On his Nov.18, 1915, Einstein presented his proof about GR correctly predicting
an advance of 43"arc/century on Mercury's perihelion.

Notwithstanding his hacks and fudges in the first part of the paper, which
deserves a separate thread for the fanatic mathematicians here, this problem
only requires an elementary knowledge of integrals and a table at hand.

The last part of the paper is devoted to develop a solution for this integral:

Φ = [1 + 1/2 α (α₁+ α₂)] ∫ dx/√[ - (x - α₁) (x – α₂) (1 – α x)] , between α₁ and α₂

or, with an approximation for the last root

Φ = [1 + 1/2 α (α₁+ α₂)] ∫ (1 – α x/2)/√[ - (x - α₁) (x – α₂)] dx , between α₁ and α₂

α₁ = 1/Ap = 1.43232E-11 = 1/69817000000 1/m (NASA)

α₂ = 1/Pe = 2.17382E-11 = 1/46002000000 1/m (NASA)

α = 2GM/c² = 2954.13 m (to be used to calculate approximations, later)

Einstein affirmed (quote):

"The integration leads to Φ = π [1 + 3/4 α (α₁+ α₂)]"

"or, if one takes α₁ and α₂ as reciprocal values of the maximal and
minimal distance from the Sun,

(Eq.12 in the 1915 paper) Φ = π [1 + 3α/[2a (1 - e²)]] "

END QUOTING.

The problem is that, solving the integral (between α₁ and α₂)

Φ = [1 + 1/2 α (α₁+ α₂)] ∫ (1 – α x/2)/√[ - (x - α₁) (x – α₂)] dx

gives

Φ = - π [1 + 5/8 (α₁+α₂) α + (α₁+α₂)2 α2/16 ]

Even if discarding the last term, for being too small,

Φ = - π [1 + 5/8 α (α₁+α₂) ]

very different from what he stated on the paper:

Φ = + π [1 + 3/4 α (α₁+ α₂)]

1) There is a - sign, which indicates a retrograde orbit

2) From this value, his famous Eq. 12 (almost Gerber, requires 1 step)
can't be derived in any possible way.

So, is the negative value of Φ a wrong calculation of a simple integral
or the fudging has been present there all the time?

To prove or disproof this, only a table with integrals is required.

P.S.: The integral with α in the numerator is the "relativistic contribution",
as the first part is pure newtonian, and gives π for half orbit.

On Tuesday, November 9, 2021 at 3:13:35 PM UTC-8, Richard Hertz wrote:
> The problem is that, solving the integral (between α₁ and α₂)
> Φ = [1 + 1/2 α (α₁+ α₂)] ∫ (1 – α x/2)/√[ - (x - α₁) (x – α₂)] dx

You have the wrong sign in the numerator of the integrand. Should be 1 + α x/2.

On Tuesday, November 9, 2021 at 8:33:29 PM UTC-3, Al Coe wrote:
> On Tuesday, November 9, 2021 at 3:13:35 PM UTC-8, Richard Hertz wrote:
> > The problem is that, solving the integral (between α₁ and α₂)
> > Φ = [1 + 1/2 α (α₁+ α₂)] ∫ (1 – α x/2)/√[ - (x - α₁) (x – α₂)] dx

> You have the wrong sign in the numerator of the integrand. Should be 1 + α x/2.

You are right, but it'd a typo. In the derivation, I used the correct + sign.

On Tuesday, November 9, 2021 at 4:14:42 PM UTC-8, Richard Hertz wrote:
> On Tuesday, November 9, 2021 at 8:33:29 PM UTC-3, Al Coe wrote:
> > On Tuesday, November 9, 2021 at 3:13:35 PM UTC-8, Richard Hertz wrote:
> > > The problem is that, solving the integral (between α₁ and α₂)
> > > Φ = [1 + 1/2 α (α₁+ α₂)] ∫ (1 – α x/2)/√[ - (x - α₁) (x – α₂)] dx
>
> > You have the wrong sign in the numerator of the integrand. Should be 1 + α x/2.
> You are right, but it'd a typo. In the derivation, I used the correct + sign.

Then you evaluated the integral incorrectly. You can find many detailed explications, line by line, of that paper, including explaining the misprints and explaining how to evaluate the integral.

I made a correction in the derivation, which gives almost the same formula, BUT with a MINUS sign.
The final derivation gives
Φ = - π [1 + α/2 (α₁ + α₂)] . [1 + α/4 (α₁ + α₂)] = - π . [1 + 3/4 α (α₁ + α₂) + 1/8 α² (α₁ + α₂)² ]
Φ ≈ - π . [1 + 3/4 α (α₁ + α₂)]
Here is the complete derivation, if anyone wants to check the origin of (-) sign, which would indicate a negative precession.
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Φ = [1 + α/2 (α₁ + α₂) ] { ∫ dx/√ [ - (x - α₁) (x - α₂) (1 - αx)] dx } , between α₁ and α₂
Φ ≈ [1 + α/2 (α₁ + α₂) ] { ∫ (1+ α/2 x)/√ [ - (x - α₁) (x - α₂)] dx } , between α₁ and α₂
Φ = [1 + α/2 (α₁ + α₂) ] { ∫ dx/√ [ - (x - α₁) (x - α₂)] dx + α/2 ∫ x dx/√ [ - (x - α₁) (x - α₂)] } , between α₁ and α₂
K = [1 + α/2 (α₁ + α₂)]
P(x) = √ [ - (x - α₁) (x - α₂)] = √ [- x2 + (α₁+α₂) x - α₁ α₂ ] = √[- a x² + b x + c]
N(x) = ∫ dx/P(x)
M(x) = ∫ x/P(x) dx
Φ = K [N(x) + α/2 M(x)] , between α₁ and α₂

Solving N(x)

N(x) = ∫ dx/P(x) = ∫ dx/√ [- x² + (α₁+α₂) x - α₁ α₂] dx = ∫ dx/√[- a x² + b x + c] , a = 1; b = (α₁ + α₂) ; c = - α₁ α₂

SOLUTION: N(x) = 1/√a arcsine (√(a/d) (x – b/2a)) + C , d = b²/4a + c
N(x) = N(α₂) – N(α₁) = arcsine [1/√ (b²/4 + c) (α₂ – b/2)] - arcsine [1/√ (b²/4 + c) (α₁ – b/2)]
1/√ (b²/4 + c) = 1/√ ((α₁ + α₂)²/4 - α₁ α₂) = 1/√ ((α₁² + 2α₁ α₂ + α₂²)/4 - α₁ α₂) = 2/√ (α₁² - 2α₁ α₂ + α₂²) = 2/(α₁ - α₂)
N(α₂) – N(α₁) = arcsine [2/(α₁ - α₂) (α₂ – (α₁ + α₂)/2)] - arcsine [2/(α₁ - α₂) (α₁ – (α₁ + α₂)/2))]
N(α₂) – N(α₁) = arcsine [1/(α₁ - α₂) (α₂ – α₁)] - arcsine [1/(α₁ - α₂) (α₁ – α₂)]
N(α₂) – N(α₁) = arcsine (-1) – arcsine (1) = −π/2 − π/2 = - π
NOTE: Here appears the first (-) sign, integrating from aphelion to perihelion, as Einstein wrote:
QUOTE: "or, if one takes α₁ and α₂ as reciprocal values of the maximal and minimal distance from the Sun,"

Solving α/2 M(x)

α/2 M(x) = α/2 ∫ x/ P(x) dx = α/2 ∫x/√[- a x² + b x + c] dx , a = - 1; b = (α₁ + α₂) ; c = - α₁ α₂
SOLUTION: α/2 M(x) = α/2 {1/a P(x) – b/(2a^3/2) ln [(2ax + b)/√a + 2 P(x) ]}
As P(x) = √ [ - (x - α₁) (x - α₂)] = 0 , for x = α₁ or x = α₂
α/2 M(x) = α/2 {– b/(2a^3/2) ln [(2ax + b)/√a ]} , between α₁ and α₂
NOTE: a^3/2 = (-1)^3/2 = √(-1)
α/2 [M(α₂) - M(α₁)] = α/2 {– (α₁ + α₂)/(2√(-1)) ln [(-2 α₂ + α₁ + α₂)/√(-1) ] + (α₁ + α₂)/(2√(-1)) ln [(-2 α₁ + α₁ + α₂)/√(-1) ] }
α/2 [M(α₂) - M(α₁)] = α/2 {– (α₁ + α₂)/(2√(-1)) ln [(- α₂ + α₁)/√(-1) ] + (α₁ + α₂)/(2√(-1)) ln [(- α₁ + α₂)/√(-1) ] }
α/2 [M(α₂) - M(α₁)] = - α/2 (α₁ + α₂)/(2√(-1)) { ln [(- α₂ + α₁)/√(-1) ] - ln [(- α₁ + α₂)/√(-1) ] }
α/2 [M(α₂) - M(α₁)] = - α/2 (α₁ + α₂)/(2√(-1)) ln { [(- α₂ + α₁)/√(-1) ] / [(- α₁ + α₂)/√(-1) ] }
α/2 [M(α₂) - M(α₁)] = - α/2 (α₁ + α₂)/(2√(-1)) ln (- 1) = - α/2 (α₁ + α₂)/(2√(-1)) √(-1) π = - α/4 (α₁ + α₂) π

N(x) + α/2 M(x)] = - π [1 + α/4 (α₁ + α₂)]
Φ = K [N(x) + α/2 M(x)]
Φ = - π [1 + α/2 (α₁ + α₂)] . [1 + α/4 (α₁ + α₂)] = - π . [1 + 3/4 α (α₁ + α₂) + 1/8 α² (α₁ + α₂)² ]
Φ ≈ - π . [1 + 3/4 α (α₁ + α₂)]

NOTE: Following the indications in the paper EXACTLY, a (-) sign appears in the first and second integrals. Should be (+),
otherwise Mercury's perihelion is receding, as it happened on his 1913 calculation with Besso, or in 2nd. Nordstrom theory (1913).
Negative values were found also by von Laue and Sommerfeld in 2010 and 2011, using Minkowski scalar SR theory.
Even Poincaré approach (1906) shows about -6"/century. Here is some info about the 1913 Einstein-Besso calculations:
https://groups.google.com/u/1/g/sci.physics.relativity/c/oLRiO1XB0fU/m/dI7HBmg2AwAJ
My interest was to find out how such equation was solved, for which I tried first its accuracy by replacing parameters
with data extracted from NASA site, to understand the absolute values of α, α₁ and α₂. Next, followed the elliptic integral.

On Wednesday, November 10, 2021 at 2:36:40 AM UTC-3, Richard Hertz wrote:
> I made a correction in the derivation, which gives almost the same formula, BUT with a MINUS sign.
>
> The final derivation gives
>
> Φ = - π [1 + α/2 (α₁ + α₂)] . [1 + α/4 (α₁ + α₂)] = - π . [1 + 3/4 α (α₁ + α₂) + 1/8 α² (α₁ + α₂)² ]
>
> Φ ≈ - π . [1 + 3/4 α (α₁ + α₂)]

<snip>

I forgot to highlight the negative precession:

For an entire orbit,

Φ ≈ - 2π . [1 + 3/4 α (α₁ + α₂)] , so

Φ + 2π ≈ - 3/2 π α (α₁ + α₂)]

Hence, the right side shows a DEFICIT in completing an entire orbit, of about - 5.026E-07" of arc (gives - 43"/ century in 415 orbits).

On Tuesday, November 9, 2021 at 9:36:40 PM UTC-8, Richard Hertz wrote:
> Here is the complete derivation, if anyone wants to check the origin of (-) sign...
> N(x) = ∫ dx/P(x)

Your nomenclature is messed up, because x is a dummy variable on the right side, so it can't be the argument on the left.

> SOLUTION: N(x) = 1/√a arcsine (√(a/d) (x – b/2a)) + C , d = b²/4a + c

This indefinite integral has a square root inside the arcsine function, and you need to select the appropriate sign for it. For purposes of forming the definite integral from α₁ to α₂, you need the other sign, so that you have α₂ - α₁ as the denominator inside the arcsine, instead of α₁ - α₂ as you have it in your later expression. That's why you get the wrong sign. But...

There's a *much* simpler way of evaluating that integral, with a substitution of variables that makes the integral trivial, and gives the correct result unambiguously.

On Tuesday, November 9, 2021 at 9:36:40 PM UTC-8, Richard Hertz wrote:
> Here is the complete derivation, if anyone wants to check the origin of (-) sign...
> SOLUTION: N(x) = 1/√a arcsine (√(a/d) (x – b/2a)) + C , d = b²/4a + c

This indefinite integral has a square root inside the arcsine function, and you need to select the appropriate sign for it. For purposes of forming the definite integral from α₁ to α₂, you need the opposite sign, so that you have α₂ - α₁ as the denominator inside the arcsine, instead of α₁ - α₂ as you have it in your later expression. That's why you get the wrong sign. But...

There is a *much* simpler way of evaluating that integral, with a substitution of variables that makes the integral trivial, and gives the correct result unambiguously.

On Wednesday, November 10, 2021 at 3:54:14 AM UTC-3, Al Coe wrote:
> On Tuesday, November 9, 2021 at 9:36:40 PM UTC-8, Richard Hertz wrote:

<snip>

> > SOLUTION: N(x) = 1/√a arcsine (√(a/d) (x – b/2a)) + C , d = b²/4a + c
> This indefinite integral has a square root inside the arcsine function, and you need to select the appropriate sign for it. For purposes of forming the definite integral from α₁ to α₂, you need the opposite sign, so that you have α₂ - α₁ as the denominator inside the arcsine, instead of α₁ - α₂ as you have it in your later expression. That's why you get the wrong sign.. But...
>
> There is a *much* simpler way of evaluating that integral, with a substitution of variables that makes the integral trivial, and gives the correct result unambiguously.

Thanks for your replies. I still have a doubt about your comment.

I tried to be as explanatory as I could, so anyone could follow the derivations. For clarity, I expressed

Φ(x) = K [N(x) + α/2 M(x)] , between α₁ and α

with solutions

N(x) = - π and M(x) = - π α/4 (α₁ + α₂)

You are telling me that solutions

N(x) = ± π and M(x) = ± π α/4 (α₁ + α₂) are equally valid, but in the resulting expression

Φ = K [ ± π ± π α/4 (α₁ + α₂)]

I have to choose one in four combinations that give the desired result Φ = + K [ π + π α/4 (α₁ + α₂)] ?

Any other of the three remaining combinations of sign (±) would give either a negative recession (two choices) or the whole
integral having being done backwards?

I'd try to study your comment more in depth. I don't question your obvious knowledge on this matter, but allow me to check
everything again, this time with more academic sources. I haven't done this kind of work for too many years to count, so I've
became a little sloppy.

I used to do this kind of things for fun in the '70s and '80s, even deriving integrals out of thin air using tricks, but obviously too
much time have passed and techniques and formalism are lost in memory. But I'm sure that I'll recover the know-how with practice.

Actually, this topic has brought to me the desire to have a new hobby solving complex integrals of this type. I have a list that
I downloaded from a university, so I'll start developing them on my own. The table only provides results, and what is "fun" is to
develop it.

Too much of a "nerd", isn't it?

On Tuesday, November 9, 2021 at 11:24:36 PM UTC-8, Richard Hertz wrote:
> You are telling me that solutions
> N(x) = ± π and M(x) = ± π α/4 (α₁ + α₂) are equally valid, but in the resulting expression
> Φ = K [ ± π ± π α/4 (α₁ + α₂)]
> I have to choose one in four combinations that give the desired result Φ = + K [ π + π α/4 (α₁ + α₂)] ?

We must choose the correct roots and values of the arcsine, but these are not arbitrary choices, there's only one correct choice for the definite integral from α₁ to α₂, as can be shown by using the more direct method of evaluation that I mentioned, in which the positive square root under the integral gives the right sign. Also note that, for any given argument, the arcsine is a multi-valued function as well, and the arcsines of +1 and -1 can be in either order, differing by ±π, so it's necessary to carefully select the appropriate values when forming the definite integral using the complicated method. That's why people preferto use the simpler method that avoids the inverse trig functions, etc.

Richard Hertz wrote:

> I forgot to highlight the negative precession:
> For an entire orbit,
> Φ ≈ - 2π . [1 + 3/4 α (α₁ + α₂)] , so
> Φ + 2π ≈ - 3/2 π α (α₁ + α₂)]
> Hence, the right side shows a DEFICIT in completing an entire orbit, of
> about - 5.026E-07" of arc (gives - 43"/ century in 415 orbits).

thank you. You are the pro writing the most correct mathematics in this
group here.

I was reasoning about the meanings of the expression

Φ ≈ ± π . [1 + 3/4 α (α₁ + α₂)]

valid for two combinations (each term with the same sign) of the solutions of

Φ = K [N(x) + α/2 M(x)] , between α₁ and α₂

It gives

Φ₁ ≈ + π . [1 + 3/4 α (α₁ + α₂)] (Einstein's paper)

or

Φ₂ ≈ - π . [1 + 3/4 α (α₁ + α₂)]

For a number of N entire orbits, it gives

Φ₁ - 2π N ≈ + 3/2 π N α (α₁ + α₂) (Einstein's paper)

Φ₂ + 2π N ≈ - 3/2 π N α (α₁ + α₂)

IMHO, in the case of Φ₁ solution, it appears that precession accumulates going BACK in time, orbit after orbit.

Again IMHO, in the case of Φ₂ solution, it appears that precession accumulates going FORWARD in time, orbit after orbit.

I'm far from being a knowledgeable person in planetary motion, so I can be deadly wrong in my interpretation.

I only imagined what was the relationship of accumulations of radians with TIME, for (+) or (-) sign, as I imagined how the
test massless particle (Mercury) swept angles from an arbitrary reference with time evolving (forward or backwards).

I'll try to find how astronomy manage this particular topic.

P.S.: In the 1915 paper, Einstein gave up with the idea of thinking in terms of t or τ time. Even more, he used t or τ arbitrarily,
as the phenomena is analyzed as it's seen from Earth. This position allowed him to use data from observational astronomy,
which were based on newtonian references of time and space, and pre-dated his theories of relativity for hundred of years.

On Wednesday, November 10, 2021 at 7:41:39 AM UTC-8, despicable kapo Richard Hertz wrote:
> I'm far from being a knowledgeable person in planetary motion, so I can be deadly wrong in my interpretation.

No, you are just a pathetic cretin desperately trying to "disprove" relativity. You are failing. Miserably.

On Wednesday, November 10, 2021 at 7:41:39 AM UTC-8, cretin Richard Hertz mused:
> I was reasoning about the meanings of the expression
>
> Φ ≈ ± π . [1 + 3/4 α (α₁ + α₂)]
>
> valid for two combinations (each term with the same sign) of the solutions of
>
> Φ = K [N(x) + α/2 M(x)] , between α₁ and α₂
>
> It gives
>
> Φ₁ ≈ + π . [1 + 3/4 α (α₁ + α₂)] (Einstein's paper)
>
> or
>
> Φ₂ ≈ - π . [1 + 3/4 α (α₁ + α₂)]
>
> For a number of N entire orbits, it gives
>
> Φ₁ - 2π N ≈ + 3/2 π N α (α₁ + α₂) (Einstein's paper)
>
> Φ₂ + 2π N ≈ - 3/2 π N α (α₁ + α₂)
>
> IMHO, in the case of Φ₁ solution, it appears that precession accumulates going BACK in time, orbit after orbit.
>
> Again IMHO, in the case of Φ₂ solution, it appears that precession accumulates going FORWARD in time, orbit after orbit.
>

Dumbestfuck,

When viewed from the direction of Polaris, precession has the opposite sign as when viwed from the direction of Southern Cross. As such, BOTH solutions are valid.

On Wednesday, November 10, 2021 at 2:32:11 PM UTC-3, Dono. wrote:

<snip>

> > For a number of N entire orbits, it gives
> >
> > Φ₁ - 2π N ≈ + 3/2 π N α (α₁ + α₂) (Einstein's paper)
> >
> > Φ₂ + 2π N ≈ - 3/2 π N α (α₁ + α₂)
> >
> > IMHO, in the case of Φ₁ solution, it appears that precession accumulates going BACK in time, orbit after orbit.
> >
> > Again IMHO, in the case of Φ₂ solution, it appears that precession accumulates going FORWARD in time, orbit after orbit.
> >
> Dumbestfuck,
>
> When viewed from the direction of Polaris, precession has the opposite sign as when viwed from the direction of Southern Cross. As such, BOTH solutions are valid.

Now tell me from where Le Verrier was "viewing" when, in 1847, he discovered the phenomenom and CALCULATED the
extra value of 38" of arc/century?.

Don't come with Newcomb, who got 43"/cy 50 years later (more know-how and data by then).

You can use Google Translate to check it in French sites. In other sites, specially US and UK, Le Verrier's theoretical model is still disputed.

On Wednesday, November 10, 2021 at 7:41:39 AM UTC-8, Richard Hertz wrote:
> I was reasoning about the meanings of the expression
> Φ ≈ ± π . [1 + 3/4 α (α₁ + α₂)]

Only the positive one is consistent with the governing equations.

> I only imagined what was the relationship of accumulations of radians with
> TIME, for (+) or (-) sign, as I imagined how the test massless particle (Mercury)
> swept angles from an arbitrary reference with time evolving (forward or backwards).

It's true that the equations are time-symmetrical, but if you run the situation backward in time, the perihelion advances in the opposite direction, but the direction of Mercury's orbit is also reversed, so the precession is still forward in the sense that the perihelion advances in the direction of the orbit, not retrograde to the orbital direction. There's no ambiguity about this.

On Wednesday, November 10, 2021 at 4:39:20 PM UTC-3, Al Coe wrote:

<snip>

> It's true that the equations are time-symmetrical, but if you run the situation backward in time, the perihelion advances in the opposite direction, but the direction of Mercury's orbit is also reversed, so the precession is still forward in the sense that the perihelion advances in the direction of the orbit, not retrograde to the orbital direction. There's no ambiguity about this.

I'd not say time-symmetrical, but time independent. This is the result of eliminating time (I mean the effect of dt) in the calculation of
the orbit of Mercury. The only presence of time is in the amount of seconds adopted as parameter for the period of the orbit.

If an equation r(Φ,t) could be developed, it would not conform the equation of a fixed ellipse whose perihelion recedes, but another
much more complex orbital path: an spiral. But, as I understand for what I read, this requires an extraordinary complex mathematical
theory, which is considered not necessary for practical applications.

If you read Gerber's paper, which was considerably respected in the years that followed his publication in 1898, you could read right
there in the text, how conscious was Gerber about this particular issue. He wrote about infinitesimal changes in orbital path during
infinitesimal changes in time (dr/dt), but he didn't go any further than write about the actual behavior.

For how the precession has been managed for more than 175 years, since Le Verrier, you could describe what slowly happens in this way:

- Take a perfect ellipse that satisfy its geometrical description: in algebra, an immutable form described by (x/a)²+(y/b)² = 1; select
the point considered as perihelion and make it rotate around one focci as time passes; IMHO, what results is a circular orbit of such
perihelion along time.

The above is a simplification, in order to conserve the geometry of an ellipse while time flows.

Now, due to the fact that the entire process of orbiting has been simplified to such extreme, now you can (from a given time to and
a coordinate pair (x,y) of the plane at which the orbit is drawn), that you could describe the motion of the perihelion forward and
backward in time, if your unit of time is the orbital period To (which is considered perennial).

So, with the above and within a limited number of centuries (say 1,000), you could calculate the position of the perihelion in instances
of time ± N.To, where N is an integer.

This is my understanding, based on that such small scale of time (± 0.1 million years) is considered in astronomy as a blip in the age
of the solar system, and that every parameter of different orbits is considered as constant.

I don't believe, of course, that such scale could be extended arbitrarily, because strange things happened in scales of 1 million years
steps, which I'm sure that we'll never know (changes in avg. radius, periods, inclinations of orbital planes, etc.).

On Wednesday, November 10, 2021 at 10:01:45 AM UTC-8, Richard Hertz wrote:
> On Wednesday, November 10, 2021 at 2:32:11 PM UTC-3, Dono. wrote:
>
> <snip>
> > > For a number of N entire orbits, it gives
> > >
> > > Φ₁ - 2π N ≈ + 3/2 π N α (α₁ + α₂) (Einstein's paper)
> > >
> > > Φ₂ + 2π N ≈ - 3/2 π N α (α₁ + α₂)
> > >
> > > IMHO, in the case of Φ₁ solution, it appears that precession accumulates going BACK in time, orbit after orbit.
> > >
> > > Again IMHO, in the case of Φ₂ solution, it appears that precession accumulates going FORWARD in time, orbit after orbit.
> > >
> > Dumbestfuck,
> >
> > When viewed from the direction of Polaris, precession has the opposite sign as when viwed from the direction of Southern Cross. As such, BOTH solutions are valid.
> Now tell me from where Le Verrier was "viewing" when, in 1847, he discovered the phenomenom and CALCULATED the
> extra value of 38" of arc/century?.

Dumbestfuck,

LeVerrier used MEASURED data coming from LaPlace and Bradley. You have been told this many times before, yet you continue to eat shit. Bon appetit!

On Wednesday, November 10, 2021 at 6:54:58 PM UTC-3, Dono. wrote:

<snip>
..
> > Now tell me from where Le Verrier was "viewing" when, in 1847, he discovered the phenomenom and CALCULATED the
> > extra value of 38" of arc/century?.
> Dumbestfuck,
>
> LeVerrier used MEASURED data coming from LaPlace and Bradley. You have been told this many times before, yet you continue to eat shit. Bon appetit!

Lame, pathetic attempt of a failed shill&troll to deviate attention from its ignorance and derail the thread. Ignorant pretender who don't
know a bit about astronomy and avoided (stupidly) to answer my simple question.

If the reptilian lifeform is so upset until now, I don't want to imagine the hate and fury after my next posts on this topic.

I'm messing with the Nov. 18, 1915 paper from Einstein, Dono, going from the end up to the beggining, checking the fudges and hacks.

So, retort, kick, scream and spit full of hate because I'm doing such thing on such "sacred paper", studied by million of scholars!

You know what? Go fuck yourself!

And wait for it, as I'm just warming up.

On Wednesday, November 10, 2021 at 4:10:44 PM UTC-8, despicable kapo Richard Hertz wrote:
> On Wednesday, November 10, 2021 at 6:54:58 PM UTC-3, Dono. wrote:
>
> <snip>
> .
> > > Now tell me from where Le Verrier was "viewing" when, in 1847, he discovered the phenomenom and CALCULATED the
> > > extra value of 38" of arc/century?.
> > Dumbestfuck,
> >
> > LeVerrier used MEASURED data coming from LaPlace and Bradley. You have been told this many times before, yet you continue to eat shit. Bon appetit!
> I'm messing with my dick
>
Yep, you keep jerking off.

On Tuesday, November 9, 2021 at 3:13:35 PM UTC-8, Richard Hertz wrote:
> On his Nov.18, 1915, Einstein presented his proof about GR correctly predicting
> an advance of 43"arc/century on Mercury's perihelion.
>
> Notwithstanding his hacks and fudges in the first part of the paper, which
> deserves a separate thread for the fanatic mathematicians here,

Einstein didn't do any "fudges".

> this problem
> only requires an elementary knowledge of integrals and a table at hand.

And not making mistakes as you do.

--
Jan

On Wednesday, November 10, 2021 at 1:15:02 PM UTC-8, Richard Hertz wrote:
> I'd not say time-symmetrical, but time independent.

The solution gives the precise geodesic path as a function of time, so it is manifestly not independent of time, but it does have the time translation symmetry along the Schwarzschild time coordinate, which is a Killing field..

> If an equation r(Φ,t) could be developed...

Again, the geodesic equations for the path are given explicitly, so we can compute t, r, and Φ as functions of the proper time.

> it would not conform the equation of a fixed ellipse whose perihelion recedes...

The axis of the perihelion doesn't "recede", it precesses, meaning it advances in the orbital direction. This geodesic for a spherically symmetrical field conforms exactly to an ellipse with precession.

> But, as I understand for what I read, this requires an extraordinary complex mathematical
> theory...

It's fairly straight-forward and simple for the spherically symmetrical case. It becomes complex (but still straight-forward) only when we consider more general fields. The equations in such general cases are ordinarily solved numerically (as they are in Newtonian physics).

On Wednesday, November 10, 2021 at 2:36:40 AM UTC-3, Richard Hertz wrote:

<snip>

WELL, THIS IS INTERESTING. I FOUND ANOTHER SOLUTION FOR N(x) WITH A DIFFERENT INTEGRAL DERIVATION:

> K = [1 + α/2 (α₁ + α₂)]
> P(x) = √ [ - (x - α₁) (x - α₂)] = √ [- x2 + (α₁+α₂) x - α₁ α₂ ] = √[- a x² + b x + c]

P(α₁) = P(α₂) = 0

> N(x) = ∫ dx/P(x)
> M(x) = ∫ x/P(x) dx
> Φ = K [N(x) + α/2 M(x)] , between α₁ and α₂
Solving N(x)

N(x) = ∫ dx/P(x) = ∫ dx/√[a x² + b x + c] = 1/√a ln [(2ax+b) /√a + 2 P(x)] = 1/√a ln [(2ax+b) /√a ]

N(α₂) - N(α₁) = 1/√-1 {ln [(-2α₂ + α₁+α₂) /√-2 ] - ln [(-2α₁+ α₁+α₂) /√-1 ]} = 1/√-1 {ln [(-α₂ + α₁) /√-2 ] - ln [(-α₁+ α₂) /√-1 ]}

N(α₂) - N(α₁) = 1/√-1 ln { [(-α₂ + α₁) /√-2 ]/[(-α₁+ α₂) /√-1 ]} = 1/√-1 ln (-1) = +π

<snip>

As for M(x) the integral uses ln X as well, both terms N(x) and M(x) are derived with the same method.

This is a change for N(x) + α/2 M(x), which now gives ] = + π [1 - α/4 (α₁ + α₂)]

Φ = + π [1 + α/2 (α₁ + α₂)] . [1 - α/4 (α₁ + α₂)] = + π . [1 + 1/4 α (α₁ + α₂) - 1/8 α² (α₁ + α₂)² ]

Φ ≈ + π . [1 + 1/4 α (α₁ + α₂)]

which differs from Φ ≈ + π . [1 + 3/4 α (α₁ + α₂)] derived by Einstein just one step before his Eq.. (12): Φ = + π . [1 + 3α/[a (1 - e²)]]

I respected the rules for integrals between limits (which get rid of any constant of integration) and used ln x, which are not ambiguous
as the previous use of arcsine.

So, what to think about this?. Did Einstein used a different table for integrals?

Now, it's more evident that the play with ± signs in the two components don't work, because it won't change the fact that N(x) and M(x)
have opposite signs, derived from the same type of solutions for elliptic integrals, using ln (x).

In combinations of the equivalence ± (+A) ± (-B), you have:

1) +A - B
2) +A + B
3) -A - B
4) -A + B

Which is the scientific reason by which I'm FORCED to accept ONLY combination (2) because it's written in an sloppy 105 years old paper?

And I even question any explanation for the need of combinations due to a new nature (that I didn't know) about integrals with ± chances
in front of them. One thing is a generic integral (indefinite), and other thing is one with specific limits of integration.

I quote:

* Indefinite Integrals: It is an integral of a function when there is no limit for integration. It contains an arbitrary constant.
* Definite Integrals: An integral of a function with limits of integration. There are two values as the limits for the interval of integration.

My conclusion, so far, is that the solution of the integral in the paper is wrong, is hacked to get (12) and it invalidates the entire paper.

P.S.: I'm studying the 1915 paper from the end to the principle in detail, and I account so far a big number of fudges and hacks.
I know that he and his partners wrote such paper in a hurry, in less than a week, and that he had experience in the subject (wrong
experience) for his work with Besso in 1913-1914, but I'm convinced that the Nov.18, 1915 is wrong and was cooked.

Never again, nor Einstein neither any other scientist which work I know by now, tried to develop an exact analytical formula (Gerber)
since then. You can read 100 different papers, with Schwarzschild's metric or whatever, but none of them provides an exact formula
like Einstein produced for such presentation before the Prussian Academy of Science (Mie, Klein, ...).

I'm using Weinstein's paper

https://www.researchgate.net/publication/221659836_From_the_Berlin_Entwurf_Field_Equations_to_the_Einstein_Tensor_IINovember_1915_until_March_1916/link/547d6f550cf27ed978623a54/download

as a good guide through the entire 1915, as well as my own judgement.

To settle this, what I'll do is to recalculate Eq. (11), after reverting the replacement x = 1/r, to integrate in r domain. I'll use:

(Einstein eq. 11): (dx/dΦ)² = 2A/B² + α/B² x −x² + αx³

as [d/dΦ (1/r)]² = 2A/B² + 2GM/c²B² . 1/r −1/r² + 2GM/c²B² . 1/r³

and will post the numerical results, once I got them.

On Wednesday, November 10, 2021 at 11:37:40 PM UTC-8, cretin Richard Hertz lied:

> Never again, nor Einstein neither any other scientist which work I know by now, tried to develop an exact analytical formula (Gerber)
> since then. You can read 100 different papers, with Schwarzschild's metric or whatever, but none of them provides an exact formula
> like Einstein produced for such presentation before the Prussian Academy of Science (Mie, Klein, ...).

Utter crank,

Symbolic (exact analytic) solutions can be found in ANY book on general relativity. Crack open a book, utter crank . You can try Missner, Thorne, Wheeler "Gravitation".

Einstein must have done something right
if the atheists had to judge him...

Mitchell Raemsch