Re: [SR] Usefulness of real velocities in accelerated relativistic frames of reference.

Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :

According to SR:
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 The rocket will pass Tau Ceti at the terrestrial time:
  t = √((d/c)²+2⋅d/a) = 12.9156 y

 Absolutely.
 

The proper time of the rocket when it passes Tau Ceti is:
  τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

 Here is the error.  Physicists use the Minkowskian metric and it is not correct.
You have to use (I understand that this is confusing) the Newtonian equation.
Physicists absolutely need to understand something:
the rocket is AT REST in its frame of reference, and it is from start to finish.
Everything happens, for her, as if the surrounding space were accelerating by 10m/s² around her.
The big complaint (because it's confusing when you don't have 40 years of thinking on the subject like me) is to say: "Yes, but the more time passes, the more the surrounding space will contract. for the rocket, and therefore Vr=a.Tr is no longer valid".
But we forget SEVERAL things.
Firstly the space to travel does not contract but expands, because we are in a longitudinal journey and if we must correctly use the =l.sqrt(1-Vo²/c²)/(1+cosµ.Vo/c) ), we see that there is dilation of the anterior space every second and according to the speed reached.
But that's not all, this would not change the fact that the equation would not be constant all the same (since the varies as a function of time).
But we forget that there will be an inverse correction due to the expansion of the apparent speeds.
We will therefore have an instantaneous distance to cover as
D'=D.sqrt[(1+Vo/c)/(1-Vo/c)]
but at the same time, the speed of approach of the star is like:
Vapp=Vo/(1+cosµ.Vo/c) or here Vapp=Vo/(1-Vo/c) at each moment of the evolution.
We know that tau=D'/Vapp (tau is the time specific to each instant necessary for the rocket to reach the star in the event of a sudden cessation of acceleration).
However, at each instant tau=To.sqrt(1-Voi²/c²).
It follows that what we gain on one side, we lose on the other, and that the acceleration is constant and that there is no relativistic correction to be made in the present case.
We can therefore keep Vr=a.Tr without any problem.

 The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:
   v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y

 No.
 Vo=0.980c
 (Vr=5.024c)

 The average speed is:
   vₘ = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t = 0.9291 ly/y

 Là, oui.

Note that  vₘ/v > 1/2

 Vom/Voi > 1/2 absolutely.
 but Vrm/Vri = 1/2
 R.H.