Math, asked by thelightning2492, 8 months ago

ਜੇਕਰ ਸਮੀਕਰਨਾਂ kx-y=2 ਅਤੇ 6x-2y=3 ਦਾ ਇੱਕ ਹੱਲ ਹੋਵੇ ਤਾਂ k ਦਾ ਮੁੱਲ ਪਤਾ ਕਰੋ

Answers

Answered by sivaprasadkotte
2

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 singularity is the set of points where a metric is undefined. It is a geometric property of a manifold that is not limited to black holes, and is not necessarily a single point.

A singularity can be just a coordinate singularity that disappears if you choose a different coordinate system for the manifold. For example, the Schwarzschild metric that describes non-rotating electrically neutral black holes has a coordinate singularity at the event horizon in Schwarzschild coordinates, but not in Kruskal–Szekeres

That singularity does not need to be a point either. For example, in the Kerr and Kerr–Newman metrics that describe rotating black holes (with angular momentum J ≠ 0) the singularity is a ring (a set of adjacent points/events).

Also, it is important to understand that singularities with black holes are not (a set of) points in space, but in space*time*: they are a set of *events*.

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.

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