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)
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.
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.
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
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.
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.
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
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
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.
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.
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
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!
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.
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)
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 → ∞
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.)
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
Andersen...). The others not only learned anything, but answer nonsense
when asked a simple problem.
Above you have demonstrated that you belong to the group that
falls into hysterical reactions.
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.
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.
And you know of course that SR is confirmed by innumerable
experiments and falsified by none.
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?
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?"
Den 14.03.2024 03:09, skrev Richard Hachel:
Contradicting fact:
-------------------
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 =====================================================================
Contradicting fact:
-------------------
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! ====================================================
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.
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.
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).
I am saying that SR does not predict that accelerated objects
will behave as you claim they do.
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
where your "correct" equations are:
Speed: Vr(t) = a⋅t
Average speed Vm(t) = a⋅t/2
:-D
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
There is not "Two ways of seeing SR".
SR is mathematical consistent and can't be made
to give contradicting answers to the same problem.
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.
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?
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
where your "correct" equations are:
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?
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.
Contradicting fact:
-------------------
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
=====================================================================
Contradicting fact:
-------------------
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?
On 3/15/2024 10:30 AM, Richard Hachel wrote:
Le 15/03/2024 à 15:11, "Paul B. Andersen" a écrit :You don't explain or even acknowledge the problem with what happens when
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
where your "correct" equations are:
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?
Tr > c/a. Besides, your statement Vr(Tr)=a*Tr is Newtonian/Galilean
physics, not SR.
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.
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
where your "correct" equations are:
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
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
Den 15.03.2024 15:39, skrev Richard Hachel:
So Richard Hachel's theory is identical to Newtonian Mechanics.
Den 15.03.2024 15:39, skrev Richard Hachel:
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.
But why is "[SR]" in the subject line when you only
are talking about Newtonian Mechanics?
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
Why should SR "explain" what doesn't exist?
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.
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.
Your comments are criminal.
When I write To=(x/c).sqrt(1+c²/ax) it 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.
SR's prediction is experimentally verified.
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.
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.
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
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.
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
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
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.
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".
<snip nonsense>
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!
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
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)
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
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
are talking about Newtonian Mechanics?
Vous plaisantez, monsieur.
C'est indigne de vous.
Je vous supplie de revenir à plus d'intelligence.
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.
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