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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
[SR] Question

<rCUg_Trt3F-OhxIpzh7WbRf0iXU@jntp>

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Mon, 9 May 2022 00:02 UTC

I see that few people react to my posts, and I understand, of course, why.

The question I am asking causes some confusion.

I recall, for those who follow, my problem.

I consider that the proper time of a uniformly accelerated rocket will be
like Tr=sqrt(2x/a).

And the improper time (observable in the terrestrial frame) will be
To=(x/c).sqrt(1+2c²/ax)

This second time is unanimous, and everyone is in agreement.

But a huge problem arises, for the proper time, because the mathematical
integration, whatever the way I go about it (by integrating the speeds, or
the times, or the distances, etc...) always falls back on the same results
as the result of the scientists.

We will say: "But then where is the problem?".

The problem is that I know that this result obtained by integration is
false.

It would seem that in this particular relativistic case, the mathematical
integrations which are obviously mathematically correct, are physically
wrong.

That's not how it works.

In short, the sum of all the small dVr or dTr, obviously gives Vr or Tr.

But a complex phenomenon seems to appear.

It is wrong to write To=∫ΔTo

In short, the sum of all the observable (improper) time parts is not the
mathematical sum of all the parts (which may seem absurd).

It is not easy to understand why.

The question is therefore: "But what is happening?".

R.H.

Re: [SR] Question

<t5asum$2jc$1@dont-email.me>

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From: mikko.le...@iki.fi (Mikko)
Newsgroups: sci.physics.relativity
Subject: Re: [SR] Question
Date: Mon, 9 May 2022 14:12:22 +0300
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 by: Mikko - Mon, 9 May 2022 11:12 UTC

On 2022-05-09 00:02:17 +0000, Richard Hachel said:

> I see that few people react to my posts, and I understand, of course, why.
>
> The question I am asking causes some confusion.

No, it does not. I reveals an already existing confusion.

> I recall, for those who follow, my problem.
>
> I consider that the proper time of a uniformly accelerated rocket will
> be like Tr=sqrt(2x/a).
>
> And the improper time (observable in the terrestrial frame) will be
> To=(x/c).sqrt(1+2c²/ax)
>
> This second time is unanimous, and everyone is in agreement.
>
> But a huge problem arises, for the proper time, because the
> mathematical integration, whatever the way I go about it (by
> integrating the speeds, or the times, or the distances, etc...) always
> falls back on the same results as the result of the scientists.

There is only one correct result that that integration can give.
Every other result is wrong.

> We will say: "But then where is the problem?".
>
> The problem is that I know that this result obtained by integration is false.

Then your knowledge is wrong.

Mikko

Re: [SR] Question

<762023b5-f582-4dd7-8679-fe52c42caedcn@googlegroups.com>

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Subject: Re: [SR] Question
From: maluwozn...@gmail.com (Maciej Wozniak)
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 by: Maciej Wozniak - Mon, 9 May 2022 11:24 UTC

On Monday, 9 May 2022 at 13:12:26 UTC+2, Mikko wrote:
> On 2022-05-09 00:02:17 +0000, Richard Hachel said:
>
> > I see that few people react to my posts, and I understand, of course, why.
> >
> > The question I am asking causes some confusion.
> No, it does not.

Nothing can ever confuse a fanatic relativistic
idiot, RH. Face it.

Re: [SR] Question

<l-a856BKyJ8pONkBJC158u_4h8w@jntp>

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Mon, 9 May 2022 11:46 UTC

Le 09/05/2022 à 13:12, Mikko a écrit :
> On 2022-05-09 00:02:17 +0000, Richard Hachel said:
>
>> I see that few people react to my posts, and I understand, of course, why.
>>
>> The question I am asking causes some confusion.
>
> No, it does not. I reveals an already existing confusion.
>
>> I recall, for those who follow, my problem.
>>
>> I consider that the proper time of a uniformly accelerated rocket will
>> be like Tr=sqrt(2x/a).
>>
>> And the improper time (observable in the terrestrial frame) will be
>> To=(x/c).sqrt(1+2c²/ax)
>>
>> This second time is unanimous, and everyone is in agreement.
>>
>> But a huge problem arises, for the proper time, because the
>> mathematical integration, whatever the way I go about it (by
>> integrating the speeds, or the times, or the distances, etc...) always
>> falls back on the same results as the result of the scientists.
>
> There is only one correct result that that integration can give.
> Every other result is wrong.
>
>> We will say: "But then where is the problem?".
>>
>> The problem is that I know that this result obtained by integration is false.
>
> Then your knowledge is wrong.
>
> Mikko

No one would think of multiplying the square root of six carrots by the
combination of ten potatoes.

I see that unfortunately a huge form of arrogance continues to prevail
(even if those who practice like that may think they are doing well).

I understood very well what the theory of relativity was, I understood
very well what integration was.

You urgently need to learn modesty in your dealings with others.

I am not saying that one should NEVER do integration, nor study Leibniz
(stolen by Newton, as Poincaré was stolen by the Anglo-Germanic school).

That's not what I'm saying.

I beg you to read me again.

I said that when we have the improper time of an object or a particle, it
may seem tempting to do an integration to find the proper time.

This will obviously work in uniform movements.

If we integrate all the small equivalent pieces of Tr, we will obtain Trf
(final proper real time).

If we integrate by adding all the small pieces of To, we will obtain To
(improper or "observable" time).

Simply, what I'm saying is that it no longer works for accelerated
movements. There is a problem that will arise in the way in which we will
believe we can carry out the integration.

The final result, well known, and presented by Jean-Pierre Messager on a
recurring basis, is false.

The expected result should be Tr=sqrt(2x/a) and nothing else.

Something is wrong with the carrots and turnips that you think you can add
up.

The possible error is perhaps this: "Can we really consider that a
uniformly accelerated movement can be considered as a sequence of uniform
movements when we divide it into infinitesimal parts?"

If we divide it by two, we see that no.

By four, by eight either, and so on.

At what moment can we think that the equation x=(1/2)a Tr² will be solved
with the stick by x=Vr.Tr?

Isn't this way of changing a carrot into a turnip responsible for a
conceptual trap of which there are many?

R.H.

Re: [SR] Question

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Subject: Re: [SR] Question
From: rot...@gmail.com (rotchm)
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 by: rotchm - Mon, 9 May 2022 13:20 UTC

> I am not saying that one should NEVER do integration,

> I said that when we have the improper time of an object or a particle, it
> may seem tempting to do an integration to find the proper time.
> This will obviously work in uniform movements.
>
> If we integrate all the small equivalent pieces of Tr, we will obtain Trf
> (final proper real time).
>
> If we integrate by adding all the small pieces of To, we will obtain To
> (improper or "observable" time).
>
> Simply, what I'm saying is that it no longer works for accelerated
> movements. There is a problem

I haven't read are really followed your thread. But I will say this:

It is well known in math that Δ is not the same as dx or del_x.
It is well known that you can't replace "adding small linear" terms by integrating;
One needs extra conditions for that procedure to be valid [ specifying the topological structure for instance].

Mathematicians know this and hopefully most physicists know this.
It is also known in the realm of special relativity (physics of space-time)..
This is why some say that there is a silent hypothesis in special relativity, namely the clock hypothesis.
This ensures that we can integrate.

https://en.wikipedia.org/wiki/Time_dilation#Clock_hypothesis

Re: [SR] Question

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Subject: Re: [SR] Question
From: maluwozn...@gmail.com (Maciej Wozniak)
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 by: Maciej Wozniak - Mon, 9 May 2022 13:34 UTC

On Monday, 9 May 2022 at 15:20:40 UTC+2, rotchm wrote:
> > I am not saying that one should NEVER do integration,
> > I said that when we have the improper time of an object or a particle, it
> > may seem tempting to do an integration to find the proper time.
> > This will obviously work in uniform movements.
> >
> > If we integrate all the small equivalent pieces of Tr, we will obtain Trf
> > (final proper real time).
> >
> > If we integrate by adding all the small pieces of To, we will obtain To
> > (improper or "observable" time).
> >
> > Simply, what I'm saying is that it no longer works for accelerated
> > movements. There is a problem
> I haven't read are really followed your thread. But I will say this:
>
> It is well known in math that Δ is not the same as dx or del_x.

Speaking of math, it's always good to remind that your
bunch of idiots had to announce its oldest, very important
and successful part false - as it didn't want to support the
Holy Postulates of your idiot guru.

Re: [SR] Question

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Date: Mon, 9 May 2022 06:50:46 -0700 (PDT)
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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Mon, 9 May 2022 13:50 UTC

On Monday, May 9, 2022 at 4:46:07 AM UTC-7, Richard Hachel wrote:
> I consider that the [elapsed coordinate] time of a uniformly accelerated
> rocket will be delta x = (x/c)sqrt(1+2c²/ax) and the [elapsed proper time
> will be delta tau = sqrt(2x/a).

That is logically self-contradictory, because it signifies that for a rocket
with constant proper acceleration the elapsed proper time between any
two events e1 and e2 along its path equals the elapsed proper time along
a uniformly moving (unaccelerated) path from e1 to e2. And you've also
agreed that the latter equals sqrt[(t2-t1)^2 - (x2-x1)^2]. The contradiction
arises when you consider three events e1,e2,e3 along the accelerated path.
According to your claim, the elapsed proper time along this accelerated path
from e1 to e2 equals the elapsed proper time of the uniform path from e1 to
e2, and furthermore, according to your claim the elapsed proper time of the
accelerating path from e2 to e3 equals the elapsed proper time of the uniform
path from e2 to e3. But according to your claim, the elapsed proper time of
the accelerating path from e1 to e3 equals the elapsed proper time along the
unaccelerated path from e1 to e3, which contradicts the first two claims.

Do you understand now that your beliefs are self-contradictory?

Re: [SR] Question

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Subject: Re: [SR] Question
From: maluwozn...@gmail.com (Maciej Wozniak)
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 by: Maciej Wozniak - Mon, 9 May 2022 13:54 UTC

On Monday, 9 May 2022 at 15:50:47 UTC+2, Stan Fultoni wrote:
> On Monday, May 9, 2022 at 4:46:07 AM UTC-7, Richard Hachel wrote:
> > I consider that the [elapsed coordinate] time of a uniformly accelerated
> > rocket will be delta x = (x/c)sqrt(1+2c²/ax) and the [elapsed proper time
> > will be delta tau = sqrt(2x/a).
>
> That is logically self-contradictory

Besides, whatever is proper according to a relativistic idiot -
sane people piss at that and (as anyone can check at GPS)
keep making improper clocks measuring t'=t, just like all
serious clocks always did.

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Subject: Re: [SR] Question
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 by: Mikko - Mon, 9 May 2022 14:19 UTC

On 2022-05-09 11:46:04 +0000, Richard Hachel said:

> Le 09/05/2022 à 13:12, Mikko a écrit :
>> On 2022-05-09 00:02:17 +0000, Richard Hachel said:
>>
>>> I see that few people react to my posts, and I understand, of course, why.
>>>
>>> The question I am asking causes some confusion.
>>
>> No, it does not. I reveals an already existing confusion.
>>
>>> I recall, for those who follow, my problem.
>>>
>>> I consider that the proper time of a uniformly accelerated rocket will
>>> be like Tr=sqrt(2x/a).
>>>
>>> And the improper time (observable in the terrestrial frame) will be
>>> To=(x/c).sqrt(1+2c²/ax)
>>>
>>> This second time is unanimous, and everyone is in agreement.
>>>
>>> But a huge problem arises, for the proper time, because the
>>> mathematical integration, whatever the way I go about it (by
>>> integrating the speeds, or the times, or the distances, etc...) always
>>> falls back on the same results as the result of the scientists.
>>
>> There is only one correct result that that integration can give.
>> Every other result is wrong.
>>
>>> We will say: "But then where is the problem?".
>>>
>>> The problem is that I know that this result obtained by integration is false.
>>
>> Then your knowledge is wrong.
>>
>> Mikko
>
> No one would think of multiplying the square root of six carrots by the
> combination of ten potatoes.

You already did.

> I see that unfortunately a huge form of arrogance continues to prevail

There are at least two possible solutions:
a. don't be arrogant
b. don't consider it unfortunate but normal and expected

> (even if those who practice like that may think they are doing well).

Looks like you indeed think so.

> I understood very well what the theory of relativity was, I understood
> very well what integration was.
>
> You urgently need to learn modesty in your dealings with others.
>
> I am not saying that one should NEVER do integration,

Nor am I. But if you don't like the result, you may want to do somenthing
else instead.

> nor study Leibniz (stolen by Newton, as Poincaré was stolen by the
> Anglo-Germanic school).
>
> That's not what I'm saying.
>
> I beg you to read me again.
>
> I said that when we have the improper time of an object or a particle,
> it may seem tempting to do an integration to find the proper time.

Indeed. But if you don't like the result then you should reconsider.

> This will obviously work in uniform movements.
>
> If we integrate all the small equivalent pieces of Tr, we will obtain
> Trf (final proper real time).

True.

> If we integrate by adding all the small pieces of To, we will obtain To
> (improper or "observable" time).

Also true.

> Simply, what I'm saying is that it no longer works for accelerated
> movements. There is a problem that will arise in the way in which we
> will believe we can carry out the integration.

Every way gives the same result. If you get something else you didn't
have a way but an error.

> The final result, well known, and presented by Jean-Pierre Messager on
> a recurring basis, is false.

It is the only correct result of the integration. If you want something
else you must do somenting else.

> The expected result should be Tr=sqrt(2x/a) and nothing else.

When the expeted result differs from the true result then either
your expectation was wrong or you don't want that integral.

> Something is wrong with the carrots and turnips that you think you can add up.

There is nothing wrong with them. You just want something else.

> The possible error is perhaps this: "Can we really consider that a
> uniformly accelerated movement can be considered as a sequence of
> uniform movements when we divide it into infinitesimal parts?"

No, it is not there.

> If we divide it by two, we see that no.
>
> By four, by eight either, and so on.
>
> At what moment can we think that the equation x=(1/2)a Tr² will be
> solved with the stick by x=Vr.Tr?

When the difference between the last and the second last approximate
result is so small that the difference does not matter for your purposes.

> Isn't this way of changing a carrot into a turnip responsible for a
> conceptual trap of which there are many?

No argument over matters of taste.

Mikko

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Mon, 9 May 2022 14:54 UTC

Le 09/05/2022 à 15:20, rotchm a écrit :

> This ensures that we can integrate.

Is it a guarantee that we can integrate?

R.H.

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 by: Python - Mon, 9 May 2022 15:33 UTC

Richard "Hachel" Lengrand (M.D.) wrote:
....
> I see that unfortunately a huge form of arrogance continues to prevail
> (even if those who practice like that may think they are doing well).

*your* arrogance. Nobody cares by the way...

> I understood very well what the theory of relativity was, I understood
> very well what integration was.

Correction: *understand ... *understand ... *is

You don't, and you don't.

> You urgently need to learn modesty in your dealings with others.

*you* need to learn modesty in your dealing with others.

> The expected result should be Tr=sqrt(2x/a) and nothing else.

No way, you don't even need algebra to demonstrate why it is
not possible... A simple though experiment and qualitative
thinking is enough.

But you will persist in putting your head in the sand and
taking all the nonsense coming out of whatever organ you
use as gold, won't you Richard?

Humanity won't lose much when you'll die, Richard. Quite
the opposite.

Re: [SR] Question

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 by: Mikko - Mon, 9 May 2022 15:38 UTC

On 2022-05-09 13:50:46 +0000, Stan Fultoni said:

> That is logically self-contradictory, because it signifies that for a rocket
> with constant proper acceleration the elapsed proper time between
> anytwo events e1 and e2 along its path equals the elapsed proper time
> alonga uniformly moving (unaccelerated) path from e1 to e2.

That by itself wouldn't be self-contradictory. In Galilean relativity
the elapsed proper time between two events is independent of the path
as long as no part of the path goes backwards in time. In order to get
a contradiction more is needed (and was given).

Mikko

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Mon, 9 May 2022 15:51 UTC

Le 09/05/2022 à 17:33, Python a écrit :

> Humanity won't lose much when you'll die, Richard.

Words coming from the obscure Python, it's crispy.

Quel déconnard, ce Jean-Pierre.

R.H.

Re: [SR] Question

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 by: Maciej Wozniak - Mon, 9 May 2022 16:53 UTC

On Monday, 9 May 2022 at 17:34:00 UTC+2, Python wrote:
> Richard "Hachel" Lengrand (M.D.) wrote:
> ...
> > I see that unfortunately a huge form of arrogance continues to prevail
> > (even if those who practice like that may think they are doing well).
> *your* arrogance. Nobody cares by the way...

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?

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Tue, 10 May 2022 10:53 UTC

Le 09/05/2022 à 15:50, Stan Fultoni a écrit :
> That is logically self-contradictory, because it signifies that for a rocket
> with constant proper acceleration the elapsed proper time between any
> two events e1 and e2 along its path equals the elapsed proper time along
> a uniformly moving (unaccelerated) path from e1 to e2. And you've also
> agreed that the latter equals sqrt[(t2-t1)^2 - (x2-x1)^2]. The contradiction
> arises when you consider three events e1,e2,e3 along the accelerated path.
> According to your claim, the elapsed proper time along this accelerated path
> from e1 to e2 equals the elapsed proper time of the uniform path from e1 to
> e2, and furthermore, according to your claim the elapsed proper time of the
> accelerating path from e2 to e3 equals the elapsed proper time of the uniform
> path from e2 to e3. But according to your claim, the elapsed proper time of
> the accelerating path from e1 to e3 equals the elapsed proper time along the
> unaccelerated path from e1 to e3, which contradicts the first two claims.

Whether the proper time of an object that joins E1 to E2 is the same
whether it is in uniform motion or in accelerated motion, I see no
apparent problem.

We have said, to be consistent, that in the observing reference frame
(earth or laboratory), the time To between A and B was deliberately the
same in both experiments.

It is therefore sufficient to calculate the real speed Vr to set Vr=AB/Tr
in the first case (displacement at constant speed), and in the second case
to calculate the acceleration a and the final speed Vrf.
AB=(1/2)aTr²
Vrf=a.Tr

Now we will continue to E3, setting Tr'(new)=2*Tr
and always setting To1=To2 in both cases.

We will then have AC=(1/2)aTr'²
Vrf=a.Tr'

But we will not have the same uniform real speed Vr to choose so that
To1(constant)=To2(accelerated).

It should be noted that the relation which joins Tr and To is, in the
mediums with constant speed is:
To=Tr.sqrt(1+Vr²/c²) or To=Tr/sqrt(1-Vo²/c²) if you want,
and in accelerated environments
To=Tr.sqrt(1+4Vrf²/c²).

I repeat this last formula which will perhaps seem strange:
To=Tr.sqrt(1+4Vrf²/c²)

N.B. If we want to work in observable speed, we must set
Vof=Vrf/sqrt(1+Vrf²/c²)

So, I repeat, the physicists' equation is very largely incorrect when they
use proper time.

Only the equation for the observable time To (or from the lab, or
terrestrial, or improper) is correct.

It is therefore necessary to use for accelerated media:
Tr=sqrt(2x/a)
To=(x/c).sqrt(1+2c²/ax)

R.H.

Re: [SR] Question

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 by: Python - Tue, 10 May 2022 11:09 UTC

Richard "Hachel" Lengrand (M.D.) wrote:
....
> Whether the proper time of an object that joins E1 to E2 is the same
> whether it is in uniform motion or in accelerated motion, I see no
> apparent problem.

You *refuse* to see the problem, which is, nevertheless, very simple to
point out:

Traveller A and B starts and end their journey at the same event
E1 and E2. A is inertial, B is uniformly accelerating.

In A frame, B leaves A then returns to A, this a "standard" twin
scenario : A "not moving", B leaving and turning back.

Time displayed on clocks cannot be the same (except if Galilean
Transformations are correct, but we *know* they are not).

Re: [SR] Question

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Subject: Re: [SR] Question
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Tue, 10 May 2022 13:41 UTC

On Tuesday, May 10, 2022 at 3:53:14 AM UTC-7, Richard Hachel wrote:
> > That is logically self-contradictory, because it signifies that for a rocket
> > with constant proper acceleration the elapsed proper time between any
> > two events e1 and e2 along its path equals the elapsed proper time along
> > a uniformly moving (unaccelerated) path from e1 to e2. And you've also
> > agreed that the latter equals sqrt[(t2-t1)^2 - (x2-x1)^2]. The contradiction
> > arises when you consider three events e1,e2,e3 along the accelerated path.
> > According to your claim, the elapsed proper time along this accelerated path
> > from e1 to e2 equals the elapsed proper time of the uniform path from e1 to
> > e2, and furthermore, according to your claim the elapsed proper time of the
> > accelerating path from e2 to e3 equals the elapsed proper time of the uniform
> > path from e2 to e3. But according to your claim, the elapsed proper time of
> > the accelerating path from e1 to e3 equals the elapsed proper time along the
> > unaccelerated path from e1 to e3, which contradicts the first two claims.
>
> Whether the proper time of an object that joins E1 to E2 is the same
> whether it is in uniform motion or in accelerated motion, I see no
> apparent problem.

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?

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Tue, 10 May 2022 15:59 UTC

Le 10/05/2022 à 13:09, Python a écrit :
> In A frame, B leaves A then returns to A, this a "standard" twin
> scenario : A "not moving", B leaving and turning back.

It's not the same scenario.

Do you really think that the Langevin traveler's problem (which I know
better than you, don't worry) is the same as the Tau Ceti traveler's
problem?
On the one hand constant speed travel, on the other uniformly accelerated
speed travel. It's not the same thing.

R.H.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Tue, 10 May 2022 16:48 UTC

Le 10/05/2022 à 15:41, Stan Fultoni a écrit :
> On Tuesday, May 10, 2022 at 3:53:14 AM UTC-7, Richard Hachel wrote:
>> > That is logically self-contradictory, because it signifies that for a rocket
>> > with constant proper acceleration the elapsed proper time between any
>> > two events e1 and e2 along its path equals the elapsed proper time along
>> > a uniformly moving (unaccelerated) path from e1 to e2. And you've also
>> > agreed that the latter equals sqrt[(t2-t1)^2 - (x2-x1)^2]. The contradiction
>> > arises when you consider three events e1,e2,e3 along the accelerated path.
>> > According to your claim, the elapsed proper time along this accelerated path
>> > from e1 to e2 equals the elapsed proper time of the uniform path from e1 to
>> > e2, and furthermore, according to your claim the elapsed proper time of the
>> > accelerating path from e2 to e3 equals the elapsed proper time of the uniform
>> > path from e2 to e3. But according to your claim, the elapsed proper time of
>> > the accelerating path from e1 to e3 equals the elapsed proper time along the
>> > unaccelerated path from e1 to e3, which contradicts the first two claims.
>>
>> Whether the proper time of an object that joins E1 to E2 is the same
>> whether it is in uniform motion or in accelerated motion, I see no
>> apparent problem.
>
> 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]

Tr1=sqrt(To1²-x²/c²)

Yes.

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

Tr2=sqrt(To2²-(x2²/c²)

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

Tr3=sqrt(To3²-(x3²/c²)

>
> but these three relations are self-contradictory. Do you understand this?

Why?

R.H.

Re: [SR] Question

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 by: Stan Fultoni - Tue, 10 May 2022 18:44 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), and yet 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. Now do you see that your claims are self-contradictory?

Re: [SR] Question

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 by: Python - Tue, 10 May 2022 22:39 UTC

Richard "Hachel" Lengrand (M.D.) wrote:
> Le 10/05/2022 à 13:09, Python a écrit :
>> In A frame, B leaves A then returns to A, this a "standard" twin
>> scenario : A "not moving", B leaving and turning back.
>
> It's not the same scenario.

Exactly not the same, because constant acceleration is assumed
instead of instaneous change of motion (which is unrealistic
by the way), but it doesn't matter.

> Do you really think that the Langevin traveler's problem (which I know
> better than you, don't worry)

I don't worry. You know knothing, this a fantasy of yours forged
by year of demencia.

> is the same as the Tau Ceti traveler's
> problem?
> On the one hand constant speed travel, on the other uniformly
> accelerated speed travel. It's not the same thing.

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) ?

(v0, acc. as seen in A's frame)

(clock setting and go)
t=0 A
B ---> v0 (initial velocity v0 Tau Ceti
<--- (constant acceleration)

middle of the trip, turning back :

A B Tau Ceti (v=0)
<--- (constant acceleration)

return trip A:

A
<---- B v (initial velocity Tau Ceti
<--- (constant acceleration)

Arrival at A:

t A
B Tau Ceti
t' <--- (constant acceleration)

When B and A will? When reunited, will their clocks show the
same elapsed times on their clocks?

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 05:05 UTC

On Wednesday, 11 May 2022 at 00:39:25 UTC+2, Python wrote:
> Richard "Hachel" Lengrand (M.D.) wrote:
> > Le 10/05/2022 à 13:09, Python a écrit :
> >> In A frame, B leaves A then returns to A, this a "standard" twin
> >> scenario : A "not moving", B leaving and turning back.
> >
> > It's not the same scenario.
> Exactly not the same, because constant acceleration is assumed

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?

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Wed, 11 May 2022 10:26 UTC

Le 11/05/2022 à 00:39, Python a écrit :
> Exactly not the same, because constant acceleration is assumed
> instead of instaneous change of motion (which is unrealistic
> by the way), but it doesn't matter.

No, it's not the same thing.

In the case of "Langevin", three standards must be used.

Only two in the case of "Tau Ceti".

If I have time, I'll make you a little diagram so that you understand that
it's not the same thing.

In the case of a "Tau Ceti", we join A and B. The two events A and the two
events B are joined.
We set To²=Tr²+Et² (with the spatial anisochrony Et=x/c).

The two proper times, if the two improper times are equivalent, and if Et
is equivalent,
are also equivalent.

For the Langevin, it's different, there are three lines. One is the
reference frame of the primary path, the other of the secondary path, and
the other the accelerated frame that joins A and B.
It is therefore necessary to use two lines to join A and B. And we do not
add the turnip line with the carrot line.

R.H.

Re: [SR] Question

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 by: Python - Wed, 11 May 2022 10:46 UTC

Richard "Hachel" Lengrand (M.D):
> Le 11/05/2022 à 00:39, Python a écrit :
>> Exactly not the same, because constant acceleration is assumed
>> instead of instaneous change of motion (which is unrealistic
>> by the way), but it doesn't matter.
>
> No, it's not the same thing.
>
> [snip evading bs]

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.

Re: [SR] Question

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From: r.hac...@tiscali.fr (Richard Hachel)
 by: Richard Hachel - Wed, 11 May 2022 10:51 UTC

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) ?

In the example you give, we're using three repositories, and we can't add
things to others without a clear idea of ​​what we're doing.

You have to limit yourself to what I said.

I said that during a transfer from A to B, whether it is done at constant
speed Vr (real speed) or at uniformly accelerated speed aTr, if the two
observable times To are the same (they leave and meet together), the two
improper times To will be the same, and that, in both cases, one can
reasonably apply To²=Tr²+Et².

But that only works if we take a single straight line to join A and B.

Let's take the example of a "Tau Ceti" where, instead of a constant speed
Vr (or Vo if you want to speak in observable), I will use two.

For example, a certain acceleration a, in the uniformly accelerated path
corresponds to two speeds Vr1 and Vr2 chosen for the first half of the
path and for the second half.

There will therefore be a total time Tr=Tr1+Tr2
and a total time To=To1+To2

With the possibility of writing To1²=Tr1²+(x/2)²/c²

But as soon as the other line appears (let's assume that Vo1=0.6c and
Vo2=0.8c), we are no longer in the same simple relationship.

The concordance that I gave, and which works for all frames of reference,
both uniform and accelerated, is only valid if the line joining A and B is
a line, and not two lines.

In this sense, Vr must always be constant, and placed in a single frame of
reference.

R.H.

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