History, asked by nbrintha2019, 11 months ago

III. True / False
1. Which of the following is not true?
a. Hyder Ali and Tipu Sultan were against British EIC.
b. Hyder Ali and Tipu Sultan were helped by the French in modernizing the army.
c. Hyder Ali and Tipu Sultan allowed EIC to trade through the Malabar Coast.​

Answers

Answered by sivaprasadkotte
0

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.

Answered by kriselledsouza25
0

Mark me as the Brainliest I will be thankful to you

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.

Similar questions