Re: Incorrect mathematical integration

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

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

>
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.

 Your assertion is in violent disagreement with the LTE:
 dx' = gamma(dx - vdt)
dt' = gamma(dt - vdx)
 For an object stationary in the unprimed frame, dx = 0:
 dx' = gamma(-vdt)
dt' = gamma(dt)
 v' = dx'/dt' = -v
 For an object moving at v in the unprimed frame, dx' = 0
 v = dx/dt = v.
 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.

But NO!
WE MUST APPLY POINCARE'S TRANSFORMATIONS!
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.
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).
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.
x' = (x-Vo.To) / sqrt (1-Vo² / c²)
x' = (9 + 0.8 * 9) / 0.6 = 27 cm.
I see a longer rod coming towards me.
The same goes for a proton launched at 0.8c on a 9 meter path. At what distance is the proton at the moment it is ejected at 0.8c from its receiver in the laboratory frame of reference? It is 9 meters.
BUT if I place myself at the level of the proton, at this moment, where is the receiver that will come towards me at high apparent speed (Vapp = 4c)?
At 27 meters in the proton frame of reference.
The dilation of distances is frightening.
But it is sad to cry to have to explain the same things over and over again (although of an incredible logic, beauty, and precision when one correctly understands the Poincaré equations and does not say anything.
My friends, I beg you to understand that the gamma factor, here 0.6, produces a dilation of distances, lengths and times.
Those who see "contractions" there are simply sick people who have not correctly understood the theory, and who, from the Poincré equations have gone into an abstract and fanciful geometry, while the good doctor Hachel has given all the correct geometry, and all the correct explanations.
R.H.