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tech / sci.physics.relativity / Re: [SR] Question

SubjectAuthor
* [SR] QuestionRichard Hachel
`* Re: [SR] QuestionMikko
 +- Re: [SR] QuestionMaciej Wozniak
 `* Re: [SR] QuestionRichard Hachel
  +* Re: [SR] Questionrotchm
  |+- Re: [SR] QuestionMaciej Wozniak
  |`- Re: [SR] QuestionRichard Hachel
  +* Re: [SR] QuestionStan Fultoni
  |+- Re: [SR] QuestionMaciej Wozniak
  |+- Re: [SR] QuestionMikko
  |`* Re: [SR] QuestionRichard Hachel
  | +* Re: [SR] QuestionPython
  | |`* Re: [SR] QuestionRichard Hachel
  | | `* Re: [SR] QuestionPython
  | |  +- Re: [SR] QuestionMaciej Wozniak
  | |  +* Re: [SR] QuestionRichard Hachel
  | |  |`* Re: [SR] QuestionPython
  | |  | `- Re: [SR] QuestionRichard Hachel
  | |  `* Re: [SR] QuestionRichard Hachel
  | |   `* Re: [SR] QuestionPython
  | |    +- Re: [SR] QuestionMaciej Wozniak
  | |    `* Re: [SR] QuestionRichard Hachel
  | |     +* Re: [SR] QuestionStan Fultoni
  | |     |`* Re: [SR] QuestionRichard Hachel
  | |     | `* Re: [SR] QuestionStan Fultoni
  | |     |  `* Re: [SR] QuestionRichard Hachel
  | |     |   +- Re: [SR] QuestionRichard Hachel
  | |     |   `* Re: [SR] QuestionStan Fultoni
  | |     |    `* Re: [SR] QuestionRichard Hachel
  | |     |     `* Re: [SR] QuestionStan Fultoni
  | |     |      +* Re: [SR] QuestionRichard Hachel
  | |     |      |+* Re: [SR] QuestionStan Fultoni
  | |     |      ||`* Re: [SR] QuestionRichard Hachel
  | |     |      || `* Re: [SR] QuestionStan Fultoni
  | |     |      ||  `* Re: [SR] QuestionRichard Hachel
  | |     |      ||   `* Re: [SR] QuestionStan Fultoni
  | |     |      ||    `* Re: [SR] QuestionRichard Hachel
  | |     |      ||     `- Re: [SR] QuestionStan Fultoni
  | |     |      |`- Re: [SR] QuestionMikko
  | |     |      `* Re: [SR] QuestionMaciej Wozniak
  | |     |       `* Re: [SR] QuestionRichard Hachel
  | |     |        `* Re: [SR] QuestionMaciej Wozniak
  | |     |         +- Re: [SR] QuestionRichard Hachel
  | |     |         +* Re: [SR] QuestionRichard Hachel
  | |     |         |`* Re: [SR] QuestionMaciej Wozniak
  | |     |         | `* Re: [SR] QuestionRichard Hachel
  | |     |         |  `- Re: [SR] QuestionStan Fultoni
  | |     |         +* Re: [SR] QuestionRichard Hachel
  | |     |         |`- Re: [SR] QuestionMaciej Wozniak
  | |     |         `* Re: [SR] QuestionPython
  | |     |          `* Re: [SR] QuestionMaciej Wozniak
  | |     |           `- Re: [SR] QuestionPython
  | |     `* Re: [SR] QuestionPython
  | |      +* Re: [SR] QuestionMaciej Wozniak
  | |      |`- Re: [SR] QuestionRichard Hachel
  | |      `* Re: [SR] QuestionRichard Hachel
  | |       `- Re: [SR] QuestionPython
  | `* Re: [SR] QuestionStan Fultoni
  |  `* Re: [SR] QuestionRichard Hachel
  |   `- Re: [SR] QuestionStan Fultoni
  +- Re: [SR] QuestionMikko
  `* Re: [SR] QuestionPython
   +- Re: [SR] QuestionRichard Hachel
   `- Re: [SR] QuestionMaciej Wozniak

Pages:123
Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Wed, 11 May 2022 11:04 UTC

Le 11/05/2022 à 12:45, Python a écrit :
> Keep your head in the sand, the FACT is that your claim
> leads to a direct contradiction.
>
> You have snipped the scenario I posted, hence did not
> answer my question. You guess I'm right and you wrong, right?
>
> There is no consideration about proper/other time there,
> only the reading of clocks at start and arrival...
>
> If you weren't a coward you would have examined the situation
> from the point of view of the uniformly moving twin in your
> tau ceti scenario. You could have noticed yourself how your
> claim leads to a contradiction.

Yes, you are right, we must limit ourselves to very specific facts.

But what do you think I'm doing when I talk about Tr and To?

I'm talking about taking departure and arrival clocks into account.

And not Santa Claus.

Tr is the time difference of the clocks in the proper frame, and To the
time of the clocks in the observing frame.

It's that simple.

I said that if the events E1 (departure) and E2 (arrival) are joint for
the two protagonists, then the proper times Tr will also be equal.

This means that we can directly use the formula To²=Tr²+Et² in the Tau
Ceti (as long as we stay within the defined framework of two reference
frames with respect to each other).

And also the formula Tr=sqrt(2x/a) to have directly, and much more simply,
the proper time of the traveler or the lab particle.

I repeat what I always say: "The theory of relativity is very simple
mathematically; it doesn't fly any further than sines, cosines, and square
roots, even the integrations we do in high school classes. But it's full
of little traps."

R.H.

Re: [SR] Question

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 by: Python - Wed, 11 May 2022 11:21 UTC

Richard "Hachel" Lengrand (M.D.) wrote:
> Le 11/05/2022 à 00:39, Python a écrit :
>> What do you think will happen is this scenario below,
>> where A is a "stay at home" twin and B starts to
>> travel to Tau Ceti and back with a given initial
>> speed and a constant accelation (directed to A) ?
>
>
> [snip evading bs]

The question is simple : will clocks display the same
elapsed time. The answer can be YES or NO.

Re: [SR] Question

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Subject: Re: [SR] Question
From: maluwozn...@gmail.com (Maciej Wozniak)
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 by: Maciej Wozniak - Wed, 11 May 2022 11:29 UTC

On Wednesday, 11 May 2022 at 13:21:09 UTC+2, Python wrote:
> Richard "Hachel" Lengrand (M.D.) wrote:
> > Le 11/05/2022 à 00:39, Python a écrit :
> >> What do you think will happen is this scenario below,
> >> where A is a "stay at home" twin and B starts to
> >> travel to Tau Ceti and back with a given initial
> >> speed and a constant accelation (directed to A) ?
> >
> >
> > [snip evading bs]
>
> The question is simple : will clocks display the same
> elapsed time. The answer can be YES or NO.

And while brainwashed morons are resolving their
gedanken scenarios, real clocks of real GPS give YES.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Wed, 11 May 2022 11:40 UTC

Le 11/05/2022 à 13:21, Python a écrit :
> The question is simple : will clocks display the same
> elapsed time. The answer can be YES or NO.

Quelles horloges?

Je le répète, s'il s'agit de comparer les horloges du voyageur de
l'espace qui évolue
à vitesse constante Vr (ou Vo si tu veux) et celles du voyageur de Tau
Ceti en mouvement accéléré,
les deux temps propres, au final, seront égaux, pour peu qu'ils soient
aussi égaux
dans le référentiel Terrestre.

Je le répète aussi : mais cela n'est valable que pour une vitesse
uniforme constante dans la cas du référentiel choisi comme référentiel
de vitesse constante.

Si je divise en deux le trajet, et que j'en fais la première partie à
Vo=0.54c et l'autre à Vo=0.62c,
il est évident que nous ne sommes plus dans le cas de figure que je
décris. Même si c'est deux référentiels de vitesse uniforme. Je ne
peux pas non plus ajouter une carotte à un navet.

Cela ne reste valable que dans le cas très précis d'une comparaison des
temps propres entre deux référentiels immuables et bien définis (même
si l'un d'entre eux est accéléré, et l'autre en vitesse constante).

R.H.

--
"Mais ne nous trompons pas. Il n'y a pas que de la violence
avec des armes : il y a des situations de violence."
Abbé Pierre.
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Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Wed, 11 May 2022 13:14 UTC

On Tuesday, May 10, 2022 at 9:48:09 AM UTC-7, Richard Hachel wrote:
> > Again, the contradiction is that for any three events e1,e2,e3 on a constantly
> > accelerating path, where the accelerating clock reads the proper time values
> > tau1,tau2,tau3, your claim is that
> >
> > tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> > tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
> > tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
> >
> > but these three relations are self-contradictory. Do you understand this?
>
> Why?

Because (tau2-tau1) + (tau3-tau2) = (tau3-tau1), but if you add the right sides
of the first two expression, it does not equal the right side of the third expression
unless the three events e1,e2,e3 are co-linear, meaning the accelerating path is
not accelerating. This proves that your claims are self-contradictory.

Re: [SR] Question

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 by: Python - Wed, 11 May 2022 13:31 UTC

Le 11/05/2022 à 13:40, Richard Hachel a écrit :
> Le 11/05/2022 à 13:21, Python a écrit :
>> The question is simple : will clocks display the same
>> elapsed time. The answer can be YES or NO.
>
> Quelles horloges?

Which clocks? Obviously the ones both travellers
carried with them, identical clocks, setted at
t=0 when the journey started at the place the
journey started.

Stop producing bullshit instead of answering simple
questions.

Re: [SR] Question

<4051f529-0add-40c2-a9bc-7fabd8733a56n@googlegroups.com>

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Subject: Re: [SR] Question
From: maluwozn...@gmail.com (Maciej Wozniak)
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 by: Maciej Wozniak - Wed, 11 May 2022 13:43 UTC

On Wednesday, 11 May 2022 at 15:31:09 UTC+2, Python wrote:
> Le 11/05/2022 à 13:40, Richard Hachel a écrit :
> > Le 11/05/2022 à 13:21, Python a écrit :
> >> The question is simple : will clocks display the same
> >> elapsed time. The answer can be YES or NO.
> >
> > Quelles horloges?
> Which clocks? Obviously the ones both travellers
> carried with them, identical clocks, setted at

Oh, stinker Python is opening its muzzle again,
and trying to pretend he knows something.
Tell me, poor stinker, what is your definition of
a "theory" in the terms of Peano arithmetic?
See: if a theorem is going to be a part of a theory,
it has to be formulable in the language of the
theory. Do you get it? Or are you too stupid even for
that, poor stinker?

And identical clocks of Giant Guru are only so
obvious for brainwashed relativistic morons;
as anyone can check in GPS, sane people from
outside of relativistic gedankens are not going
to rely on them.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Wed, 11 May 2022 13:50 UTC

Le 11/05/2022 à 15:14, Stan Fultoni a écrit :
> On Tuesday, May 10, 2022 at 9:48:09 AM UTC-7, Richard Hachel wrote:
>> > Again, the contradiction is that for any three events e1,e2,e3 on a constantly
>> > accelerating path, where the accelerating clock reads the proper time values
>> > tau1,tau2,tau3, your claim is that
>> >
>> > tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
>> > tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
>> > tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
>> >
>> > but these three relations are self-contradictory. Do you understand this?
>>
>> Why?
>
> Because (tau2-tau1) + (tau3-tau2) = (tau3-tau1),

tau2-tau1=Tr1
tau3-tau2=Tr2
tau3-tau1=Tr

So we have : Tr=Tr1+Tr2

In effect.

And also To=To1+To2 in the other trame.

> but if you add the right sides
> of the first two expression, it does not equal the right side of the third
> expression
> unless the three events e1,e2,e3 are co-linear, meaning the accelerating path is
> not accelerating. This proves that your claims are self-contradictory.

In the first trame (constant speed Vr):
Tr=x/Vr
To=Tr.sqrt(1+Vr²/c²)

In the second frame (accelerated):
Tr=sqrt(x/a)
To=Tr.sqrt(1+2c²/ax)

Tr=Tr
To=To

In the two cases, To²=Tr²+(x²/c²)

R.H.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Wed, 11 May 2022 13:52 UTC

Le 11/05/2022 à 15:31, Python a écrit :
> Le 11/05/2022 à 13:40, Richard Hachel a écrit :
>> Le 11/05/2022 à 13:21, Python a écrit :
>>> The question is simple : will clocks display the same
>>> elapsed time. The answer can be YES or NO.
>>
>> Quelles horloges?
>
> Which clocks? Obviously the ones both travellers
> carried with them, identical clocks, setted at
> t=0 when the journey started at the place the
> journey started.

Ah oui?

Et qu'est ce que tu crois que je fais? Que je m'amuse avec des tomates et
des navets?

Tu deviens "ridiculous".

R.H.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Wed, 11 May 2022 14:19 UTC

Le 11/05/2022 à 15:43, Maciej Wozniak a écrit :

> Oh, stinker Python is opening its muzzle again,
> and trying to pretend he knows something.

C'est clair que de croire que je ne sais pas que le temps mis pour faire
un voyage,
c'est l'heure d'arrivée moins l'heure du départ, c'est quand même
être un peu fêlé du bocal.

Son obstruction voulue systématique à tout ce que je dis l'a rendu fou.

R.H.

Re: [SR] Question

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 by: Python - Wed, 11 May 2022 14:29 UTC

Le 11/05/2022 à 15:52, Richard Hachel a écrit :
> Le 11/05/2022 à 15:31, Python a écrit :
>> Le 11/05/2022 à 13:40, Richard Hachel a écrit :
>>> Le 11/05/2022 à 13:21, Python a écrit :
>>>> The question is simple : will clocks display the same
>>>> elapsed time. The answer can be YES or NO.
>>>
>>> Quelles horloges?
>>
>> Which clocks? Obviously the ones both travellers
>> carried with them, identical clocks, setted at
>> t=0 when the journey started at the place the
>> journey started.
>
> Ah oui?
>
> Et qu'est ce que tu crois que je fais? Que je m'amuse avec des tomates
> et des navets?

It's worse than that. You can't answer such a simple question.

> Tu deviens "ridiculous".

Well... You've already been, and these time more than ever.

Moreover, it's disrespectful to post in French in an English
speaking group (the other way also). If you want to discuss
the issue in French do it in fr.sci.physique, if you want to
discuss the issue here to it in English, at least try.

Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Wed, 11 May 2022 18:50 UTC

On Wednesday, May 11, 2022 at 6:50:14 AM UTC-7, Richard Hachel wrote:
> >> > Again, the contradiction is that for any three events e1,e2,e3 on a constantly
> >> > accelerating path, where the accelerating clock reads the proper time values
> >> > tau1,tau2,tau3, your claim is that
> >> >
> >> > tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> >> > tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
> >> > tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
> >> >
> >> > but these three relations are self-contradictory. Do you understand this?
> >>
> >> Why?
> >
> > Because (tau2-tau1) + (tau3-tau2) = (tau3-tau1), but if you add the right sides
> > of the first two expression, it does not equal the right side of the third
> > expression unless the three events e1,e2,e3 are co-linear, meaning the
> > accelerating path is not accelerating. This proves that your claims are
> > self-contradictory.
>
> In the two cases, To²=Tr²+(x²/c²)

No, the above explains that your beliefs are self-contradictory, because you
claim that the sum of the first two square roots above equals the third square
root, which is false unless the "accelerating path" is not accelerating. Therefore,
your beliefs are self-contradictory. Do you understand this?

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Thu, 12 May 2022 11:01 UTC

Le 11/05/2022 à 20:50, Stan Fultoni a écrit :
> No, the above explains that your beliefs are self-contradictory, because you
> claim that the sum of the first two square roots above equals the third square
> root, which is false unless the "accelerating path" is not accelerating.
> Therefore,
> your beliefs are self-contradictory. Do you understand this?

I read very carefully and with great interest the answers you gave me.

I understand the difficulties that my assertions and theses may cause,
including that they may be perceived as contradictory or absurd.

But the things I say, strange as they seem, are nevertheless true.

I have long talked about the theory of realtivity in constant velocity
motion (with the cutting edge of reasoning to properly explain Langevin's
obvious traveler's paradox), and how it could be solved by understanding
things correctly.

Today, I managed to understand the error of the theoreticians concerning
proper time in accelerated media.

They miscalculate it, and I know why.

There is a big mistake on their part (the special theory of relativity is
at the mathematical level of an average 18-year-old high school student,
but it is full of little pitfalls) comes from the fact that they do not
apply correctly the relation To²=Tr²+Et²

And the integration they do is then not correct at all.

We then end up with a false Tr.

The answer should be:
Tr=sqrt(1+2x/a)
To=(x/c).sqrt(1+2c²/ax)

We must also have, for constant media:
To=Tr.sqrt(1+Vr²/c²)
or also,
Tr=To.sqrt(1-Vo²/c²)

And for accelerated media:
To=Tr.sqrt(1+(1/4)Vrf(final)²/c²
or also
To=sqrt(1+Vrm(average speed)²/c²)

Thank you for your attention.

R.H.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Thu, 12 May 2022 11:20 UTC

Le 12/05/2022 à 13:01, Richard Hachel a écrit :
> Le 11/05/2022 à 20:50, Stan Fultoni a écrit :
>> No, the above explains that your beliefs are self-contradictory, because you
>> claim that the sum of the first two square roots above equals the third square
>> root, which is false unless the "accelerating path" is not accelerating.
>> Therefore,
>> your beliefs are self-contradictory. Do you understand this?
>
> I read very carefully and with great interest the answers you gave me.
>
> I understand the difficulties that my assertions and theses may cause, including
> that they may be perceived as contradictory or absurd.
>
> But the things I say, strange as they seem, are nevertheless true.
>
> I have long talked about the theory of realtivity in constant velocity motion
> (with the cutting edge of reasoning to properly explain Langevin's obvious
> traveler's paradox), and how it could be solved by understanding things correctly.
>
> Today, I managed to understand the error of the theoreticians concerning proper
> time in accelerated media.
>
> They miscalculate it, and I know why.
>
> There is a big mistake on their part (the special theory of relativity is at the
> mathematical level of an average 18-year-old high school student, but it is full
> of little pitfalls) comes from the fact that they do not apply correctly the
> relation To²=Tr²+Et²
>
> And the integration they do is then not correct at all.
>
> We then end up with a false Tr.
>
> The answer should be:
> Tr=sqrt(1+2x/a)
> To=(x/c).sqrt(1+2c²/ax)
>
> We must also have, for constant media:
> To=Tr.sqrt(1+Vr²/c²)
> or also,
> Tr=To.sqrt(1-Vo²/c²)
>
> And for accelerated media:
> To=Tr.sqrt(1+(1/4)Vrf(final)²/c²
> or also
> To=sqrt(1+Vrm(average speed)²/c²)

Typographical error : To=Tr.sqrt(1+Vrm²/c²)

>
> Thank you for your attention.
>
> R.H.

Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Thu, 12 May 2022 18:37 UTC

On Thursday, May 12, 2022 at 4:01:26 AM UTC-7, Richard Hachel wrote:
> > The above explains that your beliefs are self-contradictory, because you
> > claim that the sum of the first two square roots above equals the third square
> > root, which is false unless the "accelerating path" is not accelerating.
> > Therefore, your beliefs are self-contradictory. Do you understand this?
>
> I understand the difficulties that my assertions and theses may cause,
> including that they may be perceived as contradictory or absurd.

It isn't a matter of perception. Your claims amount to the assertion
that 1 = 0. This is because you say the difference between the sum of
the first two square roots and the third square root is zero, but it is not
zero (unless your accelerating path is no accelerating. Your claims are
therefore self-contradictory. Do you understand this?

Re: [SR] Question

<m6UIKA3jXRODVH9x7LIsfkJC6EQ@jntp>

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Thu, 12 May 2022 19:07 UTC

Le 12/05/2022 à 20:37, Stan Fultoni a écrit :
> It isn't a matter of perception. Your claims amount to the assertion
> that 1 = 0. This is because you say the difference between the sum of
> the first two square roots and the third square root is zero, but it is not
> zero (unless your accelerating path is no accelerating. Your claims are
> therefore self-contradictory. Do you understand this?

I said that there was a fundamental relation which linked proper time and
improper (observable) time.

We have To²=Tr²+Et²

Et=x/c

Take the case of a twin who leaves the earth for 18 years of his own time
(we don't know at what constant speed he makes the journey, we don't even
need to).

Let's bypass Tau-Ceti at 12 ly and come back to earth.

Let To²=Tr²+Et²

To=sqrt(18²+24²)=30 years

But the equation still works for accelerated motion.

Physicists calculate that if the traveler accelerates from the earth at
a=10m/s² the improper (observable) terrestrial time will be To=12,915
years.

I have the same calculation as them.

But they are mistaken when they want to give proper time, and make a bad
mathematical integration.

The proper time can be calculated in several ways, the first being a
simple one:
Tr=sqrt(2x/a)

But it can also be calculated with the same formula as for displacements
at constant speed.

To²=Tr²+Et²

Let Tr=sqrt(To²-Et²)

Or To=sqrt(12.915²-12²)=~4.775 years.

I don't see where you can run into any problems.

R.H.

Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Thu, 12 May 2022 22:36 UTC

On Thursday, May 12, 2022 at 12:07:06 PM UTC-7, Richard Hachel wrote:
> > It isn't a matter of perception. Your claims amount to the assertion
> > that 1 = 0. This is because you say the difference between the sum of
> > the first two square roots and the third square root is zero, but it is not
> > zero (unless your accelerating path is no accelerating. Your claims are
> > therefore self-contradictory. Do you understand this?
>
> I don't see where you can run into any problems.

Again, the problem is that your beliefs imply 1 = 0, as explained in detail
in the previous message.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Fri, 13 May 2022 10:10 UTC

Le 13/05/2022 à 00:36, Stan Fultoni a écrit :
> On Thursday, May 12, 2022 at 12:07:06 PM UTC-7, Richard Hachel wrote:
>> > It isn't a matter of perception. Your claims amount to the assertion
>> > that 1 = 0. This is because you say the difference between the sum of
>> > the first two square roots and the third square root is zero, but it is not
>> > zero (unless your accelerating path is no accelerating. Your claims are
>> > therefore self-contradictory. Do you understand this?
>>
>> I don't see where you can run into any problems.
>
> Again, the problem is that your beliefs imply 1 = 0, as explained in detail
> in the previous message.

I don't understand this story of the sum of square roots being equal to
zero. What do you want to talk about?

R.H.

Re: [SR] Question

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Subject: Re: [SR] Question
From: maluwozn...@gmail.com (Maciej Wozniak)
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 by: Maciej Wozniak - Fri, 13 May 2022 10:33 UTC

On Friday, 13 May 2022 at 00:36:08 UTC+2, Stan Fultoni wrote:
> On Thursday, May 12, 2022 at 12:07:06 PM UTC-7, Richard Hachel wrote:
> > > It isn't a matter of perception. Your claims amount to the assertion
> > > that 1 = 0. This is because you say the difference between the sum of
> > > the first two square roots and the third square root is zero, but it is not
> > > zero (unless your accelerating path is no accelerating. Your claims are
> > > therefore self-contradictory. Do you understand this?
> >
> > I don't see where you can run into any problems.
> Again, the problem is that your beliefs imply 1 = 0, as explained in detail
> in the previous message.

So did the beliefs of your idiot guru, as explained
many times here.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Fri, 13 May 2022 12:34 UTC

Le 13/05/2022 à 12:33, Maciej Wozniak a écrit :
> So did the beliefs of your idiot guru, as explained
> many times here.

What are you talking about, Maciej?
I am explaining here something very important.
I am saying that there are, in the currently accepted theory of
relativity, paradoxes and absurdities.
I beg you to believe that when you say these things, you get a kick out of
it.
And you take it even more in the nose, when you say: "But I'm going to
show you my big cock, I'm going to explain them to you".
No need to add to the fact that there are gurus or visionaries.
I don't see the point of it.
Sheep, for sure, that's 99.99% of humanity.
The enlightened ones, the gurus, they have to be tested. But you have to
test them scientifically and honestly.
This is obviously very little done.

R.H.

Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Fri, 13 May 2022 13:31 UTC

On Friday, May 13, 2022 at 3:10:19 AM UTC-7, Richard Hachel wrote:
> I don't understand this story of the sum of square roots being equal to
> zero. What do you want to talk about?

As explained before, the contradiction is that for any three events e1,e2,e3 on
a constantly accelerating path, where the accelerating clock reads the proper
time values tau1,tau2,tau3, your claim is that

tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]

but these three relations are self-contradictory, because

(tau2-tau1) + (tau3-tau2) = (tau3-tau1)

and if you add the right sides of the first two expression, it does not equal
the right side of the third expression unless the three events e1,e2,e3 are
co-linear, meaning the accelerating path is not accelerating. This proves
that your claims are self-contradictory.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Fri, 13 May 2022 15:36 UTC

Le 13/05/2022 à 15:31, Stan Fultoni a écrit :

Let "Tr1"="Tr2"="Tr3"/2 et x1+x2=x3

> tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]

Tr1² = To1² - (x1/c)²

> tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]

? ? ?

> tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]

Tr3² = To3² - (x3/c)²
>
> but these three relations are self-contradictory, because
>
> (tau2-tau1) + (tau3-tau2) = (tau3-tau1)

Yes.

But is correct your second affirmation?

R.H.

Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Fri, 13 May 2022 18:55 UTC

On Friday, May 13, 2022 at 8:36:12 AM UTC-7, Richard Hachel wrote:
> > tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> > tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
> > tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
>
> But is correct your second affirmation?

None of those three expressions are correct (for an accelerating path), but YOU are affirming that all three of them are correct. Remember, you claim that the elapsed proper time along a uniform path (no acceleration) between two given events ei and ej is sqrt[(ti-tj)^2) - (xi-xj)^2)], and you also claim that this equals the elapsed proper time along a path undergoing constant proper acceleration between those same two events. Your statements refer to any two events, so you are affirming all three of the equations above. But the sum of the first two square roots does not equal the third square root (unless the "accelerating path" is not accelerating), so your beliefs are contradictory. Understand?

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Fri, 13 May 2022 19:01 UTC

Le 13/05/2022 à 20:55, Stan Fultoni a écrit :
> On Friday, May 13, 2022 at 8:36:12 AM UTC-7, Richard Hachel wrote:
>> > tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
>> > tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
>> > tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
>>
>> But is correct your second affirmation?
>
> None of those three expressions are correct (for an accelerating path), but YOU
> are affirming that all three of them are correct. Remember, you claim that the
> elapsed proper time along a uniform path (no acceleration) between two given
> events ei and ej is sqrt[(ti-tj)^2) - (xi-xj)^2)], and you also claim that this
> equals the elapsed proper time along a path undergoing constant proper
> acceleration between those same two events. Your statements refer to any two
> events, so you are affirming all three of the equations above. But the sum of the
> first two square roots does not equal the third square root (unless the
> "accelerating path" is not accelerating), so your beliefs are contradictory.
> Understand?

I see you don't understand what I'm saying.

I will prepare a small explanatory text that I will post maybe tomorrow if
I have time.

On the other hand, I understand your grievances.

But you can't make an omelette without at least breaking the shell of an
egg.

In the meantime, if you have time, patiently re-read what I've written
these days on accelerated repositories.

You will see that it is not so stupid.

R.H.

Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
Injection-Date: Fri, 13 May 2022 19:12:29 +0000
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 by: Stan Fultoni - Fri, 13 May 2022 19:12 UTC

On Friday, May 13, 2022 at 12:01:57 PM UTC-7, Richard Hachel wrote:
> >> > tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> >> > tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
> >> > tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
> >>
> >> But is correct your second affirmation?
> >
> > None of those three expressions are correct (for an accelerating path), but YOU
> > are affirming that all three of them are correct. Remember, you claim that the
> > elapsed proper time along a uniform path (no acceleration) between two given
> > events ei and ej is sqrt[(ti-tj)^2) - (xi-xj)^2)], and you also claim that this
> > equals the elapsed proper time along a path undergoing constant proper
> > acceleration between those same two events. Your statements refer to any two
> > events, so you are affirming all three of the equations above. But the sum of the
> > first two square roots does not equal the third square root (unless the
> > "accelerating path" is not accelerating), so your beliefs are contradictory.
>
> I see you don't understand what I'm saying.

Which of your two claims are you now retracting? To remind you, your
two claims are:

(1) The elapsed proper time along a uniform path (no acceleration) between two given
events ei and ej is sqrt[(ti-tj)^2) - (xi-xj)^2)].
(2) The elapsed proper time along a path undergoing constant proper acceleration
between two given events equals the elapsed proper time along an unaccelerated
path between those two events.

Your first claim (1) is true, but your second claim (2) is false. Agreed?

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