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.

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

Richard Hachel wrote:

> 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).

more proofs. The fucking polaker warshaw is genuine Russian lol. No
fucking ukrein anywhere on the map. That word, ukrein, doesn't even
exists. It's a stolen country configuration.

𝗧𝗵𝗲_𝗚𝗲𝗿𝗺𝗮𝗻_𝗶𝘀_𝗶𝗻_𝘀𝗵𝗼𝗰𝗸._𝗛𝗲_𝗳𝗼𝘂𝗻𝗱_𝗮_𝗺𝗮𝗽_𝗳𝗿𝗼𝗺_1908_𝗻𝗼_𝗨𝗸𝗿𝗮𝗶𝗻𝗲,_𝗪𝗮𝗿𝘀𝗮𝘄_𝗶𝘀_𝗥𝘂𝘀𝘀𝗶𝗮.
https://b%69%74%63%68%75te.com/video/jrqhFzg3wvJx

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/

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
talking about Vo.
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.

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.

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/

Den 19.03.2024 21:29, skrev Richard Hachel:
> 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
> talking about Vo.
> 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.
>

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/

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.

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.

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
is born dead.

>
> 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/

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
about absolutely inhabit a wonderland.
The observers from the real world have nothing in
common with them.

Maciej Wozniak wrote:

> 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 about absolutely
> inhabit a wonderland. The observers from the real world have nothing in
> common with them.

𝗥𝘂𝘀𝘀𝗶𝗮𝗻_𝗺𝗼𝗻𝗲𝘆_‘𝗱𝗼𝗲𝘀𝗻’𝘁_𝗯𝗲𝗹𝗼𝗻𝗴_𝘁𝗼_𝗮𝗻𝘆𝗼𝗻𝗲’_–_𝗚𝗲𝗿𝗺𝗮𝗻𝘆’𝘀_𝗦𝗰𝗵𝗼𝗹𝘇
Profits generated from Moscow’s frozen assets can be used to support the
Ukrainian military, the chancellor has argued
https://r%74.com/news/594637-russian-frozen-assets-scholz/

these 𝗸𝗵𝗮𝘇𝗮𝗿_𝗴𝗼𝘆𝘀 of gearmony are stealing money like the gypsies.
Disgusting sacks of shit, they think they are arians. Fuckin nazi gypsies
each and everyone.

China, India, get your money out of the EU/US as soon as possible.

This is the man who tell you that Germany doesn't belong to the Germans
and that he works for someone more important than the German people.

German Chancellor Olaf Scholz criminal acts are his way of distracting the
people of Germany from the covid scam.

The money "does not belong to anyone" but Brussels can use it as their
own. New European rationalism..

Something a Nazi occupier would say during WW2. Some people never change.

This is the strongest reason why foreign investors should stop investing
in western assets.

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.

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/

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.

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/

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.

Richard Hachel wrote:

> Le 27/03/2024 à 13:24, "Paul B. Andersen" a écrit :
>> 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.

it's a Law of Nature, my man, aka Newtons 3rd law of motion. That's simply
physics and mathematics. The amrica is fucked up by 𝙩𝙝𝙚_𝙨𝙢𝙚𝙡𝙡𝙚𝙣𝙨𝙠𝙞𝙚𝙨 of
ukrein. The 𝙠𝙝𝙖𝙯𝙖𝙧_𝙜𝙤𝙮𝙨 from The Bible.

compare 𝙩𝙝𝙚_𝙖𝙘𝙘𝙚𝙡𝙚𝙧𝙖𝙩𝙞𝙤𝙣, sharp 𝙨𝙝𝙞𝙛𝙩_𝙞𝙣_𝙙𝙞𝙧𝙚𝙘𝙩𝙞𝙤𝙣, to hit the bridge,
which has nothing to do with "𝙩𝙝𝙚_𝙡𝙤𝙨𝙨_𝙤𝙛_𝙥𝙤𝙬𝙚𝙧"

𝗙𝗿𝗮𝗻𝗰𝗶𝘀_𝗦𝗰𝗼𝘁𝘁_𝗞𝗲𝘆_𝗕𝗿𝗶𝗱𝗴𝗲_𝗶𝗻_𝗕𝗮𝗹𝘁𝗶𝗺𝗼𝗿𝗲_𝗱𝗲𝘀𝘁𝗿𝗼𝘆𝗲𝗱_𝘄𝗵𝗲𝗻_𝗮_𝘀𝗵𝗶𝗽_𝗹𝗼𝘀𝗲𝘀_𝗽𝗼𝘄𝗲𝗿
https://bi%74%63%68%75te.com/video/5tboc58SFodU

with this, which is nothing, except that was "𝙘𝙤𝙣𝙨𝙩𝙧𝙪𝙘𝙩𝙞𝙤𝙣_𝙬𝙤𝙧𝙠𝙚𝙧𝙨" on the
bridge, one a clock in the morning!!

𝗠𝗮𝗿𝘆𝗹𝗮𝗻𝗱_𝗯𝗿𝗶𝗱𝗴𝗲_𝗰𝗼𝗹𝗹𝗮𝗽𝘀𝗲𝗱_𝗮𝗳𝘁𝗲𝗿_𝗯𝗲𝗶𝗻𝗴_𝘀𝘁𝗿𝘂𝗰𝗸_𝗯𝘆_𝗰𝗼𝗻𝘁𝗮𝗶𝗻𝗲𝗿_𝘀𝗵𝗶𝗽
https://bi%74%63%68%75te.com/video/i6wCi3FywKqK

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.

Volney wrote:

> On 3/27/2024 8:25 AM, Paul B. Andersen wrote:
>> 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.

my friend, I love you so much, but you guys dont undrestand Newtons 1st
low of motion, which is a Law of Nature in this manifold. It's the 1st
law, not the 3rd. Here with proofs, because I'm not eating shit.

compare 𝙩𝙝𝙚_𝙖𝙘𝙘𝙚𝙡𝙚𝙧𝙖𝙩𝙞𝙤𝙣, sharp 𝙨𝙝𝙞𝙛𝙩_𝙞𝙣_heading/𝙙𝙞𝙧𝙚𝙘𝙩𝙞𝙤𝙣, to hit the bridge, which
must be 𝗶𝗻𝘁𝗲𝗻𝘁𝗶𝗼𝗻𝗮𝗹𝗹𝘆, and has nothing to do with "𝙩𝙝𝙚_𝙡𝙤𝙨𝙨_𝙤𝙛_𝙥𝙤𝙬𝙚𝙧". Myohhmy,
america hit in the face for money, by the underdeveloped 𝗸𝗵𝗮𝘇𝗮𝗿_𝗴𝗼𝘆𝘀 from
the fictious ukrein. You people are never waking the fuck up.

𝗙𝗿𝗮𝗻𝗰𝗶𝘀_𝗦𝗰𝗼𝘁𝘁_𝗞𝗲𝘆_𝗕𝗿𝗶𝗱𝗴𝗲_𝗶𝗻_𝗕𝗮𝗹𝘁𝗶𝗺𝗼𝗿𝗲_𝗱𝗲𝘀𝘁𝗿𝗼𝘆𝗲𝗱_𝘄𝗵𝗲𝗻_𝗮_𝘀𝗵𝗶𝗽_𝗹𝗼𝘀𝗲𝘀_𝗽𝗼𝘄𝗲𝗿
https://bi%74%63%68%75te.com/video/5tboc58SFodU

with this, which is nothing, except that was "𝙘𝙤𝙣𝙨𝙩𝙧𝙪𝙘𝙩𝙞𝙤𝙣_𝙬𝙤𝙧𝙠𝙚𝙧𝙨" on the
bridge, one a clock in the morning!!

𝗠𝗮𝗿𝘆𝗹𝗮𝗻𝗱_𝗯𝗿𝗶𝗱𝗴𝗲_𝗰𝗼𝗹𝗹𝗮𝗽𝘀𝗲𝗱_𝗮𝗳𝘁𝗲𝗿_𝗯𝗲𝗶𝗻𝗴_𝘀𝘁𝗿𝘂𝗰𝗸_𝗯𝘆_𝗰𝗼𝗻𝘁𝗮𝗶𝗻𝗲𝗿_𝘀𝗵𝗶𝗽
https://bi%74%63%68%75te.com/video/i6wCi3FywKqK

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.

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
>> velocity addition formula.
>
> 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.
Tell me, poor stinker, have you already read
learnt what a function is? Is the clause
"for any element of the domain" still confusing
you?

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.

Your answer is very interesting, I will just modify it slightly.
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.
You are talking about additions of observable (measurable) speeds, the
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.

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
>> velocity addition formula.
>
> 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.