Consider a bell and a microphone separated by 4.2 sound-years in a universe filled with air. In said universe one sound-year = 343 m/s x 3600 sec/hr x 24 hr/day x 365.25 days/year = 1.08 x 10^10 meters. The bell and the speaker always remain at restwrt their own co-moving coordinate system. The bell is located at the origin and the microphone is located 4.2 sound-years in the positive-x direction. If the bell begins to ring, the first clang of the bell will not be picked up by the microphone until
Now consider a distant observer located on the x-axis several more sound-years beyond the microphone in the positive-x direction. We now accelerate the bell and microphone until they are traveling at 0.867 times the speed of sound towards the distantobserver. How far will the bell have had to have been from the distant observer when its first clang sounded in order for the first clang and the distant observer to arrive at the microphone simultaneously?
We now repeat the experiment. But this time we accelerate the observer until he is traveling at 0.867 times the speed of sound towards the bell and microphone, which are at rest wrt each other. This time the observer only needs to be 4.2/0.867 = 4.84sound-years away from the microphone when the first clang sounds in order to arrive at the microphone simultaneously with the first clang. And when he does arrive there will be exactly 4.2 years worth of clangs between the bell and the microphone/
The two cases lack symmetry. The same lack of symmetry pertains in the Helical Path Paradox when Big Ben (the earth) and Proxima Centauri are accelerated versus the observer being accelerated instead.
I challenge you to find any errors or absurdities in the following:
Problem:
An observer O is racing past Proxima Centauri on her way to Big Ben
at .867c relative to the Big Ben.
Question to answer:
How many rotations will the little hand of Big-Ben make
from the observer O is passing Proxima Centauri to she hits
the Earth?
Let's call Earth's rest frame K(t,x).
We will call the position of the Earth E, and the position of
Proxima Centauri P in this frame.
O->v
K: P-----------------E
0 L
At t = t₀ = 0, the observer O is at P.
At t = t₁ the observer O is at E
L = 4.2 [ly] proper distance Earth - Proxima Centauri in K
v = 0.867c
γ = 2.0068
f₀ = 730.5 [cycles/y], proper frequency of the BB clock.
T = 1/f₀ = 0.001369 [y], proper duration of a cycle
t₁ = L/v = 4.844 y
So the answer to the question above is:
N₀ = f₀⋅t₁ = f₀⋅L/v = 3538.75 cycles
===================================
This is the same as what NM predicts, because we
have not asked what is measured in O's rest frame. ____________________________________________________
The observer's clock is moving in K:
--------------------------------------
Let K'(t',x') be O's rest frame.
There are two events of interest:
E0: The observer is at P
In K: t₀ = 0, x₀ = 0
In K': t₀' = 0, x₀' = 0
E1: The observer is at E
In K: t₁ = L/v = 4.84429 y, x₁ = L = 4.2 ly
In K':
t₁' = γ(t₁-v⋅x₁/c²) = L/γv = 2.41395 y
x₁' = γ(x₁-v⋅t₁) = 0
In K: t₂ = T = 0.001369 y, x₂ = L/v = 2.09289 ly
In K': t₂'= γ(t₂-v⋅x₂/c²) = T/γ = 0.00068215 y
f₀' = γ⋅f₀ = 1465.96 cycles/y , the frequency measured in K'
So SR predicts that O will measure (count):
N₁ = f₀'⋅t₁' = γ⋅f₀⋅L/γv = f₀⋅L/v = 3538.75 cycles
=================================================>
Note this:
The observer's clock advances the proper time:
τ' = t₁'- t₀'= 2.41395 y
while the difference between the coordinate time
t₀ at x₀ and t₁ at x₁ changes by:
(t₁ - t₀) = L/v = 4.84429 y.
The observer's moving clock appears to run slow as measured in K. ________________________________________________________________
Big Ben is moving in K':
-------------------------
t₄' = 0
O
P-----------E
0 x₄'
At Event E₄ is E at x₄' when t₄' = 0
We know that E always is at x = L in K
t₄' = γ(t₄-v⋅L/c²) = 0 => t₄ = v⋅L/c² = 3.6414 y
x₄' = γ(x₄-v⋅t₄) = γ(L-L(v²/c²)) = L/γ = 2.09289 ly
So measured in K' at the time t' = 0, E is at the position L/γ
and BB is showing the proper time τ₄ = v⋅L/c² = 3.6414 y
At Event E1, when E is at P, we have from above:
BB is showing the the proper time τ₁ = t₁ = L/v = 4.84429 y
We still have:
f₀' = γ⋅f₀ = 1465.96 cycles/y , the frequency measured in K'
t₁' = L/γv = 2.41395 y
So SR predicts that O will measure (count):
N₁ = f₀'⋅t₁' = γ⋅f₀⋅L/γv = f₀⋅L/v = 3538.75 cycles
=================================================>
Note this:
Big Ben advances the proper time:
(τ₁-τ₄) = L/v-v⋅L/c² = (L/v)(1-v²/c²) = L/γ²v = 1.20289 y
while the difference between the coordinate time t₄' at x₄'
and t₁' at x₁' changes by:
(t₁' - t₄') = L/γv = 2.41395 y
The moving Big Ben appears to run slow as measured in K'. _________________________________________________________________
On Sunday, December 31, 2023 at 9:40:53 AM UTC-8, patdolan wrote:
On Sunday, December 31, 2023 at 1:34:55 AM UTC-8, Paul B. Andersen wrote: >>>
But it doesn't matter if you consider the observer
to be stationary and the Earth-AC system to be moving,
or you consider the Earth-AC system to be stationary
and the observer to be moving.
I have showed you this before. Didn't you read it?
December 18, 2023, Paul B. Andersen wrote:
I challenge you to find any errors or absurdities in the following:
Problem:
An observer O is racing past Proxima Centauri on her way to Big Ben
at .867c relative to the Big Ben.
Question to answer:
How many rotations will the little hand of Big-Ben make
from the observer O is passing Proxima Centauri to she hits
the Earth?
Let's call Earth's rest frame K(t,x).
We will call the position of the Earth E, and the position of
Proxima Centauri P in this frame.
vK: P-----------------E
0 L
At t = t₀ = 0, the observer O is at P.
At t = t₁ the observer O is at E
L = 4.2 [ly] proper distance Earth - Proxima Centauri in K
v = 0.867c
γ = 2.0068
f₀ = 730.5 [cycles/y], proper frequency of the BB clock.
T = 1/f₀ = 0.001369 [y], proper duration of a cycle
t₁ = L/v = 4.844 y
So the answer to the question above is:
N₀ = f₀⋅t₁ = f₀⋅L/v = 3538.75 cycles
===================================
This is the same as what NM predicts, because we
have not asked what is measured in O's rest frame.
____________________________________________________
The observer's clock is moving in K:
--------------------------------------
Earth-AC system stationary, observer moving,
Let K'(t',x') be O's rest frame.
There are two events of interest:
E0: The observer is at P
In K: t₀ = 0, x₀ = 0
In K': t₀' = 0, x₀' = 0
E1: The observer is at E
In K: t₁ = L/v = 4.84429 y, x₁ = L = 4.2 ly
In K':
t₁' = γ(t₁-v⋅x₁/c²) = L/γv = 2.41395 y
x₁' = γ(x₁-v⋅t₁) = 0
In K: t₂ = T = 0.001369 y, x₂ = L/v = 2.09289 ly
In K': t₂'= γ(t₂-v⋅x₂/c²) = T/γ = 0.00068215 y
f₀' = γ⋅f₀ = 1465.96 cycles/y , the frequency measured in K'
So SR predicts that O will measure (count):
N₁ = f₀'⋅t₁' = γ⋅f₀⋅L/γv = f₀⋅L/v = 3538.75 cycles >>>> =================================================>
Note this:
The observer's clock advances the proper time:
τ' = t₁'- t₀'= 2.41395 y
while the difference between the coordinate time
t₀ at x₀ and t₁ at x₁ changes by:
(t₁ - t₀) = L/v = 4.84429 y.
The observer's moving clock appears to run slow as measured in K.
________________________________________________________________
Big Ben is moving in K':
-------------------------
Observer stationary, Earth-AC system moving
t₄' = 0
O
P-----------E
0 x₄'
At Event E₄ is E at x₄' when t₄' = 0
We know that E always is at x = L in K
t₄' = γ(t₄-v⋅L/c²) = 0 => t₄ = v⋅L/c² = 3.6414 y
x₄' = γ(x₄-v⋅t₄) = γ(L-L(v²/c²)) = L/γ = 2.09289 ly
So measured in K' at the time t' = 0, E is at the position L/γ
and BB is showing the proper time τ₄ = v⋅L/c² = 3.6414 y
At Event E1, when E is at P, we have from above:
BB is showing the the proper time τ₁ = t₁ = L/v = 4.84429 y
We still have:
f₀' = γ⋅f₀ = 1465.96 cycles/y , the frequency measured in K'
t₁' = L/γv = 2.41395 y
So SR predicts that O will measure (count):
N₁ = f₀'⋅t₁' = γ⋅f₀⋅L/γv = f₀⋅L/v = 3538.75 cycles >>>> =================================================>
Note this:
Big Ben advances the proper time:
(τ₁-τ₄) = L/v-v⋅L/c² = (L/v)(1-v²/c²) = L/γ²v = 1.20289 y >>>> while the difference between the coordinate time t₄' at x₄'
and t₁' at x₁' changes by:
(t₁' - t₄') = L/γv = 2.41395 y
The moving Big Ben appears to run slow as measured in K'.
_________________________________________________________________
You are proven wrong - again.
Which you will ignore and not even try to comment.
Probably because you are not able to read the math.
Paul, you never come to the correct answer according to SR, which is
3.747 x 2.1 years x 730.5 turns/year = 5748 turns
Ooops. Make that
3.747 x ( 2.1 light-years/0.867c ) x 730.5 turns/years = 6630 turns
How do you answer this discrepancy between SR and your own calculations, Paul?
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