Re: Accelerated frame and Tau Ceti problem

Richard Hachel wrote:

>
Le 02/06/2024 à 05:08, hitlong@yahoo.com (gharnagel) a écrit :

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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.
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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.

YES!

As obvious as it may seem, this result is false.

Nope.

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.

The astonishment is that Dr. Hachel has let his three Nobel prizes go
to
his head.  They are not in relativity nor in mathematics.  Two of my
four Nobel prizes are.

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).

Wrongo.  There must be FIVE watches, one at O and one each at A, B, C
and D.
You correctly specified that they must be synchronized.  You used slow
clock
transport which is approximately correct.  Better to use Einstein
synchronization.

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.

It's IOTTMCO.

However, this is false.

Nope.

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).

You error hugely.  The watches have no "anisochrony" because they are
stationary to one another and they have been synchronized.

We cannot subtract one deformation from another deformation in such a simplistic way, and it is a serious relativistic mistake to do
so.

The mistake is one who has three Nobel prizes in basket weaving
purports
to know relativity.

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.

And Nobel prizes which are not correct :-))

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

Velocity is irrelevant since the watches are at rest wrt earth.  They
record the time as the ship flies past.

For the observable times between games taken at random, it should be noted:
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<http://news2.nemoweb.net/jntp?uS_y7EWpgPGPmPD64dwz49GUbzs@jntp/Data.Media:1>
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Thank you for listening.
>
R.H.

Thank you for not babbling nonsense in the future.