Re: Incorrect mathematical integration

W dniu 24.07.2024 o 22:27, Python pisze:

Le 24/07/2024 à 20:45, Paul.B.Andersen a écrit :

Den 24.07.2024 15:08, skrev Richard Hachel:

Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
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Measured in the lab frame the proton is moving around
the L = 27 km long ring in T = 90.0623065140618 μs.
The very real speed of the proton relative to the lab is
v = L/T =  0.999999991·c
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γ = 7460
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Measured in the proton frame, the length of the ring is
L' = L/γ = 3.6193029490616624 m.
The proton is moving around the L' long ring in the time
 τ = T/γ = 12.072695243171824 ns
The very real speed of the lab relative to the proton is
v =  L'/τ = (L/γ)/(T/γ ) = L/T = 0.999999991·c
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This should be blazingly obvious for anybody but complete morons:
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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.

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A bit more precisely put:
In the lab frame the proton is passing a point in the ring with
the speed v = L/T =  0.999999991·c.
In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
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This is the only interesting sentence in your post.
The rest is just nonsense or tautology.
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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.

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If we transpose into real speed Vr, we have:
Vr=Vo/sqrt(1-Vo²/c)=235.7c

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Nothing is moving at the speed L/τ = 235.7c
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Likewise, this real speed is reciprocal.

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The reciprocal of L/τ is (L/γ)/T = 0.0001340c
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Equally meaningless. Not the speed of anything.
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In the proton frame, it is point A which passes near it at Vr=235.7c.

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In the proton frame the point in the ring is passing the proton with
the speed v = (L/γ)/τ = 0.999999991·c.
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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?

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Irrelevant.
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The point A is at any instant adjacent to the proton.
We consider an arbitrary instant I.
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Let K(x,t) be an inertial frame of reference which at the instant I
is momentarily comoving with the point A.
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The speed of the proton in K is v = dx/dt = L/T =  0.999999991·c
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Do you know another definition of the speed of the proton in K
than dx/dt ?
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Let K'(x',τ) be an inertial frame of reference which at the instant I
is momentarily comoving with the proton.
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The speed of the point A in K' is v' = dx'/dτ = (L/γ)/τ = 0.999999991·c
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Do you know another definition of the speed of the point A in K'
than  dx'/dτ ?
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This is a good relativistic physics question.
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Have fun answering this question...
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I hope you have a lot of fun.
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Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.

 We are dealing on fr.sci.* with this idiot for thirty years, go figure!

And whatever you say - Poincare had enough wit
to understand how idiotic rejecting Euclid
would be, and he has written it clearly
enough for anyone able to read (even if not
clearly enough for you, poor stinker).