Python wrote:

> Le 21/04/2024 à 15:53, Ross Finlayson a écrit :
>> Can you help further explain for the rest of us why this isn't
>> necessarily the usual interpretation or why it sort of doesn't arrive
>> at the same results of some of the usual thought experiments like the
>> traveling twins?
>
> In the twins scenario, twins does not share the same space-time path.

no need to. The observer just need to center its position in time and
space, where both paths would output the same.

here this guy is telling the truth. And it's not a mistake.

𝗕𝗶𝗱𝗲𝗻_𝗮𝘀𝗸𝘀_𝗔𝗺𝗲𝗿𝗶𝗰𝗮𝗻𝘀_𝘁𝗼_𝗰𝗵𝗼𝗼𝘀𝗲_𝗯𝗲𝘁𝘄𝗲𝗲𝗻_𝗳𝗿𝗲𝗲𝗱𝗼𝗺_𝗮𝗻𝗱_𝗱𝗲𝗺𝗼𝗰𝗿𝗮𝗰𝘆
The president’s latest blunder comes as a majority of Americans doubt his
mental capacity for office
https://www.r%74.com/news/596344-biden-freedom-democracy-gaffe/

He spoke the truth. The devil made him do it.

Freedom or Democracy the choice given by the Fascist government

American people have neither Freedom or Democracy what they have is a
Zionist mafia robbing the country blind.

He is so true..yankee are not free but slave of zionists

W dniu 21.04.2024 o 16:48, Python pisze:
> Le 21/04/2024 à 15:53, Ross Finlayson a écrit :
>> On 04/20/2024 09:52 AM, Tom Roberts wrote:
>>> On 4/13/24 1:36 AM, Richard Hachel wrote:
>>>> "If two different observers travel an identical path in equal
>>>> observable times, then their proper times will necessarily be equal.
>>>
>>> Yes, of course.
>>>
>>> My response differs from others because I interpret your context
>>> differently.
>>>
>>> Since you mention "proper times", your context must be relativity; it
>>> does not matter whether SR or GR, because "travel an identical path"
>>> means they travel along a single worldline through spacetime -- i.e.
>>> they are always co-located and co-moving, so of course their elapsed
>>> proper times are equal (counting from any event on their worldline).
>>>
>>>      Your "in equal observable times" is redundant. For any
>>>      observer this directly follows from them following the
>>>      same worldline through spacetime.
>>>
>>> Note this is essentially the first time I agree with Hachel. I doubt
>>> that he understands why what he wrote is actually correct, because he
>>> followed it with a bunch of obfuscatory nonsense.
>>>
>>>> [... enormous amount of gibberish ignored.]
>>>
>>> Tom Roberts
>>
>> Can you help further explain for the rest of us why
>> this isn't necessarily the usual interpretation or
>> why it sort of doesn't arrive at the same results of
>> some of the usual thought experiments like the traveling twins?
>
> In the twins scenario, twins does not share the same
> space-time path.

Samely like in Star Wars scenario Jedi knighrs wave their
lightsabers.

Le 21/04/2024 à 16:48, Python a écrit :
> Le 21/04/2024 à 15:53, Ross Finlayson a écrit :
>> On 04/20/2024 09:52 AM, Tom Roberts wrote:
>>> On 4/13/24 1:36 AM, Richard Hachel wrote:
>>>> "If two different observers travel an identical path in equal
>>>> observable times, then their proper times will necessarily be equal.
>>>
>>> Yes, of course.
>>>
>>> My response differs from others because I interpret your context
>>> differently.
>>>
>>> Since you mention "proper times", your context must be relativity; it
>>> does not matter whether SR or GR, because "travel an identical path"
>>> means they travel along a single worldline through spacetime -- i.e.
>>> they are always co-located and co-moving, so of course their elapsed
>>> proper times are equal (counting from any event on their worldline).
>>>
>>>      Your "in equal observable times" is redundant. For any
>>>      observer this directly follows from them following the
>>>      same worldline through spacetime.
>>>
>>> Note this is essentially the first time I agree with Hachel. I doubt
>>> that he understands why what he wrote is actually correct, because he
>>> followed it with a bunch of obfuscatory nonsense.
>>>
>>>> [... enormous amount of gibberish ignored.]
>>>
>>> Tom Roberts
>>
>> Can you help further explain for the rest of us why
>> this isn't necessarily the usual interpretation or
>> why it sort of doesn't arrive at the same results of
>> some of the usual thought experiments like the traveling twins?
>
> In the twins scenario, twins does not share the same
> space-time path.

We may not have the same space-time path, and have the same proper time,
the same path, and the same improper time (for other observers).
Note that "space-time path", I don't understand the geometric concept very
well.
Relativist theorists must say clear things if they want to be understood.
However, I see that their geometry is not entirely clear and easily
understandable. Above all, I see results that are as false as they are
strange appearing at the end of the race. All this is not normal and
deserves more serious reflection.

R.H.

Le 21/04/2024 à 23:07, Richard Hachel a écrit :
> Le 21/04/2024 à 16:48, Python a écrit :
>> Le 21/04/2024 à 15:53, Ross Finlayson a écrit :
>>> On 04/20/2024 09:52 AM, Tom Roberts wrote:
>>>> On 4/13/24 1:36 AM, Richard Hachel wrote:
>>>>> "If two different observers travel an identical path in equal
>>>>> observable times, then their proper times will necessarily be equal.
>>>>
>>>> Yes, of course.
>>>>
>>>> My response differs from others because I interpret your context
>>>> differently.
>>>>
>>>> Since you mention "proper times", your context must be relativity; it
>>>> does not matter whether SR or GR, because "travel an identical path"
>>>> means they travel along a single worldline through spacetime -- i.e.
>>>> they are always co-located and co-moving, so of course their elapsed
>>>> proper times are equal (counting from any event on their worldline).
>>>>
>>>>      Your "in equal observable times" is redundant. For any
>>>>      observer this directly follows from them following the
>>>>      same worldline through spacetime.
>>>>
>>>> Note this is essentially the first time I agree with Hachel. I doubt
>>>> that he understands why what he wrote is actually correct, because he
>>>> followed it with a bunch of obfuscatory nonsense.
>>>>
>>>>> [... enormous amount of gibberish ignored.]
>>>>
>>>> Tom Roberts
>>>
>>> Can you help further explain for the rest of us why
>>> this isn't necessarily the usual interpretation or
>>> why it sort of doesn't arrive at the same results of
>>> some of the usual thought experiments like the traveling twins?
>>
>> In the twins scenario, twins does not share the same
>> space-time path.
>
> We may not have the same space-time path, and have the same proper time,
> the same path, and the same improper time (for other observers).

To have the same improper time is meaningless because in a given
frame of reference there is a single time elapsed between any
two events. And there are only two events : travelers' depart
and travelers' meeting again.

Moreover to have the same path (in the spatial sense) is a
property that is only true in a single frame of reference.

It makes no sense (outside of a trivial one in Galilean Relativity)
to claim that a condition depending on a given chosen frame could
imply a property (equality of proper times) that does not depend
on that.

> Note that "space-time path", I don't understand the geometric concept
> very well.

It's not a big deal! It is the set of events "traveler is at this
position at that time".

> [snip babbling]

Your claim is directly violating the principle of Relativity. End of
Story.

Le 20/04/2024 à 19:07, Python a écrit :
>> On 4/13/24 1:36 AM, Richard Hachel wrote:
>>> "If two different observers travel an identical path in equal
>>> observable times, then their proper times will necessarily be equal.
>>
>> Yes, of course.
>>
>> My response differs from others because I interpret your context
>> differently.
>
> You interpret path as a path in space-time while Hachel is considering
> path in space only (so it is true in only ONE reference frame)
>
> For instance, he claims that if two travelers would go from Earth
> to Tau Ceti, one at constant velocity w.r.t Earth the other one
> with a constant acceleration both will measure the same (proper)
> time for their trips.

Yes, this is entirely correct.

If two different protagonists use different frames of reference to join
Tau Ceti (12al), and if we notice that in the terrestrial frame of
reference the travel times are similar and the path is similar (direct
path), then the two proper times will be similar.
This is the case of a trip by Stella in Galilean mode at constant speed
Vo=0.929c, and of a trip in accelerated mode by Bella
(10m/s²=1.052ly/y²).
The duration of the trip, which is also very well calculated by my
relativistic theoretician friends around the world, will be To=x/Vo in the
first case, and To=(x/c).sqrt(1+2c²/ax) in the second case, i.e. 12,915
years.
But this duration will also be the same for the two rockets with regard to
their own time. Tr=4.776 years.
Physicists tell me that while this is true for Stella, it is not true for
Bella.
However, here, it is they who are mistaken and misuse space-time concepts
in an accelerated environment.
On the other hand, be careful, for this to be true, and for it to remain
true, two clarifications:
1. You must only use one reference frame during the entire trip.
2. the accelerated twin must go into “native” state, i.e. at rest.

R.H.

Le 21/04/2024 à 23:31, Richard Hachel a écrit :
> Le 20/04/2024 à 19:07, Python a écrit :
>>> On 4/13/24 1:36 AM, Richard Hachel wrote:
>>>> "If two different observers travel an identical path in equal
>>>> observable times, then their proper times will necessarily be equal.
>>>
>>> Yes, of course.
>>>
>>> My response differs from others because I interpret your context
>>> differently.
>>
>> You interpret path as a path in space-time while Hachel is considering
>> path in space only (so it is true in only ONE reference frame)
>>
>> For instance, he claims that if two travelers would go from Earth
>> to Tau Ceti, one at constant velocity w.r.t Earth the other one
>> with a constant acceleration both will measure the same (proper)
>> time for their trips.
>
> Yes, this is entirely correct.
>
> If two different protagonists use different frames of reference to join
> Tau Ceti (12al), and if we notice that in the terrestrial frame of
> reference the travel times are similar and the path is similar (direct
> path), then the two proper times will be similar.

The equality of travel times in any frame is always true. There is
a single value for the time elapsed between two events. You are
ridiculous, Richard.

The path being the same is only true in a single inertial frame,
and false in all others.

It is logically unsound that a property that depends on a
chosen frame of reference could imply a property that does
not. This is basic logic.

Moreover It is obvious that your claim is violating the principle of
Relativity: just consider the situation from the frame of reference of
the inertial traveler.

This is quite pathetic that you're unable to grasp such a
simple argument.

Le 21/04/2024 à 23:31, Richard Hachel a écrit :
> [snip babbling]
> 1. You must only use one reference frame during the entire trip.

"use" a reference frame... sigh...

> 2. the accelerated twin must go into “native” state, i.e. at rest.

At rest with respect to what? Absolute space again! You are
insufferable.

Le 21/04/2024 à 23:19, Python a écrit :
>> We may not have the same space-time path, and have the same proper time,
>> the same path, and the same improper time (for other observers).
>
> To have the same improper time is meaningless because in a given
> frame of reference there is a single time elapsed between any
> two events. And there are only two events : travelers' depart
> and travelers' meeting again.
>
> Moreover to have the same path (in the spatial sense) is a
> property that is only true in a single frame of reference.
>
> It makes no sense (outside of a trivial one in Galilean Relativity)
> to claim that a condition depending on a given chosen frame could
> imply a property (equality of proper times) that does not depend
> on that.
>
>> Note that "space-time path", I don't understand the geometric concept
>> very well.
>
> It's not a big deal! It is the set of events "traveler is at this
> position at that time".
>
>> [snip babbling]
>
> Your claim is directly violating the principle of Relativity. End of
> Story.

Having the same improper time is on the contrary an important relativistic
concept, as important as having the same proper time.

For example, in the example of the two travelers from Tau Ceti, Stella and
Bella, one in Galilean mode (Vo=0.929c), the other in accelerated mode
(a=1.052ly/y²), we notice that the The observer Terrence (terrestrial)
will give them the same improper time. That is 12,915 years.

This means that they leave together, make the same journey, and arrive
together, although the journey was never joint at any point of the
journey.

This is a point that all physicists in the world accept (I hope).

It is on Bella's notion of proper time that we no longer agree because the
way physicists go about it in accelerated frames of reference does not
seem entirely correct or logical to me.

R.H.

Le 21/04/2024 à 23:39, Python a écrit :
>
> The equality of travel times in any frame is always true.

? ? ?

R.H.

Le 21/04/2024 à 23:42, Richard Hachel a écrit :
> Le 21/04/2024 à 23:19, Python a écrit :
>>> We may not have the same space-time path, and have the same proper
>>> time, the same path, and the same improper time (for other observers).
>>
>> To have the same improper time is meaningless because in a given
>> frame of reference there is a single time elapsed between any
>> two events. And there are only two events : travelers' depart
>> and travelers' meeting again.
>>
>> Moreover to have the same path (in the spatial sense) is a
>> property that is only true in a single frame of reference.
>>
>> It makes no sense (outside of a trivial one in Galilean Relativity)
>> to claim that a condition depending on a given chosen frame could
>> imply a property (equality of proper times) that does not depend
>> on that.
>>
>>> Note that "space-time path", I don't understand the geometric concept
>>> very well.
>>
>> It's not a big deal! It is the set of events "traveler is at this
>> position at that time".
>>
>>> [snip babbling]
>>
>> Your claim is directly violating the principle of Relativity. End of
>> Story.
>
> Having the same improper time is on the contrary an important
> relativistic concept, as important as having the same proper time.

Idiot! There are only two events involved! One for when the travelers
departs and another one when they meet again.

Hence a SINGLE time elapsed between these two events, as seen in any
frame of reference.

Such a triviality about a single value being equal to itself is not
"important" Richard. It is just showing how silly you are!

Le 21/04/2024 à 23:44, Richard Hachel a écrit :
> Le 21/04/2024 à 23:39, Python a écrit :
>>
>> The equality of travel times in any frame is always true.
>
> ? ? ?

Oh come on!

There are only two events involved! One for when the travelers
departs and another one when they meet again.

Hence a SINGLE time elapsed between these two events, as seen in any
frame of reference.

You seemed to have, at least, understood this recently. It appears
you don't. You are insufferable.

Le 21/04/2024 à 23:42, Python a écrit :
> Le 21/04/2024 à 23:31, Richard Hachel a écrit :
>> [snip babbling]
>> 1. You must only use one reference frame during the entire trip.
>
> "use" a reference frame... sigh...
>
>> 2. the accelerated twin must go into “native” state, i.e. at rest.
>
> At rest with respect to what? Absolute space again! You are
> insufferable.

Au repos par rapport au référentiel terrestre qui a permis de mesurer
une distance de 12 ly, entre la terre et Tau Ceti.

Je te rappelle que si tu comprends bien la RR du docteur Hachel (c'est
moi) les notions de longueurs et de distances sont relatives.

Les 12 ly sont mesurées dans le référentiel terrestre et sont une
distance propre mesurée par celui-ci.

Je te rappelle qu'une fusée qui croise la terre à cet instant, et se
dirige vers Tau Ceti à Vo=0.8c (Vapp=4c) VOIT Tau Ceti à 36 ly.

Il faut donc que la fusée qui part (Bella) ait un départ au repos pour
que l'équation soit correcte, et que les temps propres qui sont corrects
chez moi, soient égaux entre Bella et Stella (4.776 ans).

R.H.

Le 21/04/2024 à 23:47, Python a écrit :
> Oh come on!
>
> There are only two events involved! One for when the travelers
> departs and another one when they meet again.
>
> Hence a SINGLE time elapsed between these two events, as seen in any
> frame of reference.
>
> You seemed to have, at least, understood this recently. It appears
> you don't. You are insufferable.

Breathe in, exhale.

Breathe.

When Bella and Stella make their trip, they take the same amount of time.
Everyone agrees on that.

The calculated time is the same for all physicists in the world, French,
American, German, Spanish, English, and even Richard Hachel.

Here, 12,915 years old.

I call this time improper time because it does not correspond to the
proper time of the two rockets and in any system of relativistic Axelian
or Minkowskian calculations.

It goes without saying that it defines the time between the joint
departure event of the two rockets and the joint arrival of the two
rockets.

We have entered into tautology there.

R.H.

Le 22/04/2024 à 00:01, Richard Hachel a écrit :
> Le 21/04/2024 à 23:47, Python a écrit :
>> Oh come on!
>>
>> There are only two events involved! One for when the travelers
>> departs and another one when they meet again.
>>
>> Hence a SINGLE time elapsed between these two events, as seen in any
>> frame of reference.
>>
>> You seemed to have, at least, understood this recently. It appears
>> you don't. You are insufferable.
>
> [snip babbing]
> It goes without saying that it defines the time between the joint
> departure event of the two rockets and the joint arrival of the two
> rockets.
>
> We have entered into tautology there.

You are the one insisting as taking an always true condition of the
form a = a as a precondition. This is utterly stupid.