Le 19/04/2022 à 21:00, Stan Fultoni a écrit :

> That is just the special relativity equation with different symbols.

Yes, we are obliged to take symbols to write things down. You are right.

> You are
> using the symbol "Tr" to denote the elapsed proper time,

It's true.

> usually called dtau,
> and you are using the symbol "To" to denote the elapsed coordinate time,
> usually called dt,

Yes.

> and you're using "Et" to denote the coordinate spatial
> distance, usually called dx.

Yes.

> So your equation is simply dt^2 = dtau^2 + dx^2,

Yes. Or To²=Tr²+(x/c)²

Et=x/c in the vacuum.

> which is the same as the Minkowski pseudometric dtau^2 = dt^2 - dx^2.

On that, yes.

But for the rest, I don't use the same metric. I don't believe in this
geometry which is absurd if you study it carefully.

>
>> This equation remains valid in accelerated repositories.
>
> Well, in special relativity that equation remains perfectly valid for
> accelerating objects.

That what I say.

> To get the correct answer for an accelerating
> object, we simply split the distance into small segments, during which
> the speed of the object has some average value, and we compute the elapsed
> time for each segment, and then we add all the elapsed times for the
> entire journey. If we split the distance into infinitely many small
> segments, each with a specific velocity, we get the exact result. That's
> what it means to integrate dtau = sqrt(dt^2 - dx^2) to get the total.

Perhaps.
>> Take the case of the traveler from Tau Ceti.
>
> For that case the total elapsed proper time is 3.14 years.

No.

YOU, you say that the proper time will be 3.14 years.

That's not what I'm saying.

>
>> Or again: To²=Tr²+Et²
>
> That relation is true for each infinitesimal segment, and to get the
> total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
> along the path (using your symbols). This gives the correct result 3.14
> years.

You also say that the time To of the terrestrial observer will be 12 years
and 338 days; and on that I completely agree.

The equation you are using is correct.

Your equation (and mine) is : To=(x/c).sqrt(1+2c²/ax)

But for Tr, there is a confusion of concept which I have already explained
several times.

Your equation is wrong.

It's mine that is correct, you have to write:

Tr=sqrt(2x/a)

As Newtonians would do, and it's incredible and very difficult to admit.
They are the ones who are right about that. And I am not newtonian.

Moreover, LOL, Tr being false in your case, it does not check
To²=Tr²+Et² when it should.

R.H.

Le 19/04/2022 à 21:00, Stan Fultoni a écrit :

>
>> Take the case of the traveler from Tau Ceti.
>
> For that case the total elapsed proper time is 3.14 years.
>
>> Or again: To²=Tr²+Et²
>
> That relation is true for each infinitesimal segment, and to get the
> total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
> along the path (using your symbols). This gives the correct result 3.14
> years.

? ? ?

To=12,92
Tr=3.14 ? ? ?
Et=12

To²=Tr²+Et² ---> 12.92 = sqrt(12²+3.14²)=12,40

The error is several months old !!!

R.H.

On Tuesday, 19 April 2022 at 19:04:23 UTC+2, bodk...@gmail.com wrote:
> Maciej Wozniak <maluw...@gmail.com> wrote:
> > On Tuesday, 19 April 2022 at 16:59:03 UTC+2, bodk...@gmail.com wrote:
> >> Michael Moroney <mor...@world.std.spaamtrap.com> wrote:
> >>> On 4/19/2022 7:27 AM, Richard Hachel wrote:
> >>>> Le 19/04/2022 à 07:49, Michael Moroney a écrit :
> >>>>> On 4/18/2022 4:07 PM, Richard Hachel wrote:
> >>>>>> Le 18/04/2022 à 21:35, Odd Bodkin a écrit :
> >>>>>>> It will help enormously if you would refrain from connecting
> >>>>>>> judgments of
> >>>>>>> your ideas to judgements of you and your mental health.
> >>>>>>> You can have a wrong idea and still be mentally ok.
> >>>>>>> You can be mentally ill, but have some idea of what you’re talking
> >>>>>>> about.
> >>>>>>> In your case, there are problems with both, but they can be fixed
> >>>>>>> independently.
> >>>>>>
> >>>>>> You're going to spread the idea that I'm a dangerous guy soon, aren't
> >>>>>> you?
> >>>>>>
> >>>>> Why would he? Most mentally ill people are not dangerous, and I don't
> >>>>> see any reason that you would be.
> >>>>
> >>>> Yes, you are 100% correct.
> >>>>
> >>>> The mentally ill are not dangerous.
> >>>
> >>> I said most aren't. Some are obviously dangerous. Don't change what I wrote.
> >>>>
> >>>> So explain to me why, all my life, I've been considered an extremely
> >>>> dangerous guy.
> >>>
> >>> Have you? By whom? Do you have a police record for violent crimes? Are
> >>> you posting from within a prison (or secure psychiatric ward)?
> >>>
> >> Have you noticed that a common aspiration among the cranks to groups like
> >> this is to be the lone genius, the outside agitator, the dangerous threat,
> >> the crazy but crafty one?
> >
> > Bod, for instance, is deeply believing that his barking
> > is causing some horrible pain and "cranks" don't run
> > before it only when/because they're sick.
> >
> Actually, no, that is not my belief and it is not my intent.

Actually, yes, and you were dumb enough to write
it straight.

On 4/19/2022 11:21 AM, Richard Hachel wrote:
> Le 19/04/2022 à 17:01, Python a écrit :
>> They picture themselves as lions while they are roaches...
>
> I am a lion.
>
Someone turned on the lights! Run!! Hide!!

On Tuesday, April 19, 2022 at 12:45:28 PM UTC-7, Richard Hachel wrote:
>> To get the correct answer for an accelerating object, we
>> simply split the distance into small segments, during which
>> the speed of the object has some average value, and we compute
>> the elapsed time for each segment, and then we add all the
>> elapsed times for the entire journey. If we split the distance
>> into infinitely many small segments, each with a specific velocity,
>> we get the exact result. That's what it means to integrate
>> dtau = sqrt(dt^2 - dx^2) to get the total.
>
>Your equation (and mine) is : To=(x/c).sqrt(1+2c²/ax)

That's not correct. The coordinate time for the trip is 12.91 years,
and the coordinate distance traveled is 12 light-years.

>>> Or again: To²=Tr²+Et²
>>
>> That relation is true for each infinitesimal segment, and to get the
>> total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
>> along the path (using your symbols). This gives the correct result 3.14
>> years.
> To=12,92 Tr=3.14 ? ? ? Et=12
> To²=Tr²+Et² ---> 12.92 = sqrt(12²+3.14²)=12,40
>The error is several months old !!!

To get the total elapsed proper time, that you call "Tr", you must integrate
sqrt(To^2 - Et^2) along the path. This gives 3.14 years.

The same applies to Pythagorean geometry of the flat plane. If you
want to know the length of a curve whose end points are separated by
delta x and delta y, you can't just compute sqrt[(delta x)^2 + (delta y)^2],
you have to compute the integral of sqrt(dx^2 - dy^2) along the curve.

On Tuesday, April 19, 2022 at 12:45:28 PM UTC-7, Richard Hachel wrote:
>> To get the correct answer for an accelerating object, we
>> simply split the distance into small segments, during which
>> the speed of the object has some average value, and we compute
>> the elapsed time for each segment, and then we add all the
>> elapsed times for the entire journey. If we split the distance
>> into infinitely many small segments, each with a specific velocity,
>> we get the exact result. That's what it means to integrate
>> dtau = sqrt(dt^2 - dx^2) to get the total.
>
>Your equation (and mine) is : To=(x/c).sqrt(1+2c²/ax)

That is not correct. The coordinate time for the trip is 12.91 years,
and the coordinate distance traveled is 12 light-years.

>>> Or again: To²=Tr²+Et²
>>
>> That relation is true for each infinitesimal segment, and to get the
>> total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
>> along the path (using your symbols). This gives the correct result 3.14
>> years.
> To=12,92 Tr=3.14 ? ? ? Et=12
> To²=Tr²+Et² ---> 12.92 = sqrt(12²+3.14²)=12,40
>The error is several months old !!!

To get the total elapsed proper time, that you call "Tr", you must integrate
sqrt(To^2 - Et^2) along the path. This gives 3.14 years.

The same applies to Pythagorean geometry of the flat plane. If you
want to know the length of a curve whose end points are separated by
delta x and delta y, you can't just compute sqrt[(delta x)^2 + (delta y)^2]
because that works only for straight lines. In general you have to compute
the integral of sqrt(dx^2 + dy^2) along the curve.

Le 20/04/2022 à 03:16, Stan Fultoni a écrit :
> To get the total elapsed proper time, that you call "Tr", you must integrate
> sqrt(To^2 - Et^2) along the path. This gives 3.14 years.
>
> The same applies to Pythagorean geometry of the flat plane. If you
> want to know the length of a curve whose end points are separated by
> delta x and delta y, you can't just compute sqrt[(delta x)^2 + (delta y)^2],
> you have to compute the integral of sqrt(dx^2 - dy^2) along the curve.

No, in this specific case, you should not integrate.

It's very nice, the integrations, but sometimes that's not the problem.

The answer you give is wrong.

3.14 years is wrong.

Now I've given you the equation to use, do what you want with it.

As for terrestrial time, you tell me (against the relativists) that my
equation is not correct either, whereas I have always said that I had the
same as them.

I put it here:
To=(x/c).sqrt(1+2c²/ax)

This equation is correct and recognized by all.

The result for a=10m/s² and x=12ly in this equation is To=12.925 years.

Twelve years and 338 days

(one year = 365.25d)

R.H.

Le 20/04/2022 à 03:16, Stan Fultoni a écrit :

> To get the total elapsed proper time, that you call "Tr", you must integrate
> sqrt(To^2 - Et^2) along the path. This gives 3.14 years.

It is both much more complicated and much simpler.

Conceptually, it is much more complicated, because we must not forget that
the distance which remains to be covered for the rocket will constantly
change in a relativistic way according to its speed.

And it will change as I say, and there, the slap that I give to the
relativists, it will make them very funny.

They think the distance will contract according to sqrt(1-v²/c²) LOL.

The distance will stretch to become longer and longer.

That's what drives them all crazy and that's what overwhelms them when I
speak.

Suddenly, they take me for a "crank".

The equation is at each moment of the course:
x'=x.sqrt(1-v²/c²)/(1+cosµv/c²)

They are therefore very far from the true concept.

FURTHER, in the frame of reference of the rocket which can be considered
fixed from one small position to another and so on, it is Tau Ceti which
is approaching.

The approach apparent speed Vapp will be Vapp=Vo/(1-v/c)

If we take the two equations that I have given here, we realize that if
Tr=x'/Vapp we can no longer do anything by concealing these things.

The equation is therefore not the one given by realtivists, but overall,
we arrive at Tr=sqrt(2x/a)

It is extremely simple.

Note that it is of infinite logic for two reasons.

It confirms To²=Tr²+Et²

It confirms the law x=(1/2)at²
if the right measures are taken.

x is in R (but that's normal).
a is in R' (rocket)
t is in R': it is Tr.

Thank you for your attention.

R.H.

Richard "Hachel" Lengrand (M.D.) wrote:
> Le 20/04/2022 à 03:16, Stan Fultoni a écrit :
>> To get the total elapsed proper time, that you call "Tr", you must
>> integrate sqrt(To^2 - Et^2) along the path.  This gives 3.14 years.
>>
>> The same applies to Pythagorean geometry of the flat plane.  If you
>> want to know the length of a curve whose end points are separated by
>> delta x and delta y, you can't just compute sqrt[(delta x)^2 + (delta
>> y)^2], you have to compute the integral of sqrt(dx^2 - dy^2) along the
>> curve.

It's quite simple to prove Lengrand wrong here. He claims that a second
traveller moving uniformly at such a speed that he starts and arrives
at the same time as the uniformly accelerated traveller will have their
clocks displaying the same value on arrival.

But taking the point of view of this second traveller whose frame of
reference is Galilean, one can consider the trajectory of the uniformly
accelerated traveller. At first he is receding, then he slows down and
come back to arrive at the same position at the same time on Tau Ceti.

This is the "classical" twins scenario. We know that their clocks cannot
display the same value on arrival.

On Wednesday, April 20, 2022 at 4:15:51 AM UTC-7, Richard Hachel wrote:
> > To get the total elapsed proper time, that you call "Tr", you must integrate
> > sqrt(To^2 - Et^2) along the path. This gives 3.14 years.
> >
> > The same applies to Pythagorean geometry of the flat plane. If you
> > want to know the length of a curve whose end points are separated by
> > delta x and delta y, you can't just compute sqrt[(delta x)^2 + (delta y)^2],
> > you have to compute the integral of sqrt(dx^2 + dy^2) along the curve.
>
> No, in this specific case, you should not integrate.

Suppose you begin at one corner of a square that is 100 meters on each
edge, and you want to walk to the opposite corner of the square. One way
would be to follow the edges of the square, so the distance you walk
would be 200 meters. Another way would be to walk directly along the
diagonal, so the distance you walk would be 141.4 meters. Another path
you could follow would be along the perimeter of a circle that
circumscribes the square, which has the path length 222.1 meters.
In every case, the length of the path is given by integrating sqrt(dx^2 + dy^2).
According to your beliefs, all these paths have the same length. This
is logically self-contradictory. You agree that the edge length is 100
meters, so you are saying that someone who travels along the edges
goes 100 + 100 = 141.4, which is obviously false.

> As for terrestrial time, you tell me (against the relativists) that my
> equation is not correct either, whereas I have always said that I had the
> same as them. I put it here: To=(x/c).sqrt(1+2c²/ax)

Yes, I misread that equation... it is indeed true that the coordinate distance
traveled by the accelerating object is delta t = (x/c)sqrt(1 + 2c^2/ax), which,
as I said, is about 12.91 years. This is just as special relativity says. This
comes directly from the equations relating delta x and delta t with delta tau,
which gives the elapsed proper time delta tau = 3.14 years, by integrating
sqrt(dt^2 - dx^2).

Le 20/04/2022 à 15:37, Stan Fultoni a écrit :

> Yes, I misread that equation... it is indeed true that the coordinate distance
> traveled by the accelerating object is delta t = (x/c)sqrt(1 + 2c^2/ax), which,
> as I said, is about 12.91 years. This is just as special relativity says. This
> comes directly from the equations relating delta x and delta t with delta tau,
> which gives the elapsed proper time delta tau = 3.14 years, by integrating
> sqrt(dt^2 - dx^2).

Thank you for your honesty.

So here we are in agreement on the equation I gave.

I independently researched and found the relativists' equation.

They have the same.

It would therefore be very strange if this equation were false.

If on the one hand Richard Hachel affirms one thing, and on the other the
scientists of the entire world affirm the same, it would still be very
strange if it were not a scientific reality.

We can therefore consider that this equation is correct.

Now there is a very tricky problem.

Why don't scientists and Hachel agree on proper time?

A quick answer would be: "Because Hachel, he's stupid".

But I have the impression that the answer was made in a hurry.

R.H.

Le 20/04/2022 à 15:37, Stan Fultoni a écrit

> tau = 3.14 years, by integrating sqrt(dt^2 - dx^2).

Practicing good integration in bad geometry does little good.

R.H.

Richard Hachel <r.hachel@tiscali.fr> wrote:
> Le 20/04/2022 à 15:37, Stan Fultoni a écrit :
>
>> Yes, I misread that equation... it is indeed true that the coordinate distance
>> traveled by the accelerating object is delta t = (x/c)sqrt(1 + 2c^2/ax), which,
>> as I said, is about 12.91 years. This is just as special relativity says. This
>> comes directly from the equations relating delta x and delta t with delta tau,
>> which gives the elapsed proper time delta tau = 3.14 years, by integrating
>> sqrt(dt^2 - dx^2).
>
> Thank you for your honesty.
>
> So here we are in agreement on the equation I gave.
>
> I independently researched and found the relativists' equation.
>
> They have the same.
>
> It would therefore be very strange if this equation were false.
>
> If on the one hand Richard Hachel affirms one thing, and on the other the
> scientists of the entire world affirm the same, it would still be very
> strange if it were not a scientific reality.
>
> We can therefore consider that this equation is correct.
>
> Now there is a very tricky problem.
>
> Why don't scientists and Hachel agree on proper time?

Because you’re mistaken.

>
> A quick answer would be: "Because Hachel, he's stupid".

No. Don’t be so thin-skinned. Being mistaken does not make you stupid.
Repeat that a dozen times. On the other hand, you not being stupid does not
make you right.

This is the purpose of experiment in science. Very smart people come up
with elegant and well-formed ideas that make clean predictions. Most of
those elegant and well-formed ideas are wrong. This they discover when they
compare the results of their calculations with real data from real
experiments. Not thought experiments, not hypothetical situations. Real
experiments.

When a smart physicist with an elegant and well-formed idea discovers that
it is wrong, because it disagrees with experimental data, he drops the idea
and moves on. Or significantly alters the idea.

This is why you arguing about your answer vs a physicist’s answer about
proper time to Tau Ceti is a pointless boondoggle. That is not a real
experiment. Put your idea to test against real data, because that is what
real physicists do.

>
> But I have the impression that the answer was made in a hurry.
>
> R.H.
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Wednesday, April 20, 2022 at 8:36:32 AM UTC-7, Richard Hachel wrote:
> > To get the total elapsed proper time, that you call "Tr", you must integrate
> > sqrt(To^2 - Et^2) along the path. This gives 3.14 years.
> >
> > The same applies to Pythagorean geometry of the flat plane. If you
> > want to know the length of a curve whose end points are separated by
> > delta x and delta y, you can't just compute sqrt[(delta x)^2 + (delta y)^2],
> > you have to compute the integral of sqrt(dx^2 + dy^2) along the curve.
>
> No, in this specific case, you should not integrate.

Suppose you begin at one corner of a square that is 100 meters on each
edge, and you want to walk to the opposite corner of the square. One way
would be to follow the edges of the square, so the distance you walk
would be 200 meters. Another way would be to walk directly along the
diagonal, so the distance you walk would be 141.4 meters. Another path
you could follow would be along the perimeter of a circle that
circumscribes the square, which has the path length 222.1 meters.
In every case, the length of the path is given by integrating sqrt(dx^2 + dy^2).
According to your beliefs, all these paths have the same length. This
is logically self-contradictory. You agree that the edge length is 100
meters, so you are saying that someone who travels along the edges
goes 100 + 100 = 141.4, which is obviously false.

> Why don't scientists and Hachel agree on proper time?

Well, scientists believe 100 + 100 equals 200, whereas you claim it equals
141.4. If we subtract 141.4 from both sides of your claimed equality, and
then divide both sides by 58.6 you get 1 = 0. This is the quintessential
logical impossibility. Your claim is logically impossible. Agreed?

On Wednesday, 20 April 2022 at 14:39:00 UTC+2, Python wrote:
> Richard "Hachel" Lengrand (M.D.) wrote:
> > Le 20/04/2022 à 03:16, Stan Fultoni a écrit :
> >> To get the total elapsed proper time, that you call "Tr", you must
> >> integrate sqrt(To^2 - Et^2) along the path. This gives 3.14 years.
> >>
> >> The same applies to Pythagorean geometry of the flat plane. If you
> >> want to know the length of a curve whose end points are separated by
> >> delta x and delta y, you can't just compute sqrt[(delta x)^2 + (delta
> >> y)^2], you have to compute the integral of sqrt(dx^2 - dy^2) along the
> >> curve.
> It's quite simple to prove Lengrand wrong here. He claims that a second

Oh, stinker Python is opening its muzzle again,
and trying to pretend he knows something.
Tell me, poor stinker, what is your definition of
a "theory" in the terms of Peano arithmetic?
See: if a theorem is going to be a part of a theory,
it has to be formulable in the language of the
theory. Do you get it? Or are you too stupid even for
that, poor stinker?

On Wednesday, 20 April 2022 at 18:01:08 UTC+2, bodk...@gmail.com wrote:
> Richard Hachel <r.ha...@tiscali.fr> wrote:
> > Le 20/04/2022 à 15:37, Stan Fultoni a écrit :
> >
> >> Yes, I misread that equation... it is indeed true that the coordinate distance
> >> traveled by the accelerating object is delta t = (x/c)sqrt(1 + 2c^2/ax), which,
> >> as I said, is about 12.91 years. This is just as special relativity says. This
> >> comes directly from the equations relating delta x and delta t with delta tau,
> >> which gives the elapsed proper time delta tau = 3.14 years, by integrating
> >> sqrt(dt^2 - dx^2).
> >
> > Thank you for your honesty.
> >
> > So here we are in agreement on the equation I gave.
> >
> > I independently researched and found the relativists' equation.
> >
> > They have the same.
> >
> > It would therefore be very strange if this equation were false.
> >
> > If on the one hand Richard Hachel affirms one thing, and on the other the
> > scientists of the entire world affirm the same, it would still be very
> > strange if it were not a scientific reality.
> >
> > We can therefore consider that this equation is correct.
> >
> > Now there is a very tricky problem.
> >
> > Why don't scientists and Hachel agree on proper time?
> Because you’re mistaken.
> >
> > A quick answer would be: "Because Hachel, he's stupid".
> No. Don’t be so thin-skinned. Being mistaken does not make you stupid.
> Repeat that a dozen times. On the other hand, you not being stupid does not
> make you right.
>
> This is the purpose of experiment in science.

Stop fucking, poor halfbrain. Your wonders are some
tautologies of your moronic newspeak, nothing more.

Le 20/04/2022 à 18:44, Stan Fultoni a écrit :

> Well, scientists believe 100 + 100 equals 200

Do you really think I think it takes 199?

R.H.

On Wednesday, April 20, 2022 at 10:26:12 AM UTC-7, Richard Hachel wrote:
>> Scientists believe 100 + 100 equals 200, whereas you claim it equals
>> 141.4. If we subtract 141.4 from both sides of your claimed equality, and
>> then divide both sides by 58.6 you get 1 = 0. This is the quintessential
>> logical impossibility. Your claim is logically impossible. Agreed?
>
> Do you really think I think it takes 199?

No, you claim that 100 + 100 = 141.4, which implies 1 = 0, which is
obviously logically impossible. Agreed?

On Tuesday, April 19, 2022 at 4:21:17 AM UTC-7, Richard Hachel wrote:
> > The answer given by special relativity is quite simple: Just integrate
> > sqrt(dt^2 - dx^2) along the path to give the elapsed proper time.
>
> Let's pose it in the Langevin problem (the twin starts at 0.8c of constant
> speed to travel 12 ly. Its proper time (all physicists know this) will be 18 years.
> I note Tr=18. The time noted by its brother which remains in the terrestrial
> reference frame is To (observable time). The brother notes To=30. If we set Et=x/c=24
> We come across the equation that links the proper time Tr of one to the
> observable time To of the other. I repeat: To²=Tr²+Et²

That's just the special relativity equation with different symbols. You are
using the symbol "Tr" to denote the elapsed proper time, usually called dtau,
and you are using the symbol "To" to denote the elapsed coordinate time,
usually called dt, and you're using "Et" to denote the coordinate spatial
distance, usually called dx. So your equation is simply dt^2 = dtau^2 + dx^2,
which is the same as the Minkowski pseudometric dtau^2 = dt^2 - dx^2.

> This equation remains valid in accelerated repositories.

Of course, in special relativity that equation remains perfectly valid for
accelerating objects. To get the correct answer for an accelerating
object, we simply split the distance into small segments, during which
the speed of the object has some average value, and we compute the elapsed
time for each segment, and then we add all the elapsed times for the
entire journey. If we split the distance into infinitely many small
segments, each with a specific velocity, we get the exact result. That's
what it means to integrate dtau = sqrt(dt^2 - dx^2) to get the total.

> Take the case of the traveler from Tau Ceti.

For that case the total elapsed proper time is 3.14 years.

> Or again: To²=Tr²+Et²

That relation is true for each infinitesimal segment, and to get the
total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
along the path (using your symbols). This gives the correct result 3.14
years.

On 4/20/2022 11:36 AM, Richard Hachel wrote:

> If on the one hand Richard Hachel affirms one thing, and on the other
> the scientists of the entire world affirm the same, it would still be
> very strange if it were not a scientific reality.
>
> We can therefore consider that this equation is correct.
>
> Now there is a very tricky problem.
>
> Why don't scientists and Hachel agree on proper time?

Because Hachel is wrong.
>
> A quick answer would be: "Because Hachel, he's stupid".

No, the real answer is because Hachel is too stubborn to accept being
corrected. Just like how Hachel is too stubborn to communicate using
standard terminology, or, frequently, English in an English-using group.
>
> But I have the impression that the answer was made in a hurry.
>
> R.H.

Le 21/04/2022 à 03:53, Stan Fultoni a écrit :
> On Tuesday, April 19, 2022 at 4:21:17 AM UTC-7, Richard Hachel wrote:

>
>> Take the case of the traveler from Tau Ceti.
>
> For that case the total elapsed proper time is 3.14 years.
>
>> Or again: To²=Tr²+Et²
>
> That relation is true for each infinitesimal segment, and to get the
> total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
> along the path (using your symbols). This gives the correct result 3.14
> years.

There is something wrong with the way relativists calculate proper time.

The integration they are doing is not correct.

They mix carrot and turnip.

I haven't stopped saying it, the special theory of relativity is a true,
beautiful, experimental theory, but it is full of little pitfalls.

Here, the trap is not mathematical, but conceptual.

The correct equation is here:

<http://news2.nemoweb.net/jntp?UyzzGy0AB6-TDJvLN7tJBprLC7Q@jntp/Data.Media:1>

As for the proper time, it will be longer than their prediction.

They set a Tr that is much too short which does not correspond to reality.

Tr= 4.77 years and not 3.14 years.

Tr ~ 4 years and 281 days ; not 3 years and 51 days.

R.H.

Le 21/04/2022 à 03:53, Stan Fultoni a écrit :

>> Or again: To²=Tr²+Et²

For Tau Ceti (12 ly), a = 10m/s²

Tr = 4.77

To = 12.92

Et = x/c = 12

R.H.

Richard Hachel <r.hachel@tiscali.fr> wrote:
> Le 21/04/2022 à 03:53, Stan Fultoni a écrit :
>> On Tuesday, April 19, 2022 at 4:21:17 AM UTC-7, Richard Hachel wrote:
>
>>
>>> Take the case of the traveler from Tau Ceti.
>>
>> For that case the total elapsed proper time is 3.14 years.
>>
>>> Or again: To²=Tr²+Et²
>>
>> That relation is true for each infinitesimal segment, and to get the
>> total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
>> along the path (using your symbols). This gives the correct result 3.14
>> years.
>
>
> There is something wrong with the way relativists calculate proper time.
>
> The integration they are doing is not correct.
>
> They mix carrot and turnip.
>
> I haven't stopped saying it, the special theory of relativity is a true,
> beautiful, experimental theory, but it is full of little pitfalls.
>
> Here, the trap is not mathematical, but conceptual.
>
> The correct equation is here:
>
> <http://news2.nemoweb.net/jntp?UyzzGy0AB6-TDJvLN7tJBprLC7Q@jntp/Data.Media:1>
>
> As for the proper time, it will be longer than their prediction.
>
> They set a Tr that is much too short which does not correspond to reality.
>
> Tr= 4.77 years and not 3.14 years.

Experiment says otherwise. See particle lifetimes in accelerators.

A real scientist will not insist his idea is correct when it is discovered
that experimental data do not agree.

>
> Tr ~ 4 years and 281 days ; not 3 years and 51 days.
>
> R.H.
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

Le 21/04/2022 à 17:01, Odd Bodkin a écrit :

> A real scientist will not insist his idea is correct when it is discovered
> that experimental data do not agree.

C'est ce que je dis (That's what I say).

R.H.

On Thursday, 21 April 2022 at 17:01:20 UTC+2, bodk...@gmail.com wrote:
> Richard Hachel <r.ha...@tiscali.fr> wrote:
> > Le 21/04/2022 à 03:53, Stan Fultoni a écrit :
> >> On Tuesday, April 19, 2022 at 4:21:17 AM UTC-7, Richard Hachel wrote:
> >
> >>
> >>> Take the case of the traveler from Tau Ceti.
> >>
> >> For that case the total elapsed proper time is 3.14 years.
> >>
> >>> Or again: To²=Tr²+Et²
> >>
> >> That relation is true for each infinitesimal segment, and to get the
> >> total elapsed "Tr" over the journey we need to integrate sqrt(To^2 - Et^2)
> >> along the path (using your symbols). This gives the correct result 3.14
> >> years.
> >
> >
> > There is something wrong with the way relativists calculate proper time..
> >
> > The integration they are doing is not correct.
> >
> > They mix carrot and turnip.
> >
> > I haven't stopped saying it, the special theory of relativity is a true,
> > beautiful, experimental theory, but it is full of little pitfalls.
> >
> > Here, the trap is not mathematical, but conceptual.
> >
> > The correct equation is here:
> >
> > <http://news2.nemoweb.net/jntp?UyzzGy0AB6-TDJvLN7tJBprLC7Q@jntp/Data.Media:1>
> >
> > As for the proper time, it will be longer than their prediction.
> >
> > They set a Tr that is much too short which does not correspond to reality.
> >
> > Tr= 4.77 years and not 3.14 years.
> Experiment says otherwise.

YOU say otherwise, poor halfbrain. In the meantime
in the real world, however, forbidden by your insane
Shit TAI keep measuring t'=t, just like all serious clocks
always did.