Isn't there a huge bug?

R.H.

Richard "Hachel" Lengrand (M.D.) wrote:
>
> Isn't there a huge bug?

It is legitimate.

If you dared to learn this, you could know why.

By the way, your claim that times at arrival for the uniformly
accelerated traveller and the uniformly moving one will be the
same IMPLIES that, in the usual twins scenario, the times would
be the same too. I've shown you why, you don't even need a
formula to demonstrate that. Your answer was then "my bollocks".

So you're not taking seriously neither Relativity, physics nor
intellectual integrity. You only want to brag about your own
silly bunch of ill defined ideas you've pulled out of your ass.

Don't expect any benevolence, you don't deserve it.

Le 23/04/2022 à 13:03, Python a écrit :

> By the way, your claim that times at arrival for the uniformly
> accelerated traveller and the uniformly moving one will be the
> same IMPLIES that, in the usual twins scenario, the times would
> be the same too.

No.

The two problems are different.

I defined very precisely the numerical concepts and criteria so that
things are completely clear.

In the first problem, which physicists know as Langevin's traveler, a twin
must go around a star located 12 ly away and return to earth. Its speed is
constant 0.8c.
We then seek to describe the relative times throughout the course, for one
and for the other.
I gave the full explanation of that (unheard of apparently).

In the other problem, the one called the traveler of Tau Ceti, it is about
a man who will leave for a single trip to Tau Ceti (12 ly), using a
uniformly accelerated rocket at 10m/s².
It is a question of comparing the times, the speeds, the accelerations.

Obviously, the two problems cannot be confused.

The given equations are not the same.

In fine, proper times are not the same rapport with observable times.

(1) Tr=To/sqrt(1-vo²/c²) or To=Tr/sqrt(1+Vr²/c²)

(2) Tr=sqrt(2x/a) and To=(x/c).sqrt(1+2c²/ax)

R.H.

Richard Hachel wrote:

> Isn't there a huge bug?

Absolutely sure. Here's what the *russian_army* should do from the very
beginning, along the provincial region of its territory, illegally named
"ukraine".

You take out their *_lying_propaganda_machine_* from the very beginning.
Very simple, very easy thing to do. How the hell, so many generals over
there, not knowing the basics in math and physics, about the
*_lying_propaganda_machine_*, of which you have to take it out, a way or
another.

the nazi "defense" criminal organization NATO BOMBINGS IN YUGOSLAVIA - WAR
CRIMES https://www.bitchute.com/video/z5nN1KEyZ2mM/

On Saturday, 23 April 2022 at 13:04:02 UTC+2, Python wrote:
> Richard "Hachel" Lengrand (M.D.) wrote:
> >
> > Isn't there a huge bug?
> It is legitimate.
>
> If you dared to learn this, you could know why.

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?

Maciej Wozniak wrote:

>> > Isn't there a huge bug?
>> It is legitimate.
>>
>> If you dared to learn this, you could know why.
>
> Oh, stinker Python is opening its muzzle again,
> and trying to pretend he knows something.

my-ohh-my, capitalist western europe fucked up. Driving fake economies, on
energy stolen on *fake_money*, not possible anymore. A 2000 years ago,
they killed an innocent man for their *_fake_money_*

Milan, Italy...food rations
https://www.bitchute.com/video/0qrc3H2vyFgi/

On Saturday, April 23, 2022 at 4:32:07 AM UTC-7, Richard Hachel wrote:
> Le 23/04/2022 à 13:03, Python a écrit :
> > By the way, your claim that times at arrival for the uniformly
> > accelerated traveller and the uniformly moving one will be the
> > same IMPLIES that, in the usual twins scenario, the times would
> > be the same too.
>
> No. The two problems are different.

No, the two situations are identical for a traveling twin with constant
acceleration. In your case, you have one twin moving (in terms of the
given inertial coordinates) at constant speed 0.9295c a distance of 12
light years in 12.91 years, and you have another twin beginning and ending
at the same events, but subject to constant acceleration of 10 m/sec
beginning at speed zero.

Now, as Python pointed out, you can regard the first twin as stationary,
and in terms of his inertial coordinates the other twin begins with speed
0.9295c and undergoes constant acceleration toward his brother, so he
initially departs, and reaches a maximum distance of 1.627 light years
and then finally re-unites with his brother. This is the classic twins paradox
with a constantly accelerating rocket beginning at high speed. According
to your claim, the traveling twin will be the same age as the "stationary"
twin, but you know that is not true, so your beliefs are self-contradictory..
Agreed?

>> Your self-contradictory claim that the elapsed proper time is always
>> sqrt(2x/a) has been shown to be logically untenable. Agreed?
>
> No.

But there's no rational basis for your denial. You have agreed that the
elapsed proper time is the integral of sqrt(dt^2 - dx^2) along the path,
and you've been shown that for your rocket with constant acceleration
the integral is 3.14 years. So your beliefs are illogical and self-contradictory,
right?

Le 23/04/2022 à 18:16, Stan Fultoni a écrit :
> On Saturday, April 23, 2022 at 4:32:07 AM UTC-7, Richard Hachel wrote:
>> Le 23/04/2022 à 13:03, Python a écrit :
>> > By the way, your claim that times at arrival for the uniformly
>> > accelerated traveller and the uniformly moving one will be the
>> > same IMPLIES that, in the usual twins scenario, the times would
>> > be the same too.
>>
>> No. The two problems are different.
>
> No, the two situations are identical for a traveling twin with constant
> acceleration. In your case, you have one twin moving (in terms of the
> given inertial coordinates) at constant speed 0.9295c a distance of 12
> light years in 12.91 years, and you have another twin beginning and ending
> at the same events, but subject to constant acceleration of 10 m/sec
> beginning at speed zero.
>
> Now, as Python pointed out, you can regard the first twin as stationary,
> and in terms of his inertial coordinates the other twin begins with speed
> 0.9295c and undergoes constant acceleration toward his brother, so he
> initially departs, and reaches a maximum distance of 1.627 light years
> and then finally re-unites with his brother. This is the classic twins paradox
> with a constantly accelerating rocket beginning at high speed. According
> to your claim, the traveling twin will be the same age as the "stationary"
> twin, but you know that is not true, so your beliefs are self-contradictory.
> Agreed?
>
>>> Your self-contradictory claim that the elapsed proper time is always
>>> sqrt(2x/a) has been shown to be logically untenable. Agreed?
>>
>> No.
>
> But there's no rational basis for your denial. You have agreed that the
> elapsed proper time is the integral of sqrt(dt^2 - dx^2) along the path,
> and you've been shown that for your rocket with constant acceleration
> the integral is 3.14 years. So your beliefs are illogical and
> self-contradictory,
> right?

Non.

Le voyageur de Tau Ceti (à ne pas confondre avec le voyageur de Langevin)
va joindre Tau Ceti avec une accélération de a=10m/s².

Ce que je vous SUPPLIE de comprendre, c'est que ce voyageur est au repos
dans son référentiel.

Il observe le monde qui, au contraire accélère autour de lui.

Si l'on plaçait une énorme quantité de petites billes alignées autour
de lui tous les kilomètres, il verrait passer les billes de plus en plus
vite et aurait l'impression que ce sont toutes les billes qui
accélèrent.

Ce point est très important.

Pour le voyageur, qui a une distance x à avancement au départ, et qui
sait qu'il va mettre un temps Tr (il a déjà fait maintes fois le
trajet), et qui sait que son accélération est a, connait aussi la loi de
l'accélération qui s'applique ici SANS BUG.

x=1/2 a Tr²

Et sans bug aussi : Tr = sqrt(2x/a).

Sans bug aussi Tr²+Et²=To²

Et sans bug aussi Tr1=Tr2
si Vo constant =0.9295 comme vous venez de le calculer (et votre calcul
est correct).

Maintenant, parlons peu parlons bien.

Il y a un bug de concept dans la façon dont les scientifiques font
l'intégration de tous les dTo et dTr.

J'ai dénoncé leur erreur il y a bien longtemps déjà.

Je suis en train de chercher quand j'ai des instants libres, comment il
faut faire cette intégration pour qu'elle soit correcte.

On pose Vri = vitesse réelle instantanée

On pose Voi= vitesse observable instantanément.

On sait que Vri=Vro/sqrt(1-Vro²/c²)
Voi=Vri/sqrt(1+Vri²/c²)

Sur la partie de £dTo=£dTr.sqrt(1+Vri²/c²)
ou bien de
£dTo=£dTr/sqrt(1-Voi²/c²)
ou bien
£dTr=£dTo.sqrt(1-Voi²/c²)
ou bien de
£dTr=£dTo/sqrt(1+Vri²/c²)

Elle ne peut retomber que sur Tr= sqrt(2x/a) évidemment.

Mais je ne l'ai pas encore fait.

Je réfléchis à ça...

Bon, il faut que j'aille voter pour la présidentielle.

Merci de votre écoute.

R.H.

On Sunday, April 24, 2022 at 5:33:34 AM UTC-7, Richard Hachel wrote:
> Le voyageur de Tau Ceti (à ne pas confondre avec le voyageur de Langevin)
> va joindre Tau Ceti avec une accélération de a=10m/s².

The situations are the same, as explained in the previous message.
According to your "reasoning", the twins would always have the same
elapsed time, but you know they do not. So your "reasoning" is illogical
and self-contradictory.

> Ce que je vous SUPPLIE de comprendre, c'est que ce voyageur est au repos
> dans son référentiel.

Every object is instantaneously at rest in the co-moving system of inertial
coordinates at that instant, but an accelerating object is constantly
accelerating away from that inertial reference system, whereas an
unaccelerated object is continuously at rest in terms of a given inertial
reference system. In all cases, the elapsed proper time for the object
is the integral of sqrt(dt^2 - dx^2) along its path, where x,t is an _single_
system of inertial coordinates.

> Pour le voyageur, qui a une distance x à avancement au départ, et qui
> sait qu'il va mettre un temps Tr (il a déjà fait maintes fois le
> trajet), et qui sait que son accélération est a, connait aussi la loi de
> l'accélération qui s'applique ici SANS BUG.
>
> x=1/2 a Tr²

No, that equation relates the elapsed proper time along an unaccelerated
path between two events on a path with constant proper acceleration a.
It does not represent the elapsed proper time along the accelerated path.

> Je suis en train de chercher quand j'ai des instants libres, comment il
> faut faire cette intégration pour qu'elle soit correcte.
> Mais je ne l'ai pas encore fait. Je réfléchis à ça....

You have been shown the explicit integration, which is just high school
math, and you were not able to find any flaw in it. Always you say "I
will find the true explanation.... tomorrow... tomorrow...." For how many
years have you been ignoring the actual simple explanation, and promising
to provide your alternative explanation... tomorrow?

Le 24/04/2022 à 17:39, Stan Fultoni a écrit :

> and promising
> to provide your alternative explanation... tomorrow?

Yes.

Why not?

Laissez, laissez, je vais le faire...

Je suis plus à ça près.

R.H.