It is very surprising to say, for example, that the correct formula for
determining the proper time of an accelerated object is Tr=sqrt(2x/a).

With To=(x/c).sqrt(2c²/ax)

This obviously makes everyone laugh.

And everyone will laugh even more if I say that the way in which we do the
integration of the proper times is not correct, and that we must not
practice this integration.

I explained how to do it, I won't go back to it now.

What you have to understand is that when we take a small segment AB, we
can determine the speed in A, then the speed in B.

But which value to choose to make a possible high precision measurement?

Someone with an attentive eye may say to me: "The best thing is to take
the average value of the sepment", and he will be absolutely right.

Let Vrm=(Vra+Vrb)/2

We will therefore be able to ask:
ΔTo=ΔTr.sqrt(1+(Vra+Vrb)²/4c²)

Let ΔTo=ΔTr.sqrt(1+(Vra²+2.Vra.Vrb+Vrb²)/4c²)

However, if the segment is very small: Vra=Vrm=Vrb

And we can consider that: ΔTo=ΔTr.sqrt(1+(Vrm²/c²)

We come back to the equation of Galilean environments.

But this is only a very local approximation.

And we cannot integrate this approximation.

It's a relativist blunder and everyone does it.

If we want to give To as a function of Tr, we must set:
To=Tr.sqrt(1+(1/4).Vr²/c²)

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
₀₀₀
<http://news2.nemoweb.net/?DataID=nF_td_Grs5Nxu0QJuUCdBIm7NLg@jntp>

On Wednesday, September 21, 2022 at 4:16:02 PM UTC-7, Richard Hachel wrote:
> The equation (tj - ti) = (tauj - taui) sqrt[(1 + (a^2/4)(tauj+taui)^2]
> is only valid if the start is made at rest.

No, that equation is equivalent to your Δt = Δtau .sqrt[(1 + Vri² + Vri.Vra + (1/4).Vra²], since you define Vri as the initial velocity a*taui and you define Vra as the change in velocity from start to finish, which is a*(tauj - taui). Inserting these into your equation gives the result (tj - ti) = (tauj - taui) sqrt[(1 + (a^2/4)(tauj+taui)^2], which confirms that your beliefs imply 1=0, and are therefore absurd, as explained in the previous messages.

> I persist in saying that something is wrong with the integration that we
> are doing...

If there was something wrong with special relativity, you'd be able to point it out, but you can't. Remember, you've admitted that you were wrong about the unaccelerated rocket case, and to find the elapsed time on the accelerating rocket, we solve t=x*sqrt(1+2/(ax)) for x to give x = (1/a)[sqrt(1 + (at)²) - 1], so we have dx = [at/sqrt(1+(at)^2)]dt, and substituting for dx into dtau² = dt² - dx² gives dtau = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) = (1/a)invcos(1+ax).

> If the acceleration is neglected, it must also be neglected on the
> following segment, and so on, an infinity of times. In short, the
> acceleration in each segment never exists.

Right, each differential segment is essentially inertial, and the elapsed proper time along an incremental segment of a trajectory with components dx and dt is dtau = sqrt(dt^2 - dx^2), where x,t is any standard system of inertial coordinates. The total elapsed proper time along a path is the sum of the proper times along the incremental parts of the path.

> It is very surprising to say, for example, that the correct formula for
> determining the proper time of an accelerated object is Tr=sqrt(2x/a).

No, that is the elapsed proper time along the *inertial* path, as explained to you before, and you ran away. Remember?

> With t=x.sqrt(2/(ax)) This obviously makes everyone laugh.

Again, this is the relation that leads to the correct result, as explained above. Understand?

Le 27/09/2022 à 03:08, Stan Fultoni a écrit :

> whose integral is tau = (1/a)invsinh(at) = (1/a)invcos(1+ax).

The question that arises is: "Certainly, but can we integrate?"

Minkoski's geometry allows it.

Let's admit.

Does the geometry of the good doctor Hachel allow it?

So the problem is no longer there.

It is up to the geometry to apply.

Minkowski has a basic four-dimensional geometry.

Hachel's is seven.

Which is better?

Everyone is quick to say "ours".

It's not scientific.

R.H.

On Tuesday, September 27, 2022 at 7:20:30 AM UTC-4, Richard Hachel wrote:
> Le 27/09/2022 à 03:08, Stan Fultoni a écrit :
>
> > whose integral is tau = (1/a)invsinh(at) = (1/a)invcos(1+ax).
> The question that arises is: "Certainly, but can we integrate?"
>
> Minkoski's geometry allows it.

So you agree that we can integrate, good!

> So the problem is no longer there.
> It is up to the geometry to apply.

Its up to the model used. Its good that you finally realize this and that we can integrate, as you admitted above.

> Minkowski has a basic four-dimensional geometry.
> Hachel's is seven.

Irrelevant to the discussion. Stay on topic.
> Which is better?
> Everyone is quick to say "ours".

Indeed.

> It's not scientific.

But ours (SR) predicts the values obtained by exp's. Yours doesn't.

On Tuesday, September 27, 2022 at 5:17:43 AM UTC-7, rotchm wrote:

> But ours (SR) predicts the values obtained by exp's. Yours doesn't.

"Ours"? You have no clue what SR is, Stephane, you are as crank as you have ever been.

On Tuesday, 27 September 2022 at 14:17:43 UTC+2, rotchm wrote:

> But ours (SR) predicts the values obtained by exp's. Yours doesn't.

In the meantime in the real world, however, forbidden
by your bunch of idiots GPS and TAI keep measuring t'=t,
just like all serious clocks always did. And the mumble of
your idiot guru wasn't even self consistent.

Le 27/09/2022 à 14:17, rotchm a écrit :

> But ours (SR) predicts the values obtained by exp's. Yours doesn't.

J'espère que tu plaisantes.

C'est la seule théorie qui a une perfection interne (logique
mathématique complète),
et externe (preuves expérimentales tous azimuts).

Tu crois vraiment que je perdrais mon temps à expliquer des trucs
nullissimes?

R.H.

On Tuesday, September 27, 2022 at 4:20:30 AM UTC-7, Richard Hachel wrote:
> > whose integral is tau = (1/a)invsinh(at) = (1/a)invcos(1+ax).
> The question that arises is: "Certainly, but can we integrate?"

Since you don't know how to integrate, you can just split the path into several very small segments, each of which is essentially inertial, and compute the elapsed time on each segment as sqrt(dt^2 - dx^2), and then add them up. If you split up the total path into more and more smaller and smaller segments, the sum approaches a certain value. That is the total elapsed time for the accelerating trajectory. In your favorite example it is 3.14 years. Understand?

Remember, when we use your method, we get 1=0, which is absurd. So special relativity is correct, and your beliefs are absurd. Remember?

Le 27/09/2022 à 15:53, Stan Fultoni a écrit :
> Since you don't know how to integrate, you can just split the path into several
> very small segments, each of which is essentially inertial, and compute the
> elapsed time on each segment as sqrt(dt^2 - dx^2), and then add them up. If you
> split up the total path into more and more smaller and smaller segments, the sum
> approaches a certain value. That is the total elapsed time for the accelerating
> trajectory. In your favorite example it is 3.14 years. Understand?
>
> Remember, when we use your method, we get 1=0, which is absurd. So special
> relativity is correct, and your beliefs are absurd. Remember?

Cette forme n'est pas correcte.

Elle est correcte dans les vitesses constantes, car Vra=Vrm=Vrb

On peut donc simplifier en To=Tr.sqrt(1+Vr²/c²)

Mais ce n'est qu'une simplification locale.

Si l'on fait la même chose avec des référentiels accélérés, ça ne
marche plus.

Ca marche que si ΔVr=0.

Mais ΔVr=0, ça accélère pas beaucoup.

R.H.

On Tuesday, September 27, 2022 at 9:52:18 AM UTC-4, Richard Hachel wrote:
> Le 27/09/2022 à 14:17, rotchm a écrit :
>
> > But ours (SR) predicts the values obtained by exp's. Yours doesn't.
> J'espère que tu plaisantes.
>
> C'est la seule théorie qui a

<snip>

The used language does not follow the discussion.
Spam reported.

I incite others to do the same.

Le 27/09/2022 à 16:14, rotchm a écrit :

> I incite others to do the same.

Oh, noooo... Snifff....

R.H.

Richard Hachel wrote:

> It is very surprising to say, for example, that the correct formula for
> determining the proper time of an accelerated object is Tr=sqrt(2x/a).
>
> With To=(x/c).sqrt(2c²/ax)

relativity is so exact and good, they still don't even have a
*solution_manual* to it.

On Tuesday, September 27, 2022 at 7:02:01 AM UTC-7, Richard Hachel wrote:
> > Since you don't know how to integrate, you can just split the path into several
> > very small segments, each of which is essentially inertial, and compute the
> > elapsed time on each segment as sqrt(dt^2 - dx^2), and then add them up.. If you
> > split up the total path into more and more smaller and smaller segments, the sum
> > approaches a certain value. That is the total elapsed time for the accelerating
> > trajectory. In your favorite example it is 3.14 years. Understand?
>
> The equation (tj - ti) = (tauj - taui) sqrt[(1 + (a^2/4)(tauj+taui)^2]
> is only valid if the start is made at rest.

No, that equation is equivalent to your Δt = Δtau .sqrt[(1 + Vri² + Vri.Vra + (1/4).Vra²], since you define Vri as the initial velocity a*taui and you define Vra as the change in velocity from start to finish, which is a*(tauj - taui). Inserting these into your equation gives the result (tj - ti) = (tauj - taui) sqrt[(1 + (a^2/4)(tauj+taui)^2], which confirms that your beliefs imply 1=0, and are therefore absurd, as explained in the previous messages. Agreed?