• #### [SR] The traveler of Tau Ceti

From Richard Hachel@21:1/5 to All on Tue Mar 19 08:35:10 2024
The Traveler of Tau Ceti

The Tau Ceti Traveler is a relativistic problem imagined by theorist
Richard Hachel to describe what would happen to a traveler leaving to join
Tau Ceti in accelerated mode.

We assume that the Sun-Tau Ceti system is stationary.

We set x=12 light years.

Thanks to new technology, we use a comfortable acceleration of a=10m/s² (a=1.052ly/y²).

At the start, we start all the watches. The terrestrial time is noted
To=0, the rocket's own time is noted τ=0.

The problem consists first of determining what the travel time will be for
the traveler, and what will be the observable times To (in the terrestrial reference frame) and apparent Tapp (what we see in a powerful telescope)
noted by the sedentary observer.

The problem then consists of determining the instantaneous observable
velocity (Voi) at the moment when the rocket will cross Tau Ceti, and its instantaneous acceleration measured in the terrestrial reference frame.

Finally, to determine, while we know that the rocket will be 12 light
years from the earth at this instant, how far the earth will be from the
rocket in the rocket's frame of reference.

R.H.

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• From Mikko@21:1/5 to Richard Hachel on Tue Mar 19 12:30:22 2024
On 2024-03-19 08:35:10 +0000, Richard Hachel said:

The Traveler of Tau Ceti

The Tau Ceti Traveler is a relativistic problem imagined by theorist
Richard Hachel to describe what would happen to a traveler leaving to
join Tau Ceti in accelerated mode.

We assume that the Sun-Tau Ceti system is stationary.

We set x=12 light years.

Thanks to new technology, we use a comfortable acceleration of a=10m/s² (a=1.052ly/y²).

At the start, we start all the watches. The terrestrial time is noted
To=0, the rocket's own time is noted τ=0.

The problem consists first of determining what the travel time will be
for the traveler, and what will be the observable times To (in the terrestrial reference frame) and apparent Tapp (what we see in a
powerful telescope) noted by the sedentary observer.

The problem then consists of determining the instantaneous observable velocity (Voi) at the moment when the rocket will cross Tau Ceti, and
its instantaneous acceleration measured in the terrestrial reference
frame.

Finally, to determine, while we know that the rocket will be 12 light years

Not finally but next.

from the earth at this instant, how far the earth will be from the
rocket in the rocket's frame of reference.

Then calculate what part of the ships initial mass must be converted
to energy, considering that energy cannot be created from nothing.

--
Mikko

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• From Paul B. Andersen@21:1/5 to All on Tue Mar 19 20:45:48 2024
Den 19.03.2024 09:35, skrev Richard Hachel:
The Traveler of Tau Ceti

The Tau Ceti Traveler is a relativistic problem imagined by theorist
Richard Hachel to describe what would happen to a traveler leaving to
join Tau Ceti in accelerated mode.

We assume that the Sun-Tau Ceti system is stationary.

We set x=12 light years.

Thanks to new technology, we use a comfortable acceleration of a=10m/s² (a=1.052ly/y²).

At the start, we start all the watches. The terrestrial time is noted
To=0, the rocket's own time is noted τ=0.

The problem consists first of determining what the travel time will be
for the traveler, and what will be the observable times To (in the terrestrial reference frame) and apparent Tapp (what we see in a
powerful telescope) noted by the sedentary observer.

You have in another posting said that the traveller's clock
would show τ = √(2⋅d/a) = 4.7764 y , and the speed relative to
Tau Ceti would be Vr = a⋅t = 5.0279 ly/y when she passes the star.

Since it is experimentally confirmed that the speed relative
to the star never can exceed c, the theory you have used
to arrive at these predictions is obviously falsified.

The "theory" is obviously Newtonian mechanics with Galilean relativity.

The problem then consists of determining the instantaneous observable velocity (Voi) at the moment when the rocket will cross Tau Ceti,

What's the point with inventing an apparent (not real)
"observable velocity" which is less than c when you know
that the real velocity according to your theory is > c?

and
its instantaneous acceleration measured in the terrestrial reference frame.

The coordinate acceleration in the terrestrial frame is obviously a.

Finally, to determine, while we know that the rocket will be 12 light
years from the earth at this instant, how far the earth will be from the rocket in the rocket's frame of reference.

According to your theory (NM), the Earth will be 12 ly away from
the rocket measured in the rocket's instant rest frame.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Tue Mar 19 20:41:21 2024
Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :

The problem then consists of determining the instantaneous observable
velocity (Voi) at the moment when the rocket will cross Tau Ceti,

What's the point with inventing an apparent (not real)
"observable velocity" which is less than c when you know
that the real velocity according to your theory is > c?

The acceleration of the rocket is a in the rocket's frame of reference,
and it will always be the same as the rocket progresses.
Breathe, breathe out, breathe...
It's so obvious that I wonder how you can argue with it.
Imagine your rocket at rest in a frame of reference Vo=0.6c, and another
at rest in a frame of reference Vo=0.8c, accelerate the two rockets
according to a=10m/s².
There is NO difference. There is no absolute reference.
It comes from this that whatever the instantaneous speed of the rocket, it
is at rest in this frame of reference whatever the speed reached.
Always, and for itself, it will accelerate with a constant acceleration in
the frame of reference where it will be found.
It is only for the terrestrial observer that the acceleration will be
relative and that the rocket will appear to accelerate less and less
quickly (it will never exceed Vo=c).

and
its instantaneous acceleration measured in the terrestrial reference frame.

The coordinate acceleration in the terrestrial frame is obviously a.

In rocket frame : always a=1.052 ly.y²

In the terrestrial frame :

<http://news2.nemoweb.net/jntp?oaUZj2Vqx_XIyisVSLVjzhx15P8@jntp/Data.Media:1>

R.H.

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• From Richard Hachel@21:1/5 to All on Tue Mar 19 20:29:58 2024
Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :

You have in another posting said that the traveller's clock
would show τ = √(2⋅d/a) = 4.7764 y ,

That's actually what I said.

and the speed relative to
Tau Ceti would be Vr = a⋅t = 5.0279 ly/y when she passes the star.

Absolutely.

Since it is experimentally confirmed that the speed relative
to the star never can exceed c, the theory you have used
to arrive at these predictions is obviously falsified.

I beg you to understand something...

When I talk in Vr notation, I'm talking about real speeds (which can take
any value).
You are talking about speeds observable in a frame of reference which is
not that of the mobile, but that of the observer, and therefore you are
It is very obvious, and I have never said the opposite in 40 years of explanations that I wanted to be consistent, that Vo could be greater than
c.
It’s YOU who made me say it.
I never said that.
I implore you to show a little more humility when responding to me. To say "Doctor Hachel, you are an idiot, you don't know that we cannot exceed c",
is to be both very extravagant, and above all very unhumble.

I would not allow myself to make such a stupid and humiliating remark to
you.

The "theory" is obviously Newtonian mechanics with Galilean relativity.

Absolutely not.

My concepts are ultra-relativistic.

Absolutely not newtonian, nor einsteinian.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Wed Mar 20 20:19:42 2024
Den 19.03.2024 21:41, skrev Richard Hachel:

The acceleration of the rocket is a in the rocket's frame of reference,
and it will always be the same as the rocket progresses.
Breathe, breathe out, breathe...
It's so obvious that I wonder how you can argue with it.
Imagine your rocket at rest in a frame of reference Vo=0.6c, and another
at rest in a frame of reference Vo=0.8c, accelerate the two rockets
according to a=10m/s².
There is NO difference. There is no absolute reference.
It comes from this that whatever the instantaneous speed of the rocket,
it is at rest in this frame of reference whatever the speed reached.
Always, and for itself, it will accelerate with a constant acceleration
in the frame of reference where it will be found.
It is only for the terrestrial observer that the acceleration will be relative and that the rocket will appear to accelerate less and less
quickly (it will never exceed Vo=c).

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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• From Richard Hachel@21:1/5 to All on Wed Mar 20 21:06:01 2024
Le 20/03/2024 à 20:18, "Paul B. Andersen" a écrit :
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

The answers I can give you are very simple as long as you understand
correctly what I am saying.

But I repeat again and again, observable speeds are not real speeds. This
is very important to understand, because you will realize that things will logically start to go wrong.

If you use real speeds (Vr) you will no longer have any problems, and the equations will remain both simple and true.

If you use traditional observable velocities (v or Vo)
you will notice that the observable speeds can be different for various observers present in the same frame of reference. Which may seem absurd if
we do not understand that, precisely, these speeds are not real but a distortion of what is real.

I can easily give you all the equations you need.

Here you are asking me what is the instantaneous observable velocity for
each point placed on the path as the rocket passes in front of it.

We have :
Voi/c=[1+c²/2ax]^(-1/2)

That's not what the relativists say.

They give too high instantaneous speeds (like you).

But that’s THEIR problem.

Same thing for clean times. They give clean times that are too low.

tau=sqrt(2x/a)

To=(x/c).sqrt(1+2c²/ax)

R.H.

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• From Maciej Wozniak@21:1/5 to All on Thu Mar 21 10:10:33 2024
W dniu 20.03.2024 o 20:19, Paul B. Andersen pisze:
Den 19.03.2024 21:41, skrev Richard Hachel:

The acceleration of the rocket is a in the rocket's frame of
reference, and it will always be the same as the rocket progresses.
Breathe, breathe out, breathe...
It's so obvious that I wonder how you can argue with it.
Imagine your rocket at rest in a frame of reference Vo=0.6c, and
another at rest in a frame of reference Vo=0.8c, accelerate the two
rockets according to a=10m/s².
There is NO difference. There is no absolute reference.
It comes from this that whatever the instantaneous speed of the
rocket, it is at rest in this frame of reference whatever the speed
reached.
Always, and for itself, it will accelerate with a constant
acceleration in the frame of reference where it will be found.
It is only for the terrestrial observer that the acceleration will be
relative and that the rocket will appear to accelerate less and less
quickly (it will never exceed Vo=c).

Consider an inertial observer in space.
She has instruments like clocks and telescopes and computers,

And, as anyone can check in a GPS satellite,
they're ignoring your absurd religion and
assume t'=t.

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• From Paul B. Andersen@21:1/5 to Richard Hachel on Thu Mar 21 21:07:10 2024
Den 20.03.2024 22:06, skrev Richard Hachel:
Le 20/03/2024 à 20:18, "Paul B. Andersen" a écrit :
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.

This defines Doctor Hachel's theory:

On 12.03.2024, Richard Hachel wrote:
| 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.

Below are the inevitable consequences of Doctor Hachel's theory:

The speed of an inertial observer relative to the rocket is
Vᵣ = a⋅Tᵣ (1)
where a is the constant proper acceleration of the rocket
and Tᵣ is the proper time of the rocket.

This means that the observer will move a distance x relative to
the rocket:

x(Tᵣ) = ∫Vᵣ⋅dTᵣ + x(0) = a⋅∫Tᵣ⋅dTᵣ + x(0) = a⋅Tᵣ²/2 + x(0) (2)

Per definition is Vᵣ = 0 when the rocket is passing O₀,
and we define Tᵣ = 0 and x = 0 at the same event.

That means that when an observer has moved the distance x,
the time Tᵣ is:
Tᵣ = √(2⋅x/a) (3)

Since a = 1 ly/y² and it is 1 ly between two observers,
observer Oₙ at x = n ly will pass the rocket at the time
Tᵣ = √(2⋅n) y.

The speed of observer Oₙ relative to the rocket when she
passes the rockets will be:
Vᵣ = √(2⋅n) ly/y

Observer Oₙ will measure the speed of the rocket relative
to herself to be:
Vᵣ = √(2⋅n) ly/y.

Since all the observers will measure that the speed of the rocket
is faster than the speed of light, we know that Doctor Hachel's theory

The answers I can give you are very simple as long as you understand correctly what I am saying.

If they are simple, why haven't you tried to calculate
what speeds of the rocket the observers will measure?

But I repeat again and again, observable speeds are not real speeds.
This is very important to understand, because you will realize that
things will logically start to go wrong.

The observers measure obviously the real speed of the rocket!
The observers inhabit the real world, not Wonderland.

That they measure an unreal imaginary speed is indeed a weird idea.
Or is "a stupid idea" a more adequate expression?

If you use real speeds (Vr) you will no longer have any problems, and
the equations will remain both simple and true.

"No longer"? I never had any problems.
Measured by observer Oₙ the speed of the rocket is √(2⋅n) ly/y.
Simple and true.

If you use traditional observable velocities (v or Vo)
you will notice that the observable speeds can be different for various observers present in the same frame of reference. Which may seem absurd
if we do not understand that, precisely, these speeds are not real but a distortion of what is real.

Of course any measurement of anything will have a limited precision.
But that is not what you are talking about.

I can easily give you all the equations you need.

Here you are asking me what is the instantaneous observable velocity for
each point placed on the path as the rocket passes in front of it.

We have :
Voi/c=[1+c²/2ax]^(-1/2)

So you claim that the rocket passes observer O₅ at the real speed
3.1623 ly/y, but O₅ will measure the speed to be 0.9535 ly/y.

Or generally:
You claim that the speed of an object in an inertial frame
may be several times the speed of light, but will always be
measured to be less than c.

Which is utter nonsense!

Give my regards to Alice.

Case closed.

--
Paul

https://paulba.no/

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• From Maciej Wozniak@21:1/5 to All on Thu Mar 21 21:18:33 2024
W dniu 21.03.2024 o 21:07, Paul B. Andersen pisze:

The observers measure obviously the real speed of the rocket!
The observers inhabit the real world, not Wonderland.

Sorry, poor trash, the observers you're mumbling
The observers from the real world have nothing in
common with them.

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• From Richard Hachel@21:1/5 to All on Fri Mar 22 08:49:46 2024
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :
Or generally:
You claim that the speed of an object in an inertial frame
may be several times the speed of light, but will always be
measured to be less than c.

It's not just that I claim it, it's that it's logical, coherent,
mathematical.

You seem, despite your intelligence well above the average man, to have difficulty understanding WHY
Vo (the measured speed) is not equal to Vr (the real speed).

Already forty years ago, I gave the five basic equations (hundreds will
follow) of SR.

To²=Tr²+Et²
To=Tr.sqrt(1+Vr²/c²)
Tr=To.sqrt(1-Vo²/c²)
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

These are the equations that you don't make the effort to understand.

Not necessarily to accept them, but, at least, to understand them.

You do not understand this notion of anisochrony which is the basis of the entire theory of special relativity.

You tell me: “Measuring a false speed would be absurd”.

Yet this is what is done by using two separate watches.

Only the mobile, which only uses one watch, has the correct time.

Let's take the case of a medieval man who goes from Paris to Moscow and
who measures time with a very precise hourglass.

It will measure twenty days exactly (or 480 one-hour hourglasses).

He left Paris at noon, and he said to himself, I covered PM in 480 hours.
My speed is real, it's not Alice in Wonderland.

But suddenly, he observes the hourly clock present in front of the
Kremlin, and it does not mark midday, but a few more hours.

He then realizes that the Paris sundial is out of sync
relative to the Moscow sundial, and that if he uses
t(arrival)M-t(departure)P, he will get an incorrect time, and an incorrect speed.

This is a bit like how it works in relativity.

We cannot measure the real natural time of a mobile with two watches
placed in two different places and by performing a hasty subtraction.

We will observe in SR a speed Vo which will never exceed c.

While Vr will be able to take all the values without this being any
problem.

We will say: “So be it!”

We are going to place a watch C equidistant from A and B,
and we will subtract, tB{C}-tA{C} since C is just one watch.

This is what physicists do.

Except that doesn't change anything at all. It's a single smartwatch with
two different watches. Only the mobile watch is unique, and unfalsifiable
by anisochrony.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Tue Mar 26 21:47:31 2024
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You claim that the speed of an object in an inertial frame
may be several times the speed of light, but will always be
measured to be less than c.

It's not just that I claim it, it's that it's logical, coherent, mathematical.

Already forty years ago, I gave the five basic equations (hundreds will follow) of SR.

To²=Tr²+Et²
To=Tr.sqrt(1+Vr²/c²)
Tr=To.sqrt(1-Vo²/c²)
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

In the Large Hadron Collider [LHC] at Cern the measured speed
of the protons is Vo = 0.9999999896⋅c

The circumference of the LHC and the time to go around
the circuit are precisely known.

Are you claiming that the real speed of the protons in the LHC is
Vr = 6927⋅c ?

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Wed Mar 27 06:23:17 2024
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You claim that the speed of an object in an inertial frame
may be several times the speed of light, but will always be
measured to be less than c.

It's not just that I claim it, it's that it's logical, coherent,
mathematical.

Already forty years ago, I gave the five basic equations (hundreds will
follow) of SR.

To²=Tr²+Et²
To=Tr.sqrt(1+Vr²/c²)
Tr=To.sqrt(1-Vo²/c²)
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

In the Large Hadron Collider [LHC] at Cern the measured speed
of the protons is Vo = 0.9999999896⋅c

The circumference of the LHC and the time to go around
the circuit are precisely known.

Are you claiming that the real speed of the protons in the LHC is
Vr = 6927⋅c ?

Absolutely.

That's what I said.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Wed Mar 27 13:25:53 2024
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You claim that the speed of an object in an inertial frame
may be several times the speed of light, but will always be
measured to be less than c.

It's not just that I claim it, it's that it's logical, coherent,
mathematical.

Already forty years ago, I gave the five basic equations (hundreds
will follow) of SR.

To²=Tr²+Et²
To=Tr.sqrt(1+Vr²/c²)
Tr=To.sqrt(1-Vo²/c²)
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

In the Large Hadron Collider [LHC] at Cern the measured speed
of the protons is Vo = 0.9999999896⋅c

The circumference of the LHC and the time to go around
the circuit are precisely known.

Are you claiming that the real speed of the protons in the LHC is
Vr = 6927⋅c ?

Absolutely.

That's what I said.

Since it is you, Doctor Richard Hachel, i will assume you are serious.
(Nobody but you could seriously make such a ridiculous claim.)

You are obviously ignorant of how a synchrotron works.

The protons in the LHC are moving around a ring with
circumference 26659 m. The ring consist of straight
stretches and bends. In the straight stretches there
are eight RF-cavities which accelerate the protons.
In the bends there are magnets which make the path of
the protons bent. The protons will radiate some of their
kinetic energy as synchrotron radiation (light with a special
spectrum) in the bends, and when the synchrotron is in steady
state at peak power, the energy gained in the RF-cavities will
be lost in the bends.

In a RF-cavity there is an electric field which is changing
direction sinusoidally all the time. The protons are moving in
bunches, and a bunch must be at a RF-cavity exactly at the time
when the electric field is at peak value in the right direction.
Since there are many (N) bunches in the ring, and each bunch are
going around the ring many times (M) per second, the frequency
of the RF-field in the RF-cavity must be a multiple of N x M Hz.
The nominal frequency is 400.8 MHz, but this is finely tuned
depending on the speed of the protons.

The point is that the speed of the protons is very precisely known,
and the measured and real speed of the protons is the same.

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D

But maybe you were joking.
In that case you had me!

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Wed Mar 27 13:23:01 2024
Le 27/03/2024 à 13:24, "Paul B. Andersen" a écrit :
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You claim that the speed of an object in an inertial frame
may be several times the speed of light, but will always be
measured to be less than c.

It's not just that I claim it, it's that it's logical, coherent,
mathematical.

Already forty years ago, I gave the five basic equations (hundreds
will follow) of SR.

To²=Tr²+Et²
To=Tr.sqrt(1+Vr²/c²)
Tr=To.sqrt(1-Vo²/c²)
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

In the Large Hadron Collider [LHC] at Cern the measured speed
of the protons is Vo = 0.9999999896⋅c

The circumference of the LHC and the time to go around
the circuit are precisely known.

Are you claiming that the real speed of the protons in the LHC is
Vr = 6927⋅c ?

Absolutely.

That's what I said.

Since it is you, Doctor Richard Hachel, i will assume you are serious. (Nobody but you could seriously make such a ridiculous claim.)

You are obviously ignorant of how a synchrotron works.

The protons in the LHC are moving around a ring with
circumference 26659 m. The ring consist of straight
stretches and bends. In the straight stretches there
are eight RF-cavities which accelerate the protons.
In the bends there are magnets which make the path of
the protons bent. The protons will radiate some of their
kinetic energy as synchrotron radiation (light with a special
spectrum) in the bends, and when the synchrotron is in steady
state at peak power, the energy gained in the RF-cavities will
be lost in the bends.

In a RF-cavity there is an electric field which is changing
direction sinusoidally all the time. The protons are moving in
bunches, and a bunch must be at a RF-cavity exactly at the time
when the electric field is at peak value in the right direction.
Since there are many (N) bunches in the ring, and each bunch are
going around the ring many times (M) per second, the frequency
of the RF-field in the RF-cavity must be a multiple of N x M Hz.
The nominal frequency is 400.8 MHz, but this is finely tuned
depending on the speed of the protons.

The point is that the speed of the protons is very precisely known,
and the measured and real speed of the protons is the same.

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D

But maybe you were joking.
In that case you had me!

CERN physicists are doing their job.
We have accustomed them to working at classic relativistic speed (useful
but false).
So it makes sense that they find the speed they expect.
I tell them that the proton rotates 78 million times per second,
They tell me no.
I tell them that to be consistent you have to measure things with a single watch, and that measuring the departure of the proton and the arrival of
the proton with two different watches (in this case, the lab watch that
counts the revolutions is TWO watches) can only lead to measuring speeds incredibly lower than reality.
Only the proton itself has the correct watch, because it, in its frame of reference, accelerated or not, is invariant.
Now, if we want to know the real speed of the proton, we must know its
momentum in an instant of its journey.
Physicists will necessarily find p=m.Vr, and not p=m.Vo.

Vo is a decoy caused by universal anisochrony.

This is not the true particle (or rocket) speed.

R.H.

--- SoupGate-Win32 v1.05
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• From Volney@21:1/5 to Paul B. Andersen on Wed Mar 27 13:07:22 2024
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D

But maybe you were joking.
In that case you had me!

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Python@21:1/5 to All on Thu Mar 28 00:03:23 2024
Le 27/03/2024 à 18:07, Volney a écrit :
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D

But maybe you were joking.
In that case you had me!

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein velocity addition formula.

It is not even that, it is worse, far worse.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to All on Thu Mar 28 07:55:23 2024
W dniu 28.03.2024 o 00:03, Python pisze:
Le 27/03/2024 à 18:07, Volney a écrit :
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D

But maybe you were joking.
In that case you had me!

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein

It is not even that, it is worse, far worse.

Oh, stinker Python is opening its muzzle again,
and trying again to pretend he knows something.
learnt what a function is? Is the clause
"for any element of the domain" still confusing
you?

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Thu Mar 28 11:06:48 2024
Le 27/03/2024 à 23:03, Python a écrit :
Le 27/03/2024 à 18:07, Volney a écrit :
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D

But maybe you were joking.
In that case you had me!

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein

It is not even that, it is worse, far worse.

Je ne vois pas l'intérêt de ton intervention.

J'attends toujours tes excuses pour ton comportement stupide sur le description du voyageur de Langevin
en vitesses apparentes, et pourquoi la formule Vapp=Vo/(1+cosµ.Vo/c) ne fonctionne que pour l'aller et pas pour le retour.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Thu Mar 28 11:02:11 2024
Le 27/03/2024 à 17:07, Volney a écrit :
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein velocity addition formula.

You say that for Doctor Hachel the notion of real speed is not a speed.
Of course yes, and it's even the only consistent speed.
This is the usual notion of observable speed Vo which is NOT a speed but a simple decoy.
This is also why relativistic speeds do not add up as one could do in the
case of Newtonian speeds.
I have given all the formulas that must be used in relativistic physics, including in uniformly accelerated media and rotating media, including
many things different from what physicists say, but with much more clarity
and coherence than them.
formula is here:

<http://news2.nemoweb.net/jntp?imChwfC2KmQA4gQfMNTxoMdVOMA@jntp/Data.Media:1>

As for real speeds, the addition is not simpler,
because we add a real speed in one frame of reference, and a real speed in another frame of reference. The formula is here:

<http://news2.nemoweb.net/jntp?imChwfC2KmQA4gQfMNTxoMdVOMA@jntp/Data.Media:2>

I remind you that physicists who say that real speeds are of no interest
are wrong. On the contrary, these speeds are of great interest when the problems become more complicated, and when applying reality becomes more
useful than applying the observable.

R.H.

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• From Python@21:1/5 to All on Thu Mar 28 13:58:36 2024
Le 28/03/2024 à 12:06, M.D. Richard "Hachel" Lengrand a écrit :
Le 27/03/2024 à 23:03, Python a écrit :
Le 27/03/2024 à 18:07, Volney a écrit :
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D >>>>
But maybe you were joking.
In that case you had me!

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein

It is not even that, it is worse, far worse.

Je ne vois pas l'intérêt de ton intervention.
J'attends toujours tes excuses pour ton comportement stupide sur le description du voyageur de Langevin en vitesses apparentes, et pourquoi
la formule Vapp=Vo/(1+cosµ.Vo/c) ne fonctionne que pour l'aller et pas
pour le retour.

I won't apologize for having stated a basic geometric fact. The
formula applied for both *but* not for the *whole* trajectory for
the return trip :

https://gitlab.com/python_431/cranks-and-physics/-/blob/main/Hachel/divagation_lengrand.pdf

I won't apologize either for having pointing out, as many others, that
is directly violating the principle of relativity.

It is not my fault if you, M.D Lengrand, are stubborn, dishonest and
stupid.

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to All on Thu Mar 28 14:58:31 2024
Den 27.03.2024 14:23, skrev Richard Hachel:
Le 27/03/2024 à 13:24, "Paul B. Andersen" a écrit :
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You claim that the speed of an object in an inertial frame
may be several times the speed of light, but will always be
measured to be less than c.

It's not just that I claim it, it's that it's logical, coherent,
mathematical.

Right.
Newtonian Mechanics is logical and mathematical coherent.
And falsified.

Already forty years ago, I gave the five basic equations (hundreds
will follow) of SR
The "Hachel theory" has nothing to do with SR.
The speed of the object is what Newtonian Mechanics predicts.
And then you claim that the measured speed will always be
the speed divided by sqrt(1+Vr²/c²) where Vr is the speed.

"It is impossible to measure time and speed".

A genial theory! :-D

To²=Tr²+Et²
To=Tr.sqrt(1+Vr²/c²)
Tr=To.sqrt(1-Vo²/c²)
Vo=Vr/sqrt(1+Vr²/c²)
Vr=Vo/sqrt(1-Vo²/c²)

In the Large Hadron Collider [LHC] at Cern the measured speed
of the protons is Vo = 0.9999999896⋅c

The circumference of the LHC and the time to go around
the circuit are precisely known.

Are you claiming that the real speed of the protons in the LHC is
Vr = 6927⋅c ?

Absolutely.

That's what I said.

Since it is you, Doctor Richard Hachel, i will assume you are serious.
(Nobody but you could seriously make such a ridiculous claim.)

You are obviously ignorant of how a synchrotron works.

You are obviously still ignorant.
You never read what you are told.

The protons in the LHC are moving around a ring with
circumference 26659 m. The ring consist of straight
stretches and bends. In the straight stretches there
are eight RF-cavities which accelerate the protons.
In the bends there are magnets which make the path of
the protons bent. The protons will radiate some of their
kinetic energy as synchrotron radiation (light with a special
spectrum) in the bends, and when the synchrotron is in steady
state at peak power, the energy gained in the RF-cavities will
be lost in the bends.

In a RF-cavity there is an electric field which is changing
direction sinusoidally all the time. The protons are moving in
bunches, and a bunch must be at a RF-cavity exactly at the time
when the electric field is at peak value in the right direction.
Since there are many (N) bunches in the ring, and each bunch are
going around the ring many times (M) per second, the frequency
of the RF-field in the RF-cavity must be a multiple of N x M Hz.
The nominal frequency is 400.8 MHz, but this is finely tuned
depending on the speed of the protons.

The point is that the speed of the protons is very precisely known,
and the measured and real speed of the protons is the same.

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D

But maybe you were joking.
In that case you had me!

So you were not joking!

I have a hard time believing that it is possible
to be so ignorant about the real world as you have to
be to claim what you do.

But now you have convinced me!

CERN physicists are doing their job.
We have accustomed them to working at classic relativistic speed (useful
but false).
So it makes sense that they find the speed they expect.
I tell them that the proton rotates 78 million times per second,
They tell me no.
I tell them that to be consistent you have to measure things with a
single watch, and that measuring the departure of the proton and the
arrival of the proton with two different watches (in this case, the lab
watch that counts the revolutions is TWO watches) can only lead to
measuring speeds incredibly lower than reality.

If you had read what I told you, you would have known
that the speed of the protons is measured with one clock,
namely the frequency of the RF-field in the RF-cavites.
The protons are accelerated by this field, and if the field
isn't at peak value in the right direction when the proton
is in a cavity, the Accelerator wouldn't work. So the time
between each time a proton is in a certain cavity is known
with extreme precision, and the circumference of the ring
is obviously precisely known.

So you see, the speed is close to c, but will never exceed c.

To believe that a proton in an accelerator move thousands
of times faster than the speed of light is beyond ignorance.

Any sane person who know what a proton is, must understand
that you can't make a proton go around a 27 km long ring if
you do not know _exactly_ how protons behave in electric and
magnetic fields.

No accelerator would work if charged particles didn't behave
_exactly_ as predicted by SR and Maxwell.

Only the proton itself has the correct watch, because it, in its frame
of reference, accelerated or not, is invariant.
Now, if we want to know the real speed of the proton, we must know its momentum in an instant of its journey.
Physicists will necessarily find p=m.Vr, and not p=m.Vo.

Physicists measure the momentum to be m⋅v⋅γ
and the energy to be m⋅c²⋅γ
where v is the speed and γ = 1/√(1−v²/c²).

Exactly as predicted by SR.

Sorry Richard. Your "theory" is utter nonsense.

But you better stay in Wonderland where you are free
to make up how protons and other objects behave.
Learning something about the real world and realizing
that what you have believed for 40 years is nonsense
would probably kill you.

--
Paul

https://paulba.no/

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• From Richard Hachel@21:1/5 to All on Thu Mar 28 15:13:36 2024
Le 28/03/2024 à 12:58, Python a écrit :

I won't apologize

I won't apologize

Je sais.

R.H.

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• From Maciej Wozniak@21:1/5 to All on Thu Mar 28 15:31:02 2024
W dniu 28.03.2024 o 14:58, Paul B. Andersen pisze:

Sorry Richard. Your "theory" is utter nonsense.

But you better stay in Wonderland where you are free
to make up how protons and other objects behave.
Learning something about the real world and realizing
that what you have believed for 40 years is nonsense
would probably kill you.

Sorry, Paul, the same applies to you and The
Shit of your idiot guru. And in the meantime
in the real world, forbidden by idiots like you
"improper" clocks keep measuring t'=t, just like
all the serious clocks always did.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Thu Mar 28 15:45:11 2024
Le 28/03/2024 à 14:56, "Paul B. Andersen" a écrit :

Physicists measure the momentum to be m⋅v⋅γ
and the energy to be m⋅c²⋅γ
where v is the speed and γ = 1/√(1−v²/c²).

Exactly as predicted by SR.

You think I don't know that? ? ?

Paul, I beg you to be a little more serious.

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Thu Mar 28 15:41:54 2024
Le 28/03/2024 à 14:56, "Paul B. Andersen" a écrit :
If you had read what I told you, you would have known
that the speed of the protons is measured with one clock,
namely the frequency of the RF-field in the RF-cavites.
The protons are accelerated by this field, and if the field
isn't at peak value in the right direction when the proton
is in a cavity, the Accelerator wouldn't work. So the time
between each time a proton is in a certain cavity is known
with extreme precision, and the circumference of the ring
is obviously precisely known.

So you see, the speed is close to c, but will never exceed c.

To believe that a proton in an accelerator move thousands
of times faster than the speed of light is beyond ignorance.

Any sane person who know what a proton is, must understand
that you can't make a proton go around a 27 km long ring if
you do not know _exactly_ how protons behave in electric and
magnetic fields.

No accelerator would work if charged particles didn't behave
_exactly_ as predicted by SR and Maxwell.

Where did I say the opposite of that?

R.H.

--- SoupGate-Win32 v1.05
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• From Volney@21:1/5 to Richard Hachel on Thu Mar 28 16:54:19 2024
On 3/28/2024 7:02 AM, Richard Hachel wrote:
Le 27/03/2024 à 17:07, Volney a écrit :
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein

You say that for Doctor Hachel the notion of real speed is not a speed.
Of course yes, and it's even the only consistent speed.
This is the usual notion of observable speed Vo which is NOT a speed but
a simple decoy.

You have this backwards. The velocity v is real but the rapidity r is
not a speed/velocity but instead is a function of speed, specifically r=arctanh(v/c).

At low speeds (v<<c), arctanh(v/c) ~= v/c which is why adding (low)
speeds "works" even though doing so technically calculates a rapidity.

You are talking about additions of observable (measurable) speeds, the formula is here:

The observable/measurable speed v is the real speed. Its range is simply
-c < v < c. Rapidity can take on any real value, the rapidity of light
moving at c is infinite.

I remind you that physicists who say that real speeds are of no interest
are wrong.

The real speed v is of extreme interest, of course. Rapidity, as far as
I know, is little used except as a simplifier as it has gamma 'built
in'. Perhaps someone more familiar with it can provide details.

--- SoupGate-Win32 v1.05
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• From Tom Roberts@21:1/5 to Volney on Thu Mar 28 17:19:12 2024
On 3/27/24 12:07 PM, Volney wrote:
Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity.

Yes. It is the hyperbolic angle of rotation in a space-time plane due to relative velocity v (Minkowski geometry). Just like \theta is typically
the angle of rotation in a space-space plane (Euclidean geometry).

Tom Roberts

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• From Richard Hachel@21:1/5 to All on Thu Mar 28 22:34:15 2024
Le 28/03/2024 à 20:54, Volney a écrit :
On 3/28/2024 7:02 AM, Richard Hachel wrote:

Rapidity can take on any real value, the rapidity of light
moving at c is infinite.

This is absolutely obvious.

For 40 years, I have thought at length about all these problems of special relativity, because I am naturally curious, and I find it stupid to learn notions by heart without understanding them, or without being sure that
they are correct.

When we understand things correctly, after 40 years of reflection, we
realize that it is always very difficult to teach what we have understood, because of two things: the enormous a priori (that is to say stupidity and human ignorance), and intellectual violence (i.e. arrogance in front of a
man who understands things better, and who explains them differently).

It is then very very difficult to speak and explain, especially when it
seems "a little crazy".

studied by Paul B. Andersen. Is the observable speed of a mobile invariant
in a frame of reference? We will find the question absolutely absurd.
However, what is true in a Galilean frame of reference is no longer true
in an accelerated frame of reference, and this is very strange when one is
not prepared to understand it.
It seems absurd, and I myself didn't believe it for a long time, saying to myself: "It's absurd, I'm wrong." Speed cannot change depending on
position.

However, my calculations were correct, the traditional observable speed Vo VARIES depending on the position in certain frames of reference.

This is precisely because these speeds are not real, but illusions.

If I study, for example, the traveler of Tau Ceti in accelerated mode,
and in real speed there is no problem, the real speed remains perfectly consistent.

But not the observable speed, and that is NORMAL since it is only an
illusion due to measurements made by watches located in different places.

Let's ask Paul, who is one of the best posters on usenet, and ask him what
the observable speed of a rocket crossing Tau Ceti located at 12 ly will
be.

a=1.052 (10m/s²)

It will say 0.997c.

This is false Vo=0.980c

We can calculate it directly with:

<http://news2.nemoweb.net/jntp?JpQ45DGZXSvcO3qB1xXWEJE5_U4@jntp/Data.Media:1>

R.H.

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• From Richard Hachel@21:1/5 to All on Thu Mar 28 22:45:20 2024
Le 28/03/2024 à 23:19, Tom Roberts a écrit :
On 3/27/24 12:07 PM, Volney wrote:
Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity.

Yes. It is the hyperbolic angle of rotation in a space-time plane due to relative velocity v (Minkowski geometry). Just like \theta is typically
the angle of rotation in a space-space plane (Euclidean geometry).

Tom Roberts

No.

LOL.

"Rapidity is related to speed but isn't a speed" : c'est l'inverse qui est vrai.

Observable speed is a relativistic lure, only rapidity is the real speed.

p is not p=m.Vo

p is p=m.Vr

Vo is a lure.

"La vitesse de la lumière est le premier des leurres relativistes ; ce
cheval dans ce pré,
cette lune dans ce ciel, cette galaxie dans ce télescope, je les vois instantanément,
l'émission étant simultanée de l'absorption, en direct-live.
Croire le contraire, c'est porter crédit au fait que l'on peu mesurer
des instants avec des montres désynchronisées de nature".
Dr Richard Hachel.

This explains the instantaneous transfers of information in correlated
photon experiments.
Exactly the same thing happens between our retina and the star, the body,
or the animal that we observe in perfect simultaneity.

R.H.

Si on ne comprend pas cela, on court à la catastrophe conceptuelle.

R.H.

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• From Paul B. Andersen@21:1/5 to All on Fri Mar 29 13:14:57 2024
Den 28.03.2024 16:41, skrev Richard Hachel:
Le 28/03/2024 à 14:56, "Paul B. Andersen" a écrit :

No accelerator would work if charged particles didn't behave
_exactly_ as predicted by SR and Maxwell.

Where did I say the opposite of that?

R.H.

Doctor Richard Hachel, I beg you to be a little more serious.

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Fri Mar 29 12:49:18 2024
Le 29/03/2024 à 13:13, "Paul B. Andersen" a écrit :
Den 28.03.2024 16:41, skrev Richard Hachel:
Le 28/03/2024 à 14:56, "Paul B. Andersen" a écrit :

No accelerator would work if charged particles didn't behave
_exactly_ as predicted by SR and Maxwell.

Where did I say the opposite of that?

R.H.

Doctor Richard Hachel, I beg you to be a little more serious.

Oh, I'm always very serious.
The big problem is that "the others", in general, are not.
Sometimes out of stupidity, more often out of interest.
all human behavior in general.
As soon as a small interest is at stake, human beings are incapable of remaining fair and consistent.
It is a law of nature like the law of gravitation.
Your problem, Paul B. Andersen, is that you don't understand what I'm
saying. We must then ask ourselves the question: “But what is
happening?”.
Richard speaks and Paul does not understand. For what?
Two solutions are immediately considered by those who read,
according to their scientific ideological orientation.
1. Paul doesn't understand Richard, because Doctor Hachel is a moron.
2. Paul doesn't understand Richard, because Paul is a moron.
It is obvious, for anyone with a minimum of intelligence and knowledge of
the forums, both propositions are absurd.
The question remains the same: “But then? What’s going on?”

R.H.

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• From Mikko@21:1/5 to Richard Hachel on Fri Mar 29 15:26:09 2024
On 2024-03-29 12:49:18 +0000, Richard Hachel said:

Le 29/03/2024 à 13:13, "Paul B. Andersen" a écrit :
Den 28.03.2024 16:41, skrev Richard Hachel:
Le 28/03/2024 à 14:56, "Paul B. Andersen" a écrit :

No accelerator would work if charged particles didn't behave
_exactly_ as predicted by SR and Maxwell.

Where did I say the opposite of that?

R.H.

Doctor Richard Hachel, I beg you to be a little more serious.

Your problem, Paul B. Andersen, is that you don't understand what I'm saying.

No, that is not his problem but yours. Nobody needs to understand you.
You want to be understood. But that not going to happen unless uou
find out how you can make yourself understood.

--
Mikko

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• From Richard Hachel@21:1/5 to All on Fri Mar 29 13:39:59 2024
Le 29/03/2024 à 13:26, Mikko a écrit :
No, that is not his problem but yours. Nobody needs to understand you.
You want to be understood. But that not going to happen unless uou
find out how you can make yourself understood.

“Whoever wants to teach new things must do so with great caution.”

Nietszche dit la même chose : "On n'aime pas trop les nouvelles tables".

I remind you that the problem comes neither from me nor from the reader.

There are no morons in my house or in theirs.

Additionally, I only use equations that are easily understandable for
average 16 or 17 year old students.

Sines, cosines, square roots and nothing more.

This is sufficient to propose a space-time geometry, based on Poincaré's equations, but simpler and truer than that of Minkowski, and which has no paradox or contradiction.

The question is “but what happens then?”

The answer is infinitely simple and understandable for a seven-year-old
child: "We don't want Doctor Hachel to rule over us, he overshadows us."

It's stupid.

BUT it is nevertheless true.

R.H.

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• From Python@21:1/5 to All on Fri Mar 29 15:00:14 2024
Le 29/03/2024 à 14:39, Richard Hachel a écrit :
Le 29/03/2024 à 13:26, Mikko a écrit :
No, that is not his problem but yours. Nobody needs to understand you.
You want to be understood. But that not going to happen unless uou
find out how you can make yourself understood.

“Whoever wants to teach new things must do so with great caution.”

Nietszche dit la même chose : "On n'aime pas trop les nouvelles tables".

I remind you that the problem comes neither from me nor from the reader.

There are no morons in my house or in theirs.

Additionally, I only use equations that are easily understandable for
average 16 or 17 year old students.

Sines, cosines, square roots and nothing more.

This is sufficient to propose a space-time geometry, based on Poincaré's equations, but simpler and truer than that of Minkowski, and which has

The question is “but what happens then?”

The answer is infinitely simple and understandable for a seven-year-old child: "We don't want Doctor Hachel to rule over us, he overshadows us."

It's stupid.

BUT it is nevertheless true.

R.H.

https://fr.wiktionary.org/wiki/delusion_of_grandeur

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• From Mikko@21:1/5 to Richard Hachel on Sat Mar 30 08:47:06 2024
On 2024-03-29 13:39:59 +0000, Richard Hachel said:

The answer is infinitely simple and understandable for a seven-year-old child: "We don't want Doctor Hachel to rule over us, he overshadows us."

That's right. A teacher should illuminate, not keep in dark.

--
Mikko

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• From Paul B. Andersen@21:1/5 to All on Sat Mar 30 13:05:33 2024
Den 29.03.2024 13:49, skrev Richard Hachel:
Le 29/03/2024 à 13:13, "Paul B. Andersen" a écrit :
Den 28.03.2024 16:41, skrev Richard Hachel:
Le 28/03/2024 à 14:56, "Paul B. Andersen" a écrit :

No accelerator would work if charged particles didn't behave
_exactly_ as predicted by SR and Maxwell.

Where did I say the opposite of that?

The "opposite" you claim not to have said is:
"Accelerators would work even if charged particles didn't
behave as predicted by SR."

Doctor Richard Hachel, I beg you to be a little more serious.

Oh, I'm always very serious.

Since you, Doctor Richard Hachel, claim that charged particles
do not behave as predicted by SR, and you _know_ that accelerators
work, you can't seriously claim that you haven't said:
"Accelerators do work even if charged particles don't behave
as predicted by SR."

The big problem is that "the others", in general, are not.
Sometimes out of stupidity, more often out of interest.
all human behavior in general.
As soon as a small interest is at stake, human beings are incapable of remaining fair and consistent.
It is a law of nature like the law of gravitation.
Your problem, Paul B. Andersen, is that you don't understand what I'm
saying. We must then ask ourselves the question: “But what is happening?”.
Richard speaks and Paul does not understand. For what?

Richard, are you trying to divert the attention from the issue
So why don't you address the issue in stead of this irrelevant
babble?

Two solutions are immediately considered by those who read,
according to their scientific ideological orientation.
1. Paul doesn't understand Richard, because Doctor Hachel is a moron.
2. Paul doesn't understand Richard, because Paul is a moron.

3. Paul does understand what Richard says because Richard expresses
himself very clearly with words that can't be misunderstood:

| Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :

Are you claiming that the real speed of the protons in the LHC is
Vr = 6927⋅c ?

Absolutely.

That's what I said.

| Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

CERN physicists are doing their job.
We have accustomed them to working at classic relativistic speed.
So it makes sense that they find the speed they expect.
I tell them that the proton rotates 78 million times per second,

So Paul does understand that Doctor Richard Hachel claims that
protons behave very differently from what SR predicts.

Since accelerators do work, the fact that no accelerator would work
if charged particles didn't behave _exactly_ as predicted by SR,
proves that Doctor Richard Hachel's claim is wrong.

Nothing more to discuss.

--
Paul

https://paulba.no/

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• From Volney@21:1/5 to Richard Hachel on Sat Mar 30 12:08:09 2024
On 3/28/2024 7:06 AM, Richard Hachel wrote:
Le 27/03/2024 à 23:03, Python a écrit :
Le 27/03/2024 à 18:07, Volney a écrit :
On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
Den 27.03.2024 07:23, skrev Richard Hachel:
Le 26/03/2024 à 21:45, "Paul B. Andersen" a écrit :
Den 22.03.2024 09:49, skrev Richard Hachel:
Le 21/03/2024 à 21:05, "Paul B. Andersen" a écrit :

You are claiming that the protons are going around the ≈ 27 km ring
≈ 78 million times per second.
The real value is ≈ 11.25 thousand times per second.

Don't you think the physicists at CERN had noticed the difference? :-D >>>>
But maybe you were joking.
In that case you had me!

It appears "Dr." Hachel stumbled across the physics concept of
"rapidity". Rapidity is related to speed but isn't a speed. It is
related to speed by r = arctanh(v/c) where v is the velocity. The
rapidity of light is infinity. AFAIK, rapidity is used sometimes to
simplify the math. For example, rapidities of two objects in one
dimensional motion can be just added but velocities need the Einstein

It is not even that, it is worse, far worse.

Je ne vois pas l'intérêt de ton intervention.
J'attends toujours tes excuses pour ton comportement stupide sur le description du voyageur de Langevin en vitesses apparentes, et pourquoi
la formule Vapp=Vo/(1+cosµ.Vo/c) ne fonctionne que pour l'aller et pas
pour le retour.
R.H.

Why are you asking us 'Have you ever wondered why you can't taste your
tongue?'

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• From gharnagel@21:1/5 to Volney on Sat Mar 30 19:53:29 2024
Volney wrote:

On 3/28/2024 7:06 AM, Richard Hachel wrote:

Je ne vois pas l'intérêt de ton intervention.
J'attends toujours tes excuses pour ton comportement stupide sur le description du voyageur de Langevin en vitesses apparentes, et pourquoi
la formule Vapp=Vo/(1+cosµ.Vo/c) ne fonctionne que pour l'aller et pas pour le retour.
R.H.

Why are you asking us 'Have you ever wondered why you can't taste your tongue?'

:-))

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