On Thursday, December 2, 2021 at 5:06:46 PM UTC-8, Richard Hertz wrote:
> What wouldn't do students around 1900?
> 1/√(2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22) = 1/√ [- (x - α₁) (x – α₂) (1 - αx)] EXACTLY.

That's not true at all. To get a monic polynomial you factor out the coefficient of x^3, which we can call "a", and then factor the cubic like this: a(x-a1)(x-a2)(x-a3) = a(a3-x)[-(x-a1)(x-a2)]. Now, you are claiming (above) that a(a3-x) = 1-ax, but that would be true only if a*a3 = 1, which is not true. Students around 1900 would be able to tell you instantly what a*a3 equals, and how this leads immediately to the correct result. You're just making a simple algebraic mistake.

On Thursday, December 2, 2021 at 5:50:36 PM UTC-8, Townes Olson wrote:
> On Thursday, December 2, 2021 at 5:06:46 PM UTC-8, Richard Hertz wrote:
> > What wouldn't do students around 1900?
> > 1/√(2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22) = 1/√ [- (x - α₁) (x – α₂) (1 - αx)] EXACTLY.
> That's not true at all. To get a monic polynomial you factor out the coefficient of x^3, which we can call "a", and then factor the cubic like this: a(x-a1)(x-a2)(x-a3) = a(a3-x)[-(x-a1)(x-a2)]. Now, you are claiming (above) that a(a3-x) = 1-ax, but that would be true only if a*a3 = 1, which is not true. Students around 1900 would be able to tell you instantly what a*a3 equals, and how this leads immediately to the correct result. You're just making a simple algebraic mistake.
There is no mistake , you are just conversing with the number one crank here, a very sick specimen bent on "proving" Einstein wrong. You will never convince Dick that he's wrong, simply enjoy the spectacle of how low this scum gets.

To the Lone Ranger and Tonto, or Batman and Robin, depending on your sexual orientation.

Your work in tandem couldn't be more evident: Good LIAR, Bad LIAR. But this has to stop because, Townes, you are beyond ridicule
and repeats yourself pathetically and without any meaning, asshole alter ego!

Watch yourself:
***********************************
Townes Olson, Dec 2, 2021, 10:50:36 PM
On Thursday, December 2, 2021 at 5:06:46 PM UTC-8, Richard Hertz wrote:
> What wouldn't do students around 1900?
> 1/√(2954 x³ - 1.000000106525 x² + 3.60613E-11 x - 3.1136E-22) = 1/√ [- (x - α₁) (x – α₂) (1 - αx)] EXACTLY.

That's not true at all. To get a monic polynomial you factor out the coefficient of x^3, which we can call "a", and then factor the cubic like this: a(x-a1)(x-a2)(x-a3) = a(a3-x)[-(x-a1)(x-a2)]. Now, you are claiming (above) that a(a3-x) = 1-ax, but that would be true only if a*a3 = 1, which is not true.
Students around 1900 would be able to tell you instantly what a*a3 equals, and how this leads immediately to the correct result. You're just making a simple algebraic mistake.

Dono. , Dec 2, 2021, 11:10:37 PM (3 hours ago)
There is no mistake , you are just conversing with the number one crank here, a very sick specimen bent on "proving" Einstein wrong. You will never convince Dick that he's wrong, simply enjoy the spectacle of how low this scum gets.
***********************************

Townes alter ego: You don't even know what a polynomial represent, retarded.. You are in despair, looking at Google or deep into your ass
any shit that can CONTRADICT ME. Well, asshole, things aren't so complicated, even for a 15 years old.

So, stop embarrassing yourself with your paroxysm. Just get over and deal with the fact that you HAVE NOTHING.

Worse yet, having made the product [- (x - α₁) (x – α₂) (1 - αx)] = P x³ + Q x² + R x + S to identify EXACTLY the values of (P, Q, R, S)
you would found out what is the meaning of Q = - 1.000000106525, which WOULD SCARE THE SHIT OUT OF YOU (and DONO).

But as you and Dono (alter alias) are coward, deceivers and liars, I will do it for you, so you don't have excuses anymore.

- (x - α₁) (x - α₂) (1 – α x) = αx³ - [1 + α(α₁ + α₂)] x² + (α α₁ α₂ + α₁ + α₂) x - α₁ α₂ = P x³ + Q x² + R x + S

Asshole Townes, do you mean what does Q represents in the denominator, previous to the square root? NO? Of course not, idiot.

Q = [1 + α(α₁ + α₂)] represents NEGATIVE PRECESSION, which Einstein (the crook) ELIMINATED by making Q = 1 on his integral equation.

The same at which he introduced A BETTER FIX/FUDGE, K = [1 + 1/2 α (α₁ + α₂)] , to provide a final result ε = 3/2 α (α₁ + α₂).

What would happen if he HAD EXTRACTED 1/√Q and put it at the front of the integral? The coefficients of the polynomial are
BARELY affected, but what happens with 1/√Q?

Well, 1/√Q = [1 - 1/2 α(α₁ + α₂)], and would give ε = - 1/2 α (α₁ + α₂), a NEGATIVE PRECESSION OF -14,25” of arc/100 years.

But I didn't want to follow that path, being the FRAUD even MORE OUTRAGEOUS!

I just ignored K and made it EXACTLY K = 1, to avoid the scandal addressed at this forum SEVERAL TIMES between 2009 and 2012:
that the REAL RESULT is a NEGATIVE VALUE OF -14,25 arcsec/100 years.

I preferred to stay with a POSITIVE VALUE OF +14,25 arcsec/100 years, which is less painful for relativists. At least IS POSITIVE.

Please, don't play this game with me. I'm far too good at this for you to challenge me. I have every fucking equation and numbers
CHECKED 4 OR 5 TIMES, and with 10 decimal digits numerical values plus different graphs to analyze roots, integrals, errors, etc.

You both are FAR BEYOND MY LEAGUE.

Concede my result, imbeciles (or better, just one imbecile with two faces).

On Thursday, December 2, 2021 at 10:01:28 PM UTC-8, Richard Hertz wrote:

> You both are FAR BEYOND MY LEAGUE.

If only you understood in good old American English what you just wrote you would be embarrassed beyond belief!

You did NOT say what you thought you said! Priceless!

On Friday, December 3, 2021 at 3:40:19 AM UTC-3, Paul Alsing wrote:
> On Thursday, December 2, 2021 at 10:01:28 PM UTC-8, Richard Hertz wrote:
>
> > You both are FAR BEYOND MY LEAGUE.
> If only you understood in good old American English what you just wrote you would be embarrassed beyond belief!
>
> You did NOT say what you thought you said! Priceless!

I just checked with Google. Well, stupid but not too embarrassing. I'm expensive, after all. LOL!

I didn't know, for sure. I just wanted to finish such long post and that phrase come to my mind. I understand now!

On Thursday, December 2, 2021 at 10:01:28 PM UTC-8, Richard Hertz wrote:
> ...having made the product [- (x - α₁) (x – α₂) (1 - αx)] = P x³ + Q x² + R x + S to identify
> EXACTLY the values of (P, Q, R, S) you would found out what is the meaning of
> Q = - 1.000000106525...

That's another obvious mistake. Remember, you said:

> (Eq. 11, Einstein's paper) : (dx/dφ)²= 2A/B² + α/B² x −x² + αx³ ,
> where A = -3.764371725E-07 ; B² = 8.19161E+13 ; α = α = 2954.13 ;
> α1 = 1.43236E-11 ; α2 = 2.17378E-11 (geometrical units) gives
> (dx/dφ)²= 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22

but in Einstein's equation the coefficient of x^2 is -1, whereas merely substituting numerical values for A, B, and "a" you say this implies that the coefficient of x^2 is -1,000000106525. That's obviously wrong, an arithmetic mistake. So in a single post you have twice erroneously claimed that something different from 1 equals 1. The other one is that you claim a*a3 = 1, which is a simple algebraic mistake.

> I will do it for you, so you don't have excuses anymore.
> - (x - α₁) (x - α₂) (1 – α x)

Again, that's a simple algebraic mistake. You are assuming a*a3 = 1, which is false. The correct expression is a(a3-x)[-(x-a1)(x-a2)], which would equal your expression only if a*a3 = 1, which it is not. When you use the correct grade school algebra, you get the correct leading factor, resulting in the correct precession to the lowest order. You just need to eliminate your arithmetic and algebra mistakes.

On Friday, December 3, 2021 at 4:19:50 AM UTC-3, Townes Olson wrote:
> On Thursday, December 2, 2021 at 10:01:28 PM UTC-8, Richard Hertz wrote:
> > ...having made the product [- (x - α₁) (x – α₂) (1 - αx)] = P x³ + Q x² + R x + S to identify
> > EXACTLY the values of (P, Q, R, S) you would found out what is the meaning of
> > Q = - 1.000000106525...
> That's another obvious mistake. Remember, you said:
>
> > (Eq. 11, Einstein's paper) : (dx/dφ)²= 2A/B² + α/B² x −x² + αx³ ,
> > where A = -3.764371725E-07 ; B² = 8.19161E+13 ; α = α = 2954.13 ;
> > α1 = 1.43236E-11 ; α2 = 2.17378E-11 (geometrical units) gives
> > (dx/dφ)²= 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22
>
> but in Einstein's equation the coefficient of x^2 is -1, whereas merely substituting numerical values for A, B, and "a" you say this implies that the coefficient of x^2 is -1,000000106525. That's obviously wrong, an arithmetic mistake. So in a single post you have twice erroneously claimed that something different from 1 equals 1. The other one is that you claim a*a3 = 1, which is a simple algebraic mistake.
> > I will do it for you, so you don't have excuses anymore.
> > - (x - α₁) (x - α₂) (1 – α x)
> Again, that's a simple algebraic mistake. You are assuming a*a3 = 1, which is false. The correct expression is a(a3-x)[-(x-a1)(x-a2)], which would equal your expression only if a*a3 = 1, which it is not. When you use the correct grade school algebra, you get the correct leading factor, resulting in the correct precession to the lowest order. You just need to eliminate your arithmetic and algebra mistakes.

You are an idiot beyond redemption. I give up with you, miserable retarded!

On Thursday, December 2, 2021 at 11:59:35 PM UTC-8, Richard Hertz wrote:
> ax³ −x² + a/B² x + 2A/B² =
> a(x–a1)(x–a2)(x–a3) = (1–ax) [–(x–a1) (x–a2)]

The right-hand expression is incorrect. By simple algebra, we actually have
a(x–a1)(x–a2)(x–a3) = a(a3–x)[–(x–a1)(x–a2)]

which differs from your right-hand expression because a*a3 does not equal 1.. (Remember, the coefficient of x^2 is -1.) Using the correct expression, we get the correct leading factor (to the first order) on the integral, resulting in the correct value of the precession. This is not controversial, it is clearly and simply explained in widely available texts.

On Thursday, December 2, 2021 at 10:01:28 PM UTC-8, odious Richard Hertz wrote:

>
> Concede that my my result is an imbecility

Agreed/

Dick,

There are multiple errors in your postings. You will never figure them out because your imbecility is surpassed by your arrogance. Keep it up, dumbestfuck!

El viernes, 3 de diciembre de 2021 a las 4:59:35 UTC-3, Richard Hertz escribió:
> On Friday, December 3, 2021 at 4:19:50 AM UTC-3, Townes Olson wrote:
> > On Thursday, December 2, 2021 at 10:01:28 PM UTC-8, Richard Hertz wrote:
> > > ...having made the product [- (x - α₁) (x – α₂) (1 - αx)] = P x³ + Q x² + R x + S to identify
> > > EXACTLY the values of (P, Q, R, S) you would found out what is the meaning of
> > > Q = - 1.000000106525...
> > That's another obvious mistake. Remember, you said:
> >
> > > (Eq. 11, Einstein's paper) : (dx/dφ)²= 2A/B² + α/B² x −x² + αx³ ,
> > > where A = -3.764371725E-07 ; B² = 8.19161E+13 ; α = α = 2954.13 ;
> > > α1 = 1.43236E-11 ; α2 = 2.17378E-11 (geometrical units) gives
> > > (dx/dφ)²= 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22
> >
> > but in Einstein's equation the coefficient of x^2 is -1, whereas merely substituting numerical values for A, B, and "a" you say this implies that the coefficient of x^2 is -1,000000106525. That's obviously wrong, an arithmetic mistake. So in a single post you have twice erroneously claimed that something different from 1 equals 1. The other one is that you claim a*a3 = 1, which is a simple algebraic mistake.
> > > I will do it for you, so you don't have excuses anymore.
> > > - (x - α₁) (x - α₂) (1 – α x)
> > Again, that's a simple algebraic mistake. You are assuming a*a3 = 1, which is false. The correct expression is a(a3-x)[-(x-a1)(x-a2)], which would equal your expression only if a*a3 = 1, which it is not. When you use the correct grade school algebra, you get the correct leading factor, resulting in the correct precession to the lowest order. You just need to eliminate your arithmetic and algebra mistakes.
> You are an idiot beyond redemption. I give up with you, miserable retarded!

Seriously Richard, you should stop making, everyday, a fool of yourself.

On Friday, December 3, 2021 at 9:04:23 AM UTC-8, Paparios wrote:

> Seriously Richard, you should stop making, everyday, a fool of yourself.

No, no, Dick is the best entertainer of all the cranks.

What a disgrace of a person you are, Townes. Now you INSERT TEXT that I never wrote.
And I see that besides you and Dono, Paparios (MIGUEL FÉLIX RÍOS OJEDA) has joined the club of ignorant liars and imbeciles.

You wrote a text that I never used, because it involves to write α₃ = 1/α as the third root, and I NEVER WROTE THAT, because to
write α₃, which you replaced with your IDIOTIC a3, requires that I specifically find a "Subscript three: ₃" in a table and copy&paste it,
which never happened because I'm working only with the roots α₁ and α₂. It's visible that you have problems writing in that way.

Before all, to Paparios (Miguelito): Siempre supe que eras un boludo, pero ahora sos un boludo ignorante. La peor clase.

Now, Townes (AKA Monic Polynomial), this is your last post which I copy here:

-----------------------------------------------------------------------------------------------------------------------
Townes Olson , Dec 3, 2021, 1:30 AM UTC-8
On Thursday, December 2, 2021 at 11:59:35 PM UTC-8, Richard Hertz wrote:
> ax³ −x² + a/B² x + 2A/B² =
> a(x–a1)(x–a2)(x–a3) = (1–ax) [–(x–a1) (x–a2)]

The right-hand expression is incorrect. By simple algebra, we actually have
a(x–a1)(x–a2)(x–a3) = a(a3–x)[–(x–a1)(x–a2)]

which differs from your right-hand expression because a*a3 does not equal 1.. (Remember, the coefficient of x^2 is -1.)
Using the correct expression, we get the correct leading factor (to the first order) on the integral, resulting in the correct
value of the precession. This is not controversial, it is clearly and simply explained in widely available texts.
-----------------------------------------------------------------------------------------------------------------------

I could have a fest by highligting the imbecilities you posted, but I'll address only the essential.
THIS WAS INVENTED BY YOU (I rewrite using α instead of your stupid a):
****************
On Thursday, December 2, 2021 at 11:59:35 PM UTC-8, Richard Hertz wrote:
> α x³ −x² + α/B² x + 2A/B² = α (x–α₁)(x–α₂)(x–α₃) = (1–αx) [–(x–α₁) (x–α₂)]

The right-hand expression is incorrect. By simple algebra, we actually have α (x–α₁)(x–α₂)(x–α₃) =α (α₃–x)[–(x–α₁)(x–α₂)]
which differs from your right-hand expression because α*α₃ does not equal 1.
****************

Do you even realize HOW STUPID your ASSERTION α*α₃ ≠1 sounds, imbecile distorter of truth?

YOUR INVENTED a3, actually α₃, is EXACTLY α₃ = 1/α, the inverse of the Schwarzschild radius (in geometrical units).

I restaured the PHYSICAL VALUES for α₁ = 1/AP, α₂ = 1/PE and α = 2GMm/c² = Rs (for the Sun), in order to have a PHYSICAL expression

(Eq. 11, Einstein's paper) : (dx/dΦ)²= 2A/B² + α/B² x −x² + αx³ , which has units 1/(m. radian)² verified at EVERY TERM!

And my restauration from geometrical units to PHYSICAL UNITS, allowed me to write:

(dx/dΦ)²= 2A/B² + α/B² x −x² + αx³ = 2954 x³ - 1,000000106525 x² + 3.60613E-11 x - 3,1136E-22 , range (α₁, α₂) [units: 1/(m. radian)² ]

But this is TOO DIFFICULT for a retarded like you to UNDERSTAND.

This physical equation, which represent the ACTUAL MOTION of Mercury, has been verified in more than half a dozen scientific papers.

So, Washington state IDIOT, take your crap along with you monic polynomial, and GO FUCK YOURSELF.

The FRAUD is proven and, ultimately, is worse than what I wrote: The solution gives a NEGATIVE PRECESSION such as with THIS
EQUATION, Mercury MISSES to complete an entire orbit by a fraction ε = -1.6734E-07 radians.

That means that SOLVING (dΦ/dx) for a complete orbit GIVES (2π -1.6734E-07) radians instead of the expected (2π +5.0261E-07) rad.

Now, you (Townes), Dono and the newcomer Miguelito (AKA Paparios) have to return your degrees from HIGH SCHOOL, ignorants!

Richard Hertz wrote:

> What a disgrace of a person you are, Townes. Now you INSERT TEXT that I
> never wrote.
> And I see that besides you and Dono, Paparios (MIGUEL FÉLIX RÍOS OJEDA)
> has joined the club of ignorant liars and imbeciles.

no, he is a Pablo.

On Friday, December 3, 2021 at 10:01:09 AM UTC-8, Richard Hertz wrote:

> That means that SOLVING (dΦ/dx) for a complete orbit GIVES (2π -1.6734E-07) radians instead of the expected (2π +5.0261E-07) rad.
>
It only means that you are the village idiot. Thanks for the never ending entertainment, clown!

Dono. wrote:

> On Friday, December 3, 2021 at 10:01:09 AM UTC-8, Richard Hertz wrote:
>
>> That means that SOLVING (dΦ/dx) for a complete orbit GIVES (2π
>> -1.6734E-07) radians instead of the expected (2π +5.0261E-07) rad.
>>
> It only means that you are the village idiot. Thanks for the never
> ending entertainment, clown!

good observation, disregard shape, an orbit is always 2pi radians.

World Health Organization Says "No Evidence" Booster Jabs Would Offer
"Greater Protection" To The Healthy

El viernes, 3 de diciembre de 2021 a las 15:01:09 UTC-3, Richard Hertz escribió:

> Before all, to Paparios (Miguelito): Siempre supe que eras un boludo, pero ahora sos un boludo ignorante. La peor clase.
>

Keep up the good job of entertaining us. If you had any minimal research capability, you would not be writing all your nonsense here, but you will be writing a paper and submitting it to some journal (preferably cited by ISI) and you would find if some one (any one) disagree with the notion that you are nuts.
> Now, you (Townes), Dono and the newcomer Miguelito (AKA Paparios) have to return your degrees from HIGH SCHOOL, ignorants!

At least we have some degrees!!!

On Friday, December 3, 2021 at 3:28:45 PM UTC-3, Harif Kuloo wrote:
> Dono. wrote:
>
> > On Friday, December 3, 2021 at 10:01:09 AM UTC-8, Richard Hertz wrote:
> >
> >> That means that SOLVING (dΦ/dx) for a complete orbit GIVES (2π
> >> -1.6734E-07) radians instead of the expected (2π +5.0261E-07) rad..
> >>
> > It only means that you are the village idiot. Thanks for the never
> > ending entertainment, clown!
> good observation, disregard shape, an orbit is always 2pi radians.
>
> World Health Organization Says "No Evidence" Booster Jabs Would Offer
> "Greater Protection" To The Healthy

We are not talking about the real deal here, which are Newtonian orbits, based on Kepler!

You have to understand that, under GR conception, orbit of Mercury NEVER IS AN ELLIPSE, but AN SPIRAL!

The RELATIVISTIC equation dΦ/dx (in Einstein's 18 Nov. 1915 paper) is based on a cubic polynomial with
a THIRD TERM αx³ = 2954 x³ which FORCES the solution of the differential equation dΦ/dx to HAVE AN EXCESS OF
ε radians, which REPRESENTS the advance of Mercury's perihelion.

Just GET RID of αx³ (eliminate it from the differential equation) and the solution will give, EXACTLY, 2π radians.

Don't forget that, either with Newton or GR, the orbital path u(x) = u(1/r) NEVER IS COMPUTED. Only the results at
Perihelion and Aphelion.

To obtain the REAL DEAL for the motion of ANY planet, f(r, Φ, ψ) in polar coordinates, and integrate it ALONG THE
PERIMETER of the elliptical orbit has a VERY COMPLEX expression, and a formula MUCH MORE COMPLEX than
this simplification.

Is due to this problem that space agencies from US, Russia, China, India, France, Europe, Israel, etc., uses complex
polynomials of Nth. degree greater than 7 (which Le Verrier used almost 170 years ago), plus a LOT of modern
math to include almost every perturbation (difficult to model) from the rest of the celestial bodies in the Solar System.

In college, to study Newton dynamics on planets, a REDUCED LAGRANGIAN expression is used (Einstein did it also).

The orbit of a planet around the Sun is based on conservation of energy E (or A) and momentum L (or B), both taken as constants.

So, any BASIC solution for Newton or GR model is based on this general equation, in polar coordinates around Sun's Equator:

The equations of motion, [Total energy of the System, KE+U, in polar coordinates (r, ɸ)], are

1/2 m [ r². (dɸ/dt)² + (dr/dt)²] – GMm/r = - E

Being r² dɸ/dt = L/m a conserved quantity (angular moment L divided by the mass m of the planet).

The target is to derive dɸ/dr from the above equation, which is done with a couple of tricks.

But ALWAYS, under Newton, the solution of dɸ/dr for an ENTIRE ORBIT gives, EXACTLY, 2π radians.

GR claims a positive excess of ε radians on this equation, when modified under GR terms, BUT FAILS.

On Friday, December 3, 2021 at 10:57:20 AM UTC-8, Paparios wrote:
> El viernes, 3 de diciembre de 2021 a las 15:01:09 UTC-3, Richard Hertz escribió:
>
> > Before all, to Paparios (Miguelito): Siempre supe que eras un boludo, pero ahora sos un boludo ignorante. La peor clase.
> >
> Keep up the good job of entertaining us. If you had any minimal research capability, you would not be writing all your nonsense here, but you will be writing a paper and submitting it to some journal (preferably cited by ISI) and you would find if some one (any one) disagree with the notion that you are nuts.

Can you imagine the coronary the Dick would be getting when he starts getting the rejectionsto the garbage he's would be submitting?
> > Now, you (Townes), Dono and the newcomer Miguelito (AKA Paparios) have to return your degrees from HIGH SCHOOL, ignorants!
> At least we have some degrees!!!
So does the Dick. Sadly.

On Friday, December 3, 2021 at 3:57:20 PM UTC-3, Paparios wrote:
> El viernes, 3 de diciembre de 2021 a las 15:01:09 UTC-3, Richard Hertz escribió:
>
> > Before all, to Paparios (Miguelito): Siempre supe que eras un boludo, pero ahora sos un boludo ignorante. La peor clase.
> >
> Keep up the good job of entertaining us. If you had any minimal research capability, you would not be writing all your nonsense here, but you will be writing a paper and submitting it to some journal (preferably cited by ISI) and you would find if some one (any one) disagree with the notion that you are nuts.
> > Now, you (Townes), Dono and the newcomer Miguelito (AKA Paparios) have to return your degrees from HIGH SCHOOL, ignorants!
> At least we have some degrees!!!

Don't be an asshole! I started this shit about 1 month ago because I saw funny things on Einstein's paper, which didn't fit reason.

Never before I gave a shit about GR, except to prove that Einstein was a complete imbecile, cretin and crook. The Einstein-Besso
manuscript, which I analyzed very carefully months ago, gave me proofs of their complete ineptitude. I couldn't believe what I read.

And, in this way, this story started.

I don't have any interest in research about GR. I'm just learning new things or relearning things that I haven't used for decades.

I'm retired, and with a lot of time available to study whatever I want.

But studying about the shitty relativity and post here controversial truths is PRICELESS, and very entertaining when I observe reactions.

On Friday, December 3, 2021 at 11:16:00 AM UTC-8, Richard Hertz wrote:
>
> But ALWAYS, under Newton, the solution of dɸ/dr for an ENTIRE ORBIT gives, EXACTLY, 2π radians.
>

Yet, for 400 years, the observation has proven Newton wron. So, you get to eat shit once again.

> GR claims a positive excess of ε radians on this equation, when modified under GR terms, BUT FAILS.

Only in your demented mind, village clown.

Richard Hertz wrote:

>> good observation, disregard shape, an orbit is always 2pi radians.
>>
>> World Health Organization Says "No Evidence" Booster Jabs Would Offer
>> "Greater Protection" To The Healthy
>
> We are not talking about the real deal here, which are Newtonian orbits,
> based on Kepler!
>
> You have to understand that, under GR conception, orbit of Mercury NEVER
> IS AN ELLIPSE, but AN SPIRAL!
>
> The RELATIVISTIC equation dΦ/dx (in Einstein's 18 Nov. 1915 paper) is
> based on a cubic polynomial with a THIRD TERM αx³ = 2954 x³ which FORCES
> the solution of the differential equation dΦ/dx to HAVE AN EXCESS OF ε
> radians, which REPRESENTS the advance of Mercury's perihelion.

I wish could understand what you say. You may spiral all you want.
Radians are about angles, how many degrees are around a point in a plane?
360 = 2pi. How many steradians?

On Friday, December 3, 2021 at 11:23:46 AM UTC-8, Richard Hertz wrote:
> On Friday, December 3, 2021 at 3:57:20 PM UTC-3, Paparios wrote:
> > El viernes, 3 de diciembre de 2021 a las 15:01:09 UTC-3, Richard Hertz escribió:
> >
> > > Before all, to Paparios (Miguelito): Siempre supe que eras un boludo, pero ahora sos un boludo ignorante. La peor clase.
> > >
> > Keep up the good job of entertaining us. If you had any minimal research capability, you would not be writing all your nonsense here, but you will be writing a paper and submitting it to some journal (preferably cited by ISI) and you would find if some one (any one) disagree with the notion that you are nuts.
> > > Now, you (Townes), Dono and the newcomer Miguelito (AKA Paparios) have to return your degrees from HIGH SCHOOL, ignorants!
> > At least we have some degrees!!!
> I started this shit about 1 month ago

You started your shit years ago. Thanks for the show!
> I'm retired,
....and deeply demented.

On Friday, December 3, 2021 at 11:25:09 AM UTC-8, Dono. wrote:
> On Friday, December 3, 2021 at 11:16:00 AM UTC-8, Richard Hertz wrote:
> >
> > But ALWAYS, under Newton, the solution of dɸ/dr for an ENTIRE ORBIT gives, EXACTLY, 2π radians.
> >
Yet, for 400 years, the observation has proven Newton wrong. So, you get to eat shit once again.
> > GR claims a positive excess of ε radians on this equation, when modified under GR terms, BUT FAILS.
Only in your demented mind, village clown.

On Friday, December 3, 2021 at 10:01:09 AM UTC-8, Richard Hertz wrote:
> > ax³ −x² + a/B² x + 2A/B² =
> > a(x–a1)(x–a2)(x–a3) = (1–ax) [–(x–a1) (x–a2)]
>
> The right-hand expression is incorrect. By simple algebra, we actually have
> a(x–a1)(x–a2)(x–a3) = a(a3–x)[–(x–a1)(x–a2)]
> which differs from your right-hand expression because a*a3 does not equal 1.
> (Remember, the coefficient of x^2 is -1.) Using the correct expression, we get
> the correct leading factor (to the first order) on the integral, resulting in the
> correct value of the precession.
>
> YOUR INVENTED a3 is EXACTLY a3 = 1/a...

No, it is not. The quantity a3 is the third root of the cubic polynomial, and per elementary algebra it is definitely not equal to 1/a. That's the fundamental mistake that you are making, and that's why you are getting the wrong precession rate. After correcting that mistake, you get the correct precession rate.

On Friday, December 3, 2021 at 4:47:15 PM UTC-3, Townes Olson wrote:
> On Friday, December 3, 2021 at 10:01:09 AM UTC-8, Richard Hertz wrote:
> > > ax³ −x² + a/B² x + 2A/B² =
> > > a(x–a1)(x–a2)(x–a3) = (1–ax) [–(x–a1) (x–a2)]
> >
> > The right-hand expression is incorrect. By simple algebra, we actually have
> > a(x–a1)(x–a2)(x–a3) = a(a3–x)[–(x–a1)(x–a2)]
> > which differs from your right-hand expression because a*a3 does not equal 1.
> > (Remember, the coefficient of x^2 is -1.) Using the correct expression, we get
> > the correct leading factor (to the first order) on the integral, resulting in the
> > correct value of the precession.
> >
> > YOUR INVENTED a3 is EXACTLY a3 = 1/a...
>
> No, it is not. The quantity a3 is the third root of the cubic polynomial, and per elementary algebra it is definitely not equal to 1/a. That's the fundamental mistake that you are making, and that's why you are getting the wrong precession rate. After correcting that mistake, you get the correct precession rate.

I don't know from where in Washington state came you, son of Igor. Maybe from a general clinic for severely retarded poor
people, and they let use to use Internet since one month ago, poor little guy. Do your parents know that you're online late at night?

You need to have permanent custody, because your rotten brain poses a danger to you and who surround you.

The extraction of the third root of a polynomial, given that you know the other two (FORCED TO BE α₁ = 1/AP, α₂ = 1/PE ), is
a process that is known since 250 years!

Go and find out the method, cretin.

Also, if you have P(x) = [- (x - α₁) (x – α₂) (1 - αx)], the third root is at plain sight (x = 1/α).

Why do you persist with this, imbecile? That's the decomposition that Einstein himself wrote. ou need to read the translation
of the original paper. Here is a link:

Einstein's Paper:“Explanation of the Perihelion Motion of Mercury from General Relativity Theory”
Anatoli Vankov, January 2011

https://www.researchgate.net/publication/228923053_Einstein%27s_PaperExplanation_of_the_Perihelion_Motion_of_Mercury_from_General_Relativity_Theory

Read it, shut up and fuck off!