Incorrect mathematical integration in SR.

[Richard Hachel]

Incorrect mathematical integration in SR.

If we take a very small segment (but it must be extremely small on the
order of a few seconds), we can consider that a body in accelerated motion
behaves like a body in Galilean motion.

But it is of the order of only a few seconds.

If the proper velocity time segment is very small, we can consider the
true equation, which is:
ΔTo=[Tr2.sqrt(1+(1/4)Vr2²/c²]-[Tr1.sqrt(1+(1/4)Vr1²/c²] tends to
ΔTo=ΔTr.sqrt(1+Vr²/c²)

But the problem is that we cannot physically integrate this erroneous
notion.

And that the residual calculation will become totally grotesque compared
to the reality of things.

3.139 years instead of 4.776 years, in the famous problem of the traveler
of Tau Ceti, it is no longer an error.

It is a relativist drama.

R.H.

[rotchm]

On Sunday, October 2, 2022 at 9:01:44 PM UTC-4, Richard Hachel wrote:

> Relativists do not realize that they are using an infinite number of
> clocks to measure the rocket's path, and not a single clock.


That's a lie. "relativists" do realize that they are using an infinite number of
clocks to measure the rocket's path... [if I understood what you meant].

Given a coordinate system (CS) , that is, having coordinated/defined time and position, then we can
express the motion of a rocket in this CS. In our scenario, he have set up the usual (inertial)
"grid of clocks and rulers". Declaring that the rocket is at x at time t, means that as the rocket
coincides with the spatial coordinate x, the value of the clock there indicates t.
Since the path of the rocket is continuous, there are "infinitely" many clocks along the x axis.
Relativists are well aware of this, contrary to your claim.

Now physically, we can't put clocks everywhere. So clocks are put only there where they are needed
(usually at the beginning and end, at x=0 and x=D, say) .

Now, given a path x(t), SR says that we can find the value of the traveling watch as it
coincides with the spatial coordinate x (and thus at time t, the value of the clock at x): dtau² = dt² - dx².
Or, you can break it up into segments as you seem to want [using c=1 here]:

(tb - ta)√(1-va²) + (tc - tb)√(1-vb²) + ... (tz - ty)√(1-vy²).

The more sums (intervals) used, this sum becomes arbitrarily close to the result given by dtau² = dt² - dx².

[JanPB]

You keep making same mistakes over and over again. Why do
you automatically assume it physicists' mistake and not yours?

This is the weird part. I guess that's what makes crank a crank: the
lack of the immediate and instinctive feel for one's knowledge
approaching a boundary.

--
Jan

[mitchr...@gmail.com]

Why do you believe authority has everything right already jan?
Authority is in the negative instead jan...
IT shows why you are a hypocrite that
has nothing but her pride...

Mitchell Raemsch

[JanPB]

I don't believe, I have proofs. Authority doesn't enter into it.

--
Jan

[mitchr...@gmail.com]

You don't have proof you just have pride jan...
What are you claiming? Science says it can't prove anything jan.
Are you doing science. Or are you just a hypocrite that claims
she has proof she knows? Where is your proof of proof there?

Mitchell Raemsch
> --
> Jan

[JanPB]

No. I have proofs. Period.

> What are you claiming? Science says it can't prove anything jan.

Whatever.

--
Jan

[JanPB]

On Monday, October 3, 2022 at 7:04:25 PM UTC-7, Stan Fultoni wrote:
> On Monday, October 3, 2022 at 3:15:26 PM UTC-7, Richard Hachel wrote:

> > If we take a very small segment of a trajectory, we can consider that a
> > body in accelerated motion behaves like a body in inertial motion.
>
> Right, and for that small segment the body moves a small distance Δx in a small time Δt. During this segment the elapsed proper time is Δtau = sqrt(Δt^2 - Δx^2). The total elapsed time for the trajectory is the sum of the elapsed times of the small segments.
>
> In your example, this gives 3.14 years for the trajectory of constant proper acceleration. If we apply this to the unaccelerated trajectory between the same starting and ending events, we get 4.77 years. Understand?

He is going to say that the additive terms come from "different clocks",
therefore their adding up to anything means nothing :-)

--
Jan

[Stan Fultoni]

On Monday, October 3, 2022 at 3:15:26 PM UTC-7, Richard Hachel wrote:

> If we take a very small segment of a trajectory, we can consider that a
> clock in accelerated motion behaves like a clock in inertial motion.

Right, and for that small segment the clock moves a small distance Δx in a small time Δt. During this segment the elapsed proper time is Δtau = sqrt(Δt^2 - Δx^2). The total elapsed time for the clock on the whole trajectory is the sum of the elapsed times of the small segments.

In your example, this gives 3.14 years for the trajectory of constant proper acceleration, whereas if we apply this to the unaccelerated trajectory between the same starting and ending events, we get 4.77 years. Understand?

[Maciej Wozniak]

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

[Athel Cornish-Bowden]

501

--
Athel -- French and British, living mainly in England until 1987.

[Richard Hachel]

Le 04/10/2022 à 02:08, JanPB a écrit :

> You keep making same mistakes over and over again. Why do
> you automatically assume it physicists' mistake and not yours?

> Jan

I said that to be true, a theory must have internal beauty and external
beauty.

Internal beauty is its logic.

External beauty is its proof and its experimental reproducibility.

Experimentally, nothing has ever been able to contradict me, while
physicists are caught in the trap of experimental contradiction by the
instantaneous transmission of information (quantum effect).

I don't have this problem.

Now let's get to the theory. It is not even immediately acceptable to a
good scientific tribunal because of the real Langevin paradox. This
paradox is not really the one we are describing, but the fact that the
general covariance becomes absurd if we study the path in apparent
velocities (what we see in telescopes).
Physicists cannot explain how a rocket which sees the earth return to it
for 9 years of its own time, and with an apparent speed of 4c in the
telescope, can affirm that it sees our celestial body moving over 7.2
light-years.
On this point, they put the dust under the carpet to keep intact their
theory based only on the covariance and the relativity of chronotropy.

It's not that they're wrong on that, it's that on these two things, we
must also add the notion of universal anisochrony and proper present time
relating to each observer, whether mobile or at rest compared to another.

They then confuse the notion of instant, and the notion of duration.

Chronotropy, which relates to the notion of speed, measures durations.

The anisotropy, relative to the position of the object, measures the
instants.

These are two very different phenomena.

The first is a quadratic square root.

The second a cosine geometry.

We must always add the two terms.

These two terms are real and present in all the equations.

<http://news2.nemoweb.net/jntp?RebakoJHl5RMKVpJlwY-2GDQx7c@jntp/Data.Media:1>

R.H.

[Maciej Wozniak]

On Tuesday, 4 October 2022 at 12:04:18 UTC+2, Richard Hachel wrote:
> Le 04/10/2022 à 02:08, JanPB a écrit :
>
> > You keep making same mistakes over and over again. Why do
> > you automatically assume it physicists' mistake and not yours?
> > Jan
>
> I said that to be true, a theory must have internal beauty and external
> beauty.

Unlike the reality, which can be really ugly.

>
> Internal beauty is its logic.
>
> External beauty is its proof and its experimental reproducibility.
>
> Experimentally, nothing has ever been able to contradict me, while

While "improper" clocks of GPS and TAI keep measuring
t;=t in "improper" old seconds.

[Richard Hachel]

I mean by that, that the equation To=Tr.sqrt(1+Vr²/c²) used by
relativists is only correct for the moment when the rocket passes "in
front" of the watch.

It is still correct at the moment when the rocket passes "in front of"
another clock.

And so on.

But that we cannot integrate all these times which are carrots and
turnips.

The real equation being:
To=Tr.sqrt(1+(1/4).Vr²/c²)

So of course, if we shrink the segment studied into a segment of one
second for example, we will be able to simplify the accelerated movement
into a Galilean movement.

It is easily demonstrated, and I did it many posts ago.

But that means that we consider that the body, precisely, is not
accelerated. Acceleration therefore no longer exists.

And to transcribe all that, by integration, in a long accelerated journey
is a blunder.

It is true that if I pose the two equations that I gave yesterday, their
result tends to the same thing when DTr is smaller and smaller (do the
calculation), because Vra, Vrm, Vrb, all tend towards Vr .

But this is no longer true as soon as the range of initial and final
segment speeds is opened up.

Even for a single second.

R.H.

[Richard Hachel]

Le 04/10/2022 à 05:35, JanPB a écrit :

[Stan Fultoni]

On Monday, October 3, 2022 at 3:15:26 PM UTC-7, Richard Hachel wrote:

> The equation To=Tr.sqrt(1+Vr²/c²) used by relativists is only correct

> for the moment when the rocket passes "in front" of the watch.

Adults don't use your backwards equation, they use Δtau = sqrt(Δt^2 - Δx^2), and in your example this is applied to the trajectory t = x*sqrt(1+2/(ax)). We split the trajectory into many small segments, and compute Δtau for each segment, and then add these to give the total Δtau on the entire trajectory.

> The equation (tj - ti) = (tauj - taui) sqrt[(1 + (a^2/4)(tauj+taui)^2] is only
> valid if the start is made at rest.

No, that equation is equivalent to your Δt = Δtau .sqrt[(1 + Vri² + Vri.Vra + (1/4).Vra²], since you define Vri as the initial velocity a*taui and you define Vra as the change in velocity from start to finish, which is a*(tauj - taui). Inserting these into your equation gives the result (tj - ti) = (tauj - taui) sqrt[(1 + (a^2/4)(tauj+taui)^2]. This confirms that your beliefs imply 1=0, and are therefore absurd, as explained in the previous messages.

> If we take a very small segment of a trajectory, we can consider that a

> clock in accelerated motion behaves like a clock in inertial motion.

Right, and for that small segment the clock moves a small distance Δx in a small time Δt. During this segment the elapsed proper time is Δtau = sqrt(Δt^2 - Δx^2). The total elapsed time for the clock on the whole trajectory is the sum of the elapsed times of the small segments.

In your example, this gives 3.14 years for the trajectory of constant proper acceleration, whereas if we apply this to the unaccelerated trajectory between the same starting and ending events, we get 4.77 years.

[Mikko]

On 2022-10-03 22:15:23 +0000, Richard Hachel said:

> But the problem is that we cannot physically integrate this erroneous notion.

You don't need any integration. Everything you can logically infer can be
inferred without integration.

Integration can be useful if you try to find out what to infer.
However, there are other ways to find out.

Mikko

[Richard Hachel]

Le 04/10/2022 à 07:45, Stan Fultoni a écrit :
> On Monday, October 3, 2022 at 3:15:26 PM UTC-7, Richard Hachel wrote:
>> If we take a very small segment of a trajectory, we can consider that a
>> clock in accelerated motion behaves like a clock in inertial motion.
>
> Right, and for that small segment the clock moves a small distance Δx in a
> small time Δt. During this segment the elapsed proper time is Δtau = sqrt(Δt^2

> - Δx^2). The total elapsed time for the clock on this accelerating trajectory is

> the sum of the elapsed times of the small segments.
>
> In your example, this gives 3.14 years for the trajectory of constant proper

> acceleration. If we apply this to the unaccelerated trajectory between the same
> starting and ending events, we get 4.77 years. Understand?

No, precisely, the clock does not move.

Each watch is ideally placed far enough away, and perpendicular to the
segment, so as not to have to take into account the ANISOCHRONIC Doppler
effect.

We have a precise and pure measurement of the chronotropic effect (ie what
relativists have known since the beginning of the 20th century) without a
suspected Doppler problem.

It's very important to Hachel (that's me).

Each small clock will then note precisely how much the value of the proper
time that corresponds to it is worth.

We split into 94608000 small pieces established by beeps of the rocket
every second.

We place in the terrestrial reference frames, 94608000 small watches in an
ingenious way so that each watch is placed in front of a small segment.

We will then have 94608000 small observable times for each small proper
time of one second.

We will then think that there is only to add the small ends between them,
and that the sum will be the total observable time.

We will then find approximately 13,338 years (approximately) of observable
time.

Wrong answer, because it is 12,914 years.

So what's wrong?

It is the fact that the equation is not correct and that we cannot add to
each other the total of more than 94 million watches placed in different
places, and more and more far from each other others.

R.H.

[rotchm]

On Tuesday, October 4, 2022 at 6:44:08 PM UTC-4, Richard Hachel wrote:
> Le 04/10/2022 à 07:45, Stan Fultoni a écrit :

> > In your example, this gives 3.14 years for the trajectory of constant proper
> > acceleration. If we apply this to the unaccelerated trajectory between the same
> > starting and ending events, we get 4.77 years. Understand?
>
> No, precisely, the clock does not move.

The clock will be taken to be fixed objects within the main inertial frame.
The *watches* will be those clocks that travel.
So you have the watch of the accelerating rocket, and you have another watch for the inertial rocket.

> Each watch is ideally placed far enough away,

You mean the clocks?

> and perpendicular to the segment,

What does that mean??

> so as not to have to take into account the ANISOCHRONIC Doppler
> effect.

There is no Doppler except in the scenario. The watches gold from the first event to the second event.
Wherever the watch is, it compares its *value* with the value of the clock "under" it.

[Richard Hachel]

Le 05/10/2022 à 01:03, rotchm a écrit :
> There is no Doppler except in the scenario. The watches gold from the first
> event to the second event.
> Wherever the watch is, it compares its *value* with the value of the clock
> "under" it.

If you do not want to place the watch perpendicularly and far enough to
observe a small transfer in the defined segment, you can also place the
watch in the center of the segment.
The watch will have a small information delay when the rocket arrives for
example at A, but since it will have the same information delay as when
the watch arrives at B, the time dTo of this watch will be valid.

But the problem is not there.

The problem is that we believe that in a relativistic medium, we can add
all the small times dTo measured by its 94 million watches placed in
different places and do a simple integration.

A bit like if we wanted to add relativistic velocities in a simple way,
and without going through the formula that I have given many times.

<http://news2.nemoweb.net/jntp?r_9PN0wQl1cwpVKtFpkiKSwcSxQ@jntp/Data.Media:1>

R.H.



--
"Mais ne nous trompons pas.
Il n'y a pas que de la violence avec des armes : il y a des situations de
violence."
Abbé Pierre
<http://news2.nemoweb.net/?DataID=r_9PN0wQl1cwpVKtFpkiKSwcSxQ@jntp>

[Richard Hachel]

Le 05/10/2022 à 01:03, rotchm a écrit :

> On Tuesday, October 4, 2022 at 6:44:08 PM UTC-4, Richard Hachel wrote:
>> Le 04/10/2022 à 07:45, Stan Fultoni a écrit :
>
>> > In your example, this gives 3.14 years for the trajectory of constant proper
>> > acceleration. If we apply this to the unaccelerated trajectory between the
>> same
>> > starting and ending events, we get 4.77 years. Understand?
>>
>> No, precisely, the clock does not move.
>
> The clock will be taken to be fixed objects within the main inertial frame.
> The *watches* will be those clocks that travel.
> So you have the watch of the accelerating rocket, and you have another watch for
> the inertial rocket.

Absolutly.

I remind you that the traveler leaves the earth in a rocket accelerated
at 10m/s (that is a=1.052 c/ly) and that he will travel over a distance of
12 ly.

There are two watches in this adventure.

The terrestrial watch (earth-Tau Ceti frame of reference) which will
measure the observable time in this frame of reference, and the rocket
watch, which will measure the rocket's own time.

All theoreticians in the world today agree that To=(x/c).sqrt(1+2c²/ax)

It is on the proper time of the rocket that colossal differences exist in
relativistic predictions because those who use Minkowki space (4D) do not
use Hachel space (3D-space+4D-time ) which seems much simpler and more
logical to me.

R.H.

[rotchm]

On Tuesday, October 4, 2022 at 7:27:49 PM UTC-4, Richard Hachel wrote:

> If you do not want to place the watch perpendicularly

That sentence makes no sense. Trying to stated differently.

> and far enough to observe a small transfer in the defined segment,

That sentence makes no sense oh, it's gibberish.
Try to restate it differently, coherently, and clearly.

> you can also place the
> watch in the center of the segment.

No. The watches are the devices that move. The clocks are the devices that remain fixed in the main inertial frame.
Try to be consistent in your scenario and on the terminology.

> The watch will have a small information delay when the rocket arrives for
> example at A, but since it will have the same information delay as when
> the watch arrives at B, the time dTo of this watch will be valid.

Not sure what you mean by all that. You might need to restate your scenario completely and in clear terms.


> <http://news2.nemoweb.net/?DataID=r_9PN0wQl1cwpVKtFpkiKSwcSxQ@jntp>

Link doesn't work.

[rotchm]

On Tuesday, October 4, 2022 at 7:41:17 PM UTC-4, Richard Hachel wrote:
> Le 05/10/2022 à 01:03, rotchm a écrit :

> I remind you that the traveler leaves the earth in a rocket accelerated
> at 10m/s (that is a=1.052 c/ly) and that he will travel over a distance of
> 12 ly.

OK...

> There are two watches in this adventure.

Lets see about that...

> The terrestrial watch (earth-Tau Ceti frame of reference)

No, that's a clock. Again, a *watch* will represent a clock that is moving relative to the
primary inertial frame (Earth frame).

> which will
> measure the observable time in this frame of reference,

Observable time? What do you mean by that?
The value of which clock/watch?

> and the rocket
> watch, which will measure the rocket's own time.

Yes it's watch Cycles through values. It displays values.

> All theoreticians in the world today agree that To=(x/c).sqrt(1+2c²/ax)

Specify what To represents.

> Hachel space (3D-space+4D-time ) which seems much simpler and more

To you it seems simpler and more logical. Two others it doesn't. Moreover, yours does not agree with actual experiments.

[Stan Fultoni]

On Tuesday, October 4, 2022 at 3:44:08 PM UTC-7, Richard Hachel wrote:
> >> If we take a very small segment of a trajectory, we can consider that a
> >> clock in accelerated motion behaves like a clock in inertial motion.
> >
> > Right, and for that small segment the clock moves a small distance Δx in a
> > small time Δt. During this segment the elapsed proper time is Δtau = sqrt(Δt^2
> > - Δx^2). The total elapsed time for the clock on this accelerating trajectory is
> > the sum of the elapsed times of the small segments.
> >
> > In your example, this gives 3.14 years for the trajectory of constant proper
> > acceleration. If we apply this to the unaccelerated trajectory between the same
> > starting and ending events, we get 4.77 years. Understand?
>

> The clock does not move.

The elapsed proper time inside the rocket is measured by a clock that is inside the rocket, moving along with the accelerating trajectory of the rocket. The trajectory of the rocket (with constant proper acceleration "a") in terms of the coordinates x,t is defined by t = sqrt(x^2 + 2x/a)). We can choose a series of x values and compute the corresponding t values

x ... t
0 ... 0
1 ... 1.703
2 ... 2.793
3 ... 3.835
4 ... 4.858
: ... :
12 ... 12.915

Each segment is virtually inertial, so the elapsed proper time on the clock inside the rocket moving from xi,ti to xj,tj is virtually sqrt[(tj-ti)^2 - (xj-xi)^2]. Adding up these proper times gives about 3.14 years, and the more segments we use, the closer the sum approaches 3.14 years.

Question for Richard Hachel:
If a clock moves with constant velocity from xi,ti to xj,tj, how much elapsed time will show on the clock?

[Cesario De rege]

rotchm wrote:

> On Tuesday, October 4, 2022 at 7:41:17 PM UTC-4, Richard Hachel wrote:
>> Le 05/10/2022 à 01:03, rotchm a écrit :
>
>> I remind you that the traveler leaves the earth in a rocket accelerated
>> at 10m/s (that is a=1.052 c/ly) and that he will travel over a distance
>> of 12 ly.
>
> OK...
>
>> There are two watches in this adventure.
>
> Lets see about that...

shut the fuck up fool. You don't know what a ly is.