Le 19/10/2022 à 04:07, Stan Fultoni a écrit :
> On Tuesday, October 18, 2022 at 6:27:56 PM UTC-7, Richard Hachel wrote:

> Now, in one particular case, your formula actually gives the (asymptotically)
> correct answer for sufficiently small segments. This is when we choose the
> coordinate system in which the segment is initially at rest. For that system, we
> have v1=0 and v2=0.025744, and your formula gives 0.024426 years. If we do the
> same for all the other segments, and add them up, we get 3.14 years for the total
> trip.

No.

There's something you obviously don't understand somewhere.

>
> This proves all your beliefs are false, right?

But no!

R.H.

Le 19/10/2022 à 04:07, Stan Fultoni a écrit :
> Now, in one particular case, your formula actually gives the (asymptotically)
> correct answer for sufficiently small segments. This is when we choose the
> coordinate system in which the segment is initially at rest. For that system, we
> have v1=0 and v2=0.025744, and your formula gives 0.024426 years. If we do the
> same for all the other segments, and add them up, we get 3.14 years for the total
> trip.
>
> This proves all your beliefs are false, right?

? ? ?

Vo1=0c
Vo2=0.025744c

Vr1=0c
Vr2=0.025752535c

Tr1=0
Tr2=Vr2/a=0.054682years

Je ne comprends absolument rien de vos concepts géométriques.

C'est, je pense, l'espace de Minkowski qui vous fait complètement
dériver vers des doctrines abstraites et fausses.

Mikowski et son espace-temps, c'est du pur pipeau.

C'est pas comme ça que ça marche.

R.H.

On Wednesday, October 19, 2022 at 7:35:42 AM UTC-4, Richard Hachel wrote:

> The traveler problem of Tau Ceti is a very simple relativistic problem
> which brings into play the relations between the proper time of an
> accelerated rocket (Tr=tau)

OK

> and the observable time (To=t) or terrestrial time.

Ambiguous. In your scenario, you have the value of the earth clock.
You also have the values of the clocks of the earths inertial frame. This is usually labeled by "t".
This iframe also has a clock at Tau Ceti (TC)
So, you have the watch of the traveler, you have the earth clock and the clock at TC.
The earth clock and TC clock are in the same iframe, the "earth frame".

So, what did you mean by "and the observable time (To=t) or terrestrial time. " ?

> The rocket is launched on a uniformly accelerated rectilinear trajectory
> from the earth.

OK.

> The value of constant acceleration is defined as:
> a=10m/s²=0.512ly/y²

?? Did you mean proper acceleration?
If so, what do you mean by proper acceleration, what is its definition?
And how did you get the value of 0.512 ly/y² ?

Clear up these questions before we proceed,
Else we will just remain confused on what we mean.
If you want a coherent dialogue, we must be clear on the meanings of the words we use.

Le 19/10/2022 à 14:30, rotchm a écrit :

> So, what did you mean by "and the observable time (To=t) or terrestrial time. "
> ?

Le temps du référentiel terre-Tau Ceti

R.H.

Le 19/10/2022 à 14:30, rotchm a écrit :

>> The value of constant acceleration is defined as:
>> a=10m/s²=0.512ly/y²
>
> ? ? Did you mean proper acceleration?
> If so, what do you mean by proper acceleration, what is its definition?
> And how did you get the value of 0.512 ly/y² ?
>

Ouch!

J'ai tapé trop vite et inversé les chiffres...

a=10m/s²=1.052y/y²

Valeur acceptée par Stan Fultoni, ici.

R.H.

On Wednesday, October 19, 2022 at 4:35:42 AM UTC-7, Richard Hachel wrote:
> The value of constant acceleration is defined as:
> a=10m/s²=0.512ly/y²

No, a = 10 m/sec^2 = 1.052 Y^-1.
> We know that the rocket will arrive at time To=12.914 years.

Right, you borrowed this from special relativity, and the equation for the hyperbolic trajectory t = sqrt(x^2 + 2x/a) begining from rest at the origin. You do not know how to derive this relation, you just copied it.

> Stan calculated that the entry speed in this segment would be
> v1=0.970706c and the output speed v2=0.972156c
> I have v1=0.958853c and v2=0.959487c

You are mistaken. You already agreed that the trajectory (starting from rest at the origin) is t = sqrt(x^2 + 2x/a), so the velocity at any x is dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1). At x=3 and x=3.1 this gives v1=0.970620 and v2=0.972074. (My numbers above were based on a=1.054, but should be 1.052, so slightly different.) Agreed?

On Wednesday, October 19, 2022 at 8:40:55 AM UTC-4, Richard Hachel wrote:
> Le 19/10/2022 à 14:30, rotchm a écrit :
>
>
> > So, what did you mean by "and the observable time (To=t) or terrestrial time. "
> > ?
> Le temps du référentiel terre-Tau Ceti

In case you haven't noticed, the conversation was in English.
So, be respectful and remain clear in your replies: reply in English.

On Wednesday, October 19, 2022 at 8:30:55 AM UTC-4, rotchm wrote:
> On Wednesday, October 19, 2022 at 7:35:42 AM UTC-4, Richard Hachel wrote:

> > The value of constant acceleration is defined as:
> > a=10m/s²=0.512ly/y²
> ?? Did you mean proper acceleration?

No answer? Don't forget, we are trying to make sure that we understand each other.
She refused to explain what your words mean, then you are not interested in discussing.

> If so, what do you mean by proper acceleration, what is its definition?

No answer?
Are you saying that you do not know what proper acceleration means?
We must make sure you know what it means before we can proceed if we want to use it.

Le 19/10/2022 à 15:18, Stan Fultoni a écrit :
> On Wednesday, October 19, 2022 at 4:35:42 AM UTC-7, Richard Hachel wrote:
>> The value of constant acceleration is defined as:
>> a=10m/s²=0.512ly/y²
>
> No, a = 10 m/sec^2 = 1.052 Y^-1.

C'est ce que je dis (j'ai inversé les chiffres en tapant trop vite).

>> We know that the rocket will arrive at time To=12.914 years.
>
> Right, you borrowed this from special relativity, and the equation for the
> hyperbolic trajectory t = sqrt(x^2 + 2x/a) begining from rest at the origin. You
> do not know how to derive this relation, you just copied it.

No.

Si je voulais recopier, j'aurais été le roi du monde.

J'ai toujours choisi l'inverse, de m'en prendre plein la gueule, en
justement, me mettant hors jeu et ne recopiant jamais.

Les grands hommes passent toujours par deux phases:

1. Mon ennemi dit des conneries.
2. Mon ennemi dit des vérités, mais il copie
>
>> Stan calculated that the entry speed in this segment would be
>> v1=0.970706c and the output speed v2=0.972156c
>> I have v1=0.958853c and v2=0.959487c
>
> You are mistaken. You already agreed that the trajectory (starting from rest at
> the origin) is t = sqrt(x^2 + 2x/a), so the velocity at any x is dx/dt =
> a*sqrt(x^2 + 2x/a)/(ax+1). At x=3 and x=3.1 this gives v1=0.970620 and
> v2=0.972074. (My numbers above were based on a=1.054, but should be 1.052, so
> slightly different.) Agreed?

X1 = 3 years
x2 =3.1 years

Tr1=sqrt(2*x1/a)=sqrt((2*3/1.052)=2.388 years
Tr2=sqrt(2*x2/a)=sqrt(2*3.1/1.052)=2.428 years

Vr1=a.Tr1=2.512c
Vr2=a.Tr2=2.554c

Vo1=Vr1/sqrt(1+Vr1²/c²)=0.929c
Vo2=Vr2/sqrt(1+Vr2²/c²)=0.931c

It is not your mathematical calculations that are in question, but your
methodology and your religious belief in Minkowski's space-time.

It is obvious that the universe is based on seven simple dimensions. Three
of space (hypersimple) and four of time (with three temporal dimensions
conjoined to three spatial dimensions).

It is appallingly simple.

It's your geometry that's wrong, not your calculations.

R.H.

Le 19/10/2022 à 15:38, rotchm a écrit :
> On Wednesday, October 19, 2022 at 8:40:55 AM UTC-4, Richard Hachel wrote:
>> Le 19/10/2022 à 14:30, rotchm a écrit :
>>
>>
>> > So, what did you mean by "and the observable time (To=t) or terrestrial time.
>> "
>> > ?
>> Le temps du référentiel terre-Tau Ceti
>
> In case you haven't noticed, the conversation was in English.
> So, be respectful and remain clear in your replies: reply in English.

You're not going to make me believe that you don't know how to translate
this sentence?

Vous exagérez vos carences.

R.H.

On Wednesday, 19 October 2022 at 14:30:55 UTC+2, rotchm wrote:
> On Wednesday, October 19, 2022 at 7:35:42 AM UTC-4, Richard Hachel wrote:
>
> > The traveler problem of Tau Ceti is a very simple relativistic problem
> > which brings into play the relations between the proper time of an
> > accelerated rocket (Tr=tau)
> OK
> > and the observable time (To=t) or terrestrial time.
> Ambiguous. In your scenario, you have the value of the earth clock.
> You also have the values of the clocks of the earths inertial frame. This is usually labeled by "t".
> This iframe also has a clock at Tau Ceti (TC)
> So, you have the watch of the traveler, you have the earth clock and the clock at TC.
> The earth clock and TC clock are in the same iframe, the "earth frame".
>
> So, what did you mean by "and the observable time (To=t) or terrestrial time. " ?
> > The rocket is launched on a uniformly accelerated rectilinear trajectory
> > from the earth.
> OK.
> > The value of constant acceleration is defined as:
> > a=10m/s²=0.512ly/y²
> ?? Did you mean proper acceleration?
> If so, what do you mean by proper acceleration, what is its definition?
> And how did you get the value of 0.512 ly/y² ?
>
> Clear up these questions before we proceed,
> Else we will just remain confused on what we mean.
> If you want a coherent dialogue, we must be clear on the meanings of the words we use.

And in the meantime in the real world, forbidden
by your bunch of idiots GPS and TAI will keep
measuring t'=t in forbidden by your bunch of
idiots old seconds.

On Wednesday, October 19, 2022 at 10:16:48 AM UTC-4, Richard Hachel wrote:
> Le 19/10/2022 à 15:38, rotchm a écrit :

> > In case you haven't noticed, the conversation was in English.
> > So, be respectful and remain clear in your replies: reply in English.
> You're not going to make me believe that you don't know how to translate
> this sentence?

Why translate it into English when you can directly type it in English like you do for most of your posts?
Why do you keep changing from one language to another?
That is not the way to proceed to have a sensible conversation.
Do you really want to discuss your scenario or do you just want a troll?

On Wednesday, October 19, 2022 at 9:44:31 AM UTC-4, rotchm wrote:
> On Wednesday, October 19, 2022 at 8:30:55 AM UTC-4, rotchm wrote:
> > On Wednesday, October 19, 2022 at 7:35:42 AM UTC-4, Richard Hachel wrote:
>
> > > The value of constant acceleration is defined as:
> > > a=10m/s²=0.512ly/y²
> > ?? Did you mean proper acceleration?
> No answer? Don't forget, we are trying to make sure that we understand each other.

Still no answer?

> If you refuse to explain what your words mean, then you are not interested in discussing.
> If so, what do you mean by proper acceleration, what is its definition?
> No answer?

Still no answer?
Are you saying that you do not know what proper acceleration means?

Le 19/10/2022 à 17:04, rotchm a écrit :

> Are you saying that you do not know what proper acceleration means?

Your question is interesting.

In return, before answering, I would like to know, you, the definition
that you give of a proper acceleration of 10m/s² in Tau Ceti?

In the text of a commentator, page 48, I read: "proper acceleration is the
acceleration measured by the instruments of the rocket". Nothing more.

It seems very vague to me.

What do you think?

R.H.

On Wednesday, October 19, 2022 at 11:15:54 AM UTC-4, Richard Hachel wrote:
> Le 19/10/2022 à 17:04, rotchm a écrit :
>
> > Are you saying that you do not know what proper acceleration means?
> Your question is interesting.
>
> In return, before answering, I would like to know, you, the definition
> that you give of a proper acceleration of 10m/s² in Tau Ceti?

I use proper acceleration as it is unused in physics.

> In the text of a commentator, page 48, I read: "proper acceleration is the
> acceleration measured by the instruments of the rocket". Nothing more.

Irrelevant what another author says.
I am asking *you*, what definition you are using?
What formula are you using to define it?

Like speed (wrt earth frame) is v = dx/dt.
And coordinate acceleration (wrt earth frame) is dv/dt = d²x/dt².

> It seems very vague to me.
> What do you think?

Yes, the wording is vague. that's why I'm asking you what formula you are using to define proper acceleration?

On 2022-10-19 14:21:15 +0000, Maciej Wozniak said:

> On Wednesday, 19 October 2022 at 14:30:55 UTC+2, rotchm wrote:
>> On Wednesday, October 19, 2022 at 7:35:42 AM UTC-4, Richard Hachel
>> wrote:>> > The traveler problem of Tau Ceti is a very simple
>> relativistic problem> > which brings into play the relations between
>> the proper time of an> > accelerated rocket (Tr=tau)
>> OK
>>> and the observable time (To=t) or terrestrial time.
>> Ambiguous. In your scenario, you have the value of the earth clock.>
>> You also have the values of the clocks of the earths inertial frame.
>> This is usually labeled by "t".> This iframe also has a clock at Tau
>> Ceti (TC)> So, you have the watch of the traveler, you have the earth
>> clock and the clock at TC.> The earth clock and TC clock are in the
>> same iframe, the "earth frame".>> So, what did you mean by "and the
>> observable time (To=t) or terrestrial time. " ?
>>> The rocket is launched on a uniformly accelerated rectilinear
>>> trajectory> > from the earth.
>> OK.
>>> The value of constant acceleration is defined as:> > a=10m/s²=0.512ly/y²
>> ?? Did you mean proper acceleration?> If so, what do you mean by proper
>> acceleration, what is its definition?> And how did you get the value of
>> 0.512 ly/y² ?>> Clear up these questions before we proceed,> Else we
>> will just remain confused on what we mean.> If you want a coherent
>> dialogue, we must be clear on the meanings of the words we use.
>
> And in the meantime in the real world, forbidden
> by your bunch of idiots GPS and TAI will keep
> measuring t'=t in forbidden by your bunch of
> idiots old seconds.

520. I won't bother to record every instance from now -- just every 10th.

--
Athel -- French and British, living mainly in England until 1987.

On Wednesday, October 19, 2022 at 7:13:45 AM UTC-7, Richard Hachel wrote:
> > You borrowed this from special relativity, and the equation for the
> > hyperbolic trajectory t = sqrt(x^2 + 2x/a) begining from rest at the origin. You
> > do not know how to derive this relation, you just copied it.
>
> No.

Yes, that equation describes the hyperbolic trajectory of a path of constant proper acceleration in special relativity. It is not true in Galilean physics. In Galilean physics the trajectory would be t = sqrt(2x/a), which is a parabola, but the equation you are using is t = sqrt(2x/a + x^2), which is a hyperbola.

> > You are mistaken. You already agreed that the trajectory (starting from rest at
> > the origin) is t = sqrt(x^2 + 2x/a), so the velocity at any x is dx/dt =
> > a*sqrt(x^2 + 2x/a)/(ax+1). At x=3 and x=3.1 this gives v1=0.970620 and
> > v2=0.972074. Agreed?
>
> [No, I claim that v=0.929 and v2 = 0.931.]

These numbers are very different than what you claimed in your previous messge, but they are still wrong. Remember, in terms of the x,t coordinates, the velocity is defined as dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1), which gives at x=3 and x=3.1 the velocities v1=0.970620 and v2=0.972074. This is what the phrase "velocity in terms of x,t" means. This is not open to debate, it is the definition. These are the velocities you would measure with standard rulers and clocks at rest and synchronized in the x,t frame. Agreed?

Le 19/10/2022 à 20:24, Stan Fultoni a écrit :
> On Wednesday, October 19, 2022 at 7:13:45 AM UTC-7, Richard Hachel wrote:

>>> t = sqrt(x^2 + 2x/a)
>>> you just copied it.
>>
>> No.
>
> Yes

No.

R.H.

> In the text of a commentator, page 48, I read: "proper acceleration is the
> acceleration measured by the instruments of the rocket". Nothing more.

Irrelevant what another author says.
I am asking *you*, what definition you are using?
What formula are you using to define it?

No asnwers?

Like speed (wrt earth frame) is v = dx/dt.
And coordinate acceleration (wrt earth frame) is dv/dt = d²x/dt².

> It seems very vague to me.
> What do you think?

Yes, the wording is vague. that's why I'm asking you what formula you are using to define proper acceleration?

No answer? You don't know its "formula" definition?
Don't be afraid to admit it if it is the case.
It's better for you to ask ask for help rather than staying ignorant, no?

Le 19/10/2022 à 20:24, Stan Fultoni a écrit :
> On Wednesday, October 19, 2022 at 7:13:45 AM UTC-7, Richard Hachel wrote:
>> > You borrowed this from special relativity, and the equation for the
>> > hyperbolic trajectory t = sqrt(x^2 + 2x/a) begining from rest at the origin.
>> You
>> > do not know how to derive this relation, you just copied it.
>>
>> No.
>
> Yes, that equation describes the hyperbolic trajectory of a path of constant
> proper acceleration in special relativity. It is not true in Galilean physics.
> In Galilean physics the trajectory would be t = sqrt(2x/a), which is a parabola,
> but the equation you are using is t = sqrt(2x/a + x^2), which is a hyperbola.
>
>> > You are mistaken. You already agreed that the trajectory (starting from rest
>> at
>> > the origin) is t = sqrt(x^2 + 2x/a), so the velocity at any x is dx/dt =
>> > a*sqrt(x^2 + 2x/a)/(ax+1). At x=3 and x=3.1 this gives v1=0.970620 and
>> > v2=0.972074. Agreed?
>>
>> [No, I claim that v=0.929 and v2 = 0.931.]
>
> These numbers are very different than what you claimed in your previous messge,
> but they are still wrong. Remember, in terms of the x,t coordinates, the velocity
> is defined as dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1), which gives at x=3 and x=3.1 the
> velocities v1=0.970620 and v2=0.972074. This is what the phrase "velocity in
> terms of x,t" means. This is not open to debate, it is the definition. These are
> the velocities you would measure with standard rulers and clocks at rest and
> synchronized in the x,t frame. Agreed?

NOOOOOOO !!

No.

The correct direct equation is:

Vo/c=[c²/2ax + 1]^-0.5

For x=3 then Vo=0.9291c

For x=3.1 then Vo=0.9311c

Please, Stan, STOP!

R.H.

Le 19/10/2022 à 22:55, rotchm a écrit :

> It's better for you to ask ask for help rather than staying ignorant, no?

S'il te plait, un peu moins d'arrogance et de forfanterie dans tes mots.

Pour ce qui est de l'accélération propre c'est une tautologie que de
dire l'accélération mesurée dans le référentiel propre.

Si l'on prend x=(1/2)aTr², on peut alors poser a=2x/Tr².

On peut aussi poser To=(Vr/2a)sqrt(4+Vr²/c²) soit
a=(Vr/to).sqrt(1+(1/4)Vr²/c²)

Ou aussi Vr=aTr soit a=Vr/Tr

L'accélération propre est égale à la vitesse réelle Vr par unité de
temps réel Tr.

N.B. Je ne suis pas éternel, et ce que je dis ici, ce sont souvent des
pépites théoriques.
Merci pour ceux qui le comprendraient de bien vouloir les noter ; à
défaut de les comprendre, malgré leur fantastique évidence et
simplicité.

R.H.

R.H.

On Wednesday, October 19, 2022 at 5:14:40 PM UTC-4, Richard Hachel wrote:
> Le 19/10/2022 à 22:55, rotchm a écrit :
>
> > It's better for you to ask ask for help rather than staying ignorant, no?
> S'il te plait, un peu moins d'arrogance et de forfanterie dans tes mots.

You forgot to answer the folowing:

Why translate it into English when you can directly type it in English like you do for most of your posts?
Why do you keep changing from one language to another?

No answers?

> Si l'on prend x=(1/2)aTr², on peut alors poser a=2x/Tr².

That is not the definition for the proper acceleration "a" as used in physics.
So, why are you calling "a" the proper acceleration?

And, if you declare that "a" is the proper acceleration, then the formula for x as a function of
tau (your Tr here?) is not x=(1/2)aTr².

Don't forget, our goal here is to make sure that we are using the same language so that we can understand each other.
We are currently stuck on the concept of proper acceleration.
Since you invoked it, you must make sure that you know what it means in physics.
Once this notion of proper acceleration is understood by us both, then we can proceed.

> L'accélération propre est égale à la vitesse réelle Vr par unité de
> temps réel Tr.

This too is not the definition of proper acceleration as used in physics.
If you are referring to something else, do not call it proper acceleration, else you will just confuse matters.
Try again.

Le 19/10/2022 à 23:40, rotchm a écrit :

> This too is not the definition of proper acceleration as used in physics.

No, but it's mine.

Physicists do what they want.

R.H.

Le 19/10/2022 à 23:40, rotchm a écrit :

>> Si l'on prend x=(1/2)aTr², on peut alors poser a=2x/Tr².

> And, if you declare that "a" is the proper acceleration, then the formula for x
> as a function of
> tau (your Tr here?) is not x=(1/2)aTr².

No.

YOU, you say that tau (or Tr) using here is not x=1/2aTr².

And me, I said it is.

In totality.

R.H.

On Wednesday, October 19, 2022 at 6:16:32 PM UTC-4, Richard Hachel wrote:
> Le 19/10/2022 à 23:40, rotchm a écrit :
>
> > This too is not the definition of proper acceleration as used in physics.
> No, but it's mine.

No it is not.

You are using the expression 'proper acceleration' in the context of physics, in this NG.
In this NG and in physics, proper acceleration has a specific meaning. When you use that expression you are using that agreed-upon meaning. You can't just randomly change the definitions and especially without saying it. That is not how language works, that is not how efficient communication works. Unless your goal here is not to communicate your thoughts but rather to troll?

So, for your concept of "a", use a different expression than 'proper acceleration'.
Then, Define what you mean by your 'a'; what is its defining equation.

Again our goal here is to make sure we understand each other before we proceed; we need to make sure that we know the meanings of the words we use.