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Coordinate components of the transformed metric

PDF p. 10: §2.3, first display

Statement

Adopt the stationary vacuum model and regular stationary vacuum spacetime data, including the original maximal exterior section and mass-one normalization. Let q∈[0,1], and suppose the Harrison one-form and metric are supplied on an open set V where Dq>0. Thus f=−g(T,T), Dq=∣q(1+σ)+(1−σ)∣2, dq=Dq(p) at a supplied zero p of T, and Wq(T)=dq−Dq.

In an original stationary coordinate chart (x0,x1,x2,x3) on V, meaning T=∂0 there, the transformed metric has components

(gq)00=−dq2f4Dq,(gq)0i=dq4(g0i−f(Wq)iDq),

(gq)ij=14(Dqgij+g0i(Wq)j+g0j(Wq)i−f(Wq)i(Wq)jDq),i,j∈{1,2,3}.

Facts & Assumptions

[A1]

The stationary vacuum model and regular spacetime data are assumed, together with the specified maximal exterior section and mass-one convention. The component calculation is conditional on the supplied chart and fields. Stationary vacuum model and asymptotic normalization Regular non-degenerate stationary vacuum spacetime Mass-one convention and Harrison denominator

[F1]

Stationary Lorentzian data supply the metric g, its Killing field T, the metric dual T♭=g(T,⋅), and f=−g(T,T). Stationary Lorentzian data and tensor conventions

[F2]

The Harrison normalization defines Dq for the supplied potential. Mass-one convention and Harrison denominator The Harrison one-form definition sets dq=Dq(p) at the supplied zero p of T. Stationary Harrison one-form and metric

[F3]

For the supplied one-form, Wq(T)=dq−Dq, and gq=14(Dqg+T♭⊗Wq+Wq⊗T♭−fDq−1Wq⊗Wq) on V. Stationary Harrison one-form and metric

[F4]

A smooth chart supplies the coordinate frame (∂0,∂1,∂2,∂3). We write gab=g(∂a,∂b), (gq)ab=gq(∂a,∂b), and αa=α(∂a) for a one-form α. Local frame fields, coordinate frames and tetrad fields

[F5]

For covectors α,β on the same tangent space, their tensor product satisfies (α⊗β)(u,v)=α(u)β(v). The tensor product of multilinear tensors

Proof

Given: The model, data, parameter, supplied Harrison one-form and metric, positivity domain, and stationary coordinate chart in the Statement.

Goal: Calculate the three displayed component formulas.

1.1givenF1F3F4

Since T=∂0, the coordinate components satisfy g00=g(T,T)=−f, (T♭)0=−f, (T♭)i=g0i, and (Wq)0=Wq(T)=dq−Dq. Positivity of Dq permits the displayed divisions.

2.1F3F4F5step 1.1algebra

Set w=dq−Dq. Evaluating the tensor formula on (∂0,∂0) gives 4(gq)00=−f(Dq+2w+w2/Dq)=−fdq2/Dq. Evaluating it on (∂0,∂i) gives 4(gq)0i=(Dq+w)g0i−f(1+w/Dq)(Wq)i=dq(g0i−f(Wq)i/Dq).

3.1F3F4F5step 1.1step 2.1algebra∎

Evaluating the same tensor formula on (∂i,∂j) gives the displayed spatial expression directly, because (T♭)i=g0i. These calculations prove all three formulas on the chart.

Source notes

The article citation attributes the component formulas. The proof is the local tensor expansion from the supplied Harrison definition; it does not use the article's assertion as a premise.

Depends on

Used by · 3

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Sources