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Subject: Re: [SR] Usefulness of real velocities in accelerated relativistic frames 
 of reference.
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Date: Thu, 14 Mar 24 16:18:32 +0000
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From: Richard Hachel <r.hachel@tiscali.fr>
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Le 14/03/2024 à 15:02, "Paul B. Andersen" a écrit :
> Den 14.03.2024 03:09, skrev Richard Hachel:

> Contradicting fact:
> -------------------
> So this is wrong.
> You can see the correct derivation here:
> https://paulba.no/pdf/TwinsByMetric.pdf
> See chapter 2.3, equation (15)
> 
> Vr(t) = a⋅t/√(1+(a⋅t/c)²)
> 
> Note that:
>   Vr → a⋅t when t → 0
>   Vr → c   when t → ∞
> 
> 
> Your problem is that you do not understand the difference
> between proper acceleration of the rocket, and the rocket's
> coordinate acceleration in the inertial frame.
> 
> If A is the coordinate acceleration in K, we have:
> 
> A = dVr/dt = a/(√(1+(a⋅t/c)²))³
> 
> Note that:
>   A → a when t → 0
>   A → 0 when t → ∞
> 
> So  Vr(t) = ∫(from 0 to t)A⋅dt = a⋅t/√(1+(a⋅t/c)²)
> 
> You claim:
> According to SR is the average speed of the rocket Vm(t) = Vr(t)/2
> =====================================================================
> 
> Contradicting fact:
> -------------------
> This is wrong.
> 
> Vr(t) = a⋅t/√(1+(a⋅t/c)²)
> 
> The average speed Vm at the time t is:
> Vm = (integral from t=0 to t=t of Vr(t)dt)/t
> Vm = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t
> 
> Note that:
>   Vm → a⋅t/2 when t → 0
>   Vm → c     when t → ∞
> 
> So:
>   Vm/Vr  → 1/2  when t → 0
>   rm/Vr  → 1    when t → ∞
> 
> So for any t > 0   Vm > Vr/2
> 
> It is not possible to make SR predict anything else!
> ====================================================

You don't understand anything I'm telling you...

In these conditions, it is very difficult to discuss.

R.H.