On Friday, August 12, 2022 at 8:18:32 AM UTC-4, Richard Hachel wrote:
> Le 12/08/2022 à 10:20, Stan Fultoni a écrit :

> You speak of three times.
> t, tau, TAU.
> I don't mind.

Know that you are the one who invoked these times.
He is just using the conventional "words" for these times that YOU invoked, since you use unconventional Words which makes your rants gibberish.

On Friday, August 12, 2022 at 5:43:51 AM UTC-7, Richard Hachel wrote:
> Is it the same to integrate dtau=sqrt(dt²-dx²) and dt²=sqrt(dtau²+dx²) ?

Sure, but to do the latter you would need to know tau as a function of x, which is what you're trying to determine. We know x as a function of t for the path of constant proper acceleration, and we are trying to determine tau, so we integrate dtau=sqrt(dt^2 - dx^2).

> Are you sure that the way you put dx² is correct?

Yes. Remember, you already agreed that t = x.sqrt(1+2/(ax)), and solving this for x gives the relation x = (1/a)[sqrt(1 + (at)²) - 1], so we have dx = [at/sqrt(1+(at)^2]]dt, and therefore dtau = sqrt(dt^2 - dx^2) = dt/sqrt(1 + (at)^2), whose integral is tau = (1/a)invsinh(at) = 3.139 years at t=12.915 years. There is no ambiguity about this.

> But if I take the duration t in the terrestrial frame of reference, and cut it into a thousand pieces.
> It is obvious that in the first piece, I will have To=tau=TAU since the speeds are very low.

No, the value of TAU (which you call Tr) is the elapsed proper time along the inertial path from the origin to the event x=12, t=12.915 at the constant speed 0.92915, so dTAU is not the same as dt, even at the beginning.

> I only use one Tr (tau)

No, you use Tr = TAU, which is not the relevant parameter. You need to compute the elapsed proper time along the accelerated path, not along the path with constant speed 0.92915. Understand?

Your beliefs are self-contradictory, because you claim to agree that dtau^2 = dt^2 - dx^2, but then you deny it, and then you re-affirm it, and then deny, and so on.