Kepler's third law rules the Gravity
Hello all.
I would like to show something that I hope, you may like.
You will see, that Kepler was perfectly right, but in some unexpected way.
I provide references as links in text.
In whole post we assume that c=1 to simplify calculations.
We take 3rd Keppler law for some rotation in R distance where T is period. We may transform it to form of:
Let our constans for considered rotation will be equal to Schwarschild radius R_s, then:
Let us derive time dilation factor gamma for such move:
As you probably see, it is the same factor that is present in Schwarzschild metric, where for geodesics we have:
Let us differentiate our gamma factor by R. Surprise! - we obtain gravitational acceleration...
If you do not believe it is correct, look at the reference, formula (25), keeping in mind, that:
Looks interesting?
Take a look at this. Now, we derive rest of Schwarschild metric...
At first we recall and transform simply Newtonian formula for accelerated move for our case:
as it is easy to calculate if we differentiate above by "R" we obtain:
thus:
Now, we define observer resting in R distance to source of gravity (f.e. we on Earth).
His proper time formula we may denote as:
Thanks to above observer, Schwarzschild metric may be rewritten for "every particular observer" (without any bad looking "dt" in infinity). For geodesics Schwarzschild metric will be in form of:
Let us rewrite above as:
Now, we show, that difference between Schwarzschild and Minkowski is equal to t_g:
If we denote spatial increment as dx
Schwarzschild metric appears to be consequence of some Kepler rotation in Minkowski timespace...
If you are interested why it works this way, I explain it in my article draft and partly in my previous posts.
