Re: Langevin's paradox again

Liste des GroupesRevenir à p relativity 
Sujet : Re: Langevin's paradox again
De : relativity (at) *nospam* paulba.no (Paul.B.Andersen)
Groupes : sci.physics.relativity
Date : 14. Jul 2024, 00:34:44
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v6v2l5$3pri7$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11
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Den 12.07.2024 15:55, skrev Richard Hachel:
Le 12/07/2024 à 15:24, "Paul.B.Andersen" a écrit :
Den 11.07.2024 20:29, skrev Richard Hachel:
>
< snip whining and heavy breathing >
>
>
I repeat, during the U-turn, nothing happens at all on the TIME side.
>
Except that TIME is passing?
 Yes.
For Stella : tau(stella)=24 hours  tau(Terrence)=40 hours
 For Terrence : tau(Terrence)=40 hours  tau(Stella)=24 hours
Why do you write French?
But we have Google translator.

These values ​​are insignificant and useless to calculate to understand correctly the Langevin paradox, which has no connection with the acceleration phases, none.
Langevin's example was a constant speed out, instant turnaround,
and constant speed back. Instant turnaround means infinite acceleration, which is impossible, and will be a mathematical singularity.
----
Your scenario is trivially simple. Measured in Terrence's inertial
rest frame Stella is moving at the constant speed v = 0.8c, γ = 1/0.6.
Since Terrence's proper time for the whole journey is τₜ = 70 hours,
Stella's proper time must be τₛ = τₜ/γ = 42 hours.
If the rate of Terrence's clock is constant fₜ, the rate of Stella's
clock is constant fₛ = fₜ/γ (Stella's clock runs slow).
As long as Stella's speed is constant, the shape of Stella's path is irrelevant, it can be circular, elliptic, partly straight and partly
curved, or whatever shape you might like. (But no sharp corners, he path
must be an analytic function.)
Of course Stella must accelerate at some part of the journey in
order to get back to Terrence, but when the speed is constant
the acceleration is transversal and will have no effect on her
proper time.

 Everything is played out during the simple Galilean phases. All.
Galilean phase?
If there were a "Galilean phase" Terrence's and Stella's clocks
would always show the same.

 You must simply use the formula Tapp=Tr.(1+cosµ.Vo/c)/sqrt(1-Vo²/c²) each time, and for all measurements.
This is the relativistic Doppler shift equation.
We can use Doppler shift to calculate how much Terrence
and Stella will age.
See:
https://paulba.no/pdf/TwinsByDoppler.pdf

The concordance is mathematical.
We absolutely don't need anything else.
It is true that we don't need anything but Doppler shift
to calculate how much Terrence and Stella will age.
But when Stella is accelerating, the method will be rather
cumbersome.
So it is better to do it like this:
https://paulba.no/pdf/TwinsByMetric.pdf

The rest is pure nonsense from relativistic physicists, who, understanding that pouic, say anything (time-gap, watches that panic, and other joys).
Their ridiculous space-time à la Minkowski plunged them 120 years into deep theoretical darkness.
R.H.
Whining again, Richard?
--
Paul
https://paulba.no/

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