2018
DOI: 10.4236/jmp.2018.910125
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Minkowski, Schwarzschild and Kerr Metrics Revisited

Abstract: In recent papers, a few physicists studying Black Hole perturbation theory in General Relativity have tried to construct the initial part of a differential sequence based on the Kerr metric, using methods similar to the ones they already used for studying the Schwarzschild geometry. Of course, such a differential sequence is well known for the Minkowski metric and successively contains the Killing (10, order 1), Riemann (20, order 2), Bianchi (20, order 1 again) operators in the linearized framework, as a part… Show more

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Cited by 10 publications
(33 citation statements)
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“…3 ) < dim(R 3 ) < dim(R 4 ) with 11 < 12 < 15.However, we have the general Theorem 2.A.7 in ( [20]) providing the useful prolongation/ projection (PP) procedure, namely that we have ρ r (R (1) q ) = R (1) q+r , ∀r ≥ 0 whenever the symbol g q of R q is 2-acyclic. In the present case, we have indeed ρ r (R…”
Section: Example 23 Revisitedmentioning
confidence: 95%
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“…3 ) < dim(R 3 ) < dim(R 4 ) with 11 < 12 < 15.However, we have the general Theorem 2.A.7 in ( [20]) providing the useful prolongation/ projection (PP) procedure, namely that we have ρ r (R (1) q ) = R (1) q+r , ∀r ≥ 0 whenever the symbol g q of R q is 2-acyclic. In the present case, we have indeed ρ r (R…”
Section: Example 23 Revisitedmentioning
confidence: 95%
“…However, in order to prove that d 2 ξ 0,3 − d 3 ξ 0,2 = 0 or equivalently that d 2 W 3 − d 3 W 2 = 0, the previous procedure cannot work but we must never forget that U, V 2 , V 3 , W 2 , W 3 both belong to j 2 (Ω). Introducing the formal Lie derivative R = L(ξ 1 )ρ, we recall that: we have proved in ( [20]) that the third order CC d 2 W 3 − d 3 W 2 = 0 is not a generating one because it is just a differential consequence of the second order CC R 01,23 = 0.…”
Section: Example 21 Revisitedmentioning
confidence: 99%
“…However, it can be shown that, on Kerr, there is no second order operator L with symbol σ p (ϑ 2 ) such that LK 0 = 0, and hence K 0 fails to be involutive. It was rightly noted in [29,30] that constructing a full compatibility operator for an involutive version of K 0 is much easier. However, we point out that our K 0 is tied to the fixed notion of gauge symmetry and gauge invariance in linearized gravity, hence we are not free to replace it with its involutive prolongation.…”
Section: Counting Invariantsmentioning
confidence: 99%
“…Nevertheless, this paper is the first to fully demonstrate completeness for a set of gauge invariants for the Kerr spacetime. See [29,30] for work on related problems.…”
Section: Introductionmentioning
confidence: 99%
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