given a circle, what is a chord determined by 7 grain on its perimeter
Answers
Step-by-step explanation:
A singularity is NOT a (specific) place. It is a property of a function where the value of that function approaches infinity. For example, the function f: x ⟼ 1/x has a singularity at x = 0 because lim_{x → 0⁺} f(x) = +∞.
A spacetime singularity does not have to be a spacetime *curvature* singularity. For example, the Schwarzschild metric, the spacetime metric of a spherical mass distribution with total mass M, zero angular momentum, and zero electric charge, has a singularity at the Schwarzschild radius r = rₛ := 2 G M/c²:
ds² = ±(1 − rₛ/r) c²dt² ∓ 1/(1 − rₛ/r) dr² ∓ r² (dθ² + sin²θ dφ²).
Because r = rₛ ⇒ 1/(1 − rₛ/r) = 1/0 ⇒ lim_{r → rₛ} ds² = ∓∞.
However, this is NOT a spacetime *curvature* singularity because it can be avoided by using a different coordinate system. For example, the metric is in Kruskal–Szekeres coordinates:
ds² = 32G³M³/r exp(−r/(2 G M)) (±dT² ∓ dX²) ∓ r² (dθ² + sin²θ dφ²),
where c = 1. Now,
r = rₛ = 2 G M/c² ⇒ ds² = 16G²M² exp(−1) (±dT² ∓ dX²) ∓ 4G²M² (dθ² + sin²θ dφ²) ≠ ±∞,
and the spacetime singularity at r = rₛ disappears.
There is still a spacetime *curvature* singularity at r = 0 because 32G³M³/0 is not defined in these coordinates, and there are no known coordinates that can avoid that.
Finally, even a spacetime *curvature* singularity does NOT have to be point-like. The curvature singularity of the Kerr and Kerr–Newman metrics, for a black hole with non-zero angular momentum, is *ring*-shaped.