[...] Then, tK and tA are synchronized in the einstenian way.
On Wednesday, November 8, 2023 at 8:12:14 PM UTC-3, Tom Roberts
wrote:
On 11/8/23 3:56 PM, Richard Hertz wrote:
[...] Then, tK and tA are synchronized in the einstenian way.These clocks are at rest in different inertial frames, and thus
cannot be synchronized with each other.
I clearly wrote that tK and tA are synchronized as Einstein claimed
in his 1905 paper:
On 11/8/23 5:43 PM, Richard Hertz wrote:
On Wednesday, November 8, 2023 at 8:12:14 PM UTC-3, Tom Roberts
wrote:
On 11/8/23 3:56 PM, Richard Hertz wrote:
[...] Then, tK and tA are synchronized in the einstenian way.These clocks are at rest in different inertial frames, and thus
cannot be synchronized with each other.
I clearly wrote that tK and tA are synchronized as Einstein claimedYou can write whatever you want; that does not make whatever you wrote correct or possible. It is PHYSICALLY IMPOSSIBLE to synchronize clocks
in his 1905 paper:
that are at rest in different inertial frames [#], and Einstein NEVER described what you wrote as "synchronization".
[#] The GPS synchronizes satellite clocks in ECI coordinates;
while they are called "clocks", they are NOT standard clocks,
and are not at rest in inertial frames (SR). Moreover, earth's
gravitation is an essential aspect of this, required to keep
them in orbit, which is not possible in SR.
Note: you may set clocks tK and tA to the same value when they pass each other, but that does NOT synchronize them with each other, and that is
NOT how Einstein synchronized clocks in his 1905 paper (or anywhere
else). Only clocks at rest in a single inertial frame can be mutually synchronized, and Einstein's methods EXPLICITLY require this.
You REALLY need to understand SR before attempting to write about it, or criticize it.
Tom Roberts
On Wednesday, November 8, 2023 at 8:12:14 PM UTC-3, Tom Roberts wrote:
On 11/8/23 3:56 PM, Richard Hertz wrote:
[...] Then, tK and tA are synchronized in the einstenian way.
These clocks are at rest in different inertial frames, and thus cannot
be synchronized with each other.
Tom Roberts
I clearly wrote that tK and tA are synchronized as Einstein claimed in his 1905 paper:
When A (at origin of K') passes BY THE ORIGIN of K, tA = tK = 0, exactly BEFORE the photon is emitted from A to B.
On 11/8/2023 6:43 PM, Richard Hertz wrote:
On Wednesday, November 8, 2023 at 8:12:14 PM UTC-3, Tom Roberts wrote:
On 11/8/23 3:56 PM, Richard Hertz wrote:
[...] Then, tK and tA are synchronized in the einstenian way.
These clocks are at rest in different inertial frames, and thus cannot
be synchronized with each other.
Tom Roberts
I clearly wrote that tK and tA are synchronized as Einstein claimed in his 1905 paper:
When A (at origin of K') passes BY THE ORIGIN of K, tA = tK = 0, exactly BEFORE the photon is emitted from A to B.
That's not Einstein synchronization. Einstein synchronization is for two objects not located at the same point but stationary relative to each
other. (so it's impossible to set t=t'=0 since x is never equal to x'
(so x=x'=0 never happens).
On 11/8/23 5:43 PM, Richard Hertz wrote:
On Wednesday, November 8, 2023 at 8:12:14 PM UTC-3, Tom Roberts
wrote:
On 11/8/23 3:56 PM, Richard Hertz wrote:
[...] Then, tK and tA are synchronized in the einstenian way.These clocks are at rest in different inertial frames, and thus
cannot be synchronized with each other.
I clearly wrote that tK and tA are synchronized as Einstein claimedYou can write whatever you want; that does not make whatever you wrote correct or possible. It is PHYSICALLY IMPOSSIBLE to synchronize clocks
in his 1905 paper:
that are at rest in different inertial frames
My description is quite simple:
1) Mount an artifact with two mirrors located at A and B. Let it move (being a moving frame K' with origin in A) at uniform speed v =0.9c.
2) When the point A in the moving frame K' happens to be (for an infinitesimal amount of time) coincident with the
origin of frame K (at relative rest) then set both clocks (tK at frame K origin and tA at frame K' origin) to zero.
This is the same as of saying that when x = x' = 0, then make t = t' = 0.
3) Being that tK and tA are counting 1.0000000 μs ticks,...
classic physics said
SR says that, while tK shows a TRUE COUNT of 1.0000000 μs ticks,
any observer standing by tK will PERCEIVE that tA accumulates ticks at a higher rate.
BUT, any observer moving with frame K' (standing by the clock tA) will say that the
accumulation of ticks keep going at 1.0000000 μs/tick.
The best way to show the incoherence of SR is that the readings in the moving frame use digital counters
AND A GIANT DISPLAYS pointing at the origin of K,
so the REAL COUNT
at K' could be observed from the origin of K (using a telescope, if needed in this thought experiment).
By moving to the DIGITAL DOMAIN in K', all the equations of SR COLLAPSE,
because the analog math of SR is KILLED by digitalization.
So, the observer at relative rest in the origin of K FACES A DILEMMA: Because of
Lorentz, he PERCEIVES A FALLACIOUS TIME at tA,
which is in absolute conflict with the remote reading of the DIGITAL DISPLAY located at A, moving at 0.9c.
How do you assist, psychologically, to the observer at K origin, to avoid that he become crazy?
His PERCEPTION crashes with his observation of the display.
Observers, which is synonym to 'logging device'
On Thursday, November 9, 2023 at 2:52:15 AM UTC-3, Volney wrote:
On 11/8/2023 6:43 PM, Richard Hertz wrote:
On Wednesday, November 8, 2023 at 8:12:14 PM UTC-3, Tom Roberts wrote: >>>> On 11/8/23 3:56 PM, Richard Hertz wrote:That's not Einstein synchronization. Einstein synchronization is for two
[...] Then, tK and tA are synchronized in the einstenian way.
These clocks are at rest in different inertial frames, and thus cannot >>>> be synchronized with each other.
Tom Roberts
I clearly wrote that tK and tA are synchronized as Einstein claimed in his 1905 paper:
When A (at origin of K') passes BY THE ORIGIN of K, tA = tK = 0, exactly BEFORE the photon is emitted from A to B.
objects not located at the same point but stationary relative to each
other. (so it's impossible to set t=t'=0 since x is never equal to x'
(so x=x'=0 never happens).
K' origin is moving from - infinity towards + infinity, at 0.9c.
When A pass by K origin, light signaling is used to set tK = tA = 0.
c = 299792458 m/s = 299.792458 m/μs
Two perfect mirrors (A,B) are placed 149,896229 m apart, being A at the origin of the relatively moving frame K'.
A photon, in the visible light range, is emitted from the mirror A (which also contains a clock) exactly towards mirror B.
The photon bounces back at mirror B, towards mirror A. Its arrival from the roundtrip is marked in tA clock as being exactly 1.0000000 μs.
This process continues forever, without any losses, while A clock is accumulating 1.0000000 μs counts.
The arrangement in K' moves at v speed wrt the origin at K, where a clock tK is located (it also ticks at 1.0000000 μs.
Both clocks (tK and tA) are reset to zero exactly when A pass by the origin K, and it forces the emission of the photon.
Then, tK and tA are synchronized in the einstenian way.
The speed v = 0.9 c = 269813212 m/s
WHAT SR (LORENTZ) SAYS THAT HAPPENS AFTERWARDS.
γ = 2.294157339
LENGTH CONTRACTION (Horizontal light clock is perceived to run faster) http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c1
x' = γ (x - vt)
Δx' = AB AND Δx = AB/γ = 0.435889894 AB, as measured in the frame at rest. As both measurements are made simultaneously in the frame at rest, there is A PERCEIVED LENGTH CONTRACTION of AB in the moving frame.
But the speed of light IS INVARIANT wrt to any frame, so IT'S PERCEIVED that clock tA IS RUNNING FASTER.
The roundtrip of the photon is PERCEIVED to travel 2 x 0.435889894 AB = 0.871779789 x AB, which means that the clock's tick in the moving frame IS PERCEIVED TO TICK every 0.871779789 μs, instead of 1.00000000 μs.
TIME DILATION (Horizontal light clock is perceived to run slower) http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c2
t' = γ (t - vx/c)
Δt' = Δt/γ = 0.435889894 Δt (Time is PERCEIVED as running slower in the moving frame).
For every 1.00000000 μs in the frame at rest, it's PERCEIVED that only 0.435889894 μs passed in the moving frame.
The time measurements are made in the same location, so it doesn't affect the results.
CONCLUSION: Using an horizontal light clock, the results of TIME FLOW due to length contraction and time dilation are in conflict. Clock tA is PERCEIVED to run faster or slower than clock tK, in the two different transforms.
QUESTION: WHERE IS THE TRICK EMBEDDED IN THIS PARADOX?
"Length contraction is the PHENOMENON that a moving
2) Measured = CALCULATED using the first Lorentz formula.
This is very simple, and is clearly explained in the Wiki article:
Den 08.11.2023 22:56, skrev Richard Hertz:
c = 299792458 m/s = 299.792458 m/μs
Two perfect mirrors (A,B) are placed 149,896229 m apart, being A at the origin of the relatively moving frame K'.
A photon, in the visible light range, is emitted from the mirror A (which also contains a clock) exactly towards mirror B.
The photon bounces back at mirror B, towards mirror A. Its arrival from the roundtrip is marked in tA clock as being exactly 1.0000000 μs.
This process continues forever, without any losses, while A clock is accumulating 1.0000000 μs counts.OK!
The arrangement in K' moves at v speed wrt the origin at K, where a clock tK is located (it also ticks at 1.0000000 μs.
Both clocks (tK and tA) are reset to zero exactly when A pass by the origin K, and it forces the emission of the photon.OK!
Then, tK and tA are synchronized in the einstenian way.No, this has noting to do with Einstein's way of synchronizing
clocks in inertial frames.
Two clocks are set equal (to zero) when they pass each other,
that's all.
We will also assume that both clocks are placed at the origin of
their respective rest frames.
We then have:
K':-tA------B-------->x'->v
K :-tK--------------->x
The speed v = 0.9 c = 269813212 m/s
WHAT SR (LORENTZ) SAYS THAT HAPPENS AFTERWARDS.In the following we will use L as the distance
γ = 2.294157339
between the mirrors. We can put in numbers later.
To find what SR says, we use the Lorentz transform.
We have three event of interest.
E0: tA and tK are aligned and the photon is emitted from A.
E1: the photon is reflected from B.
E2: the photon is reflected from A
Let's find the coordinates of these event's in K and K':
γ = 1/√(1 −v²/c²) = 2.294157339 when v = 0.9c
EO:
Coordinates in K': x₀' = 0, t₀' = 0
Coordinates in K: x₀ = 0, t₀ = 0
E1: (this event could be skipped)
Coordinates in K': x₁' = L, t₁' = L/c
Coordinates in K:
x₁ = γ(x₁'+ v⋅t₁')=γ(L+ v⋅L/c) = L⋅√((c+v)/(c-v))
t₁ = γ(t₁'+ (v/c²)⋅x₁') = γ(L/c + (v/c²)⋅L) = (L/c)⋅√((c+v)/(c-v))
E2:
Coordinates in K': x₂' = 0, t₂' = 2L/c
Coordinates in K:
x₂ = γ(x₂'+ v⋅t₁') = γ(0 + v⋅2L/c) = (2Lv/c)/√(1 −v²/c²) t₂ = γ(t₂'+ (v/c²)⋅x₂') = γ(L/c + (v/c²)⋅L) = (2L/c)/√(1 −v²/c²)
So while the moving clock tA has advanced the proper time
t₂' = 2L/c = 1 μs ,
the coordinate time in the stationary frame has advanced
t₂ (2L/c)/√(1 −v²/c²) = 2.294157339 μs
t₂'/t₂ = √(1 −v²/c²) = 0.435889894
"Moving clocks run slow"
LENGTH CONTRACTION (Horizontal light clock is perceived to run faster) http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c1
x' = γ (x - vt)
Δx' = AB AND Δx = AB/γ = 0.435889894 AB, as measured in the frame at rest. As both measurements are made simultaneously in the frame at rest, there is A PERCEIVED LENGTH CONTRACTION of AB in the moving frame.
But the speed of light IS INVARIANT wrt to any frame, so IT'S PERCEIVED that clock tA IS RUNNING FASTER.
The roundtrip of the photon is PERCEIVED to travel 2 x 0.435889894 AB = 0.871779789 x AB, which means that the clock's tick in the moving frame IS PERCEIVED TO TICK every 0.871779789 μs, instead of 1.00000000 μs.
TIME DILATION (Horizontal light clock is perceived to run slower) http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c2
t' = γ (t - vx/c)
Δt' = Δt/γ = 0.435889894 Δt (Time is PERCEIVED as running slower in the moving frame).
For every 1.00000000 μs in the frame at rest, it's PERCEIVED that only 0.435889894 μs passed in the moving frame.
The time measurements are made in the same location, so it doesn't affect the results.
CONCLUSION: Using an horizontal light clock, the results of TIME FLOW due to length contraction and time dilation are in conflict. Clock tA is PERCEIVED to run faster or slower than clock tK, in the two different transforms.
QUESTION: WHERE IS THE TRICK EMBEDDED IN THIS PARADOX?There is no paradox.
The trick is to know how to use the Lorentz transform.
You don't.
It's mathematics which you always screw up.
--
Paul
https://paulba.no/
On Thursday, November 9, 2023 at 5:50:00 PM UTC-5, Richard Hertz wrote:
"Length contraction is the PHENOMENON that a moving<snip>. Irrelevant to the discussion.
2) Measured = CALCULATED using the first Lorentz formula.That is not what 'measured' means.
You have been systematically showing that you do not understand the meaning of the words you use.
I explain to you the meanings, and you just 'disappear' or change the subject. That is crank behavior.
This is very simple, and is clearly explained in the Wiki article:As I told you, forget what others say. Do what SR says, since that is what you want to analyze.
<rest of confusions snipped>.
I have explained to you what SR says and doesnt say. You have not given one rebuttal to my claims and calcs.
But you did divert and changed the topic(s). You are looking more and more like a crank...More and more...
On Thursday, November 9, 2023 at 5:50:00 PM UTC-5, Richard Hertz wrote:
"Length contraction is the PHENOMENON that a moving<snip>. Irrelevant to the discussion.
2) Measured = CALCULATED using the first Lorentz formula.That is not what 'measured' means.
On Thursday, November 9, 2023 at 10:43:06 PM UTC-5, Richard Hertz wrote:
I didn't reply to your post because, besides the criticism on my use of words,For two parties to understand each other, they must know the meaning of the words they use.
The words you use are ill defined and not necessary.
I didn't reply to your post because, besides the criticism on my use of words,
you changed the meaning of my post by "counting photon clicks, not time".
If 1.00000000 μs is too discrete for you, change the clock tick to 1.0000000 ps
The passage from the analog domain, in which Lorentz transforms are developed, to the digital domain, is based
on discretization of time, which has
a logical limit that applies also to the analog domain. You can't work with a fraction of a wavelength as it has no meaning.
One example of lower limit in the digital domain is to use 550 nm green photons and use AB = 1,1 μm,
The time measured is, in the digital domain, a multiple of the roundtrip time.
Lorentz transforms work with infinitesimal fractions of time,
but LOOSE
ANY MEANING below the lowest possible visible wavelength,
So, this thought experiment is about to accumulate counts of discretized time. In this case, digital math kills Lorentz math.
It's ridiculous to apply SR even for 1 second, if v is ALMOST c. The distance involved of 300,000 Km
is RIDICULOUS. And better not to talk about 10 sec.
On Thursday, November 9, 2023 at 5:18:57 PM UTC-3, Paul B. Andersen wrote:
Den 08.11.2023 22:56, skrev Richard Hertz:
c = 299792458 m/s = 299.792458 m/μsOK!
Two perfect mirrors (A,B) are placed 149,896229 m apart, being A at the origin of the relatively moving frame K'.
A photon, in the visible light range, is emitted from the mirror A (which also contains a clock) exactly towards mirror B.
The photon bounces back at mirror B, towards mirror A. Its arrival from the roundtrip is marked in tA clock as being exactly 1.0000000 μs.
This process continues forever, without any losses, while A clock is accumulating 1.0000000 μs counts.
OK!
The arrangement in K' moves at v speed wrt the origin at K, where a clock tK is located (it also ticks at 1.0000000 μs.
Both clocks (tK and tA) are reset to zero exactly when A pass by the origin K, and it forces the emission of the photon.
No, this has noting to do with Einstein's way of synchronizing
Then, tK and tA are synchronized in the einstenian way.
clocks in inertial frames.
Two clocks are set equal (to zero) when they pass each other,
that's all.
We will also assume that both clocks are placed at the origin of
their respective rest frames.
We then have:
K':-tA------B-------->x'->v
K :-tK--------------->x
The speed v = 0.9 c = 269813212 m/s
WHAT SR (LORENTZ) SAYS THAT HAPPENS AFTERWARDS.
γ = 2.294157339
In the following we will use L as the distance
between the mirrors. We can put in numbers later.
To find what SR says, we use the Lorentz transform.
We have three event of interest.
E0: tA and tK are aligned and the photon is emitted from A.
E1: the photon is reflected from B.
E2: the photon is reflected from A
Let's find the coordinates of these event's in K and K':
γ = 1/√(1 −v²/c²) = 2.294157339 when v = 0.9c
EO:
Coordinates in K': x₀' = 0, t₀' = 0
Coordinates in K: x₀ = 0, t₀ = 0
E1: (this event could be skipped)
Coordinates in K': x₁' = L, t₁' = L/c
Coordinates in K:
x₁ = γ(x₁'+ v⋅t₁')=γ(L+ v⋅L/c) = L⋅√((c+v)/(c-v))
t₁ = γ(t₁'+ (v/c²)⋅x₁') = γ(L/c + (v/c²)⋅L) = (L/c)⋅√((c+v)/(c-v))
E2:
Coordinates in K': x₂' = 0, t₂' = 2L/c
Coordinates in K:
x₂ = γ(x₂'+ v⋅t₁') = γ(0 + v⋅2L/c) = (2Lv/c)/√(1 −v²/c²) >> t₂ = γ(t₂'+ (v/c²)⋅x₂') = γ(L/c + (v/c²)⋅L) = (2L/c)/√(1 −v²/c²)
So while the moving clock tA has advanced the proper time
t₂' = 2L/c = 1 μs ,
the coordinate time in the stationary frame has advanced
t₂ (2L/c)/√(1 −v²/c²) = 2.294157339 μs
t₂'/t₂ = √(1 −v²/c²) = 0.435889894
"Moving clocks run slow"
There is no paradox.
LENGTH CONTRACTION (Horizontal light clock is perceived to run faster)
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c1
x' = γ (x - vt)
Δx' = AB AND Δx = AB/γ = 0.435889894 AB, as measured in the frame at rest. As both measurements are made simultaneously in the frame at rest, there is A PERCEIVED LENGTH CONTRACTION of AB in the moving frame.
But the speed of light IS INVARIANT wrt to any frame, so IT'S PERCEIVED that clock tA IS RUNNING FASTER.
The roundtrip of the photon is PERCEIVED to travel 2 x 0.435889894 AB = 0.871779789 x AB, which means that the clock's tick in the moving frame IS PERCEIVED TO TICK every 0.871779789 μs, instead of 1.00000000 μs.
TIME DILATION (Horizontal light clock is perceived to run slower)
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c2
t' = γ (t - vx/c)
Δt' = Δt/γ = 0.435889894 Δt (Time is PERCEIVED as running slower in the moving frame).
For every 1.00000000 μs in the frame at rest, it's PERCEIVED that only 0.435889894 μs passed in the moving frame.
The time measurements are made in the same location, so it doesn't affect the results.
CONCLUSION: Using an horizontal light clock, the results of TIME FLOW due to length contraction and time dilation are in conflict. Clock tA is PERCEIVED to run faster or slower than clock tK, in the two different transforms.
QUESTION: WHERE IS THE TRICK EMBEDDED IN THIS PARADOX?
The trick is to know how to use the Lorentz transform.
You don't.
It's mathematics which you always screw up.
--
Paul
https://paulba.no/
https://en.wikipedia.org/wiki/Length_contraction
"Length contraction is the PHENOMENON that a moving object's length is MEASURED TO BE SHORTER than its proper length, which is the length as measured in the object's own rest frame.[1] It is also known as Lorentz contraction or Lorentz–FitzGeraldcontraction (after Hendrik Lorentz and George Francis FitzGerald) and is usually only noticeable at a substantial fraction of the speed of light."
NOTES: 1) Phenomenon? ; 2) Measured = CALCULATED using the first Lorentz formula.
Paul, you should stop scrambling concepts and using Lorentz mathemagics, and stick to what I posted, without changes.
This is very simple, and is clearly explained in the Wiki article:
L = L₀/γ(v)
where
L is the length OBSERVED by an observer in motion relative to the object.
L₀ is the proper length (the length of the object in its rest frame).
γ(v) is the Lorentz factor, defined as γ(v) = 1/√(1 - v²/c²).
In my own terms,
Δx(v) = AB/γ(v) ; roundtrip time "OBSERVED/PERCEIVED" in K' from K: Δt' = 2AB/(c.γ)
Δx(0) = AB ; Δt' = 1.000000 μs
Δx(0.9c) = 0.43589 AB ; Δt' = 0.87178 μs , because c is invariant. Δx(0.999c) = 0.04471 AB ; Δt' = 0.089420 μs , because c is invariant. Δx(0.999999c) = 0.001414 AB ; Δt' = 0.002828 μs , because c is invariant.
Δx(0.999999999c) = 0.000045 AB ; Δt' = 0.000089 μs , because c is invariant.
Use two arrays AB, with a digital counter tA located at the A side of the assembly AB. Put one of them at the origin
of the K frame, with A located in the origin and put the other on the moving K' frame, with A at its origin.
These twin systems are COUNTING 1.000000 μs ticks of the horizontal light clock AB.
The observer at the origin of K frame monitor the counts of the two counters (moving and being at rest). He CAN CERTIFY that
the local and remote digital displays show THE SAME ACCOUNT OF 1.000000 μs ticks.
Yet, for his dismay and confusion, when use the Lorentz formula, it DECEIVES HIM telling that 1.000000 μs tick in his local clock/counter
is happening at an increasingly lower value in the remote system (mathemagics), BUT it's in conflict with the value of the remote digital
display, which IS EQUAL TO HIS LOCAL DISPLAY.
Here, 60 years after the 1905 paper, digital electronics DESTROY THE ANALOG LORENTZ FORMULA, proving that it's a FICTION, a fairy tale.
Now, what could happen in the head of the poor observer? He might arrive to the conclusion that SR is a flawed pseudo-science, and that all
that he believed since his conversion to relativity IS A LIE.
Blame the second postulate and the horizontal light clock, if you want.
Den 09.11.2023 23:49, skrev Richard Hertz:
In my own terms,
Δx(v) = AB/γ(v) ; roundtrip time "OBSERVED/PERCEIVED" in K' from K: Δt' = 2AB/(c.γ)
What is "round triptime "OBSERVED/PERCEIVED" in K' from K:"
supposed to mean? It is meaningless babble.
You can't "perceive" what is measured it K' from K.
The round trip time _measured_ in K' is 2L₀/c.
The round trip time _measured_ in K is (2L₀/c)/√(1 −v²/c²)
I showed this above, but you have probably not read it.
So let me remind you:
It's very simple, and should be impossible to screw up: =======================================================
When the light pulse has moved from A to B and back to A,
the coordinates of A in K' are x' = 0, t' = 2L₀/c
We must use the Lorentz transform to find the coordinates in K:
t = γ(t'+ (v/c²)x') = γ(L₀/c + (v/c²)⋅L₀) = (2L₀/c)/√(1 −v²/c²)
It is impossible to make the Lorentz transform give any other result.
You have not used the Lorentz transform at all!
On Friday, November 10, 2023 at 4:43:55 PM UTC-3, Paul B. Andersen wrote:
Den 09.11.2023 23:49, skrev Richard Hertz:
In my own terms,
Δx(v) = AB/γ(v) ; roundtrip time "OBSERVED/PERCEIVED" in K' from K: Δt' = 2AB/(c.γ)
What is "round triptime "OBSERVED/PERCEIVED" in K' from K:"
supposed to mean? It is meaningless babble.
You can't "perceive" what is measured it K' from K.
The round trip time _measured_ in K' is 2L₀/c.
The round trip time _measured_ in K is (2L₀/c)/√(1 −v²/c²)
You're full of historical shit.
Let me RECAP: You CAN'T measure the round trip time of a K' event being in the frame K, at relative rest!. In particular, being that K' is moving at almost c.
If you want to know the K' round trip time from K, read THE FUCKING DIGITAL DISPLAY IN K'!!So when you proved SR inconsistent you to read
And that is the ONLY TRUTH in your pseudo-scientific relativism.
Because mathematics (Lorentz) IS NOT PHYSICS. But you'll never learn, fanatic indoctrinated relativist!
Your early marriage with the Lorentz crap and Einstein pseudo-philosophy ruined your brain since decades ago. Now it's USELESS.
On Friday, November 10, 2023 at 4:43:55 PM UTC-3, Paul B. Andersen wrote:
The round trip time _measured_ in K' is 2L₀/c.You're full of historical shit.
The round trip time _measured_ in K is (2L₀/c)/√(1 −v²/c²)
Let me RECAP: You CAN'T measure the round trip time of a K' event being in the frame K,
at relative rest!. In particular, being that K' is moving at almost c.
If you want to know the K' round trip time from K, read THE FUCKING DIGITAL DISPLAY IN K'!!
And that is the ONLY TRUTH in your pseudo-scientific relativism.
Because mathematics (Lorentz) IS NOT PHYSICS. But you'll never learn, fanatic indoctrinated relativist!
in your journey toward full relativistic indoctrination. Again, mathematics IS NOT PHYSICS.
Den 11.11.2023 04:26, skrev Richard Hertz:
On Friday, November 10, 2023 at 4:43:55 PM UTC-3, Paul B. Andersen wrote:
Den 09.11.2023 23:49, skrev Richard Hertz:
In my own terms,
Δx(v) = AB/γ(v) ; roundtrip time "OBSERVED/PERCEIVED" in K' from K: Δt' = 2AB/(c.γ)
What is "round triptime "OBSERVED/PERCEIVED" in K' from K:"
supposed to mean? It is meaningless babble.
You can't "perceive" what is measured it K' from K.
The round trip time _measured_ in K' is 2L₀/c.
The round trip time _measured_ in K is (2L₀/c)/√(1 −v²/c²)
You're full of historical shit.Let me remind you of the subject line:
"Conflicts in SR when using horizontal light clocks in the moving frame."
Your opinion of SR is irrelevant.
The question is: Have you succeeded in proving SR inconsistent?
To prove SR inconsistent, that is self contradictory,
you must show that the math of SR can give contradictory
predictions for the same scenario.
"The math of SR" is in this case the Lorentz transform.
You claim to have proven that SR is inconsistent, but you
haven't used the Lorentz transform to show it.
Your attempt to use "the math SR" is only naive nonsense.
Let me RECAP: You CAN'T measure the round trip time of a K' event being in the frame K, at relative rest!. In particular, being that K' is moving at almost c.The Lorentz transform predicts what would be _measured_.
Not "PERCEIVED", whatever that might mean,
There is only one way to measure the round trip time of
a moving light clock.
That is to:
1. Compare it with a coordinate clock in K at the instant
the photon leave A, and:
2. Compare it with a coordinate clock in K at the instant
the photon is back at A after being reflected from B.
Let me RECAP:
1. Photon leave A, Coordinates of A:
In K': x₀' = 0, t₀' = 0
In K: x₀ = 0, t₀ = 0
2: Photon back at A, Coordinates of A:
In K': x₁' = 0, t₁' = 2L₀/c
In K:
x₁ = γ(x₁'+ v⋅t₁') = γ(0 + v⋅2L₀/c) = (2L₀v/c)/√(1 −v²/c²)
t₁ = γ(t₁'+ (v/c²)⋅x₁') = γ(L₀/c + (v/c²)⋅L₀) = (2L₀/c)/√(1 −v²/c²)
This means:
Clock tA has advanced the proper time (t₁'-t₀') = 2L₀/c
while the coordinate time of K has advanced
(t₁-t₀) = (2L₀/c)/√(1 −v²/c²)
In K the moving clock is measured to run slow by √(1 −v²/c²).
-----
Since you are so interested in the Lorentz contraction, we could ask:
Where in K is B at the time t₁ = γ2L₀/c ?
Coordinates of B:
In K': x₂' = L₀, t₂' = ?
In K:
t₁ = γ(t₂'+ (v/c²)⋅x₂') = γ2L₀/c => t₂' = 2L₀/c - vL₀/c² x₂ = γ(x₂'+ v⋅t₂') = γ(L₀ + v⋅(2L₀/c -vL₀/c²)) = γ⋅L₀(1+2v/c-v²/c²)
The distance between A and B measured in K is:
x₂-x₁ = γ⋅L₀(1+2v/c-v²/c²)- γ⋅2L₀v/c = γ⋅L₀(1-v²/c²) = L₀⋅√(1−v²/c²)
In K, the proper distance between the mirrors L₀ is Lorentz contracted (measured to be) L₀⋅√(1−v²/c²)
------
The above is what SR predicts, and if you want to prove SR inconsistent
you have to show that the Lorentz transform can produce alternative
results to the above!
If you want to know the K' round trip time from K, read THE FUCKING DIGITAL DISPLAY IN K'!!So when you proved SR inconsistent you to read
THE FUCKING DIGITAL DISPLAY IN K' which showed that
the round trip time in K' was 1μs, and then you PERCEIVED
what it would be in K? :-D
And that is the ONLY TRUTH in your pseudo-scientific relativism.
Because mathematics (Lorentz) IS NOT PHYSICS. But you'll never learn, fanatic indoctrinated relativist!
Your early marriage with the Lorentz crap and Einstein pseudo-philosophy ruined your brain since decades ago. Now it's USELESS.So now, when you have realized that you are unable to prove
the Lorentz transform inconsistent, you overturn the table
so the pieces are spread on the floor, shouting:
"I don't want to play the game 'proving SR inconsistent' anymore,
because mathematics (Lorentz) IS NOT PHYSICS. SO THERE"
I accept your surrender.
--
Paul
https://paulba.no/
Stating it as simply as I can:
1) Put a DIGITAL DISPLAY pointing to K origin, so it displays the time of the moving light clock tA.
2) Now, while CALCULATING the time elapsed in the moving frame, being at the K origin, READ AND COMPARE the remote digital display.
Two completely different values:
- THE REAL ONE: What is observed in the digital display.
c = 299792458 m/s = 299.792458 m/μs
Two perfect mirrors (A,B) are placed 149,896229 m apart, being A at the origin of the relatively moving frame K'.
A photon, in the visible light range, is emitted from the mirror A (which also contains a clock) exactly towards mirror B.
The photon bounces back at mirror B, towards mirror A. Its arrival from the roundtrip is marked in tA clock as being exactly 1.0000000 μs.
This process continues forever, without any losses, while A clock is accumulating 1.0000000 μs counts.
The arrangement in K' moves at v speed wrt the origin at K, where a clock tK is located (it also ticks at 1.0000000 μs.
Both clocks (tK and tA) are reset to zero exactly when A pass by the origin K, and it forces the emission of the photon.
Then, tK and tA are synchronized in the einstenian way.
The speed v = 0.9 c = 269813212 m/s
WHAT SR (LORENTZ) SAYS THAT HAPPENS AFTERWARDS.
γ = 2.294157339
LENGTH CONTRACTION (Horizontal light clock is perceived to run faster) http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c1
x' = γ (x - vt)
Δx' = AB AND Δx = AB/γ = 0.435889894 AB, as measured in the frame at rest. As both measurements are made simultaneously in the frame at rest, there is A PERCEIVED LENGTH CONTRACTION of AB in the moving frame.
But the speed of light IS INVARIANT wrt to any frame, so IT'S PERCEIVED that clock tA IS RUNNING FASTER.
The roundtrip of the photon is PERCEIVED to travel 2 x 0.435889894 AB = 0.871779789 x AB, which means that the clock's tick in the moving frame IS PERCEIVED TO TICK every 0.871779789 μs, instead of 1.00000000 μs.
TIME DILATION (Horizontal light clock is perceived to run slower) http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/tdil.html#c2
t' = γ (t - vx/c)
Δt' = Δt/γ = 0.435889894 Δt (Time is PERCEIVED as running slower in the moving frame).
For every 1.00000000 μs in the frame at rest, it's PERCEIVED that only 0.435889894 μs passed in the moving frame.
The time measurements are made in the same location, so it doesn't affect the results.
CONCLUSION: Using an horizontal light clock, the results of TIME FLOW due to length contraction and time dilation are in conflict. Clock tA is PERCEIVED to run faster or slower than clock tK, in the two different transforms.
QUESTION: WHERE IS THE TRICK EMBEDDED IN THIS PARADOX?The trick is that both length contraction and time dilation are merely alterations of the units of measure. Such an alteration would result in two different speeds of light in the two frames.
On Saturday, November 11, 2023 at 9:48:27 AM UTC-3, Paul B. Andersen wrote:
Let me RECAP:
1. Photon leave A, Coordinates of A:
In K': x₀' = 0, t₀' = 0
In K: x₀ = 0, t₀ = 0
2: Photon back at A, Coordinates of A:
In K': x₁' = 0, t₁' = 2L₀/c
In K:
x₁ = γ(x₁'+ v⋅t₁') = γ(0 + v⋅2L₀/c) = (2L₀v/c)/√(1 −v²/c²)
t₁ = γ(t₁'+ (v/c²)⋅x₁') = γ(L₀/c + (v/c²)⋅L₀) = (2L₀/c)/√(1 −v²/c²)
1) Put a DIGITAL DISPLAY pointing to K origin, so it displays the time of the moving light clock tA.
2) Now, while CALCULATING the time elapsed in the moving frame,
being at the K origin, READ AND COMPARE the remote digital display.
You request now that it sends out an "image" of this value (these values). Thus, clock A will sequentially display 0,1,2, ....
On Saturday, November 11, 2023 at 9:48:27 AM UTC-3, Paul B. Andersen wrote:
There is only one way to measure the round trip time of
a moving light clock.
That is to:
1. Compare it with a coordinate clock in K at the instant
the photon leave A, and:
2. Compare it with a coordinate clock in K at the instant
the photon is back at A after being reflected from B.
Let me RECAP:
1. Photon leave A, Coordinates of A:
In K': x₀' = 0, t₀' = 0
In K: x₀ = 0, t₀ = 0
2: Photon back at A, Coordinates of A:
In K': x₁' = 0, t₁' = 2L₀/c
In K:
x₁ = γ(x₁'+ v⋅t₁') = γ(0 + v⋅2L₀/c) = (2L₀v/c)/√(1 −v²/c²)
t₁ = γ(t₁'+ (v/c²)⋅x₁') = γ(L₀/c + (v/c²)⋅L₀) = (2L₀/c)/√(1 −v²/c²)
This means:
Clock tA has advanced the proper time (t₁'-t₀') = 2L₀/c
while the coordinate time of K has advanced
(t₁-t₀) = (2L₀/c)/√(1 −v²/c²)
In K the moving clock is measured to run slow by √(1 −v²/c²).
-----
Stating it as simply as I can:
1) Put a DIGITAL DISPLAY pointing to K origin, so it displays the time of the moving light clock tA.
2) Now, while CALCULATING the time elapsed in the moving frame, being at the K origin, READ AND COMPARE the remote digital display.
Two completely different values:
- THE REAL ONE: What is observed in the digital display.
- THE PSEUDO-SCIENTIFIC calculation: What the observer at rest in K origin get, when applying Lorentz.
REALITY VS. MYSTICISM.
Now tell me, Paul, what the FUCK is the value of the Lorentz's results in the real world? Do you dispute the reading of the digital display?
Of course, you do. Your credence is in danger and you have to write a tirade of shitty, retorted reasons to defend your posture.
But electronics KILLS relativity. Try something for your unpleasant butthurt.
So we can conclude that SR is consistent! Can't we?
Den 11.11.2023 17:30, skrev Richard Hertz:
So we can conclude that SR is consistent! Can't we?
The following appears very confused.
Are you really serious, or are you joking.
1) Put a DIGITAL DISPLAY pointing to K origin, so it displays the time of the moving light clock tA.
2) Now, while CALCULATING the time elapsed in the moving frame, being at the K origin, READ AND COMPARE the remote digital display.
Two completely different values:
- THE REAL ONE: What is observed in the digital display.There is no such thing as a _real_ light clock. Of obvious reasons!
And you will put a digital display on it to see what it _really_ shows!
I hope for your sake that you are joking.
(But I don't believe so!)
- THE PSEUDO-SCIENTIFIC calculation: What the observer at rest in K origin get, when applying Lorentz.
REALITY VS. MYSTICISM.
Now tell me, Paul, what the FUCK is the value of the Lorentz's results in the real world? Do you dispute the reading of the digital display?
Of course, you do. Your credence is in danger and you have to write a tirade of shitty, retorted reasons to defend your posture.
But electronics KILLS relativity. Try something for your unpleasant butthurt.
Electronics KILLS thought experiments!
One of your better! :-D
--
Paul
https://paulba.no/
On Sunday, November 12, 2023 at 11:13:30 AM UTC-3, Paul B. Andersen wrote:
So we can conclude that SR is consistent! Can't we?NO, doctrine-blinded and religion-clothed-brain-arteries, NO!
SR is a deceptive pseudo-science, which FORCES YOU TO BELIEVE OR ELSE.
The time registered in the moving frame by anything (human, ghost or instruments)
IS IMMUNE to your stupid Lorentz equations. Time flows normally there.
Throw away the light clock and use a hydrogen maser atomic clock instead.
Don't use the fucking giant digital display.
Use a 1-way microwave link to transmit to the origin the DIGITAL READOUT of the moving clock,
up to picoseconds resolution.
At origin of K compare the output of the encoded data with the LOCAL CLOCK tK: THEY ARE THE SAME!
It's the STUPID LORENTZ EQUATION THAT FORCES YOU TO BELIEVE THAT THEY ARE DIFFERENT,
On Sunday, November 12, 2023 at 5:12:21 PM UTC-5, Richard Hertz wrote:
On Sunday, November 12, 2023 at 11:13:30 AM UTC-3, Paul B. Andersen wrote:
Then show us what step, what calculation was wrong.So we can conclude that SR is consistent! Can't we?NO, doctrine-blinded and religion-clothed-brain-arteries, NO!
He (and I) showed you in very simple algebra steps what SR predicts.
No. SR makes the predictions. The actual exps "force" us to believe.
On Sunday, November 12, 2023 at 11:13:30 AM UTC-3, Paul B. Andersen wrote:
Den 11.11.2023 17:30, skrev Richard Hertz:On Saturday, November 11, 2023 at 9:48:27 AM UTC-3, Paul B. Andersen wrote:
So back to the issue of the thread shown in the subject line:
According to Richard Hertz there are "Conflicts in SR when
using horizontal light clocks in the moving frame."
Richard Hertz claims that SR can predict two different values
for the round trip time in the "stationary frame".
Let's see how he did it:
Richards starts with the equation x' = γ(x - vt) and claims that
LENGTH CONTRACTION leads to the conclusion that the round trip time measured (He uses "perceived") in the stationary frame K is (2L₀/c)/γ,
which means that the light clock is perceived to run fast, not slow.
He hasn't calculated the length contraction, though. You can't do that
from the equation x' = γ(x - vt) alone. But he knows that according
to SR, the Lorentz contraction is L'= L₀/γ.
He has made the very naive calculation that since the length is contracted by 1/γ in K, then the round trip time in K must be (2L₀/γ)/c.
This is horribly wrong, because the mirrors are moving in K.
(And this is the only calculation he has made!)
Here is the correct calculation of the round trip time based
on the correct Lorentz contraction L' = L₀/γ:
In the following is tf the transit time from A to B,
and tb is the transit time from B to A.
|<--------------c⋅tf---------------->|
|<----------L'------------->|<-v⋅tf->|
c⋅tf = L' + v⋅tf => tf = L'/(c-v)
|<-----------L'------------>|
|<--------c⋅tb----->|<-v⋅tb->|
L' = c⋅tb + v⋅tb => tb = L'/(c+v)
t₁ = tf + tb = 2L'⋅c/(c² + v²) = (2L'/c)⋅γ² = (2L₀/c)⋅γ
Which is correct, and the same as predicted by the Lorentz transform.
-------------------
Then he starts with the equation t' = γ (t - vx/c)
and claim that TIME DILATION leads to the conclusion that the round trip time measured in the stationary frame K is (2L₀/c)⋅γ.
Which is correct, and the same as predicted by the Lorentz transform.
Again, he hasn't calculated the time dilation from the equation
t' = γ (t - vx/c), but he knows that the time dilation according
to SR is Δt' = Δt/γ. And since this IS time dilation, he gets it right. >>
So we can conclude that SR is consistent! Can't we?
NO, doctrine-blinded and religion-clothed-brain-arteries, NO!
SR is a deceptive pseudo-science, which FORCES YOU TO BELIEVE OR ELSE.
The time registered in the moving frame by anything (human, ghost or instruments) IS IMMUNE to your stupid Lorentz equations. Time flows normally there.
Throw away the light clock and use a hydrogen maser atomic clock instead. Don't use the fucking giant digital display.
Use a 1-way microwave link to transmit to the origin the DIGITAL READOUT of the moving clock, up to picoseconds resolution.
At origin of K compare the output of the encoded data with the LOCAL CLOCK tK: THEY ARE THE SAME!
It's the STUPID LORENTZ EQUATION THAT FORCES YOU TO BELIEVE THAT THEY ARE DIFFERENT, BECAUSE YOU ARE FORCED TO BELIEVE!
That's the tragedy of this shitty pseudo-science: Took a mind like yours, that COULD HAVE BEEN useful before SR absorption of your reasoning,
and converted it in something small, with fewer capabilities than the brain OF A FUCKING PARROT!
I feel sorry for your wasted life. Now it's too late for you. Deal with it.
Den 12.11.2023 23:12, skrev Richard Hertz:
On Sunday, November 12, 2023 at 11:13:30 AM UTC-3, Paul B. Andersen wrote:
Den 11.11.2023 17:30, skrev Richard Hertz:On Saturday, November 11, 2023 at 9:48:27 AM UTC-3, Paul B. Andersen wrote:
So back to the issue of the thread shown in the subject line:
According to Richard Hertz there are "Conflicts in SR when
using horizontal light clocks in the moving frame."
Richard Hertz claims that SR can predict two different values
for the round trip time in the "stationary frame".
Let's see how he did it:
Richards starts with the equation x' = γ(x - vt) and claims that
LENGTH CONTRACTION leads to the conclusion that the round trip time measured (He uses "perceived") in the stationary frame K is (2L₀/c)/γ,
which means that the light clock is perceived to run fast, not slow.
He hasn't calculated the length contraction, though. You can't do that
from the equation x' = γ(x - vt) alone. But he knows that according
to SR, the Lorentz contraction is L'= L₀/γ.
He has made the very naive calculation that since the length is contracted by 1/γ in K, then the round trip time in K must be (2L₀/γ)/c.
This is horribly wrong, because the mirrors are moving in K.
(And this is the only calculation he has made!)
Here is the correct calculation of the round trip time based
on the correct Lorentz contraction L' = L₀/γ:
In the following is tf the transit time from A to B,
and tb is the transit time from B to A.
|<--------------c⋅tf---------------->|
|<----------L'------------->|<-v⋅tf->|
c⋅tf = L' + v⋅tf => tf = L'/(c-v)
|<-----------L'------------>|
|<--------c⋅tb----->|<-v⋅tb->|
L' = c⋅tb + v⋅tb => tb = L'/(c+v)
t₁ = tf + tb = 2L'⋅c/(c² + v²) = (2L'/c)⋅γ² = (2L₀/c)⋅γ
Which is correct, and the same as predicted by the Lorentz transform.
-------------------
Then he starts with the equation t' = γ (t - vx/c)
and claim that TIME DILATION leads to the conclusion that the round trip time measured in the stationary frame K is (2L₀/c)⋅γ.
Which is correct, and the same as predicted by the Lorentz transform.
Again, he hasn't calculated the time dilation from the equation
t' = γ (t - vx/c), but he knows that the time dilation according
to SR is Δt' = Δt/γ. And since this IS time dilation, he gets it right.
So we can conclude that SR is consistent! Can't we?
NO, doctrine-blinded and religion-clothed-brain-arteries, NO!
SR is a deceptive pseudo-science, which FORCES YOU TO BELIEVE OR ELSE.May I remind you: Your opinion of SR is irrelevant.
The issue is if SR is inconsistent.
You believed you had proved SR inconsistent, but as I have shown
above, you only demonstrated your poor mathematical skills.
SR is a consistent theory,
and the only way you can falsify it
is by performing an experiments which prove its predictions wrong.
Den 12.11.2023 23:12, skrev Richard Hertz:
On Sunday, November 12, 2023 at 11:13:30 AM UTC-3, Paul B. Andersen wrote:
Den 11.11.2023 17:30, skrev Richard Hertz:On Saturday, November 11, 2023 at 9:48:27 AM UTC-3, Paul B. Andersen wrote:
So back to the issue of the thread shown in the subject line:
According to Richard Hertz there are "Conflicts in SR when
using horizontal light clocks in the moving frame."
Richard Hertz claims that SR can predict two different values
for the round trip time in the "stationary frame".
Let's see how he did it:
Richards starts with the equation x' = γ(x - vt) and claims that
LENGTH CONTRACTION leads to the conclusion that the round trip time measured (He uses "perceived") in the stationary frame K is (2L₀/c)/γ,
which means that the light clock is perceived to run fast, not slow.
He hasn't calculated the length contraction, though. You can't do that
from the equation x' = γ(x - vt) alone. But he knows that according
to SR, the Lorentz contraction is L'= L₀/γ.
He has made the very naive calculation that since the length is contracted by 1/γ in K, then the round trip time in K must be (2L₀/γ)/c.
This is horribly wrong, because the mirrors are moving in K.
(And this is the only calculation he has made!)
Here is the correct calculation of the round trip time based
on the correct Lorentz contraction L' = L₀/γ:
In the following is tf the transit time from A to B,
and tb is the transit time from B to A.
|<--------------c⋅tf---------------->|
|<----------L'------------->|<-v⋅tf->|
c⋅tf = L' + v⋅tf => tf = L'/(c-v)
|<-----------L'------------>|
|<--------c⋅tb----->|<-v⋅tb->|
L' = c⋅tb + v⋅tb => tb = L'/(c+v)
t₁ = tf + tb = 2L'⋅c/(c² + v²) = (2L'/c)⋅γ² = (2L₀/c)⋅γ
Which is correct, and the same as predicted by the Lorentz transform.
-------------------
Then he starts with the equation t' = γ (t - vx/c)
and claim that TIME DILATION leads to the conclusion that the round trip time measured in the stationary frame K is (2L₀/c)⋅γ.
Which is correct, and the same as predicted by the Lorentz transform.
Again, he hasn't calculated the time dilation from the equation
t' = γ (t - vx/c), but he knows that the time dilation according
to SR is Δt' = Δt/γ. And since this IS time dilation, he gets it right.
So we can conclude that SR is consistent! Can't we?
NO, doctrine-blinded and religion-clothed-brain-arteries, NO!
SR is a deceptive pseudo-science, which FORCES YOU TO BELIEVE OR ELSE.May I remind you: Your opinion of SR is irrelevant.
The issue is if SR is inconsistent.
You believed you had proved SR inconsistent, but as I have shown
above, you only demonstrated your poor mathematical skills.
SR is a consistent theory, and the only way you can falsify it
is by performing an experiments which prove its predictions wrong.
But you have to PERFORM the experiment in the real word.
Claiming that the experiment below MUST show what you think
it will show is pathetic.
And a little funny. :-D
The time registered in the moving frame by anything (human, ghost or instruments) IS IMMUNE to your stupid Lorentz equations. Time flows normally there.
Throw away the light clock and use a hydrogen maser atomic clock instead. Don't use the fucking giant digital display.
Use a 1-way microwave link to transmit to the origin the DIGITAL READOUT of the moving clock, up to picoseconds resolution.
At origin of K compare the output of the encoded data with the LOCAL CLOCK tK: THEY ARE THE SAME!So there is no delay in the microwave link?
Or will you compensate for the unknown distance and
unknown speed of the moving clock?
It's the STUPID LORENTZ EQUATION THAT FORCES YOU TO BELIEVE THAT THEY ARE DIFFERENT, BECAUSE YOU ARE FORCED TO BELIEVE!Real several experiments with atomic clock are performed.
The designers of the experiments were smarter that you,
so the experiments worked.
https://paulba.no/paper/Ives_Stilwell.pdf https://paulba.no/paper/Hafele_Keating.pdf
https://paulba.no/paper/Alley.pdf
see pages 708-716 https://paulba.no/paper/Initial_results_of_GPS_satellite_1977.pdf https://paulba.no/paper/Vessot.pdf
Even you know what these experiment show.
You can, and will, kick and scream and claim that
the experiments are faked.
It will only make you look ignorant and pathetic.
That's the tragedy of this shitty pseudo-science: Took a mind like yours, that COULD HAVE BEEN useful before SR absorption of your reasoning,
and converted it in something small, with fewer capabilities than the brain OF A FUCKING PARROT!
I feel sorry for your wasted life. Now it's too late for you. Deal with it.Richard, you are not shouting loud enough.
Use more capitals, and more profanities.
One FUCKING PARROT won't do!
Are you getting weak in your old days?
the cesium-133 atom is 9192631770 Hz by definition.
Everywhere. Always.
Fact! Nothing to discuss.
In your website table you have the following data.
r rate
----------------------
10R 1.000000000626
9R 1.000000000618
8R 1.000000000608
7R 1.000000000596
6R 1.000000000579
5R 1.000000000556
4R 1.000000000522
3R 1.000000000464
2R 1.000000000348
1R 1.000000000000
https://paulba.no/temp/ClockRate.pdf
Is that picoseconds gained per second?
How would you calculate what frequency the rest frame
(1R) c-133 atom frequency of 9192661770 hz would be at 2R?
On Monday, 13 November 2023 at 18:36:49 UTC, Paul B. Andersen wrote:
Den 13.11.2023 14:36, skrev Lou:
In your website table you have the following data.
r rate
----------------------
10R 1.000000000626
9R 1.000000000618
8R 1.000000000608
7R 1.000000000596
6R 1.000000000579
5R 1.000000000556
4R 1.000000000522
3R 1.000000000464
2R 1.000000000348
1R 1.000000000000
https://paulba.no/temp/ClockRate.pdf
Is that picoseconds gained per second?Read the statement above the table:
“r" is the distance from the clock to the centre of the Earth.
"R" is the radius of the Earth.
The clock transmits a frequency 1 Hz,
"rate" is the frequency received on the ground.
Gravitational Doppler shift, Gravitational blue shift.
How would you calculate what frequency the rest frameIt can't be "calculated".
(1R) c-133 atom frequency of 9192661770 hz would be at 2R?
The frequency of the photon associated with
the ground-state hyperfine transition of
the cesium-133 atom is 9192631770 Hz by definition.
Everywhere. Always.
Fact! Nothing to discuss.
Yes I realise that under GR the rule is c-133 frequency never changes.
Under GR its always everywhere the same. But
Den 13.11.2023 14:36, skrev Lou:
In your website table you have the following data.
r rate
----------------------
10R 1.000000000626
9R 1.000000000618
8R 1.000000000608
7R 1.000000000596
6R 1.000000000579
5R 1.000000000556
4R 1.000000000522
3R 1.000000000464
2R 1.000000000348
1R 1.000000000000
https://paulba.no/temp/ClockRate.pdf
Is that picoseconds gained per second?Read the statement above the table:
“r" is the distance from the clock to the centre of the Earth.
"R" is the radius of the Earth.
The clock transmits a frequency 1 Hz,
"rate" is the frequency received on the ground.
Gravitational Doppler shift, Gravitational blue shift.
How would you calculate what frequency the rest frameIt can't be "calculated".
(1R) c-133 atom frequency of 9192661770 hz would be at 2R?
The frequency of the photon associated with
the ground-state hyperfine transition of
the cesium-133 atom is 9192631770 Hz by definition.
Everywhere. Always.
Fact! Nothing to discuss.
Den 13.11.2023 14:36, skrev Lou:
In your website table you have the following data.
r rate
----------------------
10R 1.000000000626
9R 1.000000000618
8R 1.000000000608
7R 1.000000000596
6R 1.000000000579
5R 1.000000000556
4R 1.000000000522
3R 1.000000000464
2R 1.000000000348
1R 1.000000000000
https://paulba.no/temp/ClockRate.pdf
Is that picoseconds gained per second?Read the statement above the table:
“r" is the distance from the clock to the centre of the Earth.
"R" is the radius of the Earth.
The clock transmits a frequency 1 Hz,
"rate" is the frequency received on the ground.
Gravitational Doppler shift, Gravitational blue shift.
How would you calculate what frequency the rest frameIt can't be "calculated".
(1R) c-133 atom frequency of 9192661770 hz would be at 2R?
The frequency of the photon associated with
the ground-state hyperfine transition of
the cesium-133 atom is 9192631770 Hz by definition.
Everywhere. Always.
Fact! Nothing to discuss.
--
Paul
https://paulba.no/
On Monday, 13 November 2023 at 18:36:49 UTC, Paul B. Andersen wrote:
Den 13.11.2023 14:36, skrev Lou:
In your website table you have the following data.
r rate
----------------------
10R 1.000000000626
9R 1.000000000618
8R 1.000000000608
7R 1.000000000596
6R 1.000000000579
5R 1.000000000556
4R 1.000000000522
3R 1.000000000464
2R 1.000000000348
1R 1.000000000000
https://paulba.no/temp/ClockRate.pdf
Read the statement above the table:
“r" is the distance from the clock to the centre of the Earth.
"R" is the radius of the Earth.
The clock transmits a frequency 1 Hz,
"rate" is the frequency received on the ground.
Gravitational Doppler shift, Gravitational blue shift.
How would you calculate what frequency the rest frame
(1R) c-133 atom frequency of 9192661770 hz would be at 2R?
It can't be "calculated".
The frequency of the photon associated with
the ground-state hyperfine transition of
the cesium-133 atom is 9192631770 Hz by definition.
Everywhere. Always.
Fact! Nothing to discuss.
Yes I realise that under GR the rule is c-133 frequency never changes.
Under GR its always everywhere the same.
But you say a broadcast
frequency can appear to be blueshifted to an observer on the ground
and I’m trying to find out what that blueshifted f would be to an
earth ground observer.
So if it were possible to observe, from the ground, a clock orbiting at 2R.
What frequency would 9192631770 Hz on a clock at 2R be blueshifted to
an observer on the ground be? Would it appear to a ground observer
to be 9192631773.48hz ?
Yes I realise that under GR the rule is c-133 frequency never changes. Under GR its always everywhere the same.Not "under GR". In the real world. Everywhere. Always. By definition!
Yes I realise that under GR the rule is c-133 frequency never changes. Under GR its always everywhere the same.Not "under GR". In the real world. Everywhere. Always. By definition!
But you say a broadcastNot "can appear to be". IS blue shifted. Measured in the real world.
frequency can appear to be blueshifted to an observer on the ground
and I’m trying to find out what that blueshifted f would be to an
earth ground observer.
https://paulba.no/paper/Pound&Rebka.pdf
If the receiver is above the sender, it is gravitational red shift.
So if it were possible to observe, from the ground, a clock orbiting at 2R.Not orbiting! Stationary in the ECI frame at 2R.
What frequency would 9192631770 Hz on a clock at 2R be blueshifted toThe frequency of the received signal on the ground would
an observer on the ground be? Would it appear to a ground observer
to be 9192631773.48hz ?
be gravitational blue shifted. That's a kind of Doppler shift.
That you hear the train-whistle change frequency when
the train passes you doesn't mean that the emitted
frequency has changed.
f_received = 1.000000000348⋅f_transmitted = 9192631773.1990 Hz
BECAUSE f_transmitted = 9192631770 Hz.
On Monday, November 13, 2023 at 3:36:49 PM UTC-3, Paul B. Andersen wrote:
First, you pollute this thread arguing nonsense around my OP.
Now you introduce more crap, playing with Einstein's 1911 crap. Why don't you take a rest?
You introduce gravitational time dilation playing with 1911 Einstein's shit:
snip fresh imbecilities
Yes I realise that under GR the rule is c-133 frequency never changes.
Under GR its always everywhere the same.
Not "under GR". In the real world. Everywhere. Always. By definition!
But you say a broadcast
frequency can appear to be blueshifted to an observer on the ground
and I’m trying to find out what that blueshifted f would be to an
earth ground observer.
Not "can appear to be". IS blue shifted. Measured in the real world.
https://paulba.no/paper/Pound&Rebka.pdf
If the receiver is above the sender, it is gravitational red shift.
So if it were possible to observe, from the ground, a clock orbiting at 2R. >> Not orbiting! Stationary in the ECI frame at 2R.
What frequency would 9192631770 Hz on a clock at 2R be blueshifted to
an observer on the ground be? Would it appear to a ground observer
to be 9192631773.48hz ?
The frequency of the received signal on the ground would
be gravitational blue shifted. That's a kind of Doppler shift.
That you hear the train-whistle change frequency when
the train passes you doesn't mean that the emitted
frequency has changed.
f_received = 1.000000000348⋅f_transmitted = 9192631773.1990 Hz
BECAUSE f_transmitted = 9192631770 Hz.
Interesting thanks. But are you sure it’s correct to convert the gps signal frequency gain of 348 pico seconds at 10.23 Mhz to the gain of 3.19 Hz
for the recieved C-133 frequency using that method?
Because if I try your same formula for the transmitted f of 10.23MHz, I
get the following:
f_received = 1.000000000348 x 10.23Mhz =10230000.0036 hz
I would have thought it should be 10230000.0038 hz.
Oh, true, you ran away from it. We have cornered you and you havent presented any counterargument whatsoever.
You havent answered one question psoed to you and havent answered one clarification posed to you.
You just run away or change subject. Now, isn't that the behavior of a crank?
Den 14.11.2023 15:01, skrev Lou:
Who wrote what? DON'T SNIP THE ATTRIBUTIONS!
Yes I realise that under GR the rule is c-133 frequency never changes. >>> Under GR its always everywhere the same.
Not "under GR". In the real world. Everywhere. Always. By definition!
But you say a broadcast
frequency can appear to be blueshifted to an observer on the ground
and I’m trying to find out what that blueshifted f would be to an
earth ground observer.
Not "can appear to be". IS blue shifted. Measured in the real world.
https://paulba.no/paper/Pound&Rebka.pdf
If the receiver is above the sender, it is gravitational red shift.
So if it were possible to observe, from the ground, a clock orbiting at 2R.Not orbiting! Stationary in the ECI frame at 2R.
What frequency would 9192631770 Hz on a clock at 2R be blueshifted to >>> an observer on the ground be? Would it appear to a ground observer
to be 9192631773.48hz ?
The frequency of the received signal on the ground would
be gravitational blue shifted. That's a kind of Doppler shift.
That you hear the train-whistle change frequency when
the train passes you doesn't mean that the emitted
frequency has changed.
f_received = 1.000000000348⋅f_transmitted = 9192631773.1990 Hz
BECAUSE f_transmitted = 9192631770 Hz.
Interesting thanks. But are you sure it’s correct to convert the gps signalThis statement is meaningless.
frequency gain of 348 pico seconds at 10.23 Mhz to the gain of 3.19 Hz
for the recieved C-133 frequency using that method?
A GPS satellite is at r = 4.187883R
and is moving in the ECI frame at the speed 3.874 km/s.
https://paulba.no/pdf/GPS_clock_rate.pdf
Because if I try your same formula for the transmitted f of 10.23MHz, I get the following:A more precise value is:
f_received = 1.000000000348 x 10.23Mhz =10230000.0036 hz
f_received = 1.000000000347674 x 10.23Mhz = 10230000.0035567 Hz
I would have thought it should be 10230000.0038 hz.OK, so that's what you believe.
On Tuesday, 14 November 2023 at 19:33:59 UTC, Paul B. Andersen wrote:
A GPS satellite is at r = 4.187883R
and is moving in the ECI frame at the speed 3.874 km/s.
https://paulba.no/pdf/GPS_clock_rate.pdf
Using your formula I get approximately 10230000.0054hz
recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
I can’t find any reference to confirm this. Do you know if that
approx frequency is observed at earth ground stations?
But a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.
It is however impossible to directly measure this frequency on the ground, because the Doppler shift due
to the orbital speed of the satellite is order of magnitudes greater than the shift due to GR, and it
is changing all the time.
But the ground stations can read what the satellite clock shows, and can compare it to clocks on the ground (GPS-time, UTC).
The difference will increase with time.
On Wednesday, November 15, 2023 at 3:20:28 PM UTC-3, Paul B. Andersen wrote:
<snip>
But a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.
It is however impossible to directly measure this frequency on the ground, because the Doppler shift due
to the orbital speed of the satellite is order of magnitudes greater than the shift due to GR, and it
is changing all the time.
But the ground stations can read what the satellite clock shows, and can compare it to clocks on the ground (GPS-time, UTC).
The difference will increase with time.
https://paulba.no/pdf/GPS_clock_rate.pdf
Relativist, you are so full of shit that it's extremely disgusting.
So, relativistic effects can't be measured in any way because Doppler effect bury them 1,000 times under its values.
But EVERYONE have to trust you about the readings of the ground clock, because YOU HAD A JOB measuring it.
LIAR, DECEIVER, HYPOCRITE!
Den 15.11.2023 21:53, skrev Richard Hertz:
On Wednesday, November 15, 2023 at 3:20:28 PM UTC-3, Paul B. Andersen
wrote:
<snip>
But a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.
It is however impossible to directly measure this frequency on the
ground, because the Doppler shift due
to the orbital speed of the satellite is order of magnitudes greater
than the shift due to GR, and it
is changing all the time.
But the ground stations can read what the satellite clock shows, and
can compare it to clocks on the ground (GPS-time, UTC).
The difference will increase with time.
Relativist, you are so full of shit that it's extremely disgusting.
So, relativistic effects can't be measured in any way because Doppler
effect bury them 1,000 times under its values.
The satellite is moving, the ground is moving, so the Doppler
shift of the carrier signals from a satellite will be anything
in the range (1 ± 1E-7). The receiver will obviously not measure
this frequency. Why should it?
The bandwidth of the receiver is bigger that that.
Receivers can receive the signals of up to 12 satellites at
the same time, and all the signals will be differently Doppler shifted.
They are separated by their PRN code. Phase locked loops.
Didn't you know this, Richard?
Did you believe that the receiver could measure the frequency
to a precision of 1E-12?
But EVERYONE have to trust you about the readings of the ground clock,
because YOU HAD A JOB measuring it.
LIAR, DECEIVER, HYPOCRITE!
Mind the blood pressure, Richard. :-D
This was done in the first GPS-satellite with a clock
which was not corrected for relativistic effects.
Den 15.11.2023 14:29, skrev Lou:
On Tuesday, 14 November 2023 at 19:33:59 UTC, Paul B. Andersen wrote:You can find the answer to your question below here.
A GPS satellite is at r = 4.187883R
and is moving in the ECI frame at the speed 3.874 km/s.
https://paulba.no/pdf/GPS_clock_rate.pdf
Using your formula I get approximately 10230000.0054hzThe received frequency from a stationary satellite
recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
I can’t find any reference to confirm this. Do you know if that
approx frequency is observed at earth ground stations?
at GPS altitude which transmits the frequency 10.23 MHz
would be 10.230000005408 MHz.
But a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.
It is however impossible to directly measure this
frequency on the ground, because the Doppler shift due
to the orbital speed of the satellite is order of
magnitudes greater than the shift due to GR, and it
is changing all the time.
But the ground stations can read what the satellite clock shows,
and can compare it to clocks on the ground (GPS-time, UTC).
The difference will increase with time.
This was done in the first GPS-satellite with a clock
which was not corrected for relativistic effects.
https://paulba.no/paper/Initial_results_of_GPS_satellite_1977.pdf
On Wednesday, 15 November 2023 at 18:20:28 UTC, Paul B. Andersen wrote:
Den 15.11.2023 14:29, skrev Lou:
Using your formula I get approximately 10230000.0054hz
recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
I can’t find any reference to confirm this. Do you know if that
approx frequency is observed at earth ground stations?
The received frequency from a stationary satellite
at GPS altitude which transmits the frequency 10.23 MHz
would be 10.230000005408 MHz.
But a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.
Interesting facts thanks for clarifying things. Still a bit of discrepancy though.The current NIST reference says the gps sat clock f is designated
as 10229999.99543Hz.
If I assume gps at 4.12R ** then it should have an approx picosecond gain
of 1.000000000542s
So using your formula I get:
10229999.99543 × 1.000000000542 = 10230000.001hz.
Small error but maybe still significant.
To get exactly 10.23 Mhz for GR clock gains at receiver I need a calculation using closer to a 3R altitude clock gain of only 450 picoseconds.
Not 542 picoseconds
10229999.99543 × 1.00000000045 = 10.23 Mhz.
That’s equivelent to one whole radius r distance based on your table
below
As you can see below the sat clock f setting isn’t
matching GR predictions. Is that because the sat f
also has to take account the various Doppler speeds
you mentioned? I would have thought that the average
Doppler shift over time would be zero seeing as 1/2 the
time it’s(gps sat) coming towards reciever, 1/2 time away.
**
10R 1.000000000626 0.0100
9R 1.000000000618 0.0123
8R 1.000000000608 0.0156
7R 1.000000000596 0.0204
6R 1.000000000579 0.0278
5R 1.000000000556 0.0400
4.12 1.000000000542. *
4R 1.000000000522 0.0625
3R 1.000000000464 0.1111 *
2R 1.000000000348 0.2500
1R 1.000000000000 1.0000
https://paulba.no/temp/ClockRate.pdf
Den 16.11.2023 13:58, skrev Lou:
On Wednesday, 15 November 2023 at 18:20:28 UTC, Paul B. Andersen wrote:
Den 15.11.2023 14:29, skrev Lou:
Using your formula I get approximately 10230000.0054hz
recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
I can’t find any reference to confirm this. Do you know if that
approx frequency is observed at earth ground stations?
As I told you and you have ignored:The received frequency from a stationary satellite
at GPS altitude which transmits the frequency 10.23 MHz
would be 10.230000005408 MHz.
THE SATELLITE IS MOVING IN THE ECI-FRAME ========================================
Note that I didn't say a GPS satellite, I said a satelliteBut a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.
in GPS orbit which transmits a 10.23 Mhz signal.
Then the signal received on the ground would be:
f = (1 + 4.4647E-10)⋅10.23 MHz = 10.230000004567389 MHz
Interesting facts thanks for clarifying things. Still a bit of discrepancy though.The current NIST reference says the gps sat clock f is designated as 10229999.99543Hz.Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
The frequency is adjusted (the famous "GR-correction") to:
f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
This frequency would observed on the ground be:
f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
(The frequency f₀ isn't transmitted at all, but it is used as reference for the satellite clock, to make it stay in sync with GPS-time (UTC).)Yes OK
If I assume gps at 4.12R ** then it should have an approx picosecond gain of 1.000000000542sThe satellite is moving!!!!
So using your formula I get:
10229999.99543 × 1.000000000542 = 10230000.001hz.
https://paulba.no/pdf/GPS_clock_rate.pdf
equation (8)
Read it!
Small error but maybe still significant.
To get exactly 10.23 Mhz for GR clock gains at receiver I need a calculation
using closer to a 3R altitude clock gain of only 450 picoseconds.
Not 542 picoseconds
10229999.99543 × 1.00000000045 = 10.23 Mhz.
That’s equivelent to one whole radius r distance based on your table below
As you can see below the sat clock f setting isn’t
matching GR predictions. Is that because the sat f
also has to take account the various Doppler speeds
you mentioned? I would have thought that the average
Doppler shift over time would be zero seeing as 1/2 the
time it’s(gps sat) coming towards reciever, 1/2 time away.
**The GPS-satellites are moving!
10R 1.000000000626 0.0100
9R 1.000000000618 0.0123
8R 1.000000000608 0.0156
7R 1.000000000596 0.0204
6R 1.000000000579 0.0278
5R 1.000000000556 0.0400
4.12 1.000000000542. *
4R 1.000000000522 0.0625
3R 1.000000000464 0.1111 *
2R 1.000000000348 0.2500
1R 1.000000000000 1.0000
https://paulba.no/temp/ClockRate.pdf
================================
Den 16.11.2023 13:58, skrev Lou:
On Wednesday, 15 November 2023 at 18:20:28 UTC, Paul B. Andersen wrote:
Den 15.11.2023 14:29, skrev Lou:
Using your formula I get approximately 10230000.0054hz
recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
I can’t find any reference to confirm this. Do you know if that
approx frequency is observed at earth ground stations?
As I told you and you have ignored:The received frequency from a stationary satellite
at GPS altitude which transmits the frequency 10.23 MHz
would be 10.230000005408 MHz.
THE SATELLITE IS MOVING IN THE ECI-FRAME ========================================
Note that I didn't say a GPS satellite, I said a satelliteBut a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.
in GPS orbit which transmits a 10.23 Mhz signal.
The satellite is moving!!!!
On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:
Den 16.11.2023 13:58, skrev Lou:
Interesting facts thanks for clarifying things. Still a bit of discrepancy >>> though.The current NIST reference says the gps sat clock f is designated >>> as 10229999.99543Hz.
Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
Yes. I just told you it was 10.229999995432612 MHz I got that from NIST.
The frequency is adjusted (the famous "GR-correction") to:
f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
This frequency would observed on the ground be:
f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
Where did you get 1 + 4.4647E-10 ?
The GPS-satellites are moving!
================================
Got it now?
“They are moving”. Could you elaborate.
Den 16.11.2023 21:16, skrev Lou:
On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:
Den 16.11.2023 13:58, skrev Lou:
Interesting facts thanks for clarifying things. Still a bit of discrepancy
though.The current NIST reference says the gps sat clock f is designated >>> as 10229999.99543Hz.
Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
Yes. I just told you it was 10.229999995432612 MHz I got that from NIST.
The frequency is adjusted (the famous "GR-correction") to:
f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
This frequency would observed on the ground be:
f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
Where did you get 1 + 4.4647E-10 ?The correction is (1 - 4.4647E-10)
so f₁ = (1/(1 - 4.4647E-10))⋅f₀ ≈ (1 + 4.4647E-10)⋅f₀
The correction is specified in the
"INTERFACE SPECIFICATION DOCUMENT" for GPS:
https://www.gps.gov/technical/icwg/IS-GPS-200N.pdf
From 3.3.1.1 Frequency Plan:
"The carrier frequencies for the L1 and L2 signals shall be
coherently derived from a common frequency source within the SV.
The nominal frequency of this source -- as it appears to an observer
on the ground -- is 10.23 MHz. The SV carrier frequency and clock
rates -- as they would appear to an observer located in the SV --
t' = γ (t - vx/c)
On Friday, 17 November 2023 at 12:10:04 UTC+1, Paul B. Andersen wrote:
Den 16.11.2023 21:16, skrev Lou:
On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:The correction is (1 - 4.4647E-10)
Den 16.11.2023 13:58, skrev Lou:
Interesting facts thanks for clarifying things. Still a bit of discrepancy
though.The current NIST reference says the gps sat clock f is designated >>>>> as 10229999.99543Hz.
Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
Yes. I just told you it was 10.229999995432612 MHz I got that from NIST. >>>
The frequency is adjusted (the famous "GR-correction") to:
f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
This frequency would observed on the ground be:
f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
Where did you get 1 + 4.4647E-10 ?
so f₁ = (1/(1 - 4.4647E-10))⋅f₀ ≈ (1 + 4.4647E-10)⋅f₀
The correction is specified in the
"INTERFACE SPECIFICATION DOCUMENT" for GPS:
https://www.gps.gov/technical/icwg/IS-GPS-200N.pdf
From 3.3.1.1 Frequency Plan:
"The carrier frequencies for the L1 and L2 signals shall be
coherently derived from a common frequency source within the SV.
The nominal frequency of this source -- as it appears to an observer
on the ground -- is 10.23 MHz. The SV carrier frequency and clock
rates -- as they would appear to an observer located in the SV --
Paul, poor halfbrain, who cares what "would appear" to a
non-existing gedanken person?
It's the measurement result
that counts, andthis is 10.23, both measured with
ground or satellite clock.
On 11/17/2023 2:14 PM, Maciej Wozniak wrote:
On Friday, 17 November 2023 at 12:10:04 UTC+1, Paul B. Andersen wrote:
Den 16.11.2023 21:16, skrev Lou:
On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote: >>>> Den 16.11.2023 13:58, skrev Lou:The correction is (1 - 4.4647E-10)
Interesting facts thanks for clarifying things. Still a bit of discrepancy
though.The current NIST reference says the gps sat clock f is designated
as 10229999.99543Hz.
Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
Yes. I just told you it was 10.229999995432612 MHz I got that from NIST. >>>
The frequency is adjusted (the famous "GR-correction") to:
f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
This frequency would observed on the ground be:
f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
Where did you get 1 + 4.4647E-10 ?
so f₁ = (1/(1 - 4.4647E-10))⋅f₀ ≈ (1 + 4.4647E-10)⋅f₀
The correction is specified in the
"INTERFACE SPECIFICATION DOCUMENT" for GPS:
https://www.gps.gov/technical/icwg/IS-GPS-200N.pdf
From 3.3.1.1 Frequency Plan:
"The carrier frequencies for the L1 and L2 signals shall be
coherently derived from a common frequency source within the SV.
The nominal frequency of this source -- as it appears to an observer
on the ground -- is 10.23 MHz. The SV carrier frequency and clock
rates -- as they would appear to an observer located in the SV --
Paul, poor halfbrain, who cares what "would appear" to aNIST does, obviously, since they wrote that.
non-existing gedanken person?
It's the measurement resultSo now it is you invoking the "would appear" since you are talking about
that counts, andthis is 10.23, both measured with
ground or satellite clock.
the satellite clock without an astronaut checking. But regardless of
whether there is an astronaut there or not, it still runs at 10.229999995432612 MHz
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