Re: Incorrect mathematical integration

Den 24.07.2024 20:55, skrev Richard Hachel:

Le 24/07/2024 à 20:44, "Paul.B.Andersen" a écrit :

Den 24.07.2024 15:08, skrev Richard Hachel

>
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?

 Irrelevant.
 The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
 Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
 The speed of the proton in K is v = dx/dt = L/T =  0.999999991·c
 Do you know another definition of the speed of the proton in K
than dx/dt ?
 Let K'(x',τ) be an inertial frame of reference which at the instant I
is momentarily comoving with the proton.
 The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ = 0.999999991·c
 Do you know another definition of the speed of the point A in K'
than  dx'/dτ ?

To the very slow reader R.H:
At the instant in question the point A and the momentarily colocated
proton can be at any arbitrary point in the circuit, and K and K'
are momentarily comoving with the point A and the proton respectively.
This means that K and K' are momentarily colocated, and their relative speed is 0.999999991·c.
Even the shape of the accelerator (it's not a circle!) is irrelevant.

 It's not a joke, it's a question.
 We take a very large particle accelerator of several kilometers.
 On this particle accelerator, we fix a point A, coordinates (0,0,0) and we ask to draw the trajectory of the proton in R.
 We assume z=0.
 We then have a large circle.
 We then request a change of reference frame, and we request the trajectory of point A in the proton reference frame.
 I wish good luck to whoever answers, it's not high school level.
 R.H.

And your point is?
--
Paul
https://paulba.no/