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tech / sci.physics.relativity / Re: Relativistic accelerated frames

SubjectAuthor
* Relativistic accelerated framesRichard Hachel
`- Re: Relativistic accelerated framesmitchr...@gmail.com

1
Relativistic accelerated frames

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From: r.hac...@frite.fr (Richard Hachel)
 by: Richard Hachel - Tue, 21 Feb 2023 09:48 UTC

I had said that if the equation giving the improper times (observable
times To in Hachel writing),
was correct, i.e. To=(x/c).sqrt(1+2c²/ax), the equation relating to the
proper times (real time Tr in Hachel writing) was completely false, and
that the equation correct was Tr=sqrt(2x/a).
In fact, you're going to feel bad my friends, I was only telling half the
truth.
The first equation is true, but it is only true for a standing start.
I have spoken a lot since the beginning of the year on fr.sci.physics and
on sci.physics.relativity.
I say more.
I thank Samuel Devulder who gave the correct numerical answer of
integrating my formula,
in perfect agreement with the equation given above.
I recall my formula giving the instantaneous speed which is the one that
should be used by physicists to the exclusion of any other:
Vo/c = [1+c²/2ax]^(-1/2)
I recall my formula for proper times Tr=sqrt(2x/a)
As for the formula To=(x/c).sqrt(1+2c²/ax), it is only valid for a
segment x₂-x₁ where x₁=0.
For all the other segments, you have to take the general formula that I
write here:
ΔTo=(Δx/c).sqrt(1+2c²[sqrt(x₂)-sqrt(x₁)]²/aΔx²)
Thank you for your attention.
Attention, it is obvious that people familiar with the theory of
relativity, will point out to me that the results are absurd, because the
time difference between two trips over different distances with return to
earth are incorrect.
They will be right.
But that's not what I'm talking about.
I am not speaking for this case, which only takes up stopped starts, which
are subtracted from each other, I am speaking of the general evolution,
segment by segment, and there is no contradiction in what I say.
Whoever sees a contradiction in this must convince himself that the fault
is his, and the fact that he did not fully understand what I am trying to
say about the geometry of space-time, and the fact that it is not a
Minkowskian geometry that must be taken, but such as I give it, me.

R.H.

Re: Relativistic accelerated frames

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Subject: Re: Relativistic accelerated frames
From: mitchrae...@gmail.com (mitchr...@gmail.com)
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 by: mitchr...@gmail.com - Wed, 22 Feb 2023 22:57 UTC

On Tuesday, February 21, 2023 at 1:48:29 AM UTC-8, Richard Hachel wrote:
> I had said that if the equation giving the improper times (observable
> times To in Hachel writing),
> was correct, i.e. To=(x/c).sqrt(1+2c²/ax), the equation relating to the
> proper times (real time Tr in Hachel writing) was completely false, and
> that the equation correct was Tr=sqrt(2x/a).
>
> In fact, you're going to feel bad my friends, I was only telling half the
> truth.
>
> The first equation is true, but it is only true for a standing start.
>
> I have spoken a lot since the beginning of the year on fr.sci.physics and
> on sci.physics.relativity.
>
> I say more.
>
> I thank Samuel Devulder who gave the correct numerical answer of
> integrating my formula,
> in perfect agreement with the equation given above.
>
> I recall my formula giving the instantaneous speed which is the one that
> should be used by physicists to the exclusion of any other:
>
> Vo/c = [1+c²/2ax]^(-1/2)
>
> I recall my formula for proper times Tr=sqrt(2x/a)
>
> As for the formula To=(x/c).sqrt(1+2c²/ax), it is only valid for a
> segment x₂-x₁ where x₁=0.
>
> For all the other segments, you have to take the general formula that I
> write here:
>
> ΔTo=(Δx/c).sqrt(1+2c²[sqrt(x₂)-sqrt(x₁)]²/aΔx²)
>
> Thank you for your attention.
>
> Attention, it is obvious that people familiar with the theory of
> relativity, will point out to me that the results are absurd, because the
> time difference between two trips over different distances with return to
> earth are incorrect.
> They will be right.
> But that's not what I'm talking about.
> I am not speaking for this case, which only takes up stopped starts, which
> are subtracted from each other, I am speaking of the general evolution,
> segment by segment, and there is no contradiction in what I say.
> Whoever sees a contradiction in this must convince himself that the fault
> is his, and the fact that he did not fully understand what I am trying to
> say about the geometry of space-time, and the fact that it is not a
> Minkowskian geometry that must be taken, but such as I give it, me.
>
> R.H.

How is there two times before the event horizon?
What time ends what time begins?
How is there more than one?

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