MIT Dumbest physicist Walter Lewin

By first physicist Joe Nahhas; joenahhas1958@yahoo.com

Greetings: My name is Joe Nahhas founder of real time physics and astronomy

500 Years of western civilization physics laws and physicists claims are expression of artifacts. Western physicists measure the motion around one object (planet) around another object (sun) from a third place (Earth) and end up in relativistic measurements. Relativistic measurements are not expressions of physical reality but expressions of physical reality artifacts or measurement errors when measuring the motion of one object around another object from a third place. Artifacts equation is a 5th grade arithmetic equation (below). I can get Einstein's numbers using 5th grade arithmetic or use any physics concept written by any physicist from any period of time in human history and get the same numbers as Einstein and the majority of Nobel Prize winner's numbers because the measurement error has the same value. I am not only

saying physicist Walter Lewin is a Darwin's ape to say the least but what I can prove without a least of doubt that the entire western civilization physicists are Alfred Nobel dumb apes to be exact. I am not only the greatest physicist of all time but only physicist since the beginning of time! Here it is.

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Professor of Physics, Emeritus

Name: Walter H.G. Lewin

Title(s): Professor of Physics, Emeritus

Email: lewin@mit.edu

Phone: (617) 253-4282

Assistant: Teresa Santiago (617) 253-7078

Address:

Massachusetts Institute of Technology 77 Massachusetts Avenue, Bldg. 37-627Cambridge, MA 02139

Related Links:

Chandra X-ray Center Rossi X-Ray Timing Explorer Project (RXTE) The XMM-Newton Observatory Hubble Space Telescope (HST) Integral Swift

Walter Lewin: You are here and the proof is below with my tow middle fingers up Nobel

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If a car moves from point A to point B its visual changes from C to D.

A C

B D

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

B = B is self evident; add and subtract A

B = A + (B - A); divide by A

(B/A) = 1 + (B - A)/A; multiply by C

(B/A) C = C + [(B - A)/B] C --------------------- Nahhas 5th grade equation- 1

D = D is self evident; add and subtract C

D = C + (D - C) ----------------------------------- Nahhas 5th grade equation- 2

Compare equation - 1 with equation - 2

D - C = [(B - A)/B)] C = measurement errors

= "Macdonald" physics and Physicists = Nobel physics and physicists

If an object moves from point A to point B measured by a human from a third place its visual changes from size C to size D

Humans measure a planet orbit (1) around the Sun (2) from Earth (3)

The measurement error is: D - C = [(B - A)/B)] C = measurement error = Nobel

Or (C - D)/D = (A - B)/B =

Or Δ D/D = Δ B/B; Divide by Δ t

(1/D) (Δ D/ Δ t) = (1/B) (Δ B/ Δ t)

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Limit [(1/D) (Δ D/ Δ t)] = Limit [(1/B) (Δ B/ Δ t)] = +/- (λ + í ω)

Δ t 0 Δ t 0

Or, d B/B = (λ + í ω) d t and B = B0 e (λ + í ω) t = A e - / + (λ + í ω) t

B = A e - / + (λ + í ω) t

Distance is A; real time distance is B = A e - / + (λ + í ω) t

With λ = 0; B = A e - /+ í ω t

In general if an object moves from A to B, the visual of A is B = A e - /+ í ω t

In general real time distance r = r0 e - /+ í ω t

To calculate Einstein's error numbers

I can use this 5th grade arithmetic D - C = [(B - A)/B)] C

Or use Newton's Gravitational equation F = G m M/r2.

Relativistic measurements = error

SUN

Earth

Mercury

In Picture:

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What is Mercury's perihelion?

Kepler said Planets revolve around the Sun in an ellipse

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Newton's said the reason Planets revolve around the Sun is due to gravity expressed by Newton's gravitational equation

F = - G m M/r2; r > 0

Where m is Planet Mercury's mass; M = Sun mass; r = distance between the Sun and Planet Mercury; G = a constant number = gravitational constant

Or, r = a (1 - ε2)/ (1 + ε cosine θ); definition of an ellipse -------- I Newton's solution

French astronomer Le Verrier said the ellipse axis rotates and the rate of rotation is 43 seconds of an arc per century

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Einstein said to have a different look at space and time to explain the visual artifacts or planet Mercury's perihelion rotation and his thoughts resemble a sand clock

The Sand inside the glass is sand up and sand down

Sand up = sand down

Space (sand up) = velocity x time (sand down)

If velocity is constant c; then Space = c x time

Or space r = velocity c x time t

Or space (r) = constant (light velocity c) x time (t)

Or if you can travel in space, then you can travel in time but at a different rate and Nature is space - time and not just space analyzed by time

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Or nature is space and time and the only difference is that space and time runs by a different rate and if you can travel in space you could travel in time

Planet Mercury travel in space 70.75 arc seconds per century

Also planet Mercury travel in time 43 arc seconds per century

For that reason a time force in space form should be added to Newton's equation

Einstein's added a force and modified Newton's equation:

Newton's force space equation: F = - G m M/r2

Einstein's force space - time equation: F = - G m M/r2 + k1/ r4

The solution to this equation: r = 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 constant

M =2x1030kg = mass of the Sun

Eccentricity of Mercury’s orbit = ε = 0.206

T= 88 days = period of Planet mercury’s rotation around the Sun

And c = 299792.458 km/sec =light speed/sec

And 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 x

[1– (0.206)²]} x [180/ π] [36526/88] [3600] = 43 seconds of an arc per century.

In Short:

Einstein added a time force in space form k1/ r4

Newton's force space equation: F = - G m M/r2

Newton's Ellipse r = a (1 - ε2)/ (1 + ε cosine θ)

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Einstein's force space - time equation: F = - G m M/r2 + k1/ r4

Einstein's rotating ellipse: r = a (1 - ε2)/ [1 + ε cosine (θ - Ψ)

Ψ = {6 π G M/a c² (1 - ε²)} [180/ π] [36526/T (days)] [3600]

Newton's space ----------------Nobel - Einstein's space - time

In July 21, 1969, and after graduating from 5th grade I watched Apollo 11 and came up with the visual difference in size of Apollo 11: D - C = [(B - A)/B)] C

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A = Sun - Mercury distance = 5.82 x 109 meters;

B = Sun Earth distance = 149.6 x 109 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 θ0'= 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 δθ0'

= (2 x π /88 x 24 x 60 x 60) (180/ π) (36526/88) (3600)

= 70.75 arc sec per century.

If C = δθ0' = 70.75 arc sec per century measured from the Sun, then how much it is diminished if measured from Earth?

A = 5.82 x 109; B = 149.6 x 109; C = 70.75

And the answer is [(A - B)/B] C = [(5.82 x 109 - 149.6 x 109)/ 149.6 x 109] 70.75 = 43 arc sec per/100 years same numbers as Einstein's numbers

Take any physics definition written by any physicists from any period of time and calculate (D - C) and the answer shall be = 43 arc sec per/100 years same numbers as Einstein's errors numbers

Le Verrier historical mistake is

The angular velocity of Mercury around the Sun is:

Is: θ ' = v /r = Mercury's orbital velocity/ mercury's - Sun orbital distance

If plant Mercury's angular velocity is measured from the sun, then θ ' = v /r

But planet Mercury's angular velocity is not measured from the Sun

Planet Mercury angular around the sun is measured from earth

If planet Mercury angular around the sun measured from earth, then the

angular velocity of Planet mercury around the Sun θ m' measured is: : θ m' (Earth) = (v + v 0)/r

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Where v = Mercury's orbital speed around the sun = 48.14 km/sec

And v0 = Earth's orbital speed around the sun = 29.78 km/sec

And r = Mercury's - Sun distance

Then θ m' (Earth) = (v +/- v0)/r

And θ m' (Earth) = [(v /r) + (v0 /r 0)], and not v0 /r; Le Verrier error

is: (v0 /r)

= (Earth's orbital speed around the sun/ Mercury's - Sun distance)

Le Verrier mistake was/is not taking into account Earth's motion/speed.

The angular speed shift: v0 /r; taking into account Earth rotation v º

Le Verrier historical error is: = [(v0 - vº) /r] radians

= (Earth's orbital speed around the sun - Earth's spin speed/ Mercury's - Sun distance)

In arc second per century multiplying by (180/ π) (3600) (100 τ / τ 0)

Changing radians to degrees (180/ π) to seconds (3600) per century (100 τ / τ 0)

And τ = 1 year = 365.2526; 1 century = 100 years = 100 τ = 36525.26 days

And τ 0 = Planet Mercury's period around the Sun = 88 days

And v0 = Earth's speed around the Sun = 29.78 km/sec

And v º = Earth's spin speed = 0.4651 km/sec

Le Verrier error in arc second per century is:

= [(v0 - v º) /r)] (180/ π) (3600) (100 τ / τ 0)

= [(29.8 - 0.465) /48.14)] (180/ π) (3600) (36525.26/ 88)

= 43 arc sec/100y

Planet Mercury's Perihelion is a measurements error that can at least be calculated 100 different ways from 100 different physics concepts using 100 different formulas from 100 different periods of time in human history and get the same numbers as Einstein 43 seconds of an arc per century using the formula: D - C = [(B - A)/B)] C = Visual artifact = measurement error = modern + Nobel

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Newton's equation is solved wrong for 350 years and the correct solution deletes 112 of Nobel science and scientists:

F = -G m M/r2; r > 0

Physics Faculty wrong solution of ellipse, r (θ, 0) = a (1 - ε2)/ (1 + ε cosine θ)

Correct solution is a rotating ellipse, r (θ, t) = [a (1-ε²)/ (1+ ε cosine θ)] ℮(λ + ì ω) t

Newton's equation in polar coordinates

F = m γ; γ= [r" - r θ'²] r1 + [2 r' θ' + r θ"] θ1

With m (r" - r θ'²) = - Gm M/r2 Eq-1

And 2 r' θ' + r θ"= 0 Eq-2

A - Real numbers or time independent solution

Eq-2: 2 r' θ' + r θ"= 0

Multiply by r> 0

Then 2 r r' θ' + r2 θ"= 0

Or, d (r²θ')/d t = 0

And integrating: r²θ' = h = constant

With m (r" - r θ'²) = - Gm M/r2

Then, (r" - r θ'²) = - GM/r2

Let u = 1/r; r = 1/u; r²θ' = h = θ’ /u²

And r' = d r/d t = (d r/ d u) (d u /d θ) (d θ/ d t)

= (- /u ²) (d u /d θ) θ'

= (-θ'/u ²) (d u /d θ)

= - h (d u/ d θ)

And r' = d r/d t = - h (d u/ d θ)

And r" = d² r/ d t² = d (d r'/ d t)/ d t

= d [- h (d u/ d θ)]/ d t

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Multiply (d θ/ d θ)

Then r" = d² r/ d t² = {d [- h (d u/ d θ)]/ d t} (d θ/ d θ)

= θ' {d [- h (d u/ d θ)]/ d θ}

= - h θ’ (d² u/ d θ ²)

= (- h²/r²) (d² u/ d θ ²)

= - h² u² (d² u/ d θ ²)

And r" = d² r/ d t² = - h² u² (d² u/ d θ ²)

With d² r/dt² - r θ'² = - G M/r2 E q – 1

And - h² u² (d² u/ d θ ²) – (1/u) (h u²) ² = - G M u2

Then (d² u/ d θ ²) + u = G M/h2

And u = G M/h2 + A cosine θ

The r = 1/u = 1/ (G M/h2 + A cosine θ); divide by G M/h2

And r = (h2/G M)/ [1 + (A h2/G M) cosine θ]

With; h2/G M = a (1 - ε2); (A h2/G M) = ε

This is Newton's equation classical solution

Or, r = a (1 - ε2)/ (1 + ε cosine θ); definition of an ellipse ------------- I

Newton's time independent solution

B - Real time or complex numbers solution:

Newton's equation in polar coordinates

F = m γ; γ= [r" - r θ'²] r1 + [2 r' θ' + r θ"] θ1

With m (r" - r θ'²) = - Gm M/r2 Eq-1

And 2 r' θ' + r θ"= 0 Eq-2

Eq-2: 2 r' θ' + r θ"= 0

Separate the variables: 2 r' θ' = - r θ"

Or 2(r'/r) = - (θ"/θ') = - 2 (λ + í ω)

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Then: (r'/r) = λ + í ω

Or d r/r = (λ + í ω) d t

Then r = r 0 ℮ (λ + ì ω) t

And r = r (θ, 0) r (0, t); r 0 = r (θ, 0)

And r = r (θ, 0) ℮ (λ + ì ω) t

And r (0, t) = ℮ (λ + ì ω) t

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

Then, r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ℮ (λ + ì ω) t ------------- I

Newton's time dependent solution = quantum mechanics

If time is frozen that is t = 0

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

Relativistic is the difference between I and Real II

With - (θ"/θ') = - 2 (λ + í ω)

Then θ' = θ'0 ℮ -2 (λ + ì ω) t

With θ'0 = h/ [r (θ, 0)] 2

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

And, θ'(θ, t) = θ' (θ, 0) θ' (0, t)

And θ' (0, t) = ℮ -2 (λ + ì ω) t

At Perihelion:

We Have θ' (0, 0) = h (0, 0)/r² (0, 0) = 2πab/ τ0 a² (1- ε) ²; τ0 = orbital period

= 2πa² [√ (1- ε²)]/ τ0a² (1- ε) ²]

= 2π [√ (1- ε²)]/ τ0 (1- ε) ²]

Then θ'(0, t) = 2 π √ [(1- ε²)/ τ0 (1- ε) ²] ℮ -2 (λ + ì ω) t

With λ= 0

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Then θ'(0, t) = 2 π √ [(1-ε²)/ τ0 (1-ε) ²] ℮ -2 (λ + ì ω) t

= 2 π √ [(1-ε²)/ τ0 (1- ε) ²] (cosine 2 ω t - ỉ sine 2 ω t)

Real θ'(0, t) = 2 π √ [(1- ε²)/ τ0 (1-ε) ²] cosine 2 ω t

Real θ'(0, t) = 2 π √ [(1-ε²)/ τ0 (1-ε) ²] (1 - 2sine² ω t)

Naming θ' = θ'(0, t); θ'0 = 2 π √ [(1-ε²)/ τ0 (1-ε) ²]

Then θ' = 2 π √ [(1- ε²)/ τ0 (1- ε) ²] (1 - 2 sine² ω t)

And θ' = θ'0 (1 - 2 sine² ω t)

And θ' - θ'0 = - 2 θ'0 sine² ω t

= -2{2 π √ [(1-ε²)/ τ0 (1-ε) ²]} sine² ω t

And θ' - θ'0 = -4 π √ [(1-ε²)/ τ0 (1-ε) ²] sine² ω t

With, v ° = spin velocity; v0 = orbital velocity; τ0 = orbital period

And ω τ0= tan-1 [(v° + v0)/c]; light aberrations

Δ θ' = θ' - θ'0

= - 4 π √ [(1-ε²)]/ τ0 (1-ε) ²] sine² tan-1 [(v° + v0)/c] radians per τ0

In degrees per period is multiplication by 180/ π

Δ θ' = (-720) √ [(1-ε²)/ τ0 (1-ε) ²] sine² tan-1 [(v° + v0)/c]

The angle difference in degrees per period is:

Δ θ = (Δ θ') τ0

Δ θ = (-720) √ [(1-ε²)/ (1-ε) ²] sine² tan-1 [(v° + v0)/c] calculated in degrees per

century is multiplication = 100 τ; τ = Earth orbital period = 100 x 365.26 = 36526

days and dividing by using τ0 in days

Δ θ (100 τ / τ0) = Δ θ in degrees per century

= (-72000 τ / τ0) √ [(1-ε²)/ (1-ε) ²] sine² tan-1 [(v° + v0)/c]

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In arc second per century is multiplying by 3600

Δ θ = - 3600 x 720 (100 τ / τ0) √ [(1-ε²)/ (1-ε) ²] x Sine² tan-1 [(v° + v0)/c]

Approximations I

With v° << c and v* << c

Then Sine² tan-1 [(v°+ v0)/c] ≈ (v° + v0)/c

Δ θ (Calculated in arc second per century)

= (-720x36526x3600/ τ0 days) √ [(1-ε²)/ (1-ε) ²] [(v° + v0)/c] ²

Approximations II

The circumference of an ellipse

Is: 2 π a (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2 π a (1- ε²/4); r0 = a (1- ε²/4)

From Newton's laws for a circular orbit:

F = [M/m F = - Gm M/r02 = m v0²/ r0

Then v0² = GM/ r0

For planet Mercury

And v0 = √ [GM/ r] = √ [GM/a (1-ε²/4)]

G = 6.673 x 10 -11; M = 2 x1030 kg; a = 58.2 x 109 meters; ε = 0.206

Then v0 = √ [6.673 x 10 -11 x 2 x1030 /58.2 x 109 (1- 0.206 ²/4)]

And v0 = 48.14 km/sec [Mercury]; c = 300,000

Δ θ (Calculated in arc second per century)

= (-720x36526x3600/ τ0 days) √ [(1-ε²)/ (1-ε) ²] [(v° + v0)/c] ²

With ε = 0.206; √ [(1-ε²)/ (1-ε) ²] = 1.552; v° = 3 meters per second

Δ θ = (-720x36526x3600/88) 1.552 (48.143/300,000) ²

Δ θ = 43 arc second per century

Summary

= (-720x36526x3600/ τ0 days) √ [(1-ε²)/ (1-ε) ²] [(v° + v0)/c] ²

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= (-720x36526x3600/88) 1.552 (48.143/300,000) ²

= 43 arc second per century

Or, r = a (1 - ε2)/ (1 + ε cosine θ); definition of an ellipse

Rotating ellipse, r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ℮ (λ + ì ω) t ------------- I

Newton's equation is: F = -G m M/r2; r > 0

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

Correct: r (θ, t) = [a (1-ε²)/ (1+ε cosine θ)] ℮ (λ + ì ω) t ------------- II

Read my article: 50 Solutions of Mercury's Perihelion www.scribd/joenahhas

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