How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit :
The notion of universal anisochrony means that each watch will lag
behind the other with an anisochrony Et=x/c, a reciprocal phenomenon
that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you
miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Given that your stubbornness in not wanting to understand what you don't
get at the first reading is even stronger than your stupidity (which is saying something!), I doubt you'll even try to comprehend. However, here
are a few intermediate exercises to help you understand what most people grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
The notion of universal anisochrony means that each watch will lag
behind the other with an anisochrony Et=x/c, a reciprocal phenomenon
that will affect all the watches in the universe.
How naive is it possible to be?
You don't sync two clocks to each other, you sync one clockto another clock.
You still don't understand.
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit :
The notion of universal anisochrony means that each watch will lag
behind the other with an anisochrony Et=x/c, a reciprocal phenomenon
that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you
miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Given that your stubbornness in not wanting to understand what you don't
get at the first reading is even stronger than your stupidity (which is
saying something!), I doubt you'll even try to comprehend. However, here
are a few intermediate exercises to help you understand what most people
grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie,[snip whining] -
the hypothesis was no way confirmed.
But it was a self-denying absurd instead.
Le 16/08/2024 à 12:56, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit :
The notion of universal anisochrony means that each watch will lag
behind the other with an anisochrony Et=x/c, a reciprocal phenomenon
that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you
miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ >>>
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Given that your stubbornness in not wanting to understand what you don't >>> get at the first reading is even stronger than your stupidity (which is
saying something!), I doubt you'll even try to comprehend. However, here >>> are a few intermediate exercises to help you understand what most people >>> grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie,[snip whining] -
the hypothesis was no way confirmed.
A Review of One-Way and Two-Way Experiments to Test the Isotropy of the
Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
https://arxiv.org/abs/1011.1318
But it was a self-denying absurd instead.
Because you say so?
speed two-way experiments to confirm it to be invariant.
W dniu 16.08.2024 o 13:05, Python pisze:
Le 16/08/2024 à 12:56, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit :
The notion of universal anisochrony means that each watch will lag
behind the other with an anisochrony Et=x/c, a reciprocal
phenomenon that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you
miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ >>>>
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can >>>> refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received. >>>>
Given that your stubbornness in not wanting to understand what you
don't
get at the first reading is even stronger than your stupidity (which is >>>> saying something!), I doubt you'll even try to comprehend. However,
here
are a few intermediate exercises to help you understand what most
people
grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie,[snip whining] -
the hypothesis was no way confirmed.
A Review of One-Way and Two-Way Experiments to Test the Isotropy of
the Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
https://arxiv.org/abs/1011.1318
I could as well write those experiments
are testing and confirming the advantage
of communism over rotten capitalism. But
they don't.
But it was a self-denying absurd instead.
Because you say so?
Because that's a VERY simple consequence
of a definition, in the time you're talking
about - in the time when your idiot guru
lived and mumbled - valid also for his
moronic church.
Unfortunately there is nothing absurd into light
speed two-way experiments to confirm it to be invariant.
Oh, yes, it is. You're an idiot, so
you're unable to notice.
Le 16/08/2024 à 12:56, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit :
The notion of universal anisochrony means that each watch will lag
behind the other with an anisochrony Et=x/c, a reciprocal phenomenon
that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you
miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ >>>
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Given that your stubbornness in not wanting to understand what you don't >>> get at the first reading is even stronger than your stupidity (which is
saying something!), I doubt you'll even try to comprehend. However, here >>> are a few intermediate exercises to help you understand what most people >>> grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie,[snip whining] -
the hypothesis was no way confirmed.
A Review of One-Way and Two-Way Experiments to Test the Isotropy of the
Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
https://arxiv.org/abs/1011.1318
But it was a self-denying absurd instead.
Because you say so? Unfortunately there is nothing absurd into light
speed two-way experiments to confirm it to be invariant.
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit :
The notion of universal anisochrony means that each watch will lag
behind the other with an anisochrony Et=x/c, a reciprocal phenomenon
that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you
miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Given that your stubbornness in not wanting to understand what you don't
get at the first reading is even stronger than your stupidity (which is
saying something!), I doubt you'll even try to comprehend. However, here
are a few intermediate exercises to help you understand what most people
grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie, as expected from a relativistic idiot -
the hypothesis was no way confirmed. But it
was a self-denying absurd instead.
Le 16/08/2024 à 14:05, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 13:05, Python pisze:
Le 16/08/2024 à 12:56, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit :
The notion of universal anisochrony means that each watch will lag >>>>>> behind the other with an anisochrony Et=x/c, a reciprocal
phenomenon that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you
miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ >>>>>
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can >>>>> refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is
received.
Given that your stubbornness in not wanting to understand what you
don't
get at the first reading is even stronger than your stupidity
(which is
saying something!), I doubt you'll even try to comprehend. However,
here
are a few intermediate exercises to help you understand what most
people
grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie,[snip whining] -
the hypothesis was no way confirmed.
A Review of One-Way and Two-Way Experiments to Test the Isotropy of
the Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
https://arxiv.org/abs/1011.1318
I could as well write those experiments
are testing and confirming the advantage
of communism over rotten capitalism. But
they don't.
But it was a self-denying absurd instead.
Because you say so?
Because that's a VERY simple consequence
of a definition, in the time you're talking
about - in the time when your idiot guru
lived and mumbled - valid also for his
moronic church.
Unfortunately there is nothing absurd into light
speed two-way experiments to confirm it to be invariant.
Oh, yes, it is. You're an idiot, so
you're unable to notice.
So all you have is incoherent babbling and insults, as usual.
W dniu 16.08.2024 o 14:08, Python pisze:
Le 16/08/2024 à 14:05, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 13:05, Python pisze:
Le 16/08/2024 à 12:56, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit : >>>>>>>
The notion of universal anisochrony means that each watch will
lag behind the other with an anisochrony Et=x/c, a reciprocal
phenomenon that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you >>>>>> miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ >>>>>>
If the meaning of t_A, t_B, and t'_A are still unknown to you, you >>>>>> can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is
received.
Given that your stubbornness in not wanting to understand what you >>>>>> don't
get at the first reading is even stronger than your stupidity
(which is
saying something!), I doubt you'll even try to comprehend.
However, here
are a few intermediate exercises to help you understand what most
people
grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie,[snip whining] -
the hypothesis was no way confirmed.
A Review of One-Way and Two-Way Experiments to Test the Isotropy of
the Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
https://arxiv.org/abs/1011.1318
I could as well write those experiments
are testing and confirming the advantage
of communism over rotten capitalism. But
they don't.
But it was a self-denying absurd instead.
Because you say so?
Because that's a VERY simple consequence
of a definition, in the time you're talking
about - in the time when your idiot guru
lived and mumbled - valid also for his
moronic church.
Unfortunately there is nothing absurd into light
speed two-way experiments to confirm it to be invariant.
Oh, yes, it is. You're an idiot, so
you're unable to notice.
So all you have is incoherent babbling and insults, as usual.
And a proof of inconsistency of incoherent
[SR]
poor stinker
A Review of One-Way and Two-Way Experiments to Test the Isotropy of the
Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
Because you say so? Unfortunately there is nothing absurd into light
speed two-way experiments to confirm it to be invariant.
I could as well write those experiments
are testing and confirming the advantage
of communism over rotten capitalism. But
they don't.
Le 16/08/2024 à 13:05, Python a écrit :
A Review of One-Way and Two-Way Experiments to Test the Isotropy of
the Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
Il y a des rigolos qui ont voulu tester des conneries pareilles.
C'est quoi, c'est des bédouins?
Le 16/08/2024 à 14:15, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 14:08, Python pisze:
Le 16/08/2024 à 14:05, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 13:05, Python pisze:
Le 16/08/2024 à 12:56, Maciej Wozniak a écrit :
W dniu 16.08.2024 o 12:47, Python pisze:
Le 15/08/2024 à 21:38, M.D. Richard "Hachel" Lengrand a écrit : >>>>>>>>
The notion of universal anisochrony means that each watch will >>>>>>>> lag behind the other with an anisochrony Et=x/c, a reciprocal
phenomenon that will affect all the watches in the universe.
How naive is it possible to be?to another clock.
You don't sync two clocks to each other, you sync one clock
You still don't understand.
You completely messed up your quotes above. Anyway...
You're probably a bit intellectually challenged to understand a
procedure that is fairly simple, just as you were in 2007 when you >>>>>>> miserably demonstrated it back then:
https://groups.google.com/g/fr.sci.physique/c/KgqI9gqTkR8/m/oMc9X0XjCWMJ
If the meaning of t_A, t_B, and t'_A are still unknown to you,
you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted; >>>>>>>
t_B is the time shown by clock B when the signal is received and >>>>>>> re-emitted;
t'_A is the time shown by clock A when the returned signal is
received.
Given that your stubbornness in not wanting to understand what
you don't
get at the first reading is even stronger than your stupidity
(which is
saying something!), I doubt you'll even try to comprehend.
However, here
are a few intermediate exercises to help you understand what most >>>>>>> people
grasp on the first try:
1. Using the hypothesis (confirmed by experiment) that:
A lie,[snip whining] -
the hypothesis was no way confirmed.
A Review of One-Way and Two-Way Experiments to Test the Isotropy of
the Speed of Light
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer >>>>>
https://arxiv.org/abs/1011.1318
I could as well write those experiments
are testing and confirming the advantage
of communism over rotten capitalism. But
they don't.
But it was a self-denying absurd instead.
Because you say so?
Because that's a VERY simple consequence
of a definition, in the time you're talking
about - in the time when your idiot guru
lived and mumbled - valid also for his
moronic church.
Unfortunately there is nothing absurd into light
speed two-way experiments to confirm it to be invariant.
Oh, yes, it is. You're an idiot, so
you're unable to notice.
So all you have is incoherent babbling and insults, as usual.
And a proof of inconsistency of incoherent
[SR]
You have nothing but complete baloney, as it has, it vain,
been explained to you.
poor stinker
Nice signature though.
Md. Farid Ahmed, Brendan M. Quine, Stoyan Sargoytchev, A. D. Stauffer
Il y a des rigolos qui ont voulu tester des conneries pareilles?
Ce sont des scientifiques
C'est quoi, c'est des bédouins?
Pharos or not pharos? Zat iz ze kwestion.
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
t1, t2, tB, AB, c, and 0 are numbers. Numbers are the same for
everyone, including A and B, and therefore the realtions between
the numbers are the same, too.
Mikko
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
On Sat, 17 Aug 2024 8:56:06 +0000, Richard Hachel wrote:
Le 17/08/2024 à 10:26, Mikko a écrit :
Richard Hachel wrote:
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
t1, t2, tB, AB, c, and 0 are numbers. Numbers are the same for
everyone, including A and B, and therefore the realtions between
the numbers are the same, too.
Mikko
No, the numbers are not the same for everyone.
Let's take the case of the Langevin traveler, Vo=0.8c. D=2*12al.
Richard, it seems to me that you're conflating two different
situations, which causes monumental confusion. The case that
was under discussion was synchronizing two clocks at rest with
respect to each other, not the case where the clocks are in
relative motion, n'est-ce pas?
Le 17/08/2024 à 10:26, Mikko a écrit :
Richard Hachel wrote:
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
t1, t2, tB, AB, c, and 0 are numbers. Numbers are the same for
everyone, including A and B, and therefore the realtions between
the numbers are the same, too.
Mikko
No, the numbers are not the same for everyone.
Let's take the case of the Langevin traveler, Vo=0.8c. D=2*12al.
This means that, for an observer on Earth, the duration of the journey,
which is a number, will be To=D/Vo.
Or To=24/0.8=30 years.
Problem: this is not an invariant. Stella on board the rocket does not
have the same number as Terrence.
She is 18 years old.
The same is true here, in special relativity explained by Dr. Hachel,
which is fundamentally different from that taught by Minkowski.
Fundamentally different.
If we believe in the flat earth theory (today it is the belief in flat
time and the absolute isochrony of the universe), we will say that for A
the outward journey is AB/c and the same for the return journey.
And the same for B.
But this way of thinking, even if it is intuitive, even if it seems of
the most colossal logic, is false.
R.H.
Le 17/08/2024 à 10:26, Mikko a écrit :
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
t1, t2, tB, AB, c, and 0 are numbers. Numbers are the same for
everyone, including A and B, and therefore the realtions between
the numbers are the same, too.
Mikko
No, the numbers are not the same for everyone.
Let's take the case of the Langevin traveler, Vo=0.8c. D=2*12al.
This means that, for an observer on Earth, the duration of the journey,
which is a number, will be To=D/Vo.
Den 16.08.2024 14:24, skrev Richard Hachel:
Le 16/08/2024 à 12:47, Python a écrit :
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
Le 16/08/2024 à 12:47, Python a écrit :
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Here's the clown continuing.
If we look closely, what he says seems sensible.
This is why all of humanity is wrong about the theory of special
relativity as taught.
Because the more we rub our eyes, the more sensible it seems.
Except that... breathe, blow...
That clock A notes a time t1 when the signal leaves A (we don't care
about the value by the way,
it could be t1=0 or it could be t1=153), I'm willing.
We place B at 3.10^8m, and time is measured in seconds.
That clock A notes a time t2 when the signal returns, I'm willing again,
and as Jean-Pierre
Messager says (sometimes he says intelligent things, although it's
rare), I'm going to say that t2=2 or that t2=155.
How will Jean-Pierre achieve this prophetic feat?
Let's not get carried away, I know how to do it too.
Here's how I do it, breathe, blow.
t2 = t1 + 2AB/c
And this is so true that it applies to the entire universe, and all
inertial frames of reference.
But unlike Jean-Pierre, Henri or Albert, I'll stop here.
Because this is where the vast ocean of relativistic science begins.
What time is it in B when B receives the information? I don't know at all, and first of all, depending on how I synchronized B, it could be t=4532
or t=-12.
So I don't know at all.
Jean-Pierre is intelligent enough to understand that it is therefore necessary to first synchronize B with A,
to have something coherent, because saying that tA=0 tB=4532 and tA'=2
is always feasible, we are not lying, but it is very unhelpful.
Except that Jean-Pierre still has not understood Hachel's thinking, and
he remains in the hypothesis of a flat present (the horizontal plane of
the present time), as others remain in the hypothesis of the flat earth.
Nature is not made like that, that's not how it works.
So what time is it in B?
Jean-Pierre does not bother with embellishments: "We only have to artificially set tB=(t2-t1)/2 and thus, everything will be very simple
and very practical".
Except that it is an artificial synchronization.
And except that it will not be true for A, nor for B.
It will only be true for M, a point placed at an equal distance from A
and B, and the synchronization will be called M synchronization.
Because in the universe of A, this M synchronization is completely
false, everything that is part of the "3D present time" of M is not
part, and we are infinitely far from it, of the present time of A, and
ditto for B.
Each chosen point, A, B, or M have the same 3D inertial frame, but they
are not part of the same 4D frame, and each can only have its own
(because of anisochrony, and the fourth component t).
The synchronization of Einstein, Poincaré, physicists, is therefore only
an abstract synchronization,
which represents a point M, placed very far perpendicularly, in an
imaginary fourth dimension,
and which apprehends all the points of the 3D universe at the same
distance and at the same present moment of M.
It is obviously totally imaginary, but it is very useful.
For this point, indeed, we can say that tB=(t2-t1)/2 but it is a
convention M.
For A as for B, it is absolutely impossible to synchronize these two
watches between them FOR them.
As it is also impossible to synchronize A or B with the imaginary point M.
Always, always, always, there will remain a universal anisochrony.
And always, always, always, in the reality of things, if we have
practiced a synchronization M:
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
I don't know if it will take Lengruche four years to understand that (a+b)(a-b)=a²-b²
but it is certain that in 30 years Ybmuche will still not have
understood what I have just detailed here.
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
In our very big, inertial lab, we have two points A and B which are
separated by some distance. Let's call the transit time for light
to go from A to B is x seconds. We will _define_ that the transit time
is the same from B to A. (This follows from Einstein's definition
of simultaneity).
At point A we have:
Clock C_A, a light-detector, a flash-light and a computer.
The computer can register the time shown by C_A when
the flash-light is flashing, and when the light-detector
registers a light-flash.
At point B we have:
Clock C_B, a light-detector, a mirror and a computer.
The computer can register the time shown by C_B when
the light-detector registers a light-flash.
In the following we will synchronise clock C_B to clock C_A.
That is, we will adjust clock C_B so it become synchronous
with clock C_A.
Now we let the flash-light at point A flash.
At this instant, C_A is showing tA = n seconds.
tA is measured by C_A at A.
When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
A short time later the light detector at A registers
the light reflected by the mirror at B.
At this instant Clock C_A shows t'A = n + 2x seconds.
t'A is measured by C_A at A.
Einstein:
"The two clocks synchronise if tB − tA = t'A − tB."
Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
That is, to be synchronous clock C_B must show a time midway
between tA and t'A when the light is reflected by the mirror.
So tB should show (n + x) seconds when the light is reflected
by the mirror.
But at that instant tB is showing m seconds, so to make the two
clocks synchronous, we must adjust clock C_B by:
δ = (n-m) + x seconds.
After this correction, we have:
tB − tA = (m - n) seconds + δ = x seconds
t'A − tB = (n + 2x - m) seconds - δ = x seconds
The clocks are now synchronised.
Please explain what in the above you find impossible
to do in your lab.
I have explained these things a hundred times.
And your explanation has been refuted millions of times :-))
For this point, indeed, we can say that tB=(t2-t1)/2 but it is a
convention M.
It is a man made _definition_.
Einstein's _definition_ of simultaneity is based on symmetry.
The transit time is the same in both direction.
It is possible to define simultaneity in other ways, but
it would be very inconvenient if the speed of light were
not isotropic.
For A as for B, it is absolutely impossible to synchronize these two
watches between them FOR them.
As it is also impossible to synchronize A or B with the imaginary point M. >>
Always, always, always, there will remain a universal anisochrony.
And always, always, always, in the reality of things, if we have
practiced a synchronization M:
FOR A:
tB-t1=2AB/c
t2-tB=0
FOR B:
tB-t1=0
t2-tB=2AB/c
What an idiot!
tB is a time measured by B.
t1 and t2 are measured by A.
They are proper times and can't be
different for A and B.
If the clocks are synchronous:
tB = (t1+t2)/2
t2 = t1 + 2AB/c
For both!
I don't know if it will take Lengruche four years to understand that
(a+b)(a-b)=a²-b²
but it is certain that in 30 years Ybmuche will still not have
understood what I have just detailed here.
If Ybmuche is a sane person, you are probably right. :-D
Le 19/08/2024 à 22:32, "Paul.B.Andersen" a écrit :
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
In our very big, inertial lab, we have two points A and B which are separated by some distance. Let's call the transit time for light
to go from A to B is x seconds. We will _define_ that the transit time
is the same from B to A. (This follows from Einstein's definition
of simultaneity).
At point A we have:
Clock C_A, a light-detector, a flash-light and a computer.
The computer can register the time shown by C_A when
the flash-light is flashing, and when the light-detector
registers a light-flash.
At point B we have:
Clock C_B, a light-detector, a mirror and a computer.
The computer can register the time shown by C_B when
the light-detector registers a light-flash.
In the following we will synchronise clock C_B to clock C_A.
That is, we will adjust clock C_B so it become synchronous
with clock C_A.
Now we let the flash-light at point A flash.
At this instant, C_A is showing tA = n seconds.
tA is measured by C_A at A.
When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
A short time later the light detector at A registers
the light reflected by the mirror at B.
At this instant Clock C_A shows t'A = n + 2x seconds.
t'A is measured by C_A at A.
Einstein:
"The two clocks synchronise if tB − tA = t'A − tB."
Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
That is, to be synchronous clock C_B must show a time midway
between tA and t'A when the light is reflected by the mirror.
So tB should show (n + x) seconds when the light is reflected
by the mirror.
But at that instant tB is showing m seconds, so to make the two
clocks synchronous, we must adjust clock C_B by:
δ = (n-m) + x seconds.
After this correction, we have:
tB − tA = (m - n) seconds + δ = x seconds
t'A − tB = (n + 2x - m) seconds - δ = x seconds
The clocks are now synchronised.
Please explain what in the above you find impossible
to do in your lab.
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located
in different places.
It is not a technical problem, it is not a problem offinding a round square, or a natural number between 5
intelligence, it is simply that it is as impossible as
and 6. Because nature is not made like that.
Anyone who wants to synchronize watches with each other does
not realize that he is looking for rabbit horns.
The only thing we can do is to find a kind of abstract,
imaginary synchronization procedure, called the Einstein
procedure where tAB is supposed to be equal to tBA
not only for the point M which is the only real origin of
the synchronization, but for everyone.
We then have a useful procedure, but false.
[Rest of misinformed baloney deleted]
How can tAB not be equal to tBA for light is vacuum?
How can tAB not be equal to tBA for light is vacuum?
not only for the point M which is the only real origin of
the synchronization, but for everyone.
If tAB = tBA, then Paul's procedure works. If tAB <> tBA,
then your point M method won't work either.
We then have a useful procedure, but false.
I set my watch to my wall clock months ago and it's still
in sync. You're spouting baloney, Richard.
[Rest of misinformed baloney deleted]
On Mon, 19 Aug 2024 23:15:28 +0000, Richard Hachel wrote:
If tAB = tBA, then Paul's procedure works. If tAB <> tBA,
then your point M method won't work either.
On Tue, 20 Aug 2024 0:35:42 +0000, Richard Hachel wrote:
Even a child knows that the present tense can exist elsewhere.
Even Wozniak and the GPS agree that "t' = t." It's not stupid,
but even if it were stupid but it works, it's not stupid. It's
real.
On Tue, 20 Aug 2024 0:35:42 +0000, Richard Hachel wrote:
Le 20/08/2024 à 01:33, hitlong@yahoo.com (gharnagel) a écrit :
If tAB = tBA, then Paul's procedure works. If tAB <> tBA,
then your point M method won't work either.
If we synchronize on an abstract point M placed at an equal
distance from A, B, C, D, E, we obtain for E a plane of the
present time where A, B,C,D and E are simultaneously located.
Richard, Richard, Richard! If tAB <> tBA, your method won't
work Breath, blow, and THINK! Given
A ___________ M ___________ B
If tAB <> tBA, then tAM <> tMB and A and B won't be synchronized.
If, OTOH tMA = tMB, then tAB = tBA.
Le 20/08/2024 à 01:33, hitlong@yahoo.com (gharnagel) a écrit :
If tAB = tBA, then Paul's procedure works. If tAB <> tBA,
then your point M method won't work either.
If we synchronize on an abstract point M placed at an equal
distance from A, B, C, D, E, we obtain for E a plane of the
present time where A, B,C,D and E are simultaneously located.
....
The universal present does not exist!!!
I repeat again, even if it must be said with knees in the testicles,
and big slaps in the snout.
"Your stupid present tense doesn't exist." I will never understand
this blockage for a concept that I thought was understandable for
an eleven-year-old child, and which seems to block everyone.
R.H.
Den 16.08.2024 14:24, skrev Richard Hachel:
Le 16/08/2024 à 12:47, Python a écrit :
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can
refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
Am Montag000019, 19.08.2024 um 22:33 schrieb Paul.B.Andersen:
Den 16.08.2024 14:24, skrev Richard Hachel:
Le 16/08/2024 à 12:47, Python a écrit :
If the meaning of t_A, t_B, and t'_A are still unknown to you, you can >>>> refer to Einstein 1905 article.
t_A is the time shown by clock A when a light signal is emitted;
t_B is the time shown by clock B when the signal is received and
re-emitted;
t'_A is the time shown by clock A when the returned signal is received.
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
This is a self-contradicting statement!
'Synchronization' does not mean 'to turn the clocks to the same shown
time once'.
Instead 'Synchronization' means 'adjust both clocks, that they are maintaining to show the same time'.
One part of the procedure would be to adjust the readings.
But the other part would be to adjust the tick-rate, hence the length of
the second.
We could simply discard 'adjust the reading' from our todo list, at
least for while, because upon our home-planet we have also time-zones.
That would leave 'adjustment of the tick-rate' as main requirement for synchronization.
But you have already preassumed, that this adjustment was already made.
(with " The clocks run at the same rate.")
This would violate your statement, that both clocks are not synchronized.
Am Montag000019, 19.08.2024 um 22:33 schrieb Paul.B.Andersen:
Instead 'Synchronization' means 'adjust both clocks, that they are maintaining to show the same time'.
Le 19/08/2024 à 22:32, "Paul.B.Andersen" a écrit :
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
In our very big, inertial lab, we have two points A and B which are
separated by some distance. Let's call the transit time for light
to go from A to B is x seconds. We will _define_ that the transit time
is the same from B to A. (This follows from Einstein's definition
of simultaneity).
At point A we have:
Clock C_A, a light-detector, a flash-light and a computer.
The computer can register the time shown by C_A when
the flash-light is flashing, and when the light-detector
registers a light-flash.
At point B we have:
Clock C_B, a light-detector, a mirror and a computer.
The computer can register the time shown by C_B when
the light-detector registers a light-flash.
In the following we will synchronise clock C_B to clock C_A.
That is, we will adjust clock C_B so it become synchronous
with clock C_A.
Now we let the flash-light at point A flash.
At this instant, C_A is showing tA = n seconds.
tA is measured by C_A at A.
When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
A short time later the light detector at A registers
the light reflected by the mirror at B.
At this instant Clock C_A shows t'A = n + 2x seconds.
t'A is measured by C_A at A.
Einstein:
"The two clocks synchronise if tB − tA = t'A − tB."
Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
That is, to be synchronous clock C_B must show a time midway
between tA and t'A when the light is reflected by the mirror.
So tB should show (n + x) seconds when the light is reflected
by the mirror.
But at that instant tB is showing m seconds, so to make the two
clocks synchronous, we must adjust clock C_B by:
δ = (n-m) + x seconds.
After this correction, we have:
tB − tA = (m - n) seconds + δ = x seconds
t'A − tB = (n + 2x - m) seconds - δ = x seconds
The clocks are now synchronised.
Please explain what in the above you find impossible
to do in your lab.
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
I have explained these things a hundred times.
And your explanation has been refuted millions of times :-))
That's what I say.
But refuted with prehistoric clubs.
You will never be able to synchronize two watches placed in different
places with each other.
It is impossible.
But between them the relativity of the notion of simultaneity will meanthat for A, B will have started running later,
and, reciprocally B will accuse A of having started running later.
The reciprocal delay between A and B will be t=AB/c.
The notion of present time is relative.
It is still incredible that you cannot understand this.
I didn't say that synchronization was stupid,
nor that we couldn't make satellites or GPS work with this
synchronization.
I simply said that this Einsteinian synchronization based
on M was an abstract, unreal, imaginary synchronization.
Le 20/08/2024 à 13:13, Mikko a écrit :
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
You have to understand the words and concepts, and in relativistic
theory, this is very poorly done. There is no theory in the world that
is so poorly explained, and the fact that some people think they
understand it, when they only possibly understand how to do the
equations (and again, I am amazed at the enormous errors made on proper
times and instantaneous observable speeds in accelerated frames of
reference, the misunderstandings in Langevin in apparent speeds and the madness of rotating frames of reference) does not really help the dissemination of a theory that is both beautiful and true, which I have
been trying to do (under idiotic laughter) for 40 years, with,
nevertheless, some great personal successes.
As for Paul's or Python's synchronization procedure, I'm not saying it's wrong, or useless. On the contrary, it's very useful for giving a
coherent and practicable universal time.
It's just that it's only true for an imaginary, abstract observer, who, placed far away and in an ideal fourth dimension, synchronizes the
entire universe on itself, with the particularity, for him, of
considering that all movements can ONLY be transversal, including the movements of information which then take the value c in all directions.
What I mean is that two watches set according to this procedure each
advance on the other by an artificial and false value which is AB/c.
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
Le 20/08/2024 à 14:01, Python a écrit :
This means that your synchronization criteria is broken. That's all.
While Einstein-Poincaré's criteria is NOT broken.
End Of Story.
This is just the beginning of the story.
You know Jesus Christ: "Heaven and earth will pass away, my words will absolutely not pass away".
Well, it's the same for Hachel's theory of relativity.
Einstein will pass away, Hachel will absolutely not pass away (I'm not talking about men, I'm talking about principles and equations).
The great truths are eternal "water is wet", "two joints of seven make
nine", "snow is white".
The doctored truths only please for a time, they are rivers of Babylon.
This means that your synchronization criteria is broken. That's all.
While Einstein-Poincaré's criteria is NOT broken.
End Of Story.
Le 20/08/2024 à 14:01, Python a écrit :
This means that your synchronization criteria is broken. That's all.
While Einstein-Poincaré's criteria is NOT broken.
End Of Story.
This is just the beginning of the story.
You know Jesus Christ: "Heaven and earth will pass away, my words will absolutely not pass away".
Well, it's the same for Hachel's theory of relativity.
Einstein will pass away, Hachel will absolutely not pass away (I'm not talking about men, I'm talking about principles and equations).
The great truths are eternal "water is wet",
"two joints of seven make nine",
"snow is white".
The doctored truths only please for a time, they are rivers of Babylon.
R.H.
It is impossible to synchronize two watches A and B located in different places.
It is not a technical problem, it is not a problem of intelligence, it
is simply that it is as impossible as finding a round square, or a
natural number between 5 and 6.
Because nature is not made like that.
Anyone who wants to synchronize watches with each other does not realize
that he is looking for rabbit horns.
The only thing we can do is to find a kind of abstract, imaginary synchronization procedure, called the Einstein procedure where tAB is supposed to be equal to tBA not only for the point M which is the only
real origin of the synchronization, but for everyone.
We then have a useful procedure, but false.
There is no global synchronization, but a synchronization of type M, as
I have explained.
A and B, as I explained, will never "live" in the same global, real, and reciprocal present moment.
Several points.
(1) What you said was indeed simple, but incorrect.
(2) Einstein synchronization is NOT based on M.
(3) If it's "unreal" or "imaginary" it doesn't work.
(4) If it doesn't work, it's stupid.
(5) But it works, so it not stupid, it's not unreal
and it's not imaginary.
You, Richard, are the one who is "not making an effort" to
understand. Signal time delay is irrelevant because those
doing the synchronization aren't stupid.
Look Richard, you say there is an inherent nonsynchronization
due to time delay of the signal. If that were true, it would
depend on the speed of the signal. You say it's AB/c. But
the speed of light is NOT c on the earth: it's c/n where n is
the refractive index of air. If the signal were sent over a
wire connecting A and B, it would be even slower.
And if tachyons exist, it would be SMALLER than AB/c! I can
see why you vociferously denounce the existence of tachyons,
even if they haven't been refuted (and when neutrinos may be
tachyons for all we know). It would demolish your mistaken
belief about the relativity of simultaneity.
Den 20.08.2024 01:25, skrev Richard Hachel:...
I see that you are not making any effort to understand what I am saying.
What else can I do, you keep repeating the same nonsense based on
ignorance or the refusal of the notion of universal anisochrony?
You can address what I write in stead of snipping it,
probably without reading it.
Einstein:
"The two clocks synchronise if tB − tA = t'A − tB."
Le 20/08/2024 à 14:43, Richard "Hachel" Lengrand a écrit :
Le 20/08/2024 à 14:01, Python a écrit :
This means that your synchronization criteria is broken. That's all.
While Einstein-Poincaré's criteria is NOT broken.
End Of Story.
This is just the beginning of the story.
You know Jesus Christ: "Heaven and earth will pass away, my words will
absolutely not pass away".
Well, it's the same for Hachel's theory of relativity.
Einstein will pass away, Hachel will absolutely not pass away (I'm not
talking about men, I'm talking about principles and equations).
The great truths are eternal "water is wet", "two joints of seven make
nine", "snow is white".
The doctored truths only please for a time, they are rivers of Babylon.
Something proven wrong and contradictory in 2024 will still be wrong
and contradictory in 2124.
Den 20.08.2024 01:15, skrev Richard Hachel:
It is impossible to synchronize two watches A and B located in
different places.
It is not a technical problem, it is not a problem of intelligence, it
is simply that it is as impossible as finding a round square, or a
natural number between 5 and 6.
Because nature is not made like that.
Anyone who wants to synchronize watches with each other does not
realize that he is looking for rabbit horns.
--------------------
The only thing we can do is to find a kind of abstract, imaginary
synchronization procedure, called the Einstein procedure where tAB is
supposed to be equal to tBA not only for the point M which is the only
real origin of the synchronization, but for everyone.
So the only thing we can do is to do what Einstein did,
namely to define a synchronisation procedure which
makes it possible to synchronise two clocks at different
places in an inertial frame of reference.
W dniu 20.08.2024 o 15:08, Paul.B.Andersen pisze:
Den 20.08.2024 01:15, skrev Richard Hachel:
It is impossible to synchronize two watches A and B located in
different places.
It is not a technical problem, it is not a problem of intelligence,
it is simply that it is as impossible as finding a round square, or a
natural number between 5 and 6.
Because nature is not made like that.
Anyone who wants to synchronize watches with each other does not
realize that he is looking for rabbit horns.
--------------------
The only thing we can do is to find a kind of abstract, imaginary
synchronization procedure, called the Einstein procedure where tAB is
supposed to be equal to tBA not only for the point M which is the
only real origin of the synchronization, but for everyone.
So the only thing we can do is to do what Einstein did,
namely to define a synchronisation procedure which
makes it possible to synchronise two clocks at different
places in an inertial frame of reference.
That's for sure the only thing [you]
can do. Professionals of GPS could do more.
Note that clocks showing UTC are synchronous in the non-rotating
Earth centred frame of reference (ECI-frame), but they are NOT
synchronous in the ground frame.
Note the rather peculiar phenomenon that the UTC clocks
are synchronous in the frame were they are moving, but
not synchronous in the frame where they are stationary.
I see that you are not making any effort to understand what I am saying.
What else can I do, you keep repeating the same nonsense based on
ignorance or the refusal of the notion of universal anisochrony?
BTW. You are posting here because you love to be spanked, aren't you?
No.
Le 20/08/2024 à 16:01, Maciej Wozniak a écrit :
W dniu 20.08.2024 o 15:08, Paul.B.Andersen pisze:
Den 20.08.2024 01:15, skrev Richard Hachel:
It is impossible to synchronize two watches A and B located in
different places.
It is not a technical problem, it is not a problem of intelligence,
it is simply that it is as impossible as finding a round square, or
a natural number between 5 and 6.
Because nature is not made like that.
Anyone who wants to synchronize watches with each other does not
realize that he is looking for rabbit horns.
--------------------
The only thing we can do is to find a kind of abstract, imaginary
synchronization procedure, called the Einstein procedure where tAB
is supposed to be equal to tBA not only for the point M which is the
only real origin of the synchronization, but for everyone.
So the only thing we can do is to do what Einstein did,
namely to define a synchronisation procedure which
makes it possible to synchronise two clocks at different
places in an inertial frame of reference.
That's for sure the only thing [you]
can do. Professionals of GPS could do more.
Sure ! Thanks to GR.
BTW. You are posting here because you love to be spanked, aren't you?
On 2024-08-19 23:15:28 +0000, Richard Hachel said:
Le 19/08/2024 à 22:32, "Paul.B.Andersen" a écrit :
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
In our very big, inertial lab, we have two points A and B which are
separated by some distance. Let's call the transit time for light
to go from A to B is x seconds. We will _define_ that the transit time
is the same from B to A. (This follows from Einstein's definition
of simultaneity).
At point A we have:
Clock C_A, a light-detector, a flash-light and a computer.
The computer can register the time shown by C_A when
the flash-light is flashing, and when the light-detector
registers a light-flash.
At point B we have:
Clock C_B, a light-detector, a mirror and a computer.
The computer can register the time shown by C_B when
the light-detector registers a light-flash.
In the following we will synchronise clock C_B to clock C_A.
That is, we will adjust clock C_B so it become synchronous
with clock C_A.
Now we let the flash-light at point A flash.
At this instant, C_A is showing tA = n seconds.
tA is measured by C_A at A.
When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
A short time later the light detector at A registers
the light reflected by the mirror at B.
At this instant Clock C_A shows t'A = n + 2x seconds.
t'A is measured by C_A at A.
Einstein:
"The two clocks synchronise if tB − tA = t'A − tB."
Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
That is, to be synchronous clock C_B must show a time midway
between tA and t'A when the light is reflected by the mirror.
So tB should show (n + x) seconds when the light is reflected
by the mirror.
But at that instant tB is showing m seconds, so to make the two
clocks synchronous, we must adjust clock C_B by:
δ = (n-m) + x seconds.
After this correction, we have:
tB − tA = (m - n) seconds + δ = x seconds
t'A − tB = (n + 2x - m) seconds - δ = x seconds
The clocks are now synchronised.
Please explain what in the above you find impossible
to do in your lab.
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
Hachel now pretends that tB − tA = t'A − tB can be true or false depending on the observer.
Le 20/08/2024 à 13:13, Mikko a écrit :
On 2024-08-19 23:15:28 +0000, Richard Hachel said:
Le 19/08/2024 à 22:32, "Paul.B.Andersen" a écrit :
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
In our very big, inertial lab, we have two points A and B which are
separated by some distance. Let's call the transit time for light
to go from A to B is x seconds. We will _define_ that the transit time >>>> is the same from B to A. (This follows from Einstein's definition
of simultaneity).
At point A we have:
Clock C_A, a light-detector, a flash-light and a computer.
The computer can register the time shown by C_A when
the flash-light is flashing, and when the light-detector
registers a light-flash.
At point B we have:
Clock C_B, a light-detector, a mirror and a computer.
The computer can register the time shown by C_B when
the light-detector registers a light-flash.
In the following we will synchronise clock C_B to clock C_A.
That is, we will adjust clock C_B so it become synchronous
with clock C_A.
Now we let the flash-light at point A flash.
At this instant, C_A is showing tA = n seconds.
tA is measured by C_A at A.
When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
A short time later the light detector at A registers
the light reflected by the mirror at B.
At this instant Clock C_A shows t'A = n + 2x seconds.
t'A is measured by C_A at A.
Einstein:
"The two clocks synchronise if tB − tA = t'A − tB."
Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
That is, to be synchronous clock C_B must show a time midway
between tA and t'A when the light is reflected by the mirror.
So tB should show (n + x) seconds when the light is reflected
by the mirror.
But at that instant tB is showing m seconds, so to make the two
clocks synchronous, we must adjust clock C_B by:
δ = (n-m) + x seconds.
After this correction, we have:
tB − tA = (m - n) seconds + δ = x seconds
t'A − tB = (n + 2x - m) seconds - δ = x seconds
The clocks are now synchronised.
Please explain what in the above you find impossible
to do in your lab.
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
It's much more complicated than that.
We can accept it for a Galilean frame of reference,
for example the Earth frame of reference.
But for an accelerated frame of reference, for example, it doesn't work anymore.
If we ask a relativistic physicist, for example Paul who is still an
educated and intelligent person (compared to Python the clown)
to give
me the time taken by Bella to reach Tau Ceti (12 ly; a=1.052 ly/y²) he
will answer me correctly and set To=(x/c).sqrt(1+2c²/ax)=12.9156 years.
The problem is, if I ask him for Bella's proper time, everything will
sink into horror, because he will give me an incredibly low proper time,
by performing an abstract integration adding abstract times.
A clown is someone pretending that 3 can be 4 for some obervers.
Le 20/08/2024 à 17:02, Python a écrit :
A clown is someone pretending that 3 can be 4 for some obervers.
This is not exactly what I am saying.
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least 40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because 2. because you are tied up with relativistic thoughts all learned, but false.
[irrelevant multi-posted content]Abuse report sent to usenet at bofh dot team.
Le 20/08/2024 à 17:32, Python a écrit :
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Numbers like t_B or t'_A cannot be "simultaneous". Events
like e1 = "B sends back a light signal to A" or e2 = "A receives
this signal" can be.
If this is what you meant, which is likely, then, according
to Einstein-Poincaré *definition* of simultaneity:
e1 and e2 are NOT simultaneous for A (nor for B)
No.
[snip whining]
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Numbers like t_B or t'_A cannot be "simultaneous". Events
like e1 = "B sends back a light signal to A" or e2 = "A receives
this signal" can be.
If this is what you meant, which is likely, then, according
to Einstein-Poincaré *definition* of simultaneity:
e1 and e2 are NOT simultaneous for A (nor for B)
Le 20/08/2024 à 18:18, M.D. Richard "Hachel" Lengrand a écrit :
Le 20/08/2024 à 17:32, Python a écrit :
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Numbers like t_B or t'_A cannot be "simultaneous". Events
like e1 = "B sends back a light signal to A" or e2 = "A receives
this signal" can be.
If this is what you meant, which is likely, then, according
to Einstein-Poincaré *definition* of simultaneity:
e1 and e2 are NOT simultaneous for A (nor for B)
No.
According to Einstein-Poincaré definition of simultaneity
they are not simultaneous.
However you like or not this *definition* this is something
you CANNOT deny.
A definition cannot be false.
W dniu 20.08.2024 o 18:22, Python pisze:
Le 20/08/2024 à 18:18, M.D. Richard "Hachel" Lengrand a écrit :
Le 20/08/2024 à 17:32, Python a écrit :
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Numbers like t_B or t'_A cannot be "simultaneous". Events
like e1 = "B sends back a light signal to A" or e2 = "A receives
this signal" can be.
If this is what you meant, which is likely, then, according
to Einstein-Poincaré *definition* of simultaneity:
e1 and e2 are NOT simultaneous for A (nor for B)
No.
According to Einstein-Poincaré definition of simultaneity
they are not simultaneous.
However you like or not this *definition* this is something
you CANNOT deny.
A definition cannot be false.
If your idiot guru defined a cow as a 6 leg
small animal with chitine armour - would it
be "not false" to you as well, poor halfbrain?
Le 20/08/2024 à 18:38, Maciej Wozniak a écrit :
W dniu 20.08.2024 o 18:22, Python pisze:
Le 20/08/2024 à 18:18, M.D. Richard "Hachel" Lengrand a écrit :
Le 20/08/2024 à 17:32, Python a écrit :
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Numbers like t_B or t'_A cannot be "simultaneous". Events
like e1 = "B sends back a light signal to A" or e2 = "A receives
this signal" can be.
If this is what you meant, which is likely, then, according
to Einstein-Poincaré *definition* of simultaneity:
e1 and e2 are NOT simultaneous for A (nor for B)
No.
According to Einstein-Poincaré definition of simultaneity
they are not simultaneous.
However you like or not this *definition* this is something
you CANNOT deny.
A definition cannot be false.
If your idiot guru defined a cow as a 6 leg
small animal with chitine armour - would it
be "not false" to you as well, poor halfbrain?
Who knows?
[snip irrelevant abuse]
W dniu 20.08.2024 o 18:42, Python pisze:
Le 20/08/2024 à 18:38, Maciej Wozniak a écrit :
W dniu 20.08.2024 o 18:22, Python pisze:
Le 20/08/2024 à 18:18, M.D. Richard "Hachel" Lengrand a écrit :
Le 20/08/2024 à 17:32, Python a écrit :
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Numbers like t_B or t'_A cannot be "simultaneous". Events
like e1 = "B sends back a light signal to A" or e2 = "A receives
this signal" can be.
If this is what you meant, which is likely, then, according
to Einstein-Poincaré *definition* of simultaneity:
e1 and e2 are NOT simultaneous for A (nor for B)
No.
According to Einstein-Poincaré definition of simultaneity
they are not simultaneous.
However you like or not this *definition* this is something
you CANNOT deny.
A definition cannot be false.
If your idiot guru defined a cow as a 6 leg
small animal with chitine armour - would it
be "not false" to you as well, poor halfbrain?
Who knows?
You surely don't. Or maybe you do and you
will answer
poor stinker
Nym-shifting troll aka Fehmi Bezrukov wrote :
[snip irrelevant abuse]
Note to sci.* participants: do not sent abuse reports to OVH.
paganini.bofh.team administrator is addressing the issue of
the nym-shifting troll. I'll sent updates.
On 2024-08-20 22:59:08 +0000, Python said:
Nym-shifting troll aka Fehmi Bezrukov wrote :
[snip irrelevant abuse]
Note to sci.* participants: do not sent abuse reports to OVH.
paganini.bofh.team administrator is addressing the issue of
the nym-shifting troll. I'll sent updates.
That's great. I did send one abuse report, but I found it wanted much moreeffort than I expected. At the age that I am things are more
difficult than one might guess.
Anyway, I've noticed that the troll has abandoned any pretence of giving
his nyms plausible names:
Bunnie Belogubov <beb@ublb.ru> -- how likely is it that a Russian would
be called that?
Zasko Mihalkov <ksd@zoaal.ru> -- likewise
Mubarak Schitov <uaka@acikt.ru> -- likewise
Balabaanoff Bibitinsky <yyinsy@krtikb.ru> -- crazier still
Yemill Karkampasis <ly@srak.gr> -- how many Greeks are called Yemill?
Dereck Moraitopoulos <loar@aeteo.gr> -- or Dereck?
Hanoi Bagdasaroff <faa@fgs.ru> -- is Hanoi used as a given name in Russia? Guadalupe Hankoev <ogpa@uauvv.ru> -- not impossible, I suppose, for a
person of mixed Spanish-Russian parents, but not very likely
Fehmi Bezrukov <efbh@eror.ru> -- etc., etc.
Le 20/08/2024 à 19:46, Maciej Wozniak a écrit :
W dniu 20.08.2024 o 18:42, Python pisze:
Le 20/08/2024 à 18:38, Maciej Wozniak a écrit :
W dniu 20.08.2024 o 18:22, Python pisze:
Le 20/08/2024 à 18:18, M.D. Richard "Hachel" Lengrand a écrit :
Le 20/08/2024 à 17:32, Python a écrit :
In short, for A, tB and tA' are simultaneous.
I repeat, for A, tB and tA' are simultaneous.
Numbers like t_B or t'_A cannot be "simultaneous". Events
like e1 = "B sends back a light signal to A" or e2 = "A receives >>>>>>> this signal" can be.
If this is what you meant, which is likely, then, according
to Einstein-Poincaré *definition* of simultaneity:
e1 and e2 are NOT simultaneous for A (nor for B)
No.
According to Einstein-Poincaré definition of simultaneity
they are not simultaneous.
However you like or not this *definition* this is something
you CANNOT deny.
A definition cannot be false.
If your idiot guru defined a cow as a 6 leg
small animal with chitine armour - would it
be "not false" to you as well, poor halfbrain?
Who knows?
You surely don't. Or maybe you do and you
will answer
I could. This would mean that I take you seriously.
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least 40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because 2. because you are tied up with relativistic thoughts all learned, but false.
R.H.
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least 40
years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in R
(where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because 2.
because you are tied up with relativistic thoughts all learned, but false. >>
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least
40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in
R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because
2. because you are tied up with relativistic thoughts all learned, but
false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Le 20/08/2024 à 13:13, Mikko a écrit :
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
You have to understand the words and concepts, and in relativistic
theory, this is very poorly done.
Le 20/08/2024 à 13:13, Mikko a écrit :
On 2024-08-19 23:15:28 +0000, Richard Hachel said:
Le 19/08/2024 à 22:32, "Paul.B.Andersen" a écrit :
Below I show how two real clocks in the real world can be
synchronised, strictly according to Einstein's method.
We have to equal clocks C_A and C_B. They are not synced in any way,
but they are using the same time unit, let's call it second.
The clocks run at the same rate.
In our very big, inertial lab, we have two points A and B which are
separated by some distance. Let's call the transit time for light
to go from A to B is x seconds. We will _define_ that the transit time >>>> is the same from B to A. (This follows from Einstein's definition
of simultaneity).
At point A we have:
Clock C_A, a light-detector, a flash-light and a computer.
The computer can register the time shown by C_A when
the flash-light is flashing, and when the light-detector
registers a light-flash.
At point B we have:
Clock C_B, a light-detector, a mirror and a computer.
The computer can register the time shown by C_B when
the light-detector registers a light-flash.
In the following we will synchronise clock C_B to clock C_A.
That is, we will adjust clock C_B so it become synchronous
with clock C_A.
Now we let the flash-light at point A flash.
At this instant, C_A is showing tA = n seconds.
tA is measured by C_A at A.
When the flash hits the light-detector at B,
Clock C_B shows tB = m seconds.
tB is measured by C_B at B.
A short time later the light detector at A registers
the light reflected by the mirror at B.
At this instant Clock C_A shows t'A = n + 2x seconds.
t'A is measured by C_A at A.
Einstein:
"The two clocks synchronise if tB − tA = t'A − tB."
Or: tB = (tA + t'A)/2 = (n+n+2x)/2 = (n + x)
That is, to be synchronous clock C_B must show a time midway
between tA and t'A when the light is reflected by the mirror.
So tB should show (n + x) seconds when the light is reflected
by the mirror.
But at that instant tB is showing m seconds, so to make the two
clocks synchronous, we must adjust clock C_B by:
δ = (n-m) + x seconds.
After this correction, we have:
tB − tA = (m - n) seconds + δ = x seconds
t'A − tB = (n + 2x - m) seconds - δ = x seconds
The clocks are now synchronised.
Please explain what in the above you find impossible
to do in your lab.
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
It's much more complicated than that.
We can accept it for a Galilean frame of reference,
for example the Earth frame of reference.
But for an accelerated frame of reference, for example, it doesn't work anymore.
Le 20/08/2024 à 15:34, Paul.B.Andersen a écrit :
Den 20.08.2024 01:25, skrev Richard Hachel:...
I see that you are not making any effort to understand what I am saying. >>> What else can I do, you keep repeating the same nonsense based on
ignorance or the refusal of the notion of universal anisochrony?
You can address what I write in stead of snipping it,
probably without reading it.
This is Hachel's (aka Richard Lengrand) habits :
- Don't read and make stupid answers
- Read in a very lazy way (he is utterly lazy in addition to be
stupid and egomaniac)
- Answer in an even more stupid manner
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least 40 years.
Now, yes, obviously I assume it.
Le 20/08/2024 à 17:02, Python a écrit :
A clown is someone pretending that 3 can be 4 for some obervers.
This is not exactly what I am saying.
On 2024-08-20 11:27:56 +0000, Richard Hachel said:
Le 20/08/2024 13:13, Mikko a crit :
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
You have to understand the words and concepts, and in relativistic
theory, this is very poorly done.
What words or concepts in Paul B. Andersen's prodedure did you consider poorly undestood?
If you consider "doable" as I used it in my response then I may clarify
that nothing is doable if it is not clear from the specification what
the intended action should be. But as far as I can see there is nothing
in that procedure that could be regarded as "poorly understood".
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least 40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in
R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because
2. because you are tied up with relativistic thoughts all learned, but
false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent opinion
on the matter.
Mikko <mikko.levanto@iki.fi> wrote:
On 2024-08-20 11:27:56 +0000, Richard Hachel said:
Le 20/08/2024 à 13:13, Mikko a écrit :
I have explained these things a hundred times.
It is impossible to synchronize two watches A and B located in
different places.
So you agree that Paul B. Andersen's prodedure is doable and achieves
what you call "impssible".
You have to understand the words and concepts, and in relativistic
theory, this is very poorly done.
What words or concepts in Paul B. Andersen's prodedure did you consider
poorly undestood?
If you consider "doable" as I used it in my response then I may clarify
that nothing is doable if it is not clear from the specification what
the intended action should be. But as far as I can see there is nothing
in that procedure that could be regarded as "poorly understood".
Yes, and not just doable, routinely done in billionfold.
All clock comparisons between all standard clocks world-wide
(which is needed to establish TAI)
is done by exchanging signals through the GPS system,
taking travel times of signals into account of course,
He's still convinced that GPS satellites clocks are
synchronized against an infinitely far imaginary clock on a
fourth spatial dimension.
Den 21.08.2024 22:20, skrev Richard Hachel:
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false >>>>> depending on the observer.
This is what I have always said for at least
40 years.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read off YOUR watch.
Or is your deeper and more intelligent opinion that the time YOU
read off YOUR watch depend on the observer?
Can I have the opinion that you read something else off your watch
than you did?
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent opinion
on the matter.
Does that mean that your deeper and more intelligent opinion is
that it is NOT a fact that according to Einstein's definition
the clocks are synchronous in the inertial frame?
I am surprised by the stupidity (I do not say this maliciously but with
sadness) of those who read me, and who, surprised, do not understand
anything at all of what I explain to them.
See? You don't even try to address what I write, you flee,
whining about why nobody acknowledge your genius.
You never EXPLAIN anything. You only CLAIM a lot of nonsense.
But now you have the opportunity to EXPLAIN why the clocks
according to Einstein's definition are NOT synchronous in
the inertial frame.
Can you do that?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
--------------
I bet you will flee the challenge yet again. Prove me wrong!
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
This is what I have always said for at least
40 years.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read off YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light, >>
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent opinion
on the matter.
I am surprised by the stupidity (I do not say this maliciously but with sadness) of those who read me, and who, surprised, do not understand
anything at all of what I explain to them.
Le 22/08/2024 à 12:11, Python a écrit :
He's still convinced that GPS satellites clocks are
synchronized against an infinitely far imaginary clock on a
fourth spatial dimension.
<http://news2.nemoweb.net/jntp?omIg9w0Iy0ZX5hSc9cLG74AlKY0@jntp/Data.Media:1>
I never said that, LOL.
You're a buffoon.
<http://news2.nemoweb.net/jntp?omIg9w0Iy0ZX5hSc9cLG74AlKY0@jntp/Data.Media:2>
R.H.
Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least
40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers
in R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because
2. because you are tied up with relativistic thoughts all learned,
but false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
Therefore, you have read something, that should be there, but wasn't.
In fact I have spent a lot of time to verify, that 'delay' or anything equaivalent was actually missing in Einstein's 1905 paper.
Now you have invented in your own mind something, what should be there
(but wasn't).
To verify my statement yourself, you need to go carefully through the
paper and identify the statement, where you think, that Einstein had
delay (or anything equivalent) in mind.
But I was unsuccesful in this realm, because Einstein simply forgot delay.
That's why you can search as long as you like for 't_d' or 'delay' or 'transit time', because they are not present.
Also no equation or any other statement can possibly be interpreted as calculation of transit time.
It's simply not there!
Nym-shifting troll aka Fehmi Bezrukov wrote :
[snip irrelevant abuse]
Note to sci.* participants: do not sent abuse reports to OVH.
paganini.bofh.team administrator is addressing the issue of
the nym-shifting troll. I'll sent updates.
Le 22/08/2024 à 12:27, "Paul.B.Andersen" a écrit :
Den 21.08.2024 22:20, skrev Richard Hachel:
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false >>>>>> depending on the observer.
This is what I have always said for at least 40 years.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read off YOUR watch.
Or is your deeper and more intelligent opinion that the time YOU
read off YOUR watch depend on the observer?
Can I have the opinion that you read something else off your watch
than you did?
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent opinion
on the matter.
Does that mean that your deeper and more intelligent opinion is
that it is NOT a fact that according to Einstein's definition
the clocks are synchronous in the inertial frame?
I am surprised by the stupidity (I do not say this maliciously but with
sadness) of those who read me, and who, surprised, do not understand
anything at all of what I explain to them.
See? You don't even try to address what I write, you flee,
whining about why nobody acknowledge your genius.
You never EXPLAIN anything. You only CLAIM a lot of nonsense.
But now you have the opportunity to EXPLAIN why the clocks
according to Einstein's definition are NOT synchronous in
the inertial frame.
Can you do that?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
--------------
I bet you will flee the challenge yet again. Prove me wrong!
I am not avoiding debate, and on the contrary, I have already explained dozens of times what the notion of universal anisochrony is and how
things should be understood and taught.
But each time, and I do not understand why, no one makes the effort to integrate what I say.
... the Germans (Einstein and Minkowski) ...
... the French spirit (Poincaré, Hachel)
On 2024-08-22 10:52:42 +0000, Richard Hachel said:
Le 22/08/2024 à 12:27, "Paul.B.Andersen" a écrit :
Den 21.08.2024 22:20, skrev Richard Hachel:
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false >>>>>>> depending on the observer.
This is what I have always said for at least 40 years.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read off YOUR watch.
Or is your deeper and more intelligent opinion that the time YOU
read off YOUR watch depend on the observer?
Can I have the opinion that you read something else off your watch
than you did?
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected >>>>> light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent
opinion on the matter.
Does that mean that your deeper and more intelligent opinion is
that it is NOT a fact that according to Einstein's definition
the clocks are synchronous in the inertial frame?
I am surprised by the stupidity (I do not say this maliciously but
with sadness) of those who read me, and who, surprised, do not
understand anything at all of what I explain to them.
See? You don't even try to address what I write, you flee,
whining about why nobody acknowledge your genius.
You never EXPLAIN anything. You only CLAIM a lot of nonsense.
But now you have the opportunity to EXPLAIN why the clocks
according to Einstein's definition are NOT synchronous in
the inertial frame.
Can you do that?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light, >>>
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
--------------
I bet you will flee the challenge yet again. Prove me wrong!
I am not avoiding debate, and on the contrary, I have already
explained dozens of times what the notion of universal anisochrony is
and how things should be understood and taught.
But each time, and I do not understand why, no one makes the effort to
integrate what I say.
What you post here is not important. If you were a genius willing to
tell something important you would do it in better place, e.g. in a
book.
Le 22/08/2024 à 12:52, Richard "Hachel" Lengrand a écrit :
... the Germans (Einstein and Minkowski) ...
... the French spirit (Poincaré, Hachel)
This is a bogus dichotomy.
The real one is:
Poincaré, Einstein, Minkowski: brilliant scientists
Hachel, Wozniak, ...: crooks
W dniu 22.08.2024 o 13:49, Mikko pisze:
On 2024-08-22 10:52:42 +0000, Richard Hachel said:
Le 22/08/2024 à 12:27, "Paul.B.Andersen" a écrit :
Den 21.08.2024 22:20, skrev Richard Hachel:
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :Or is your deeper and more intelligent opinion that the time YOU
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false >>>>>>>> depending on the observer.
This is what I have always said for at least 40 years.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read off YOUR watch. >>>>
read off YOUR watch depend on the observer?
Can I have the opinion that you read something else off your watch
than you did?
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected >>>>>> light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according >>>>>> to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition >>>>>> the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent
opinion on the matter.
Does that mean that your deeper and more intelligent opinion is
that it is NOT a fact that according to Einstein's definition
the clocks are synchronous in the inertial frame?
I am surprised by the stupidity (I do not say this maliciously but
with sadness) of those who read me, and who, surprised, do not
understand anything at all of what I explain to them.
See? You don't even try to address what I write, you flee,
whining about why nobody acknowledge your genius.
You never EXPLAIN anything. You only CLAIM a lot of nonsense.
But now you have the opportunity to EXPLAIN why the clocks
according to Einstein's definition are NOT synchronous in
the inertial frame.
Can you do that?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light, >>>>
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
--------------
I bet you will flee the challenge yet again. Prove me wrong!
I am not avoiding debate, and on the contrary, I have already
explained dozens of times what the notion of universal anisochrony is
and how things should be understood and taught.
But each time, and I do not understand why, no one makes the effort
to integrate what I say.
What you post here is not important. If you were a genius willing to
tell something important you would do it in better place, e.g. in a
book.
The problem with idiots like you is: you
have no slightest clue what "good" means,
but you still feel an invincible expert
of what is good and what is better.
W dniu 22.08.2024 o 14:23, Python pisze:...
Le 22/08/2024 à 14:21, Maciej Wozniak a écrit :
W dniu 22.08.2024 o 13:49, Mikko pisze:
The problem with idiots like you is: you
have no slightest clue what "good" means,
but you still feel an invincible expert
of what is good and what is better.
As a self-proclamed "one the best logician Humanity ever had" (you still
owe me a keyboard for that one!) don't you think that your genial
thoughts would deserve a book Maciej?
No, I don't,
Le 22/08/2024 à 14:21, Maciej Wozniak a écrit :
W dniu 22.08.2024 o 13:49, Mikko pisze:
On 2024-08-22 10:52:42 +0000, Richard Hachel said:
Le 22/08/2024 à 12:27, "Paul.B.Andersen" a écrit :
Den 21.08.2024 22:20, skrev Richard Hachel:
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false >>>>>>>>> depending on the observer.
This is what I have always said for at least 40 years.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read off YOUR
watch.
Or is your deeper and more intelligent opinion that the time YOU
read off YOUR watch depend on the observer?
Can I have the opinion that you read something else off your watch
than you did?
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the
reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame >>>>>>> independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according >>>>>>> to Einstein's _definition_ are synchronous in the inertial frame. >>>>>>>
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition >>>>>>> the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent
opinion on the matter.
Does that mean that your deeper and more intelligent opinion is
that it is NOT a fact that according to Einstein's definition
the clocks are synchronous in the inertial frame?
I am surprised by the stupidity (I do not say this maliciously but >>>>>> with sadness) of those who read me, and who, surprised, do not
understand anything at all of what I explain to them.
See? You don't even try to address what I write, you flee,
whining about why nobody acknowledge your genius.
You never EXPLAIN anything. You only CLAIM a lot of nonsense.
But now you have the opportunity to EXPLAIN why the clocks
according to Einstein's definition are NOT synchronous in
the inertial frame.
Can you do that?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected
light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
--------------
I bet you will flee the challenge yet again. Prove me wrong!
I am not avoiding debate, and on the contrary, I have already
explained dozens of times what the notion of universal anisochrony
is and how things should be understood and taught.
But each time, and I do not understand why, no one makes the effort
to integrate what I say.
What you post here is not important. If you were a genius willing to
tell something important you would do it in better place, e.g. in a
book.
The problem with idiots like you is: you
have no slightest clue what "good" means,
but you still feel an invincible expert
of what is good and what is better.
As a self-proclamed "one the best logician Humanity ever had" (you still
owe me a keyboard for that one!) don't you think that your genial
thoughts would deserve a book Maciej?
When I talk about type M (or Einstein type) synchronization,
I am talking about the need to represent a coherent universal
simultaneity for the entire terrestrial frame of reference.
However, this absolute and universal simultaneity, even
in a simple, simple inertial reference frame, cannot be used,
because it DOES NOT EXIST.
We must therefore find a "neutral" point, for which a certain
form of simultaneity would exist, but all the points of
the universe would have to be at an equal distance from it,
in order to synchronize all the watches, on its notion of proper
present .
We must therefore imagine a point ideally placed very far away,
in an imaginary fourth dimension, and imagine that it is this
which simultaneously gives the start to all the watches in
the universe.
This is what Einstein synchronization does if you look closely,
and this is why it is mathematically coherent and easily usable.
But that does not mean that two watches A and B marking the same
time for M and "existing in perfect simultaneity with M" mark
together and reciprocally the same time.
This is a false and abstract thought from a person who does
not understand the basis of the theory of relativity.
The two watches are actually offset from each other by dt=AB/c
anisochrony value valid for the entire universe.
The notion of absolute universal present does not exist,
and it
is this special synchronization which creates it on what
a distant, idealized and “neutral” observer would perceive.
Of course "simultaneity" and "synchronism" are man made,
theoretical notions, but they are very practical, and
the world would be even more chaotic than it is without it.
Le 22/08/2024 à 12:27, "Paul.B.Andersen" a écrit :
Den 21.08.2024 22:20, skrev Richard Hachel:
Le 21/08/2024 à 20:41, "Paul.B.Andersen" a écrit :
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false >>>>>> depending on the observer.
This is what I have always said for at least 40 years.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read off YOUR watch.
Or is your deeper and more intelligent opinion that the time YOU
read off YOUR watch depend on the observer?
Can I have the opinion that you read something else off your watch
than you did?
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected
light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Note this:
-----------
It is an indisputable FACT that according to Einstein's definition
the clocks are synchronous in the inertial frame.
It is not possible to have different opinions about this.
Yes, it is possible to have a much deeper and more intelligent
opinion on the matter.
Does that mean that your deeper and more intelligent opinion is
that it is NOT a fact that according to Einstein's definition
the clocks are synchronous in the inertial frame?
I am surprised by the stupidity (I do not say this maliciously but
with sadness) of those who read me, and who, surprised, do not
understand anything at all of what I explain to them.
See? You don't even try to address what I write, you flee,
whining about why nobody acknowledge your genius.
You never EXPLAIN anything. You only CLAIM a lot of nonsense.
But now you have the opportunity to EXPLAIN why the clocks
according to Einstein's definition are NOT synchronous in
the inertial frame.
Can you do that?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light, >>
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
--------------
I bet you will flee the challenge yet again.
I am not avoiding debate, and on the contrary, I have already explained dozens of times what the notion of universal anisochrony is and how
things should be understood and taught.
But each time, and I do not understand why, no one makes the effort to integrate what I say. I think it is out of conformity. I do not think it
is out of laziness or lack of intelligence, because there are posters
like you who are courageous (you have to be courageous to write pdfs
rather than watch television) and who are intelligent, even curious.
The reason therefore comes from conformity and the fear of shaking up
ideas, even if the ideas are ugly and false (ridiculous integration of improper times in your pdf, bad equations for instantaneous observable
speeds and proper times of accelerated objects, delirium about rotating disks).
Yet EVERYTHING I say should be clear and obvious to someone who would
detach himself from what the Germans (Einstein and Minkowski) said to
get closer to the French spirit (Poincaré, Hachel). You just have to understand, and everything becomes clearer, more beautiful, truer and
more obvious to teach.
R.H.
Den 22.08.2024 13:09, skrev Richard Hachel:
| But Einstein showed that there is no absolute simultaneity,
| and clocks can't be absolutely synchronised.
If time is a local phenomenon, you cannot assume, that perceived delay
(or 'transit time') would be independent of movement.
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least
40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in
R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because
2. because you are tied up with relativistic thoughts all learned, but
false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected
light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
But he write down two equations that implies directly that a delay
is taken into account.
Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least 40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in
R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because
2. because you are tied up with relativistic thoughts all learned, but
false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light, >>
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:
Den 20.08.2024 17:12, skrev Richard Hachel:
Le 20/08/2024 à 15:39, Python a écrit :
Hachel now pretends that tB − tA = t'A − tB can be true or false
depending on the observer.
You are lying.
I do not claim it "now". This is what I have always said for at least 40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also
for all the stationary points of the inertial frame of reference of A
and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference.
On the other hand the value tB-tA (go) will vary for most observers in
R (where A and B are stationary), as will the value tA'-tB (return).
But you cannot understand this, because 1. You are stupid and because
2. because you are tied up with relativistic thoughts all learned, but
false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
This is not, what 'invariant' means in the context of relativity.
Meant is, that time would not change, if you switch from one frame of reference to another.
Whether this is actually true or not is irrelevant here, because you
need to compensate the delay somehow. Otherwise you would get a 'mutual
time stretching' effect, what cannot be a real physical effect, because
it is visible only at the far side by the remote observer.
Since both of these observers are of equal rights, both could claim the
other time reading invalid, hence both readings ARE invalid.
Therefore you must compensate the delay 'by hand'.
...
TH
This is not, what 'invariant' means in the context of relativity.
Yes, it is.
Meant is, that time would not change, if you switch from one frame of
reference to another.
No, it means that whatever is called "invariant" is the same for all
frames. In the current case, the number wirtten on the paper is invariant.
Mikko
https://paulba.no/paper/Electrodynamics.pdf
Quote from § 1. Definition of Simultaneity: -------------------------------------------
"If at the point A of space there is a clock, an observer at
A can determine the time values of events in the immediate
proximity of A by finding the positions of the hands which
are simultaneous with these events.
If there is at the point B of space another clock in all
respects resembling the one at A, it is possible for an observer
at B to determine the time values of events in the immediate
neighbourhood of B.
But it is not possible without further assumption to compare,
in respect of time, an event at A with an event at B.
We have so far defined only an “A time” and a “B time.”
We have not defined a common “time” for A and B, for
the latter cannot be defined at all unless we establish
by definition that the “time” required by light to travel
from A to B equals the “time” it requires to travel from B to A.
"
If you can read, you will see that Einstein did say what I said.
I am quite certain that Einstein wasn't talking about
the same thing as Richard Hachel.
It is remarkable that a person who pride himself of having studied
relativity issues for 40 years is ignorant of the most basic concepts
in the Special Theory of Relativity.
I am not going to teach you SR (or GR).
If you really want to learn, read a book.
I would be very happy to understand you and for you to be able to define
your words.
All the "words" used in SR are defined in any book about relativity,
but you have to read it yourself.
I will remind you:
In the world there are millions of clocks which are synchronous
in the ECI frame, and the world would be even more chaotic than
it is without them. Think if it was no way to tell you when your
train or aeroplane would go, and there was no way to tell you
when you would arrive at the destination.
The world is _very_ dependent on synchronous clocks.
The civilisation as we know it couldn't exist without them.
And you say it is impossible to synchronise clocks? :-D
Paul
Le 22/08/2024 à 20:19, "Paul.B.Andersen" a écrit :
It is no absolute and universal simultaneity.
Since you still seem to think that it was Richard Hachel
who discovered this, it is obvious that you do not read
what I and others write to you.
Den 22.08.2024 Paul B. Andersen wrote:
|
| And you believe it is YOU that have discovered that? 😂
|
| Before 1905 everybody believed it was a "universal, present now",
| that simultaneity was absolute, and that clocks could be
| absolutely synchronised. Newton took it for granted!
|
| But Einstein showed that there is no absolute simultaneity,
| and clocks can't be absolutely synchronised.
|
| https://paulba.no/paper/Electrodynamics.pdf
| See: § 1. Definition of Simultaneity
|
| Did you really not know that it was Einstein who discovered this? 😂
Let's admit it, Paul.
You're wrong, Einstein didn't say anything at all, and always just
repeated what Poincaré said, but hey, it doesn't matter, we'll admit
that you're right.
There's just one thing I don't understand in your grievances and mockery. What do you mean by: "I, Paul B. Andersen, think that there is no
absolute simultaneity, and that not all watches can be synchronized"?
There is no _absolute_ simultaneity, but Einstein _defined_
what he meant by simultaneity _in an inertial frame_.
And with this _definition_, we can make two clocks at different
locations in the inertial frame simultaneously show the same;
the clocks are synchronous _in said frame of reference_.
But they are NOT synchronous in a frame of reference which
is moving relative to the first frame of reference.
Python said that the terms had to be clear (for once, he's not lying).
The terms have to be clear.
In the introduction to my pdf, I wrote two A4 pages (in a compressed handwriting) and without putting a single equation, just to talk about
the notion of simultaneity, proof that it's not so obvious (even if it
makes you laugh).
I too am talking about the notion of simultaneity, but am I sure we are talking about the same thing?
Is Einstein talking about the same thing?
Can you explain to me, in the greatest clarity, as Python recommends,
what you mean, what you understand by the following words: "In special relativity, the notion of simultaneity is relative"?
This is very important, and it is the very basis of the theory as it
actually exists in nature.
What do you mean by these words?
Do you mean that the internal chronotropy of watches is relative?
Do you mean that the notion of instant is relative to position (spatial anisochrony)?
Do you mean something else?
Personally, I define my words, and I explain what I consider clearly.
I would be very happy to understand you and for you to be able to define
your words.
Le 23/08/2024 à 07:41, Thomas Heger a écrit :
Whether this is actually true or not is irrelevant here, because you
need to compensate the delay somehow. Otherwise you would get a 'mutual
time stretching' effect, what cannot be a real physical effect, because
it is visible only at the far side by the remote observer.
Since both of these observers are of equal rights, both could claim the
other time reading invalid, hence both readings ARE invalid.
Therefore you must compensate the delay 'by hand'.
...
TH
Beaucoup de choses intéressantes dans ce post.
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
https://paulba.no/paper/Electrodynamics.pdf
Quote from § 1. Definition of Simultaneity:
-------------------------------------------
"If at the point A of space there is a clock, an observer at
A can determine the time values of events in the immediate
proximity of A by finding the positions of the hands which
are simultaneous with these events.
If there is at the point B of space another clock in all
respects resembling the one at A, it is possible for an observer
at B to determine the time values of events in the immediate
neighbourhood of B.
But it is not possible without further assumption to compare,
in respect of time, an event at A with an event at B.
We have so far defined only an “A time” and a “B time.”
We have not defined a common “time” for A and B, for
the latter cannot be defined at all unless we establish
by definition that the “time” required by light to travel
from A to B equals the “time” it requires to travel from B to A.
"
If you can read, you will see that Einstein did say what I said.
Here is finally a solid basis.
And that is very well said.
The small drawback that remains is that Einstein proposes a definition,
but without explaining which observer will be able to consider the proposition as true.
Le 23/08/2024 à 10:55, Mikko a écrit :
This is not, what 'invariant' means in the context of relativity.
Yes, it is.
Meant is, that time would not change, if you switch from one frame of
reference to another.
No, it means that whatever is called "invariant" is the same for all
frames. In the current case, the number wirtten on the paper is invariant. >>
Mikko
Here is yet another proof of what I am saying, and of the need to
re-explain things correctly.
When I say "I bought a white horse, and I gave it to Father François;
he will take care of it because he is retired, and he owns a field",
everyone understands what I am saying.
But if I say: "All the watches will be desynchronized", it is clear
that no one will clearly understand what that means, and so on for a
great many terms used.
Den 22.08.2024 21:12, skrev Richard Hachel:
Le 22/08/2024 à 20:19, "Paul.B.Andersen" a écrit :
It is no absolute and universal simultaneity.
Since you still seem to think that it was Richard Hachel
who discovered this, it is obvious that you do not read
what I and others write to you.
Den 22.08.2024 Paul B. Andersen wrote:
|
| And you believe it is YOU that have discovered that? 😂
|
| Before 1905 everybody believed it was a "universal, present now",
| that simultaneity was absolute, and that clocks could be
| absolutely synchronised. Newton took it for granted!
|
| But Einstein showed that there is no absolute simultaneity,
| and clocks can't be absolutely synchronised.
|
| https://paulba.no/paper/Electrodynamics.pdf
| See: § 1. Definition of Simultaneity
|
| Did you really not know that it was Einstein who discovered this? 😂 >>>
Let's admit it, Paul.
You're wrong, Einstein didn't say anything at all, and always just
repeated what Poincaré said, but hey, it doesn't matter, we'll admit
that you're right.
https://paulba.no/paper/Electrodynamics.pdf
Quote from § 1. Definition of Simultaneity: -------------------------------------------
"If at the point A of space there is a clock, an observer at
A can determine the time values of events in the immediate
proximity of A by finding the positions of the hands which
are simultaneous with these events.
If there is at the point B of space another clock in all
respects resembling the one at A, it is possible for an observer
at B to determine the time values of events in the immediate
neighbourhood of B.
But it is not possible without further assumption to compare,
in respect of time, an event at A with an event at B.
We have so far defined only an “A time” and a “B time.”
We have not defined a common “time” for A and B, for
the latter cannot be defined at all unless we establish
by definition that the “time” required by light to travel
from A to B equals the “time” it requires to travel from B to A.
"
Of course "simultaneity" and "synchronism" are man made,
theoretical notions, but they are very practical, and
the world would be even more chaotic than it is without it.
W dniu 23.08.2024 o 07:41, Thomas Heger pisze:
If time is a local phenomenon, you cannot assume, that perceived delay
(or 'transit time') would be independent of movement.
He can always assume whatever idiocy he wants.
but time is neither local nor a phenomenon.
Am Donnerstag000022, 22.08.2024 um 13:11 schrieb Python:
...
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected >>>> light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
But he write down two equations that implies directly that a delay
is taken into account.
The equation on page 3 COULD be interpeted as calculation of the delay.
But Einstein wrote, that would be the definition of the speed of light.
I would miss the word 'delay' in this context (or something similar).
Also the remainder of this paper does not mention delay neither.
So: where have YOU found any use of delay or transit time in this paper?
I 'combed' the text very carefully and could not find any statement,
which eventually would match this discription.
Le 23/08/2024 à 07:59, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 07:41, Thomas Heger pisze:
If time is a local phenomenon, you cannot assume, that perceived
delay (or 'transit time') would be independent of movement.
He can always assume whatever idiocy he wants.
but time is neither local nor a phenomenon.
So what is it?
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
It is remarkable that a person who pride himself of having studied
relativity issues for 40 years is ignorant of the most basic concepts
in the Special Theory of Relativity.
I am not going to teach you SR (or GR).
If you really want to learn, read a book.
Please, a little more seriousness and dignity in your answers.
W dniu 23.08.2024 o 15:41, Python pisze:
Le 23/08/2024 à 07:59, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 07:41, Thomas Heger pisze:
If time is a local phenomenon, you cannot assume, that perceived
delay (or 'transit time') would be independent of movement.
He can always assume whatever idiocy he wants.
but time is neither local nor a phenomenon.
So what is it?
An universal/global coordinate.
Having
all the usual properties of a coordinate,
including being a human made mental
construct.
On 2024-08-20 22:59:08 +0000, Python said:
Nym-shifting troll aka Fehmi Bezrukov wrote :
[snip irrelevant abuse]
Note to sci.* participants: do not sent abuse reports to OVH.
paganini.bofh.team administrator is addressing the issue of
the nym-shifting troll. I'll sent updates.
Thank you. I would like to know whem (if ever) it is time to
remove paganini from my kill file.
W dniu 23.08.2024 o 15:59, Python pisze:...
Le 23/08/2024 à 15:57, Maciej Wozniak a écrit :
An universal/global coordinate.
This is a postulate that such a universal/global can exist and
be consistent.
No, [snip profanity], any function defined
from the set of events into R will do.
..furthermore, i don't understand how anyone thinks
they can sync two clocks if Time Dialation will always
UN-sync...a clock?
Le 23/08/2024 à 15:57, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 15:41, Python pisze:
Le 23/08/2024 à 07:59, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 07:41, Thomas Heger pisze:
If time is a local phenomenon, you cannot assume, that perceived
delay (or 'transit time') would be independent of movement.
He can always assume whatever idiocy he wants.
but time is neither local nor a phenomenon.
So what is it?
An universal/global coordinate.
This is a postulate that such a universal/global can exist and
be consistent.
Having
all the usual properties of a coordinate,
including being a human made mental
construct.
This is right. Bravo !
Le 23/08/2024 à 17:03, The Starmaker a écrit :
...
..furthermore, i don't understand how anyone thinks
they can sync two clocks if Time Dialation will always
UN-sync...a clock?
Synchronization is about at least two clocks by definition.
Co-moving clocks do not "unsynchronize"
Le 23/08/2024 à 16:27, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 15:59, Python pisze:...
Le 23/08/2024 à 15:57, Maciej Wozniak a écrit :
An universal/global coordinate.
This is a postulate that such a universal/global can exist and
be consistent.
No, [snip profanity], any function defined
from the set of events into R will do.
Including the constant function?
W dniu 23.08.2024 o 16:29, Python pisze:
Le 23/08/2024 à 16:27, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 15:59, Python pisze:...
Le 23/08/2024 à 15:57, Maciej Wozniak a écrit :
An universal/global coordinate.
This is a postulate that such a universal/global can exist and
be consistent.
No, [snip profanity], any function defined
from the set of events into R will do.
Including the constant function?
It's not a good coordinate, but still.
Le 23/08/2024 à 13:57, M.D. Richard "Hachel" Lengrand a écrit :
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
It is remarkable that a person who pride himself of having studied
relativity issues for 40 years is ignorant of the most basic concepts
in the Special Theory of Relativity.
I am not going to teach you SR (or GR).
If you really want to learn, read a book.
Please, a little more seriousness and dignity in your answers.
He is perfectly right. You've stuffed your own mind with idiotic
nonsense for 40 years, intoxicating your brain with silly unsound
idiocies. If you want to understand SR you HAVE TO empty all this
sh*t out from your mind and start from scratch.
Le 23/08/2024 à 18:15, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 16:29, Python pisze:
Le 23/08/2024 à 16:27, Maciej Wozniak a écrit :
W dniu 23.08.2024 o 15:59, Python pisze:...
Le 23/08/2024 à 15:57, Maciej Wozniak a écrit :
An universal/global coordinate.
This is a postulate that such a universal/global can exist and
be consistent.
No, [snip profanity], any function defined
from the set of events into R will do.
Including the constant function?
It's not a good coordinate, but still.
Why? What are your criteria for a "good candidate"?
See, Wozniak, this is actually quite an interesting question. Instead
of insulting people why would you not start a discussion on this point
here.
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
https://paulba.no/paper/Electrodynamics.pdf
Quote from § 1. Definition of Simultaneity:
-------------------------------------------
"If at the point A of space there is a clock, an observer at
A can determine the time values of events in the immediate
proximity of A by finding the positions of the hands which
are simultaneous with these events.
If there is at the point B of space another clock in all
respects resembling the one at A, it is possible for an observer
at B to determine the time values of events in the immediate
neighbourhood of B.
But it is not possible without further assumption to compare,
in respect of time, an event at A with an event at B.
We have so far defined only an “A time” and a “B time.”
We have not defined a common “time” for A and B, for
the latter cannot be defined at all unless we establish
by definition that the “time” required by light to travel
from A to B equals the “time” it requires to travel from B to A.
"
If you can read, you will see that Einstein did say what I said.
Here is finally a solid basis.
And that is very well said.
The small drawback that remains is that Einstein proposes a definition,
but without explaining which observer will be able to consider the proposition as true.
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :|
Den 22.08.2024 21:12, skrev Richard Hachel:
Can you explain to me, in the greatest clarity,
as Python recommends, what you mean, what you understand
by the following words: "In special relativity,
the notion of simultaneity is relative"?
https://paulba.no/paper/Electrodynamics.pdf
Read:
§ 1. Definition of Simultaneity
§ 2. On the Relativity of Lengths and Times
It is explained with the greatest clarity.
Einstein proposes an interesting synchronization, and that I take upGobbledegook!
again by speaking of synchronization of type M,
based on an imaginary observer placed in M in a teletransverse way in an abstract fourth dimension.
The problem is that he does not say it or at worst, he does not know it. Saying "Between A and B, the speed of light is c, we know it, because we
have measured it" does not make sense. Who measures this speed? A? No.
B? Neither. We must therefore define things.
Saying:
"My dear Jane, I bought an animal", is ridiculous.
We must say "My dear Jane, I bought for your birthday this white horse
that you wanted".
This is why, for 40 years, I have been saying that this introduction
needs to be rewritten in a clearer, more understandable and more obvious
way.
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
Den 22.08.2024 21:12, skrev Richard Hachel:
I too am talking about the notion of simultaneity,I am quite certain that Einstein wasn't talking about
but am I sure we are talking about the same thing?
Is Einstein talking about the same thing? >>
the same thing as Richard Hachel.
It's obvious.
I have already made many complaints to you, and among them the rotation
that you make by changing the frame of reference, I explained to you
that there was no rotation, but translation in x (y and z remaining invariant).
Your new point in R' is therefore erroneous each time. In addition, you
keep an invariant time To since your distance d becomes invariant. All
this is not serious, but it is not your fault.
It is that of German thought (Einstein, Minkowski).
R.H.
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
In the world there are millions of clocks which are synchronous
in the ECI frame, and the world would be even more chaotic than
it is without them. Think if it was no way to tell you when your
train or aeroplane would go, and there was no way to tell you
when you would arrive at the destination.
The world is _very_ dependent on synchronous clocks.
The civilisation as we know it couldn't exist without them.
And you say it is impossible to synchronise clocks? :-D
Paul
That's not really what I said.
I said that in the common world, we could use a synchronization in
hours, minutes and seconds, to qualify various events.
But that a synchronization in nanoseconds was only possible under
certain conditions, and in particular the creation of a universal time capable of governing all of this.
However, it is impossible for such a universal time to exist, since
according to Hachel (too bad if I contradict Einstein on that) the
notion of a plane of universal present time does not exist, it is a
powerful thought (like the flat earth) but abstract from reality.
We will therefore never be able to agree even two watches placed in the
same frame of reference. Of course, they will beat at the same speed,
that is to say they will have the same internal chronotropy, but each
will have its own notion of the surrounding simultaneity, that is to say
its own hyperplane of present time.
R.H.
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
Den 22.08.2024 21:12, skrev Doctor Richard Hachel:
Can you explain to me, in the greatest clarity, as Python recommends,
what you mean, what you understand by the following words:
"In special relativity, the notion of simultaneity is relative"?
It is remarkable that a person who pride himself of having studied
relativity issues for 40 years is ignorant of the most basic concepts
in the Special Theory of Relativity.
I am not going to teach you SR (or GR).
If you really want to learn, read a book.
Please, a little more seriousness and dignity in your answers.
Le 23/08/2024 à 16:04, Python a écrit :
Le 23/08/2024 à 13:57, M.D. Richard "Hachel" Lengrand a écrit :
Le 23/08/2024 à 13:23, "Paul.B.Andersen" a écrit :
Den 22.08.2024 21:12, skrev Doctor Richard Hachel:
Can you explain to me, in the greatest clarity, as Python recommends, what you mean, what you understand by the following words: "In special relativity, the notion of simultaneity is relative"?It is remarkable that a person who pride himself of having studied
relativity issues for 40 years is ignorant of the most basic concepts
in the Special Theory of Relativity.
I am not going to teach you SR (or GR).
If you really want to learn, read a book.
Please, a little more seriousness and dignity in your answers.
He is perfectly right. You've stuffed your own mind with idiotic
nonsense for 40 years, intoxicating your brain with silly unsound
idiocies. If you want to understand SR you HAVE TO empty all this
sh*t out from your mind and start from scratch.
No, it's not "perfectly right".
It's not up to me to pick up a book and read what relativists
have said and written (I did forty years ago).
It is up to the relativists to read what I have written,
and to see if it is coherent or not.
I pointed out to Paul where the errors were,
and although the language barrier may play a role,
it is impossible for him not to understand what I am
saying if it takes the effort to understand.
He makes no effort and tirelessly repeats "Albert is God,
and I am his prophet; Hachel is the suppository of Satan". >
This is not scientific behavior.
Den 23.08.2024 13:57, skrev Doctor Richard Hachel:
Please, a little more seriousness and dignity in your answers.
More dignity? Should I address you with "Doctor" or "Sir"?
So: where have YOU found any use of delay or transit time in this paper?
Equations there implies t'_A = t_B - (AB)/c, (AB)/c is the light
propagation delay between A and B or B and A.
Le 22/08/2024 à 13:24, Mikko a écrit :
Thank you. I would like to know whem (if ever) it is time to remove
paganini from my kill file.
He has been kicked of from paganini, he's now using Eternal September,
from where he will be banned too.
Python wrote:
But he write down two equations that implies directly that a delay is
taken into account.
then you don't understand physics. That's exactly NOT a delay in
relativity, which is all about NOT having delays, exactly. This
uneducated troll don't know what a delay is.
'Delay' is a VERY common phenomenon.
E.g. if you use satellites for phone-calls, the long distance from
ground station A to satellite and from there to ground station B causes audible delays. All sorts of other devices or situations are also
influenced by the finite speed of light, too.
It is simply everywhere and all around us!
Den 23.08.2024 13:57, skrev Doctor Richard Hachel:
Le 23/08/2024 13:23, "Paul.B.Andersen" a écrit :
Den 22.08.2024 21:12, skrev Doctor Richard Hachel:
Can you explain to me, in the greatest clarity, as Python recommends,
what you mean, what you understand by the following words:
"In special relativity, the notion of simultaneity is relative"?
"The notion of simultaneity" is a very basic concept in SR.
So I seriously think that:
It is remarkable that a person who pride himself of having studied
relativity issues for 40 years is ignorant of the most basic concepts
in the Special Theory of Relativity.
You can't expect me to teach you the basic concepts of SR in this forum.
So:
I am not going to teach you SR (or GR).
If you really want to learn, read a book.
This is my serious advice, I am not joking.
Please, a little more seriousness and dignity in your answers.
More dignity? Should I address you with "Doctor" or "Sir"?
Python wrote:
Le 22/08/2024 à 13:24, Mikko a écrit :
Thank you. I would like to know whem (if ever) it is time to remove
paganini from my kill file.
He has been kicked of from paganini, he's now using Eternal September,
from where he will be banned too.
... They blocked the posts from
everybody, you included.
W dniu 23.08.2024 o 07:41, Thomas Heger pisze:
If time is a local phenomenon, you cannot assume, that perceived delay
(or 'transit time') would be independent of movement.
He can always assume whatever idiocy he wants.
but time is neither local nor a phenomenon.
Am Freitag000023, 23.08.2024 um 07:59 schrieb Maciej Wozniak:
W dniu 23.08.2024 o 07:41, Thomas Heger pisze:Well, depends on the definition of 'phenomenon'.
If time is a local phenomenon, you cannot assume, that perceived
delay (or 'transit time') would be independent of movement.
He can always assume whatever idiocy he wants.
but time is neither local nor a phenomenon.
The concept of time is actually based on counint repeaded events, about
which we assume, they would alway occur at the same frequency.
I do not claim it "now". This is what I have always said for at
least 40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but
also for all the stationary points of the inertial frame of
reference of A and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference. >>>>
On the other hand the value tB-tA (go) will vary for most observers
in R (where A and B are stationary), as will the value tA'-tB (return). >>>>
But you cannot understand this, because 1. You are stupid and
because 2. because you are tied up with relativistic thoughts all
learned, but false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light, >>>
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
Einstein used tB - tA and similar expressions. Nothing else needs be
said about delays. The equation tB − tA = t'A − tB and the text that describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.
On 2024-08-23 10:58:35 +0000, Richard Hachel said:
Le 23/08/2024 à 07:41, Thomas Heger a écrit :
Whether this is actually true or not is irrelevant here, because you
need to compensate the delay somehow. Otherwise you would get a
'mutual time stretching' effect, what cannot be a real physical
effect, because it is visible only at the far side by the remote
observer.
Since both of these observers are of equal rights, both could claim
the other time reading invalid, hence both readings ARE invalid.
Therefore you must compensate the delay 'by hand'.
...
TH
Beaucoup de choses intéressantes dans ce post.
If you believe a fool you are a fool.
Am Freitag000023, 23.08.2024 um 10:51 schrieb Mikko:
...
I do not claim it "now". This is what I have always said for at least 40 years.
Now, yes, obviously I assume it.
The value (tA'-tA) = 2AB/c is the same not only for A and B, but also >>>>> for all the stationary points of the inertial frame of reference of A >>>>> and B.
Better, if I change frame of reference it will remain true, by
invariance of the transverse speed of light in any frame of reference. >>>>>
On the other hand the value tB-tA (go) will vary for most observers in >>>>> R (where A and B are stationary), as will the value tA'-tB (return). >>>>>
But you cannot understand this, because 1. You are stupid and because >>>>> 2. because you are tied up with relativistic thoughts all learned, but >>>>> false.
R.H.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
Einstein used tB - tA and similar expressions. Nothing else needs be
said about delays. The equation tB − tA = t'A − tB and the text that
describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.
But this has nothing to do with synchronicity, but with a process to
turn remote clocks to the same time value.
Time is not equal to what clocks say, because clocks are measuring
devices, which measure time, but do not determine, what the measured
quantity is.
The process to synchronize clocks require technical means, which are
here light signals:
The clock at some point A emmits a timing signal, which a remote clock receives a little time later, because such signals have finite speed.
Now it should be obvious, that the remote clock had to compensate this
delay, because otherwise it would not show the time of the master
clock, but an asynchronous value.
For uncertain reasons Einstein had not mentioned this requirement at
all, even if transit time per se was actually mentionend.
Am Freitag000023, 23.08.2024 um 14:18 schrieb Mikko:
On 2024-08-23 10:58:35 +0000, Richard Hachel said:But he believes me and not you!
Le 23/08/2024 à 07:41, Thomas Heger a écrit :
Whether this is actually true or not is irrelevant here, because you
need to compensate the delay somehow. Otherwise you would get a 'mutual >>>> time stretching' effect, what cannot be a real physical effect, because >>>> it is visible only at the far side by the remote observer.
Since both of these observers are of equal rights, both could claim the >>>> other time reading invalid, hence both readings ARE invalid.
Therefore you must compensate the delay 'by hand'.
...
TH
Beaucoup de choses intéressantes dans ce post.
If you believe a fool you are a fool.
Le 22/08/2024 à 13:24, Mikko a écrit :
On 2024-08-20 22:59:08 +0000, Python said:
Nym-shifting troll aka Fehmi Bezrukov wrote :
[snip irrelevant abuse]
Note to sci.* participants: do not sent abuse reports to OVH.
paganini.bofh.team administrator is addressing the issue of
the nym-shifting troll. I'll sent updates.
Thank you. I would like to know whem (if ever) it is time to
remove paganini from my kill file.
He has been kicked of from paganini, he's now using Eternal
September, from where he will be banned too.
Le 23/08/2024 à 10:55, Mikko a écrit :
On 2024-08-23 05:41:50 +0000, Thomas Heger said:
Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
This is not, what 'invariant' means in the context of relativity.
Yes, it is.
Meant is, that time would not change, if you switch from one frame of
reference to another.
No, it means that whatever is called "invariant" is the same for all
frames. In the current case, the number wirtten on the paper is invariant. >>
Mikko
Here is yet another proof of what I am saying, and of the need to
re-explain things correctly.
Den 23.08.2024 13:30, skrev Richard Hachel:
Le 23/08/2024 à 10:55, Mikko a écrit :
On 2024-08-23 05:41:50 +0000, Thomas Heger said:
Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch.
This is not, what 'invariant' means in the context of relativity.
Yes, it is.
Meant is, that time would not change, if you switch from one frame of
reference to another.
No, it means that whatever is called "invariant" is the same for all
frames. In the current case, the number wirtten on the paper is invariant. >>>
Mikko
Here is yet another proof of what I am saying, and of the need to
re-explain things correctly.
Do you mean that the fact that Tomas Heger doesn't know what
"invariant" means, is a proof of the need to re-explain
my statement correctly?
My statement was:
" Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch."
Both "proper time" and "invariant" are explained in the text.
Exactly what do you not understand?
What is needed to be re-explained correctly?
J'ai signalé à Paul où se trouvaient les erreurs, et même si la barrière de la langue peut jouer, il est impossible qu'il ne comprenne pas ce que
je dis s'il faut l'effort de comprendre.
I pointed out to Paul where the errors were,
and although the language barrier may play a role,
it is impossible for him not to understand what I am
saying if it takes the effort to understand.
That is the same thing. Two clocks are sychronous if and only if the
show the same at the same time.
Python wrote:
Le 22/08/2024 à 13:24, Mikko a écrit :
Thank you. I would like to know whem (if ever) it is time to remove
paganini from my kill file.
He has been kicked of from paganini, he's now using Eternal September,
from where he will be banned too.
... They blocked the posts from everybody, you included.
Nope ;-)
Here is yet another proof of what I am saying, and of the need to
re-explain things correctly.
Do you mean that the fact that Tomas Heger doesn't know what
"invariant" means, is a proof of the need to re-explain
my statement correctly?
My statement was:
" Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch."
Both "proper time" and "invariant" are explained in the text.
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected >>>>> light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according
to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
Einstein used tB - tA and similar expressions. Nothing else needs be
said about delays. The equation tB − tA = t'A − tB and the text that >>> describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.
But this has nothing to do with synchronicity, but with a process to
turn remote clocks to the same time value.
That is the same thing. Two clocks are sychronous if and only if the
show the same at the same time.
Time is not equal to what clocks say, because clocks are measuring
devices, which measure time, but do not determine, what the measured
quantity is.
The intent is to adjust the clocks so that the value shown is the time,
or at least to find out the difference so that the time can be inferred
from the shown value.
The process to synchronize clocks require technical means, which are
here light signals:
The clock at some point A emmits a timing signal, which a remote clock
receives a little time later, because such signals have finite speed.
And also a light signal to the opposite direction.
Now it should be obvious, that the remote clock had to compensate this
delay, because otherwise it would not show the time of the master
clock, but an asynchronous value.
And in order to do that, the delay needs be known. Therefore the
requirement
to adjust so that the apparent delay is the same in both directions.
For uncertain reasons Einstein had not mentioned this requirement at
all, even if transit time per se was actually mentionend.
For obvious reason, Einstein required what he required.
But the necessary step was missing, that the remote station had to add
the transit time to the received timing value.
No such such statement can be found in Einstein's paper, hence we are
forced to beleive, that he didn't wanted to compensate the delay.
Le 24/08/2024 à 12:08, "Paul.B.Andersen" a écrit :
Den 23.08.2024 13:30, skrev Richard Hachel:
Le 23/08/2024 à 10:55, Mikko a écrit :
On 2024-08-23 05:41:50 +0000, Thomas Heger said:
Am Mittwoch000021, 21.08.2024 um 20:42 schrieb Paul.B.Andersen:
This is not, what 'invariant' means in the context of relativity.
Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch. >>>>>
Yes, it is.
Meant is, that time would not change, if you switch from one frame
of reference to another.
No, it means that whatever is called "invariant" is the same for all
frames. In the current case, the number wirtten on the paper is
invariant.
Mikko
Here is yet another proof of what I am saying, and of the need to
re-explain things correctly.
Do you mean that the fact that Tomas Heger doesn't know what
"invariant" means, is a proof of the need to re-explain
my statement correctly?
My statement was:
" Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch." >>
Both "proper time" and "invariant" are explained in the text.
Exactly what do you not understand?
What is needed to be re-explained correctly?
What you say is quite obvious, and that is not the problem.
We all say, even the buffoon Python, that when the event e1 occurs (A
beeps), A starts his watch.
At A, we note tA(e1)=0
e2 is the capture of the beep by B...
e3 is the event that characterizes the return of the signal to A.
We note tA(e3)=2
We know that AB=3.10^8m/s
This leads to tA(e3)-tA(e1)=2AB/c
That's what I say, and I see with sadness (don't laugh friends), that my intelligence seems to surpass the entire scientific community, and that
for having taken, what I say is distorted.
Am Samstag000024, 24.08.2024 um 10:02 schrieb Mikko:
...
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2?td when it is hit by the reflected
light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB ? tA = t'A ? tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according >>>>> to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
Einstein used tB - tA and similar expressions. Nothing else needs be
said about delays. The equation tB ? tA = t'A ? tB and the text that
describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.
But this has nothing to do with synchronicity, but with a process to
turn remote clocks to the same time value.
That is the same thing. Two clocks are sychronous if and only if the
show the same at the same time.
Well, I would agree on that.
But what do you mean with 'at the same time'?
As I see it, we need to adress the so called 'hyperplane of the present'
with 'at the same time'.
This is the set of events, which would require an infinetely fast
signal, to recognize them at the same time.
Since no such signal exists, the hyperplane of the present is mainly invisible.
What we actually see, like in the nicht sky and call 'universe', is
visible, hence does not belong to the hyperplane of the present.
Therefore, synchronization with light signals isn't a very good idea,
because it is light what we see and light would not allow infinite fast communication.
That's why we need some means, to compensate the delay, caused by the
finite speed of light.
This compensation isn't that difficult, because we could easily measure
the delay and use this value for compesation.
BUT: Einstein didn't do this nor even mentioned this requirement.
Time is not equal to what clocks say, because clocks are measuring
devices, which measure time, but do not determine, what the measured
quantity is.
The intent is to adjust the clocks so that the value shown is the time,
or at least to find out the difference so that the time can be inferred from the shown value.
The 'value shown' isn't time, neither.
In physics I would distinguish between a quantity we like to measure,
the measuring device and the value shown by that device.
In case of time we have a clock as device to measure that quantity and
the positions of its hand as shown value.
But the positions of the hands are not time, but the outcome of a
certain measurement (contrary to what Einstein had written).
The process to synchronize clocks require technical means, which are
here light signals:
The clock at some point A emmits a timing signal, which a remote clock
receives a little time later, because such signals have finite speed.
And also a light signal to the opposite direction.
Sure, the process should produce the same setting, if it is initiated
from the other side.
But Einstein's process would not fullfil this requirement, because it
was based on one frame of reference and the observations from there.
The obvious step would be, that the delay was measured and the measured
value used to compensate this delay.
Since Einstein didn't mention anything like this, his method would not produce symmetric snychronization.
Now it should be obvious, that the remote clock had to compensate this
delay, because otherwise it would not show the time of the master
clock, but an asynchronous value.
And in order to do that, the delay needs be known. Therefore the requirement
to adjust so that the apparent delay is the same in both directions.
For uncertain reasons Einstein had not mentioned this requirement at
all, even if transit time per se was actually mentionend.
For obvious reason, Einstein required what he required.
Well, yes, but we are not discussing what Einstein wanted, but what he
wrote in this particular article.
If he forgot to mention his requirements, then they are not there.
And what is not there where it should be, that does not exist in the
context of this paper.
Am Samstag000024, 24.08.2024 um 10:02 schrieb Mikko:
...
How is it possible to fail to understand this?
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected light,
then tA, tB, tA', t1 and td are all proper times which are frame
independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according >>>>>> to Einstein's _definition_ are synchronous in the inertial frame.
You introduced t_d or 'transit time' (aka 'delay'), while Einstein
didn't use any of these terms.
Einstein used tB - tA and similar expressions. Nothing else needs be
said about delays. The equation tB − tA = t'A − tB and the text that >>>> describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.
But this has nothing to do with synchronicity, but with a process to
turn remote clocks to the same time value.
That is the same thing. Two clocks are sychronous if and only if the
show the same at the same time.
Well, I would agree on that.
But what do you mean with 'at the same time'?
As I see it, we need to adress the so called 'hyperplane of the
present' with 'at the same time'.
This is the set of events, which would require an infinetely fast
signal, to recognize them at the same time.
Since no such signal exists, the hyperplane of the present is mainly invisible.
What we actually see, like in the nicht sky and call 'universe', is
visible, hence does not belong to the hyperplane of the present.
Therefore, synchronization with light signals isn't a very good idea,
because it is light what we see and light would not allow infinite fast communication.
On 2024-08-25 06:55:16 +0000, Thomas Heger said:
Am Samstag000024, 24.08.2024 um 10:02 schrieb Mikko:
...
How is it possible to fail to understand this?You introduced t_d or 'transit time' (aka 'delay'), while Einstein >>>>>> didn't use any of these terms.
If we have two stationary clocks in an inertial frame,
and clock A shows tA = t1 when it emits light,
and clock B shows tB = t1 + td when the light hits it,
and clock A shows tA'= t1 + 2⋅td when it is hit by the reflected >>>>>>> light,
then tA, tB, tA', t1 and td are all proper times which are frame >>>>>>> independent (invariants) and "the same for all".
tB − tA = t'A − tB = td
The transit time td is a frame independent invariant and
the same in both directions, which means that the clocks according >>>>>>> to Einstein's _definition_ are synchronous in the inertial frame. >>>>>>
Einstein used tB - tA and similar expressions. Nothing else needs be >>>>> said about delays. The equation tB − tA = t'A − tB and the text that >>>>> describes the situation and defines what tA, t'A and tB mean define
clearly and unambiguously what simultaneity and synchronity mean.
But this has nothing to do with synchronicity, but with a process to
turn remote clocks to the same time value.
That is the same thing. Two clocks are sychronous if and only if the
show the same at the same time.
Well, I would agree on that.
But what do you mean with 'at the same time'?
Basically it means that the time coordinates of the events have the
same value.
As Einstein noted, these words don't mean anything until a defintion
is given. Then he gave his definition. That is a reasonable definition
Am Samstag000024, 24.08.2024 um 12:09 schrieb Paul.B.Andersen:
A piece of paper containing some time value has nothing to do with
Here is yet another proof of what I am saying, and of the need to
re-explain things correctly.
Do you mean that the fact that Tomas Heger doesn't know what
"invariant" means, is a proof of the need to re-explain
my statement correctly?
My statement was:
" Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch."
Both "proper time" and "invariant" are explained in the text.
time, let alone 'proper time', because the value written gets
immediatly out of synch with the clock, from which that value was
copied.
On 2024-08-25 07:07:58 +0000, Thomas Heger said:
Am Samstag000024, 24.08.2024 um 12:09 schrieb Paul.B.Andersen:
A piece of paper containing some time value has nothing to do with
Here is yet another proof of what I am saying, and of the need to
re-explain things correctly.
Do you mean that the fact that Tomas Heger doesn't know what
"invariant" means, is a proof of the need to re-explain
my statement correctly?
My statement was:
" Richard, read your watch NOW. Write down the time nn:nn:nn.
The time nn:nn:nn is a proper time (read off a clock), it is
invariant, not depending on frame of reference.
Nobody can have another opinion of what time YOU read of YOUR watch." >>>
Both "proper time" and "invariant" are explained in the text.
time, let alone 'proper time', because the value written gets
immediatly out of synch with the clock, from which that value was copied.
It is the proper time of the watch at the event of reading it. Nobody
claimed it be useful, just that it is invariont.
Le 24/08/2024 à 08:50, Thomas Heger a écrit :
...
But the necessary step was missing, that the remote station had to add
the transit time to the received timing value.
No such such statement can be found in Einstein's paper, hence we are
forced to beleive, that he didn't wanted to compensate the delay.
THis is silly. You are not "forced" to believe such a idiotic thing.
Basic algebra shows that what you call "delay" is embedded in the
very definition of synchronization. This is something that you
are not "forced" to "believe", but forced to acknowledge.
Am Sonntag000025, 25.08.2024 um 09:24 schrieb Python:
Le 24/08/2024 à 08:50, Thomas Heger a écrit :
...
But the necessary step was missing, that the remote station had to
add the transit time to the received timing value.
No such such statement can be found in Einstein's paper, hence we are
forced to beleive, that he didn't wanted to compensate the delay.
THis is silly. You are not "forced" to believe such a idiotic thing.
Basic algebra shows that what you call "delay" is embedded in the
very definition of synchronization. This is something that you
are not "forced" to "believe", but forced to acknowledge.
Where have you found the delay-corrections 'embedded' in Einstein's
paper???
Einstein's method is actually based on the idea to take the perceived
time from a hypothical clock at a point in the middle between A and B.
This is why Einstein had no time-measure, which is valid throughout an
entire coordinate system, but regarded time as dependent on the position
of the clocks.
But 'position' means 'point within that particular coordinate system'.
So Einstein excluded uniform time within a certain coordinate system.
This is a strong indication, that he didn't plan to compensate delay.
If you think otherwise, than you should quote the statement, which I had apparently overlooked.
Bt_A
Le 26/08/2024 à 08:13, Thomas Heger a écrit :
Am Sonntag000025, 25.08.2024 um 09:24 schrieb Python:
Le 24/08/2024 à 08:50, Thomas Heger a écrit :
From (*) you can get : t'_A - t_A = 2AB/c so another way to
describe the same delay : twice the distance AB divided by c.
So there such a delay is present in paragraph I.1. THREE times
as a term in an equation and ONCE as a term you can obtain in
ONE step of basic algebra.
It is difficult to believe you've "overlooked" this and continue
to do so for YEARS.
At first it could have been a symptom of your complete inability
to understand a single sentence of the article (i.e. sheer stupidity),
since you've published your idiotic comments and got some clues from
numerous people here it is definitely a symptom of your dishonesty
Thomas.
Le 26/08/2024 à 12:46, Python a écrit :
Le 26/08/2024 à 08:13, Thomas Heger a écrit :
Am Sonntag000025, 25.08.2024 um 09:24 schrieb Python:
Le 24/08/2024 à 08:50, Thomas Heger a écrit :
From (*) you can get : t'_A - t_A = 2AB/c so another way to
describe the same delay : twice the distance AB divided by c.
So there such a delay is present in paragraph I.1. THREE times
as a term in an equation and ONCE as a term you can obtain in
ONE step of basic algebra.
It is difficult to believe you've "overlooked" this and continue
to do so for YEARS.
At first it could have been a symptom of your complete inability
to understand a single sentence of the article (i.e. sheer stupidity),
since you've published your idiotic comments and got some clues from
numerous people here it is definitely a symptom of your dishonesty
Thomas.
Dishonesty Thomas?
You're talking nonsense, nonsense...
Plus you insult the posters who answer you.
You're crazy.
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