LemmaStatement: AI-adaptedProof: AI-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart

Statement

Let $(M, g, N)$ be the Schwarzschild exterior datum with Schwarzschild radius $s > 0$ and $f (r) = 1 - s / r$ (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation). Let $M_{e x t} : = R \times (0, \infty) \times S^{2}$ carry the product smooth structure and let $ι_{e x t} : M_{e x t} \to S^{2}$ be the third projection. Its product atlas includes ordinary sphere charts at the poles (Products of smooth manifolds have a canonical product smooth structure). Define the tortoise function

$r_{*} (r) : = r + s \log ∣ \frac{r}{s} - 1 ∣, r > 0, r \neq s,$

on each of the two intervals $(0, s)$ and $(s, \infty)$ , and define the advanced-chart metric globally on $M_{e x t}$ by

$g_{e x t} = - f (r) d v \otimes d v + d v \otimes d r + d r \otimes d v + r^{2} ι_{e x t}^{*} γ_{S^{2}} .$

In spherical-polar product charts its components are

$g_{e x t} = - f (r) d v^{2} + 2 d v d r + r^{2} d Ω^{2},$

namely $g_{v v} = - f$ , $g_{v r} = g_{r v} = 1$ , $g_{ϑ ϑ} = r^{2}$ , $g_{φ φ} = r^{2} \sin^{2} ϑ$ and all other components zero. In another sphere chart the angular block is $r^{2} (γ_{i j})$ . Then:

(a) Smooth Lorentzian metric. $g_{e x t}$ is a smooth symmetric covariant $2$ -tensor field on $M_{e x t}$ , and at every point it is nondegenerate with inertia $(3, 1, 0)$ ; hence $(M_{e x t}, g_{e x t})$ is a smooth Lorentzian four-manifold of signature $(-, +, +, +)$ on all of $r > 0$ , including $r = s$ .

(b) Time orientation. The smooth vector field

$N_{e x t} : = A (r) \partial_{v} - \partial_{r}, A (r) : = \frac{f (r) + \sqrt{f (r)^{2} + 4}}{2} > 0,$

satisfies $g_{e x t} (N_{e x t}, N_{e x t}) < 0$ everywhere and $g_{e x t} (N_{e x t}, \partial_{v}) < 0$ everywhere; hence $N_{e x t}$ is a time orientation of $g_{e x t}$ , and on the region $r > s$ it determines the same future cone as the datum's time orientation $\partial_{v} = \partial_{0}$ .

(c) Isometric embedding of the exterior. The map $Φ : M \to M_{e x t}$ , $(x^{0}, r, ω) \mapsto (x^{0} + r_{*} (r), r, ω)$ restricts to a diffeomorphism of $M$ onto the open subset ${r > s}$ of $M_{e x t}$ , and $Φ^{*} g_{e x t} = g$ ; the exterior datum is therefore isometric to the region $r > s$ of the extension.

(d) Vacuum and Kretschmann scalar of the extension. The metric-selected connection of $g_{e x t}$ exists and is unique, its Ricci tensor, scalar curvature and Einstein tensor vanish identically on $M_{e x t}$ , and its Kretschmann scalar equals

$R_{a b c d} R^{a b c d} = \frac{12 s^{2}}{r^{6}}$

on all of $M_{e x t}$ . In particular the scalar is smooth and finite on the horizon $r = s$ , where it takes the value $12 / s^{4}$ .

(e) The horizon is a null hypersurface. The level set ${r = s}$ is a smooth null hypersurface of $(M_{e x t}, g_{e x t})$ whose null generator direction is $\partial_{v}$ ; every smooth product-chart frame remains a basis across $r = s$ . In spherical-polar charts $\det (g_{e x t}) = - r^{4} \sin^{2} ϑ$ ; in an arbitrary sphere chart the determinant is $- r^{4} \det (γ_{i j})$ , which is nonzero.

(f) Scope. The extension constructed here is a smooth Lorentzian manifold containing the exterior isometrically and regular across $r = s$ ; it is not claimed to be maximal, and no uniqueness or global structure theorem is asserted. The region $r \leq 0$ and the surfaces relevant to the Kruskal model of this page are outside this statement.

Facts & Assumptions

Given: The Schwarzschild exterior datum with $s > 0$ , $f = 1 - s / r$ ; the product manifold $M_{e x t} = R \times (0, \infty) \times S^{2}$ ; the function $r_{*}$ of the statement; the assigned components of $g_{e x t}$ ; and the maps $Φ$ and $N_{e x t}$ .

[L1]

On each of $(0, s)$ and $(s, \infty)$ the positive function $∣ r / s - 1 ∣$ is smooth. The natural logarithm is smooth on $(0, \infty)$ because it is the inverse of the smooth exponential with nonzero derivative, and its derivative is $1 / x$ ; hence $r_{*} (r) = r + s \log ∣ r / s - 1 ∣$ is smooth on each branch and $r_{*}^{'} = 1 + s / (r - s) = r / (r - s) = 1 / f (r)$ there (The natural logarithm as the inverse of the exponential function, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The real exponential function and the number $e$ by a power series, The exponential function is smooth and $(\exp)^{'} = \exp$ , Choice-free smooth inverse function theorem in Euclidean space, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Rational powers $a^{r}$ of a positive base). The function $f$ is smooth on $(0, \infty)$ and is positive exactly for $r > s$ .

[L2]

Globally $g = - f d (x^{0}) \otimes d (x^{0}) + f^{- 1} d r \otimes d r + r^{2} ι^{*} γ_{S^{2}}$ ; in a spherical-polar product chart its angular components are $γ_{22} = 1$ , $γ_{33} = \sin^{2} ϑ$ . The map $Φ$ is the identity on the sphere factor, so its pullback preserves $r^{2} γ_{S^{2}}$ (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation clauses (b)-(c), The Schwarzschild exterior datum is a smooth time-oriented Lorentzian four-manifold, The round metric on the sphere as an induced metric).

[L3]

The base block $(\begin{array}{cc} - f & 1 \\ 1 & 0 \end{array})$ has determinant $- 1$ , inverse $(\begin{array}{cc} 0 & 1 \\ 1 & f \end{array})$ , and one positive and one negative eigenvalue; the sphere block is $r^{2} γ_{S^{2}}$ , positive definite at every sphere point. Their orthogonal product therefore has inertia $(3, 1, 0)$ everywhere. In a polar chart the full determinant is $- r^{4} \sin^{2} ϑ$ , while in any angular chart it is $- r^{4} \det (γ_{i j})$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, The round metric on the sphere as an induced metric).

[L4]

A smooth vector field $N$ with $g (N, N) < 0$ everywhere is a time orientation; if two timelike fields $N, N^{'}$ satisfy $g (N, N^{'}) < 0$ at every point of a connected set, they determine the same future cone there (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model clause (c), Two time orientations agree or are opposite on each connected set).

[L5]

The global tensor $g_{e x t} = - f d v \otimes d v + d v \otimes d r + d r \otimes d v + r^{2} ι_{e x t}^{*} γ_{S^{2}}$ is smooth because all scalar functions, one-forms and the round metric are smooth; thus the product chart atlas, including pole charts, gives continuous Ricci and curvature-scalar components (Smooth covariant 2-tensor fields on a smooth manifold, Chart independence and the transformation law of a smooth covariant 2-tensor field, $C^{r}$ and smooth maps between smooth manifolds, The round metric on the sphere as an induced metric).

[L6]

If two continuous maps from a topological space into a Hausdorff space agree on a dense subset, they agree everywhere (Two continuous maps into a Hausdorff space that agree on a dense subset are equal).

[L7]

The Ricci tensor, scalar curvature and Einstein tensor of the metric $g_{f} = - f d (x^{0})^{2} + f^{- 1} d r^{2} + r^{2} d Ω^{2}$ with $f = 1 - s / r$ vanish identically on $r > s$ ; the computation of (a)-(c) of Curvature components and the vanishing of the Ricci and Einstein tensors of the Schwarzschild exterior metric is algebraic in $f, f^{'}, f^{''}$ and the identities $f^{''} / 2 + f^{'} / r = 0$ , $1 - f - r f^{'} = 0$ hold for $f = 1 - s / r$ at every $r \neq 0$ (Curvature components and the vanishing of the Ricci and Einstein tensors of the Schwarzschild exterior metric, Christoffel symbols of the Schwarzschild exterior metric in product charts).

[L8]

The Kretschmann scalar of the exterior metric is $12 s^{2} / r^{6}$ on $r > s$ (Orthonormal-frame curvature components and the Kretschmann scalar 12s^2/r^6 of the Schwarzschild exterior metric); a fully contracted curvature scalar of a smooth nondegenerate metric is a smooth function in every chart.

Proof

technique · verify smoothness and signature from the explicit component matrix, verify the pullback identity by the chain rule, obtain the vacuum property on $r\ne s$ algebraically and extend by continuity, and verify the horizon and orientation claims directly

1.1L3L5L1algebra

(Smoothness and signature.) The global tensor expression in [L5] is smooth on the product manifold. At any point its base block on the $(v, r)$ factor is $(\begin{array}{cc} - f & 1 \\ 1 & 0 \end{array})$ , with determinant $- 1$ and inertia $(1, 1, 0)$ , while its angular block is $r^{2} γ_{S^{2}}$ , positive definite on the entire sphere including the poles. The product splitting therefore gives a smooth nondegenerate metric of inertia $(3, 1, 0)$ everywhere. In a spherical-polar chart the determinant is $- r^{4} \sin^{2} ϑ$ ; in an arbitrary sphere chart it is $- r^{4} \det (γ_{i j}) \neq 0$ . This proves (a) and the determinant claims of (e).

1.2L1L3L4algebra

(Time orientation.) For $N_{e x t} = A \partial_{v} - \partial_{r}$ with $A = (f + \sqrt{f^{2} + 4}) / 2$ , one has $g_{e x t} (N_{e x t}, N_{e x t}) = - f A^{2} - 2 A = - A (f A + 2)$ and $g_{e x t} (N_{e x t}, \partial_{v}) = - f A - 1$ . By [L1], $A > 0$ because $\sqrt{f^{2} + 4} > ∣ f ∣ \geq - f$ . Next $f A + 1 = \frac{f^{2} + f \sqrt{f^{2} + 4} + 2}{2} > 0$ : for $f \geq 0$ its numerator is at least $2$ ; for $f < 0$ , writing $u = - f > 0$ , the inequality $u \sqrt{u^{2} + 4} < u^{2} + 2$ follows by squaring the two positive sides, $u^{2} (u^{2} + 4) < (u^{2} + 2)^{2}$ , so the numerator is positive again. Hence $f A + 2 > f A + 1 > 0$ , and therefore $g_{e x t} (N_{e x t}, N_{e x t}) = - A (f A + 2) < 0$ and $g_{e x t} (N_{e x t}, \partial_{v}) = - (f A + 1) < 0$ . Thus $N_{e x t}$ is a smooth nowhere-zero timelike field; by [L4] it is a time orientation of $g_{e x t}$ , and on the region $r > s$ , where $f > 0$ and $\partial_{v} = \partial_{0}$ , the negative pairing $g_{e x t} (N_{e x t}, \partial_{v}) < 0$ shows by [L4] that it determines the same future cone as the datum's orientation. This proves (b).

1.3L1L2algebra

(Pullback identity.) The map $Φ$ and its inverse $(v, r, ω) \mapsto (v - r_{*} (r), r, ω)$ are smooth on $r > s$ by [L1], so $Φ$ is a diffeomorphism onto that open subset. Pull back the base terms using $d v = d (x^{0}) + f^{- 1} d r$ : the cross terms cancel and the $d r^{2}$ coefficient becomes $- f (f^{- 1})^{2} + 2 f^{- 1} = f^{- 1}$ . The map is the identity on the sphere factor, so it pulls $r^{2} ι_{e x t}^{*} γ_{S^{2}}$ back to $r^{2} ι^{*} γ_{S^{2}}$ . Thus $Φ^{*} g_{e x t} = - f d (x^{0}) \otimes d (x^{0}) + f^{- 1} d r \otimes d r + r^{2} ι^{*} γ_{S^{2}} = g$ by [L2], proving (c).

2.1L6L7step 1.1step 1.3algebra

(Vacuum property on $r \neq s$ .) On $r > s$ the pullback identity of step 1.3 expresses $g_{e x t}$ as the datum metric of [L7], which is Ricci-flat there. On $0 < r < s$ the same coordinate calculation applies on the other branch of $r_{*}$ ; its Christoffel and Ricci formulas are algebraic in $f, f^{'}, f^{''}$ and $r$ and do not use the sign of $f$ , so the metric is Ricci-flat there too. Thus the Ricci tensor vanishes on the dense open set $D = R \times ((0, s) \cup (s, \infty)) \times (S^{2} ∖ {two poles})$ . The global metric is smooth and nondegenerate by step 1.1, so its Ricci tensor is smooth. In any smooth product chart, each Ricci component is continuous and vanishes on the dense subset $D$ of that chart; it therefore vanishes on the chart, including at $r = s$ and at the sphere poles. The scalar curvature and Einstein tensor vanish by their definitions. This proves the first part of (d).

3.1L6L8step 2.1algebra

(Kretschmann scalar of the extension.) By [L8] and the algebraic calculation on the two non-horizon branches, the global scalar $K = R_{a b c d} R^{a b c d}$ equals $12 s^{2} / r^{6}$ on $D$ defined in step 2.1. Both sides are smooth on all of $M_{e x t}$ : the left by the smooth nondegenerate global metric and the right because $r > 0$ . They therefore agree everywhere by continuity and density, including at the horizon, the sphere poles, and $r = s$ , where the value is $12 / s^{4}$ . This proves the remaining part of (d).

4.1step 1.1L3L5given∎

(The horizon and the conclusion.) At $r = s$ one has $f = 0$ , so the inverse base block of [L3] is $(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ and the covector $d r$ has squared norm $g_{e x t}^{- 1} (d r, d r) = 0$ . Since $d r \neq 0$ and the level set ${r = s}$ has codimension one, it is a null hypersurface. Its tangent space is spanned by $\partial_{v}$ and the sphere tangent space; $\partial_{v}$ is null at $r = s$ because $g_{e x t} (\partial_{v}, \partial_{v}) = - f = 0$ , and it is the null generator direction. The global block decomposition of step 1.1 is nondegenerate at the horizon, so every smooth chart frame there is a basis; in polar charts the determinant is $- s^{4} \sin^{2} ϑ$ , and in pole charts it is $- s^{4} \det (γ_{i j}) \neq 0$ . Clause (f) records that no maximality or uniqueness is claimed. This proves (e) and completes the proof.

Remarks

What the extension achieves and what it does not. The construction shows that the apparent breakdown of the Schwarzschild chart at $r = s$ is a property of that chart: the same metric is smooth and nondegenerate in the advanced chart across the horizon, and its curvature invariant is finite there. It does not identify the maximal extension of the model; that involves the Kruskal model of this page, and it does not show that every geodesic extends, a statement requiring global arguments not made here.
The two branches. The tortoise function is defined separately on $(0, s)$ and $(s, \infty)$ and is not continuous across $r = s$ ; it is only its derivative $1 / f$ that is used in the pullback computation, and the metric components of $g_{e x t}$ do not involve $r_{*}$ at all. This is why the extension is smooth even though the auxiliary coordinate $x^{0}$ is not defined across the horizon as a single-valued function (only its combinations $v$ and $u = v - 2 r_{*}$ are).

Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources