ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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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.

The isotropic Schwarzschild end has energy Mc squared

Example

Example references

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources