Re: Incorrect mathematical integration

Le 24/07/2024 à 22:55, Maciej Wozniak a écrit :

W dniu 24.07.2024 o 22:45, Python pisze:

Le 24/07/2024 à 22:41, Maciej Wozniak a écrit :

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.

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

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And whatever you say - Poincare

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... would kick your asses: yours and Lengrand's.

 Could be.

No doubt about that. Glad to see you admit it. Is that because
you *know* you're posting bs?