Re: Incorrect mathematical integration

Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :

If the proton is passing a point in the ring with the speed v
relative to the point, then the point in the ring is passing
the proton a the speed v relative to the proton.

This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
Indeed, if the proton passes at Vo=0.999991 c (for example) at a point A of the device, then the laws of physics state that point A passes at Vo=0.999991c.
If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Likewise, this real speed is reciprocal.
In the proton frame, it is point A which passes near it at Vr=235.7c.
Now what does the global ring look like in the proton frame of reference, and above all what is the trajectory of point A during one revolution?
This is a good relativistic physics question.
Have fun answering this question...
I hope you have a lot of fun.
R.H.