Re: Space-time interval...

Le 13/08/2024 à 13:42, Python a écrit :

Le 13/08/2024 à 13:38, Mikko a écrit :

 Until very very recently (you can check on fr.sci.physique) he firmly
believed that ds^2 is always zero, go figure!

That's not what I actually said.
I was talking about an event occurring in a frame of reference
and whose information reached any observer present in this frame of reference.
For example, a terrestrial observer who observes the explosion of a supernova.
If the explosion took place 15,000 years ago, the observer will note (dl,dt)=(15,000,-15,000)
and therefore ds²=0
I can go further, and ask for a more precise notation than dl.
I then set E=(x,y,z,To,t) in Hachel notation that I do not explain,
because those who read me are intelligent enough to decode me easily without me giving them the bottle.
So, for example, we have E=(12000, 9000, 0, -15000.0)
I said that the space-time interval will be noted ds²=0 for all the joint observers who will cross the solar system at this precise moment, whatever their speed and direction.
This is very obvious, and it even becomes ridiculous to talk about it too much.
Let's take a single case: a rocket passes on the Earth's Ox axis, at two hundred and forty thousand km/s (Vo=0.8c).
A well-understood Lorentz transformation immediately gives me.
E'=(40000, 9000, 0, -41000, 0)
We see that here again ds²=0
This is trivial.
Although it allows me to emphasize a remark that I have often made: If one or more observers are conjoined,
whatever their relativistic speeds or their direction, they all observe the same present universe.
Very deformed in x (Poincaré-Lorentz transformation)
x'=(x+Vo.To)/sqrt(1-Vo²/c²)
but always with t"=t'=t=0.
Do you understand these things better than the buffoon Python, who says anything and everything.
R.H.