Le 13/11/2022 à 17:57, Volney a écrit :
> On 11/13/2022 11:51 AM, Richard Hachel wrote:
>> Le 13/11/2022 à 17:38, Volney a écrit :
>>> On 11/13/2022 7:34 AM, Richard Hachel wrote:
>>>
>>>> I repeat it once again (even I have a feeling it's going to be useless):
>>>>
>>>> They will cross the same space, and arrive at the same time.
>>>>
>>>> I didn't say "the same space-time".
>>>
>>> And that's your mistake.
>>>
>>> They don't follow the same paths through spacetime.
>>
>> Et mes couilles, elles suivent le même chemin, dans l'espace-temps?
>>
>> J'ai beau les supplier d'arrêter.
>>
>> Ils continuent.
>>
>> Que puis-je faire?
>>
>> Je les supplie, une fois, deux fois, trois fois, d'arrêter...
>>
>> Ils continuent...
>>
>> Mille fois, sic mille ans...
>> Quoi faire?
>>
>> Suis-je Superman?
>> Non, je ne suis pas Superman. Je ne viens pas de Krypton, et je ne vole
>> pas dans les airs.
>>
>> Cela, ils ne le comprennent pas.
>> Merde, c'est pourtant pas con à comprendre.
>> Ils veulent quoi, les hommes, un Messie qui les soumets à grandes
>> claques dans la gueule, à coup de Knout dans les caves nazies?
>> Oui, je le dis avec des larmes dans les yeux, c'est ça qu'ils veulent,
>> les bandes de pauvres cons.
>>
>> Bien sûr que oui.
>> R.H.
>>
>>
>>
> So you agree they don't take the same paths through spacetime.

Yes my love.

That's what I keep saying.

They follow the same path, in space.

But not in space-time, since I begged millions of physicists to tell the
difference (but they're too dumb, it's horrible).

Shit!

I have been begging people for years to be very careful with the words,
and the dogmas they use.

They spit in my face.

So yes, my love, that's what I say, the same path.

Not the same "4D trajectory" that I don't understand, because I don't care
about your bullshit of bullshit.

I am talking about the same spatial trajectory in the observing frame of
reference.

The other bullshit and stupidity, I don't care.

R.H.

Richard Hachel wrote:
>
> Le 13/11/2022 à 06:05, Tom Roberts a écrit :
> > On 11/12/22 3:42 AM, Richard Hachel wrote:
> >> "If two relativistic mobiles travel, in equal observable times,
> >> identical paths, their proper times will remain equal."
> >>
> >> It's a very simple sentence that I think is quite correct.
> >
> > No. There are too many undefined terms in it.
>
> In Einstein description, yes.
>
> In mine no.
>
> "If two relativistic mobiles travel, in equal observable times,
> identical paths, their proper times will remain equal."
>
> R.H.

but you began the sentence with "If..."

that means everything after the word "If" is meaningless. It does not
agree with Nature.

Any 3 year old can observe objects traveling on a stream of water and
see there is no "identical paths".

The paper boat travels alone
On it's own path
At it's own time.

Even my two watches travel on different paths in time.

--
The Starmaker -- To question the unquestionable, ask the unaskable,
to think the unthinkable, mention the unmentionable, say the unsayable,
and challenge
the unchallengeable.

Den 13.11.2022 13:26, skrev Richard Hachel:
> Le 13/11/2022 à 09:45, "Paul B. Andersen" a écrit :
>>
>> Another, not equally vacuous, "Hachel's principle":
>>
>> On Saturday, November 5, Richard Hachel wrote:
>>
>>   "For the trajectory t=sqrt(x^2 + 2x/a)
>>    we have  dx/dt = [1+c²/2ax]^(-1/2)"
>>
>> :-D
>
> I didn't write that. >
> For these two equations, I wrote, and specifying well:
>
> To=(x/c).sqrt(1+2c²/ax)

Quite.
t = (x/c).sqrt(1+2c²/ax) or with c = 1: t=sqrt(x^2 + 2x/a)

>
> I am talking about the terrestrial observable time for an object
> accelerated according to a, and over a chosen distance x.
>
> I then wrote that the instantaneous observable speed of the object
> (rocket or particle) was, at any point of the journey:
> dx/dt = [1+c²/2ax]^(-1/2)

And with c = 1
dx/dt = [1+1/2ax]^(-1/2)

So according to Hachel:
"For the trajectory t = (x/c).sqrt(1+2c²/ax)
we have dx/dt = [1+c²/2ax]^(-1/2)"

or if you prefer:
"For the trajectory t=sqrt(x^2 + 2x/a)
we have dx/dt = [1+1/2ax]^(-1/2)"

This alone proves that you are talking nonsense.
There is nothing more to discuss.

And you know why, you have been told many times,
this is elementary calculus!

--
Paul

https://paulba.no/

On Sunday, November 13, 2022 at 4:09:33 AM UTC-8, Richard Hachel wrote:
> Le 13/11/2022 à 06:05, Tom Roberts a écrit :
> > On 11/12/22 3:42 AM, Richard Hachel wrote:
> >> "If two relativistic mobiles travel, in equal observable times,
> >> identical paths, their proper times will remain equal."
> >>
> >> It's a very simple sentence that I think is quite correct.
> >
> > No. There are too many undefined terms in it.
> In Einstein description, yes.
>
> In mine no.

Yes, you used the following undefined terms:

"equal observable times"
"identical paths"

Your sentence has no meaning until you state precisely what those
terms mean.

> "If two relativistic mobiles travel, in equal observable times,
> identical paths, their proper times will remain equal."

Yes, as stated it's gobbledygook. You need to define the
undefined terms.

--
Jan

On Sunday, November 13, 2022 at 9:13:07 AM UTC-8, Richard Hachel wrote:
> Le 13/11/2022 à 17:57, Volney a écrit :
> > On 11/13/2022 11:51 AM, Richard Hachel wrote:
> >> Le 13/11/2022 à 17:38, Volney a écrit :
> >>> On 11/13/2022 7:34 AM, Richard Hachel wrote:
> >>>
> >>>> I repeat it once again (even I have a feeling it's going to be useless):
> >>>>
> >>>> They will cross the same space, and arrive at the same time.
> >>>>
> >>>> I didn't say "the same space-time".
> >>>
> >>> And that's your mistake.
> >>>
> >>> They don't follow the same paths through spacetime.
> >>
> >> Et mes couilles, elles suivent le même chemin, dans l'espace-temps?
> >>
> >> J'ai beau les supplier d'arrêter.
> >>
> >> Ils continuent.
> >>
> >> Que puis-je faire?
> >>
> >> Je les supplie, une fois, deux fois, trois fois, d'arrêter...
> >>
> >> Ils continuent...
> >>
> >> Mille fois, sic mille ans...
> >> Quoi faire?
> >>
> >> Suis-je Superman?
> >> Non, je ne suis pas Superman. Je ne viens pas de Krypton, et je ne vole
> >> pas dans les airs.
> >>
> >> Cela, ils ne le comprennent pas.
> >> Merde, c'est pourtant pas con à comprendre.
> >> Ils veulent quoi, les hommes, un Messie qui les soumets à grandes
> >> claques dans la gueule, à coup de Knout dans les caves nazies?
> >> Oui, je le dis avec des larmes dans les yeux, c'est ça qu'ils veulent,
> >> les bandes de pauvres cons.
> >>
> >> Bien sûr que oui.
> >> R.H.
> >>
> >>
> >>
> > So you agree they don't take the same paths through spacetime.
> Yes my love.
>
> That's what I keep saying.
>
> They follow the same path, in space.
>
> But not in space-time,

OK, so your statement from your first post is false then.

> since I begged millions of physicists to tell the
> difference (but they're too dumb, it's horrible).

Stop fantasising about experts' stupidity. You'll get
nowhere using this method of discourse.

--
Jan

Le 13/11/2022 à 21:55, "Paul B. Andersen" a écrit :
> Den 13.11.2022 13:26, skrev Richard Hachel:

>> To=(x/c).sqrt(1+2c²/ax)

I confirm.

This is the observable time (or improper time, or terrestrial time) that
we will note according to the length of the path of a rocket or a particle
and according to its acceleration.

That's what I found when writing what I thought was right, and it's
exactly the same formula that I found in books dealing with relativity.

Which I think, makes it credible since through different ideas and paths,
we find an identical formula.

On the other hand, for proper tenses my ideas and mametry are
fundamentally different.

> So according to Hachel:
> "For the trajectory t = (x/c).sqrt(1+2c²/ax)
> we have dx/dt = [1+c²/2ax]^(-1/2)"

That's not exactly what I said.

I said that the instantaneous observable speed in x, was given by the
relation dx/dt = Vo/c(inst)=[1+c²/2ax]^(-1/2)

Which is a much lower speed than predicted by relativists.

Of course, it is clear that if I take any segment of x, and divide it by
the corresponding observable time To, I obtain an instantaneous speed much
greater than those of the two edges of the segment.

So there is something weird here.

I'm not saying something that's wrong, but something that sounds weird.

R.H.

On 11/13/2022 12:13 PM, Richard Hachel wrote:
> Le 13/11/2022 à 17:57, Volney a écrit :

>> So you agree they don't take the same paths through spacetime.
>
> That's what I keep saying.
>
> They follow the same path, in space.
>
> But not in space-time, since I begged millions of physicists to tell the
> difference (but they're too dumb, it's horrible).

Since you agree they don't follow the same path through spacetime, the
two travelers will experience different proper times. And the proper
time of one traveler should not be using the formula of the other
traveler, and your original claim is wrong.

On Monday, 14 November 2022 at 06:29:41 UTC+1, Volney wrote:
> On 11/13/2022 12:13 PM, Richard Hachel wrote:
> > Le 13/11/2022 à 17:57, Volney a écrit :
>
> >> So you agree they don't take the same paths through spacetime.
> >
> > That's what I keep saying.
> >
> > They follow the same path, in space.
> >
> > But not in space-time, since I begged millions of physicists to tell the
> > difference (but they're too dumb, it's horrible).
> Since you agree they don't follow the same path through spacetime, the
> two travelers will experience different proper times. And the proper
> time of one traveler should not be using the formula of the other
> traveler, and your original claim is wrong.

Still believing that adjusting clocks to your ISO
idiocy means putting them into "Newton mode",
stupid Mike?

Richard Hachel <r.hachel@frite.fr> wrote:

> Hachel's principle.
>
> "If two relativistic mobiles travel, in equal observable times, identical
> paths, their proper times will remain equal."
>
> It's a very simple sentence that I think is quite correct.
>
> Yet this principle deeply shocks all relativist physicists in the world.

Would it shock you deeply to learn that no physicist in the world
knows about it, or will ever know about it?

FYI, you are not a physicist,

Jan

On 2022-11-13 12:34:50 +0000, Richard Hachel said:

> Le 13/11/2022 à 11:49, Mikko a écrit :
>> On 2022-11-12 23:42:52 +0000, Richard Hachel said:
>>
>>> I take the well-known example of Tau Ceti.
>>>
>>> We're going to arrange for two rockets (it's a thought experiment),
>>> leaving at the same time, from the same place.
>>>
>>> They start from the ground, and will connect Tau Ceti (12ly).
>>>
>>> One starts according to the accelerated example that we have long
>>> treated here, and which is 10m/s².
>>>
>>> Let a=1.052 ly/y² as Stan Fultoni rightly points out.
>>>
>>> The other will start with a constant real speed of 2.5122c.
>>>
>>> They will cross the same space, and arrive at the same time.
>>>
>>> Don't forget what Hachel says (that's me): "events coinjoint in one
>>> frame of reference are cojoint in all".
>>>
>>> If they leave at the same time, they leave at the same time in all the
>>> observing reference frames.
>>>
>>> If they arrive at the same time, they arrive at the same time in all
>>> the observing reference frames.
>>
>> However, their paths between those two events is not the same in all
>> reference frames. The first rocket travels along a curved path in most
>> reference frames. The second rocent travels along a straight path in
>> every reference frame.
>>
>> Mikko
>
> When I talk about the same path, you will notice that I didn't use the
> confusing term "trajectory".

I did notice that. Did you notice that I didn't use "trajectory", either?

> The term trajctoire can be confused with the term trajectory in space-time.
> I'm not talking about that.

Neither am I.

> I beg you to read carefully what I wrote.

I did. Can you read carefully what I wrote?

> Nobody reads right, although I write right, and very careful what I say
> because I know it's hard on your indoctrinated brains.

You can't know whether nobody reads right. Most readers don't tell how
they read. But you don't write carefylly enough. If you later feel that
you need to explain that your "path" does not mean "trajectory" then
your feeling is a proof that your original text should be more clear.

> You don't want to get out of it, and for that, you make me say what I
> didn't say.

I didn't make you say anything. I said myself what you failed to say.

> I repeat it once again (even I have a feeling it's going to be useless):
> They will cross the same space, and arrive at the same time.

As seen by some observers. Most observers observe one traveller going
through a location that the other does not.

> I didn't say "the same space-time".

You didn't need, as there is no other space-time.

Mikko

On Monday, 14 November 2022 at 12:14:01 UTC+1, Mikko wrote:

> > I repeat it once again (even I have a feeling it's going to be useless):
> > They will cross the same space, and arrive at the same time.
> As seen by some observers. Most observers observe one traveller going
> through a location that the other does not.

There are no observers. It's a gedanken, poor halfbrain.
You've fabricated it.

Den 14.11.2022 01:59, skrev Richard Hachel:
> Le 13/11/2022 à 21:55, "Paul B. Andersen" a écrit :
>> Den 13.11.2022 13:26, skrev Richard Hachel:
>
>>> To=(x/c).sqrt(1+2c²/ax)
>
> I confirm.
>
>  This is the observable time (or improper time, or terrestrial time)
> that we will note according to the length of the path of a rocket or a
> particle and according to its acceleration.
>
>
>> So according to Hachel:
>>     "For the trajectory t = (x/c).sqrt(1+2c²/ax)
>>      we have  dx/dt = [1+c²/2ax]^(-1/2)"
>
> That's not exactly what I said.
>
> I said that the instantaneous observable speed in x, was given by the
> relation dx/dt = Vo/c(inst)=[1+c²/2ax]^(-1/2)

So you confirm that you said:
"For the trajectory t = (x/c).sqrt(1+2c²/ax)
we have dx/dt = [1+c²/2ax]^(-1/2)"

And you pretend not to understand what's wrong with it!

Can you solve the following problems?

Given the function: t = (x/c).sqrt(1+2c²/ax) = sqrt(x²/c²+2x/a)

Question #1
what is dt/dx?

Question #2
what is dx/dt?

If you are ignorant of calculus:
https://www.derivative-calculator.net

--
Paul

https://paulba.no/

Le 14/11/2022 à 06:29, Volney a écrit :
> On 11/13/2022 12:13 PM, Richard Hachel wrote:
>> Le 13/11/2022 à 17:57, Volney a écrit :
>
>>> So you agree they don't take the same paths through spacetime.
>>
>> That's what I keep saying.
>>
>> They follow the same path, in space.
>>
>> But not in space-time, since I begged millions of physicists to tell the
>> difference (but they're too dumb, it's horrible).
>
> Since you agree they don't follow the same path through spacetime, the
> two travelers will experience different proper times. And the proper
> time of one traveler should not be using the formula of the other
> traveler, and your original claim is wrong.

That's not what I'm saying.

This is what relativistic physicists say, based on a poor understanding of
Lorentz transformations, and have imagined a space-time block in four
dimensions, one of which is more or less "complex".

What I say myself, and what I ask that we not distort (it is very tempting
to distort what an "opponent" says to try to have a hold), is that if the
space crossed (simple threeD) is the same, and if the departures and the
arrivals take place simultaneously, not only the observable times will be
equal to each other (tautology), but also, the proper times will be equal
to each other.

I repeat, I do not use the same space-time as other physicists, and in the
end, I sometimes say very different things than experimentation
will prove theoretical consistency, I am sure.

R.H.

Le 14/11/2022 à 10:47, nospam@de-ster.demon.nl (J. J. Lodder) a écrit :
> Would it shock you deeply to learn that no physicist in the world
> knows about it, or will ever know about it?
>
> FYI, you are not a physicist,

It does not matter.

The Beatles couldn't recognize a C from a D.

That didn't stop them from becoming the biggest band in the universe.

R.H.

Le 14/11/2022 à 13:29, "Paul B. Andersen" a écrit :

> If you are ignorant of calculus

Please, don't be arrogant.

R.H.

On Monday, 14 November 2022 at 15:13:13 UTC+1, Richard Hachel wrote:
> Le 14/11/2022 à 06:29, Volney a écrit :
> > On 11/13/2022 12:13 PM, Richard Hachel wrote:
> >> Le 13/11/2022 à 17:57, Volney a écrit :
> >
> >>> So you agree they don't take the same paths through spacetime.
> >>
> >> That's what I keep saying.
> >>
> >> They follow the same path, in space.
> >>
> >> But not in space-time, since I begged millions of physicists to tell the
> >> difference (but they're too dumb, it's horrible).
> >
> > Since you agree they don't follow the same path through spacetime, the
> > two travelers will experience different proper times. And the proper
> > time of one traveler should not be using the formula of the other
> > traveler, and your original claim is wrong.
> That's not what I'm saying.
>
> This is what relativistic physicists say, based on a poor understanding of
> Lorentz transformations, and have imagined a space-time block in four
> dimensions, one of which is more or less "complex".
>
> What I say myself, and what I ask that we not distort (it is very tempting
> to distort what an "opponent" says to try to have a hold), is that if the
> space crossed (simple threeD) is the same, and if the departures and the
> arrivals take place simultaneously, not only the observable times will be
> equal to each other (tautology), but also, the proper times will be equal
> to each other.
>
> I repeat, I do not use the same space-time as other physicists, and in the
> end, I sometimes say very different things than experimentation
> will prove theoretical consistency, I am sure.

And in the meantime in the real world - forbidden by your bunch
of idiots, "improper" GPS and TAI keep measuring t'=t, just like all
serious clocks always did.

On Monday, November 14, 2022 at 6:13:13 AM UTC-8, Richard Hachel wrote:
> if the space crossed (simple threeD) is the same, and if the
> departures and the arrivals take place simultaneously, not
> only the observable times will be equal to each other (tautology),
> but also, the proper times will be equal to each other.

The concept of "space crossed" is coordinate-dependent. For example, when described in terms of one frame of reference, one twin moves in the positive x direction at a constant speed of 10 mph for two hours, and another twin moves in the same direction at 15 mph for one hour and then 5 mph for one hour. They both traveled 20 miles in 2 hours, following the same spatial path, so you claim they have the same elapsed proper times.

However, in terms of a different coordinate system, the first twin was stationary for 2 hours, and the second twin moved at +5 mph for one hour, and -5 mph for 1 hour. You have agreed that these twins do *not* have the same elapsed proper time (Langevin). So, you contradict yourself, because they are the same twins, merely described in terms of different coordinates. So your beliefs are self-contradictory and hence absurd.

Le 14/11/2022 à 21:16, Stan Fultoni a écrit :
> The concept of "space crossed" is coordinate-dependent. For example, when
> described in terms of one frame of reference, one twin moves in the positive x
> direction at a constant speed of 10 mph for two hours, and another twin moves in
> the same direction at 15 mph for one hour and then 5 mph for one hour. They both
> traveled 20 miles in 2 hours, following the same spatial path, so you claim they
> have the same elapsed proper times.
>
> However, in terms of a different coordinate system, the first twin was
> stationary for 2 hours, and the second twin moved at +5 mph for one hour, and -5
> mph for 1 hour. You have agreed that these twins do *not* have the same elapsed
> proper time (Langevin). So, you contradict yourself, because they are the same
> twins, merely described in terms of different coordinates. So your beliefs are
> self-contradictory and hence absurd.

No, I'm not saying that.

In the example that you take, and which is that of the Langevin twins, the
twins use several reference frames.

It's not a good example, and you're right to say it's nonsense.

I am not talking about this example, but about the traveler of Tau Ceti,
in uniformly accelerated movement (a=1.052ly/y²).

This traveler has only one frame of reference and it remains in the same
throughout the journey.

We could then take another twin, and make it start in uniform motion from
Vr=2.5122c.

They leave together, they arrive together (one catching up with the other
at the end of the race).

Their observable (terrestrial) times will obviously be the same since it
is postulated.

But also their own time: Tr=4.776 years.

Yet they don't have the same space-time path, I agree.

But they will have the same spatial path, and the same starting and ending
spatial coordinates in all the reference frames of the universe.

R.H.

On Monday, November 14, 2022 at 3:16:29 PM UTC+1, Richard Hachel wrote:
> Le 14/11/2022 à 10:47, nos...@de-ster.demon.nl (J. J. Lodder) a écrit :
> > Would it shock you deeply to learn that no physicist in the world
> > knows about it, or will ever know about it?
> >
> > FYI, you are not a physicist,
> It does not matter.
>
> The Beatles couldn't recognize a C from a D.
>
> That didn't stop them from becoming the biggest band in the universe.

Science is different. If you cannot recognize a C from a D, you're out.

--
Jan

On Monday, November 14, 2022 at 12:29:54 PM UTC-8, Richard Hachel wrote:
> the example that you take is that of the Langevin twins…
> I am not talking about this example, but about the traveler
> of Tau Ceti,

The two example are the same, merely described in terms of different coordinate systems. In both cases we have twins who separate at the departure event, and one of them moves without acceleration to the reunion event, whereas the other moves with some acceleration. Look at these two examples:

Langevin: One twin is stationary at the origin for 2 hours of coordinate time, and the other twin moves at -5 mph for 1 hour of coordinate time and then abruptly accelerates and moves at +5 miles for 1 hour of coordinate time.

Tau Ceti: One twin moves at 10 mph for 2 hours of coordinate time, and the other twin moves at 5 mph for 1 hour of coordinate time, and then abruptly accelerates and moves at 15 mph for 1 hour of coordinate time.

Note that these two scenarios are exactly the same, merely described in terms of two different coordinate systems. However, according to your beliefs, in the first description the twins have unequal proper times whereas in the latter description they have equal proper times. This is self-contradictory, so your beliefs are absurd.

You will complain: No! My rocket accelerates continuously, not abruptly! But that is an invalid complaint, because we can just as well use a smoothly accelerating twin as an abruptly accelerating one, both for Langevin and for Tau Ceti. Thus, your beliefs are absurd, and none of your pointless attempts to evade logic have any validity.

Le 15/11/2022 à 00:54, Stan Fultoni a écrit :
> On Monday, November 14, 2022 at 12:29:54 PM UTC-8, Richard Hachel wrote:
>> the example that you take is that of the Langevin twins…
>> I am not talking about this example, but about the traveler
>> of Tau Ceti,
>
> The two example are the same, merely described in terms of different coordinate systems. In both cases we have twins who separate at the departure event, and one of them moves without acceleration to the reunion event, whereas the other moves with some acceleration. Look at these two examples:
>
> Langevin: One twin is stationary at the origin for 2 hours of coordinate time, and the other twin moves at -5 mph for 1 hour of coordinate time and then abruptly accelerates and moves at +5 miles for 1 hour of coordinate time.
>
> Tau Ceti: One twin moves at 10 mph for 2 hours of coordinate time, and the other twin moves at 5 mph for 1 hour of coordinate time, and then abruptly accelerates and moves at 15 mph for 1 hour of coordinate time.
>
> Note that these two scenarios are exactly the same, merely described in terms of two different coordinate systems. However, according to your beliefs, in the first description the twins have unequal proper times whereas in the latter description they have equal proper times. This is self-contradictory, so your beliefs are absurd.
>
> You will complain: No! My rocket accelerates continuously, not abruptly! But that is an invalid complaint, because we can just as well use a smoothly accelerating twin as an abruptly accelerating one, both for Langevin and for Tau Ceti. Thus, your beliefs are absurd, and none of your pointless attempts to evade logic have any validity.

I explained exactly this to Richard months ago on fr.sci.physique.

Guess what was his answer then? "My bollocks". Sigh.

On Tuesday, 15 November 2022 at 00:19:36 UTC+1, JanPB wrote:
> On Monday, November 14, 2022 at 3:16:29 PM UTC+1, Richard Hachel wrote:
> > Le 14/11/2022 à 10:47, nos...@de-ster.demon.nl (J. J. Lodder) a écrit :
> > > Would it shock you deeply to learn that no physicist in the world
> > > knows about it, or will ever know about it?
> > >
> > > FYI, you are not a physicist,
> > It does not matter.
> >
> > The Beatles couldn't recognize a C from a D.
> >
> > That didn't stop them from becoming the biggest band in the universe.
> Science is different. If you cannot recognize a C from a D, you're out.

Science, and poor idiot Jan especially, is, of course, free of
weaknesses of ordinary mortal worms.

Den 14.11.2022 15:21, skrev Richard Hachel:
> Le 14/11/2022 à 13:29, "Paul B. Andersen" a écrit :
>
>> If you are ignorant of calculus
>
> Please, don't be arrogant.

Does that mean that you are not ignorant of calculus?

In that case the following questions should be easy to solve:

Given the function: t = (x/c).sqrt(1+2c²/ax) = sqrt(x²/c²+2x/a)

Question #1
what is dt/dx?

Question #2
what is dx/dt?

Even if you are good at calculus, this is
an easy way out. I use it!

https://www.derivative-calculator.net

--
Paul

https://paulba.no/

On 2022-11-14 20:29:52 +0000, Richard Hachel said:

> In the example that you take, and which is that of the Langevin twins,
> the twins use several reference frames.

...

> Yet they don't have the same space-time path, I agree.
>
> But they will have the same spatial path, and the same starting and
> ending spatial coordinates in all the reference frames of the universe.

In almost every coordinate system one of the twins has a bent or curved
path and the other has a straight path.

How can a bent or curved path be the same as a straight path?

Mikko

Le 15/11/2022 à 00:54, Stan Fultoni a écrit :
> On Monday, November 14, 2022 at 12:29:54 PM UTC-8, Richard Hachel wrote:
>> the example that you take is that of the Langevin twins…
>> I am not talking about this example, but about the traveler
>> of Tau Ceti,
>
> The two example are the same, merely described in terms of different coordinate
> systems. In both cases we have twins who separate at the departure event, and one
> of them moves without acceleration to the reunion event, whereas the other moves
> with some acceleration. Look at these two examples:
>
> Langevin: One twin is stationary at the origin for 2 hours of coordinate time,
> and the other twin moves at -5 mph for 1 hour of coordinate time and then abruptly
> accelerates and moves at +5 miles for 1 hour of coordinate time.
>
> Tau Ceti: One twin moves at 10 mph for 2 hours of coordinate time, and the
> other twin moves at 5 mph for 1 hour of coordinate time, and then abruptly
> accelerates and moves at 15 mph for 1 hour of coordinate time.
>
> Note that these two scenarios are exactly the same, merely described in terms of
> two different coordinate systems. However, according to your beliefs, in the
> first description the twins have unequal proper times whereas in the latter
> description they have equal proper times. This is self-contradictory, so your
> beliefs are absurd.
>
> You will complain: No! My rocket accelerates continuously, not abruptly! But
> that is an invalid complaint, because we can just as well use a smoothly
> accelerating twin as an abruptly accelerating one, both for Langevin and for Tau
> Ceti. Thus, your beliefs are absurd, and none of your pointless attempts to evade
> logic have any validity.

Stan, please keep your seriousness.

These are not the same examples.

Langevin's traveler is classically defined as an astronaut who will leave
the earth at a constant speed of 0.8c, from departure to arrival, will do
a quick half-turn in one day (thirty-six terrestrial hours), then return.
We then compare the times and we realize that the star twin has aged only
18 years, while the terrestrial twin has aged 30 years.

The earth twin used a single frame of reference, the star twin three.

As for the traveler of Tau Ceti, this one uses only one referential, and
his earthly companion too.

In the "Tau Ceti II" variant, a third twin accompanies the second, but at
a constant speed of Vr=2.5122c,
i.e. Vo=0.9291c.

The two rockets leave together, arrive together at Tau Ceti, (we call this
conjunctions).

They traveled the same spatial path (a straight line cers Ta Ceti).

In this case, what I'm saying is that their terrestrial times To will be
the same (since that's what we wanted), but still that their proper times
will be the same for this precise example.

That's what I'm saying, and that's how I would like people to read me.

If you read me otherwise, you risk, of course, having me say grotesque
nonsense.

R.H.