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Asymptotic ends and ADM surface integrals — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

The isotropic Schwarzschild end has energy Mc squared

Physical scope

A fixed-chart calculation on the standard isotropic Schwarzschild slice data; no positive-mass theorem or empirical statement is asserted.

Example

Let $M > 0$ be a mass in kilograms and put $m = G_{N} M / c^{2}$ , a length. On the end $E = {x \in R^{3} : r = ∣ x ∣ > m / 2}$ , take the standard isotropic Schwarzschild slice data $u = 1 + \frac{m}{2 r}, h = u^{4} δ, K = 0.$ Here “time-symmetric” means precisely $K = 0$ . With the fixed Cartesian end chart and unit asymptotic time normalization $x^{0} = c t$ , the ADM energy is $M c^{2}$ joules.

Example references

A specified asymptotically Euclidean ADM end

ADM energy of a conformally flat end

Facts & Assumptions

Given: The stated isotropic Schwarzschild initial data with $M > 0$ , $m = G_{N} M / c^{2}$ , and the fixed-chart ADM convention.

[F1]

A fixed-chart asymptotically Euclidean end has $h_{i j} = δ_{i j} + a_{i j} / r + O_{2} (r^{- 1 - ϵ})$ , $K_{i j} = O_{1} (r^{- 2})$ , and the ADM energy is the limit of the specified surface integral multiplied by $c^{4} / (16 π G_{N})$ (A specified asymptotically Euclidean ADM end).

[F2]

If $h = u^{4} δ$ and $u = 1 + m / (2 r) + O_{2} (r^{- 2})$ with $u > 0$ , then the fixed-chart ADM limit exists and equals $c^{4} m / G_{N}$ (ADM energy of a conformally flat end).

[A1]

$M > 0$ , $m = G_{N} M / c^{2} > 0$ , $u > 0$ on $E$ , and the supplied initial data have $K = 0$ ; the metric form is the standard isotropic Schwarzschild slice representation recorded in the cited full texts.

Verification

Given: [F1], [F2], [A1], and the data in the Example.

1.1F1F2A1algebra

The remainder in $u = 1 + m / (2 r) + ρ$ is $ρ = 0$ , so $ρ = O_{2} (r^{- 2})$ . Also $u > 1$ on $E$ and $K = 0$ satisfies $K = O_{1} (r^{- 2})$ . Hence the end meets the hypotheses of [F1] and [F2].

2.1F2step 1.1algebra∎

By [F2], $E_{A D M} = c^{4} m / G_{N}$ . Substituting $m = G_{N} M / c^{2}$ gives $E_{A D M} = c^{4} (G_{N} M / c^{2}) / G_{N} = M c^{2}$ . The result is the exact model calculation for this fixed end and time normalization.

Remarks

The source texts use geometric units in the surface formula; the factor $c^{4} / G_{N}$ restores SI energy units. This calculation does not prove a positivity theorem.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

A compact interior change preserves the ADM limit but changes scalar curvature

Physical scope

A smooth Riemannian metric example on a fixed asymptotic end; no Einstein constraint, completeness, or physical-solution claim is made.

Example

Fix a length $R_{0} > 0$ , set $E = {x \in R^{3} : ∣ x ∣ > R_{0}}$ , and let $x_{*} = (3 R_{0}, 0, 0)$ . Define $ϕ (x) = {\begin{cases} \exp ⁣ (- \frac{1}{1 - ∣ x - x_{*} ∣^{2} / R_{0}^{2}}), & ∣ x - x_{*} ∣ < R_{0}, \\ 0, & ∣ x - x_{*} ∣ \geq R_{0} . \end{cases}$ For $0 < ε < 1$ , set $u_{0} = 1$ , $u_{ε} = 1 + ε ϕ$ , $h_{0} = u_{0}^{4} δ$ , $h_{ε} = u_{ε}^{4} δ$ , and $K_{0} = K_{ε} = 0$ on $E$ . Both data sets have the same fixed-chart ADM energy, $0$ , but their scalar curvatures differ at $x_{*}$ .

Example references

A specified asymptotically Euclidean ADM end

ADM energy of a conformally flat end

Compact interior changes leave a fixed ADM integral unchanged

Scalar curvature of a conformally Euclidean three-metric

Facts & Assumptions

Given: The Euclidean end $E$ , the smooth bump candidate $ϕ$ , $0 < ε < 1$ , and the conformal metrics and tensors in the Example.

[F1]

The fixed-chart ADM energy depends only on $h$ and equals the prescribed limit of its Cartesian surface integrand; $K = 0$ satisfies the required decay (A specified asymptotically Euclidean ADM end).

[F2]

For a positive smooth conformal factor $u$ , $S_{u^{4} δ} = - 8 u^{- 5} Δ_{δ} u$ (Scalar curvature of a conformally Euclidean three-metric).

[F3]

A positive conformal factor $u = 1 + m / (2 r) + O_{2} (r^{- 2})$ has $E_{A D M} = c^{4} m / G_{N}$ (ADM energy of a conformally flat end).

[F4]

If two metrics agree outside a compact subset of the same fixed end chart and both limits exist, their ADM energies are equal (Compact interior changes leave a fixed ADM integral unchanged).

[A1]

$R_{0} > 0$ , $x_{*} = (3 R_{0}, 0, 0)$ , and $0 < ε < 1$ ; all coordinates are Euclidean length coordinates.

Verification

Given: [F1]–[F4] and [A1].

1.1A1constructalgebra

Put $y = x - x_{*}$ and $q = 1 - ∣ y ∣^{2} / R_{0}^{2}$ . For $q > 0$ , each derivative of $e^{- 1 / q}$ is a polynomial in $q^{- 1}$ times $e^{- 1 / q}$ , and $q^{- N} e^{- 1 / q} \to 0$ as $q ↓ 0$ for every integer $N \geq 0$ . Therefore extension by zero at $q \leq 0$ is $C^{\infty}$ , so $ϕ$ is a smooth bump supported in $\overline{B_{R_{0}} (x_{*})}$ . Since $∣ x_{*} ∣ = 3 R_{0}$ , this compact support lies in ${∣ x ∣ \geq 2 R_{0}} \subset E$ . Also $0 \leq ϕ \leq e^{- 1}$ , hence both $u_{0}$ and $u_{ε}$ are positive.

2.1F1F3F4step 1.1algebra

The metrics $h_{0}$ and $h_{ε}$ agree outside the compact support of $ϕ$ , and both have $K = 0$ . The conformal end result [F3] with $m = 0$ gives existence of the ADM limit for $h_{0}$ and $E_{A D M} (h_{0}) = 0$ . For $r > 4 R_{0}$ , $ϕ = 0$ on an open neighborhood, so $h_{ε} = h_{0} = δ$ there and their first Cartesian derivatives agree. Thus their surface integrals in [F1] agree for every $R > 4 R_{0}$ , which proves existence of the $h_{ε}$ limit. Both metrics satisfy the end decay because they equal $δ$ there. Now [F4] applies and gives $E_{A D M} (h_{ε}) = E_{A D M} (h_{0}) = 0$ .

3.1F2step 1.1algebra∎

At $x_{*}$ , $q = 1$ , $ϕ (x_{*}) = e^{- 1}$ , $\partial_{i} q = 0$ , and $\partial_{i} \partial_{j} q = - 2 δ_{i j} / R_{0}^{2}$ . Differentiating $ϕ = e^{- 1 / q}$ at this point gives $\partial_{i} ϕ (x_{*}) = 0$ and $\partial_{i} \partial_{j} ϕ (x_{*}) = - 2 e^{- 1} δ_{i j} / R_{0}^{2}$ , so $Δ_{δ} ϕ (x_{*}) = - 6 e^{- 1} / R_{0}^{2}$ . The flat metric $h_{0} = δ$ has scalar curvature $0$ , while [F2] gives $S_{h_{ε}} (x_{*}) = - 8 (1 + ε e^{- 1})^{- 5} ε Δ_{δ} ϕ (x_{*}) = \frac{48 ε e^{- 1}}{R_{0}^{2} (1 + ε e^{- 1})^{5}} > 0.$ Thus the scalar curvatures differ at $x_{*}$ although the ADM limits agree.

Remarks

The example changes a metric on an exterior manifold with an inner boundary and does not impose the Einstein constraint equations or completeness. It is not a counterexample to a positive-mass theorem with its own hypotheses.

Asymptotic ends and ADM surface integrals — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

The isotropic Schwarzschild end has energy Mc squared

Example

Example references

Facts & Assumptions

Verification

Remarks

A compact interior change preserves the ADM limit but changes scalar curvature

Example

Example references

Facts & Assumptions

Verification

Remarks

Sources