The problem of relativistic synchronisation

[Richard Hachel]

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.
The theory of relativity is sorely lacking an article on the bases of the
theory and the great problem of synchronization, that is to say the
relations between space and time.
Einstein does not speak of the causes of the invariance of the speed of
light, and he does not even explain that the measurement of the speed of
light is in fact only a transversal decoy for a teletransverse observer.
What he does is just think and give a famous postulate, but written in
haste, and without explanation. A postulate moreover taken once again from
Poincaré.
All this is very insufficient.
It is urgent, I think, to rewrite an article, at least two pages long (and
not three lines) on the very foundations of the theory.
The most difficult thing is not to understand things (I have a fairly
clear understanding) but to write them in a simple and universal way, and
especially by getting rid of this idea of ​​a FUCKING FLAT PLANE OF
PRESENT TIME that clutters the thoughts of men from Methuselah to even
today's physicists.

R.H.

[Mikko]

Post by Richard Hachel
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.

[Richard Hachel]

Post by Mikko

Post by Richard Hachel
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.

R.H.

[Python]

Post by Richard Hachel

Post by Mikko

Post by Richard Hachel
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.

[Richard Hachel]

Post by Python

Post by Richard Hachel

Post by Mikko

Post by Richard Hachel
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.

En parlant de ça, perso, je baisse un peu les bras, même si je suis
convaincu de l'utilité
d'un petit article de quelques lignes sur la notion de simultanéité et
de synchronisation (la base de la RR).

Je pense qu'en moyenne (tant pis si ça leur fait mal) les habitués sont
trop cons, je veux surtout dire trop crétins de côté de la bite, la
bite, toujours la bite.

C'est très malheureux, mais on n'en sort pas plus ici sur les forums
anglo-saxons, que sur les forums francophones. Ce sera à qui est le plus
con avec la bite la plus grosse.

C'est dommage, il y a pourtant matière à réflexion, et certaines
réflexions sont parfois intéressantes,
comme les posts sur la synchronisation relativiste entre deux points A et
B.

Ca part parfois bien (comme tes explications des événement e1, e2, e3)
et le fait qu'on puisse déjà
proposer des évidences PRUDENTES avant d'aller plus loin.

On peut alors poser sans crainte : tA(e3)-tA(e1)=2AB/c

Puis, en admettant que A prévienne de e1 et de e3, soit avec des
photons, soit avec des limaces de même vitesse, n'importe quel point M du
référentiel stationnaire, on a encore une autre tautologie :

tM(e3)-tM(e1)= tA(e3)-tA(e1) = 2AB/c

On ne peut pour l'instant en dire plus sur la vitesse de la lumière
entre A et B dans le sens AB,
ni dans le sens BA.

Sur cela, on respire on souffle, Einstein ne semble pas d'accord avec
Hachel. Pour Einstein, la question ne se pose pas, et il semble acquis que
t(AB)=t(BA).

Sauf que ce n'est plus vrai en milieu anisochrone, et que notre univers
n'est pas "un hyperplan 4D de simultanéité absolu, même pour un simple
référentiel inertiel".

On peut alors proposer UNE synchronisation basée sur UN hyperplan de
simultanéité, mais il faut proposer LE candidat adéquat, et ce ne peut
être, évidement, ni A, ni B.

Donc on continue à partir de là.

On peut alors proposer une synchronisation de A et B par M (et on aura
une synchronisation de type M).

Puis-je le faire sans rire, et comment?

A noter que si M est purement en position perpendiculaire sur la sécante
du milieu AB, alors quelque soit la vitesse de l'information (c dans les
deux sens chez Einstein, à la fois c/2 ou ∞ selon le sens chez Hachel),
la réception du signal de synchro émis par M sera simultané en
réception A et B pour M,
et simultané aussi en retour pour M.

On peut donc dans risque, poser que, pour M après synchronisation A et B
de sa part (mais QUE pour M):
tM(e1)=0 tM(e2)=1 et tM(e3)=2

La synchronisation est parfaite pour M.

On peut, en imaginant un point M imaginaire, placé très loin et
perpendiculairement à tous els points de l'univers stationnaire étudié,
avec une synchronisation parfaite de type To, et parfaite pour tout
le référentiel.

Par changement de référentiel inertiel, M devient M' et To devient To',
car la chronotropie devient réciproquement relative.

Tout est dit. Les bases sont données. La RR devient alors très simple
et les équations qui vont avec évidentes.

Reste à le faire comprendre, et ce n'est pas en faisant le singe qu'on
va me comprendre, ou en me faisant des réponses à la con du genre : "La
vitesse de la lumière met une seconde pour venir de la lune jusqu'ici".

C'est d'une profonde débilité de relativistes bédouins, ça.

Mais il semble que ça plaise, et qu'on peut même agrémenter ça de
smileys ridicules.

R.H.

[Paul.B.Andersen]

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

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.
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.

[Python]

Post by Paul.B.Andersen
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.
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
"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

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.
  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
"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."
 "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.

This is exactly what I tried to explain him on fr.sci.physique.

In vain.

[Richard Hachel]

Post by Paul.B.Andersen
"You" is probably Python. Richard never quote what he is
referring to, and he doesn't define the events e1, e2, e3.
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
"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

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.
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
"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."
"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
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.

[Thomas Heger]

Am Freitag000030, 30.08.2024 um 18:31 schrieb Richard Hachel:
...

Post by Paul.B.Andersen
"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."
  "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



...

[Richard Hachel]

Post by Thomas Heger
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.

This is what I have been explaining to Python for thirty years.

Python has always remained at the anal stage of the isochrony of the
universe,
and of the generalized Cartesian hyperplane.

Damn, thirty years...

Forty years for the others...

What should we do to make them understand?

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, and that if there
is a problem, it is because the photons do not run fast enough.

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?

R.H.

[Paul.B.Andersen]

Post by 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?

Your wife, who is in Berlin see in the timetable that the train
will arrive in Berlin at the time 20:41. Which clock would she
use to meet you at the railway station at the right time,
a solar clock in Berlin, or her wristwatch which is showing UTC+2h?

When the train leaves Nantes, you see the clock on the railway station
showing 8:32, and you start your stop watch. When you arrive Berlin,
you see the watch on the railway station showing 20:41. You stop your
stop watch which show that the duration of the journey is 12h 9m.

How is it possible that the difference between the Berlin clock
at arrival and the Nantes clock at departure is equal to the duration
of your journey, 20:41 - 8:32 = 12h 9m ?

Could it be that the clock at the railway station in Berlin
and the clock at the railway station in Nantes are synchronous?

You will of course not respond to this.

You will flee, as you always do.

Won't you? :-D

Post by Richard Hachel
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.

Why are you stating the bleeding obvious?
Solar clocks can't be synchronised.

That's why we use clocks showing UTC+2h in Nantes and Berlin.
They do mark the time 12:00 (which is not the noon) at the same
time in Nantes and Berlin.

Same time = simultaneously as defined by Einstein.

Post by Richard Hachel
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

Le 24/08/2024 à 21:12, "Paul.B.Andersen" a écrit :
| If we use three weeks on the transfer, and we ignore the gravitational
| blue shift, and pretend that the ECI frame is a true inertial frame,
| then the lunar clock will lag 0.45 μs on the Earth clock.

So the clocks are not synchronous to within ns,
but they are synchronous within 1 μs which will be acceptable
in most cases.

In your example 1 second was the resolution of the timing.
And you agreed that the clock at your table and the lunar clock
would be synchronous.

| 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.

so when you see in the telescope the lunar watch showing 00:00:07",
the lunar watch will simultaneously show 00:00:08, the same
as the watch at your table.

Post by Richard Hachel
He does not understand that there is a real procedure of natural
desynchronization by spatial position,

If the clocks are moved relative to each other in an asymmetric way,
they will not stay synchronous.
In this case the clocks are 0.45 μs out of sync.

Post by Richard Hachel
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).

An awkward way of explaining mutual time dilation.

https://paulba.no/pdf/Mutual_time_dilation.pdf

Post by Richard Hachel
What should we do to make him understand?

As shown above, I understand everything.

[Maciej Wozniak]

Post by Paul.B.Andersen

Post by 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?

Or maybe it will be some gedanken clock of
your idiot guru, showing god-only-know-what?

[Richard Hachel]

Post by Paul.B.Andersen
https://paulba.no/pdf/Mutual_time_dilation.pdf

? ? ?

<http://nemoweb.net/jntp?***@jntp/Data.Media:1>


R.H.

[Python]

Post by Paul.B.Andersen
https://paulba.no/pdf/Mutual_time_dilation.pdf

? ? ?

Nemo is quite unstable these days (this very link leads to "access
denied").

The image is available there : https://ibb.co/7C0S4nJ

Just in case Paul would want to answer to idiotic annotations on his
artice (been there, done that :-) ).

This time, concerning Hachel's idioticy: the sky is the limit.

[Paul.B.Andersen]

Post by Paul.B.Andersen
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

The spatial coordinate was irrelevant, but it would be:

x₂ = γ⋅(x₂' + v⋅t₂') = γ⋅(-d + v⋅d/v) = 0

which is a trivial result since clock A is stationary at x = 0.

--------------

Your d' is probably the distance between the clocks in K measured
in frame K', and vice versa.
But this 'contacted' distance is never used, and is irrelevant.

In the Lorentz transform you never use contacted distances and
dilated times. You _only_ use proper distances and proper times.

Didn't you know that?

[Maciej Wozniak]

Post by Paul.B.Andersen

Post by Paul.B.Andersen
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
| 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.
-----------------------------------------------
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
t₂ = γ⋅(t₂' + (v/c²)⋅x₂') = equation (1) in the paper

And angels on a pinhead will dance foxtrot.
Anyone can check GPS, your delusions have nothing
in common with real clocks, real observations
or real whatever.

[Richard Hachel]

Post by Python
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)= ?

R.H.

[Paul.B.Andersen]

Post by Richard Hachel

Post by Python
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

Post by Richard Hachel
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

[Richard Hachel]

Post by Paul.B.Andersen

Post by Richard Hachel

Post by Python
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

Post by Richard Hachel
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

Je n'accepte pas ta réponse.

Je prends pour base v=0.8c.

c=10^8m/s, c'est à dire, dans mon exemple, c=3.0000000000.10^8m/s

Please, respect.

t₂ = ?



R.H.

[Paul.B.Andersen]

Post by Paul.B.Andersen

Post by 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.

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

c=10^8m/s, that is to say, in my example, c=3.0000000000.10^8m/s

What the hell is your point?

You didn't give a speed of light 'in your example',
but that's OK because I know the speed of light.

Please, respect.

Why? I don't respect you.

t₂ = ?

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?

[Richard Hachel]

Post by Paul.B.Andersen

Post by Paul.B.Andersen

Post by 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

No. My question. YOUR problem.

c = 300000000 m/s
d = 3e8 m
v = 0.8⋅c = 240000000 m/s
t₂ = (d/v)⋅√(1−v²/c²) = 0.75 second

R.H.

[Paul.B.Andersen]

Post by Richard Hachel

Post by Paul.B.Andersen
c = 299792458 m/s

No.
c = 300000000 m/s

'nuff said!

[Richard Hachel]

Post by Paul.B.Andersen
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.

[Ross Finlayson]

Post by Richard Hachel

Post by Paul.B.Andersen
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".

[Ross Finlayson]

Post by Ross Finlayson

Post by Richard Hachel

Post by Paul.B.Andersen
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"

[Ross Finlayson]

Post by Ross Finlayson

Post by Ross Finlayson

Post by Richard Hachel

Post by Paul.B.Andersen
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"

"Ornstein-Uelenbech"? Ulen-beck, Ornstein Uelen-bech.

Ornstein-Eulenbech models stationary processes in
Gaussian boxes, Uhlenbeck, points out a same kind
of example, what results a centralizing tendency,
results reducing or reversing model invariants.

Mathematically, ....

"___ models stationary processes in ___, ...."


See, generative is just "show me where the blanks are
then show what makes fill-in-the-blanks".

Mathematically, ....

In physics then there's that the light-like,
has these various corners, their cases for
mathematical induction and model induction,
about that mathematically light is both
geometric, a straight ray, and optical, focussing.

Now, everybody knows that writing the tenses
of focus the verb and plurals of focus the noun,
focus the noun focus the verb to focus the infinitive,
focussing the gerund and focusses or focuses, verb noun,
"foci", hard or soft c, light is geometric, a
straight ray, and optical, focussing.

Then here for example, in the near field the light's
optical, is independent the near field the charge's
"optical", as with regards to "the model invariant",
what is "optical" behavior, or what is "model splitting"
or "model breaking", symmetry breaking"and model breaking
in the near field, and in the far field, and from and to
the point field, and from and to the global field.

Then this "quadratic gradient-positive", "flow", in a
space, that it's the flow over the space, the gradient,
is that "gradient-positive" means the same as "a-diabatic".

There are lots of non-adiabatic models breakings in the
adiabatic, often for example it's the first or only
assumption in a formalism, what establishes the directrix
of the current term, "flow", its gradient.

Anyways such as an idea about root-mean square, or,
Clausius, Fresnel or you know about Fresnel,
when sources move the optical, you can find wherever
there's "1/2" or "squared", "inverse square", then
giving that the triangle rule that the triangle inequality
holds for the metric in each space, so that outer 0.25
falls out of taking square then its root, is the usual
sort of case that define what also of course perfectly
holds, for all factors in the quadratic, what they do.

I.e. it's like "the discarded term", yet in whatever
gradient it's so, "the definite term", establishing
the scalar distances Cauchy-Schwarz, triangle rule,
then "synchronization" here is about "sources and
moving sources", then besides that the optical
of light and the "optical" of "electromagnetic
energy", is that in the interfaces that the light
is variously absorbed or reflected, while the
usual linear instantaneous electromotive and shock
potential of the "electromagnetic" in terms of
"power optically" is quite about zero with respect
to "power optically", while it's electromotive potential
or electrostatic potential, is a lot different,
with free reflection in optics in the light-like,
as with regards to potential in the electromagnetic.

So, it makes sense to just write in algebra what
happens to look like identity, to square the term
then take its root, that in terms of all the equations
in that as a quantity, that equations together define
a quantity, to square the term and take the root of the
square, _is_ the same quantity, "the" root of "the" square.


Or, it's as simple to make something like that out of "0.25",
as just a sort of "symmetrical" term then in what attenuates
as according their centralizing tendency or gradient.

"Root-square and power-root: not inverses,
to take the root and square,
to take the square and root",

these are among most usual of all manipulations in algebra,
and the point is that equations together define a quantity,
so that then when it's attached "the root or power would
attenuate these centralizing tendencies' terms, or reverse
them", simply and quite directly attaches that, a usual
model of an indicator of flow.



Then, here is that there's electron physics and there's
the matter of frequency and energy, as with regards to
"electricity's energy" and "electricity's frequency",
is that's either resulting or reducing potential in
both the electrostatic and electromotive, passively,
and always passively if actively.

While, light just reflects and expands, ....



So, it's a most usual sort of "the corner case" in terms.

[Python]

Post by Ross Finlayson

Post by Ross Finlayson

Post by 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.

[Ross Finlayson]

Post by Python

Post by Ross Finlayson

Post by Ross Finlayson

Post by Richard Hachel

Post by Paul.B.Andersen
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".

[Python]

Post by Ross Finlayson

Post by Python

Post by Ross Finlayson

Post by Ross Finlayson

Post by 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".

:-)

Nice.

[Paul.B.Andersen]

Post by Paul.B.Andersen

Post by 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.

In stead of making a fool of yourself like this,
you can rather respond to the more interesting part of
the post you were responding to.

Please, answer the simple question below:

When the train leaves Nantes, you see the watch on the railway station
showing 8:32, and you start your stop watch. When you arrive in Berlin,
you see the watch on the railway station showing 20:41. You stop your
stop watch which show that the duration of the journey was 12h 9m.

The question:
-------------
Why is the difference between the Berlin clock at arrival and
the Nantes clock at departure equal to the duration of your journey,
20:41 - 8:32 = 12h 9m ?

You will keep fleeing, won't you? Chicken!

[Richard Hachel]

Post by 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

But now, we have to go further. Do you want to remain a chicken?

No.

Why, Paul?

Because I don't want to be a chicken anymore. I want to be a brave man.

OK, Paul.

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.

But, however, Paul is a chicken. And he's going to run away (prophecy of
Nostradamus).
"Great King Hachel speak usenet
Afraid chicken run away".

For 0.75 seconds, B' rushes toward A.

Two questions that today represent two of the biggest questions in the
entire existence of sci.physics?relativity

1. At what apparent (i.e. APPARENT) speed does A apprehend B' rushing
toward him?

Chicken!

2. What is the apparent distance traveled by B' during the interval noted
by the watch?

Chicken! Chicken!

R.H.

[Python]

Post by Richard Hachel

Post by 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

If you insist, but this in no way a requisite of Einstein-Poincaré's
synchronization procedure. t_A = 451 is another possible value :-)

Post by Richard Hachel
tA'(e1)= 0

LOL!!! Whaaat the Hell is that???

Post by Richard Hachel
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 poulet piteux Lengrand est tout déplumé :-)

[Richard Hachel]

Post by Richard Hachel
tA (e1)= 0
tA'(e1)= 0

LOL!!! Whaaat the Hell is that? ? ?

Post by Richard Hachel
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.

R.H.

[Python]

Post by Richard Hachel

Post by Richard Hachel
tA (e1)= 0
tA'(e1)= 0

LOL!!! Whaaat the Hell is that? ? ?

Post by Richard Hachel
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!
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.

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.

[Python]

Post by Python
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.

[Paul.B.Andersen]

Post by Python

Post by Richard Hachel
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 :-)

Post by Richard Hachel
tA'(e1)= 0

LOL!!! Whaaat the Hell is that???

Post by Richard Hachel
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.

You are misunderstanding.
Richard never quote what he is responding to, so the context
is lost.

This is not about Einstein's synchronisation method,
but about this paper:

https://paulba.no/pdf/Mutual_time_dilation.pdf

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

[Richard Hachel]

Post by Paul.B.Andersen
tA(e1) = 0
tA'(e1)= 0
by definition
c ≈ 3e8 m/s
d = 3e8 m
v = 0.8c
tA(e2) = (d/v)⋅√(1−v²/c²) ≈ 0.75 seconds

Absolutely.

But it's not over, contrary to what Jean-Pierre is asking.
He wants us to stop this "ridiculous" discussion and surrender.
We will not surrender.
And all the tanks and planes that are going to Ukraine, we will burn them.

Now, let's go further.

We have:

tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75


And what's missing:
tA'(e2) = 2.25

Maintenant, ce n'est pas tout, les relativistes idiots, qui ne comprennent
rien à rien (comme Python),
on va tous les brûler, faut pas vous inquiéter.

Nous allons maintenant parler du B.

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.

Il y a deux façons principales de synchroniser B et B', soit les
synchroniser sur A et A' en notant tB(e1)=-1 et tB'(e1)=-1 ; ou soit en
pratiquant une synchronisation de type M (synchronisation Einstein),
en notant tB(e1)=0 et tB'(e1)=0.

Nous allons utiliser la synchronisation Einstein pour faire plaisir à
Jean-Pierre.

Pour la synchronisation sur AA', et non sur M(R) et M'(R'), il suffira
d'ajouter Δt=-1.

On a alors:
tB(e1)=0 {-1}
tB'(e1)=0 {-1}
tB(e2)=0.75 {-0.25}
tB'(e2)=0.25 {-0.75)

Nous allons maintenant pour ennuyer Jean-Pierre, ajouter l'événement e3
qui est la conjonction BB'
qui va se faire à un certain moment.

On a apparemment :
tA(e3)= 0.50
tA'(e3)= 1.50
tB(e3)= -1.50
tB'(3)= -0.50

Pour ces deux dernières données, on remarque qu'elles sont négatives,
mais c'est normal.

Pour B et pour B', la synchronisation est faite après la conjonction BB'.

Je vous remercie de votre attention.

R.H.

[Richard Hachel]

Post by Paul.B.Andersen
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s

OK.

Post by Paul.B.Andersen
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
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

This is perfectly correct, and I am happy that we say the same thing, even
if I give other values, because I speak in H (Hachel) synchronization and
you in M ​​(Einstein) synchronization.
With your method of synchronization, that is to say (I beg you to
understand what I am saying, because it is very important), you place
yourself at the level of an observer M as I have defined it.
I remind you that each observer, even stationary, has his own hypercone of
present time, and that only an imaginary observer placed OUTSIDE the 3D
universe can synchronize the universe in his "absolute" fashion.
Everything then depends on the present moment considered FOR WHOM.
In Einstein synchronization, everything you have just said is true.
In HAchel synchronization (that is to say by putting oneself in the place
not of M, but of A, or B, or A' or B', everything I say is also true.
My notation is however MORE true, because it describes exactly the time on
the watch at the moment when, for it, the event occurs (live).
It is not a fantasy, on my part. If I do it, attracting contempt, hatred,
misunderstandings, it is because it is necessary for the song.
tA(e1)=0
tA'(e1)=0
tA(e2)=0.75
tA'(e2)=2.25

You, you note in synchronization M.
I, I note in synchronization H, according to what the watch SEES for
itself, on itself when the event occurs.
You, you note in synchronization M, according to what M perceives of the
event (abstract absolute synchronization).

tM(e1)=0
tM(e2)=0.75
tM'(e1)=0
tM'(e2)=1.25

Note :
tM(e1)=tM'(e1).sqrt(1-Vo²/c²)
tM(e2)=tM'(e2).sqrt(1-Vo²/c²)

This is classic chronotropic dilation.

R.H.

[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.

At e2, tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s
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 ?

You need to add the anisochronous delay.
There is one second between A and B in R, and one second between A' and B'
in R'.

R.H.

[Paul.B.Andersen]

Post by 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.

This is so idiotic that I have had enough!

What's the matter with you?
Are you drunk, or just insane?

[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'.

Vo=0.8c

Vapp=Vo/(1+cosµ.Vo/c)

Vapp'=(4/9)c

Vapp"=4c

R.H.

[Paul.B.Andersen]

Post by 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?

----------------------

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))

[Richard Hachel]

Post by Paul.B.Andersen

Post by 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.
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?

Post by Paul.B.Andersen
Note that v_app > c when v > c/(sin(μ)+cos(μ))

What's the matter with you?
Are you drunk, or just insane?

R.H.

[Paul.B.Andersen]

Post by 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?

How confused is it possible to be? 😂

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?

[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?

Tu vas finir par devenir aussi fou que Python pour garder ton pré carré
et grillagé.

Tu ne te rends même plus compte que, pour me tenir tête (ce qui est
particulièrement idiot dans le marasme actuel), tu assures que les
vitesses apparentes n'existent pas, parce que les vitesses classiques
existent.

MAIS NON!

BORDEL DE MERDE, réveillez vous, usenet est en train de courir à la
catastrophe, et tout le monde est en train de devenir fou on ne sait pas
pourquoi...

IL Y A des vitesses classiques, dites vitesses OBSERVABLES. C'est une
chose.

Par exemple Vo=0.8c.

Mais il y a AUSSI ce qu'on appelle une notion de vitesses apparentes en
physique, et en astrophysique.

Cette notion EXISTE.

Vapp=v/(1+cosµ.v/c)

C'est évident. On l'explique soit par l'anisochronie (Hachel), soit pas
la vitesse de la lumière (Einstein). Mais ne dites pas que ça n'existe
pas!!!

R.H.

[Python]

Post by Richard Hachel
...
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

[Python]

Post by Python

Post by Richard Hachel
...
Cette notion EXISTE.
Vapp=v/(1+cosµ.v/c)

https://gitlab.com/python_431/cranks-and-physics/-/blob/main/Hachel/divagation_lengrand.pdf

By the way, invoking a cosine here is an utter pathetic pedantry: by
a trivial change of variables you can end up with -1 or 1 for the
v/c coefficient.

Do you know how idiot and pedant at the same time you look (at every
of your posts)?

[Mikko]

Post by Richard Hachel

Post by Paul.B.Andersen

Post by 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.
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?

Post by Paul.B.Andersen
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.

[Richard Hachel]

Post by Mikko

Post by Richard Hachel

Post by Paul.B.Andersen

Post by 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.
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?

Post by Paul.B.Andersen
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.

I am willing for Paul to know and understand certain simple things.
For example, he knows and understands perfectly the notion of reciprocal
relativity of internal chronotropies.
He is not a thug, he is not a moron, he is not a criminal.

But there are things that he does not understand, and, not understanding
them, assumes that those who say them do not have all their mental
faculties.

I explained to him correctly why he was making a colossal relativistic
blunder when he thought he could integrate (Leibniz) improper times into
accelerated frames of reference.

The blunder is of the same type as the poorly understood addition of
relativistic speeds, and which some consider very simple, and of the type
W=v+u.

I explained the blunder clearly, and showed that by ricochet, an
observable time measured too large would induce in the opposite direction,
a proper time measured too small.

Paul takes this as a joke, because "that's not what relativists say".

It's a shame that he's stubborn, because a good understanding of things
could necessarily earn him a better judgment.

When two parties confront each other, they must be heard both clearly and
completely. We can then judge.

Problem: Paul doesn't understand anything I'm telling him.

So he only judges with one hand.

It's not scientific.

For those who want to understand: what is observable time?

If we set To²=Tr²+Et², we see that it is the diagonal of proper time
and anisochrony.

It's visible to everyone.

However, this diagonal will stretch as proper time increases, or as the
distance traveled increases.

What Paul does not understand (and neither do physicists) is that we must
constantly take into account the length of this diagonal,
and not the path taken by the end of this diagonal (which forms a curve).

If we take the path, we have too large a value of the improper time.

This is not serious as it stands, since we easily have
To=(x/c).sqrt(2c²/ax)

But the problem will arise when we want to calculate Tr (the proper time)
with respect to this curve. We will inevitably imagine a proper time that
is much too short, whereas if we refer to the simple progression of the
length of the diagonal without taking into account its progressive
rotation, but simply its value,
there is absolutely no possible error.

I am providing the little diagram explaining the difference between the
blue curve and the red diagonal, and the dramatic confusion that we make
by confusing the two values.

<http://nemoweb.net/jntp?***@jntp/Data.Media:1>

R.H.

[Richard Hachel]

Post by Paul.B.Andersen
Vapp = v⋅sin(μ)/(1 - (v/c)⋅cos(μ))

This equation is correct, but it is not the equation of the apparent
speed.
The apparent speed is given as follows, if we take the angle µ, formed by
the direction of the mobile and the projection of the sight.

Vapp= v/(1+cosµ.v/c) Your equation, my dear Paul, and which is this one:
Vapp(perp)=sinµ.v/(1+cosµ.v/c) and which is the perpendicular projection
of the apparent speed, is not the apparent speed, but its transverse
projection on the background of the sky.

This projection, which is not really an apparent speed, but the apparent
transverse speed of the apparent speed, can, as you say, and the
phenomenon may seem surprising, be faster than c.

It is indeed strange, but logical if we know the RR well, that the
transverse projection of an apparent speed can be greater than this same
purely transverse speed.

This is a particular relativistic effect indeed.

I simply ask you not to confuse apparent transverse velocity (Vapp=v), and
transverse projection of an apparent velocity (Vapp" = sinµ.Vapp =
sinµ.v/(1+cosµ.v/c)

<http://nemoweb.net/jntp?***@jntp/Data.Media:1>

R.H.
Clic here if you can't see the draw ---> <https://www.nemoweb.net/?DataID=***@jntp>

[Thomas Heger]

Post by Paul.B.Andersen

Post by 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.

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.

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?

...


TH

[Maciej Wozniak]

Post by Thomas Heger

Post by Paul.B.Andersen

Post by 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.

No, usually you have no observer and you just
gedanke/imagine/fabricate. And assert.

[Richard Hachel]

Post by Thomas Heger
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.

R.H.

[Thomas Heger]

Post by Richard Hachel

Post by Thomas Heger
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.

Well, yes, but we had two objects: A and B.

A moves in the eyes of B and B in the eyes of A.

Since both are of equal rights, A does not move in respect to A himself,
same as B, while simulataniously move with 1,6 c away from the other.


TH

[Richard Hachel]

Post by Thomas Heger
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.

These values ​​remain systematically reciprocal between them according
to the laws of relativity.

R.H.

[Thomas Heger]

Post by Richard Hachel

Post by Thomas Heger
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.

We are talking here in the context of SRT.

But SRT assumes inertial frames of refence, hence assumes a background
space, which is not the real universe we live in.

In SRT there exist only unaccelearted streight worldlines, along which
any object would float.

This is not how the real universe operates, where practically nothing
moves in a streight line.

Now we can only use the SRT background here, because otherwise SRT would
not work, anhow.

This SRT-space would allow to regard any movement along a streight line
as being equivalent to an object at rest.

Now we could repeat the same trick and apply 0.8 c once, stop that to
zero, apply 0.8c a second time and still are allowed to halt this to zero.

But how would you introduce a 'speedlimit' in a space with such features?



TH

[Richard Hachel]

Post by Thomas Heger
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?
...

Oh my god!

This is where I see the criminal behavior and efficiency of all the
anti-Hachel morons who are on the forums and try to rot everything he has
to say.

If I were allowed to speak, and if what I say was accepted, not only would
the RR be simpler and fairer for the Nobel Prize winners (unable to speak
as well as I do), but in addition, such sentences full of
misunderstandings would not exist.

Saint Paul already said it in the time of Jesus Christ:
"By rejecting Hachel, physicists are therefore fully responsible for the
crimes and offenses committed against science, because having been able to
recognize Richard Hachel as their God, they did not want to recognize him
as God, preferring to lose the normal use of women to fuck each other,
with lousy concepts, and imaginary and false equations".

It is always the same waters that flow.

Man remains man, sunk in these hominid ideologies.

R.H.

[Richard Hachel]

Post by Richard Hachel
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.

But I never said that.

We really live in a crazy world.

I never said that!!!!

R.H.

[Richard Hachel]

Above I asked you to read my paper again.

Oui, c'est un bon conseil.

https://paulba.no/pdf/Mutual_time_dilation.pdf

C'est effectivement un papier intéressant (ainsi que d'autre pdf de
votre composition.

You are clearly too ignorant and incompetent to understand it.

Ce n'est pas très gentil, et ce n'est pas ainsi que les lecteurs de ce
forum vont progresser
dans l'intérêt scientifique et la convivialité.

R.H.

[Paul.B.Andersen]

Post by Paul.B.Andersen
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s

OK.

And we could add:
tA(e2) = (d/v)⋅√(1−v²/c²) = 0.75 s
tB'(e2) = d/v = 1.25 s

tA'(e3) = (d/v)⋅√(1−v²/c²) = 0.75 s
tB(e3) = d/v = 1.25 s

Post by Paul.B.Andersen
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.

I think you should read the paper again:
https://paulba.no/pdf/Mutual_time_dilation.pdf

Post by Paul.B.Andersen
The problem with B is that B is in B, that is to say somewhere other
than in the conjunction AA'.

Is the "conjunction of AA'" the same as event E1?
Why is it a problem that B isn't present at this event?

Post by Paul.B.Andersen
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.

Post by Paul.B.Andersen
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.

Yes, there are idiots, and you are one of them. Python is not.

Post by Paul.B.Andersen
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;

Good grief!
What the hell is tB(e1)=-1 and tB'(e1)=-1 supposed to mean?

LEARN WHAT AN EVENT IS.

The rest is mindless babble!

Post by Paul.B.Andersen
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.
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,

I think Jean-Pierre is pleased to see that all he
has said about you is correct.

Post by Paul.B.Andersen
add the e3 event
which is the conjunction BB'
which will happen at a certain time.
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.

Above I asked you to read my paper again.

https://paulba.no/pdf/Mutual_time_dilation.pdf

FORGET IT!

You are clearly too ignorant and incompetent to understand it.
As you so convincingly have demonstrated!

[Richard Hachel]

Post by Paul.B.Andersen
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s

OK.

Post by Paul.B.Andersen
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

Attention, je parle de l'heure exacte où A' perçoit réellement e2 en
direct-live, et qui représente l'heure réelle des choses.

Si l'on veut juger, et compter en fonction d'une synchronisation
abstraite, basée sur "la vitesse transversale de le lumière", il faut
ôter une seconde d'anisochrnie entre A' et B'. Ce qui donne un temps
supposé (mais faux) de 1.25 sec.

<http://nemoweb.net/jntp?***@jntp/Data.Media:2>

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.

R.H.

[Paul.B.Andersen]

Post by Paul.B.Andersen
tA(e1) = 0
tA'(e1)= 0
tA(e2) = 0.75 s

OK.

Post by Paul.B.Andersen
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
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

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'.

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 ?

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.

Please note, I am talking about the exact time when Paul
actually perceives that Richard snap his fingers 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.

Well said!

I have always admired your ability to clearly explain things.
Of course we have to remove a second when we are talking
about "the transverse speed of light".

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.

I have understood that we have to remove a second based on
"the transverse speed of light".
Based on "the longitudinal speed of light" we would have to
add a second, obviously.

[Paul.B.Andersen]

Post by Richard Hachel
OK, Paul.
So we're going to continue, because it's very important.

OK. Let's play.

https://paulba.no/pdf/Mutual_time_dilation.pdf
c ≈ 3e8 m/s
d = 3e8 m
v = 0.8c
tA(e1) = 0
tA'(e1)= 0
tA(e2) = (d/v)⋅√(1−v²/c²) ≈ 0.75 seconds.

Post by Richard Hachel
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?

I have no idea of what your "apparent speed" is, and I don't care.

I can however tell you what A will _measure_ the speed of B'
relative to A will be.

A knows that his clock shows 0 at e1, when B' is at
the position x = (-3e8m + (v/c²)⋅0)/√(1−v²/c²) = -3e8m/√(1−v²/c²)
A knows that his clock shows 0.75 seconds when B' is at
the position x = 0.
So B' has moved the distance 3e8m/√(1−v²/c²) in 0.75 s
v = (3e8m/0.75s)/√(1−v²/c²)
v = 3e8m/√((0.75s)²+(3e8m)²/(3e8m/s)²) = 240000000 m/s = 0.8c

Post by Richard Hachel
2. What is the apparent distance traveled by B' during the interval
noted by the watch?

The distance B' has travelled in the rest frame of A is shown above.
d' = 3e8m/√(1−v²/c²) = 500000000 m
It is nothing 'apparent' about this distance.


Now that I have answered your questions, you can possibly answer mine:

When the train leaves Nantes, you see the watch on the railway station
showing 8:32, and you start your stop watch. When you arrive in Berlin,
you see the watch on the railway station showing 20:41. You stop your
stop watch which show that the duration of the journey was 12h 9m.

The question:
-------------
Why is the difference between the Berlin clock at arrival and
the Nantes clock at departure equal to the duration of your journey,
20:41 - 8:32 = 12h 9m ?

[Paul.B.Andersen]

Post by Paul.B.Andersen
https://paulba.no/pdf/Mutual_time_dilation.pdf

? ? ?

You point out that you can't even read and understand
a paper with simple math, but you flee from the real challenge
in the post you are responding to.

So please, answer the simple question below:

When the train leaves Nantes, you see the watch on the railway station
showing 8:32, and you start your stop watch. When you arrive in Berlin,
you see the watch on the railway station showing 20:41. You stop your
stop watch which show that the duration of the journey was 12h 9m.

The question:
-------------
Why is the difference between the Berlin clock at arrival and
the Nantes clock at departure equal to the duration of your journey,
20:41 - 8:32 = 12h 9m ?


You will flee again, won't you?

[Maciej Wozniak]

Post by Paul.B.Andersen

Post by Paul.B.Andersen
https://paulba.no/pdf/Mutual_time_dilation.pdf

? ? ?

You point out that you can't even read and understand
a paper with simple math

And speaking of simple math - it's always
good to remind that your idiot guru had
to announce it false, as it didn't want
to fit his madness.

[Thomas Heger]

Post by Richard Hachel

Post by Thomas Heger
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.

I used a term called 'time domaine'.

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.

Timezones on planet Earth has a conncetion to the sun and where it is in
the sky at a certain place.

But we could actually use concepts like GMT and use one abstract time
standard around the globe, which does not refer to sunsets.

Only required is, that a second in Paris is also a second in Sydney, for
instance.

This is the case for all points around the globe at sea-level.

This set of points is what I call a 'time-domaine'.

But other times are thinkable, which includes a 'time-domaine' where
time flows 'sideways' or backwards.

This is so, because I regard time as imaginary pseudo-scalar and the
perpendicular spacelike axes as real.

This picutre can be rotated by multiplying it with i (or multiples or
fractions of i).

This is a very odd thing and not so easy to understand.

But apparently the universe allows this and uses local time only.

Now, to all such local time axes exists a world, where that time is an
imaginary pseudo-scalar, which defines a perpendicular hyperplane, that
beings there would call 'space'.

Iow: there exists a multitude of spaces, which all have their own axis
of time.

And thes axes can have an angle towards other universes' times.


TH

see here (this is a link to my 'book')
https://docs.google.com/presentation/d/1Ur3_giuk2l439fxUa8QHX4wTDxBEaM6lOlgVUa0cFU4/edit?usp=sharing

[Richard Hachel]

Post by Thomas Heger
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.

R.H.

[Mikko]

Post by Richard Hachel

Post by Thomas Heger
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.

[Richard Hachel]

Post by Mikko

Post by Richard Hachel

Post by Thomas Heger
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.

At the beginning, in the time of Newton and Descartes, there was no
problem using this term, because we did not go too far, and
above all, because we did not travel very fast.

Today, telescopes see very far, much better (apparent speeds of supernovae
extensions), and above all, electrons and protons go very fast in particle
accelerators.

The word reference frame then becomes a dubious term, because if at the
beginning, we could represent a reference frame as four axes perpendicular
to each other, we know today from Dr. Hachel, that this notion is no
longer possible because of the time component. It is no longer possible to
conceive of the universe as a vast four-dimensional hypervolume, since
each point of the universe that composes it is at the base (at the origin
O) of its own hypervolume (because of the time component).

Thus, if the term "stationary frame" remains correct, the term referential
is only correct if we admit that we are talking about the referential
centered on a single object, and that a simple little metric translation
from wherever it starts destroys the referential to create another one.

R.H.

[Mikko]

Post by Richard Hachel

Post by Mikko

Post by Richard Hachel

Post by Thomas Heger
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.

[Richard Hachel]

Post by Mikko

Post by Richard Hachel

Post by Mikko

Post by Richard Hachel

Post by Thomas Heger
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.

Le terme pose problème en physique pure, et j'ai expliqué pourquoi.

De plus, la barrière des langues n'aide pas.

Je rappelle, en français, en physique courante, on appelle repère une
structure spatiale en 3D; et on appelle référentiel une structure
spatio-temporelle en 4D.

En milieu relativiste, le terme référentiel devient inutilisable tel que
les physiciens l'emploient.

Il ne peut servir, en temps que strucutre 4D aux axes perpendiculaires,
que pour une seule entité (cette fusée, cet observateur, ce cheval dans
ce pré) à l'exclusion de tout autre.

Autant un repère ne change pas ses axes, autant un référentiel a un
gros problème d'angularité des axes x,y,z, par rapport à l'axe t dès
qu'on effectue le moindre mouvement.

Roméo sur ce banc, Juliette sur cet autre, ne font PAS partie du MÊME
référentiel.

Chacun a le sien.

Alors qu'ils font partie de même repère.

R.H.

[Thomas Heger]

Post by Richard Hachel

Post by Thomas Heger
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.

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.

This is actually a continuus effect, which extends 'outwards' in space.

In space we see past events only and the further away the longer ago.

The future is not visible, mainly because future comes later than what
we call 'now'.

This 'now' is called 'hyperplane of the present' in relativity lingo.

In a typical spacetime diagram it is a horizontal plane, while the axis
of time points up vertically.

This axis of time is surrounded by our own futur light cone.

Now we have a consistent picture, which describes, what we mean by 'now'.

To reach such events now along the hyperplane of the present from our
own position is, of course, not possible.

But we could assume something hypothetical, which could and moves with
infinite velocity.

Since light isn't THAT fast (though fast), we need to compensate the
difference 'by hand'.

This is simple: since our signals actually move along our future light
cone, we need to subtract the delay and would reach a point on that
hyperplane of the present.

If we receive a signal, we would need to add the delay, because that
would 'raise' a point from our own past light-cone to that hyperplane.


TH

[Richard Hachel]

Post by Thomas Heger
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).

R.H.

[Maciej Wozniak]

Post by Richard Hachel

Post by Thomas Heger
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.

Depends of the details of the construction.
But doesn't really matter.

What matters is that it's going to either
indicate t'=t with the precision of an
acceptable error - or land in a trash can.

And that's because - apart of some religious
maniacs worshipping some insane crazie - nobody
wants clocks not indicating t'=t and nobody is
going to tolerate such clocks. And your precious
experiments can do nothing about that.

[Thomas Heger]

Post by Richard Hachel

Post by Thomas Heger
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).

There exist no 'center of the universe', because everything moves.

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.


TH

[Thomas Heger]

Post by Thomas Heger

Post by Richard Hachel

Post by Thomas Heger
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).

There exist no 'center of the universe', because everything moves.
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.

This is actually the reason, why 'big-bang-theory' must be wrong.
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.

Sure, but 'everywhere' was a single point.

I compare 'big-bang' to other sigularities, like a black hole at the
center of a galaxy.

This singularity isn't a single point at aall, but looks like.

Look at this webpage:

https://en.wikipedia.org/wiki/Light_cone

If you exchange 'observer' for 'big-bang' you would get a consistent
picture of the alleged beginning of time, if you would regard the
obersers future light cone as 'universe'.


TH

[Python]

Post by Thomas Heger

Post by Richard Hachel

Post by Thomas Heger
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).

There exist no 'center of the universe', because everything moves.
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.

This is actually the reason, why 'big-bang-theory' must be wrong.
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.

[snip demented nonsense]

[Richard Hachel]

Post by Thomas Heger
This 'now' is called 'hyperplane of the present' in relativity lingo.

This is a Hachel's term.

Except that for Hachel, it is not global, it is not universal to a given
stationary frame, and for the entirety of the frame.
Romeo and Juliet, each on their own bench, are in the same stationary
frame (3D), but they each have their own frame of reference (4D), that is
to say their own hyperplane of present time, of universal simultaneity.

Two simultaneous events in Romeo's universe are no longer simultaneous in
Juliet's universe.

It is this obviousness and this conceptual beauty that physicists refuse
to admit, while they easily admit the effects of the second degree (but
not the first) of the theory of relativity.

When Dr. Richard Hachel speaks to them of dyschronotropy by change of
inertial frame of reference (the three terms are semantic exact), they
understand him, and he acclaims its mathematical clarity and physical
relevance.

Hachel then speaks to them of universal anisochrony. They drown very
quickly as if they had fallen into the icy waters of the North Atlantic,
off the coast of Terre-Neuve.

R.H.

[Richard Hachel]

Post by Thomas Heger
In a typical spacetime diagram it is a horizontal plane, while the axis
of time points up vertically.

Absolutely.

Horizontal hyperplane.

HORIZONTAL HYPERPLANE OF PRESENT TIME. (Thomas Heger copyrights).

Thomas Heger Président 2024 !!!

Doctor Richard Hachel fourth Nobel prize !

"Horizontal hyperplane of proper simultaneity" est la base même de la
théorie de la relativité bien compris.

Et si l'on comprend que cette notion est relative entre Roméo et
Juliette, assis dans la même cour, mais pas sur le même banc, on a tout
COMPRIS de la théorie. TOUT en découle.

Les plus grands physiciens du monde n'y ont cependant compris que pouic.

La théorie, très belle en elle-même (peut être la plus belle de
l'histoire de l'humanité) a alors subi la catastrophe allemande
(Einstein, Minkowski) et le désastre scientifique qui s'en est suivi.

R.H.

[Python]

[nonsense answering nonsense]
see here (this is a link to my 'book')
https://docs.google.com/presentation/d/1Ur3_giuk2l439fxUa8QHX4wTDxBEaM6lOlgVUa0cFU4/edit?usp=sharing

This is an atrocious bunch of bullshit.

[Python]

Post by Thomas Heger
...

Post by Paul.B.Andersen
"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."
  "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.

You have no more excuses for this lie Thomas.

[Richard Hachel]

Post by Paul.B.Andersen
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)?

Si un jour je dis : "L'un des moutons du père François s'est
probablement égaré,
il se trouve dans le bosquet du lieu-dit "les Primevères".

Et que là, un crétin vient dire : "C'est quoi un mouton, c'est quoi un
bosquet, c'est quoi une primevère?"

Franchement, même avec une petite bite, on n'a plus envie de répondre.

Je laisse ça à YBM, il fait ça très bien.

R.H.

[Paul B. Andersen]

Post by Paul.B.Andersen
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)?

"One of Father François' sheep has probably got lost,
it is in the grove in the place called "les Primevères".
“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.

It may well be that your dick is bigger than mine,
but that doesn't answer my questions.

Post by Paul.B.Andersen
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.
tM(e1)=0 tM(e2)=1 and tM(e3)=2

Without proper definition of the terms, this equation is meaningless.
So please explain:
What are e1, e2 and e3?
What are tM(e1), tM(e2), and tM(e3)?

Your big dick won't help you.
You have to use the other end of your body.

[Mikko]

Post by Richard Hachel

Post by Mikko

Post by Richard Hachel
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.

That depends on your purposes. For some putposes it is sufficient, for
others a good start.

Post by Richard Hachel
Chacun lit ce que d'autres ont lu, et on tourne en rond.

Of course. Whenever possible, the author should be the first person to
read the text, then the co-authors if there are any, and the editors
and reviewers.

About our current topic, the best texts are well tested even before they
are written. Their authors have presented the material many times to their
students and have answered their students questions and that way have
learned how to present what they want to present, and also what to present
in the elementary book and what in an advanced book.

Post by Richard Hachel
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.

Of course, and that is done, even about topics that are covered by
ancient books; for all important and many other topics.

[J. J. Lodder]

Post by Richard Hachel
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.

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

[Richard Hachel]

Post by J. J. Lodder
Insufficient -for you-.

Yes

Post by J. J. Lodder
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.

Post by J. J. Lodder
Jan

R.H.

[gharnagel]

Post by Richard Hachel

Post by J. J. Lodder
Insufficient -for you-.

Yes

Post by J. J. Lodder
For Einstein and Poincare it sufficed,
because it is so blindingly obvious.
No competent reader has had any problem with it, ever since,

I was required to write two high school term papers, one in English
and one in Physics. They were both on relativity and each earned
an A, but they were pretty naive. I've never taken a relativity
class and I never got straightened out about it until I joined a
discussion group several years ago. Tom Roberts was a big help,
but relativity deniers were also (I had to study and learn to be
able to answer their objections).

Post by Richard Hachel
Which is still a big problem in the history of humanity, but it
seems obvious to me that it will necessarily be corrected.

Although I found Saint Albert's papers a bit obtuse in my naive
years, I don't find them so now. I understand what he was trying
to say. The answer to the problem is to read copiously and try
to understand the criticism you receive. IOW, be humble (i.e.,
teachable).

“Education isn’t something you can finish.” – Isaac Asimov

And if you think you've finished, you're finished.

“Changelessness is decay.” – Isaac Asimov

Post by Richard Hachel
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.

“spacetime is likely to be an approximate description of something quite
different.” – Steven Carlip

[Richard Hachel]

Post by gharnagel

Post by Richard Hachel
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.

R.H.

[gharnagel]

Post by Richard Hachel

Post by gharnagel

Post by Richard Hachel
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.

Absolutely so, just like Newtonian physics is an approximation to
reality, just as SR is an approximation to reality, just as GR is.

Post by Richard Hachel
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.

That has nothing to do with Minkowski. It comes from SR whole cloth.

https://en.wikipedia.org/wiki/Space_travel_under_constant_acceleration#:~:text=The%20distance%20traveled%2C%20under%20constant%20proper%20acceleration%2C%20from,9%5D%20where%20c%20is%20the%20speed%20of%20light.

That is, it comes from the LT equations applied to the problem.
I must regretfully inform you that your calculation is incorrect.

[Paul.B.Andersen]

Post by 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?

In any case, this is obviously impossible in the real world,
and can never be verified.

So why not use a scenario which is possible in the real word?

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.

Ginette is always stationary on the Equator.
Elise is travelling eastwards at low altitude in an aeroplane.
Wanda is travelling westwards at low altitude in an aeroplane.
Both are travelling once around Earth at equator.

Note that the altitude is so low that the gravitational
blue shift can be ignored.

From the time they are co-located, to they again are co-located
after Elise's and Wanda's journey, Ginette's clock shows that
the duration of their journey is τ_G = two sidereal days.

Please find what the duration of the journey will be
measured by Elise and Wanda, τ_E and τ_W.

Some data:
Circumference of Earth at equator L = 40075 km
Sidereal day Tday = 86164.0905 s
Ginette's speed in the non rotating Earth centred frame of reference
(ECI frame), v = L/Tday = 465.1 m/s

SR predicts: τ_E − τ_G = −259.2 ns, τ_W − τ_G = +155.5 ns

Note that τ_W - τ_E = 414.7 ns, Wanda ages 414.7 ns more than Elise
during their journey around the Earth

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.

[Richard Hachel]

Post by Paul.B.Andersen

Post by 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.

[Paul.B.Andersen]

Post by Richard Hachel

Post by 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 ","
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.

Ginette is always stationary on the Equator.
Elise is travelling eastwards at low altitude in an aeroplane.
Wanda is travelling westwards at low altitude in an aeroplane.
Both are travelling once around Earth at equator.

Note that the altitude is so low that the gravitational
blue shift can be ignored.

From the time they are co-located, to they again are co-located
after Elise's and Wanda's journey, Ginette's clock shows that
the duration of their journey is τ_G = two sidereal days.
(Ginette's clock is a UTC clock on the geoid)

Please find what the duration of the journey will be
measured by Elise and Wanda, τ_E and τ_W.

Some data:
Circumference of Earth at equator L = 40075 km
Sidereal day Tday = 86164.0905 s
Ginette's speed in the non rotating Earth centred frame of
reference (ECI frame) is v = L/Tday = 465.1 m/s

SR predicts: τ_E − τ_G = −259.2 ns, τ_W − τ_G = +155.5 ns

Note that τ_W - τ_E = 414.7 ns, Wanda ages 414.7 ns more than Elise
during their journey around the Earth

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.

[Maciej Wozniak]

Post by Paul.B.Andersen

Post by Richard Hachel

Post by 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 ","
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!
------------------------------------------------
The triplets Ginette, Elise and Wanda are co-located on
the equator. They all have an atomic clock.

Fortunately, we have GPS now. With its
atomic clocks it's demonstrating very
clearly that your delusional scenarios
have nothing in common with real
observations, real atomic clocks or
real anything.

[Python]

Post by Richard Hachel

Post by gharnagel

Post by Richard Hachel
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.

Just because you said so? As a matter of fact your claim about
accelerated travelers can be shown in contradiction with the Principle
of Relativity.

Post by Richard Hachel
There can be no photography.

This is not, afaik, a English proper expression. The French "Il n'y a
pas photo" can be translated as "There is no possible contest".

You're right: there is no possible contest that your claims are WRONG.

[Richard Hachel]

Post by gharnagel
Perhaps, but Minkowski spacetime will remain an approximation to
reality.
“spacetime is likely to be an approximate description of something quite
different.” – Steven Carlip

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.

Everything is connected.

R.H.

[Paul.B.Andersen]

Post by Richard Hachel
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.
"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.

Since Doctor Hachel claims that Python and Paul claim that addition
and multiplication are the same, Doctor Hachel must obviously be
much smarter than Python and Paul.

Elementary Hachelian logic!

Post by Richard Hachel
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.

Quite.
SR and GR don't work in the Hachelian fantastic relativistic world.
SR even claims that clocks can be synchronous in an inertial frame,
and that the speed of light is invariant and not depending on
the position of Richard Hachel!

In the real world, OTOH, is SR and GR thoroughly tested and
never falsified.

https://paulba.no/paper/index.html

But of course, experimental evidence in the real world has nothing
to do with what can and can't happen in the Hachelian world.

[Richard Hachel]

Post by Paul.B.Andersen
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.

[Paul.B.Andersen]

Post by Richard Hachel

Post by Paul.B.Andersen
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.

Joking about your divinity won't make the experimental evidence go away.

https://paulba.no/div/index.html

Don't read it. You might loose your faith!

[Mikko]

Post by Richard Hachel

Post by J. J. Lodder
Insufficient -for you-.

Yes

Post by J. J. Lodder
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.

Do you find any problem with Augustinian belief?

[Athel Cornish-Bowden]

Post by Richard Hachel
When I read the contributors to the French and Anglo-Saxon

You can write "anglosaxon" in French if you insist, but the name of the
language in English is "English".

Post by Richard Hachel
forums, when I read Einstein (three lines of explanation) or Poincaré
(one line of explanation), I realize that it is very insufficient.

[ … ] (skip the usual claptrap)

Post by Richard Hachel
--

athel -- biochemist, not a physicist, but detector of crackpots