Re: Incorrect mathematical integration

Le 26/07/2024 à 22:20, Ross Finlayson a écrit :

On 07/25/2024 01:30 PM, Richard Hachel wrote:

You mean the distance _in_ the space _in_ the frame?

We must be careful about our understanding of relativistic things.
Physicists make things too simple.
They say D'=D.sqrt(1-Vo²/c²).
Then they rub their hands.
However, this is completely false, it all depends on where we stand and in which frame of reference.
I have already said a thousand times that D'=D.sqrt(1-Vo²/c²), applied hastily and haphazardly, is pure nonsense.
The true equation being D'=D.sqrt(1-Vo²/c²)/(1+cosµ.Vo/c)
So we start again:
We have a particle with a constant speed Vo=0.8c that goes from A to B.
In the lab frame of reference, AB is 3 meters.
Whether I place myself at A or B, it is logical that AB is three meters.
Except that I beg you to understand something important.
I am inertial with A and B when I measure AB.
Now let's place ourselves at the level of the proton for example.
What is the distance AB?
Physicists answer me, insulting me when possible, threatening me or hating me when they can: D'=3*0.6=1.8m.
Except that having said that, they have not said anything coherent at all, and they make me laugh, they who believe, because they have studied a adulterated SR,
that it is me who is making fun.
No, to say that is to say an abstract, incoherent sentence, and no more real than "I like round squares" or "I would like to drink dehydrated water", or "I prefer the color scarlet white".
It means NOTHING.
We come back to the proton, what is the distance AB for it?
Well, it all depends on its POSITION.
And this is what physicists have trouble understanding (I still have 40 SR, and it is logical that I am stronger than them).
When the proton passes through A, the distance AB is 9 meters.
When the proton passes through the center of AB (in the lab frame of reference) AB measures 5 m (0.5+4.5).
When the proton arrives at B, AB is 1 meter.
Space is a reference mollusk.
R.H.