W dniu 17.03.2024 o 14:44, Paul B. Andersen pisze:

> SR's prediction is experimentally verified.

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

Le 17/03/2024 à 14:42, "Paul B. Andersen" a écrit :
> Den 16.03.2024 15:26, skrev Richard Hachel:
>> Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
>>> Den 15.03.2024 15:39, skrev Richard Hachel:
>>
>>> So Richard Hachel's theory is identical to Newtonian Mechanics.
>>
>> Absolutely not.
>
> Since the equations:
> Speed of rocket in inertial frame: Vr=a.Tr
> Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
> are valid _only_ in Newtonian Mechanics with Galilean relativity,
> then the theory which is consistent with said equations
> is Newtonian Mechanics.

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

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

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

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

Except I never wrote that.

R.H.

Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
> Den 16.03.2024 15:29, skrev Richard Hachel:

> So the coordinate transformation equations are:
> x' = x - v⋅t
> y' = y
> z' = z
> t' = t
>
> Which IS Galilean relativity.

Absolutly, this is galilean relativity.

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

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

I never said anything else.

R.H.

Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
> Don't be ridiculous.
> An intelligent Doctor and scientist like you will obviously
> understand that since the equations:
> Speed of rocket in inertial frame: Vr=a.Tr
> Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
> are valid _only_ in Newtonian Mechanics with Galilean relativity,
> then the theory which is consistent with said equations
> is Newtonian Mechanics.
>
> Or don't you? :-D

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

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

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

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

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

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

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

R.H.

Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
> Den 16.03.2024 15:16, skrev Richard Hachel:
>> Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
>>
>>> Richard Hachel's equations:
>>>   Speed of rocket in inertial frame:             Vr=a.Tr
>>>   Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
>>>
>>> Are correct only in Newtonian Mechanics with Galilean relativity.
>>>
>>> So Richard Hachel's theory is identical to Newtonian Mechanics.
>>>
>>> Paul
>>
>> The greatness of a man is taking serious things seriously.
>>
>> There you are joking.
>>
>> That's not what I expect from a man like you.
>
> Maybe this is what you expect?
>
> Since the equations:
> Speed of rocket in inertial frame: Vr=a.Tr
> Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
> are valid _only_ in Newtonian Mechanics with Galilean relativity,
> then the theory which is consistent with said equations
> is _only_ Newtonian Mechanics.

No.

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

R.H.

Den 18.03.2024 10:27, skrev Richard Hachel:
> Le 17/03/2024 à 14:42, "Paul B. Andersen" a écrit :
>>
>> Since the equations:
>>   Speed of rocket in inertial frame:             Vr=a.Tr
>>   Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
>>   are valid _only_  in Newtonian Mechanics with Galilean relativity,
>>   then the theory which is consistent with said equations
>>   is Newtonian Mechanics.
>
> The equations I give, if written correctly, are valid in both systems.
> But you have to write them correctly.
> For example if I write, in the Newtonian system,
> v=a.t
> This is valid.
> In the same Newtonian system, we can also write:
> v_m=(1/2)v_i
>
> We agree on this, and I don't think, even regarding the craziest posters
> (Python example), anyone will come and contradict.

Quite.

We can now review the journey to Tau Ceti.

Both Earth and Tau Ceti are considered to be inertial.

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

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

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

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

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

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

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

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

Note that d/vₘ = 4.5403 y < t Why :-D

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

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

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

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

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

Note that vₘ/v > 1/2

--
Paul

https://paulba.no/

Den 18.03.2024 10:38, skrev Richard Hachel:
> Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
>> Den 16.03.2024 15:29, skrev Richard Hachel:
>
>> So the coordinate transformation equations are:
>>   x' = x - v⋅t
>>   y' = y
>>   z' = z
>>   t' = t
>>
>> Which IS Galilean relativity.
>
> Absolutly, this is galilean relativity.
> But that's not what I'm talking about.

Yes, that's what you are talking about.

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

We can now review the journey to Tau Ceti.

Both Earth and Tau Ceti are considered to be inertial.

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

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

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

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

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

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

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

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

Note that d/vₘ = 4.5403 y < t Why :-D

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

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

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

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

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

Note that vₘ/v > 1/2

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

The equations below are not consistent with your your equations
v = a⋅t and vₘ = a⋅t/2.

> I'm talking about special relativity (Hachel sauce).
> The transformations of Hachel (that's me) concerning the Galilean
> environments are as follows:
> x'=(x-Vo.To)/sqrt(1-Vo²/c²)
> y'=y
> z'=z
> To'=[To-(x.Vo/c²)]/sqrt(1-Vo²/c²)]
>
> The opposite becoming:
>  x=(x'+Vo.To')/sqrt(1-Vo²/c²)
> y=y'
> z=z'
> To=[To'+(x'.Vo/c²)]/sqrt(1-Vo²/c²)]
>
> I never said anything else.
>
> R.H.

--
Paul

https://paulba.no/

Den 18.03.2024 10:52, skrev Richard Hachel:
> Le 17/03/2024 à 14:43, "Paul B. Andersen" a écrit :
>> Don't be ridiculous.
>> An intelligent Doctor and scientist like you will obviously
>> understand that since the equations:
>>   Speed of rocket in inertial frame:             Vr=a.Tr
>>   Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
>>   are valid _only_  in Newtonian Mechanics with Galilean relativity,
>>   then the theory which is consistent with said equations
>>   is Newtonian Mechanics.
>>
>> Or don't you? :-D

>
> I don't understand why a man like you, who has repeatedly shown in his
> speeches that he has a good command of his subject, cannot listen to
> what I say.

I am listening to what you say.

You say:
Speed of rocket in inertial frame: Vr=a.Tr
Average speed of rocket in the inertial frame: Vrm=(1/2)Vr

These statements are valid _only_ in Newtonian Mechanics
with Galilean relativity, so the theory which is consistent
with said equations is Newtonian Mechanics.

We can now review the journey to Tau Ceti.

Both Earth and Tau Ceti are considered to be inertial.

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

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

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

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

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

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

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

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

Note that d/vₘ = 4.5403 y < t Why :-D

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

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

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

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

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

Note that vₘ/v > 1/2

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

> I will never question your intellectual qualities.
> I have begged you many times to breathe calmly as you read me.
> What I blame you for is your impatience and your lack of listening.
> You tell me, Paul, that my equations are Newtonian physics, which is
> both untruthful and at the same time proof that you are not reading me
> correctly.

The point is that with the starting point v = a⋅t and vₘ = a⋅t/2
you can't get anything but Newtonian mechanics.

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

I see a number of equations with no definition of the entities in
the equations. So they are meaningless.

But i guess they all are inconsistent with your equations
v = a⋅t and vₘ = a⋅t/2
>
> <http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:1>
>
> <http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:2>
>
> <http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:3>
>
> <http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:4>
>
> <http://news2.nemoweb.net/jntp?68amBojevNcrI6dyBbapMP-hJb8@jntp/Data.Media:5>
>
> R.H.
>

--
Paul

https://paulba.no/

Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
> Quite.
>
> We can now review the journey to Tau Ceti.
>
> Both Earth and Tau Ceti are considered to be inertial.
>
> A rocket is stationary on Earth, When its clock show τ = 0 and
> the Earth clock show t = 0 the rocket engine starts an give
> the rocket a constant proper acceleration a = 10 m/s².
>
> a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly

Tout cela est entièrement exact.

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

Please note : Vr=a.Tr
Vrm=a.Tr/2

> ===================================================
>
> d = ∫a⋅t⋅dt + 0 ly = a⋅t²/2 => t = √(2⋅d/a)
>
> The rocket will pass Tau Ceti at the terrestrial time:
> t = √(2⋅d/a) = 4.7764 y

No, no, no, no !

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

To=12.92 years.

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

Absolutely.

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

Yes, Vr=5.0245c

BUT: Vo=0.980c

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

Vrm=2.51c if Vr=5.0245c

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

R.H.

Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
> According to SR:
> ================
>
> The rocket will pass Tau Ceti at the terrestrial time:
> t = √((d/c)²+2⋅d/a) = 12.9156 y

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

Here is the error.

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

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

No.

Vo=0.980c

(Vr=5.024c)

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

Là, oui.

> Note that vₘ/v > 1/2

Vom/Voi > 1/2 absolutely.

but Vrm/Vri = 1/2

R.H.

Richard Hachel wrote:

> Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
>> The proper time of the rocket when it passes Tau Ceti is:
>> τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
>
> Here is the error. Physicists use the Minkowskian metric and it is not
> correct. You have to use (I understand that this is confusing) the
> Newtonian equation. Physicists absolutely need to understand something:
> the rocket is AT REST in its frame of reference, and it is from start to
> finish.

you guys don't even know what a nazi is. A nazi is 𝗮_𝗸𝗵𝗮𝘇𝗮𝗿_𝗴𝗼𝘆 thinking he
is arian lol.

𝗔𝘇𝗼𝘃_𝗡𝗮𝘇𝗶𝘀_𝗺𝗮𝗸𝗲_𝘃𝗶𝗱𝗲𝗼_𝗶𝗻𝘃𝗶𝘁𝗶𝗻𝗴_𝗙𝗿𝗲𝗻𝗰𝗵_𝗙𝗼𝗿𝗲𝗶𝗴𝗻_𝗟𝗲𝗴𝗶𝗼𝗻_𝘁𝗼_𝗵𝗲𝗹𝗽_𝗮𝗴𝗮𝗶𝗻𝘀𝘁_𝗥𝘂𝘀𝘀𝗶𝗮_𝗶𝗻_𝗨𝗸𝗿𝗮𝗶𝗻𝗲
https://b%69%74%63%68%75te.com/video/k6vuaKzLIQ8V

Den 19.03.2024 10:17, skrev Richard Hachel:
> Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
>>
>> A rocket is stationary on Earth, When its clock show τ = 0 and
>> the Earth clock show t = 0 the rocket engine starts an give
>> the rocket a constant proper acceleration a = 10 m/s².
>>
>> a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly
>> >> According to SR:
>> ================
>>
>> The rocket will pass Tau Ceti at the terrestrial time:
>>   t = √((d/c)²+2⋅d/a) = 12.9156 y
>
> Absolutely.
>
>> The proper time of the rocket when it passes Tau Ceti is:
>>   τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
>
> Here is the error.

No error!

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

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

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

Facts. Indisputable!

> Physicists use the Minkowskian metric and it is not correct.

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

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

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

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

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

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

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

Think of it. The answer is rather obvious.

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

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

> <snip nonsense>

--
Paul

https://paulba.no/

Den 19.03.2024 09:50, skrev Richard Hachel:
> Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
>>
>> We can now review the journey to Tau Ceti.
>>
>> Both Earth and Tau Ceti are considered to be inertial.
>>
>> A rocket is stationary on  Earth, When its clock show  τ = 0 and
>> the Earth clock show t = 0 the rocket engine starts an give
>> the rocket a constant proper acceleration a = 10 m/s².
>>
>> a = 10 m/s² = 1.05265 ly/y/y   c = 1 ly/y  d = 12 ly

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

>>
>> According to your equations v = a⋅t and vₘ = a⋅t/2:
>> ===================================================
>>
>> d =  ∫a⋅t⋅dt + 0 ly = a⋅t²/2  => t = √(2⋅d/a)
>>
>> The proper time of the rocket when it passes Tau Ceti is:
>> τ = √(2⋅d/a) = 4.7764 y

> Absolutely.

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

My typo: v = a⋅t = 5.0279 ly/y

> Yes, Vr=5.0245c

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

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

>
> BUT: Vo=0.980c
>
> Observable speeds Vo is not real speeds Vr,

:-D

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

--
Paul

https://paulba.no/

W dniu 19.03.2024 o 20:45, Paul B. Andersen pisze:
> Den 19.03.2024 10:17, skrev Richard Hachel:
>> Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
>>>
>>> A rocket is stationary on  Earth, When its clock show  τ = 0 and
>>> the Earth clock show t = 0 the rocket engine starts an give
>>> the rocket a constant proper acceleration a = 10 m/s².
>>>
>>> a = 10 m/s² = 1.05265 ly/y/y   c = 1 ly/y  d = 12 ly
>>>  >> According to SR:
>>> ================
>>>
>>> The rocket will pass Tau Ceti at the terrestrial time:
>>>   t = √((d/c)²+2⋅d/a) = 12.9156 y
>>
>> Absolutely.
>>
>>> The proper time of the rocket when it passes Tau Ceti is:
>>>   τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
>>
>> Here is the error.
>
> No error!
>
> What SR predicts is not a matter of opinion,
> it is a matter of fact.

Your opinion about it is worthless, because you're
stupid.

>
> So _ACCORDING TO SR_:
> ======================
> The proper time of the rocket when it passes Tau Ceti is:
>  τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
>
> The speed of the rocket in the terrestrial frame
> when it passes Tau Ceti is:
>   v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y
>
> Facts. Indisputable!
>
>> Physicists use the Minkowskian metric and it is not correct.
>
> So you are saying that the predictions of SR are wrong
> because the metric is wrong.
> You are of course free to have that opinion, but remember that
> SR is confirmed by innumerable experiments, and falsified by none.
>
>> You have to use (I understand that this is confusing) the Newtonian
>> equation.
>> Physicists absolutely need to understand something:
>> the rocket is AT REST in its frame of reference, and it is from start
>> to finish.
>
>> Everything happens, for her, as if the surrounding space were
>> accelerating by 10m/s² around her.
>
> Ah! This is an old, rather stupid misconception:
> The twins must age equally because of the symmetry,
> both the speed and acceleration of the other twin
> are equal for both twins.
>
> This is obviously nonsense. SR do predict that
> the twin's age differently. Math don't lie!

And speaking of math - it's always good to remind
that your bunch of idiots had to announce its
oldest part false, as it didn't want to cooperate
with the madness of your insane guru.

Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :
> Den 19.03.2024 10:17, skrev Richard Hachel:
>> Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
>>>
>>> A rocket is stationary on Earth, When its clock show τ = 0 and
>>> the Earth clock show t = 0 the rocket engine starts an give
>>> the rocket a constant proper acceleration a = 10 m/s².
>>>
>>> a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly
>>> >> According to SR:
>>> ================
>>>
>>> The rocket will pass Tau Ceti at the terrestrial time:
>>>   t = √((d/c)²+2⋅d/a) = 12.9156 y
>>
>> Absolutely.
>>
>>> The proper time of the rocket when it passes Tau Ceti is:
>>>   τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
>>
>> Here is the error.
>
> No error!
>
> What SR predicts is not a matter of opinion,
> it is a matter of fact.
>
> So _ACCORDING TO SR_:
> ======================
> The proper time of the rocket when it passes Tau Ceti is:
> τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
>
> The speed of the rocket in the terrestrial frame
> when it passes Tau Ceti is:
> v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y
>
> Facts. Indisputable!

Vous vous trompez.

Doublement.

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

Je gagne toujours, ne l'oubliez pas.

R.H.

Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :
>
>> Physicists use the Minkowskian metric and it is not correct.
>
> So you are saying that the predictions of SR are wrong
> because the metric is wrong.
> You are of course free to have that opinion, but remember that
> SR is confirmed by innumerable experiments, and falsified by none.
>
>> You have to use (I understand that this is confusing) the Newtonian

I have already answered this many times.

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

You are going around in circles all alone.

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

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

R.H.

Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :
> Den 19.03.2024 09:50, skrev Richard Hachel:
>> Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :
>>>
>>> We can now review the journey to Tau Ceti.
>>>
>>> Both Earth and Tau Ceti are considered to be inertial.
>>>
>>> A rocket is stationary on  Earth, When its clock show  τ = 0 and
>>> the Earth clock show t = 0 the rocket engine starts an give
>>> the rocket a constant proper acceleration a = 10 m/s².
>>>
>>> a = 10 m/s² = 1.05265 ly/y/y   c = 1 ly/y  d = 12 ly
>
> Richard Hackel uses 10 m/s² = 1.052 ly/y²
> but my value above is more precise.
> (A year is ≈ 356.25 days, not 365 days)

If you want.

>>> According to your equations v = a⋅t and vₘ = a⋅t/2:
>>> ===================================================
>>>
>>> d =  ∫a⋅t⋅dt + 0 ly = a⋅t²/2  => t = √(2⋅d/a)
>>>
>>> The proper time of the rocket when it passes Tau Ceti is:
>>> τ = √(2⋅d/a) = 4.7764 y
>
>> Absolutely.
>
>>> The speed of the rocket in the terrestrial frame
>>> when it passes Tau Ceti is:  v = a⋅t = 5.2860 ly/y
>
> My typo: v = a⋅t = 5.0279 ly/y
>
>> Yes, Vr=5.0245c
>
> So your "theory" is identical to NM and predicts
> that there is no limit to the speed of the rocket
> in an inertial frame of reference.
>
> You know of course that experimental evidence
> show that the speed of an object can't exceed c,
> so why do you promote a "theory" you know is false?
>
>>
>> BUT: Vo=0.980c
>>
>> Observable speeds Vo is not real speeds Vr,
>
> :-D
>
> How do you think that inventing a speed that isn't real
> can change the fact that your "theory" predicts that it
> is no limit to the speed of the rocket in an inertial
> frame of reference, and therefore is falsified?
>
> --
> Paul

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

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

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

But this is only a local illusion.

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

R.H.

On 3/16/2024 10:29 AM, Richard Hachel wrote:
> Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
>> Den 15.03.2024 15:39, skrev Richard Hachel:
>
>> Richard Hachel's equations:
>>   Speed of rocket in inertial frame:             Vr=a.Tr
>
> Absolutely.
>>   Average speed of rocket in the inertial frame: Vrm=(1/2)Vr
>
> Absolutely.
>
>>
>> Are correct only in Newtonian Mechanics with Galilean relativity.
>
> Sure.
>
>>
>> So Richard Hachel's theory is identical to Newtonian Mechanics.
>
> Absolutely not.
>
> Try again.
>
>> But why is "[SR]" in the subject line when you only
>> are talking about Newtonian Mechanics?
>
> Vous plaisantez, monsieur.
> C'est indigne de vous.
>
> Je vous supplie de revenir à plus d'intelligence.
>
Why are you telling Paul that his father smelt of elderberries?

Den 19.03.2024 21:09, skrev Richard Hachel:
> Le 19/03/2024 à 20:44, "Paul B. Andersen" a écrit :
>> What SR predicts is not a matter of opinion,
>> it is a matter of fact.
>>
>> So _ACCORDING TO SR_:
>> ======================
>> The proper time of the rocket when it passes Tau Ceti is:
>>   τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
>>
>> The speed of the rocket in the terrestrial frame
>> when it passes Tau Ceti is:
>>    v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y
>>
>> Facts. Indisputable!
>
> Vous vous trompez.
>
> Doublement.
> 1. Cette optique relativiste est incohérente et théoriquement incorrecte.
> 2. Je suis persuadé que les faits expérimentaux iront contre.
> Je gagne toujours, ne l'oubliez pas.
> R.H.

"Goddag mann - økseskaft" er en norsk historie som passer bra til
dette tullet. Gagga drutto totto libre abba setruska babbell og
hip urra.

Farmatiskitabletull!

So there!

--
Paul

https://paulba.no/

Den 19.03.2024 21:59, skrev Richard Hachel:
>
> I have told you many times that if we use low speeds, the observable
> speeds are the real speeds.
> This trotting horse, this motorcycle, this passing train.
> What I measure is the reality of things.
>
> I also said that the nature of space is such that if we want to have a
> correct notion of speed, we absolutely must use only one watch and NEVER
> two watches placed in different places.
>
> The best measure of time will therefore be the mobile's own time, which
> goes from A to B, and which clicks during both events. There can
> therefore be no measurement error.
> It notes tau (or Tr).
> If I want the real speed of a mobile, in the reference frame where I am,
> I therefore need the AB measurement in this reference frame, BUT the
> mobile's own time.
>
> If I make the mistake of taking time A noted by watch A, and time B
> noted by watch B, I am using two watches placed in two different
> theaters, and which will never be naturally in tune, because the
> universe is not “done like that”. The notion of universal present time
> is as abstract an notion as the notion of a flat earth.
> Now, in relativity, this is what we do every day, and we find a
> measurement that is false, and therefore a speed that is false.
> It doesn't matter for low speeds.
> But for relativistic speeds, the measurement errors are considerable,
> and it "seems" to us that nothing can exceed c.
> But it is only an abstract idea that is very difficult to disengage from
> the human mind.
> Vo=Vr/sqrt(1+Vr²/c²)
> Vr=Vo/sqrt(1-Vo²/c²)
>
> We then understand that all real speeds are permitted, but that, by the
> way we use distinct watches,
> an impression of speed Vo appears, and it cannot exceed c.
>
> But this is only a local illusion.

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

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

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

c = 1 light year per year.

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

--
Paul

https://paulba.no/

Volney wrote:

> n 3/16/2024 10:29 AM, Richard Hachel wrote:
>> Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
>>> But why is "[SR]" in the subject line when you only are talking about
>>> Newtonian Mechanics?
>>
>> Vous plaisantez, monsieur. C'est indigne de vous.
>> Je vous supplie de revenir à plus d'intelligence.
>>
> Why are you telling Paul that his father smelt of elderberries?

your sincerity my friend.But they had 𝗺𝗲𝗲𝘁𝗶𝗻𝗴𝘀_𝗮𝗳𝘁𝗲𝗿_𝗺𝗲𝗲𝘁𝗶𝗻𝗴𝘀_𝗶𝗻_𝗘𝗨, about
balkanizing Russia, stealing their resources etc. With the motherfucker
𝗡𝗮𝘃𝗮𝗹𝗻𝘆_𝗮𝗻𝗱_𝗬𝘂𝗹𝗹𝗮𝗻𝗱𝗮 in place, to take over and everything. My butt, almost
incredible. Here the proof, because proofs are essential in Physics. The
fucking EU are corrupt like shit, obvious.

(18:00) thank you my friend
𝗙𝗿𝗲𝗻𝗰𝗵_𝗺𝗲𝗱𝗶𝗮_𝘄𝗮𝗿𝗴𝗮𝗺𝗶𝗻𝗴_20𝗞_𝘁𝗿𝗼𝗼𝗽𝘀.
𝗥𝘂𝘀𝘀𝗶𝗮_𝗜𝗻𝘁𝗲𝗹,_2𝗞_𝗙𝗿𝗲𝗻𝗰𝗵_𝘁𝗿𝗼𝗼𝗽𝘀_𝗨𝗸𝗿𝗮𝗶𝗻𝗲.
𝗘𝗨_𝗽𝗮𝗻𝗶𝗰,_𝗸𝗲𝗲𝗽_𝘄𝗮𝗿_𝗴𝗼𝗶𝗻𝗴
https://youtu.be/07yaxhmvFJA

Volney wrote:

> On 3/16/2024 10:29 AM, Richard Hachel wrote:
>> Le 16/03/2024 à 14:18, "Paul B. Andersen" a écrit :
>>> But why is "[SR]" in the subject line when you only are talking about
>>> Newtonian Mechanics?
>>
>> Vous plaisantez, monsieur. C'est indigne de vous.
>> Je vous supplie de revenir à plus d'intelligence.
>>
> Why are you telling Paul that his father smelt of elderberries?

they had 2,3 days events and conferences in EU to kill Russia. What is it
you don't undrestand?? Come on, you have to undrestand. Probably not even
the gay actor 𝘀𝗺𝗲𝗹𝗹𝗲𝗻𝘀𝗸𝘆 knew about this plan, speaking the witch.

(from 18:00)
𝗙𝗿𝗲𝗻𝗰𝗵_𝗺𝗲𝗱𝗶𝗮_𝘄𝗮𝗿𝗴𝗮𝗺𝗶𝗻𝗴_20𝗞_𝘁𝗿𝗼𝗼𝗽𝘀. 𝗥𝘂𝘀𝘀𝗶𝗮_𝗜𝗻𝘁𝗲𝗹,_2𝗞_𝗙𝗿𝗲𝗻𝗰𝗵_𝘁𝗿𝗼𝗼𝗽𝘀_𝗨𝗸𝗿𝗮𝗶𝗻𝗲.
𝗘𝗨_𝗽𝗮𝗻𝗶𝗰,_𝗸𝗲𝗲𝗽_𝘄𝗮𝗿_𝗴𝗼𝗶𝗻𝗴
https://youtu.be/07yaxhmvFJA

so fucking sad with these despicable satanists.