Den 31.12.2023 08:24, skrev patdolan:

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 rest

wrt 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
4.2 years have passed.

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 distant

observer. 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 = 4.84

sound-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/
observer.

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.

It is obviously nonsensical to _accelerate_ the rest frame of
Earth-Proxima Centauri.

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.

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:
--------------------------------------

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

https://paulba.no/

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Den 31.12.2023 19:02, skrev patdolan:

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.

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:
--------------------------------------

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

I suppose this is a very naive attempt to calculate the
the number of turns BB make using Doppler shift.

But you failed again! Twice! :-D

Do you really not understand that when O is at AC
and visually observe BB, then he will see the light
that was emitted a time L/c before O reached AC?

I have showed you the correct derivation before.
Here it is again.

We are talking about visual observation of the BB clock.

O-v
P-----------------E
0 L

Since O is approaching the Earth, he will measure
(see above) the frequency of BB to be f₀' = γ⋅f₀.

He will visually observe this frequency to be Doppler shifted:
f = sqrt((1+v/c)/(1-v/c))f₀' = f₀/(1-v/c)

He will observe this frequency for the time L/v.
But when he is at P, he will see the light emitted from BB
at a time L/c before he arrived at P, so he must subtract
the counts he received the first time L/c.
That means that he must count the cycles received during
the time Δt = L/v - L/c = (L/v)(1-v/c)

The number of counts emitted from BB during this time is:
N = f⋅Δt = (f₀/(1-v/c))(L/v)(1-v/c) = f₀⋅L/v = 3538.75 cycles ==============================================================

How do you answer this discrepancy between SR and your own calculations, Paul?

There is no discrepancy between SR and my calculations.

The discrepancy between SR and your calculations are easy to explain:
The simple math of SR is beyond your very limited abilities.

--
Paul

https://paulba.no/

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