Le 09/08/2022 à 03:42, Stan Fultoni a écrit :
> On Monday, August 8, 2022 at 12:38:52 PM UTC-7, Richard Hachel wrote:

>
>> Your equation tau = (1/g) asinh(g t) is not correct.
>
> It is correct, where tau is the elapsed proper time along the path of constant
> proper acceleration from rest at the origin, nd t is the coordinate time, and g is
> the proper acceleration. The simple derivation was provided to you, simply
> integrating dtau = sqrt(dt^2 -dx^2).

You see, I'm not as stupid as people say on the forums.

Worse, I find this equation of the first.

Whereas among the relativists, they find theirs of the second.

I start from Tr = tau = sqrt(2x/a)

and I come to
To = (x/c).sqrt(1+2c²/ax)

They start from To = (x/c).sqrt(1+2c²/ax)

and arrive at tau = (1/g) asinh(g t)

There must be something hugely wrong between them and me.

How come their first answer is right, and my second answer is right (the
same)?

But that we don't have the same them on their SECOND, and me, on my first?

Can an error be retroactive? ? ?

It's absurd.

So the error can only come from them.

A child would understand that.

R.H.

Le 09/08/2022 à 03:42, Stan Fultoni a écrit :
> On Monday, August 8, 2022 at 12:38:52 PM UTC-7, Richard Hachel wrote:

> simply integrating dtau = sqrt(dt^2 -dx^2).

Cette manière d'agir n'est pas correcte. Il y a un biais conceptuel.

Il se trouve dans dx².

Il faut procéder ainsi pour avoir l'équation relativiste correcte.

dtau = sqrt(dt² -dx²)

dtau = sqrt(dt² - (1/2.a.dtau²)²/c²)

dtau = sqrt(dt² - dtau²(1/4.a².dtau²/c²)

dtau = dt /sqrt (1+(1/4)Vr²/c²)

Attention, cela ne fonctionne que départ arrêté. Si le départ est
"lancé", c'est plus compliqué.

Dans ce cas il faut utiliser x=(1/2)aTr²+Vri.Tr

Vri (vitesse réelle initiale)

R.H.

Le 09/08/2022 à 03:42, Stan Fultoni a écrit :
> On Monday, August 8, 2022 at 12:38:52 PM UTC-7, Richard Hachel wrote:

> Your "system" is logically inconsistent,

More consistent.

>because you agree that dtau = sqrt(dt^2 - dx^2),

Yes.

Obviously.

But I define better dx². Is not, in this case, dx=x/c like in initial
frames.

>
>> When you tell me that the rocket will very quickly reach the speed of
>> light, it proves that you do not understand everything I say correctly.
>
> No, you claimed that x = (1/2)at²,

That's what I say.

> which would imply that dx/dt exceeds the speed of light after about 1 year.

No.

That's not what I say. Never.

Never.

I beg you to be careful.

When the rocket reaches for example Tau Ceti, its speed is Vr=5.0245c.

That's what I said.

You are confusing real speed Vr with observable speed Vo.

If you want to speak in observable speed (traditional), you have to set
Vo=Vr/sqrt(1+Vr²/c²)

You will find that nothing can exceed c.

Here, Vo=0.980c

> You are confusing constant coordinate acceleration for constant proper
> acceleration (a parabola versus a hyperbola).

Yes, but I do it voluntarily.

> You even admitted later that the correct relation is x = (1/a) [sqrt(1 +
> (at)²) - 1]. Remember?

Yes.

x=(c²/a).[sqrt(1+a²To²/c²)-1]

R.H.

On Tuesday, August 9, 2022 at 5:11:42 AM UTC-7, Richard Hachel wrote:
> I start from tau = sqrt(2x/a)
> and I come to
> t = x.sqrt(1+2/(ax))

Your mistake is failing to distinguish between the proper time "tau" along the accelerated path versus the proper time along the inertial path, which we can call "TAU". Correct equations are

x = (1/2)a TAU^2
t = TAU sqrt[1 + (1/4)a^2 TAU^2]

and if we eliminate TAU from these equations we get t = x sqrt(1+2/(ax)) .. You see, this is all you are doing. The problem is that you are confusing TAU with tau.

> They start from To = (x/c).sqrt(1+2c²/ax)
> and arrive at tau = (1/g) asinh(g t)

Well, no, the correct result begins with dtau = sqrt[dt^2 - dx^2], and integrates along the accelerating path to give tau = (1/g) asinh(g t). This is equivalent to t = (1/a) sinh(a tau), and we also have x = (1/a)[cosh(a tau) - 1].

> There must be something hugely wrong between them and me.

Yes, you are making the huge error of confusing TAU (proper time along inertial path) with tau (proper time along accelerating path).

Le 09/08/2022 à 15:49, Stan Fultoni a écrit :
> On Tuesday, August 9, 2022 at 5:11:42 AM UTC-7, Richard Hachel wrote:
>> I start from tau = sqrt(2x/a)
>> and I come to
>> t = x.sqrt(1+2/(ax))
>
> Your mistake is failing to distinguish between the proper time "tau" along the
> accelerated path versus the proper time along the inertial path, which we can call
> "TAU". Correct equations are
>
> x = (1/2)a TAU^2
> t = TAU sqrt[1 + (1/4)a^2 TAU^2]
>
> and if we eliminate TAU from these equations we get t = x sqrt(1+2/(ax)) .

Yes.

This is what I wrote in the equations I posted several weeks ago here.

To=(x/c).sqrt(1+2c²/ax)

No problem.

<http://news2.nemoweb.net/jntp?7VrmUd2e4_0PhHyhfbvm7bjIqOI@jntp/Data.Media:1>

<http://news2.nemoweb.net/?DataID=7VrmUd2e4_0PhHyhfbvm7bjIqOI@jntp>

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

On 2022-08-09 12:34:27 +0000, Richard Hachel said:

> Il faut procéder ainsi pour avoir l'équation relativiste correcte.
> dtau = sqrt(dt² -dx²)
>
> dtau = sqrt(dt² - (1/2.a.dtau²)²/c²)
>
> dtau = sqrt(dt² - dtau²(1/4.a².dtau²/c²)
>
> dtau = dt /sqrt (1+(1/4)Vr²/c²)
>
> Attention, cela ne fonctionne que départ arrêté. Si le départ est
> "lancé", c'est plus compliqué.

A clock does not know how fast it is moving or how long it has been
moving or even whethet it is moving. Therefore its behaviour is best
described with equations that don't care whether it was moveing in
the beginning. The equation dtau = sqrt(dt² - dx²) satifies this
requirement.

Mikko

Le 09/08/2022 à 16:49, Mikko a écrit :
> A clock does not know how fast it is moving or how long it has been
> moving or even whethet it is moving. Therefore its behaviour is best
> described with equations that don't care whether it was moveing in
> the beginning. The equation dtau = sqrt(dt² - dx²) satifies this
> requirement.

Yes.

Obviously.

dtau = sqrt(dt² - dx²)

Or To²=Tr²+Et²

BUT!

But, but, but, but!

How do you calculate dx in this case?

If the rocket leaves when it is at rest, it is not the same thing as if it
already has an acquired speed.

The equation is not the same, but becomes x=(1/2)aTr²+Vri.Tr

With Vri=real initial speed.

Warning: Not to be confused with Voi=observable initial speed.

R.H.

On Tuesday, August 9, 2022 at 7:04:57 AM UTC-7, Richard Hachel wrote:
> No problem.

No problem? I just explained why all your beliefs are wrong, and you happily reply "no problem"?

What we have shown is that you are making the huge error of confusing TAU (proper time along inertial path) with tau (proper time along accelerating path). That's why everything you type is senseless gibberish. Agreed?

> I start from tau = sqrt(2x/a)
> and I come to
> t = x.sqrt(1+2/(ax))

Again, your mistake is failing to distinguish between the proper time "tau" along the accelerated path versus the proper time along the inertial path, which we can call "TAU". Correct equations are

.... x = (1/2)a TAU^2
.... t = TAU sqrt[1 + (1/4)a^2 TAU^2]

and if we eliminate TAU from these equations we get t = x sqrt(1+2/(ax)) , so this is all you are doing. The problem is that you are confusing TAU with tau. Understand your mistake now?

> They start from To = (x/c).sqrt(1+2c²/ax)
> and arrive at tau = (1/g) asinh(g t)

Again, no, the correct result begins with dtau = sqrt[dt^2 - dx^2], and integrates along the accelerating path to give tau = (1/g) asinh(g t). This is equivalent to t = (1/a) sinh(a tau), and we also have x = (1/a)[cosh(a tau) - 1].

> There must be something hugely wrong between them and me.

Yes, there issomething hugely wrong... you are making the huge error of confusing TAU (proper time along inertial path) with tau (proper time along accelerating path). Do you understand now?

On 2022-08-09 15:00:02 +0000, Richard Hachel said:

> Le 09/08/2022 à 16:49, Mikko a écrit :
>> A clock does not know how fast it is moving or how long it has been
>> moving or even whethet it is moving. Therefore its behaviour is best
>> described with equations that don't care whether it was moveing in
>> the beginning. The equation dtau = sqrt(dt² - dx²) satifies this
>> requirement.

> How do you calculate dx in this case?

Here t and x are time and space coordinates in a rectilinear orthogonal
isometric coordinate system that is aligned so that the clock'q motion
is along x-axis.

If the clocks motion is not along the x-axis then a more complete
formula dtau = sqrt(dt² - dx² - dy² - dz²) must be used.

> If the rocket leaves when it is at rest, it is not the same thing as if
> it already has an acquired speed.

It is. If the rocket is initially standing on Earth's surface it is
at rest in a coordiate system that is attached to the ground but
moving in a coordinate system that is attached to Sun. The latter
is important when the rocket is navigating in the Solar system.

Mikko

On Tuesday, August 9, 2022 at 7:04:57 AM UTC-7, Richard Hachel wrote:
> No problem.

No problem? We've shown that you are confusing TAU (proper time along inertial path) with tau (proper time along accelerating path). That's why everything you type is senseless.

> I start from tau = sqrt(2x/a)
> and I come to
> t = x.sqrt(1+2/(ax))

Again, you fail to distinguish between the proper time "tau" along the accelerated path versus the proper time along the inertial path, which we can call "TAU". The following equations are correct:

x = (1/2)a TAU^2
t = TAU sqrt[1 + (1/4)a^2 TAU^2]

If we eliminate TAU from these equations we get t = x sqrt(1+2/(ax)). The problem is that you are confusing TAU with tau.

> They start from To = (x/c).sqrt(1+2c²/ax)
> and arrive at tau = (1/g) asinh(g t)

Again, the correct result begins with dtau = sqrt[dt^2 - dx^2], and integrates along the accelerating path to give tau = (1/g) asinh(g t). This is equivalent to t = (1/a) sinh(a tau), and we also have x = (1/a)[cosh(a tau) - 1].

> There must be something hugely wrong between them and me.

Yes, there is something hugely wrong... you are making the huge error of confusing TAU (proper time along inertial path) with tau (proper time along accelerating path).

> Obviously dtau = sqrt(dt² - dx²), BUT how do you calculate dx in this case?

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).

Le 10/08/2022 à 03:17, Stan Fultoni a écrit :

> 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).

Le problème est là.

Il faut montrer que c'est là qu'il y a un truc qui cloche.

Comment d'une équation correcte, on en arrive à un résultat faux.

R.H.

On Wednesday, August 10, 2022 at 4:27:11 AM UTC-7, Richard Hachel wrote:
> > 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).
>
> Le problème est là. Il faut montrer que c'est là qu'il y a un truc qui cloche.
> Comment d'une équation correcte, on en arrive à un résultat faux.

You contradict yourself, because you have agreed that dtau = dt/sqrt(1 + (at)^2), which implies tau = (1/a)invsinh(at). This is simple and irrefutable. Learn calculus. In contrast, your claims are erroneous because you conflate the proper time along the accelerating path with the proper time along the inertial path.

Le 09/08/2022 à 17:33, Mikko a écrit :
> Here t and x are time and space coordinates in a rectilinear orthogonal
> isometric coordinate system that is aligned so that the clock'q motion
> is along x-axis.
>
> If the clocks motion is not along the x-axis then a more complete
> formula dtau = sqrt(dt² - dx² - dy² - dz²) must be used.
>
>> If the rocket leaves when it is at rest, it is not the same thing as if
>> it already has an acquired speed.
>
> It is. If the rocket is initially standing on Earth's surface it is
> at rest in a coordiate system that is attached to the ground but
> moving in a coordinate system that is attached to Sun. The latter
> is important when the rocket is navigating in the Solar system.
>
> Mikko

Yes, dear Mikko, you are totally right.

But that's not what we're talking about.

R.H.

Le 10/08/2022 à 15:49, Stan Fultoni a écrit :
> On Wednesday, August 10, 2022 at 4:27:11 AM UTC-7, Richard Hachel wrote:
>> > 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).
>>
>> Le problème est là. Il faut montrer que c'est là qu'il y a un truc qui
>> cloche.
>> Comment d'une équation correcte, on en arrive à un résultat faux.
>
> You contradict yourself, because you have agreed that dtau = dt/sqrt(1 +
> (at)^2), which implies tau = (1/a)invsinh(at). This is simple and irrefutable.
> Learn calculus. In contrast, your claims are erroneous because you conflate the
> proper time along the accelerating path with the proper time along the inertial
> path.

Can you explain your equation more precisely?

This is the distance traveled by the rocket in the terrestrial frame as a
function of terrestrial time.

Not for you?

But what exactly do you mean by: x = (1/a) [sqrt(1 + (at)²) - 1] ?

1) x = (c²/a) [sqrt(1 + (a.tau/c)²) - 1]

or 2) x = (c²/a) [sqrt(1 + (a.To/c)²) - 1]

To me, is the second equation who is correct.

R.H.

On 2022-08-10 15:20:42 +0000, Richard Hachel said:

> Yes, dear Mikko, you are totally right.
>
> But that's not what we're talking about.

Don't say "proper time" if you are talking about something else.

Mikko

On Wednesday, August 10, 2022 at 8:28:23 AM UTC-7, Richard Hachel wrote:
> >> > 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).
> >>
> >> Le problème est là. Il faut montrer que c'est là qu'il y a un truc qui
> >> cloche.
> >> Comment d'une équation correcte, on en arrive à un résultat faux.
> >
> > You contradict yourself, because you have agreed that dtau = dt/sqrt(1 +
> > (at)^2), which implies tau = (1/a)invsinh(at). This is simple and irrefutable.
> > Learn calculus. In contrast, your claims are erroneous because you conflate the
> > proper time along the accelerating path with the proper time along the inertial
> > path.
>
> Can you explain your equation more precisely?

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).

> What exactly do you mean by: x = (1/a) [sqrt(1 + (at)²) - 1] ?

That is the equation describing the position x of the rocket as a function of the coordinate time t for constant proper acceleration "a" beginning at rest at the origin. It describes the hyperbolic path. Note that I'm using units so c=1. The proper time along that path is given by integrating sqrt(dt^2 - dx^2) along that path, as explained above.

Le 10/08/2022 à 20:32, Stan Fultoni a écrit :
> On Wednesday, August 10, 2022 at 8:28:23 AM UTC-7, Richard Hachel wrote:
>> >> > 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).
>> >>
>> >> Le problème est là. Il faut montrer que c'est là qu'il y a un truc qui
>> >> cloche.
>> >> Comment d'une équation correcte, on en arrive à un résultat faux.
>> >
>> > You contradict yourself, because you have agreed that dtau = dt/sqrt(1 +
>> > (at)^2), which implies tau = (1/a)invsinh(at). This is simple and irrefutable.
>>
>> > Learn calculus. In contrast, your claims are erroneous because you conflate
>> the
>> > proper time along the accelerating path with the proper time along the
>> inertial
>> > path.
>>
>> Can you explain your equation more precisely?
>
> 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).
>
>> What exactly do you mean by: x = (1/a) [sqrt(1 + (at)²) - 1] ?
>
> That is the equation describing the position x of the rocket as a function of
> the coordinate time t for constant proper acceleration "a" beginning at rest at
> the origin. It describes the hyperbolic path. Note that I'm using units so c=1.
> The proper time along that path is given by integrating sqrt(dt^2 - dx^2) along
> that path, as explained above.

x = (c²/a) [sqrt(1 + (a.To/c)²) - 1]

OK.

Well if you put a=10m/s² and To=12.91years then x=12ly.

It's correct.

Our problem is on tau (Tr, proper time) who is very different to you and
to me.

We don't use the same concept, the same space-time geometry.

R.H.

On Wednesday, August 10, 2022 at 11:54:00 AM UTC-7, Richard Hachel wrote:
> > 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).
>
> OK. Well if you put a=10m/s² and t=12.91years then x=12ly.
> Our problem is on tau (proper time) who is very different to you and
> to me.

The proper time, tau, along a given path is defined as the elapsed time on a standard clock following that path. There is no ambiguity about this. The elapsed proper time along the path of constant proper acceleration "a" from rest at the origin to the event x,t is tau = (1/a)invsinh(at). This is what the clock reads, as has been confirmed experimentally. Your claims to the contrary are logically self-contradictory and empirically falsified..

Le 10/08/2022 à 03:17, Stan Fultoni a écrit :

> 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.

For accelerated baselines on departure:

x=(1/2).a.Tr
To=Tr.sqrt(1+(1/4)a².Tr²/c²)

If a.Tr=Vrf (Real final speed), we obtain
x=(1/2).a.Tr
To=Tr.sqrt(1+(1/4)Vrf²/c²)

For constant velocity repositories:
x=Vr.Tr
To=Tr.sqrt(1+Vr²/c²)

Now what happens if the rocket (or the particle) is in accelerated motion
and has, in addition, an initial real velocity Vri in the reference frame?

We then have:
To²=Tr²+Et²

To²=Tr²+(x²/c²)

With x=(1/2)aTr²+Vri.Tr

To²=Tr²+[Tr.((1/2)a.Tr+Vri)]²/c²

To²=Tr²(1+(1/2)a.Tr+Vri)²/c²

To=Tr.sqrt[1+((1/2)a.Tr+Vri)²/c²]

Indeed if a=0 (constant movement) then To=Tr.sqrt(1+Vr²/c²)

and if Vri=0 (start stopped):
To=Tr.sqrt(1+(1/4)Vrf²/c²)

R.H.

On Thursday, August 11, 2022 at 5:54:23 AM UTC-7, Richard Hachel wrote:
> Le 10/08/2022 à 03:17, Stan Fultoni a écrit :
> > 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).

Le 11/08/2022 à 16:28, Stan Fultoni a écrit :
> On Thursday, August 11, 2022 at 5:54:23 AM UTC-7, Richard Hachel wrote:
>> Le 10/08/2022 à 03:17, Stan Fultoni a écrit :
>> > 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:
To=Tr.sqrt(1+0.25Vr²/c²) Hachel notation <=> tau =
t/sqrt(1+0.25Vr²/c²) Fultoni notation

and not tau = (1/a)invsinh(at).

I repeat that : "il y a quelque chose qui cloche".

R.H.

R.H.

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.

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?

R.H.

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

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?

He says genius things to you.

As it is holy written: "What a man! What a dick! What a lover!"

Otherwise, listen to me.

There is an error not in the fact that one cannot try to integrate, but it
is not necessary.

But if you integrate, you have to integrate with the right equation.

And it's not the one given here by Stan Fultoni.

The right equation is more complex, because it takes into account Vri,
which is the initial real velocity in the small time fragment.

This is what must be integrated if one wants to integrate.

We then have:
To²=Tr²+Et²
To²=Tr²+(x²/c²)
With x=(1/2)aTr²+Vri.Tr
To²=Tr²+[Tr.((1/2)a.Tr+Vri)]²/c²
To²=Tr²(1+(1/2)a.Tr+Vri)²/c²
To=Tr.sqrt[1+((1/2)a.Tr+Vri)²/c²]

I'd be surprised if Stan practiced like that.

Hence the fact that even if its integration is correct, the fact that it
is based on a misunderstood concept by taking an incorrect equation, the
result is false.

R.H.