Paul B. Andersen :
https://paulba.no/pdf/TwinsByDoppler.pdf
Well.
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.
[snip already debunked claims]
Le 05/09/2024 à 23:45, M.D. Richard "Hachel" Lengrand a écrit :
Paul B. Andersen :
https://paulba.no/pdf/TwinsByDoppler.pdf
Well.
There are often few numerical applications in the
developments of relativistic theorists, because this can pose some
problems of contradiction, or even absurdity.
You can be sure that we have no problems with what you call
"numerical applications" (funny gallicism). This is a fetish of
yours, as you fight with basic equations and run like a chicken
when confronted :-)
Still, contradiction and absurdity are on YOUR side Richard.
Paul B. Andersen :
https://paulba.no/pdf/TwinsByDoppler.pdf
Well.
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.
It is clear that, for Stella, on the way there, as on the way back, she
will beep
108 times (216 times in total). >
SO, Terrence will receive 216 beeps.
For his part, Terrence will beep for 30 years. He will therefore send
360 beeps.
Except that if Terrence receives the same number of beeps on the way
there as on the way back (108 each time), Stella receives many more
beeps on the way back than on the way there.
It is then interesting to know how many beeps Stella receives on the way there, and how many she receives on the way back.
Once this is done, we can still progress in understanding things.
We just have to not be afraid, and have a playful spirit.
R.H.
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
But since you never read anything, I will repeat it here.
It is clear that, for Stella, on the way there, as on the way back, she
will beep
108 times (216 times in total). >
SO, Terrence will receive 216 beeps.
OK. Stella has aged Tₛ = 2L/vγ = 18 y
Number of received beeps Ns = Tₛ⋅f = 216
For his part, Terrence will beep for 30 years. He will therefore send
360 beeps.
OK
Except that if Terrence receives the same number of beeps on the way
there as on the way back (108 each time), Stella receives many more
beeps on the way back than on the way there.
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
Doppler shift when Stella is moving back Db = √((c+v)/(c-v)) = 3
So number of received beeps Nt = Tₛ⋅f⋅Do/2 + Tₛ⋅f⋅Db/2 = 36 + 324 =
360
Once this is done, we can still progress in understanding things.
We just have to not be afraid, and have a playful spirit.
Terrence will age C = 2L/v = 30y
Am I guessing wrong when I guess that you will present us with
the paradox that Terrence will receive
Tₜₛ⋅f⋅Do/2 + Tₜₛ⋅f⋅Db/2 = 60 + 540 = 600 beeps ? :-D
On Thu, 5 Sep 2024 21:45:22 +0000, Richard Hachel wrote:
We just have to not be afraid, and have a playful spirit.
There is no contradiction.
Terrance receives the exact number of beeps that Stella has transmitted.
Stella receives the exact number of beeps that Terrance has transmitted.
This is TRIVIAL to demonstrate using a spacetime diagram.
No matter what path through spacetime Stella follows, this must be true.
Don't bother with my table. Just LOOK at my Figure 4-4. https://en.wikipedia.org/wiki/Special_relativity#Twin_paradox
For this, you will have to check the math in my table.
There is no such thing as a 4c approach.
You've made some stupid math errors.
Really, really STOOPID math errors.
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.
Doppler shift when Stella is moving back Db = √((c+v)/(c-v)) = 3
Absolutely.
So number of received beeps Nt = Tₛ⋅f⋅Do/2 + Tₛ⋅f⋅Db/2 = 36 + 324 = 360
Wonderfull.
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?
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?
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.
Is the Langevin paradox solved?
Le 07/09/2024 à 21:14, "Paul.B.Andersen" a écrit :
Is the Langevin paradox solved?
Absolutely not.
The key to the problem is not explained by the time gap, but not the space-zoom.
< snip >
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).
What distance will he measure? x=0.4444c*27=12al.
On the way back, he will measure x=4c*3=12
It is undeniable and mathematical.
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.
All we have to do to make it solvable is to introduce
a high but finite acceleration for a short time at turnaround.
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...
Where did you see an infinite acceleration?
Stella goes into the stars for a journey of 24 light years.(12 ly).
The speed is 0.8c on the way there (12 ly), and 0.8c on the way back
In the example I gave, the U-turn takes 40 hours in Terrence's frame of reference (24 in Stella's).
It's certainly a huge acceleration, but it's not infinite.
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?
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.
Le 11/09/2024 à 22:19, Paul.B.Andersen a écrit :
Den 09.09.2024 20:54, skrev Richard Hachel:
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)
Le 11/09/2024 à 22:18, "Paul.B.Andersen" a écrit :
Den 09.09.2024 20:54, skrev Richard Hachel:
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.
Same as for the previous post.
You use observers M1 and M2 and not Terrence himself.
Here, Terrence always remaining in the same inertial frame of reference,
M1 and M2 are the same external reference point.
But it is not Terrence.
Of course, I may give you the impression that I am quibbling, but that
is not the case.
Things must be said so that it is both infinitely beautiful and
infinitely true.
The measurements that you give here, with this concept M (which is not
known since you really think that you are talking about Stalla, then Terrence, and not very distant lateral points) are not beautiful, even
if, for the moment, they are true.
The blow of the baton can only come later.
When you are going to talk about the same things, but with accelerated
frames of reference, it will no longer be beautiful or true.
But for Stella, and this is the great key to the paradox, and thebrilliant explanation, the distance traveled by the earth will NOT be 12
This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation". If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
As I said:
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.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".
Le 12/09/2024 à 16:00, "Paul.B.Andersen" a écrit :
This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation". >> If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
As I said:
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.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper
times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".
Le 12/09/2024 à 16:41, Richard Hachel a écrit :
Le 12/09/2024 à 16:00, "Paul.B.Andersen" a écrit :
This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation".
If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
As I said:
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.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper >>> times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".
t'es vraiment con comme un balai : tu as foutu toute ta "réponse" en ... signature.
you are stupid as stone: you put your entire "answer" in ... your signature.
Trump is inspiring you, I'd say :-)
Le 12/09/2024 à 17:32, Python a écrit :
Le 12/09/2024 à 16:41, Richard Hachel a écrit :
Le 12/09/2024 à 16:00, "Paul.B.Andersen" a écrit :
This is the statement I responded to.
It was YOU who introduced Stella's view ("for Stella").
You even called it "the key to the paradox" and "the brilliant explanation".
If Stella shall return to Terrence, she _must_ accelerate during
her journey, so Stella's frame is always an accelerated frame of
reference (in flat spacetime).
But it isn't beautiful, even if it true.
Stella's view is rather pointless.
As I said:
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.
If you change Stella's path in Terrence frame a little,
Stella's view of Terrence may change it a lot, because
Stella's idea of what is simultaneous changes.
(You can change the path without changing Stella's and Terrence's proper >>>> times.)
If we describes the "twin paradox" in Terrence's inertial frame,
Stella's view adds nothing of interest.
So we can do without Stella's view of Terrence.
It is certainly not "the key to the paradox".
t'es vraiment con comme un balai : tu as foutu toute ta "réponse" en ...
signature.
you are stupid as stone: you put your entire "answer" in ... your signature. >>
Trump is inspiring you, I'd say :-)
Trump 2024 !!!
R.H.
Le 12/09/2024 à 16:00, "Paul.B.Andersen" a écrit :
I have shown you Terrence's, speed, position and proper time
measured in Stella's rest frame, as a function of Stella's time.
It is quite simple:
During Stella's time 0y -> 9y, Stella is inertial and so is her frame.
MEASURED IN STELLA'S FRAME, Terrence is moving away at the speed 0.8c, his clock runs slow by 1/γ, and his position is x = 0.8c⋅t_Stella'
So when Stella's clock show 9h, t_Terrence = 5.4y and x = 7.2 ly
During Stella's time 9y -> 9y + 24h, Stella is accelerating towards
Terrence, so Stella's frame is an accelerated frame.
MEASURED IN STELLA'S ACCELERATED FRAME, Terrence will move away
from Stella at high speed, and his clock will run thousands of times
faster than Stella's.
So when Stella's clock shows 9h 12h, t_Terrence = 15y and x = 12ly
When Stella's clock shows 9h 24h, t_Terrence = 24.6y 25.6h
and x = 7.2ly + 19.2lh
During Stella's time 9y 24h - 18y, Stella is inertial and so is her frame. >> MEASURED IN STELLA'S FRAME, Terrence is moving away at the speed 0.8c, his clock runs slow by 1/γ, and his position is
x = 7.2ly + 19.2lh + 0.8c⋅t_Stella'
At Stella-s time 18y + 24h Terrence is back.
What is it that you don't understand?
Physicists don't realize that they are practicing artificially.
The frame of reference that they attribute to Stella is firstly not the real one. It's not hers, but the frame of reference of a comoving observer placed very far away and observing things transversally.
This is the first problem, but the second will be worse. For the return, we are forced to do the same thing, but here, it's no longer like with Terrence where M1=M2.
In Stella's frame of reference, M1 is imaginarily infinitely far from M2. We "jump" from frame of reference, which is neither pretty nor credible. We then "jump" from clocks. And all this is nothing more than a vulgar patch-up to "be even more right".
If we finally wanted to understand things (but we don't want to),
we would realize that a frame of reference can only be attributed, in relativity, to a single observer.
This observer is at point O of the frame of reference.respective space-time. We "jump" frames of reference.
In Stella's problem; Stella is NOT in her correct relativistic place, and we are forced to apply non-existent subterfuges (time-gap) to validate the passage from M1 (go) to M2 (return) while ideally these two points are infinitely far away in their
I don't practice like that at all.
I always stay in Stella's frame of reference, WITH Stella as center and origin. So she NEVER jumps frames of reference. She is always in her frame of reference and at point O.
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