Re: Langevin and Doppler effects...

Den 06.09.2024 22:57, skrev Richard Hachel:

Le 06/09/2024 à 21:03, "Paul.B.Andersen" a écrit :

Den 05.09.2024 23:45, skrev Richard Hachel:

 

There are often few numerical applications in the
developments of relativistic theorists, because this can pose some problems of contradiction, or even absurdity.
>
We will return to the eternal problem, which has scared away all the contributors I have dealt with for 40 years, although the opposite is always said, and that I am the one who would be a wimp.
>
So we return to the problem, which is nevertheless very simple.
>
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).
>
We quickly obtain the following data by the Doppler effect.
>
Stella and Terrence agree to beep every month.

>
>
You will find the answer here:
https://paulba.no/pdf/TwinsByDoppler.pdf
L = 12 ly
c = 1 ly/y
f = 12/y
v = 0.8c
γ = 5/3

 Yes. Very good.
 

It is then interesting to know how many beeps Stella receives on the way there, and how many she receives on the way back.

>
>
Doppler shift when Stella is moving out  Do = √((c−v)/(c+v)) = 1/3

 Absolutely.

Remember that you know this:
The Doppler shift is given by the Relative speed Terrence-Stella.

 

Doppler shift when Stella is moving back Db = √((c+v)/(c-v)) = 3

 Absolutely.

Remember that you know this:
The Doppler shift is given by Relative speed Terrence-Stella.

 

So number of received beeps Nt = Tₛ⋅f⋅Do/2 +  Tₛ⋅f⋅Db/2 = 36 + 324 = 360

 Wonderfull.

Remember that you know this:
Stella receives 36 beeps on her way out and  324 beeps on her way back,
total 360 beeps.

 

Den 07.09.2024 01:18, skrev Richard Hachel:

My question is: Stella sees Terrence in her ultra-powerful telescope moving away from her with an apparent speed of 0.4444c, and for 9 years.
Is it true, or is it not true?

It is irrelevant.
Terrence is moving away from her at the speed v = 0.8c.
That's why Stella will see the Doppler shifted frequency to be
  f₁ = f⋅Do = f⋅√((c−v)/(c+v)) = f/3 = 4 beeps/y
Remember?
What Stella might think Terrence speed appears to be
can have no physical consequences whatsoever.
Only the real speed v has a consequence for the Doppler shift.

On the way back, for 9 years, she sees Terrence coming back to her for 9 years, with a speed of 4c.
Is it true or is it not true?

It is irrelevant.
Terrence is approaching her at the speed v = 0.8c.
That's why Stella will see the Doppler shifted frequency to be
f₂ = f⋅Db = f⋅√((c+v)/(c-v)) = f⋅3 = 36 beeps/y as you agreed to above.
Remember?
What Stella might think Terrence speed appears to be
can have no physical consequences whatsoever.
Only the real speed v has a consequence for the Doppler shift.
------
Remember that you know:
Stella receives 36 beeps on her way out and  324 beeps on her way back,
total 360 beeps, _because_ her speed relative to Terrence is  0.8c,
and _because_ Stella's proper time each way  is 9 y, and _because_
Terrence is sending 12 beeps each year.
If Stella has some confused idea of what Terrence speed appears
to be, then this is a fantasy  with no physical consequences
whatsoever. It is utterly irrelevant.
So what was your point with your "apparent speeds"?

If it is true why not say it?
In his human trickery, man is very intelligent, he intuitively KNOWS that he cannot say it.
If he says: it is false. He makes a fool of himself.
If he says: it is true, he knows that we will answer, and what distance does the earth travel in both cases?
And this is frankly intolerable for relativistic believers who see their God naked, and his two laughable German prophets.

Gobbledegook.
I see you have failed to see the point with the Doppler method.
It is that neither Stella nor Terrence have to know anything
about the speed or times of the other twin. All they have to know
is that both are sending 12 peeps a year.
All they have to do is to count the peeps.
So far, we have not calculated how many beeps
Terrence will receive. We have only said it must be
216 beeps because Stella is emitting 12 beeps per year for 18 years.
This is true, but it is circular, and doesn't prove that Doppler
shift method give this result.
Here we will show that it does:
After 9 years Stella emits the last beep before turnaround.
Terrence time is L/v = 15 years when Stella emits the last beep.
Terrence will receive this beep a time L/c  = 12 years later.
So Terrence will receive the Doppler shifted frequency 4 peeps/y
for L/v + L/c = 27 y, so he will receive 27⋅4 beeps = 108 beeps
emitted from Stella on her way out.
It is then obvious that Terrence must receive the Doppler shifted
frequency 36 beeps/y for L/v - L/c = 3 y, so he will receive
3⋅36 beeps = 108 beeps emitted from Stella on her way back.
To sum it up:
All Terrence and Stella have to do, is to count the received beeps.
Stella will receive 360 beeps.
 From this she can deduce that Terrence must
have aged 360/12 y = 30 years.
Terrence will receive 216 beeps.
 From this he can deduce that Stella must
have aged 216/12 y = 18 years.
Is the Langevin paradox solved?
--
Paul
https://paulba.no/