Langevin and Doppler effects...

[Richard Hachel]

Paul B. Andersen :

https://paulba.no/pdf/TwinsByDoppler.pdf

Well.

There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.

We will return to the eternal problem, which has scared away all the
contributors I have dealt with for 40 years, although the opposite is
always said, and that I am the one who would be a wimp.

So we return to the problem, which is nevertheless very simple.

Stella goes into the stars for a journey of 24 light years.

The speed is 0.8c on the way there (12 ly), and 0.8c on the way back (12
ly).

We quickly obtain the following data by the Doppler effect.

Stella and Terrence agree to beep every month.

It is clear that, for Stella, on the way there, as on the way back, she
will beep
108 times (216 times in total).

SO, Terrence will receive 216 beeps.

For his part, Terrence will beep for 30 years. He will therefore send
360 beeps.

Except that if Terrence receives the same number of beeps on the way there
as on the way back (108 each time), Stella receives many more beeps on the
way back than on the way there.

It is then interesting to know how many beeps Stella receives on the way
there, and how many she receives on the way back.

Once this is done, we can still progress in understanding things.

We just have to not be afraid, and have a playful spirit.

R.H.

[Python]

Post by Richard Hachel
https://paulba.no/pdf/TwinsByDoppler.pdf
Well.
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.

You can be sure that we have no problems with what you call
"numerical applications" (funny gallicism). This is a fetish of
yours, as you fight with basic equations and run like a chicken
when confronted :-)

Still, contradiction and absurdity are on YOUR side Richard.

Post by Richard Hachel
[snip already debunked claims]

[Richard Hachel]

Post by Python

Post by Richard Hachel
https://paulba.no/pdf/TwinsByDoppler.pdf
Well.
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.

You can be sure that we have no problems with what you call
"numerical applications" (funny gallicism). This is a fetish of
yours, as you fight with basic equations and run like a chicken
when confronted :-)
Still, contradiction and absurdity are on YOUR side Richard.

Bah, le problème tout le monde le connait : "Le bon Docteur Hachel est
parvenu à énoncer une théorie de la relativité d'une rare beauté
conceptuelle et d'une rare évidence mathématique."

Seulement ça emmerde bien des gens, et on se demande bien pourquoi.

Je te rappelle que, question application numérique, tu ne m'as toujours
pas dit comment il se fait qu'au retour (c'est là le problème le plus
manifeste et le plus percutant) Stella vieillit de 9 ans (ce que tout le
monde admet), et voit la terre revenir sur elle à 4c (ce qui est la
vitesse apparente d'une fusée se déplaçant à 0.8c dans l'énoncé).

Tu es incapable de l'expliquer.

Les physiciens eux mêmes (TOUS!!!) sont incapables de l'expliquer.

Alors s'il te plait, avec tant d'incompétence, tais-toi donc un peu et
laisse parler le maître.

Si possible, sereinement.

Je te remercie par avance.

R.H.

[Paul.B.Andersen]

Post by Richard Hachel
https://paulba.no/pdf/TwinsByDoppler.pdf
Well.
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.
We will return to the eternal problem, which has scared away all the
contributors I have dealt with for 40 years, although the opposite is
always said, and that I am the one who would be a wimp.
So we return to the problem, which is nevertheless very simple.
Stella goes into the stars for a journey of 24 light years.
The speed is 0.8c on the way there (12 ly), and 0.8c on the way back (12
ly).
We quickly obtain the following data by the Doppler effect.
Stella and Terrence agree to beep every month.

You will find the answer here:
https://paulba.no/pdf/TwinsByDoppler.pdf
L = 12 ly
c = 1 ly/y
f = 12/y
v = 0.8c
γ = 5/3

But since you never read anything, I will repeat it here.

Post by Richard Hachel
It is clear that, for Stella, on the way there, as on the way back, she
will beep
108 times (216 times in total). >
SO, Terrence will receive 216 beeps.

OK. Stella has aged Tₛ = 2L/vγ = 18 y
Number of received beeps Ns = Tₛ⋅f = 216

Post by Richard Hachel
For his part, Terrence will beep for 30 years. He will therefore send
360 beeps.

OK

Post by Richard Hachel
Except that if Terrence receives the same number of beeps on the way
there as on the way back (108 each time), Stella receives many more
beeps on the way back than on the way there.
It is then interesting to know how many beeps Stella receives on the way
there, and how many she receives on the way back.

Doppler shift when Stella is moving out Do = √((c−v)/(c+v)) = 1/3
Doppler shift when Stella is moving back Db = √((c+v)/(c-v)) = 3
So number of received beeps Nt = Tₛ⋅f⋅Do/2 + Tₛ⋅f⋅Db/2 = 36 + 324 = 360

Post by Richard Hachel
Once this is done, we can still progress in understanding things.
We just have to not be afraid, and have a playful spirit.

Terrence will age C = 2L/v = 30y

Am I guessing wrong when I guess that you will present us with
the paradox that Terrence will receive
Tₜₛ⋅f⋅Do/2 + Tₜₛ⋅f⋅Db/2 = 60 + 540 = 600 beeps ? :-D

Hint:
https://paulba.no/pdf/TwinsByDoppler.pdf

Post by Richard Hachel
R.H.

[Richard Hachel]

Post by Richard Hachel

Post by Richard Hachel
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.
We will return to the eternal problem, which has scared away all the
contributors I have dealt with for 40 years, although the opposite is
always said, and that I am the one who would be a wimp.
So we return to the problem, which is nevertheless very simple.
Stella goes into the stars for a journey of 24 light years.
The speed is 0.8c on the way there (12 ly), and 0.8c on the way back (12
ly).
We quickly obtain the following data by the Doppler effect.
Stella and Terrence agree to beep every month.

https://paulba.no/pdf/TwinsByDoppler.pdf
L = 12 ly
c = 1 ly/y
f = 12/y
v = 0.8c
γ = 5/3

Yes. Very good.

Post by Richard Hachel
But since you never read anything, I will repeat it here.

I read, I read, don't worry

Post by Richard Hachel

Post by Richard Hachel
It is clear that, for Stella, on the way there, as on the way back, she
will beep
108 times (216 times in total). >
SO, Terrence will receive 216 beeps.

OK. Stella has aged Tₛ = 2L/vγ = 18 y
Number of received beeps Ns = Tₛ⋅f = 216

Good!

Post by Richard Hachel

Post by Richard Hachel
For his part, Terrence will beep for 30 years. He will therefore send
360 beeps.

OK

I am magic.

Post by Richard Hachel

Post by Richard Hachel
Except that if Terrence receives the same number of beeps on the way
there as on the way back (108 each time), Stella receives many more
beeps on the way back than on the way there.
It is then interesting to know how many beeps Stella receives on the way
there, and how many she receives on the way back.

Doppler shift when Stella is moving out Do = √((c−v)/(c+v)) = 1/3

Absolutely.

Post by Richard Hachel
Doppler shift when Stella is moving back Db = √((c+v)/(c-v)) = 3

Absolutely.

Post by Richard Hachel
So number of received beeps Nt = Tₛ⋅f⋅Do/2 + Tₛ⋅f⋅Db/2 = 36 + 324 = 360

Wonderfull.

Post by Richard Hachel

Post by Richard Hachel
Once this is done, we can still progress in understanding things.
We just have to not be afraid, and have a playful spirit.

Terrence will age C = 2L/v = 30y
Am I guessing wrong when I guess that you will present us with
the paradox that Terrence will receive
Tₜₛ⋅f⋅Do/2 + Tₜₛ⋅f⋅Db/2 = 60 + 540 = 600 beeps ? :-D

No, not at all.

Terrence will receive the 216 beeps sent by Stella for 18 years.

He can't receive more (otherwise, it's absurd).

R.H.

[Paul.B.Andersen]

Post by Richard Hachel

Post by Richard Hachel

Post by Richard Hachel
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.
We will return to the eternal problem, which has scared away all the
contributors I have dealt with for 40 years, although the opposite is
always said, and that I am the one who would be a wimp.
So we return to the problem, which is nevertheless very simple.
Stella goes into the stars for a journey of 24 light years.
The speed is 0.8c on the way there (12 ly), and 0.8c on the way back
(12 ly).
We quickly obtain the following data by the Doppler effect.
Stella and Terrence agree to beep every month.

https://paulba.no/pdf/TwinsByDoppler.pdf
L = 12 ly
c = 1 ly/y
f = 12/y
v = 0.8c
γ = 5/3

Yes. Very good.

Post by Richard Hachel

Post by Richard Hachel
It is then interesting to know how many beeps Stella receives on the
way there, and how many she receives on the way back.

Doppler shift when Stella is moving out  Do = √((c−v)/(c+v)) = 1/3

Absolutely.

Remember that you know this:
The Doppler shift is given by the Relative speed Terrence-Stella.

Post by Richard Hachel

Post by Richard Hachel
Doppler shift when Stella is moving back Db = √((c+v)/(c-v)) = 3

Absolutely.

Remember that you know this:
The Doppler shift is given by Relative speed Terrence-Stella.

Post by Richard Hachel

Post by Richard Hachel
So number of received beeps Nt = Tₛ⋅f⋅Do/2 +  Tₛ⋅f⋅Db/2 = 36 + 324 = 360

Wonderfull.

Remember that you know this:
Stella receives 36 beeps on her way out and 324 beeps on her way back,
total 360 beeps.

Post by Richard Hachel
My question is: Stella sees Terrence in her ultra-powerful telescope moving away from her with an apparent speed of 0.4444c, and for 9 years.
Is it true, or is it not true?

It is irrelevant.
Terrence is moving away from her at the speed v = 0.8c.
That's why Stella will see the Doppler shifted frequency to be
f₁ = f⋅Do = f⋅√((c−v)/(c+v)) = f/3 = 4 beeps/y

Remember?

What Stella might think Terrence speed appears to be
can have no physical consequences whatsoever.
Only the real speed v has a consequence for the Doppler shift.

Post by Richard Hachel
On the way back, for 9 years, she sees Terrence coming back to her for 9 years, with a speed of 4c.
Is it true or is it not true?

It is irrelevant.
Terrence is approaching her at the speed v = 0.8c.
That's why Stella will see the Doppler shifted frequency to be
f₂ = f⋅Db = f⋅√((c+v)/(c-v)) = f⋅3 = 36 beeps/y as you agreed to above.

Remember?

What Stella might think Terrence speed appears to be
can have no physical consequences whatsoever.
Only the real speed v has a consequence for the Doppler shift.

------

Remember that you know:

Stella receives 36 beeps on her way out and 324 beeps on her way back,
total 360 beeps, _because_ her speed relative to Terrence is 0.8c,
and _because_ Stella's proper time each way is 9 y, and _because_
Terrence is sending 12 beeps each year.

If Stella has some confused idea of what Terrence speed appears
to be, then this is a fantasy with no physical consequences
whatsoever. It is utterly irrelevant.

So what was your point with your "apparent speeds"?

Post by Richard Hachel
If it is true why not say it?
In his human trickery, man is very intelligent, he intuitively KNOWS that he cannot say it.
If he says: it is false. He makes a fool of himself.
If he says: it is true, he knows that we will answer, and what distance does the earth travel in both cases?
And this is frankly intolerable for relativistic believers who see their God naked, and his two laughable German prophets.

Gobbledegook.

I see you have failed to see the point with the Doppler method.
It is that neither Stella nor Terrence have to know anything
about the speed or times of the other twin. All they have to know
is that both are sending 12 peeps a year.
All they have to do is to count the peeps.

So far, we have not calculated how many beeps
Terrence will receive. We have only said it must be
216 beeps because Stella is emitting 12 beeps per year for 18 years.
This is true, but it is circular, and doesn't prove that Doppler
shift method give this result.

Here we will show that it does:
After 9 years Stella emits the last beep before turnaround.
Terrence time is L/v = 15 years when Stella emits the last beep.
Terrence will receive this beep a time L/c = 12 years later.
So Terrence will receive the Doppler shifted frequency 4 peeps/y
for L/v + L/c = 27 y, so he will receive 27⋅4 beeps = 108 beeps
emitted from Stella on her way out.
It is then obvious that Terrence must receive the Doppler shifted
frequency 36 beeps/y for L/v - L/c = 3 y, so he will receive
3⋅36 beeps = 108 beeps emitted from Stella on her way back.

To sum it up:
All Terrence and Stella have to do, is to count the received beeps.

Stella will receive 360 beeps.
From this she can deduce that Terrence must
have aged 360/12 y = 30 years.

Terrence will receive 216 beeps.
From this he can deduce that Stella must
have aged 216/12 y = 18 years.

Is the Langevin paradox solved?

[Richard Hachel]

Post by Paul.B.Andersen
Is the Langevin paradox solved?

Absolutely not.

We had a very interesting discussion, you and I, on the relativistic
Doppler effect.

We are neither you nor I, cretins, bandits, thugs, war criminals.

Each of us agrees that we wish to understand, master, and teach Henri
Poincaré's theory of relativity (the most beautiful theory ever to have
germinated in a human mind) in the most correct way possible.

We have just, by a very simple example, shown the relativity of pulsatile
frequencies, and there is no doubt, for me, as for you, that if these
things can one day be tested, they will give strictly the results that we
have proposed, and which are perfectly in agreement in your vision of
things, and in mine.

But this is only the first approach. This is NOT the complete resolution
of the Langevin paradox, because we talked about the relativity of the
pulsatile frequency, but we never
talked about the relative distances traveled by the two protagonists.

And that is the main point of the subject (the Doppler effect is just the
appetizer).

I have never stopped saying, for forty years, that we must respect the
word of our old physicists, and the difficulty they had at the time to
believe in the theory of relativity because we could not say on the one
hand that there was no absolute frame of reference, and on the other,
say that Stella, in her own frame of reference, came back younger than
Terrence, which is absurd, since constantly, AS MUCH on the way out,
DURING the U-turn, and AS MUCH on the way back, second after second,
for her, Terrence's time passes less quickly.

There must be something wrong.

Because that's the reality of things, always, always, always, chronotropy
is of the type To'=To/sqrt(1-Vo²/c²) for both participants. Always,
always, always, the relative speed will be v=0.8c (even during the
half-circle of the U-turn) between the two, and always, always, always,
Stella will observe Terrence's chronotropy lower than hers, and Terrence
will observe that it is Stella's chronotropy that is lower.

So what's going on?

The Doppler effect solves part of the things, but not ALL of the things,
and in this, a problem persists: the Langevin paradox.

It is ONLY partially clarified.

Because there will persist, on Stella's side, a theoretical enormity that
physicists cannot explain, and which will then be denied (by human
reflex).

We know that Terrence will observe Stella moving at Vapp=0.4444c,
on the way there, and for 27 years (he receives half of the beeps during
this period).

What distance will he measure? x=0.4444c*27=12al.

On the way back, he will measure x=4c*3=12 al

It is undeniable and mathematical.

But for Stella, and this is the great key to the paradox, and the
brilliant explanation, the distance traveled by the earth will NOT be 12
al, neither on the way there, nor on the way back, because it is not in
the same frame of reference.

It is on this incredible relativity of distances that the paradox is
based, BUT she does not travel a shorter distance (which is absurd because
she would have a longer time) but a longer one.

That is why in total she has a shorter time than Terrence.

I beg you to UNDERSTAND what is really happening.

This "time-gap" story is just dust swept under the carpet.

There is NO time-gap in Hachel, because we do not need it, and it is
totally false.

In Stella's frame of reference the earth will travel x=0.4444*9=4ly on the
way out.

And x=4c*9=36al on the way back.

This is the phenomenon of elasticity of relative distances well
understood.

During the U-turn, nothing changes for Stella in terms of time.

I beg you to understand this.

She notes 9 years on her clock, at the entrance to the U-turn, and 9 years
at the exit.

She sees the earth, over there, at 4ly, which marks 3 years, when she
enters her U-turn (which she will do in 24 hours of proper time), and at
the exit, the earth still marks 3 years.

There is no time gap.

She will therefore see the earth come back to her in 9 years of proper
time, but with an earth clock that turns three times faster.

The key to the problem is not explained by the time gap, but not the
space-zoom.

During her U-turn, she sees the earth go from 4 ly, to 36 ly, in her frame
of reference.

THIS is what explains the Langevin paradox.

Not the time-gap, which is false, ridiculous, and a filthy attempt to
sweep dust under the carpet.

R.H.

[Richard Hachel]

Post by Richard Hachel

Post by Paul.B.Andersen
Is the Langevin paradox solved?

Absolutely not.
The key to the problem is not explained by the time gap, but not the space-zoom.

Correction.

The key to the problem is not explained by the time gap, but by the
space-zoom.

R.H.

[Paul.B.Andersen]

< snip >
We know that Terrence will observe Stella moving at Vapp=0.4444c,
on the way there, and for 27 years (he receives half of the beeps during
this period).

READ THIS, AND THINK!

The only thing Terrence must know about Stella is that she is
emitting 12 beeps/y.

A year after Stella left, Terrence will know that he receives 4 beeps/y.
Since Stella is emitting 12 beeps/y, the Doppler shift is 1/3.
So Terrence knows that Stella is moving away at the speed 0.8c.

Why would Terrence say that Stella appears to move away
at the speed 0.4444c when he knows that her speed is 0.8c? :-D

When Terrence sees that the received beep frequency suddenly changes,
Terrence clock shows 27 y, so he has received 4*27 = 108 beeps.
Stella's clock shows 108/12 = 9 year at the time of emission of
the beep Terrence received when his clock showed 27 y.

What distance will he measure? x=0.4444c*27=12al.

Since Stella is moving away at the speed 0.8c, and her moving
clock shows 9 y, Terrence will know that his own stationary
clock simultaneously must have shown 9h⋅γ = 9⋅5/3 y = 15 y.
("moving clocks run slow" , remember?)

Terrence will obviously, as opposed to Hachel, understand
that his clock didn't show 27h at the instant when Stella's
clock showed 9h. He knows that his clock showed 15 y, and
the light used 12 y to reach him when his clock showed 27 y.

Since Stella is travelling at 0.8c in Terrence's rest frame,
and his stationary clock in said frame shows 15 h at the instant
when Stella's clock shows 9h, the distance to Stella is
15h⋅0.8c = 12 ly.

So Terrence knows that Stella's turning point is 12 ly from him
in his rest frame.

On the way back, he will measure x=4c*3=12

(Is Stella moving at 4 times the speed of light for 3h? :-D)

One year after the instant when Stella's clock showed 9h,
when Terrence's clock shows 28 y, Terrence will see that he is
receiving 36 peeps/y and that the Doppler shift thus is 3.
So Terrence knows that Stella is approaching at the speed 0.8c.
And since the turning point is 12 ly from him, he knows that
his clock will advance 12ly/0.8c = 15 y during Stella's travel back.

Since his clock showed 15h at turnaround, it will show 30h
when Stella is back.

It is undeniable and mathematical.

Indeed.
All I said above is irrefutable true according to
The Special Theory of Relativity.
No paradox, no mystery.

-----------------------------

But for Stella, and this is the great key to the paradox, and the
brilliant explanation, the distance traveled by the earth will NOT be 12
al, neither on the way there, nor on the way back, because it is not in
the same frame of reference.

I will come back to the scenario seen from Stella's rest frame
in a separate post, this is long enough. (It probably won't be today.)

But I will say this:
The scenario described here is physically impossible to do
in the real world. The problem is the abrupt change of velocity
at turnaround. Stella will then have an infinite high acceleration
for zero time. This is a mathematical singularity, which make
it impossible to solve exactly. This is a problem when seen
from Stella's point of view, because what happens to her view
of Terrence's clock when her acceleration is infinite?

All we have to do to make it solvable is to introduce
a high but finite acceleration for a short time at turnaround.

Back tomorrow.

[Richard Hachel]

Post by Paul.B.Andersen
All we have to do to make it solvable is to introduce
a high but finite acceleration for a short time at turnaround.

But no!

You don't pay any attention to what I'm telling you.

You've decided to say that in RR everything is fine, that everything is
acceptable, that everything is true, and that you should definitely not
say anything.

From then on, it turns into religious fundamentalism.

I explained RR like no one had explained it before, and instead of being
interested, everyone is playing the monkey by opposing me with
anything....

Pffff...

Where did you see an infinite acceleration?

In the example I gave, the U-turn takes 40 hours in Terrence's frame of
reference (24 in Stella's).

It's certainly a huge acceleration, but it's not infinite.

You've decided not to listen to me.

So we're inevitably going around in circles.

This is not new, I have been explaining for forty years how to correctly
understand this theory which is not so complicated (we do not even need an
integral or a tangent for the whole theory).

The problem is not that "Richard Hachel's theory is wrong", it is "We do
not want Richard Hachel to talk to us about that".

We live in a world of sick people.

Furthermore, you do like Python, you criticize but without ever answering
the question.
How far is the earth from Stella, at the moment when, after 9 years, and
after having seen it flee behind her at 0.444c, Stella will start to brake
to begin her U-turn?
Is it 4 ly, or is it not 4 ly?
How far is the earth from Stella, quad, the U-turn completed, she is again
at 0.8c, but towards the earth?
Is it 36 ​​ly, or is it not 36 ly?
We must still take responsibility and say things?
How far does Stella see the earth?
What are these distances?



R.H.

[Paul.B.Andersen]

Post by Richard Hachel

Post by Paul.B.Andersen
I will come back to the scenario seen from Stella's rest frame
in a separate post, this is long enough.
The scenario described here is physically impossible to do
in the real world. The problem is the abrupt change of velocity
at turnaround. Stella will then have an infinite high acceleration
for zero time. This is a mathematical singularity, which make
it impossible to solve exactly. This is a problem when seen
from Stella's point of view, because what happens to her view
of Terrence's clock when her acceleration is infinite?
All we have to do to make it solvable is to introduce
a high but finite acceleration for a short time at turnaround.

But no!
You don't pay any attention to what I'm telling you.
You've decided to say that in RR everything is fine, that everything is
acceptable, that everything is true, and that you should definitely not
say anything.
From then on, it turns into religious fundamentalism.
I explained RR like no one had explained it before, and instead of being
interested, everyone is playing the monkey by opposing me with anything....
Pffff...

What are you whining about?

Post by Richard Hachel
Where did you see an infinite acceleration?

| Den 05.09.2024 23:45, skrev Richard Hachel:>
|> Stella goes into the stars for a journey of 24 light years.
|> The speed is 0.8c on the way there (12 ly), and 0.8c on the way back
(12 ly).

Let's call this example #1.

Post by Richard Hachel
In the example I gave, the U-turn takes 40 hours in Terrence's frame of
reference (24 in Stella's).

That was a different example.
In Terrence frame Stella travels with the constant speed
0.8 c for 30 years and 40 hours. When Terrence's clock
shows 15 years, Stella goes into a sharp U-turn which
last 40 hours.

Let's call this example #2.

Post by Richard Hachel
It's certainly a huge acceleration, but it's not infinite.

Yes.
But this acceleration is only transverse, and doesn't affect the speed.
But it does make it possible to calculate what Terrence's proper time
and position in Stella's frame is, as a function of Stella's time.

----

In both examples is the speed constant, so it trivially simple to find
Stella's proper time.

In example #1:
Stella's clock at return = 30y/γ = 18y

In example #2:
Stella's clock at return = (30y 40h)/γ = 18y 24h

-----

Note:
We will calculate what SR say will be measured in Stella's frame.
NOT what will be observed in a telescope.

Example #1:
------------
Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9- y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y

When Stella's clock shows 9+ y, Stella is on her way back.
Terrence will still be at the position x = 7.2 ly,
but will now move at the speed 0.8c towards Stella.
Terrence's clock will now show 30y-5.4y = 24.6y

When Stella's clock shows 18y, Stella is back.
Terrence will now be at the position x = 7.2ly - 0.8c⋅9y = 0 ly.
Terrence clock will now show 24.6y + 5.4y = 30 y

Note this:
When Stella is moving away from Terrence and her clock shows 9y,
Terrence clock will _simultaneously_ in Stella,s frame show 5.4y
When Stella is moving towards Terrence and her clock shows 9y,
Terrence clock will _simultaneously_ in Stella,s frame show 24.6y

Both clocks run at a steady rate, 1 second per second, but
what the two clocks _simultaneously_ will show in Stella's frame
depend on Stella's velocity relative to Terrence.
When Stella abruptly change her direction, her idea of what time is
simultaneous in Terrence frame changes from 5.4y to 22.8y,

But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that changes.

Example #2:
------------
Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y

When Stella's clock show 9y 12h, Terrence will be at the position
x = 12 ly in Stella's frame. Terrence clock will now show 15 y.

When Stella's clock shows 9y 24h, Terrence will now be at the position
x = 0.8c⋅(9y 24h) = (7.2 ly + 19.2 lh) in Stella's frame.
Terrence clock will now show (30y 40h) - (9y 24h)/γ = 24.6y 25.6h

When Stella's clock shows 18y 24h, Stella is back.
Terrence will now be at the position x = 0 ly.
Terrence clock will now show (24.6y 25.6h) + (9y 24h)/γ = 30y 40h

--------------

Note that when Stella's clock show 9y, Terrence clock simultaneously
in Stella's frame show 5.4y, and when Stella's clock show 9y 12h,
Terrence clock simultaneously in Stella's frame show 15y.

That means that Terrence clock runs (15-5.5)y/12h = 7012.8 times
faster than Stella's clock during the first part of the U-turn.

But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.

(It may be typos)

[Python]

Post by Paul.B.Andersen

Post by Richard Hachel

Post by Paul.B.Andersen
I will come back to the scenario seen from Stella's rest frame
in a separate post, this is long enough.
The scenario described here is physically impossible to do
in the real world. The problem is the abrupt change of velocity
at turnaround. Stella will then have an infinite high acceleration
for zero time. This is a mathematical singularity, which make
it impossible to solve exactly. This is a problem when seen
from Stella's point of view, because what happens to her view
of Terrence's clock when her acceleration is infinite?
All we have to do to make it solvable is to introduce
a high but finite acceleration for a short time at turnaround.

But no!
You don't pay any attention to what I'm telling you.
You've decided to say that in RR everything is fine, that everything
is acceptable, that everything is true, and that you should definitely
not say anything.
 From then on, it turns into religious fundamentalism.
I explained RR like no one had explained it before, and instead of
being interested, everyone is playing the monkey by opposing me with
anything....
Pffff...

What are you whining about?

https://ibb.co/9yZCxzy

[Richard Hachel]

Quand tu auras fini tes conneries, tu posteras peut-être des trucs
intéressants, voire tu répondra aux concepts et exercices qui t'ont
été posés.

Au moins Paul il fait l'effort de dialoguer intelligemment et avec
recherche d'exemples numériques possibles.

Tu es la honte de la physique française.

R.H.

[Richard Hachel]

Post by Paul.B.Andersen
------------
Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9- y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y
When Stella's clock shows 9+ y, Stella is on her way back.
Terrence will still be at the position x = 7.2 ly,
but will now move at the speed 0.8c towards Stella.
Terrence's clock will now show 30y-5.4y = 24.6y
When Stella's clock shows 18y, Stella is back.
Terrence will now be at the position x = 7.2ly - 0.8c⋅9y = 0 ly.
Terrence clock will now show 24.6y + 5.4y = 30 y
When Stella is moving away from Terrence and her clock shows 9y,
Terrence clock will _simultaneously_ in Stella,s frame show 5.4y
When Stella is moving towards Terrence and her clock shows 9y,
Terrence clock will _simultaneously_ in Stella,s frame show 24.6y
Both clocks run at a steady rate, 1 second per second, but
what the two clocks _simultaneously_ will show in Stella's frame
depend on Stella's velocity relative to Terrence.
When Stella abruptly change her direction, her idea of what time is
simultaneous in Terrence frame changes from 5.4y to 22.8y,
But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that changes.

Yes, I understood very well.
I have known about SR for forty years, and I have thought about it, I
think, much more than a simple man who, having studied a few hours,
convinced himself that it was very pretty and that Einstein and Minkowski
were probably right.
I have thought about it for thousands of hours, and I have always
understood that something would go wrong, and WHY it would go wrong.
I will tell you what is wrong with your demonstration which is
mathematically impeccable, but physically useless and false if we scratch
much deeper. You are not talking to me here about Stella's frame of
reference but about two different frames of reference of type M (i.e. M1
for the outward journey, and M2 for the return journey).
It is NOT Syella's frame of reference but something useless, which is
found in Stella's frame 1, then in Stella's frame 2.
This explains the time jump during the U-turn, because we go from what is
measured by the watch M1 in the outbound frame of reference, to what is
measured by the watch M2 in the return frame of reference. That is to say
a carrot and a turnip.

This way of seeing things is not correct. For the sake of great clarity
(and it is infinitely clearer when we master Hachel's concepts) we must
place ourselves in Stella's frame of reference, and never leave it.

Not only do you go from Stella to a hypothetical and mathematical observer
M1, but suddenly, M1 no longer following you at the time of the U-turn,
you go to another infinitely different observer M2.

This is not correct, and this is not how we should see things.

Even if what you say about M1 and M2 is true.

R.H.

[Richard Hachel]

Post by Paul.B.Andersen
------------
Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y
When Stella's clock show 9y 12h, Terrence will be at the position
x = 12 ly in Stella's frame. Terrence clock will now show 15 y.
When Stella's clock shows 9y 24h, Terrence will now be at the position
x = 0.8c⋅(9y 24h) = (7.2 ly + 19.2 lh) in Stella's frame.
Terrence clock will now show (30y 40h) - (9y 24h)/γ = 24.6y 25.6h
When Stella's clock shows 18y 24h, Stella is back.
Terrence will now be at the position x = 0 ly.
Terrence clock will now show (24.6y 25.6h) + (9y 24h)/γ = 30y 40h
--------------
Note that when Stella's clock show 9y, Terrence clock simultaneously
in Stella's frame show 5.4y, and when Stella's clock show 9y 12h,
Terrence clock simultaneously in Stella's frame show 15y.
That means that Terrence clock runs (15-5.5)y/12h = 7012.8 times
faster than Stella's clock during the first part of the U-turn.
But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.
(It may be typos)

Same as for the previous post.

You use observers M1 and M2 and not Terrence himself.

Here, Terrence always remaining in the same inertial frame of reference,
M1 and M2 are the same external reference point.

But it is not Terrence.

Of course, I may give you the impression that I am quibbling, but that is
not the case.

Things must be said so that it is both infinitely beautiful and infinitely
true.

The measurements that you give here, with this concept M (which is not
known since you really think that you are talking about Stalla, then
Terrence, and not very distant lateral points) are not beautiful, even if,
for the moment, they are true.

The blow of the baton can only come later.

When you are going to talk about the same things, but with accelerated
frames of reference, it will no longer be beautiful or true.

And there, the loss is double and the predictions catastrophic.

R.H.

[Python]

This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation".
If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
Nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper
times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".

t'es vraiment con comme un balai : tu as foutu toute ta "réponse" en ...
signature.
you are stupid as stone: you put your entire "answer" in ... your signature.
Trump is inspiring you, I'd say :-)

Trump 2024 !!!
R.H.

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

https://ibb.co/YXQbVth

[Richard Hachel]

This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation".
If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
Nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper
times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".

t'es vraiment con comme un balai : tu as foutu toute ta "réponse" en ...
signature.
you are stupid as stone: you put your entire "answer" in ... your signature.
Trump is inspiring you, I'd say :-)

Trump 2024 !!!

R.H.

[Richard Hachel]

This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation".
If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
Nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper
times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".

--
Paul

According to physicists, the key to the paradox is to add a time gap, like
adding dust under a carpet.
It's not pretty, and it's not true.
Physicists don't realize that they are practicing artificially.
The frame of reference that they attribute to Stella is firstly not the
real one. It's not hers, but the frame of reference of a comoving observer
placed very far away and observing things transversally.
This is the first problem, but the second will be worse. For the return,
we are forced to do the same thing, but here, it's no longer like with
Terrence where M1=M2.

In Stella's frame of reference, M1 is imaginarily infinitely far from M2.
We "jump" from frame of reference, which is neither pretty nor credible.
We then "jump" from clocks. And all this is nothing more than a vulgar
patch-up to "be even more right".

If we finally wanted to understand things (but we don't want to),
we would realize that a frame of reference can only be attributed, in
relativity, to a single observer.

This observer is at point O of the frame of reference.

In Stella's problem; Stella is NOT in her correct relativistic place, and
we are forced to apply non-existent subterfuges (time-gap) to validate the
passage from M1 (go) to M2 (return) while ideally these two points are
infinitely far away in their respective space-time. We "jump" frames of
reference.

I don't practice like that at all.

I always stay in Stella's frame of reference, WITH Stella as center and
origin. So she NEVER jumps frames of reference. She is always in her frame
of reference and at point O.

We then realize that there is never a clock jump to imagine.
Never. Terrence observes things with a clock that is smooth for him,
and fluid for her.
For her part, it is the same thing for Stella, she observes her own clock
in a fluid way, and Terrence's in a fluid way.

I am repeating this because it is important, too bad if the reader finds
that I am repeating myself.

Terrence (that is to say HIM and not M) observes this in his telescope and
live.
It is Ta=0 (his watch) and Tb=0 (Stella's watch) at the start.
When Stella approaches her aphelion, her clock marks Tb=9 years.
Terrence's FOR Terrence marks 27 years!
Stella turns, and comes back. She reappears there, this time rushing
towards the earth. Terrence always marks 27 years (+40 hours) and Stella
always marks 9 years (+24 hours).
No time jump on Terrence's part, neither for him (that would be absurd),
nor for her (that would also be absurd).

From Stella's point of view:
At the start Ta = 0 Tb = 0.
When she arrives there, just before starting her half-turn.
Her clock marks 9 years and she sees the terrestrial clock which marks 3
years. According to the laws of relativity.
When she has finished turning, she sees her watch always marks 9 years,
and Terrence's marks THREE YEARS.
I beg you to understand this.
There is NO time gap.
Stella's watch for Stella indicates 9 years and 24 hours. And Terrence's
(seen by Stella in HER frame of reference) marks 3 years and 40 hours.

The paradox is therefore not explained by the time-gap.

We need to put an end to this patching up.

But then how is it explained?

Hold on tight my friends, breathe, blow. I'm going to tell you something
brilliant that you don't want to understand because of the finiteness of
your neurons and even more because of your arrogance towards me.

Space is a mollusk of reference, and if we understand Poincaré's
magnificent transformations, we realize that, for Stella when she turns,
all her space will be transformed.

The earth which is at 4 ly (and not 7.2 ly) will be carried in its own
frame of reference, centered on it, and not on M1 or M2, at 36 ly (and not
at 7.2 ly).

We call this the relativistic zoom effect.

It is such an effect that is not understood, that is refused, and on which
we spit.

Yet this is what is true, and our universe is wrong as I say it, and not
as thousands of physicists who think they are stronger than me will say
it.

Always, always, always, they will contradict, until the day when too much
experimental evidence will go on my side.

The situation will become untenable for them.

And they will be obliged to make the effort to understand me by crying.

Because they will realize that if the concept is brilliant, the equations
that follow are much simpler, and at the level of a good high school
student, or even middle school student.

Hence my sadness when I read the forums or the relativistic websites.

No one makes the real effort to understand.

R.H.
Ce message a été posté avec Nemo : <https://www.nemoweb.net/?DataID=zgxExgejF-***@jntp>

[Python]

This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation".
If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
Nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper
times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".

t'es vraiment con comme un balai : tu as foutu toute ta "réponse" en ...
signature.

you are stupid as stone: you put your entire "answer" in ... your
signature.

Trump is inspiring you, I'd say :-)

[Paul.B.Andersen]

Post by Richard Hachel

Post by Paul.B.Andersen
I have shown you Terrence's, speed, position and proper time
measured in Stella's rest frame, as a function of Stella's time.
During Stella's time 0y -> 9y, Stella is inertial and so is her frame.
MEASURED IN STELLA'S FRAME, Terrence is moving away at the speed 0.8c, his clock runs slow by 1/γ, and his position is x = 0.8c⋅t_Stella'
So when Stella's clock show 9h, t_Terrence = 5.4y and x = 7.2 ly
During Stella's time 9y -> 9y + 24h, Stella is accelerating towards
Terrence, so Stella's frame is an accelerated frame.
MEASURED IN STELLA'S ACCELERATED FRAME, Terrence will move away
from Stella at high speed, and his clock will run thousands of times
faster than Stella's.
So when Stella's clock shows 9h 12h, t_Terrence = 15y and x = 12ly
When Stella's clock shows 9h 24h, t_Terrence = 24.6y 25.6h
and x = 7.2ly + 19.2lh
During Stella's time 9y 24h - 18y, Stella is inertial and so is her frame.
MEASURED IN STELLA'S FRAME, Terrence is moving away at the speed 0.8c, his clock runs slow by 1/γ, and his position is
x = 7.2ly + 19.2lh + 0.8c⋅t_Stella'
At Stella-s time 18y + 24h Terrence is back.
What is it that you don't understand?

This is how Doctor Richard Hachel understands the above:

Or should I say:
This is how Doctor Richard Hachel doesn't understand the above?

Post by Richard Hachel
Physicists don't realize that they are practicing artificially.
The frame of reference that they attribute to Stella is firstly not the real one. It's not hers, but the frame of reference of a comoving observer placed very far away and observing things transversally.

So Stella's rest frame is not the real one, and it isn't hers,
but the frame of a distant comoving observer who is observing
things transversely.

Post by Richard Hachel
This is the first problem, but the second will be worse. For the return, we are forced to do the same thing, but here, it's no longer like with Terrence where M1=M2.
In Stella's frame of reference, M1 is imaginarily infinitely far from M2. We "jump" from frame of reference, which is neither pretty nor credible. We then "jump" from clocks. And all this is nothing more than a vulgar patch-up to "be even more right".

Ooops. You lost me there!

Post by Richard Hachel
If we finally wanted to understand things (but we don't want to),
we would realize that a frame of reference can only be attributed, in relativity, to a single observer.

So Terrence can't have a speed and position in Stella's rest frame?

Post by Richard Hachel
This observer is at point O of the frame of reference.
In Stella's problem; Stella is NOT in her correct relativistic place, and we are forced to apply non-existent subterfuges (time-gap) to validate the passage from M1 (go) to M2 (return) while ideally these two points are infinitely far away in their respective space-time. We "jump" frames of reference.

And because Stella is NOT in her correct relativistic place, we are
forced to apply non-existing subterfuges to validate the passage between
two points which are infinitely far away.

As always, you express yourself crystal clear, Richard. Well done!

Post by Richard Hachel
I don't practice like that at all.
I always stay in Stella's frame of reference, WITH Stella as center and origin. So she NEVER jumps frames of reference. She is always in her frame of reference and at point O.

So Stella's rest frame is not the real one, and it isn't hers,
but the frame of a distant comoving observer who is observing
things transversely.
But YOU always stay in Stella's frame of reference with Stella as
centre of origin.

---------------------

I think it is enough now.

[Paul.B.Andersen]

Post by Richard Hachel

Post by Paul.B.Andersen
------------
Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y
When Stella's clock show 9y 12h, Terrence will be at the position
x = 12 ly in Stella's frame. Terrence clock will now show 15 y.
When Stella's clock shows 9y 24h, Terrence will now be at the position
x = 0.8c⋅(9y 24h) = (7.2 ly + 19.2 lh)  in Stella's frame.
Terrence clock will now show (30y 40h) - (9y 24h)/γ = 24.6y 25.6h
When Stella's clock shows 18y 24h, Stella is back.
Terrence will now be at the position x = 0 ly.
Terrence clock will now show (24.6y 25.6h) + (9y 24h)/γ = 30y 40h
--------------
Note that when Stella's clock show 9y, Terrence clock simultaneously
in Stella's frame show 5.4y, and when Stella's clock show 9y 12h,
Terrence clock simultaneously in Stella's frame show 15y.
That means that Terrence clock runs (15-5.5)y/12h = 7012.8 times
faster than Stella's clock during the first part of the U-turn.
But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.

Same as for the previous post.
You use observers M1 and M2 and not Terrence himself.
Here, Terrence always remaining in the same inertial frame of reference,
M1 and M2 are the same external reference point.
But it is not Terrence.
Of course, I may give you the impression that I am quibbling, but that
is not the case.

You are not quibbling, you are babbling.

According to SR:

I have shown you Terrence's, speed, position and proper time
measured in Stella's rest frame, as a function of Stella's time.

It is quite simple:

During Stella's time 0y -> 9y, Stella is inertial and so is her frame.
MEASURED IN STELLA'S FRAME, Terrence is moving away at the speed 0.8c,
his clock runs slow by 1/γ, and his position is x = 0.8c⋅t_Stella'
So when Stella's clock show 9h, t_Terrence = 5.4y and x = 7.2 ly

During Stella's time 9y -> 9y + 24h, Stella is accelerating towards
Terrence, so Stella's frame is an accelerated frame.
MEASURED IN STELLA'S ACCELERATED FRAME, Terrence will move away
from Stella at high speed, and his clock will run thousands of times
faster than Stella's.
So when Stella's clock shows 9h 12h, t_Terrence = 15y and x = 12ly
When Stella's clock shows 9h 24h, t_Terrence = 24.6y 25.6h
and x = 7.2ly + 19.2lh

During Stella's time 9y 24h - 18y, Stella is inertial and so is her frame.
MEASURED IN STELLA'S FRAME, Terrence is moving away at the speed 0.8c,
his clock runs slow by 1/γ, and his position is
x = 7.2ly + 19.2lh + 0.8c⋅t_Stella'

At Stella-s time 18y + 24h Terrence is back.

What is it that you don't understand?

Post by Richard Hachel
Things must be said so that it is both infinitely beautiful and
infinitely true.
The measurements that you give here, with this concept M (which is not
known since you really think that you are talking about Stalla, then
Terrence, and not very distant lateral points) are not beautiful, even
if, for the moment, they are true.
The blow of the baton can only come later.
When you are going to talk about the same things, but with accelerated
frames of reference, it will no longer be beautiful or true.

| Den 07.09.2024 23:42, skrev Richard Hachel:
|> But for Stella, and this is the great key to the paradox, and the
brilliant explanation, the distance traveled by the earth will NOT be 12
al, neither on the way there, nor on the way back, because it is not in
the same frame of reference.

This is the statement I responded to.

It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation".
If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).

But it isn't beautiful, even if it true.

Stella's view is rather pointless.

As I said:
Nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.

If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper
times.)

If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.

So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".

[ProkaryoticCaspaseHomolog]

Post by Richard Hachel
We just have to not be afraid, and have a playful spirit.

There is no contradiction.

Terrance receives the exact number of beeps that Stella has transmitted.
Stella receives the exact number of beeps that Terrance has transmitted.

This is TRIVIAL to demonstrate using a spacetime diagram.
No matter what path through spacetime Stella follows, this must be true.

Don't bother with my table. Just LOOK at my Figure 4-4.
https://en.wikipedia.org/wiki/Special_relativity#Twin_paradox

[Richard Hachel]

Post by ProkaryoticCaspaseHomolog

Post by Richard Hachel
We just have to not be afraid, and have a playful spirit.

There is no contradiction.
Terrance receives the exact number of beeps that Stella has transmitted.
Stella receives the exact number of beeps that Terrance has transmitted.

Absolutely.

In 18 years, Stella has pulsed 18*12=216 times.

In 30 years, Terrence has pulsed 30*12=360 times

During the trip, it is impossible for beeps to have been lost, neither for
Terrence nor for Stella.

Everything is of a sumptuous obviousness and a great relativistic clarity.

Post by ProkaryoticCaspaseHomolog
This is TRIVIAL to demonstrate using a spacetime diagram.

Oh no please, please no, I beg you...

No satanic spacetime diagram... Please...pleasa... No....

Post by ProkaryoticCaspaseHomolog
No matter what path through spacetime Stella follows, this must be true.
Don't bother with my table. Just LOOK at my Figure 4-4.
https://en.wikipedia.org/wiki/Special_relativity#Twin_paradox

Irrelevant.

Is not my question.

My question is: Stella sees Terrence in her ultra-powerful telescope
moving away from her with an apparent speed of 0.4444c, and for 9 years.
Is it true, or is it not true?
On the way back, for 9 years, she sees Terrence coming back to her for 9
years, with a speed of 4c.
Is it true or is it not true?
If it is true why not say it?
In his human trickery, man is very intelligent, he intuitively KNOWS that
he cannot say it.
If he says: it is false. He makes a fool of himself.
If he says: it is true, he knows that we will answer, and what distance
does the earth travel in both cases?
And this is frankly intolerable for relativistic believers who see their
God naked, and his two laughable German prophets.


R.H.

[ProkaryoticCaspaseHomolog]

Post by Richard Hachel

Post by ProkaryoticCaspaseHomolog

Post by Richard Hachel
We just have to not be afraid, and have a playful spirit.

There is no contradiction.
Terrance receives the exact number of beeps that Stella has transmitted.
Stella receives the exact number of beeps that Terrance has transmitted.

Absolutely.
In 18 years, Stella has pulsed 18*12=216 times.
In 30 years, Terrence has pulsed 30*12=360 times
During the trip, it is impossible for beeps to have been lost, neither for
Terrence nor for Stella.
Everything is of a sumptuous obviousness and a great relativistic clarity.

Post by ProkaryoticCaspaseHomolog
This is TRIVIAL to demonstrate using a spacetime diagram.

Oh no please, please no, I beg you...
No satanic spacetime diagram... Please...pleasa... No....

Post by ProkaryoticCaspaseHomolog
No matter what path through spacetime Stella follows, this must be true.
Don't bother with my table. Just LOOK at my Figure 4-4.
https://en.wikipedia.org/wiki/Special_relativity#Twin_paradox

Irrelevant.
Is not my question.
My question is: Stella sees Terrence in her ultra-powerful telescope
moving away from her with an apparent speed of 0.4444c, and for 9 years.
Is it true, or is it not true?
On the way back, for 9 years, she sees Terrence coming back to her for 9
years, with a speed of 4c.
Is it true or is it not true?
If it is true why not say it?
In his human trickery, man is very intelligent, he intuitively KNOWS that
he cannot say it.
If he says: it is false. He makes a fool of himself.
If he says: it is true, he knows that we will answer, and what distance
does the earth travel in both cases?
And this is frankly intolerable for relativistic believers who see their
God naked, and his two laughable German prophets.

For this, you will have to check the math in my table.
There is no such thing as a 4c approach.
You've made some stupid math errors.
Really, really STOOPID math errors.

[Richard Hachel]

Post by ProkaryoticCaspaseHomolog
For this, you will have to check the math in my table.
There is no such thing as a 4c approach.
You've made some stupid math errors.
Really, really STOOPID math errors.

If it is to intervene on subjects as obvious as relativistic apparent
speeds, by having alcoholic remarks, you should not feel obliged to post.
We are basically telling you that apparent speeds do not exist, or are not
what I say.
So please tell me at what apparent speed you observe a rocket or any other
mobile coming towards you with a speed of v=0.8c.
I am beginning to wonder if the level of the posters of
sci.physics.relativity is not even lower than that of the French posters
of fr.sci.physique

R.H.