Re: Incorrect mathematical integration

Den 26.07.2024 22:36, skrev Richard Hachel:

Le 26/07/2024 à 21:54, "Paul.B.Andersen" a écrit : >>>> Den 25.07.2024 21:50, skrev Richard Hachel:

>
The proton only goes around once, and the time it takes, measured by the laboratory clock (which is actually TWO clocks A and B combined into one) is T = 90.0623 μs.

Right.
So the proton crossed the distance L = 27 km in the laboratory
reference frame in T = 90.0623 μs, and the speed in
the laboratory reference frame is v = L/T = 0.999999991·c

But if I measure with the watch that the proton wears on his left wrist, I will measure a time of τ = 12.0727 ns.

 Right again.
This is a trivial fact, disputed by no one.

This is experimentally confirmed in the real world.
(With satellites and aeroplanes.)

 It is not a question of discussing what is of rare evidence in both theories.
We must remain simple.
We have here two theories, and it is infinitely probable that one of the two is correct.
The problem is that on certain concepts, we do not agree. It is therefore certain that if there is an error, the error is there.
It cannot be found where we agree as here, or when each one poses:
To=(x/c).sqrt(1+2c²/ax) to determine the observable time of an accelerated object.

What are you talking about? :-D

>
Thus, for the proton, the distance AB (in the laboratory reference frame) was crossed 7460 times faster

!!!!!! :-D

You say that the proton crossed the distance L = 27 km in
_the laboratory reference frame_ in τ = 12.0727 ns, thus is
the speed in the laboratory reference frame v = L/τ ≈ 7460·c
 How do you think that the proton can have two different speeds
in the laboratory frame?
 Of course it can't in the real world.

We must say simple things, and we must say true things.
It is very difficult in special relativity because of the frequent conceptual errors. Sometimes when I read certain things here or elsewhere, I have the impression that everything is sinking into horror.
We must be careful about the confusion of words.
You say, a body can only have one speed, and you seem to think that I am an idiot.
But no, I am not an idiot, and it is precisely because of morons like Python that I can pass for an idiot.

Do you think that I am so stupid to say that a moving body can have two different speeds at the same time?

I did think so, but now you have corrected me.
So you know that the proton can have only one speed in the lab frame.

When I say that a body can have, in the same frame of reference, many different speeds, that is OBVIOUSLY not what I am talking about.
Let us assume a speed Vo=0.8c.
It is quite obvious that I cannot have at the same time, at the risk of being absurd, I who claim to describe the most beautiful, the simplest and the most logical theory, a life that is Vo=0.8/c, Vo=0.9c, Vo=0.5c and Vo=0.999c.
It would be absurd, and it would be dishonest to make me say what I did not say.
Now, I can still write Vo=0.8c, Vr=(4/3)c, Vapp'=0.4444c and Vapp"=4c.
You just have to understand what I write, why I write it, and validate it without spitting on it.

Why did you use so many words to say that you agree:
"The proton can only have one speed in the lab frame."
You are right of course, because:
It is _one_ proton going once around the circuit.
The speed of that _one_ proton in the lab frame is
obviously v = dx/dt = L/T = 0.999999991·c.
Nothing else!
That the clock on the _one_ proton measures the proper time
τ = T/γ = 12.0727 ns doesn't change the fact that
the speed of the _one_ proton in the lab frame is
v = L/T = 0.999999991·c.
But it means that the speed of the lab in the proton frame is:
v = dx'/dτ = (L/γ)/τ = L/T = 0.999999991·c.

 Only an ignoramus like you will fail to understand that if
the speed of the proton in the lab frame is 0.999999991·c,
then the speed of the lab in the proton frame is 0.999999991·c.

 C'est ce que je dis.

No, that's not what you are saying.
You are saying:
"Thus, for the proton, the distance AB (in the laboratory
  reference frame) was crossed 7460 times faster."
And since I now know that you agree that the proton can't
have but one speed in the laboratory frame, I have to
interpret your statement a bit differently:
Since you say "for the proton" you must mean "in the proton frame",
and in this frame the proton doesn't move, so it must be
the distance AB which has crossed the proton at the speed 7460·c.
So you are saying:
"The speed of the proton in the lab frame is 0.999999991·c,
  but the speed of the lab in the proton frame is 7460·c.
--
Paul
https://paulba.no/