Re: Langevin and Doppler effects...

Den 09.09.2024 20:54, skrev Richard Hachel:

Le 09/09/2024 à 19:42, "Paul.B.Andersen" a écrit :
 

I will come back to the scenario seen from Stella's rest frame
in a separate post, this is long enough.
 But I will say this:
The scenario described here is physically impossible to do
in the real world. The problem is the abrupt change of velocity
at turnaround. Stella will then have an infinite high acceleration
for zero time. This is a mathematical singularity, which make
it impossible to solve exactly. This is a problem when seen
from Stella's point of view, because what happens to her view
of Terrence's clock when her acceleration is infinite?
 All we have to do to make it solvable is to introduce
a high but finite acceleration for a short time at turnaround.

 But no!
 You don't pay any attention to what I'm telling you.
 You've decided to say that in RR everything is fine, that everything is acceptable, that everything is true, and that you should definitely not say anything.
  From then on, it turns into religious fundamentalism.
 I explained RR like no one had explained it before, and instead of being interested, everyone is playing the monkey by opposing me with anything....
 Pffff...

What are you whining about?

 Where did you see an infinite acceleration?

| Den 05.09.2024 23:45, skrev Richard Hachel:>
|> Stella goes into the stars for a journey of 24 light years.
|> The speed is 0.8c on the way there (12 ly), and 0.8c on the way back (12 ly).
Let's call this example #1.

 In the example I gave, the U-turn takes 40 hours in Terrence's frame of reference (24 in Stella's).

That was a different example.
In Terrence frame Stella travels with the constant speed
0.8 c for 30 years and 40 hours. When Terrence's clock
shows 15 years, Stella goes into a sharp U-turn which
last 40 hours.
Let's call this example #2.

 It's certainly a huge acceleration, but it's not infinite.

Yes.
But this acceleration is only transverse, and doesn't affect the speed.
But it does make it possible to calculate what Terrence's proper time
and position in Stella's frame is, as a function of Stella's time.
----
In both examples is the speed constant, so it trivially simple to find
Stella's proper time.
In example #1:
  Stella's clock at return = 30y/γ = 18y
In example #2:
  Stella's clock at return = (30y 40h)/γ = 18y 24h
-----
Note:
We will calculate what SR say will be measured in Stella's frame.
NOT what will be observed in a telescope.
Example #1:
------------
Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9- y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y
When Stella's clock shows 9+ y, Stella is on her way back.
Terrence will still be at the position x = 7.2 ly,
but will now move at the speed 0.8c towards Stella.
Terrence's clock will now show 30y-5.4y = 24.6y
When Stella's clock shows 18y, Stella is back.
Terrence will now be at the position x = 7.2ly - 0.8c⋅9y = 0 ly.
Terrence clock will now show 24.6y + 5.4y = 30 y
Note this:
When Stella is moving away from Terrence and her clock shows 9y,
Terrence clock will _simultaneously_ in Stella,s frame show 5.4y
When Stella is moving towards Terrence and her clock shows 9y,
Terrence clock will _simultaneously_ in Stella,s frame show 24.6y
Both clocks run at a steady rate, 1 second per second, but
what the two clocks _simultaneously_ will show in Stella's frame
depend on Stella's velocity relative to Terrence.
When Stella abruptly change her direction, her idea of what time is
simultaneous in Terrence frame changes from  5.4y to 22.8y,
But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that changes.
Example #2:
------------
Terrence will move away at the speed 0.8c in Stella's frame.
When Stella's clock shows 9y, Terrence will be at the position
x = 0.8c⋅9h = 7.2 ly in Stella frame. (Stella is at x = 0ly)
Terrence clock will run at the rate 1/γ so it will show 9h/γ = 5.4y
When Stella's clock show 9y 12h, Terrence will be at the position
x = 12 ly in Stella's frame. Terrence clock will now show 15 y.
When Stella's clock shows 9y 24h, Terrence will now be at the position
x = 0.8c⋅(9y 24h) = (7.2 ly + 19.2 lh)  in Stella's frame.
Terrence clock will now show (30y 40h) - (9y 24h)/γ = 24.6y 25.6h
When Stella's clock shows 18y 24h, Stella is back.
Terrence will now be at the position x = 0 ly.
Terrence clock will now show (24.6y 25.6h) + (9y 24h)/γ = 30y 40h
--------------
Note that when Stella's clock show 9y, Terrence clock simultaneously
in Stella's frame show 5.4y, and when Stella's clock show 9y 12h, Terrence clock simultaneously in Stella's frame show 15y.
That means that Terrence clock runs (15-5.5)y/12h = 7012.8 times
faster than Stella's clock during the first part of the U-turn.
But nothing happens to Terrence clock, it runs normally.
It is _only_ Stella's idea of what is simultaneous that
make it appear that Terrence clock run fast.
(It may be typos)
--
Paul
https://paulba.no/