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Subject: Re: Accelerated frame and Tau Ceti problem
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Date: Sun, 02 Jun 24 07:20:05 +0000
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From: Richard Hachel <richard.hachel@invalid.fr>
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Le 02/06/2024 à 05:08, hitlong@yahoo.com (gharnagel) a écrit :
> Richard Hachel wrote:
> 
>> > When we move to level 3 ly, the local watch marks: 3.8345 years
>> > When we move to level 6 ly, the local watch marks: 6.8852 years
>> > When we move to level 9 ly, the local watch marks: 9.9050 years
>> > When we cross Tau Ceti (12 ly), the local watch says: 12.9156 years
>> >
>> > We have therefore just verified experimentally that the equation is 
>> > correct.
>> >
>> > It seems very simple, but we now know Richard Hachel, and we know what
>> 
>> > kind of madness attacked his neurons.
>> >
>> > We will then say: What is the observable time that exists between the
>> 
>> > passage in (3ly) and the passage in (6ly)?
> 
> Well, 6.8852 - 3.8345 = 3.0507 

 No.
As obvious as it may seem, this result is false.
I understand the immense astonishment I cause by saying this, because I 
say that the whole is not equal to the sum of the parts.
But your result, which you think is quite simple, is false.
I explain why.
When you use, and correctly, To=(x/c).sqrt(1+2c²/ax) an equation which 
should be known by heart by all students from the age of 17, you do so 
under two specific conditions.
1. The departure is in O (on land) and at rest.
2. Measuring time requires two watches, one placed at O ​​and the 
other placed at A (3 ly), B (6 ly), C (9 ly), D (12 ly).

The measurements are therefore correct.

Now, I want to measure the time taken between A and B. And you try to give 
it to me by doing a simple subtraction, as this may seem so obvious to 
you.

However, this is false.

As incredible as it may seem if we don't understand what we are doing, 
that is to say placing ourselves at the level of two other watches which 
are not in O.

It is not the same thing, whereas the observable times are deformations of 
the proper times (which are the real times because they are not distorted 
by the natural anisochrony present between any two watches, whatever they 
may be). We cannot subtract one deformation from another deformation in 
such a simplistic way, and it is a serious relativistic mistake to do so.

Yet this is what physicists do, who then obtain times which are not 
correct, instantaneous speeds which are not correct, and proper times 
which are not correct either.

For instantaneous observable velocities the correct equation is:
Voi/c=[1+c²/2ax]^(-1/2)

For the observable times between games taken at random, it should be 
noted:

<http://news2.nemoweb.net/jntp?uS_y7EWpgPGPmPD64dwz49GUbzs@jntp/Data.Media:1>


Thank you for listening.

R.H.