On Thursday, August 11, 2022 at 11:58:39 AM UTC-4, Richard Hachel wrote:

> If I integrate the equation,

Here, we are talking about the eqs: dtau = dt/sqrt(1 + (at)^2)

> I find: tau = t/sqrt(1+0.25Vr²/c²)

Then you are flunking basic calculus, and flunking reading skills.
Do you agree that there is no There is no "Vr" in dtau = dt/sqrt(1 + (at)^2)?

And its integral is tau = (1/a)invsinh(at). Do you agree that that is its integral?

> > you already agreed that x = (1/a) [sqrt(1 + (at)²) - 1],

Do you remember having agreed to this?

>> so dx = [at/sqrt(1+(at)^2]]dt,

Do you agree that this follows from the above?

> t/sqrt(1+0.25Vr²/c²) Fultoni notation

No, that is not his notation; it is yours. Don't bestow your gibberish on to others.

On Thursday, August 11, 2022 at 9:44:30 AM UTC-7, Richard Hachel wrote:
> >> >> > The following equations are correct:
> >> >> > x = (1/2)a TAU^2
> >> >> > t = TAU sqrt[1 + (1/4)a^2 TAU^2]
> >> >>
> >> >> That's what I've been saying for some time now.
> >> >
> >> > No, it is not. In those equations, TAU represents the elapsed proper time
> >> along
> >> > the inertial path from the origin to x,t, but you have been fantasizing that
> >> it
> >> > represents the elapsed proper time along the accelerating path. That's why
> >> > everyting you type is gibberish.
> >> >
> >> > You already agreed that x = (1/a) [sqrt(1 + (at)²) - 1], so dx =
> >> > [at/sqrt(1+(at)^2]]dt, and if you then integrate dtau = sqrt(dt^2 - dx^2) you
> >> get
> >> > tau = (1/a)invsinh(at).
> >>
> >> If I integrate the equation, I find: tau = t/sqrt(1+0.25Vr²/c²)
> >
> > Then you are flunking high school calculus. Again, you already agreed that x =
> > (1/a) [sqrt(1 + (at)²) - 1], so dx = [at/sqrt(1+(at)^2]]dt, so we have dtau =
> > sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at).
> >
> > There is no ambiguity here. The integral of dt/sqrt(1 + (at)^2) is simply
> > (1/a)invsinh(at). This is simple high school calculus.
>
> So there's something wrong in high school calculus.

No, there is nothing wrong with calculus, or with rational thought in general. The problem is that you are confusing TAU with tau, as explained previously. Again, TAU is the proper time along the unaccelerated path, whereas tau is the proper time along the accelerated path. The tau is what a clock following the path would show, and how much a person would age, etc. Understand?

On Thursday, August 11, 2022 at 11:06:08 AM UTC-7, Richard Hachel wrote:
> Le 11/08/2022 à 19:55, "Dono." a écrit :
> > On Thursday, August 11, 2022 at 9:44:30 AM UTC-7, Richard Hachel wrote:
> >> Le 11/08/2022 à 18:41, Stan Fultoni a écrit :
> >> > On Thursday, August 11, 2022 at 8:58:39 AM UTC-7, Richard Hachel wrote:
> >> >> >> > The following equations are correct:
> >> >> >> > x = (1/2)a TAU^2
> >> >> >> > t = TAU sqrt[1 + (1/4)a^2 TAU^2]
> >> >> >>
> >> >> >> That's what I've been saying for some time now.
> >> >> >
> >> >> > No, it is not. In those equations, TAU represents the elapsed proper time
> >> >> along
> >> >> > the inertial path from the origin to x,t, but you have been fantasizing
> >> that
> >> >> it
> >> >> > represents the elapsed proper time along the accelerating path. That's why
> >> >> > everyting you type is gibberish.
> >> >> >
> >> >> > You already agreed that x = (1/a) [sqrt(1 + (at)²) - 1], so dx =
> >> >> > [at/sqrt(1+(at)^2]]dt, and if you then integrate dtau = sqrt(dt^2 - dx^2)
> >> you
> >> >> get
> >> >> > tau = (1/a)invsinh(at).
> >> >>
> >> >> If I integrate the equation, I find: tau = t/sqrt(1+0.25Vr²/c²)
> >> >
> >> > Then you are flunking high school calculus. Again, you already agreed that x =
> >>
> >> > (1/a) [sqrt(1 + (at)²) - 1], so dx = [at/sqrt(1+(at)^2]]dt, so we have dtau =
> >>
> >> > sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau =
> >> (1/a)invsinh(at).
> >> >
> >> > There is no ambiguity here. The integral of dt/sqrt(1 + (at)^2) is simply
> >> > (1/a)invsinh(at). This is simple high school calculus.
> >> So there's something wrong in high school calculus.
> >>
> >> What? ? ?
> >>
> >> Why?
> >>
> > Not wrong in high school calculus. Wrong in your demented brain
> Do you know what the demented brain says to you?
>
Imbecilities

Le 11/08/2022 à 20:36, rotchm a écrit :
> On Thursday, August 11, 2022 at 11:58:39 AM UTC-4, Richard Hachel wrote:
>
>> If I integrate the equation,
>
> Here, we are talking about the eqs: dtau = dt/sqrt(1 + (at)^2)
>
>> I find: tau = t/sqrt(1+0.25Vr²/c²)
>
> Then you are flunking basic calculus, and flunking reading skills.
> Do you agree that there is no There is no "Vr" in dtau = dt/sqrt(1 + (at)^2)?
>
>
> And its integral is tau = (1/a)invsinh(at). Do you agree that that is its
> integral?
>
>> > you already agreed that x = (1/a) [sqrt(1 + (at)²) - 1],
>
> Do you remember having agreed to this?
>
>>> so dx = [at/sqrt(1+(at)^2]]dt,
>

I sometimes have trouble with scientific notations, and I use mine which I
try to define better (for example I use in relativity three kinds of time
(Tapp, Tr, To) and three kinds of speeds ( Vapp, Vo, Vr).

I try to be careful not to make a mistake, as it is written in the Holy
Bible of reality: Richard Hachel chap 15 verse 3:
"The theory of relativity is a very simple theory, mathematically it does
not often exceed square roots, sines and cosines, but it is full of small
pitfalls".

The notions of Vapp, Vo, Tapp, and To are the same as the usual notions.

Tr and Vr are new notions, but which I think are essential.

Now we come back to the equation:

> Do you agree that there is no There is no "Vr" in dtau = dt/sqrt(1 + (at)^2)?

I do not write the equation in this form where we give tau as a function
of t and (at)². I don't see the immediate benefit.

I pose, clearly specifying the terms:

To²=Tr²+Et²

Hence if the movement is uniformly accelerated with a start at rest:

To²=Tr²+((1/2)a.Tr²)²/c²

Let To=Tr.sqrt(1+(1/4)Vr²/c²)

Here, there is no need to integrate.

The answer is direct.

If we seek Tr, and if we know Vr and To, we invert the equation
Tr=To/sqrt(1+(1/4)Vr²/c²)

R.H.

> Do you agree that this follows from the above?
>
>> t/sqrt(1+0.25Vr²/c²) Fultoni notation
>
> No, that is not his notation; it is yours. Don't bestow your gibberish on to
> others.

Le 11/08/2022 à 20:38, Stan Fultoni a écrit :

> The problem is that you are confusing TAU with tau, as explained previously.
> Again, TAU is the proper time along the unaccelerated path, whereas tau is the
> proper time along the accelerated path.

I call To the time observed in the terrestrial reference frame.

I call Tr (tau for you) the real time observed in the reference frame of
the object (whether in uniform or accelerated motion).

I call Tapp the apparent time measured in the terrestrial reference frame,
and which takes into account universal anosochrony.

These three notions, I understand them very well.

On the other hand, I do not see, here, the interest of a fourth time which
one would call TAU, but would not be tau, but another proper time quite
other than the proper time.

I don't want to rise to this kind of speech which brings nothing to the
real problem.

R.H.

On Thursday, August 11, 2022 at 1:04:14 PM UTC-7, Richard Hachel wrote:
> Le 11/08/2022 à 20:38, Stan Fultoni a écrit :
> > The problem is that you are confusing TAU with tau, as explained previously.
> > Again, TAU is the proper time along the unaccelerated path, whereas tau is the
> > proper time along the accelerated path.
>
> I call To the time observed in the terrestrial reference frame.

That is the coordinate time, which grown-ups call t.

> I call Tr (tau for you) the real time observed in the reference frame of
> the object (whether in uniform or accelerated motion).

The elapsed proper time between the origin and the final event is different along the accelerated path than along the inertial path, so you need to distinguish between them. The time of interest is the time along the accelerating path, which we can call tau, and then if you insist on talking about the elapsed proper time along the inertial path, you can call that by capitalized letters TAU. If you prefer, you could call them tau_accel and tau_inert. It doesn't matter what you call them.

The problem is that you're confusing TAU with tau, as explained previously. Again, TAU is the proper time along the unaccelerated path, whereas tau is the proper time along the accelerated path.

Le 11/08/2022 à 23:12, Stan Fultoni a écrit :
> The elapsed proper time between the origin and the final event is different
> along the accelerated path than along the inertial path, so you need to
> distinguish between them. The time of interest is the time along the accelerating
> path, which we can call tau, and then if you insist on talking about the elapsed
> proper time along the inertial path, you can call that by capitalized letters TAU.
> If you prefer, you could call them tau_accel and tau_inert. It doesn't matter
> what you call them.
>
> The problem is that you're confusing TAU with tau, as explained previously.
> Again, TAU is the proper time along the unaccelerated path, whereas tau is the
> proper time along the accelerated path.

I call proper time of an object the time it measures between two events.

For example, in Langevin's traveler problem, I posit an event which is the
departure, and an event which is the return to earth.
The subject must go around Tau Ceti with a constant speed of 0.8c. I note
a proper time (or real time) Tr for this subject.

On the way back, I am Tr=18 years old.

His brother, who remained on earth, will also measure the time interval
between the two events. He notes To = 30 years.

I don't see where we can add tau or TAU times which would be different
from Tr and timed by who knows who.

It is the same in the accelerated traveler problem of Tau Ceti. It leaves
the ground with a= 10m/s², and will evolve on 12ly.

Arrived there (Hachel calculation) his watch marks Tr=4 years and 284
days, in the Terrestre-Tau Ceti repository, watches mark 12 years 338
days. (slight variations if we count 365d or 365.25d per year).

But in no case exists a watch that notes TAU or tau, placed who knows
where.

The rocket, it notes only one clean hour.

Otherwise, it's nonsense.

R.H.

On Thursday, August 11, 2022 at 2:35:29 PM UTC-7, Richard Hachel wrote:
> > The elapsed proper time between the origin and the final event is different
> > along the accelerated path than along the inertial path, so you need to
> > distinguish between them. The time of interest is the time along the accelerating
> > path, which we can call tau, and then if you insist on talking about the elapsed
> > proper time along the inertial path, you can call that by capitalized letters TAU.
> > If you prefer, you could call them tau_accel and tau_inert. It doesn't matter
> > what you call them.
> >
> > The problem is that you're confusing TAU with tau, as explained previously.
> > Again, TAU is the proper time along the unaccelerated path, whereas tau is the
> > proper time along the accelerated path.
>
> I call proper time of an object the time it measures between two events.

Yes, that is correct.

> The subject must go around Tau Ceti with a constant speed of 0.8c.

You are switching to a different problem. If, in terms of the earth's inertial coordinate system, someone goes a distance of x=12 LY at a constant speed 0.8c, it will take then t=15 years of coordinate time, and he will undergo 9 years of elapsed proper time. But that is not the question you are trying to answer. You are considering someone who begins at rest and undergoes a constant proper acceleration of 10 m/sec^2 = 1.052 LY^-1. His journey outward will take 12.915 years of coordinate time, and he will undergo 3.139 years of elapsed proper time. Now, if someone went at constant speed from the origin to that arrival event, he would undergo 7.602 years of elapsed proper time. We covered this before. Remember?

On Thursday, August 11, 2022 at 3:59:14 PM UTC-4, Richard Hachel wrote:
> Le 11/08/2022 à 20:36, rotchm a écrit :

> > And its integral is tau = (1/a)invsinh(at). Do you agree that that is its
> > integral?

No answer?

> >> > you already agreed that x = (1/a) [sqrt(1 + (at)²) - 1],
> > Do you remember having agreed to this?

No answer?

> I sometimes have trouble with scientific notations,

If you have trouble with that, then this shows you are limited, that you do not have what it takes to discuss math and physics.

> and I use mine which I try to define better

You may try but you have not succeeded. You just introduce gibberish. So, don't use your notation. Learn, and adopt the conventional notations.

> Tr and Vr are new notions, but which I think are essential.

Will those Notions change the values displayed on the clocks?
No? Then they are useless.

> Now we come back to the equation:
> > Do you agree that there is no There is no "Vr" in dtau = dt/sqrt(1 + (at)^2)?
> I do not write the equation in this form where we give tau as a function
> of t and (at)². I don't see the immediate benefit.

Your answer does not addressed the above question. The question has a simple yes or no answer.
Try again.

<gibberish snipped>

On Thursday, August 11, 2022 at 5:35:29 PM UTC-4, Richard Hachel wrote:

> I call proper time of an object the time it measures between two events.

Proper time of an object (clock) is the value displayed on that clock (typically set to zero initially).
It is typically symbolized as 't' or tau in physics.

> For example, in Langevin's traveler problem, I posit an event which is the
> departure, and an event which is the return to earth.
> The subject must go around Tau Ceti with a constant speed of 0.8c. I note
> a proper time (or real time) Tr for this subject.

Proper time, real-time, Tr.... Why not just use the conventional notation t !!
It's that simple. It's the value displayed by the given clock. There is no need to invent three other words or notations for it.

> On the way back, I am Tr=18 years old.

On the way back, the value displayed by the clock (me, my watch, my age) is 18 units.

> His brother, who remained on earth, will also measure the time interval
> between the two events. He notes To = 30 years.

Value on his clock is 30 units.

> I don't see where we can add tau or TAU times which would be different
> from Tr and timed by who knows who.

The value displayed by the given clock is such and such.

> Arrived there (Hachel calculation) his watch marks Tr=4 years and 284
> days,

You mean, the value of his clock displays 4 years and 284 days.
Etc.

Triad hearing to the expression 'the value displayed by the given clock'.
This removes many ambiguities.

> But in no case exists a watch that notes TAU or tau, placed who knows
> where.

Again the proper time on a given clock is the value displayed by that Clock.

On Thursday, August 11, 2022 at 2:35:29 PM UTC-7, Richard Hachel wrote:
> > The elapsed proper time between the origin and the final event is different
> > along the accelerated path than along the inertial path, so you need to
> > distinguish between them. The time of interest is the time along the accelerating
> > path, which we can call tau, and then if you insist on talking about the elapsed
> > proper time along the inertial path, you can call that by capitalized letters TAU.
> > If you prefer, you could call them tau_accel and tau_inert. It doesn't matter
> > what you call them.
> >
> > The problem is that you're confusing TAU with tau, as explained previously.
> > Again, TAU is the proper time along the unaccelerated path, whereas tau is the
> > proper time along the accelerated path.
>
> I call proper time of an object the time it measures between two events.

Yes, that's correct.

> The subject must go around Tau Ceti with a constant speed of 0.8c.

You're switching to a different problem. If, in terms of the earth's inertial coordinate system, someone goes a distance of x=12 LY at a constant speed 0.8c, it will take then t=15 years of coordinate time, and he will undergo 9 years of elapsed proper time. But that is not the question you are trying to answer. You are considering someone who begins at rest and undergoes a constant proper acceleration of 10 m/sec^2 = 1.052 LY^-1. His journey outward will take 12.915 years of coordinate time, and he will undergo 3..139 years of elapsed proper time. Now, if someone went at constant speed from the origin to that arrival event, he would be traveling at constant speed 0.92915c, and undergo 4.77 years of elapsed proper time. We covered this before. Remember?

Le 12/08/2022 à 01:59, Stan Fultoni a écrit :

> If, in terms of the earth's inertial coordinate system, someone goes a distance
> of x=12 LY at a constant speed 0.8c, it will take then t=15 years of coordinate
> time, and he will undergo 9 years of elapsed proper time.

Perfectly.

> But that is not the question you are trying to answer. You are considering
> someone who begins at rest and undergoes a constant proper acceleration of 10
> m/sec^2.

Yes.

> His journey outward will take 12.915 years of coordinate time,

Absolutly.

To=(x/c).sqrt(1+2c²/ax)

>and he will undergo 3.139 years of elapsed proper time.

No.

I think your equation is not correct, and you just have to put
Tr=sqrt(2x/a).

Tr=4.776 years

> Now, if someone went at constant speed from the origin to that arrival event, he
> would undergo 7.602 years of elapsed proper time. We covered this before.
> Remember?

To=12.915

If Vo=cte then Vo=x/To=0.92915c

Then Vr=2.513c
Tr=x/Vr=12/2.513=4.776 years

R.H.

Le 12/08/2022 à 03:41, rotchm a écrit :
> On Thursday, August 11, 2022 at 3:59:14 PM UTC-4, Richard Hachel wrote:
>> Le 11/08/2022 à 20:36, rotchm a écrit :
>
>> > And its integral is tau = (1/a)invsinh(at). Do you agree that that is its
>> > integral?
>
> No answer?

I have answered many times already.

You must follow.

I said that for me, this equation is false.
>
>
>> >> > you already agreed that x = (1/a) [sqrt(1 + (at)²) - 1],
>> > Do you remember having agreed to this?
>
> No answer?

J'ai déjà repondu plusieurs fois que je considérais cette équation
comme correcte, et que je l'écrivais ainsi:
x = (c²/a).[sqrt(1+(aTo)²)-1]

You must follow.

>
>> I sometimes have trouble with scientific notations,
>
> If you have trouble with that, then this shows you are limited, that you do not
> have what it takes to discuss math and physics.
>
>> and I use mine which I try to define better
>
> You may try but you have not succeeded. You just introduce gibberish. So, don't
> use your notation. Learn, and adopt the conventional notations.
>
>> Tr and Vr are new notions, but which I think are essential.
>
> Will those Notions change the values displayed on the clocks?
> No? Then they are useless.
>
>> Now we come back to the equation:
>> > Do you agree that there is no There is no "Vr" in dtau = dt/sqrt(1 + (at)^2)?
>> I do not write the equation in this form where we give tau as a function
>> of t and (at)². I don't see the immediate benefit.
>
> Your answer does not addressed the above question. The question has a simple yes
> or no answer.
> Try again.

I don't see what this equation could want to show.
She has no interest whatsoever for me.

If I want to give the observable time as a function of the real (proper)
time in the accelerated reference frames starting at rest I ask:
To=Tr.sqrt(1+(1/4)Vr²/c²)

If there is a rolling start, I set To=Tr.sqrt[1+((1/2)aTr+Vri)²/c²)]

R.H.

On Thursday, August 11, 2022 at 11:34:45 PM UTC-7, Richard Hachel wrote:
> > If, in terms of the earth's inertial coordinate system, someone goes a distance
> > of x=12 LY at a constant speed 0.8c, it will take then t=15 years of coordinate
> > time, and he will undergo 9 years of elapsed proper time.
>
> Perfectly.
>
> > But that is not the question you are trying to answer. You are considering
> > someone who begins at rest and undergoes a constant proper acceleration of 10
> > m/sec^2. His journey outward will take 12.915 years of coordinate time,
>
> Yes.
>
> >and he will undergo 3.139 years of elapsed proper time.
>
> No.

You are mistaken. That is the elapsed proper time of the accelerating traveler. This has been explained to you in detail. Remember?

> t = x.sqrt(1+2/(ax))

Right, that is the relationship between the coordinates t and x for the accelerating (hyperbolic) trajectory. Remember, you already agreed that x = (1/a) [sqrt(1 + (at)²) - 1], so dx = [at/sqrt(1+(at)^2]]dt, so we have dtau = sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) = 3.139 years.

> I think your equation is not correct, and you just have to put Tr=sqrt(2x/a).

No, what you call "Tr" is the proper time along the *inertial* path from the origin to the arrival event at x=12, t=12.915. This has been explained to you before. The elapsed time along that inertial path is 4.77 years, but the elapsed proper time along the accelerated path is 3.139 years. Both of those are given by integrating along the respective path, although for the inertial path the integration is trivial. Remember:

tau_inertial = sqrt(2x/a)
tau_accelerated = (1/a)invsinh(at)

Your mistake is confusing tau_inertial with tau_accelerated. Understand?

Le 12/08/2022 à 09:08, Stan Fultoni a écrit :
> On Thursday, August 11, 2022 at 11:34:45 PM UTC-7, Richard Hachel wrote:
>> > If, in terms of the earth's inertial coordinate system, someone goes a
>> distance
>> > of x=12 LY at a constant speed 0.8c, it will take then t=15 years of
>> coordinate
>> > time, and he will undergo 9 years of elapsed proper time.
>>
>> Perfectly.
>>
>> > But that is not the question you are trying to answer. You are considering
>> > someone who begins at rest and undergoes a constant proper acceleration of 10
>> > m/sec^2. His journey outward will take 12.915 years of coordinate time,
>>
>> Yes.
>>
>> >and he will undergo 3.139 years of elapsed proper time.
>>
>> No.
>
> You are mistaken. That is the elapsed proper time of the accelerating traveler.
> This has been explained to you in detail. Remember?
>
>> t = x.sqrt(1+2/(ax))
>
> Right, that is the relationship between the coordinates t and x for the
> accelerating (hyperbolic) trajectory. Remember, you already agreed that x = (1/a)
> [sqrt(1 + (at)²) - 1], so dx = [at/sqrt(1+(at)^2]]dt, so we have dtau = sqrt(dt^2
> - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) = 3.139
> years.
>
>> I think your equation is not correct, and you just have to put Tr=sqrt(2x/a).
>
> No, what you call "Tr" is the proper time along the *inertial* path from the
> origin to the arrival event at x=12, t=12.915. This has been explained to you
> before. The elapsed time along that inertial path is 4.77 years, but the elapsed
> proper time along the accelerated path is 3.139 years. Both of those are given by
> integrating along the respective path, although for the inertial path the
> integration is trivial. Remember:
>
> tau_inertial = sqrt(2x/a)
> tau_accelerated = (1/a)invsinh(at)
>
> Your mistake is confusing tau_inertial with tau_accelerated. Understand?

It's not a mistake.

I think that's the reality of things.

In my opinion, the integration we are doing is not a correct geometric
concept. One integrates carrots and turnips inconsiderately.

Tu as essayé d'intégrer l'équation correcte?

To=ΔTr.∫sqrt[1+((1/2)aΔTr+Vri)²/c²)]

R.H.

On Friday, August 12, 2022 at 12:34:30 AM UTC-7, Richard Hachel wrote:
> Le 12/08/2022 à 09:08, Stan Fultoni a écrit :
> > On Thursday, August 11, 2022 at 11:34:45 PM UTC-7, Richard Hachel wrote:
> >> > If, in terms of the earth's inertial coordinate system, someone goes a
> >> distance
> >> > of x=12 LY at a constant speed 0.8c, it will take then t=15 years of
> >> coordinate
> >> > time, and he will undergo 9 years of elapsed proper time.
> >>
> >> Perfectly.
> >>
> >> > But that is not the question you are trying to answer. You are considering
> >> > someone who begins at rest and undergoes a constant proper acceleration of 10
> >> > m/sec^2. His journey outward will take 12.915 years of coordinate time,
> >>
> >> Yes.
> >>
> >> >and he will undergo 3.139 years of elapsed proper time.
> >>
> >> No.
> >
> > You are mistaken. That is the elapsed proper time of the accelerating traveler.
> > This has been explained to you in detail. Remember?
> >
> >> t = x.sqrt(1+2/(ax))
> >
> > Right, that is the relationship between the coordinates t and x for the
> > accelerating (hyperbolic) trajectory. Remember, you already agreed that x = (1/a)
> > [sqrt(1 + (at)²) - 1], so dx = [at/sqrt(1+(at)^2]]dt, so we have dtau = sqrt(dt^2
> > - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) = 3.139
> > years.
> >
> >> I think your equation is not correct, and you just have to put Tr=sqrt(2x/a).
> >
> > No, what you call "Tr" is the proper time along the *inertial* path from the
> > origin to the arrival event at x=12, t=12.915. This has been explained to you
> > before. The elapsed time along that inertial path is 4.77 years, but the elapsed
> > proper time along the accelerated path is 3.139 years. Both of those are given by
> > integrating along the respective path, although for the inertial path the
> > integration is trivial. Remember:
> >
> > tau_inertial = sqrt(2x/a)
> > tau_accelerated = (1/a)invsinh(at)
> >
> > Your mistake is confusing tau_inertial with tau_accelerated. Understand?
>
> It's not a mistake.

It is plainly a mistake, because what you want is the elapsed timefor the accelerating rocket, but what you are calculating is the elapsed time along the inertial path between the starting and ending events.

> I think that's the reality of things.

Reality can;t be logically self-contradictory. 1 = 0 is not reality.

> In my opinion, the integration we are doing is not a correct geometric concept.

It isn't a matter of opinion. Remember, you have actually agreed that t = x.sqrt(1+2/(ax)), and solving this for x gives the relation x = (1/a)[sqrt(1 + (at)²) - 1], so we have dx = [at/sqrt(1+(at)^2]]dt, and therefore dtau = sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) = 3.139 years. There is no ambiguity or "opinion" about this.

> Tu as essayé d'intégrer l'équation correcte?
> To=ΔTr.∫sqrt[1+((1/2)aΔTr+Vri)²/c²)]

That is not the correct equation, it is infantile drivel. The correct equation to be integrated is dtau = sqrt(dt^2 - dx^2) along the hyperbolic path of constant proper acceleration, as explained above. This is logically correct, and it yields the empirically verified result. You have agreed with each logical step... so there is no excuse for you to revert back to your infantile drivel. Agreed?

On Sunday, August 7, 2022 at 12:19:17 AM UTC-7, Richard Hachel wrote:
> Le 06/08/2022 à 22:52, Stan Fultoni a écrit :
> > On Saturday, August 6, 2022 at 12:26:04 PM UTC-7, Richard Hachel wrote:
> >> > You claim both "X" and "not X",
> >>
> >> No.
> >
> > Again, X is the proposition that the elapsed time along any path between any two
> > events ei and ej is tauj - taui = sqrt[(tj-ti)^2 - (xj-xi)^2/c^2]. ) You claim
> > both X and not X, so you are contradicting yourself. That's why your beliefs
> > entail 1=0. Remember?
> It is more complicated than that.

In this case, no.

> Because this equation is correct for all inertial frames in uniform
> motion, and for all accelerated frames with a stopped start.

It is an equation in terms of an inertial frame. You are confusing the
movement of the object (an accelerated rocket) with that of the
observer frame (which is inertial) used for coordinate tracking the
rocket.

> But if we take the case of an accelerated frame of reference with a
> non-stop start, it does not work.

The formula for the elapsed time is not in terms of the (curvilinear)
accelerated frame. It's in terms of an inertial frame. That's why the
formula for length is so simple (the determinant of the Minkowski
metric is -1).

> As with relativistic velocity additions, you can't do what you want
> without the correct equations.

True but you are forever confused by the entire subject. I suggested
multiple times that you pick a hobby that's better suited for your
innate talents. Less frustration for you.

--
Jan

On Friday, August 12, 2022 at 2:58:20 AM UTC-4, Richard Hachel wrote:

> >> > And its integral is tau = (1/a)invsinh(at). Do you agree that that is its
> >> > integral?
> >
> > No answer?
> I have answered many times already.

And your answer was yes sometimes and no sometimes.
You must be consistent which you are not. Try again.

> I said that for me, this equation is false.

And that is irrelevant since it is not the issue, It is not the topic.
The topic was the integral of a certain function. No matter what that initial function represents or if it is true or not.
Try to focus and remain on topic.

> >> >> > you already agreed that x = (1/a) [sqrt(1 + (at)²) - 1],
> >> > Do you remember having agreed to this?
> >
> > No answer?
> J'ai déjà repondu plusieurs fois que je considérais cette équation

We were discussing in English.
Try to remain consistent in the discussion.
Since your act was deliberate, I report you as spam. I incite others to do the same.

> >> Now we come back to the equation:
> >> > Do you agree that there is no There is no "Vr" in dtau = dt/sqrt(1 + (at)^2)?

> >> I do not write the equation in this form where we give tau as a function
> >> of t and (at)². I don't see the immediate benefit.

It gives you the email relation between the coordinate time and the proper time.
That is, it gives you an immediate relation between the value displayed on the "fixed inertial clock on a ground" and the value of the watch of The Traveler...

> If I want to give the observable time as a function of the real (proper)
> time in the accelerated reference frames starting at rest I ask:
> To=Tr.sqrt(1+(1/4)Vr²/c²)
>
> If there is a rolling start, I set To=Tr.sqrt[1+((1/2)aTr+Vri)²/c²)]

Do you realize that the values your predict are not the same as relativity's prediction, and your values contradict actual results of experiments? Do you realize this?

On Friday, August 12, 2022 at 3:34:30 AM UTC-4, Richard Hachel wrote:
> Le 12/08/2022 à 09:08, Stan Fultoni a écrit :

> > tau_inertial = sqrt(2x/a)
> > tau_accelerated = (1/a)invsinh(at)
> >
> > Your mistake is confusing tau_inertial with tau_accelerated. Understand?
> It's not a mistake.

Yes it is.

> I think that's the reality of things.

Don't just think it's reality, compare it to actual results of experiments. That is "reality", agreed?

> In my opinion, the integration we are doing is not a correct geometric
> concept.

Irrelevant if it's a correct geometric concept or not. It gives you a formula, a prediction. And this prediction agrees with results of experiments.

> One integrates carrots and turnips inconsiderately.

But it works, Contrary to your equations.

> Tu as essayé d'intégrer l'équation correcte?

Spam recorded. I incite others to do the same.

On Tuesday, August 9, 2022 at 6:17:02 PM UTC-7, Stan Fultoni wrote:
> sqrt(dt^2 - dx^2) you get tau = (1/a)invsinh(at).

Not that it's the end of the world or anything but the inverse of the sinh
function is called "ar sinh" or "arsinh" (short for "area hyperbolic sine",
note it's "ar", not "arc").

Same for "arcosh", "artanh", etc.

--
Jan

Le 12/08/2022 à 10:20, Stan Fultoni a écrit :
> On Friday, August 12, 2022 at 12:34:30 AM UTC-7, Richard Hachel wrote:
>> Le 12/08/2022 à 09:08, Stan Fultoni a écrit :
>> > On Thursday, August 11, 2022 at 11:34:45 PM UTC-7, Richard Hachel wrote:
>> >> > If, in terms of the earth's inertial coordinate system, someone goes a
>> >> distance
>> >> > of x=12 LY at a constant speed 0.8c, it will take then t=15 years of
>> >> coordinate
>> >> > time, and he will undergo 9 years of elapsed proper time.
>> >>
>> >> Perfectly.
>> >>
>> >> > But that is not the question you are trying to answer. You are considering
>> >> > someone who begins at rest and undergoes a constant proper acceleration of
>> 10
>> >> > m/sec^2. His journey outward will take 12.915 years of coordinate time,
>> >>
>> >> Yes.
>> >>
>> >> >and he will undergo 3.139 years of elapsed proper time.
>> >>
>> >> No.
>> >
>> > You are mistaken. That is the elapsed proper time of the accelerating
>> traveler.
>> > This has been explained to you in detail. Remember?
>> >
>> >> t = x.sqrt(1+2/(ax))
>> >
>> > Right, that is the relationship between the coordinates t and x for the
>> > accelerating (hyperbolic) trajectory. Remember, you already agreed that x =
>> (1/a)
>> > [sqrt(1 + (at)²) - 1], so dx = [at/sqrt(1+(at)^2]]dt, so we have dtau =
>> sqrt(dt^2
>> > - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) =
>> 3.139
>> > years.
>> >
>> >> I think your equation is not correct, and you just have to put Tr=sqrt(2x/a).
>>
>> >
>> > No, what you call "Tr" is the proper time along the *inertial* path from the
>> > origin to the arrival event at x=12, t=12.915. This has been explained to you
>> > before. The elapsed time along that inertial path is 4.77 years, but the
>> elapsed
>> > proper time along the accelerated path is 3.139 years. Both of those are given
>> by
>> > integrating along the respective path, although for the inertial path the
>> > integration is trivial. Remember:
>> >
>> > tau_inertial = sqrt(2x/a)
>> > tau_accelerated = (1/a)invsinh(at)
>> >
>> > Your mistake is confusing tau_inertial with tau_accelerated. Understand?
>>
>> It's not a mistake.
>
> It is plainly a mistake, because what you want is the elapsed timefor the
> accelerating rocket, but what you are calculating is the elapsed time along the
> inertial path between the starting and ending events.
>
>> I think that's the reality of things.
>
> Reality can;t be logically self-contradictory. 1 = 0 is not reality.
>
>> In my opinion, the integration we are doing is not a correct geometric concept.
>
> It isn't a matter of opinion. Remember, you have actually agreed that t =
> x.sqrt(1+2/(ax)), and solving this for x gives the relation x = (1/a)[sqrt(1 +
> (at)²) - 1], so we have dx = [at/sqrt(1+(at)^2]]dt, and therefore dtau =
> sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at)
> = 3.139 years. There is no ambiguity or "opinion" about this.
>
>> Tu as essayé d'intégrer l'équation correcte?
>> To=ΔTr.∫sqrt[1+((1/2)aΔTr+Vri)²/c²)]
>
> That is not the correct equation, it is infantile drivel. The correct equation
> to be integrated is dtau = sqrt(dt^2 - dx^2) along the hyperbolic path of constant
> proper acceleration, as explained above. This is logically correct, and it yields
> the empirically verified result. You have agreed with each logical step... so
> there is no excuse for you to revert back to your infantile drivel. Agreed?

Is it the same to integrate dtau=sqrt(dt²-dx²) and
dt²=sqrt(dtau²+dx²) ?

Are you sure that the way you put dx² is correct?

Something else.

You speak of three times.

t, tau, TAU.

I don't mind.

But if I take the duration t (To in the terrestrial frame of reference),
and cut it into a thousand pieces.

It is obvious that in the first piece, I will have To=tau=Tau since the
speeds are very low.

But in the last piece (Vr=5.0245c), it goes without saying that dTr will
be very different from dtau and from TAU.

However, you tell me (and it's true) that dtau and dTAU (the two proper
times you describe) will be very different from each other.

This means that in the earlier pieces tau and TAU are the same thing, but
not so in the later ones.

However, there is no absolute frame of reference, no absolute
acceleration, wherever I find myself on the journey, I can consider myself
at rest, and consider that I have just accelerated.

And so that tau=TAU.

I only use one Tr (tau) and I find it difficult to understand the
usefulness of another tau which you call TAU and which I don't know what
to do with.

R.H.

Le 12/08/2022 à 13:45, rotchm a écrit :

> Do you realize that the values your predict are not the same as relativity's
> prediction, and your values contradict actual results of experiments? Do you
> realize this?

No.

What resultes? What experiences?

I think that specific experiments could one day be done on this specific
subject.

But that's not the case yet.

In the meantime, I am trying to find where the calculation error comes
from (which is a relativistic geometry error and a confusion between
carrots and turnips).

I remind you that this is not the first time that I suspect such errors..

In the history of the traveler of Langevin, it is proven that I was right
by the simple theoretical deduction in apparent speed (manifest absurdity
which I explained well).

For the accelerated traveler of Tau Ceti too, there is a bug.

R.H.

Le 12/08/2022 à 13:50, rotchm a écrit :

> And this prediction agrees with results of experiments.

Be careful what you say.

R.H.

On Friday, August 12, 2022 at 5:18:32 AM UTC-7, Richard Hachel wrote:
> Is it the same to integrate dtau=sqrt(dt²-dx²) and
> dt²=sqrt(dtau²+dx²) ?

Sure, but to do the latter you would need to know tau as a function of x, which is what you are trying to determine. We know x as a function of t for the path of constant proper acceleration, and we are trying to determine tau, so we integrate dtau=sqrt(dt^2 - dx^2).

> Are you sure that the way you put dx² is correct?

Yes. Remember, you already agreed that x = (1/a)[sqrt(1 + (at)²) - 1], so dx = [at/sqrt(1+(at)^2]]dt, and hence we have dtau = sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) = 3.139 years.
> > It isn't a matter of opinion. Remember, you have actually agreed that t =
> > x.sqrt(1+2/(ax)), and solving this for x gives the relation x = (1/a)[sqrt(1 +
> > (at)²) - 1], so we have dx = [at/sqrt(1+(at)^2]]dt, and therefore dtau =
> > sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at)
> > = 3.139 years. There is no ambiguity or "opinion" about this.
> >
> >> Tu as essayé d'intégrer l'équation correcte?
> >> To=ΔTr.∫sqrt[1+((1/2)aΔTr+Vri)²/c²)]
> >
> > That is not the correct equation, it is infantile drivel. The correct equation
> > to be integrated is dtau = sqrt(dt^2 - dx^2) along the hyperbolic path of constant
> > proper acceleration, as explained above. This is logically correct, and it yields
> > the empirically verified result. You have agreed with each logical step.... so
> > there is no excuse for you to revert back to your infantile drivel. Agreed?

> You speak of three times. t, tau, TAU.

Well, YOU speak of three times. The value of TAU is not actually relevant, but TAU is what you call Tr, and that is what you are computing. What you should be computing is tau, which is the proper time along the accelerating path.

> But if I take the duration t in the terrestrial frame of reference, and cut it into a thousand pieces.
> It is obvious that in the first piece, I will have To=tau=TAU since the speeds are very low.

No, the value of TAU (which you call Tr) is the elapsed proper time along the inertial path from the origin to the event x=12, t=12.915 at the constant speed 0.92915, so dTAU is not the same as dt, even at the beginning.

> I only use one Tr (tau)

No, you use Tr = TAU, which is not the relevant parameter. You need to compute the elapsed proper time along the accelerated path, not along the path with constant speed 0.92915. Understand?

On Friday, August 12, 2022 at 8:42:17 AM UTC-4, Richard Hachel wrote:
> Le 12/08/2022 à 13:45, rotchm a écrit :
>
> > Do you realize that the values your predict are not the same as relativity's
> > prediction, and your values contradict actual results of experiments? Do you
> > realize this?
> No.

Then you are a reality denier, and you even deny that.

> What resultes? What experiences?

Google is your friend.

> I think that specific experiments could one day be done on this specific
> subject.

Such experiments have been done numerous times. Google is your friend.

> But that's not the case yet.

You are either very confused or a reality denier.

> In the meantime, I am trying to find where the calculation error comes
> from (which is a relativistic geometry error

Perhaps you made calculation errors, but there are no such errors in relativity.

> For the accelerated traveler of Tau Ceti too, there is a bug.

No, there is no bug. And you have failed to show that there is a bug, And your reasonings are wrong. But you are unable to accept that fact because you are a reality denier.