Top ten wrong Physics and wrong Physicists
By Professor Joe Nahhas;
Zewail ChuNewton
ChadwickFeynmanLe VerrierReinesSoldner
Einstein
Modern physics is based on optical illusions and these are the formulas of optical Illusions that explains it all
And mass m = m(0)e ỉ ω t; distance r = r(0)e ỉ ω t; time Γ = te ỉ ω t
And ω t = arc tan (v/c)
Real time
We can not see or measure something that did not happen.
We can only see or measure something that had happened.
What we measure is not what happened.
We measure in present time an event that happened in past time.
That is we measure past events in present time
Present time = present time
Present time = past time + [present time - past time]
Present time = past time + time delay (difference)
Real time physics = event time physics + real time delay physics
Quantum = classical + time delays (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 delays
Γ = 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 delays
Δ Γ (y) is perpendicular to the line of sight time delays
Page 1
Γ = t (Γ/t)
Γ = t e ỉ ω t;Γ/t = e ỉ ω t; ω t = arc tan (v/c)
Γ = t e ί θ; θ = arc tan (v/c)
Γ = t [cosine ω t + ỉ sine ω t]
Γ = Γ (x) + ỉ Γ (y) = t cosine ω t + ỉ t sine ω t
Γ (x) = t cosine ω t = t {1 - 2 sine² [(1/2) arc tan (v/c)]}
Δ Γ (x) = Γ (x) - t = - 2 t sine ² [(1/2) arc tan (v/c)]
Δ Γ (x) = - 2 t sine ² [(1/2) arc tan (v/c)]
Γ (y) = sine ω t = sine arc tan (v/c)
Δ Γ (x) = - 2 t sine ² [(1/2) arc tan (v/c)]
Δ Γ (x) /2 = - t sine ² [(1/2) arc tan (v/c)] seconds
And using the 1/2 cycle measurements
Where t = 100 years = 100 x 24 x 3600 seconds
Δ Γ (x) = - (100 x 365.26 x 24 x 3600) sine ² [(1/2) arc tan (v/c)]
With v* = orbital speed and vº = spin speed
And v = [v* (m) +/- vº (m)](Mercury)–[v (e) +/- vº](Earth)
= (47.9 – 0.002) km/sec –(29.8 km/sec – 0.465km/sec) = 18.565 km/sec
And c = light velocity = 300,000 km/sec
In ½ period and in arc per second
Or W" = (15) Δ Γ (x) = (7.5) t sine ² [(1/2) arc tan (v/c)] arc seconds
W" = 15 x 100 x 365.2624x 24 x 3600 x sine ² [(1/2) arc tan (18.565/300,000)]
= 43.0" arc seconds per 100 years
Δ Γ (x) = - 2 t sine ² [(1/2) arc tan (v/c)] in seconds
A century t = 100 (years) x 365.26 (days) x 24 (hours) x 3600 (seconds)
In second per century
Δ Γ (x) = - 2 (100 x 24 x 365.26 x 24 x 3600) sine ² [(1/2) arc tan (v/c)]
arc second per century
Δ Γ (x) = - 30(100 x 24 x 365.26 x 24 x 3600) t sine ² [(1/2) arc tan (v/c)]
Page 2
In arc per century using ½ cycle
And v = 47.9 – 29.335 = 18.565 km/sec
With c = 300,000 km/sec
Δ Γ (x) = - 15 sine ² [(1/2) arc tan (v/c)] (100 x 24 x 365.26 x 24 x 3600)
Δ Γ (x) = - 15 sine ² [(1/2) arc tan (18.435/300,000)] x
(100 x 24 x 365.26 x 24 x 3600)
= 43.0 arc sec per century
For Venus v (Mercury) = 35.1 km/sec
And [v* (Orbital speed of Earth) - v° (Spin speed of Earth)] = 29.8 - 0.465
Or,[v* (Orbital speed of Earth) - v° (Spin speed of Earth)] = 29.335 km/sec
And v = 35.1 – 29.335 = 5.765 km/sec
With c = 300,000 km/sec
Δ Γ (x) = - 15 sine ² [(1/2) arc tan (v/c)] (100 x 24 x 365.26 x 24 x 3600)
Δ Γ (x) = - 15 sine ² [(1/2) arc tan (5.765 /300,000)] x
(100 x 24 x 365.26 x 24 x 3600)
= 4.37 arc sec per century
Using full cycle
Δ Γ (x) = - 2 x 15 sine ² [(1/2) arc tan (5.765 /300,000)] x
(100 x 24 x 365.26 x 24 x 3600)
= 8.74 arc sec per century
It published literature both numbers, 4.37 arc sec per century for ½ cycle and 8.74 arc sec per century for full cycle are given without an explanation
Astronomers use 1/2 cycle
Then (Γ - t)/2 = - t sine² {[arc tan (v/c)]/2}; If v < c
Then (Γ - t)/2 = - (t /4) (v/c) ² seconds per 100 years in 1/2 cycles
And (Γ - t)/2 = - 15 (t /4) (v/c) ² arc seconds per 100 years in 1/2 cycle
= 3.75 t (v/c) ² arc seconds per 100 years in 1/2 cycle
With t = 36526 x 24 x 3600 seconds in 100 years
Δ Γ (x) = (Γ - t)/2 = 3.75 x 36526 x 24 x 3600 (v/c) ²
If we are using full cycle
Δ Γ (x) = - 2 t sine² {[arc tan (v/c)]/2}; If v < c
Then (Γ - t) = - 2 (t /4) (v/c) ² seconds per 100 years in 1/2 cycles
And (Γ - t) = - 30 (t /4) (v/c) ² arc seconds per 100 years in 1/2 cycle
= 7.5 t (v/c) ² arc seconds per 100 years in 1/2 cycle
Page 3
With t = 36526 x 24 x 3600 seconds in 100 years
For Mercury: (Γ - t)/2 = - 3.75 x [36526 x 24 x 3600] x (18.1/300,000) ²
= - 43.0 arc sec per century
(Γ - t) = - 7.5 x [36526 x 24 x 3600] x (18.1/300,000) ²
= - 86.0 arc sec per century
For Venus Velocity instead of Earth's Velocity
That is: 35.1[Venus orbital speed] + 6.52 [Venus spin speed] - 29.8
= 11.82 km/sec
Half - (Γ - t)/2 = - 3.75 x [36526 x 24 x 3600] x (11.82/300,000) ²
= 4.592821605
Of arc sec per century
Full (Γ - t) = - 2 x 3.75 x [36526 x 24 x 3600] x (11.82/300,000) ²
= 9.18654321arc sec per century
If a bicycle is ridden in a circular orbitof origin O and radius r making an angle θ from a fixed starting point A and going counter clock wise direction at a circular speed v then the angular speed of the bicycle is the angle change measured d θ the bicycle traces from the fixed starting point Adivided by time interval dt it travels or the angular speed is:
The angular speed is θ'= d θ/ d t =orbital velocity/ radius = v/r
Page 4
If Planet Mercury is at a distance r (m) from the Sun and rotating around the Sun with orbital speed v (m) and making an angle θ(m) from fixed point A, then planet Mercury angular velocity around the Sun:
Is θ’ (m) = v (m)/r (m)
With r = planet average distancefrom the Sun
And v = planet average speed around the sun
And θ = the angle traced by a planet from a fixed point when moving around the Sun.
And θ’ = angular speed of a planet moving around the Sun
With r (e) = Earth – Sun distance and r (m) = Mercury – Sun distance
With v (e) = Earth orbital speedaround the Sun
And v (m) = Mercury orbital speedaround the Sun
If we to measure the angular velocity of Planet Mercury from the Sun while planet Mercury moving around then Sun then, Planet Mercury Angularspeed
Is:θ' = v (m)/r (m)
If Planet mercury, m, orbital speed is to be measured from the Sun, S, then planet mercury orbital speed is θ’ (m) = v (m)/ r (m)
If Planet mercury, m, orbital speed is to be measured from the Earth, e, then planet mercury orbital speed is θ’ (m) = [v (m) + V (e)]/ r (m)
Page 5
Real time mechanics = absolute time mechanics + time delaysmechanics
Then, θ’ = [v (m) + v (e)]/r (m) = v (m)/r (m) + v (e)/r (m)
And not v (m)/ r (m)
The angular speed delay is: d θ' = v (e)/r (m)
Taking into account Earth rotation vº (e) then
With v (m) = 49.7 km/sec and v (e) = 29.8 km/sec
And taking into account Earth rotation vº (e) then
Then the angular speed delay is
W = d θ' = [v* (e) +/- v º/r] = [(29.8 km/sec) – 0.465km/sec]/ (58.2 x 106km)
W = (29.335km/sec)/ (58.2 x 106km) radian/second
In arc second per century multiplying by [(180/π) (3600) (26526/T)]
W (arc – sec /century) = [v* (e) +/- v º (e) /r (m)] X [(180/π) (3600) (26526/T)]
Or
W (arc – sec /century) = [v* (m) +/- v º (m) /r (e)] X [(180/π) (3600) (26526/T)]
X [r (e) - r (m)]/r (m)
W (arc – sec/century) = [(29.335km/sec)/ (58.2 x 106km)]
X [(180/π) (3600) (26526/T)]
= 43.1 arc second per century
This angular speed delay is a real time angular delay due to motion. Physicists call the 43.0 seconds of an arc delay is caused by time travel and it is presented as the first experimental proof of general relativity theory.
Planet / Distance rX 106km / Planet
Orbit T / Orbit speed v in km/sec / Less Earth speed / Spin speed
km/sec / Angular velocity; v/r
arc sec/ century
Mercury / 58.2 / 88 / 47.9 / 18.1 / .002 / 70.29
Venus / 108.2 / 224.7 / 35.05 / 5.7 / 6.52 / 10.86
Earth / 149.6 / 365.26 / 29.8 / .46511 / 4.1
Mars / 227.936 / 687 / 24.14 / 0.2411
Jupiter / 778.412 / 4333 / 13.06 / 12.6
Saturn / 1,426.725 / 10760 / 9.65 / 9.87
Uranus / 2,870.97 / 30690 / 6.80 / 2.59
Neptune / 4,498 / 60180 / 5.43 / 2.68
Pluto / 5906.4 / 90730 / 4.74
Page 6
For planet Venus
T (Venus) = 224.7days
For planet Venus
W"(v) = [v (e)/ r (v)] [36528/T (v)] (180/π) (3600) = 9.0"/century
The Angular velocity delay/ Planet Venus is v (e)/r (v) = [v* (e) +/- v° (e)]/r (Venus)
With T (v) = 224.7, v (e) = [29.8 km/sec - 0.465 km/sec]; r (v) = 108.2 x 10 9m
W"(m) = [v (e)/ r (v)] [36528/T (v)] (180/π) (3600) = 9.0"/century
W"(v) = [(29.8 – 0.465)/108.2 x 10 6] [36526/224.7] (180/π) (3600) = 9.09"/century
The number 43.0"/centuy is taught and advertised as the First experimental proofs of space –time confusions of modern physics and dearest to Einstein's general relativity theory.
If this visual Illusion is to be transferred to the Sun it will be seen as follows
Actual value is θ' = v/r = [v (m) + v (M)]/r
Where m = mass of primary; M = mass of the secondary
And v (m) = √ [GM²/ (m + M) a]
And v (M) = √ [GM²/ (m + M) a]
And v (m) = √ [GM²/ (m + M) a] + v (M) = √ [Gm²/ (m + M) a]
= √ [G (m + M)/ a]
And θ' = v/r = [v (m) + v (M)]/r
= √ [G (m + M)/ a³] Radians per seconds
In degree per century
Then θ' = {√ [G (m + M)/ a³]} x (180/π) (36526/T)
Mercury Advance of perihelion motion solution
With m = 0.23 x 10-6 M (0) and M = 1 M (0); G = 6.673 x 10- 11; a = 58.2 x 109 m
And M (0) = 2 x 1030 kg; R (0) = 0.696 x 109 meters; T = 88 days
With θ' = {√ [G (m + M)/ a³]} x (180/π) (36526/T) (3600) arc sec/century
Then θ' = {√ [6.673 x 10- 11 x 2 x 1030]/ (58.2x109)³]} (180/π) (36526/88) (3600)
= 43.0 arc sec per century
Venus Advance of perihelion motion solution
With m = 4.868 x 10-6 M (0) and M = 1 M (0); G = 6.673 x 10- 11; a = 108.2 x 109 m
And M (0) = 2 x 1030 kg; R (0) = 0.696 x 109 meters; T = 224.7 days
With θ' = {√ [G (m + M)/ a³]} x (180/π) (36526/T) (3600) arc sec/century
Then θ' = {√ [6.673 x 10- 11 x 2 x 1030]/ (108.2x109)³]} (180/π) (36526/88) (3600)
= 9.0 arc sec per century
Page 7
Kepler's speed laws told us about this mistake
Or, r (1)θ'²(1) = r (2)θ'²(2) = location x speed = constant
= Areal velocity
Or, θ' (1) = {√[r (2)/ r (1)]} θ' (2)
And θ' (1) - θ' (2) = {{√[r (2)/ r (1)]} - 1} θ' (2)
Δ θ' = {{√[r (2)/ r (1)]} - 1} θ' (2)
This is the angular time delay and will be seen as angular visual Illusion
The angular speed is θ' = v/r
For 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", then
And 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" = {{√[r (2)/ r (1)]} - 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" = {{√[r (2)/ r (1)]} - 1} W" (2)
Δ W" = {{√ [149.6/ 58.2]} - 1} [-70.29] = 43.0" arc per century
For Planet Venus
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
The angular speed is θ' = v/r
For Venus: θ' = v/r = (35.1km/sec)/108,200,000 km = 0.000000324 radians/sec
If you want the accumulation value in arc sec /century W", then
And W" = (v/r) (180/π) (3600) (26526/T) = angular velocity in arc sec per century. If it is measured for planet Venus then
W" = (35.1/108,200,000) (180/π) (3600) (26526/88)
W"= - 10.87687234 arc second per century
Or, Δ W" = {{√[r (2)/ r (1)]} - 1} W" (2)
What is the angular visual Illusion for planet Venus that would be seen when measured from Earth with Earth location r (1) = Earth = 149.6 x 106
And r (2) = Venus = 108.2x 106
And W" (2) = 10.87687234arc sec /century
Δ W" = {{√[r (2)/ r (1)]} - 1} W" (2)
Δ W" = {{√ [149.6/108.2]} - 1} [-10.87687234]= 4.16 arc per century
Page 8
Another way is: To T²/a³ = 4 π²/G (M + m)
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)
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
W " (calculated) = [- π m/MT] (180/π degrees) (3600 seconds) (36526 century); T = days; m = 0.32 x1024 kg; M = 2.0 x1030 kg; T = 88 days
W' (calculated) = (-180 x 36526 x 3600/T) (m/M) = 43.0" seconds of arc /100 years
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 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
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 = acceleration
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)
= [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
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) + rθ'[- 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 Ĵ
Then 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)
Defineγ = 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 = - 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] Kepler's findings
Frames are related by the quotient of their velocities θ = arc tan (v m/ v n) and the advance of perihelion is the opposite of the Michelson experiment or thequantity tan θ = (v m/ v n)= [(v* +/- v°)/ (v**)] = Earth speed and spin/Mercury speed and spin
Advance of perihelion Period is given by:
Kepler's equation:
And ω² = [GM/r³] [(v* +/- v°)/ (v**)]²
Page 10
In arc sec / century
Then ω = {[GM/r³]} 1/2 {[(v* +/- v°)/ (v**)]} [(180/π) (3600) (36526/T) = 43"/century
G = gravitational constant; M = sun mass; r = Mercury - Sun distance
Where v* = orbital speed of earth; v° = spin speed of earth
And v** = orbital speed of observed Planet
Then ω = {[GM/r³]} 1/2 {[(v* +/- v°)/ (v**)]} [(180/π) (3600) (36526/T) = 43"/century
For Planet Mercury
ω = {[6.673x x10 -11x 2 x 1030/ (58.2x109)³]} 1/2 {[(29.8-0.465)/ 47.9]} = [(180/π) (3600) (36526/88)
= 43.0 “/century
For planet Venus
ω = {[6.673x 2 x 1030/ (108.2x106)³]} 1/2 {[(29.8-0.465)/ 35.1]} [(180/π) (3600) (36526/224.7)
= 9.0”/century
The conclusion is
With ω = 2 π f = 2 π/T angular frequency in event time
And ω (real time) = ω (event time) + Δ ω
Δ ω = ω (real time) - ω (event time)
= 2 π f [v* (e) +/- vº (e)] / [v* (m) +/- vº (m)]
= 2 π f Z; Z = [v* (e) +/- vº (e)] / [v* (m) +/- vº (m)] = red shift
And ω = 2 π f (1 + Z)
The Advance of Planet Mercury Perihelion is2 π f Z = 2 π Z/ T
Δ ω = (2 π/T) [v* (e) +/- vº (e)] / [v* (m) +/- vº (m)]
In arc seconds per century: Multiply by: (180/π) [36526/T (days)] (3600)
And Δ ω = [2 π / T (seconds)][v*(e) +/- vº (e)] / [v* (m) +/- vº (m)] x
(180/π) [36526/T (days)] (3600);
T (seconds) = T days x 24x 3600
Δ ω = [2 π / T days x 24x 3600][v*(e) +/- vº (e)] / [v* (m) +/- vº (m)] x
(180/π) [36526/T (days)] (3600)
Δ ω = [15 x 36526 / T² (days)] [v*(e) +/- vº (e)] / [v* (m) +/- vº (m)]
With Z = (29.8 – 0.465)/ (48.2) =29.335/48.2; T = 88 days
Δ ω = 15 (36526/ 88²) (29.335/48.2) = [547890/88²] (29.335/58.2) = 43
Δ ω (sec) = [36526 / T² (days)] [v*(e) +/- vº (e)] / [v* (m) +/- vº (m)]
Page 11
We went to the Lab and found the force F between two objects to be
F = - G mM/r²
Where G = Universal constant = 6.673 x 10-11; m = primary mass = planet
M = secondary mass = Sun
And we solved the equation using Newton’s law F = mγ
The solution came out to be an elliptic motion
When applied to planetary motion Newton’s solution in event time came out to be; r (θ, t) = [a (1-ε²)/ (1+ ε cosine θ)]
With Sun-Planet distance r and Sun is at the focus of the ellipse and θ is the angle of rotation.
When astronomers turned their telescopes to the skies they did not see an ellipse but an optical illusion of a rotating ellipse. They found a rotating ellipse with axial rotation rate of 43.0 seconds of an arc per century or Planet Mercury appears that to rotate one extra time every 3,013,953.488 without explanation. Although this rate of Mercury’s apparent axial rotation rate is tiny and insignificant to science like all of relativity theory and quantum mechanics. Einstein and all other 100,000 living physicists and 100,000 dead physicists accepted Einstein time travel solution
Page 12
With rotating angle ψ = 43 arc sec per century
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) r
Velocity = 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
F = m d² r/d t² + 2 (d m/d t) (d r/d t) + (d² m/d t) ² r
Force = Change of momentum
F = m a + 2 m ' v + m" r
F = - GmM/r²
Page 13
Or, Newton's Kepler's equation: F = - GmM/r²
With 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/dt² - r θ'²] = -GM/r² Newton's Gravitational Equation (1)
And d (r²θ')/d t = 0 Kepler's force law (2)
Newton’s space solution r (θ, 0) = [a (1-ε²)/ (1+ ε cosine θ)]
Proof:
(2): d (r²θ')/d t = 0 <=> r²θ' = h = constant
Let r =1/u
With d r/ d t = -u'/u²
= -(1/u²)(θ')d u/d θ
= (- θ'/u²)d u/d θ
And d r/ d t = - h d u/d θ
And d²(m r)/dt² = - hθ'd²u/dθ²
= - h u²[d²u/dθ²]
With [d² r/dt² - r θ'²] = -GM/r²
Then– h u² [d²u/dθ²] - (1/u) (hu²)² = -G M u²
And [d²u/ dθ²] + u = GM/h²
The solution u = G M/h² + A cosine θ
Then r (θ, 0) = 1/u = 1/ [G M /h² + A cosine θ]
= [h²/GM]/ {1 + [Ah²/ G M] [cosine θ]}
= [h²/G M]/ (1 + ε cosine θ)
And r (θ, 0) = [a (1-ε²)/ (1+ε cosine θ)]
Gives r (θ, 0) = [a (1-ε²)/ (1+ε cosine θ)] Newton's classical solution
And it is the equation of an ellipse{a, b = √ [1 - a²], c = ε a}
Einstein came and added time travel equation is and a new force k/r 4
With d² (m r)/dt² - (m r) θ'² = -GmM/r² + k/r 4 Einstein's Space-time (1)
And d (m²r²θ')/d t = 0 Central force law (2)
Einstein’s solution r (θ, 0) = a (1 – ε²)/ [1 + ε cosine (θ – φ)]
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And φ = [6πGM/ac² (1 – ε²)] radians/second x (180/π) (36526/T) (3600) = 43.03''
Is our measurement of the gravitational force F = -GmM/r² is wrong?
Or, Einstein Gravitational force F = -GmM/r² + k/r 4 is wrong?
Real time Universal Mechanics solution
With d²(mr)/dt² - (mr)θ'² = -GmM/r² Newton's Gravitational Equation (1)
And d(m²r²θ')/dt = 0 Central force law (2)
(2): d (m²r²θ')/d t = 0 <=> m²r²θ' = H (0, 0) = constant
= m² (0, 0) h (0, 0)
= m² (0, 0) r² (0, 0) θ'(0, 0); h (0, 0)
= [r² (θ, 0)] [θ'(θ, 0)]
= [m² (θ, 0)] [r² (θ, 0)] [θ'(θ, 0)]
= [m² (θ, 0)] h (θ, 0); h (θ, 0) = [r² (θ, 0)] [θ'(θ, 0)]
= [m² (θ, t)] [r² (θ, t)] [θ'(θ, t)]
= [m² (θ, 0) m² (0, t)] [r² (θ, 0) r² (0, t)] [θ'(θ, t)]
Now d (m²r²θ')/d t = 0
Or 2mm'r²θ' + 2m²rr'θ' + m²r²θ" = 0
Dividing by m²r²θ' to get 2(m'/m) + 2(r'/r) + (θ"/θ') = 0
This differential equation has a solution:
With 2(m'/m) = 2[λ (m) + ì ω (m)]
And λ (m) + ì ω (m) = constant complex number
And λ (m) and ω (m) are real numbers
Then (m'/m) = λ (m) + ì ω (m)
And dm/m = [λ (m) + ì ω (m)] d t
Integrating both sides
Then m = m [θ (t = 0), 0) m (0, t)
= m (θ, 0) e[λ (m) + ì ω (m)] t
And m (0, t) = e[λ (m) + ỉ ω (m)] t
And m = m (0) e[λ (m) + ỉ ω (m)] t
This Equation (3) is Kepler's time dependent mass equation
B- 2(r'/r) = 2[λ (r) + ì ω (r)]; λ (r) + ì ω (r) = constant complex number; λ (r) and ω (r)
Are real numbers
Now r (θ, t) = r (θ, 0) r (0, t) = r (θ, 0) e[λ (r) + ỉ ω (r)] t----- I
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And r (0, t) =e[λ (r) + ỉ ω (r)] t
And this Equation (4) is Kepler's time dependent location equation
C- Then θ'(θ, t) = {H(0, 0)/[m²(θ,0)r(θ,0)]}e{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}}
And θ'(θ, t) = θ' (θ, 0) e{-2{[λ (m) + λ (r)] t + ì [ω (m) + ω (r)] t}}--II
This is angular velocity time dependent equation
And θ'(θ, t) = θ' (θ, 0)θ' (0, t)
Then θ'(0, t) = θ'(0, 0) e{-2{[λ (m) + λ (r)] t + ì [ω (m) + ω (r)] t}}}
This is Angular velocity time dependent equation
Now
(1): d² (m r)/dt² - (m r) θ'² = -GmM/r² = -Gm³M/m²r²
d² (m r)/dt² - (m r) θ'² = -Gm³ (θ, 0) m³ (0, t) M/ (m²r²)
Let m r =1/u
With d (m r/ d t) = -u'/u²
= -(1/u²)(θ')d u/d θ
= (- θ'/u²)d u/d θ
And d (m r/ d t) = -H d u/d θ
And d²(m r)/dt² = -Hθ'd²u/dθ²
= - Hu²[d²u/dθ²]
With -Hu² [d²u/dθ²] - (1/u) (Hu²)² = -Gm³ (θ, 0) m³ (0, t) M u²
And [d²u/ dθ²] + u = Gm³ (θ, 0) m³ (0, t) M/H²
At t = 0; m³ (0, 0) = 1
[d²u/ dθ²] + u = Gm³ (θ, 0) M/H²
[d²u/ dθ²] + u = Gm (θ, 0) M/h² (θ, 0)
The solution u = Gm (θ, 0) M/h² (θ, 0) + A cosine θ
Then m (θ, 0) r (θ, 0) = 1/u = 1/ [Gm (θ, 0) M (θ, 0)/h² (θ, 0) + A cosine θ]
= [h²/Gm (θ, 0) M (θ, 0)]/ {1 + [Ah²/ Gm (θ, 0) M (θ, 0)] [cosine θ]}
= [h² (θ, 0)/Gm (θ, 0) M (θ, 0)]/ (1 + ε cosine θ)
And m (θ, 0) r (θ, 0) = [a (1-ε²)/ (1+ε cosine θ)] m (θ, 0)
Gives r (θ, 0) = [a (1-ε²)/ (1+ε cosine θ)] this is the classical Newton's equation (5)
And it is the equation of an ellipse{a, b = √ [1 - a²], c = ε a}
We Have m r = m (θ, t) r (θ, t)
= m (θ, 0) m (0, t) r (θ, 0) r (0, t)
And r (θ, t) = r (θ, 0) r (0, t)
With r (0, t) = e[λ(r) + ỉ ω (r)] t
And r (θ, 0) = [a (1-ε²)/ (1+ε cosine θ)]
Then r = (θ, t) = [a (1-ε²)/ (1+εcosθ)] e[λ(r) + ỉ ω (r)] t
This is the new real time solution of Newton's equation
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Classical Newton's Equation solution:
Is: r = r (θ) = r (θ, 0) = a (1-ε²)/ (1+εcosθ)
Real time solution r = (θ, t) = [a (1-ε²)/ (1+εcosθ)] e[λ(r) + ỉ ω (r)] t
We have r (θ, t) = r (θ, 0) r (0, t)
With r (0, t) = e[λ(r) + ỉ ω (r)] t
And r (θ, 0) = [a (1-ε²)/ (1+ε cosine θ)]
If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit
By fixed mass we mean no matter (constant mass) added or subtracted
By fixed orbit we mean that:
These quantities are constant {a, b = √ [1 - a²], c = ε a}
Then r (θ, t) = r (θ, 0) r (0, t) = [a (1-ε²)/ (1+ε cosine θ)] e ỉ ω (r) t
And m = m (θ, 0) eỉ ω (m)] t
We Have θ'(0, 0) = h (0, 0)/r² (0, 0) = 2πab/ Ta² (1-ε) ²
= 2πa² [√ (1-ε²)]/T a² (1-ε) ²
= 2π [√ (1-ε²)]/T (1-ε) ²
We get θ'(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²
Then θ'(0, t) = {2π [√ (1-ε²)]/ T (1-ε) ²} e{-2[ω (m) + ω (r)] t
= {2π [√ (1-ε²)]/ (1-ε) ²} {cosine 2[ω (m) + ω (r)] t - ỉ sin 2[ω (m) + ω (r)] t}
And θ'(0, t) = θ'(0, 0) {1 - 2sine² [ω (m) t + ω (r) t]}
- 2ỉ θ'(0, 0) sin [ω (m) + ω(r)] t cosine [ω (m) + ω(r)] t
Δ θ' (0, t) = Real Δ θ' (0, t) + Imaginary Δ θ (0, t)
Real Δ θ' (0, t) = θ'(0, 0) {1 - 2 sine² [ω (m) t ω(r) t]}
Let W (ob) = Δ θ' (0, t) (observed) = Real Δ θ (0, t) - θ'(0, 0)
= -2θ'(0, 0) sine² [ω (m) t + ω(r) t]
= -2[2π [√ (1-ε²)]/T (1-ε) ²] sine² [ω (m) t + ω(r) t]
If this apsidal motion is to be found as visual effects, then
With, v ° = spin velocity; v* = orbital velocity; v°/c = tan ω (m) T°; v*/c = tan ω (r) T*
Where T° = spin period; T* = orbital period
And ω (m) T° = Inverse tan v°/c; ω (r) T*= Inverse tan v*/c
W (ob) = -4 π [√ (1-ε²)]/T (1-ε) ²] sine² [Inverse tan v°/c + Inverse tan v*/c] radians
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Multiplication by 180/π
W° (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} sine² {Inverse tan [v°/c + v*/c]/ [1 - v° v*/c²]}
Degrees and multiplication by 1 century = 36526 days and using T in days
Where Inverse tan [v°/c + v*/c]/ [1 - v° v*/c²] = Inverse tan v°/c + Inverse tan v*/c
W° (ob) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x
sine² {Inverse tan [v°/c + v*/c]/ [1 - v° v*/c²]} degrees/100 years
Approximations I
With v° < c and v* < c, then v° v* < c² and [1 - v° v*/c²] ≈ 1
Then W° (ob) ≈ (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x sine² Inverse tan [v°/c + v*/c] degrees/100 years
Approximations II
With v° < c and v* < c,
Then sine Inverse tan [v°/c + v*/c] ≈ (v° + v*)/c
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W° (calculated) = (-720x36526/Tdays) {[√ (1-ε²)]/ (1-ε) ²} x [(v° + v*)/c] ² degrees/100 years
This 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)
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/r²
Planet --- r (cm) ----- Center of mass ------r (CM) ------Mother Sun
Planet ------r ------Mother Sun
Center of mass law m r (cm) = M r (CM); m = planet mass; M = sun mass
And r (cm) = distance of planet to the center of mass
And r (CM) = distance of sun to center of mass
And 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:
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And v* = v (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 48.14 km for planet Mercury
Velocity of secondary or Mother Sun velocity
And v* (M) = √ [Gm² / (m + M) a (1-ε²/4)]
Applications: mercury ellipse and its axis rotation of 43 " /century
1- Planet Mercury axial "apparent" rotation rate Einstein and Harvard MIT Cal-Tech and all of Modern physicists and NASA call time travel
W (cal (-720x36526x3600/T) {[√ (1-ε²]/ (1-ε) ²} (v* + v°/c) ² seconds of arc per century
In planetary motion planets do no emit light and their spin rotations are very small
The 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=2x10^30kg; m=.32x10^24kg
ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec; v° = 0.002km/sec
Calculations yield: v* =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552
W (calculated) = (-720x36526x3600/88) x (1.552) (48.14/299792)²
=43.0”/century
2- Venus Advance of perihelion solution:
W" (ob) = (-720x36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² seconds/100 years
Data: T=244.7days v°= v° (p)] = 6.52km/sec; ε = 0.0.0068; v*(p) = 35.12
Calculations
With 1-ε = 0.0068; (1-ε²/4) = 0.99993; [√ (1-ε²)] / (1-ε) ² = 1.00761
G=6.673x10^-11; M (0) = 1.98892x19^30kg; R= 108.2x10^9m
V* (p) = √ [GM²/ (m + M) a (1-ε²/4)] = 41.64 km/sec
Advance of perihelion of Venus motion is given by this formula:
W" (ob) = (-720x36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²]} [(v° + v*)/c] ² seconds/100 years
W" (ob) = (-720x36526x3600/T) {[√ (1-ε²)]/ (1-ε) ²} sine² [Inverse tan 41.64/300,000] = -720x36526/10.55) (1.00762) (41.64/300,000)²
W" (observed) = 8.2"/100 years; observed 8.4"/100years
This is an excellent result within the scientific errors
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Double throwing one stone
Inverse 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 = (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) r
And d² r/ dθ² - r [1 - (k/mh²)] = 0
And r (θ, 0) = r (0, 0) eỉ {√ [1 - (k/mh²)]} θ
From (2) d (r²θ')/d t = 0; r²θ' = h
Then 2rr'θ' + r²θ'' = 0
Dividing by r²θ'
We get 2 (r'/r) + (θ''/θ') = 0
And 2 (r'/r) = - θ''/θ' = 2ỉ ω t
And r = r (0, 0) eỉ {√ [1 - (k/mh²)]}θeỉ ω t
And θ' = θ' (θ, 0) e- 2ỉ ω t
Or r = r (0, 0) eỉ {√ [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/h²)]} θ + ω t]
- ỉ sine 2 [{√ [1 - (k/h²)]} θ + ω t]}
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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; h = 2π a b/T
Then δ θ' = - 2 θ' (0, 0)] sine² [{√ [1 - (GMαT²/4π²a²b²)]} θ + ω t]
Taking Kepler's: T²/4π²a³ = 1/GM
And GM = 4π²a³/ T²; GM α T² = 4π²a³ α
And (GMαT²/4π²a²b²) = a α/b²
Then δ θ' = - 2 θ' (0, 0) sine² [{√ (1 - a α/b²)} θ + ω t]
If α = mb²/aM
Then δ θ' = - 2 θ' (0, 0)] sine² [{√ [1 - (m/ M)]} θ + ω t]
If θ = 0
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 t
With ω T = arc tan v/c < 1
Then δ θ' = - 4π/T√ (1 - ε²)/ (1 - ε) ² sine² arc tan (v/c) radians per T
Or 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 years
Or δ θ' = - 720 [36526/T] (3600) (1.552) (48.2/c) ² = 43.11 " arc second/100 years
Or If ω = 0
Then δ θ' = - 2 θ' (0, 0)] sine² {√ [1 - (m/ M)]} θ
And (m/ M) < 1
Then δ θ' = - 2 θ' (0, 0) sine² θ
If θ = arc tan (v/c)
Then δ θ' = - 2 θ' (0, 0) sine² arc tan (v/c)
With θ' (0, 0) = 2π√ (1 - ε²)/T (1 - ε) ²
Then δ θ' = - 720 [36526/T] (3600) √ (1 - ε²)/ (1 - ε) ²] (v/c) ² arc second/100 years
Or r= r (0, 0) Exp ỉ [{√ [1 - (k/mh²)]} θ + ω t]
Or r = r (0, 0) Exp ỉ [{√ [1 - (m/M)]} θ + ω t]
If m/ M < 1
Then r≈ r (0, 0) e ỉ [{[1 - (m/ 2M)]} θ + ω t]
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And r≈ r (0, 0) eỉ [(θ - φ) + ω t]
With φ = m/ 2M θ
Taking θ = 2π
Then φ = π m/ M radians
And φ = π m/ (M) [180/π] [36526/T] [3600] arc second/100 years
Or φ = (m/ M) (180) [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
Or and extra Newton's φ = π m/(m+ M)[180/π] [36526/T] [3600] arc second/100 years
Either one of the three formulas works and gives 43 seconds of an arc for planet Mercury.
Why Einstein’s formula works on Mercury and not anywhere else?
Einstein rigged everything to come up with the 43.0 seconds of arc per century and what made this formula work is [√ (1-ε²]/ (1-ε) ²} = 1.552
And vº = 0
Einstein’s φ = [6πGM/ac² (1 – ε²)] radians/second x (180/π) (36526/T) (3600) = 43.03''
GM/a = v²
And φ = [6πv²/c² x (180/π) (36526/T) (3600) = 43.03''
= 1.5 x (-720x36526x3600/T) (v/c) ² seconds of arc
W (cal) = (-720x36526x3600/T) {[√ (1-ε²]/ (1-ε) ²} (v/c) ²
{[√ (1-ε²]/ (1-ε) ²} = 1.552
Nuclear Gravity F = (-GmM/r²) ek/r
With d²(mr)/dt² – (mr)θ'² = [-GmM/r²] ek/rNuclear gravity Equation (1)
And (m²r²θ')/dt = 0 Kepler's Areal Velocity (2)
(2): d (m²r²θ')/d t = 0
Then m²r²θ' = constant; if m is taken as constant then r²θ' = h
And (1): d² r/dt² - r θ'² = [-GmM/r²]ek/r
Let m r =1/u
Then d r/d t = -u'/u²
= - (1/u²) (θ') d u/d θ
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= (- θ'/u²) d u/d θ