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Le 24/07/2024 à 14:47, "Paul.B.Andersen" a écrit :A bit more precisely put:
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.
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:Nothing is moving at the speed L/τ = 235.7c
Vr=Vo/sqrt(1-Vo²/c)=235.7c
Likewise, this real speed is reciprocal.The reciprocal of L/τ is (L/γ)/T = 0.0001340c
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
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.
This is a good relativistic physics question.Quite.
Have fun answering this question...
I hope you have a lot of fun.
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