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Schwarzschild as the zero-spin member of Kerr

PDF p. 1: §1.1, sentence spanning pp.1–2; PDF p. 2: §1.1, sentence spanning pp.1–2

Example

For $M > 0$ , the $a = 0$ member of the Kerr exterior is the Schwarzschild exterior of mass parameter $M$ . Its angular momentum is $J = 0$ and its radial domain is $r > 2 M$ . Here the Schwarzschild exterior is the tensor on $R_{t} \times (2 M, \infty) \times S^{2}$ with line element $- (1 - 2 M / r) d t^{2} + (1 - 2 M / r)^{- 1} d r^{2} + r^{2} (d θ^{2} + \sin^{2} θ d ϕ^{2})$ away from the spherical coordinate poles.

Example references

Kerr metric, parameters, exterior and extremality

Kerr exterior tensor including the extremal parameter range

Facts & Assumptions

[F1]

For $M > 0$ and $∣ a ∣ \leq M$ , the Kerr definition gives $J = M a$ , $r_{+} = M + \sqrt{M^{2} - a^{2}}$ , and the exterior $r > r_{+}$ , together with its displayed Boyer–Lindquist tensor. Kerr metric, parameters, exterior and extremality

[F2]

The Kerr exterior tensor is smooth on the sphere, and its $a = 0$ line element is $- (1 - 2 M / r) d t^{2} + (1 - 2 M / r)^{- 1} d r^{2} + r^{2} (d θ^{2} + \sin^{2} θ d ϕ^{2})$ . Kerr exterior tensor including the extremal parameter range

Proof

Given: $M > 0$ and the Kerr exterior with $a = 0$ .

Goal: Identify its domain, angular momentum, and tensor with the stated Schwarzschild exterior.

1.1givenF1

Since $∣ 0 ∣ \leq M$ , [F1] applies. The positive square root satisfies $\sqrt{M^{2}} = M$ , so $r_{+} = 2 M$ and the exterior domain is $r > 2 M$ . The angular momentum is $J = M \cdot 0 = 0$ .

2.1givenF2step 1.1∎

By [F2], setting $a = 0$ gives exactly the displayed Schwarzschild line element. Its smooth extension over the spherical coordinate poles is the same sphere tensor supplied by [F2]. Together with the domain from step 1.1, this identifies the $a = 0$ Kerr exterior with the stated Schwarzschild exterior.

Source notes

The article identifies the $a = 0$ member as Schwarzschild in §1.1. This is attribution only; the calculation above uses the completed geometric suppliers.

Depends on

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Sources

The uniqueness of smooth stationary black holes · §1.1, PDF pp. 1–2, sentence ending on p. 2