On Sunday, August 7, 2022 at 1:00:30 AM UTC-7, Richard Hachel wrote:
> > 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.
>
> I can add all the small pieces of Tr (proper time tau) to have the
> total proper time.

Right, but on the basis of your claims, this leads to the logically impossible result 1=0, which proves that your claims are logically inconsistent. Remember?

> The scientist will then commit a terrible and effective blunder.
> And also [sum the incremental coordinate times to give the total
> coordinate time].

You are mistaken, and that is another of your logically inconsistent claims. If three given events have the coordinate times t1, t2 and t3 such that t1 < t2 < t3, then (t2-t1) + (t3-t2) = (t3-t1). This is an algebraic tautology, remember?

> Yeah, that's wrong.

Your claim is logically self-contradictory, since the algebraic tautology is not wrong. You admitted this before, remember?

Le 07/08/2022 à 18:04, Richard Hachel a écrit :
> 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.

What is invariable is the formula To²=Tr²+Et²

If the start is at rest, the two formulas meet.

If we take a fragment of the evolution of the acceleration, there is
already a "launched velocity" and the equation Tr=sqrt(2x/g) is no longer
correct.

To²=Tr²+Et² on the other hand, it remains correct, but the calculation
is more complex.

R.H.

On Sunday, August 7, 2022 at 9:04:32 AM UTC-7, Richard Hachel wrote:
> Le 07/08/2022 à 11:29, Ufonaut a écrit :
> > "This is valid only from a standing start. " ? ? ? ? ? ? ? ?
>
> Yes.

Again, you contradicted yourself by saying that the relation is both true and not true for the same two events. When considering 1,2,3 you said the relations is valid for 1,2, but when considering 1,q,2 where q is an event between 1 and 2 you said the relation is not valid for events 1,2. That is "X" and "not X", so you are blatantly contradicting yourself. Understand?

On Sunday, August 7, 2022 at 9:10:55 AM UTC-7, Richard Hachel wrote:
> Le 07/08/2022 à 18:04, Richard Hachel a écrit :
> > Le 07/08/2022 à 11:29, Ufonaut a écrit :
> >
> >> "This is valid only from a standing start. " ? ? ? ? ? ? ? ?
> >
> > Yes. If the start is at rest, the two formulas meet.

Again, you contradict yourself by saying that the relation is both true and not true for the same two events. When considering 1,2,3 you said the relations is valid for events 1,2, but when considering 1,q,2 where q is an event between 1 and 2 you said the relation is not valid for events 1,2. That is "X" and "not X" for the very same starting and stopping events, so you are blatantly contradicting yourself. (Of course, "starting at rest" is relative, so your claim can also be debunked on those grounds, but that is not necessary, since your claim is already logically self-contradictory even for intervals beginning "at rest".)

Are you able to understand this?

Le 07/08/2022 à 18:05, Stan Fultoni a écrit :
> On Sunday, August 7, 2022 at 1:00:30 AM UTC-7, Richard Hachel wrote:
>> > 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.
>>
>> I can add all the small pieces of Tr (proper time tau) to have the
>> total proper time.
>
> Right, but on the basis of your claims, this leads to the logically impossible
> result 1=0, which proves that your claims are logically inconsistent. Remember?
>
>> The scientist will then commit a terrible and effective blunder.
>> And also [sum the incremental coordinate times to give the total
>> coordinate time].
>
> You are mistaken, and that is another of your logically inconsistent claims. If
> three given events have the coordinate times t1, t2 and t3 such that t1 < t2 < t3,
> then (t2-t1) + (t3-t2) = (t3-t1).

It's obsiously always true for tau (Tr). Pure logic.

But for "To" (improper time), it's false. Because without realizing it,
the watch that measures To is constantly changing position and adding up
all this falsifies the reality of things.

The error consists in believing that the uniformly accelerated reference
frame is like the simple addition of small uniform reference frames.

Do you understand this extremely difficult thing to understand?

To=To1+To2+To3+To4+To5

Let's admit.

But not by setting To1=Tr1.sqrt(1-v²/c²) etc... and in addition the
whole.

Yeah, that's wrong.

R.H.

Le 07/08/2022 à 18:05, Stan Fultoni a écrit :
> On Sunday, August 7, 2022 at 1:00:30 AM UTC-7, Richard Hachel wrote:
>> > 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.
>>
>> I can add all the small pieces of Tr (proper time tau) to have the
>> total proper time.
>
> Right, but on the basis of your claims, this leads to the logically impossible
> result 1=0, which proves that your claims are logically inconsistent. Remember?
>
>> The scientist will then commit a terrible and effective blunder.
>> And also [sum the incremental coordinate times to give the total
>> coordinate time].
>
> You are mistaken, and that is another of your logically inconsistent claims. If
> three given events have the coordinate times t1, t2 and t3 such that t1 < t2 < t3,
> then (t2-t1) + (t3-t2) = (t3-t1).

It's obsiously always true for tau (Tr). Pure logic.

But for "To" (improper time), it's false. Because without realizing it,
the watch that measures To is constantly changing position and adding up
all this falsifies the reality of things.

The error consists in believing that the uniformly accelerated reference
frame is like the simple addition of small uniform reference frames.

Do you understand this extremely difficult thing to understand?

To=To1+To2+To3+To4+To5

Let's admit.

But not by setting To1=Tr1/sqrt(1-v²/c²) etc... and in addition the
whole.

Yeah, that's wrong.

R.H.

On Sunday, August 7, 2022 at 9:28:53 AM UTC-7, Richard Hachel wrote:
> > If three given events have the coordinate times t1, t2 and t3 such that t1 < t2 < t3,
> > then (t2-t1) + (t3-t2) = (t3-t1). This is an algebraic tautology.
>
> ... it's false.

Algebraic tautologies cannot be false. For any three values t1,t2,t3 we have the relation (t2-t1) + (t3-t2) = (t3-t1). You admitted this previously. Now you deny it, adding to your long list of self-contradictions.

Le 07/08/2022 à 18:35, Stan Fultoni a écrit :
> On Sunday, August 7, 2022 at 9:28:53 AM UTC-7, Richard Hachel wrote:
>> > If three given events have the coordinate times t1, t2 and t3 such that t1 <
>> t2 < t3,
>> > then (t2-t1) + (t3-t2) = (t3-t1). This is an algebraic tautology.
>>
>> ... it's false.
>
> Algebraic tautologies cannot be false. For any three values t1,t2,t3 we have
> the relation (t2-t1) + (t3-t2) = (t3-t1). You admitted this previously. Now you
> deny it, adding to your long list of self-contradictions.

You are right, Stan.

For any values tau1, tau2, tau3, tau4, tau, of the rocket
(Tr1,Tr2,Tr3,Tr4,Tr in Hachel's notation) we have a relation and an unique
relation with a To.

Well. To1, To2, To3, To4, To.

x_1 = 0.526ly (Alice) ---> Tr1=1year
x_2 = 2.104ly (Bob) ---> Tr2=2years
x_3 = 4.734ly (Charles) ---> Tr3=3years
x_4 = 8.416ly (Dave) ---> Tr4=4years
D = 12ly (final) ---> Tr= 4.776 years

The real relation between Tr(tau) and x, is Tr=sqrt(2x/g)

The relation (accepted by all) between To and x, is
To=(x/c).sqrt(1+2c²/gx)

To find, for example, To4-To3 we can do To4-To3 =
[(x_4/c).sqrt(1+2c²/gx_4)] - [(x_3/c).sqrt(1+2c²/gx_3)]

But not (To4-To3)=(Tr4-Tr3)/sqrt(1-v²/c²) . This is completly false in
accelerated frames.

It is wrong to assume that if the part is very small, we can consider
that we are in a Galilean frame of reference.

If the start is at rest, we can give the relation, and the true equation
as a function of x, is not To=Tr.sqrt(1+Vr²/c²) BUT
To=Tr.sqrt(1+(1/4)Vr²/c²)

With Vr=Vo/sqrt(1-Vo²/c²)
or, it's the same, Vo=Vr/sqrt(1+Vr²/c²)

R.H.

On Sunday, August 7, 2022 at 2:44:35 PM UTC-7, Richard Hachel wrote:
> The real relation between tau and x is tau = sqrt(2x/g)

That's the relation between x and the elapsed proper time along the *inertial* path between the two given events, not the accelerated path. The elapsed proper time (with c=1) along the path of constant proper acceleration g from the origin to x,t is tau = (1/g)acosh(ax+1).

> The relation (accepted by all) between t and x, is
> t = x sqrt[1 + 2/(gx)]

Well, that's a somewhat obtuse way of writing it, but yes.

> To find, for example, t4 - t3 ...

To find those values you need to define events 3 and 4. If they are supposed to be the events at which the elapsed proper time along the path accelerating from rest at the origin with constant proper acceleration g has the values tau3 and tau4, then we have t3 = (1/g)sinh(g tau3) and t4 = (1/g)sinh(g tau4).

On the other hand, if you define events 3 and 4 as events at which x equal x3 and x4, then
t3 = x3 sqrt[1 + 2/(gx3)]
t4 = x4 sqrt[1 + 2/(gx4)]

> But (t4 - t3) = (tau4 - tau3)/sqrt(1 - v²) is completly false in accelerated frames.

First, no one is talking about accelerated frames. The coordinates x,t are a single standard inertial coordinate system. Second, everyone knows that the equation you typed is not correct for the elapsed time between two separate events along an accelerating path, because you must take the integral, meaning that we have dtau/dt = sqrt(1-v^2) at every event along the path, where v is the speed of the path at that event. Carrying out the integration along a path with constant proper acceleration g from rest at the origin we get t = (1/g)sinh(g tau).

> If the start is at rest, we can give the relation, and the true equation
> as a function of x, is not t = tau sqrt(1 + v²) BUT t = tau sqrt[1+(1/4)v²].

Nope, the correction relation is t = (1/g)sinh(g tau), and also x = (1/g)[cosh(g tau) - 1]. Again, you must integrate the absolute interval along an accelerating path. This is just Relativity 101.

Le 08/08/2022 à 00:53, Stan Fultoni a écrit :
>> The real relation between tau and x is tau = sqrt(2x/g)
>
> That's the relation between x and the elapsed proper time along the *inertial*
> path between the two given events, not the accelerated path. The elapsed proper
> time (with c=1) along the path of constant proper acceleration g from the origin
> to x,t is tau = (1/g)acosh(ax+1).

I think you are wrong.

Only experimental proof can decide between us.

R.H.

Le 08/08/2022 à 00:53, Stan Fultoni a écrit :
>> The relation (accepted by all) between t and x, is
>> t = x sqrt[1 + 2/(gx)]
>
> Well, that's a somewhat obtuse way of writing it, but yes.

I didn't write that, but:

To = (x/c).sqrt(1+2c²/gx)

R.H.

Le 08/08/2022 à 00:53, Stan Fultoni a écrit :
>> To find, for example, t4 - t3 ...
>
> To find those values you need to define events 3 and 4. If they are supposed to
> be the events at which the elapsed proper time along the path accelerating from
> rest at the origin with constant proper acceleration g has the values tau3 and
> tau4, then we have t3 = (1/g)sinh(g tau3) and t4 = (1/g)sinh(g tau4).
>
> On the other hand, if you define events 3 and 4 as events at which x equal x3
> and x4, then
> t3 = x3 sqrt[1 + 2/(gx3)]
> t4 = x4 sqrt[1 + 2/(gx4)]

I cannot endorse what you are saying.

R.H.

Le 08/08/2022 à 00:53, Stan Fultoni a écrit :
> First, no one is talking about accelerated frames. The coordinates x,t are a
> single standard inertial coordinate system. Second, everyone knows that the
> equation you typed is not correct for the elapsed time between two separate events
> along an accelerating path, because you must take the integral, meaning that we
> have dtau/dt = sqrt(1-v^2) at every event along the path, where v is the speed of
> the path at that event. Carrying out the integration along a path with constant
> proper acceleration g from rest at the origin we get t = (1/g)sinh(g tau).

I am killing myself to tell you that, precisely, one must not take the
integral and that this integral is false.

It is doubly wrong:
1. If the start is not in a stopped position. IMPORTANT !!!
2.If we take this equation by setting:
To=ΔTr∫(1/sqrt(1+Vr²/c²)) instead of
To=ΔTr∫(1/sqrt(1+(1/4)Vr²/c²))

R.H.

Le 08/08/2022 à 00:53, Stan Fultoni a écrit :
>> If the start is at rest, we can give the relation, and the true equation
>> as a function of x, is not t = tau sqrt(1 + v²) BUT t = tau sqrt[1+(1/4)v²].
>
> Nope, the correction relation is t = (1/g)sinh(g tau), and also x = (1/g)[cosh(g
> tau) - 1]. Again, you must integrate the absolute interval along an accelerating
> path. This is just Relativity 101.

No. You don't must integrate the absolute interval along an accelerating
path.

It is a bad concept.

Because your equation is not the good equation to use.

I repeat : To=Tr/sqrt(1+Vr²/c²) is false in the accelerated frames.

You must use To=Tr/sqrt((1+0.25.Vr²/c²)

R.H.

On Sunday, August 7, 2022 at 2:44:35 PM UTC-7, Richard Hachel wrote:
> I repeat : t = tau/sqrt(1 + v²) is false in the accelerated frames..

That is not only false, it is gibberish, but you have made it yourself. The correct relation is dt = dtau/sqrt(1 - v^2), which is equivalent to dtau = sqrt(dt^2 - dx^2). (You must learn that t and dt are not the same things.) We integrate dtau along the path to get the elapsed tau along the path.

> You must use t = tau/sqrt((1 + 0.25v²)

That is complete gibberish. Look, if you draw a curve on a flat surface with Cartesian coordinate system x,y, the length of the curve from one point p1 on the curve to another point p2 is given by integrating ds = sqrt(dx^2 + dy^2) along the curve. If the curve happens to be a straight line, then this integral gives the length sqrt[(x2-x1)^2 + (y2-y1)^2], but if the curve is not straight, you have the integrate. It's the same for elapsed proper time, i.e., you have to integrate the metric along the path. You are confusing the distance between the end points with the length of the (curved) path. Those are different. Understand?

Le 08/08/2022 à 03:34, Stan Fultoni a écrit :

> That is not only false, it is gibberish, but you have made it yourself. The
> correct relation is dt = dtau/sqrt(1 - v^2), which is equivalent to dtau =
> sqrt(dt^2 - dx^2). (You must learn that t and dt are not the same things.) We
> integrate dtau along the path to get the elapsed tau along the path.

I do the reverse. I take the set of d_tau, and I integrate them to find
To.

It's easier.

Still it is necessary to integrate correctly, and to take the correct
equation.

Physicists use To=ΔTau.∫sqrt(1+Vr²/c²) but this is not correct.

We have to take To=ΔTau.∫sqrt(1+(1/4)Vr²/c²)

R.H.

Le 08/08/2022 à 03:34, Stan Fultoni a écrit :
> That is complete gibberish. Look, if you draw a curve on a flat surface with
> Cartesian coordinate system x,y, the length of the curve from one point p1 on the
> curve to another point p2 is given by integrating ds = sqrt(dx^2 + dy^2) along the
> curve. If the curve happens to be a straight line, then this integral gives the
> length sqrt[(x2-x1)^2 + (y2-y1)^2], but if the curve is not straight, you have the
> integrate. It's the same for elapsed proper time, i.e., you have to integrate the
> metric along the path. You are confusing the distance between the end points with
> the length of the (curved) path. Those are different. Understand?

No. YOU think and say it's gibberish.

I think you don't understand the structure of space-time.

You then use bad mathematical tools, and you integrate a false equation.

The integrations, you do them as well as I do, but I repeat, you integrate
a wrong equation.

The correct equation is To²=Tr²+Et².

Which easily gives To=Tr.sqrt(1+Vr²/c²) for uniform media, BUT, for
accelerated media, with start at standstill, this gives To=Tr.sqrt(1+(1/4)
Vr²/c²)
If the start is made with a speed already acquired, the equations become
even more complicated.
I'll let you find out if you want. My job is to dust off the problem;
yours to finalize it if you want.
If you don't want to, I'll do it myself.
And shame, I will leave it on you all, to which I give pearls of all
kinds, but it makes you laugh.

R.H.

On Monday, August 8, 2022 at 7:17:04 AM UTC-7, Richard Hachel wrote:
> The correct equation is t² = tau² + x².

That's the equation (with c=1) relating x and t to the elapsed proper time tau along an inertial path from the orign to x,t, but to apply this to accelerating paths we must use the differential form (dt)² = (dtau)² + (dx)², which implies dtau = sqrt[(dt)² - (dx)²]. We integrate this for a path of constant proper acceleration g to give the result tau = (1/g) asinh(g t).

> Which easily gives t = tau.sqrt(1 + v²) for uniform media...

No, that equation makes no sense at all.

> BUT, for accelerated media, with start at standstill, this gives
> t = tau.sqrt(1 + (1/4)v²)

That makes no sense either. The correct derivation is as given above, based on the fundamental relation (dtau)² = (dt)² - (dx)². Do you dispute this?

Le 08/08/2022 à 20:42, Stan Fultoni a écrit :
> On Monday, August 8, 2022 at 7:17:04 AM UTC-7, Richard Hachel wrote:
>> The correct equation is t² = tau² + x².
>
> That's the equation (with c=1) relating x and t to the elapsed proper time tau
> along an inertial path from the orign to x,t, but to apply this to accelerating
> paths we must use the differential form (dt)² = (dtau)² + (dx)², which implies
> dtau = sqrt[(dt)² - (dx)²]. We integrate this for a path of constant proper
> acceleration g to give the result tau = (1/g) asinh(g t).

Yes, that's what I'm saying. But this result is wrong. You don't have to
integrate like that.

x=(1/2)at²

To²=Tr²+Et²

To²=Tr²+[(1/2)aTr²]²/c²

To²=Tr²+Tr²[(1/2)aTr]²/c²

To=Tr.sqrt(1+(1/4)a²Tr²/c²)

a.Tr=Vr

To=Tr.sqrt(1+(1/4)Vr²/c²)

To=ΔTr.∫sqrt(1+(1/4)Vr²/c²)

>
>> Which easily gives t = tau.sqrt(1 + v²) for uniform media...
>
> No, that equation makes no sense at all.

No. YOU, you say that equation makes no sense at all.
>
>> BUT, for accelerated media, with start at standstill, this gives
>> t = tau.sqrt(1 + (1/4)v²)
>
> That makes no sense either.

No. YOU, you say that equation makes no sense at all.

> The correct derivation is as given above, based on the fundamental relation
> (dtau)² = (dt)² - (dx)². Do you dispute this?

No, obviously.

Tr²=To²-(x/c)²

Et=x/c

Then To²=Tr²+Et²

No problem.

R.H.

On Monday, August 8, 2022 at 12:38:52 PM UTC-7, Richard Hachel wrote:
> >> The correct equation is t² = tau² + x².
> >
> > That's the equation (with c=1) relating x and t to the elapsed proper time tau
> > along an inertial path from the orign to x,t, but to apply this to accelerating
> > paths we must use the differential form (dt)² = (dtau)² + (dx)², which implies
> > dtau = sqrt[(dt)² - (dx)²]. We integrate this for a path of constant proper
> > acceleration g to give the result tau = (1/g) asinh(g t).
>
> x=(1/2)at²

No, that would be the relation between x and t for constant *coordinate* acceleration, but that would result in exceeding the speed of light after about one year, and would be 3c after about 3 years. That is not possible. What we are discussing is a rocket undergoing constant *proper* acceleration, so the actual relation is

x = (1/a) [sqrt(1 + (at)²) - 1]

For small values of "at" this is approximately (1/2)at², but for larger values it deviates very significantly. This is a crucial mistake you are making. Now, it so happens that x = (1/2)aT² where T is the elapsed time along the inertial path from the origin to x,t, so T is neither the coordinate time nor the proper time along the accelerating path.

> > The correct derivation is as given above, based on the fundamental relation
> > (dtau)² = (dt)² - (dx)². Do you dispute this?
>
> No, obviously.

Great, then it is simple math to derive the fact that tau = (1/g) asinh(g t), proving that all your claims to the contrary are nonsense.

Le 09/08/2022 à 01:07, Stan Fultoni a écrit :

> No, that would be the relation between x and t for constant *coordinate*
> acceleration, but that would result in exceeding the speed of light after about
> one year, and would be 3c after about 3 years. That is not possible.

Please! Stop !!!

Stan, please, stop!

R.H.

Le 09/08/2022 à 01:07, Stan Fultoni a écrit :

> x = (1/a) [sqrt(1 + (at)²) - 1]

Yes.

Here is correct.

You are right.

x=(c²/a).[sqrt(1+a²To²/c²)-1]

R.H.

Le 09/08/2022 à 01:07, Stan Fultoni a écrit :

> Great, then it is simple math to derive the fact that tau = (1/g) asinh(g t),
> proving that all your claims to the contrary are nonsense.

Tr = tau = sqrt(2x/a)

To = (x/c).sqrt(1+2c²/ax)

Your equation Tr = tau = (1/g) asinh(g t) is not correct.

I will advise you something.

First check the consistency of my system based on thousands of hours of
reflection.

You will realize that if my system is not the same as yours, it has its
internal consistency (theoretical beauty).

It is even more coherent than the current system which does not hold water
for a single second if the apparent relativistic velocities are used, as I
have already shown dozens of times.

Next, we must move on to the experimental facts. There again, I explain
very easily the phenomena of quantum transmissions.

So I have a theoretical external perfection also superior to yours).

I repeat, if you don't want to make the effort to understand me, it's not
scientific, it's just human self-defense behavior.

I do not understand this behavior which seems to me rather to be of the
political or religious type rather than scientific.

On the other hand, we must do things correctly and honestly.

When you tell me that the rocket will very quickly reach the speed of
light, exceed it, and arrive at Tau Ceti at 5.0245c, and that this is
impossible, it proves that you do not understand everything I say
correctly.

You are confusing real speeds Vr, traditional observable speeds Vo, and
apparent speeds Vapp.

These are three different concepts.

Vo and Vapp are commonly accepted notions. Physicists have the same as me.

Vr is a new concept that I introduced in 1986.

It is the real speed in an anisochronous space.

The equation that links this real speed with the observable speed Vo is
Vr=Vo/sqrt(1-Vo²/c²)

The inverse being Vo=Vr/sqrt(1+Vr²/c²)

Attention again: do not confuse Vr with a kind of absolute speed.

There are obviously no absolute speeds or absolute references.

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
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On Monday, August 8, 2022 at 12:38:52 PM UTC-7, Richard Hachel wrote:
> > Given that (dtau)² = (dt)² - (dx)², it is simple math to derive the fact
> > that tau = (1/g) asinh(g t), proving that all your claims to the contrary
> > are nonsense.
>
> tau = sqrt(2x/a)

The "tau" you have written there is not the elapsed proper time along the accelerating path, it is the elapsed proper time along the inertial path from the origin to the event x,t. You keep confusing these things.

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

Yes, that is a correct equation relating xand t along the path of constant proper acceleration.

> Your equation tau = (1/g) asinh(g t) is not correct.

It is correct, where tau is the elapsed proper time along the path of constant proper acceleration from rest at the origin, nd t is the coordinate time, and g is the proper acceleration. The simple derivation was provided to you, simply integrating dtau = sqrt(dt^2 -dx^2).

> First check the consistency of my system based on thousands of hours of
> reflection.

Your "system" is logically inconsistent, because you agree that dtau = sqrt(dt^2 - dx^2), from which the correct expression for tau follows by simple integration, and yet you deny this. Didn't you ever take calculus?

> When you tell me that the rocket will very quickly reach the speed of
> light, it proves that you do not understand everything I say correctly.

No, you claimed that x = (1/2)at², which would imply that dx/dt exceeds the speed of light after about 1 year. You are confusing constant coordinate acceleration for constant proper acceleration (a parabola versus a hyperbola). You even admitted later that the correct relation is x = (1/a) [sqrt(1 + (at)²) - 1]. Remember?

Le 09/08/2022 à 03:42, Stan Fultoni a écrit :
> On Monday, August 8, 2022 at 12:38:52 PM UTC-7, Richard Hachel wrote:
>> > Given that (dtau)² = (dt)² - (dx)², it is simple math to derive the fact
>> > that tau = (1/g) asinh(g t), proving that all your claims to the contrary
>> > are nonsense.
>>
>> tau = sqrt(2x/a)
>
> The "tau" you have written there is not the elapsed proper time along the
> accelerating path, it is the elapsed proper time along the inertial path from the
> origin to the event x,t. You keep confusing these things.

Yes, I understand where you are coming from.

I just think it's real proper time.

Now, it's true that I don't give the concrete mathematical explanation, I
admit it.

I'll think about the problem, and see how I can prove it.

I had known for a long time that with regard to Langevin's description of
the voyageur, there was a "shell" in the relativists, until the day when I
was able to actually prove it using the covariance of the apparent
velocities.

This I have proven.

So I will also have to prove that you are completely wrong with a
mathematical description.

Although I know the human being. Even if I do, I'm not sure I'll be heard.

Relativity as enacted has become a dogma.

Dogma with its truths and its errors.

But dogma.

R.H.