I am a real person in real space in real time in real fight with modern and Nobel

Transcription

1 I am a real person in real space in real time in real fight with modern and Nobel Modern and Nobel physicists and astronomers measure in present time (real time = measured time) and calculate in past time (event time = actual time) and puzzled by the difference between present time and past time (shift time = measured time - actual time). Shift time is the time difference between when an event happens and when an event is measured. Shift time in labs is a measurement error or visual effects that make up make - believe modern and Nobel physics and astronomy. I am not saying that modern physics and astronomy is all wrong and can be deleted but modern physics and astronomy were deleted and replaced with real time physics and astronomy on July I am not only the greatest physicist of all time but the only physicist since the beginning of time because I know/knew what time it is and all others do/did not. Humans since the beginning of time measure/measured in wrong time. Page 1

2 Introduction In July 21, 1969 at age 11 and after graduating from 5th grade and on the same day a man landed a foot on the moon I watched Apollo 11 take off and disappear into the skies on its way to the moon and then saw Neil Armstrong land human's first step on the moon. I wondered how someone sees and measures distances in space and how someone sees and measures sizes of objects in space. Apollo 11 rocket looked like it is shrinking in size while moving up into the skies; not as if the rocket shrunk in size but as if the rocket visual changed in size indication a different location and not a different size. Apollo 11 looked similar to a moving car moving away and shrinking in size. I realized that objects location and size has to do with how we see things (eye as an instrument). I imagined two snap shot of Apollo 11 at different distances A and B. At snap shot distance A Apollo 11 looked like it has a size C, and at snaps shot distance B Apollo 11 looked like it has a shrunk size D as shown below: C Eye A B I asked myself the question: how A, B, C, and D are related 1 = 1 is self evident; 2 = 2 is self evident A = A is self evident If A = A; add and subtract B; then A = B + (A - B) Divide by B; then (A/B) = 1 + (A - B)/B Multiply by C; then (A/B) C = C + [(A - B)/B] C Equation - 1 Also D = D; add and subtract C; then D = C + (D - C) Equation - 2 Comparing Equation - 1 and Equation - 2 (A/B) C = D; or, AC = BD C = C D - C = [(A - B)/B] The answer is AC = BD = is how distances A and B related to sizes C and D And D - C = [(A - B)/B] C = visual contraction of C when moved from A to B AC = BD = actual distance x actual size = visual distance x visual size Page 2 D

3 The initial condition solution to AC = BD is A = B and C = D The general solution to AC = BD = constant = actual distance x actual size AC = BD = constant = actual distance x actual size (Cosine ω t + í sine ω t) = e i ω t ; (Cosine ω t - í sine ω t) = e - i ω t (Cosine ω t + í sine ω t) (Cosine ω t - í sine ω t) = e i ω t e - i ω t = 1 AC = BD = k Taking AC = k Differentiating with respect to time Then d A/ d t + d C/ d t = d k/ d t = 0 And d A/ d t = - d C/ d t = λ + í ω; method of separation of variables A = e (λ + i ω t) = e λ e i ω t = C e i ω t C = e - (λ + i ω t) = e - λ e - i ω t = D e - i ω t A = B e i ω t C = D e - i ω t AC = B e i ω t D e - i ω t = B D e i ω t e - i ω t = BD x 1 = BD A real number C has a visual complex number D = C e i ω t D along the line of sight = D x = C cosine ω t = C - 2 C sine 2 (ω t/2) D x - C = - 2 C sine 2 (ω t/2) D x = C (1 - sine 2 ω t); with ω t = tan -1 (v/c) = aberration angle D x = C [1 - sine 2 tan -1 (v/c)]; (v/c) << 1 D x = C [1 - (v/c) 2 ] D x = C cosine ω t D x / C = cosine ω t And ω t= cosine -1 (D x / C) = 1-2 sine 2 cosine -1 (D x / C) D x / C = cosine ω t = 1-2 sine 2 cosine -1 (D x / 2 C) [(D x / C) -1] = - 2 sine 2 cosine (D x / 2 C) D - C = [(A - B)/B] C D - C = - 2 C sine 2 [cosine -1 (A / 2 B)] Page 3

4 In practice: physicists and astronomers measure orbits of planets around the Sun not from the Sun (distance A) but from Earth (distance B) A = B e i ω t C = D e - i ω t D - C = [(A - B)/B] C D - C = - 2 C sine 2 [cosine -1 (A / 2 B)] D - C = Einstein's numbers without Einstein = Illusions Astronomers measure planet Mercury orbit around the Sun (distance A) not from the Sun but from Earth (distance B) and that means the orbit has visually shrunk and not actually shrunk by the quantity [(A - B)/B] C A = Sun - Mercury distance = 5.82 x 10 9 meters; B = Sun Earth distance = x 10 9 meters Sun - Mercury Period in seconds = 88 days x 24 hours x 60 minutes x 60 seconds Planet Mercury angular velocity around the Sun Is θ 0 '= 2 x π/88 x 24 x 60 x 60 radians per period Planet Mercury angular velocity around the Sun in arc second per century δθ 0 ' = (2 x π /88 x 24 x 60 x 60) (180/ π) (36526/88) (3600) = arc sec per century. If C = δθ 0 ' = arc sec per century measured from the Sun, then how much it is diminished if measured from Earth? A = 5.82 x 10 9 ; B = x 10 9 ; C = And the answer is [(A - B)/B] C = [(5.82 x x 10 9 )/ x 10 9 ] = 43 arc sec per/100 years same numbers as Einstein's numbers Defining distance r = r x + í r y = r 0 e í ω t And r = r 0 [cosine ω t + sine í ω t] and r x = r 0 [cosine ω t And r x - r 0 = r 0 [cosine ω t - 1] = - 2 r 0 sine 2 ω t/2; ω t = cosine -1 (r x/r 0 ) And [(r x - r 0 )/r 0 ] = - 2 sine 2 {[cosine -1 (r x/r 0 )]/2} And [(r x - r 0 )/r 0 ] δθ 0 '= - 2 sine 2 {[cosine -1 (58.2/149.6)]/2} = A real object (classical) of size C has a visual (quantum) of size D = C e í ω t 2 - D - C = visual illusions = (Einstein's relativity theory) The fight is about this Everything modern and Nobel said in physics and astronomy is all wrong! Modern and Nobel deletion is in progress and this book is 1st replacement Page 4

5 Chapter One: light signal processing Light signal wave length shift Naming: r = Visual distance And r 0 = actual distance Then, r 0 = r 0 ; add and subtract visual distance r And, r 0 = r + (r 0 - r); divide by r Then (r 0 / r) = 1 + [(r 0 - r)/ r] multiply by δθ 0 ' And (r 0 / r) δθ 0 ' = δθ 0 ' + [(r 0 - r)/ r] δθ 0 ' Modern and Nobel error # 1 is: [(r 0 - r)/ r] δθ 0 ' = [(r 0 - r)/ r] = [(58,200,000, ,600,000)/ 149,600,000] = 43 arc sec/ 100y Light signal period shift: T 0 = r 0 /c = 58,200,000,000/300,000 (light velocity km/sec) = 194 seconds T = r/c = 149,600,000/300,000 = seconds Naming: T = visual time And T 0 = actual time Then, T 0 = T 0 ; add and subtract real time T T 0 = T + (T 0 - T); divide by T (T 0 / T) = 1 + [(T 0 - T)/ T]; multiply by δθ' 0 (T 0 / T) θ' 0 = θ' 0 + [(T 0 - T)/ T] θ' 0 (T 0 / T) δθ' 0 = δθ' 0 + [(T 0 - T)/ T] δθ' 0 Modern and Nobel error # 2 is: [(T 0 - T)/ T] δθ' 0 = [(T 0 - T)/ T] = [( )/ ] = 43 arc second per century Light signal argument shift; tan θ 0 = v 0 /c; tan θ = v/c Naming: v = visual velocity And v 0 = actual velocity Then, v 0 = v 0 ; add and subtract v And, v 0 = v + (v 0 - v); divide by v Then (v 0 / v) = 1 + [(v 0 - v)/ v] multiply by δθ' 0 And (v 0 / v) δθ' 0 = δθ' 0 + [(v 0 - v)/ v] δθ' 0 Modern and Nobel error # 3 is: Is: [(v 0 /c) - (v/c)]/ (v/c) = (tan θ 0 - tan θ)/ tan θ = [(v 0 - v)/ v] δθ' 0 = [(v 0 - v)/ v] = [( )/ 29.8] = 43 arc sec/100y Light signal surface acceleration frequency shift; f 0 = γ 0/c; f = γ /c Naming: γ = g = Earth gravitational acceleration = 9.8 And γ 0 = g 0 = Mercury gravitational acceleration = 3.8 Then: g 0 = g 0 And g 0 = g + (g 0 - g) Page 5

6 And (g 0 / g) = 1 + [(g 0 - g)/ g] multiply by δθ' 0 And (g 0 / g) δθ' 0 = δθ' 0 + [(g 0 - g)/ g] δθ' 0 And [(g 0 /c)/ (g/c) δθ' 0 = δθ' 0 + {(g 0 /c) - (g/c)]/ (g/c)} δθ' 0 And [(f 0 - f)]/ f] δθ' 0 Modern and Nobel error # 4 is: [(f 0 - f)]/ f] δθ' 0 [(g 0 - g)/ g] δθ' 0 = [( )/ 9.8] = 43 arc sec/100y 1.5 Light signal momentum P shift With λ p = h; differentiating d λ/λ = - d p/p = [(r 0 - r)/ r] Alfred Nobel Prize winner's and Einstein's error # 5 is: [(f 0 - f)]/ f] δθ' 0 = (70.75) x 0.61 = 43 arc sec/100y 1.6 Light signal Energy E shift E 0 = h f 0 ; E = h f [(E 0 - E)/E] δθ' 0 = [(f 0 - f)/f] δθ' 0 Modern and Nobel error # 6 is: [(E 0 - E)/E] δθ' 0 = [(f 0 - f)/f] δθ' 0 = (0.61) = 43 arc sec/100y Planet Distance/km Signal period Orbital period Spin velocity Mass Eccentricity Mercury 58,200, sec 88days.003km/sec 0.33 x Earth 149,600, sec km/sec 5.97X Sun radius 0.696x10 6 km Sun mass 2X Page 6

7 Chapter two: Orbit processing Distance shift r = Visual distance; r 0 = actual distance Then, r 0 = r 0 ; add and subtract visual distance r And, r 0 = r + (r 0 - r); divide by r Then (r 0 / r) = 1 + [(r 0 - r)/ r] multiply by δθ 0 ' And (r 0 / r) δθ 0 ' = δθ 0 ' + [(r 0 - r)/ r] δθ 0 ' Modern and Nobel error # 7 is: [(r 0 - r)/ r] δθ 0 ' = [(r 0 - r)/ r] = [(58,200,000, ,600,000)/ 149,600,000] = 43 arc sec/ 100y Velocity shift Light signal argument shift; tan θ 0 = v 0 /c; tan θ = v/c Naming: v = visual velocity; v 0 = actual velocity Then, v 0 = v 0 ; add and subtract v And, v 0 = v + (v 0 - v); divide by v Then (v 0 / v) = 1 + [(v 0 - v)/ v] multiply by δθ' 0 And (v 0 / v) δθ' 0 = δθ' 0 + [(v 0 - v)/ v] δθ' 0 Modern and Nobel error # 8 is: [(v 0 - v)/ v] δθ' 0 = [(v 0 - v)/ v] = [( )/ 29.8] = 43 arc sec/100y Acceleration shift Naming: γ = g = Earth gravitational acceleration = 9.8 And γ 0 = g 0 = Mercury gravitational acceleration = 3.8 Then: g 0 = g 0 And g 0 = g + (g 0 - g) And (g 0 / g) = 1 + [(g 0 - g)/ g] multiply by δθ' 0 And (g 0 / g) δθ' 0 = δθ' 0 + [(g 0 - g)/ g] δθ' 0 Modern and Nobel error # 9 is: [(g 0 - g)/ g] δθ' 0 = [( )/ 9.8] = 43 arc sec/100y Linear distance shift Naming: d = Real time linear distance And d 0 = event time linear distance Then, d 0 = d 0; add and subtract real time distance d And, d 0= d + (d 0 - d); divide by d Then (d 0/ d) = 1 + [(d 0 - d)/ d]; multiply by δθ' 0 And (d 0 / d) δθ' 0 = δθ' 0 + [(d 0 - d)/ d] δθ' 0 Modern and Nobel error # 10 is: [(d 0 - d)/ d] δθ' 0 = [(d 0 - d)/ d] With d 0 = v 0 τ 0= 47.9 x 88, and d = v τ = 29.8 x = [(v 0 τ 0 - v τ /)/ v τ] = [(47.9 x x )/ 29.8 x ] = 43 arc sec/100y Page 7

8 Chapter 3: light aberrations processing Light signal period Aberration: Modern and Nobel error # 11is: = {[Tan -1 (T 0 / τ 0)/tan -1 (T/ τ)] -1} δθ' 0 = {[Tan -1 (194/ 88 x 24 x 3600)/tan -1 (498.67/ 88 x 24 x 3600)] -1} = x 0.61 = 43 arc sec/100 y Light signal distance Aberration: Modern and Nobel error # 12 is: = {[tan -1 (r 0 / c τ 0)/tan -1 (r/ c τ)] -1} δθ' 0 = {[Tan -1 (58,200,000/ 300,000x88 x 24 x 3600)/tan -1 (149,600,000/ 88 x 24 x 3600)] -1} = = x 0.61 = 43 arc sec/100 y Light signal velocity Aberration: Modern and Nobel error # 13 is: = {[tan -1 (v 0 / c)/tan -1 (v/ c)] -1} δθ' 0 = {[Tan -1 (48.1/ 300,000)/tan -1 (29.8/300,000)] -1} = x 0.61 = 43 arc sec/100 y Light signal frequency Aberration: Modern and Nobel error # 14 is: = [(tan -1 f 0 /tan -1 f) -1] δθ' 0 = {[tan -1 (γ 0 / c)/tan -1 (γ 0 / c)] -1} δθ' 0 = {[Tan -1 (3.8/3x10 8 )/tan -1 (9.8/3x10 8 )] -1} = x 0.61 = 43 arc sec/100 y 3.5 Image processing Orbit and planet sizes are governed by AC = BD = constant (A/B) -1 = (D/C) - 1 Radius of Earth is 6731 km and radius of Mercury is 2440 km The fact we measure from Earth then velocity of earth is 29.8 km/sec The fact that we measure Earth orbit and confused with Mercury orbit the velocity is = km/sec The mistake astronomers are doing is taking Mercury's orbit as Earth orbit but with less orbital speed of when they use Sun as reference point and measure from Earth. They are looking at Earth orbit and not mercury's orbit Modern and Nobel error # 15 is: [(D /C)(29.8/29.335) - 1] δθ' 0 [(D x 29.8/C x ) - 1] δθ' 0 = [(2440 x 29.8/6371x ) - 1] δθ' 0 = 0.61 x = 43 arc sec/ century 8

9 Chapter four: Mistakes of Kepler, Newton Kepler's Period historical mistake Kepler s law: a³/t² = k = constant Or, a 1 ³/ T 1 ² = a 2 ³/ T 2 ² = k Or, a 1 / a 2 = (T 1 / T 2 ) 2/3 And (a 1 - a 2 )/ a 2 = (τ 0 / τ) 2/3 1 Or (a 0 a)/ a = (τ 0/ τ) 2/3 1 And [(a m a e )/ a e ] δ θ 0 = [(a 0 a)/ a] (v 0 /r 0 ) δθ' 0 = [(a 0 a)/ a] δθ' 0 Modern and Nobel error # 16 is: [(τ 0/ τ) 2/3 1] δθ' 0 = [(88/365.26) 2/3-1] = 43 arc sec/100y Newton's distance historical mistake Newton force is F = k/r 2 = G m M/r 2 = m v 2 /r And velocity of a celestial object measured from the sun is = (GM/r) 1/2 With r 3 0 = (GM/v 2 0); r 0 = (GM/v 2 0) And r 3 = (GM/v 2 ); r = (GM/v 2 ) Modern and Nobel error # 17 [(r 0 - r)/r] δθ' 0 = {[(GM/v 2 0)/ (GM/v 2 )] - 1} δθ' 0 = [(v/v 0 ) 2-1] δθ' 0 = [(29.8/47.9) 2-1] x = 43 arc sec/100y Newton's velocity historical mistake Newton force is F = k/r 2 = G m M/r 2 = m v 2 /r And v 0 = (GM/r 0 ) ½ ; v = (GM/r) 1/2 Modern and Nobel error # 18 [(v 0 - v)/v] δθ' 0 = [(GM/r 0 ) 1/2 / (GM/r) ½ -1] δθ' 0 = [(r/r 0 ) 1/2-1] δθ' 0 = [(149.6/58.2) 1/2-1] x = 43 arc sec/100y Newton's surface gravity historical mistake Newton force is F = k/r 2 = G m M/R 2 = m g And g 0 = Gm 0 /R 2 0; g = Gm/R 2 Modern and Nobel error # 19 = [(g 0 - g)/ g] δθ' 0 = {[(Gm 0 /R 02 ) - (Gm/R 2 )]/ (Gm/R 2 )} δθ' 0 = {[(m 0 / R 02 ) - (m/r 2 )]/ (m/r 2 )} δθ' 0 = [(g 0 - g)/ g] δθ' 0 = [(3.8/9.8) -1] = 43 arc sec/100 years Note: [(m 0 / R 02 )/ (m/r 2 ) (29.8/29.335) 2-1] δθ' 0 =43 arc sec/100y Newton's signal time historical mistake Newton force is F = k/r 2 = G m M/r 2 = m v 2 /r; v 0 = (GM/r 0 ) ½ ; v = (GM/r) 1/2 Modern and Nobel error # 20 [(r/r 0 ) 1/2-1] δθ' 0 = {[(r/c)/(r 0 /c) ½ ] - 1]} δθ' 0 = [(T/T 0 ) 1/2-1] δθ' 0 = [(498/194) 1/2-1] = 43 arc sec/100y Page 9

10 Chapter five: mistakes of Le Verrier 5.1- Le Verrier velocity historical mistake The angular velocity of Mercury around the Sun is: θ 0 ' = v 0 /r 0 If it is measured for planet Mercury from the sun, then θ 0 ' = v 0 /r 0 If planet Mercury around the sun measured from earth Then θ m ' (Earth) = (v 0 + v)/r 0 And θ m ' (Earth) = v 0 /r 0 + v /r 0 ; not v 0 /r 0 Le Verrier mistake is: v /r 0 The angular speed shift: v /r 0 ; taking into account Earth rotation v º Le Verrier mistake: = [(v +/- v º ) /r 0 ] In arc second per century multiplying by (180/ π) (3600) (100 τ / τ 0) = [(v +/- v º) /r 0 ] [(180/ π) (3600) (100 τ / τ 0)] = [(v +/- v º) /v 0 ) (v 0 /r 0 )] (180/ π) (3600) (100 τ / τ 0) Modern and Nobel error # 21 is: [(v +/- v º) /v 0 )] δθ' 0 [( ) /47.9)] = 43 arc sec/100y 5.2- Le Verrier distance mistake And v= [G M 2 / (m + M) r] 1/2 = (G M / r) 1/2 ; m <<M; Solar system And v 0 = [G M 2 / (m 0 + M) r] 1/2 = (G M / r 0 ) 1/2 ; m 0 <<M; Solar system And v = (G M / r) ½ ; v 0 = (G M / r 0 ) 1/2 And (v/ v 0 ) = (G M / r) ½ / (G M / r 0 ) ½ = (r 0 / r) 1/2 With [(v - v º) /v 0 )] δθ' 0 = (r 0 / r) 1/2 [1- (v º/v)] δθ' 0 Modern and Nobel error # 22 is: (r 0 /r) 1/2 [1- (v º/v)] δθ' 0 = (58.2/149.6) 1/2 x 0.98 x = 0.61 x = 43 arc sec/100y Le Verrier signal time mistake With (r/r 0 ) 1/2 [1- (v º/v)] δθ' 0 = [(r 0 /c) /(r /c)] 1/2 [(1- (v º/v)] δθ' 0 Modern and Nobel error # 23 is: (T 0 /T) 1/2 [(1- (v º/v)] δθ' 0 = (194 /498.67) 1/2 x 0.98 x = 43 arc sec/100y Le Verrier orbital period mistake Kepler s law: a³/t² = k = constant Or, r 0 ³/ τ 0 ² = r³/ τ ² = k; r 0 / r = (τ 0 / τ) 2/3 With (r 0 /r) 1/2 [(1- (v º/v)] δθ' 0 Modern and Nobel error # 24 is: (τ 0 / τ) 1/3 [(1- (v º/v)] δθ' 0 = (88/365.26) 1/3 x 0.98 x = 43 arc sec/100y Le Verrier gravity mistake And γ 0 = g 0 = Gm 0 /r 2 0; γ = g = Gm/r 2 And (γ / γ 0 ) = (g/g 0 ) = (Gm/r 2 )/ (Gm 0 /r 2 0) = (m r 2 0/ m 0 r 2 ) And (r 0 /r) = (m 0 g/g 0 m) ½ ; (r 0 /r) 1/2 = (m 0 g/g 0 m) 1/4 With (r/r 0 ) 1/2 [1- (v º/v)] δθ' 0 Modern and Nobel error # 25 is: (m 0 g/g 0 m) 1/4 [1- (v º/v)] δθ' 0 = (0.33 x 9.8/3.8 x5.97) 1/ x = 43 arc sec/100y Page 10

11 Chapter six: Real time Vision 6.1 Distance: AC = BD = k Differentiation with respect to time of BD = k Gives D (db/d t) + B (d D/d t) = 0 And D (db/d t) = - B (d D/d t) And [(db/d t)/b] = - [(d D/d t)/d] = - í ω B = A e - í ω t D = C e í ω t Or, r = r 0 e í ω t ; r 0 = r e - í ω t And r 0 = r x + í r y = r (cosine ω t + í sine ω t); ω t = cosine -1 (r x /r) And r x = r cosine ω t = r [1-2 sine 2 (ω t/2)] And (r x - r)/r = - 2 sine 2 (ω t/2) And (r x - r)/r = - 2 sine 2 {[cosine -1 (r x / r)]/2} = [(r x - r)/r] δθ' 0 = - 2 sine 2 {[cosine -1 (r x / r)]/2} δθ' 0 Modern and Nobel error # 26 is: - 2 sine 2 {[cosine -1 (r x / r)]/2} δθ' 0 = - 2 sine 2 {[cosine -1 (58.2/ 149.6)]/2} x = Time r = r 0 e í ω t and r = c T and r 0 = c T 0 Then T = T 0 e í ω t ; T 0 = T And T = T x + í T y = T 0 (cosine ω t + í sine ω t); ω t = cosine -1 (T/ T 0 ) And T x = T 0 cosine ω t = T 0 [1-2 sine 2 (ω t/2)] And (T x - T 0 )/T 0 = - 2 sine 2 (ω t/2) And (T x - T 0 )/T 0 = - 2 sine 2 {[cosine -1 (T/ T 0 )]/2} = [(T x - T 0 )/T 0 ] [(v 0 /r 0 )] [(180/π) (3600) (100 τ/ τ 0 )] Modern and Nobel error # 27 is: - 2 sine 2 {[cosine -1 (T x / T 0 )]/2} δθ' 0 = - 2 sine 2 {[cosine -1 (194/ )]/2} x = Velocity v = 2 π r / τ ; and v 0 = 2 π r 0 / τ 0 Then r = (v τ /2 π) and r 0 = (v 0 τ 0/2 π) And (r /r 0 ) = (v τ /2 π)/ (v 0 τ 0/2 π) = (v τ)/ (v 0 τ 0) = [(r - r 0 )/r 0 ] [(v 0 /r 0 )] [(180/π) (3600) (100 τ/ τ 0 )] = - 2 sine 2 {[cosine -1 (r/ r 0 )]/2} [(v 0 /r 0 )] [(180/π) (3600) (100 τ/ τ 0 )] Modern and Nobel error # 28 is - 2 sine 2 {[cosine -1 (v τ)/ (v 0 τ 0)]/2} δθ' 0 = - 2 sine 2 {[cosine -1 (48.1 x 88/ ( ) x ]/2} x = 43 Page 11

12 6.4 Areal velocity r v = h = r 0 v 0 The r/r 0 = v 0 /v; taking Earth spin into account Then r/r 0 = (v 0 - v 0 )/v Modern and Nobel error # 29 is: - 2 sine 2 {[cosine -1 (v 0 - v 0 )/v]/2} δθ' 0 =- 2 sine 2 {[cosine -1 ( )/48.14]/2} δθ' Acceleration g = GM/r 2 ; g 0 = GM/r 0 The r/r 0 = (g 0 /g) 1/2 ; taking Earth spin into account Then r/r 0 = (g 0 /g) ½ (1 - v 0 / v) Modern and Nobel error # 30 : - 2 sine 2 {[cosine -1 (g 0 /g) ½ (1 - v 0 / v)/2} δθ' 0 =- 2 sine 2 {cosine -1 {(3.8/9.8)[1 - (0.451/29.8]}/2} = 43 arc sec/100y 6.6 Argument r/r 0 = (v 0 /c)/ (v/c) same as Le Verrier The r/r 0 = (v 0 /c)/ (v/c); taking Earth spin into account Then r/r 0 = [(v 0 - v 0 )/c]/ (v/c) Modern and Nobel error # 31: - 2 sine 2 {[cosine -1 {[(v 0 - v 0 )/c]/ (v/c)}/2} δθ' 0 =- 2 sine 2 {[cosine -1 {{[( )]/ 3x10 8 }/ (48.14/3x10 8 )]}/2} = 43 arc sec/100y 6.7 Frequency [(g 0 /c)/ (g/c)] ½ (1 - v 0 / v) The r/r 0 = v 0 /v; taking Earth spin into account Then r/r 0 = (v 0 - v 0 )/v Modern and Nobel error # 32: - 2 sine 2 {{cosine -1 [(g 0 /c)/ (g/c)] ½ (1 - v 0 / v)}/2} δθ' 0 =- 2 sine 2 {{cosine -1 {[(3.8/3x10 8 )/ (9.8/3x10 8 )] 1/2 x0.98}/2}} = 43 arc sec/100y Kepler's Period historical mistake II Kepler s law: a³/t² = k = constant Or, a 1 ³/ T 1 ² = a 2 ³/ T 2 ² = k Or, a 1 / a 2 = (T 1 / T 2 ) 2/3 And (a 1 - a 2 )/ a 2 = (τ 0 / τ) 2/3 1 Or (a 0 a)/ a = (τ 0/ τ) 2/3 1 And [(a m a e )/ a e ] δ θ 0 = [(a 0 a)/ a] (v 0 /r 0 ) δθ' 0 = [(a 0 a)/ a] δθ' 0 Modern and Nobel error # 16 is: [(τ 0/ τ) 2/3 1] δθ' 0 = [(88/365.26) 2/3-1] = 43 arc sec/100y Modern and Nobel error # 33 is: - 2 sine 2 {[cosine -1 (τ 0/ τ) 2/3 ]/2} δθ' 0 = - 2 sine 2 {[cosine -1 (88/365.26) 2/3 ]/2} = 43 arc sec/100y Page 12

13 6.9 - Newton's distance historical mistake II Newton force is F = k/r 2 = G m M/r 2 = m v 2 /r And velocity of a celestial object measured from the sun is = (GM/r) 1/2 With r 3 0 = (GM/v 2 0); r 0 = (GM/v 2 0) And r 3 = (GM/v 2 ); r = (GM/v 2 ) Modern and Nobel error # 17 = [(r 0 - r)/r] δθ' 0 = {[(GM/v 2 0)/ (GM/v 2 )] - 1} δθ' 0 = [(v/v 0 ) 2-1] δθ' 0 = [(29.8/47.9) 2-1] x = 43 arc sec/100y Modern and Nobel error # 34: = - 2 sine 2 {[cosine -1 (v/v 0 ) 2 ]/2} δθ' 0 = - 2 sine 2 {[cosine -1 (v/v 0 ) 2 ]/2} δθ' Newton's velocity historical mistake II Newton force is F = k/r 2 = G m M/r 2 = m v 2 /r And v 0 = (GM/r 0 ) ½ ; v = (GM/r) 1/2 Modern and Nobel error # 18 = [(v 0 - v)/v] δθ' 0 = [(GM/r 0 ) 1/2 / (GM/r) ½ -1] δθ' 0 = [(r/r 0 ) 1/2-1] δθ' 0 = [(149.6/58.2) 1/2-1] x = 43 arc sec/100y Modern and Nobel error # 35: = - 2 sine 2 {[cosine -1 (r/r 0 ) 1/2 ]/2} δθ' 0 = - 2 sine 2 {[cosine -1 (149.6/58.2) 1/2 ]/2} = 43 arc sec/100y Newton's surface gravity historical mistake II Newton force is F = k/r 2 = G m M/R 2 = m g And g 0 = Gm 0 /R 2 0; g = Gm/R 2 Modern and Nobel error # 19 = [(g 0 - g)/ g] δθ' 0 = {[(Gm 0 /R 02 ) - (Gm/R 2 )]/ (Gm/R 2 )} δθ' 0 = {[(m 0 / R 02 ) - (m/r 2 )]/ (m/r 2 )} δθ' 0 = [(g 0 - g)/ g] δθ' 0 = [(3.8/9.8) -1] = 43 arc sec/100 years Note: [(m 0 / R 02 )/ (m/r 2 ) (29.8/29.335) 2-1] δθ' 0 = 43 arc sec/100y Modern and Nobel error # 36: = - 2 sine 2 {{cosine -1 [(m 0 / R 02 )/ (m/r 2 )]}/2} δθ' 0 = - 2 sine 2 {{cosine -1 [(0.33/ )/ (5.97/ )]}/2} = 43 arc sec/100y Le Verrier mistake II Modern and Nobel error # 21 is: [(v +/- v º) /v 0 )] δθ' 0 = (v /v 0 ) [1- (v º/v)] δθ' 0 Modern and Nobel error # 37: - 2 sine 2 {{cosine -1 (v /v 0 ) [1- (v º/v)]}/2} δθ' 0 = - 2 sine 2 {{cosine -1 (29.8 /48.14) [1- (0.4651/29.8)]}/2} = 43 arc sec/100y Page 13

14 Le Verrier distance mistake II And v = [G M 2 / (m + M) r] 1/2 = (G M / r) 1/2 ; m <<M; Solar system And v 0 = [G M 2 / (m 0 + M) r] 1/2 = (G M / r 0 ) 1/2 ; m 0 <<M; Solar system And v = (G M / r) ½ ; v 0 = (G M / r 0 ) 1/2 And (v/ v 0 ) = (G M / r) ½ / (G M / r 0 ) ½ = (r 0 / r) 1/2 With [(v - v º) /v 0 )] δθ' 0 = (r 0 / r) 1/2 [1- (v º/v)] δθ' 0 Modern and Nobel error # 22 is: (r 0 /r) 1/2 [1- (v º/v)] δθ' 0 = (58.2/149.6) 1/2 x 0.98 x = 0.61 x = 43 arc sec/100y = 43 arc sec/100y Modern and Nobel error # 38:- 2 sine 2 {{cosine -1 (r 0 /r) 1/2 [1- (v º/v)]}/2} δθ' 0 = - 2 sine 2 {{cosine -1 (58.2/149.6) 1/2 [1- (0.4651/29.8)]}/2} = 43 arc sec/100y Le Verrier signal time mistake II With (r/r 0 ) 1/2 [1- (v º/v)] δθ' 0 = [(r 0 /c) /(r /c)] 1/2 [(1- (v º/v)] δθ' 0 Modern and Nobel error # 23 is: (T 0 /T) 1/2 [(1- (v º/v)] δθ' 0 = (194 /498.67) 1/2 x 0.98 x = 43 arc sec/100y Modern and Nobel error # 39:- 2 sine 2 {{cosine -1 (v /v 0 ) [1- (v º/v)]}/2} δθ' 0 = - 2 sine 2 {{cosine -1 (194 /498.67) 1/2 x 0.98}/2} x Le Verrier orbital period mistake Kepler s law: a³/t² = k = constant Or, r 0 ³/ τ 0 ² = r³/ τ ² = k; r 0 / r = (τ 0 / τ) 2/3 With (r 0 /r) 1/2 [(1- (v º/v)] δθ' 0 Modern and Nobel error # 24 is: (τ 0 / τ) 1/3 [(1- (v º/v)] δθ' 0 = (88/365.26) 1/3 x 0.98 x = 43 arc sec/100y Modern and Nobel error # 40: = - 2 sine 2 {{cosine -1 (τ 0 / τ) 1/3 [(1- (v º/v)}}/2} = - 2 sine 2 {{cosine -1 (88 / 149.6) 1/3 [(1- (0.4561/29.8)}}/2} = 43 arc sec/100y Le Verrier gravity mistake And γ 0 = g 0 = Gm 0 /r 2 0; γ = g = Gm/r 2 And (γ / γ 0 ) = (g/g 0 ) = (Gm/r 2 )/ (Gm 0 /r 2 0) = (m r 2 0/ m 0 r 2 ) And (r 0 /r) = (m 0 g/g 0 m) ½ ; (r 0 /r) 1/2 = (m 0 g/g 0 m) 1/4 With (r/r 0 ) 1/2 [1- (v º/v)] δθ' 0 Modern and Nobel error # 24 is: (m 0 g/g 0 m) 1/4 [1- (v º/v)] δθ' 0 = (0.33 x 9.8/3.8 x5.97) 1/ x = 43 arc sec/100y Modern and Nobel error # 41: =- 2 sine 2 {{cosine -1 (m 0 g/g 0 m) 1/4 [1- (v º/v)]/2} δθ' 0 = - 2 sine 2 {{cosine -1 (0.33 x 9.8/ /3.8 x5.97) 1/4 [(1- (0.4561/29.8)}}/2} = 43 arc sec/100y Page 14

15 Chapter 7: Fourier Light motion visual analysis r = r 0 e í ω t i (θ + ω t) Distance r in real time is r = r 0 e With r 2 θ' = h = r 2-2 i (θ + ω t) 0 θ' 0 ; θ' = θ' 0 e At θ = 0; Perihelion r = r 0 e í ω t Taking r = c t and r 0 = c t 0 Then (t/t 0 ) = e i ω t And the Fourier transform is: Γ = (1/2 π) { - e i ω t d t} And Γ = (1/2 π) 0 τ 0 e i ω t d t Γ = (1/2 π) [e i ω τ 0-1]/ i ω = (1/2 π) [cosine ω τ 0 + i sine ω τ 0-1]/ i ω Along the line of sight Γ x = (1/2 π) [sine ω τ 0 ]/ ω Γ x [100 τ / τ 0 ] = (100 τ /2 π) [sine ω τ 0 / ω τ 0 ] per 100 τ With ω τ 0 = arc tan (v 0/c); 100 τ = 1 century = days Γ x [100 τ / τ 0 ] per century = (100 τ /360) {sine [arc tan (v 0/c)] / [arc tan (v 0/c)]} Γ x [100 τ / τ 0 ] per century = (36526days /360degrees) {sine [arc tan (v 0/c)] / [arc tan (v 0/c)]} In arc seconds per century: Γ x [100 τ / τ 0 ] per century = (36526x 24 x 3600 /360 x 3600) {sine [arc tan (v 0/c)] / [arc tan (v 0/c)]} Γ x [100 τ / τ 0 ] per century = (36526x 24 x 3600/360 x 3600) {sine [arc tan (v 0/c)] / [arc tan (v 0/c)]} Where v 0 = 47.9 km/second; c = 300,000 km/second Modern and Nobel error # 42: Γ x [100 τ / τ 0] = (36526 /15) {sine [arc tan (47.9/300,000)] / [arc tan (47.9/300,000)]} = 42.5 arc seconds per century Modern and Nobel error # 43: Γ x [100 τ/ τ 0] = (36526 /15) {sine [arc tan (29.8/300,000)] / [arc tan (29.8/300,000)]} = 42.5 arc seconds per century Page 15

16 Angular velocity θ' in real time is θ' = θ' 0 e Taking θ'= 2 π f and θ' 0 = 2 π f 0-2i (θ + ω t) Then (f/f 0 ) = e At θ = 0; Perihelion And (f/f 0 ) = e - 2i ω t And the Fourier transform is: Γ = (1/2 π) { - e - 2i ω t d t} - 2i (θ + ω t) And Γ = (1/2 π) 0 τ 0 e - 2i ω t d t Γ = (1/2 π) [e - 2i ω t -1]/ -2 í ω = (1/2 π) [cosine 2 ω τ 0 + í sine 2 ω τ 0-1]/ -2 í ω Along the line of sight Γ x = - (1/2 π) [sine 2ω τ 0 ]/ 2ω Γ x [100 τ / τ 0 ] = (100 τ /2 π) [sine 2 ω τ 0 / 2 ω τ 0 ] per 100 τ With ω τ 0 = arc tan (v 0/c); 100 τ = 1 century = days Γ x [100 τ / τ 0 ] per century = (100 τ /360) {sine 2 [arc tan (v 0/c)] / [arc tan (v 0/c)]} Γ x [100 τ / τ 0 ] per century = (36526days /360degrees) {sine2 [arc tan (v 0/c)] / 2 [arc tan (v 0/c)]} In arc seconds per century: Γ x [100 τ / τ 0 ] per century = (36526x 24 x 3600 /360 x 3600) {sine 2[arc tan (v 0/c)] / 2[arc tan (v 0/c)]} Where v 0 = 47.9 km/second; c = 300,000 km/second Modern and Nobel error # 44: Γ x [100 τ /τ 0 ] = (36526 /15) {sine 2[arc tan (47.9/300,000)] / 2[arc tan (47.9/300,000)]} = 42.5 arc seconds per century Modern and Nobel error # 45: Γ x [100 τ / τ 0 ] = (36526 /15) {sine 2[arc tan (29.8/300,000)] / 2[arc tan (29.8/300,000)]} = 42.5 arc seconds per century Page 16

17 Chapter 8: Axial tilt processing In Chapter 7 Fourier analysis said that planet Mercury's Perihelion has nothing to do with planet Mercury but has something to do with Earth. Astronomers measure with an axial tilt of because they measure from Earth and use the Sun as reference point Parallaxes are shifted by an angle of Meaning measurements are shifted by a factor of: (1 - sine ) Modern and Nobel error # 46: (1 - sine ) δθ' 0 = (1 - sine ) = 43 arc sec/ 100y Page 17

18 Chapter ten: Descartes and LaGrange failures 10.1 Γ = Γ x + Γ y = Γ 0 e í ω t Along the line of sight Γ x = Γ 0 cosine ω t Then Γ x (seconds) = Γ x - t = - 2 t sine² {[arc tan (V r m - V r e )/ ( 2c)]/ 2} = 43 And 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 Modern and Nobel error # 47: Γ x (Arc second) = Γ x - Γ 0 = - 30 t sine² {[arc tan (V r m - V r e )/ ( 2c)]/ 2} = 43 arc second /century 10.2 La Grange historical mistake Angular velocity with respect to center of mass θ' c m = v c m /r Angular velocity with respect to Sun θ' s = v s /r And Astronomers observed: θ' s = v s /r Angular velocity with to the sun And v c m = [GM²/ (m + M) a]; m = 0.32 x10 24 kg; and M = 2.0 x10 30 kg 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)] / { [(m + M) a/ GM²]} - 1 = [(m + M)/M] - 1 = [1 + (m/m)] (m/2 M) 1 m/2 M And (θ' cm - θ s ) /θ s (m/2 M); and (θ' cm - θ s ) = (2 π /T s ) (m/2 M) And (θ' cm - θ s ) T s = (2 π + δ θ - 2 π) = π m/ M; δ θ = π m/ M Multiplying by [(180/π) (3600) (26526/ τ 0 )] Modern and error # 48 = π (m/ M) [(180/π) (3600) (26526/ τ 0 )] = 43 = π (0.32 x /2x ) [(180/π) (3600) (26526/ τ 0 )] = 43 Page 18

19 Chapter 9: Newton's equation was/is solved wrong for 350 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 Y - Axis θ 1 r 1 r Angle θ X - Axis First r = r r 1 = r [cosine θ î + sine θ Ĵ] Define r 1 = cosine θ î + sine θ Ĵ Define v = d r/d t = (d r/d t) r' r 1 + r (d r 1 /d t) = r' r 1 + r (d r 1 /d t) = r' r 1 + r θ'[- sine θ î + cosine θ Ĵ] = 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 Define γ = d (r' r 1 + r θ' θ 1 ) /d t = r" r 1 + r' (d r 1 /d t) + r' θ' θ 1 + r θ" θ 1 + r θ' (d θ 1 /d t) = (r" - rθ'²) r 1 + (2r'θ' + r θ") θ (1) 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 m'(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 Page 19

20 With d² (m r)/dt² - (m r) θ'² = F (r) And d (m²r²θ')/d t = 0 If light mass m = constant, then With m [d² r/dt² - r θ'²] = F (r) Eq-1 And d (r²θ')/d t = 0 Eq-2 A - Newton's gravitational classical or past time solution Newton took m = constant Then F = m γ With d² (m r)/dt² - (m r) θ'² = - G m M/r² Newton's Gravitational Equation (1) And d (m²r²θ')/d t = 0 Kepler's law (2) Then m (d² r/dt² - r θ'²) = - G m M/r² Eq-1 And d (r²θ')/d t = 0 Eq-2 From Eq-2: d (r²θ')/d t = 0 Then r²θ' = h = constant 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) 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 = - G M/r 2 E q 1 And - h² u² (d² u/ d θ ²) (1/u) (h u²) ² = - G M u 2 Then (d² u/ d θ ²) + u = G M/h 2 And u = G M/h 2 + A cosine θ The r = 1/u = 1/ (G M/h 2 + A cosine θ); divide by G M/h 2 And r = (h 2 /G M)/ [1 + (A h 2 /G M) cosine θ] With; h 2 /G M = a (1 - ε 2 ); (A h 2 /G M) = ε Or, r = a (1 - ε 2 )/ (1 + ε cosine θ); definition of an ellipse This is Newton's equation classical solution Measuring planetary orbit in real time using Newton's equation classical solution does not match Newton's equation classical solution but solving Newton's equation in real time or solving Newton's equation in present time will match measurements of planetary orbits in real time. Solving an equation in real or present time is solving it in complex numbers system. Solving Newton's equation in complex numbers produces quantum mechanics solution. The difference between real numbers classical solution and real time or complex numbers solution will produce relativistic effects as visual effects. In short: Real (Complex numbers solution) = real numbers solution + relativistic effects Page 19

21 B - Real time solution or complex numbers solution of Newton's equation is: F = m [(r"-rθ'²) r 1 + (2r'θ' + r θ") θ m'(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 With d² (m r)/dt² - (m r) θ'² = F (r) = - G m M/r 2 E q - 1 And d (m²r²θ')/d t = 0 E q - 2 From E q - 2; d (m²r²θ')/d t = 0 Then m²r²θ' = constant = H = m² h; h = r² θ' With m²r²θ' = constant Differentiate with respect to time Then 2mm'r²θ' + 2m²rr'θ' + m²r²θ" = 0 Divide by m²r²θ' Then 2 (m'/m) + 2(r'/r) + θ"/θ' = 0 This equation will have a solution 2 (m'/m) = 2(λ m + ì ω m ) And 2(r'/r) = 2(λ r + ì ω m ) And θ"/θ' = -2[λ m + λ r + ỉ [ω m + ω r )] Then (m'/m) = (λ m + ì ω m ) Or d m/m d t = (λ m + ì ω m ) And dm/m = (λ m + ì ω m ) d t Then m = m 0 ( λ m + ì ω m) t And m = m (θ, 0) m (0, t); m 0 = m (θ, 0) And m = m (θ, 0) ( λ m + ì ω m) t And m (0, t) = ( λ m + ì ω m) t Finally, m = m 0 ( λ m + ì ω m) t Similarly we can get (r'/r) = (λ r + ì ω r ) Or d r/r d t = (λ r + ì ω r ) And d r/r = (λ r + ì ω r ) d t Then r = r 0 ( λ r + ì ω r) t And r = r (θ, 0) r (0, t); r 0 = r (θ, 0) And r = r (θ, 0) ( λ r + ì ω r) t And r (0, t) = ( λ r + ì ω r) t Finally, r = r 0 ( λ r + ì ω r) t Page 20

22 And θ'(θ, t) = θ' (θ, 0)] -2 [(λ m + ỉ ω m ) + (λ r + ỉ ω r )] t And, θ'(θ, t) = θ' (θ, 0) θ' (0, t) And θ' (0, t) = -2 [(λ m + λ r ) + ỉ (ω m + ω r )] t Also θ' = H / m² r² From (1): d² (m r)/dt² - (m r) θ'² = - G m M/r² = -G M m³/m²r² = - G M m 3 u 2 Let m r =1/u Then d (m r)/d t = -u'/u² = - (1/u²) (θ') d u/d θ = (- θ'/u²) d u/d θ = - H d u/d θ And d² (m r)/dt² = -H θ'd²u/dθ² = - Hu² [d²u/dθ²] - Hu² [d²u/dθ²] - (1/u) (Hu²)² = -G M m³ u² And (d²u/ dθ²) + u = G M m³/ H² And [d²u (θ, 0)/ dθ²] + u (θ, 0) = G M (θ, 0) m³ (θ, 0)/ H² (θ, 0) Then u (θ, 0) = G M m³ (θ, 0)/ H² (θ, 0) + A cosine θ = G m 0 M 0 / h² + A cosine θ And m 0 r = 1/u (θ, 0) = 1/ [G m 0 M 0 / h² + A cosine θ] Or, r = 1/ [G M 0 / h² + A cosine θ] And r = h²/ G M 0 / [1 + (Ah²/ G M 0 ) cosine θ] Then r (θ, 0) = a (1-ε²)/ (1+ ε cosine θ) This is Newton's gravitational law classical solution of two body problem which is the equation of an ellipse of semi-major axis of length a and semi minor axis b = a (1 - ε²) and focus length c = ε a Then, r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ( λ r + ì ω r) t I This is real time solution or present solution of Newton's equation It is the math formula that matches astronomical measurements If time is frozen that is t = 0 Then r (θ, 0) = a (1-ε²)/ (1+ε cosine θ) or classical or event time solution -- II Relativistic is the difference between Real I and II And it is the visual difference motion and motion measurement Page 20

23 The difference between and event and its measurement in real time Real time solution = Event time solution + time shift solution Real of a complex orbit solution = real numbers orbit solution + shift solution We Have θ' (0, 0) = h (0, 0)/r² (0, 0) = 2πab/ τ 0 a² (1- ε) ² = 2πa² [ (1- ε²)]/ τ 0 a² (1- ε) ²] = 2π [ (1- ε²)]/ τ 0 (1- ε) ²] Then θ'(0, t) = 2 π [(1- ε²)/ τ 0 (1- ε) ²] -2 [(λ m + λ r ) + ỉ (ω m + ω r )] t Assuming that λ m + λ r = 0; or λ m = λ r = 0 Then θ'(0, t) = 2 π [(1-ε²)/ τ 0 (1-ε) ²] -2 ỉ (ω m + ω r ) t = 2 π [(1-ε²)/ τ 0 (1- ε) ²] [cosine 2 (ω m + ω r ) t - ỉ sine 2 (ω m + ω r ) t] Real θ'(0, t) = 2 π [(1- ε²)/ τ 0 (1-ε) ²] cosine 2 (ω m + ω r ) t Real θ'(0, t) = 2 π [(1-ε²)/ τ 0 (1-ε) ²] [1-2sine² (ω m + ω r ) t] Naming θ' = θ'(0, t); θ' 0 = 2 π [(1-ε²)/ τ 0 (1-ε) ²] Then θ' = 2 π [(1- ε²)/ τ 0 (1- ε) ²] [1-2 sine² (ω m + ω r ) t] And θ' = θ' 0 [1-2sine² (ω m + ω r ) t] And θ' - θ' 0 = - 2 θ' 0 sine² (ω m + ω r ) t] = -2{2 π [(1-ε²)/ τ 0 (1-ε) ²]} sine² (ω m + ω r ) t] And θ' - θ' 0 = -4 π [(1-ε²)/ τ 0 (1-ε) ²] sine² (ω m + ω r ) t] If this apsidal motion is to be found as visual effects, then With, v = spin velocity; v 0 = orbital velocity; τ 0 = orbital period And ω m τ 0 = tan -1 (v /c); ω r τ 0 = tan -1 (v 0 /c) Δ θ' = θ' - θ' 0 = - 4 π [(1-ε²)]/ τ 0 (1-ε) ²] sine² [tan -1 (v /c) + tan -1 (v 0 /c)] radians per τ 0 In degrees per period is multiplication by 180/ π Δ θ' = (-720) [(1-ε²)/ τ 0 (1-ε) ²] sine² {tan -1 [(v + v 0 )/c]/ [1 - v v 0 /c²]} The angle difference in degrees per period is: Δ θ = (Δ θ') τ 0 Δ θ = (-720) [(1-ε²)/ (1-ε) ²] sine² {tan -1 [(v + v 0 )/c]/ [1 - v v 0 /c²]} calculated in degrees per century is multiplication = 100 τ; τ = Earth orbital period = 100 x = days and dividing by using τ 0 in days Page 21

24 Δ θ (100 τ / τ 0 ) = Δ θ in degrees per century = ( τ / τ 0 ) [(1-ε²)/ (1-ε) ²] sine² {tan -1 (v + v 0 )/c]/ [1 - v v 0 /c²]} In arc second per century is multiplying by 3600 Δ θ = x 720 (100 τ / τ 0 ) [(1-ε²)/ (1-ε) ²] x Sine² {tan -1 [(v + v 0 )/c]/ [1 - v v 0 /c²]} Approximations I With v << c and v 0 << c, then v v 0 <<< c² and [1 - v v*/c²] 1 Δ θ x 720 (100 τ / τ 0 ) [ (1-ε²)]/ (1-ε) ²] Sine² tan -1 [(v + v 0 )/c] Arc second per century Approximations II With v << c and v* << c Then Sine² tan -1 [(v + v 0 )/c] (v + v 0 )/c Δ θ (Calculated in arc second per century) = (-720x36526x3600/ τ 0 days) [(1-ε²)/ (1-ε) ²] [(v + v 0 )/c] ² Approximations III The circumference of an ellipse Is: 2 π a (1 - ε²/4 + 3/16(ε²)²- --.) 2 π a (1- ε²/4); r 0 = a (1- ε²/4) From Newton's laws for a circular orbit: F = [M/m F = - Gm M/r 02 = m v 0 ²/ r 0 Then v 0 ² = GM/ r 0 For planet Mercury And v 0 = [GM/ r] = [GM/a (1-ε²/4)] G = x ; M = 2 x10 30 kg; a = 58.2 x 10 9 meters; ε = Then v 0 = [6.673 x x 2 x10 30 /58.2 x 10 9 ( ²/4)] And v 0 = km/sec [Mercury]; c = 300,000 Δ θ (Calculated in arc second per century) = (-720x36526x3600/ τ 0 days) [(1-ε²)/ (1-ε) ²] [(v + v 0 )/c] ² With ε = 0.206; [(1-ε²)/ (1-ε) ²] = 1.552; v = 3 meters per second Δ θ = (-720x36526x3600/88) (48.143/300,000) ² Δ θ = 43 arc second per century Modern Nobel error # 49: = (-720x36526x3600/ τ 0 days) [(1-ε²)/ (1-ε) ²] [(v + v 0 )/c] ² = (-720x36526x3600/88) (48.143/300,000) ² = 43 arc second per century With θ' = θ' 0 e - 2i ω t Then τ = τ 0 e 2i ω t ; Δ τ = - 2 τ 0 sine 2 arc tan (v/c) Modern and Nobel error # 50 is Δ θ (per century) = - 2 x 15 τ 0 sine 2 arc tan (v/c); τ 0 = 100 years = - 30 (100 x x 24 x 3600) sine 2 tan -1 (48.14/300,000) = 43 arc sec/100y Page 22 All rights reserved