x=12
a=10m/s²
1 year=365.25 days
Tr=4.776 years
To= 12.915 years

We are going to place four observers on the path who will detect the
proper time of the rocket when it passes close to them.

x_1 = 0.526ly
x_2 = 2.104ly
x_3 = 4.734ly
x_4 = 8.416ly

x = 12ly

we will then have Tr_1 = 1 year
Tr_2 = 2 years
Tr_3 = 3 years
Tr_4 = 4 years
Tr = 4.776 years

To find To (the earth's improper time), you must very simply use the
equation To²=Tr²+€t²

Let To=sqrt[Tr²+(x/c)²]

This is not only true, but remarkably simple.

Attention! I use these formulas in this specific case, it is valid, but it
does not work if we take other coordinates, and we can fall into the
relativistic trap that I denounced in several posts yesterday.

R.H.

On Thursday, August 4, 2022 at 4:42:30 AM UTC-7, Richard Hachel wrote:
> x=12 , a=10m/s²
> We are going to place four observers on the path who will detect the
> proper time of the rocket when it passes close to them. We will then have
> Tr_1 = 1 years .... x_1 = 0.526 ly
> Tr_2 = 2 years .... x_2 = 2.104 ly
> Tr_3 = 3 years .... x_3 = 4.734 ly
> Tr_4 = 4 years .... x_4 = 8.416 ly
> Tr = 4.776 years .... x = 12 ly

If a rocket begins at rest at the origin of inertial coordinate system x,t and undergoes constant proper acceleration a = 10 m/s^2, then the rocket reaches x = 12 LY at the rocket's proper time tau = 3.139 years and the coordinate time t = 12.915, and at that event the magnitude of the timelike interval from the origin is 7.682 years. The latter represents the proper time along an inertial path from the origin to this event.

Le 05/08/2022 à 00:59, Stan Fultoni a écrit :
> On Thursday, August 4, 2022 at 4:42:30 AM UTC-7, Richard Hachel wrote:
>> x=12 , a=10m/s²
>> We are going to place four observers on the path who will detect the
>> proper time of the rocket when it passes close to them. We will then have
>> Tr_1 = 1 years .... x_1 = 0.526 ly
>> Tr_2 = 2 years .... x_2 = 2.104 ly
>> Tr_3 = 3 years .... x_3 = 4.734 ly
>> Tr_4 = 4 years .... x_4 = 8.416 ly
>> Tr = 4.776 years .... x = 12 ly
>
> If a rocket begins at rest at the origin of inertial coordinate system x,t and
> undergoes constant proper acceleration a = 10 m/s^2, then the rocket reaches x =
> 12 LY at the rocket's proper time tau = 3.139 years and the coordinate time t =
> 12.915, and at that event the magnitude of the timelike interval from the origin
> is 7.682 years. The latter represents the proper time along an inertial path from
> the origin to this event.

Thank you for your intervention which says exactly the same thing as me
from a general point of view, that is to say that relativists calculate
very well the observable (terrestrial) time To.

Let To=12.9156years.

But I repeat it tirelessly, there is an error in the calculation of the
proper time of the rocket.

Tr=tau=4.7765 years

The relativistic geometry used to calculate the proper times is not
correct.

For observable times (here terrestrial), yes, it is perfectly correct.

R.H.

On Thursday, August 4, 2022 at 4:28:10 PM UTC-7, Richard Hachel wrote:
> >> x=12 , a=10m/s²
> >> We are going to place four observers on the path who will detect the
> >> proper time of the rocket when it passes close to them. We will then have
> >> Tr_1 = 1 years .... x_1 = 0.526 ly
> >> Tr_2 = 2 years .... x_2 = 2.104 ly
> >> Tr_3 = 3 years .... x_3 = 4.734 ly
> >> Tr_4 = 4 years .... x_4 = 8.416 ly
> >> Tr = 4.776 years .... x = 12 ly
> >
> > If a rocket begins at rest at the origin of inertial coordinate system x,t and
> > undergoes constant proper acceleration a = 10 m/s^2, then the rocket reaches x =
> > 12 LY at the rocket's proper time tau = 3.139 years and the coordinate time t =
> > 12.915, and at that event the magnitude of the timelike interval from the origin
> > is 7.682 years. The latter represents the proper time along an inertial path from
> > the origin to this event.
>
> [scientists] calculate very well the observable (terrestrial) time...

Yes, and scientists also calculate very well the spatial and temporal coordinates, and the proper times along every path. For the rocket proper times tau=0, 1, 2, 3, and 3.139 years the rocket is at locations x=0, 0.576, 3.007, 10.242, and 12 lightyears, at the coordinate times t=0, 1.195, 3.841, 11.151, and 12.915 years, respectively, and the magnitudes of the intervals from the origin to those events are tauI = 0, 1.047, 2.390, 4.411, and 7.602 years. So, every one of your calculated values were wrong.

> ... there is an error in the calculation of the proper time of the rocket..

There is an error in your calculation of all the quantities, including the proper time of the rocket, and this is why your beliefs are self-contraidctory, because they imply 1=0. When this was proven, you ran away (again). Remember?

> [the proper time of the rocket is] 4.7765 years

Nope, it is 3.139 years, as noted above. This is elementary Relativity 101..

On Thursday, August 4, 2022 at 9:42:30 PM UTC+10, Richard Hachel wrote:
> x=12
> a=10m/s²
> 1 year=365.25 days
> Tr=4.776 years
> To= 12.915 years
>
> We are going to place four observers on the path who will detect the
> proper time of the rocket when it passes close to them.
>
> x_1 = 0.526ly
> x_2 = 2.104ly
> x_3 = 4.734ly
> x_4 = 8.416ly
>
> x = 12ly
>
> we will then have Tr_1 = 1 year
> Tr_2 = 2 years
> Tr_3 = 3 years
> Tr_4 = 4 years
> Tr = 4.776 years
>
> To find To (the earth's improper time), you must very simply use the
> equation To²=Tr²+€t²
>
> Let To=sqrt[Tr²+(x/c)²]
>
> This is not only true, but remarkably simple.
>
> Attention! I use these formulas in this specific case, it is valid, but it
> does not work if we take other coordinates, and we can fall into the
> relativistic trap that I denounced in several posts yesterday.
>
>
> R.H.

Great, so as our traveller accelerates towards Tau Ceti at 10m/s² , each year (by his clock) he passes 4 stationary (relative to the Earth) observers - I'm still going to call them Alice, Bob, Charles, Dave , because if we call both the observers and the equations "1", "2", "3" and "4", that'll lead to confusion !

You say he passes Observer 1 Alice at x_A = 0.526ly and David at x_D = 8.416ly - so, you're using your own equations instead of the standard equations found in text books. OK, we'll worry about that later, but first let's bring in your other equations to see how they apply :

From your opening post in your "Special Relativity" thread :
On Wednesday, August 3, 2022 at 9:13:13 AM UTC+10, Richard Hachel wrote:
>
> (tau2 - tau1) + (tau3 - tau2) = (tau3 - tau1)
> Yes
> tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> Yes, all rigth.
> tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
> Yes OK absolutly.
> tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
> Your mistake is here.

So, for any three events "1", "2", "3" along the traveller's path :

you AGREE with EQUATION 1 : tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
(you said "Yes, all rigth. " ..... although you do also say : "There is a conceptual error here: sqrt[(t2-t1)² - (x2-x1)²/c²]" )

you AGREE with EQUATION 2 : tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]

You DISAGREE with EQUATION 3 : tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]

Right ?

Le 06/08/2022 à 14:50, Ufonaut a écrit :
> On Thursday, August 4, 2022 at 9:42:30 PM UTC+10, Richard Hachel wrote:
>> x=12
>> a=10m/s²
>> 1 year=365.25 days
>> Tr=4.776 years
>> To= 12.915 years
>>
>> We are going to place four observers on the path who will detect the
>> proper time of the rocket when it passes close to them.
>>
>> x_1 = 0.526ly
>> x_2 = 2.104ly
>> x_3 = 4.734ly
>> x_4 = 8.416ly
>>
>> x = 12ly
>>
>> we will then have Tr_1 = 1 year
>> Tr_2 = 2 years
>> Tr_3 = 3 years
>> Tr_4 = 4 years
>> Tr = 4.776 years
>>
>> To find To (the earth's improper time), you must very simply use the
>> equation To²=Tr²+€t²
>>
>> Let To=sqrt[Tr²+(x/c)²]
>>
>> This is not only true, but remarkably simple.
>>
>> Attention! I use these formulas in this specific case, it is valid, but it
>> does not work if we take other coordinates, and we can fall into the
>> relativistic trap that I denounced in several posts yesterday.
>>
>>
>> R.H.
>
> Great, so as our traveller accelerates towards Tau Ceti at 10m/s² , each year
> (by his clock) he passes 4 stationary (relative to the Earth) observers - I'm
> still going to call them Alice, Bob, Charles, Dave , because if we call both the
> observers and the equations "1", "2", "3" and "4", that'll lead to confusion !

Absolutly.

x_1 = 0.526ly (Alice)
x_2 = 2.104ly (Bob)
x_3 = 4.734ly (Charles)
x_4 = 8.416ly (Dave)

> You say he passes Observer 1 Alice at x_A = 0.526ly and David at x_D = 8.416ly -
> so, you're using your own equations instead of the standard equations found in
> text books. OK, we'll worry about that later, but first let's bring in your other
> equations to see how they apply :
>
> From your opening post in your "Special Relativity" thread :
> On Wednesday, August 3, 2022 at 9:13:13 AM UTC+10, Richard Hachel wrote:
>>
>> (tau2 - tau1) + (tau3 - tau2) = (tau3 - tau1)
>> Yes
>> tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
>> Yes, all rigth.
>> tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
>> Yes OK absolutly.
>> tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
>> Your mistake is here.
>
> So, for any three events "1", "2", "3" along the traveller's path :
>
> you AGREE with EQUATION 1 : tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]

Absolutly.

> you AGREE with EQUATION 2 : tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]

Absolutly.

>
> You DISAGREE with EQUATION 3 : tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]

Yes, I desagree. It's a concept error.

> Right ?

Absolutly.

You translate my thoughts quite correctly.

Absolutly.

R.H.

On Saturday, 6 August 2022 at 14:50:18 UTC+2, Ufonaut wrote:
> On Thursday, August 4, 2022 at 9:42:30 PM UTC+10, Richard Hachel wrote:
> > x=12
> > a=10m/s²
> > 1 year=365.25 days
> > Tr=4.776 years
> > To= 12.915 years
> >
> > We are going to place four observers on the path who will detect the
> > proper time of the rocket when it passes close to them.
> >
> > x_1 = 0.526ly
> > x_2 = 2.104ly
> > x_3 = 4.734ly
> > x_4 = 8.416ly
> >
> > x = 12ly
> >
> > we will then have Tr_1 = 1 year
> > Tr_2 = 2 years
> > Tr_3 = 3 years
> > Tr_4 = 4 years
> > Tr = 4.776 years
> >
> > To find To (the earth's improper time), you must very simply use the
> > equation To²=Tr²+€t²
> >
> > Let To=sqrt[Tr²+(x/c)²]
> >
> > This is not only true, but remarkably simple.
> >
> > Attention! I use these formulas in this specific case, it is valid, but it
> > does not work if we take other coordinates, and we can fall into the
> > relativistic trap that I denounced in several posts yesterday.
> >
> >
> > R.H.
> Great, so as our traveller accelerates towards Tau Ceti at 10m/s² , each year (by his clock) he passes 4 stationary (relative to the Earth) observers - I'm still going to call them Alice, Bob, Charles, Dave , because if we call both the observers and the equations "1", "2", "3" and "4", that'll lead to confusion !

Remember, however, that this gedanken bullshit
has nothing in common with real clocks, real
observers, real measurements.

On Saturday, August 6, 2022 at 5:58:52 AM UTC-7, Richard Hachel wrote:
> >> x=12 , a=10m/s² We are going to place four observers on the path who will
> >> detect the proper time of the rocket when it passes close to them. We will
> >> then have
> >> Tr_1 = 1 years .... x_1 = 0.526 ly
> >> Tr_2 = 2 years .... x_2 = 2.104 ly
> >> Tr_3 = 3 years .... x_3 = 4.734 ly
> >> Tr_4 = 4 years .... x_4 = 8.416 ly
> >> Tr = 4.776 years .... x = 12 ly
> >
> > If a rocket begins at rest at the origin of inertial coordinate system x,t and
> > undergoes constant proper acceleration a = 10 m/s^2, then the rocket reaches x > > 12 LY at the rocket's proper time tau = 3.139 years and the coordinate time t > > 12.915, and at that event the magnitude of the timelike interval from the origin
> > is 7.682 years. The latter represents the proper time along an inertial path from
> > the origin to this event.
>
> [scientists] calculate very well the observable (terrestrial) time...

Yes, and scientists also calculate very well the spatial and temporal coordinates, and the proper times along every path. For the rocket proper times tau = 0, 1, 2, 3, and 3.139 years the rocket is at locations x = 0, 0.576, 3.007, 10.242, and 12 lightyears, at the coordinate times t = 0, 1.195, 3.841, 11.151, and 12.915 years, respectively, and the magnitudes of the intervals from the origin to those events are tauI = 0, 1.047, 2.390, 4.411, and 7.602 years. So, every one of your calculated values were wrong.

> [the proper time of the rocket is] 4.7765 years

No, it is 3.139 years, as noted above. Remember, we've shown that your beliefs are self-contradictory. Each time this is explained, you just run away.

Le 06/08/2022 à 17:37, Stan Fultoni a écrit :
> On Saturday, August 6, 2022 at 5:58:52 AM UTC-7, Richard Hachel wrote:
>> >> x=12 , a=10m/s² We are going to place four observers on the path who will
>> >> detect the proper time of the rocket when it passes close to them. We will
>> >> then have
>> >> Tr_1 = 1 years .... x_1 = 0.526 ly
>> >> Tr_2 = 2 years .... x_2 = 2.104 ly
>> >> Tr_3 = 3 years .... x_3 = 4.734 ly
>> >> Tr_4 = 4 years .... x_4 = 8.416 ly
>> >> Tr = 4.776 years .... x = 12 ly
>> >
>> > If a rocket begins at rest at the origin of inertial coordinate system x,t and
>> > undergoes constant proper acceleration a = 10 m/s^2, then the rocket reaches x
>> =
>> > 12 LY at the rocket's proper time tau = 3.139 years and the coordinate time t
>> =
>> > 12.915, and at that event the magnitude of the timelike interval from the
>> origin
>> > is 7.682 years. The latter represents the proper time along an inertial path
>> from
>> > the origin to this event.
>>
>> [scientists] calculate very well the observable (terrestrial) time...
>
> Yes, and scientists also calculate very well the spatial and temporal
> coordinates, and the proper times along every path. For the rocket proper times
> tau = 0, 1, 2, 3, and 3.139 years the rocket is at locations x = 0, 0.576, 3.007,
> 10.242, and 12 lightyears, at the coordinate times t = 0, 1.195, 3.841, 11.151,
> and 12.915 years, respectively, and the magnitudes of the intervals from the
> origin to those events are tauI = 0, 1.047, 2.390, 4.411, and 7.602 years. So,
> every one of your calculated values were wrong.
>
>> [the proper time of the rocket is] 4.7765 years
>
> No, it is 3.139 years,

No.

YOU, you say it is 3.139.

It is 4.7765 years.

<http://news2.nemoweb.net/jntp?OKc_vlnD9vVzcVHT-Gz1_rLvj4w@jntp/Data.Media:1>

R.H.

On Saturday, August 6, 2022 at 9:49:35 AM UTC-7, Richard Hachel wrote:
> >> >> x=12 , a=10m/s² We are going to place four observers on the path who will
> >> >> detect the proper time of the rocket when it passes close to them. We will
> >> >> then have
> >> >> Tr_1 = 1 years .... x_1 = 0.526 ly
> >> >> Tr_2 = 2 years .... x_2 = 2.104 ly
> >> >> Tr_3 = 3 years .... x_3 = 4.734 ly
> >> >> Tr_4 = 4 years .... x_4 = 8.416 ly
> >> >> Tr = 4.776 years .... x = 12 ly
> >> >
> >> > If a rocket begins at rest at the origin of inertial coordinate system x,t and
> >> > undergoes constant proper acceleration a = 10 m/s^2, then the rocket reaches
> >> > x = 12 LY at the rocket's proper time tau = 3.139 years and the coordinate time
> >> > t = 12.915, and at that event the magnitude of the timelike interval from the
> >> > origin is 7.682 years. The latter represents the proper time along an inertial path
> >> > from the origin to this event.
> >>
> >> [scientists] calculate very well the observable (terrestrial) time...
> >
> > Yes, and scientists also calculate very well the spatial and temporal
> > coordinates, and the proper times along every path. For the rocket proper times
> > tau = 0, 1, 2, 3, and 3.139 years the rocket is at locations x = 0, 0.576, 3.007,
> > 10.242, and 12 lightyears, at the coordinate times t = 0, 1.195, 3.841, 11.151,
> > and 12.915 years, respectively, and the magnitudes of the intervals from the
> > origin to those events are tauI = 0, 1.047, 2.390, 4.411, and 7.602 years. So,
> > every one of your calculated values were wrong.
> >
> >> [No, the proper time of the rocket is] 4.7765 years.
No, the correct and clear calculation of the proper time of the rocket, 3.139 years, has been provided to you. Your calculation method yielding 4.7765 years is self-contradictory gibberish, implying 1=0. We covered this before. Again, the elapsed proper time along the path of the rocket is the integral of sqrt[(dt)^2 - (dx/c)^2] along that path, and this gives 3.139 years.

Le 06/08/2022 à 19:31, Stan Fultoni a écrit :
> On Saturday, August 6, 2022 at 9:49:35 AM UTC-7, Richard Hachel wrote:
>> >> >> x=12 , a=10m/s² We are going to place four observers on the path who will
>>
>> >> >> detect the proper time of the rocket when it passes close to them. We will
>>
>> >> >> then have
>> >> >> Tr_1 = 1 years .... x_1 = 0.526 ly
>> >> >> Tr_2 = 2 years .... x_2 = 2.104 ly
>> >> >> Tr_3 = 3 years .... x_3 = 4.734 ly
>> >> >> Tr_4 = 4 years .... x_4 = 8.416 ly
>> >> >> Tr = 4.776 years .... x = 12 ly
>> >> >
>> >> > If a rocket begins at rest at the origin of inertial coordinate system x,t
>> and
>> >> > undergoes constant proper acceleration a = 10 m/s^2, then the rocket
>> reaches
>> >> > x = 12 LY at the rocket's proper time tau = 3.139 years and the coordinate
>> time
>> >> > t = 12.915, and at that event the magnitude of the timelike interval from
>> the
>> >> > origin is 7.682 years. The latter represents the proper time along an
>> inertial path
>> >> > from the origin to this event.
>> >>
>> >> [scientists] calculate very well the observable (terrestrial) time...
>> >
>> > Yes, and scientists also calculate very well the spatial and temporal
>> > coordinates, and the proper times along every path. For the rocket proper
>> times
>> > tau = 0, 1, 2, 3, and 3.139 years the rocket is at locations x = 0, 0.576,
>> 3.007,
>> > 10.242, and 12 lightyears, at the coordinate times t = 0, 1.195, 3.841,
>> 11.151,
>> > and 12.915 years, respectively, and the magnitudes of the intervals from the
>> > origin to those events are tauI = 0, 1.047, 2.390, 4.411, and 7.602 years. So,
>>
>> > every one of your calculated values were wrong.
>> >
>> >> [No, the proper time of the rocket is] 4.7765 years.
>
> No, the correct and clear calculation of the proper time of the rocket, 3.139
> years, has been provided to you. Your calculation method yielding 4.7765 years is
> self-contradictory gibberish, implying 1=0. We covered this before. Again, the
> elapsed proper time along the path of the rocket is the integral of sqrt[(dt)^2 -
> (dx/c)^2] along that path, and this gives 3.139 years.

No.

You have to take the integral of the correct equation.

And that's not the equation to use when it comes to uniformly accelerated
repositories.

R.H.

On Saturday, August 6, 2022 at 10:43:24 AM UTC-7, Richard Hachel wrote:
> > The elapsed proper time along the path of the rocket is the integral of
> > sqrt[(dt)^2 - (dx/c)^2] along that path, and this gives 3.139 years.
>
> And that's not the equation to use when it comes to uniformly accelerated
> repositories.

The rocket is stipulated to be undergoing constant proper acceleration, so it follows a hyperbolic trajectory in terms of any standard system of inertial coordinates x,t, and the elapsed proper time along the path is the integral of sqrt[(dt)^2 - (dx/c)^2] along that path, which gives 3.139 years.

Remember, your proposed alternative is self-contradictory, implying 1=0, so your beliefs are obviously wrong. The correct result is as stated above..

Le 06/08/2022 à 20:20, Stan Fultoni a écrit :
> The rocket is stipulated to be undergoing constant proper acceleration, so it
> follows a hyperbolic trajectory in terms of any standard system of inertial
> coordinates x,t, and the elapsed proper time along the path is the integral of
> sqrt[(dt)^2 - (dx/c)^2] along that path, which gives 3.139 years.
>
> Remember, your proposed alternative is self-contradictory, implying 1=0, so your
> beliefs are obviously wrong. The correct result is as stated above.

If you want to use an integral (I don't need one, but hey), it's better to
use the correct one.

To=(Tr/N).∫sqrt(1+0.25Vr²n²/c²N²)

Je dis ça, je dis rien.

A la limite, je m'en fous.

R.H.

Le 06/08/2022 à 20:20, Stan Fultoni a écrit :

> implying 1=0

In science as elsewhere, things can only be based on facts.

R.H.

On Saturday, August 6, 2022 at 11:36:30 AM UTC-7, Richard Hachel wrote:
> The rocket undergoes constant proper acceleration, so it follows a hyperbolic
> trajectory in terms of any standard system of inertial coordinates x,t, and the
> elapsed proper time along the path is the integral of sqrt[(dt)^2 - (dx/c)^2] along
> that path, which gives 3.139 years.
>
> Remember, your proposed alternative is self-contradictory, implying 1=0, so your
> beliefs are obviously wrong. The correct result is as stated above.
>
> In science as elsewhere, things can only be based on facts.

Yes, and since your claim that 1=0 is obviously not a fact, your beliefs are false, as well as self-contradictory. You claim both "X" and " not X", so your claims are incoherent nonsense.

Le 06/08/2022 à 20:46, Stan Fultoni a écrit :

> You claim both "X" and " not X",

No, that, no.

R.H.

On Saturday, August 6, 2022 at 12:26:04 PM UTC-7, Richard Hachel wrote:
> > You claim both "X" and "not X",
>
> No.

Again, X is the proposition that the elapsed time along any path between any two events ei and ej is tauj - taui = sqrt[(tj-ti)^2 - (xj-xi)^2/c^2]. ) You claim both X and not X, so you are contradicting yourself. That's why your beliefs entail 1=0. Remember?

Le 06/08/2022 à 22:52, Stan Fultoni a écrit :
> On Saturday, August 6, 2022 at 12:26:04 PM UTC-7, Richard Hachel wrote:
>> > You claim both "X" and "not X",
>>
>> No.
>
> Again, X is the proposition that the elapsed time along any path between any two
> events ei and ej is tauj - taui = sqrt[(tj-ti)^2 - (xj-xi)^2/c^2]. ) You claim
> both X and not X, so you are contradicting yourself. That's why your beliefs
> entail 1=0. Remember?

It is more complicated than that.

Because this equation is correct for all inertial frames in uniform
motion, and for all accelerated frames with a stopped start.

But if we take the case of an accelerated frame of reference with a
non-stop start, it does not work.

As with relativistic velocity additions, you can't do what you want
without the correct equations.

One thing is important to understand:
all physicists have looked at the fact that, if we take a very short time
Tr (tau), we can estimate that the mobile is in uniform motion, and by
simply integrating all this, well we can find To ( improper time in the
terrestrial frame of reference).

It seems of infinite logic.

And yet, it is completely false, we cannot add all the small To in a
simple arrhythmic way.

The relation To=Tr.sqrt(1+Vr²/c²) being false for accelerated media in
the theory of relativity.

That is to say that setting tau=To.sqrt(1-v²/c²) is not correct, as
obvious as it may seem if we are not careful to fall into a terrible trap:
the clock which measures To is not fixed as it is in uniform reference
frames, it constantly changes places, and the real sum of all the small To
is not a simple mathematical integration, since it is the sum of an
infinity of ways of measuring To by an infinity of clocks and not a single
coherent clock.

Comprenez-vous (un peu) où je veux en venir et pourquoi je dis que la
façon dont les relativistes procèdent, sur ça, est totalement erronée?

R.H.

On Saturday, August 6, 2022 at 10:58:52 PM UTC+10, Richard Hachel wrote:
> Le 06/08/2022 à 14:50, Ufonaut a écrit :

> > So, for any three events "1", "2", "3" along the traveller's path :
> >
> > you AGREE with EQUATION 1 : tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> Absolutly.
> > you AGREE with EQUATION 2 : tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
> Absolutly.
> >
> > You DISAGREE with EQUATION 3 : tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
> Yes, I desagree. It's a concept error.
>
> > Right ?
>
> Absolutly.
>
> You translate my thoughts quite correctly.
>
> Absolutly.
>
> R.H.

Great - so we agree that at each of the 4 events A, B, C, D, we have fixed concrete values for that event's :

- Tau (being the time on the traveller's clock in years, so tau_A=1, tau_B=2, tau_C=3, and tau_D=4), AND

- x (being the location of the A, B, C or D along the x axis as measured in ly by those observers, so x_A=0.526ly , x_B=2.104ly , x_C=4.734ly and x_D=8.416ly ), AND

- t (being the time on A, B, C or D's clock in years as the traveller passes them. We haven't calculated what that value is yet, but suffice to say t_A, t_B, t_C and t_D are fixed concrete values just as tau_ and t_ are)

So we can plug those values into one of the equations - for example :
tau_C - tau_A = sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]

now that equation is must be EITHER true OR false, depending on those fixed values above. Obviously it's a non-sensical contradiction if for those fixed concrete set of values of tau_C, tau_A, t_C, t_A, x_C and x_A, we were to say that that equation is true one in one statement, and false in another statement - right ?

So let's make it a logical statement "X", so :
X is "tau_C - tau_A DOES EQUAL sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]

and therefore of course :
"NOT X" is "tau_C - tau_A DOES NOT EQUAL sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]

So all well and good :).

Now let's imagine that you've been explaining this whole scenario to two of your friends - let's call them "Harry" and "Meghan" :)

So you go on to say to them :
"So we have 3 Equations : "EQUATION 1", "EQUATION 2" and "EQUATION 3" that can be applied to any three events along the traveller's path. So I want each of you to choose 3 events from the A,B,C,D for us to apply those equations"

"OK", says Harry, "I choose events A, C and D"

"Right", you reply, "So setting event 1 is A, event 2 is C and event 3 is D, so therefore :

STATEMENT 1) AGREE with EQUATION 1 : tau_C - tau_A = sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]

STATEMENT 2) AGREE with EQUATION 2 : tau_D - tau_A = sqrt[(t_D - t_A)^2 - (x_C - x_A)^2/c^2]

STATEMENT 3) but DISAGREE with EQUATION 3 : tau_D - tau_C = sqrt[(t_D - t_C)^2 - (x_D - x_C)^2/c^2]
"

Meghan then says "Well, I am going to choose events A, B and C"

"Right", you reply, "So setting event 1 is A, event 2 is B and event 3 is C, so therefore :

STATEMENT 4) AGREE with EQUATION 1 : tau_B - tau_A = sqrt[(t_B - t_A)^2 - (x_B - x_A)^2/c^2]

STATEMENT 5) AGREE with EQUATION 2 : tau_C - tau_A = sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]

STATEMENT 6) but DISAGREE with EQUATION 3 : tau_C - tau_A = sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]
"

Agree ?

However, STATEMENT 1 is a statement that you are claiming that tau_C - tau_A DOES EQUAL sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2] - in other words, that you are claiming that logical statement X above.

Likewise, STATEMENT 6 is a statement that you are claiming that tau_C - tau_A DOES NOT EQUAL sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2] - in other words, that you are claiming "NOT X" above.

So for this scenario of the traveller, you are claiming both X and NOT X - right ?

On Sunday, August 7, 2022 at 12:19:17 AM UTC-7, Richard Hachel wrote:
> > Again, X is the proposition that the elapsed time along any path between any two
> > events ei and ej is tauj - taui = sqrt[(tj-ti)^2 - (xj-xi)^2/c^2]. ) You claim
> > both X and not X, so you are contradicting yourself.
>
> This equation is correct for all inertial frames in uniform motion...

Right, and the proof of the self-contradiction in your claim is carried out entirely in terms of a single standard system of inertial coordinates, x,t.. In that single standard system of inertial coordinates, you claim both "X" and "not X", so you are contradicting yourself.

> If we take the case of an accelerated frame of reference...

We do not use any accelerating frames of reference. We are using only a single inertial frame of reference with x,t coordinates. Also, you cannot make any distinction because we showed that you declared the same proposition with i=1, j=2 to be both true and false. These are just two events, and when you are asked if the equation gives the elapsed proper time you first answer yes, and then when this leads to 1=0, you go back and answer no.. So you are blatantly contradicting yourself.

Le 07/08/2022 à 09:36, Ufonaut a écrit :

> Great - so we agree that at each of the 4 events A, B, C, D, we have fixed
> concrete values for that event's :
>
> - Tau (being the time on the traveller's clock in years, so tau_A=1, tau_B=2,
> tau_C=3, and tau_D=4),

Yes
x_1 = 0.526ly (Alice) ---> Tr1=1year
x_2 = 2.104ly (Bob) ---> Tr2=2years
x_3 = 4.734ly (Charles) ---> Tr=3years
x_4 = 8.416ly (Dave) ---> Tr=4years

R.H.

Le 07/08/2022 à 09:36, Ufonaut a écrit :
> So we can plug those values into one of the equations - for example :
> tau_C - tau_A = sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]

No.

because, here, tau_A is different of zero.

Equation becomes false.

This is valid only from a standing start.

Here is the error.

R.H.

Le 07/08/2022 à 09:36, Stan Fultoni a écrit :
> Right, and the proof of the self-contradiction in your claim is carried out
> entirely in terms of a single standard system of inertial coordinates, x,t. In
> that single standard system of inertial coordinates, you claim both "X" and "not
> X", so you are contradicting yourself.
>
>> If we take the case of an accelerated frame of reference...
>
> We do not use any accelerating frames of reference. We are using only a single
> inertial frame of reference with x,t coordinates. Also, you cannot make any
> distinction because we showed that you declared the same proposition with i=1, j=2
> to be both true and false. These are just two events, and when you are asked if
> the equation gives the elapsed proper time you first answer yes, and then when
> this leads to 1=0, you go back and answer no. So you are blatantly contradicting
> yourself.

No, I'm not contradicting myself.

Simply things are not easy to understand and may seem counter-intuitive.

For example I can add all the small pieces of Tr (proper time tau) to have
the total proper time.

That is to say that the whole is the sum of all the parts.

Tr=Tr1+Tr2+Tr3+Tr4+Tr5

The scientist will then commit a terrible and effective blunder.

And also set (why not) To=To1+To2+To3+To4+To5

It seemed absolutely obvious.

And yet, therein lies the catch.

Yeah, that's wrong.

We think we add times measured by the same clock, and we add carrots and
turnips.

We must therefore do as the good Doctor Hachel suggests.

The traditional relativistic equation relating to proper times is
shattered.

R.H.

On Sunday, August 7, 2022 at 5:48:23 PM UTC+10, Richard Hachel wrote:
> Le 07/08/2022 à 09:36, Ufonaut a écrit :
> > So we can plug those values into one of the equations - for example :
> > tau_C - tau_A = sqrt[(t_C - t_A)^2 - (x_C - x_A)^2/c^2]
> No.
>
> because, here, tau_A is different of zero.
>
> Equation becomes false.
>
> This is valid only from a standing start.
>
> Here is the error.
>
> R.H.

"This is valid only from a standing start. " ????????

This is the first time you have said that Event 1 MUST be the standing start.

On the contrary, I **EXPLICITLY** confirmed with you :

On Saturday, August 6, 2022 at 10:58:52 PM UTC+10, Richard Hachel wrote:
> Le 06/08/2022 à 14:50, Ufonaut a écrit :

> > So, for any three events "1", "2", "3" along the traveller's path :
> >
> > you AGREE with EQUATION 1 : tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> Absolutly.

ANY three events. "ANY"

To which you replied "Absolutly". Not a hint of "Oh, only if event 1 is the standing start, so all its associated values are zero".

You started the original thread with the post :

> (tau2 - tau1) + (tau3 - tau2) = (tau3 - tau1)
> Yes
> tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2]
> Yes, all rigth.
> tau3-tau1 = sqrt[(t3-t1)^2 - (x3-x1)^2/c^2]
> Yes OK absolutly.
> tau3-tau2 = sqrt[(t3-t2)^2 - (x3-x2)^2/c^2]
> Your mistake is here.

Again, not a hint of "only valid from a starting start, so I'm pre-assuming that tau1 =0, x1=0, t1=0". No, just an obscure "Your mistake is here" on the only equation that does not happen to mention that Event 1, which is your pre-assumed assumption that Event 1 is the standing start.

So those equations above assume tau1 =0, x1=0, t1=0 do they ? So you just introduced zero three times, giving it three different names to look like they meant something ?
When you said "tau2-tau1 = sqrt[(t2-t1)^2 - (x2-x1)^2/c^2] ", all that you really meant was "tau2 = sqrt[t2^2 - x2^2/c^2]" ????

Is that right ????

Le 07/08/2022 à 11:29, Ufonaut a écrit :

> "This is valid only from a standing start. " ? ? ? ? ? ? ? ?

Yes.

<http://news2.nemoweb.net/jntp?r-Ko8QAUC5Fnb9MV7eVMb7n5DdQ@jntp/Data.Media:1>

R.H.