Re: [SR] David's little pebble against Goliath

Dr Richard "Hachel" Lengrand à écrit:

Le 31/07/2022 à 15:16, "Paul B. Andersen" a écrit :

d'=d.sqrt(1-v²/c²)/(1+cosΦ.v/c)

>
What is Φ?

  ? ? ?
  ? ? ? ? ?
   ? ? ? ? ? ? ? ? ?
  Je te supplie d'arrêter tes conneries.

Tu ne définis, comme d'habitude, aucun des termes que tu utilises et qui
ne sont pas parmi les conventions usuelles (genre x, c ou v) et tu
t'étonnes que ton charabia soit incompréhensible?
Et quand tu te prends une réponse circonstanciée tu n'as d'autre
réaction que de parler de ta bite et du fait que tu ne la liras
pas (entre autres grossièretés ridicules), bref tu te débines, comme
d'habitude.
la réponse complète de Paul Andersen, pour info:

   Den 31.07.2022 11:51, skrev Richard Hachel:

David's little pebble against Goliath
>
And it can be summarized like this:
- the theory is that the star traveler will age 9 years on the way out and 9 years on the way back.
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- We assume that at 0.8c, the distance of 12 bar will be contracted and will measure 7.2 bar.

 bar = light year?
 

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- We assume that if the traveler has a good telescope, when he returns, he will see the earth approaching him with an apparent speed of 4c.

 This is rather irrelevant.
There is no way the traveller can visually observe the Earth
approaching at 4c.
 It is true that you can define an apparent speed = v/(1-v/c) = 4c.
But you would never be able to observe this in
the real world because it usually is impossible to
observe the apparent speed of the approaching object.
 And it should be blatantly obvious to you that the speed
of the approaching object is v, and not v/(1-v/c).
 There is however possible to observe an apparent transverse speed:
 Vapp = v⋅sinθ/(1−(v/c)⋅cosθ)
where v is the speed of the object and θ is the angle of
the object's velocity relative to you.
(See superluminal jets.)
 

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On that, everyone agrees.
>
How can a traveler who observes for nine years the earth returning towards him at an apparent speed of 4c, can he see it evolve on 7.2al?

  What he _can_ visually observe is this:
(Hard to observe with a telescope, but it could be
a transmitter on the Earth sending video of the Earth clock.)
 On the way out he will see the Earth clock run slow
by sqrt((1-0.8)/(1+0.8)) = 1/3.
So when he turns around when his clock shows 9 years,
he will see the Earth clock shows 3 years.
 On the way back he will see the Earth clock run fast
by sqrt((1+0.8)/(1-0.8)) = 3.
So when he is back after another 9 years on his clock,
he will see the Earth clock showing (3+3*9) years = 30 years.
 Which is in accordance with the Lorentz transform.
2*9/sqrt(1-0.8*0.8) years = 30 years
 

>
Doctor Hachel explained that there was a dilation of the distances in this precise case. Logical dilation and predicted by the Poincaré-Lorentz equations, and that this dilation was like d'=d.sqrt(1-v²/c²)/(1+cosΦ.v/c)

 What is Φ?
 

>
Either here: 36al.

 al?
 Let us place a body at the turning point.
 This body is stationary in the Earth frame (x,t) at x = x₁ = 12 ly.
The Earth is at x = x₀ = 0.
The distance between Earth and the body is 12 ly in the Earth frame.
 Now we want to find the distance in a frame (x',t') that is moving
at v relative to the Earth frame. Let x = x' = 0 at t = t' = 0.
 We want to find the distance between the Earth and the body
at the time t' = 0 in the moving frame.
 t' = γ(t₁ -  x₁⋅v/c²) = 0  ->  t₁ = x₁⋅v/c²
x'₁ = γ(x₁ - v⋅t₁) = γ⋅x₁(1 - v²/c²) = x₁⋅√(1 − v²/c²)
 with x₁ = 12 and v = 0.8c , x'₁ = 7.2 ly, t₁ = 9.6 y
 The distance between Earth and the body is 7.2 ly in the moving frame.
 Note that when t' = 0 in the moving frame,
a clock on the body which is in sync with the a clock
on the Earth would show 9.6 years.
 Time dilation and Lorentz contraction are two sides
of the same coin.
  -- Paul
 https://paulba.no/