Richard Hertz

2021-11-20 18:27:45 UTC

This post specifically address the paper that made Einstein famous, not General Relativity, and is not a critic about GR in general (for now).

On Nov. 18, 1915, Einstein made his third presentation in November to the Prussian Academy of Science. The final paper, with the solution of the gravitational field equation, would be made on Nov. 18, 1915. The Nov.18 paper would be celebrated as a triumph of General Relativity, and would made him famous immediately in the scientific community and a worldwide celebrity in 1919, thanks to Eddington’s expedition outcome.

In its 10 pages, it delivered two numbers: An explanation for the missing +43”/century in the total advance of the perihelion of Mercury, and a brief comment about a new value of 1.75” for the deflection of stars light by the Sun, when it passes by its surface. It was twice the value he announced in 1911.

The paper, “Explanation of the Perihelion Motion of Mercury from General Relativity Theory”, contains 44 formulae and less than a half are numbered, for references within the paper. The equation 11, only 9 before the end of the paper, divide it in two parts: the application of GR for the gravitational field in a vacuum for a suitably chosen system of coordinates and the calculations for the extra Mercury’s perihelion advance, ε.

The last part contains two fudging and cooking actions, in order to obtain the final Eq. 14 (43”/cy) from Eq. 11.

Going backward from Eq. 14, allows to understand the process done to reverse engineering the 1898 Gerber’s formula, which was known to include as a parameter (not an unknown), ε = 43”/century.

From Eq. 14: ε = 24π³a²/[T²c² (1 – e²)], and using Kepler’s identity a³/T² = GM/4π² (error 0.07%).

Knowing that the relativistic gravitational potential α =rS = 2GM/c², the equation 14 can be expressed as Eq. 13:

(Equation 13) ε = + 3π α/[a (1 – e²)] , which is the difference per orbit (radians) respect to Newton’s theory.

The problem is that ε should have a value ε = + 5.0186E-07 rad/orbit to obtain ε = 42.94”/cy ≈ 43”/cy.

Given α = 2953.25 m ; a = 5.7909E10 m ; e = 0.2056 ; T = 87.9694 days or T = 7600556.16 sec (NASA)

The real calculated values (as values in this paper only) are:

ε = -1.6729E-07 rad/orbit , which gives to obtain ε = -14.31”/century.

Then, the paper contains two errors: a negative sign for precession and a 2/3 times higher real value of ε.

Real ε = -8π³a²/[T²c² (1 – e²)], as recalculated using correct results of the integral and avoiding fudging.

Mathematical proofs

(Equation 11) (dx/dφ)²= 2A/B² + α/B² x −x² + αx³

Is the relativistic equation for the orbital motion of a planet around the Sun, in geometrical units (G, M, m, c and T equal to 1).

The newtonian equivalent has the form:

1/r⁴ . (dr/dɸ)² = 2A.m/L² + (2GM.m²/L²) . 1/r – 1/r² , which after a substitution x = 1/r has the form

(dx/dɸ)² = 2A.m/B² + (2GM.m²/B²) . x – x² , where A: Total energy of the system ; B: angular moment

This equation has two real roots: α₁ = 1/AP = 2.17378E-11 m-1 and α₂ = 1/PE = 1.43236E-11 m-1

These are the same two newtonian roots that are contained in the cubic polynomial of Equation 11.

The third root of Eq. 11, never detailed because α is the variable variable, is α3 = 1/α = 1/rS = 1/2953.25 m-1.

Going one step back from Eq. 13, and considering that the integral only cover half an orbit (from α₁ to α₂), the angular sweep in half orbit is given by Equation 12:

(Equation 12) Φ = π [ 1+ 3α/[2a (1 – e²)]] , and replacing with the identity 1/[2a (1 – e²)] = (α₁ + α₂)/2

Φ = π [ 1+ 3/4 α (α₁ + α₂)] , which is the equation previous to Eq. 12, as given in the paper.

The problem is that going from Eq. 11 to this point, actually Φ = π [1 – 1/4 α (α₁ + α₂)].

This produces a value 2/3 times lower in the final Equation 14, plus a (-) sign that indicates a retrograde orbital difference with Newton, not an advance.

And due to these facts, two actions of fudging and cooking happens between Eq. 11 and Eq. 12.

1) Einstein equated the polynomial in Eq. 11 to a new one, with a fudge parameter K introduced from nowhere:

Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + K) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]

K = α/2 (α₁ + α₂). Now proceeds to extract 1/√(1 - αx) ≈ 1 + α/2 x from the polynomial to make it quadratic:

Φ ≈ [(1 + α/2 (α₁ + α₂)] ∫ [1 + α/2 x)] dx /√[ - (x - α₁) (x - α₂)] (between α₁ and α₂), the next equation.

And Einstein simply writes an expression before writing the Eq. 12, commenting “The integration leads to”

Φ = π [ 1+ 3/4 α (α₁ + α₂)]

What happens in one single step is two fold:

1) Take (1 + K) = [(1 + α/2 (α₁ + α₂)] = 1 + 5.3265E-05 ≈ 1, is deceiving, not innocent at all.

This is because the value that counts is K = 5.3265E-05, which constitutes 2/3 of the final value for ε, because the value 1 present in the equation vanishes

when a 2π value for an entire orbit is subtracted from Φ.

So, this is cheating, fudging.

2) The result of the integral: ∫ [1 + α/2 x)] dx /√[ - (x - α₁) (x - α₂)] (between α₁ and α₂), is literally replaced.

The real result +π [1 - 1/4 α(α₁ + α₂)] is replaced by +π [1 + 1/4 α(α₁ + α₂)].

This is needed to make

Φ ≈ π [(1 + α/2 (α₁ + α₂)] . [1 + 1/4 α(α₁ + α₂)] ≈ π [ 1+ 3/4 α (α₁ + α₂)] , discarding the product 1/8 α² (α₁ + α₂)²

3/4 α (α₁ + α₂) = 7.9874E-08 ; 1/8 α² (α₁ + α₂)² = 1.4177E-15

Introducing K falsely gives the desired value for Φ. If discarded making K = 0, as should be, the REAL VALUE of Φ:

Φ = π [1 – 1/4 α (α₁ + α2) ] , which makes ε = - 1/2 π α (α₁ + α₂) = - π α/[a (1 – e2)] (only 1/3 of the published value).

Solving the integral ∫ [1 + α/2 x)] dx /√[ - (x - α₁) (x - α₂)]

It has two parts:

P1(x) = ∫ dx /√[ - (x - α₁) (x - α₂)] , (between α₁ and α₂)

P2(x) = ∫ [α/2 x dx /√[ - (x - α₁) (x - α₂)] , (between α₁ and α₂)

For simplicity, P(x) = - (x - α₁) (x - α₂) = - x² + (α₁ + α₂) x - α₁ α₂ = ax² + bx + c ; P(α₁) = P(α₂) = 0

P1(x) = ∫ dx /√P(x) = 1/√a ln [(2ax+b)/√a + 2 P(x)] = 1/√a ln [(2ax+b)/√a], between α₁ and α₂ ; P(x) = 0

P1(α₂) – P1(α₁) = 1/√(-1) ln [(-2α₂+b)/(-2α₁+b)] = 1/√(-1) ln [(-α₂+ α₁)/(-α₁+ α₂)] = +π

P2(x) = α/2 ∫ x/√P(x) dx = α/2 . 1/a P(x) – α/2. b/(2a^3/2) ln {[(2a x +b)/√a ] + 2 P(x) }

P2(x) = – α/2. b/(2a^3/2) ln [(2a x +b)/√a ] between α₁ and α₂ ; P(x) = 0

replacing a = -1 ; b = (α₁ + α₂)

P2(α₂) – P2(α₁) = – α/2. (α₁ + α₂)/(2√(-1)) { ln (- α₂ + α₁)/(- α₁ + α₂)} = - 1/4π α (α₁ + α₂)

Finally,

H(α₂) - H(α₁) = π - 1/4π α (α₁ + α₂) = π [1 - 1/4 α (α₁ + α₂)] , which produces a difference

ε = - π α/[a(1 – e²)]

Which is very different by the value required by Einstein to reach Eq. 13 and Eq. 14.

The final result Φ, using K = 0 (it was planted with no reason) and H(α2) - H(α1) gives

Φ = π [1 - 1/4 α (α₁ + α₂)] = - π [1 - 1/2 α/[a(1 – e²)]] , as calculated here.

Φ = π [1 + 3/4 α (α₁ + α₂)] = + π [1 + 3/2 α/[a(1 – e²)] , as written by Einstein on his Nov. 18, 1915 paper.

The new and real value of Φ causes that the final result be:

(Equation 14) ε = - 8π3a²/[T²c² (1 – e²)], negative and 3 times lower.

ε(Einstein)/ε(Real) = -3 ; ε(Real) = -1/3 ε(Einstein)

On Nov. 18, 1915, Einstein made his third presentation in November to the Prussian Academy of Science. The final paper, with the solution of the gravitational field equation, would be made on Nov. 18, 1915. The Nov.18 paper would be celebrated as a triumph of General Relativity, and would made him famous immediately in the scientific community and a worldwide celebrity in 1919, thanks to Eddington’s expedition outcome.

In its 10 pages, it delivered two numbers: An explanation for the missing +43”/century in the total advance of the perihelion of Mercury, and a brief comment about a new value of 1.75” for the deflection of stars light by the Sun, when it passes by its surface. It was twice the value he announced in 1911.

The paper, “Explanation of the Perihelion Motion of Mercury from General Relativity Theory”, contains 44 formulae and less than a half are numbered, for references within the paper. The equation 11, only 9 before the end of the paper, divide it in two parts: the application of GR for the gravitational field in a vacuum for a suitably chosen system of coordinates and the calculations for the extra Mercury’s perihelion advance, ε.

The last part contains two fudging and cooking actions, in order to obtain the final Eq. 14 (43”/cy) from Eq. 11.

Going backward from Eq. 14, allows to understand the process done to reverse engineering the 1898 Gerber’s formula, which was known to include as a parameter (not an unknown), ε = 43”/century.

From Eq. 14: ε = 24π³a²/[T²c² (1 – e²)], and using Kepler’s identity a³/T² = GM/4π² (error 0.07%).

Knowing that the relativistic gravitational potential α =rS = 2GM/c², the equation 14 can be expressed as Eq. 13:

(Equation 13) ε = + 3π α/[a (1 – e²)] , which is the difference per orbit (radians) respect to Newton’s theory.

The problem is that ε should have a value ε = + 5.0186E-07 rad/orbit to obtain ε = 42.94”/cy ≈ 43”/cy.

Given α = 2953.25 m ; a = 5.7909E10 m ; e = 0.2056 ; T = 87.9694 days or T = 7600556.16 sec (NASA)

The real calculated values (as values in this paper only) are:

ε = -1.6729E-07 rad/orbit , which gives to obtain ε = -14.31”/century.

Then, the paper contains two errors: a negative sign for precession and a 2/3 times higher real value of ε.

Real ε = -8π³a²/[T²c² (1 – e²)], as recalculated using correct results of the integral and avoiding fudging.

Mathematical proofs

(Equation 11) (dx/dφ)²= 2A/B² + α/B² x −x² + αx³

Is the relativistic equation for the orbital motion of a planet around the Sun, in geometrical units (G, M, m, c and T equal to 1).

The newtonian equivalent has the form:

1/r⁴ . (dr/dɸ)² = 2A.m/L² + (2GM.m²/L²) . 1/r – 1/r² , which after a substitution x = 1/r has the form

(dx/dɸ)² = 2A.m/B² + (2GM.m²/B²) . x – x² , where A: Total energy of the system ; B: angular moment

This equation has two real roots: α₁ = 1/AP = 2.17378E-11 m-1 and α₂ = 1/PE = 1.43236E-11 m-1

These are the same two newtonian roots that are contained in the cubic polynomial of Equation 11.

The third root of Eq. 11, never detailed because α is the variable variable, is α3 = 1/α = 1/rS = 1/2953.25 m-1.

Going one step back from Eq. 13, and considering that the integral only cover half an orbit (from α₁ to α₂), the angular sweep in half orbit is given by Equation 12:

(Equation 12) Φ = π [ 1+ 3α/[2a (1 – e²)]] , and replacing with the identity 1/[2a (1 – e²)] = (α₁ + α₂)/2

Φ = π [ 1+ 3/4 α (α₁ + α₂)] , which is the equation previous to Eq. 12, as given in the paper.

The problem is that going from Eq. 11 to this point, actually Φ = π [1 – 1/4 α (α₁ + α₂)].

This produces a value 2/3 times lower in the final Equation 14, plus a (-) sign that indicates a retrograde orbital difference with Newton, not an advance.

And due to these facts, two actions of fudging and cooking happens between Eq. 11 and Eq. 12.

1) Einstein equated the polynomial in Eq. 11 to a new one, with a fudge parameter K introduced from nowhere:

Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + K) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]

K = α/2 (α₁ + α₂). Now proceeds to extract 1/√(1 - αx) ≈ 1 + α/2 x from the polynomial to make it quadratic:

Φ ≈ [(1 + α/2 (α₁ + α₂)] ∫ [1 + α/2 x)] dx /√[ - (x - α₁) (x - α₂)] (between α₁ and α₂), the next equation.

And Einstein simply writes an expression before writing the Eq. 12, commenting “The integration leads to”

Φ = π [ 1+ 3/4 α (α₁ + α₂)]

What happens in one single step is two fold:

1) Take (1 + K) = [(1 + α/2 (α₁ + α₂)] = 1 + 5.3265E-05 ≈ 1, is deceiving, not innocent at all.

This is because the value that counts is K = 5.3265E-05, which constitutes 2/3 of the final value for ε, because the value 1 present in the equation vanishes

when a 2π value for an entire orbit is subtracted from Φ.

So, this is cheating, fudging.

2) The result of the integral: ∫ [1 + α/2 x)] dx /√[ - (x - α₁) (x - α₂)] (between α₁ and α₂), is literally replaced.

The real result +π [1 - 1/4 α(α₁ + α₂)] is replaced by +π [1 + 1/4 α(α₁ + α₂)].

This is needed to make

Φ ≈ π [(1 + α/2 (α₁ + α₂)] . [1 + 1/4 α(α₁ + α₂)] ≈ π [ 1+ 3/4 α (α₁ + α₂)] , discarding the product 1/8 α² (α₁ + α₂)²

3/4 α (α₁ + α₂) = 7.9874E-08 ; 1/8 α² (α₁ + α₂)² = 1.4177E-15

Introducing K falsely gives the desired value for Φ. If discarded making K = 0, as should be, the REAL VALUE of Φ:

Φ = π [1 – 1/4 α (α₁ + α2) ] , which makes ε = - 1/2 π α (α₁ + α₂) = - π α/[a (1 – e2)] (only 1/3 of the published value).

Solving the integral ∫ [1 + α/2 x)] dx /√[ - (x - α₁) (x - α₂)]

It has two parts:

P1(x) = ∫ dx /√[ - (x - α₁) (x - α₂)] , (between α₁ and α₂)

P2(x) = ∫ [α/2 x dx /√[ - (x - α₁) (x - α₂)] , (between α₁ and α₂)

For simplicity, P(x) = - (x - α₁) (x - α₂) = - x² + (α₁ + α₂) x - α₁ α₂ = ax² + bx + c ; P(α₁) = P(α₂) = 0

P1(x) = ∫ dx /√P(x) = 1/√a ln [(2ax+b)/√a + 2 P(x)] = 1/√a ln [(2ax+b)/√a], between α₁ and α₂ ; P(x) = 0

P1(α₂) – P1(α₁) = 1/√(-1) ln [(-2α₂+b)/(-2α₁+b)] = 1/√(-1) ln [(-α₂+ α₁)/(-α₁+ α₂)] = +π

P2(x) = α/2 ∫ x/√P(x) dx = α/2 . 1/a P(x) – α/2. b/(2a^3/2) ln {[(2a x +b)/√a ] + 2 P(x) }

P2(x) = – α/2. b/(2a^3/2) ln [(2a x +b)/√a ] between α₁ and α₂ ; P(x) = 0

replacing a = -1 ; b = (α₁ + α₂)

P2(α₂) – P2(α₁) = – α/2. (α₁ + α₂)/(2√(-1)) { ln (- α₂ + α₁)/(- α₁ + α₂)} = - 1/4π α (α₁ + α₂)

Finally,

H(α₂) - H(α₁) = π - 1/4π α (α₁ + α₂) = π [1 - 1/4 α (α₁ + α₂)] , which produces a difference

ε = - π α/[a(1 – e²)]

Which is very different by the value required by Einstein to reach Eq. 13 and Eq. 14.

The final result Φ, using K = 0 (it was planted with no reason) and H(α2) - H(α1) gives

Φ = π [1 - 1/4 α (α₁ + α₂)] = - π [1 - 1/2 α/[a(1 – e²)]] , as calculated here.

Φ = π [1 + 3/4 α (α₁ + α₂)] = + π [1 + 3/2 α/[a(1 – e²)] , as written by Einstein on his Nov. 18, 1915 paper.

The new and real value of Φ causes that the final result be:

(Equation 14) ε = - 8π3a²/[T²c² (1 – e²)], negative and 3 times lower.

ε(Einstein)/ε(Real) = -3 ; ε(Real) = -1/3 ε(Einstein)