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

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Sujet : Re: [SR] Usefulness of real velocities in accelerated relativistic frames of reference.
De : relativity (at) *nospam* paulba.no (Paul B. Andersen)
Groupes : sci.physics.relativity
Date : 18. Mar 2024, 22:14:04
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <utaao2$2akmq$1@i2pn2.org>
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User-Agent : Mozilla Thunderbird
Den 18.03.2024 10:27, skrev Richard Hachel:
Le 17/03/2024 à 14:42, "Paul B. Andersen" a écrit :
>
Since the equations:
  Speed of rocket in inertial frame:             Vr=a.Tr
  Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
  are valid _only_  in Newtonian Mechanics with Galilean relativity,
  then the theory which is consistent with said equations
  is Newtonian Mechanics.
 The equations I give, if written correctly, are valid in both systems.
But you have to write them correctly.
For example if I write, in the Newtonian system,
v=a.t
This is valid.
In the same Newtonian system, we can also write:
v_m=(1/2)v_i
 We agree on this, and I don't think, even regarding the craziest posters (Python example), anyone will come and contradict.
Quite.
We can now review the journey to Tau Ceti.
Both Earth and Tau Ceti are considered to be inertial.
A rocket is stationary on  Earth, When its clock show  τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².
a = 10 m/s² = 1.05265 ly/y/y   c = 1 ly/y  d = 12 ly
According to your equations v = a⋅t and vₘ = a⋅t/2:
===================================================
d =  ∫a⋅t⋅dt + 0 ly = a⋅t²/2  => t = √(2⋅d/a)
The rocket will pass Tau Ceti at the terrestrial time:
  t = √(2⋅d/a) = 4.7764 y
The proper time of the rocket when it passes Tau Ceti is:
  τ = √(2⋅d/a) = 4.7764 y
The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:  v = a⋅t = 5.2860 ly/y
The average speed is: vₘ = a⋅t/2 = 2.6430 ly/y
Note that d/vₘ = 4.5403 y < t    Why :-D
According to SR:
================
The rocket will pass Tau Ceti at the terrestrial time:
  t = √((d/c)²+2⋅d/a) = 12.9156 y
The proper time of the rocket when it passes Tau Ceti is:
  τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
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
The average speed is:
   vₘ = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t = 0.9291 ly/y
Note that  vₘ/v > 1/2
--
Paul
https://paulba.no/

Date Sujet#  Auteur
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