Re: Langevin and Doppler effects...

Den 07.09.2024 23:42, skrev Richard Hachel:

 < snip >

Here we go again:

We know that Terrence will observe Stella moving at Vapp=0.4444c,
on the way there, and for 27 years (he receives half of the beeps during this period).

READ THIS, AND THINK!
The only thing Terrence must know about Stella is that she is
emitting  12 beeps/y.
A year after Stella left, Terrence will know that he receives 4 beeps/y.
Since Stella is emitting 12 beeps/y, the Doppler shift is 1/3.
So Terrence knows that Stella is moving away at the speed 0.8c.
Why would Terrence say that Stella appears to move away
at the speed 0.4444c when he knows that her speed is 0.8c? :-D
When Terrence sees that the received beep frequency suddenly changes,
Terrence clock shows 27 y, so he has received 4*27 = 108 beeps.
Stella's clock shows 108/12 = 9 year at the time of emission of
the beep Terrence received when his clock showed 27 y.

What distance will he measure? x=0.4444c*27=12al.

Since Stella is moving away at the speed 0.8c, and her moving
clock shows 9 y, Terrence will know that his own stationary
clock simultaneously must have shown 9h⋅γ = 9⋅5/3 y = 15 y.
("moving clocks run slow" , remember?)
Terrence will obviously, as opposed to Hachel, understand
that his clock didn't show 27h at the instant when Stella's
clock showed 9h. He knows that his clock showed 15 y, and
the light used 12 y to reach him when his clock showed 27 y.
Since Stella is travelling at 0.8c in Terrence's rest frame,
and his stationary clock in said frame shows 15 h at the instant
when Stella's clock shows 9h, the distance to Stella is
15h⋅0.8c = 12 ly.
So Terrence knows that Stella's turning point is 12 ly from him
in his rest frame.

 On the way back, he will measure x=4c*3=12

(Is Stella moving at 4 times the speed of light for 3h? :-D)
One year after the instant when Stella's clock showed 9h,
when Terrence's clock shows 28 y, Terrence will see that he is
receiving 36 peeps/y and that the Doppler shift thus is 3.
So Terrence knows that Stella is approaching at the speed 0.8c.
And since the turning point is 12 ly from him, he knows that
his clock will advance 12ly/0.8c = 15 y during Stella's travel back.
Since his clock showed 15h at turnaround, it will show 30h
when Stella is back.

 It is undeniable and mathematical.

Indeed.
All I said above is irrefutable true according to
The Special Theory of Relativity.
No paradox, no mystery.
-----------------------------

But for Stella, and this is the great key to the paradox, and the brilliant explanation, the distance traveled by the earth will NOT be 12 al, neither on the way there, nor on the way back, because it is not in the same frame of reference.

I will come back to the scenario seen from Stella's rest frame
in a separate post, this is long enough. (It probably won't be today.)
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.
Back tomorrow.
--
Paul
https://paulba.no/