• #### [SR] Usefulness of real velocities in accelerated relativistic frames o

From Richard Hachel@21:1/5 to All on Tue Mar 12 09:00:29 2024
We know that in accelerated frames of reference the average speed is proportional to the instantaneous speed.

Let Vrm=(1/2)Vri

We are talking about real relativistic speeds (Richard Verret's hobby).

But we know that relativistic physicists do not use this notion, and it is important to remind them of the relationship between real speed and
observable speed.

Vr=Vo/sqrt(1-Vo²/c²)

Vom/sqrt(1-Vom²/c²)=(1/2)Voi/sqrt(1-Voi²/c²)

and therefore to:

Vom²(1-Voi²/c²)=(1/4)Voi²(1-Vom²/c²)

Let Vom²=(1/4)Voi²+(3/4)Voi².Vom²/c²

Hence Vom=(1/2)Voi/sqrt(1-(3/4)Vom²/c²) and, conversely, Voi=2.Vom/sqrt(1+3Vom²/c²)

Thank you for your kind attention.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Tue Mar 12 20:20:43 2024
Den 12.03.2024 10:00, skrev Richard Hachel:
We know that in accelerated frames of reference the average speed is proportional to the instantaneous speed.

Let Vrm=(1/2)Vri

This is meaningless without definition of the entities.

But I can guess:
You are not talking about "speed in an accelerated frame".

You are talking about the speed of a stationary object
in an accelerated frame, measured in an inertial frame.

If the object is instantly at rest at t = 0, and
the coordinate acceleration in the inertial frame is constant a,
Then the speed Vri(t) = at
and the average speed from t=0 to t=t is Vrm(t) = at/2.

So Vrm=(1/2)Vri

We are talking about real relativistic speeds (Richard Verret's hobby).

According to SR:

For the above to be true, the coordinate acceleration must
be constant. This means that the proper acceleration must
increase with time.

Let 1g = 1 c per year = 9.4998 m/s².

To keep the coordinate acceleration a equal to 1 g
the proper acceleration A must be:
at t = 0 year A = 1.00 g
at t = 0.8 year A = 1.65 g
at t = 0.9 year A = 3.33 g
at t = 0.99 year A = 33.4 g
at t = 0.999 year A = 333.5 g
at t = 1.000 year A = infinite

when t ≥ 1.0 year the coordinate acceleration can't
be kept equal to 1 g.

Generally:
The coordinate acceleration a can't be kept equal to n g
when t ≥ 1/n year.

It is obviously normal to keep the proper acceleration constant,
and then Vrm ≠ (1/2)Vri

But we know that relativistic physicists do not use this notion, and it
is important to remind them of the relationship between real speed and observable speed.

Vr=Vo/sqrt(1-Vo²/c²)

This is nonsense.
According to "relativistic physicists" there is no difference
between the "real speed" and the "observable (measurable) speed"
in the inertial frame.

This immediately leads us to Vom/sqrt(1-Vom²/c²)=(1/2)Voi/sqrt(1-Voi²/c²)

and therefore to:

Vom²(1-Voi²/c²)=(1/4)Voi²(1-Vom²/c²)

Let Vom²=(1/4)Voi²+(3/4)Voi².Vom²/c²

Hence Vom=(1/2)Voi/sqrt(1-(3/4)Vom²/c²) and, conversely, Voi=2.Vom/sqrt(1+3Vom²/c²)

Thank you for your kind attention.

R.H.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Tue Mar 12 19:42:31 2024
Le 12/03/2024 à 20:19, "Paul B. Andersen" a écrit :
Den 12.03.2024 10:00, skrev Richard Hachel:
We know that in accelerated frames of reference the average speed is
proportional to the instantaneous speed.

Let Vrm=(1/2)Vri

This is meaningless without definition of the entities.

But I can guess:
You are not talking about "speed in an accelerated frame".

You are talking about the speed of a stationary object
in an accelerated frame, measured in an inertial frame.

If the object is instantly at rest at t = 0, and
the coordinate acceleration in the inertial frame is constant a,
Then the speed Vri(t) = at
and the average speed from t=0 to t=t is Vrm(t) = at/2.

So Vrm=(1/2)Vri

That is what I am saying.

Vr=a.Tr

Vrm=(1/2)Vri

We are talking about real relativistic speeds (Richard Verret's hobby).

According to SR:

For the above to be true, the coordinate acceleration must
be constant. This means that the proper acceleration must
increase with time.

Let 1g = 1 c per year = 9.4998 m/s².

To keep the coordinate acceleration a equal to 1 g
the proper acceleration A must be:
at t = 0 year A = 1.00 g
at t = 0.8 year A = 1.65 g
at t = 0.9 year A = 3.33 g
at t = 0.99 year A = 33.4 g
at t = 0.999 year A = 333.5 g
at t = 1.000 year A = infinite

when t ≥ 1.0 year the coordinate acceleration can't
be kept equal to 1 g.

Generally:
The coordinate acceleration a can't be kept equal to n g
when t ≥ 1/n year.

It is obviously normal to keep the proper acceleration constant,
and then Vrm ≠ (1/2)Vri

I don't understand what you are saying.

I think that between you and me, there is a different understanding of
things.

But we know that relativistic physicists do not use this notion, and it
is important to remind them of the relationship between real speed and
observable speed.

Vr=Vo/sqrt(1-Vo²/c²)

This is nonsense.
According to "relativistic physicists" there is no difference
between the "real speed" and the "observable (measurable) speed"
in the inertial frame.

That is what I am saying.
Physicists do not differentiate between Vo and Vr.

This immediately leads us to Vom/sqrt(1-Vom²/c²)=(1/2)Voi/sqrt(1-Voi²/c²)

and therefore to:

Vom²(1-Voi²/c²)=(1/4)Voi²(1-Vom²/c²)

Let Vom²=(1/4)Voi²+(3/4)Voi².Vom²/c²

Hence Vom=(1/2)Voi/sqrt(1-(3/4)Vom²/c²) and, conversely,
Voi=2.Vom/sqrt(1+3Vom²/c²)

Thank you for your kind attention.

It is really a shame that a large majority of physicists are not very intelligent, they understand nothing of what they are saying,
which is dramatic since their number serves as truth.
A bit like in Nazi Germany everyone believed they were right because
everyone held out their arm in front of Hitler.
Physicists are educated, but very few are intelligent.

On usenet, there are perhaps five or six people with whom we can seriously discuss relativity, without immediately falling into hysterical reactions. (Julien Arlandis, Michel Talon, Paul B Anderson...). The others not only learned anything, but answer nonsense when asked a simple problem.

A good example of this madness here is Python, as stupid as he is
aggressive.

R.H.

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• From Richard Hachel@21:1/5 to All on Tue Mar 12 19:54:01 2024
Le 12/03/2024 à 20:19, "Paul B. Andersen" a écrit :
Den 12.03.2024 10:00, skrev Richard Hachel:

Let 1g = 1 c per year = 9.4998 m/s².

To keep the coordinate acceleration a equal to 1 g
the proper acceleration A must be:
at t = 0 year A = 1.00 g
at t = 0.8 year A = 1.65 g
at t = 0.9 year A = 3.33 g
at t = 0.99 year A = 33.4 g
at t = 0.999 year A = 333.5 g
at t = 1.000 year A = infinite

No.

It's not the current problem.

at t = 0 year A = 1.00 g
at t = 0.8 year A = 1.00 g
at t = 0.9 year A = 1.00 g
at t = 0.99 year A = 1.00 g
at t = 0.999 year A = 1.00 g
at t = 1.000 year A = 1.00 g

I beg you to understand something:
in the rocket's frame of reference, the acceleration is constant.
The motors are manufactured to give a constant acceleration of 10m/s² (a=1.052ly/y²) and there is no need to look for rabbit crones.
Certainly, in the terrestrial reference frame, this acceleration will
appear to gradually decrease over time and this will give the following equation for the acceleration observed in the terrestrial reference frame:

<http://news2.nemoweb.net/jntp?xw8jiaP3Bbvt3aHF_FHQ1Rt5-Jc@jntp/Data.Media:1>

But in the framework of the rocket, built for that, there is no need to
make any unnecessary modifications.

<http://news2.nemoweb.net/?DataID=xw8jiaP3Bbvt3aHF_FHQ1Rt5-Jc@jntp>

R.H.

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• From Richard Hachel@21:1/5 to All on Tue Mar 12 19:55:39 2024
Le 12/03/2024 à 20:19, "Paul B. Andersen" a écrit :
Den 12.03.2024 10:00, skrev Richard Hachel:
We know that in accelerated frames of reference the average speed is
proportional to the instantaneous speed.

Let Vrm=(1/2)Vri

This is meaningless without definition of the entities.

But I can guess:
You are not talking about "speed in an accelerated frame".

You are talking about the speed of a stationary object
in an accelerated frame, measured in an inertial frame.

If the object is instantly at rest at t = 0, and
the coordinate acceleration in the inertial frame is constant a,
Then the speed Vri(t) = at
and the average speed from t=0 to t=t is Vrm(t) = at/2.

So Vrm=(1/2)Vri

That is what I am saying.

Vr=a.Tr

Vrm=(1/2)Vri

We are talking about real relativistic speeds (Richard Verret's hobby).

According to SR:

For the above to be true, the coordinate acceleration must
be constant. This means that the proper acceleration must
increase with time.

Let 1g = 1 c per year = 9.4998 m/s².

To keep the coordinate acceleration a equal to 1 g
the proper acceleration A must be:
at t = 0 year A = 1.00 g
at t = 0.8 year A = 1.65 g
at t = 0.9 year A = 3.33 g
at t = 0.99 year A = 33.4 g
at t = 0.999 year A = 333.5 g
at t = 1.000 year A = infinite

when t ≥ 1.0 year the coordinate acceleration can't
be kept equal to 1 g.

Generally:
The coordinate acceleration a can't be kept equal to n g
when t ≥ 1/n year.

It is obviously normal to keep the proper acceleration constant,
and then Vrm ≠ (1/2)Vri

I don't understand what you are saying.

I think that between you and me, there is a different understanding of
things.

But we know that relativistic physicists do not use this notion, and it
is important to remind them of the relationship between real speed and
observable speed.

Vr=Vo/sqrt(1-Vo²/c²)

This is nonsense.
According to "relativistic physicists" there is no difference
between the "real speed" and the "observable (measurable) speed"
in the inertial frame.

That is what I am saying.
Physicists do not differentiate between Vo and Vr.

This immediately leads us to Vom/sqrt(1-Vom²/c²)=(1/2)Voi/sqrt(1-Voi²/c²)

and therefore to:

Vom²(1-Voi²/c²)=(1/4)Voi²(1-Vom²/c²)

Let Vom²=(1/4)Voi²+(3/4)Voi².Vom²/c²

Hence Vom=(1/2)Voi/sqrt(1-(3/4)Vom²/c²) and, conversely,
Voi=2.Vom/sqrt(1+3Vom²/c²)

Thank you for your kind attention.

It is really a shame that a large majority of physicists are not very intelligent, they understand nothing of what they are saying,
which is dramatic since their number serves as truth.
A bit like in Nazi Germany everyone believed they were right because
everyone held out their arm in front of Hitler.
Physicists are educated, but very few are intelligent.

On usenet, there are perhaps five or six people with whom we can seriously discuss relativity, without immediately falling into hysterical reactions. (Julien Arlandis, Michel Talon, Paul B Andersen...). The others not only learned anything, but answer nonsense when asked a simple problem.

A good example of this madness here is Python, as stupid as he is
aggressive.

R.H.

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• From Maciej Wozniak@21:1/5 to All on Wed Mar 13 07:09:33 2024
W dniu 12.03.2024 o 20:42, Richard Hachel pisze:

On usenet, there are perhaps five or six people with whom we can
seriously discuss relativity, without immediately falling into
hysterical reactions. (Julien Arlandis, Michel Talon, Paul B
Anderson...). The others not only learned anything, but answer nonsense

Nope, no relativistic idiot can be discussed. The
Shit is not training its doggiesd for discussing,
it's training them for barking.

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to All on Wed Mar 13 22:00:30 2024
Den 12.03.2024 20:42, skrev Richard Hachel:
Le 12/03/2024 à 20:19, "Paul B. Andersen" a écrit :
Den 12.03.2024 10:00, skrev Richard Hachel:
We know that in accelerated frames of reference the average speed is
proportional to the instantaneous speed.

Let Vrm=(1/2)Vri

This is meaningless without definition of the entities.

But I can guess:
You are not talking about "speed in an accelerated frame".

You are talking about the speed of a stationary object
in an accelerated frame, measured in an inertial frame.

If the object is instantly at rest at t = 0, and
the coordinate acceleration in the inertial frame is constant a,
Then the speed Vri(t) = at
and the average speed from t=0 to t=t is Vrm(t) = at/2.

So  Vrm=(1/2)Vri

That is what I am saying.

Yes. And I agree.

NOTE THIS:
If the object is instantly at rest at t = 0, and
the coordinate acceleration in the inertial frame is constant a,
Then the speed Vri(t) = at

Vr=a.Tr

Where Vr (Vri(t)) is the speed in the inertial frame,
and Tr (t) is the time coordinate in the inertial frame, and ----------------------------------------------------------------
a is the constant coordinate acceleration in the inertial frame! ================================================================

The consequence of this is that when t > c/a v > c
which is impossible in SR.

That means that the coordinate acceleration a can't be constant! ==============================================================

According to SR:

For the above to be true, the coordinate acceleration must
be constant. This means that the proper acceleration must
increase with time.

Let 1g = 1 c per year = 9.4998 m/s².

when t ≥ 1.0 year the coordinate acceleration can't
be kept equal to 1 g.

Generally:
The coordinate acceleration a can't be kept equal to n g
when t ≥ 1/n year.

It is obviously normal to keep the proper acceleration constant,
and then Vrm  ≠ (1/2)Vri

As you said yourself:
"I beg you to understand something:
in the rocket's frame of reference, the acceleration is constant."

That is the proper acceleration A is constant, and then
the coordinate acceleration a is NOT constant, it is
decreasing with time, Vri < at and Vrm > (1/2)Vri.

Se exact calculation below.

I don't understand what you are saying.

I think that between you and me, there is a different understanding of things.

Indeed!

Understand this:
What SR predicts is not a matter of opinion,
it is a matter of fact.

https://paulba.no/pdf/TwinsByMetric.pdf
see equation (38)

It is a FACT that according to SR, the speed of
an object with constant proper acceleration A is:

Vri(t) = A⋅t/√(1+(A⋅t/c)²)

Note that:
Vri → A⋅t when t → 0
Vri → c when t → ∞

The average speed Vrm at the time t is:
Vrm = (integral from t=0 to t=t of Vri(t)dt)/t
Vrm = c²⋅(√(1+(A⋅t/c)²)-1)/A⋅t

Note that:
Vrm → A⋅t/2 when t → 0
Vrm → c when t → ∞

So:
Vrm/Vri → 1/2 when t → 0
Vrm/Vri → 1 when t → ∞

So for any t > 0 Vrm > (1/2)Vri

It is not possible to make SR predict anything else! ====================================================

The coordinate acceleration a is:
a = dVri/dt = A/(√(1+(A⋅t/c)²))³
where A is the proper acceleration

Note that:
a → A when t → 0
a → 0 when t → ∞

But we know that relativistic physicists do not use this notion, and
it is important to remind them of the relationship between real speed
and observable speed.

Vr=Vo/sqrt(1-Vo²/c²)

This is nonsense.
According to "relativistic physicists" there is no difference
between the "real speed" and the "observable (measurable) speed"
in the inertial frame.

That is what I am saying.
Physicists do not differentiate between Vo and Vr.

Because there is no such difference.

You are claiming that when you measure the speed of a passing object,
then you will always measure the real speed divided by γ.

Can you please explain how you arrived at this conclusion?

It is really a shame that a large majority of physicists are not very intelligent, they understand nothing of what they are saying,
which is dramatic since their number serves as truth.

:-D

The "large majority of physicists" are obviously much more
intelligent than average. They have to be to pass the exams.
(I am not a physicist.)

A bit like in Nazi Germany everyone believed they were right because
everyone held out their arm in front of Hitler.
Physicists are educated, but very few are intelligent.

A hysterical reaction!

On usenet, there are perhaps five or six people with whom we can
seriously discuss relativity, without immediately falling into
hysterical reactions. (Julien Arlandis, Michel Talon, Paul B
Anderson...). The others not only learned anything, but answer nonsense

Above you have demonstrated that you belong to the group that
falls into hysterical reactions.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Thu Mar 14 02:00:57 2024
Le 13/03/2024 à 21:59, "Paul B. Andersen" a écrit :
Den 12.03.2024 20:42, skrev Richard Hachel:

It is obviously normal to keep the proper acceleration constant,
and then Vrm  ≠ (1/2)Vri

We know that in accelerated frames of reference the average speed is proportional to the instantaneous speed.

Let Vrm=(1/2)Vri

We are talking about real relativistic speeds (Richard Verret's hobby).

But we know that relativistic physicists do not use this notion, and it is important to remind them of the relationship between real speed and
observable speed.

Vr=Vo/sqrt(1-Vo²/c²)

Vom/sqrt(1-Vom²/c²)=(1/2)Voi/sqrt(1-Voi²/c²)

and therefore to:

Vom²(1-Voi²/c²)=(1/4)Voi²(1-Vom²/c²)

Let Vom²=(1/4)Voi²+(3/4)Voi².Vom²/c²

Hence
Vom=(1/2)Voi/sqrt(1-(3/4)Vom²/c²)
and, conversely,
Voi=2.Vom/sqrt(1+3Vom²/c²)

R.H.

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• From Richard Hachel@21:1/5 to All on Thu Mar 14 01:57:38 2024
Le 13/03/2024 à 21:59, "Paul B. Andersen" a écrit :
Den 12.03.2024 20:42, skrev Richard Hachel:

So  Vrm=(1/2)Vri

That is what I am saying.

Yes. And I agree.

NOTE THIS:
If the object is instantly at rest at t = 0, and
the coordinate acceleration in the inertial frame is constant a,
Then the speed Vri(t) = at

Vr=a.Tr

Where Vr (Vri(t)) is the speed in the inertial frame,
and Tr (t) is the time coordinate in the inertial frame, and ----------------------------------------------------------------
a is the constant coordinate acceleration in the inertial frame! ================================================================

The consequence of this is that when t > c/a v > c
which is impossible in SR.

That means that the coordinate acceleration a can't be constant!

In the rocket frame, a is constant.
Always.
a=10m/s² (a=1.052ly/y²)
The rocket is at rest in its frame of reference, and the speed Vr of the surrounding space becomes Vr=a.Tr
There is no problem, the speed of the rocket, that is to say the real
speed of movement of the terrestrial frame of reference, is indeed Vr.
But you know the conversion equation that I gave forty years ago: Vo=Vr/sqrt(1+Vr²/c²)

Everything is remarkably clear and of great theoretical beauty if you understand what I am saying.

R.H.

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• From Richard Hachel@21:1/5 to All on Thu Mar 14 02:09:10 2024
Le 13/03/2024 à 21:59, "Paul B. Andersen" a écrit :
Den 12.03.2024 20:42, skrev Richard Hachel:

Understand this:
What SR predicts is not a matter of opinion,
it is a matter of fact.

Above all, we must prioritize experimentation.
And if two theories face each other, we must take the one which has the approval of the experimenters.

If we take Einstein's RR, which is taken from the Poincaré equation and
the Minkowski metric, and which is called "local", we realize that two
enormous problems appear.
- The Langevin paradox (which we camouflaged, but which resurfaces using apparent speeds, that is to say what we would observe if we had very good telescopes and sufficient rocket technology).
- The principle of non-locality of Aspect and its contradiction with instantaneous transfers of information.

These contradictions do not exist with me, and I have never seen a single
fact contradicting everything I have said for 40 years.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Thu Mar 14 02:35:35 2024
Le 13/03/2024 à 21:59, "Paul B. Andersen" a écrit :
Den 12.03.2024 20:42, skrev Richard Hachel:

It is a FACT that according to SR, the speed of
an object with constant proper acceleration A is:

Vri(t) = A⋅t/√(1+(A⋅t/c)²)

? ? ?

Vri(t)=a.Tr

Voi(Tr!!!)=Vri/sqrt(1+Vri²/c²)

Voi=[1+c²/2ax]^(-1/2)

To=Tr.sqrt[1+(1/4)Vri²/c²]

Note that:
Vri → A⋅t when t → 0
Vri → c when t → ∞

The average speed Vrm at the time t is:
Vrm = (integral from t=0 to t=t of Vri(t)dt)/t
Vrm = c²⋅(√(1+(A⋅t/c)²)-1)/A⋅t

Note that:
Vrm → A⋅t/2 when t → 0
Vrm → c when t → ∞

No.

Vrm → ∞

Vr is not Vo.

So:
Vrm/Vri → 1/2 when t → 0
Vrm/Vri → 1 when t → ∞

No.

Physicists do not differentiate between Vo and Vr.

Because there is no such difference.

? ? ?

The "large majority of physicists" are obviously much more
intelligent than average. They have to be to pass the exams.
(I am not a physicist.)

Yes, they have to take exams.

Like everyone.

Personally, I'm happy to be theoretically logical and experimental in
racing.

This is NOT the current condition of physicists.

Their theory is imperfect and contradictory (Vapp in Langevin), it is
refuted by experimentation (their Sr is local, but not mine).

A bit like in Nazi Germany everyone believed they were right because
everyone held out their arm in front of Hitler.
Physicists are educated, but very few are intelligent.

A hysterical reaction!

No. In SR, is fact.

Ils ne comprennent pas ce qu'ils disent, ce n'est PAS normal.

On usenet, there are perhaps five or six people with whom we can
seriously discuss relativity, without immediately falling into
hysterical reactions. (Julien Arlandis, Michel Talon, Paul B
Andersen...). The others not only learned anything, but answer nonsense

Above you have demonstrated that you belong to the group that
falls into hysterical reactions.

Simply ask yourself: “Is this man lying, or is he telling the truth?”

R.H.

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• From Paul B. Andersen@21:1/5 to All on Thu Mar 14 15:04:13 2024
Den 14.03.2024 03:09, skrev Richard Hachel:
Le 13/03/2024 à 21:59, "Paul B. Andersen" a écrit :

Understand this:
What SR predicts is not a matter of opinion,
it is a matter of fact.

The subject line is:
"SR: Usefulness of real velocities in accelerated relativistic frames of reference."

I take this to mean that you are stating what you claim SR predicts.

Above all, we must prioritize experimentation.
And if two theories face each other, we must take the one which has the approval of the experimenters.

Indeed.
And you know of course that SR is confirmed by innumerable
experiments and falsified by none.
Some of them:
https://paulba.no/paper/index.html

But the issue is:
Does SR predict what you claim it predicts?

If we take Einstein's RR, which is taken from the Poincaré equation and
the Minkowski metric, and which is called "local", we realize that two enormous problems appear.
- The Langevin paradox (which we camouflaged, but which resurfaces using apparent speeds, that is to say what we would observe if we had very
good telescopes and sufficient rocket technology).
- The principle of non-locality of Aspect and its contradiction with instantaneous transfers of information.

Your opinion of SR is irrelevant.

SR is a consistent theory, and the issue is:
"Does SR predict that accelerated objects will behave as
you claim they do?"

These contradictions do not exist with me, and I have never seen a
single fact contradicting everything I have said for 40 years.

This is probably true, because it is obvious that you never

The scenario as defined by you:
"In the rocket frame, a is constant. The rocket is at rest
in its frame of reference, and the speed Vr of the surrounding
space becomes Vr=a.Tr."

A bit more precisely put:
A rocket is accelerating at the constant proper acceleration a.
An inertial frame of reference K(x,t) is at the time t = 0
instantly co-moving with the rocket.

You claim:
According to SR the speed of the rocket in K is Vr(t) = a⋅t ===========================================================

Note that this means that Vr > c when t > c/a
which according to SR is impossible.

-------------------
So this is wrong.
You can see the correct derivation here: https://paulba.no/pdf/TwinsByMetric.pdf
See chapter 2.3, equation (15)

Vr(t) = a⋅t/√(1+(a⋅t/c)²)

Note that:
Vr → a⋅t when t → 0
Vr → c when t → ∞

Your problem is that you do not understand the difference
between proper acceleration of the rocket, and the rocket's
coordinate acceleration in the inertial frame.

If A is the coordinate acceleration in K, we have:

A = dVr/dt = a/(√(1+(a⋅t/c)²))³

Note that:
A → a when t → 0
A → 0 when t → ∞

So Vr(t) = ∫(from 0 to t)A⋅dt = a⋅t/√(1+(a⋅t/c)²)

You claim:
According to SR is the average speed of the rocket Vm(t) = Vr(t)/2 =====================================================================

-------------------
This is wrong.

Vr(t) = a⋅t/√(1+(a⋅t/c)²)

The average speed Vm at the time t is:
Vm = (integral from t=0 to t=t of Vr(t)dt)/t
Vm = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t

Note that:
Vm → a⋅t/2 when t → 0
Vm → c when t → ∞

So:
Vm/Vr → 1/2 when t → 0
rm/Vr → 1 when t → ∞

So for any t > 0 Vm > Vr/2

It is not possible to make SR predict anything else! ====================================================

--
Paul

https://paulba.no/

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• From Maciej Wozniak@21:1/5 to All on Thu Mar 14 15:54:14 2024
W dniu 14.03.2024 o 15:04, Paul B. Andersen pisze:

And you know of course that SR is confirmed by innumerable
experiments and falsified by none.

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

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• From Richard Hachel@21:1/5 to All on Thu Mar 14 16:03:46 2024
Le 14/03/2024 à 15:02, "Paul B. Andersen" a écrit :
The subject line is:
"SR: Usefulness of real velocities in accelerated relativistic frames of reference."

I take this to mean that you are stating what you claim SR predicts.

Above all, we must prioritize experimentation.
And if two theories face each other, we must take the one which has the
approval of the experimenters.

Indeed.
And you know of course that SR is confirmed by innumerable
experiments and falsified by none.
Some of them:
https://paulba.no/paper/index.html

But the issue is:
Does SR predict what you claim it predicts?

Here is the problem:
We now know that Newtonian physics is out of the running, and that we must
use a relativistic theory.
This is certain, and the evidence is so abundant that no one disputes it anymore, except a few crazy people.
But here we have, face to face, two ways of seeing the SR.
That of Einstein and Minkowski, and that of Hachel and Aspect.
Both theories predict the same things.
Except that Einstein and Minkowski's theory is ridiculous and
contradictory on paper (internal imperfection), and that it is
contradicted by the non-locality experiment (external refutation).
I am the only one to propose a coherent theory, theoretically unassailable
and flawless, and with all the experiments on my side (including the instantaneous longitudinal transmission of information and the
impossibility of transverse transmissions greater than c).

R.H.

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• From Richard Hachel@21:1/5 to All on Thu Mar 14 16:13:46 2024
Le 14/03/2024 à 15:02, "Paul B. Andersen" a écrit :
Your opinion of SR is irrelevant.

SR is a consistent theory, and the issue is:
"Does SR predict that accelerated objects will behave as
you claim they do?"

No, YOU are saying that my concepts are irrelevant.

As for accelerated frames of reference, I said that things are poorly
explained by physicists and that certain equations are false.

Some are correct, like
To=(x/c).sqrt(1+2c²/ax)
Or
x=(c²/a)[sqrt(1+a²To²/c²)-1]

But many others are incorrect.

Same thing for rotating frames of reference where we sink into conceptual madness by proposing a crazy geometry where a rotating disk becomes an
unknown figure on the earth (LOL).

R.H.

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• From Richard Hachel@21:1/5 to All on Thu Mar 14 16:18:32 2024
Le 14/03/2024 à 15:02, "Paul B. Andersen" a écrit :
Den 14.03.2024 03:09, skrev Richard Hachel:

-------------------
So this is wrong.
You can see the correct derivation here: https://paulba.no/pdf/TwinsByMetric.pdf
See chapter 2.3, equation (15)

Vr(t) = a⋅t/√(1+(a⋅t/c)²)

Note that:
Vr → a⋅t when t → 0
Vr → c when t → ∞

Your problem is that you do not understand the difference
between proper acceleration of the rocket, and the rocket's
coordinate acceleration in the inertial frame.

If A is the coordinate acceleration in K, we have:

A = dVr/dt = a/(√(1+(a⋅t/c)²))³

Note that:
A → a when t → 0
A → 0 when t → ∞

So Vr(t) = ∫(from 0 to t)A⋅dt = a⋅t/√(1+(a⋅t/c)²)

You claim:
According to SR is the average speed of the rocket Vm(t) = Vr(t)/2 =====================================================================

-------------------
This is wrong.

Vr(t) = a⋅t/√(1+(a⋅t/c)²)

The average speed Vm at the time t is:
Vm = (integral from t=0 to t=t of Vr(t)dt)/t
Vm = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t

Note that:
Vm → a⋅t/2 when t → 0
Vm → c when t → ∞

So:
Vm/Vr → 1/2 when t → 0
rm/Vr → 1 when t → ∞

So for any t > 0 Vm > Vr/2

It is not possible to make SR predict anything else! ====================================================

You don't understand anything I'm telling you...

In these conditions, it is very difficult to discuss.

R.H.

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• From Maciej Wozniak@21:1/5 to All on Thu Mar 14 18:38:11 2024
W dniu 14.03.2024 o 17:03, Richard Hachel pisze:
Le 14/03/2024 à 15:02, "Paul B. Andersen" a écrit :
The subject line is:
"SR: Usefulness of real velocities in accelerated relativistic frames
of reference."

I take this to mean that you are stating what you claim SR predicts.

Above all, we must prioritize experimentation.
And if two theories face each other, we must take the one which has
the approval of the experimenters.

Indeed.
And you know of course that SR is confirmed by innumerable
experiments and falsified by none.
Some of them:
https://paulba.no/paper/index.html

But the issue is:
Does SR predict what you claim it predicts?

Here is the problem:
We now know that Newtonian physics is out of the running, and that we
must use a relativistic theory.
This is certain, and the evidence is so abundant that no one disputes it anymore, except a few crazy people.

And we also know communism must win,
as it's the best.
We know it, of course, except a few
crazy people.

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• From Paul B. Andersen@21:1/5 to All on Fri Mar 15 15:13:31 2024
Den 14.03.2024 17:03, skrev Richard Hachel:
Le 14/03/2024 à 15:02, "Paul B. Andersen" a écrit :
The subject line is:
"SR: Usefulness of real velocities in accelerated relativistic frames
of reference."

I take this to mean that you are stating what you claim SR predicts.

Above all, we must prioritize experimentation.
And if two theories face each other, we must take the one which has
the approval of the experimenters.

Indeed.
And you know of course that SR is confirmed by innumerable
experiments and falsified by none.
Some of them:
https://paulba.no/paper/index.html

But the issue is:
Does SR predict what you claim it predicts?

Here is the problem:
We now know that Newtonian physics is out of the running, and that we
must use a relativistic theory.
This is certain, and the evidence is so abundant that no one disputes it anymore, except a few crazy people.
But here we have, face to face, two ways of seeing the SR.

There is not "Two ways of seeing SR".
SR is mathematical consistent and can't be made

(c⋅dτ)² = (c⋅dt)² − dx² − dy² − dz²
and everything follows from that.

The equations relating to our discussion are:

Rocket with constant proper acceleration a.

Entities measured in the inertial frame K(x,t):

Speed: Vr(t) = a⋅t/√(1+(a⋅t/c)²)
Average speed Vm(t) = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t
Coordinate acceleration A = dVr/dt = a/(√(1+(a⋅t/c)²))³

It is not possible to make SR predict anything else! ====================================================

show the math that leads to the alternative equations
you think are right!

I am the only one to propose a coherent theory, theoretically
unassailable and flawless, and with all the experiments on my side
(including the instantaneous longitudinal transmission of information
and the impossibility of transverse transmissions greater than c).

If you have a theory which give different answers from SR,
then it will be experimentally falsified.

I am not interested in discussing yet another crackpot theory.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Fri Mar 15 14:20:15 2024
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :
I am saying that SR does not predict that accelerated objects
will behave as you claim they do.

We agree.
The SR does not predict things the way I predict them.
The problem, although enormous, is that experimentally I am credible and
not the SR which cannot explain instantaneous transfers of information,
and that theoretically I do not have the enormous concern of completely inconsistent apparent speeds. if only in a simple Langevin paradox.

R.H.

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• From Richard Hachel@21:1/5 to All on Fri Mar 15 14:30:21 2024
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :
I suppose you are referring to my "false" equations:
Speed: Vr(t) = a⋅t/√(1+(a⋅t/c)²)
Average speed Vm(t) = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t

Speed: Vr(t) = a⋅t
Average speed Vm(t) = a⋅t/2

:-D

The speed of the accelerated mobile or particle
is a function of time.
The more time passes, the greater the speed.
Everyone agrees on that.
Now we have to give the correct equation.
I wrote that the correct equation is:
Vo(Tr)=a.Tr/sqrt(1+(a.Tr)²/c²)
I don't see what your problem is.
Vr(Tr)=a.Tr
Vrm=Vri/2
What do you not understand?

R.H.

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• From Richard Hachel@21:1/5 to All on Fri Mar 15 14:39:18 2024
Le 15/03/2024 à 15:12, "Paul B. Andersen" a écrit :
I do indeed understand that you telling me:
Vr(t) = a⋅t
and:
Vm(t) = Vr(t)/2

And I do indeed understand that what you are telling me is wrong.
And it is a very naive and elementary blunder!

In these conditions, it is very difficult to discuss.

I do understand that you find it difficult to defend your own words.
So that's why you don't even try, right?

Paul

No, you simply understand that what I say, on certain points, is different
from what the usual SR says.
You conclude, without any serious examination, that it is therefore false.
Your behavior is then not scientific, it is just based on a kind of religiosity.

R.H.

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• From Maciej Wozniak@21:1/5 to All on Fri Mar 15 15:52:47 2024
W dniu 15.03.2024 o 15:13, Paul B. Andersen pisze:

There is not "Two ways of seeing SR".
SR is mathematical consistent and can't be made

No it is not, you've got a proof
and pretending you haven't noticed
won't change anything, poor trash.

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• From Maciej Wozniak@21:1/5 to All on Fri Mar 15 15:54:40 2024
W dniu 15.03.2024 o 15:39, Richard Hachel pisze:
Le 15/03/2024 à 15:12, "Paul B. Andersen" a écrit :
I do indeed understand that you telling me:
Vr(t) = a⋅t
and:
Vm(t) = Vr(t)/2

And I do indeed understand that what you are telling me is wrong.
And it is a very naive and elementary blunder!

In these conditions, it is very difficult to discuss.

I do understand that you find it difficult to defend your own words.
So that's why you don't even try, right?

Paul

No, you simply understand that what I say, on certain points, is
different from what the usual SR says.
You conclude, without any serious examination, that it is therefore false. Your behavior is then not scientific, it is just based on akind of religiosity.

Of course; why wouldn't behaviour of
a fanatic religious maniac be based
on a kind of religiosity?

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• From Richard Hachel@21:1/5 to All on Fri Mar 15 15:47:09 2024
Le 15/03/2024 à 15:54, Maciej Wozniak a écrit :
W dniu 15.03.2024 o 15:39, Richard Hachel pisze:
Le 15/03/2024 à 15:12, "Paul B. Andersen" a écrit :
I do indeed understand that you telling me:
Vr(t) = a⋅t
and:
Vm(t) = Vr(t)/2

And I do indeed understand that what you are telling me is wrong.
And it is a very naive and elementary blunder!

In these conditions, it is very difficult to discuss.

I do understand that you find it difficult to defend your own words.
So that's why you don't even try, right?

Paul

No, you simply understand that what I say, on certain points, is
different from what the usual SR says.
You conclude, without any serious examination, that it is therefore false. >> Your behavior is then not scientific, it is just based on akind of
religiosity.

Of course; why wouldn't behaviour of
a fanatic religious maniac be based
on a kind of religiosity?

The theory of relativity has two important enemies.
Those who believe in it too much and make it their religion without even understanding the basics, and those who don't believe in it at all.
A guitar string must be tensioned correctly. If you don't tighten it
enough, it makes a deep and unpleasant sound. If you stretch it too much,
it will break.

R.H.

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• From Volney@21:1/5 to Richard Hachel on Fri Mar 15 13:45:59 2024
On 3/15/2024 10:30 AM, Richard Hachel wrote:
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :
I suppose you are referring to my "false" equations:
Speed:            Vr(t) = a⋅t/√(1+(a⋅t/c)²)
Average speed     Vm(t) = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t

Speed:  Vr(t) = a⋅t
Average speed  Vm(t) = a⋅t/2

:-D

The speed of the accelerated mobile or particle
is a function of time.
The more time passes, the greater the speed.
Everyone agrees on that.
Now we have to give the correct equation.
I wrote that the correct equation is:
Vo(Tr)=a.Tr/sqrt(1+(a.Tr)²/c²)
I don't see what your problem is.
Vr(Tr)=a.Tr
Vrm=Vri/2
What do you not understand?

You don't explain or even acknowledge the problem with what happens when
Tr > c/a. Besides, your statement Vr(Tr)=a*Tr is Newtonian/Galilean
physics, not SR.

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• From Python@21:1/5 to All on Sat Mar 16 03:56:46 2024
Le 15/03/2024 à 15:13, Paul B. Andersen a écrit :
Den 14.03.2024 17:18, skrev Richard Hachel:
Le 14/03/2024 à 15:02, "Paul B. Andersen" a écrit :
A rocket is accelerating at the constant proper acceleration a.
An inertial frame of reference K(x,t) is at the time t = 0
instantly co-moving with the rocket.

You claim:
According to SR the speed of the rocket in K is Vr(t) = a⋅t
===========================================================

Note that this means that Vr > c when t > c/a
which according to SR is impossible.

A rocket is accelerating at the constant proper acceleration a.
An inertial frame of reference K(x,t) is at the time t = 0
instantly co-moving with the rocket.

You claim:
According to SR the speed of the rocket in K is Vr(t) = a⋅t
===========================================================

Note that this means that Vr > c when t > c/a
which according to SR is impossible.

-------------------
So this is wrong.
You can see the correct derivation here:
https://paulba.no/pdf/TwinsByMetric.pdf
See chapter 2.3, equation (15)

Vr(t) = a⋅t/√(1+(a⋅t/c)²)

Note that:
Vr → a⋅t when t → 0
Vr → c   when t → ∞

Your problem is that you do not understand the difference
between proper acceleration of the rocket, and the rocket's
coordinate acceleration in the inertial frame.

If A is the coordinate acceleration in K, we have:

A = dVr/dt = a/(√(1+(a⋅t/c)²))³

Note that:
A → a when t → 0
A → 0 when t → ∞

So  Vr(t) = ∫(from 0 to t)A⋅dt = a⋅t/√(1+(a⋅t/c)²)

You claim:
According to SR is the average speed of the rocket Vm(t) = Vr(t)/2
=====================================================================

-------------------
This is wrong.

Vr(t) = a⋅t/√(1+(a⋅t/c)²)

The average speed Vm at the time t is:
Vm = (integral from t=0 to t=t of Vr(t)dt)/t
Vm = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t

Note that:
Vm → a⋅t/2 when t → 0
Vm → c     when t → ∞

So:
Vm/Vr  → 1/2  when t → 0
rm/Vr  → 1    when t → ∞

So for any t > 0   Vm > Vr/2

It is not possible to make SR predict anything else!
====================================================

You don't understand anything I'm telling you...

I do indeed understand that you telling me:
Vr(t) = a⋅t
and:
Vm(t) = Vr(t)/2

And I do indeed understand that what you are telling me is wrong.
And it is a very naive and elementary blunder!

In these conditions, it is very difficult to discuss.

I do understand that you find it difficult to defend your own words.
So that's why you don't even try, right?

As he did on fr.sci.physique for decades.

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• From Richard Hachel@21:1/5 to All on Sat Mar 16 07:56:00 2024
Le 15/03/2024 à 18:46, Volney a écrit :
On 3/15/2024 10:30 AM, Richard Hachel wrote:
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :
I suppose you are referring to my "false" equations:
Speed:            Vr(t) = a⋅t/√(1+(a⋅t/c)²)
Average speed     Vm(t) = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t

Speed:  Vr(t) = a⋅t
Average speed  Vm(t) = a⋅t/2

:-D

The speed of the accelerated mobile or particle
is a function of time.
The more time passes, the greater the speed.
Everyone agrees on that.
Now we have to give the correct equation.
I wrote that the correct equation is:
Vo(Tr)=a.Tr/sqrt(1+(a.Tr)²/c²)
I don't see what your problem is.
Vr(Tr)=a.Tr
Vrm=Vri/2
What do you not understand?

You don't explain or even acknowledge the problem with what happens when
Tr > c/a. Besides, your statement Vr(Tr)=a*Tr is Newtonian/Galilean
physics, not SR.

You are absolutely right :
when I say Vr(Tr)=a.Tr, it IS Newtonian relativity.
However, the equation is correct.
Warning! Tr is a proper time, and not a observable time.
We must therefore inhale and breathe, that is to say : try to understand.
It is the same when I say: "There will therefore exist an impassable speed which will extend to all the particles and all the properties of physics"
which is RR, and which I say, at the same time "This horse in this meadow,
this moon in the sky, this galaxy in this telescope, I see them live, instantly, absolutely indicative of themselves. The emission of the photon
and its reception by my retina occur in the same instant", which is
quantum theory.
We will say: it is absurd and contradictory.
Absolutely not, all this is of prodigious theoretical beauty, mathematical simplicity and manifest experimental evidence (Alain Aspect).
The only problem is just the neurons of those who read me,
formatted with an incorrect vision of the world, and that was imposed on
them from a very young age.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Sat Mar 16 14:20:00 2024
Den 15.03.2024 15:20, skrev Richard Hachel:
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :
I am saying that SR does not predict that accelerated objects
will behave as you claim they do.

We agree.
The SR does not predict things the way I predict them.

Right. You predict them like Newton did.

Richard Hachel's equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Are correct only in Newtonian Mechanics with Galilean relativity.

So Richard Hachel's theory is identical to Newtonian Mechanics.

The problem, although enormous, is that experimentally I am credible and
not the SR which cannot explain instantaneous transfers of information.

Why should SR "explain" what doesn't exist?

You know of course that SR is thoroughly experimentally verified
and never falsified, while NM is falsified.

and that theoretically I do not have the enormous concern of completely inconsistent apparent speeds. if only in a simple Langevin paradox.

Are there "inconsistent apparent speeds" in the "twin paradox"? :-D

--
Paul

https://paulba.no/

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• From Paul B. Andersen@21:1/5 to All on Sat Mar 16 14:19:54 2024
Den 15.03.2024 15:30, skrev Richard Hachel:
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :
I suppose you are referring to my "false" equations:
Speed:            Vr(t) = a⋅t/√(1+(a⋅t/c)²)
Average speed     Vm(t) = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t

Speed:  Vr(t) = a⋅t
Average speed  Vm(t) = a⋅t/2

:-D

The speed of the accelerated mobile or particle
is a function of time.
The more time passes, the greater the speed.
Everyone agrees on that

Indeed! Vr increases with time.

Den 12.03.2024, skrev Richard Hachel:
| In the rocket frame, a is constant. Always.
| The rocket is at rest in its frame of reference,
| and the speed Vr of the surrounding space becomes Vr=a.Tr
| There is no problem, the speed of the rocket, that is to
| say the real speed of movement of the terrestrial frame
| of reference, is indeed Vr.

This means that Vr is the relative speed between the rocket
and the terrestrial frame. Since the terrestrial frame is
inertial while the rocket is accelerating this means that
Vr is the speed of rocket in the inertial terrestrial frame.

So there is no problem: Vr = a.Tr
where a is the proper acceleration of the rocket
and Vr is the speed of the rocket in the inertial frame.

Den 16.03.2024 skrev Richard Hachel:
| When I say Vr(Tr)=a.Tr, it IS Newtonian relativity.
| However, the equation is correct.
| Warning! Tr is a proper time.

OK.
This is Newtonian (Galilean) relativity.
Tr is the proper time of the rocket and the time coordinate
in the inertial (terrestrial) frame.

There is no limit for the speed of the rocket.

We can conclude:
================
Richard Hachel's equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Are correct only in Newtonian Mechanics with Galilean relativity.

So Richard Hachel's theory is identical to Newtonian Mechanics.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Sat Mar 16 14:16:51 2024
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :

Richard Hachel's equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Are correct only in Newtonian Mechanics with Galilean relativity.

So Richard Hachel's theory is identical to Newtonian Mechanics.

Paul

The greatness of a man is taking serious things seriously.

There you are joking.

That's not what I expect from a man like you.

R.H.

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• From Richard Hachel@21:1/5 to All on Sat Mar 16 14:26:51 2024
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
Den 15.03.2024 15:39, skrev Richard Hachel:

So Richard Hachel's theory is identical to Newtonian Mechanics.

Absolutely not.

When I write To=(x/c).sqrt(1+c²/ax) it is not SR.

When I give you the correct equations to apply for rotating frames of reference, that is not SR.

When I describe a Langevin paradox to you, it is not SR.

All my equations relating to accelerated frames of reference are not SR.

Do not lie.

You don't get anything very scientific from lies.

Only lies.

Want to look like Python?

I don't recommend it.

You are worth, I am not kidding, much better than that.

R.H.

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• From Richard Hachel@21:1/5 to All on Sat Mar 16 14:29:25 2024
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
Den 15.03.2024 15:39, skrev Richard Hachel:

Richard Hachel's equations:
Speed of rocket in inertial frame: Vr=a.Tr

Absolutely.

Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Absolutely.

Are correct only in Newtonian Mechanics with Galilean relativity.

Sure.

So Richard Hachel's theory is identical to Newtonian Mechanics.

Absolutely not.

Try again.

But why is "[SR]" in the subject line when you only

Vous plaisantez, monsieur.

C'est indigne de vous.

Je vous supplie de revenir à plus d'intelligence.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Sat Mar 16 14:19:45 2024
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
Den 15.03.2024 15:20, skrev Richard Hachel:
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :
I am saying that SR does not predict that accelerated objects
will behave as you claim they do.

We agree.
The SR does not predict things the way I predict them.

Right. You predict them like Newton did.

Richard Hachel's equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Are correct only in Newtonian Mechanics with Galilean relativity.

So Richard Hachel's theory is identical to Newtonian Mechanics.

Absolutely not.

The problem, although enormous, is that experimentally I am credible and
not the SR which cannot explain instantaneous transfers of information.

Why should SR "explain" what doesn't exist?

You know of course that SR is thoroughly experimentally verified
and never falsified, while NM is falsified.

and that theoretically I do not have the enormous concern of completely
inconsistent apparent speeds. if only in a simple Langevin paradox.

Are there "inconsistent apparent speeds" in the "twin paradox"? :-D

Je vous supplie de comprendre ce que je dis avant de juger.

Vous ne le faites pas.

Ce n'est pas à la hauteur d'un homme tel que vous, que je reconnais,
moi, pour un très bon posteur.

R.H.

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• From Maciej Wozniak@21:1/5 to All on Sat Mar 16 17:45:33 2024
W dniu 16.03.2024 o 14:20, Paul B. Andersen pisze:

Why should SR "explain" what doesn't exist?

And why it does. Simple - because it was
created by an idiot mystician unable even
to mumble consistently.

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to All on Sun Mar 17 14:44:36 2024
Den 16.03.2024 15:16, skrev Richard Hachel:
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :

Richard Hachel's equations:
Speed of rocket in inertial frame:             Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Are correct only in Newtonian Mechanics with Galilean relativity.

So Richard Hachel's theory is identical to Newtonian Mechanics.

Paul

The greatness of a man is taking serious things seriously.

There you are joking.

That's not what I expect from a man like you.

Maybe this is what you expect?

Since the equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
are valid _only_ in Newtonian Mechanics with Galilean relativity,
then the theory which is consistent with said equations
is _only_ Newtonian Mechanics.

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to then the theory which is consistent on Sun Mar 17 14:44:26 2024
Den 16.03.2024 15:26, skrev Richard Hachel:
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
Den 15.03.2024 15:39, skrev Richard Hachel:

So Richard Hachel's theory is identical to Newtonian Mechanics.

Absolutely not.

Since the equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
are valid _only_ in Newtonian Mechanics with Galilean relativity,
then the theory which is consistent with said equations
is Newtonian Mechanics.

A very lethal argument! :-D

When I write To=(x/c).sqrt(1+c²/ax) it is not SR.

And it is not mathematical consistent with the equations:
Vr=a.Tr
Vrm=(1/2)Vr
so it is not Newtonian Mechanics.

But maybe you don't think that a theory of physics
has to be mathematically consistent?

When I describe a Langevin paradox to you, it is not SR.

In 1911 Langevin discovered that SR predicted that
two twins could age differently after different journeys.

And you say the "Langevin paradox" is NOT SR? :-D

I find it rather strange that it has been so many
strange attempts to "resolve" the "twin paradox",
because it follows straight forward from SR.

You can see SR's prediction for the "Langevin paradox" here: https://paulba.no/pdf/TwinsByMetric.pdf https://paulba.no/pdf/TwinsByDoppler.pdf

SR's prediction is experimentally verified. https://paulba.no/paper/Hafele_Keating.pdf
https://paulba.no/pdf/H&K_like.pdf
Two twins travelling in opposite directions
around the Earth will age differently.

All my equations relating to accelerated frames of reference are not SR.

And they are NOT mathematical consistent with
your equations Vr=a.Tr and Vrm=(1/2)Vr so
they are not NM.

Your equations which are not consistent
with neither NM nor SR are nonsense.

--
Paul

https://paulba.no/

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• From Maciej Wozniak@21:1/5 to All on Sun Mar 17 15:37:30 2024
W dniu 17.03.2024 o 14:44, Paul B. Andersen pisze:

SR's prediction is experimentally verified.

Sure, anyone can check GPS, the delusions
of your idiot guru have nothing in common
with the real clocks. No surprise, since
theyy were not even consistent.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Mon Mar 18 09:38:41 2024
Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
Den 16.03.2024 15:29, skrev Richard Hachel:

So the coordinate transformation equations are:
x' = x - v⋅t
y' = y
z' = z
t' = t

Which IS Galilean relativity.

Absolutly, this is galilean relativity.

But that's not what I'm talking about.
I'm talking about special relativity (Hachel sauce).
The transformations of Hachel (that's me) concerning the Galilean
environments are as follows:
x'=(x-Vo.To)/sqrt(1-Vo²/c²)
y'=y
z'=z
To'=[To-(x.Vo/c²)]/sqrt(1-Vo²/c²)]

The opposite becoming:
x=(x'+Vo.To')/sqrt(1-Vo²/c²)
y=y'
z=z'
To=[To'+(x'.Vo/c²)]/sqrt(1-Vo²/c²)]

I never said anything else.

R.H.

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• From Richard Hachel@21:1/5 to All on Mon Mar 18 09:54:20 2024
Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
Den 16.03.2024 15:16, skrev Richard Hachel:
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :

Richard Hachel's equations:
Speed of rocket in inertial frame:             Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Are correct only in Newtonian Mechanics with Galilean relativity.

So Richard Hachel's theory is identical to Newtonian Mechanics.

Paul

The greatness of a man is taking serious things seriously.

There you are joking.

That's not what I expect from a man like you.

Maybe this is what you expect?

Since the equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
are valid _only_ in Newtonian Mechanics with Galilean relativity,
then the theory which is consistent with said equations
is _only_ Newtonian Mechanics.

No.

Relisez mieux ce que j'ai écrit, vous allez trop vite pour railler des opinions que vous ne comprenez pas.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Mon Mar 18 09:52:23 2024
Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
Don't be ridiculous.
An intelligent Doctor and scientist like you will obviously
understand that since the equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
are valid _only_ in Newtonian Mechanics with Galilean relativity,
then the theory which is consistent with said equations
is Newtonian Mechanics.

Or don't you? :-D

I don't understand why a man like you, who has repeatedly shown in his
speeches that he has a good command of his subject, cannot listen to what
I say.
I will never question your intellectual qualities.
I have begged you many times to breathe calmly as you read me.
What I blame you for is your impatience and your lack of listening.
You tell me, Paul, that my equations are Newtonian physics, which is both untruthful and at the same time proof that you are not reading me
correctly.

I beg you, once again, to look at this, and tell me if these are Newtonian equations.

<http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:1>

<http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:2>

<http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:3>

<http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:4>

<http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:5>

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Mon Mar 18 09:27:41 2024
Le 17/03/2024 à 14:42, "Paul B. Andersen" a écrit :
Den 16.03.2024 15:26, skrev Richard Hachel:
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
Den 15.03.2024 15:39, skrev Richard Hachel:

So Richard Hachel's theory is identical to Newtonian Mechanics.

Absolutely not.

Since the equations:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
are valid _only_ in Newtonian Mechanics with Galilean relativity,
then the theory which is consistent with said equations
is Newtonian Mechanics.

The equations I give, if written correctly, are valid in both systems.
But you have to write them correctly.
For example if I write, in the Newtonian system,
v=a.t
This is valid.
In the same Newtonian system, we can also write:
v_m=(1/2)v_i

We agree on this, and I don't think, even regarding the craziest posters (Python example), anyone will come and contradict.

Now let's go further and talk about special relativity.
Vrm=a.Tr
and Vrm=(1/2)Vri
These equations remain true, as p=mv remains true in classical physics and p=m.Vr in relativistic physics (Hachel notation).

If now, you, Paul B. Andersen, claim that it is false to ask:
Vo=a.To
Vom=(1/2)Voi
p=m.Vo
you will obviously be absolutely right.

Except I never wrote that.

R.H.

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to All on Mon Mar 18 22:14:04 2024
Den 18.03.2024 10:27, skrev Richard Hachel:
Le 17/03/2024 à 14:42, "Paul B. Andersen" a écrit :

Since the equations:
Speed of rocket in inertial frame:             Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
are valid _only_  in Newtonian Mechanics with Galilean relativity,
then the theory which is consistent with said equations
is Newtonian Mechanics.

The equations I give, if written correctly, are valid in both systems.
But you have to write them correctly.
For example if I write, in the Newtonian system,
v=a.t
This is valid.
In the same Newtonian system, we can also write:
v_m=(1/2)v_i

We agree on this, and I don't think, even regarding the craziest posters (Python example), anyone will come and contradict.

Quite.

We can now review the journey to Tau Ceti.

Both Earth and Tau Ceti are considered to be inertial.

A rocket is stationary on Earth, When its clock show τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².

a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly

According to your equations v = a⋅t and vₘ = a⋅t/2: ===================================================

d = ∫a⋅t⋅dt + 0 ly = a⋅t²/2 => t = √(2⋅d/a)

The rocket will pass Tau Ceti at the terrestrial time:
t = √(2⋅d/a) = 4.7764 y

The proper time of the rocket when it passes Tau Ceti is:
τ = √(2⋅d/a) = 4.7764 y

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is: v = a⋅t = 5.2860 ly/y

The average speed is: vₘ = a⋅t/2 = 2.6430 ly/y

Note that d/vₘ = 4.5403 y < t Why :-D

According to SR:
================

The rocket will pass Tau Ceti at the terrestrial time:
t = √((d/c)²+2⋅d/a) = 12.9156 y

The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:
v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y

The average speed is:
vₘ = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t = 0.9291 ly/y

Note that vₘ/v > 1/2

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Tue Mar 19 08:50:25 2024
Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
Quite.

We can now review the journey to Tau Ceti.

Both Earth and Tau Ceti are considered to be inertial.

A rocket is stationary on Earth, When its clock show τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².

a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly

Tout cela est entièrement exact.

According to your equations v = a⋅t and vₘ = a⋅t/2:

Vrm=a.Tr/2

===================================================

d = ∫a⋅t⋅dt + 0 ly = a⋅t²/2 => t = √(2⋅d/a)

The rocket will pass Tau Ceti at the terrestrial time:
t = √(2⋅d/a) = 4.7764 y

No, no, no, no !

In terrestrial time, To=(x/c)/sqrt(1+2c²/ax)

To=12.92 years.

The proper time of the rocket when it passes Tau Ceti is:
τ = √(2⋅d/a) = 4.7764 y

Absolutely.

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is: v = a⋅t = 5.2860 ly/y

Yes, Vr=5.0245c

BUT: Vo=0.980c

The average speed is: vₘ = a⋅t/2 = 2.6430 ly/y

Vrm=2.51c if Vr=5.0245c

Observable speeds Vo is not real speeds Vr, apparent speeds Vapp is not observable speeds Vo.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Tue Mar 19 09:17:07 2024
Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
According to SR:
================

The rocket will pass Tau Ceti at the terrestrial time:
t = √((d/c)²+2⋅d/a) = 12.9156 y

Absolutely.

The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

Here is the error.

Physicists use the Minkowskian metric and it is not correct.
You have to use (I understand that this is confusing) the Newtonian
equation.
Physicists absolutely need to understand something:
the rocket is AT REST in its frame of reference, and it is from start to finish.
Everything happens, for her, as if the surrounding space were accelerating
by 10m/s² around her.
The big complaint (because it's confusing when you don't have 40 years of thinking on the subject like me) is to say: "Yes, but the more time
passes, the more the surrounding space will contract. for the rocket, and therefore Vr=a.Tr is no longer valid".
But we forget SEVERAL things.
Firstly the space to travel does not contract but expands, because we are
in a longitudinal journey and if we must correctly use the =l.sqrt(1-Vo²/c²)/(1+cosµ.Vo/c) ), we see that there is dilation of the anterior space every second and according to the speed reached.
But that's not all, this would not change the fact that the equation would
not be constant all the same (since the varies as a function of time).
But we forget that there will be an inverse correction due to the
expansion of the apparent speeds.
We will therefore have an instantaneous distance to cover as D'=D.sqrt[(1+Vo/c)/(1-Vo/c)]
but at the same time, the speed of approach of the star is like: Vapp=Vo/(1+cosµ.Vo/c) or here Vapp=Vo/(1-Vo/c) at each moment of the evolution.
We know that tau=D'/Vapp (tau is the time specific to each instant
necessary for the rocket to reach the star in the event of a sudden
cessation of acceleration).
However, at each instant tau=To.sqrt(1-Voi²/c²).
It follows that what we gain on one side, we lose on the other, and that
the acceleration is constant and that there is no relativistic correction
to be made in the present case.
We can therefore keep Vr=a.Tr without any problem.

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:
v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y

No.

Vo=0.980c

(Vr=5.024c)

The average speed is:
vₘ = c²⋅(√(1+(a⋅t/c)²)-1)/a⋅t = 0.9291 ly/y

Là, oui.

Note that vₘ/v > 1/2

Vom/Voi > 1/2 absolutely.

but Vrm/Vri = 1/2

R.H.

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to All on Tue Mar 19 20:45:43 2024
Den 19.03.2024 10:17, skrev Richard Hachel:
Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :

A rocket is stationary on Earth, When its clock show τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².

a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly
According to SR:
================

The rocket will pass Tau Ceti at the terrestrial time:
t = √((d/c)²+2⋅d/a) = 12.9156 y

Absolutely.

The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

Here is the error.

No error!

What SR predicts is not a matter of opinion,
it is a matter of fact.

So _ACCORDING TO SR_:
======================
The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:
v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y

Facts. Indisputable!

Physicists use the Minkowskian metric and it is not correct.

So you are saying that the predictions of SR are wrong
because the metric is wrong.
You are of course free to have that opinion, but remember that
SR is confirmed by innumerable experiments, and falsified by none.

You have to use (I understand that this is confusing) the Newtonian
equation.
Physicists absolutely need to understand something:
the rocket is AT REST in its frame of reference, and it is from start to finish.

Everything happens, for her, as if the surrounding space were
accelerating by 10m/s² around her.

Ah! This is an old, rather stupid misconception:
The twins must age equally because of the symmetry,
both the speed and acceleration of the other twin
are equal for both twins.

This is obviously nonsense. SR do predict that
the twin's age differently. Math don't lie! https://paulba.no/pdf/TwinsByMetric.pdf

Consider this simple thought experiment:
Two rockets A and B are instantly side by side.
A is accelerating at 10m/s², B is inertial.

Will "everything happen to the astronaut in A as if
she were inertial, and B and the surrounding space were
accelerating by 10m/s² around her"?

Think of it. The answer is rather obvious.

The big complaint (because it's confusing when you don't have 40 years
of thinking on the subject like me) is to say: "Yes, but the more time passes, the more the surrounding space will contract. for the rocket,
and therefore Vr=a.Tr is no longer valid".

It's pathetic, and a bit sad, to have thought in 40 years
to come up with utter nonsense.

<snip nonsense>

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
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• From Maciej Wozniak@21:1/5 to All on Tue Mar 19 20:49:56 2024
W dniu 19.03.2024 o 20:45, Paul B. Andersen pisze:
Den 19.03.2024 10:17, skrev Richard Hachel:
Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :

A rocket is stationary on  Earth, When its clock show  τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².

a = 10 m/s² = 1.05265 ly/y/y   c = 1 ly/y  d = 12 ly
According to SR:
================

The rocket will pass Tau Ceti at the terrestrial time:
t = √((d/c)²+2⋅d/a) = 12.9156 y

Absolutely.

The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

Here is the error.

No error!

What SR predicts is not a matter of opinion,
it is a matter of fact.

stupid.

So _ACCORDING TO SR_:
======================
The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:
v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y

Facts. Indisputable!

Physicists use the Minkowskian metric and it is not correct.

So you are saying that the predictions of SR are wrong
because the metric is wrong.
You are of course free to have that opinion, but remember that
SR is confirmed by innumerable experiments, and falsified by none.

You have to use (I understand that this is confusing) the Newtonian
equation.
Physicists absolutely need to understand something:
the rocket is AT REST in its frame of reference, and it is from start
to finish.

Everything happens, for her, as if the surrounding space were
accelerating by 10m/s² around her.

Ah! This is an old, rather stupid misconception:
The twins must age equally because of the symmetry,
both the speed and acceleration of the other twin
are equal for both twins.

This is obviously nonsense. SR do predict that
the twin's age differently. Math don't lie!

And speaking of math - it's always good to remind
oldest part false, as it didn't want to cooperate

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Richard Hachel@21:1/5 to All on Tue Mar 19 20:09:52 2024
Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :
Den 19.03.2024 10:17, skrev Richard Hachel:
Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :

A rocket is stationary on Earth, When its clock show τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².

a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly
According to SR:
================

The rocket will pass Tau Ceti at the terrestrial time:
t = √((d/c)²+2⋅d/a) = 12.9156 y

Absolutely.

The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

Here is the error.

No error!

What SR predicts is not a matter of opinion,
it is a matter of fact.

So _ACCORDING TO SR_:
======================
The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:
v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y

Facts. Indisputable!

Vous vous trompez.

Doublement.

1. Cette optique relativiste est incohérente et théoriquement
incorrecte.
2. Je suis persuadé que les faits expérimentaux iront contre.

Je gagne toujours, ne l'oubliez pas.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Tue Mar 19 20:14:34 2024
Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :

Physicists use the Minkowskian metric and it is not correct.

So you are saying that the predictions of SR are wrong
because the metric is wrong.
You are of course free to have that opinion, but remember that
SR is confirmed by innumerable experiments, and falsified by none.

You have to use (I understand that this is confusing) the Newtonian

If you want to have a coherent discussion with me, it is necessary, at
least, to understand what I am saying.

You are going around in circles all alone.

When you write, I read you, and I make the effort to understand you; I
usually explain where you are wrong, and why you are wrong.

But the opposite is not true. You don't read me, and you constantly repeat
what you learned at school, and not what I added.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Tue Mar 19 20:59:10 2024
Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :
Den 19.03.2024 09:50, skrev Richard Hachel:
Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :

We can now review the journey to Tau Ceti.

Both Earth and Tau Ceti are considered to be inertial.

A rocket is stationary on  Earth, When its clock show  τ = 0 and
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².

a = 10 m/s² = 1.05265 ly/y/y   c = 1 ly/y  d = 12 ly

Richard Hackel uses 10 m/s² = 1.052 ly/y²
but my value above is more precise.
(A year is ≈ 356.25 days, not 365 days)

If you want.

According to your equations v = a⋅t and vₘ = a⋅t/2:
===================================================

d =  ∫a⋅t⋅dt + 0 ly = a⋅t²/2  => t = √(2⋅d/a)

The proper time of the rocket when it passes Tau Ceti is:
τ = √(2⋅d/a) = 4.7764 y

Absolutely.

The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:  v = a⋅t = 5.2860 ly/y

My typo: v = a⋅t = 5.0279 ly/y

Yes, Vr=5.0245c

So your "theory" is identical to NM and predicts
that there is no limit to the speed of the rocket
in an inertial frame of reference.

You know of course that experimental evidence
show that the speed of an object can't exceed c,
so why do you promote a "theory" you know is false?

BUT: Vo=0.980c

Observable speeds Vo is not real speeds Vr,

:-D

How do you think that inventing a speed that isn't real
can change the fact that your "theory" predicts that it
is no limit to the speed of the rocket in an inertial
frame of reference, and therefore is falsified?

--
Paul

My Dear Paul, I have begged you many times to try to understand what I
was saying, and I believe that is an impossible task because you do not
WANT to understand.

I have told you many times that if we use low speeds, the observable
speeds are the real speeds.
This trotting horse, this motorcycle, this passing train.
What I measure is the reality of things.
Paul! Breathe! Exhale!
I also said that the nature of space is such that if we want to have a
correct notion of speed, we absolutely must use only one watch and NEVER
two watches placed in different places.
Paul! Breathe! Exhale!
The best measure of time will therefore be the mobile's own time, which
goes from A to B, and which clicks during both events. There can therefore
be no measurement error.
It notes tau (or Tr).
If I want the real speed of a mobile, in the reference frame where I am, I therefore need the AB measurement in this reference frame, BUT the
mobile's own time.
Paul! Breathe! Exhale!
If I make the mistake of taking time A noted by watch A, and time B noted
by watch B, I am using two watches placed in two different theaters, and
which will never be naturally in tune, because the universe is not “done
like that”. The notion of universal present time is as abstract an
notion as the notion of a flat earth.
Now, in relativity, this is what we do every day, and we find a
measurement that is false, and therefore a speed that is false.
It doesn't matter for low speeds.
But for relativistic speeds, the measurement errors are considerable, and
it "seems" to us that nothing can exceed c.
But it is only an abstract idea that is very difficult to disengage from
the human mind.
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

We then understand that all real speeds are permitted, but that, by the
way we use distinct watches,
an impression of speed Vo appears, and it cannot exceed c.

But this is only a local illusion.

In reality, instantaneous transfers of information are evident (Aspect's experience) if one understands what one is doing.

R.H.

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• From Volney@21:1/5 to Richard Hachel on Wed Mar 20 13:31:10 2024
On 3/16/2024 10:29 AM, Richard Hachel wrote:
Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
Den 15.03.2024 15:39, skrev Richard Hachel:

Richard Hachel's equations:
Speed of rocket in inertial frame:             Vr=a.Tr

Absolutely.
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

Absolutely.

Are correct only in Newtonian Mechanics with Galilean relativity.

Sure.

So Richard Hachel's theory is identical to Newtonian Mechanics.

Absolutely not.

Try again.

But why is "[SR]" in the subject line when you only

Vous plaisantez, monsieur.
C'est indigne de vous.

Je vous supplie de revenir à plus d'intelligence.

Why are you telling Paul that his father smelt of elderberries?

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• From Paul B. Andersen@21:1/5 to All on Wed Mar 20 20:19:35 2024
Den 19.03.2024 21:59, skrev Richard Hachel:

I have told you many times that if we use low speeds, the observable
speeds are the real speeds.
This trotting horse, this motorcycle, this passing train.
What I measure is the reality of things.

I also said that the nature of space is such that if we want to have a correct notion of speed, we absolutely must use only one watch and NEVER
two watches placed in different places.

The best measure of time will therefore be the mobile's own time, which
goes from A to B, and which clicks during both events. There can
therefore be no measurement error.
It notes tau (or Tr).
If I want the real speed of a mobile, in the reference frame where I am,
I therefore need the AB measurement in this reference frame, BUT the
mobile's own time.

If I make the mistake of taking time A noted by watch A, and time B
noted by watch B, I am using two watches placed in two different
theaters, and which will never be naturally in tune, because the
universe is not “done like that”. The notion of universal present time
is as abstract an notion as the notion of a flat earth.
Now, in relativity, this is what we do every day, and we find a
measurement that is false, and therefore a speed that is false.
It doesn't matter for low speeds.
But for relativistic speeds, the measurement errors are considerable,
and it "seems" to us that nothing can exceed c.
But it is only an abstract idea that is very difficult to disengage from
the human mind.
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

We then understand that all real speeds are permitted, but that, by the
way we use distinct watches,
an impression of speed Vo appears, and it cannot exceed c.

But this is only a local illusion.

Let's leave the illusionary world and revert to the real world:

Consider an inertial observer in space.
She has instruments like clocks and telescopes and computers,
so she can measure the speed of a passing rocket relative
to herself.
Please don't say that this in principle is impossible in the real world.

Eleven such observers (O_0 ..O_10) are stationary relative to each
other, and are arranged along a straight line with 1 light year
between them.
A rocket which is accelerating at the constant proper acceleration
a = 1 c per year is instantly at rest relative to O_0.
The rocket is moving along a line parallel to the line of observers.

c = 1 light year per year.

Please show what you think the observers O_1 to O_10 would
measure the speed of the rocket to be relative to themselves.

--
Paul

https://paulba.no/

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