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Subject: Re: Langevin's paradox again
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Date: Mon, 15 Jul 24 12:23:32 +0000
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From: Richard Hachel <r.hachel@wanadou.fr>
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Le 15/07/2024 à 13:27, "Paul.B.Andersen" a écrit :
> Den 14.07.2024 23:11, skrev Richard Hachel:

> What does your theory predict the speed of the protons in the LHC is?
> 
> What is the measured speed of the protons in the LHC?

But I have already answered you a thousand times.
The observable, i.e. measurable, speed of the proton in the LHC is 
0.999999989c.
I set Vo=0.999999989c.
It's still very simple to understand, and if you made more effort to 
achieve coherent discussions, we wouldn't be talking such banalities.

I just pointed out, even if it may seem very destabilizing, that this is 
only the observable, measured speed, but that it is not the deep reality 
of things.

You know very well that the measurement of the shadow of a building is not 
the measurement of the building itself. It depends on the position of the 
sun.

We have the same thing in relativity.

When we measure, in our frame of reference, the distances to be covered, 
we have a precise measurement to note.

But when we measure the time it takes to cover this distance, a huge 
illusion appears because apart from the rocket or the particle, which only 
have one watch, we are obliged to use two watches (even if we come back to 
the same physical watch, after a long journey, it is no longer really "the 
same watch").

Paul, breathe, exhale.

Measuring time with two watches placed in different places can only lead 
to temporal aberrations.

Speed ​​being the quotient of distance over time, we will no longer 
have the same notion of speed. An illusion will appear since our way of 
measuring things becomes incorrect.

When you measure the observable speed Vo of your proton, you find 
Vo=0.999999989c but if you want to know the real speed
in the lab reference frame, you must convert (i.e. remove the bias from 
the measurement).
Vr=Vo/sqrt(1-Vo²/c²)

Conversely, if you know Vr, you use the reciprocal equation to find Vo.

Vo=Vr/sqrt(1+Vr²/c²)

The real speed of your proton (not the one you measure) is therefore 
Vr=6947c.

This is enormous speed.

If you know the mass of your proton, and if you know its momentum at this 
instant, you just need to use p=m.v

You will see that v, in reality, is Vr=6947c and not Vo=0.999999989c.

R.H.