• #### Proper time in SR.

From Richard Hachel@21:1/5 to All on Tue Oct 17 13:43:39 2023
I have posted several times, here or elsewhere, details on the equality of proper tenses in certain cases.

I said, speaking of the two travelers from Tau Ceti (variant of the
problem where the traveler was alone), that if the observable
(terrestrial) times of the two protagonists were the same (they leave
together, cross the same space of 12 ly, and arrive at the same time, then their own times would be equal.

One of the travelers is in uniform Galilean motion (Vo=0.9291c), the
other is in accelerated motion (a=1.052 ly/y²).

This seems to surprise relativistic physicists.

But if we take the following example: we imagine a large circle of 24
light years, and we make a rocket travel the right half-circumference at
speed 0.9291c, and the left half-circumference
at the same speed speed by another rocket.

The two rockets will meet at the bottom.

To=12,914 years.

Tr=4,776 years.

The answer is obvious, and it will be the same for both rockets since they
do the same thing.

However, between them, the rockets undergo separations, rapprochements, accelerations, changes in speed, and constantly, the two rockets, each,
will have stronger chronotropies than the other rocket.

We can also do the same thing with rockets accelerated on the same course.
Each rocket has an acceleration a=1.052ly/y² and they are at the bottom
of the diagram at the same time.
Here again the proper times will be the same (there is no need to
demonstrate this since they do the same thing with each other).
But throughout the journey, there will be a constantly positive
chronotropy for one rocket looking at the other, with reciprocal effect.

It is only a variation of Langevin's traveler where a paradox seems to
arise at the end.

But that's not the question.

What happens if we take a rocket from the first example on the left, and a rocket from the second example on the right?

Doctor Hachel affirms that the proper times will again be equal, and he
still masters relativistic kinematics much better than the average person.

He says that the way scientists measure improper times in accelerated
frames is correct To=(x/c).sqrt(1+2c²/ax) but NOT the way they measure
proper times in this kind of frame.

Hence the fact that the proper times will not be equal for the two
rockets, according to physicists.

Who is lying? Who tells the truth?

R.H.

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• From Paul B. Andersen@21:1/5 to All on Tue Oct 17 20:53:09 2023
Den 17.10.2023 15:43, skrev Richard Hachel:
I have posted several times, here or elsewhere, details on the equality
of proper tenses in certain cases.

I said, speaking of the two travelers from Tau Ceti (variant of the problem where the traveler was alone), that if the observable
(terrestrial) times of the two protagonists were the same (they leave together, cross the same space of 12 ly, and arrive at the same time,
then their own times would be equal.

I suppose "their own time" means their proper time.

One of the travelers is in uniform Galilean motion (Vo=0.9291c), the
other is in accelerated motion (a=1.052 ly/y²).

c = 1 ly/y
d = 12 ly

Accelerated traveller:
a = 1.052 ly/y²
terrestrial time: t₁ = √((d/c)²+2d/a) = 12.9156 y
proper time: τ₁ = (c/a)⋅arsinh(at₁/c) = 3.1403 y

Inertial traveller:
v = 0.9291 ly/y
γ = √(1−v²/c²) = 2.703955
terrestrial time: t₂ = d/v = 12.9157 y
proper time: τ₂ = t₂/γ = 4.7766 y

This seems to surprise relativistic physicists.

The proper times are not equal.

No point in going on.

Who is lying? Who tells the truth?

R.H.

You are not lying because you don't know better.
But you are wrong.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Tue Oct 17 19:09:35 2023
Le 17/10/2023 à 20:52, "Paul B. Andersen" a écrit :
Den 17.10.2023 15:43, skrev Richard Hachel:
I have posted several times, here or elsewhere, details on the equality
of proper tenses in certain cases.

I said, speaking of the two travelers from Tau Ceti (variant of the
problem where the traveler was alone), that if the observable
(terrestrial) times of the two protagonists were the same (they leave
together, cross the same space of 12 ly, and arrive at the same time,
then their own times would be equal.

I suppose "their own time" means their proper time.

One of the travelers is in uniform Galilean motion (Vo=0.9291c), the
other is in accelerated motion (a=1.052 ly/y²).

c = 1 ly/y

Yes.

d = 12 ly

Absolutely.

Accelerated traveller:
a = 1.052 ly/y²

Sure.

terrestrial time: t₁ = √((d/c)²+2d/a) = 12.9156 y

It's that I said.

proper time: τ₁ = (c/a)⋅arsinh(at₁/c) = 3.1403 y

This formula is not correct.

This is the one that is taught.

But it is not correct.This formula is not correct.

This is the one that is taught.

But it is not correct.

Inertial traveller:
v = 0.9291 ly/y

Yes.

γ = √(1−v²/c²) = 2.703955
terrestrial time: t₂ = d/v = 12.9157 y

Yes.

proper time: τ₂ = t₂/γ = 4.7766 y

Yes.

This seems to surprise relativistic physicists.

The proper times are not equal.

The proper times, in this particular case, must be equal.

I specify, IN THIS PARTICULAR CASE.

Tr=4.776 in both cases; for both rockets.

No point in going on.

Who is lying? Who tells the truth?

R.H.

You are not lying because you don't know better.
But you are wrong.

I refer the accusation to you.

Je vous remercie de votre réponse, qui clarifie les choses, et montre où
nous ne sommes plus d'accord du tout.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Tue Oct 17 22:04:26 2023
Den 17.10.2023 21:09, skrev Richard Hachel:
Le 17/10/2023 à 20:52, "Paul B. Andersen" a écrit :
Den 17.10.2023 15:43, skrev Richard Hachel:
I have posted several times, here or elsewhere, details on the
equality of proper tenses in certain cases.

I said, speaking of the two travelers from Tau Ceti (variant of the
problem where the traveler was alone), that if the observable
(terrestrial) times of the two protagonists were the same (they leave
together, cross the same space of 12 ly, and arrive at the same time,
then their own times would be equal.

I suppose "their own time" means their proper time.

One of the travelers is in uniform Galilean motion (Vo=0.9291c),
the other is in accelerated motion (a=1.052 ly/y²).

c = 1 ly/y

Yes.

d = 12 ly

Absolutely.

Accelerated traveller:
a = 1.052 ly/y²

Sure.
terrestrial time:  t₁ = √((d/c)²+2d/a) = 12.9156 y

It's that I said.

proper time:  τ₁ = (c/a)⋅arsinh(at₁/c)  = 3.1403 y

This formula is not correct.

This is the one that is taught.

But it is not correct.This formula is not correct.

This is the one that is taught.

But it is not correct.

It is what SR predicts.

What SR predicts is not a matter of opinion,
it is a matter of fact.

And the fact is that "this formula" follows with
mathematical necessity from the postulates of SR.

Inertial traveller:
v = 0.9291 ly/y

Yes.
γ = √(1−v²/c²) = 2.703955
terrestrial time: t₂ = d/v = 12.9157 y

Yes.
proper time: τ₂ = t₂/γ = 4.7766 y

Yes.

This seems to surprise relativistic physicists.

The proper times are not equal.

The proper times, in this particular case, must be equal.

Not according to SR.

If you are talking about the predictions of a different
theory than SR, I am not interested.

So don't tell me what your theory is.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Tue Oct 17 21:12:16 2023
Le 17/10/2023 à 22:03, "Paul B. Andersen" a écrit :
Den 17.10.2023 21:09, skrev Richard Hachel:

But it is not correct.

It is what SR predicts.

That's what I said.

SR predictions (in this precise case) are not correct.

The proper times are not equal.

The proper times, in this particular case, must be equal.

Not according to SR.

That's what I said.

SR predictions (in this precise case) are not correct.

If you are talking about the predictions of a different
theory than SR, I am not interested.

I'm not talking about a different theory, but a different understanding of things.

So don't tell me what your theory is.

:))

You do not have to be afraid, Paul.

I don't bite anyone.

R.H.

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• From Volney@21:1/5 to Richard Hachel on Tue Oct 17 16:31:57 2023
On 10/17/2023 3:09 PM, Richard Hachel wrote:

Je vous remercie de votre réponse, qui clarifie les choses, et montre où nous ne sommes plus d'accord du tout.

Why are you telling Paul that his hovercraft is full of eels?

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• From Richard Hachel@21:1/5 to All on Tue Oct 17 21:25:20 2023
Le 17/10/2023 à 22:31, Volney a écrit :
On 10/17/2023 3:09 PM, Richard Hachel wrote:

Je vous remercie de votre réponse, qui clarifie les choses, et montre où >> nous ne sommes plus d'accord du tout.

Why are you telling Paul that his hovercraft is full of eels?

I won't go that far.

I'm not saying that Paul is wrong about everything, or about many things.

Je trouve ses interventions très pertinentes et le plus souvent très intéressantes.

I just noticed some errors in the way he was taught the theory of
relativity, which he reproduced in some pages of his pdfs.

For example when it performs a rotation instead of a translation, or when
it gives the instantaneous speed of an accelerated object, or the proper
time of an accelerated object.

He uses the theory as it is taught in universities, but not as I think it should be taught.

Far too many things are imprecise or very poorly understood in this
theory.

I see with amazement that the big names themselves do not have clear ideas
on this and that they apply a half-crazy geometry, while the one that I
believe to be true is simpler, more logical, and more predictive.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Wed Oct 18 10:00:01 2023
Den 17.10.2023 23:25, skrev Richard Hachel:

He uses the theory as it is taught in universities, but not as I think
it should be taught.

R.H.

Shame on him!

--
Paul

https://paulba.no/

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• From Dono.@21:1/5 to Richard Hachel on Wed Oct 18 01:29:01 2023
On Tuesday, October 17, 2023 at 2:25:23 PM UTC-7, Richard Hachel wrote:

He uses the theory as it is taught in universities, but not as I think it should be taught.

....says one of the village idiots (the others being Dick Hertz , LooLoo and Patycakes Dolan)

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• From mitchrae3323@gmail.com@21:1/5 to All on Mon Oct 23 14:15:36 2023
Time can change in two ways. If it slows down
it can speed back up...
If a space ship speeds up its time slows down.
Then it must slow back down to land on the other world.
So its time speeds back up by being again slow in space...

Mitchell Raemsch

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• From Tom Roberts@21:1/5 to mitchr...@gmail.com on Mon Oct 23 22:54:27 2023
On 10/23/23 4:15 PM, mitchr...@gmail.com wrote:
Time can change in two ways.

No,it cannot.

If it slows down it can speed back up...

No, it does not.

If a space ship speeds up its time slows down.

Nope.

Why do you keep wasting time posting nonsense to the Net? -- your time
would be MUCH better spent studying and learning about physics.

Tom Roberts

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• From Maciej Wozniak@21:1/5 to Tom Roberts on Mon Oct 23 23:20:55 2023
On Tuesday, 24 October 2023 at 05:54:41 UTC+2, Tom Roberts wrote:

Why do you keep wasting time posting nonsense to the Net? -- your time
would be MUCH better spent studying and learning about physics.

Yeah!! Learning, that we're FORCED!!! To THE BEST WAY!!!

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