Re: Incorrect mathematical integration

On Fri, 26 Jul 2024 0:54:30 +0000, Richard Hachel wrote:

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Le 26/07/2024 à 01:29, hitlong@yahoo.com (gharnagel) a écrit :

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On Thu, 25 Jul 2024 20:30:09 +0000, Richard Hachel wrote:

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In the case you are proposing, there is no contraction of the

distances,

because the particle is heading TOWARDS its receptor.
>
The equation is no longer D'=D.sqrt(1-Vo²/c²) and to believe this is

to

fall into the trap of ease, but D'=D.sqrt[(1+Vo/c)/ (1-Vo/c)] since
cosµ=-1.

>
You are conflating Doppler effect with length contraction.  LC is

ALWAYS

D'=D.sqrt(1-Vo²/c²).
>

For the particle the distance to travel (or rather that the receiver
travels towards it) is extraordinarily greater than in the

laboratory

reference frame.
>
R.H.

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Your assertion is in violent disagreement with the LTE:
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dx' = gamma(dx - vdt)
dt' = gamma(dt - vdx)
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For an object stationary in the unprimed frame, dx = 0:
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dx' = gamma(-vdt)
dt' = gamma(dt)
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v' = dx'/dt' = -v
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For an object moving at v in the unprimed frame, dx' = 0
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v = dx/dt = v.
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There is no "extraordinarily greater" speed in either frame.  This
is true in Galilean motion also.  Galileo described it perfectly
with his ship and dock example and blows your assertion out of the
water, so to speak.

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But NO!
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WE MUST APPLY POINCARE'S TRANSFORMATIONS!
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It took years to find them, and without Poincaré, it is likely that they
would have been found only ten or fifteen years later, when they already
had them in 1904.
>
That is why I am almost certain that Einstein copied them from Poincaré
despite his period denials (which he would later contradict by saying
that he had read Poincaré and that he had been captivated by the
intellectual
power of this man, considered the best mathematician in the world at
that time).
>
We must apply Poincaré.

If Einstein copied Poincaré, then Einstein's equations are Poincaré's.

What does Poincaré say?
>
If an observer moves towards me, at speed Vo=v, and crosses me at
position 0, then for me, he is at (0,0,0,0) and for him, I am at
(0,0,0,0).

Not necessarily. "Position 0" is insufficient  for being at (0,0,0,0).

But let's assume that it is only a piece of rod 9 cm long
that crosses me, and that the other end has not yet passed.
At what distance will I see the other end of the rod? Let Vo = 0.8c.

You are going off on a tangent, not sticking to the problem you posed.
Furthermore, you haven't defined what you believe Poincaré's equations
are.  Consequently, your deflection is merely buzz words.