Re: Incorrect mathematical integration

Am Freitag000026, 26.07.2024 um 22:11 schrieb Richard Hachel:

Le 26/07/2024 à 21:54, "Paul.B.Andersen" a écrit :

Den 26.07.2024 18:14, skrev Richard Hachel:

 

Right.
So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the speed in
the laboratory reference frame is v = L/T = 0.999999991·c

 There you go.
Your sentence is almost perfect.
It is perfect in traditional relativity, and almost perfect in RH-style relativity.
I just add the word "observable" not because I like to brag, but because I find it useful for the song.
  So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the observable speed in
the laboratory reference frame is v = L/T = 0.999999991·c
 

But which speed does the proton have?
If the proton compares its own speed in respect to itself, this would be zero, because the proton does not move in respect to itself.
This is actually a valid and possible 'inertial' frame, too, hence 'speedlimit c' is nonsense, because the proton could not move in repect to itself at all.
'Speedlimit c' would require something, which Einstein had explicitly excluded: some 'absolute' space (like e.g. 'the universe').
But since the relativity principle allows to regard all inertial frames as equally valid, I could choose the comoving frame, too.
Now, this frame does not move in respect to the proton, because it is 'attatched' to it.
The consequence: the proton does not move at all, hence cannot possibly reach anything close to c.
TH