Den 01.01.2024 21:58, skrev patdolan:
There has never been a mathematical calculation offered on any platform to refute the BBP. Not a single one. The only counter response has been, and will continue to be, taking it down off the platform or closing it to all further discussion. That
is the only response imaginable by its critics. No other theory in the entire history of science has been met with this treatment. This attests to the BBP's everlasting potency to destroy SR and GR in one swell scoop.
______________________________________
Consider a distant observer traveling at 0.867c ( 𝛾 = 2 ) relative to the solar system along the line that is collinear with the sun's axis of rotation. As the clockwork solar system spins beneath him, the distant observer peers through his powerful
telescope at Big Ben in London. After taking relativistic doppler into account, the distant observer measures Big Ben's little hand to make one revolution for every two revolutions of his own wristwatch's little hand, in accordance with relativistic time
dilation. He also observes that Big Ben's little hand still makes 730.5 revolutions for every revolution that the earth makes around the sun. From these two observations the distant observer concludes that in his inertial frame of reference the earth's
orbital velocity is only half the velocity necessary to keep the earth in stable orbit around the sun, given the invariant spacetime curvature in the vicinity of the sun through which the earth’s geodesic passes.
Will the earth spiral into the sun? If not, why not?
Because what SR predicts is not a matter of opinion,
it is a matter of fact.
The fact is that your claim of what SR predicts
is proved wrong below:
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'. _________________________________________________________________
Calculation with Doppler shift.
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 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 ==============================================================
Keep ignoring that you are proved wrong,
and I will keep reminding you.
--
Paul
https://paulba.no/
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