Most people look to the world with a straight mathematical XYZ-geometry, as Nugatory mentioned. That is the reference for what we "see". The problem is that they think this is equal to space, so get confused when space is curved.
Returning to my first statement: I notice that space is curved...
Indeed I suppose that this is what I mean. I notice a curvature of (existing) space against what "I see"', which is probably a mathematical (non existing) straight XYZ "space". Perhaps this is not the curvature of GT calculations, but still results in a curvature in what I see.
Indeed then my understanding of spacetime curvature was wrong. So I suppose that if a space is curved extrinsic, an observers Euclidean space is curved in the same way, so he does see this curvature. Or perhaps better: the observers Euclidean space does not have (does not see) the dimension in...
Allright, I understand that because spacetime is an intrinsic curvature it does not need a reference. However basically both extrinsic and intrinsic curvature are curved against what an observer sees, its mathematical Euclidean space. This probably is not useful for GR theory and calculation...
I think PauloConstantino is looking for a more fundamental answer, about against which reference space is curved.
My answer would be: against an mathematical Euclidean space, which is absolute straight, connected to the masses causing the curvature.
Measuring is easy: sit on the mass (earth)...
You cannot "observe my friend's clocks" while they move. That is only possible when those clocks are at the same location. So suppose A and B return back to you. Then A will see you 10 years older and B will see you 18 years older. Indeed A and B will conclude afterwards that your clock was...
That is what you wrote. That math is there. Then you are not interested or not familiar with this math
There are three formula which together give a wrong result. Then one of the formula must wrong. Unless math also changes in GR...
If you don't know the answer, please wait for someone who...
The classic calculation of an orbit is based on: s = 0.5g.t2 and y = x2/2r (circle at x ≈ 0) which is : s = v2t2/2r
together: 0.5g.t2 = v2t2/2r, resulting in: g = v2/r
In classic the orbit formula is v2 = G.m/r with gives in above: g = G.m/r2
In GR the orbit formula is v2 = G.m/(r-rs) with...
That is right.
My formula was not correct indeed, but when I calculate it I get (and is until rs):
$$
g = \frac{Gm}{r^2 (1 - \frac{r_s}{r})}
$$
If that is not the case, then other formula does change.
So summing up: the formula outside a BH is (r - rs) = G.m/v2, which means that g = G.m/(r - rs)2. This applies to both photons and mass, although for mass r cannot be smaller then 6rs (v = c/2), because of reasons not explained here the orbit becomes unstable.
Indeed, although he would be please when informed that information in a BH is proportional with the surface of the sphere rs in stead of its content (but I know, still for the wrong reason)
But now about mass: is your formule the same (r - rs) = G.m/v2? Then in the smallest stable orbit v is...
If I transfer this to a more familiar form it becomes: (r - rs) = G.m/v2, so standard Newton, only he would think that in a BH all mass is concentrated in a shell with radius rs (instead of a singularity in the center).
For photons (v = c): r = G.m/c2 + rs = G.m/c2 + 2G.m/c2 = 3G.m/c2
Are...
No, I measure the diameters of the black holes with the telescope and subtract them. By definition of the GR formula I know that the radius is half that size. No reference needed.
I do the same with objects orbeting the BH on large distances where Newton can be applied.