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Subject: Re: [SR] Their proper times will necessarily be equal
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Date: Sun, 21 Apr 24 21:31:16 +0000
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From: Richard Hachel   <bts.orange@licorne-bleue.fr>
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Le 20/04/2024 à 19:07, Python a écrit :
>> On 4/13/24 1:36 AM, Richard Hachel wrote:
>>> "If two different observers travel an identical path in equal 
>>> observable times, then their proper times will necessarily be equal.
>> 
>> Yes, of course.
>> 
>> My response differs from others because I interpret your context
>> differently.
> 
> You interpret path as a path in space-time while Hachel is considering
> path in space only (so it is true in only ONE reference frame)
> 
> For instance, he claims that if two travelers would go from Earth
> to Tau Ceti, one at constant velocity w.r.t Earth the other one
> with a constant acceleration both will measure the same (proper)
> time for their trips.

Yes, this is entirely correct.

If two different protagonists use different frames of reference to join 
Tau Ceti (12al), and if we notice that in the terrestrial frame of 
reference the travel times are similar and the path is similar (direct 
path), then the two proper times will be similar.
This is the case of a trip by Stella in Galilean mode at constant speed 
Vo=0.929c, and of a trip in accelerated mode by Bella 
(10m/s²=1.052ly/y²).
The duration of the trip, which is also very well calculated by my 
relativistic theoretician friends around the world, will be To=x/Vo in the 
first case, and To=(x/c).sqrt(1+2c²/ax) in the second case, i.e. 12,915 
years.
But this duration will also be the same for the two rockets with regard to 
their own time. Tr=4.776 years.
Physicists tell me that while this is true for Stella, it is not true for 
Bella.
However, here, it is they who are mistaken and misuse space-time concepts 
in an accelerated environment.
On the other hand, be careful, for this to be true, and for it to remain 
true, two clarifications:
1. You must only use one reference frame during the entire trip.
2. the accelerated twin must go into “native” state, i.e. at rest.

R.H.