Re: The problem of relativistic synchronisation

Den 30.08.2024 00:15, skrev Richard Hachel:

"You" is probably Python. Richard never quote what he is
referring to, and he doesn't define the events e1, e2, e3.
But we know:
e1 is the event that light is emitted from A
e2 is the event that light is reflected from B
e3 is the event that the reflected light hits A
 Note e1 and e3 are happening at A, e2 is happening at B.
 

 We can then pose without fear: tA(e3)-tA(e1)=2AB/c

 tA(e3) is the reading of a clock at A at the event e3
tA(e1) is the reading of a clock at A at the event e1
 Einstein says:
"We assume the quantity 2AB/(tA(e3)-tA(e1)) = c
  to be a universal constant—the velocity of light in empty space."
 It is a postulate in SR that say the speed of light
in vacuum is constant and invariant, and his paper is about
the consequence of the postulates, so of course he assumes that.
 It is thoroughly experimentally verified that the speed
of light indeed is constant and invariant.
So we _know_ that tA(e3)-tA(e1)= 2AB/c
 

Then, admitting that A warns of e1 and e3, either with photons or with slugs of the same speed, any point M of the stationary frame of reference, we have yet another tautology:

 This statement doesn't parse.
 

 tM(e3)-tM(e1)= tA(e3)-tA(e1) = 2AB/c

 tM can only be the reading of a clock at the point M,
which is an arbitrary stationary point in the frame where
A and B are stationary.
 But the events e1 and e3 happen at A, and not on M,
so tM(e3)-tM(e1) is meaningless.
 Richard doesn't seem to know what an event is.
 

For the moment we cannot say more about the speed of light between A and B in the direction AB,
nor in the BA sense.

 Above Richard say the posts on relativistic synchronization
between two points A and B are interesting, but so far he
has said nothing about synchronization.
 The equation:
   tB(e2) - tA(e1) = tA(e3) - tB(e2)
is Einstein's _definition_ of synchronism and simultaneity.
If this equation is true, then the clocks are synchronous
in the frame where A and B are stationary.
 

 On this, we breathe we breathe, Einstein does not seem to agree with Hachel. For Einstein, the question does not arise, and it seems certain that t(AB)=t(BA).

 What are you saying? :-D
 Einstein says:
"We have not defined a common “time” for A and B, for the latter
  cannot be defined at all unless we establish _by definition_
  that the “time” required by light to travel from A to B equals
  the “time” it requires to travel from B to A."
 And:
  "In accordance with definition the two clocks synchronize if
      tB(e2) - tA(e1) = tA(e3) - tB(e2)"
   tB(e2) - tA(e1) is the time the light uses to go from A to B
  tA(e3) - tB(e2) is the time the light uses to go from B to A
 Einstein _defines_ that the clocks simultaneously show
the same (are synchronous in the stationary system)
if the time the light uses to go from A to B equals
the time the light uses to go from B to A.
 

 Except that this is no longer true in an anisochronous environment, and that our universe is not "a 4D hyperplane of absolute simultaneity, even for a simple inertial frame of reference".
We can then propose A synchronization based on A hyperplane of simultaneity, but we must propose THE appropriate candidate, and it can obviously be neither A nor B.
So we continue from there.
We can then propose a synchronization of A and B by M (and we will have a synchronization of type M).

 SR doesn't depend on the definition of simultaneity
(but everything would be very awkward without it)
so another definition is possible.
 But in the real world no other than Einstein's definition
would work, so no one would use your alternative definition.
 

 Can I do it without laughing, and how?
Note that if M is purely in a perpendicular position on the secant of the middle AB, then whatever the speed of the information (c in both directions for Einstein, both c/2 or ∞ depending on the meaning for Hachel), reception of the sync signal transmitted by M will be simultaneous in reception A and B for M, and also simultaneous in return for M.
 We can therefore at risk, pose that, for M after synchronization A and B on its part (but ONLY for M):
tM(e1)=0 tM(e2)=1 and tM(e3)=2

 Without a proper definition of the terms, this is nonsense.
What are e1, e2 and e3?
What are tM(e1), tM(e2), and tM(e3)?
 

 The timing is perfect for Mr.
 We can, by imagining an imaginary point M, placed very far away and perpendicular to all the points of the stationary universe studied, with perfect synchronization of type To, and perfect for all
the repository.

 Richard, this is mindless babble.
 _In the real world_ we have two equal clocks at two points A and B.
Please explain _exacly_ how you will go about to synchronise them
according to _your_ definition.
 Imaginary points won't do. Everything must be _real_.
 Einstein could do it. Can you?
 

 By change of inertial reference frame, M becomes M' and To becomes To', because the chronotropy becomes reciprocally relative.
Everything is said. The basics are given. RR then becomes very simple and the equations that go with it obvious.

 

It remains to be made understood, and it is not by acting like a monkey that anyone will understand me, or by giving me stupid answers like: "The speed of light takes one second to come from the moon to 'here".

 You mean like the stupid answer you gave me?
 |Den 26.08.2024 13:02, skrev Richard Hachel:>
|> There is a one-second time difference between 00:00:08 and 00:00:07",
|> between my watch and the watch on the moon.
|> This is because of the speed of light, which is quite slow,
|> and takes at least a second to reach me.
|>
 Shot yourself in the foot, Richard?
 

 This is profoundly stupid for Bedouin relativists, that.
 But it seems that people like it, and that we can even decorate it with ridiculous smileys.

 

R.H.