Discussion:
Proofs about Einstein fudging and cooking the paper that made him famous worldwide
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Richard Hertz
2021-11-20 18:27:45 UTC
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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)
Dono.
2021-11-20 18:42:56 UTC
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Snip Richard Hertz' imbecilities<
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Richard Hertz
2021-11-20 18:53:17 UTC
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On Saturday, November 20, 2021 at 3:42:57 PM UTC-3, Dono. wrote:

<snip>
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong or shut the fuck up, liar cretin. I NEVER did this before. It's a work that took me 20 hours to complete,
with an study and representations in the computer of BOTH EQUATIONS for the orbital motion of planets.

I don't make crap when I become serious on a subject. And I'm not prone to make mistakes, if I analyze things many times,
like in this case. I'm not going to expose myself so naively, fucking retarded!

Now, instead of talking, do the walking or SHUT UP!
Dono.
2021-11-20 18:58:02 UTC
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Post by Richard Hertz
<snip>
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong
You already calculated the integral incorrectly a few days ago, lying piece of shit.
Dono.
2021-11-20 19:24:35 UTC
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Post by Richard Hertz
<snip>
Post by Dono.
Post by Richard Hertz
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong
You already calculated the integral incorrectly a few days ago, lying piece of shit.
What I did two weeks ago was to use TWO different sources for the solution of both integrals.
.....and you made elementary mistakes on BOTH of them. You are repaeting the same mistakes, Thank you for the continued entertainment.
Richard Hertz
2021-11-20 20:28:05 UTC
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On Saturday, November 20, 2021 at 4:24:37 PM UTC-3, Dono. wrote:

<snip>
Post by Dono.
Post by Richard Hertz
Post by Dono.
Post by Richard Hertz
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong
You already calculated the integral incorrectly a few days ago, lying piece of shit.
What I did two weeks ago was to use TWO different sources for the solution of both integrals.
.....and you made elementary mistakes on BOTH of them. You are repaeting the same mistakes, Thank you for the continued entertainment.
I'm truly convinced that a fanatic relativist like you, also an ignorant imbecile depending on a SW to do math,
can't deal with the FACTS that I specifically detailed.

If you don't like the integrals part, then try with the other fudge:

How he used 1 = 1 + α/2 (α₁ + α₂) so he could FUDGE, COOK the final formula.

Just explain this or justifiy this FRAUD, fudge, hack our of many others in the paper, that I didn't tell.

But of course, you are a cretin FANATIC and will never concede anything against your pagan religion of science.

Because you are a cheater like him, and you proved it here infinite times.

How come you are not going to endorse a cheater that TRIPLED HIS RESULT.
Dono.
2021-11-20 20:34:59 UTC
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Post by Richard Hertz
<snip>
Post by Richard Hertz
Post by Dono.
Post by Richard Hertz
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong
You already calculated the integral incorrectly a few days ago, lying piece of shit.
What I did two weeks ago was to use TWO different sources for the solution of both integrals.
.....and you made elementary mistakes on BOTH of them. You are repeating the same mistakes, Thank you for the continued entertainment.
How he used 1 = 1 + α/2 (α₁ + α₂) so he could FUDGE, COOK the final formula.
Cretinoid


The same exact approximation can be seen on page 243 in the book that I just recommended (Rindler). Keep up the entertainment, clown.
JanPB
2021-11-28 20:27:04 UTC
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Post by Richard Hertz
<snip>
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong or shut the fuck up, liar cretin.
One can never prove to an ignoramus that he is wrong. This has been noted
for centuries, Newton and Goethe commented on this phenomenon.
Post by Richard Hertz
I NEVER did this before. It's a work that took me 20 hours to complete,
with an study and representations in the computer of BOTH EQUATIONS for the orbital motion of planets.
I don't make crap when I become serious on a subject.
You do. Sorry but you simply don't have what it takes to work in this.

--
Jan
Richard Hertz
2021-11-28 21:03:39 UTC
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<snip>
Post by JanPB
Post by Richard Hertz
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong or shut the fuck up, liar cretin.
One can never prove to an ignoramus that he is wrong. This has been noted
for centuries, Newton and Goethe commented on this phenomenon.
Show me and others at this forum where there is any mathematical error, genius. Honor your BsC in math, if it's true.

If not, you also shut the fuck up, ignorant. This is not post-post graduated math. It's only High School math, so it should
be very easy for you to find any error. Isn't it, imbecile?

Or it's that you CAN'T FIND ANY?
Post by JanPB
Post by Richard Hertz
I NEVER did this before. It's a work that took me 20 hours to complete,
with an study and representations in the computer of BOTH EQUATIONS for the orbital motion of planets.
I don't make crap when I become serious on a subject.
You do. Sorry but you simply don't have what it takes to work in this.
You don't have any idea of what you're talking about. It looks like the reply of an angry and stupid child.

DO THE MATH AND PROVE ME WRONG, PEACOCK. It's your chance to show me as an idiot and you as a winner.
Chason Aceta
2021-11-28 21:30:33 UTC
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Post by Richard Hertz
Post by JanPB
You do. Sorry but you simply don't have what it takes to work in this.
You don't have any idea of what you're talking about. It looks like the
reply of an angry and stupid child.
DO THE MATH AND PROVE ME WRONG, PEACOCK. It's your chance to show me as
an idiot and you as a winner.
his math is like his moon landing, fake.
Richard Hertz
2021-11-28 22:09:15 UTC
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On Sunday, November 28, 2021 at 6:30:39 PM UTC-3, Chason Aceta wrote:

<snip>
Post by Chason Aceta
Post by Richard Hertz
Post by JanPB
You do. Sorry but you simply don't have what it takes to work in this.
You don't have any idea of what you're talking about. It looks like the
reply of an angry and stupid child.
DO THE MATH AND PROVE ME WRONG, PEACOCK. It's your chance to show me as
an idiot and you as a winner.
his math is like his moon landing, fake.
Do you know what's really sad? That JabPB degree was a gift from USA to Poland for helping in the fall of the USSR.

After that event, USA was determined to conquer minds, souls and bodies of everyone living at Eastern Europe and, in particular,
Russia, Ukraine and Poland (geopolitical game plus NATO interests).

So, during all the '90s and among many other things, a line of grants were established to poor students at those countries, in order
to came to USA and get a degree at several first class universities, so they could go back to Europe highly brainwashed with the
American Way of Life. Of course, US gov. mandated that bright people be offered a job to stay at US. The rest was sent home.

JanPB is the kind of USA brainwashed type that was sent back (not good enough) but, as Indians while UK ruling there, he was converted.

Back in Poland with a BsC on Math (C+ grade), he only could find a job as a PIC/DSP programmer. But it wasn't to throw down his dream
to belong to the western community of mathematician, which was very though. So he did choose Applied Math, and selected GR, where
the competition was scarce and the complexity high enough to keep busy someone for years. But this was his pet project.

Once JanPB gained confidence with GR, he started to show off his knowledge at different forums by early 2000's. But he was repeteadly
kicked off for being a presumptuous amateur and, finally, he arrived here and claimed a position of "GR top dog" at first, and of
"any shit involving math Top Dog", which is his current state of mind.

And what's sad is that he need to downplay ANY OTHER that touch the domains of his own realm, in a fierce way, being highly derogatory.

He needs that, in his fucking and tortured mind, in order to avoid facing that HE FAILED AS A PROFESSIONAL.

So he come here to VOMIT his hate, disguised by comments like:

* You don't even understand what are you writing about.
* Why do you mess with things that are beyond your comprehension?
* You'll never understand this.
* Why do you persist doing things beyond your capabilities?
* Etc, etc.

He's not bad and dark as Bodkin or Dono, or a fucking cretin like Dork (Dirk). He's a child-like naive person who have to show his teeth
in self-defence, because he lives in fear like a chihuahua.

The chihuahua doesn't mean to be a bad boy. It respond aggressively just to be left alone, undisturbed.

JanPB is like a traumatized chihuahua. If you pet him and feed his ego, he´ll respond nicely, because he's a good boy, after all.

But, is a chihuahua equivalent of a mathematician like JanPB a TOP DOG?

Please, I can't even think about it. I'm laughing alone here, so badly that hurts. LOL!
Chason Aceta
2021-11-28 22:24:06 UTC
Reply
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Post by Richard Hertz
<snip>
Post by Chason Aceta
Post by Richard Hertz
Post by JanPB
You do. Sorry but you simply don't have what it takes to work in this.
You don't have any idea of what you're talking about. It looks like
the reply of an angry and stupid child.
DO THE MATH AND PROVE ME WRONG, PEACOCK. It's your chance to show me
as an idiot and you as a winner.
his math is like his moon landing, fake.
Do you know what's really sad? That JabPB degree was a gift from USA to
Poland for helping in the fall of the USSR.
wow, I hate these traitors. Good post.
Post by Richard Hertz
After that event, USA was determined to conquer minds, souls and bodies
of everyone living at Eastern Europe and, in particular, Russia, Ukraine
and Poland (geopolitical game plus NATO interests).
So, during all the '90s and among many other things, a line of grants
were established to poor students at those countries, in order to came
to USA and get a degree at several first class universities, so they
could go back to Europe highly brainwashed with the American Way of
Life. Of course, US gov. mandated that bright people be offered a job to
stay at US. The rest was sent home.
JanPB is the kind of USA brainwashed type that was sent back (not good
enough) but, as Indians while UK ruling there, he was converted.
Back in Poland with a BsC on Math (C+ grade), he only could find a job
as a PIC/DSP programmer. But it wasn't to throw down his dream to belong
to the western community of mathematician, which was very though. So he
did choose Applied Math, and selected GR, where the competition was
scarce and the complexity high enough to keep busy someone for years.
But this was his pet project.
Once JanPB gained confidence with GR, he started to show off his
knowledge at different forums by early 2000's. But he was repeteadly
kicked off for being a presumptuous amateur and, finally, he arrived
here and claimed a position of "GR top dog" at first, and of "any shit
involving math Top Dog", which is his current state of mind.
And what's sad is that he need to downplay ANY OTHER that touch the
domains of his own realm, in a fierce way, being highly derogatory.
He needs that, in his fucking and tortured mind, in order to avoid
facing that HE FAILED AS A PROFESSIONAL.
* You don't even understand what are you writing about.
* Why do you mess with things that are beyond your comprehension?
* You'll never understand this.
* Why do you persist doing things beyond your capabilities?
* Etc, etc.
He's not bad and dark as Bodkin or Dono, or a fucking cretin like Dork
(Dirk). He's a child-like naive person who have to show his teeth in
self-defence, because he lives in fear like a chihuahua.
The chihuahua doesn't mean to be a bad boy. It respond aggressively just
to be left alone, undisturbed.
JanPB is like a traumatized chihuahua. If you pet him and feed his ego,
he´ll respond nicely, because he's a good boy, after all.
But, is a chihuahua equivalent of a mathematician like JanPB a TOP DOG?
Please, I can't even think about it. I'm laughing alone here, so badly that hurts. LOL!
Paul Alsing
2021-11-29 02:45:37 UTC
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Post by Richard Hertz
Do you know what's really sad? That JabPB degree was a gift from USA to Poland for helping in the fall of the USSR.
Do you actually think that making shit up as you go along elevates anyone's opinion of you? You are even more delusional than I thought! Jan has forgotten more physics than you will ever know!

You should just slink off into the sunset before the embarrassment you suffer becomes fatal... or even worse!
Richard Hertz
2021-11-29 03:50:56 UTC
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On Sunday, November 28, 2021 at 11:45:38 PM UTC-3, Paul Alsing wrote:

<snip>
Post by Paul Alsing
Post by Richard Hertz
Do you know what's really sad? That JabPB degree was a gift from USA to Poland for helping in the fall of the USSR.
Do you actually think that making shit up as you go along elevates anyone's opinion of you? You are even more delusional than I thought! Jan has forgotten more physics than you will ever know!
You should just slink off into the sunset before the embarrassment you suffer becomes fatal... or even worse!
1) You repeat yourself too often the shit "forgotten more physics than you will ever know". You applied it to you few days ago. Senility?

2) Ask JanPB IF and WHEN studied at USA (I don't want to tell where). Ask him if he tried hard to enhance his English there. Also
ask him when and why returned to Poland. Finally, you can fact check the old post of Jan Piotr Bielawski (he doesn't like Piotr).
Finally, ask him WHO got a degree by 2012 and where was he at that moment.

Pedantic idiot, good for nothing.
Paul Alsing
2021-11-29 04:26:29 UTC
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Post by Richard Hertz
<snip>
Post by Paul Alsing
Post by Richard Hertz
Do you know what's really sad? That JabPB degree was a gift from USA to Poland for helping in the fall of the USSR.
Do you actually think that making shit up as you go along elevates anyone's opinion of you? You are even more delusional than I thought! Jan has forgotten more physics than you will ever know!
You should just slink off into the sunset before the embarrassment you suffer becomes fatal... or even worse!
1) You repeat yourself too often the shit "forgotten more physics than you will ever know". You applied it to you few days ago. Senility?
Many people here have forgotten more physics than you will ever know... so it is not 'shit', it is a fact! You may hear it many times more before it sinks into your shriveled brain...

You have probably seen this before, too, and will probably see it again since it is so apropos in your case...

"In an anti-intellectual society, people who know nothing about a complex subject are emboldened to ridicule experts who have spent a lifetime studying it."
- George Kiser
Richard Hertz
2021-11-29 05:15:47 UTC
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On Monday, November 29, 2021 at 1:26:31 AM UTC-3, Paul Alsing wrote:

<snip>
Post by Paul Alsing
"In an anti-intellectual society, people who know nothing about a complex subject are emboldened to ridicule experts who have spent a lifetime studying it."
- George Kiser
Thank you. It serves me perfectly to demonstrate what a pompous and ridicule sentence is, and yet imbeciles quote it.

Complex subject: The Bible, Evolution of Species, Universe, the atom, etc.

Experts who have spent a lifetime studying it:

The Bible: Theologians. Do they understand the existence of God? NO! They give up and adopt a credo
or resign. A life spent FOR NOTHING.

Anthropology: the study of human biological and physiological characteristics and their evolution.
Evolution of Species: Anthropologists. Do they understand how, when and from what the actual human specie evolved? NO.
Worse than with Theologians because, day after day, a new piece of the puzzle appears and makes to fall apart previous
understanding. A life spent FOR NOTHING.

Universe: Cosmologists. Better for you that I don't address this subject, which is sinking into madness and ridiculous fixes
every day. A life spent FOR NOTHING.


GR specialists. I've read about people who spent 30 years of his life studying Ricci-Civita complexities and solutions, having to assign a
physical meaning or discard each that couldn't be explained IN HIS FUCKING MIND!. A life spent FOR NOTHING.

Human psyche: 100+ years of accumulated studies, 2000000000 hours invested at it, and yet it's not known why humans kill humans
in different circumstances, or humans cheat and or lie to other humans. A life spent FOR NOTHING.

I could keep going forever: Cubism, Beethoven music, molecular cuisine, etc.

So, the IDIOT you quote blindly (I'm sure that you bought his thought as you bought relativity) is just a presumptuous ass. Could be Bodkin,
the Supreme Thinker at this forum.

You are the inflexible with a fossilized mind and full of resentment upon free thinkers. You are the prisoner of your own mind.

But don't worry. It's not your blame. Nature just designed you in this way, as it does with EVERY single human. The problem is the
Hubris mental disease that affects you and many other assholes here.
Paul Alsing
2021-11-29 05:30:16 UTC
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Post by Richard Hertz
You are the inflexible with a fossilized mind and full of resentment upon free thinkers. You are the prisoner of your own mind.
“It pays to keep an open mind, but not so open your brains fall out.”

― Author unclear...
Michael Moroney
2021-11-29 05:42:18 UTC
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Post by Richard Hertz
full of resentment upon free thinkers.
"Free thinkers". I think that should be included in my kook vocabulary list.
Odd Bodkin
2021-11-29 14:21:42 UTC
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Post by Richard Hertz
<snip>
Post by Paul Alsing
"In an anti-intellectual society, people who know nothing about a
complex subject are emboldened to ridicule experts who have spent a
lifetime studying it."
- George Kiser
Thank you. It serves me perfectly to demonstrate what a pompous and
ridicule sentence is, and yet imbeciles quote it.
Complex subject: The Bible, Evolution of Species, Universe, the atom, etc.
The Bible: Theologians. Do they understand the existence of God? NO!
They give up and adopt a credo
or resign. A life spent FOR NOTHING.
Anthropology: the study of human biological and physiological
characteristics and their evolution.
Evolution of Species: Anthropologists. Do they understand how, when and
from what the actual human specie evolved? NO.
Worse than with Theologians because, day after day, a new piece of the
puzzle appears and makes to fall apart previous
understanding. A life spent FOR NOTHING.
Universe: Cosmologists. Better for you that I don't address this subject,
which is sinking into madness and ridiculous fixes
every day. A life spent FOR NOTHING.
GR specialists. I've read about people who spent 30 years of his life
studying Ricci-Civita complexities and solutions, having to assign a
physical meaning or discard each that couldn't be explained IN HIS
FUCKING MIND!. A life spent FOR NOTHING.
Human psyche: 100+ years of accumulated studies, 2000000000 hours
invested at it, and yet it's not known why humans kill humans
in different circumstances, or humans cheat and or lie to other humans. A
life spent FOR NOTHING.
I could keep going forever: Cubism, Beethoven music, molecular cuisine, etc.
So, the IDIOT you quote blindly (I'm sure that you bought his thought as
you bought relativity) is just a presumptuous ass. Could be Bodkin,
the Supreme Thinker at this forum.
You are the inflexible with a fossilized mind and full of resentment upon
free thinkers. You are the prisoner of your own mind.
But don't worry. It's not your blame. Nature just designed you in this
way, as it does with EVERY single human. The problem is the
Hubris mental disease that affects you and many other assholes here.
More in a long train of “If it’s not engineering, it’s a waste of time.”
--
Odd Bodkin -- maker of fine toys, tools, tables
JanPB
2021-11-29 04:49:26 UTC
Reply
Permalink
Post by Richard Hertz
<snip>
Post by JanPB
Post by Richard Hertz
Post by Dono.
You tried this before only to prove that you are unable to calculate a simple integral correctly.
Prove me wrong or shut the fuck up, liar cretin.
One can never prove to an ignoramus that he is wrong. This has been noted
for centuries, Newton and Goethe commented on this phenomenon.
Show me and others at this forum where there is any mathematical error, genius.
As I said, one cannot explain to an arrogant ignoramus that he is wrong, ever. This
is well-known and I won't belabor the point any further. If you're interested, you can
find a lot of resources on the Internet regarding this phenomenon.
Post by Richard Hertz
Honor your BsC in math, if it's true.
If not, you also shut the fuck up, ignorant.
No, I will keep pointing out your made up idiocies regarding Einstein whenever
I am in the mood. It's a very easy thing to do as your output here is both
voluminous and virtually fully incorrect. It's amazing an adult (I'm assuming
you are an adult?) would waste his life away on pursuing a completely
nonsensical quest like this.
Post by Richard Hertz
This is not post-post graduated math. It's only High School math, so it should
be very easy for you to find any error. Isn't it, imbecile?
But you don't understand this math, let alone the physics. Your posts are
full of idiocies, errors, and non-sequiturs.
Post by Richard Hertz
Or it's that you CAN'T FIND ANY?
I can find them very easily. They've been pointed out to you many times,
naturally this didn't work since all ignoramuses like you are like that: they never
accept that they can be wrong. It's infantilism that's amazing to see in an adult
(assuming you are one?)
Post by Richard Hertz
Post by JanPB
Post by Richard Hertz
I NEVER did this before. It's a work that took me 20 hours to complete,
with an study and representations in the computer of BOTH EQUATIONS for the orbital motion of planets.
I don't make crap when I become serious on a subject.
You do. Sorry but you simply don't have what it takes to work in this.
You don't have any idea of what you're talking about.
Yes, I do. You'll have to spend a few decades to catch up to me in that
department. There is nothing we can do about it.

--
Jan
Richard Hertz
2021-11-20 19:37:57 UTC
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This topic was addressed once, 15 years ago, and a couple of forum members reached the same
conclusion, except that they used a different approach, analyzing the Lagrangian of orbits.

The whole Einstein paper is dubious and that's why Schwarzschild, who was present at the Prussian
Academy on Nov. 18, 1915, criticized Einstein's approach and proceeded to write his paper about an
exact analytical solution of the problem that Einstein posted in the first part of the paper.

Obviously, being one of three that recommended Einstein for Berlin's professorship, he was not going
to accuse Einstein of plagiarism or fudging, but his letter on Dec. 1915 to Einstein put the things in
perspective, with a careful diplomatic language:

1) I don't know what you did to reach Eq. 11, but here you have A NEW ONE.
Maciej Wozniak
2021-11-21 07:16:27 UTC
Reply
Permalink
Post by Dono.
Post by Richard Hertz
This topic was addressed once, 15 years ago, and a couple of forum members reached the same
conclusion, except that they used a different approach, analyzing the Lagrangian of orbits.
Cretinoid
Einstein eq (13) is exactly the same as the solution obtained by using the Euler-Lagrange equations derived from Schwarzschild metric. It is well known (not by cranks like you) that the Euler-Lagrange equations derived from the Schwarzschild metric are IDENTICAL to the geodesics equations obtained from the EFEs. Therefore, the eq (13) from the Einstein paper is the SAME solution as the one that once can see for example on page 243 on Rindler's book (Relativity, Special, General and Cosmological) whereby \alpha=2m. You being a crank, would never learn that
In the meantime in the real world, however, forbidden by your moronic
religion GPS clocks keep measuring t'=t, just like all serious clocks
always did.
Richard Hertz
2021-11-20 20:36:50 UTC
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On Saturday, November 20, 2021 at 5:18:37 PM UTC-3, Townes Olson wrote:

<snip>
Your comments are not correct at all. First, the extra degree of freedom that confused Schwarzschild was nothing but the free choice of origin for the radial coordinate, which has no physical significance. Second, Schwarzschild was just pointing out that the essentially unique exact solution of Einstein's field equations for the metric can be used, in place of Einstein's first-order approximation. Third, Schwarzschild was not criticizing the evaluation of the elementary integral (well known to astronomers) used by Einstein, nor was he saying that this well-known integral is ambiguous. And for good reason, because (fourth) the integral is not ambiguous at all, as was explained in a recent thread. There is a simple substitution of variables that makes the integral trivial. You can find this (and the subsequent developments of Droste, etc.) explained in detail in any good book on the foundations of relativity.
Schwarzschild criticized the GR part, not 1st. year calculus.

About the integral, one of the two FRAUDS, forget about it.

Tell me why IS NOT CHEATIN to make 1 = 1 + α/2 (α₁ + α₂)

being that this FUDGE contributes with 67% of the final value.

Go ahead, and justify how he pulled out of his ass the factor α/2 (α₁ + α₂). It's calling FRAUD, cheating.

And another thing: Stop lecturing me, because for me you are an important asshole, and I could put you
down on any physical or mathematical problem.

I'm smarter that you, and you hate it, like the fucking reptilian lifeform.

Deal with it every day, asshole.
Richard Hertz
2021-11-20 23:33:03 UTC
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On Saturday, November 20, 2021 at 7:40:13 PM UTC-3, Townes Olson wrote:

<snip>
Good, so we can dispense with your erroneous claim about the evaluation of the integral being ambiguous.
Post by Richard Hertz
Tell me why IS NOT CHEATIN to make 1 = 1 + α/2 (α₁ + α₂)
The paper does not claim 1 equals 1 + α/2 (α₁ + α₂). It asserts, correctly, that, to the first order, 1/sqrt[1-a(a1+a2)] equals
1 + α/2 (α₁ + α₂). Understand?

Still patronizing, Wilson? You never learn, don't you?

You now, TO SAVE FACE or because you are an IDIOT, pretend to impose to me the rationalization of the cheating done by Einstein?

Stop fucking around, and show some intelligence, please!

In this IDENTITY:

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

where 1 + K =1 + α/2 (α₁ + α₂) or, 1 = 1 + K ≈ 1 + α/2 (α₁ + α₂) and α/2 (α₁ + α₂) = 5.3265E-05

you have the stone face to affirm that α/2 (α₁ + α₂) = 5.3265E-05 OR α/2 (α₁ + α₂) = 0, DEPENDING ON the circumstances.

And, specially, because K contributes with 67% to the final result of 43"/century.

Just write about THAT IDENTITY, and don't INVENT. There are only 6 formulae between the first integral and the end of the paper.

FOCUS on my topic. Don't derail it because you CAN'T or DON'T WANT TO understand the fudge.
Townes Olson
2021-11-21 00:10:33 UTC
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Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + α/2 (α₁ + α₂)) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]
That isn't an *identity*, it is "established with the precision demanded of us", which is to say, to the first significant order of approximation necessary to show the non-Newtonian effects to the precision of observation. If you have trouble seeing how Einstein goes from the left side to the right side of that expression, it's because the high school curriculum in those days included things like the algebraic theory of polynomials that make it obvious, but modern students don't have that background, so it looks mysterious. (Likewise the simple substitution of variables that trivializes the integral is fairly obvious to any mathematically literate person, but looks mysterious to most students today.) Fortunately you can find a detailed explanation of the intermediate steps (which are omitted from Einstein's paper as being obvious to his readers) in any good book on the foundations of relativity.
Dono.
2021-11-21 00:14:10 UTC
Reply
Permalink
Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + α/2 (α₁ + α₂)) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]
That isn't an *identity*, it is "established with the precision demanded of us", which is to say, to the first significant order of approximation necessary to show the non-Newtonian effects to the precision of observation. If you have trouble seeing how Einstein goes from the left side to the right side of that expression, it's because the high school curriculum in those days included things like the algebraic theory of polynomials that make it obvious, but modern students don't have that background, so it looks mysterious. (Likewise the simple substitution of variables that trivializes the integral is fairly obvious to any mathematically literate person, but looks mysterious to most students today.) Fortunately you can find a detailed explanation of the intermediate steps (which are omitted from Einstein's paper as being obvious to his readers) in any good book on the foundations of relativity.
Richard Hertz is an 67 year old fart crank, he's not a young student. He never had the mathematical ability to understand Einstein's paper. He is just parroting the Vankov paper
Richard Hertz
2021-11-21 00:33:02 UTC
Reply
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Post by Dono.
Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + α/2 (α₁ + α₂)) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]
That isn't an *identity*, it is "established with the precision demanded of us", which is to say, to the first significant order of approximation necessary to show the non-Newtonian effects to the precision of observation. If you have trouble seeing how Einstein goes from the left side to the right side of that expression, it's because the high school curriculum in those days included things like the algebraic theory of polynomials that make it obvious, but modern students don't have that background, so it looks mysterious. (Likewise the simple substitution of variables that trivializes the integral is fairly obvious to any mathematically literate person, but looks mysterious to most students today.) Fortunately you can find a detailed explanation of the intermediate steps (which are omitted from Einstein's paper as being obvious to his readers) in any good book on the foundations of relativity.
Richard Hertz is an 67 year old fart crank, he's not a young student. He never had the mathematical ability to understand Einstein's paper. He is just parroting the Vankov paper
Hello, trolling reptilian lifeform! Greetings from Earth.

I quoted Vankov at my convenience. This is another fucking relativist who spent the whole paper to kiss Einstein's ass.

Read it, so you have an einsteingasm.
Richard Hertz
2021-11-21 03:40:57 UTC
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I should have known better, checking who Townes Olson is.

The shill/troll has participated only in 13 threads since he showed up on last September 8th.

He/she is particularly picked by any mention about Mercury, and engaged with Paul Andersen on his last post
about his new work on the advance of perihelion of planets and his downloadable application.

Paul hasn't showed again since his last thread.

What happened to him since his last post 40 days ago? He used to contribute with 7 to 10 posts per month, lately.

Everyone on this thread was a little jealous, and showed it up, criticizing his simulation.

What GR predicts for the perihelion advance of planets
https://groups.google.com/u/1/g/sci.physics.relativity/c/Anb8KZYi2Lk/m/S4KZtBWiAgAJ
Richard Hertz
2021-11-21 00:29:34 UTC
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On Saturday, November 20, 2021 at 9:10:34 PM UTC-3, Townes Olson wrote:

<snip>
Φ = ∫dx/√(2A/B² + α/B² x - x² + α x³) = (1 + α/2 (α₁ + α₂)) ∫ dx/√ [- (x - α₁) (x - α₂) (1 - αx)]
That isn't an *identity*, it is "established with the precision demanded of us", which is to say, to the first significant order of approximation necessary to show the non-Newtonian effects to the precision of observation. If you have trouble seeing how Einstein goes from the left side to the right side of that expression, it's because the high school curriculum in those days included things like the algebraic theory of polynomials that make it obvious, but modern students don't have that background, so it looks mysterious. (Likewise the simple substitution of variables that trivializes the integral is fairly obvious to any mathematically literate person, but looks mysterious to most students today.) Fortunately you can find a detailed explanation of the intermediate steps (which are omitted from Einstein's paper as being obvious to his readers) in any good book on the foundations of relativity.
Poor Olson. You are so full of shit and yet downplay people around.

Your polite cockiness aligned with your patronizing play make you the pathetic shadow of an "anal retentive" high school teacher.

I'm sorry for you, really. I can't believe how many sophist comments are you pulling out just to be on the other side.

We have a personal problem, you and me. You don't like me and it's reciprocal.

Making an elementary cost/benefit analysis, it turns out that you don't worth one more minute of my attention. You are a clone of JanPB.

In Argentina, we eat alive assholes like you: at the job, in the school/college, in social meetings. We just make them cry and run.

We don't like pretentious idiots that find delight in make corrections to other persons just to show what? More intelligence?

Quite the opposite of an intelligent behavior is yours. I like intelligence + flexibility of thought + prudence. You seem to have few or none.

Lucky me that you're not a grammar teacher, so you don't fuck with my poor english. We should debate in chinese, in the next life.

For now, thanks for showing some interest on this thread. You and the other retarded, Dono, were the only ones.
Richard Hertz
2021-11-20 21:46:40 UTC
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https://en.wikipedia.org/wiki/Sch%C3%B6n_scandal

The Schön scandal concerns German physicist Jan Hendrik Schön (born August 1970 in Verden an der Aller, Lower Saxony, Germany) who briefly rose to prominence after a series of apparent breakthroughs with semiconductors that were later discovered to be fraudulent.[1] Before he was exposed, Schön had received the Otto-Klung-Weberbank Prize for Physics and the Braunschweig Prize in 2001, as well as the Outstanding Young Investigator Award of the Materials Research Society in 2002, both of which were later rescinded.

The scandal provoked discussion in the scientific community about the degree of responsibility of coauthors and reviewers of scientific articles. The debate centered on whether peer review, traditionally designed to find errors and determine relevance and originality of articles, should also be required to detect deliberate fraud.

Julia Hsu and Lynn Loo originally noticed issues with Schön's paper describing the assembly of molecular transistors whilst attempting to patent research on lithography, realising by accident that Schön had DUPLICATED figures.

********************

The physicist Jan Schön was caught duplicating figures in one paper and lost everything.

Einstein, on his Nov. 18, 1915, TRIPLICATED ONE FIGURE, DUPLICATED OTHER in the same paper and become
the icon of science.

Corruption + fraud, and managed to get free of charges?

My, my, how times are changed. Einstein should have represented powerful interests of the establishment to get unpunished.

Tell me that everyone is equal under the law. Some are much more equal than others.

Nobody in his sane mind can justify a double fudging done to FIT the formula of a dead physicist.
Not Gerber or von Soldner, not even Poincaré or the poor Bose, etc.

To be a crook and get clean, you need powerful friends, like it happens today in finance.






And regarding elliptic integrals, if one pair of solutions provide the required answer to a problem, and poses no doubts: WHY
should one select OTHER solution that get what OTHERS are looking for?

The solutions using natural logaritms ARE PERFECT, no doubt about it.

F(x) = ax² + bx + c

∫ dx /√F(x) = (1/√a) ln [(2ax+b)/√a + 2 F(x)] + C

∫ x √F(x) dx = 1/a √F(x) – b/(2a^3/2) {ln [(2a x +b)/√a ] +2 F(x) } + C

What is SO HARD to understand or SO DIFFICULT to accept? Only that these integrals prove, without any doubt, Einstein WRONG.

And that is the problem: proving Einstein WRONG is forbidden, due to 100 years of worship and advantages saying NO, NO, NO!
Richard Hertz
2021-11-20 21:49:42 UTC
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Permalink
Post by Richard Hertz
https://en.wikipedia.org/wiki/Sch%C3%B6n_scandal
The Schön scandal concerns German physicist Jan Hendrik Schön (born August 1970 in Verden an der Aller, Lower Saxony, Germany) who briefly rose to prominence after a series of apparent breakthroughs with semiconductors that were later discovered to be fraudulent.[1] Before he was exposed, Schön had received the Otto-Klung-Weberbank Prize for Physics and the Braunschweig Prize in 2001, as well as the Outstanding Young Investigator Award of the Materials Research Society in 2002, both of which were later rescinded.
The scandal provoked discussion in the scientific community about the degree of responsibility of coauthors and reviewers of scientific articles. The debate centered on whether peer review, traditionally designed to find errors and determine relevance and originality of articles, should also be required to detect deliberate fraud.
Julia Hsu and Lynn Loo originally noticed issues with Schön's paper describing the assembly of molecular transistors whilst attempting to patent research on lithography, realising by accident that Schön had DUPLICATED figures.
********************
The physicist Jan Schön was caught duplicating figures in one paper and lost everything.
Einstein, on his Nov. 18, 1915, TRIPLICATED ONE FIGURE, DUPLICATED OTHER in the same paper and become
the icon of science.
Corruption + fraud, and managed to get free of charges?
My, my, how times are changed. Einstein should have represented powerful interests of the establishment to get unpunished.
Tell me that everyone is equal under the law. Some are much more equal than others.
Nobody in his sane mind can justify a double fudging done to FIT the formula of a dead physicist.
Not Gerber or von Soldner, not even Poincaré or the poor Bose, etc.
To be a crook and get clean, you need powerful friends, like it happens today in finance.
And regarding elliptic integrals, if one pair of solutions provide the required answer to a problem, and poses no doubts: WHY
should one select OTHER solution that get what OTHERS are looking for?
The solutions using natural logaritms ARE PERFECT, no doubt about it.
F(x) = ax² + bx + c
∫ dx /√F(x) = (1/√a) ln [(2ax+b)/√a + 2 F(x)] + C
∫ x √F(x) dx = 1/a √F(x) – b/(2a^3/2) {ln [(2a x +b)/√a ] +2 F(x) } + C
What is SO HARD to understand or SO DIFFICULT to accept? Only that these integrals prove, without any doubt, Einstein WRONG.
And that is the problem: proving Einstein WRONG is forbidden, due to 100 years of worship and advantages saying NO, NO, NO!
Typo. Should say ∫ x/√F(x) dx and not ∫ x √F(x) dx
Dono.
2021-11-20 23:25:06 UTC
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What the FUCK has to be with this thread HIS almost newtonian demonstration (11.51) with this discussion?
Rindler's demonstration is GR, not Newtonian , cretinoid.
His Δ = 2π/(1- 3 m²/h²) ≈ 6πm/[a (1 – e²)] comes from a completely different equation, with his "wiggling" and use of "u".
Comes from the EFE geodesics, utter crank. Same equations used by Einstein in his 1915 paper but you are too cretin to understand.
Keep up the entertainment, old fart clown.
Dono.
2021-11-21 00:02:27 UTC
Reply
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Post by Dono.
What the FUCK has to be with this thread HIS almost newtonian demonstration (11.51) with this discussion?
Rindler's demonstration is GR, not Newtonian , cretinoid.
His Δ = 2π/(1- 3 m²/h²) ≈ 6πm/[a (1 – e²)] comes from a completely different equation, with his "wiggling" and use of "u".
Comes from the EFE geodesics, utter crank. Same equations used by Einstein in his 1915 paper but you are too cretin to understand.
Keep up the entertainment, old fart clown.
It comes from (dx/dφ)² = 2A/B² + α/B² x −x² + αx³ , LIAR!
(Equation 11) (dx/dφ)² = 2A/B² + α/B² x −x² + αx³
"we are led without approximation to the following differential equation:"
These are the geodesics, crank

Keep it up, dumbestfuck!
Richard Hertz
2021-11-21 00:04:38 UTC
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The liar and deceiver knew when he fucked it up beyond any explanation or correction.
With E=mc2, at least he tried six more times to prove it, before giving up in 1942.


Einstein’s Paper: “Explanation of the Perihelion Motion of Mercury from General Relativity Theory”
Anatoli Andrei Vankov
IPPE, Obninsk, Russia; Bethany College, KS, USA; ***@hotmail.com


2.1.1 Historical remarks

Einstein’s paper devoted to the GR prediction of Mercury’s perihelion advance,
Doc.24 (see Notes), is the only one among his publications that contains
the explanation of the GR effect. In his following paper The Foundations
of the General Theory of Relativity, 1916, Doc.30, Einstein presents
his new (he called it “correct”) calculation of the bending of light

while the Mercury perihelion is only mentioned by referring it as in Doc.24, along
with Schwarzshild’s work on “the exact solution”.

Since then, to our knowledge, he never returned to the methodology of the GR perihelion advance problem.
Richard Hertz
2021-11-21 06:42:17 UTC
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Permalink
This is how cheating is rationalized and applied in science?

It needs a second definition.

This is the first one, "legalized" and widely used in science. It talks about HIDDEN VALUES, not formulas.
----------------------------------------------------------------------------------------------------------------------------------
https://en.wikipedia.org/wiki/Fudge_factor

Fudge factor

A fudge factor is an ad hoc quantity or element introduced into a calculation, formula or model in order to make it fit observations or expectations. Also known as a "Correction Coefficient" which is defined by:

Kc = Experimental value/Theoretical value

Examples include Einstein's Cosmological Constant, dark energy, the initial proposals of dark matter and inflation.
----------------------------------------------------------------------------------------------------------------------------------

This could be a second definition, to justify Einstein's paper:

Given two formulae: E(obtained) and G(desired), a fudging factor CH is that

CH(cheat) = E(obtained)/G(desired)

If

E = π [1 - 1/4 α (α₁ + α₂)]
G = π [1 + 3/4 α (α₁ + α₂)]

with CH = G/E = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]

Then risk a fudging factor (1 + K) so that

(1 + K) = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]

then, find K as the desired fudging factor, calling X = (α₁ + α₂) to simplify:

(1 + K) = [1 + 3/4 α X]/[1 - 1/4 α X]

1 -1/4 α X + K - 1/4 α X K = 1 + 3/4 α X

-1/4 α X + K - 1/4 α X K = 3/4 α X

K = α X/(1 - 1/4 α X) = α X.(1 - 1/4 α X)/(1 - 1/16 α² X²) = (α X - 1/4 α² X²)/(1 - 1/16 α2 X2)

Now, as α X = 1.065E-04 ; - 1/16 α² X² = - 0.709E-15 ; - 1/4 α² X² = -2.836E-15

Dismiss all small values, so K = α X = α (α₁ + α₂)

CH = 1 + α (α₁ + α₂)

Works like charm, isn’t it?

Still is cheating. It fudged the equation to get what was desired BY BRUTE FORCE. Einstein did this, twice in the same paper.

He was a winner! Solved all the problems in the final part and everyone looked in the other way.

Still, Schwarzschild has saved Einstein's ass for 106 years and counting. Never again the method used in the paper was repeated.


.
Richard Hertz
2021-11-21 06:47:35 UTC
Reply
Permalink
Post by Richard Hertz
This is how cheating is rationalized and applied in science?
It needs a second definition.
This is the first one, "legalized" and widely used in science. It talks about HIDDEN VALUES, not formulas.
----------------------------------------------------------------------------------------------------------------------------------
https://en.wikipedia.org/wiki/Fudge_factor
Fudge factor
Kc = Experimental value/Theoretical value
Examples include Einstein's Cosmological Constant, dark energy, the initial proposals of dark matter and inflation.
----------------------------------------------------------------------------------------------------------------------------------
Given two formulae: E(obtained) and G(desired), a fudging factor CH is that
CH(cheat) = E(obtained)/G(desired)
If
E = π [1 - 1/4 α (α₁ + α₂)]
G = π [1 + 3/4 α (α₁ + α₂)]
with CH = G/E = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]
Then risk a fudging factor (1 + K) so that
(1 + K) = [1 + 3/4 α (α₁ + α₂)]/[1 - 1/4 α (α₁ + α₂)]
(1 + K) = [1 + 3/4 α X]/[1 - 1/4 α X]
1 -1/4 α X + K - 1/4 α X K = 1 + 3/4 α X
-1/4 α X + K - 1/4 α X K = 3/4 α X
K = α X/(1 - 1/4 α X) = α X.(1 - 1/4 α X)/(1 - 1/16 α² X²) = (α X - 1/4 α² X²)/(1 - 1/16 α2 X2)
Now, as α X = 1.065E-04 ; - 1/16 α² X² = - 0.709E-15 ; - 1/4 α² X² = -2.836E-15
Dismiss all small values, so K = α X = α (α₁ + α₂)
CH = 1 + α (α₁ + α₂)
Works like charm, isn’t it?
Still is cheating. It fudged the equation to get what was desired BY BRUTE FORCE. Einstein did this, twice in the same paper.
He was a winner! Solved all the problems in the final part and everyone looked in the other way.
Still, Schwarzschild has saved Einstein's ass for 106 years and counting. Never again the method used in the paper was repeated.
.
Typo. It has to be:

CH(cheat) = G(desired)/E(obtained)
Dono.
2021-11-21 17:51:21 UTC
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Permalink
snip repeated imbecilities<
Does your family know?
Richard Hertz
2021-11-21 19:02:19 UTC
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Post by Dono.
Does your family know?
It's very SIMPLE, imbecile.

As a mathematician, just PROVE (to save Einstein reputation) that

∫ G(x) dx = 1 + 1.0E-07 EXACTLY!

G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) , between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11

If you don't try it, you are just an inept CHARLATAN, like the fucker.

But, probably, you are one of many retarded with a college degree that PESTER this world, lying and cheating as most of them.

Faggot commie! You are just a RINO (You know what it means: Republican In Name Only). A fucking commie in disguise
and, worse, an worshiper of another imbecile.
Dono.
2021-11-21 19:26:34 UTC
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Post by Dono.
Does your family know?
It's very SIMPLE, I am an imbecile.
Yes
.
Richard Hertz
2021-11-21 17:33:46 UTC
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Straight from the integral of equation of motion 11, in the paper, and without any simplification using approximations


ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where

G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) , between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11

Whoever solve it, must know that if ε is negative, the lie of Einstein have been around for 106 years without any condemnation,
because the relativistic aditional precession predicted by his 1915 GR is FALSE, and he committed FRAUD.

There are solutions available for numerical computation of the integral, using software.


OR, just make a graphic of G(x) and inspect the AREA within limits. This could be the first step.

ε comes from the equation that solves (dx/dΦ)² = 2A/B² + α/B² x −x² + αx³

Φ = ∫ dx/√(2A/B² + α/B² x −x² + αx³) = π + ε/2 (here, ε is an assumption of GR. Could be 0, positive or negative).
Townes Olson
2021-11-21 19:15:12 UTC
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ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
Whoever solve it, must know that if ε is negative...
It's easy to see that the integrand is always positive. Remember, you are evaluating
∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.

Of course, we can say much more than just "it's positive". It's quite easy to explicitly evaluate the integral in closed form, which confirms that it is positive.
Richard Hertz
2021-11-21 20:24:26 UTC
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ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
Whoever solve it, must know that if ε is negative...
It's easy to see that the integrand is always positive. Remember, you are evaluating
∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
Of course, we can say much more than just "it's positive". It's quite easy to explicitly evaluate the integral in closed form, which confirms that it is positive.
You didn't understand the challenge that I posted. NO APPROXIMATIONS.

It's a matter of NUMERICALLY PROVE exactly the integral of G(x), which is the inverse of the square root of a cubic polynomial
and HAS NOT an analytical solution.

And, solving NUMERICALLY this integral with the limits have to give an EXACT RESULT of 1 + 1.0E-07, no more and no less.

Otherwise, Einstein's paper is wrong and it's not true that the formula gives 43" of arc per century.

It seems that you, for being so knowledgeable and perfectionist, FAILS TO understand an very specific and detailed problem.

I'm more clear than crystal water in my challenge. Don't change it. Just solve it or shut up.
Townes Olson
2021-11-21 20:49:29 UTC
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Post by Richard Hertz
ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
Whoever solve it, must know that if ε is negative...
It's easy to see that the integrand is always positive. Remember, you are evaluating
∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
You didn't understand the challenge that I posted. NO APPROXIMATIONS.
The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.
Richard Hertz
2021-11-23 21:59:39 UTC
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Post by Richard Hertz
ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
Whoever solve it, must know that if ε is negative...
∫ G(x) dx can be 0 (Newton), or < 2π (negative ε). It has to be proven that 2π [ ∫ G(x) dx - 1] = ε = 10E-08 EXACTLY, or the paper is wrong.
Post by Richard Hertz
It's easy to see that the integrand is always positive. Remember, you are evaluating
∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
Of course that the area under G(x) HAS TO BE POSITIVE!. If NOT, G(x) would be imaginary between α₁ and α₂, which would have no sense.
Post by Richard Hertz
You didn't understand the challenge that I posted. NO APPROXIMATIONS.
The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.
I repeat. The integral HAS TO BE POSITIVE, because if not it would be IMAGINARY. There is an square root there, that dictates the rules.

Now, I have to make a clarification: When I wrote "no approximations", I was referring to the extraction of one inverse term and using series
expansion, to obtain a simple numerator plus the inverse of the square root of a quadratic polynomial (like Einstein did).

I've searched about the topic of integrals of the inverse square root of cubic polynomials, and the results are conclusive: many researchers
tried in the past to find a solution, even using Jacobi functions or similar theories. There is NO ANALYTICAL SOLUTION for the problem.

It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.

I tried a numerical solution, with N = 30,000, for the expression ∫ G(x) dx, with steps Δx = 2.4717E-16, which are 1:10E+08 respect ε.

The formula, for Δx = (α₂ - α₁)/30000 = 2.4717E-16, is:

∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1.

Σᴷ [G(α₁ +KΔx) . Δx] = 3.12473257 , with an error of -0.01686008 to reach π.

It also can be seen, graphically, how the error decreases from - 2.8691 (step 20) to -0.01686 (step 30,000).

You can prove it in Excel, as I did.

The problem is that for this kind of functions, being integrated between to consecutive roots, is that the result diverge to infinity on them.
And it doesn't matter how small do you try the first and last Δx (α₁+Δx and α₂-Δx), the result increases up to infinity.

It made me wonder of the real value of any elliptic integral with inverse square root of quadratic polynomials, between consecutive roots.

All of them are infinite, because the AREA UNDER THE CURVE is infinite.

And this is a problem with GR on Mercury, for ANY solution I've seen until now (many papers). The solution is not credible, and the
singularities are FIXED, as in QED (that "renormalized" by eliminating infinities and planting desired values there).

If any has some idea or comment about this case of integrals, I'll appreciate.

But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
divergence, as I proved with other numbers for N.

In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.
Dono.
2021-11-23 22:12:23 UTC
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Post by Richard Hertz
But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
divergence, as I proved with other numbers for N.
Cretinoid,

You obviously failed the section of calculus that deals with the issue of integrals with singularities. You are dead wrong, the integral in the Einstein paper is not infinite, you are as incompetent at math as you are at physics. Eat (some more) shit: https://mathworld.wolfram.com/SingularIntegral.html
Richard Hertz
2021-11-23 22:35:34 UTC
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Post by Dono.
Post by Richard Hertz
But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
divergence, as I proved with other numbers for N.
Cretinoid,
You obviously failed the section of calculus that deals with the issue of integrals with singularities. You are dead wrong, the integral in the Einstein paper is not infinite, you are as incompetent at math as you are at physics. Eat (some more) shit: https://mathworld.wolfram.com/SingularIntegral.html
Fucking commie illiterate, it's "perseveres", not "perseberes".

Plus, I'm learning/re-learning things about orbits of planets, which I find fascinating. And this problem finally gave me the chance
to study something about celestial mechanics that I always wanted to learn a bit but didn't have the time or the spirit.

So, and using my philosophy of refinement by successive steps, I'm doing OK in the learning curve. And this attitude is much better
that your stupid fanaticism that have frozen your mind.

And one more comment: I don't depend on a Mathematica software to help me, SOB. And I don't want it.

I'm fine doing the learning process with Excel and Word, as I decided not to write in paper anymore. And, for developing mathematical
formulations, Word has proven to be a much more useful tool that a notepad, a pencil and a rubber. You can highlight parts at convenience
and can have a working board with numbers, equations and graph all of them in one screen, which you design at will.

Having to deal with dozen of parameters at once, Excel proves to be very flexible FOR MY PURPOSES, which are modest and amateur.
Dono.
2021-11-23 23:24:04 UTC
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Post by Richard Hertz
So, and using my philosophy of refinement by successive steps, I'm doing OK in the learning curve.
No, you are lying, you are not learning anything. Because you are a crank, blinded by your hatred for Einstein.
Integrals with singularities are taught in second year calculus, you obviously flunked the subject. Because you keep insisting that the integral in Einstein paper is "infinite". Keep up the enetertainment, clown.
Richard Hertz
2021-11-23 23:40:57 UTC
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On Tuesday, November 23, 2021 at 8:24:06 PM UTC-3, Dono. wrote:

<snip>
Post by Dono.
No, you are lying, you are not learning anything. Because you are a crank, blinded by your hatred for Einstein.
Integrals with singularities are taught in second year calculus, you obviously flunked the subject. Because you keep insisting that the integral in Einstein paper is "infinite". Keep up the enetertainment, clown.
Demential LIAR! What a miserable piece of shit you are!

My OP is far from the derail of the thread. I stated there that he FUDGED the formuale TWICE to get Gerber's one, but he FAILED,
and his REAL result is negative (FORMULA) and 1/3 of Gerber's formula.

If he was an IMBECILE. You couldn't expect less from him. He had the luck to have Schwarzschild's back up, who HELPED HIM by
providing his exact analytical solution, that put the fucking paper into oblivion!

No one used Einstein's "approximations" or "high cuisine". Every further reinterpretation of Mercury's problem is under the light
of Scwarzschild metric, and NONE of the IMBECILE!
Dono.
2021-11-23 23:46:36 UTC
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Post by Richard Hertz
Post by Dono.
No, you are lying, you are not learning anything. Because you are a crank, blinded by your hatred for Einstein.
Integrals with singularities are taught in second year calculus, you obviously flunked the subject. Because you keep insisting that the integral in Einstein paper is "infinite". Keep up the entertainment, clown.
Demential LIAR! What a miserable piece of shit you are!
My OP is far from the derail of the thread. I stated there that he FUDGED the formuale TWICE to get Gerber's one,
All you have proved, once again, is that you are a crank. Keep it up, dumbestfuck.
Townes Olson
2021-11-23 22:25:27 UTC
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Post by Richard Hertz
There is NO ANALYTICAL SOLUTION for the problem.
I think you mean there is no closed-form solution in terms of "simple" functions (which is an arbitrary classification). That is obviously true, but the integral has a unique real value, which can easily be explicitly evaluated to any desired precision.
Post by Richard Hertz
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
to obtain any number or expression for the area.
That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies to regions with infinite height.
Post by Richard Hertz
You can prove it in Excel, as I did.
We can easily evaluate the integral analytically to any desired precision, so playing with Excel is pointless and not leading you to any insight.
Post by Richard Hertz
If any has some idea or comment about this case of integrals, I'll appreciate.
Anagin, the integral is easily evaluated analytically to any desired precision. You can find this is any good book on relativity.
Post by Richard Hertz
In the case of the Gerber's paper, he didn't need to work with algebraic polynomials
but with trigonometric equations, so it was error free.
The problem with Gerber's paper was not trivial math errors, it was simply that his conclusions don't follow from his premises. He smuggled in an extra factor into the potential without justification, other than to match the known precession rate. Again, you can read a detailed analysis of Gerber's paper (and Besso's and Einstein's discussion of Gerber's paper) in any good book on the foundations of relativity.
Richard Hertz
2021-11-23 22:50:57 UTC
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Post by Richard Hertz
There is NO ANALYTICAL SOLUTION for the problem.
I think you mean there is no closed-form solution in terms of "simple" functions (which is an arbitrary classification). That is obviously true, but the integral has a unique real value, which can easily be explicitly evaluated to any desired precision.
Post by Richard Hertz
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
to obtain any number or expression for the area.
That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies to regions with infinite height.
Post by Richard Hertz
You can prove it in Excel, as I did.
We can easily evaluate the integral analytically to any desired precision, so playing with Excel is pointless and not leading you to any insight.
Post by Richard Hertz
If any has some idea or comment about this case of integrals, I'll appreciate.
Anagin, the integral is easily evaluated analytically to any desired precision. You can find this is any good book on relativity.
Post by Richard Hertz
In the case of the Gerber's paper, he didn't need to work with algebraic polynomials
but with trigonometric equations, so it was error free.
The problem with Gerber's paper was not trivial math errors, it was simply that his conclusions don't follow from his premises. He smuggled in an extra factor into the potential without justification, other than to match the known precession rate. Again, you can read a detailed analysis of Gerber's paper (and Besso's and Einstein's discussion of Gerber's paper) in any good book on the foundations of relativity.
Understood. Read the paper: ELLIPTIC INTEGRALS AND SOME APPLICATIONS, which I posted before. This is one of the many papers
that I have read. Like this other:

The Jacobi elliptic functions and their applications in the advance of mercury’s perihelion
Revista Mexicana de Fısica

https://www.researchgate.net/publication/281748195_The_Jacobi_elliptic_functions_and_their_applications_in_the_advance_of_mercury%27s_perihelion

I'll study it with substitutions of the kind suggested in the first paper I quoted here, to see how it goes.

Regarding "the degree of precision you want", I ACK that, as I saw several methods in other domain than algebraic polynomials.
Richard Hertz
2021-11-23 23:10:18 UTC
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On Tuesday, November 23, 2021 at 7:25:29 PM UTC-3, Townes Olson wrote:

<snip>
Post by Richard Hertz
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
to obtain any number or expression for the area.
That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies to regions with infinite height.
Yes it is, Townes.

Just try to analyze, by approximations and the limit of the FPU of your computer (64 bits, 15 decimal digits mantissa):

Analyze ∫ G(x) dx between x = α₁= 1.43232E-11 and x = α₁ + 1.00E-11, being

G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) , Δx = 1.00E-11/N, and

∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1, under the new limits of integration.

The formula Σᴷ [G(α₁ +KΔx) . Δx] would diverge to infinity the faster as the greater N is. Prove it with N: 10, 100, 1000, 10000, etc.

Actually, you have to start a little bit ahead of α₁ , even 10E-15 apart, so you avoid the infinity in the formula of G(x) if you use α₁.

It doesn't converge at all, unless Excel starts to break with such small values (It happened, but I don't this that in this case).
Townes Olson
2021-11-24 00:37:24 UTC
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Post by Richard Hertz
That is not true. The integrand is infinite at the boundaries of integration, but the
integral is finite.
Yes it is, Townes.
No, for example the integral from x=0 to 1 of 1/sqrt(x) is 2, even though the integrand is infinite at x=0.
Post by Richard Hertz
Just try to analyze, by approximations and the limit of the FPU of your computer...
The analytical solution is trivial.. you can evaluate it without using a computer. If you are getting infinite numerical results, you are making mistakes.
Post by Richard Hertz
It doesn't converge at all, unless Excel starts to break with such small values...
You are losing numerical significance. The only way to evaluate the integral is to do so analytically to give an expression for the result that can be computed trivially.
Post by Richard Hertz
The problem with Gerber's paper was not trivial math errors, it was simply that his conclusions don't follow from his premises. He smuggled in an extra factor into the potential without justification, other than to match the known precession rate. Again, you can read a detailed analysis of Gerber's paper (and Besso's and Einstein's discussion of Gerber's paper) in any good book on the foundations of relativity.
Besso's and Einstein's discussion of Gerber's paper? But this is IMPOSSIBLE!
Nope. You can read about this in any good book on the history of relativity.
Post by Richard Hertz
And regarding "smuggling" a factor to delay potentials, as done by Gerber, what
about Einstein's kitchen?
Gerber smuggled, Einstein did not.
Post by Richard Hertz
Remember (Eq. 8a) Φ = - α/2r , which is written TWO PAGES BEFORE 7c.
You can read a detailed explanation of every step in Einstein's analysis in any good book on the foundations of relativity. Each step is valid. In contrast, the steps in Gerber's paper are not valid. This has all been studied very carefully, and is quite well known.
Richard Hertz
2021-11-23 23:33:09 UTC
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On Tuesday, November 23, 2021 at 7:25:29 PM UTC-3, Townes Olson wrote:

<snip>
The problem with Gerber's paper was not trivial math errors, it was simply that his conclusions don't follow from his premises. He smuggled in an extra factor into the potential without justification, other than to match the known precession rate. Again, you can read a detailed analysis of Gerber's paper (and Besso's and Einstein's discussion of Gerber's paper) in any good book on the foundations of relativity.
Besso's and Einstein's discussion of Gerber's paper? But this is IMPOSSIBLE! Einstein firmly sworn that he DIDN'T KNOW about Gerber's
paper until the accusations of plagiarism started in 1916.

And regarding "smuggling" a factor to delay potentials, as done by Gerber, what about Einstein's kitchen?:

"Finally, one chooses s√(1 − 2A) as the second variable, and AGAIN CALLS IT s, so that one has a SLIGHTLY CHANGED
meaning of the constant B:"

(Eq. 7c) Φ = - α/2r (1+ B²/r²) "

Remember (Eq. 8a) Φ = - α/2r , which is written TWO PAGES BEFORE 7c. Why he planted there what he needed for a cubic polynomial?

Cooking with the MasterCheff, at plain sight and no remorse.

Now, of course you or Dono will rationalize and justify this fudging. I'll wait.
Dono.
2021-11-23 23:41:20 UTC
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Post by Richard Hertz
Now, of course you or Dono will rationalize and justify this fudging. I'll wait.
You are clearly a cretin when it comes to calculating integrals, a more advanced subject. Now, you demonstrate that you are an even bigger cretin by not understanding simple expression substitutions. You see, Dick, your hatred for Einstein really blocks your brain. Whatever you have left inside your skull.
Richard Hertz
2021-11-24 06:11:34 UTC
Reply
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On Tuesday, November 23, 2021 at 7:25:29 PM UTC-3, Townes Olson wrote:

<snip many>
Post by Richard Hertz
There is NO ANALYTICAL SOLUTION for the problem.
I think you mean there is no closed-form solution in terms of "simple" functions (which is an arbitrary classification). That is obviously true, but the integral has a unique real value, which can easily be explicitly evaluated to any desired precision.
Townes: I insist: There is NO ANALYTICAL SOLUTION for the problem.
Post by Richard Hertz
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
to obtain any number or expression for the area.
That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies to regions with infinite height.
Townes: I insist: There is NO ANALYTICAL SOLUTION for the problem.
Post by Richard Hertz
You can prove it in Excel, as I did.
We can easily evaluate the integral analytically to any desired precision, so playing with Excel is pointless and not leading you to any insight.
Townes: I insist: There is NO ANALYTICAL SOLUTION for the problem.
Post by Richard Hertz
If any has some idea or comment about this case of integrals, I'll appreciate.
Anagin, the integral is easily evaluated analytically to any desired precision. You can find this is any good book on relativity.
Post by Richard Hertz
In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.
Townes: I insist: There is NO ANALYTICAL SOLUTION for the problem. Einstein's paper is FULL OF SHIT. He cooked it with
approximations that replaces the cubic polynomial with a quadratic one. So, he CHEATED!

The error margin for the substitution, which I've calculated is HUGE. He fudged and cooked the paper to get Gerber's formula.

I'm giving up finding an analytical solution for Φ = ∫ dx/√(2A/B² + α/B² x −x² + αx³) = K ∫ dx/√[- (x - α₁) (x - α₂) (1 - αx)]

1) A first substitution of variable x = = α₁ + u² gives new limits u₁ = 0 and u₂ = √(α₂ – α₁) and an integral

Φ = 2K/√α ∫ du/√ [ (u² + α₁ - α₂) (u² + α₁ – 1/α) ] = ∫ P(u) du

While P(u₁) = 0, P(u₂) = ∞ , so GOOD BYE for a solution with this substitution.

Following the paper "ELLIPTIC INTEGRALS AND SOME APPLICATIONS ", devoted to the Mercury's problem and GR , the author state
that "Many integrals are reducible to elliptic type.". MANY, NOT ALL OF THEM.

I continue with the following step, as adviced.

2) A second substitution of variable u = tan ϴ, ; A = (α₁ - α₂) ; B = (α₁ – 1/α) , gives new limits ϴ₁ = 0 and ϴ₂ = √(α₂ – α₁) and an integral

Φ = 2K/√α ∫ dϴ/√[sin⁴ ϴ + (A + B) sin² ϴ cos² ϴ + A . B cos⁴ ϴ] = ∫ P(ϴ) dϴ , between ϴ₁ and ϴ₂

P(ϴ₁) = K/√[(α₁ - α₂) (α₁ – 1/α)] , which is finite and > 0.

But P(ϴ₂) = ∞ , so GOOD BYE for ANY SOLUTION for this integral.

However, after using a battery of tricks and limits on infinite series decomposition, the author manages to obtain a DUBIOUS solution,
because again, APPROXIMATIONS ARE INTRODUCED in limit theorems.

He uses the substitution x = α₁ + 1/t² and 1/α = α₃ , to obtain a substitution:

Φ = K ∫ dx/√[(x - α₁) (x - α₂) (x - α₃)] = C ∫ dt/√[ (t² - a²) (t² - b²) ]

where a² = 1/(α₂ - α₁) and b² = 1/(α₃ - α₁)

only to finish in a quotient of TRUNCATED expansions of series UP TO second order.

ε = 2κ/η - 2π = A QUOTIENT OF TRUNCATED SERIES = 3𝜋α (in the limit).

What is funny in THIS FINE PAPER, PRO-RELATIVITY, is that α = 5.09×E−08 (extra relativistic precession), has been used
all the time and PRESUMED TO EXIST before the end.

α was introduced early on, on the solution of an elliptic integral 30 steps before, as θ √α = desired solution. (THIS IS FUDGING, COOKING, CHEATING).

Infinite series WITH THREE PARAMETERS estimated on the limit of many recursions, an error function 𝑂(α²) conveniently
forgotten in the final expression (even when it's declared 5 steps before, and α = 5.09×E−08 previously known IS NOT SERIOUS.

BTW, this is the paper that Dono called as, QUOTE, "A very good paper. Contradicts the imbecilities you have been posting."

But of course it is, because it fits the narrative: Keep proving Einstein is right, no matter what.

Fucking cretin. Get it for once and al: THERE IS NO ANALYTICAL SOLUTION FOR EINSTEIN'S INTEGRAL, AND ANY ATTEMPT TO SOLVE
IT BY APPROXIMATIONS IS FALSE. The objective is to reach to Gerber's formula AT ANY COST.

And the next 100 years worshipers will try to fill the gaps, so the fucking paper can be protected. What is it? A CULT?
Dono.
2021-11-24 06:22:42 UTC
Reply
Permalink
BTW, this is the paper that Dono called as, QUOTE, "A very good paper. Contradicts the imbecilities you have been posting."
...in stark contrast with the cretinisms that you have been posting. The only consolation is that you will continue to post (even bigger) cretinisms until the end of your life. For our amusement and for your embarrassment. Keep it up, dumbestfuck.
Townes Olson
2021-11-24 14:25:54 UTC
Reply
Permalink
Post by Townes Olson
I think you mean there is no closed-form solution in terms of "simple" functions
(which is an arbitrary classification). That is obviously true, but the integral has a
unique real value, which can easily be explicitly evaluated to any desired precision.
I insist: There is NO ANALYTICAL SOLUTION for the problem.
You are mistaken. There is no "closed form solution in simple functions", but that's an arbitrary classification, and there is a simple analytic solution.
Post by Townes Olson
Post by Richard Hertz
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility
to obtain any number or expression for the area.
That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why, consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies to regions with infinite height.
I insist: There is NO ANALYTICAL SOLUTION for the problem.
Repeating your error doesn't make it correct. Remember, you said the reason there can be no analytic solution is that the integrand is infinite at the boundary, but I just gave you a trivial example of a finite integral that is infinite at the boundary. This illustrates that your reasoning is invalid.
[Einstein] cooked it with approximations that replaces the cubic polynomial with a
quadratic one. So, he CHEATED!
No, he was expressly evaluating the precession to just the lowest order of approximation, which is the only order that is large enough to be perceptible in this circumstance. That is not cheating. Of course, we can work out the solution analytically to all higher orders as well.
I'm giving up finding an analytical solution...
That is wise... you will never find it on your own.
BTW, this is the paper that Dono called as, QUOTE, "A very good paper..."
If it doesn't show you the simple and well-known analytic solution, it isn't a very good paper.
The objective is to reach to Gerber's formula AT ANY COST.
No, the objective is to account for the known non-Newtonian precession of orbits in a rational way that is not ad hoc. Gerber just smuggled in the known answer, which does not follow rationally from his (frankly wacky) premises, whereas the result follows quite naturally and unambiguously from the very plausible premises of general relativity (equivalence principle, etc.).
Richard Hertz
2021-11-24 15:44:06 UTC
Reply
Permalink
THIS IS WHAT A DIALOG BETWEEN TWO DEAF PERSONS LOOKS LIKE:

One is an retired EE, very good at math in general, who doesn't give up easily, who's relearning things 45 years in the past.
The other one is a physicist or mathematician, good at his skills, but highly biased toward relativity, who doesn't want to concede one of many Einstein's fudges.


On Wednesday, November 24, 2021 at 11:25:56 AM UTC-3, Townes Olson wrote:

<snip>
Post by Richard Hertz
I insist: There is NO ANALYTICAL SOLUTION for the problem.
You are mistaken. There is no "closed form solution in simple functions", but that's an arbitrary classification, and there is a simple analytic solution.
Pure sophism and dialectics. A self-contradictory statement: "No closed form solution" but there is "a simple analytic solution".
Remember that we are talking here with an elliptic integral with THREE singularities: two very close one to each other, and a third 1,000 time far away
WITHOUT ANY SUBSTITUTION that may change that fact. You can eliminate 2 out of 3 singularities with change of variables, BUT NOT ALL OF THEM!
Post by Richard Hertz
Post by Richard Hertz
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.
That is not true. The integrand is infinite at the boundaries of integration, but the integral is finite. See above. This is not unusual. To understand why,
consider the area under 1/x^2 as x ranges from 1 to infinity. Just because the region has infinite width doesn't mean the area is infinite. The same applies
to regions with infinite height.
Again a deaf person's dialog: You can get rid of ONE out of TWO singularities at the boundaries of integration, but not BOTH. It doesn't matter how
sophisticated is your variable's replacement. The only solution that remains is:

1) Remove the farthest singularity from the inverse polynomial (Einstein's solution). Prove that substitution introduces a negligible error (less than 0.0001).
2) Solve the remaining inverse quadratic polynomial by integrating within its TWO singularities. Apply here substitution of values (Newton's remains).
Post by Richard Hertz
I insist: There is NO ANALYTICAL SOLUTION for the problem.
Repeating your error doesn't make it correct. Remember, you said the reason there can be no analytic solution is that the integrand is infinite at the
boundary, but I just gave you a trivial example of a finite integral that is infinite at the boundary. This illustrates that your reasoning is invalid.
Exactly a TRIVIAL example. Now try with the integral of the inverse square roots of a quadratic polynomial, EXACTLY between its singularities.
After all, it's all about HOW TO CALCULATE an area of an expression that diverges at infinity in both ends. Newton's methods for planet orbitals are problems
of THIS KIND. But you did an APPROXIMATION to the real problem with an inverse cubic polynomial.

Given that the DIFFERENCE that is been sought has a positive value and is about 1/1000000 of the numerical solution, I'd like to see how do you operate
with accuracy in such a difference on ε = (FINAL VALUE - π), being ε = +10E08. This is not a mathematical problem. This is a precision engineering problem.
Post by Richard Hertz
[Einstein] cooked it with approximations that replaces the cubic polynomial with a quadratic one. So, he CHEATED!
No, he was expressly evaluating the precession to just the lowest order of approximation, which is the only order that is large enough to be perceptible
in this circumstance. That is not cheating. Of course, we can work out the solution analytically to all higher orders as well.
You seem to forget that Einstein's paper has ONLY ONE NUMBER: the famous 43" at the final. Not a single clue about the values of α, α₁, α₂ or ε are provided.
That would have been very interesting to know for what they were valued 105 years ago. A lot of data has changed since then (except for ε > 0).
Post by Richard Hertz
I'm giving up finding an analytical solution...
That is wise... you will never find it on your own.
Don't be so sure. In the end, ALL THE PROBLEM REDUCES TO calculate the length of HALF the perimeter of an ellipse, or his equivalent area.
And let me say this, under this new light: IT HAS NOT AN ANALYTICAL PROBLEM. You have: Euler's approximation, Ramanujan's approximation, etc.

P ≈ π [ 3 (a + b) - √[(3a + b) (a + 3b) ]] P ≈ π (a + b) [ 1 + (3h) / (10 + √(4 - 3h) ) ], where h = (a - b)2/(a + b) (SEE? APPROXIMATION TO THE PERIMETER)
Post by Richard Hertz
BTW, this is the paper that Dono called as, QUOTE, "A very good paper..."
If it doesn't show you the simple and well-known analytic solution, it isn't a very good paper.
Because the author introduces ε as a KNOWN PARAMETER, when start the development, which is about to TRY TO PROVE that such PLANT OF ε is OK
within certain limits of error after truncating TWO INFINITE SERIES. This is not the problem. The problem is TO DERIVE ε.
Post by Richard Hertz
The objective is to reach to Gerber's formula AT ANY COST.
No, the objective is to account for the known non-Newtonian precession of orbits in a rational way that is not ad hoc. Gerber just smuggled in the
known answer, which does not follow rationally from his (frankly wacky) premises, whereas the result follows quite naturally and unambiguously from
the very plausible premises of general relativity (equivalence principle, etc.).
You are misguided on purpose here. You are denying the truth: Gerber used ε as a KNOWN PARAMETER, because his paper is about to find the value of c,
the speed of gravity, which resulted to be very close to the speed of light.

His goal was to prove that delayed gravitational potentials caused ε, which is (after all) what GR causes to the limits of "action at a distance" effects, that
forbid them to being instantaneous.

At any case, here we are discussing a fucking integral, to prove or disprove Einstein. Incredible how the things reduce to the elementary, isn't it?
Dono.
2021-11-24 16:05:23 UTC
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One is an imbecile retired EE, totally inept at math in general and at physics in specific, who has an insane fixation on Einstein
Right Dickie-boy, right
Richard Hertz
2021-11-24 17:18:35 UTC
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On Wednesday, November 24, 2021 at 1:05:25 PM UTC-3, Dono. wrote:

You, disgrace of a human being, has become an reincarnation of Paul Joseph Goebbels. You have succeeded on it, SOB!

A jew using nazi tacticts: LIE, THAT SOMETHING WILL REMAIN.

What a disgusting piece of shit you are. I feel pity for those who live around you, cretin.

Paul Joseph Goebbels: a German Nazi politician who was the Gauleiter of Berlin, chief propagandist for the Nazi Party,
and then Reich Minister of propaganda from 1933 to 1945

Goebbels quotes:

"A lie told once remains a lie but a lie told a thousand times becomes the truth"

"If you repeat a lie often enough, people will believe it, and you will even come to believe it yourself."

My contributions:

"If I, Dono, change what other people said to arrange it at my convenience, some idiots will believe it's true"

"I, Dono, have for highest purpose on life, to lie, to distort true and to make a living, in order to preserve the narrative about my pagan god, Einstein".

"I, Dono, believe that there are no enough low blows when it comes to defend my patrons, who feed me. I'm the rottweiler of hell in this forum, and
I live 24x7 permanent watch to attack who dares to question."

See, Dono? You are a fucking nazi jew. And this is a conflict that rotten your life and your soul. I called you a reptilian lifeform, because you have not
a single trace of humanity.

Life takes care of SOB like you: It's called KARMA, and never ever fails.

My purposes on this thread are clear since the OP: To prove Einstein wrong in this paper. It's a legitimate challenge, and should be dealt with respect.
You, on the contrary, are a dark person, who's hiding his agenda while monitoring this forum and reacting like the fucking piece of shit that you have being
for more than 15 years here. No less than 40 threads have been dedicated to show how despicable you are, but you never cease, because you need to
put your evil being at work. And, as this forum is, somehow, a sample of the character of a person, it pictures you as a miserable human being. You was born
that way, lives being such and will die being a miserable and despicable being (not human at all).
Post by Dono.
One is an imbecile retired EE, totally inept at math in general and at physics in specific, who has an insane fixation on Einstein
Right Dickie-boy, right
Dono.
2021-11-24 19:31:54 UTC
Reply
Permalink
I, kapo Richard Hertz, am a disgrace of a human being and I am trying to become a reincarnation of Paul Joseph Goebbels. Nazi Party, and then Reich Minister of propaganda from 1933 to 1945
You are right again, Dick
Dirk Van de moortel
2021-11-24 19:28:59 UTC
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Post by Richard Hertz
One is an retired EE,
Say no more.

Dirk Vdm
Townes Olson
2021-11-24 19:54:10 UTC
Reply
Permalink
Post by Richard Hertz
A self-contradictory statement: "No closed form solution" but there is "a simple analytic solution".
Not at all. Suppose that you are seeking the value of f(x) and you determine that f(x) = x - x^3/3! + x^5/5! - ... This is an analytic function that converges for all x. You might say this is not a legitimate solution or "only approximate", but that isn't true in any meaningful sense, because it is a well-define analytic function. Now, if someone tells you that f(x)=sin(x), suddenly you are happy, and you say "that's an exact solution!". But that is just silly, because sin(x) is just a name given to that power series. They represent the same thing. Likewise we can analytically define the solution of the perihelion problem, as an analytic function of m/r. If you want to call that function Fred, then you can say the solution is Fred(m/r), just like sin(x).
Post by Richard Hertz
You can eliminate 2 out of 3 singularities with change of variables, BUT NOT ALL OF THEM!
The solution does not involve "eliminating" any singularities. The analytic function that solves the problem is well-behaved.
The point is that an integral with any number of singularities can have a finite value (I gave you a trivial example, which you can splice together into as many singularities as you like), so your argument is obviously specious.
Post by Richard Hertz
Exactly a TRIVIAL example.
The more trivial the counter-example to your claim, the more ridiculous is your claim.
Post by Richard Hertz
Now try with the integral of the inverse square roots of
a quadratic polynomial, EXACTLY between its singularities.
That has already been done. We (meaning scientists) have the exact analytic function that is the solution of the perihelion integral. It doesn't happen to be a named function, like sine(m/r), but you can name it Fred(m/r) if you like.
Post by Richard Hertz
But you did an APPROXIMATION to the real problem with an inverse cubic polynomial.
No, it is solved exactly and analytically. It is no more an approximation than x - x^3/3! + x^5/5! - ... is an approximation of sin(x). Note the three dots.
Post by Richard Hertz
Not a single clue about the values of α, α₁, α₂ or ε are provided.
Huh? Those are just a orbital elements for Mercury.
Dirk Van de moortel
2021-11-24 21:01:03 UTC
Reply
Permalink
Post by Townes Olson
Post by Richard Hertz
A self-contradictory statement: "No closed form solution" but there is "a simple analytic solution".
Not at all. Suppose that you are seeking the value of f(x) and you determine that f(x) = x - x^3/3! + x^5/5! - ... This is an analytic function that converges for all x. You might say this is not a legitimate solution or "only approximate", but that isn't true in any meaningful sense, because it is a well-define analytic function. Now, if someone tells you that f(x)=sin(x), suddenly you are happy, and you say "that's an exact solution!". But that is just silly, because sin(x) is just a name given to that power series. They represent the same thing. Likewise we can analytically define the solution of the perihelion problem, as an analytic function of m/r. If you want to call that function Fred, then you can say the solution is Fred(m/r), just like sin(x).
I can't believe this. You are lowering yourself to arguing
about such elementary mathematics. With an engineer of all
people. Good grief.

Dirk Vdm
Richard Hertz
2021-11-24 21:43:00 UTC
Reply
Permalink
On Wednesday, November 24, 2021 at 6:01:06 PM UTC-3, Dirk Van de moortel wrote:

<snip>
I can't believe this. You are lowering yourself to arguing > about such elementary mathematics. With an engineer of all people. Good grief.
Dirk Vdm
At least he's not posting about what happens with twins and imaginary rockets that fly at 800,000,000 Km/hour like you do, imbecile and asshole!

Your superiority complex should pay dearly in your fucking life, of which you WASTED 20% of the your past 20 years or more at this forum.

Your psychological profile has been deteriorating SEVERELY in your last 6 years here, becoming more and more a perfect imbecile asshole at this forum.

Dirk, know it: You are more the Inspector Closeau type of asshole than Mr. Bean type of asshole. At least Mr. Bean know that he's an asshole and
don't give a shit about it. But Closeau embrace his hubris with conviction.

I'll persist in the work about the topic of this thread because I CAN. And, regarding precission and accuracy, I have an entire professional life working
at the limits of state of the art instrumentation, so suffer and shut the fuck up.

To increase your pain, now I'm searching about the REAL ORIGIN of integrals for the inverse square root of quadratic polynomials, and I'm checking
the accuracy of some special integrals NUMERICALLY, because I can. And I have more SW tools than a fucking Excel, with precision well beyond 30
digits, if I want. Now, eat your soup and go to bed, imbecile.
Michael Moroney
2021-11-24 22:54:58 UTC
Reply
Permalink
Post by Richard Hertz
<snip>
I can't believe this. You are lowering yourself to arguing > about such elementary mathematics. With an engineer of all people. Good grief.
Your superiority complex should pay dearly in your fucking life, of which you WASTED 20% of the your past 20 years or more at this forum.
And how many years of your life have you wasted by barking at the 1905
SR paper and your obsession of hating Einstein for no other reason than
'just because' ?
Dirk Van de moortel
2021-11-24 22:59:13 UTC
Reply
Permalink
Post by Michael Moroney
Post by Richard Hertz
<snip>
I can't believe this. You are lowering yourself to arguing > about
such elementary mathematics. With an engineer of all people. Good grief.
Your superiority complex should pay dearly in your fucking life, of
which you WASTED 20% of the your past 20 years or more at this forum.
And how many years of your life have you wasted by barking at the 1905
SR paper and your obsession of hating Einstein for no other reason than
'just because' ?
Hold on. Not 'just because'.
For no other reason than because 'I really, honestly don't get it,
and I SHOULD SHOULD SHOULD! Because I am an ENGINEER GODDAMMIT!'

Dirk Vdm
Dono.
2021-11-24 23:18:24 UTC
Reply
Permalink
Post by Dirk Van de moortel
Post by Michael Moroney
Post by Richard Hertz
<snip>
I can't believe this. You are lowering yourself to arguing > about
such elementary mathematics. With an engineer of all people. Good grief.
Your superiority complex should pay dearly in your fucking life, of
which you WASTED 20% of the your past 20 years or more at this forum.
And how many years of your life have you wasted by barking at the 1905
SR paper and your obsession of hating Einstein for no other reason than
'just because' ?
Hold on. Not 'just because'.
For no other reason than because 'I really, honestly don't get it,
and I SHOULD SHOULD SHOULD! Because I am an ENGINEER GODDAMMIT!'
Dirk Vdm
What makes Richard Hertz stand out:

1. He dabbles in GR (like Koobee Wublee)
2. He's strident, he gets quickly violent when proven wrong
3. He dabbles in calculus (and fails miserably)
4. He is very conceited
5. He is Jewish but he's an antisemite, with a deep hatred for Einstein in specific (just like "hanson")
6. He is an exponent of the Nazi doctrine of Deutsche Physik (cannot like enough ass the Gerber, Soldner, etc)
Dono.
2021-11-24 23:19:37 UTC
Reply
Permalink
Post by Dono.
Post by Dirk Van de moortel
Post by Michael Moroney
Post by Richard Hertz
<snip>
I can't believe this. You are lowering yourself to arguing > about
such elementary mathematics. With an engineer of all people. Good grief.
Your superiority complex should pay dearly in your fucking life, of
which you WASTED 20% of the your past 20 years or more at this forum.
And how many years of your life have you wasted by barking at the 1905
SR paper and your obsession of hating Einstein for no other reason than
'just because' ?
Hold on. Not 'just because'.
For no other reason than because 'I really, honestly don't get it,
and I SHOULD SHOULD SHOULD! Because I am an ENGINEER GODDAMMIT!'
Dirk Vdm
1. He dabbles in GR (like Koobee Wublee)
2. He's strident, he gets quickly violent when proven wrong
3. He dabbles in calculus (and fails miserably)
4. He is very conceited
5. He is Jewish but he's an antisemite, with a deep hatred for Einstein in specific (just like "hanson")
6. He is an exponent of the Nazi doctrine of Deutsche Physik (cannot lick enough ass to the likes of Gerber, Soldner, etc)
Richard Hertz
2021-11-25 01:21:56 UTC
Reply
Permalink
On Wednesday, November 24, 2021 at 8:19:39 PM UTC-3, Dono. wrote:

<snip>
Post by Dono.
1. He dabbles in GR (like Koobee Wublee)
I returned here on July 2021, after a trial by 2019. I never liked GR or SR just because are
fairy tails (SR) and products of sick MINDS (GR).
Space is Euclidean, the Universe is infinite and eternal, no length contraction exist, t' = t,
the only BB that happened was when you farted after eating a burrito, and the only thing
that has severe time delay and is twisted is your fucking rotten brain.
I started to meddle with GR, months ago, only to make all of you cry foul, faggots.
Post by Dono.
2. He's strident, he gets quickly violent when proven wrong
And the fucking nazi jew accuse me of written violence. Just you, piece of shit,
with your 15 years of recorded insults in this forum only. What a cretin.
I can be violent with words, but is up to others to counteract. You can't, because I word
every phrase in the most possible hurting way to make you cry. Obviously, I succeed and
you don't have the INTELLIGENCE to reply accordingly, because YOU ARE AN IMBECILE.
Don't play if you are not apt enough for the wording game.
Post by Dono.
3. He dabbles in calculus (and fails miserably)
I'm relearning calculus after decades of oblivion. Rusty at the start, but gaining traction
day after day. It's called REFINEMENT OF LEARNING CURVE BY SUCCESSIVE STEPS.
A valuable lesson for any of you. What is more, now I'm willing to use the PC up to the
limit of precision in any mathematical expression, and I will not tolerate less than six sigma
of errors to accept something. I can do it, and I will do it.

For instance, TODAY I verified that the error replacing 1/√(1 - αx) ≈ 1 + α/2 x is lower than
2E-15 for the integration range between α₁ and α₂, what allows me to continue with quadratic
polynomials. Also, reversing the development of the solution for the integral ∫ [dx/√ [- (x - α₁) (x - α₂)
allowed me to find substitutions by exponentials.

See how it works, retarded? Refinement by successive steps, plus the PLEASURE to confirm things by myself,
without using any book (except to fact-check). And I enjoy this game, and will continue until satisfaction.
Post by Dono.
4. He is very conceited
I'm not going around like you, Dork, Jan and others, showing off knowledge that you BORROWED. I'm developing
know how on myself, but I'm not the type of person who is intellectually racist like you all, cretins.
Post by Dono.
5. He is Jewish but he's an antisemite, with a deep hatred for Einstein in specific (just like "hanson")
Fuck you. You are more a nazi than a jew. You would have fit perfectly with Ukrainian jews that sided with Hitler's SS in WWII.
There is a lot still there, where you should live.
Post by Dono.
6. He is an exponent of the Nazi doctrine of Deutsche Physik (cannot lick enough ass to the likes of Gerber, Soldner, etc)
Von Soldner died 180 years ago. Gerber died in 1908. Are you so in despair that uses whatever you can trying to construct a LIE?

Not only a nazi jew, but a FAILED nazi jew. That's you, nazi Dono (pathetic Dono sounds similar to me).

AND I'LL CONTINUE TRYING TO PROVE NUMERICALLY THAT EINSTEIN'S PAPER ON MERCURY IS FRAUDULENT.
I already did, theoretically in the OP here. Now, I'll try to do numerically.

And, finally, fuck you, cretin.
Dono.
2021-11-25 01:32:11 UTC
Reply
Permalink
I never liked GR or SR just because are fairy tails (SR) and products of sick MINDS (GR).
Space is Euclidean, the Universe is infinite and eternal, no length contraction exist, t' = t,
the only BB that happened was when you farted after eating a burrito, and the only thing
that has severe time delay and is twisted is your fucking rotten brain.
Very entertaining, keep it up, dumbestfuck!
Odd Bodkin
2021-11-25 13:46:44 UTC
Reply
Permalink
Post by Richard Hertz
<snip>
Post by Dono.
1. He dabbles in GR (like Koobee Wublee)
I returned here on July 2021, after a trial by 2019. I never liked GR or
SR just because are
fairy tails (SR) and products of sick MINDS (GR).
Space is Euclidean, the Universe is infinite and eternal, no length
contraction exist, t' = t,
the only BB that happened was when you farted after eating a burrito, and the only thing
that has severe time delay and is twisted is your fucking rotten brain.
I started to meddle with GR, months ago, only to make all of you cry foul, faggots.
Post by Dono.
2. He's strident, he gets quickly violent when proven wrong
And the fucking nazi jew accuse me of written violence. Just you, piece of shit,
with your 15 years of recorded insults in this forum only. What a cretin.
I can be violent with words, but is up to others to counteract. You can't, because I word
every phrase in the most possible hurting way to make you cry. Obviously, I succeed and
you don't have the INTELLIGENCE to reply accordingly, because YOU ARE AN IMBECILE.
Don't play if you are not apt enough for the wording game.
Post by Dono.
3. He dabbles in calculus (and fails miserably)
I'm relearning calculus after decades of oblivion. Rusty at the start, but gaining traction
day after day. It's called REFINEMENT OF LEARNING CURVE BY SUCCESSIVE STEPS.
A valuable lesson for any of you. What is more, now I'm willing to use the PC up to the
limit of precision in any mathematical expression, and I will not
tolerate less than six sigma
of errors to accept something. I can do it, and I will do it.
For instance, TODAY I verified that the error replacing 1/√(1 - αx) ≈ 1 +
α/2 x is lower than
2E-15 for the integration range between α₁ and α₂, what allows me to
continue with quadratic
polynomials. Also, reversing the development of
the solution for the integral ∫ [dx/√ [- (x - α₁) (x - α₂)
allowed me to find substitutions by exponentials.
See how it works, retarded? Refinement by successive steps, plus the
PLEASURE to confirm things by myself,
without using any book (except to fact-check). And I enjoy this game, and
will continue until satisfaction.
Post by Dono.
4. He is very conceited
I'm not going around like you, Dork, Jan and others, showing off
knowledge that you BORROWED. I'm developing
know how on myself, but I'm not the type of person who is intellectually
racist like you all, cretins.
Post by Dono.
5. He is Jewish but he's an antisemite, with a deep hatred for Einstein
in specific (just like "hanson")
Fuck you. You are more a nazi than a jew. You would have fit perfectly
with Ukrainian jews that sided with Hitler's SS in WWII.
There is a lot still there, where you should live.
Post by Dono.
6. He is an exponent of the Nazi doctrine of Deutsche Physik (cannot
lick enough ass to the likes of Gerber, Soldner, etc)
Von Soldner died 180 years ago. Gerber died in 1908. Are you so in
despair that uses whatever you can trying to construct a LIE?
Not only a nazi jew, but a FAILED nazi jew. That's you, nazi Dono
(pathetic Dono sounds similar to me).
AND I'LL CONTINUE TRYING TO PROVE NUMERICALLY THAT EINSTEIN'S PAPER ON
MERCURY IS FRAUDULENT.
I already did, theoretically in the OP here. Now, I'll try to do numerically.
And, finally, fuck you, cretin.
My, my, my. Don Quixote hones his rusty lance to tilt at windmills.
Cursing the whole while.
--
Odd Bodkin -- maker of fine toys, tools, tables
Dono.
2021-11-25 14:56:43 UTC
Reply
Permalink
Post by Richard Hertz
<snip>
Post by Dono.
1. He dabbles in GR (like Koobee Wublee)
I returned here on July 2021, after a trial by 2019. I never liked GR or
SR just because are
fairy tails (SR) and products of sick MINDS (GR).
Space is Euclidean, the Universe is infinite and eternal, no length
contraction exist, t' = t,
the only BB that happened was when you farted after eating a burrito, and the only thing
that has severe time delay and is twisted is your fucking rotten brain.
I started to meddle with GR, months ago, only to make all of you cry foul, faggots.
Post by Dono.
2. He's strident, he gets quickly violent when proven wrong
And the fucking nazi jew accuse me of written violence. Just you, piece of shit,
with your 15 years of recorded insults in this forum only. What a cretin.
I can be violent with words, but is up to others to counteract. You can't, because I word
every phrase in the most possible hurting way to make you cry. Obviously, I succeed and
you don't have the INTELLIGENCE to reply accordingly, because YOU ARE AN IMBECILE.
Don't play if you are not apt enough for the wording game.
Post by Dono.
3. He dabbles in calculus (and fails miserably)
I'm relearning calculus after decades of oblivion. Rusty at the start, but gaining traction
day after day. It's called REFINEMENT OF LEARNING CURVE BY SUCCESSIVE STEPS.
A valuable lesson for any of you. What is more, now I'm willing to use the PC up to the
limit of precision in any mathematical expression, and I will not
tolerate less than six sigma
of errors to accept something. I can do it, and I will do it.
For instance, TODAY I verified that the error replacing 1/√(1 - αx) ≈ 1 +
α/2 x is lower than
2E-15 for the integration range between α₁ and α₂, what allows me to
continue with quadratic
polynomials. Also, reversing the development of
the solution for the integral ∫ [dx/√ [- (x - α₁) (x - α₂)
allowed me to find substitutions by exponentials.
See how it works, retarded? Refinement by successive steps, plus the
PLEASURE to confirm things by myself,
without using any book (except to fact-check). And I enjoy this game, and
will continue until satisfaction.
Post by Dono.
4. He is very conceited
I'm not going around like you, Dork, Jan and others, showing off
knowledge that you BORROWED. I'm developing
know how on myself, but I'm not the type of person who is intellectually
racist like you all, cretins.
Post by Dono.
5. He is Jewish but he's an antisemite, with a deep hatred for Einstein
in specific (just like "hanson")
Fuck you. You are more a nazi than a jew. You would have fit perfectly
with Ukrainian jews that sided with Hitler's SS in WWII.
There is a lot still there, where you should live.
Post by Dono.
6. He is an exponent of the Nazi doctrine of Deutsche Physik (cannot
lick enough ass to the likes of Gerber, Soldner, etc)
Von Soldner died 180 years ago. Gerber died in 1908. Are you so in
despair that uses whatever you can trying to construct a LIE?
Not only a nazi jew, but a FAILED nazi jew. That's you, nazi Dono
(pathetic Dono sounds similar to me).
AND I'LL CONTINUE TRYING TO PROVE NUMERICALLY THAT EINSTEIN'S PAPER ON
MERCURY IS FRAUDULENT.
I already did, theoretically in the OP here. Now, I'll try to do numerically.
And, finally, fuck you, cretin.
My, my, my. Don Quixote hones his rusty lance to tilt at windmills.
Cursing the whole while.
Exactly. He's losing more of the plot with every post.
Maciej Wozniak
2021-11-25 07:28:32 UTC
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Post by Dirk Van de moortel
Post by Richard Hertz
A self-contradictory statement: "No closed form solution" but there is "a simple analytic solution".
Not at all. Suppose that you are seeking the value of f(x) and you determine that f(x) = x - x^3/3! + x^5/5! - ... This is an analytic function that converges for all x. You might say this is not a legitimate solution or "only approximate", but that isn't true in any meaningful sense, because it is a well-define analytic function. Now, if someone tells you that f(x)=sin(x), suddenly you are happy, and you say "that's an exact solution!". But that is just silly, because sin(x) is just a name given to that power series. They represent the same thing. Likewise we can analytically define the solution of the perihelion problem, as an analytic function of m/r. If you want to call that function Fred, then you can say the solution is Fred(m/r), just like sin(x).
I can't believe this. You are lowering yourself to arguing
about such elementary mathematics. With an engineer of all
people. Good grief.
Speaking of mathematics, it's always good to remind that
your moronic cult had to announce its oldest part false,
as it didn't want to cooperate with the Holy Postulates.
Dono.
2021-11-23 22:27:47 UTC
Reply
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Post by Richard Hertz
Post by Richard Hertz
ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
Whoever solve it, must know that if ε is negative...
∫ G(x) dx can be 0 (Newton), or < 2π (negative ε). It has to be proven that 2π [ ∫ G(x) dx - 1] = ε = 10E-08 EXACTLY, or the paper is wrong.
Post by Richard Hertz
It's easy to see that the integrand is always positive. Remember, you are evaluating
∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
Of course that the area under G(x) HAS TO BE POSITIVE!. If NOT, G(x) would be imaginary between α₁ and α₂, which would have no sense.
Post by Richard Hertz
You didn't understand the challenge that I posted. NO APPROXIMATIONS.
The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.
I repeat. The integral HAS TO BE POSITIVE, because if not it would be IMAGINARY. There is an square root there, that dictates the rules.
Now, I have to make a clarification: When I wrote "no approximations", I was referring to the extraction of one inverse term and using series
expansion, to obtain a simple numerator plus the inverse of the square root of a quadratic polynomial (like Einstein did).
I've searched about the topic of integrals of the inverse square root of cubic polynomials, and the results are conclusive: many researchers
tried in the past to find a solution, even using Jacobi functions or similar theories. There is NO ANALYTICAL SOLUTION for the problem.
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.
I tried a numerical solution, with N = 30,000, for the expression ∫ G(x) dx, with steps Δx = 2.4717E-16, which are 1:10E+08 respect ε.
∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1.
Σᴷ [G(α₁ +KΔx) . Δx] = 3.12473257 , with an error of -0.01686008 to reach π.
It also can be seen, graphically, how the error decreases from - 2.8691 (step 20) to -0.01686 (step 30,000).
You can prove it in Excel, as I did.
The problem is that for this kind of functions, being integrated between to consecutive roots, is that the result diverge to infinity on them.
And it doesn't matter how small do you try the first and last Δx (α₁+Δx and α₂-Δx), the result increases up to infinity.
It made me wonder of the real value of any elliptic integral with inverse square root of quadratic polynomials, between consecutive roots.
All of them are infinite, because the AREA UNDER THE CURVE is infinite.
And this is a problem with GR on Mercury, for ANY solution I've seen until now (many papers). The solution is not credible, and the
singularities are FIXED, as in QED (that "renormalized" by eliminating infinities and planting desired values there).
If any has some idea or comment about this case of integrals, I'll appreciate.
But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
divergence, as I proved with other numbers for N.
In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.
I forgot to mention this paper, which deals with elliptic integrals and substitutions for polynomials. It's devoted to prove GR in the case
of Mercury's perihelion. Very interesting, yet...
ELLIPTIC INTEGRALS AND SOME APPLICATIONS Jay Villanueva Florida Memorial University
https://www.pearson.com/content/dam/one-dot-com/one-dot-com/us/en/files/Jay-Villanuevaictcm3013.pdf
A very good paper. Contradicts the imbecilities you have been posting.
Richard Hertz
2021-11-23 22:40:28 UTC
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Permalink
Post by Dono.
Post by Richard Hertz
Post by Richard Hertz
ε = 2π [ ∫ G(x) dx - 1] , in its purest expression, where
G(x) = 1/√(29159 x³ – 9.8696 x² + 3.559E-10 x – 3.073E-21) ,
between x = α₁= 1.43232E-11 and x = α₂ = 2.17382E-11
Whoever solve it, must know that if ε is negative...
∫ G(x) dx can be 0 (Newton), or < 2π (negative ε). It has to be proven that 2π [ ∫ G(x) dx - 1] = ε = 10E-08 EXACTLY, or the paper is wrong.
Post by Richard Hertz
It's easy to see that the integrand is always positive. Remember, you are evaluating
∫ (1+a/2 x)/√[-(x-a1)(x-a2)] dx
as x ranges from a1 to a2, so aside from the singular boundaries, one of the factors (x-a1) and (x-a2) in the denominator is always positive and the other is always negative, so the product is always negative, and it is negated in the integrand, so the integrand is always positive, and the numerator is also positive, so the integral is positive.
Of course that the area under G(x) HAS TO BE POSITIVE!. If NOT, G(x) would be imaginary between α₁ and α₂, which would have no sense.
Post by Richard Hertz
You didn't understand the challenge that I posted. NO APPROXIMATIONS.
The above explanation applies to the exact integral as well. The integrand is 1/sqrt(f(x)) where f(x) is a cubic polynomial which factors as 29159(x-a1)(x-a2)(x-a3) where a1, a2, a3 are the real roots, and this is being integrated as x ranges from a1 to a2. We know that x-a3 is negative, and we know that one of (x-a1) and (x-a2) is positive and the other is negative, so the cubic is strictly positive for x between a1 and a2. Hence the integral is positive.
I repeat. The integral HAS TO BE POSITIVE, because if not it would be IMAGINARY. There is an square root there, that dictates the rules.
Now, I have to make a clarification: When I wrote "no approximations", I was referring to the extraction of one inverse term and using series
expansion, to obtain a simple numerator plus the inverse of the square root of a quadratic polynomial (like Einstein did).
I've searched about the topic of integrals of the inverse square root of cubic polynomials, and the results are conclusive: many researchers
tried in the past to find a solution, even using Jacobi functions or similar theories. There is NO ANALYTICAL SOLUTION for the problem.
It happens that at α₁ and α₂, G(x) diverges to infinity, which indicates the impossibility to obtain any number or expression for the area.
I tried a numerical solution, with N = 30,000, for the expression ∫ G(x) dx, with steps Δx = 2.4717E-16, which are 1:10E+08 respect ε.
∫ G(x) dx ≈ Σᴷ [G(α₁ +KΔx) . Δx] , from K=1 to K=N-1.
Σᴷ [G(α₁ +KΔx) . Δx] = 3.12473257 , with an error of -0.01686008 to reach π.
It also can be seen, graphically, how the error decreases from - 2.8691 (step 20) to -0.01686 (step 30,000).
You can prove it in Excel, as I did.
The problem is that for this kind of functions, being integrated between to consecutive roots, is that the result diverge to infinity on them.
And it doesn't matter how small do you try the first and last Δx (α₁+Δx and α₂-Δx), the result increases up to infinity.
It made me wonder of the real value of any elliptic integral with inverse square root of quadratic polynomials, between consecutive roots.
All of them are infinite, because the AREA UNDER THE CURVE is infinite.
And this is a problem with GR on Mercury, for ANY solution I've seen until now (many papers). The solution is not credible, and the
singularities are FIXED, as in QED (that "renormalized" by eliminating infinities and planting desired values there).
If any has some idea or comment about this case of integrals, I'll appreciate.
But, a solid technique like numerical approximation FAILS even when 1,000,000 intervals are computed. In fact, with more intervals, more
divergence, as I proved with other numbers for N.
In the case of the Gerber's paper, he didn't need to work with algebraic polynomials but with trigonometric equations, so it was error free.
I forgot to mention this paper, which deals with elliptic integrals and substitutions for polynomials. It's devoted to prove GR in the case
of Mercury's perihelion. Very interesting, yet...
ELLIPTIC INTEGRALS AND SOME APPLICATIONS Jay Villanueva Florida Memorial University
https://www.pearson.com/content/dam/one-dot-com/one-dot-com/us/en/files/Jay-Villanuevaictcm3013.pdf
A very good paper. Contradicts the imbecilities you have been posting.
Had you followed my challenge, I said NO SUBSTITUTIONS. Townes suggested since time zero to use substitutions, but my
challenge (a game) was to keep the solution within polynomials, which I prove that's impossible. For not being my specialty, I'm OK
with my fucking LEARNING CURVE. It allowed me to study something using Newton and the fucker.
Dono.
2021-11-23 23:29:16 UTC
Reply
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Post by Richard Hertz
Post by Dono.
https://www.pearson.com/content/dam/one-dot-com/one-dot-com/us/en/files/Jay-Villanuevaictcm3013.pdf
A very good paper. Contradicts the imbecilities you have been posting.
Had you followed my challenge, I said NO SUBSTITUTIONS.
Cretinoid,

One of the standard methods of calculating integrals is "change of variable", i.e. "substitution". So, you cannot impose your demented rules.
Athel Cornish-Bowden
2021-11-25 18:32:10 UTC
Reply
Permalink
Post by Richard Hertz
This post specifically address the paper that made Einstein famous, not
General Relativity, and is not a critic about GR in general (for now).
The paper that got Einstein the Nobel Prize was not that one, but the
one on the photoelectric effect. Do you have any opinions about whether
that work was fraudulent?
Post by Richard Hertz
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,
(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)
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).
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
(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
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
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)]
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 vanisheswhen 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
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)]
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.
(Equation 14) ε = - 8π3a²/[T²c² (1 – e²)], negative and 3 times lower.
ε(Einstein)/ε(Real) = -3 ; ε(Real) = -1/3 ε(Einstein)
--
Athel -- French and British, living mainly in England until 1987.
Richard Hertz
2021-11-26 04:27:07 UTC
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On Thursday, November 25, 2021 at 3:32:15 PM UTC-3, Athel Cornish-Bowden wrote:

<snip most of the reposting of my OP>
Post by Athel Cornish-Bowden
Post by Richard Hertz
This post specifically address the paper that made Einstein famous, not
General Relativity, and is not a critic about GR in general (for now).
The paper that got Einstein the Nobel Prize was not that one, but the
one on the photoelectric effect. Do you have any opinions about whether
that work was fraudulent?
<snip>
Post by Athel Cornish-Bowden
Athel -- French and British, living mainly in England until 1987.
If you read carefully the title of this thread, it says: Proofs about Einstein fudging and cooking the paper that made him famous worldwide,
and I made it very clear that this thread was devoted ONLY to prove FALSE his famous Nov. 18, 1915, on Mercury's missing 43"/cy.

Many scientists before him had failed: Minkowski, Sommerfeld, von Laue, Nordstrom, Abraham, Mie, etc.

The only who got the right formula was Gerber, in 1898. And this because he used ψ = 4.789 × 10-7 rad/orbit as an entry parameter
in his search for the speed c of gravity.

Einstein failed again (first time with Entwurf in 1913) to find a correct formula, so he committed FRAUD in this promised paper to the
Prussian Academy of Science. He was VERY concerned.

In the same paper, let the PAS that he achieved to obtain twice the deflection of light by the Sun than in his 1911 paper, or 1.75" of arc.

Einstein went public to announce the triumph of his theory of GR, with the first proof (1 out of 3) for which GR was going to be tested.

Such publicity made him immediately famous in Europe and USA, but had to wait 3 more years for Eddington's dubious proof of light
deflection by the Sun's surface in 1.75" of arc. Then, his fame exploded worldwide overnight.

I've demonstrated here that his fame was undue, because his development was fraudulent, fudged and cooked in high cuisine.

No intention, AT THIS THREAD, to go against GR, SR or photoelectric effect. I did something here to prove how useless is SR in earthly
applications, including Moon and neighborhood.

But, thanks for reposting my OP. Too long and boring, so I'll make a new thread with a simplified explanation.
Paul Alsing
2021-11-26 04:45:29 UTC
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Post by Richard Hertz
I've demonstrated here that his fame was undue, because his development was fraudulent, fudged and cooked in high cuisine.
No, you have not. You have only *claimed* that his fame was undue but you have failed to prove it.

You are a dumbfuck now and will remain a dumbfuck for a very long time. You should retire before you embarrass yourself any further...but, of course, this is just my opinion...
Richard Hertz
2021-11-26 05:15:24 UTC
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Post by Paul Alsing
Post by Richard Hertz
I've demonstrated here that his fame was undue, because his development was fraudulent, fudged and cooked in high cuisine.
No, you have not. You have only *claimed* that his fame was undue but you have failed to prove it.
You are a dumbfuck now and will remain a dumbfuck for a very long time. You should retire before you embarrass yourself any further...but, of course, this is just my opinion...
I had to use a dictionary on the dumbfuck insult. I found that (urban slang):

"A dumb fuck is someone that is undeniably stupid, they can't get their life together because of their epic level of stupidity."

Your opinion of me matches mine about you, but a little bit less. After all, you can type your imbecile "contributions" in this forum
and post them, but with the risk to appear as a fucking peacock, who thinks that's above something on Earth.

Let me tell you what I think of you, for a start: I think that you are a pathetic relativist with a fossilized brain, that are opinionated
with BORROWED KNOWLEDGE, because you are unable to generate a single original thought by yourself, even if your life is at stake.

I've seen lots of assholes like you, who didn't want the effort to discuss something with.

In your case, your mind is severely impaired by indoctrination, which prevent you from questioning NOTHING.

No difference between you and a robot working with Windows 3.1 (1991). Prone to fail and an OS wannabe. It applies to you, in
the intellectual activities that I can read in your history here. As I said: pathetic, void, a true peacock (which is dumb as fuck).

Don't miss your pagan church's meeting, with other retarded like you, in a near location to be disclosed soon. Fucking cretin.
Paul Alsing
2021-11-26 06:09:38 UTC
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Permalink
Post by Richard Hertz
Post by Paul Alsing
Post by Richard Hertz
I've demonstrated here that his fame was undue, because his development was fraudulent, fudged and cooked in high cuisine.
No, you have not. You have only *claimed* that his fame was undue but you have failed to prove it.
You are a dumbfuck now and will remain a dumbfuck for a very long time. You should retire before you embarrass yourself any further...but, of course, this is just my opinion...
"A dumb fuck is someone that is undeniably stupid, they can't get their life together because of their epic level of stupidity."
Your opinion of me matches mine about you, but a little bit less. After all, you can type your imbecile "contributions" in this forum
and post them, but with the risk to appear as a fucking peacock, who thinks that's above something on Earth.
Let me tell you what I think of you, for a start: I think that you are a pathetic relativist with a fossilized brain, that are opinionated
with BORROWED KNOWLEDGE, because you are unable to generate a single original thought by yourself, even if your life is at stake.
I've seen lots of assholes like you, who didn't want the effort to discuss something with.
In your case, your mind is severely impaired by indoctrination, which prevent you from questioning NOTHING.
No difference between you and a robot working with Windows 3.1 (1991). Prone to fail and an OS wannabe. It applies to you, in
the intellectual activities that I can read in your history here. As I said: pathetic, void, a true peacock (which is dumb as fuck).
Don't miss your pagan church's meeting, with other retarded like you, in a near location to be disclosed soon. Fucking cretin.
I hope you feel better after this rant... it is probably not good for you to get so worked up... but I must tell you that I have forgotten more physics than you have ever known... and there isn't a thing that you can do about that...
Dono.
2021-11-26 05:45:03 UTC
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Permalink
I've demonstrated here that I am a cretin with a huge hard-on for Einstein
Paul B. Andersen
2021-11-26 12:56:39 UTC
Reply
Permalink
Post by Richard Hertz
If you read carefully the title of this thread, it says: Proofs about Einstein fudging and cooking the paper that made him famous worldwide,
and I made it very clear that this thread was devoted ONLY to prove FALSE his famous Nov. 18, 1915, on Mercury's missing 43"/cy.
Many scientists before him had failed: Minkowski, Sommerfeld, von Laue, Nordstrom, Abraham, Mie, etc.
The only who got the right formula was Gerber, in 1898. And this because he used ψ = 4.789 × 10-7 rad/orbit as an entry parameter
in his search for the speed c of gravity.
It is however now known that Gerber's theory was wrong,
the equation he found doesn't even follow from his theory.
Post by Richard Hertz
Einstein went public to announce the triumph of his theory of GR, with the first proof (1 out of 3) for which GR was going to be tested.
Such publicity made him immediately famous in Europe and USA, but had to wait 3 more years for Eddington's dubious proof of light
deflection by the Sun's surface in 1.75" of arc. Then, his fame exploded worldwide overnight.
I've demonstrated here that his fame was undue, because his development was fraudulent, fudged and cooked in high cuisine.
So you will have us to believe that:
1. Einstein didn't know, and was unable to calculate
what GR predicts for the perihelion advance of Mercury.
2. Einstein found Gerber's equation and since it gave
the correct answer, he claimed that it followed from GR,
which he didn't believe was true. Thus it was a fraud.
Post by Richard Hertz
No intention, AT THIS THREAD, to go against GR,
I take that to mean that you know what we all know:
GR predicts the perihelion advance of a planet is:
ε = 24π³a²/T²c²(1−e²) radians/orbit.

This is calculated many times by many physicists
and astronomers.

So when Einstein in his 1915 paper:
https://tinyurl.com/yzzbratp
fraudulently claimed that GR predicted:
ε = 24π³a²/T²c²(1−e²) equation (14)
he was very lucky when his fudged and cooked
equation later proved to be what GR predicts.
--
Paul

https://paulba.no/
Maciej Wozniak
2021-11-26 15:01:35 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Richard Hertz
If you read carefully the title of this thread, it says: Proofs about Einstein fudging and cooking the paper that made him famous worldwide,
and I made it very clear that this thread was devoted ONLY to prove FALSE his famous Nov. 18, 1915, on Mercury's missing 43"/cy.
Many scientists before him had failed: Minkowski, Sommerfeld, von Laue, Nordstrom, Abraham, Mie, etc.
The only who got the right formula was Gerber, in 1898. And this because he used ψ = 4.789 × 10-7 rad/orbit as an entry parameter
in his search for the speed c of gravity.
It is however now known that Gerber's theory was wrong,
the equation he found doesn't even follow from his theory.
Post by Richard Hertz
Einstein went public to announce the triumph of his theory of GR, with the first proof (1 out of 3) for which GR was going to be tested.
Such publicity made him immediately famous in Europe and USA, but had to wait 3 more years for Eddington's dubious proof of light
deflection by the Sun's surface in 1.75" of arc. Then, his fame exploded worldwide overnight.
I've demonstrated here that his fame was undue, because his development was fraudulent, fudged and cooked in high cuisine.
1. Einstein didn't know, and was unable to calculate
what GR predicts for the perihelion advance of Mercury.
In the meantime in the real world, however, forbidden by
your moronic religion GPS clocks keep measuring t'=t,
just like all serious clocks always did.
Richard Hertz
2021-11-26 17:16:54 UTC
Reply
Permalink
On Friday, November 26, 2021 at 9:56:42 AM UTC-3, Paul B. Andersen wrote:

<snip>
Post by Richard Hertz
Einstein went public to announce the triumph of his theory of GR, with the first proof (1 out of 3) for which GR was going to be tested.
Such publicity made him immediately famous in Europe and USA, but had to wait 3 more years for Eddington's dubious proof of light
deflection by the Sun's surface in 1.75" of arc. Then, his fame exploded worldwide overnight.
I've demonstrated here that his fame was undue, because his development was fraudulent, fudged and cooked in high cuisine.
1. Einstein didn't know, and was unable to calculate what GR predicts for the perihelion advance of Mercury.
NO! What I'm saying that he DIDN'T UNDERSTAND YET the Grossmann-Hilbert theory (it took one more year for him, and it's written).
So, he FAILED from the first part of the paper to manage equations for orbits of planets. Schwarzschild (an astronomer) did know, and
let it known to Einstein in personal letters and in HIS PUBLISHED PAPER. I quote from it:

"§2. Mr. Einstein showed that this problem, in first approximation, leads to Newton’s law and that the second approximation correctly
reproduces the known anomaly in the motion of the perihelion of Mercury. The following calculation yields the exact solution of the
problem. It is always pleasant to avail of exact solutions of simple form. More importantly, the calculation proves also the uniqueness of
the solution, about which Mr. Einstein’s treatment STILL LEFT DOUBT, and which could have been PROVED ONLY WITH GREAT
DIFFICULTY, in the way shown below, through SUCH AN APPROXIMATION METHOD The following lines THEREFORE let Mr. Einstein’s
result shine with increased clearness."

I wrote here many times that Schwarzschild HAD CRITICIZED Einstein's paper, in a VERY POLITE WAY, given that Germany was at war,
and was not a time to make the fight a big issue, and that Schwarzschild HAD RESPONSABILITIES UPON Einstein. He had been one
of the three persons responsible for giving Einstein a Berlin University's professorship by 1914 (because he believed on his ideas on GR),
but he couldn't be supportive of the Nov. 18 paper, because he KNEW THAT IT WAS PLAIN WRONG. Read carefully §2 , and tell me that's
not a SEVERE critic to his protegee (he was present at the Prussian Academy that day).

Also, read Schwarzschild FIRST paper (published Feb 1916) with extreme care, and will find more critics, and how he showed the way
to address the problem.

Also, read about his Equation 18, which contrast with Einstein's Equation 11:

( Schwarzschild Eq. 18) (dx/dΦ)² = (1 –h)/c² + hα/c² x −x² + αx³ = 2A/B + α/B x - x² + α x³

(Einstein Eq. 11) (dx/dΦ)² = 2A/B² + α/B² x - x² + α x³

DO YOU SEE THE DIFFERENCES? Both produces different values for roots α₁ and α₂. Root 1/α remains the same.

And that differences are the cause of the FAILED PAPER, so Einstein had to cheat to get Gerber's expression.

Read the rest of the paper, and find out WHERE Einstein went wrong. In particular, because he should have used

1/x = R = (r³ + α³)^1/3 instead 1/x = r.

Schwarzschild too made a mistake on the coordinate origin, because he wasn't using the latest Hilbert's theory.

The solution known as Schwarzschild metric is not the one for Feb. 1916 paper, but a MODIFICATION made by Hilbert in 1917,
which was named as Schwarzschild's solution by Hilbert to HONOR his late colleague.

And using the CURRENT Schwarzschild metric is that 1,000+ proofs in the last 90+ years provide the correct Gerber's formula.
2. Einstein found Gerber's equation and since it gave the correct answer, he claimed that it followed from GR, which he didn't
believe was true. Thus it was a fraud.
NO. Einstein knew that Gerber's formula was structurally correct, with disregard of the theory. He had to fudge his development
to match the final result, by introducing the dirty tricky K-1 ≈ 0 before solving the integral. In the ≈ 0 is hidden 2/3 of the 43" of arc/orbit.
Post by Richard Hertz
No intention, AT THIS THREAD, to go against GR,
I take that to mean that you know what we all know: GR predicts the perihelion advance of a planet is: ε = 24π³a²/T²c²(1−e²)
radians/orbit.
Using the Hilbert-Schwarzschild metric YES. In the Einstein's paper NO.
This is calculated many times by many physicists and astronomers.
ε = 24π³a²/T²c²(1−e²) equation (14)
he was very lucky when his fudged and cooked equation later proved to be what GR predicts.
It's not a matter of having luck. It's a matter of being a SOB cheater, as he always was. GR can "predict" many things,
but the mathematically INEPT fraudster couldn't. Inept, cheater and street-smart: enough for success with the PR around GR, which
was developed BY OTHER PERSONS, starting with Grossmann and ending with Hilbert.
Odd Bodkin
2021-11-26 18:07:27 UTC
Reply
Permalink
Post by Paul B. Andersen
Post by Richard Hertz
If you read carefully the title of this thread, it says: Proofs about
Einstein fudging and cooking the paper that made him famous worldwide,
and I made it very clear that this thread was devoted ONLY to prove
FALSE his famous Nov. 18, 1915, on Mercury's missing 43"/cy.
Many scientists before him had failed: Minkowski, Sommerfeld, von Laue,
Nordstrom, Abraham, Mie, etc.
The only who got the right formula was Gerber, in 1898. And this because
he used ψ = 4.789 × 10-7 rad/orbit as an entry parameter
in his search for the speed c of gravity.
It is however now known that Gerber's theory was wrong,
the equation he found doesn't even follow from his theory.
Post by Richard Hertz
Einstein went public to announce the triumph of his theory of GR, with
the first proof (1 out of 3) for which GR was going to be tested.
Such publicity made him immediately famous in Europe and USA, but had to
wait 3 more years for Eddington's dubious proof of light
deflection by the Sun's surface in 1.75" of arc. Then, his fame exploded
worldwide overnight.
I've demonstrated here that his fame was undue, because his development
was fraudulent, fudged and cooked in high cuisine.
1. Einstein didn't know, and was unable to calculate
what GR predicts for the perihelion advance of Mercury.
2. Einstein found Gerber's equation and since it gave
the correct answer, he claimed that it followed from GR,
which he didn't believe was true. Thus it was a fraud.
Post by Richard Hertz
No intention, AT THIS THREAD, to go against GR,
ε = 24π³a²/T²c²(1−e²) radians/orbit.
This is calculated many times by many physicists
and astronomers.
Hertz would have to claim that NO physicist has done the calculations after
Gerber. That either they just cited Einstein and did nothing themselves, or
that they copied Gerber themselves. Heh.
Post by Paul B. Andersen
https://tinyurl.com/yzzbratp
ε = 24π³a²/T²c²(1−e²) equation (14)
he was very lucky when his fudged and cooked
equation later proved to be what GR predicts.
--
Odd Bodkin — Maker of fine toys, tools, tables
Dirk Van de moortel
2021-11-26 18:22:29 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Paul B. Andersen
Post by Richard Hertz
If you read carefully the title of this thread, it says: Proofs about
Einstein fudging and cooking the paper that made him famous worldwide,
and I made it very clear that this thread was devoted ONLY to prove
FALSE his famous Nov. 18, 1915, on Mercury's missing 43"/cy.
Many scientists before him had failed: Minkowski, Sommerfeld, von Laue,
Nordstrom, Abraham, Mie, etc.
The only who got the right formula was Gerber, in 1898. And this because
he used ψ = 4.789 × 10-7 rad/orbit as an entry parameter
in his search for the speed c of gravity.
It is however now known that Gerber's theory was wrong,
the equation he found doesn't even follow from his theory.
Post by Richard Hertz
Einstein went public to announce the triumph of his theory of GR, with
the first proof (1 out of 3) for which GR was going to be tested.
Such publicity made him immediately famous in Europe and USA, but had to
wait 3 more years for Eddington's dubious proof of light
deflection by the Sun's surface in 1.75" of arc. Then, his fame exploded
worldwide overnight.
I've demonstrated here that his fame was undue, because his development
was fraudulent, fudged and cooked in high cuisine.
1. Einstein didn't know, and was unable to calculate
what GR predicts for the perihelion advance of Mercury.
2. Einstein found Gerber's equation and since it gave
the correct answer, he claimed that it followed from GR,
which he didn't believe was true. Thus it was a fraud.
Post by Richard Hertz
No intention, AT THIS THREAD, to go against GR,
ε = 24π³a²/T²c²(1−e²) radians/orbit.
This is calculated many times by many physicists
and astronomers.
Hertz would have to claim that NO physicist has done the calculations after
Gerber. That either they just cited Einstein and did nothing themselves, or
that they copied Gerber themselves. Heh.
Imagine that this marks the point in history where all future
generations of retired engineers just cite Hertz and do
nothing themselves. Huh.

Dirk Vdm
Maciej Wozniak
2021-11-26 18:33:12 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Paul B. Andersen
Post by Richard Hertz
If you read carefully the title of this thread, it says: Proofs about
Einstein fudging and cooking the paper that made him famous worldwide,
and I made it very clear that this thread was devoted ONLY to prove
FALSE his famous Nov. 18, 1915, on Mercury's missing 43"/cy.
Many scientists before him had failed: Minkowski, Sommerfeld, von Laue,
Nordstrom, Abraham, Mie, etc.
The only who got the right formula was Gerber, in 1898. And this because
he used ψ = 4.789 × 10-7 rad/orbit as an entry parameter
in his search for the speed c of gravity.
It is however now known that Gerber's theory was wrong,
the equation he found doesn't even follow from his theory.
Post by Richard Hertz
Einstein went public to announce the triumph of his theory of GR, with
the first proof (1 out of 3) for which GR was going to be tested.
Such publicity made him immediately famous in Europe and USA, but had to
wait 3 more years for Eddington's dubious proof of light
deflection by the Sun's surface in 1.75" of arc. Then, his fame exploded
worldwide overnight.
I've demonstrated here that his fame was undue, because his development
was fraudulent, fudged and cooked in high cuisine.
1. Einstein didn't know, and was unable to calculate
what GR predicts for the perihelion advance of Mercury.
2. Einstein found Gerber's equation and since it gave
the correct answer, he claimed that it followed from GR,
which he didn't believe was true. Thus it was a fraud.
Post by Richard Hertz
No intention, AT THIS THREAD, to go against GR,
ε = 24π³a²/T²c²(1−e²) radians/orbit.
This is calculated many times by many physicists
and astronomers.
Hertz would have to claim that NO physicist has done the calculations after
Gerber. That either they just cited Einstein and did nothing themselves, or
that they copied Gerber themselves. Heh.
And in the meantime - forbidden by your moronic
religion GPS clocks keep measuring t'=t, just
like all serious clocks always did.
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