792.458 kilometer/second =light speed in kilometers/secondAnd a = 58.2 x 106 km/sec = semi major axis of planet Mercury orbit

The grand moron of the west said Planets travel in space and time

The Grand moron said planets travel in space around the Sun in an elliptical orbit

The Grand moron said planets travel in time around the Sun rotating the elliptical orbit

21

Planet

r

Sun

This is space travel And the motion of a planet m around the Sun M measured as a rotating ellipse

This is time travelPage 22

θ

r

Sun

Mercury

M

ψ

m

θ

Unless this moron can prove time travel he can keep his German stupidity to himself and Alfred Nobel Institution because time travel is Einstein’s and Alfred Nobel Institution stupidity

I am not saying Einstein is a moron but what I am saying is all western physicists are morns to say the least and can be deleted without loss of subject to be exact. Practicing again If 1= 1

2 = 23 = 3

Then we can write anything that is equal to itself as equal to itself like: Visual = Visual

Actual = actual- Actual = - actual

----------------------- AddVisual = Actual + (Visual – actual)

Divide by actual Visual/actual = Actual/actual + (Visual – actual)/actual

OrVisual/actual = 1 + (Visual – actual)/actualImage/object = 1 + (Image – object)/object

I/O = 1+ (I – O)/OWe do not see object O but we see Image I. If we to look at planet Mercury from Earth and not from the Sun, then we are not looking at θ’ m= 2 π /T m = (v m /r0) but we are looking:At θ’0 I/O = 1 θ’0+ [(I – O)/O] θ’0Or θ’0 I/O = θ’0+ [(I – O)/O] θ’0The visual illusion and Modern Physicists confusion is: [(I – O)/O] θ’0Where I is Mercury – Sun distance = 58,200,000 kilometers = r mAnd O is Earth – Sun distance = 149,600,000 kilometers = r eThe Visual Illusions and modern Physicists confusion of Planet Mercury is: [(I – O)/O] θ’0 = [(r m – r e)/ r e] θ’0

Page 23

Multiplying by (t/t) = 1

[(I – O)/O] θ’0 = [(r m/t – r e/t)/ r e/t] θ’0Then [(I – O)/O] θ’0 = [(v m – v e)/ v e] θ’0 [(I – O)/O] θ’0 = [(r m – r e)/ r e] θ’0 = [(r m – r e)/ r e] (v0/r0)[(I – O)/O] θ’0 = [(v m – v e)/ v e] θ’0 = [(v m – v e)/ v e] (v0/r0)With v0 = v m; and r0 = r mWith r m = 58.2 x 109 m; r e = 149.6 x 109 m; v m = 47.9 km/s; T m = 88 days

With (v m /r m) [(180/π) (3600) (26526/T m) = 70.75 arc second per century

Copernicus Distance Illusion

[(I – O)/O] δ θ’0 = [(r m – r e)/ r e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43 = [(58.2 x 109 - 149.6 x 109)/ 149.6 x 109] (70.75) = [(58.2 - 149.6)/ 149.6] (70.75) = 43 arc second per century Nicklaus Copernicus Distance Illusion

[(I – O)/O] δ θ’0 = [(r m – r e)/ r e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43

M r Copernicus you are wrong and you are deletedWith [(I – O)/O] δ θ’0 = [(r m – r e)/ r e] (v m /r m) [(180/π) (3600) (26526/T m)]

Next

Is Galileo

Page 24

With r m = v m T m; r e = v e T e

Or Galileo’s linear motion visual Illusion

[(I – O)/O] δ θ’0 = [(v m T m – v e T e)/ v e T e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43

[(I – O)/O] δ θ’0= [(47.9x88 – 365.26x29.8)/ 365.26x29.8] (70.75) = 43

M r Galileo you are wrong and you are deleted With [(I – O)/O] δ θ’0 = [(v m T m – v e T e)/ v e T e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43

Page 25

Tyco Brahe Velocity Illusion: [(r m – r e)/ r e] = [(r m /t) – (r e/ t)]/ (r e /t) = (v m – v e)/ v e

[(I – O)/O] δ θ’0 = [(v m – v e)/ v e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43

[(I – O)/O] δ θ’0 = [(47.9– 29.8)/ 29.8] (70.75) = 43

M r Tyco Brahe you are wrong and you are deletedWith

[(I – O)/O] δ θ’0 = [(v m – v e)/ v e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43

Page 26

(I/O) q = q + [(I – O)/O] q

M r Lippershey you can even read the telescope and you are deleted

Hans Lippershey the person who attempted to get a patent for the telescope in 1609

Modern Physics is wrong because Hans and the entire Western civilization still have no clue how to read the telescope

(I/O) q = q + [(I – O)/O] q

Page 27

Rene Descartes Cartesian failurest = t0 e í ω (t) t

Γ = t e í ω (t) t

Along the line of sight

Γ x = t cosine ω tΓ x = t + (Γ x – t) = t + ∆ Γ x; ∆ Γ x = (Γ x – t)Γ x /t = t/t + (Γ x – t)/tΓ x /t = 1 + (Γ x – t)/t

What is [(Γ x – t)/t]

Then ∆ Γ x (seconds) = Γ x - t = - 2 t sine² {[arc tan (V r m - V r e)/ (√ 2c)]/ 2} = 43And in arc ∆ Γ x (arc seconds) = Γ x - t = - 30 t sine² {[arc tan (V r m - V r e)/ (√ 2c)]/ 2} = 43

Where r’ = V r; and r θ’ = V θ; and V r ²= 2 γ r r; and V θ ² = r γ θ

The confusion is and was γ r = γ θ; V r ²= 2 V θ ²; and taking V r = (√ 2) V θ And taking V r = V r (Mercury) - V r (Earth) = V r m - V r e

And V θ = V r / (√ 2∆ Γ x Arc second) = Γ x - t = - 30 t sine² {[arc tan (V r m - V r e)/ (√ 2c)]/ 2} = 43 arc second /century

Page 28

Chapter 3: Arabs real time light motion





Observer line: …………………Observer ………………

Visual objects seen through emitted or reflected light follow this law Light measurement of visual motion of objects law: (I/O) q = q + [(I – O)/O] q

The proof:Page 29

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location

r = r (x, y, z). The state of any object in the Universe can be expressed as the product S = m r; State = mass x location: P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rate


r = r r (1) ;v = r' r(1) + r θ' θ(1) ; γ = (r" - rθ'²)r(1) + (2r'θ' + r θ")θ(1)r = location; v = velocity; γ = accelerationF = m γ + 2m'v +m" rF = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + (m" r) r (1) = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/m r) [d (m²r²θ')/d t] θ (1) = [-Gm M/r²] r (1) ------------------------------- Newton's Gravitational LawProof:First r = r [cosine θ î + sine θ Ĵ] = r r (1)Define r (1) = cosine θ î + sine θ Ĵ Define v = d r/d t = r' r (1) + r d[r (1)]/d t = r' r (1) + r θ'[- sine θ î + cosine θĴ] = r' r (1) + r θ' θ (1)


Define γ = d [r' r (1) + r θ' θ (1)] /d t = r" r (1) + r'd [r (1)]/d t + r' θ' r (1) + r θ" r (1) +r θ'd [θ (1)]/d t γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)

With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 With m = constant, then With d² r/dt² - r θ'² = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2 With d² r/dt² - r θ'² = 0 E q – 1 And d (r²θ')/d t = 0 Eq-2

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From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant Differentiate with respect to timeThen 2rr'θ' + r²θ" = 0

Divide by r²θ' Then 2(r'/r) + θ"/θ' = 0And 2(r'/r) = - θ"/θ' = 2[λ (r) + ỉ ω (r)]Also, r = r0 (θ, 0) e [λ (r) + ỉ ω (r)] t

And θ' = θ' 0 (θ, 0) e -2[λ (r) + ỉ ω (r)] t


And θ' = θ' 0 (θ, 0) e -2ỉ ω t

With d² r/dt² - r θ'² = 0Let u = 1/r; r = 1/u; r²θ' = h = θ’ /u² And d r/d t = (d r/ d u) (d u /d θ) (d θ/ d t) = (- 1/u ²) (θ’) (d u/ d θ) = - h (d u/ d θ)

And d² r/ d t² = - h (θ’) (d² u/ d θ ²) = [- h²/r²] (d² u/ d θ ²) = - h² u² (d² u/ d θ ²)With d² r/dt² - r θ'² = 0 E q – 1 And - h² u² (d² u/ d θ ²) – (1/u) (h u²) ² = 0Then (d² u/ d θ ²) + u = 0

Then u = A e - í θ

Then r 0 (θ, 0) = (1/A) e í θ And r 0 (θ, 0) = r 0 (0, 0) e í θ And r = r 0 (0, 0) e í θ e ỉ ω t Or, r = r 0 e í (θ + ω t) This is quantum mechanics

Arabs real time astronomy is actually quantum mechanics r = r0 e í ω (r) t

Light is emitted and not an emitter and humans have no means of measuring light in motion and at best they can take a snap shot.In mathematical languageThen r (θ, 0) = r 0 e í θ at t = 0And r (θ, t) = r 0 e í (θ + ω t) = r 0 e í θ e í ω t = r (θ, 0) e í ω t

And r (θ, t) = r (θ, 0) [cosine ω t + í sine ω t] And r x (θ, t) = r x (θ, 0) cosine ω t And ω t = cosine-1 [r x (θ, t) / r x (θ, 0)]

31

Then ∆ r = r x (θ, t) - r x (θ, 0) = r x (θ, 0) [(cosine ω t) – 1]

And [r x (θ, t) - r x (θ, 0)]/ r x (θ, 0) = [(1 - 2 sine² ω t/2) – 1] = - 2 sine ² ω t/2; ω t = cosine-1 [r x (θ, t) / r x (θ, 0)]

Arabs real time astronomy:Image = Image = IObject = Object = O- Object = - Object = - O

Image = object + (Image – Object)I = O + (I – O)(I/O) = O/O + [(I – O)/O](I/O) = 1 + [(I – O)/O] (I/O) θ' 0 = θ' 0 + [(I – O)/O] θ' 0

Quantum = classical + relativistic Illusion I = r; O = r 0

Or, (r / r 0) θ' 0 = θ' 0 + [(r – r 0)/ r 0] θ' 0

And [(r – r 0)/ r 0] θ' 0Or [- 2sine ² ω t/2] θ' 0; ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r/ r 0)For planet Mercury:

The distance visual effect Error is: [(r m – re)/re] δ θ’ m

The quantity: δ θ’ m = (v m/r m) (180/ π) [36526/Tm (days)] (3600) = 70.75

With v m = 47.9 km/sec; r m = 58.2 x 106 km; Tm = 88 days

With r = 58.2 x 109 m = Sun – Mercury distance

And r e = 149.6 x 109 m = Sun Earth distance

With [(r m – re)/re] = [- 2sine ² ω t/2] = 0.61 And ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r/ r 0)Or ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r m/ r e)

And [(r m – re)/re] δ θ’ m = [- 2sine ² ω t/2] (v m/r m) (180/ π) [36526/Tm (days)] (3600) = [- 2sine ² {[cosine-1 (r m/ r e)]/2}] (v m/r m) (180/ π) [36526/Tm (days)] (3600) = 0.61 x 70.75 = 43

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Chapter five: Arabs real time light Astronomy

A- Kepler’s force laws F = - ω² a and F = - GmM/a²

Or, F = - ω² a = - GmM/a²And T²/a³ = 4 π GM

Mr Kepler you are wrong and you are deleted

Johannes Keplers: Kepler’s law: a³/T² = k = constant

Isaac Newton space Inverse square law γ = k/r² Hooke’s time Inverse square law γ = r/t² or γ = r/T²

Kepler’s Space – time Law: k/r² = r/ T²; or, a³/T² = k = constantWhere this law came from?Acceleration = γ = r/t² = k/r² γ = time coordinates r/t² = space coordinates k/r²; r substituted for t

γ = r/t² = k/r² γ = - k/r² Newton’s theft = Newtonγ = - r/t² = - r/T² = - k r Hooke’s theft = Hooke’ law Of: a³/T² = k = constantOr, a1³/ T1² = a2³/ T2² Or, a1/ a2 = (T1/ T2)2/3

And (a1 - a2)/ a2 = (T1/ T2)2/3 – 1Or (am – a e)/ a e = (Tm/ Te) 2/3 – 1Nicklaus Copernicus Distance Illusion [(I – O)/O] δ θ’0 = [(r m – r e)/ r e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43Become: Kepler’s vision of idiot Einstein: Nicklaus Copernicus Distance Illusion[(I – O)/O] δ θ’0 = [(Tm/ Te) 2/3 – 1] (v m /r m) [(180/π) (3600) (26526/T m)] = 43

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When Newton came with F = - GmM/r² Physicists calculated: θ' c m = v c m/r; Angular velocity with respect to center of mass

And Astronomers observed: θ's = v s/r Angular velocity with to the sun

And v c m = √ [GM²/ (m + M) a]

And v s = √ (GM/a)

And (θ' cm - θ’s) /θ’ s = [(2 π/T c m) - (2 π/T s)]/ (2 π/T s) = (T c m/T s) – 1 = (v s/v c m) – 1 = [√ (GM/a)] x {√ [(m + M) a/ GM²]} - 1 = √ [(m + M)/M] - 1 = √ [1 + (m/M)] - 1 ≈ 1 + (m/2 M) – 1 ≈ m/2 M

And (θ' cm - θ’ s) /θ’ s ≈ (m/2 M) And (θ' cm - θ’ s) / θ' s = (2 π/T s) (m/2 M)And [(θ' cm - θ’ s)] T s = π (m/ M) Multiplying by [(180/π) (3600) (26526/T m)]

[(I – O)/O] δ θ’0 = (π m/M) [(180/π) (3600) (26526/T m)] = 43

T²/a³ = 4 π²/GM with respect to the sunTo T²/a³ = 4 π²/G (M + m) with respect to the center of mass That is Kepler's measurements data are centered from the sun and Newton's data are centered at the center of mass and this would explain this mistake like this:T² (1)/a³ = 4 π²/GM and T² (2) /a³ = 4 π²/G (M + m)When Newton changed Kepler's equation he solved the advance of perihelion as shown:

The advance of Mercury perihelion by Newton

T (1) = T (2) √ [M/ (M + m)] And 2π/ T (1) = [2 π/ T (2)] √ (1 + m/M)And θ' (1) = θ' (2) √ (1 + m/M) And θ' (2) = θ' (1) /√ [1 + (m/M)] ≈ [1 - m/ (2M)]This approximation was not on the original work And θ' (2) - θ' (1) = θ' (1) [1 /√ [1 + (m/M)] - 1]And θ' (2) - θ' (1) ≈ - θ' (1) (m/2M) = - [2 π/T] [m/2M) = - π m/MT radians/T

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W " (calculated) = [- π m/MT] (180/π degrees) (3600 seconds) (36526 century); T = days; with m = 0.32 x1024 kg; and M = 2.0 x1030 kg; T = 88 days

[(I – O)/O] δ θ’0 = (-180 x 36526 x 3600/T) (π m/M) = 43.0" seconds of arc /100 years

Mr Newton you are wrong and you are deleted

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Hooke’s for law F = - ω²x; 3 – Newton’s law F = - GmM/x²If Hooke’s law is taken the F = - k r in one dimensionThen r’’ +ω²r = 0; then r (θ, 0) = r0 e ỉ ω t

Then r = r0 e ỉ ω t; and r = r0 [1- 2 sine² ω t/2]And (r - r0)/ r0 = - 2 sine² ω t/2 The distance visual effect Error is: [(r m – re)/re] δ θ’ m





With [(r m – re)/re] = [- 2sine ² ω t/2] = 0.61 And ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r/ r 0)Or ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r m/ r e)

And [(r m – re)/re] δ θ’ m = [- 2sine ² ω t/2] (v m/r m) (180/ π) [36526/Tm (days)] (3600) = [- 2sine ² {[cosine-1 (r m/ r e)]/2}] (v m/r m) (180/ π) [36526/Tm (days)] (3600) = 0.61 x 70.75 = 43

Mr Hooke I really like you but you can be deleted

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Chapter Five: The all time wrong AstronomerThen Urbain Jean Joseph Le Verrier cameThe angular velocity of Mercury around the Sun is: θ m' = v m /r m

And δ θ’0 = (v m /r m) (180/π) (3600) (26526/Tm); Tm = 88 days

If it is measured for planet Mercury from the sun then Then it is θ m' = v m /r m

If planet Mercury around the sun measured from earth then =Then θ m' (Earth) = (v m + v e)/r m

And θ m' (Earth) = v m /r m + v e /r mAnd not v m /r m Le Verrier [1] mistake is: v e /r mThe angular speed delay is: v* e /r mOr, [(v e/v m) (v m /r m)]Taking into account Earth rotation vº e Le Verrier mistake: Then the angular speed delay:Is: v e /r m = [v* e /r m +/- vº e /r m] In arc second per century multiplying by [(180/π) (3600) (26526/T)]

[(I – O)/O] δ θ’0 = [(v* e +/- vº e) /r m] [(180/π) (3600) (26526/T)] [(I – O)/O] δ θ’0 = [(v* e +/- vº e) /v m) (v m /r m)] [(180/π) (3600) (26526/T)] = 43 [(I – O)/O] δ θ’0 = [(v* e +/- vº e) /v m) (v m /r m)] [(180/π) (3600) (26526/Tm)]

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SM

E

Idiot Le Verrier[(I – O)/O] δ θ’0 = [(r m – r e)/ r e] (v m /r m) [(180/π) (3600) (26526/T m)] = 43[(I – O)/O] δ θ’0 = [(v m T m – v e T e)/ v e T e] (v m /r m) [(180/π) (3600) (26526/T m)]

Arabs real time Astronomy is 5 lawsAnd saying (I/O) q = q + [(I – O)/O] q; q measured quantity Eq - 1And real time distance r = r0 e í ω (r) t Eq - 2And real time t = t0 e í ω (t) t Eq - 3And real time mass m = m0 e í ω (m) t Eq - 4And S = m r, the state equation Eq - 5

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Chapter six

Newton equation is a solved wrong for 350 yearsNewton’s unprecedented stupidity

Being a good mathematician is solving own equation wrong?Newton said there is gravity Force F = -GmM/r² whose solution is this

Fig1. Newton’s gravitational lawAnd r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] But you idiot this solution is wrongReal time solution is: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ℮ [λ (r) + ỉ ω (r)] t

"Apparent advance of perihelion"[(I – O)/O] δ θ’0 = (-720x36526x3600/T) {[√ (1- ε²)]/ (1-ε) ²} x [(v°+ v*)/c] ² = 43 arc second per century

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Planet

r

Sun

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z) = r [length, width, height].

The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location:

P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rate

F = d P/d t = d²S/dt² = Total force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rate


Location = r = r r (1)

Velocity = v = r' r (1) + r θ' θ (1)

Acceleration = γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)

F = m γ + 2m'v +m" r

F = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + + (m" r) r (1)

F = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/mr) [d (m²r²θ')/d t] θ (1)

F = F1 + F2

F1 = [d² (m r)/dt² - (m r) θ'²] r (1)

F2 = (1/m r) [d (m²r²θ')/d t] θ (1)

Newton's Gravitational Equation is: F = -GmM/r² With d² (m r)/dt² - (m r) θ'² = -GmM/r² (1)And d (m²r²θ')/d t = 0 (2)

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The solution is notAnd r (θ, t) = [a (1-ε²)/ (1+εcosθ)]But this solution

Real time solution is: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ℮ [λ (r) + ỉ ω (r)] t

"Apparent advance of perihelion"[(I – O)/O] δ θ’0 = (-720x36526x3600/T) {[√ (1- ε²)]/ (1-ε) ²} x [(v°+ v*)/c] ² arc second per centuryProof:

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z) = r [length, width, height]. The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location: P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rateIn polar coordinates systemLocation = r = r r (1) Velocity = v = r' r (1) + r θ' θ (1) Acceleration = γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)F = m γ + 2m'v +m" rF = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + + (m" r) r (1)F = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/mr) [d (m²r²θ')/d t] θ (1) F = F1 + F2 F1 = [d² (m r)/dt² - (m r) θ'²] r (1)F2 = (1/mr) [d (m²r²θ')/d t] θ (1)

Newton’s force law is F1 = [-GmM/r²] r (1) Kepler's force law is:F2 = (1/mr) [d (m²r²θ’)/d t] θ (1) = 0

If m is constant then d (m²r²θ’)/d t = 0; and d (r²θ’)/d t = 0Or, r²θ’ = h = 2 π a b/T; a = mean distance from sun and is called semi major axis and b is the semi minor axis.

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And the motion of a planet m around the Sun M measured in real time is a rotating ellipse

Tyco Brahe Logged observational Data of Planets P motion around the Sun S and then Kepler stated the areal velocity law: If Planet p observed from the sun then the trajectory of planet p will cut equal areas in equal times.

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θ

r

Sun

Mercury

M

ψ

m

θ

A

AA

A

S

A

A

A

S

T

T

TTT

T

T

P

P

When the areas size A are sliced equally it was found that the times spent by planets orbiting around the Sun and making areas A each are equal also. Or, r² (0) θ' (0) = r² (1) θ' (1) = r² (2) θ' (2) = r² (3) θ' (3) = -- = location² x [angular speed] = constant = Areal velocityIf r² θ' = hThen differentiating with respect to timeThen d (r² θ')/ d t = d h/ d t And 2 r r’ θ’ + r² θ’’ = 0Or 2 (r’ /r) = - (θ’’/θ’)The r = r0 e í ω t

And θ’ = θ’0 e – 2 í ω t

With r = r0 e í ω t

In real time r (n) = r (0) e í ω t

With r² (0) θ' (0) = r² (1) θ' (1)Then θ' (1) = [r² (0)/ r² (1)] θ' (0) And θ' (1) = {[e í 2 ω t] θ' (0) And θ’ (1) = {[cosine 2 ω t + í sine 2 ω t] – 1} θ' (0) Δ θ’ (1) = Δ θ’ (x) + í Δ θ’ (y) = [cosine 2 ω t + í sine 2 ω t] θ' (0) Δ θ’ (x) = [cosine 2 ω t] θ' (0) Δ θ’ (x) = [1 – 2 sine² ω t] θ' (0) Δ θ’ (x) - θ' (0) = - 2 θ' (0) sine² ω tΔ W = Δ θ’ (x) - θ' (0) = - 2 θ' (0) sine² ω t = -2(2π) [√ (1-ε²)]/T (1-ε) ²] sine² ω t And θ' (0) = h/r² (0) = 2 π a b/T a² = 2π√ (1-ε²)]/T (1-ε) ² Δ W = - 4 π) [√ (1-ε²)]/T (1-ε) ²] sine² ω tIf this apsidal motion is to be found as visual effects, then With, v ° = spin velocity; v* = orbital velocityAnd v/c = (v* + v°)/c = tan ω tΔ W = -4 π [√ (1-ε²)]/T (1-ε) ²] sine² {Inverse tan [(v* + v°)/c]} radians Multiplication by 180/π to change to degrees Δ W ° = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [(v* + v°)/c]} Degrees And multiplication by 1 century = 36526 days and using T in days W° (ob) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x sine² {Inverse tan [(v* + v°)/c]} degrees/100 yearsApproximations With v° << c and v* << c, Then sine Inverse tan [v°/c + v*/c] ≈ (v° + v*)/cΔ W ° (calculated) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x [(v° + v*)/c] ² degrees/100 yearsThis is the equation for axial rotations rate of planetary and binary stars or any two body problem.The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)

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Finding orbital velocities

From Newton's inverse square law of an ellipse motion applied to a circular orbit gives the following: m v²/ r (cm) = GmM/r²Planet --- r (cm) ----- Center of mass ------- r (CM) --------- Mother Sun Planet ------------------- r -------------------------------------- Mother SunCenter of mass law m r (cm) = M r (CM); m = planet mass; M = sun massAnd r (cm) = distance of planet to the center of massAnd r (CM) = distance of sun to center of massAnd r (cm) + r (CM) = r = distance between sun and planet Solving to get: r (cm) = [M/ (m + M)] r And r (CM) = [m/ (m + M)] r Then v² = [GM r (cm)/ r²] = GM²/ (m + M) r And v = √ [GM²/ (m + M) r = a (1-ε²/4)] Planet orbital velocity or primary velocity:And v* = v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 48.14 km for planet MercuryVelocity of secondary or Mother Sun velocityAnd v* (M) = √ [Gm² / (m + M) a (1-ε²/4)] Applications: mercury ellipse and its axis rotation of 43 " /century1- Planet Mercury axial "apparent" rotation rate

Δ W’’ = (-720x36526x3600/T) {[√ (1-ε²]/ (1-ε) ²} (v* + v°/c) ² seconds of arc per centuryThe circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)Where v* (p) =√ [G M² / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system data: G =6.673x10-11; M=2x1030kg; m=.32x10^24kg; ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec; v° = 0.002km/secCalculations yield: v* =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552 Δ W = (-720x36526x3600/88) x (1.552) (48.14/299792)² =43.0”/centuryΔ W’’ = (-720x36526x3600/T) {[√ (1-ε²]/ (1-ε) ²} [(v* + v°)/c] ² = 43 seconds of arc per century [(I – O)/O] δ θ’0= (-720x36526x3600/T) {[√ (1-ε²]/ (1-ε) ²} [(v*m + v° m)/c] ² seconds of arc per centuryOr, Newton is plain stupid? There is no proof of action at a distance?F = -GmM/r² is not a proof of gravitational pull between planetary objects but rather the interaction of earth with its atmosphere G = 1/ (2/5) (4π/3) p T² = √ (2π)/ (4π/3)p T²; ρ = air density =1.2045kg/m³; and T = Earth rotation period = 23.9333x3600; Then G = 6.6747 x 10-11

G = 1/ (radius of gyration) (sphere) (air density) (Earth spin) ²G = √ (2 π) / (4π/3) p T² = √ (2π)/ (4π/3)p T²Suggesting that gravity measured on Earth is local to earth surface caused by Earth spin motionAlso a rotating Ellipse can be derived from many different solutions and not necessarily from an inverse square law

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Chapter sevenInverse Cube equations F = m γ = - k/r³ r (1), then in polar coordinates With m [d² r/dt² - θ'²r] = - k /r³ Inverse Cube Gravitational law (1)And d (r²θ')/d t = 0 Kepler's Areal Velocity Equation (2)

These two equations give an axial rotation rate: One: φ = π (m/ M) (180) [36526/T] [3600] arc second/100 years = 43.0344 seconds of arc / century for Mercury Two: δ θ' = - 720 [36526/T] (3600) √ (1 - ε²)/ T (1 - ε) ² (v/c) ² arc second/100 years = 43.0" seconds of arc /century for Mercury Solution:With m = constant Then d² r/dt² - θ'²r = - k/ r³ (1)And d (r²θ')/d t = 0 (2)From (2) d (r²θ')/d t = 0; r²θ' = h From (1), θ'² d² r/ dθ² - θ'²r = - k/ mr³ And θ'² [d² r/ dθ² - r] = - k/ mr³ And d² r/ dθ² - r = - (k/mh²) rAnd d² r/ dθ² - r [1 - (k/mh²)] = 0And r (θ, 0) = r (0, 0) e ỉ {√ [1 - (k/mh²)]} θ

From (2) d (r²θ')/d t = 0; r²θ' = h Then 2rr'θ' + r²θ'' = 0Dividing by r²θ' We get 2 (r'/r) + (θ''/θ') = 0And 2 (r'/r) = - θ''/θ' = 2ỉ ω tAnd r = r (0, 0) e ỉ {√ [1 - (k/mh²)]} θ e ỉ ω t

And θ' = θ' (θ, 0) e - 2ỉ ω t Or r = r (0, 0) e ỉ {√ [1 - (k/mh²)]} θ + ỉ ω t

And θ' = θ' (0, 0) e -2ỉ [{√ [1 - (k/mh²)]} θ + ω t]

And θ' = θ' (0, 0) e -2ỉ [{√ [1 - (k/mh²)]} θ + ω t]

And θ' = (θ' (0, 0) {cosine 2 [{√ [1 - (k/mh²)]} θ + ω t] - ỉ sine 2 [{√ [1 - (k/mh²)]} θ + ω t]} And θ' - θ' (0, 0) = - 2 θ' (0, 0) sine² [{√ [1 - (k/mh²)]} θ + ω t] And δ θ' = - 2 θ' (0, 0) sine² [{√ [1 - (k/mh²)]} θ + ω t] If k = Gm M α (1 - ε²); h = 2π a b/TThen: k/mh² = Gm M a (1 - ε²)/m 4 π ² a² b²/T²And k/mh² = GM T² (1 - ε²)/4 π ² a b²; multiply by (a²/a²)Then [a² (1 - ε²)/ b²] [GM T²/4 π ² a³] = [GM T²/4 π ² a³]

Then δ θ' = - 2 θ' (0, 0)] sine² [{√ [1 - (GM T²/4 π ² a³)]} θ + ω t] Taking Kepler's: GM T²/4π²a³ = 1

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Then δ θ' = - 2 θ' (0, 0) sine² ω t And θ' (0, 0) = h/r² = 2πab/Ta² (1 - ε) ² = 2πa²√ (1 - ε²)/Ta² (1 - ε) ² = 2π√ (1 - ε²)/T (1 - ε) ² And δ θ' = - 4π√ (1 - ε²)/ T (1 - ε) ² sine² ω tWith ω T = arc tan v/c << 1 Then δ θ' = - 4π/T√ (1 - ε²)/ (1 - ε) ² sine² arc tan (v/c) radians per TOr δ θ' = - 4π/T√ (1 - ε²)/ (1 - ε) ² (v/c) ² radians per T

And δ θ' = - 4π/T√ (1 - ε²)/ (1 - ε) ² (v/c) ² [180/π] [36526] [3600] arc second/100 years

Or δ θ' = - 720 [36526/T] (3600) [√ (1 - ε²)]/ (1 - ε) ²] (v/c) ² arc second/100 yearsOr δ θ' = - 720 [36526/T] (3600) (1.552) (48.2/c) ² = 43.11 " arc second/100 yearsIf k = Gm (M + m) α (1 - ε²); h = 2π a b/TIf we take Newton’s G (M + m) T²/4π²a³ = 1

Then δ θ' = - 2 θ' (0, 0)] sine² [{√ [1 – [G (M + m) T²/4 π ² a³)]} θ + ω t] And G (M + m) T²/4π²a³ = 1And 1- G (M + m) T²/4π²a³ = 0Then δ θ' = - 2 θ' (0, 0) sine² ω t Then δ θ' = - 2 2 π √ (1 - ε²)/T (1 - ε) ² sine² ω t = - 4 π √ (1 - ε²)/T (1 - ε) ² sine² ω tOr δ θ' = - 720 [36526/T] (3600) [√ (1 - ε²)]/ (1 - ε) ²] (v/c) ² arc second/100 yearsIf k = mh²; then δ θ' = - 2 θ' (0, 0) sine² ω t Or δ θ' = - 720 [36526/T] (3600) [√ (1 - ε²)]/ (1 - ε) ²] (v/c) ² arc second/100 yearsIf k = m²h²/M; then δ θ' = - 2 θ' (0, 0) sine² [{√ [1 - (k/mh²)]} θ + ω t] Become δ θ' = - 2 θ' (0, 0) sine² [{√ [1 - (m/M)]} θ + ω t] And (m/ M) << 1; t = 0 Or r = r (0, 0) e ỉ {√ [1 - (m/M)]} θ

Then r ≈ r (0, 0) e ỉ [1 - (m/ 2M)]} θ

And r ≈ r (0, 0) e ỉ [(θ - φ)]

With φ = m/ 2M θTaking θ = 2 π Then φ = π m/ M radians And φ = π m/ (M) [180/π] [36526/T] [3600] arc second/100 yearsIf k = m²h²/ (M + m)Then φ = π m/ (M + m) radians And φ = π m/ (M + m) [180/π] [36526/T] [3600] arc second/100 yearsOr φ = π (m/ M) (180/ π) [36526/T] [3600] arc second/100 years = 43.0344"/100 yearsOr φ = π m/(M +m) (180/ π) [36526/T] [3600] arc second/100 years = 43.0344"/100 years

Or δ θ' = - 720 [36526/T] (3600) √ (1 - ε²)/ (1 - ε) ²] (v/c) ² arc second/100 years = 43.0"/100 years

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Chapter eight: Nuclear gravityOr Yukawa’s: [(I – O)/O] δ θ’0: M r Yukawa travel is not physics but Alfred Nobel physicists’ stupidity

Nuclear Gravity F = (-GmM/r²) e k/r

With d² (m r)/dt² – (m r) θ'² = [-GmM/r²] e k/ r Nuclear gravity Equation (1)And (m²r²θ')/d t = 0 Kepler's Areal Velocity (2)

(2) : d (m²r²θ')/d t = 0 Then m²r²θ' = constant; if m is taken as constant then r²θ' = hAnd (1): d² r/dt² - r θ'² = [-GmM/r²] e k/r Let m r =1/uThen d r/d t = -u'/u² = - (1/u²) (θ') d u/d θ = (- θ'/u²) d u/d θ = - h d u/d θAnd d² r/dt² = -hθ'd²u/dθ² = - hu² [d²u/dθ²]With d² (m r)/dt² - (m r) θ'² = [-GmM/r²] e k/r Nuclear Gravity (1)

With e k/r ≈ 1+ k/r; k/r <<1 And -hu² (d²u/dθ²) - (1/u) (hu²)² = -GMu² [1 + k u]; α m/M r << 1And (d²u dθ²) + (1 – GMk/h²) u = GM/h²And u = [GM/ h²]/ {(1 – GMk/h²) + A cosine [√ (1 - GMk/h²)] θ} And r = 1/u = 1/ {[GM/ h²]/ {[1 – GMk/h²] + A cosine {√ [1 – GMk/h²]} θ} = [1 – GMk/h²]/ (M/ h²)]/ {1 + ε cosine {√ [1 - GMk/h²]} θ Where [1 – GMk/h²]/ (GM/h²) = a (1 – ε²)

48

And [1 – GMk/h²] = (GM/h²) a (1-ε²)And h² - GM k = G M a (1- ε²)

Then h²/GM – a (1- ε²) = k

And √ [1 – GMk/h²] =√ {[1 – (GM/h²) [(h²/GM) – a (1 - ε²)]}

= √ {[1 – 1 + (GM/h²) a (1 - ε²)]} = √ [(GM/h²) a (1 - ε²)] = √ {[GMT²/4π²a4 (1 - ε²)] a (1 - ε²)]} = √ (GMT²/4π²a³) = 1; T²/ a³ = 4π²/ GM Kepler’s Then, r = 1/u = (GM/ h²) + A cosine θIf, T²/ a³ = 4π²/ G (M +m) Newton’sThen, √ [1 – GMk/h²] = √ (GMT²/4π²a³) = √ [M/ (M + m)]= 1/ √ [1 + (m/M)] ≈ 1/ [1 + (m/2M)] = 1 – (m/ 2M)And [1 – GMk/h²]/ (M/ h²)]/ {1 + ε cosine {√ [1 - GMk/h²]} θ = a (1- ε²)/ [1 + ε cosine {√ [1 – (GMk/h²)]} θ= a (1- ε²)/ {[1 + ε cosine [(1 – m/2M) θ]} = a (1- ε²)/ [1 + ε cosine (θ – Φ)] And 2 π Φ = 2 π (m/ 2M) = π (m/M)If Newton’s law were to be F = - Gm (M + m)/r²The 2 π Φ = 2 π [m/ 2(M + m)] = π [m/ (m + M)]

Nuclear Gravity (-GmM/r²) e (m α/ M r)

Abstract: Yukawa Gravity (-GmM/r²) e (m α/ M r) or the nuclear gravity force is the crudely approximated Newtonian gravity force (-GmM/r²) explains planetary motion around the sun as a rotating ellipse with a rotation rate φ = [π m/(m + M)](180/π)(36526/T)(3600) = 43.03 seconds of an arc per century for the most talked about planet of mercury; m = 3.2 x 1024 kg; M = 2x1030 kg; T = 88days

With d² (m r)/dt² – (m r) θ'² = (-GmM/r²) e (m α/ M r) Gravity Force (1)And d (m²r²θ')/d t = 0 Kepler's Areal Velocity (2)

Then m²r²θ' = constant; if m is taken as constant then r²θ' = hAnd (1): d² r/dt² - r θ'² = -GmM/r² + Gm²α/r³

Let m r =1/uThen d r/d t = -u'/u² = - (1/u²) (θ') d u/d θ = (- θ'/u²) d u/d θ = - h d u/d θAnd d² r/dt² = -hθ'd²u/dθ² = - hu² [d²u/dθ²]-hu² [d²u/dθ²] - (1/u) (hu²)² = -G M u² [1 – (α m /M) u]; α m /M r << 1Or (d²u/ dθ²) + (1 – Gmα/h²) u = GM/h²And u = [GM/ h²]/[1 – Gmα/h²] + A cosine {√ [1 - Gmα/h²]}θ

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And r = 1/u = 1/ {[GM/ h²]/ [1 – Gmα/h²] + A cosine {√ [1 – Gmα/h²]} θ} = [1 – Gmα/h²]/M/ h²]/ {1 + ε cosine {√ [1 - Gmα/h²]} θ

Where [1 – Gmα/h²]/GM/h² = a (1 – ε²)And [1 – Gmα/h²] = (GM/h²) a (1-ε²)

If α = a (1 – ε²), the h² = G (m + M) a (1- ε²)Then Gmα/h² = G ma (1 - ε²) /G (m + M) a (1- ε²) = m/ (m + M)

With r (θ, 0) = [1 – Gmα/h²]/M/ h²]/ {1 + ε cosine {√ [1 – Gmα/h²]} θThen r (θ, 0) = a (1 – ε²)/ {1 + ε cosine {√ [1 – Gmα/h²]} θ

And r (θ, 0) = a (1 – ε²)/ {1 + ε cosine {√ [1 – m/ (m + M)]} θ And m << M; √ [1 – m/ (m + M)] ≈ 1 - m/2(m + M)

With {√ [1 – Gmα/h²]} θ ≈ [1 - m/ 2(m + M)] θIf θ = 2π, Then 2π [1- m/2(m + M)] = 2π - 2π [m/2(m + M)]

With r (θ, 0) = a (1 – ε²)/ {1 + ε cosine [θ – mθ/2(m + M)]}Then r (θ, 0) = a (1 – ε²)/ [1 + ε cosine (θ – φ)]And φ = mθ/2(m + M)

With θ = 2π, then φ = 2πm/2(m + M) = π m/ (m + M) radians/second

Multiplication by 180/π, then φ = 180m/ (m + M) degrees/second

Multiplication by (36526/T), then φ = [m/ (m + M)] (180)(36526/T) degrees/century

Multiplication by 3600, then φ = [m/ (m + M)] (180) (36526/T)(3600) seconds/century

With Planet Mercury: m = 3.2x1024kg; M = 2x1030kg

Then φ = [3.2x1024/ 2x 1030] (180) (36526/T)(3600) seconds of an arc per century

And φ = 43.03426909''/century φ = 2 π m/2(m + M) = π m/ (m + M) And [(I – O)/O] δ θ’0 = [π m/ (m + M)] (180/ π) (36526/T) (3600)]

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Chapter NineFrequency Games

What is the accumulated value of 2 π f per century for planet mercury seen from Earth?The angular frequency is ω = 2 π f How I would see ω of planet mercury turning around the sun from earth?The answer is there will be a frequency change ofW = 2 π f [v* m - v*e]/v* e; radians per second Where v*e = Earth orbital velocity around the Sun = 29.8 km/secAnd vº e = Earth spin speed = 0.465 km/sec

And v* m = Mercury orbital velocity around the Sun = 47.9 km/secWith f = 1/T; f = frequency; T = Period = 88 daysIf W is wanted in degrees multiply by: 180/π If W is wanted in degree per century multiply by (180/π) x (36526 days/ T)If W is wanted in arc second per century multiply by (180/π) x (36526 days/ T) x 3600W = [2 π f [(v* m - v*e)/v* e] x (180/π) x [36526 days/ (T days)] x 3600With f = 1/T (seconds) W = [360 x 3600 x (36526/T (days)] [1/T (seconds)] [(v* m - v*e)/v* e]

W = [360 x 3600 x (36526/T (days)] [1/ 24 x 3600 x T (days)] [(v* m - v*e)/v* e]W = 15 x (36526/T² (days)] [(v* m - v*e)/v* e] [(I – O)/O] δ θ’0 =15 x (36526/T² (days)] [(v* m - v*e)/v* e]For MercuryW = 15 x [36526/ (88)²] [(47.9 – 29.8)/29.8] = 43.0 arc second per century

The conclusion isWith ω = 2 π f = 2 π /T angular frequency in event timeAnd ω (real time) = ω (event time) + Δ ωΔ ω = ω (real time) - ω (event time) = 2 π f [(v* m - v*e)/v* e]= 2 π f Z; Z = [(v* m - v*e)/v* e]

= red shift

And Δ ω = 2 π f (1 + Z) The Advance of Planet Mercury Perihelion is 2 π f Z = 2 π Z/ TΔ ω = (2 π/T) [(v* m - v*e)/v* e]In arc seconds per century: Multiply by: (180/π) [36526/T (days)] (3600) And Δ ω = [2 π / T (seconds)] [(v* m - v*e)/v* e] x (180/π) [36526/T (days)] (3600); T (seconds) = T days x 24x 3600 Δ ω = [2 π / T days x 24x 3600] [(v* m - v*e)/v* e] x (180/π) [36526/T (days)] (3600)Δ ω = [15 x 36526 / T² (days)] [(v* m - v*e)/v* e] = 43

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Chapter Ten: Space – time unlimited stupidity Visual force: F = - Gm M/r² - Gm Mk/r³

With m [d² r/dt² - θ'²r] = - Gm M/r² - Gm Mk/r³ Visual Gravitational law (1)And d (r²θ')/d t = 0 Kepler's Areal Velocity Equation (2)

Gives an axial rotation rate of W = 15 x (36526/T² (days)] [(v* m - v*e)/v* e]

Boundary value LawWith d² (m r)/dt² - (m r) θ'² = - k/ r² Inverse square Gravitational (1) And d (m²r²θ')/d t = 0 Kepler's law (2)

At Perihelion: d² r/d t² - r θ'² = - GM/r² = - r θ'²; d² r/d t² = 0 Then r θ'² = GM/r² A quick answer by Newton would be: First θ' ² = [GM/r³]

And ω² = [GM/r³] [(v* m - v*e)/v* e] ²

In arc sec / century Then ω = {[GM/r³]} 1/2 [(v* m - v*e)/v* e] [(180/π) (3600) (36526/Tm)

= 43"/century[(I – O)/O] δ θ’0 = {[GM/r³]} 1/2 [(v* m - v*e)/v* e] [(180/π) (3600) (36526/Tm)

Constant acceleration lawOr, r 1 /θ'² 1= r 2 /θ'²2 = constant Or, θ' 2= {√[r 2/ r 1]} θ' 1

And θ' 2 - θ' 1 = {[√ (r 2/ r 1)] - 1} θ' 1Δ θ' = {[√ (r 2/ r 1)] - 1} θ' 1This is the angular time delay and will be seen as angular visual IllusionThe angular speed is θ' = v/rFor Mercury: θ' = v/r = (47.9km/sec)/58,200,000 km = 0.000000843 radians/secIf you want the accumulation value in arc sec /century W", thenAnd W" = (v/r) (180/π) (3600) (26526/T) = angular velocity in arc sec per century. If it is measured for planet Mercury then W" = (47.9/58,200,000) (180/π) (3600) (26526/88)W"= 70.75 arc second per centuryOr, Δ W" = {[√ (r 2/ r 1)] - 1} W" (1)What is the angular visual Illusion for planet Mercury that would be seen when measured from Earth with Earth location r (1) = Earth = 149.6 x 106

And r (2) = Mercury = 58.2 x 106

And W" (2) = - 70.75 arc sec /centuryΔ W" = {[√ (r 2/ r 1)] - 1} W" 1Δ W" = {{√ [149.6/58.2]} - 1} [-70.29] = 43.0" arc per century[(I – O)/O] δ θ’0 = {{√[r e/ r m]} - 1} (v m /r m)] [(180/π) (3600) (26526/T)

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Constant Areal velocity Law

Or, r² 1 θ'1= r² 2 θ'2 = location x speed = constant = Areal velocity

Or, r 1 v 1= r 2 v2

Or, θ' 1= (r 2/ r 1)² θ' 2

And θ' 1 - θ' 2 = [(r 2/ r 1)² - 1] θ' 2

Δ θ' = θ' 1 - θ' 2 = [(r 2/ r 1)² - 1] θ' 2

Δ θ' = [(v 1/ v 2)² - 1] θ' 2

This is the angular time delay and will be seen as angular visual Illusion

The angular speed is θ' = v/rFor Mercury: θ' = v/r = (47.9km/sec)/58,200,000 km = 0.000000843 radians/sec

If you want the accumulation value in arc sec /century W", thenAnd W" = (v/r) (180/π) (3600) (26526/T) = angular velocity in arc sec per century.

If it is measured for planet Mercury then W" = (47.9/58,200,000) (180/π) (3600) (26526/88)W"= 70.29 arc second per century

Or, Δ W" = [(v 1/ v 2)² - 1] W" (2)

What is the angular visual Illusion for planet Mercury that would be seen when measured from Earth with Earth location r (1) = Earth = 149.6 x 106

And r (2) = Mercury = 58.2 x 106

And W" (2) = - 70.29 arc sec /century

Δ W" = [(v 1/ v 2)² - 1] W" (2)

Δ W" = [(v 1/ v 2)² - 1] (v m /r m)] [(180/π) (3600) (26526/T)

Δ W" = [(29.8/ 47.9)² - 1] [-70.75] = 43.0" arc per century

[(I – O)/O] δ θ’0

= [(v e/ v m) ² - 1] (v m /r m)] [(180/π) (3600) (26526/T)

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No Force solution Newton said there is gravity Force F = -GmM/r² whose solution is this

Fig 1. Newton’s gravitational lawWith m [d² r/dt² - θ'²r] = 0 Zero force law equation (1)And d (r²θ')/d t = 0 Kepler's Areal Velocity Equation (2)Equation (2) r = r0 e í ω t; θ' = θ'0 e - 2 í ω t

Equation (1) - h² d² θ/ d u² - h² u³ = 0

Planet

r

Sun

Or d² θ/ d u² + u = 0Then u = u0 e - í θ And r0 = r0 (0) e + í θ Or, r = r0 (0) e í (θ + ω t) [h/ θ'0]½ e í (θ + ω t)

And θ' = θ'0 e - 2 í ω t

Δ θ' = - 2 θ'0 sine² ω t = - 2[2π/T] x [180/ π] [38526/T] [3600] sine² ω t = - 2 [2π/88 x 24 x 3600] x [180/ π] [38526/T] [3600] sine² ω t = - 141.5 sine² ω tAnd ω t = arc tan [(r m - r e)/ (r m + r e)] ½ = arc tan [(149.6 – 58.2)/ (149.6 + 58.2)] ½

Δ θ' = - 141.5 sine² arc tan [(149.6 – 58.2)/ (149.6 + 58.2)] ½

= 43 arc seconds per century

Newton’s mathematically wrong [(I – O)/O] δ θ’0 =(-720x36526x3600/88) (1.552) (48.14/300,000) ² arc sec per century

= 43This particular equation structure is the basis of stars studying. Originally, I derived it from Arabic real time physics literature. A visual Illusion along the line of sight is constructed as follows: Instead of seeing r0 we see r0 cosine arc tan (v/c) and perpendicular to the line of sight we see r0 sine arc tan (v/c)Or r = r0 [cosine arc tan (v/c) + í sine arc tan (v/c)]Or r = r0 e í arc tan (v/c)

Or r = r0 e í ω t

And areal velocity law gives the same resultsIf r² θ' = hThen differentiating with respect to timeThen d (r² θ')/ d t = d h/ d t And 2 r r’ θ’ + r² θ’’ = 0Or 2 (r’ /r) = - (θ’’/θ’)The r = r0 e í ω t

And θ’ = θ’0 e – 2 í ω t



PlanetOrbit T


Less Earth speed

Spin speedkm/sec



4,498 60180 5.43 2.68

Pluto 5906.4 90730 4.74


With r² (0) θ' (0) = r² (1) θ' (1)Then θ' (1) = [r² (0)/ r² (1)] θ' (0) And θ' (1) = {[e í 2 ω t] θ' (0) And θ’ (1) = {[cosine 2 ω t + í sine 2 ω t] – 1} θ' (0) Δ θ’ (1) = Δ θ’ (x) + í Δ θ’ (y) = [cosine 2 ω t + í sine 2 ω t] θ' (0) Δ θ’ (x) = [cosine 2 ω t] θ' (0) Δ θ’ (x) = [1 – 2 sine² ω t] θ' (0) Δ θ’ (x) - θ' (0) = - 2 θ' (0) sine² ω tΔ W = Δ θ’ (x) - θ' (0) = - 2 θ' (0) sine² ω t = -2(2π) [√ (1-ε²)]/T (1-ε) ²] sine² ω t And θ' (0) = h/r² (0) = 2 π a b/T a² = 2π√ (1-ε²)]/T (1-ε) ² Δ W = - 4 π) [√ (1-ε²)]/T (1-ε) ²] sine² ω tIf this apsidal motion is to be found as visual effects, then With, v ° = spin velocity; v* = orbital velocityAnd v/c = (v* + v°)/c = tan ω tΔ W = -4 π [√ (1-ε²)]/T (1-ε) ²] sine² {Inverse tan [(v* + v°)/c]} radians Multiplication by 180/π to change to degrees Δ W ° = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [(v* + v°)/c]} Degrees And multiplication by 1 century = 36526 days and using T in days W° (ob) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x sine² {Inverse tan [(v* + v°)/c]} degrees/100 years

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Abstract: light measurements are done along the line of sight and we measure light projections to our eye. Meaning, that we do not measure distance r but we measure r cosine arc tan (v/c) when light is used and as a consequence instead of measuring time T we measure T cosine arc tan (v/c) and here is the experimental proof of it.

Your eyes

A000 00000 00 0 0 0 00 0 0 0 0 0 0 0 0 00 0 θ (2) 0 0 00 θ (1) 0 0 θ (3) 0 00 0 0 0 00 0 0 0 0 0 0 0 0 00 0 0 0 00 0 0 0 0B------------------------- 0 --------------- 0 ------------- 0 1 2 3

The projections of light is what we measure and the projection of 1 and 2 and 3 is ABIf the angle between ray 1 and AB is θ (1)If the angle between ray 1 and AB is θ (2)If the angle between ray 1 and AB is θ (3) Then C (1) cosine θ (1) = C (2) cosine θ (2) = C (3) cosine θ (3) = c = AB You want experimental proof?Here is my iron clad proof of my theory: This is using the light projections to be equal c then the advance of perihelion and stars to solve mercury Venus and DI Herculis binary stars better than all is

If an object at location A sends light signals in all directions at time at t = 0 where B1 is; And B1 is moving to B2 then the projection of AB2 of AB1 is measured. AB1 is theorized and AB2 is measured; or AB2 = AB1 cosine θAnd θ = arc tan (v/c)B1B2 /AB1 = sine θ; AB2/AB1 = cosine θ; B1B2/AB2 = tan θ Instead of measuring time t we theorize time t and we measure time Γ = t cosine θ

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A Light V E L O C I T Y B2O O O O O O O O O O

O θ C OO O

O OO V O

O OO O

O OO O

OB1

Γ = t cosine arc tan (v/c) = t {1- 2 sine² {[arc tan (v/c)]/2}}Γ - t = - 2 t sine² {[arc tan (v/c)]/2}With r2θ’ = h and 1/T= θ’’/θ’ = - 2r’/r = 2/t

Or T = t/2 or the time measurements changes are twice period changes Or (Γ – t)/2 = - t sine² {[arc tan (v/c)]/2} is the time delays for periods

With t = 1 century and t in arc seconds is 15 times time seconds then

(Γ – t)/2 = - 15 T sine² {[arc tan (v/c)]/2}; T = One century

With v = 47.9 (mercury) – 29.8 (Earth) =18.1 km/sec

(Γ – t)/2 = - 15 (36526) sine² {[arc tan (18.1/300,000)]/2] = 43

= 15 x (36526/T² (days)] [(r m – r e)/ r e] = 43

= 15 x (36526/T² (days)] [(v m T m – v e T e)/ v e T e] = 43

= 15 x (36526/T² (days)] [(v m – v e)/ v e] = 43

= 15 x (36526/T² (days)] [(Tm/ Te) 2/3 – 1] = 43

Or, Center of mass gravity equation

[(I – O)/O] δ θ’0 = [π m/ (m + M)] (180/ π) (36526/ T m) (3600)] = 43

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Mr Laplace yes you are deleted

The solution to F = -GmM/r² is not r = [a (1-ε²)/ (1+ε cosine θ)] and it is time you get correctedReal time solution is: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ℮ [λ (r) + ỉ ω (r)] t

"Apparent advance of perihelion"

Or, Boundary value law

[(I – O)/O] δ θ’0 == {[GM/r³]} 1/2 [(r m – r e)/ r e] [(180/ π) (3600) (36526/ T m) = 43

= {[GM/r³]} 1/2 [(v m – v e)/ v e] [(180/ π) (3600) (36526/ T m) = 43

= {[GM/r³]} 1/2 [(Tm/ Te) 2/3 – 1] [(180/π) (3600) (26526/T m)] = 43

= {[GM/r³]} 1/2 [(v m T m – v e T e)/ v e T e] [(180/π) (3600) (26526/T m) = 43

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Chapter ElevenArabs real time physics in outer Space

The problems Einstein and all others can not solveApparent Apsidal motion of binary (two) stars

This is the solution to the 150 years apsidal motion puzzle that is not solvable by space-time physics or any said or published physics including 500 years of Modern physics and 500,000 modern physicists. Binary stars apsidal motion or "Apparent" rate of orbital axial rotation is a visual effect along the line of sight of moving objects applied to the angular velocity at Apses. From the thousands of close binary stars astronomers picked a dozen sets of binary stars systems that would be a good test of relativity theory and modern physics collected data for all of past century and relativity theory failed every one of them. This rate of "apparent" axial rotation is given by this new equation W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} (v* + v°/c) ² degrees/100 yearsT = period; ε = eccentricity; v° = spin velocity effect; v*= orbital velocity effects The simplest problem in physics that all 500 years modern physics and 500,000 modern physicists could not solve by any physics:

1 – As Camelopardalis2 – DI Herculis3 – V1143 Cygni4 – V 541 Cygni5 - AI Hydra6 - V 731 Cephei7 - SW Canis Majoris8 - NV Canis Majoris9 - GG Orion10- CM Draconis11- VV PYX12 – EW Oriomis13- V1147 Sagittari

14 – PSR 1913 + 1612 – PSR 0737 – 3039Solution: Location r = r r (1) ; Velocity v = r' r (1) + r θ' θ (1)Acceleration γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)S = m r; State = mass x distance P = d S/ d t = d (m r)/d t = m (d r/d t) + (d m/d t) rVelocity = v = (d r/d t); mass rate change = m' = (d m/d t) P = m v + m' r; Momentum = change of state = change in location or change in mass F = d P/d t = d² S/d t² = d [m (d r/d t) + (d m/d t)]/d t = m d² r/d t² + (d m/d t) (d r/d t) + (d m/d t) (d r/d t) + (d² m/d t) ² r

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F = m d² r/d t² + 2 (d m/d t) (d r/d t) + (d² m/d t) ² rForce = Change of momentumF = m a + 2 m ' v + m" rF = - GmM/r² Or, Newton's Kepler's equation: F = - GmM/r² ThenWith d² (m r)/dt² - (m r) θ'² = -GmM/r² Newton's Gravitational Equation (1)And d (m²r²θ')/d t = 0 Kepler's force law (2)With m = constant, then m can be taken out from both equations (1) and (2)With d² r/d t² - r θ'² = - GM/r² Newton's Gravitational Equation (1)And d (r²θ')/d t = 0 Kepler's force law (2)From 2: With m = constant; then d (m²r²θ')/d t = 0 And m² d (r²θ')/d t = 0And d (r²θ')/d t = 0And r²θ' = h With (1): d² (m r)/dt² - (m r) θ'² = -GmM/r²With m = constantThen m [d² r/ d t² - r θ'²] = - Gm M/ r² And [d² r/ d t² - r θ'²] = - G M/ r² Let r =1/uThen d r/d t = -u'/u² = - (1/u²) (θ') d u/d θ = (- θ'/u²) d u/d θ = - h d u/d θAnd d² r /d t² = - h θ'd²u/dθ² = - h u² [d²u/dθ²]And - h u² [d²u/dθ²] - (1/u) (hu²)² = - G Mu²Or, [d²u/ dθ²] + u = G M/ h ²u = G M/ h² + A cosine θ And r = 1/u = 1/ [G M / h² + A cosine θ] = (h²/ G M)/ [1 + (Ah²/ GM) cosine θ] = (h²/GM)/ (1 + ε cosine θ)

Then r (θ, 0) = a (1-ε²)/ (1+ ε cosine θ)

This is Newton's Classical Equation solution of two body problem. We solved this equation and we got the motion equation:Is: r (θ, t) = [a (1-ε²)/ (1+ ε cosine θ)] which is the equation of an ellipse with eccentricity ε and semi - major axis a, and a semi- minor axis whose value is b = a √ [1 - ε²] and two foci one equals to c = ε a, and the other foci location equals to - c = - ε a. Or the motion of one ball around the other ball should be an ellipse with the

other ball at one of the foci +/- ε a, of the ellipse with semi major axes (- a, a) on x-axis and semi minor axes (-b, b) on y- axis and foci (- c, c) on x- axis again with angle of rotation θ. If this law to work on two planets instead of two balls like planetary motion around the sun, then Astronomers should see this motion of a planet moving around the sun in an ellipse with the sun at one of its foci.

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What astronomers saw was not an ellipse but a rotating ellipse like the ellipse below wth rotating angle ψ

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M

m

θ

M

ψ

m

θ

When the areas size A are sliced equally it was found that the times spent by planets orbiting around the Sun and making areas A each are equal also. Or, r² (0) θ' (0) = r² (1) θ' (1) = r² (2) θ' (2) = r² (3) θ' (3) = -- = location² x [angular speed] = constant = Areal velocityIf r² θ' = hThen differentiating with respect to timeThen d (r² θ')/ d t = d h/ d t And 2 r r’ θ’ + r² θ’’ = 0Or 2 (r’ /r) = - (θ’’/θ’)The r = r0 e í ω t

And θ’ = θ’0 e – 2 í ω t



With r² (0) θ' (0) = r² (1) θ' (1)Then θ' (1) = [r² (0)/ r² (1)] θ' (0) And θ' (1) = {[e í 2 ω t] θ' (0) And θ’ (1) = {[cosine 2 ω t + í sine 2 ω t] – 1} θ' (0) Δ θ’ (1) = Δ θ’ (x) + í Δ θ’ (y) = [cosine 2 ω t + í sine 2 ω t] θ' (0) Δ θ’ (x) = [cosine 2 ω t] θ' (0) Δ θ’ (x) = [1 – 2 sine² ω t] θ' (0)

Δ θ’ (x) - θ' (0) = - 2 θ' (0) sine² ω tΔ W = Δ θ’ (x) - θ' (0) = - 2 θ' (0) sine² ω t = -2(2π) [√ (1-ε²)]/T (1-ε) ²] sine² ω t And θ' (0) = h/r² (0) = 2 π a b/T a² = 2π√ (1-ε²)]/T (1-ε) ²

Δ W = - 4 π) [√ (1-ε²)]/T (1-ε) ²] sine² ω t

If this apsidal motion is to be found as visual effects, then With, v ° = spin velocity; v* = orbital velocityAnd v/c = (v* + v°)/c = tan ω t

Δ W = -4 π [√ (1-ε²)]/T (1-ε) ²] sine² {Inverse tan [(v* + v°)/c]} radians

Multiplication by 180/π to change to degrees

Δ W ° = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [(v* + v°)/c]} Degrees And multiplication by 1 century = 36526 days and using T in days

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W° (ob) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x sine² {Inverse tan [(v* + v°)/c]} degrees/100 years

Astronomy re - written

Primary →Secondary ↓

v°(p) ↑ v* (p)↑ v° (p) ↑v* (p)↓ v° (p) ↓ v* (p) ↑ v° (p) ↓V* (p) ↓

v°(s) ↑ v* (s)↑ Spin=[↑,↑][↑,↑]=orbit

[↑,↑][↓,↑] [↓,↑][↑,↑] [↓,↑][↓,↑]

Spin results v°(p) + v°(s) v°(p) + v°(s) - v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples AS CAMv° (s) ↑v* (s)↓ [↑,↑][↑,↓] [↑,↑][↓,↓] [↓,↑][↑,↓] [↓,↑][↓,↓]Spin results v°(p) + v°(s) v°(p) + v°(s) -v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples

v° (p) ↓ v*(s) ↑ [↑,↓][↑,↑] [↑,↓][↓,↑] [↓,↓][↑,↑] [↓,↓][↓,↑]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examples AS CAM

v° (s) ↓V*(s) ↓ [↑,↓][↑,↓] [↑,↓][↓,↓] [↓,↓][↑,↓] [↓,↓][↓,↓]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples AS CAM

W° (calculated) = (-720x36526/T) {[√ (1- ε²)]/ (1- ε) ²]} [(v° + v*)/c] ² degrees/100 years

1- As Cameloppardalis: Binary stars System AS Cam Data T=3.431; r (m) =0.1499; m=3.3 M (0); M=2.5 M (0)R (m) =2.57 R (0); [v° (m), v° (M)] = [40, 30]; ε = 0.1695; 1- ε = 00.8305R (M) = 2.5 R (0); r (M) =0.1111; m + M=5.8 M (0); G=6.673x10-11

M (0) = 2 x 1030 kg; R (0) = 0.696x109m; The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)Finding orbital velocitiesFrom Newton's inverse square law of an ellipse motion applied to a circular orbit gives the following: m v²/ r (cm) = GmM/r²Planet --- r (cm) ----- Center of mass ------- r (CM) --------- Mother Sun Planet ------------------- r -------------------------------------- Mother SunCenter of mass law m r (cm) = M r (CM); m = planet mass; M = sun massAnd r (cm) = distance of planet to the center of mass

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And r (CM) = distance of sun to center of massAnd r (cm) + r (CM) = r = distance between sun and planet Solving to get: r (cm) = [M/ (m + M)] r And r (CM) = [m/ (m + M)] r Then v² = [GM r (cm)/ r²] = GM²/ (m + M) r And v = √ [GM²/ (m + M) r = a (1-ε²/4)] Planet orbital velocity or primary velocity:

And v* (m) = v (m) = √ [GM²/ (m + M) a (1-ε²/4)] R =a (1-ε²/4)

Calculations: (1- ε²/4) = 0.9928 [√ (1-ε²)]/ (1-ε) ² = 1.43With a = [R (m)/r (m)] = (2.57/0.1499) (0.696x109) = 11.9327x109

And v* (m) = √ [GM²/ (m + M) a (1- ε²/4)] = 110km/sec= √ [6.673x10-11 (2.5)² (2 x 1030) / (5.8) 11.932x109 (0.9928)] = 110.178km/sec

And v* (M) = √ [GM²/ (m + M) a (1- ε²/4)] = √ [6.673x10-11 (3.3)² (2 x 1030) / (5.8) 11.932x109 (0.9928)] = 145.435 km/secSpin: v° = 40 + 30=70km/secAnd v* (e) = v* Earth = 29.8 km/secAnd vº (e) = 0.465 km/secThen v* + v °= v* (m) + v* (M) + v* (e) + vº (m) + v° (M) + vº (e) = 110.178 + 145.435 + 70 + 29.8 – 0.465 = 355 m/sec[√ (1-ε²)]/ (1-ε) ² = 1.43; T = 3.431days

W°= (-720x36526/T) x {√ [(1-ε²)] (1-ε) ²} {[v* + v°]/c} ²

W°= (-720x36526/3.431) x (1.43) (355/ 300,000) ² = 15.0°/century Dr Guinan and DR Maloney 1989: W°= 15°/century 1989DR Khailullin and Dr Kozyreva: 1983 W º= 14.6 °/century Einstein: 44.3º/ century

2 - Binary stars System: DI Her Apsidal Motion SolutionDI Her Apsidal motion solution: Data: T=10.55days r(m) = 0.0621 m=5.15M(0) R(m)=2.68R(0) [v°(m),v°(M)]=[45,45] And ε = 0.4882; r (M) = 0.0574 M=4.52 M (0) R (M) =2.48; m + M=9.67 M (0)L = 2000 +/- 200 Ly

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Calculations1- ε = 0.5118; (1-ε²/4) = 0.94; [√ (1-ε²)] / (1-ε) ² = 3.33181; 1 + ε = 1.4882; G=6.673x10-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x109mV* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 99.88 km/sec V* (s) = √ [Gm²/ (m + M) a (1-ε²/4)] = 113.8km/sec

A- Apsidal motion is given by this formula: W° (cal) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years And v° = - 45 km/s - 45 km/sec = 90km/sec Now let us calculate v* (cm) = ∑m v/∑m = 106.38km/secWith v* = 2 v*(cm) = 212.76 km/sec And v° = -90 km/secThen v* + v° = 212.76 - 90 = 122.76 km/sec and W° (cal) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} sine² [Inverse tan 122.76/300,000] = (-720x36526/10.55) (3.33181) (122.76/300,000)²W° (cal) = 1.39°/centuryWith σ = √ {∑ [v* - v* (cm)] ²/2} = √ {[106.36 - 99.88]²/2 + [106.36 - 113.8]²/2} = 6.975 km/secThen W° (cal) = 1.39°/century +/- 0.16Observed 2004: 1.39° +/- 0.3°/century; Relativity: 4.27°/century

B - W° (cal) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 yearsWith v° = - 45 km/s + 45 km/sec = 0 km/sec And v* = v* (cm) = ∑m v/∑m = 106.38km/sec

W° (cal) = (-720x36526/10.55) (106.38.300, 000) ² degrees/100 yearsW° (cal) = 1.044 degrees/centuryC - V* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 99.88 km/sec

With v* x √ (1-ε²/4)/√ (1-ε) = 99.88 √ (.94)/√ (0.5118) = 135.36And v* x √ (1-ε²/4)/√ (1+ ε) = 99.88 √ (.94)/√ (1.4882) = 79.38

K (A) = [(135.36 + 79.38)/2] = 107.37 km/sec V* (s) = √ [Gm²/ (m + M) a (1-ε²/4)] = 113.8 km/sec

With v* x √ (1-ε²/4)/√ (1-ε) = 113.8 √ (.94)/√ (0.5118) = 154.2254 km/secAnd v* x √ (1-ε²/4)/√ (1+ ε) = 113.8 √ (.94)/√ (1.4882) = 90.44312 km/secK (B) = 122.3342 km/sec

V* = [m K (A) + M K (B)]/ (m + M) = 114.3646 km/secOr v* = [K (A) + K (B)]/2 = 114. 85 km/sec; with v* = 114.85

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W° (cal) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} sine² [Inverse tan 114.85/300,000]

= (-720x36526/10.55) (3.33181) (114.85/300,000)² = 1.22° degrees/ centuryObserved: 1.24° degree/century +/- 0.05°

References: Go to Smithsonian/NASA website SAO/NASA and type: 1- Apsidal motion of DI Her: Dr Edward Guinan and Dr Frank Maloney; 1985.2- New Apsidal Motion of DI Her: Dr Edward Guinan and Dr Frank Maloney; 1994. 3- D. YA. Martynov; and KH. F. Khaliulullin 1980 4- Petrie et al.19675- Petrova - Ovlav Apsidal motion catalogue6- Riazi 20037- Maloney Guinan 2004

3 - V1143 Cgyni Apsidal Motion Solution

V1143 Cgyni data T= 7.641days; r (m) = 0.059; m =1.391 M (0); R (1) =1.346R (0); ε = 0.54 And [v ° (m), v° (M)] = [18, 28]; r (M) = 0.058; M=1.347 M (0) Distance [38 +/- 2 parsec] = 123.956 +/- 6.524 Ly CalculationsWe have 1- ε = 0.46 1- ε²/4=0.9721 R (0) = 0.696x109m With a = [R (1)/r (m)] R (0) = 15.87823729x10^9m; 1+ ε = 1.54With v (p -perihelion) = √ [GM²/ (m + M) a (1- ε)] = 110 km/secAnd v (p- aphelion) = √ [GM²/ (m + M) a (1+ε)] = 60 km/secK (A) = (110 + 60)/2 = 170/2 = 85km/secWith v (s - perihelion) = √ [Gm²/ (m + M) a (1-ε)] = 113.6 km/secAnd v (s - aphelion) = √ [Gm²/ (m + M) a (1+ε)] = 62 km/secK (B) = (113.6 + 62)/2 = 175.6/2 = 87.8km/sec

With v (1) = √ [GM²/ (m + M) a] = 74.632 km/sec And v (2) = √ [Gm²/ (m + M) a] = 77.0699 km/secWhen spinning on opposite directions

1- With v° [21, 28] = 28 - 21 = 72- With v° [18, 28] = 28 - 18 = 103- Taking average 10 + 7/2 = 8.5

With v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 77.5126 km/sAnd v (M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 80.00448 km/s Also, [√ (1-ε²)]/ (1-ε) ² = 3.977622971

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Now: With 1- v° + v* = 157.51648km/sec - 10 km/sec = 147.51648km/sec And 2- v° + v* = 157.51648km/sec - 8.5 km/sec = 149.01648km/sec And 3- v° + v* = 157.51648km/sec - 7 km/sec = 150.51648km/sec

W° (obo) = (-720x36526/T) x {[√ (1-ε²)]/ (1-ε) ²} {[v* + v°]/c} ²

1- W°/century= (-720x36526/7.641) (3.977622951) (147.51648/300,000)²=3.31°/century2- W°/century= (-720x36526/7.641) (3.977622951) (149.01/300,000)²=3.3778°/century3- W°/century= (-720x36526/7.641) (3.977622951) (150.51648/300,000)² = 3.44614561°/century

With v* = 2 v* (cm) = 157.4770 km/secAnd v ° = v° (p) - v° (s) = 21 - 28 = -7 km/secThen v* + v ° (p) = 157.477 - 7 = 150.477 km/sec

W° /century= (-720x36526/7.641) (3.977622951) (150.477/300,000)²=3.44°/century Observed values are: W° = 3.393987698°/century; W° = =3.489592985Average observed: 3.44°/ century

References: 1-Geminez and Margrave, 1985 [0.00071°/cycle] = [1 century = 36526days/7.641days] = 3.393987698°/century 2- Anderson and Nordstrom and Garcia and Geminez 1987: 0.00073°/cycle [0.00073°/cycle] = [1 century = 36526days/7.641days] = 3.489592985°/century

Relativity theory: 4.254435283°/ century = 0.00089°/cycle

4- V541Solution: Apsidal motion catalogue

W° (cal) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years T = 15.3379days; r (m) = 0.0440 m = 2.4 M (0) R (m) = 1.88 R (0)

And [v°(m),v°(M)]= [24±2,24±2]And ε = 0.479 r (M) =0.0425 M=2.4 M (0) R (M) =1.79 R (0) With 1- ε = 0.521 1-ε²/4=0.94263975; [√ (1-ε²)]/ (1-ε) ² = 3.2339

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And a = [R (m)/r (m)] R (0) = (1.88/0.0440)0.696x10^9m = 29.73818182x10^9mThen a (1-ε²/4) = 28.03x10^9mAnd v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 75.5883km/sec; v° (m) =24 And v (M) = √ [Gm²/a (m + M) (1-ε²/4)] = 75.883km/sec; v° (M) =24With v°=24 + 24= 44km/secAnd v* = 151.1766km/secWith v* + v°= 151.1766 - 48 = 103.1766km/secW° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} {[v* + v°]/c} ²

W° (ob) = (- 720 x 365226/15.3379) (3.2339) (103.1766/300,000)²W° (ob) = 0.65°Notice: [v° (m), v° (M)] = [24 ± 2, 24 ± 2]If v° (m) = v° (M) = 24 + 2 = 26 Then v* + v°= 151.1766 - 52 = 99.1766km/secAnd W° (ob) = (- 720 x 365226/15.3379) (3.2339) (99.1766/300,000)²W° (ob) = 0.60°/centuryObserved is W°= 0.60° +/ -0.1/century Lacy = [0.5°; 0.7°]

Relativity: W° = 0.97°/century

1- Apsidal motion of V541Cgyni Lacy 1989

5 - AI Hydra apsidal motion puzzle solution

W° (cal) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years

AI Hydra Apsidal motion solution: Data: T=8.29days r(m) = 0.1418 m=2.15M(0) R(m)=3.92R(0) [v°(m),v°(M)]=[28,27]; and ε = 0.0.23; r (M) = 0.1002 M=1.98 M (0) R (M) =2.77(0); m + M=4.13 M (0); L = 575 +/- 15 Ly

Calculations1-ε = 0.77; (1-ε²/4) = 0.986775; [√ (1-ε²)] / (1-ε) ² = 1.6414 G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mThen a = [R (m) / r (m)] = 19.24062059 x 10 ^ 9 m

V (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 81.11439578 km/sec V (M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 88.11km/sec

Apsidal motion is given by this formula:

W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years

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With v* = v (m) + v (M) = 81.11439578km/sec + 88.11km/sec = 169.2243958km/secAnd v° = v° (m) + v° (M) = 28 + 27 = 55 km/sec Then (v* + v°) = 224.2243958 km/secW° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} sine² [Inverse tan 224.2243958/300,000] = (-720x36526/8.29) (1.6414) (224.2243958/300,000)² = 2.90°/ century W° (observed) = 2.90°/century as measured Observed is 2.90°/centurySpace-Time Relativity theory 6.8° / century

Now let us calculate v* (cm) = ∑m v/∑m = 2.15 x 81.11439578 + 1.98 x88.11= 84.46822 km/secWith v* = 2 v*(cm) = 168.936411 km/sec; And v° = 55 km/secThen v* + v° = 223.9364411 km/sec and

W° (observed) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} sine² [Inverse tan 223. 936/300,000]= (-720x36526/10.55) (3.33181) (223.9364411/300,000)²W° (observed) = 2.9°0/centuryWith σ = √ {∑ [v* - v* (cm)] ²/2} = √ {[88.11 -84.46822]²/2 + [81.11439578 - 84.46822]²/2} = √ {[3.64178]²/2 + [3.35382422]²/2} = 3.5 km/sec Then W° (ob) = 2.9°0/century; observed is W° = 2.9°/per centuryAnd Einstein's 100,000 space-timers 6.8° / centuryReferences: Go to Smithsonian/NASA website SAO/NASA and type: 1- Apsidal motion of AI Hydra Popper 1985 2- KH. F. Khaliulullin and V.S Kozyreva 1988 3- Petrova - Ovlav Apsidal motion catalogue 19996 - V731Cehpei Apsidal Motion Solution



v°(s) ↑ v* (s)↑ Spin=[↑,↑][↑,↑]=orbit

[↑,↑][↓,↑] [↓,↑][↑,↑] [↓,↑][↓,↑]

Spin results v°(p) + v°(s) v°(p) + v°(s) - v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples V731Cepheiv° (s) ↑v* (s)↓ [↑,↑][↑,↓] [↑,↑][↓,↓] [↓,↑][↑,↓] [↓,↑][↓,↓]Spin results v°(p) + v°(s) v°(p) + v°(s) -v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examplesv° (p) ↓ v*(s) ↑ [↑,↓][↑,↑] [↑,↓][↓,↑] [↓,↓][↑,↑] [↓,↓][↓,↑]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examplesv° (s) ↓V*(s) ↓ [↑,↓][↑,↓] [↑,↓][↓,↓] [↓,↓][↑,↓] [↓,↓][↓,↓]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples V731Cephei

Next the same equation will be used to find the advance of Periastron or "apparent" apsidal motion of V731 binary stars system. V731Cehpei Apsidal Motion Solution

W° (ob) = (-720x36526/T) x {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years

V731data [see below]

T= 6.068567days; m= 2.577 M (0); M = 2.577 M (0)And [v° (m), v° (M)] = [19+/-3, 18+/-3]

And ε = 0.0165 a = 23.27x R (0)

CalculationsM + m = 2.738; 1-ε=0.9835 1-ε²/4=0.9999 R (0) = .696x10^9m

With v° [21, 28] = [19 +/- 3] + [18 +/- 3] = 37 +/- 6

With v* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 85.6111965km/sec

And v* (M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 109.38km/sec

Also, [√ (1-ε²)]/ (1-ε) ² = 1.033694356

With v* (cm) = 2∑m v∑/m = 96.46688km/sec; 2 v* (cm) = 192.9337619

And σ = √ {∑ [v*-v* (cm)] ²/2} = √ {[96.46688 - 85.6] ²/2} + {[109.38 - 96.46688] ²/2} = 11.9288422 km/sec

With v* (p) = 85.6111965km/sec +/- 11.9288422 km/sec

And v* (s) = 109.38km/sec +/- 11.9288422 km/sec

Then v* (p) + v* (s) = [192.9337619 +/-] x 2 = 23.8567844 km/sec

Then v* + v° = 229.9288422km/sec +/- 29.8567844 km/sec

Now: Taking the upper limitThen v* + v° = 229.9288422km/sec + 29.8567844 km/sec = 259.7856266 km/sec

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W° (obo) = (-720x36526/T) x {[√ (1-ε²)]/ (1-ε) ²} {[v* + v°]/c} ²W°/century= (-720x36526/6.068567) (1.033694356) (259.7856266 /300,000)²=2.91°/centuryW°/century = 3.35914177°/century = 0.0335914177°/year

U = 360/0.0335914177° = 10717 years

Observed values are U = 10000 +/- 2500

References: 1- Absolute dimensional and apsidal motion of V731CepV. Batkis; M.Zejda; I. Bulut; M.Wolf; S. Bilir; H. Bakis; O.Demircan: J.w.Lee: M.Slechta: B. Kucerova. 2008

7 - NV CMa Binary stars apsidal motion table

8 - NV CMa apsidal motion solution:Data: T=1.885159 days; ε = 0; v* (p) = 128.55 km/sec; v* (s) = 130.87 km/sec[√ (1-ε²)] / (1-ε) ² = 3.33181; v° (p) = 51.7 km/sec and v° (s) = 52.4 km/secApsidal motion is given by this formula: W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years

With v* = v* (p) + v*(s) = 259.42 km/sec and v° = v° (p) + v° (s) = 104.1And v* + v° = 363.52 km/sec

W° (observed) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} sine² [Inverse tan 363.52/300,000] = (-720x36526/1.885159) (1) (363.52/300,000)² = 20.48333818°/century = 0.2048333818°/year

U = 360°/0.2048333818°/year;

U = 1757. 5 years

References: Go to Smithsonian/NASA website SAO/NASA and type: Absolute dimensions NV CMa; Kaluzny, J; Pych, W; Rucinski, S. M; Thompson, I.B

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Relativity theory coffin nail # 8 SW Canis Majoris

W° (cal) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years

T = period = 10.09 days; ε = eccentricity = 0.3179

And v° = spin velocity effect = v° (p) + v°(s) = 57km/sec

And v*= orbital velocity effect = v*(p) + v* (s) = 80.5 + 87.8 = 168.3 km/sec

For SW Canis Majoris: v* + v° = 225.3km/sec

W° (observed) = 2.99565967°/century = 0.0299565967°

U = 360/ 0.0299565967= 12017years

U (observed) = 12,000 years Einstein and space-timers 14,000 years SW Canis Majoris Binary stars



v°(s) ↑ v* (s)↑ Spin=[↑,↑][↑,↑]=orbit

[↑,↑][↓,↑] [↓,↑][↑,↑] [↓,↑][↓,↑]

Spin results v°(p) + v°(s) v°(p) + v°(s) - v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples Sw Canis Majorisv° (s) ↑v* (s)↓ [↑,↑][↑,↓] [↑,↑][↓,↓] [↓,↑][↑,↓] [↓,↑][↓,↓]Spin results v°(p) + v°(s) v°(p) + v°(s) -v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examplesv° (p) ↓ v*(s) ↑ [↑,↓][↑,↑] [↑,↓][↓,↑] [↓,↓][↑,↑] [↓,↓][↓,↑]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examplesv° (s) ↓V*(s) ↓ [↑,↓][↑,↓] [↑,↓][↓,↓] [↓,↓][↑,↓] [↓,↓][↓,↓]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples SW Canis majoris

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SW CMa apsidal motion solution: Data: T=10.09 days; r (m) = 0.0942; m = 2.22 M (0); R (m) = 3.01R (0); ε = 0.3179 And r (M) = N/A M = 2.03 M (0) R (M) =2.46 R (0); m + M = 4.25 M (0)And [v° (m), v° (M)] = [30+/-2, 27+/-3] K (1) = 80.5; K (2) = 87.8Calculations

1-ε = 0.6821; (1-ε²/4) = 0.974734898 [√ (1-ε²)] / (1-ε) ² = 2.037835646

G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mThen a = [R (m) / r (m)] = 22.23949045 x 10 ^ 9 m

V* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 77.26298km/sec

V* (s) = √ [Gm²/ (m + M) a (1-ε²/4)] = 84.4944913km/sec

And v* = v* (p) + v* (s) = 161.7574713 km/secWith v° = v° (p) + v° (s) = 30 + 27 = 57 km/sec

And, v* + v° = 218.7574713km/sec

Apsidal motion is given by this formula: W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years

W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} sine² [Inverse tan 218.7574713/300,000] = (-720x36526/10.09) (2.037835646) (218.7574713/300,000)² = 2.8242°/ century = 0.026242/yrU = 360/ 0.026242 = 12747 years Taking: v* + v° = 80.5 + 87.8 +57 = 225.3 km/secW° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} Sine² [Inverse tan 225.3 /300,000] = (-720x36526/10.09) (2.037835646) (218.7574713/300,000)² = 2.995659677°/ century = 0.0299565967°/yr

U = 360/ 0.0299565967 = 12017 years NahhasU (observed) = 12,000 years

References: Go to Smithsonian/NASA website SAO/NASA and type: 1- Lacy Apsidal motion Canis Majoris 1997

73

9 - GG Orion Apsidal motion puzzle solutionPrimary →Secondary ↓


v°(s) ↑ v* (s)↑ Spin=[↑,↑][↑,↑]=orbit [↑,↑][↓,↑] [↓,↑][↑,↑] [↓,↑][↓,↑]Spin results v°(p) + v°(s) v°(p) + v°(s) - v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples GG Orionv° (s) ↑v* (s)↓ [↑,↑][↑,↓] [↑,↑][↓,↓] [↓,↑][↑,↓] [↓,↑][↓,↓]Spin results v°(p) + v°(s) v°(p) + v°(s) -v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples

v° (p) ↓ v*(s) ↑ [↑,↓][↑,↑] [↑,↓][↓,↑] [↓,↓][↑,↑] [↓,↓][↓,↑]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examples

v° (s) ↓V*(s) ↓ [↑,↓][↑,↓] [↑,↓][↓,↓] [↓,↓][↑,↓] [↓,↓][↓,↓]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examples GG OrionData: T=6.6314948; m = 2.342 M (0); M = 0.2338 M (0); R (1) = 1.852 R (0); R (2) =1.830

ε = 0.2218; 1 - ε = 0.7782; r (1) = 0.0746; r (2) =.988 r (1); m + M = 4.68 M (0) And [v° (p); v° (s)] = [16 +/- 1; 16 +/- 1]; [v° (p); v° (s)] = [25 +/- 3; 24 +/- 3]; U = 10700 +/- 4500 yearsCalculations (1-ε²/4) = 0.9877; [√ (1-ε²)] / (1-ε) ² = 1.57 G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mCalculationsWith v* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 95.6 km/secAnd v* (s) = √ [Gm²/ (m + M) a (1-ε²/4)] = 95.735 km/secAnd v° (p) = 16 km/sec; v° (s) = 16 km/secThen v* (p) + v* (s) + v° (p) + v° (s) = 223.335 km/secW° (ob) = (-720x36526/6.6314948) (1.57) [223.335/300,000] ² degrees/100 yearsW° (ob) = 3.45°/century = 0.0345°/yearU [years] = 360/[0.0345°/year]U = 10,432 years NahhasU (observed) = 10700+/-4500years

References: Absolute dimensions and apsidal motion of eclipsing binary GG Orion Dr Lacy; Dr Torres; Dr Claret; Dr Sabby: 2000 The time has come to send relativity theories and all four-dimensional space-time confusion of physics to the...

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Chapter TwelveArabs real time physics in outer Space II

A- Arabs Binary stars and Binary Pulsar apsidal motion2009 total collapse of general relativity theory; CD Draconis Binary stars



v°(s) ↑ v* (s)↑ Spin=[↑,↑][↑,↑]=orbit

[↑,↑][↓,↑] [↓,↑][↑,↑] [↓,↑][↓,↑]

Spin results v°(p) + v°(s) v°(p) + v°(s) - v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v* (p) + v*(s) -v* (p) + v* (s)Examples CD Draconisv° (s) ↑v* (s)↓ [↑,↑][↑,↓] [↑,↑][↓,↓] [↓,↑][↑,↓] [↓,↑][↓,↓]Spin results v°(p) + v°(s) v°(p) + v°(s) -v°(p) + v°(s) -v°(p) + v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)Examplesv° (p) ↓ v*(s) ↑ [↑,↓][↑,↑] [↑,↓][↓,↑] [↓,↓][↑,↑] [↓,↓][↓,↑]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) + v*(s) -v*(p) + v*(s) v*(p) + v*(s) -v* (p) + v* (s)Examplesv° (s) ↓V*(s) ↓ [↑,↓][↑,↓] [↑,↓][↓,↓] [↓,↓][↑,↓] [↓,↓][↓,↓]Spin results v°(p) - v°(s) v°(p) - v°(s) -v°(p) - v°(s) -v°(p) - v°(s)Orbit results v*(p) - v*(s) -v*(p) - v*(s) v*(p) - v*(s) -v* (p) - v* (s)

Examples CD Draconis

10 – 2009 CD Draconis apsidal motion Table:Data: T=1.268389985days; m = 0.231 M (0); M = 0.2141 M (0); a = 3.7634 R (0); ε = 0.0051And [v° (p); v° (s)] = [9.5 +/- 1; 10.0 +/- 1]Einstein and space – timers: W° = 1.91x10-3/day Calculations m + M = 0.4451 M (0)1-ε = 0.9949; (1-ε²/4) = 0.99993498; [√ (1-ε²)] / (1-ε) ² = 1.01 G=6.673x10-11; M (0) = 1.98892x1030kg; R (0) = 0.696x10 9mCalculationsWith v* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 72.436 km/secAnd v* (s) = √ [Gm²/ (m + M) a (1-ε²/4)] = 78.153 km/secAnd v° (p) = 9.5 km/sec; v° (s) = 10 km/secThen v* (p) + v* (s) + v° (p) + v° (s) = 170.117 km/sec

Page 75

Apsidal motion is given by this formula:

W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² degrees/100 years

W° (ob) = (-720x36526/1.238389985) (1.01) [170.09/300,000] ² degrees/100 yearsW° (ob) = 6.731598944°/century = 0.06731598944°/year

U [years] = 360/[0.06731598944°/year]

U = 5348 years Nahhas

U (observed) = 5400+/-3200yearsEinstein's and space-timers U = 360/ [0.00191x365.26] = 516 yearsCan it get any better?It is not just about dumping relativity it is dumping relativity and Alfred Nobel institution with it.

References: Absolute properties of the low-mass eclipsing binary CM Draconis; 2009By : Juan Carlos Morales; Ignasi ribas; carme jordi; Guillermo Toress; Jose Gallardo; Edward F. Guinan; David Chardonneau; Marek wolf; David w.latham; Guillem Angalada Escude; David H.Bradstreet; Mark E.Everett; Francis T. O, Donavan; Georgi Mandushev; Robert D. Mathieu and other 15

11 - V 1147 SagittariV 1147 Sagittari: Data: U = 592.5 +/- 6 years

With K (p) = 117.4 km/sec; K(s) = 130.8 km/secGiving v* (p) = 117.4 + 29.8 - 0.465 = 146.735 km/secAnd v* (s) = 130.8 + 29.8 – 0.465 =160.135 km/secOrbital speed of earth = 29.8 km/sec; spin speed of earth is 0.465 km/secAnd vº (p) = 80 + 29.88 – 0.465 km/sec = 109.335 km/secAnd vº(s) =75 + 29.88 – 0.465 km/sec = 104.335 km/secThen v* + vº = 146.735 + 160.135 + 109.335 + 104.335 = 520.54km/sec[√ (1- ε²)] (1-ε) ² = 2.62 W (cal) = (-720x36526/3.28) x (2.62) (520.541/300,000)²= 63º/centuryU = [63/100]-1x 360 = 570 years.Observed values; 592.5 ReferencesAbsolute dimensions of solar-type eclipsing binaries V 1147SagittariJ. Anderson, A. Gimenez 1985

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12- EW Orionis EW Orionis: Data: U =16300 +/- 3900With K (p) = 73.18km/sec; K(s) = 75.46 km/secGiving v* (p) = 73.18 + 29.8 - 0.465 = 102.515 km/secAnd v* (s) = 75.46 + 29.8 – 0.465 =104.795 km/secAnd vº (p) = 9 + 29.8 - 0.465 = 102.515 km/secAnd vº (s) = 73.18 - 29.8 + 0.465 = 102.515 km/sec

Orbital speed of earth = 29.8 km/sec; spin speed of earth is 0.465 km/secAnd vº (p) = 9km/sec; vº(s) =8.8km/sec Then v* + vº = 102.515 + 104.795 +9.0 + 8.8 =225.11km/sec[√ (1- ε²)] (1-ε) ² = 1.167

W (cal) = (-720x36526/6.93) x (1.167) (225.11/300,000)²= 2.1367º/centuryU = [2.1367/100]-1x 360 = 16848 years. 13 - The Apsidal motion of VV PYX

Data: T = 4.596 days; m = 2.098 M (0); M = 2.098 M (0)With r (p) = 0.1156 and r (s) = 0.1156 and R (m) = 2.167 R (0) and R (M) = 2.167 R (0)And ε = 0.0956; 1 - ε = .9044; U = 3200 +/- 1000And [v° (p); v° (s)] = [23; 23] V = 381.03095km/sec

Calculations

With m + M = 4.196 M (0)And (1-ε²/4) = 0.99771516; [√ (1-ε²)] / (1-ε) ² = 1.216984877G=6.673x10^-11; M (0) = 1.98892x19^30kg; R (0) = 0.696x10^9mAnd a = [R (m)/r (m)] = [2.06/ 0.097] R (0) = [2.167/ 0.1156] [0.696x109 m]

Then a = 13.04698962 x 109 mAnd a (1-ε²/4) = [13.04698962 x10^9m] [0.99771516] = 13.01717934 x109m

K (A) = {√ [GM²/ (m + M) a (1- ε)] + √ GM²/ (m + M) a (1+ ε)]}/2

K (A) = {√ [6.673x2.098²x2x10-19/ (4.196) 13.04698962x10^9x0.9044] + √ [6.673x2.098²x2x10-19/ (4.196) 13.04698962x10^9x1.0956]}/2 = [108.85 + 98.96511511]/2 = 103.9km/sec

K (B) = {√ [Gm²/ (m + M) a (1- ε)] + √ Gm²/ (m + M) a (1+ ε)]}/2 = 103.9km/sec

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With v* (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 103.70 km/sec

And v* (M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 103.70 km/secAnd v* (cm) = [m v* (p) + M v* (s)]/ (m + M) = 103.70 km/sec

And v° (p) = 23vkm/sec; v° (s) = 23 km/sec

W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]} (v* + v°/c) ² degrees/100 years

T = period; ε = eccentricity; v° = spin velocity effect; v*= orbital velocity effects

With v* (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 103.70 km/secPrimary orbital speed contribution to apsidal motion

Is v* (m) = 103.7 + 29.8 – 0.465 = 133 km/secAnd v* (M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 103.70 km/sec

Secondary orbital speed contribution to apsidal motionIs v* (M) = 103.7 + 29.8 – 0.465 = 133 km/secAnd v° (m) = 23 km/sec

Primary spin speed contribution to apsidal motionIs vº (m) = 23 + 29.8 – 0.465 = 52.335 km/secAnd v° (M) = 23 km/sec

Secondary spin speed contribution to apsidal motionIs vº (m) = 23 + 29.8 – 0.465 = 52.335 km/secThen v * + vº = 370.67

With ε = 0.17; T = 4.596 days[√ (1-ε²)] / (1-ε) ² = 1.216984877

W° (cal) = (-720x36526/4.596) (1.216984877) [370.67/300,000] ²

U = 3386 years Nahhas'U (observed) = 3200 +/- 1000years

ReferencesAbsolute dimensions of solar-type eclipsing binaries EW OrionisJ.V. Clausen, H. Bruntt, E.H. Olsen, B.E. Helt, and A. Claret 2009

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Chapter 13: Alfred Nobel prize winner for less than a $1993 Nobel Prize Winner Dr Joseph H. Taylor

Binary Pulsar 1913 + 16For PSR 1913 + 16 Data: Joseph Taylor and Joel M. Weisberg 2004With a =2.3417725 R (0); R (0) = 0.696 x 10 9metersAnd ε = 0.6171338T = 0.322997448930And θ' = 4.226595 °/yearWith mass m = 1.4414 M (0) and M = 1.3867 M (0); M (0) = 2 x 1030 kgV* (p) = √ [GM²/ (m + M) a] = 235.9574664 km/sec V* (s) = √ [Gm²/ (m + M) a] = 245.2640841km/secW° (cal) = (-720x365.26/Tdays) x sine² [Inverse tan (v/c)] degrees/100 years= (-720x365.26/0.329974489) x sine² [Inverse tan (481.22/300, 00)] deg/100 yearsW = 4.2°/year

15- PSR J0737 – 3039Data: U =21.3 yearsWith m = 1.34 M (0); M = 1.25 M (0); M (0) = 2 x 1030

And ε = 0.0878; [vº (p), vº (s)] = [22.7 km/sec, 2km/sec]With G = 6.673 x 10 – 11; T =2.4 hours; a = 1.145 x 0.069 x 109

With K (p) = v* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 303 km/secAnd K(s) = v* (m) = √ [Gm²/ (m + M) a (1-ε²/4)] = 325 km/secGiving v* (p) = 303 + 29.8 - 0.465 = 332.335 km/secAnd v* (s) = 325+ 29.8 – 0.465 =354.335 km/secAnd vº (p) = 22.7 + 29.8 - 0.465 km/secAnd vº (s) = 2 - 29.8 + 0.465 km/secOrbital speed of earth = 29.8 km/sec; spin speed of earth is 0.465 km/sec

Then v* + vº = 332.335 + 354.335 +22.7 + 2 =711.37km/sec[√ (1- ε²)] (1-ε) ² = 1.197

W (cal) = (-720x36526/6.93) x (1.197) (711/300,000)²= 1770º/centuryU = [1770/100]-1x 360 = 20.3389 years.

ReferencesThe Double pulsar PSR J0737 - 3039Burgay, M; D’Amico,N.; A.Manchester,R.N; Lyne, A.G.; Kramer, M; Mclaughlin, M.A; Lorimer, D.R; Carmilo, F.; Stairs, I.H: Freire, P.C.C; Joshi,

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Chapter 14B- Arabs Real time GPS or Real time Global positioning systems

B- Arabs real time Global Positioning System: GPS

Abstract: The Global Positioning System or GPS 45 micro seconds per day time delays have nothing to do with Einstein's relativity theory time travels confusions of physics and they are a consequence of Satellite orbital speed and Earth rotational speed given by this formula below. Even if Einstein's formulas were correct for all practical purposes they are insignificant to the performance to the GPS system. Earth - Satellite distance is a variable that Engineers account for and the tiny "relativistic" effect has no significance whatsoever because distance adjustment is far more than any relativistic adjustment. For the president of the United States President to ask for Innovations like that in an AAAS meeting is a sign of a bankrupt scientific community and for those who knew the facts must have laughed to their teeth at the president.

NASA's and AAAS laughing at the president requires a reply with laugh at AAAS and NASA and all other including the person who wrote the speech for the president

GPS time delays of 45 micro seconds per day have nothing to do with relativity theory or Einstein or AAAS or NASA or silly time travel physics or any theory. This time delay is due to Earth rotation and satellite orbital speed that Scientists turn their heads away from because without Einstein's silly magic sock of time travel physics, Physicists will have less tricks that allows them to say anything publish anything based on nothing for jobs money prestige Nobel Prizes and silly physics.

W" (ob) = (-720x3600x15) [(v° +/- v*)/c] ² arc sec /day

T = period; ε = eccentricity; v° = spin velocity of earth; v*= orbital velocity of satelliteAnd v* = 14000km/hr = 3.88888888889 km/s; ε = 0; T = 0.5 days and v° = 0.465km/s

U = W" x (24/360) = 45.016microsecond per day Nahhas'

Relativity theory silly professor of time travel accounted for 38 Micro seconds and blamed the other 7 Micro seconds on weather

1 – Arabs Global Positioning Systems in arc second per century:W" (calculated) = [-720x36526x3600/T] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

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2 – Arabs Global Positioning Systems in arc degree /centuryW" (calculated) = [-720x36528/T] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

3 – Arabs Global Positioning Systems in arc second per year:W" (calculated) = [-720x365.26x3600/T] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

4- Arabs Global Positioning Systems in arc degree per year:W" (calculated) = [-720x365.26/T] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

5 - Arabs Global Positioning Systems delays in arc degree per cycle:W" (calculated) = [-720/T] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

6 Arabs Global Positioning Systems delays in arc second per cycle:W" (calculated) = [-720x3600/T] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

7- Arabs Global Positioning Systems delays in arc second per day:W" (calculated) = [-720x3600] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

8- Arabs Global Positioning Systems delays in second per cycle:W" (calculated) = [24/360] [-720x3600/T] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² arc sec/ century

9 - Arabs Global Positioning Systems delays in second per day:U (seconds/day) = [-720x 3600/15] {[√ (1 - ε²)]/ (1 - ε) ²} (v/c) ² sec/day

10 - Arabs Global Positioning System Circular Satellite Orbits time delays in seconds/dayU (seconds/day) = [-720x 3600/15] (v/c) ² sec/day

Application of Arabs Global positioning system

GPS Data: T = 0.5 days satellite orbital Period; ε = 0And v° = 0.465km/sec Earth spin speed; And v* = 14,000 km/hr = 35/9 km/secondThen v* +/- v° = 35/9 = 3.88888889km/sec - 0.465km/secondWe subtracted because satellite and motion and spin orientations are opposite GPS time delays are given by this formula per day in seconds of an arc W" (ob) = (-720x3600/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° +/- v*)/c] ² seconds/dayW" (ob) = (-720x36/0.5) (1) [3.423888889/300,000] ² seconds of arc /1 dayW" (ob) = 0.000675246 arc seconds per day to get this answer in seconds divide by 24hours/360degress = 15

U [seconds] = 0.000675246 x [24/360] seconds/day

U = 0.000045016 seconds/day = 45 micro seconds /day

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Chapter 15 Arabs Real time Interplanetary Telecommunications systems

C - Interplanetary telecommunications around the sun

Harvard Irwin Shapiro historical mistake

Relativity stupidity of Harvard Physics department Universal Constant Γ0 =16πGM/C³= 247.597μs

Abstract: Interplanetary time delays around the moving sun derived from three dimensional time-dependent Newton - Kepler's equations solution gives a solar round trip time delay rate of: ΔΓ= 16πGM/c³ [1 + (v°/v)] ² = ΔΓ0 [1 + (v°/v)] ² ΔΓ0 = 16πGM/c³= 247.597μsG = Gravitational constant; M=Sun mass; a=mean distance from Sun. And eccentricity; c = light speed; a = mean distance

And v = Planet speed; v°= Sum/Difference in spin between Earth and planets. When applied to actual data it gives extremely accurate results better than Shapiro's Space-time-delay analysis and without space-time fictional forces or space-time fiction.

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W° (ob) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x [(v° + v*)/c] ² degrees/100 years

The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)

Where v (m) = √ [GM²/ (m + M) a (1-ε²/4)] And v (M) = √ [Gm² / (m + M) a (1-ε²/4)]

Then W (ob) = -4π [√ (1-ε²)]/T (1-ε) ² sine² [ω (m) t + ω(r) t] Δ θ = T W (ob) = - 4π {[√ (1-ε) ²]/ (1-ε) ²} (v° + v*/c) ²} radians; and with ε = 0

Δ θ = - 4π (v° + v*/c) ² Sun-Photon; and with v° = 0Δ θ = -4π (v*/c) ²Sun-Photon: 0 = ε [Sun - Photon] ≠ ε [Earth - Mars] = 0.2075The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²---) ≈ 2 π a (1-ε²/4); R =a (1-ε²/4) v=√ [Gm M/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m<<M; Solar system ΔΓ = 2 arc length/c = 2[Δ θ] d/c = 2[- 4π (v/c) ²] 2R/c; ΔΓ = -16π/c (v/c) ²;ΔΓ = 8πd/c³ [GM/a (1-ε²/4)] =16πGM/c³ (1-ε²/4) = Γ0 (1 - ε²/4)

ε = [a (planet 1) - a (planet 2)]/ [a (planet 1) + a (planet 2)] =0.2075 Mars-Earth Γ0 = 16 πGM/c³= 247.5974607μs=universal constant

ΔΓ = 250μs Mars-Earth. x 1000 x 1000 x 1000

These data compared to Shapiro's time delay from NASA 1977 Vikings 6, 7 Earth - Mars Telecommunications mission are more accurate because the actual value is 250μs and the value published by Doctor Irwin Shapiro of Harvard is 247.597μs Although this formula works the correct formula is Δ θ = -4π [(v° + v*)/c] ² Sun-Photon; and with v° ≠ 0 v = √ [GM/a] = 24.1 km/sec; vº = 0.46511 – 0.241 = 0.224 km/secΔΓ = 2 arc length/c = 2[Δ θ] d/c = 16πGM/c³ [1 + (vº/v)] ² = 247.597 x [1 + 0.224/24.1)]²Δ Γ = 250 µs For Mars; 0.4651 = Earth rotation; 0.241 = mars rotation; v = mars speed

Planet Distant Planet- Earth

Planet+Earth

Eccentricity 1-ε²/4 ΔΓ0 ΔΓμ s

Mercury

57,910 91,690 207,510 0.441858224

0.951190328 247.5974607 260.3

Venus 108,200 41,400 257,800 0.160589604

0.993552745 247.5794607 249.2

Earth 149,600 0 299,200 0 1 247.5794607 247.597Mars 227,940 78,340 377,540 0.20750119

20.989235814 247.5794607 250.273

Jupiter 778,330 628,730 927,930 0.677561885

0.885227473 247.5794607 279.6789

Saturn 1,429,400 1,279,800 1,579,000 0.810512983

0.835767176 247.5794607 296.230

Uranus 2,870,990 2,721,390 3,020,590 0.900946504

0.797073849 247.5794607 310.61

Neptune 4,504,300 4,354,700 4,653,900 0.935709835

0.781111776 247.5794607 316.98

Pluto 5,913,520 5,763,920 6,063,120 0.950652469

0.774064971 247.5974607 319.86650

Harvard physics department is where physics changed from science to stupidity

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Physics in a capitalist country is a business and physics is not necessarily science or scientific and if it is a western invention it is to be

at least 88.88 % fraud After World War II and the emergence of USA as a global superpower and the shift of power from traditional old empires to the new world “superpower” American Universities rushed for gold and gold is “reputable” physics department that can say anything publish anything based on nothing and nothing is Robert Pound and Glen Rebka confusion of an experiment . Harvard strikes again with fraud Robert Pound and Glen Rebka (top)

22.5 meters apart

To start with gravity is a spin effect of a planet with atmosphereThe gravitational and electric and magnetic constants all are derived from two constants p and T which are air density and Earth spin

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If we took a look at the most famous physics fundamental constant G from Newton's 350 years old Equation F = - GmM/r², or, Coulomb electric force equation F = - Qq/4π ε 0, one can see that, for example, G’s value has unit dimension of [G] = [1/ρ T²]; ρ is density and T is time rotational period. If we can show that the gravitational constant G = G (ρ, T) is a variable air density and Earth rotational period dependent then we can say what Earth is saying that the most famous physics fundamental constants are not so universal constants because all three fundamental constant are air density and Earth spin dependent.

Proof: G = G (ρ, T) = 1/ [(2/5) (4 π/3) ρT²] = √ (2π)/ (4 π/3) ρT²Nothing that: 8 π /15 = (2/5) (4π/3); 4π/3 and 2/5 come from spherical aspect Where: ρ = air density; T = Earth rotation periodThe 2/5 is the ratio for radius of gyration of a sphereTaking T = 24 hours and ρ = 1.2045 kg/m³ G = 1/ (8 π/15) ρT² = 6.674 x 10 - 11

And ε 0 = 1/4πρT² = 8.85 x 10 -12

And μ 0 = 4πρT²/c²When Newton’s equation F = - mM/r² was solved it produced the angular velocity r’ = r’0 e í ω t

Divide by cAnd r’/c = (r’0/c) e í ω t

T = T0 e í ω t

And f = f0 e - í ω t

And f = f x + f y = f 0[cosine ω t - í sine ω t]And f x = f 0 cosine ω tAnd f x - f 0 = f 0 [1 – 2 sine² (ω t/2)] - f 0And δ f = f x - f 0 = – 2 f 0 sine² (ω t/2)And δ f/f 0 = - 2 sine² (ω t/2) UpAnd δ f/f 0 = 2 sine² (ω t/2) DownAnd δ f/f 0 (total) = [2 sine² (ω t/2) Down] – [- 2 sine² (ω t/2)]And δ f/f 0 (total) = 4 sine² (ω t/2) With ω t = arc tan (v/c)With v = δ f/f 0 (total) = 4 sine² [(arc tan √2gh) /2] With v << c

Then δ f/f 0 (total) ≈ 2gh/c² = 2x9.81x22.5/ (299,792,458)² = 4.91 x 10-15

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Chapter 16 Handymen in white ropesIn Cleveland Ohio there were two handymen in white robes named

Michelson Morley

Set up this round rotating table in 1887 with a lamb, two mirrors beam splitter, telescope

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In Modern times an animation of the experiment would look like

Their reasoning was that if they rotate the table 90 degrees they would see the interference pattern should change looking through the telescope.

When the interference pattern did not change they claim that physical quantities data would not change when rotated 90 degrees or 1 = 1 +i

A. Michelson and E. Morley Handymen in white ropes confusions of an Experiment

In General

V = v = Velocity of Earth assuming that emitter - receiver line-up with earth line travel at start and during this whole experiment. That would be the x-x line X' ------Earth moving in this direction with velocity v -----------XAnd C = c at start What is turning is the table that has the lamb and the mirrors and the telescope on top of it.

To turn a quantity you multiply by e î θ And that means light c and distance dC = c e î θ = velocity of light And d = d e î θ

Then: V = v; C = c e î θ; d = d e î θ; e î θ = cosine θ + î sine θFor example:

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1-Assume θ = 0° Arm AThen: V = v; C = c; d = d

Forward is C-V = c - v Backward C+V = c + vAnd t = d/ (c-v) + d/ (c + v) 2- Assume θ = 90° then e î 90 = î Arm BThen V = v; C = c î; d = d îForward is C-V = -v + c î; and its measure is = √ (c² + v²)V is along the x-x and has no component on y-y axis and î means along the y-y axis Backward C+V = v + c î; and its measure is = √ (c² + v²)And comes to this is Position IPosition 1: Right-UpperRight: t (1) = d/ (c -v) +d/ (c +v) = 2dc/ (c² - v²) = (2d/c) [1/ (1- v²/c²)]Upper: t (2) = d/√ (c² + v²) + d√ (c² + v²) = 2d/ √ (c² + v²) = (2d/c) [1/√ (1 + v²/c²)]Now Position II3-Assume θ = 90° Arm AThen: V = v; C = ỉ c; d = d Forward is C-V = -v + c î; and its measure is = √ (c² + v²) MMX mistake √ (c² - v²)Backward C+V = v + c î; and its measure is = √ (c² + v²)And t' (1) = 2d/ √ (c² + v²) = (2d/c) [1/√ (1 + v²/c²)]4- Assume θ = 90° + 90° =180 then e î 180 = -î Arm BThen V = v; C = -c; d = d Forward is C-V = -c - v; V is along the x-x and has no component on y-y axis and î means along the y-y axis Backward C+V = -c + v Then t = d/-c+ v forward and in absolute value t = d/ (c - v) And t = d/-c -v backward and in absolute value t = d/ (c + v) And t' (2) = d/ (c - v) + d/ (c + v) = 2dc/ (c² - v²) = (2d/c) [1/ (1- v²/c²)]With t' (1) = 2d/ √ (c² + v²) = (2d/c) [1/√ (1 + v²/c²)]Conclusions is turning the table 90° has no effect what so ever on time length differenceIn GeneralIf we twist the table by an angle θ then V = v; C = c e î θ; d = d e î θ; e î θ = cosine θ + î sine θ Now position I V = v; C = c cosine θ + î c sine θ; d = d cosine θ + î d sine θForward: C - V = c cosine θ -v + î c sine θ and its measure is √ [(c cosine θ -v) ² + c² sine ² θ] Backward: C + V = c cosine θ +v + î c sine θ and its measure is √ [(c cosine θ + v) ² + c² sine ² θ]

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And t (1) = d / √ [(c cosine θ -v) ² + c² sine ² θ] + d / √ [(c cosine θ + v) ² + c² sine ² θ] And Position IIV = v; C = c cosine [θ + 90°] + î c sine [θ + 90°] = c sine θ - ỉ c cosine θ d = d cosine θ + î d sine θForward: C - V = c sine θ -v - î c cosine θ and its measure is √ [(c sine θ -v) ² + c² cosine ² θ]

Backward: C + V = c sine θ + v - î c cosine θ and its measure is √ [(c sine θ + v) ² + c² cosine ² θ] And t (2) = d / √ [(c sine θ -v) ² + c² cosine ² θ] + d / √ [(c sine θ + v) ² + c² cosine ² θ] As General rule: And t (1) = d / √ [(c cosine θ -v) ² + c² sine ² θ] + d / √ [(c cosine θ + v) ² + c² sine ² θ] And t (2) = d / √ [(c sine θ -v) ² + c² cosine ² θ] + d / √ [(c sine θ + v) ² + c² cosine ² θ] If θ = 0 then, we get position I Right: t (1) = d/ (c -v) +d/ (c +v) = 2dc/ (c² - v²) = (2d/c) [1/ (1- v²/c²)]Upper: t (2) = d/√ (c² + v²) + d√ (c² + v²) = 2d/ √ (c² + v²) = (2d/c) [1/√ (1 + v²/c²)]If θ = 90°, then we get position IIWith t' (1) = 2d/ √ (c² + v²) = (2d/c) [1/√ (1 + v²/c²)]And t' (2) = d/ (c - v) d/ (c + v) = 2dc/ (c² - v²) = (2d/c) [1/ (1- v²/c²)]There is no Path difference changeIf θ = 45 degrees, then Upper: t (2 = d/ (c -v) +d/ (c +v) = 2dc/ (c² - v²) = (2d/c) [1/ (1- v²/c²)]Right: t (1) = d/√ (c² + v²) + d√ (c² + v²) = 2d/ √ (c² + v²) = (2d/c) [1/√ (1 + v²/c²)]

And the difference is still the sameNow Let us turn the Table 30°

A- Assume θ = 30° then Exp [30° î] = √ 3/2 + 1/2îThen: V = v; C = c Exp [î 30°]; d = d Exp [î 30°]; Exp [î 30°] = cosine 30° + î sine 30°And t (1) = d / √ [(c cosine 30° -v) ² + c² sine ² 30°] + d / √ [(c cosine 30° + v) ² + c² sine ² 30°] And t (2) = d / √ [(c sine30° -v) ² + c² cosine ² 30°] + d / √ [(c sine 30° + v) ² + c² cosine ² 30°] And t (1) = d / √ [(c √ 3/2 -v) ² + c² /4] + d / √ [(c √ 3/2 + v) ² + c² /4] And t (2) = d / √ [(c/2 -v) ² + c² (√ 3/2) ²] + d / √ [(c /2 + v) ² + c² (√ 3/2) ²] Now let us use v = 30km/sec and c = 300,000km/secIf θ = 0 then,

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Right: t (1) = d/ (c -v) +d/ (c +v) = 2dc/ (c² - v²) = (2d/c) [1/ (1- v²/c²)]Upper: t (2) = d/√ (c² + v²) + d√ (c² + v²) = 2d/ √ (c² + v²) = (2d/c) [1/√ (1 + v²/c²)]If θ = 30° then, And t (1) = d / c √ [(√ 3/2 -v/c) ² + 1 /4] + d / c √ [(√ 3/2 + v/c) ² + 1 /4] And t (2) = d / c √ [(1/2 -v/c) ² + (√ 3/2) ²] + d / c √ [(1 /2 + v/c) ² + (√ 3/2) ²] We can take (d/c) Then Multiply that at the end If θ = 0 then, Right: t (1)/ (d/c) = 2 / (1- v²/c²)] = 2.00000002 Upper: t (2)/ (d/c) =2/√ (1 + v²/c²)] = 1.99999999Then δ t = 3 x 10-8 (d/c) = 3 x 10-8 (d/300,000,000) = 10 -16 d

Then δ t = 10 -16 dIf θ = 30° then, And t (1)/ (d / c) = √ [(√ 3/2 -v/c) ² + 1 /4] + √ [(√ 3/2 + v/c) ² + 1 /4] = 0.99999134 + 1.00000866 = 2And t (2)/ (d / c) = √ [(1/2 -v/c) ² + (√ 3/2) ²] + √ [(1 /2 + v/c) ² + (√ 3/2) ²] = 0.999995 + 1.000005 = 2 Then δ t' = 0So the difference isΔΓ = δ t - δ t' = 10 -16 dWith fringe shift = c ΔΓ/ δ r = 3x10-8d/λ; λ = 5.9 x 10^-5 c m; d = 11m =1100c m With (d/λ) = 1100/ (5.9x 10-5)Then fringe shift = (3300x10-8)/ (5.9x 10-5) = .559322034

Well if fringe shift does not happen at angle 90° and fringe shift happens at 30° why no shift seen?

The initial problem was and is the lineup of apparatus with earth velocity which is impossible because of the infinite numbers of possibilities of earth motion in space and if it happened it will be momentarily. MMX will never work and all similar experiments will not work unless it is designed as a 3-dimensional set-up and not a one-dimensional wish washy set-up An ideal setup is a spherical set-up but a more realistic setup will be discussed below how to find earth path in space

The conclusion was 1 = 1 +i

Or E = mc²/2 = (mc²/2).1 = (mc²/2) (1 + i) ² = 2i (mc²/2) = i mc²

Or E = mc²; i in absolute value = 1

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Chapter 17 Earth Path in real space in real timeEarth path in space is possible and Einstein can keep his relativistic

trash to himself: Self – referenced motionThere is a serious problem in all of physics and that is timing of experiments and physics experiments theory.Earth absolute motion is a circle that is going to tie spin to orbit motion and waiting experimental verifications Finding new sources of energy requires the elimination of bad physics and bad physics is E = mc² because it is based on space - time physics and space time physics is wrong because I can prove it wrong as follows. What is wrong with space - time is that it is a wrong idea and it is without proofs. Time is a human made scale and making it a dimension is scientifically stupid and the reasoning behind accepting time as a dimension and not as a scale is the assumption that it matches experiment. Time as a dimension allowing the exchange of time for space or squeezing space to expand time is not acceptable to me regardless what all Nobel prize winners have said about it and regardless what the 100, 000 living physicists are saying about it and regardless what all 100, 000 dead physicists had said about it because it is wrong and proving it wrong is what I can do and did the first time I read what started it and that would be the original dumb idea and experiment in Cleveland Ohio in 1887 by Michelson and Morley. What dumb Michelson and Morley experiment did is it started stupidity as accepted science.The stupidity of Michelson and Morley experiment is the following.

We can not measure something that did not happen. We can only measure something that had happened. We can measure something in present time that had happened in past time.Michelson and Morley experiment is stupid because two handymen in white robes tried to measure something in real time and wondered why they did not see it in past time.

Present time = present timePresent time = past time + [present time - past time]Present time = past time + time delays

Real time = event time + time delays

What we measure = what happened + what changed till things are measured.Real time physics = event time physics + time delays physicsWhat we see or measure is relativistic = Absolute event + relativistic time delaysWhat happened is absolute = real time relativistic event - real time relativistic effects

Michelson and Morley mistake is the attempt:

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Real time = event time [this is the wrong idea that started space - time stupidity]What are missing are time delays

[Known as quantum theory and relativity theory]

Real time = event time + time delays

This formula is my 1973 discovery that not only Nobel Prize winners are wrong but all of physics and physicists are wrong because physics solution is based on the idea that:

Real time = event time

In one dimension:

How would I know if I am going to the right or the left?

The problem is very simple, I would send a signal to 2 signals to two equidistant repeaters that send back the signal to me and the signal that comes first indicate a motion in opposite direction. Installing a multi opposite directional repeaters system and one signal and for each tow opposite direction have on timer will produce Earth path in absolute space and may produce the greatest formula in all of Physics and that would be mass motion and spin and better than that it will erase stupidity from classrooms know as space - time.

Abstract: Finding new sources of energy require the elimination of bad physics and bad physics is Einstein's space-time (x, y, z, ict) confusion of physics that lead to energy fraud symbol E = mc² because energy spent is based on distance traveled and not relative distance and distance traveled is self referenced distance or absolute motion and finding this motion is waiting on funding for discovery since the beginning of time and experimentally is possible. Discovering Earth absolute motion in space could be one of the greatest discoveries in physics of all time because it will answer the trillion dollar question if mass m spin s and absolute motion r are related by a mathematical formula that expresses a physical law and if this law is to be found can it explains the structural design of the solar system and if so what applications of such a law is useful inside the nucleus to produce new sources of energy?

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The experimental set - up

Experimental set-up d <<<<<<<<<<<<<<<<<<<E >>>>>>>>>>>>>>>>>>>> dSensor x' Emitter Ex Lead Sensor x

Timer T x

And d <<<<<<<<<<<<<<<<<<<E>>>>>>>>>>>>>>>>>>>>> dSensor y' Emitter E y Lead Sensor y

E

Timer T y And d <<<<<<<<<<<<<<<<<<<E>>>>>>>>>>>>>>>>>>>>> d

Sensor z' Emitter E z Lead Sensor z E

Timer T z 1-Assume there is a pulse emitter E shooting pulses in six different directions of the Euclidean system of axes per unit time τ; x'-x and y'-y and z'-z axes at six equidistant receivers of distance d and has three wire leads Ex, E y, E z connected to three Timers Named T x , T y, and T z.

2 - The six equidistant receivers are named x', x, y', y, z', z.

3- Three timers are named timer T x with three wires one to receiver x' and one wire to receiver x and one wire to emitter wire lead E x and timer T y with three wires one to receiver y' and one wire to receiver y and one wire to emitter wire lead E y and timer T z with three wires one to receiver z' and one wire to receiver z and one wire to emitter wire lead E z are used 4- These six pulses has their directions grouped in three sets of pairs x'-x, y'-y, z'-z.5- Three timers T x; T y, and T z are associated with the three axes respectivelya - Timer T x takes measurements from Emitter lead E x; receiver x' and receiver xb - Timer T y takes measurements from Emitter lead E y; receiver y' and receiver yc - Timer T z takes measurements from Emitter lead E z; receiver z' and receiver z

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Now d <<<<<<<<<<<<<<<<<<<E>>>>>>>>>>>>>>>>>>>>> d

Sensor x' Emitter E x Lead Sensor xE

t x' Timer T x t x1- Timer T x receives 3 signals

a- From lead E x a signal departures from emitter E to T x register E [T x, n = 1, 2, 3,]b - From receiver x' a signal departures to timer T x [t x', n =1, 2, 3,]

c - From receiver x a signal departures to timer T x [t x, n =1, 2, 3,]If t x' - T x is the travel time from emitter E to receiver x'

And t x - T x is the travel time from emitter E to receiver x

If t x' < t x Then t x' - T x < t x - T xAnd motion Earth is in the x - direction

Let C = average velocity of the pulses and v = velocity of earth in x'-x direction

ThenRIGHT (C - V x'-x) = d/ t (x-direction) = V x = measurable quantity; t (x-direction)

measured because d is knownLEFT (C + V x'-x) = d/ t (x'-direction)/d = V x' = measurable quantity; t (x'-direction) measured

because d in knownThen C = (V x' + V x)/2

And V x'- x = (V x' - V x)/2 Equation X-1 Measurable quantitySimilarly we can get

V y'- y = (V y' - V y)/2 Equation Y-1 Measurable quantityV z'- z = (V z' - V z)/2 Equation Z-1 Measurable quantity

V (earth) = √ [V² x'-x + V² y'-y + V² z'-z]If we take unit of times τ = period of a light source (or a high speed signal)

Wave #

V x'-x

V y'-y

V z'-z

V = √ [V²x + V²y + V²z] X =τ V x

Y =τ V y

Z=τ V z

r = X + Y + Z

1 Vx1

Vy1 Vz1

V1=√ [V²x1 + V²y1 + V²z1]

X1 Y1 Z1 r (1) = X1 + Y1 + Z1

2 Vx2

Vy2 Vz2

V2=√ [V²x2 + V²y2 + V²z2]

X2 Y2 Z2 r (2) = X2 + Y2 + Z2

3 Vx3

Vy3

Vz3

V3=√ [V²x3 + V²y3 + V²z3]

X3 Y3 Z3 r (1) = X3 + Y3 + Z3

4 Vx4

Vy4 Vz4

V4=√ [V²x4 + V²y4 + V²z4]

X4 Y4 Z4 r (4) = X4 + Y4 + Z4

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Then

Conclusion: The idea of this experiment is to find distance travel segments r (1); r (2); r (3); r (4); ...ETC; then join them together. With [r (1); τ] O -------------- O [r (4); τ] - - O ------------------ [r (2); τ] - - - - [r (3); τ] - O

r (1) = X1 + Y1 + Z1r (2) = X2 + Y2 + Z2r (1) = X3 + Y3 + Z3r (4) = X4 + Y4 + Z4

The smaller the signal time interval or period τ then the more accurate the experiment. Such a set-up is not only capable of finding absolute motion but it will give clues to finding the relation between mass m spin s and its orbit r of planet Earth. If mass m motion r and spin s are found to be related it will the greatest formulas of all time because all there is in the Universe is mass m space r and spin s. Earth mass m and spin s are know but mass r is Earth absolute motion that had been waiting on funding for discovery since the beginning of time.

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Chapter 18 Johann Georg Von Soldner 1801 light bending historical mistake

Abstract: Newton proposed F = - GmM/r² as gravitational law

In 1801 Johann Georg Van Soldner was the first person to calculate the gravitational bending of light using Newtonian Mechanics and he got:Johann Georg Van Soldner ς (Johann) = 2 {cosine -1 [v²/ (-c² + v²)} - π With v² = GM /R where G = gravitational constant = 6.673 x 10-11; C = 3 x108m/sec

And M = Sun mass = 2 x 1030 kg; R = sun radius = 0.695 x 109 m; v = 437.89 Einstein said if make – believe time travel and new forces added:Then: ς (Einstein) = 4 (v/c) ²; ς (Johann) = 0.8789 arc sec; ς (Einstein) = 2(0.8789) Johann Georg Van Soldner derivation was incomplete and when completed and approximated it produces Einstein’s formula without Einstein’s space – time fiction and as light aberration and not light bending.Proof:

Johann Georg Van Soldner wrong derivation of angle of light aberration around the Sun

With d² r/d t² - r θ'² = -GM/r² Newton's Gravitational equation (1)And d (r²θ')/d t = 0 Kepler's force law (2)Assuming mass m = constant

Proof:With (2): d (r²θ')/d t = 0 Then r²θ' = constant = hDifferentiate with respect to time

Then 2rr'θ' + r²θ" = 0Divide by r²θ' Then 2(r'/r) + θ"/θ' = 0And 2(r'/r) = - (θ"/θ') = 2[λ (r) + ì ω (r)]And 2(r'/r) = 2[λ (r) + ì ω (r)]And (θ"/θ') = - 2[λ (r) + ì ω (r)]Solving for r = r (θ, t) = r (θ, 0) r (0, t) = r (θ, 0) ℮ [λ (r) + ỉ ω (r)] t

With r (0, t) = ℮ [λ (r) + ỉ ω (r)] t

Then θ'(θ, t) = [h/ r² (θ, 0)] ℮ -2[λ (r) + ỉ ω (r)] t And, θ'(θ, t) = θ' (θ, 0) θ' (0, t)

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And θ' (0, t) = ℮ -2[λ (r) + ỉ ω (r)] t

Also θ'(θ, 0) = [h/ r² (θ, 0)] And θ'(0, 0) = [h/ r² (0, 0)]

With (1): d² r/d t² - r θ'² = - GM/r² Let r =1/uThen d r/d t = -u'/u² = - (1/u²) (θ') d u/d θ = (- θ'/u²) d u/d θ = - h d u/d θAnd d² r/d t² = - hθ'd²u/dθ² = - h u² [d²u/dθ²]

And - hu² [d²u/dθ²] - (1/u) (hu²)² = - G Mu²[d²u/ dθ²] + u = G M/ h²And u = G M/ h² + A cosine θ

And du/ d θ = 0 = - A sine θ; θ = 0Then u (0) = 1/ r (0) = GM/h² + A; h = RCC = light velocity of 300,000km/sec; And A = 1/R – GM/ (RC) ² And u = G M/ h² + A cosine θ = GM/ (RC) ² + [1/R – GM/ (RC) ²] cosine θ

And r = 1/u = 1/ {GM/ (RC) ² + [1/R – GM/ (RC) ²] cosine θ} If r --- ∞; GM/ (RC) ² + [1/R – GM/ (RC) ²] cosine θ = 0 Divide by GM/ (RC) ²

Then 1 + [R²C²/ GM R – 1] cosine θ = 0And cosine θ = -1/ [C²/ (GM/ R) – 1] Or cosine θ = 1/ [1 – (C²/V²)]; GM/R = V² Or cosine θ = v²/ (v² - c²)And θ = cosine -1 [v²/ (v² - c²)]

And ς (Johann) = 2 {cosine -1 [v²/ (-c² + v²)} - π ≈ 2 [π/2 + (v/c) ²] – π = 2 (v/c) ²; v/c <<1

Einstein invented many things to come up with double the amounts:Or, ς (Einstein) = 4 (v/c) ²

Here is Johann Georg Van Soldner 1801 Historical mistakeWe have u (θ) = G M/ h² + A cosine θAnd r (θ, t) = r (θ, 0) r (0, t) = r (θ, 0) ℮ [λ (r) + ỉ ω (r)] t

And r (θ, 0) = 1/ u (θ, 0) = 1/ [G M/ h² + A cosine θ]Or, r (θ, 0) = (h²/GM)/ [1 + (h²/GM) A cosine θ]Or, r (θ, 0) = (h²/GM)/ [1 + ε cosine θ]

Real time orbit: r (θ, t) = [a (1-ε²)/ (1+ ε cosine θ)] ℮ [λ (r) + ỉ ω (r)] t

This equation is real time Universal mechanics solution

LightLight

Sun

θς

This: r (θ, t) = [a (1-ε²)/ (1+ ε cosine θ)] ℮ [λ (r) + ỉ ω (r)] t --------------------------------- IIt is the math formula that matches a physical experimentIf time is frozen that is t = 0Then r (θ, t) = [a (1-ε²)/ (1+ ε cosine θ)] we get the classical or event time solution ----------- IIRelativistic is the difference between I and II

And it is the visual illusion between motion II and Visual motion I The difference between an event and its measurement in real time

With θ’ (θ, t) = [h/ r² (θ, 0)] ℮ -2[λ (r) + ỉ ω (r)] t With (θ”/ θ’) = - 2[λ (r) + ì ω (r)]Then θ” (θ, t) = - 2[λ (r) + ì ω (r)] [h/ r² (θ, 0)] ℮ -2[λ (r) + ỉ ω (r)] t

With λ (r) = 0Then θ” (θ, t) = - 2 ì ω (r) [h/ r² (θ, 0)] ℮ -2 ỉ ω (r) t Or, θ” (θ, t) = - 2 ì ω (r) [h/ r² (θ, 0)] [cosine 2 ω (r) t + ί sine 2 ω (r) t]

The real part or along the line of sightIs Real θ” (θ, t) = 2 ω (r) [h/ r² (θ, 0)] sine 2 ω (r) t] (t/t) Or, Real θ” (θ, t) = 2 t ω (r) [h/ t r² (θ, 0)] sine 2 (r) ω t]

Or, Real θ” (θ, t) / [h/ t r² (θ, 0)] = 2 t ω (r) sine 2 ω (r) t At t = T; light aberration angle in real time is confused for light bending. With ω T = arc tan (v/c) Then ψ = visual illusion angle = θ” (θ, T) / [h/ T r² (θ, 0)]Or, ψ =2 T ω (r) sine 2 ω (r) T

Johann Georg Van Soldner 1801 historical mistakeIs: ψ = [2 arc tan (v/c)] sine [2 arc tan (v/c)]

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With (v/c) << 1; 2 arc tan (v/c) ≈ 2 (v/c) And sine 2 arc tan (v/c) ≈ sine 2 (v/c) ≈ 2 (v/c)

And ψ = [2 arc tan (v/c)] sine [2 arc tan (v/c)] ≈ [2 (v/c)] [2 (v/c)]

Or ψ ≈ 4 (v/c) ² radians Or, ψ ≈ 4 (v/c) ² x (180/π) degreesOr, ψ ≈ 4 (v/c) ² x (180/π) x 3600 secondsWith v² = GM/R Then ψ ≈ 4 GM/R c² x (180/π) x 3600 secondsEinstein with the help of others rigged eternity to come up with 4 (v/c) ² and not 2 (v/c) ² to justify experimental illusions or 2[2(v/c) ²]Ψ = 7200 [arc tan (437.89 / 300,000)] sine 2 [arc tan (437.89/300,000)] = 1.757855865 arc second

MR Eddington: keep the trash to England Royals

101

Chapter 19: 20th century greatest mistake is E = mc²

The difference between old physics and new physics is doubleThat is one Einstein equal two Newton.

Newton energy definition is En = mc²/2 Einstein doubled that amount or E e = 2 En = 2[mc²/2] = mc²

It is not just Einstein and Newton are wrong but physics is 500 years wrong Page 1Abstract: The elimination of 500 years of Physics including relativity theory is a matter of time and not a matter of science and the time is 1900.

E = mc² death certificate is now available

Real time Physics: We can only measure past events. We can not measure something that did not happen. We can only measure things that had happened. What we measure is not what happened. We measure in present time an event that happened in past time.

Present time = present timePresent time = past time + [present time - past time]Present time = past time + real time delaysReal time physics = event time physics + real time relativistic delays

What one sees is relativistic = what happened in an absolute event + relativistic effectsWhat happened in an event is absolute = real time physics - real time relativistic effects.

Observer time = observed traveler time + time delays Real time = absolute time + time delays Real time = Event time + time delays Real time Physics = event time Physics + time delays Physics

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Quantum = classical + relativistic

Naming Γ as real time and t as event time Γ = t + (Γ - t) = t [1 + (Γ – t)/t] = t (Γ/t) Γ = t + Δ Γ; Δ Γ = Γ - t

If an event happens on Planet Mercury and the event is seen from the Sun at event time t, then this same event would be seen from Earth in real time as time Γ = t + Δ Γ

Γ = t + (Γ - t) Real time = Event time + time delaysΓ = t + Δ Γ (x) + ỉ Γ (y) = t + Δ Γ

Δ Γ = Δ Γ (x) + ỉ Γ (y) Δ Γ (x) is along the line of sight time delaysΔ Γ (y) is perpendicular to the line of sight time delays

Γ = t (Γ/t)Γ = t e ỉ ω t; Γ/t = e ỉ ω t

Γ = t e ί ω t Multiplying by c, then r = c Γ = c t e ί ω t = r0 e ί ω t; ω T = arc tan (v/c)

Or r = r0 e ί ω t

A – Light aberration ω T= arc tan (v 0/c) B – Michelson and Morley ω T = [π/2 + arc tan (v 0/c)]

Real time location is r = r 0 ℮ ỉ ω t

Real time velocity v = [v 0 + í ω r 0] ℮ ỉ ω t

The definition of energy E = m (V. V)/2 = (m/2) [v 0 + í ω² r 0]. [v 0 + í ω r 0] ℮ 2 ỉ ω t

= (m/2) [(v0² - ω² r 0²) + 2 í ω (r 0 v 0)] [cosine 2ω t + í sine 2ω t] = (m/2) {[(v0² - ω² r 0²) cosine 2ω t - 2 ω (r 0 v 0) sine 2ω t] + í [(v0² - ω r 0²) sine 2ω t + 2 ω (r 0 v 0) cosine 2ω t]} = Ex + i E yEx = (m/2) [(v0² - ω² r 0²) cosine 2ω t - 2 ω (r 0 v 0) sine 2ω t]

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Case I Constant velocity of light principle: v 0 = ω r 0 = cThenEx = (m/2) [(c² - c²) cosine 2ω t - 2 c² sine 2ω t]Ex = (m/2) [- 2 c² sine 2ω t]Ex = - m c² sine 2ω tA – Light aberration ω T= arc tan (v 0/c) Ex = - m c² sine 2ω tEx = - m c² sine 2 [arc tan (v 0/c)] Ex = - m c² sine 2 [arc tan (c/c)]Ex = - m c² B – Michelson and Morley ω T = [π/2 + arc tan (v 0/c)] Ex = - m c² sine 2ω tEx = - m c² sine 2[π/2 + arc tan (v 0/c)]

Ex = - m c² sine 2[π/2 + arc tan (c/c)] Ex = - m c² sine 2[π/2 + π/4] Ex = - m c² sine [3π/2] Ex = m c²

Case IIRest Energy Principle: v 0 = 0; ω r 0 = cThenEx = (m/2) [(0² - c²) cosine 2ω t - 2 c (0) sine 2ω t]Ex = (m/2) [- 2 c² cosine 2ω t]Ex = - m c² cosine 2ω tA – Light aberration ω T= arc tan (v 0/c) Ex = - m c² cosine ω T= arc tan (v 0/c) Ex = - m c² cosine 2 [arc tan (0/c)] Ex = - m c² cosine 2 (0)Ex = - m c² B – Michelson and Morley ω T = [π/2 + arc tan (v 0/c)] Ex = - m c² cosine 2ω tEx = - m c² cosine 2[π/2 + arc tan (0/c)] Ex = - m c² cosine 2(π/2) Ex = - m c² cosine π Ex = - m c² (-1) Ex = m c²

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Michelson Morley This 1907 Nobel Prize winner Dude along with this Dude put a Lamp two mirrors and a telescope on a round rotating table.

This was the set –up.

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Michelson and Morley looked at the virtual image from the two mirrors of light reflections through the telescope. They claim that if they to rotate the table then the image should change. When Michelson and Morley rotated the table in 1887 and the image did not change they claim that you can rotate data without change in results!

Then conclusion of this wacky claim that won a Nobel Prize is that The number 1 can be replaced by 1 + i; i is the complex number

Or; in Cartesian coordinates

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If we take E = (mc²/2) (1)²And replace (1) with (1 + 1i)Then E = (mc²/2) (1)² = (mc²/2) (1 + 1i) ² E = (mc²/2) (1 + 1i) ² = (mc²/2) (1 – 1 + 2i) ²E = 2i (mc²/2) = imc²|E| = |imc²| = mc²Case III: Classical Energy Principle: v 0 = c; ω r 0 = 0: Then ωEx = (m/2) [(c² - 0²) cosine 2 ω t - 2 c (0) sine 2ω t]Ex = (m/2) (c² cosine 2 ω t)Ex = (m/2) [c² cosine 2 (0) t]Ex = m c²/2

E = mc²/2 and not E = mc² but the visual illusion ∆ E = mc²

The crooks of “University” are using visual effects to claim Dark energy or an expansion of 500 years of western stupidity and fraud in physics

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Chapter 20: special relativity stupidity

1

i

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z). The state of any object in the Universe can be expressed as the product S = m r; State = mass x location: P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rate


r = r r (1) ;v = r' r(1) + r θ' θ(1) ; γ = (r" - rθ'²)r(1) + (2r'θ' + r θ")θ(1)r = location; v = velocity; γ = accelerationF = m γ + 2m'v +m" rF = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + (m" r) r (1) = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/mr) [d (m²r²θ')/d t] θ (1) = [-GmM/r²] r (1) ------------------------------- Newton's Gravitational LawProof:First r = r [cosine θ î + sine θ Ĵ] = r r (1)Define r (1) = cosine θ î + sine θ Ĵ Define v = d r/d t = r' r (1) + r d[r (1)]/d t = r' r (1) + r θ'[- sine θ î + cosine θĴ] = r' r (1) + r θ' θ (1)


Define γ = d [r' r (1) + r θ' θ (1)] /d t = r" r (1) + r'd [r (1)]/d t + r' θ' r (1) + r θ" r (1) +r θ'd [θ (1)]/d t γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)

With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 With m = constant, then With d² r/dt² - r θ'² = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2 With d² r/dt² - r θ'² = - ω²s r E q – 1 Hooke’s LawAnd d (r²θ')/d t = 0 Eq-2

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From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant Differentiate with respect to timeThen 2rr'θ' + r²θ" = 0

Divide by r²θ' Then 2(r'/r) + θ"/θ' = 0And 2(r'/r) = - θ"/θ' = 2[λ (r) + ỉ ω (r)]Also, r = r0 (θ, 0) e [λ (r) + ỉ ω (r)] t

And θ' = θ' 0 (θ, 0) e -2[λ (r) + ỉ ω (r)] t

For a fixed orbit: λ r= 0Also, r = r 0 (θ, 0) e ỉ ω t

An object of length r could be seen longer or shorter if it extended or shortened e λ t

or twisted e í ω t making it in general r e [λ + í ω] t. If the length of an object is not extended or shortened then λ = 0 and it could be seen as a twisted visual effect. Then r -------------------------- Visual r = S = r e í ω t

1- The illusion of length contraction: Length contraction is just a visual effect of projected light aberration and it is an "apparent" visual effect and not real

Alfred Nobel Institution IdiotAn object located at r ----- light sensing -----measured as S = r e í ω t with ω t = arc tan (v/c); tan (v/c) = light aberrations angle = ω tS = r e í ω t caused by light aberrations visual effects as follows:If e í ω t = [cosine ω t + ỉ sine ω t]; From S = r e í ω t

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It changes to: S = r {√ [1- sine² arc tan (v/c)] - ỉ sine arc tan (v/c)} = r {√ [1- (v/c) ²] - ỉ (v/c)}; v/c <<< 1 = S x + ỉ S y

Where S x = √ [1- sine² arc tan (v/c)]; And S y = cosine arc tan (v/c)With v/c << 1 then; Where S x = √ [1- sine² arc tan (v/c)]; And S y = cosine arc tan (v/c)In absolute value S = rAlong the line of measurement: S x = √ [1- sine² arc tan (v/c)] ≈ r √ [1-(v/c) ²]; v/c << 1 This the equation for length contraction of Lorentz's used in Einstein's theoriesBut it is the light aberrations visual effects and it is "apparent and not real

2 - Constant velocity of light leading to Time DilationsProjected light aberrations S x = r cosine ω tHypotenuse = S x = [c t x] = c t x √ [1- sine² arc tan (v/c)]

S x ≈ c t √ [1-(v/c) ²]; from constant velocity of light and c is constant in all reference frames

Where t = local self time; t x = time by observer

t x = t √ [1-(v/c) ²]; andt = {1/√ [1-(v/c) ²]} t x absolute math

These are time dilatation equations given by Einstein’s special relativity theory.

t x' = t' √ [1-(v'/c) ²]; andt' = {1/√ [1-(v'/c) ²]} t x' absolute math

Two observers observing the same thing the time dilations are

Then, t x = t √ [1-(v/c) ²]; t = {1/√ [1-(v/c) ²]} t x absolute math; Lab purposesAnd, t x' = t √ [1-(v'/c) ²]; t' = {1/√ [1-(v'/c) ²]} t x' absolute math; Lab purposes

However; two observers looking at each other S (A) = r exp [ỉ ω t] S (B) = r exp [- ỉ ω t]S x (A) = S x = c t x ≈ c t x' √ [1-(v/c) ²]S x (B) = S x = c t x' ≈ c t x √ [1-(v/c) ²]And t x ≈ t x' √ [1-(v/c) ²]And t x' ≈ t x √ [1-(v/c) ²]That is why there no twin Paradox except on science fictions books because it is all about aberrations and nothing real.

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3 – Momentum

S x = Visual location along the line of sight = r [√ [1-(v/c) ²]P x = v [√ [1-(v/c) ²]; v =constant; P x = d [S x]/d tAnd m P x = m v [√ [1-(v/c) ²] = m (0) v

4 – Mass Then m = m (0) / [√ [1-(v/c) ²]

Also; m = m (0) / [1-1/2(v/c) ²]

5- Energy

mc² = m (0) c² / [1-1/2(v/c) ²]

E = m (0) c²; v = 0

Also m ≈ m (0) [1+ 1/2(v/c) ²]

Hence m c² ≈ m (0) c² + m v ²/2

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Chapter 21 Quantum mechanics Copenhagen Cyclic Interpretation

Ape Man Modern Physicist Nobel Prize Ape Ass.

Then came the Copenhagen Interpretation

The Copenhagen idiots with Nobel prizes That new physics is needed!This new physics is quantum mechanics!Quantum mechanics is a purely western stupidity:

I - In 950 AD with Arabs visual location law is quantum mechanics: Is r = r 0 e í (θ + ω t)

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z). The state of any object in the Universe can be expressed as the product S = m r; State = mass x location: P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rateIn polar coordinates systemr = r r (1) ;v = r' r(1) + r θ' θ(1) ; γ = (r" - rθ'²)r(1) + (2r'θ' + r θ")θ(1)r = location; v = velocity; γ = accelerationF = m γ + 2m'v +m" rF = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + (m" r) r (1) = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/mr) [d (m²r²θ')/d t] θ (1) = [-GmM/r²] r (1) ---------------------------- Newton's Gravitational Law

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Proof:First r = r [cosine θ î + sine θ Ĵ] = r r (1)Define r (1) = cosine θ î + sine θ Ĵ Define v = d r/d t = r' r (1) + r d[r (1)]/d t = r' r (1) + r θ'[- sine θ î + cosine θĴ] = r' r (1) + r θ' θ (1)

Define θ (1) = -sine θ î +cosine θ Ĵ;

And with r (1) = cosine θ î + sine θ ĴThen d [θ (1)]/d t= θ' [- cosine θ î - sine θ Ĵ= - θ' r (1) And d [r (1)]/d t = θ' [-sine θ î + cosine θ Ĵ] = θ' θ (1)

Define γ = d [r' r (1) + r θ' θ (1)] /d t = r" r (1) + r'd [r (1)]/d t + r' θ' r (1) + r θ" r (1) +r θ'd [θ (1)]/d t γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 With m = constant, then With d² r/dt² - r θ'² = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2 With d² r/dt² - r θ'² = 0 E q – 1 And d (r²θ')/d t = 0 Eq-2

From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant Differentiate with respect to timeThen 2rr'θ' + r²θ" = 0Divide by r²θ' Then 2(r'/r) + θ"/θ' = 0And 2(r'/r) = - θ"/θ' = 2[λ (r) + ỉ ω (r)]Also, r = r0 (θ, 0) e [λ (r) + ỉ ω (r)] t

And θ' = θ' 0 (θ, 0) e -2[λ (r) + ỉ ω (r)] t


And θ' = θ' 0 (θ, 0) e -2ỉ ω t

With d² r/dt² - r θ'² = 0Let u = 1/r; r = 1/u; r²θ' = h = θ’ /u² And d r/d t = (d r/ d u) (d u /d θ) (d θ/ d t) = (- 1/u ²) (θ’) (d u/ d θ) = - h (d u/ d θ)And d² r/ d t² = - h (θ’) (d² u/ d θ ²) = [- h²/r²] (d² u/ d θ ²) = - h² u² (d² u/ d θ ²)

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With d² r/dt² - r θ'² = 0 E q – 1 And - h² u² (d² u/ d θ ²) – (1/u) (h u²) ² = 0Then (d² u/ d θ ²) + u = 0


Then r 0 (θ, 0) = (1/A) e í θ And r 0 (θ, 0) = r 0 (0, 0) e í θ And r = r 0 (0, 0) e í θ e ỉ ω t Or, r = r 0 e í (θ + ω t) This is quantum mechanics

Arabs real time astronomy is actually quantum mechanics r = r0 e í ω (r) t

Light is emitted and not an emitter and humans have no means of measuring light in motion and at best they can take a snap shot.In mathematical languageThen r (θ, 0) = r 0 e í θ at t = 0And r (θ, t) = r 0 e í (θ + ω t) = r 0 e í θ e í ω t = r (θ, 0) e í ω t

And r (θ, t) = r (θ, 0) [cosine ω t + í sine ω t] And r x (θ, t) = r x (θ, 0) cosine ω t And ω t = cosine-1 [r x (θ, t) / r x (θ, 0)]Then ∆ r = r x (θ, t) - r x (θ, 0) = r x (θ, 0) [(cosine ω t) – 1]And [r x (θ, t) - r x (θ, 0)]/ r x (θ, 0) = [(1 - 2 sine² ω t/2) – 1] = - 2 sine ² ω t/2; ω t = cosine-1 [r x (θ, t) / r x (θ, 0)]rabs real time astronomy:Image = Image = IObject = Object = O- Object = - Object = - O

Image = object + (Image – Object)I = O + (I – O)(I/O) = O/O + [(I – O)/O](I/O) = 1 + [(I – O)/O] (I/O) θ' 0 = θ' 0 + [(I – O)/O] θ' 0

Quantum = classical + relativistic Illusion I = r; O = r 0

Or, (r / r 0) θ' 0 = θ' 0 + [(r – r 0)/ r 0] θ' 0

And [(r – r 0)/ r 0] θ' 0Or [- 2sine ² ω t/2] θ' 0; ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r/ r 0)For planet Mercury:

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The distance visual effect Error is: [(r m – re)/re] δ θ’ m





With [(r m – re)/re] = [- 2sine ² ω t/2] = 0.61 And ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r/ r 0)Or ω t = cosine-1 [r x (θ, t) / r x (θ, 0)] = cosine-1 (r m/ r e)[(r m – re)/re] δ θ’ m = 0.61 x 70.75 = 43

Chapter 22: Earth Absolute Motion is circularCopernicus said that Planets move around the sun in an ellipseGalileo affirmed that planets move around the sun in an ellipseTycho Brahe collected the astronomical data and confirmed that planets move around

the Sun in an ellipse

Figure - 1

Kepler wrote the acceleration law: a/T2 = k/a2

And deduced that planets move around the sun in an ellipse and that the sun is at the focus of the ellipse like in figure above

Newton and Hooke reiterated the same thing as Kepler and wrote: r/ t2 = k/r2

Astronomers found that the motion is not an ellipse but a rotating ellipse

Einstein wrote another acceleration law r/t2 = k/ r2 + k1/r4 to describe a rotating ellipse

7 billion people were told by Alfred Nobel prize winner physicists that Planets rotate around the Sun in a rotating ellipse

What if I can prove all wrong?

I am Joe Nahhas And I am better than good and I am after Alfred Nobel Institution head

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θ

r

Sun

Mercury

Newton gave the equation of the ellipse:As: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] Einstein said he corrected Newton and wrote the new equation:Is: r (θ, t) = a (1-ε²)/ [1+ε cosine (θ - Ψ)] It is a fact that Astronomers see this axis tilt and Einstein got a formula for it: ψ = {6 π G M/a c² (1 - ε²)} [180/ π] [36526/T (days)][3600] G = 6.673 x 10-11= gravitational acceleration constantM =2x1030kg = mass of the Sun

Eccentricity of Mercury’s orbit = ε = 0.206T= 88 days = period of Planet mercury’s rotation around the SunAnd c = 299792.458 km/sec =light speed/secAnd a = 58.2 x 106 km/sec = semi major axis of planet Mercury orbit

ψ = {6 (π) 6.673 x 10-11 (2 x 1030)/58.2 x 106 (299792.458) ² [1– (0.206)²]} x [180/ π] [36526/88] [3600] = 43 seconds of an arc per century

Sun

ψ

Mercury

θ

Newton used magical powers or action at a distance force F = G m M/r2

Einstein said if Newton can add magical power then I can do magic powers of Newton and add Einstein’s own magical powers of time travel force and the forceIs F = G m M/r2 (space force) + k/r4 (time force)

These two morons claim that what we see is due to magic where the Sun is pulling on the planets to keep them around? Did anyone see the Sun Pulling?Did anyone see a damn robe not the Sun hanging the planets to revolve around it but to hang these two idiots with Alfred Nobel physicists?!

There is nothing Nobel about Alfred Nobel institution that gave prizes for magical forces and time travel and garbage for physics.

I am not saying that Alfred Nobel physicists are idiots to say the least but what I am saying is that Alfred Nobel prize winner physics is stupidity itself and Alfred Nobel prize winner physicists are crooks to be exact. Time is not a structure like space to allow Alfred Nobel Institution to dictate stupidity in classrooms on the rest of us and Alfred Nobel Physicists can keep stupidity to themselves. Alfred Nobel prize is worth 1001 times in money. Alfred Nobel prize is a 1 billion dollar business each. Big Institutions with massive funding needs backed bad physics and idiots for physicists for 1 billion dollars prize and the world stuck with garbage for physics.What is modern physics? Did anyone hear the big bang? Did anyone saw a black hole? Did anyone find Einstein’s gravity waves? Did anyone hook up a hose to the sky and funnel dark energy to verify E = mc2? Does anyone have a proof of Newton’s action at a distance? Does anyone have a proof of Einstein time travel? Does anyone figure out why humans have to accept Alfred Nobel prize winners nuclear claims that the universe is made up of particles that dies and rebirth billion billion times a second? Hey dude I am a redneck one grandfather was a damn farmer and my other grandfather was a damn Carpenter and my father was a much damn redneck I sometimes feel reluctant to claim him but I never knew what a dumb ass really is till I met my science teacher talking about the greatest idiots of physics starting With the seven idiots in order Copernicus Galileo Newton and Einstein!

What did Alfred Nobel honor? My answer to that is: “University” Crooks in Alfred Nobel SuitThe difference between a crook and a Nobel Prize winner is a flip of a coinThe difference between me and all others I can prove them crooks and I dare all to prove me wrong. Starting with Einstein and Newton to make Galileo irrelevant to the subject of planetary motion around the Sun because of now and now is July 4th 1973 there is no Galileo and there is No Einstein and there is no Newton because I can prove them all wrong including NASA and Europeans space agency and all others!Newton’s space force: F = G m M/r2

Einstein’s – Newton space - time:

F = G m M/r2 (Newton’s space force) + k/r4 (Einstein’s time force)

The force is with Joe Nahhas and in real time space F = 0Nahhas said the only magic is an act of kindness of a human to all others!My act of kindness to humanity is to make humanity aware of the criminals with Nobel prizes and the idiots with Intelligent portfolios!

What I am saying is that there is no Newton’s gravitational force and Einstein’s space – time force is silly to say the least and I can prove that there is no gravitational force between the Sun and the Planets force, or, F = o solution is a rotating ellipse Illusion.

What I am saying dear Pope John Paul is that those with “knowledge” and sick intentions to be arrogant and claim magical forces and mocks the believers have absolutely no clue what they see in their telescopes and have nothing to offer beside stupidity that started with Newton and perfected with Alfred Nobel Institution’s dumbest Einstein. To start with let me solve this equation F = 0. The solution of F = 0 is the equation of a real time circle Is r = r 0 e í (θ + ω t)

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location:

P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rate

In polar coordinates systemr = r r (1) ;v = r' r(1) + r θ' θ(1) ; γ = (r" - rθ'²)r(1) + (2r'θ' + r θ")θ(1)r = location; v = velocity; γ = accelerationF = m γ + 2m'v +m" r

F = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + (m" r) r (1) = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/m r) [d (m²r²θ')/d t] θ (1) = [-G m M/r²] r (1) -------------------- Newton's Gravitational Law

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Proof:First r = r [cosine θ î + sine θ Ĵ] = r r (1)Define r (1) = cosine θ î + sine θ Ĵ Define v = d r/d t = r' r (1) + r d[r (1)]/d t = r' r (1) + r θ'[- sine θ î + cosine θĴ] = r' r (1) + r θ' θ (1)Define θ (1) = -sine θ î +cosine θ Ĵ;And with r (1) = cosine θ î + sine θ ĴThen d [θ (1)]/d t= θ' [- cosine θ î - sine θ Ĵ= - θ' r (1) And d [r (1)]/d t = θ' [-sine θ î + cosine θ Ĵ] = θ' θ (1) Define γ = d [r' r (1) + r θ' θ (1)] /d t = r" r (1) + r'd [r (1)]/d t + r' θ' r (1) + r θ" r (1) +r θ'd [θ (1)]/d t γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 With m = constant, then With d² r/dt² - r θ'² = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2 With d² r/dt² - r θ'² = 0 E q – 1 And d (r²θ')/d t = 0 Eq-2

From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant Differentiate with respect to timeThen 2rr'θ' + r²θ" = 0Divide by r²θ' Then 2(r'/r) + θ"/θ' = 0And 2(r'/r) = - θ"/θ' = 2[λ (r) + ỉ ω (r)]Also, r = r0 (θ, 0) e [λ (r) + ỉ ω (r)] t

And θ' = θ' 0 (θ, 0) e -2[λ (r) + ỉ ω (r)] t


And θ' = θ' 0 (θ, 0) e -2ỉ ω t

With d² r/dt² - r θ'² = 0Let u = 1/r; r = 1/u; r²θ' = h = θ’ /u² And d r/d t = (d r/ d u) (d u /d θ) (d θ/ d t) = (- 1/u ²) (θ’) (d u/ d θ) = - h (d u/ d θ)

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And d² r/ d t² = - h (θ’) (d² u/ d θ ²) = [- h²/r²] (d² u/ d θ ²)

= - h² u² (d² u/ d θ ²)

With d² r/dt² - r θ'² = 0 E q – 1 And - h² u² (d² u/ d θ ²) – (1/u) (h u²) ² = 0Then (d² u/ d θ ²) + u = 0


Then r 0 (θ, 0) = (1/A) e í θ And r 0 (θ, 0) = r 0 (0, 0) e í θ And r = r 0 (0, 0) e í θ e ỉ ω t

Or, r = r 0 e í (θ + ω t) What I am saying Dear Pope is this is the equation of motion of any planet! It is also the equation of motion of the smallest particle light!

Now Planet Mercury motion is r m = r m 0 e í [θ (m) + ω (m) t]

The Sun Motion is r s = r s 0 e í [θ (s) + ω (s) t]

Sun r r s Mercury r1 r0 r m X1 r2 X2 Earth

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In mathematical form: r1 = x1 + r sAnd r2 = x2 + r m

Then r2 - r1 = x2 - x1 + r m – r sOr, r = r0 + (r m – r s)To find the extreme values of r,Then: d r/d t = d r0/d t+ (d r m/d t – d r s/ d t) = 0

And (d r m/d t – d r s/ d t) = 0; d r0/d t = 0

And d r m/d t = d r s/ d tNow Planet Mercury motion is r m = r m 0 e í [θ (m) + ω (m) t]


And d r m/d t = r m 0 [θ’ (m) + ω (m)] e í [θ (m) + ω (m) t]

And d r s/ d t = r s 0 [θ’ (s) + ω (s)] e í [θ (s) + ω (s) t]

With r m 0 [θ’ (m) + ω (m)] Cosine [θ (m) + ω (m) t]= r s 0 [θ’ (s) + ω (s)] Cosine [θ (s) + ω (s) t]And r m 0 [θ’ (m) + ω (m)] sine [θ (m) + ω (m) t]= r s 0 [θ’ (s) + ω (s)] sine [θ (s) + ω (s) t]Then dividing we get: tan [θ (n) + ω (n) t] = tan [θ (s) + ω (s) t]Or, [θ (m) + ω (m) t] = [θ (s) + ω (s) t] + [n π = 0; n =0]

And θ (m) - θ (s) + = [ω (s) - ω (m)] tThis is the time delay IllusionAnd, r m 0 [θ’ (m) + ω (m)] = r s 0 [θ’ (s) + ω (s)] And [θ’ (s) + ω (s)] = (r m 0 /r s 0) [θ’ (m) + ω (m)]

And [θ’ (s) + ω (s)] - [θ’ (m) + ω (m)] = [(r m 0 /r s 0) -1] [θ’ (m) + ω (m)]

In arc second per century:

Is: [θ’ (s) + ω (s)] - [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600)= [(r m 0 /r s 0) -1] [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600)

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And, r m 0 [θ’ (m) + ω (m)] = r s 0 [θ’ (s) + ω (s)]

And r m 0 [θ’ (m) + ω (m)] = r e 0 [θ’ (e) + ω (e)]

And [θ’ (s) + ω (s)] - [θ’ (e) + ω (e)] = [(r m 0 /r e 0) -1] [θ’ (m) + ω (m)]


Is: [θ’ (e) + ω (e)] - [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600)= [(r m 0 /r e 0) -1] [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600) = 43

r m 0

r s 0 M M Mercury S Sunr m0 r 0

Is: r 0 = distance between centersIs: r max = r 0 + (r m 0 + r s 0)And r min = r 0 - (r m 0 + r s 0)And c = (r max - r min)/2 = r m 0 + r s 0And a = (r max + r min)/2 = r 0

With c = (r max - r min)/2 = r m 0 + r s 0And a = (r max + r min)/2 = r 0 This establishes visual ellipse of semi major axis a and a semi minor axis b = √ (a² - c²) = a √ [1 – (c/a) ²] = a √ (1 – ε ²)

And h = 2 π a b = 2 π a² √ (1 – ε ²) Areal Velocity is: 2 π a² √ (1 – ε ²)/T

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Figure - 1

Page 126

And Farthest distance = r max = r 0 + (r m 0 + r s 0) = a + c

θ

r

Sun

Mercury

And closest distance = r min = r 0 - (r m 0 + r s 0) = a - c

At closest approach: h =2 π a² √ (1 – ε ²)/TAnd dividing h / (a – c) ² = 2 π a² √ (1 – ε ²)/T a² (1 – ε) ² Or r²θ' = h will give θ' (a) = 2 π √ (1 – ε ²)/T (1 – ε) ²

And for any motion θ' = θ' 0 (θ, 0) e -2ỉ ω t

And θ’ = {[cosine 2 ω t + í sine 2 ω t] – 1} θ' 0 (θ, 0) And θ’ = θ’ (x) + í θ’ (y) = [cosine 2 ω t + í sine 2 ω t] θ' 0 (θ, 0)And θ’ (x) = [cosine 2 ω t] θ' 0 (θ, 0) = [1 – 2 sine² ω t] θ' 0 (θ, 0)

And θ’ (x) - θ' (0) = - 2 θ' 0 (θ, 0) sine² ω t = -2(2π) [√ (1-ε²)]/T (1-ε) ²] sine² ω t = - 4 π [√ (1-ε²)]/T (1-ε) ²] sine² ω t

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If this apsidal motion is to be found as visual effects, then

With, v ° = spin velocity; v* = orbital velocityAnd v/c = (v* + v°)/c = tan ω t

= -4 π [√ (1-ε²)]/T (1-ε) ²] sine² {Inverse tan [(v* + v°)/c]} radians

Multiplication by 180/π to change to degrees

Ψ ° = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [(v* + v°)/c]} Degrees And multiplication by 1 century = 36526 days and using T in days

Ψ’’ (calculated) = (-720x36526/T days) {[√ (1-ε²)]/ (1-ε) ²} x sine² {Inverse tan [(v* + v°)/c]} degrees/100 years

And in arc seconds per century:Ψ’’ (calculated) = (-720x36526x3600/T days) {[√ (1-ε²)]/ (1-ε) ²} x sine² {Inverse tan [(v* + v°)/c]} degrees/100 years = 43

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Sun

Ψ

Mercury

θ

The confrontation between physics and physicists By Professor Joe Nahhas 1978

Greetings: my name is Joe Nahhas and I am the greatest physicist of all time since July 4th 1973I am unknown to the world of physics but I have 1001 new real time physics formulas to change physics and history to delete all physics and wipe out written history and this is a revolt!

Abstract: There is no solar system and Earth does not move around the Sun. I can say that there is no solar system and Earth does not move around the Sun because I can prove that there is no solar system and that Earth does not move around the Sun. There is a solar group of objects humans call one star few planets and few moons. Each object of the solar group has its own independent orbit (circle) and all are in the same neighborhood looking like a solar “system” where Earth is visually looking like it moves around the Sun in an ellipse and the Sun at the focus (not center) of that ellipse. Gravity is a spin effect found inside the atmosphere. There is no gravity outside the atmosphere. There is no gravity between Earth and Sun. Taking that there is no gravity between Earth and Sun leads to a circular motion for the Sun and a circular motion for Earth. The difference between two circular motions is visual motion and not an actual motion of an ellipse. An ellipse motion requires an inverse square law that is given by Newton’s equation that says the gravitational force F between Earth of mass m and the Sun of mass M is given by F = - G m M/r² where r is the distance between Earth and sun and G is a constant number. A rotating ellipse requires Newton’s law of space motion - G m M/r² and Einstein’s time motion k/r4 to a total force of F = - G m M/r² + k/r4 in contradiction to Qur’an F = 0

In Short Planets Moons Sun all swim freely and no gravity restrictions. No gravity means the Force F = 0. The solution of F = 0 leads to a circular orbit for all

. Sun r r s Mercury r1 r0 r m X1 r2 X2 Earth

r s 0 M M M Mercury S Sun

r m 0 r 0

The Sun S moves in a circle and Planet mercury moves in a circle.And Farthest distance = r max = r 0 + (r m 0 + r s 0) = a + c


Gives a visual ellipse and visual (not actual gravity) meaning planets seems to move around the Sun and not actually moving around the Sun

Figure - 1

If we took a look at the most famous physics fundamental constant G from Newton’s 350 years old Equation F = - G m M/r² G’s value has unit dimension of [G] = [1/ρ T²]; ρ is atmospheric density and T is time rotational period. If we can show that the gravitational constant G = G (ρ, T) is a variable air density and Earth rotational period dependent then we can say what Earth is saying that the most famous physics fundamental constants are not so universal constants because all physics fundamental constant are air density and Earth spin dependent. For example: G = G (ρ, T) = 1/ [(2/5) (4 π/3) ρT²]; 5/2 ~ √ (2 π)Nothing that: 8 π /15 = (2/5) (4π/3); 4π/3 and 2/5 come from spherical aspect Where: ρ = air density; T = Earth rotation period; T = 23.9333 hours and ρ = 1.2045 kg/m³ G = 1/ (8 π/15) ρT² = 6.674 x 10 – 11

[G] = [1/ρ T²] ~ 0 if spin period is very long or atmospheric density is very highThis fact applies to electromagnetic fundamental constants:

Like ε (0) = 1/4πρT² = 8.85 x 10 -12; μ (0) = 4πρT²/c²There is no “gravity” between the “sun” and “planets” and that means: The force F = 0And not F = - G m M/r²

θ

r

Sun

Mercury

Yes I am saying that astronomers have no clue what they see using their telescopesWhat I am saying is that there is no F = - G m M/r² and there is no solar system where the motion of a planet around the sun is an ellipse with the sun at the focus of the ellipse.

Figure - 1

Kepler wrote the acceleration law: a/T2 = k/a2

And said that the motion of a planet around the sun is an ellipse like in figure - 1 And wrote the equation of motion law As: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)]

Newton and Hooke reiterated the same thing as Kepler and wrote: r/ t2 = k/r2

Astronomers found that the motion is not an ellipse but a rotating ellipseAnd wrote the equation of motion law As: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)]

Einstein wrote another acceleration law r/t2 = k/ r2 + k1/r4 to describe a rotating ellipse

θ

r

Sun

Mercury

Newton gave the equation of the ellipse:As: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] Einstein said he corrected Newton and wrote the new equation:Is: r (θ, t) = a (1-ε²)/ [1+ε cosine (θ - Ψ)]It is a fact that Astronomers see this axis tilt and Einstein got a formula for it: ψ = {6 π G M/a c² (1 - ε²)} [180/ π] [36526/T (days)][3600] G = 6.673 x 10-11= gravitational acceleration constantM =2x1030kg = mass of the Sun Eccentricity of Mercury’s orbit = ε = 0.206T= 88 days = period of Planet mercury’s rotation around the SunAnd c = 299792.458 km/sec =light speed/secAnd a = 58.2 x 106 km/sec = semi major axis of planet Mercury orbitψ = {6 (π) 6.673 x 10-11 (2 x 1030)/58.2 x 106 (299792.458) ² [1– (0.206)²]} x [180/ π] [36526/88] [3600] = 43 seconds of an arc per century

Newton used magical powers or action at a distance force F = G m M/r2

Einstein said if Newton can add magical power then I can do magic powers of Newton and add Einstein’s own magical powers of time travel force and the forceIs F = G m M/r2 (space force) + k/r4 (time force)

Sun

ψ

Mercury

θ

These two morons claim that what we see is due to magic where the Sun is pulling on the planets to keep them around? Did anyone see the Sun Pulling?Did anyone see a damn robe not the Sun hanging the planets to revolve around it but to hang these two idiots with Alfred Nobel physicists?!

There is nothing Nobel about Alfred Nobel institution that gave prizes for magical forces and garbage for physics.

Page 6I am not saying that Alfred Nobel physicists are idiots to say the least but what I am saying is that Alfred Nobel prize winner physics is stupidity itself and Alfred Nobel prize winner physicists are crooks to be exact. Time is not a structure like space to allow Alfred Nobel Institution to dictate stupidity on the rest of us and Alfred Nobel Physicists can keep stupidity to themselves. Alfred Nobel prize is worth 1001 times in money. Alfred Nobel prize is a 1 billion dollar business each. Big Institutions with massive funding needs backed bad physics and idiots for physicists for 1 billion dollars prize and the world stuck with garbage for physics.What is modern physics? Did anyone hear the big bang? Did anyone saw a black hole? Did anyone find Einstein’s gravity waves? Did anyone hook up a hose to the sky and funnel dark energy to verify E = mc2? Does anyone have a proof of Newton’s action at a distance? Does anyone have a proof of Einstein time travel? Does anyone

figure out why humans have to accept Alfred Nobel prize winners nuclear claims that the universe is made up of particles that dies and rebirth billion billion times a second? Hey dude I am a redneck one grandfather was a damn farmer and my other grandfather was a damn Carpenter and my father was a much damn redneck I sometimes feel reluctant to claim my dad but I never knew what a dumb ass is really is till I met my science teacher talking about the greatest idiots of physics Newton and Einstein! What did Alfred Nobel honor? My answer to that is: “University” Crooks in Alfred Nobel SuitThe difference between a crook and a Nobel Prize winner is a flip of a coinThe difference between me and all others I can prove them crooks and I dare all to prove me wrong. Starting with Einstein and Newton to make Galileo irrelevant to the subject of planetary motion around the Sun because of now and now is July 4th 1973 there is no Galileo and there is No Einstein and there is no Newton because I can prove them all wrong including NASA and Europeans space agency and all others!Newton’s space force: F = G m M/r2

Einstein’s – Newton space - time: F = G m M/r2 (Newton’s space force) + k/r4 (Einstein’s time force)The force is with Joe Nahhas and in real time space F = 0

Nahhas said the only magic is an act of kindness of a human to all others!My act of kindness to humanity is to make humanity aware of the criminals with Nobel prizes and the idiots with intelligent portfolios!

What I am saying is that there is no Newton’s gravitational force and Einstein’s space – time force is silly to say the least and I can prove that there is no gravitational force between the Sun and the Planets force, or, F = o solution is a rotating ellipse Illusion.

What I am saying dear Pope John Paul is that those with “knowledge” and sick intentions to be arrogant and claim magical forces and mocks the believers have absolutely no clue what they see in their telescopes and have nothing to offer beside stupidity that started with Newton and perfected with Alfred Nobel Institution’s dumbest Einstein. To start with let me solve this equation F = 0. The solution of F = 0 is the equation of a real time circle Is r = r 0 e í (θ + ω t)

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location r = r (x, y, z). The state of any object in the Universe can be expressed as the product S = m r; State = mass x location: P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment = change of location + change of mass = m v + m' r; v = velocity = d r/d t; m' = mass change rateF = d P/d t = d²S/dt² = Total force

= m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r = m γ + 2m'v +m" r; γ = acceleration; m'' = mass acceleration rate

In polar coordinates systemr = r r (1) ;v = r' r(1) + r θ' θ(1) ; γ = (r" - rθ'²)r(1) + (2r'θ' + r θ")θ(1)r = location; v = velocity; γ = accelerationF = m γ + 2m'v +m" r

F = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m'[r' r (1) + r θ' θ (1)] + (m" r) r (1) = [d² (m r)/dt² - (m r) θ'²] r (1) + (1/m r) [d (m²r²θ')/d t] θ (1) = [-G m M/r²] r (1) -------------------- Newton's Gravitational Law

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Proof:First r = r [cosine θ î + sine θ Ĵ] = r r (1)Define r (1) = cosine θ î + sine θ Ĵ Define v = d r/d t = r' r (1) + r d[r (1)]/d t = r' r (1) + r θ'[- sine θ î + cosine θĴ] = r' r (1) + r θ' θ (1)Define θ (1) = -sine θ î +cosine θ Ĵ;And with r (1) = cosine θ î + sine θ ĴThen d [θ (1)]/d t= θ' [- cosine θ î - sine θ Ĵ= - θ' r (1) And d [r (1)]/d t = θ' [-sine θ î + cosine θ Ĵ] = θ' θ (1) Define γ = d [r' r (1) + r θ' θ (1)] /d t = r" r (1) + r'd [r (1)]/d t + r' θ' r (1) + r θ" r (1) +r θ'd [θ (1)]/d t γ = (r" - rθ'²) r (1) + (2r'θ' + r θ") θ (1)With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 With m = constant, then With d² r/dt² - r θ'² = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2 With d² r/dt² - r θ'² = 0 E q – 1 And d (r²θ')/d t = 0 Eq-2 From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant Differentiate with respect to timeThen 2rr'θ' + r²θ" = 0Divide by r²θ' Then 2(r'/r) + θ"/θ' = 0And 2(r'/r) = - θ"/θ' = 2[λ (r) + ỉ ω (r)]Also, r = r0 (θ, 0) e [λ (r) + ỉ ω (r)] t

And θ' = θ' 0 (θ, 0) e -2[λ (r) + ỉ ω (r)] t

For a fixed orbit: λ r = 0

Also, r = r 0 (θ, 0) e ỉ ω t

And θ' = θ' 0 (θ, 0) e -2ỉ ω t

With d² r/dt² - r θ'² = 0

Let u = 1/r; r = 1/u; r²θ' = h = θ’ /u² And d r/d t = (d r/ d u) (d u /d θ) (d θ/ d t) = (- 1/u ²) (θ’) (d u/ d θ) = - h (d u/ d θ)

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And d² r/ d t² = - h (θ’) (d² u/ d θ ²) = [- h²/r²] (d² u/ d θ ²) = - h² u² (d² u/ d θ ²)With d² r/dt² - r θ'² = 0 E q – 1 And - h² u² (d² u/ d θ ²) – (1/u) (h u²) ² = 0Then (d² u/ d θ ²) + u = 0Then u = A e - í θ

Then r 0 (θ, 0) = (1/A) e í θ And r 0 (θ, 0) = r 0 (0, 0) e í θ And r = r 0 (0, 0) e í θ e ỉ ω t Or, r = r 0 e í (θ + ω t) What I am saying Dear Pope is this is the equation of motion of any planet! It is also the equation of motion of the smallest particle light!Now Planet Mercury motion is r m = r m 0 e í [θ (m) + ω (m) t]


Sun r r s Mercury r1 r0 r m X1 r2 X2 Earth

In mathematical form: r1 = x1 + r sAnd r2 = x2 + r mThen r2 - r1 = x2 - x1 + r m – r sOr, r = r0 + (r m – r s)

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To find the extreme values of r,Then: d r/d t = d r0/d t+ (d r m/d t – d r s/ d t) = 0

And (d r m/d t – d r s/ d t) = 0; d r0/d t = 0

And d r m/d t = d r s/ d tNow Planet Mercury motion is r m = r m 0 e í [θ (m) + ω (m) t]


And d r m/d t = r m 0 [θ’ (m) + ω (m)] e í [θ (m) + ω (m) t]

And d r s/ d t = r s 0 [θ’ (s) + ω (s)] e í [θ (s) + ω (s) t]

With r m 0 [θ’ (m) + ω (m)] Cosine [θ (m) + ω (m) t]= r s 0 [θ’ (s) + ω (s)] Cosine [θ (s) + ω (s) t]And r m 0 [θ’ (m) + ω (m)] sine [θ (m) + ω (m) t]= r s 0 [θ’ (s) + ω (s)] sine [θ (s) + ω (s) t]Then dividing we get: tan [θ (n) + ω (n) t] = tan [θ (s) + ω (s) t]Or, [θ (m) + ω (m) t] = [θ (s) + ω (s) t] + [n π = 0; n =0]

And θ (m) - θ (s) + = [ω (s) - ω (m)] tThis is the time delay IllusionAnd, r m 0 [θ’ (m) + ω (m)] = r s 0 [θ’ (s) + ω (s)] And [θ’ (s) + ω (s)] = (r m 0 /r s 0) [θ’ (m) + ω (m)]

And [θ’ (s) + ω (s)] - [θ’ (m) + ω (m)] = [(r m 0 /r s 0) -1] [θ’ (m) + ω (m)]


Is: [θ’ (s) + ω (s)] - [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600)= [(r m 0 /r s 0) -1] [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600)

And, r m 0 [θ’ (m) + ω (m)] = r s 0 [θ’ (s) + ω (s)]

And r m 0 [θ’ (m) + ω (m)] = r e 0 [θ’ (e) + ω (e)]

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And [θ’ (s) + ω (s)] - [θ’ (e) + ω (e)] = [(r m 0 /r e 0) -1] [θ’ (m) + ω (m)]


Is: [θ’ (e) + ω (e)] - [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600)= [(r m 0 /r e 0) -1] [θ’ (m) + ω (m)] x (180/π) x (36526/88) x (3600) = 43

r m 0

r s 0 M M M Mercury S Sunr m 0 r 0

Is: r 0 = distance between centers

Is: r max = r 0 + (r m 0 + r s 0)

And r min = r 0 - (r m 0 + r s 0)

And c = (r max - r min)/2 = r m 0 + r s 0

And a = (r max + r min)/2 = r 0

With c = (r max - r min)/2 = r m 0 + r s 0And a = (r max + r min)/2 = r 0

This establishes visual ellipse of semi major axis a, and a semi minor axis;

b = √ (a² - c²) = a √ [1 – (c/a) ²] = a √ (1 – ε ²)

And h = 2 π a b = 2 π a² √ (1 – ε ²) Areal Velocity is: 2 π a² √ (1 – ε ²)/T

Figure - 1

θ

r

Sun

Mercury

And Farthest distance = r max = r 0 + (r m 0 + r s 0) = a + c


At closest approach: h =2 π a² √ (1 – ε ²)/TAnd dividing h / (a – c) ² = 2 π a² √ (1 – ε ²)/T a² (1 – ε) ² Or r²θ' = h will give θ' (a) = 2 π √ (1 – ε ²)/T (1 – ε) ²

And for any motion θ' = θ' 0 (θ, 0) e -2ỉ ω t

And θ’ = {[cosine 2 ω t + í sine 2 ω t] – 1} θ' 0 (θ, 0) And θ’ = θ’ (x) + í θ’ (y) = [cosine 2 ω t + í sine 2 ω t] θ' 0 (θ, 0)And θ’ (x) = [cosine 2 ω t] θ' 0 (θ, 0) = [1 – 2 sine² ω t] θ' 0 (θ, 0)

And θ’ (x) - θ' (0) = - 2 θ' 0 (θ, 0) sine² ω t = -2(2π) [√ (1-ε²)]/T (1-ε) ²] sine² ω t = - 4 π [√ (1-ε²)]/T (1-ε) ²] sine² ω t

If this apsidal motion is to be found as visual effects, then

With, v ° = spin velocity; v* = orbital velocityAnd v/c = (v* + v°)/c = tan ω t

= -4 π [√ (1-ε²)]/T (1-ε) ²] sine² {Inverse tan [(v* + v°)/c]} radians Multiplication by 180/π to change to degrees

Ψ ° = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [(v* + v°)/c]} Degrees And multiplication by 1 century = 36526 days and using T in days

Ψ’’ (calculated) = (-720x36526/T days) {[√ (1-ε²)]/ (1-ε) ²} x sine² {Inverse tan [(v* + v°)/c]} degrees/100 yearsAnd in arc seconds per century:Ψ’’ (calculated) = (-720x36526x3600/T days) {[√ (1-ε²)]/ (1-ε) ²} x

Sun

Ψ

Mercury

θ

sine² {Inverse tan [(v* + v°)/c]} degrees/100 years = 43

Big Banger Georges Lemaitre and Time traveler Albert Einstein and Real Timer Joe Nahhas

Time Travel Bus Driver Einstein said that his proudest space – time travel achievement is the proof that planet Mercury travels with space angular velocity of 70.75 and time angular velocity of 43.0Big Banger Georges Lemaitre said that The Universe came from a Big Bang ExplosionJoe Nahhas said Buzz off:

Einstein greatest achievement in space said Planet Mercury angular speed travel in space is 70.75 and in time 43The Advance of Planet Mercury’s Perihelion given by space formula of:

[2 π/T] [180/π][36526/88][3600] = [2 π /88 x 24x3600] [180/π][36526/88][3600] = 70.75

And Einstein’s formula of: ψ = - 6 π GM/a c² (1 - ε²) [180/ π] [36526/T] [3600] = 43 angular velocity = angular speed of time travel of planet Mercury

G = 6.673 x 10-11= gravitational acceleration constantM =2x1030 kilogram = mass of the Sun Eccentricity of Mercury’s orbit = ε = 0.206T= 88 days = period of Planet mercury’s rotation around the SunAnd c = 299792.458 kilometer/second =light speed in kilometers/secondAnd a = 58.2 x 106 km/sec = semi major axis of planet Mercury orbit

In my 5th grade I did this:

Angular velocity = θ’0 = θ’0 line - 1 Visual angular velocity = θ’ = θ’ line - 2 - Visual angular velocity = - θ’ = - θ’ line - 3

Add line - 1 and Line – 2 and line - 3

Then θ’0 = θ’ + (θ’0 - θ’); Divide by θ’

And (θ’0/ θ’) = 1 + [(θ’0 – θ’)/ θ’]; multiply by θ’0

Then (θ’0/ θ’) θ’0 = θ’0 + [(θ’0 – θ’)/ θ’] θ’0

Quantum = classical + relativisticAnd [(θ’0 – θ’)/ θ’] θ’0 = [(2πR0 /T– 2πR/T)/ 2πR/T] θ’0

Or, [(θ’0 – θ’)/ θ’] θ’0 = [(R0 - R)/R] θ’0R0 = Mercury’s distance from the sun = 5.82 x 106 kmR = Earth’s distance from the sun = 149.6 x 106 kmAnd Mercury’s angular velocity;Is θ’0 = 2πR0/T = 2π x 5.82 x 106 /88x 24x3600 radians /period

And Mercury’s angular velocity accumulation in arc seconds per centuryIs: δθ’0 = (2π x 5.82 x 106 /88x 24x3600) x (36526/88) (180/ π) (3600) = 70.75[(R0 - R)/R] δθ’0 = [(5.82 x 106 - 149.6 x 106)/ 149.6 x 106] x(2π x 5.82 x 106 /88x 24x3600) x (36526/88) (180/ π) (3600) = 43.0 arc second per centuryWhere this came from?

It is called Arabs real time astronomy?





Observer line: …………………Observer ………………

If we stand in the middle of the street and watch a car moving directly away from us we see as if the size of the car shrunk. If we take a picture of the car at different distances then we see that the pictures of the car look like as if the car contracted in size. The further the car away from us the smaller image of the car in the picture. The size of the

car did not change but real time light data reflected from the car and ran through air and into our eyes changes the real time image of the car to indicate a different location of the car and not a different size of the car. Our eyes see real time images of a moving car. If the car stops, then the image of the car freezes. A picture or a snap shot of a moving car is a frozen image of the moving car. When a car moves its real time image change size but the car does not change size. The car as a moving object at different distances has a variable real time image that our eyes see in real time and depends on distance and motion as perceived by our eyes. In short statementsOur eyes see objects in real time as different objects at different distances and different speeds. Or, our eyes see real time objects we call images that change when measured from different distances or when moving. Our eyes do not see actual objects but Images of actual objects and our eyes see different images of same actual objects when at different distances of moving In mathematical form: 1 = 1 self evident2 = 2 is self evidentImage = image = I is self evident line - 1 Object = object = O = self evident line - 2And, - Object = - objects = - O = self evident line - 3I = I line - 1O = O line - 2And, - O = - O line - 3 Adding: line - 1 and line - 2 and line - 3I + O – O = O + (I – O) line - 4I = O + (I – O); O – O = 0 one left side (I/O) = (O/O) + (I – O)/O; dividing by on both sides line - 5 Or, (I/O) = 1 + (I – O)/O; O/O = 1 line - 6 (I/O) θ’0 = θ’0 + [(I – O)/O] θ’0 Line - 7 If we try to measure the angular velocity of planet Mercury around the sun θ’0 from planet Earth and not from the Sun a visual effect (error) of [(I – O)/O] θ’0 is measured that Einstein’s call time travel(I/O) θ’0 = θ’0 + [(I – O)/O] θ’0Quantum = classical + relativisticWhat we see = what is + visual effectsReal time object = event time Object + time delay objectReal time astronomy = event time astronomy + time delay astronomyIf we try to measure an object θ’0 we see it as (I/O) θ’0 and the difference accumulation:Is [(I – O)/O] θ’0 = visual effectsAnd the accumulation is [(R0 - R)/R] δθ’0And Planet Mercury angular velocity space travel Is δθ’0 = (2π x 5.82 x 106 /88x 24x3600) x (36526/88) (180/ π) (3600) = 70.75 And Arabs 10th century correction to idiot man of the 20th century Einstein

Is: [(R0 - R)/R] δθ’0 = [(5.82 x 106 - 149.6 x 106)/ 149.6 x 106] x(2π x 5.82 x 106 /88x 24x3600) x (36526/88) (180/ π) (3600) = 43.0 arc second per century

Another method is the product methodIf two objects going in two different orbital paths around the Sun and one with speed v* m = planet Mercury Orbital speed around the Sun and one with speed v* e = Planet Earth orbital speed around the Sun; then the cross multiply is: (v* e - v° e/v m + v° m) δθ’0 = [(29.8 – 0.465)/47.9 + 0.465] (70.75) = 43Buzz off EinsteinThen came the big BangerThe big Banger said that: r = rAnd r = r + 0And r = r + (r0 - r0)And r = (r - r0) + r0; dividing by r0Then: r/r0 = 1 +(r - r0)/r0 And: n λ /n λ 0= 1 + (n λ – n λ0)/ λ0 Or, λ/ λ0 = 1 + (λ – λ0)/ λ0 = 1 + ZThis quantity [(λ - λ 0)/ λ 0] = Z = Red Shift is a measure of a different is an indication of a different location and not of an expanding Universe to claim big bang Distant r, λ

Lab r0, λ0 …………………………………………………………………………………………………………………….

Lab λ0 will have a different value Lab λ0 when measured from a distant.

That is: (r/r0) λ 0= λ 0 + [(r – r0)/ r0] λ0Or, (λ / λ 0) r 0= r 0 + [(λ – λ 0)/ λ 0] r0

The conclusion is that if we use distance scale to measure wavelength λ0We will measure it as λ and a wavelength distance Illusion delay by the quantity of [(r – r0)/ r0] λ0 will be added or subtracted to λ0And if we use wavelength scale to measure distance r0; then a distance wavelength Illusion delay by the quantity of [(λ – λ 0)/ λ 0] r0 will be added or subtracted to distance r0

What Lemaitre confusion is distance Illusion expansion of [(λ – λ 0)/ λ 0] r0 when he tried to measure r0

Nahhas confession to these two: “I am real timer and I am a proud American redneck and yall Idiots; Big Bang and Time Travel is a purely western stupidity”

Stanford University most stupid professor Leonard Susskind

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Transcript of Stanford University most stupid professor Leonard Susskind