<http://news2.nemoweb.net/jntp?H3tLDulCXBCiM8mncPciqquvlbA@jntp/Data.Media:1>

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
<http://news2.nemoweb.net/?DataID=H3tLDulCXBCiM8mncPciqquvlbA@jntp>

Le 28/08/2022 à 12:28, Richard Hachel a écrit :
>
> <http://news2.nemoweb.net/jntp?H3tLDulCXBCiM8mncPciqquvlbA@jntp/Data.Media:1>
>
>
> R.H.

To=Tr.sqrt(1+Vr²/c²) is correct.
To=Tr.sqrt(1+(1/4)Vr²/c²) is correct.
It is an extraordinary paradox so great that one is obliged to say: "there
is no paradox".
Concept of psychological denial.
Well yes, there is a paradox, since we don't know how to explain it, at
least correctly, as in this Langevin twin story where the use of apparent
speeds becomes absurd.
Of course yes, it's absurd, this story of relativistic acceleration and
this problem of the "traveler from Tau Ceti".
But you need the key.
Once the key is given, all is logic and light.

R.H.

On Sunday, August 28, 2022 at 3:37:52 AM UTC-7, Richard Hachel wrote:
> To=Tr.sqrt(1+Vr²) is correct.

A correct relation is t = TAU sqrt(1 + [x/TAU]^2). Here you are defining "Vr" as x/TAU, and you are defining "Tr" as the elaped proper time along the inertial path from the origin to x,t, and you are using T0 to denote t. This relation does not involve the proper time tau along the accelerated path.

> To=Tr.sqrt(1+(1/4)Vr²) is correct.

A correct relation is t = TAU sqrt(1 + (1/4)[a*TAU]^2). Here you are changing the definition of "Vr" to a*TAU, but this is different from your previous definition of x/TAU, because if you set them equal you would have x = a*TAU^2, but in fact the correct relation is x = (1/2)a*TAU^2. So, your definitions imply 1/2 = 1, and hence 2=1 and hence 1=0.

> C'est un paradoxe extraordinaire si grand qu'on est obligé de dire : "il
> n'y a pas de paradoxe".

There is no paradox or inconsistency in the correct relations. There is an inconsistency in your relations, because you are using two different definitions for Vr that together imply 1=0. Also, note again that none of your relations involve the elapsed proper time tau along the accelerated path.. That requires integration, as explained previously.

On Sunday, 28 August 2022 at 17:59:49 UTC+2, Stan Fultoni wrote:
> On Sunday, August 28, 2022 at 3:37:52 AM UTC-7, Richard Hachel wrote:
> > To=Tr.sqrt(1+Vr²) is correct.
>
> A correct relation is t = TAU sqrt(1 + [x/TAU]^2). Here you are defining "Vr" as x/TAU, and you are defining "Tr" as the elaped proper time along the inertial path from the origin to x,t, and you are using T0 to denote t. This relation does not involve the proper time tau along the accelerated path.

In the meantime in the real world, of course,
forbidden by your bunch of idiots GPS and TAI
keep measuring t'=t, just like all serious clocks
always did.

Le 28/08/2022 à 17:59, Stan Fultoni a écrit :
> On Sunday, August 28, 2022 at 3:37:52 AM UTC-7, Richard Hachel wrote:
>> To=Tr.sqrt(1+Vr²) is correct.
>
> A correct relation is t = TAU sqrt(1 + [x/TAU]^2). Here you are defining "Vr"
> as x/TAU, and you are defining "Tr" as the elaped proper time along the inertial
> path from the origin to x,t, and you are using T0 to denote t. This relation
> does not involve the proper time tau along the accelerated path.

? ? ?

>> To=Tr.sqrt(1+(1/4)Vr²) is correct.
>
> A correct relation is t = TAU sqrt(1 + (1/4)[a*TAU]^2). Here you are changing
> the definition of "Vr" to a*TAU, but this is different from your previous
> definition of x/TAU, because if you set them equal you would have x = a*TAU^2, but
> in fact the correct relation is x = (1/2)a*TAU^2. So, your definitions imply 1/2
> = 1, and hence 2=1 and hence 1=0.

? ? ?

>
>> C'est un paradoxe extraordinaire si grand qu'on est obligé de dire : "il
>> n'y a pas de paradoxe".
>
> There is no paradox or inconsistency in the correct relations. There is an
> inconsistency in your relations, because you are using two different definitions
> for Vr that together imply 1=0. Also, note again that none of your relations
> involve the elapsed proper time tau along the accelerated path. That requires
> integration, as explained previously.

? ? ?

Well.

There are initially two equations.

<http://news2.nemoweb.net/jntp?EVJGLtt1YvAFe1fN8fqNRCS5SJ0@jntp/Data.Media:1>

Sir, I beg you to understand them, and to understand why they can be used
in some cases and not in others.

I beg you to understand my position before refuting it without
understanding it.

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<http://news2.nemoweb.net/?DataID=EVJGLtt1YvAFe1fN8fqNRCS5SJ0@jntp>

On Sunday, August 28, 2022 at 9:37:50 AM UTC-7, Richard Hachel wrote:
> >> To=Tr.sqrt(1+Vr²) is correct.
> >
> > A correct relation is t = TAU sqrt(1 + [x/TAU]^2). Here you are defining "Vr"
> > as x/TAU, and you are defining "Tr" as the elaped proper time along the inertial
> > path from the origin to x,t, and you are using T0 to denote t. This relation
> > does not involve the proper time tau along the accelerated path.
>
> ? ? ?

Which part of that do you not understand?

> >> To=Tr.sqrt(1+(1/4)Vr²) is correct.
> >
> > A correct relation is t = TAU sqrt(1 + (1/4)[a*TAU]^2). Here you are changing
> > the definition of "Vr" to a*TAU, but this is different from your previous
> > definition of x/TAU, because if you set them equal you would have x = a*TAU^2, but
> > in fact the correct relation is x = (1/2)a*TAU^2. So, your definitions imply 1/2
> > = 1, and hence 2=1 and hence 1=0.
>
> ? ? ?

Which part do you not understand?

> >> C'est un paradoxe extraordinaire si grand qu'on est obligé de dire : "il
> >> n'y a pas de paradoxe".
> >
> > There is no paradox or inconsistency in the correct relations. There is an
> > inconsistency in your relations, because you are using two different definitions
> > for Vr that together imply 1=0. Also, note again that none of your relations
> > involve the elapsed proper time tau along the accelerated path. That requires
> > integration, as explained previously.
>
> ? ? ?

Which part do you not understand?

Le 28/08/2022 à 19:00, Stan Fultoni a écrit :
> On Sunday, August 28, 2022 at 9:37:50 AM UTC-7, Richard Hachel wrote:
>> >> To=Tr.sqrt(1+Vr²) is correct.
>> >
>> > A correct relation is t = TAU sqrt(1 + [x/TAU]^2). Here you are defining "Vr"
>> > as x/TAU, and you are defining "Tr" as the elaped proper time along the
>> inertial
>> > path from the origin to x,t, and you are using T0 to denote t. This relation
>> > does not involve the proper time tau along the accelerated path.
>>
>> ? ? ?
>
> Which part of that do you not understand?
>
>> >> To=Tr.sqrt(1+(1/4)Vr²) is correct.
>> >
>> > A correct relation is t = TAU sqrt(1 + (1/4)[a*TAU]^2). Here you are changing
>> > the definition of "Vr" to a*TAU, but this is different from your previous
>> > definition of x/TAU, because if you set them equal you would have x = a*TAU^2,
>> but
>> > in fact the correct relation is x = (1/2)a*TAU^2. So, your definitions imply
>> 1/2
>> > = 1, and hence 2=1 and hence 1=0.
>>
>> ? ? ?
>
> Which part do you not understand?
>
>> >> C'est un paradoxe extraordinaire si grand qu'on est obligé de dire : "il
>> >> n'y a pas de paradoxe".
>> >
>> > There is no paradox or inconsistency in the correct relations. There is an
>> > inconsistency in your relations, because you are using two different
>> definitions
>> > for Vr that together imply 1=0. Also, note again that none of your relations
>> > involve the elapsed proper time tau along the accelerated path. That requires
>> > integration, as explained previously.
>>
>> ? ? ?
>
> Which part do you not understand?

The first equation is that of a time relation between the proper time and
the observable time during a Galilean displacement.

You have to say things clearly and give clear and correct equations of
things.

The second is the equation linking the proper times Tr and the times
observable in a uniformly accelerated medium.

It's very simple and very understandable.

Now, you have to be careful when you want to go further, and make the sign
of the cross three times, and repeat three times: "The theory of special
relativity is basic mathematics, but it's stuffed little traps.

So there, be careful, a terrible trap will come back to us all in the
face.

We are going to say that if we could pass two small objects in uniform
Galilean motion for one, uniformly accelerated for the other, for a very
short time at the same speed, the proper times will be equal, the
observable times will be equal, the instantaneous velocities will be
equal.

And that's true!!!

I demonstrated it again this week here during a post with the successive
equations.

Of course it's true but the punch in the face, as we say in French,
happens very quickly.

The physicist then thinks, in his blind madness: therefore one can
practice integration without any problem.

The problem is that the physicist confuses two things!!!

1. When the two mobiles are in the same place in my repository, THEY ARE
NOT AT THE SAME SPEED.

2. And when they are at the same Vr speed, they are not in the same
place!!!

When they are for example at Vr=(1/2)Vrf

Vrf=Final real speed.

In Galilean displacement, the object is in the middle of the path.

In the accelerated displacement; the object is only a quarter of the way.

So same speed Vr but not in the same place.

Do you understand these things?

From where the error which I had already announced for the proper times
badly calculated and given too low.

You have to apply Hachel's equations, and dare to do so (the problem being
human and of the style: "we're still not going to apply the teachings of
this little French shit").

<http://news2.nemoweb.net/jntp?4Np5p2oCAHZlZhpw6Fh5vEyV2dA@jntp/Data.Media:1>

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=4Np5p2oCAHZlZhpw6Fh5vEyV2dA@jntp>

On Sunday, August 28, 2022 at 10:02:36 AM UTC-7, Richard Hachel wrote:
> >> >> To=Tr.sqrt(1+Vr²) is correct.
> >> >
> >> > A correct relation is t = TAU sqrt(1 + [x/TAU]^2). Here you are defining "Vr"
> >> > as x/TAU, and you are defining "Tr" as the elaped proper time along the
> >> inertial path from the origin to x,t, and you are using T0 to denote t.. This
> >> relation does not involve the proper time tau along the accelerated path.
>
> >> >> To=Tr.sqrt(1+(1/4)Vr²) is correct.
> >> >
> >> > A correct relation is t = TAU sqrt(1 + (1/4)[a*TAU]^2). Here you are changing
> >> > the definition of "Vr" to a*TAU, but this is different from your previous
> >> > definition of x/TAU, because if you set them equal you would have x = a*TAU^2,
> >> but in fact the correct relation is x = (1/2)a*TAU^2. So, your definitions imply
> >> 1/2 = 1, and hence 2=1 and hence 1=0.
>
> The first equation is that of a time relation between the proper time and
> the observable time during a Galilean displacement. The second is the
> equation linking the proper times Tr and the times observable in a
> uniformly accelerated medium.

No, the first equation is true only if you define your symbols To = t and Tr = TAU and Vr = x/TAU, but your second equation is true only if you define To = t and Tr = TAU and Vr = a*TAU. But those two definitions of Vr imply x = a*TAU^2, which contradicts the relation x = (1/2)a*TAU^2. So, your two equations together imply 1=0, and therefore are gibberish. Understand?

Also, none of your equations involve tau, the elapsed proper time along the accelerated path, which is supposedly what you are trying to find. The value of tau has already been derived for you. Remember?

Le 28/08/2022 à 19:22, Stan Fultoni a écrit :
> On Sunday, August 28, 2022 at 10:02:36 AM UTC-7, Richard Hachel wrote:
>> >> >> To=Tr.sqrt(1+Vr²) is correct.
>> >> >
>> >> > A correct relation is t = TAU sqrt(1 + [x/TAU]^2). Here you are defining
>> "Vr"
>> >> > as x/TAU, and you are defining "Tr" as the elaped proper time along the
>> >> inertial path from the origin to x,t, and you are using T0 to denote t. This
>> >> relation does not involve the proper time tau along the accelerated path.
>>
>> >> >> To=Tr.sqrt(1+(1/4)Vr²) is correct.
>> >> >
>> >> > A correct relation is t = TAU sqrt(1 + (1/4)[a*TAU]^2). Here you are
>> changing
>> >> > the definition of "Vr" to a*TAU, but this is different from your previous
>> >> > definition of x/TAU, because if you set them equal you would have x =
>> a*TAU^2,
>> >> but in fact the correct relation is x = (1/2)a*TAU^2. So, your definitions
>> imply
>> >> 1/2 = 1, and hence 2=1 and hence 1=0.
>>
>> The first equation is that of a time relation between the proper time and
>> the observable time during a Galilean displacement. The second is the
>> equation linking the proper times Tr and the times observable in a
>> uniformly accelerated medium.
>
> No, the first equation is true only if you define your symbols To = t and Tr =
> TAU and Vr = x/TAU, but your second equation is true only if you define To = t and
> Tr = TAU and Vr = a*TAU. But those two definitions of Vr imply x = a*TAU^2, which
> contradicts the relation x = (1/2)a*TAU^2. So, your two equations together imply
> 1=0, and therefore are gibberish. Understand?
>
> Also, none of your equations involve tau, the elapsed proper time along the
> accelerated path, which is supposedly what you are trying to find. The value of
> tau has already been derived for you. Remember?

What I would like you to understand is that there are not two TAUs.

But you are so attached to yours (which justifies miscalculation) that you
believe in this abstract existence.

There is only one tau, one Tr, one proper time. And not two.

All this means the time marked by the departure of the rocket and the
arrival of the rocket.

You don't seem to understand that the rest is just abstraction and false
abstraction.

Your second tau, simply does not exist.

It is a useless geometric invention, and which gives a false result.

We must have tau=Tr=proper time=4.7765 years.

Period. Everything else is absolutely useless.

R.H.

On Sunday, August 28, 2022 at 10:52:39 AM UTC-7, Richard Hachel wrote:
>> The first equation is that of a time relation between the proper time and
>> the observable time during a Galilean displacement. The second is the
>> equation linking the proper times Tr and the times observable in a
>> uniformly accelerated medium.
>
> No, the first equation is true only if you define your symbols To = t and Tr =
> TAU and Vr = x/TAU, but your second equation is true only if you define To = t and
> Tr = TAU and Vr = a*TAU. But those two definitions of Vr imply x = a*TAU^2, which
> contradicts the relation x = (1/2)a*TAU^2. So, your two equations together imply
> 1=0, and therefore are gibberish. Understand?

You forgot to respond to this, which shows that your equations imply 1=0.

> > Also, none of your equations involve tau, the elapsed proper time along the
> > accelerated path, which is supposedly what you are trying to find. The value of
> > tau has already been derived for you. Remember?
>
> What I would like you to understand is that there are not two TAUs.

There are infinitely many proper times between the origin at x=0, t=0 and the final arrival event x=xf, t=tf at the star, one for each trajectory. For the inertial path, moving at constant velocity xf/tf the proper time is TAU = sqrt(tf^2 - xf^2), but for a trajectory that begins at rest and undergoes constant proper acceleration the elaped proper time is tau = (1/a)invcosh(a*xf + 1) = (1/a)invsinh(a*t). These are the proper times for two different trajectories.

Remember, you already 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 at t=12.915 years. There is no ambiguity about this.

> All this means the time marked by the departure of the rocket and the
> arrival of the rocket.

Well, if thye rocket goes at constant speed 0.929c it has the proper time TAU, but if it begins at rest and undergoes constant proper acceleration a = 1.052 year^-1 it has proper time tau, which is about 3.139 years in your example.

Le 28/08/2022 à 20:37, Stan Fultoni a écrit :

> There are infinitely many proper times between the origin at x=0, t=0 and the
> final arrival event x=xf, t=tf at the star, one for each trajectory. For the
> inertial path, moving at constant velocity xf/tf the proper time is TAU =
> sqrt(tf^2 - xf^2), but for a trajectory that begins at rest and undergoes constant
> proper acceleration the elaped proper time is tau = (1/a)invcosh(a*xf + 1) =
> (1/a)invsinh(a*t). These are the proper times for two different trajectories.

I read your answer, and thank you for it.

You tell me, dear Stan, that there is an infinity of trajectories.

I don't mind you saying it and studying it, but I don't think it's of
great interest here and for our readers.

For me, there is only one trajectory, one proper time.

The rocket goes from earth to Tau Ceti and travels x=12 ly.

It does so in an observable time in the terrestrial reference frame which
is To=12.915 years

It does so in an apparent terrestrial time of Tapp=24.915 years.

If I place myself in place of Tau Ceti, the travel time will be Tapp=0.915
years.

For now, it's very simple.

>
> Remember, you already agreed that t = x.sqrt(1+2/(ax)),

Absolutly.

To = (x/c).sqrt(1+2c²/ax))

To= 12,915 ans

> and solving this for x gives the relation x = (1/a)[sqrt(1 + (at)²) - 1],

Yes, it"s what I said.

Absolutly.

In Hachel notation : x=(c²/a).sqrt[(1+a².To²/c²)-1]

x=12ly.

R.H.

On Sunday, August 28, 2022 at 12:08:29 PM UTC-7, Richard Hachel wrote:
> The rocket goes from earth to Tau Ceti and travels x=12 ly.
> It does so in an observable time in the terrestrial reference frame which
> is To=12.915 years

Right, but there are infinitely many ways of doing that. For example, the rocket could go at constant speed 0.929c, in which case its elapsed proper time is 4.77 years. On the other hand, a rocket could begin at rest and undergo constant proper acceleration of 10 m/sec^2, in which case its elaped proper time is 3.139 years. For each trajectory, the elapsed proper time is just the integral of dtau = sqrt(dt^2 - dx^2) along that trajectory. That's why Langevin's twins have different ages when re-united.

Remember, you 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 at t=12.915 years.

Le 28/08/2022 à 20:37, Stan Fultoni a écrit :
> Remember, you already 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)

Yes, or (it is the same thing) : To²=Tr²+Et²

Absolutly.

But here, I feel that it will turn sour.

> dtau = dt/sqrt(1 + (at)^2),

? ? ?

NO!

To²=Tr²+Et²

Warning!

Et=x/c=(1/2)a.Tr²

To²=[(Tr²+(1/4)a²Tr²Tr²]=Tr²(1+(1/4)a.Tr)²

To=Tr.sqrt[1+(1/4)Vr²/c²]

Tr=To/sqrt[1+(1/4)Vr²/c²]

Note To=12.915 years Vr=5.0245c

Tr=4.776 ans.

But we could have the answer from the start, by setting x=(1.2)aTr²

Then Tr=sqrt(2x/a)

Tr=tau=proper time=4.776 years

> whose integral is tau = (1/a)invsinh(at) = 3.139 years at t=12.915 years. There
> is no ambiguity about this.

Houston, we have a problem with correcting by Hachel!

R.H.

On Sunday, August 28, 2022 at 12:31:48 PM UTC-7, Richard Hachel wrote:
> > Remember, you already 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)
>
> Yes.
>
> > dtau = dt/sqrt(1 + (at)^2),
>
> ? ? ?

This is simple grade school algebra. You agreed that
(1) .. .. .. dx = [at/sqrt(1+(at)^2]]dt
and you agreed that
(2) .. .. .. dtau = sqrt(dt^2 - dx^2)

Substitute the expression (1) for dx into expression (2), and simplify. What do you get? [Hint: You get dtau = dt/sqrt(1 + (at)^2).]