PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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ADM energy of a conformally flat end

Physical scope

A fixed-chart calculation for a conformally flat end; no constraint equation, positivity result, or empirical claim is inferred.

Statement

On a fixed-chart end $E = {x \in R^{3} : r = ∣ x ∣ > R_{0}}$ with the ADM convention stated below, let $u \in C^{\infty} (E, (0, \infty))$ satisfy $u = 1 + m / (2 r) + ρ$ , where $m \in R$ has units of metres and $ρ = O_{2} (r^{- 2})$ . Set $h = u^{4} δ$ and choose any $C^{1}$ symmetric tensor $K = O_{1} (r^{- 2})$ . Then the ADM limit exists in the specified chart and $E_{A D M} (h; x) = \frac{c^{4} m}{G_{N}} .$

Statement references

A specified asymptotically Euclidean ADM end

Facts & Assumptions

Given: The fixed Cartesian end and ADM convention of A specified asymptotically Euclidean ADM end, a smooth positive $u = 1 + m / (2 r) + ρ$ with $ρ = O_{2} (r^{- 2})$ , and $h = u^{4} δ$ .

[F1]

On the fixed chart, $E_{A D M}$ is $c^{4} / (16 π G_{N})$ times the limit of $\int_{S_{R}} (\partial_{j} h_{i j} - \partial_{i} h_{j j}) n^{i} d S$ , where $d S$ is induced by $δ ∣_{S_{R}}$ and a local parametrization $X$ has density $\sqrt{\det [δ (\partial_{y^{a}} X, \partial_{y^{b}} X)]_{a, b = 1}^{2}} d y^{1} d y^{2}$ ; $O_{2} (r^{- q})$ controls Cartesian derivatives of order $k \leq 2$ by $O (r^{- q - k})$ uniformly in angle; $c$ and $G_{N}$ have the SI units stated in the definition (A specified asymptotically Euclidean ADM end).

[A1]

$m$ is a real length parameter, $u$ is strictly positive on $E$ , and the remainder bounds are uniform in angle, as required by the Statement.

Proof

Given: [F1], [A1], and the conformally flat metric in the Statement.

1.1F1A1algebra

Set $w : = m / (2 r) + ρ$ . Derivatives of $r^{- 1}$ satisfy $\partial^{α} (r^{- 1}) = O (r^{- 1 - ∣ α ∣})$ for $∣ α ∣ \leq 2$ , since each derivative is homogeneous of degree $- 1 - ∣ α ∣$ and bounded on the unit sphere. Together with $ρ = O_{2} (r^{- 2})$ , this gives $w = O_{2} (r^{- 1})$ ; the product rule gives $w^{2}, w^{3}, w^{4} = O_{2} (r^{- 2})$ . Also $4 w = 2 m / r + O_{2} (r^{- 2})$ , so the binomial expansion $u^{4} = (1 + w)^{4} = 1 + 4 w + 6 w^{2} + 4 w^{3} + w^{4}$ yields $h_{i j} = δ_{i j} + (2 m δ_{i j}) / r + O_{2} (r^{- 2})$ . Since $u > 0$ , $h$ is positive definite and $C^{2}$ , and the supplied symmetric $K$ is $C^{1}$ with $O_{1} (r^{- 2})$ decay. Thus $(h, K)$ satisfies the definition of a fixed-chart end with $a_{i j} = 2 m δ_{i j}$ and $ϵ = 1$ . Now $h_{i j} = u^{4} δ_{i j}$ gives $\partial_{j} h_{i j} = \partial_{i} (u^{4})$ and $h_{j j} = 3 u^{4}$ , hence $\partial_{j} h_{i j} - \partial_{i} h_{j j} = - 2 \partial_{i} (u^{4}) = - 8 u^{3} \partial_{i} u$ . Contracting with $n^{i} = x^{i} / R$ on $S_{R}$ replaces $\partial_{i} u$ by $\partial_{r} u$ .

2.1F1A1step 1.1algebra

The radial derivative bounds in [F1] and $ρ = O_{2} (r^{- 2})$ give $u ∣_{S_{R}} = 1 + m / (2 R) + O (R^{- 2})$ and $\partial_{r} u ∣_{S_{R}} = - m / (2 R^{2}) + O (R^{- 3})$ , uniformly in angle. Thus $u^{3} = 1 + O (R^{- 1})$ , so step 1.1's contracted integrand is $4 m / R^{2} + O (R^{- 3})$ , uniformly on $S_{R}$ .

3.1F1step 2.1algebra

Parameterize the complement of the poles and one meridian in $S_{R}$ by $X_{R} (θ, φ) = R (\sin θ \cos φ, \sin θ \sin φ, \cos θ)$ , with coordinates $y^{1} = θ \in (0, π)$ and $y^{2} = φ \in (0, 2 π)$ . Its coordinate tangent vectors are $\partial_{θ} X_{R} = R (\cos θ \cos φ, \cos θ \sin φ, - \sin θ)$ and $\partial_{φ} X_{R} = R (- \sin θ \sin φ, \sin θ \cos φ, 0)$ , so the induced-metric Gram matrix is $[δ (\partial_{y^{a}} X_{R}, \partial_{y^{b}} X_{R})]_{a, b = 1}^{2} = (\begin{array}{cc} R^{2} & 0 \\ 0 & R^{2} \sin^{2} θ \end{array})$ , with determinant $R^{4} \sin^{2} θ$ . The area-measure definition in [F1] therefore gives $d S = R^{2} \sin θ d θ d φ$ . The omitted poles and meridian have area zero, so integrating the uniform expansion in step 2.1 over this chart yields $\int_{S_{R}} (\partial_{j} h_{i j} - \partial_{i} h_{j j}) n^{i} d S = 16 π m + O (R^{- 1})$ , since $\int_{0}^{2 π} \int_{0}^{π} \sin θ d θ d φ = 4 π$ . This proves the limit exists.

4.1F1step 3.1algebra∎

Multiplying the limit in step 3.1 by $c^{4} / (16 π G_{N})$ from [F1] gives $E_{A D M} = c^{4} m / G_{N}$ . The dimensions are $(c^{4} / G_{N}) \times m = J$ , as specified in [F1].

Remarks

The proof uses the fixed chart and the exact stated falloff. It proves neither coordinate invariance for broader asymptotic classes nor a positivity theorem.

Depends on

A specified asymptotically Euclidean ADM end