When I read the contributors to the French and Anglo-Saxon forums, when
I read Einstein (three lines of explanation) or Poincaré (one line of explanation), I realize that it is very insufficient.
On 2024-08-23 11:15:38 +0000, Richard Hachel said:
When I read the contributors to the French and Anglo-Saxon forums, when
I read Einstein (three lines of explanation) or Poincaré (one line of
explanation), I realize that it is very insufficient.
You are free to read other authors if you think those two are insufficient. After all, they have had more time to think how to explain, probably also more experience about explaining.
Le 23/08/2024 à 14:03, Mikko a écrit :
On 2024-08-23 11:15:38 +0000, Richard Hachel said:
When I read the contributors to the French and Anglo-Saxon forums, when
I read Einstein (three lines of explanation) or Poincaré (one line of
explanation), I realize that it is very insufficient.
You are free to read other authors if you think those two are
insufficient.
After all, they have had more time to think how to explain, probably also
more experience about explaining.
Lire ne suffit pas.
Chacun lit ce que d'autres ont lu, et on tourne en rond.
Il faut ré-écrire les choses, et si possible en langage universellement compréhensible, afin que cela soit clair pour tous, et qu'on arrête un
peu ce grand bluff minkowskien qui pourrit l'histoire de l'humanité.
When I read the contributors to the French and Anglo-Saxon forums, when I read Einstein (three lines of explanation) or Poincar (one line of explanation), I realize that it is very insufficient.
Insufficient -for you-.
For Einstein and Poincare it sufficed,
because it is so blindingly obvious.
No competent reader has had any problem with it, ever since,
Jan
Le 23/08/2024 à 21:51, nospam@de-ster.demon.nl (J. J. Lodder) a écrit :
Insufficient -for you-.
Yes
For Einstein and Poincare it sufficed,
because it is so blindingly obvious.
No competent reader has had any problem with it, ever since,
Which is still a big problem in the history of humanity, but it
seems obvious to me that it will necessarily be corrected.
We will not be able to always remain in Minkowskian belief, the
absurdity will fall.
R.H.
Le 23/08/2024 à 14:03, Mikko a écrit :
On 2024-08-23 11:15:38 +0000, Richard Hachel said:
When I read the contributors to the French and Anglo-Saxon forums, when
I read Einstein (three lines of explanation) or Poincaré (one line of
explanation), I realize that it is very insufficient.
You are free to read other authors if you think those two are insufficient. >> After all, they have had more time to think how to explain, probably also
more experience about explaining.
Lire ne suffit pas.
Chacun lit ce que d'autres ont lu, et on tourne en rond.
Il faut ré-écrire les choses, et si possible en langage universellement compréhensible, afin que cela soit clair pour tous, et qu'on arrête un
peu ce grand bluff minkowskien qui pourrit l'histoire de l'humanité.
We will not be able to always remain in Minkowskian belief, the
absurdity will fall.
R.H.
Perhaps, but Minkowski spacetime will remain an approximation to
reality.
Le 24/08/2024 à 04:26, hitlong@yahoo.com (gharnagel) a écrit :
Richard Hachel wrote:
We will not be able to always remain in Minkowskian belief, the
absurdity will fall.
R.H.
Perhaps, but Minkowski spacetime will remain an approximation to
reality.
Absolutely not.
When we see the stupidity and deformity of this interpretation of
Poincaré's equations, there will not be enough left to fill the shard of
a bottle.
I recall the terrible deformities of Minkowskian relativity in the face
of what Dr. Hachel says, if we take for example the example of Hachel's
Tau Ceti traveler (we send a rocket with an acceleration of 10m/s² to
Tau Ceti to travel the 12 light years that separate us from the star),
we have in Minkowski mode a proper time of about 3.5 years, against
4,776 years in Hachel.
I recall the terrible deformities of Minkowskian relativity
in the face of what Dr. Hachel says, if we take for example
the example of Hachel's Tau Ceti traveler (we send a rocket
with an acceleration of 10m/s² to Tau Ceti to travel
the 12 light years that separate us from the star),
we have in Minkowski mode a proper time of about 3.5 years,
against 4,776 years in Hachel.
Den 24.08.2024 13:28, skrev Richard Hachel:
I recall the terrible deformities of Minkowskian relativity
in the face of what Dr. Hachel says, if we take for example
the example of Hachel's Tau Ceti traveler (we send a rocket
with an acceleration of 10m/s² to Tau Ceti to travel
the 12 light years that separate us from the star),
we have in Minkowski mode a proper time of about 3.5 years,
against 4,776 years in Hachel.
SR: τ = 3.304 years
Newton: τ = 11.400 years
Hachel: τ = 4,776 years? Did you mean 4.776 years?
Le 24/08/2024 à 04:26, hitlong@yahoo.com (gharnagel) a écrit :
We will not be able to always remain in Minkowskian belief, the
absurdity will fall.
R.H.
Perhaps, but Minkowski spacetime will remain an approximation to
reality.
Absolutely not.
When we see the stupidity and deformity of this interpretation of
Poincaré's equations, there will not be enough left to fill the shard of
a bottle.
I recall the terrible deformities of Minkowskian relativity in the face
of what Dr. Hachel says, if we take for example the example of Hachel's
Tau Ceti traveler (we send a rocket with an acceleration of 10m/s² to
Tau Ceti to travel the 12 light years that separate us from the star),
we have in Minkowski mode a proper time of about 3.5 years, against
4,776 years in Hachel.
There can be no photography.
Le 23/08/2024 à 21:51, nospam@de-ster.demon.nl (J. J. Lodder) a écrit :
Insufficient -for you-.
Yes
For Einstein and Poincare it sufficed,
because it is so blindingly obvious.
No competent reader has had any problem with it, ever since,
Which is still a big problem in the history of humanity, but it seems
obvious to me that it will necessarily be corrected.
We will not be able to always remain in Minkowskian belief, the
absurdity will fall.
Le 24/08/2024 à 22:10, "Paul.B.Andersen" a écrit :
Den 24.08.2024 13:28, skrev Richard Hachel:
I recall the terrible deformities of Minkowskian relativity in the
face of what Dr. Hachel says, if we take for example the example of
Hachel's Tau Ceti traveler (we send a rocket with an acceleration of
10m/s² to Tau Ceti to travel the 12 light years that separate us from
the star), we have in Minkowski mode a proper time of about 3.5
years, against 4,776 years in Hachel.
SR: τ = 3.304 years
Newton: τ = 11.400 years
Hachel: τ = 4,776 years? Did you mean 4.776 years?
Yes, in french numerical notation we don't use "." but ","
In american notation:
SR: τ = 3.304 years
Newton: τ = 11.400 years
Hachel: τ = 4.776 years
Yes, that's exactly it, and it proves that you have much greater
relativistic knowledge than most French or foreign speakers.
So we have three opposing positions, the Newtonians, the Einsteinians,
the Hachettes.
We must call things by their name, and it is one of the greatness of
science to be able to do it with freedom and knowledge like you.
That's exactly it.
I am happy to be able to read and discuss with a correspondent like you
who thinks and who says things clearly.
R.H.
<Richard snipped the challenge>
I bet that Doctor Richard Hachel is unable to find τ_E and τ_W.
He can't calculate anything in the real world.
So he will flee the challenge as he always does.
Den 24.08.2024 22:26, skrev Richard Hachel:
Le 24/08/2024 à 22:10, "Paul.B.Andersen" a écrit :
Den 24.08.2024 13:28, skrev Richard Hachel:
I recall the terrible deformities of Minkowskian relativity in the
face of what Dr. Hachel says, if we take for example the example of
Hachel's Tau Ceti traveler (we send a rocket with an acceleration of
10m/s² to Tau Ceti to travel the 12 light years that separate us
from the star), we have in Minkowski mode a proper time of about 3.5
years, against 4,776 years in Hachel.
SR: τ = 3.304 years
Newton: τ = 11.400 years
Hachel: τ = 4,776 years? Did you mean 4.776 years?
Yes, in french numerical notation we don't use "." but ","
In american notation:
SR: τ = 3.304 years
Newton: τ = 11.400 years
Hachel: τ = 4.776 years
Yes, that's exactly it, and it proves that you have much greater
relativistic knowledge than most French or foreign speakers.
So we have three opposing positions, the Newtonians, the Einsteinians,
the Hachettes.
We must call things by their name, and it is one of the greatness of
science to be able to do it with freedom and knowledge like you.
That's exactly it.
I am happy to be able to read and discuss with a correspondent like
you who thinks and who says things clearly.
R.H.
<Richard snipped the challenge>
I bet that Doctor Richard Hachel is unable to find τ_E and τ_W.
He can't calculate anything in the real world.
So he will flee the challenge as he always does.
Wasn't I right? Or was I right? :-D
But I can always repeat the challenge, so you yet again
can demonstrate your incompetence by fleeing.
Chicken!
Consider this variant of the Langewin's paradox: ------------------------------------------------
The triplets Ginette, Elise and Wanda are co-located on
the equator. They all have an atomic clock.
Le 23/08/2024 à 14:20, M.D. Richard "Hachel" Lengrand a écrit :
Le 23/08/2024 à 14:03, Mikko a écrit :
On 2024-08-23 11:15:38 +0000, Richard Hachel said:
When I read the contributors to the French and Anglo-Saxon forums, when >>>> I read Einstein (three lines of explanation) or Poincaré (one line of >>>> explanation), I realize that it is very insufficient.
You are free to read other authors if you think those two are
insufficient.
After all, they have had more time to think how to explain, probably also >>> more experience about explaining.
Lire ne suffit pas.
Chacun lit ce que d'autres ont lu, et on tourne en rond.
Il faut ré-écrire les choses, et si possible en langage universellement
compréhensible, afin que cela soit clair pour tous, et qu'on arrête un
peu ce grand bluff minkowskien qui pourrit l'histoire de l'humanité.
It is quite contradictory to praise for the use of a comprehensible
language in *French* into a English-speaking group.
Speaking of that, personally, I give up a little, even if I am convinced of the usefulness of a short article of a few lines on the notion of simultaneity
and synchronization (the basis of RR).
I think that on average (too bad if it hurts them) the regulars are too stupid,
I especially mean too stupid aside from the dick, the dick, always the dick. It's very unfortunate, but we don't come out any more here on the Anglo-Saxon forums
than on the French-speaking forums. It will be about who is the stupidest with
the biggest dick.
It's a shame, there is nevertheless food for thought,
and certain reflections are sometimes interesting,
like the posts on relativistic synchronization
between two points A and B.
It sometimes goes well (like your explanations of events e1, e2, e3)
and the fact that we can already
offer CAREFUL evidence before going any further.
We can then pose without fear: tA(e3)-tA(e1)=2AB/c
Then, admitting that A warns of e1 and e3, either with photons or
with slugs of the same speed, any point M of the stationary frame
of reference, we have yet another tautology:
tM(e3)-tM(e1)= tA(e3)-tA(e1) = 2AB/c
For the moment we cannot say more about the speed of light
between A and B in the direction AB,
nor in the BA sense.
On this, we breathe we breathe, Einstein does not seem to agree with Hachel. For Einstein, the question does not arise, and it seems certain that t(AB)=t(BA).
Except that this is no longer true in an anisochronous environment,
and that our universe is not "a 4D hyperplane of absolute simultaneity,
even for a simple inertial frame of reference".
We can then propose A synchronization based on A hyperplane of simultaneity, but we must propose THE appropriate candidate, and it can obviously be neither A nor B.
So we continue from there.
We can then propose a synchronization of A and B by M (and we will have a synchronization of type M).
Can I do it without laughing, and how?
Note that if M is purely in a perpendicular position on the secant of the middle AB,
then whatever the speed of the information
(c in both directions for Einstein, both c/2 or ∞ depending on the meaning for Hachel),
reception of the sync signal transmitted by M will be simultaneous in reception A and B for M,
and also simultaneous in return for M.
We can therefore at risk, pose that, for M after synchronization A and B on its part (but ONLY for M):
tM(e1)=0 tM(e2)=1 and tM(e3)=2
The timing is perfect for Mr.
We can, by imagining an imaginary point M, placed very far away and perpendicular to all the points of the stationary universe studied, with perfect synchronization of type To, and perfect for all
the repository.
By change of inertial reference frame, M becomes M' and To becomes To', because the chronotropy becomes reciprocally relative.
Everything is said. The basics are given. RR then becomes very simple and the equations that go with it obvious.
It remains to be made understood, and it is not by acting like a monkey that anyone will understand me, or by giving me stupid answers like: "The speed of light takes one second to come from the moon to 'here".
There is a one-second time difference between 00:00:08 and 00:00:07", between my watch and the watch on the moon.
This is because of the speed of light, which is quite slow,
and takes at least a second to reach me.
This is profoundly stupid for Bedouin relativists, that.
But it seems that people like it, and that we can even decorate it with ridiculous smileys.
R.H.
Den 30.08.2024 00:15, skrev Richard Hachel:
Let's analyse Richard's post.
Speaking of that, personally, I give up a little, even if I am
convinced of the usefulness of a short article of a few lines on the
notion of simultaneity
and synchronization (the basis of RR).
I think that on average (too bad if it hurts them) the regulars are
too stupid, I especially mean too stupid aside from the dick, the
dick, always the dick.
It's very unfortunate, but we don't come out any more here on the
Anglo-Saxon forums than on the French-speaking forums. It will be
about who is the stupidest with the biggest dick.
I have never seen anybody but Richard Hachel boast of his big dick,
so who are the "regulars" he is accusing of doing so?
It's a shame, there is nevertheless food for thought, and certain
reflections are sometimes interesting,
like the posts on relativistic synchronization between two points A
and B.
So Richard is talking about Einstein's synchronisations method.
It sometimes goes well (like your explanations of events e1, e2, e3)
and the fact that we can already
offer CAREFUL evidence before going any further.
"You" is probably Python. Richard never quote what he is
referring to, and he doesn't define the events e1, e2, e3.
But we know:
e1 is the event that light is emitted from A
e2 is the event that light is reflected from B
e3 is the event that the reflected light hits A
Note e1 and e3 are happening at A, e2 is happening at B.
We can then pose without fear: tA(e3)-tA(e1)=2AB/c
tA(e3) is the reading of a clock at A at the event e3
tA(e1) is the reading of a clock at A at the event e1
Einstein says:
"We assume the quantity 2AB/(tA(e3)-tA(e1)) = c
to be a universal constant—the velocity of light in empty space."
It is a postulate in SR that say the speed of light
in vacuum is constant and invariant, and his paper is about
the consequence of the postulates, so of course he assumes that.
It is thoroughly experimentally verified that the speed
of light indeed is constant and invariant.
So we _know_ that tA(e3)-tA(e1)= 2AB/c
Then, admitting that A warns of e1 and e3, either with photons or with
slugs of the same speed, any point M of the stationary frame of
reference, we have yet another tautology:
This statement doesn't parse.
tM(e3)-tM(e1)= tA(e3)-tA(e1) = 2AB/c
tM can only be the reading of a clock at the point M,
which is an arbitrary stationary point in the frame where
A and B are stationary.
But the events e1 and e3 happen at A, and not on M,
so tM(e3)-tM(e1) is meaningless.
Richard doesn't seem to know what an event is.
For the moment we cannot say more about the speed of light between A
and B in the direction AB,
nor in the BA sense.
Above Richard say the posts on relativistic synchronization
between two points A and B are interesting, but so far he
has said nothing about synchronization.
The equation:
tB(e2) - tA(e1) = tA(e3) - tB(e2)
is Einstein's _definition_ of synchronism and simultaneity.
If this equation is true, then the clocks are synchronous
in the frame where A and B are stationary.
On this, we breathe we breathe, Einstein does not seem to agree with
Hachel. For Einstein, the question does not arise, and it seems
certain that t(AB)=t(BA).
What are you saying? :-D
Einstein says:
"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."
And:
"In accordance with definition the two clocks synchronize if
tB(e2) - tA(e1) = tA(e3) - tB(e2)"
tB(e2) - tA(e1) is the time the light uses to go from A to B
tA(e3) - tB(e2) is the time the light uses to go from B to A
Einstein _defines_ that the clocks simultaneously show
the same (are synchronous in the stationary system)
if the time the light uses to go from A to B equals
the time the light uses to go from B to A.
Except that this is no longer true in an anisochronous environment,
and that our universe is not "a 4D hyperplane of absolute
simultaneity, even for a simple inertial frame of reference".
We can then propose A synchronization based on A hyperplane of
simultaneity, but we must propose THE appropriate candidate, and it
can obviously be neither A nor B.
So we continue from there.
We can then propose a synchronization of A and B by M (and we will
have a synchronization of type M).
SR doesn't depend on the definition of simultaneity
(but everything would be very awkward without it)
so another definition is possible.
But in the real world no other than Einstein's definition
would work, so no one would use your alternative definition.
Den 30.08.2024 00:15, skrev Richard Hachel:
Without a proper definition of the terms, this is nonsense.
What are e1, e2 and e3?
What are tM(e1), tM(e2), and tM(e3)?
Den 30.08.2024 00:15, skrev Richard Hachel:
"You" is probably Python. Richard never quote what he is
referring to, and he doesn't define the events e1, e2, e3.
But we know:
e1 is the event that light is emitted from A
e2 is the event that light is reflected from B
e3 is the event that the reflected light hits A
Note e1 and e3 are happening at A, e2 is happening at B.
We can then pose without fear: tA(e3)-tA(e1)=2AB/c
tA(e3) is the reading of a clock at A at the event e3
tA(e1) is the reading of a clock at A at the event e1
Einstein says:
"We assume the quantity 2AB/(tA(e3)-tA(e1)) = c
to be a universal constant—the velocity of light in empty space."
It is a postulate in SR that say the speed of light
in vacuum is constant and invariant, and his paper is about
the consequence of the postulates, so of course he assumes that.
It is thoroughly experimentally verified that the speed
of light indeed is constant and invariant.
So we _know_ that tA(e3)-tA(e1)= 2AB/c
Then, admitting that A warns of e1 and e3, either with photons or
with slugs of the same speed, any point M of the stationary frame
of reference, we have yet another tautology:
This statement doesn't parse.
tM(e3)-tM(e1)= tA(e3)-tA(e1) = 2AB/c
tM can only be the reading of a clock at the point M,
which is an arbitrary stationary point in the frame where
A and B are stationary.
But the events e1 and e3 happen at A, and not on M,
so tM(e3)-tM(e1) is meaningless.
Richard doesn't seem to know what an event is.
For the moment we cannot say more about the speed of light
between A and B in the direction AB,
nor in the BA sense.
Above Richard say the posts on relativistic synchronization
between two points A and B are interesting, but so far he
has said nothing about synchronization.
The equation:
tB(e2) - tA(e1) = tA(e3) - tB(e2)
is Einstein's _definition_ of synchronism and simultaneity.
If this equation is true, then the clocks are synchronous
in the frame where A and B are stationary.
On this, we breathe we breathe, Einstein does not seem to agree with Hachel. >> For Einstein, the question does not arise, and it seems certain that
t(AB)=t(BA).
What are you saying? :-D
Einstein says:
"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."
And:
"In accordance with definition the two clocks synchronize if
tB(e2) - tA(e1) = tA(e3) - tB(e2)"
tB(e2) - tA(e1) is the time the light uses to go from A to B
tA(e3) - tB(e2) is the time the light uses to go from B to A
Einstein _defines_ that the clocks simultaneously show
the same (are synchronous in the stationary system)
if the time the light uses to go from A to B equals
the time the light uses to go from B to A.
Except that this is no longer true in an anisochronous environment,
and that our universe is not "a 4D hyperplane of absolute simultaneity,
even for a simple inertial frame of reference".
We can then propose A synchronization based on A hyperplane of simultaneity, >> but we must propose THE appropriate candidate, and it can obviously be neither A
nor B.
So we continue from there.
We can then propose a synchronization of A and B by M (and we will have a
synchronization of type M).
SR doesn't depend on the definition of simultaneity
(but everything would be very awkward without it)
so another definition is possible.
But in the real world no other than Einstein's definition
would work, so no one would use your alternative definition.
Can I do it without laughing, and how?
Note that if M is purely in a perpendicular position on the secant of the middle
AB,
then whatever the speed of the information
(c in both directions for Einstein, both c/2 or ∞ depending on the meaning for
Hachel),
reception of the sync signal transmitted by M will be simultaneous in reception
A and B for M,
and also simultaneous in return for M.
We can therefore at risk, pose that, for M after synchronization A and B on its
part (but ONLY for M):
tM(e1)=0 tM(e2)=1 and tM(e3)=2
Without a proper definition of the terms, this is nonsense.
What are e1, e2 and e3?
What are tM(e1), tM(e2), and tM(e3)?
The timing is perfect for Mr.
We can, by imagining an imaginary point M, placed very far away and
perpendicular to all the points of the stationary universe studied, with perfect
synchronization of type To, and perfect for all
the repository.
Richard, this is mindless babble.
_In the real world_ we have two equal clocks at two points A and B.
Please explain _exacly_ how you will go about to synchronise them
according to _your_ definition.
Imaginary points won't do. Everything must be _real_.
Einstein could do it. Can you?
By change of inertial reference frame, M becomes M' and To becomes To',
because the chronotropy becomes reciprocally relative.
Everything is said. The basics are given. RR then becomes very simple and the
equations that go with it obvious.
It remains to be made understood, and it is not by acting like a monkey that >> anyone will understand me, or by giving me stupid answers like: "The speed of light
takes one second to come from the moon to 'here".
You mean like the stupid answer you gave me?
|Den 26.08.2024 13:02, skrev Richard Hachel:>
There is a one-second time difference between 00:00:08 and 00:00:07", between my watch and the watch on the moon.
This is because of the speed of light, which is quite slow,
and takes at least a second to reach me.
Shot yourself in the foot, Richard?
This is profoundly stupid for Bedouin relativists, that.
But it seems that people like it, and that we can even decorate it with
ridiculous smileys.
R.H.
Le 30/08/2024 à 15:19, "Paul.B.Andersen" a écrit :
Den 30.08.2024 00:15, skrev Richard Hachel:
Without a proper definition of the terms, this is nonsense.
What are e1, e2 and e3?
What are tM(e1), tM(e2), and tM(e3)?
If one day I say:
"One of Father François' sheep has probably got lost,
it is in the grove in the place called "les Primevères".
And then, some idiot comes and says:
“What is a sheep, what is a grove, what is a primrose?”
Frankly, even with a small dick, you don't want to answer anymore.
I leave that to YBM, he does that very well.
R.H.
Den 30.08.2024 00:15, skrev Richard Hachel:transmitted by M will be simultaneous in reception A and B for M, and also simultaneous in return for M.
Except that this is no longer true in an anisochronous environment, and that our universe is not "a 4D hyperplane of absolute simultaneity, even for a simple inertial frame of reference".
We can then propose A synchronization based on A hyperplane of simultaneity, but we must propose THE appropriate candidate, and it can obviously be neither A nor B.
So we continue from there.
We can then propose a synchronization of A and B by M (and we will have a synchronization of type M).
Can I do it without laughing, and how?
Note that if M is purely in a perpendicular position on the secant of the middle AB, then whatever the speed of the information (c in both directions for Einstein, both c/2 or ∞ depending on the meaning for Hachel), reception of the sync signal
We can therefore at risk, pose that, for M after synchronization A and B on its part (but ONLY for M):
tM(e1)=0 tM(e2)=1 and tM(e3)=2
Einstein says:
"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."
And:
"In accordance with definition the two clocks synchronize if
tB(e2) - tA(e1) = tA(e3) - tB(e2)"
tB(e2) - tA(e1) is the time the light uses to go from A to B
tA(e3) - tB(e2) is the time the light uses to go from B to A
Einstein _defines_ that the clocks simultaneously show
the same (are synchronous in the stationary system)
if the time the light uses to go from A to B equals
the time the light uses to go from B to A.
What had actually the reading of a clock to do with how fast light moves
from A to B?
This would be a non sequitur, if time would not be what clocks say.
In fact light is here used to synchronize distant clocks.
But that does not say, that clocks would use light to measure time.
Usually clocks use other means than light, like pendulums or Quartz
crystals.
To syncronize distant clocks, we would need to adjust them in a way,
that they tick at the same rate and show the same time.
To measure this 'in synch' we would need to measure the delay and add
this to the observed time from the remote system.
But this step cannot be found anywhere in Einstein's paper.
...
TH
Am Freitag000030, 30.08.2024 um 18:31 schrieb Richard Hachel:
...
Einstein says:
"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."
And:
"In accordance with definition the two clocks synchronize if
tB(e2) - tA(e1) = tA(e3) - tB(e2)"
tB(e2) - tA(e1) is the time the light uses to go from A to B
tA(e3) - tB(e2) is the time the light uses to go from B to A
Einstein _defines_ that the clocks simultaneously show
the same (are synchronous in the stationary system)
if the time the light uses to go from A to B equals
the time the light uses to go from B to A.
What had actually the reading of a clock to do with how fast light moves
from A to B?
This would be a non sequitur, if time would not be what clocks say.
In fact light is here used to synchronize distant clocks.
But that does not say, that clocks would use light to measure time.
Usually clocks use other means than light, like pendulums or Quartz
crystals.
To syncronize distant clocks, we would need to adjust them in a way,
that they tick at the same rate and show the same time.
To measure this 'in synch' we would need to measure the delay and add
this to the observed time from the remote system.
But this step cannot be found anywhere in Einstein's paper.
It would not occur to anyone to place two solar clocks in their garden.
If it is well oriented, the time displayed is the right one.
No need for two solar clocks.
On the other hand, if I am no longer in Berlin, but in Nantes, I cannot
be satisfied with taking continual selfies of my Berlin solar clock to
have the solar time at Nantes. Everyone understands the ridiculousness
of the situation (except Python and Paul.B Andersen). If I am in Nantes,
I have to look at the time on a clock in a Nantes garden, and I cannot
say to my wife: "What time is it now? Please send me a selfie of my clock."
We cannot, even by slapping them in the nose, or kicking them in the
balls, synchronize solar clocks with each other. They will NEVER mark
noon at the same time in Berlin, and in Nantes.
Paul B Andersen has just informed us that a watch slowly transferred to
the moon shows the same time as a watch left on my desk
There is a one-second time difference between 00:00:08 and 00:00:07", between my watch and the watch on the moon.
This is because of the speed of light, which is quite slow,
and takes at least a second to reach me.
He does not understand that there is a real procedure of natural desynchronization by spatial position,
just as there is also a real
procedure of reciprocal dyschrnotopia when we make watches evolve
between them at relativistic speeds (they do not even beat at the same
rhythm anymore, and conversely, each beats faster than the other).
What should we do to make him understand?
Den 31.08.2024 09:05, skrev Richard Hachel:
https://paulba.no/pdf/Mutual_time_dilation.pdf
Den 31.08.2024 09:05, skrev Richard Hachel:
It would not occur to anyone to place two solar clocks in their
garden. If it is well oriented, the time displayed is the right one.
No need for two solar clocks.
On the other hand, if I am no longer in Berlin, but in Nantes, I
cannot be satisfied with taking continual selfies of my Berlin solar
clock to have the solar time at Nantes. Everyone understands the
ridiculousness of the situation (except Python and Paul.B Andersen).
If I am in Nantes, I have to look at the time on a clock in a Nantes
garden, and I cannot say to my wife: "What time is it now? Please send
me a selfie of my clock."
If you are in Nantes and want to go to Berlin by train,
and you see in the timetable that your train will leave Nantes
at the time 8:32, which clock would you use to be at the railway station
at the right time, the solar clock in Nantes, or your wristwatch
which is showing UTC+2h?
On Fri, 23 Aug 2024 21:13:10 +0000, Richard Hachel wrote:
Perhaps, but Minkowski spacetime will remain an approximation to
reality.
“spacetime is likely to be an approximate description of something quite different.” – Steven Carlip
Le 31/08/2024 à 22:10, "Paul.B.Andersen" a écrit :
Den 31.08.2024 09:05, skrev Richard Hachel:
https://paulba.no/pdf/Mutual_time_dilation.pdf
? ? ?
<http://nemoweb.net/jntp?pheofpwVPcOT3RCuNcESEqS47x0@jntp/Data.Media:1>
Le 31/08/2024 à 08:30, Thomas Heger a écrit :
What had actually the reading of a clock to do with how fast light
moves from A to B?
This would be a non sequitur, if time would not be what clocks say.
In fact light is here used to synchronize distant clocks.
But that does not say, that clocks would use light to measure time.
Usually clocks use other means than light, like pendulums or Quartz
crystals.
To syncronize distant clocks, we would need to adjust them in a way,
that they tick at the same rate and show the same time.
To measure this 'in synch' we would need to measure the delay and add
this to the observed time from the remote system.
But this step cannot be found anywhere in Einstein's paper.
...
TH
You are getting closer to the truth (a little more than Paul B. Andersen
who is drowning in it.
But it is not quite there yet.
It is not far.
For two watches to be synchronized, on MY DESK, two things are necessary. That they turn at the same speed (equal chronotropy).
That they mark the same time at the same instant (isochrony).
But hey. The beautiful thing that two watches mark the same time on my
table. One is enough for me. The second one is not much use to me.
A bit like two solar clocks.
It would not occur to anyone to place two solar clocks in their garden.
If it is well oriented, the time displayed is the right one.
No need for two solar clocks.
On the other hand, if I am no longer in Berlin, but in Nantes, I cannot
be satisfied with taking continual selfies of my Berlin solar clock to
have the solar time at Nantes. Everyone understands the ridiculousness
of the situation (except Python and Paul.B Andersen). If I am in Nantes,
I have to look at the time on a clock in a Nantes garden, and I cannot
say to my wife: "What time is it now? Please send me a selfie of my clock."
The same goes for the relativistic universe, where the time displayed on watches depends on POSITION, and where the chronotropy of watches (the
speed of the hands) depends on SPEED.
What is absolutely frightening, even highly ridiculous, is to see even
the greatest experts in world physics (Stefen Hawking himself)
completely drown in concepts that, once well explained, are nevertheless
of a college simplicity.
We cannot, even by slapping them in the nose, or kicking them in the
balls, synchronize solar clocks with each other. They will NEVER mark
noon at the same time in Berlin, and in Nantes.
Den 31.08.2024 23:14, skrev Richard Hachel:
Le 31/08/2024 à 22:10, "Paul.B.Andersen" a écrit :
Den 31.08.2024 09:05, skrev Richard Hachel:
https://paulba.no/pdf/Mutual_time_dilation.pdf
? ? ?
Thanks to Python, I see that you fail to understand
the very first equation in the paper.
https://ibb.co/7C0S4nJ
It probably won't help, but I will point out what is clearly
defined in the paper.
https://paulba.no/pdf/Mutual_time_dilation.pdf
Quote:
| Event E₂: clock A and clock B' are adjacent
| In frame K', A will be at the position -d when B' shows t₂'= d/v
I considered this to be obvious for a reasonable knowledgeable reader.
Explanation for the less knowledgeable reader: -----------------------------------------------
In frame K' the clock A has moved from the position X' = 0
to the position x' = -d with the speed v.
Since t' = 0 when A was at x' = 0, t' = d/v when A is at x' = -d.
So the coordinates of event E₂ in K' are x₂'= -d, t₂' = d/v
In frame K the temporal coordinate will be:
t₂ = γ⋅(t₂' + (v/c²)⋅x₂') = equation (1) in the paper
Le 31/08/2024 à 22:10, "Paul.B.Andersen" a écrit :
Den 31.08.2024 09:05, skrev Richard Hachel:
https://paulba.no/pdf/Mutual_time_dilation.pdf
? ? ?
Le 31/08/2024 à 22:10, "Paul.B.Andersen" a écrit :
https://paulba.no/pdf/Mutual_time_dilation.pdf
? ? ?
Am Samstag000031, 31.08.2024 um 09:05 schrieb Richard Hachel:
This is the set of all places, which can use the same kind of clocks.
They do not need to show the same time, however, but need to tick at the
same rate.
Den 31.08.2024 23:14, skrev Richard Hachel:
Le 31/08/2024 à 22:10, "Paul.B.Andersen" a écrit :
https://paulba.no/pdf/Mutual_time_dilation.pdf
? ? ?
You point out that you can't even read and understand
a paper with simple math
Den 31.08.2024 23:14, skrev Richard Hachel:
https://ibb.co/7C0S4nJ
| Event E₂: clock A and clock B' are adjacent
To say that Minkowski spacetime is an approach to reality is a huge bluff. This is like saying that addition is an approach to multiplication. >
A fanatic of the theory of the equality of two operations will tell you
that it is very close, because 1+2 makes three, and 1x2 makes 2; and two
is not far from three.
So multiplying or adding is the same reality.
This is what a fanatic will tell you.
Then, things will get worse, and Python will come to support:
"But yes, it is the same reality, it is the same operation because 0+0
makes zero; and if I do 0x0 I still have zero, the equivalence is perfect."
And then Paul will come and support: "But yes, multiplying and adding
are the same thing, we will show that 2+2=4, and that 2x2=4."
Well you see, you do the same thing by saying: "Minkowski is very close
to reality, because on Galilean frames of reference, it works".
Except that Doctor Hachel, who worked for forty years on the subject,
says that maybe, but that it no longer works for accelerated frames of reference, nor for rotating frames of reference.
It no longer works, but NOT AT ALL!!!
As 4+9 gives 13, and 4x9 gives 36.
13 and 36, it's not the same at all.
Well with Richard Hachel's relativity, we obtain a real relativistic perfection. With that of physicists, based on Minkowski, we only obtain bullshit.
Poincaré and Hachel: two Frenchmen. A formidable advance in the understanding of the universe.
Einstein and Minkowski, two Germans: a formidable derailment of the relativistic theory.
Le 01/09/2024 à 11:57, "Paul.B.Andersen" a écrit :
Den 31.08.2024 23:14, skrev Richard Hachel:
https://ibb.co/7C0S4nJ
| Event E₂: clock A and clock B' are adjacent
Yes.
Goooood!
Event (e1) : clock A and clock A' are adjacent.
Event (e2) : clock A end clock B' are ajacent.
Well.
Nous allons maintenant donner des précisions numériques.
AB (at rest) = 3.10^8m
A'B' (at rest) = 3.10^8m
v=0.8c (240000km/s)
tA(e1)= 0
tA'(e1)= 0
tA(e2)= ?
[nonsense answering nonsense]
see here (this is a link to my 'book') https://docs.google.com/presentation/d/1Ur3_giuk2l439fxUa8QHX4wTDxBEaM6lOlgVUa0cFU4/edit?usp=sharing
But of course, experimental evidence in the real world has nothing
to do with what can and can't happen in the Hachelian world.
Le 01/09/2024 à 08:27, Thomas Heger a écrit :
Am Samstag000031, 31.08.2024 um 09:05 schrieb Richard Hachel:
This is the set of all places, which can use the same kind of clocks.
They do not need to show the same time, however, but need to tick at
the same rate.
This is called a relativistic reference frame.
All clocks there have the same chronotropy: that is to say, they turn
at the same speed.
I do not use the term reference frame like physicists,
Den 01.09.2024 13:25, skrev Richard Hachel:
Le 01/09/2024 à 11:57, "Paul.B.Andersen" a écrit :
Den 31.08.2024 23:14, skrev Richard Hachel:
https://ibb.co/7C0S4nJ
| Event E₂: clock A and clock B' are adjacent
Yes.
Goooood!
Event (e1) : clock A and clock A' are adjacent.
Event (e2) : clock A end clock B' are ajacent.
Well.
Nous allons maintenant donner des précisions numériques.
AB (at rest) = 3.10^8m
A'B' (at rest) = 3.10^8m
v=0.8c (240000km/s)
No. v = 239833966.4 m/s
tA(e1)= 0
tA'(e1)= 0
tA(e2)= ?
https://paulba.no/pdf/Mutual_time_dilation.pdf
Still a problem with equation (1)?
t₂ = 0.75052 second
Den 01.09.2024 16:13, skrev Richard Hachel:
AB (at rest) = 3.10^8m
A'B' (at rest) = 3.10^8m
v=0.8c
https://paulba.no/pdf/Mutual_time_dilation.pdf
Still a problem with equation (1)?
t₂ = 0.75052 second
I do not accept your answer.
Your problem.
c = 299792458 m/s
d = 3e8 m
v = 0.8⋅c = 239833966.4 m/s
t₂ = (d/v)⋅√(1−v²/c²) = 0.75052 second
Den 01.09.2024 16:13, skrev Richard Hachel:
Why didn't you calculate calculate t₂ with the wrong speed of light?
Is the equation t₂ = (d/v)⋅√(1−v²/c²) beyond your mathematical capabilities?
Le 01/09/2024 à 14:27, "Paul.B.Andersen" a écrit :
Den 01.09.2024 13:25, skrev Richard Hachel:
Nous allons maintenant donner des précisions numériques.
AB (at rest) = 3.10^8m
A'B' (at rest) = 3.10^8m
v=0.8c
tA(e2)= ?
https://paulba.no/pdf/Mutual_time_dilation.pdf
Still a problem with equation (1)?
t₂ = 0.75052 second
I do not accept your answer.
Le 01/09/2024 à 14:27, "Paul.B.Andersen" a écrit :
Den 01.09.2024 13:25, skrev Richard Hachel:
Nous allons maintenant donner des précisions numériques.
AB (at rest) = 3.10^8m
A'B' (at rest) = 3.10^8m
v=0.8c
https://paulba.no/pdf/Mutual_time_dilation.pdf
Still a problem with equation (1)?
t₂ = 0.75052 second
I do not accept your answer.
c=10^8m/s, that is to say, in my example, c=3.0000000000.10^8m/s
Please, respect.
t₂ = ?
Le 01/09/2024 à 19:20, "Paul.B.Andersen" a écrit :
c = 299792458 m/s
No.
c = 300000000 m/s
Le 01/09/2024 à 13:45, "Paul.B.Andersen" a écrit :
But of course, experimental evidence in the real world has nothing
to do with what can and can't happen in the Hachelian world.
YOU said it.
Personally, I have always had a religious spirit.
I believe what the good Lord tells me to believe (and not necessarily men).
R.H.
Den 01.09.2024 16:13, skrev Richard Hachel:
AB (at rest) = 3.10^8m
A'B' (at rest) = 3.10^8m
v=0.8c
tA(e2)= ?
tA(e2)=0.75 sec
Chicken!
No.Why, Paul?
Because I don't want to be a chicken anymore. I want to be a brave man.
Le 01/09/2024 à 08:27, Thomas Heger a écrit :
Am Samstag000031, 31.08.2024 um 09:05 schrieb Richard Hachel:
This is the set of all places, which can use the same kind of clocks.
They do not need to show the same time, however, but need to tick at
the same rate.
This is called a relativistic reference frame.
All clocks there have the same chronotropy: that is to say, they turn
at the same speed.
I do not use the term reference frame like physicists,
That's OK but you should not use it any other way, at least not here
or any other physics group.
Mikko
On 09/01/2024 11:22 AM, Ross Finlayson wrote:
On 09/01/2024 10:38 AM, Richard Hachel wrote:
Le 01/09/2024 à 19:20, "Paul.B.Andersen" a écrit :
Den 01.09.2024 16:13, skrev Richard Hachel:
Why didn't you calculate calculate t₂ with the wrong speed of light? >>>> Is the equation t₂ = (d/v)⋅√(1−v²/c²) beyond your mathematical >>>> capabilities?
Vr=Vo/sqrt(1-Vo²/c²)
Vr=4c/3
t₂=x/Vr=0.75sec
R.H.
Yet, if you square that, then take the root,
is it not that triangle inequality replaces 0.25?
The difference?
This is a large part of when things are squared
or stored in roots for no reason then as with
regards to that as an indicator, or "dimension",
where a "dimension" only needs one bit an "indicator",
triangle rule the bit indicator for the quadrant,
that in these things making numerical emergence
for continuity, it's often the most usual rule
any matters of direction, "what 0.75 means".
It's like "power law" or "normal distribution",
"sure, it fits".
"Centralizing tendency"
Le 01/09/2024 à 19:41, "Paul.B.Andersen" a écrit :
Den 01.09.2024 16:13, skrev Richard Hachel:
AB (at rest) = 3.10^8m
A'B' (at rest) = 3.10^8m
v=0.8c tA(e2)= ?
tA(e2)=0.75 sec
Chicken!
No YOU chicken !
tA (e1)= 0
tA'(e1)= 0
tA (e2)= 0.75
On 09/01/2024 04:41 PM, Python wrote:
Le 01/09/2024 à 20:56, Ross Finlayson a écrit :
On 09/01/2024 11:22 AM, Ross Finlayson wrote:
On 09/01/2024 10:38 AM, Richard Hachel wrote:
Le 01/09/2024 à 19:20, "Paul.B.Andersen" a écrit :
Den 01.09.2024 16:13, skrev Richard Hachel:
Why didn't you calculate calculate t₂ with the wrong speed of light? >>>>>> Is the equation t₂ = (d/v)⋅√(1−v²/c²) beyond your mathematical >>>>>> capabilities?
Vr=Vo/sqrt(1-Vo²/c²)
Vr=4c/3
t₂=x/Vr=0.75sec
R.H.
Yet, if you square that, then take the root,
is it not that triangle inequality replaces 0.25?
The difference?
This is a large part of when things are squared
or stored in roots for no reason then as with
regards to that as an indicator, or "dimension",
where a "dimension" only needs one bit an "indicator",
triangle rule the bit indicator for the quadrant,
that in these things making numerical emergence
for continuity, it's often the most usual rule
any matters of direction, "what 0.75 means".
It's like "power law" or "normal distribution",
"sure, it fits".
"Centralizing tendency"
You want to play ?
"Globalizing serendipity"
Your turn.
Well, there's nothing to follow serendipity,
the serendipity is the emergent what's found,
"global" is a bit redundant, and, "globalizing"
doesn't necessarily apply, as not all find it.
As a universal and an ideal and transcendental,
though, serendipity itself shares with other
universal, ideal, transcendental concepts.
The "centralizing tendency" or "polarizing tendencies"
or "uniformizing tendencies" or "agitating tendencies",
then for tendencies and propensities, mostly result
reflecting in tendencies and propensities,
the oscillation and restitution, though
the attenuation and dissipation.
The "serendipity" is a nice place to visit.
I don't know if the book about Serendipity
in the children's literature is still very
widely read, there's a picture book that
introduces the concept through the lens of
a character, Serendip.
The concept then is given as a happy place
and an alignment, or as like that "a serendipity
is like a syzygy, a happenstance emergence in
order as what's yet sublime".
Le 01/09/2024 à 08:27, Thomas Heger a écrit :
Am Samstag000031, 31.08.2024 um 09:05 schrieb Richard Hachel:
This is the set of all places, which can use the same kind of clocks.
They do not need to show the same time, however, but need to tick at
the same rate.
This is called a relativistic reference frame.
All clocks there have the same chronotropy: that is to say, they turn at
the same speed.
I do not use the term reference frame like physicists, because the term reference frame is an abstract invention, a verbal nothing, a pure nothingness due to the fact that each point of the reference frame has
its own clock, certainly isochronotropic with respect to neighboring
clocks, which will never "absolutely" mark the same time as the
neighboring time on which it will always advance by t=x/c (and vice
versa) during a universal synchronization of type M.
Am Sonntag000001, 01.09.2024 um 12:18 schrieb Richard Hachel:
This 'now' is called 'hyperplane of the present' in relativity lingo.
Le 02/09/2024 à 08:25, Thomas Heger a écrit :
I use the observation, that clocks around the Earth surface tick at
the same rate, while they don't tick at the same rate at different
altitudes.
There is already a bias here.
If a watch is placed at altitude, it does not evolve at the same speed
as a fixed watch placed at the level of our local mass reference center
that we could put the sun, or even the galactic center.
Am Sonntag000001, 01.09.2024 um 12:18 schrieb Richard Hachel:
In a typical spacetime diagram it is a horizontal plane, while the axis
of time points up vertically.
OK, Paul.OK. Let's play.
So we're going to continue, because it's very important.
A sees the segment AB coming towards him, and when A' crosses A, which
is event e1, A starts his watch. tA(e1)=0
Then A observes that B' is approaching him at high speed, and stops his
watch when B crosses him, this is event e2, and we note tA(e2)=0.75.
There is an interval of 0.75 seconds between e1 and e2.
For 0.75 seconds, B' rushes toward A.
1. At what apparent (i.e. APPARENT) speed does A apprehend B' rushing
toward him?
2. What is the apparent distance traveled by B' during the interval
noted by the watch?
I use the observation, that clocks around the Earth surface tick at the
same rate, while they don't tick at the same rate at different altitudes.
tA (e1)= 0
tA'(e1)= 0
LOL!!! Whaaat the Hell is that? ? ?
tA (e2)= 0.75
Le 01/09/2024 à 22:29, M.D. Richard "Hachel" Lengrand a écrit :
tA (e1)= 0
If you insist, but this in no way a requisite of Einstein-Poincaré's synchronization procedure. t_A = 451 is another possible value :-)
tA'(e1)= 0
LOL!!! Whaaat the Hell is that???
tA (e2)= 0.75
Irrelevant. The time marked by clock A "when" B received the first
signal is undefined at that time (i.e. "when" is undefined for remote events). It does play any role in the procedure. Fortunately.
Le 02/09/2024 à 01:39, Python a écrit :
tA (e1)= 0
tA'(e1)= 0
LOL!!! Whaaat the Hell is that? ? ?
tA (e2)= 0.75
Bon, t'arrête de faire le crétin, toi?
Il faut te faire suivre à coups de genoux dans les couilles ou quoi?
Bouffon! Crétin! Guignol!
On répète, ici:
tA (e1) par définition : tA(e1)=0 Paul B.Andersen copyrighs.
tA'(e1) = 0 (Paul B.Andersen copyrights).
tA(e2) =0.75 (Paul B. Andersen and Richard Hachel copyrighs).
Cela fait trois données.
Dans un tel problème, il y a huit données.
tA(e1), tA(e2), tA'(e1), tA'(e2), tB(e1), tB'(e1), tB(e2), tB'(e2).
Ici, bien que Paul B. Andersen et moi même figurons parmi les plus
grands théoriciens relativistes de notre époque, nous n'avons voulu révéler au monde que trois données.
Mais les cinq autres, on va les donner, ne t'inquiète pas.
C'est comme les chars Leclerc français et les chars Leopard allemand, t'inquiète pas, qu'ils viennent sur les plaines d'Ukraine, on va les brûler.
On va tous les bruler.
Le 02/09/2024 à 08:25, Thomas Heger a écrit :
I use the observation, that clocks around the Earth surface tick at
the same rate, while they don't tick at the same rate at different
altitudes.
There is already a bias here.
If a watch is placed at altitude, it does not evolve at the same speed
as a fixed watch placed at the level of our local mass reference center
that we could put the sun, or even the galactic center. The effects of
these reference frames are perhaps negligible. I do not know. But at
least, the effects of the revolution of the object around the center of
the earth are not the same as the effects
on an object placed on the surface of the ground. Worse, for the object placed on the surface of the ground, it is the center of the earth that rotates around it; and also for the other. These effects are no longer
really Galilean, but effects of rotating reference frames for which I
have given the equations, and which cause some surprises (it is the
object that goes the fastest that has the time that passes the fastest, contrary to Galilean effects).
Sorry Richard, but the notation tX(eN) is, according to yourself,
supposed to mean "time for clock X of event eN" (which is out of the
scope of a synchronization procedure if the event eN took place
far from clock X anyway).
So when you write tA'(e2) your referencing a third clock named A'
which is definitely not a part of the synchronization procedure
that only involves TWO clocks. Worse now, tB' refer to a fourth
clocks.
What the hell is happening in your mind, Richard?
You are not even trying to make sense anymore. Anyway you've
already failed at that.
Le 03/09/2024 à 01:04, Python a écrit :
Sorry Richard, but the notation tX(eN) is, according to yourself,
supposed to mean "time for clock X of event eN" (which is out of the
scope of a synchronization procedure if the event eN took place
far from clock X anyway).
So when you write tA'(e2) your referencing a third clock named A'
which is definitely not a part of the synchronization procedure
that only involves TWO clocks. Worse now, tB' refer to a fourth
clocks.
What the hell is happening in your mind, Richard?
You are not even trying to make sense anymore. Anyway you've
already failed at that.
But no, I did not fail.
On the contrary, I told you, we will burn all the tanks, don't worry,
and the relativistic data on the problem of Paul B. Andersen's synchronization will be written down.
Don't worry, Jean-Pierre, don't worry.
It's even worse than that, I will list not only the eight data, but all twelve, I will invent an additional event e3 which is the conjunction
BB' in the problem.
Let tA(e3), tA'(e3), tB(e3), tB'(e3).
Don't worry Jean-Pierre, we will burn all your tanks, and the planes too.
Le 03/09/2024 à 11:07, M.D. Richard "Hachel" Lengrand a écrit :
Le 03/09/2024 à 01:04, Python a écrit :
Sorry Richard, but the notation tX(eN) is, according to yourself,
supposed to mean "time for clock X of event eN" (which is out of the
scope of a synchronization procedure if the event eN took place
far from clock X anyway).
So when you write tA'(e2) your referencing a third clock named A'
which is definitely not a part of the synchronization procedure
that only involves TWO clocks. Worse now, tB' refer to a fourth
clocks.
What the hell is happening in your mind, Richard?
You are not even trying to make sense anymore. Anyway you've
already failed at that.
But no, I did not fail.
On the contrary, I told you, we will burn all the tanks, don't worry,
and the relativistic data on the problem of Paul B. Andersen's
synchronization will be written down.
Don't worry, Jean-Pierre, don't worry.
It's even worse than that, I will list not only the eight data, but all
twelve, I will invent an additional event e3 which is the conjunction
BB' in the problem.
Let tA(e3), tA'(e3), tB(e3), tB'(e3).
Don't worry Jean-Pierre, we will burn all your tanks, and the planes too.
You are loosing your marbles Richard.
Den 02.09.2024 01:39, skrev Python:
tA(e1) = 0
tA'(e1)= 0
by definition
With Richards (unrealistic as always) numbers:
c ≈ 3e8 m/s
d = 3e8 m
v = 0.8c
tA(e2) = (d/v)⋅√(1−v²/c²) ≈ 0.75 seconds
When I read the contributors to the French and Anglo-Saxon
forums, when I read Einstein (three lines of explanation) or Poincaré
(one line of explanation), I realize that it is very insufficient.
--athel -- biochemist, not a physicist, but detector of crackpots
Le 01/09/2024 à 15:21, Mikko a écrit :
Le 01/09/2024 à 08:27, Thomas Heger a écrit :
Am Samstag000031, 31.08.2024 um 09:05 schrieb Richard Hachel:
This is the set of all places, which can use the same kind of clocks.
They do not need to show the same time, however, but need to tick at
the same rate.
This is called a relativistic reference frame.
All clocks there have the same chronotropy: that is to say, they turn
at the same speed.
I do not use the term reference frame like physicists,
That's OK but you should not use it any other way, at least not here
or any other physics group.
Mikko
The term reference frame carries within itself a huge bias in
relativistic physics, and explains all by itself all the problems that
will arise in the history of modern aphysics.
On 2024-09-01 21:47:38 +0000, Richard Hachel said:
Le 01/09/2024 à 15:21, Mikko a écrit :
Le 01/09/2024 à 08:27, Thomas Heger a écrit :
Am Samstag000031, 31.08.2024 um 09:05 schrieb Richard Hachel:
This is the set of all places, which can use the same kind of clocks. >>>>>
They do not need to show the same time, however, but need to tick at >>>>> the same rate.
This is called a relativistic reference frame.
All clocks there have the same chronotropy: that is to say, they turn
at the same speed.
I do not use the term reference frame like physicists,
That's OK but you should not use it any other way, at least not here
or any other physics group.
Mikko
The term reference frame carries within itself a huge bias in
relativistic physics, and explains all by itself all the problems that
will arise in the history of modern aphysics.
If that is what you want then you shoid wse the tern, otherwise you should not.
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
And what's missing:
tA'(e2) = 2.25
The problem with B is that B is in B, that is to say somewhere other
than in the conjunction AA'.
We have synchronized the watches, A and A', but how do we do it for B?
Of course there are idiots, like Python, who will say, we just have to synchronize anyhow. But they are crazy. Don't worry, we are waiting for
them, we will burn them all.
There are two main ways to synchronize B and B', namely the
synchronize on A and A' by noting tB(e1)=-1 and tB'(e1)=-1;
or either in
practicing M type synchronization (Einstein synchronization),
noting tB(e1)=0 and tB'(e1)=0.
We are going to use Einstein synchronization to please Jean-Pierre.
For synchronization on AA', and not on M(R) and M'(R'),
it will suffice to add Δt=-1.
We then have:
tB(e1)=0 {-1}
tB'(e1)=0 {-1}
tB(e2)=0.75 {-0.25}
tB'(e2)=0.25 {-0.75)
We are now going to annoy Jean-Pierre,
add the e3 event
which is the conjunction BB'
which will happen at a certain time.
We apparently have:
tA(e3)= 0.50
tA'(e3)= 1.50
tB(e3)= -1.50 tB'(3)= -0.50
For these last two data, we notice that they are negative,
but that's normal.
For B and for B', the synchronization is done after the conjunction BB'.
Thank you for your attention.
R.H.
Den 03.09.2024 12:03, skrev Richard Hachel:
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
OK.
We have synchronized the watches, A and A', but how do we do it for B?
? ? ? ? ? ? ? ? !!!!!!!!!!!!!!
How confused is it possible to be?
A and A' are moving at the speed v relative to each other,
so of course A and A' can never be synchronous.
Setting two clocks to the same value as they pass each other
isn't to make them synchronous, they will only show the same
time at the instant they are adjacent.
Le 03/09/2024 à 17:20, "Paul.B.Andersen" a écrit :
Den 03.09.2024 12:03, skrev Richard Hachel:
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
OK.
<http://nemoweb.net/jntp?nMk86znPr22YJUAQ8K6Ws3GBg9o@jntp/Data.Media:1>
Slowly please.
Our readers need to understand correctly and be able to judge correctly.
We have determined, by simple synchronization procedure AA'
which we have called event e1, that tA(e1)=0, and that tA'(e1)=0.
We have then very easily concluded that tA(e2)=0.75.
So far, so good.
We won't come back to this.
Now, we are moving forward, we are looking, at the time of e2,
what time will be displayed on clock A' when B' crosses A.
It already seems that there is a difficulty here.
You say with certainty that tA'(e2)= ? ? ? , and I say with the same certainty
that tA'(e2)=2.25.
One of us is necessarily wrong.
We must understand why.
R.H.
Den 03.09.2024 12:03, skrev Richard Hachel:
Above I asked you to read my paper again.
https://paulba.no/pdf/Mutual_time_dilation.pdf
You are clearly too ignorant and incompetent to understand it.
Den 03.09.2024 12:03, skrev Richard Hachel:
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
OK.
And what's missing:
tA'(e2) = 2.25
This is nonsense, and demonstrates that you
don't know what an event is.
e2 is the event that clock A and clock B' are adjacent
tA'(e2) is meaningless.
Den 03.09.2024 12:03, skrev Richard Hachel:
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
OK.
And what's missing:
tA'(e2) = 2.25
This is nonsense, and demonstrates that you
don't know what an event is.
e2 is the event that clock A and clock B' are adjacent
tA'(e2) is meaningless.
Am Montag000002, 02.09.2024 um 14:16 schrieb Richard Hachel:
Le 02/09/2024 à 08:25, Thomas Heger a écrit :There exist no 'center of the universe', because everything moves.
I use the observation, that clocks around the Earth surface tick at
the same rate, while they don't tick at the same rate at different
altitudes.
There is already a bias here.
If a watch is placed at altitude, it does not evolve at the same speed
as a fixed watch placed at the level of our local mass reference
center that we could put the sun, or even the galactic center. The
effects of these reference frames are perhaps negligible. I do not
know. But at least, the effects of the revolution of the object around
the center of the earth are not the same as the effects
on an object placed on the surface of the ground. Worse, for the
object placed on the surface of the ground, it is the center of the
earth that rotates around it; and also for the other. These effects
are no longer really Galilean, but effects of rotating reference
frames for which I have given the equations, and which cause some
surprises (it is the object that goes the fastest that has the time
that passes the fastest, contrary to Galilean effects).
If we define a certer of our own local frame of reference, we do this
for pratical purposes, even if no such thing as a center would exist.
I personally prefer a setting, where the observer in question rests in
the center of his own frame of reference.
I call this perspective 'subjectivism', because this is the view we have
from the world around us.
We could use any other point, however, if we decide to do so.
But this wouldn't make this point the center of the world, but the
center of our frame of reference.
But none of these 'centers' is actually real, because the universe has
no center.
Am Dienstag000003, 03.09.2024 um 07:53 schrieb Thomas Heger:
Am Montag000002, 02.09.2024 um 14:16 schrieb Richard Hachel:This is actually the reason, why 'big-bang-theory' must be wrong.
Le 02/09/2024 à 08:25, Thomas Heger a écrit :There exist no 'center of the universe', because everything moves.
I use the observation, that clocks around the Earth surface tick at
the same rate, while they don't tick at the same rate at different
altitudes.
There is already a bias here.
If a watch is placed at altitude, it does not evolve at the same
speed as a fixed watch placed at the level of our local mass
reference center that we could put the sun, or even the galactic
center. The effects of these reference frames are perhaps negligible.
I do not know. But at least, the effects of the revolution of the
object around the center of the earth are not the same as the effects
on an object placed on the surface of the ground. Worse, for the
object placed on the surface of the ground, it is the center of the
earth that rotates around it; and also for the other. These effects
are no longer really Galilean, but effects of rotating reference
frames for which I have given the equations, and which cause some
surprises (it is the object that goes the fastest that has the time
that passes the fastest, contrary to Galilean effects).
If we define a certer of our own local frame of reference, we do this
for pratical purposes, even if no such thing as a center would exist.
I personally prefer a setting, where the observer in question rests in
the center of his own frame of reference.
I call this perspective 'subjectivism', because this is the view we
have from the world around us.
We could use any other point, however, if we decide to do so.
But this wouldn't make this point the center of the world, but the
center of our frame of reference.
But none of these 'centers' is actually real, because the universe has
no center.
The big bang would be, in a way, the center of the universe and the
beginning of time.
[snip demented nonsense]
Le 03/09/2024 à 17:20, "Paul.B.Andersen" a écrit :
Den 03.09.2024 12:03, skrev Richard Hachel:
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
OK.
And what's missing:
tA'(e2) = 2.25
This is nonsense, and demonstrates that you
don't know what an event is.
e2 is the event that clock A and clock B' are adjacent
tA'(e2) is meaningless.
It is obvious that tA'(e2) has a meaning for A'.
It is the time at which in his frame of reference (A'), the event E2
exists.
Paul, Paul, you can't say it's meaningless. A little more consideration
for the other posters, and please a little more practical intelligence:
there is indeed a moment, where, for A, the event e2 exists in his frame
of reference, and if A' was synchronized at the start, there will be a
time, and only one time of e2 that will be written on his watch.
We can easily, if we correctly master the notion of relativistic
simultaneity and the notions of relative chronotropies,
reveal what this time written on the clock A' thus synchronized during
e1 will be: tA'(e1)=0.
I beg you not to say that it is absurd or meaningless.
How to proceed?
We KNOW that the travel time of A in A'B'
will be equal to the distance
A'B' in R' divided by the apparent escape velocity of an object moving
at v=0.8c.
This is unavoidable and it is mathematical.
Let tA'(e2)=tA'(e1)+(A'B'/Vapp')
Let tA'(e2)= 0 + 3.10^8/(4/9)c
tA'(2)=2.25 sec
Please note, I am talking about the exact time when A'
actually perceives e2 in direct-live, and which represents
the real time of things.
If we want to judge, and count based on synchronization
abstract, based on "the transverse speed of light", it is necessary
remove a second of anisochrony between A' and B'. Which gives a time
supposed (but false) of 1.25 sec.
Please confirm that you have understood and that you validate, which
will allow us to go further and explain all the predictive values that I have already given, but without explaining yet.
Den 04.09.2024 02:32, skrev Richard Hachel:
Le 03/09/2024 à 17:20, "Paul.B.Andersen" a écrit :
Den 03.09.2024 12:03, skrev Richard Hachel:
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
OK.
And what's missing:
tA'(e2) = 2.25
This is nonsense, and demonstrates that you
don't know what an event is.
e2 is the event that clock A and clock B' are adjacent
tA'(e2) is meaningless.
It is obvious that tA'(e2) has a meaning for A'.
It is the time at which in his frame of reference (A'), the event E2
exists.
Paul, Paul, you can't say it's meaningless. A little more
consideration for the other posters, and please a little more
practical intelligence: there is indeed a moment, where, for A, the
event e2 exists in his frame of reference, and if A' was synchronized
at the start, there will be a time, and only one time of e2 that will
be written on his watch.
We can easily, if we correctly master the notion of relativistic
simultaneity and the notions of relative chronotropies,
reveal what this time written on the clock A' thus synchronized during
e1 will be: tA'(e1)=0.
e2 is short for "the event that clock A and clock B' are adjacent"
At this event, tA = (d/v)⋅√(1−v²/c²) = 0.75 s
and clock B will simultaneously in K show tB = 0.75 s
At this event, tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s
Den 04.09.2024 02:32, skrev Richard Hachel:
Le 03/09/2024 à 17:20, "Paul.B.Andersen" a écrit :
Den 03.09.2024 12:03, skrev Richard Hachel:
We have:
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s
OK.
And what's missing:
tA'(e2) = 2.25
This is nonsense, and demonstrates that you
don't know what an event is.
e2 is the event that clock A and clock B' are adjacent
tA'(e2) is meaningless.
It is obvious that tA'(e2) has a meaning for A'.
It is the time at which in his frame of reference (A'), the event E2
exists.
Paul, Paul, you can't say it's meaningless. A little more consideration
for the other posters, and please a little more practical intelligence:
there is indeed a moment, where, for A, the event e2 exists in his frame
of reference, and if A' was synchronized at the start, there will be a
time, and only one time of e2 that will be written on his watch.
We can easily, if we correctly master the notion of relativistic
simultaneity and the notions of relative chronotropies,
reveal what this time written on the clock A' thus synchronized during
e1 will be: tA'(e1)=0.
e2 is short for "the event that clock A and clock B' are adjacent"
At this event, tA = (d/v)⋅√(1−v²/c²) = 0.75 s
and clock B will simultaneously in K show tB = 0.75 s
At this event, tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s
I beg you not to say that it is absurd or meaningless.
How to proceed?
We KNOW that the travel time of A in A'B'
At event e1 tA = 0, at event e2 tA = (d/v)⋅√(1−v²/c²) = 0.75 s
so the travel time for A to go from e1 to e2 is = 0.75 s
Den 04.09.2024 02:32, skrev Richard Hachel:
This is unavoidable and it is mathematical.
Let tA'(e2)=tA'(e1)+(A'B'/Vapp')
Let tA'(e2)= 0 + 3.10^8/(4/9)c
tA'(2)=2.25 sec
Nonsense.
At e2, tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s
Understand this:
A and B are always simultaneously showing the same in K
A' and B' are always simultaneously showing the same in K'
So how can you imagine that at event e2, when tB' = 1.25 s.
then clock A' should simultaneously show 2.25 s ?
Den 04.09.2024 02:32, skrev Richard Hachel:
will be equal to the distance
A'B' in R' divided by the apparent escape velocity of an object moving
at v=0.8c.
And what will you escape from?
A and B are moving with the speed v = 0.8c in K' <-
A' and B' are moving with the speed v = 0.8c in K ->
Nothing is moving with any other speed than v.
There are no 'apparent speeds'.
Le 04/09/2024 à 09:24, Thomas Heger a écrit :
Am Dienstag000003, 03.09.2024 um 07:53 schrieb Thomas Heger:
Am Montag000002, 02.09.2024 um 14:16 schrieb Richard Hachel:This is actually the reason, why 'big-bang-theory' must be wrong.
Le 02/09/2024 à 08:25, Thomas Heger a écrit :There exist no 'center of the universe', because everything moves.
I use the observation, that clocks around the Earth surface tick at
the same rate, while they don't tick at the same rate at different
altitudes.
There is already a bias here.
If a watch is placed at altitude, it does not evolve at the same
speed as a fixed watch placed at the level of our local mass
reference center that we could put the sun, or even the galactic
center. The effects of these reference frames are perhaps
negligible. I do not know. But at least, the effects of the
revolution of the object around the center of the earth are not the
same as the effects
on an object placed on the surface of the ground. Worse, for the
object placed on the surface of the ground, it is the center of the
earth that rotates around it; and also for the other. These effects
are no longer really Galilean, but effects of rotating reference
frames for which I have given the equations, and which cause some
surprises (it is the object that goes the fastest that has the time
that passes the fastest, contrary to Galilean effects).
If we define a certer of our own local frame of reference, we do this
for pratical purposes, even if no such thing as a center would exist.
I personally prefer a setting, where the observer in question rests
in the center of his own frame of reference.
I call this perspective 'subjectivism', because this is the view we
have from the world around us.
We could use any other point, however, if we decide to do so.
But this wouldn't make this point the center of the world, but the
center of our frame of reference.
But none of these 'centers' is actually real, because the universe
has no center.
The big bang would be, in a way, the center of the universe and the
beginning of time.
You are again making up silly stuff. In the b-b-theory there is NO
center. The Big Band happened everywhere.
Le 04/09/2024 à 21:37, "Paul.B.Andersen" a écrit :
Den 04.09.2024 02:32, skrev Richard Hachel:
This is unavoidable and it is mathematical.
Let tA'(e2)=tA'(e1)+(A'B'/Vapp')
Let tA'(e2)= 0 + 3.10^8/(4/9)c
tA'(2)=2.25 sec
Nonsense.
But no!
I simply use a synchronization system based not on M but on the watches themselves, and, moreover, I involve universal anisochrony.
Once well understood, my logic is quite simple, but the brain has been
so clouded by the notion of universal and absolute present time for centuries, that it is difficult to convince those who read me.
Den 05.09.2024 01:23, skrev Richard Hachel:
Vo=0.8c
Vapp=Vo/(1+cosµ.Vo/c)
Vapp'=(4/9)c
Vapp"=4c
R.H.
BTW, your equation above is wrong.
It should be:
Vapp = v⋅sin(μ)/(1 - (v/c)⋅cos(μ))
where μ is the angle between the observed object's
velocity and the line of sight.
That is because we can only observe the transverse
component of the object's velocity.
If the object is coming right at us, μ = 0⁰, and Vapp = 0.
Note that v_app > c when v > c/(sin(μ)+cos(μ))
It should be:
Vapp = v⋅sin(μ)/(1 - (v/c)⋅cos(μ))
Le 05/09/2024 à 14:24, "Paul.B.Andersen" a écrit :
Den 05.09.2024 01:23, skrev Richard Hachel:
We KNOW that the travel time of A in A'B' will be equal to the distance A'B' in R' divided by the apparent escape velocity of an object moving at v=0.8c. >>>
Vo=0.8c
Vapp=Vo/(1+cosµ.Vo/c)
Vapp'=(4/9)c
Vapp"=4c
R.H.
What do you say, Paul?
Den 05.09.2024 16:15, skrev Richard Hachel:
A and B are moving with the speed v = 0.8c in K' <-
A' and B' are moving with the speed v = 0.8c in K ->
Nothing is moving with any other speed than v.
There are no 'apparent speeds'.
Is this too hard for you to understand?
...
Cette notion EXISTE.
Vapp=v/(1+cosµ.v/c)
Le 05/09/2024 à 22:31, M.D. Richard "Hachel" Lengrand a écrit :
...
Cette notion EXISTE.
Vapp=v/(1+cosµ.v/c)
Yep. As shown there :
https://gitlab.com/python_431/cranks-and-physics/-/blob/main/Hachel/divagation_lengrand.pdf
Any velocity is between an object and a point of reference.
Am Donnerstag000005, 05.09.2024 um 14:25 schrieb Paul.B.Andersen:
Den 05.09.2024 01:23, skrev Richard Hachel:
Vo=0.8c
Vapp=Vo/(1+cosµ.Vo/c)
Vapp'=(4/9)c
Vapp"=4c
R.H.
How confused is it possible to be? :-D
You must know that this 'apparent speed' is a visual
observation (telescope).
From whence did you get the idiotic idea that somebody
is visually observing any of the clocks in this paper?
https://paulba.no/pdf/Mutual_time_dilation.pdf
A and B are moving with the speed v = 0.8c in K' <-
A' and B' are moving with the speed v = 0.8c in K ->
Nothing is moving with any other speed than v.
There are no 'apparent speeds'.
Is this too hard for you to understand?
Any velocity is between an object and a point of reference.
Usually we have an observer (say 'A') at a certain spot (also called
'A'), who measures distances from his own position.
Den 05.09.2024 01:23, skrev Richard Hachel:
Vo=0.8c
Vapp=Vo/(1+cosµ.Vo/c)
Vapp'=(4/9)c
Vapp"=4c
R.H.
How confused is it possible to be? :-D
You must know that this 'apparent speed' is a visual
observation (telescope).
From whence did you get the idiotic idea that somebody
is visually observing any of the clocks in this paper?
https://paulba.no/pdf/Mutual_time_dilation.pdf
A and B are moving with the speed v = 0.8c in K' <-
A' and B' are moving with the speed v = 0.8c in K ->
Nothing is moving with any other speed than v.
There are no 'apparent speeds'.
Is this too hard for you to understand?
The relevant coordinate system can now be attatched to A or B, depending
on where the observer is placed.
So: the coordinate system K is placed, that its center coincides with
'A', while the certer of K' coincides with B.
(the primed version of A and B make no sense, hence could be left away).
Now we have two systems K and K' which both receed from an imaginary
point in the center (called 'M', for simpility), which is assumed to be
not moving.
This would mean, that A would receed from M by v=0.8c and from B with
1,6 c, hence drops out of the realm of visiblity, because the image of A
gets redshifted below the value 0 Hz, if seen from the remote side (here B).
But, nevertheless, both (A and B) could remain well and good, because
who cares about distant observers, which you cannot see?
...
Usually we have an observer (say 'A') at a certain spot (also called
'A'), who measures distances from his own position.
The measured object (say 'B') moves - say- away at a certain speed v.
But seen from B the point A moves away with the same speed, though into
the opposite direction.
Le 05/09/2024 à 14:24, "Paul.B.Andersen" a écrit :
Den 05.09.2024 01:23, skrev Richard Hachel:
Vo=0.8c
Vapp=Vo/(1+cosµ.Vo/c)
Vapp'=(4/9)c
Vapp"=4c
R.H.
BTW, your equation above is wrong.
It should be:
Vapp = v⋅sin(μ)/(1 - (v/c)⋅cos(μ))
where μ is the angle between the observed object's
velocity and the line of sight.
That is because we can only observe the transverse
component of the object's velocity.
If the object is coming right at us, μ = 0⁰, and Vapp = 0.
What do you say, Paul?
Note that v_app > c when v > c/(sin(μ)+cos(μ))What's the matter with you?
Are you drunk, or just insane?
On 2024-09-05 14:15:28 +0000, Richard Hachel said:
Le 05/09/2024 à 14:24, "Paul.B.Andersen" a écrit :
Den 05.09.2024 01:23, skrev Richard Hachel:
Vo=0.8c
Vapp=Vo/(1+cosµ.Vo/c)
Vapp'=(4/9)c
Vapp"=4c
R.H.
BTW, your equation above is wrong.
It should be:
Vapp = v⋅sin(μ)/(1 - (v/c)⋅cos(μ))
where μ is the angle between the observed object's
velocity and the line of sight.
That is because we can only observe the transverse
component of the object's velocity.
If the object is coming right at us, μ = 0⁰, and Vapp = 0.
What do you say, Paul?
Note that v_app > c when v > c/(sin(μ)+cos(μ))What's the matter with you?
Are you drunk, or just insane?
No, Paul just knows and understands certain simple things.
Le 06/09/2024 à 12:31, Thomas Heger a écrit :
Any velocity is between an object and a point of reference.
This means that all speed is relative.
If you remove the reference point, there is no speed.
An isolated object in an empty universe cannot move, nor be moved.
Le 06/09/2024 à 12:31, Thomas Heger a écrit :
Usually we have an observer (say 'A') at a certain spot (also called
'A'), who measures distances from his own position.
The measured object (say 'B') moves - say- away at a certain speed v.
But seen from B the point A moves away with the same speed, though
into the opposite direction.
Here, already, there arises a problem that physicists do not see, or
even deny for forty years, that is to say since I told them.
How is this speed characterized in a relativistic universe?
There are three ways to propose what A sees of B.
-The real speed (which is a primordial concept but is never used, except
in a roundabout way by posing m'=m/sqrt(1-v²/c²), which is an absurdity imposing the relativity of masses. A bus becoming two buses by change of reference frame.
-The observable speed, measured with two separate watches A and B, and
which is the classic speed used, but false.
- The apparent speed (what we see with the naked eye in telescopes
because of anisochrony, that is to say the inverse of the speed of light).
We therefore have three simple notations of what a speed is.
Vr, Vo, Vapp.
We will note that the absence of a preferred reference frame in the
universe,
means that these three speeds remain constant by permutation of observer.
Let us pose an entity B which moves away from A at 0.8c (speed in
classical notation). We have Vo=0.8c, Vr=(4/3)c, Vapp=(4/9)c.
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