Re: Challenge for Paul; Probe that with Mercury ds^2>0 and the solution is spacelike

Le 12/02/2025 à 23:27, hertz778@gmail.com (rhertz) a écrit :

To get you out of boredom, I have this problem:
 Given the metric ds^2 = (ct)^2-x^2+-y^2-z^2 > 0 for Mercury's perihelion
analysis, prove that S gives the additional 43"/cy.
 Also explain why this problem is treated as spacelike and not as
timelike.
  Enjoy!

The advance of the perihelion of Mercury can probably be given by simple RR.
I think I did it a long time ago, and found an exact result.
We must first consider one thing, there is a contraction of the circumference as a function of the speed, and therefore of the orbit of Mercury.
The effects of the contraction are not enormous, but over a century, we can easily observe them.
Just as there is a contraction of the circumference (the orbit), there is an equal contraction of the radius (pi is an invariant in Hachel).
I think that if someone were to stick to it, they could easily find the exact advance of the perihelion of Mercury (Personally, I get tired).
I give the equations relating to transformations in rotating media here.
<http://nemoweb.net/jntp?bSU98M6KIUBzLsrpIgbePkDLQD8@jntp/Data.Media:1>
C'=C.sqrt(1-v²/c²)   Circonférence
R'=R.sqrt(1-v²/c²)    Rayon
R.H.