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From: "Paul.B.Andersen" <relativity@paulba.no>
Newsgroups: sci.physics.relativity
Subject: Re: Incorrect mathematical integration
Date: Wed, 24 Jul 2024 20:45:02 +0200
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Den 24.07.2024 15:08, skrev Richard Hachel:
> Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :
> 
>> 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
>> 
>> γ = 7460
>> 
>> 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
>> 
>> This should be blazingly obvious for anybody but complete morons:
>> 
>> 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.

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.

> 
> 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

Nothing is moving at the speed L/τ = 235.7c

> 
> Likewise, this real speed is reciprocal.

The reciprocal of L/τ is (L/γ)/T = 0.0001340c

Equally meaningless. Not the speed of anything.

> 
> In the proton frame, it is point A which passes near it at Vr=235.7c.

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

> 
> 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 ?

> This is a good relativistic physics question.
> 
> Have fun answering this question...
> 
> I hope you have a lot of fun.
> 

Quite.
But your jokes aren't funny the umpteenth time they are told,
It is getting boring.

-- 
Paul

https://paulba.no/