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Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: Python <python@invalid.org>
Newsgroups: sci.physics.relativity
Subject: Re: Incorrect mathematical integration
Date: Wed, 24 Jul 2024 23:01:21 +0200
Organization: CCCP
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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 :
>>>>>>
>>>>>>> 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.
>>>>
>>>> We are dealing on fr.sci.* with this idiot for thirty years, go figure!
>>>
>>>
>>> And whatever you say - Poincare 
>>
>> ... 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?