Den 06.02.2023 20:57, skrev patdolan:
> On Monday, February 6, 2023 at 11:37:02 AM UTC-8, Paul B. Andersen wrote:
>> Den 06.02.2023 01:01, skrev patdolan:
>>> On Sunday, February 5, 2023 at 11:20:49 AM UTC-8, Paul B. Andersen wrote:
>>>>
>>>>
https://paulba.no/paper/Vessot.pdf
https://paulba.no/paper/Vessot.pdf
Look at fig,3.
The 'predicted effect' is GR's prediction for the gravitational
frequency shift.
The 'residuals' are the difference between the measured and
the predicted gravitational frequency shift.
In the first 10 minutes, the residuals are rather large while
the system is settling down after the acceleration,
and something dramatically is happening at the time 13:50.
But during the 80 minutes between 12:00 and 13:20,
the residuals |Δf/f| < 2e-13, most of the time |Δf/f| < 1e-13.
Which means that the error most of the time is < 1e-3 of the prediction.
If you are serious about reading the paper and trying to understand
how the measurements are done, it may help to know the basic principle
of the measurements:
1:
The frequency f₀ from a maser in the rocket is sent to a receiver
on the ground. The received frequency will be both gravitational
Doppler shifted and Doppler shifted due to the speed of the rocket.
The MEASURED received frequency is:
fᵣₑ = (1 + Δf/f₀ + Δfₛ/f₀)⋅f₀
or:
(fᵣₑ/f₀ - 1) = Δfᵣₑ/f₀ = Δf/f₀ + Δfₛ/f₀ (1)
where Δf/f₀ is the gravitational shift and Δfₛ/f₀ is the speed shift.
2:
The frequency f₀ from a maser on the ground is sent up to the rocket
where it is sent back (transponder) to the ground.
This signal will be gravitational red shifted on its way up , and
the signal received by the rocket will in addition be Doppler shifted
due to the speed of the rocket.
On its way down the signal will be gravitational blue shifted, and the
signal received on the ground will in addition be Doppler shifted due to
the speed of ground relative to the rocket.
The red shift of the signal on the way up will always exactly cancel
the blue shift on its way down, so the two way Doppler shift of
the signal depends only on the speed of the rocket.
The MEASURED two way Doppler shift:
f₂ = (1 + v/c)/(1 − v/c) ≈ (1 + 2v/c) when v/c << 1
or:
Δf₂/f₀ ≈ 2v/c (2)
One way Doppler shift:
Δfₛ = √((1 + v/c)/(1 − v/c)) ≈ (1 + v/c) when v/c << 1
or:
Δfₛ/f₀ = v/c = (Δf₂/f₀)/2 (3)
We have two measured values, Δfᵣₑ/f₀ (1) and Δf₂/f₀ (2).
Combining (1),(2) and (3) yields the gravitational Doppler shift:
Δf/f₀ = Δfᵣₑ/f₀ - (Δf₂/f₀)/2
The calculation above is made in true time by hardware.
See Fig. 1.
What I have not mentioned in the explanation of the basics
principle is that the frequencies must be offset so that
the tree sent frequencies (one up and two down) are different
so that the three receivers will receive the correct signal only.
--
Paul
https://paulba.no/