Vo(m)=x/To

Vo(m)/c=(x/c)/To

To=(x/c).sqrt(1+2c²/ax)

Vo(m)/c=(x/c)/[(x/c).sqrt(1+2c²/ax)]

Vr(m)=Vo(m)/sqrt(1-Vo(m)²/c²)

Vr=2.Vr(m)

Vo=Vr/sqrt(1+Vr²/c²)

Then:

<http://news2.nemoweb.net/jntp?LdiqhYxzbreqNbtE8CJtElfK_-E@jntp/Data.Media:1>

R.H.

On Sunday, November 20, 2022 at 5:54:03 AM UTC-8, Richard Hachel wrote:
> v = x/t

No, v=dx/dt. You are conflating x/t with dx/dt. Look, if you begin at x=0,t=0 and you drive for one hour at 30 km/hr, you will be at x=30,t=1, and then you abruptly accelerate to 60 km/hr. After one hour traveling at 60 km/hr you will be x=90,t=2, and according to your insane claim this means that while your velocity is dx/dt = 60 km/hr your velocity is also x/t = 45 km/hr. As has been patiently explained to you countless times, the value of x/t is actually the velocity of a vehicle that moved at constant speed from x=0,t=0 to x=90,t=2, but it is not the velocity of a vehicle that moved at constant speed from x=30,t=1 to x=90,t=2. The velocity of a vehicle moving at 60 km/hr is 60 km/hr. The only turnips and carrots here are the ones between your ears.

> t = x*sqrt(1 + 2/(ax))

This is the equation of a hyperbolic trajectory t=sqrt(x^2+2x/a) that begins at rest at the origin and moves with constant proper acceleration "a". You copied this from special relativity, and you have no idea how to derive this (because you don't even know what proper acceleration is). Then you write this in the form x/t = [1 + 2/(ax)]^(-1/2), and then you insanely conflate x/t with dx/dt by claiming that v=x/t (which is equivalent to claiming the vehicle moving at 60 km/hr is moving at 45 km/hr, so 1=0), so you claim v=[1 + 2/(ax)]^(-1/2), which implies 1=0. Your beliefs are absurd. Again, the correct expression for the velocity of the trajectory in terms of these coordinates is v = dx/dt = sqrt[1 - 1/(1+ax)^2].

Remember, Langevin and Tau Ceti are essentially the same:

- Example 1: One twin is stationary at the origin for 2 hours, and the other twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1 hour. Do the clocks have the same elapsed proper time?

- Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do the clocks have the same elapsed proper time?

Le 20/11/2022 à 17:35, Stan Fultoni a écrit :

> You are conflating
> You copied this from special relativity

LOL.

R.H.

Le 20/11/2022 à 17:35, Stan Fultoni a écrit :
>> v = x/t
>
> No, v=dx/dt. You are conflating x/t with dx/dt.

Vo(m)= x/To

La vitesse moyenne d'un objet accéléré est égale à sa distance
parcourue par le temps mis à la parcourir.

Vo(i)=dx/dTo

La vitesse instantanée d'un mobile est égale à son incrément dx de
distance par son incrément dt de temps.

> Look, if you begin at x=0,t=0 and you drive for one hour at 30 km/hr, you will
> be at x=30,t=1, and then you abruptly accelerate to 60 km/hr. After one hour
> traveling at 60 km/hr you will be x=90,t=2,

Evidemment.

> and according to your insane claim this means that while your velocity is dx/dt
> = 60 km/hr your velocity is also x/t = 45 km/hr.

Ce n'est pas ce que je dis. Je sais bien qu'une vitesse de 60 km/h n'est
pas une vitesse de 45 km/h.

Je sais bien que ce n'est pas la même chose.

Dans ce cas, nous aurons Vo(m)=45km/h et Vo(i)=60km/h

Je n'ai aucun souci avec ça.

R.H.

On Sunday, November 20, 2022 at 4:19:03 PM UTC-8, Richard Hachel wrote:
> The average [velocity] of an ... object is equal to its
> distance traversed by the time taken to traverse it [delta x/delta t].
> The instantaneous [velocity] is equal to [dx/dt].

Right, and the usual term for "instantaneous velocity" is just "velocity", and if (for some reason) we want to talk about the average velocity of an object, which is delta x/delta t, we refer to it as the "average velocity" between the specified events on the trajectory.

>> Look, if you begin at x=0,t=0 and you drive for one hour at 30 km/hr, you will
>> be at x=30,t=1, and then you abruptly accelerate to 60 km/hr. after one hour
>> traveling at 60 km/hr you will be x=90,t=2, and according to your insane claim
>> this means that while your velocity is dx/dt = 60 km/hr your velocity is also
>> x/t = 45 km/hr.
>
> I know very well that a speed of 60 km/h is not a speed of 45 km/h.

Great. This is real progress.

> In this case, we will have Vo(m)=45km/h and Vo(i)=60km/h

No, in this case we have v=60 km/hr, and the average velocity of this object between x=30,t=1 and x=90,t=2 is also 60 km/hr, and the average velocity between any other event xe,te on the object's trajectory and x=90,t=2 is (90-xe)/(2-te). You cannot use a symbol like v(m) to signify average velocity, because that is ambiguous. You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two specified events on the trajectory of the object. An object has a unique velocity at any given event, but it does not have a unique average velocity at a given event.

> I have no problem with that.

Sure you do. All this time you have been conflating a tacitly assumed average velocity over some tacitly assumed interval with the actual velocity of an object at a specified event. You kept saying "scientists are wrong when they say the velocity of this trajectory is 0.97, because that is much too high..blah blah blah". Well, now you confess that the scientists were correct all along. It was your mistake in conflating average velocity with velocity, etc. And, again, the average velocity is the velocity of an unaccelerated trajectory over that specified interval. You kept denying this, but now you say "I have no problem with that". Well, excellent. This is great progress.

> t = x*sqrt(1 + 2/(ax))

This is the equation of a hyperbolic trajectory t=sqrt(x^2+2x/a) that begins at rest at the origin and moves with constant proper acceleration "a". You copied this from special relativity, and you have no idea how to derive this (because you don't even know what proper acceleration is). Then you write this in the form x/t = [1 + 2/(ax)]^(-1/2), and then you insanely conflate x/t with dx/dt by claiming that v=x/t (which is equivalent to claiming the vehicle moving at 60 km/hr is moving at 45 km/hr, so 1=0), so you claim v=[1 + 2/(ax)]^(-1/2), which implies 1=0. Your beliefs are absurd. Again, the velocity of the trajectory in terms of these coordinates is v = dx/dt = sqrt[1 - 1/(1+ax)^2].

Remember, Langevin and Tau Ceti are essentially the same examples:

-- Example 1: One twin is stationary at the origin for 2 hours, and the other twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1 hour. Do the clocks have the same elapsed proper time?

-- Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do the clocks have the same elapsed proper time?

Le 21/11/2022 à 02:11, Stan Fultoni a écrit :

> You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two
> specified events on the trajectory of the object.

And I beg you to understand that if the x2-x1 value is correct, the t2-t1
value is not, and that we are, WITHOUT KNOWING IT, deducting a carrot from
a turnip.
The watch that notes t1 is not the same as the one that notes t2, and it
is not the same one that will note dt either.
When you draw an orthonormal aspec-time frame, you must take a stable
frame, with the distances on the abscissa, and the REAL TIMES on the
ordinate.

I beg you to understand (I appeal to your intelligence) that if
To²=Tr²+Et²
To represents, in the frame, a simple hypotenuse, and that this hypotenuse
will progress on something other than a fixed axis (like Tr and x).

Can you understand that?

Can you make sure to teach what I say, and not the fantasies of
relativistic physicists?

<http://news2.nemoweb.net/jntp?bPz5t2nQJrSRTm8QnjMhzkKOFmk@jntp/Data.Media:1>

R.H.

--
"Mais ne nous trompons pas.
Il n'y a pas que de la violence avec des armes : il y a des situations de
violence."
Abbé Pierre
₀₀₀
<http://news2.nemoweb.net/?DataID=bPz5t2nQJrSRTm8QnjMhzkKOFmk@jntp>

On Monday, November 21, 2022 at 8:09:08 AM UTC-8, Richard Hachel wrote:
> > You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two
> > specified events on the trajectory of the object.
>
> If the x2-x1 value is correct, the t2-t1 value is not...

You contradict yourself. You already agreed that a car moving from (30,1) to (90,2) has an average velocity of 60 km/hr during that interval. That is given by (90-30)/(2-1) = 60. Likewise if it goes on to move at constant speed from (90,2) to (120,3) then it's average velocity between (30,1) and (120,3) is 45 km/hr. Also, it's average veliocity from (90,2) to (120,3) is 30 km/hr. You assured me that you have no problem with this. Were you lying?

Le 21/11/2022 à 21:08, Stan Fultoni a écrit :
> On Monday, November 21, 2022 at 8:09:08 AM UTC-8, Richard Hachel wrote:
>> > You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two
>> > specified events on the trajectory of the object.
>>
>> If the x2-x1 value is correct, the t2-t1 value is not...
>
> You contradict yourself. You already agreed that a car moving from (30,1) to
> (90,2) has an average velocity of 60 km/hr during that interval. That is given by
> (90-30)/(2-1) = 60. Likewise if it goes on to move at constant speed from (90,2)
> to (120,3) then it's average velocity between (30,1) and (120,3) is 45 km/hr.
> Also, it's average veliocity from (90,2) to (120,3) is 30 km/hr. You assured me
> that you have no problem with this. Were you lying?

I have no lack of understanding between the different types of speeds.

Perhaps that is why I am in a better position to talk about the theory of
relativity than many men.

Do you yourself understand what you are saying?

To put it simply, I remind you that I speak, and that I try to define
things, so that you understand them.

It takes effort for you, I admit.

A real speed (Vr) is a speed which really exists in a frame of reference,
and which is not deformed by the notions of chronotropy and anisochrony.
Physicists don't seem to know that notion, and I don't know why.

An observable speed (Vo) is the speed measured by those who observe the
phenomena, but with the relativistic deformations which are fundamental.
Physicists generally denote this speed v.

Apparent velocities (Vapp). They are the same ones that physicists use
when talking about the added Doppler effect.

Instantaneous speeds, which physicists also use for Voi (instantaneous
observable speed) and which they denote v(i).

I only add that you can also use Vri (real instantaneous speed).

It's not hard to understand.

Except when you refuse dialogue, convincing yourself that the other is an
enemy, and that he is stupid.

R.H.

On Monday, November 21, 2022 at 12:25:09 PM UTC-8, Marceline wrote:
> >> > You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two
> >> > specified events on the trajectory of the object.
> >>
> >> If the x2-x1 value is correct, the t2-t1 value is not...
> >
> > You contradict yourself. You already agreed that a car moving from (30,1) to
> > (90,2) has an average velocity of 60 km/hr during that interval. That is given by
> > (90-30)/(2-1) = 60. Likewise if it goes on to move at constant speed from (90,2)
> > to (120,3) then it's average velocity between (30,1) and (120,3) is 45 km/hr.
> > Also, it's average veliocity from (90,2) to (120,3) is 30 km/hr. You assured me
> > that you have no problem with this. Were you lying?
>
> I have no lack of understanding between the different types of speeds.

Great. So you agree that in terms of the standard inertial coordinates x,t (readings on a grid of standard rulers and clocks at rest and inertially synchronized in a given frame) the average velocity of a trajectory x(t) between t1 and t2 is [x(t2)-x(t1)]/(t2-t1). Excellent. You've made real progress. Do you now understand why this shows that all your beliefs are absurd?

Le 22/11/2022 à 01:27, Stan Fultoni a écrit :
> On Monday, November 21, 2022 at 12:25:09 PM UTC-8, Marceline wrote:
>> >> > You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two
>> >> > specified events on the trajectory of the object.
>> >>
>> >> If the x2-x1 value is correct, the t2-t1 value is not...
>> >
>> > You contradict yourself. You already agreed that a car moving from (30,1) to
>> > (90,2) has an average velocity of 60 km/hr during that interval. That is given
>> by
>> > (90-30)/(2-1) = 60. Likewise if it goes on to move at constant speed from
>> (90,2)
>> > to (120,3) then it's average velocity between (30,1) and (120,3) is 45 km/hr.
>> > Also, it's average veliocity from (90,2) to (120,3) is 30 km/hr. You assured
>> me
>> > that you have no problem with this. Were you lying?
>>
>> I have no lack of understanding between the different types of speeds.
>
> Great. So you agree that in terms of the standard inertial coordinates x,t
> (readings on a grid of standard rulers and clocks at rest and inertially
> synchronized in a given frame) the average velocity of a trajectory x(t) between
> t1 and t2 is [x(t2)-x(t1)]/(t2-t1). Excellent. You've made real progress. Do
> you now understand why this shows that all your beliefs are absurd?

It's not my beliefs that are absurd, it's yours.

I haven't stopped telling you, for a few weeks now, that when it comes to
special relativity, we can't afford everything.

One cannot, for example, add speeds to speeds without taking precautions.

There is a very clear and precise law that I have given you.

You cannot contradict it, because physicists have the same one, even if
they write it differently.

The same goes for the simplistic subtraction you do when you want to
calculate the instantaneous observable velocity of an accelerating body.

You set Vo=x/(t2-t1) without taking care that the two times t1 and t2 are
not times taken by the same clock.

And so you get too much instantaneous speed.

It's the same thing with the proper times of accelerated objects, the
times you calculate are too short.

If you use a Galilean average speed in place of the accelerated movement
from 0 to 3ly, then a Galilean average speed from 0 to 3.1ly, you will
notice that the average speed is greater for the 3.1ly course.

And if you look closely, by launching the two mobiles at the same time,
they will therefore not pass together at the point x=3ly.

They have neither my real speed nor the same observable speed.

Neither the same travel time for the 0 to 3ly portion.

And when you subtract the time of one time from the other, you make the
mistake of believing that the dTo obtained is the subtraction of two times
based on the same system.

I therefore give the correct equation to find the instantaneous velocities
as a function of x.

Vr=sqrt(2ax)
Vo=[1+c²/2ax]^(-1/2)

You will notice the relativistic logic which wants that
Vr=Vo/sqrt(1-Vo²/c²)
and Vo=Vr/sqrt(1+Vr²/c²)

I can't explain it any better.

It is not by saying that it is absurd without having understood what I am
explaining that you will get there.

R.H.

On Tuesday, November 22, 2022 at 12:30:13 AM UTC-8, Richard Hachel wrote:
> >> >> > You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two
> >> >> > specified events on the trajectory of the object.
> >> >>
> >> >> If the x2-x1 value is correct, the t2-t1 value is not...
> >> >
> >> > You contradict yourself. You already agreed that a car moving from (30,1) to
> >> > (90,2) has an average velocity of 60 km/hr during that interval. That is given
> >> > by (90-30)/(2-1) = 60. Likewise if it goes on to move at constant speed from
> >> > (90,2) to (120,3) then it's average velocity between (30,1) and (120,3) is 45 km/hr.
> >> > Also, it's average veliocity from (90,2) to (120,3) is 30 km/hr. You assured
> >> > me that you have no problem with this. Were you lying?
> >>
> >> I have no lack of understanding between the different types of speeds.
> >
> > Great. So you agree that in terms of the standard inertial coordinates x,t
> > (readings on a grid of standard rulers and clocks at rest and inertially
> > synchronized in a given frame) the average velocity of a trajectory x(t) between
> > t1 and t2 is [x(t2)-x(t1)]/(t2-t1). Excellent. You've made real progress. Do
> > you now understand why this shows that all your beliefs are absurd?
>
> One cannot, for example, add speeds to speeds...

Of course we can. Remember, you mistakenly thought that 0.5c + 0.5c = 0.8c, but I explained to you that, in fact, 0.5c + 0.5c = 1.0c. Your belief that special relativity invalidated arithmetic or rational thought is incorrect. You are conflating addition with composition. Furthermore, the velocities appearing in the relativistic composition formula are defined precisely as I have explained to you, i.e., the velocity of a trajectory x(t) in terms of coordinates x,t is v=dx/dt. So you cannot rationally appeal to these velocities in support of your absurd beliefs. Remember, when your car is moving at 65 km/hr in terms of a given system of coordinates it is moving at 65 km/hr in terms of that system of coordinates. You already agreed to this. You said you had no problem with it. Remember?

> You set Vo=x/(t2-t1) without taking care that the two times t1 and t2 are
> not times taken by the same clock.

No, in terms of standard inertial coordinates x,t, defined as the readings of standard rulers and clocks all at rest and inertially synchronized in a given frame, the average velocity of a trajectory x(t) between t1 and t2 is given by [x(t2)-x(t1)]/(t2-t1). Obviously the times t1 and t2 are read on different clocks, because the object is moving relative to the grid of rulers and clocks, but these clocks are all mutually at rest and inertially synchronized, as this is the meaning of a standard inertial coordinate system..

Remember, Langevin and Tau Ceti are the same:

- Example 1: One twin is stationary at the origin for 2 hours, and the other twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1 hour. Do these twins have the same elapsed proper time?

- Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do these twins have the same elapsed proper time?

Le 22/11/2022 à 14:22, Stan Fultoni a écrit :

> Of course we can. Remember, you mistakenly thought that 0.5c + 0.5c = 0.8c, but
> I explained to you that, in fact, 0.5c + 0.5c = 1.0c.

Vous avez raison.

0.5+0.5=1

0.5+0.5 = pas du tout 0.8.

Et puis j'ai pas envie d'y passer la nuit.

J'ai d'autres choses que ça à foutre.

R.H.

On Tuesday, November 22, 2022 at 5:36:56 AM UTC-8, Richard Hachel wrote:
> > Remember, you mistakenly thought that 0.5c + 0.5c = 0.8c, but
> > I explained to you that, in fact, 0.5c + 0.5c = 1.0c.
>
> You are right.

Yes, so you agree that in terms of standard inertial coordinates x,t, defined as the readings of standard rulers and clocks all at rest and inertially synchronized in a given frame, the average velocity of a trajectory x(t) between t1 and t2 is given by [x(t2)-x(t1)]/(t2-t1). Good. Now, to show that your other beliefs are absurd, remember that Langevin and Tau Ceti are the same:

- Example 1: One twin is stationary at the origin for 2 hours, and the other twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1 hour. Do these twins have the same elapsed proper time?

- Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do these twins have the same elapsed proper time?

The reason you cannot answer these simple questions is because you know your beliefs are absurd, so all you can do is tun away. Right?

Le 22/11/2022 à 14:53, Stan Fultoni a écrit :

> inertially synchronized in a given frame

Nan, écoute, Coco, tu es bien gentil, mais ma patience à des limites.

Ca devient infernal vos conneries.

C'est épouvantable.

R.H.

On Sunday, November 20, 2022 at 5:54:03 AM UTC-8, Richard Hachel wrote:
> v = x/t

No, v=dx/dt. You are conflating x/t with dx/dt. Look, if you begin at x=0,t=0 and you drive for one hour at 30 km/hr, you will be at x=30,t=1, and then you abruptly accelerate to 60 km/hr. After one hour traveling at 60 km/hr you will be x=90,t=2, and according to your claim this means that while your velocity is dx/dt = 60 km/hr your velocity is also x/t = 45 km/hr. As has been patiently explained to you countless times, the value of x/t is actually the velocity of a vehicle that moved at constant speed from x=0,t=0 to x=90,t=2, but it is not the velocity of a vehicle that moved at constant speed from x=30,t=1 to x=90,t=2. The velocity of a vehicle moving at 60 km/hr is 60 km/hr. The only turnips and carrots here are the ones between your ears.

> t = x*sqrt(1 + 2/(ax))

This is the equation of a hyperbolic trajectory t=sqrt(x^2+2x/a) that begins at rest at the origin and moves with constant proper acceleration "a". You copied this from special relativity, and you have no idea how to derive this (because you don't even know what proper acceleration is). Then you write this in the form x/t = [1 + 2/(ax)]^(-1/2), except you mistakenly place the "2" in the denominator, and then you insanely conflate x/t with dx/dt by claiming that v=x/t (which is equivalent to claiming the vehicle moving at 60 km/hr is moving at 45 km/hr, so 1=0), so you claim v=[1 + 2/(ax)]^(-1/2). This is all based on your claim that 65=45, which implies 1=0. Your beliefs are absurd. Again, the correct expression for the velocity of the trajectory in terms of these coordinates is v = dx/dt = sqrt[1 - 1/(1+ax)^2].

Remember, Langevin and Tau Ceti are the same, as explained in terms of standard inertial coordinates:

- Example 1: One twin is stationary at the origin for 2 hours, and the other twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1 hour. Do the twins have the same elapsed proper time?

- Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do the twins have the same elapsed proper time?

You cannot answer these questions, right?

On Tuesday, November 22, 2022 at 9:08:13 AM UTC-5, Richard Hachel wrote:
> Le 22/11/2022 à 14:53, Stan Fultoni a écrit :
>
> > inertially synchronized in a given frame
> Nan, écoute, Coco, tu es bien gentil, mais ma patience à des limites.

Can you state that in the appropriate language?
And whats wrong with Stans comment?
Are you unable to elaborate?

On Tuesday, November 22, 2022 at 12:30:13 AM UTC-8, Richard Hachel wrote:
> Le 22/11/2022 à 01:27, Stan Fultoni a écrit :
> > On Monday, November 21, 2022 at 12:25:09 PM UTC-8, Marceline wrote:
> >> >> > You need to specify v_ave[1,2] = (x2-x1)/(t2-t1) where 1 and 2 denote two
> >> >> > specified events on the trajectory of the object.
> >> >>
> >> >> If the x2-x1 value is correct, the t2-t1 value is not...
> >> >
> >> > You contradict yourself. You already agreed that a car moving from (30,1) to
> >> > (90,2) has an average velocity of 60 km/hr during that interval. That is given
> >> by
> >> > (90-30)/(2-1) = 60. Likewise if it goes on to move at constant speed from
> >> (90,2)
> >> > to (120,3) then it's average velocity between (30,1) and (120,3) is 45 km/hr.
> >> > Also, it's average veliocity from (90,2) to (120,3) is 30 km/hr. You assured
> >> me
> >> > that you have no problem with this. Were you lying?
> >>
> >> I have no lack of understanding between the different types of speeds.
> >
> > Great. So you agree that in terms of the standard inertial coordinates x,t
> > (readings on a grid of standard rulers and clocks at rest and inertially
> > synchronized in a given frame) the average velocity of a trajectory x(t) between
> > t1 and t2 is [x(t2)-x(t1)]/(t2-t1). Excellent. You've made real progress. Do
> > you now understand why this shows that all your beliefs are absurd?
> It's not my beliefs that are absurd, it's yours.
>
> I haven't stopped telling you, for a few weeks now, that when it comes to
> special relativity, we can't afford everything.
>
> One cannot, for example, add speeds to speeds without taking precautions.

You can if they are measured in the same frame. Relativity does not suspend the
laws of arithmetic.

--
Jan

Le 22/11/2022 à 20:25, JanPB a écrit :

>> One cannot, for example, add speeds to speeds without taking precautions.
>
> You can if they are measured in the same frame.

> Relativity does not suspend the
> laws of arithmetic.
>
> --
> Jan

Exactly, no.

This is further proof that I have been right for a long time already.

It's been a very long time since I pointed out, as the first physicists
did around 1905, that something was not clear in this theory, and that if
there was perfect covariance, there was an inextricable paradox that we
called "Langevin's paradox".

Do not take this opposition of physicists against Poincaré and Einstein
at the time as something stupid.

These physicists were right, as on the other side, Poincaré was right.

Today, the one who is most right is me. Because I'm the only one in the
world to explain all the paradoxes and why everyone is right.

Yes, there is always a huge paradox called Langevin's paradox, and yes, I
was the first to really lift this paradox (or I missed an episode).

The paradox consists in the non-symmetry, in the non-covariance of the
apparent velocities, which is absurd.

I was the first to say that the covariance was entirely maintained, and
that the defect of the relativists came from a total misunderstanding of
the Lorentz transformations which are exact, entirely exact, perfectly
exact, but that they had to be "listened to" and "understand" them to the
end.

This induces not a dilation of times, and a contraction of lengths, BUT
well understood, an ELASTICITY of times, lengths and distances.

It's not the same thing.

And there to render everything in order with a symmetry and a covariance
of great beauty and great logic.

But that's not all. With the same weapons, I tackled accelerated
repositories.

And there, there are even more problems and misunderstandings, but always
linked to the same phenomenon that I denounce: the ignorance of spatial
anisochrony, which is the key, the base, of the whole theory of
arelativity, and of which the relativity of chronotropy is only a
secondary epiphenomenon.

I now answer your question:
NOPE! Contrary to what may seem like mathematical and physical evidence to
you, we are not in the same frame of reference, only in the same frame.
And it's not the same thing.
There is indeed a benchmark in which the rocket evolves from 0 to 3 ly,
then from 3 to 3.1 ly, as Stan Fultoni pointed out in one of his examples.
BUT, and this again I beg you to understand, IT IS NOT the same frame of
reference.

There is a benchmark that calculates the time taken to go from 0 to 3ly.

But to calculate the time taken to go from 0 to 3.1 ly, we are dealing
with something else, another frame of reference, another watch.

And if I want to know what happens if I want to measure real (Vr) or
observable (Vo) velocities in this small segment, I can't use observable
velocities, nor observable times, I CANNOT ask t2-t1 like Stan does.

If I do, I get vastly overrated speeds.

Just as if I take the equations of relativists to calculate the proper
times of accelerated particles and accelerated rockets, I will find proper
times that are shorter than I should.

What is sad is that the equations I give are more beautiful, simpler,
truer and more direct.

But people prefer not to believe me.

These very simple equations for accelerated repositories are:

Tr=sqrt(2x/a)
To=(x/c).sqrt(1+2c²/ax)

Vr=sqrt(2ax)
Vo=[1+c²/2ax]^(-1/2)

Everything enters then, for the "Tau Ceti" as in my "Langevin" in a
description of a perfect logic, and without paradox.

Thank you for your attention.

I have no doubt that the future will prove me right.

Doctor Richard Hachel.

On Tuesday, November 22, 2022 at 2:21:28 PM UTC-8, Richard Hachel wrote:
> Le 22/11/2022 à 20:25, JanPB a écrit :
>
> >> One cannot, for example, add speeds to speeds without taking precautions.
> >
> > You can if they are measured in the same frame.
>
> > Relativity does not suspend the
> > laws of arithmetic.
> >
> > --
> > Jan
> Exactly, no.
>
> This is further proof that I have been right for a long time already.
>
> It's been a very long time since I pointed out, as the first physicists
> did around 1905, that something was not clear in this theory, and that if
> there was perfect covariance, there was an inextricable paradox that we
> called "Langevin's paradox".

You simply have problem grasping the theory. There is nothing in
*the theory* that's not clear. Of course there are questions surrounding the
theory which deserve attention, e.g. *why* the universe appears locally
Lorentz invariant? This question ultimately rests on "what is space"
question, etc. For now this is dealt with in philosophy but not in
physics. This may change, we don't know yet.

> Do not take this opposition of physicists against Poincaré and Einstein
> at the time as something stupid.
>
> These physicists were right, as on the other side, Poincaré was right.
>
> Today, the one who is most right is me.

No. You are labouring under an illusion.

> Because I'm the only one in the
> world to explain all the paradoxes and why everyone is right.

Nonsense. The sooner you stop fantasising, the better off you'll be.
This present dreamland of youyrs is juyst a towal waste of your time
and life.

> Yes, there is always a huge paradox called Langevin's paradox, and yes, I
> was the first to really lift this paradox (or I missed an episode).

There is a paradox but it's not anything beyond that (look up the actual
meaning of the word).

> The paradox consists in the non-symmetry, in the non-covariance of the
> apparent velocities, which is absurd.

No, there is no such paradox. You misunderstood it.

Just stop fantasising, it leads you nowhere.

--
Jan

Le 23/11/2022 à 00:37, JanPB a écrit :

> No, there is no such paradox. You misunderstood it.

No paradox?

Well, as always, you're kidding.

Everyone is joking.

Everyone laughs holding their stomachs.

But the problem is that you have no idea of ​​your moral limits.

The moral limit would be to accept losing, and it is absolutely impossible
for minds that have been indoctrinated for decades.

The problem is right there.

So what will happen?

You will constantly tell me that I am wrong (in really believing it),
except that at a certain moment contradictions and evidence will come to
light, and that you will not be able to face them morally.

Scientifically yes, you could, but you don't know the power of arrogance
and dogma, and where it can lead when you don't want to see clearly.

Want us to try it out?

You will see that not only what I say scientifically is perfectly true,
but that everything I say about the power of dogmas is also true.

Always always, rather than being wrong, the subject will withdraw into
himself, and save himself by insulting.

I know all this perfectly now.

Want to give it a try?

R.H.

> Jan

On Tuesday, November 22, 2022 at 2:21:28 PM UTC-8, Richard Hachel wrote:
> And if I want to know what happens if I want to measure real (Vr) or
> observable (Vo) velocities in this small segment, I can't use observable
> velocities, nor observable times, I CANNOT ask t2-t1 like Stan does.

Place two identically constructed clocks beside the road, 1 km apart, and synchronize them by shooting identical bullets from identical guns at rest at the mid point between them, setting each clocks to 12:00 noon when the respective bullet arrives. Now, a bicycle moves along the road, and it passes one clock when it reads 1:00pm, and it passes the next clock when it reads 1:06 pm. What is the bicycle's average speed between those clocks?

> Everything enters then, for the "Tau Ceti" as in my "Langevin" in a
> description of a perfect logic, and without paradox.

You know that is not true, because you can't even answer these two simple questions, described in terms of standard inertial coordinates:

- Example 1: One twin is stationary at the origin for 2 hours, and the other twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1 hour. Do the twins have the same elapsed proper time?

- Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do the twins have the same elapsed proper time?

> These very simple equations for accelerated repositories are:
> To = x.sqrt(1 + 2/ax)
> Vo = [1 + 1/2ax]^(-1/2)

You contradict yourself, because you claimed that Vo=x/To, but then your first equation implies Vo = [1 + 2/(ax)]^(-1/2), whereas your second equation is Vo = [1 + 1/(2ax)]^(-1/2), which implies 1/2 = 2. Of course, you failed to inser parentheses so your expressions are ambiguous (not to mention senseless), but this just illustrtates one absurdity compatible with what you typed. Agreed?

On Tuesday, November 22, 2022 at 3:49:36 PM UTC-8, Richard Hachel wrote:
> Le 23/11/2022 à 00:37, JanPB a écrit :
>
> > No, there is no such paradox. You misunderstood it.
> No paradox?

There is a paradox which is a true statement which only appears false.

> Well, as always, you're kidding.
>
> Everyone is joking.
>
> Everyone laughs holding their stomachs.
>
> But the problem is that you have no idea of ​​your moral limits.

Whatever. Get real.

--
Jan

Le 23/11/2022 à 00:37, JanPB a écrit :

> No, there is no such paradox.
If you say so...

> --
> Jan

R.H.

Le 23/11/2022 à 01:03, Stan Fultoni a écrit :
> On Tuesday, November 22, 2022 at 2:21:28 PM UTC-8, Richard Hachel wrote:
> > And if I want to know what happens if I want to measure real (Vr) or
>> observable (Vo) velocities in this small segment, I can't use observable
>> velocities, nor observable times, I CANNOT ask t2-t1 like Stan does.
>
> Place two identically constructed clocks beside the road, 1 km apart, and
> synchronize them by shooting identical bullets from identical guns at rest at the
> mid point between them, setting each clocks to 12:00 noon when the respective
> bullet arrives. Now, a bicycle moves along the road, and it passes one clock when
> it reads 1:00pm, and it passes the next clock when it reads 1:06 pm. What is the
> bicycle's average speed between those clocks?

Dear Stan.

The problem is that it seems that you don't understand at all what I'm
trying to tell you.

Two possible causes: where I express myself very badly, or you do not have
the intellectual capacity to understand me.

I tell you, I tell you again, and I tell you again (and to anyone who
reads and is happy to remain in crass ignorance and arrogance) that
physicists are wrong when they BELIEVE that we can synchronize "absolutely
"two watches placed in two different places.
You can make an adjustment, yes. But never a synchronization.
If you stand at O, in the center of AB, you will synchronize the watches
on O.
But neither on A nor on B.

Two events which will occur in A and B, can be considered as simultaneous
FOR O.

But in this case, invariably, they will not be simultaneous for A nor for
B.

A will assert that event A preceded B by AB/c.

And the reverse for B.

There is no possible simultaneity.

Let's say, I'm going to set A to B, so that for A, tA=tB. We can do it,
but B will look at things with astonishment, and assert that henceforth A
really lags behind by 2AB/c.

This is called spatial anisochrony.

To affirm that there is a dyschronotropy by change of inertial frame of
reference, and that the watches "elsewhere" reciprocally turn less
quickly, is good.

But this is not enough.

This oversight is then a source of terrible errors and total
misunderstandings, as when you subtract carrots from turnips, to find an
erroneous observable speed.

I beg you to understand this notion of anisochrony without which I advise
you to abandon all search for physical truth.

R.H.

Le 23/11/2022 à 01:03, Stan Fultoni a écrit :
> You know that is not true, because you can't even answer these two simple
> questions, described in terms of standard inertial coordinates:
>
> - Example 1: One twin is stationary at the origin for 2 hours, and the other
> twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph
> for 1 hour. Do the twins have the same elapsed proper time?
>
> - Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5
> mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do
> the twins have the same elapsed proper time?
>
>> These very simple equations for accelerated repositories are:
>> To = x.sqrt(1 + 2/ax)
>> Vo = [1 + 1/2ax]^(-1/2)
>
> You contradict yourself, because you claimed that Vo=x/To, but then your first
> equation implies Vo = [1 + 2/(ax)]^(-1/2), whereas your second equation is Vo = [1
> + 1/(2ax)]^(-1/2), which implies 1/2 = 2. Of course, you failed to inser
> parentheses so your expressions are ambiguous (not to mention senseless), but this
> just illustrtates one absurdity compatible with what you typed. Agreed?

In your Tau Ceti example, when you take the small segment between 3 and
3.1 ly,
you calculate dTo=0.102953 years.

This calculation is not good, the real calculation is dTo=0.107515 years.

So you find an average observable speed of 0.7913c.

It's too high.

The average speed in this segment is 0.9301c

You do not want to admit that the watch which will measure To1=2.388 is
not the same as the one which will measure To2=2.427

And that we cannot logically do T2-T1 without falling into a major
blunder.

Why do you think we can do it? Because you have noticed that we are in the
same frame and that the chronotropy in A is the same as in B.

But you forget the anisotropy.

Always, always, always relativists make the same mistake.

It is this colossal error that has rotted the theory and caused everyone
to be insulted and torn apart for more than a century.

R.H.