Le 23/11/2022 à 20:44, Stan Fultoni a écrit :
> On Wednesday, November 23, 2022 at 11:04:06 AM UTC-8, Richard Hachel wrote:
> [evasions]
>
> Please answer the questions:
>
> Example 1: One twin is stationary at the origin for 2 hours, and the other twin
> moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1
> hour. Do the twins have the same elapsed proper time?
>
> Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5
> mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do
> the twins have the same elapsed proper time?

I already answered that.

In the first case, we are dealing with the classic Langevin traveler.

It is obvious that the specific times of the speakers will not be the
same.

In Langevin's traveler, the proper times are fairly well calculated by
physicists.

On the other hand, it is at the level of the distances that they
understand nothing at all of what they are doing, and even less of what
Henri Poncaré told them with his Lorentz transformations.

For the other case, are you talking about a sudden acceleration?

In this case, we are dealing with several repositories, and things get
complicated.

R.H.
>
>> I SAY: Vo/c=[1+1/2ax]^(-1/2) [is] the instantaneous velocity...
>> [but] when I write Vo/c = [1+2c²/ax]^(-1/2), it is the average velocity
>> between [the origin and] x.
>
> Again, you contradict yourself. You're saying the symbol "Vo" represents both
> the instantaneous velocity of the trajectory at x and the average velocity over
> the interval from the origin to x. Those values are different, and you are using
> the same symbol to refer to both of them, so you are contradicting yourself.

No, I'm not contradicting myself.

Vo stands for observable velocity.

We have Vo(i) and Vo(m).

Vo(i) instantaneous observable velocity.

Vo(m) Average observable velocity.
>
> Moreover, you are contradicting yourself still further, because even if you give
> those expressions different names, such as Vo and Vo_ave[0,x], we can directly
> calculate the average value of Vo from your equation over the range from the
> origin to x and it does not equal your exporession for Vo_ave[0,x]. So, your
> beliefs imply 1=0, which is absurd. Right?

No, you are wrong.

You're wrong because you don't understand what I'm saying (or you don't
want to understand it, but it comes to the same thing).

R.H.

On Wednesday, November 23, 2022 at 2:06:40 PM UTC-5, rotchm wrote:
> On Wednesday, November 23, 2022 at 2:04:06 PM UTC-5, Richard Hachel wrote:
> > Le 23/11/2022 à 19:03, rotchm a écrit :
> >
> > > Irrelevant. The defintion of average speed is (x2-x1)/(t2-t1), where the
> > > variables are
> > > the *values* (of the ruler and clocks) located at the events. This is the
> > > definition, agree?
> > Je vous ai dit plusieurs fois que non.
> >
> > Cette égalité
> Can you state that in the language we were using, so that I may understand better?
>
> The language we were using (English) was understood by us both.
> Now you changed language. Why?

No answers? Stumped?

On Wednesday, November 23, 2022 at 2:24:07 PM UTC-5, rotchm wrote:
> On Wednesday, November 23, 2022 at 10:33:41 AM UTC-5, rotchm wrote:
> > On Wednesday, November 23, 2022 at 9:14:07 AM UTC-5, Richard Hachel wrote:
>
> > > >> Vo/c=(x/c)/To=(x/c)/[(x/c).sqrt(1+2c²/ax)]
> > > >
> > > > So if x = 0, what is the average velocity?
> > > If x=0 : no move.
> > So if someone leaves from x = 0, goes away and comes back to x=0, then you declare that he did not move?
> No answer?

Still no answer?
Is it because you're realize your blunder and that you are too dishonest to admit it?

On Wednesday, November 23, 2022 at 12:06:26 PM UTC-8, Richard Hachel wrote:
> > Please answer the questions:
> >
> > Example 1: One twin is stationary at the origin for 2 hours, and the other twin
> > moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1
> > hour. Do the twins have the same elapsed proper time?
> >
> > Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5
> > mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do
> > the twins have the same elapsed proper time?
>
> In the first case, we are dealing with the classic Langevin traveler.
> It is obvious that the [proper] times of the speakers will not be the
> same.

Right! Bravo. And now for the moment of truth...

> For the other case, are you talking about a sudden acceleration?

Yes, a sudden acceleration, just as in Example 1.
> In this case, we are dealing with several repositories, and things get
> complicated.

No, the answer is quite simple, and there are no accelerating systems of coordinates involved. Everything is described in terms of a single system of inertial coordinates. So, what is the answer? Are you saying you do not know the answer to this simple question? After 36 years?

> >> I SAY: Vo/c=[1+1/2ax]^(-1/2) [is] the instantaneous velocity...
> >> [but] when I write Vo/c = [1+2c²/ax]^(-1/2), it is the average velocity
> >> between [the origin and] x.
> >
> > Again, you contradict yourself. You're saying the symbol "Vo" represents both
> > the instantaneous velocity of the trajectory at x and the average velocity over
> > the interval from the origin to x. Those values are different, and you are using
> > the same symbol to refer to both of them, so you are contradicting yourself.
>
> No, I'm not contradicting myself.

You most certainly are. You define the symbol Vo to mean two different things, and even more seriously your definitions are self-contradictory because you claim the average velocity is the average of the instantaneous velocities, but your expression for the average is not actually the average of your expression for the instantaneous values over the specified range. You can check this for yourself. Compute the average of the instantaneous value, and you will see it does not equal your equations for the average. Your equations are self-contradictory gibberish. Agreed?

Le 24/11/2022 à 01:23, Stan Fultoni a écrit :
> On Wednesday, November 23, 2022 at 12:06:26 PM UTC-8, Richard Hachel wrote:
>> > Please answer the questions:
>> >
>> > Example 1: One twin is stationary at the origin for 2 hours, and the other
>> twin
>> > moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph
>> for 1
>> > hour. Do the twins have the same elapsed proper time?
>> >
>> > Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5
>>
>> > mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour.
>> Do
>> > the twins have the same elapsed proper time?
>>
>> In the first case, we are dealing with the classic Langevin traveler.
>> It is obvious that the [proper] times of the speakers will not be the
>> same.
>
> Right! Bravo. And now for the moment of truth...
>
>> For the other case, are you talking about a sudden acceleration?
>
> Yes, a sudden acceleration, just as in Example 1.
>
>> In this case, we are dealing with several repositories, and things get
>> complicated.
>
> No, the answer is quite simple, and there are no accelerating systems of
> coordinates involved. Everything is described in terms of a single system of
> inertial coordinates. So, what is the answer? Are you saying you do not know the
> answer to this simple question? After 36 years?
>
>> >> I SAY: Vo/c=[1+1/2ax]^(-1/2) [is] the instantaneous velocity...
>> >> [but] when I write Vo/c = [1+2c²/ax]^(-1/2), it is the average velocity
>> >> between [the origin and] x.
>> >
>> > Again, you contradict yourself. You're saying the symbol "Vo" represents both
>> > the instantaneous velocity of the trajectory at x and the average velocity
>> over
>> > the interval from the origin to x. Those values are different, and you are
>> using
>> > the same symbol to refer to both of them, so you are contradicting yourself.
>>
>> No, I'm not contradicting myself.
>
> You most certainly are. You define the symbol Vo to mean two different things,
> and even more seriously your definitions are self-contradictory because you claim
> the average velocity is the average of the instantaneous velocities, but your
> expression for the average is not actually the average of your expression for the
> instantaneous values over the specified range. You can check this for yourself.
> Compute the average of the instantaneous value, and you will see it does not equal
> your equations for the average. Your equations are self-contradictory gibberish.
> Agreed?

I use the term Vo more accurate than the term v of doctors.

I always differentiate, because it is of importance in the theory of
relativity of the real velocities of observable capitals.

I remind you that I introduced two fundamental notions, the notion of
anisochrony and the notion of dyschronotropy.

These notions mean that we never observe real speeds, but only observable
speeds.

Hence Vo and Vr.

Now you're right, an average speed over a course is not an instantaneous
speed, but I don't remember saying such a thing.

When I pose:
Vo/c=[1+c²/2ax]^(-1/2)
it is, as you have understood, the instantaneous speed over part of the
path at x.

For example, if you replace in the equation x by 3ly you have an
instantaneous speed of Vo=0.929c and if you replace by x=12ly, you have an
instantaneous speed of 0.980c

We can also say that the instantaneous speed is the average speed over an
infinitesimal or very small segment.

On what I have just said, two things separate us.

1. You think my equation is wrong and that this speed is significantly
higher. I deny this accusation.

2. You say my expression Vo is badly chosen for instantaneous speed, I
admit it, but how do you want me to write it without writing a big
illegible block?
It is very important to specify whether I am talking about real speeds or
observable speeds. So I have to write Vo or Vr.

Vo/c=[1+c²/2ax]^(-1/2)

Do you want me to write Vo(instantaneous speed)/c=[1+c²/2ax]^(-1/2)

Where

Vo(inst)/c=[1+c²/2ax]^(-1/2)

or Vo(i)/c=[1+c²/2ax]^(-1/2)

I chose the option to specify it earlier saying the instantaneous speed
will be:
Vr=sqrt(2ax)
Vo=[1+c²/2ax]^(-1/2)

I am of course open to any change to write if it is true and if it is
useful.

What would you suggest?

R.H.

On Thursday, November 24, 2022 at 5:45:47 AM UTC-5, Richard Hachel wrote:
> Le 24/11/2022 à 01:23, Stan Fultoni a écrit :

> I use the term Vo more accurate than the term v of doctors.

A lie on your part.
And 'doctors' don't use v.
You are very confused.

> I always differentiate, because it is of importance in the theory of
> relativity of the real velocities of observable capitals.

What are 'observable capitals' ?
You need to learn to write clearly else you will not be understood, agreed?

And what's the difference between velocity, real velocity, and observable velocity?
What are their (operational) definitions? We both need to know this if we want to discuss sensibly about those Concepts.

> Hence Vo and Vr.

You haven't defined them yet. Try again.

> When I pose:
> Vo/c=[1+c²/2ax]^(-1/2)
> it is, as you have understood, the instantaneous speed over part of the
> path at x.

So whats its instantaneous speed at x=0 according to your formula?

> 2. You say my expression Vo is badly chosen for instantaneous speed, I
> admit it,

So, fix it!

> but how do you want me to write it without writing a big
> illegible block?

Are you saying that you do not have the brains to figure that out on your own?
Are you saying that you do not have the brains to write clearly?

> Do you want me to write Vo(instantaneous speed)/c=[1+c²/2ax]^(-1/2)

Yes that is more clear.

But you still need to define the distinction between speed, instantaneous speed, observable speed, real speed Etc.

> I chose the option to specify it earlier saying the instantaneous speed
> will be:
> Vr=sqrt(2ax)
> Vo=[1+c²/2ax]^(-1/2)
>
> I am of course open to any change to write if it is true and if it is
> useful.

We have been telling you this for the past year. You are only waking up now?

> > > >> Vo/c=(x/c)/To=(x/c)/[(x/c).sqrt(1+2c²/ax)]
> > > >
> > > > So if x = 0, what is the average velocity?
> > > If x=0 : no move.
> > So if someone leaves from x = 0, goes away and comes back to x=0, then you declare that he did not move?
> No answer?

Still no answer?
Is it because you're realize your blunder and that you are too dishonest to admit it?
Still no answer?

> The language we were using (English) was understood by us both.
> Now you changed language. Why?

Still no answer?

On Thursday, November 24, 2022 at 2:45:47 AM UTC-8, Richard Hachel wrote:
> > Example 1: One twin is stationary at the origin for 2 hours, and the other
> > twin moves at -5 mph for 1 hour and then abruptly accelerates and moves
> > at +5 mph for 1 hour. Do the twins have the same elapsed proper time?
>
> It is obvious that the [proper] times of the speakers will not be the same.
>
> > Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves
> > at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for
> > 1 hour. Do the twins have the same elapsed proper time?
>
> In this case, we are dealing with several repositories, and things get
> complicated.

So, you cannot answer this simple question, and you don't care. This amounts to an admission that you have no idea what you are talking about. Agreed?

> For the trajectory t=sqrt(x^2 + 2x/a) [copied from special relativity for constant proper acceleration]
> I say the instantaneous velocity at any given x is [1+1/2ax]^(-1/2)
> and the average velocity from the origin to x is [1+2c²/ax]^(-1/2)

Again, you contradict yourself, in multiple ways. First, neither of your expressions agrees with (x2-x1)/(t2-t1) for the specified trajectory, which you already agreed is the average velocity of a bicycle going from x1,t1 to x2,t2 in terms of the x,t coordinates. Second, neither of your expressions equals dx/dt for the given trajectory, which you already agreed is the instantaneous velocity in terms of those coordinates. Third, your expression for average velocity does not equal the average of your expression for the instantaneous velocity over the specified range from the origin to x. You can verify this youself. So, you have contradicted yourself in three ways..

> You say my expression Vo is badly chosen for instantaneous speed, I
> admit it, but how do you want me to write it...

In terms of the x,t coordinates, the instantaneous velocity for the specified trajectory is dx/dt = sqrt[1 - 1/(1+ax)^2], and the average velocity between any two events x1,t2 and x2,t2 on the trajectory is (x2-x1)/(t2-t1).

> What would you suggest?

I suggest you answer the simple question: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do the twins have the same elapsed proper time?

Le 24/11/2022 à 17:46, Stan Fultoni a écrit :
> I suggest you answer the simple question: One twin moves at 10 mph for 2 hours,
> and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and
> moves at 15 mph for 1 hour. Do the twins have the same elapsed proper time?

They will cover the same distance in the same final time.

But what is the relationship with a relativistic problem?

R.H.

On Thursday, November 24, 2022 at 12:16:35 PM UTC-8, Richard Hachel wrote:
> > One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for
> > 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do the
> > twins have the same elapsed proper time?
>
> What is the relationship with a relativistic problem?

Again, please answer the question: Do the twins have the same total elapsed proper time? (This is a relativistic question.) Remember, you answered that the twins in Example 1 have different elapsed times. What about these twins in Example 2?

What are you unable to answer this simple question?

On Thursday, November 24, 2022 at 3:16:35 PM UTC-5, Richard Hachel wrote:
> Le 24/11/2022 à 17:46, Stan Fultoni a écrit :
> > I suggest you answer the simple question: One twin moves at 10 mph for 2 hours,
> > and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and
> > moves at 15 mph for 1 hour. Do the twins have the same elapsed proper time?
> They will cover the same distance in the same final time.

That does not answer his question. Try again.
he is asking, at the end of their respective journeys, what are the values on their respective watches?

> But what is the relationship with a relativistic problem?

Irrelevant.
The setup is clear. The question is clear. What will the *values* be of the watches?