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Spherical Vacuum Model — Examples

1 · Prerequisites

2 · Summary

This page collects six worked computations and counterexamples of gr-spherical-vacuum-model, each an exact evaluation of a supplied formula at an explicit value or an explicit refutation. No item reports a measurement. The Kruskal computation sets s=1 m as an exact normalization of the model parameter, not as a measured value; no numerical value of c, GN or m∗ is asserted.

Coordinates and clocks. Explicit isotropic-to-areal conversion and the second-order scalar-coefficient comparison of the Schwarzschild exterior verifies the isotropic conversion R(ρ)=ρ+s/2+s2/(16ρ), expands the inverse as ρ0(r)=r−s/2−s2/(16r)+O(s3/r2) and shows that after the conversion the areal and isotropic coefficients agree through first order and differ at second order by 18(s/r)2, while the unconverted comparison at ρ=r is already wrong at that order by 58(s/r)2. Explicit static clock-rate ratios of the Schwarzschild exterior at r = 2s and r = 4s evaluates the static clock identity at r1=2s and r2=4s, obtaining the exact ratio 2/3 and the acceleration coefficient 1/(8s) at r1; Explicit circular-orbit data of the Schwarzschild exterior at the ISCO radius 3s and the photon-sphere radius 3s/2 computes the orbit data at the two thresholds: ℓ2=3c2s2, e=22c/3 and (dφ/dt)2=c2/(54s2) at r0=3s with Vℓ′′(3s)=0, the values f(3s/2)=1/3, (dφ/dt)2=4c2/(27s2) at the photon sphere, and the stable comparison radius 4s with Vℓ′′(4s)=c2/(320s2).

The horizon. Explicit metric, orientation and curvature data of the advanced-chart extension at the horizon r = s evaluates the advanced-chart extension at the horizon: the metric block (0110), determinant −s4sin⁡2ϑ0, the orientation field Next=∂v−∂r at r=s with g(Next,Next)=−2<0, generator pairing g(Next,∂v)=−1, and Kretschmann value 12/s4, next to the divergent exterior component g11. Explicit causal-cone and radial-derivative checks in the Kruskal model at the horizon for s = 1 m instantiates the Kruskal cone criterion at (U,V)=(0,1) with s=1 m as a model normalization, where r=s and A=4s2/e: the vectors ∂U+∂V, ∂U, ∂V are future causal with the first timelike and the last two null, ∂U−∂V is spacelike, the angular cut-off is γ(ω,ω)≤4/e, and ∂Ur=−e−1, ∂Vr=0 agree with implicit differentiation.

A counterexample. The exterior Schwarzschild charts do not cover the horizon-crossing extension: the regular horizon point (v, r) = (0, s) refutes the statement that the product charts of the exterior datum cover the horizon-crossing extension (equivalently, that extension points are reached at bounded exterior coordinate x0): the point (v,r)=(0,s) is a regular point with finite nonzero Kretschmann scalar 12/s4 and nonvanishing determinant, lies outside the image of Φ because Φ preserves r>s, and any exterior sequence mapping to it has x0→+∞.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Explicit isotropic-to-areal conversion and the second-order scalar-coefficient comparison of the Schwarzschild exterior

Example

Let (M,g,N) be the Schwarzschild exterior datum with Schwarzschild radius s>0, f(r)=1−s/r and g11=1/f (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation, The Schwarzschild exterior datum is a smooth time-oriented Lorentzian four-manifold). Let

R(ρ):=ρ(1+s4ρ)2,ρ>s4,

be the isotropic-to-areal radius map of The isotropic coordinate form of the Schwarzschild exterior metric and the isothermal radius shift and let ρ0:(s,∞)→(s/4,∞) be its inverse, so that the isotropic coefficient of the spatial metric is the function ρ↦(1+s/(4ρ))4 and the areal coefficient is r↦(1−s/r)−1. The two displayed functions are coefficients in different radial bases. The following scalar-function comparisons do not identify their tensor components after a change of basis; that transformation is included in clause 5.

  1. Exact form and positivity. R(ρ)=ρ+s2+s216ρ and R′(ρ)=1−s216ρ2>0 on (s/4,∞); the inverse is ρ0(r)=(r−s2)+r(r−s)2.

  2. Conversion and areal expansion. With ρ=ρ0(r), ρ0(r)=r−s2−s216r+O ⁣(s3r2),1ρ0(r)=1r+s2r2+O ⁣(s2r3). These are expansions as r/s→∞ with fixed s>0. The areal coefficient satisfies, exactly for r>s, (1−s/r)−1=1+sr+(s/r)21−s/r, so its expansion is 1+sr+s2r2+O ⁣(s3r3).

  3. Isotropic coefficient after argument substitution. With ρ=ρ0(r) one has s4ρ0(r)=s4r+s28r2+O(s3/r3) and (1+s4ρ0(r))4=1+sr+78s2r2+O ⁣(s3r3). Hence these scalar functions agree through first order and differ at second order by 18s2r2.

  4. Failure of the unconverted comparison. Evaluating the isotropic coefficient at ρ=r instead of at ρ=ρ0(r) gives (1+s4r)4=1+sr+38s2r2+O ⁣(s3r3), which differs from the areal coefficient of clause 2 by 58s2r2 at second order: the consistency of the leading Newtonian-type term is unaffected, but the comparison of coefficients in the unchanged chart is already wrong at relative order (s/r)2.

  5. Radial tensor transformation. Put a(r)=s/(4ρ0(r)). Differentiating R(ρ0(r))=r gives ρ0′(r)=(1−a(r)2)−1. Consequently the radial coefficient in the areal basis is exactly grr=(1+a(r))4(ρ0′(r))2=(1−s/r)−1. Moreover (ρ0′)2=1+s2/(8r2)+O(s3/r3), so this coefficient is 1+s/r+s2/r2+O(s3/r3): the apparent 1/8 discrepancy of clause 3 disappears when the basis is transformed.

Facts & Assumptions

Given: The exterior datum with Schwarzschild radius s>0, the functions R, ρ0, the areal coefficient 1/f and the isotropic coefficient (1+s/(4ρ))4.

[L1]

R(ρ)=ρ(1+s/(4ρ))2 is a smooth bijection of (s/4,∞) onto (s,∞) with smooth inverse ρ0(r)=(r−s/2)+r(r−s)2, R′(ρ)=1−s2/(16ρ2)>0, the exact algebraic identity R(ρ)=ρ+s2+s216ρ, and the asymptotic expansion ρ0(r)=r−s2−s216r+O(s3/r2) for fixed s>0 as r/s→∞ (The isotropic coordinate form of the Schwarzschild exterior metric and the isothermal radius shift clauses (a) and (c), The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set, Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0, Rational powers ar of a positive base).

[L3]

In the isotropic chart of the datum the spatial coefficient is gρρ=(1+s/(4ρ))4 and in a spherical-polar product chart g11=(1−s/r)−1 (The isotropic coordinate form of the Schwarzschild exterior metric and the isothermal radius shift clause (b), The Schwarzschild exterior datum is a smooth time-oriented Lorentzian four-manifold clause (a), The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation clause (c)).

[L5]

The chain rule for differentiable functions on open real intervals gives R′(ρ0(r))ρ0′(r)=1 from R∘ρ0=id (The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c)); the radial term of the metric in [L3] is (1+s/(4ρ))4dρ2, so dρ=ρ0′(r)dr in the areal chart.

Proof

technique · use the inverse-radius expansion proved in the isotropic-coordinate lemma, expand the reciprocal and fourth power with explicit remainders, and compare the coefficients after the coordinate conversion
1.1L1algebra

(Exact form and inverse.) By [L1], R(ρ)=ρ+s2+s216ρ, so R′(ρ)=1−s216ρ2, which is positive exactly because ρ2>s2/16, that is ρ>s/4; [L1] also gives the displayed inverse ρ0, and the identity R(ρ0(r))=r holds on (s,∞). This is clause 1.

1.2L1algebra

(Expansion of the inverse.) By [L1], for fixed s>0 and r/s→∞, ρ0(r)=r−s2−s216r+O ⁣(s3r2). Write ρ0(r)=r(1+δ(r)), where δ(r)=−s/(2r)−s2/(16r2)+O(s3/r3)=O(s/r). For all sufficiently large r/s, ∣δ(r)∣≤1/2, and the exact identity (1+δ)−1=1−δ+δ2/(1+δ) gives 1ρ0(r)=1r+s2r2+O ⁣(s2r3). All remainder estimates here are for fixed s>0 and r/s→∞. This proves the two expansions of clause 2.

1.3L2L4algebra

(Areal expansion.) For r>s the identity (1−s/r)−1=1+s/r1−s/r=1+sr+(s/r)21−s/r is exact by [L2]; since 0<s/r<1 by [L4], the last term is positive and satisfies (s/r)21−s/r=(s/r)2(1+O(s/r)), so (1−s/r)−1=1+sr+s2r2+O(s3/r3) with remainder at most s3/r31−s/r, which is O(s3/r3) on r≥2s. This is the second display of clause 2.

2.1step 1.2L3L4algebra

(Isotropic coefficient after conversion.) By step 1.2, s4ρ0(r)=s4r+s28r2+O(s3/r3), so with x:=s4ρ0(r), [L4] gives (1+x)4=1+4x+6x2+O(x3), hence (1+x)4=1+sr+s22r2+6s216r2+O(s3/r3)=1+sr+78s2r2+O(s3/r3). Combining with step 1.3, the two coefficients agree in their constant and 1/r terms and differ in their 1/r2 terms by 18s2r2; this is clause 3.

2.2step 1.3L3L4algebra

(Unconverted comparison.) Evaluating at ρ=r gives x=s4r and by [L4] (1+s4r)4=1+sr+6s216r2+O(s3/r3)=1+sr+38s2r2+O(s3/r3); subtracting the areal expansion of step 1.3 termwise gives the mismatch (1−38)s2r2=58s2r2 at second order, while the constant and first-order terms continue to agree. This is clause 4 as a comparison of scalar coefficient functions, not a metric transformation.

3.1L1L3L5step 1.2step 1.3step 2.1algebra∎

(Transforming the radial basis.) By [L1] and [L5], ρ0′=1/(1−a2) with 0<a<1. Since f(r)=(1−a)2/(1+a)2 for r=R(ρ0(r)), the transformed radial coefficient is (1+a)4/(1−a2)2=(1+a)2/(1−a)2=f(r)−1. Step 1.2 gives a2=s2/(16r2)+O(s3/r3); for u=a2, the exact remainder identity (1−u)−2−1−2u=u2(3−2u)/(1−u)2 gives (ρ0′)2=1+s2/(8r2)+O(s3/r3). Multiplication by the expansion of step 2.1 changes its second-order coefficient from 7/8 to 7/8+1/8=1, exactly as in step 1.3. This proves clause 5.

Remarks

  • What the example checks. The computation is exact finite algebra with remainder bounds as r/s→∞: after argument substitution alone, the second-order isotropic coefficient is 1/2+3/8=7/8, where 1/2 comes from the coordinate shift in 1/ρ0 and 3/8 from the fourth-power expansion. The areal coefficient has second-order coefficient 1, so their difference is 1/8; the O(s2/r3) remainder of 1/ρ0 contributes only at third order. The radial tensor coefficient additionally includes the squared Jacobian of clause 5 and agrees exactly with the areal coefficient. No numerical value of s, no observed system and no measurement occurs.
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Explicit static clock-rate ratios of the Schwarzschild exterior at r = 2s and r = 4s

Example

Let (M,g,N) be the Schwarzschild exterior datum with Schwarzschild radius s>0, f(r)=1−s/r (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation) and static observer congruence n=f−1/2∂0 (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (a), Observer congruences, their observer curves and adapted charts). By The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (c), a static observer worldline at radius r0∈(s,∞) has proper-time increment

Δτ(r0)=f(r0) Δt=1−sr0  Δt

over a coordinate-time increment Δt=Δx0/c>0. Then:

  1. Exact values at the two radii. For r1=2s and r2=4s one has f(r1)=12, f(r2)=34, so Δτ(r1)=Δt2,Δτ(r2)=32 Δt,Δτ(r1)Δτ(r2)=23=63<1.
  2. Monotonicity in the radius. The function r↦f(r)=1−s/r is smooth and strictly increasing on (s,∞) with derivative ddrf(r)=s2r2f(r)>0; consequently a static observer at the larger radius accumulates strictly more proper time over the same coordinate interval: Δτ(r2)>Δτ(r1) for s<r1<r2, and the ratio of clause 1 is the exact instance at (r1,r2)=(2s,4s).
  3. Large-radius limit. f(r)→1 as r→+∞; the ratios Δτ(r1)/Δτ(r2) are invariant under the common scaling of Δt and tend to 1 as min⁡(r1,r2)→+∞ with r2>r1.
  4. The static observers are accelerated. At r1=2s the acceleration of the congruence is a=s2r2∂r (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (e)), whose coefficient at r1=2s is s2(2s)2=18s>0: the observer held at r1=2s has nonzero acceleration pointing in the direction of increasing r, so it is not freely falling.

Facts & Assumptions

Given: The exterior datum with s>0 and f=1−s/r, a fixed Δt>0, the radii r1=2s, r2=4s, and the static observer congruence n=f−1/2∂0.

[L1]

The static congruence n=f−1/2∂0 is a smooth future-directed unit timelike field, its integral curves are the worldlines at constant r, and along each such worldline Δτ=f(r0) Δt with Δt=Δx0/c; for two radii over the same nonzero coordinate interval, Δτ1/Δτ2=f(r1)/f(r2) (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clauses (a)-(d), Observer congruences, their observer curves and adapted charts, The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation).

[L2]

The acceleration of the congruence is a=∇nn=s2r2∂r (The static observer congruence of the Schwarzschild exterior model: unit field, integral curves, acceleration and clock-rate ratios clause (e)).

[L5]

A function continuous on [a,b] and differentiable on (a,b), for a<b, satisfies h(b)−h(a)=h′(ξ)(b−a) for some ξ∈(a,b) (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

Proof

technique · direct; evaluate the clock identity at the two explicit radii, differentiate the square root to get monotonicity, and substitute $r_1=2s$ in the acceleration coefficient
1.1L1L3algebra

(Values at 2s and 4s.) By [L3], f(2s)=1−s2s=12 and f(4s)=1−s4s=34, both positive, so [L1] gives Δτ(r1)=1/2 Δt=Δt2 and Δτ(r2)=3/4 Δt=32Δt. The ratio is 1/23/4=23=63, and 6/3<1 because 6<9. This is clause 1.

2.1L1L3L4L5step 1.1algebra

(Monotonicity.) On (s,∞) the chain rule gives ddrf(r)=f′(r)2f(r)=s/r22f(r)=s2r2f(r), using f′(r)=s/r2 and f>0 by [L3]. The numerator and denominator are positive, so the derivative is positive. On every [a,b]⊂(s,∞) with a<b, f is continuous and differentiable by [L3], and [L5] gives f(b)−f(a)=(b−a)s/(2ξ2f(ξ))>0 for some ξ∈(a,b). Thus f, hence r↦Δτ(r) at fixed Δt>0, is strictly increasing; with f(r1)<f(r2) by [L4], r1<r2 gives Δτ(r1)<Δτ(r2) and the ratio of clause 1 is the instance r1=2s, r2=4s. This is clause 2 and the inequality Δτ(r2)>Δτ(r1) of clause 1.

2.2L1L3step 1.1algebra

(Large-radius limit.) As r→+∞, s/r→0, so f(r)→1 and f(r)→1 by [L3]; hence Δτ(r)=f(r) Δt→Δt and the ratio Δτ(r1)/Δτ(r2)=f(r1)/f(r2)→1 whenever min⁡(r1,r2)→+∞ with the radii positive. The ratio does not involve Δt, since the common factor cancels. This is clause 3.

3.1L1L2L3algebra∎

(Acceleration at 2s.) By [L2] the acceleration field is a=s2r2∂r, whose coefficient function is smooth and positive on (s,∞); at r1=2s it is s2(2s)2=18s>0. Since the coefficient is nonzero, the worldlines of the congruence do not satisfy the geodesic equation, so the observer held at radius 2s is accelerated with the stated outward-coefficient 1/(8s) in the coordinate basis (∂0,∂r,∂ϑ,∂φ). This is clause 4.

Remarks

  • What the example computes. All four clauses are exact evaluations of the clock identity and the acceleration formula of the cited lemma at explicit radii; the coordinate-time increment Δt is an arbitrary positive coordinate difference, not a measurement, and no value of c, GN, m∗ or s in any unit system is asserted. In particular clause 1 does not report a clock comparison or a redshift observation; it is the conditional identity of the model evaluated at two radii.
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Explicit circular-orbit data of the Schwarzschild exterior at the ISCO radius 3s and the photon-sphere radius 3s/2

Example

Let (M,g,N) be the Schwarzschild exterior datum with Schwarzschild radius s>0, f(r)=1−s/r (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation), with the circular-orbit data of Circular timelike geodesics, their radial-potential stability threshold, and the photon sphere at r = 3s/2 and the radial potential Wℓ(r)=f(r)(c2+ℓ2/r2) of Conserved quantities and equatorial radial reduction of timelike geodesics of the Schwarzschild exterior model clause (c). Then the following exact evaluations hold.

In every Vℓ′′ evaluation below, differentiate with ℓ fixed as a constant and substitute the circular-orbit value only after differentiation.

  1. The innermost stable circular orbit at r0=3s. With ℓ2=c2sr022r0−3s one has ℓ2=3c2s2, e2=89c2 and e=223c, and the second derivative of Vℓ=12Wℓ vanishes: Vℓ′′(3s)=c2s(3s−3s)(3s)3(6s−3s)=0. The circular angular velocity at this radius satisfies (dφdt)2=GNm∗(3s)3=c254s2.
  2. The photon sphere at r0=3s2. For ℓ≠0, with e0=∣ℓ∣f(r0)r0=2∣ℓ∣33 s one has f(3s2)=13 and (dφdt)2=c2f(r0)r02=4c227s2; the null potential Wℓn(r)=f(r)ℓ2/r2 has (Wℓn)′(3s2)=0 and (Wℓn)′′(3s2)=−32ℓ281s4<0.
  3. A stable comparison radius. At r0=4s one has ℓ2=c2s⋅16s28s−3s=165c2s2 and Vℓ′′(4s)=c2s(4s−3s)(4s)3(8s−3s)=c2320s2>0, a strict local minimum: the two threshold values 3s/2 (existence) and 3s (stability) separate the unstable circular orbits 3s2<r0<3s from the stable ones r0>3s.
  4. Blow-up at the existence threshold. As r0↓3s2 the required angular momentum diverges, ℓ2=c2sr022r0−3s→+∞, so there is no finite angular momentum at the limiting radius; for r0≤3s2 no circular timelike geodesic exists (Circular timelike geodesics, their radial-potential stability threshold, and the photon sphere at r = 3s/2 clause (b)).

Facts & Assumptions

Given: The exterior datum with s>0, and the function f=1−s/r, the potential Wℓ(r)=f(r)(c2+ℓ2/r2), its half Vℓ, and the null potential Wℓn(r)=f(r)ℓ2/r2, with ℓ≠0 in the photon-sphere clause.

[L1]

For r0>3s2 the equatorial circular timelike geodesic constructed in Circular timelike geodesics, their radial-potential stability threshold, and the photon sphere at r = 3s/2 clause (a) has ℓ2=c2sr022r0−3s, e=c(r0−s)2r0(2r0−3s), and its angular velocity satisfies (dφdt)2=GNm∗r03; conversely every future timelike geodesic with r≡r0 constant on an interval of positive length has r0>3s2, and if equatorial it has this ℓ2 (that supplier's clause (b), The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation).

[L2]

Differentiating at fixed ℓ and then imposing ℓ2=c2sr2/(2r−3s) gives the circular stability value Vℓ′′(r)=c2s(r−3s)r3(2r−3s) for r>3s2 (Circular timelike geodesics, their radial-potential stability threshold, and the photon sphere at r = 3s/2 clause (c), Conserved quantities and equatorial radial reduction of timelike geodesics of the Schwarzschild exterior model clause (c)).

[L3]

For ℓ≠0, the null circular orbit at r0=3s2 has e0=∣ℓ∣f(r0)r0, angular velocity (dφdx0)2=f(r0)r02, and the null potential satisfies (Wℓn)′(3s2)=0, (Wℓn)′′(3s2)=−32ℓ281s4 (Circular timelike geodesics, their radial-potential stability threshold, and the photon sphere at r = 3s/2 clause (d)).

Proof

technique · direct; substitute the two threshold radii into the conserved-quantity and potential formulas of the cited lemma, and evaluate with the exact identities $f(3s)=2/3$, $f(3s/2)=1/3$
1.1L1L2L4algebra

(ISCO data at 3s.) By [L4] and [L1], at r0=3s the denominator is 2r0−3s=3s, so ℓ2=c2s⋅9s23s=3c2s2. Then c2+ℓ2/r02=c2+3c2s29s2=c2+c23=43c2, so e2=f(3s)⋅43c2=23⋅43c2=89c2 and e=223c>0. The angular velocity is (dφdt)2=GNm∗(3s)3=sc2/227s3=c254s2. By [L2], Vℓ′′(3s)=c2s⋅0(3s)3⋅3s=0. This is clause 1.

1.2L3L4algebra

(Photon-sphere data at 3s/2.) By [L4], f(3s2)=13, so e0=∣ℓ∣1/33s/2=2∣ℓ∣33 s; the angular velocity satisfies (dφdx0)2=f(3s/2)(3s/2)2=1/39s2/4=427s2, hence (dφdt)2=c2⋅427s2. The two potential identities are those of [L3] evaluated at the stated radius. This is clause 2.

1.3L1L2L4algebra

(Stable comparison radius 4s.) At r0=4s the denominator is 2r0−3s=5s, so ℓ2=c2s⋅16s25s=165c2s2; by [L2], Vℓ′′(4s)=c2s(4s−3s)(4s)3⋅5s=c2s264s3⋅5s=c2320s2>0, a strict local minimum of Vℓ at the circular-orbit radius. Combined with clause 1, the sign of Vℓ′′ changes at r=3s, separating the strict local maxima on 3s2<r0<3s from the strict local minima on r0>3s; this is clause 3.

2.1L1L4algebra∎

(Threshold divergence.) For r0>3s2 the denominator 2r0−3s is positive with 2r0−3s→0+ as r0↓3s2, while the numerator c2sr02→c2s⋅9s24>0; hence ℓ2→+∞. By the converse clause of [L1], no future timelike geodesic in any angular orientation has r≡r0 constant on an interval of positive length for s<r0≤3s2, since that would force r0>3s2. This is clause 4.

Remarks

  • What the example computes. The quantities 3c2s2, 22c/3, c2/(54s2), 4c2/(27s2) and c2/(320s2) are exact evaluations of the cited formulas at the stated radii; the factor 22/3 in e/c is irrational. The longitude velocities and angular momenta refer to the constructed equatorial curves. The example reports no orbit of any observed body and no measurement, and the word "stable" refers only to the sign of the second derivative of the stated effective potential of the restricted model.
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Explicit metric, orientation and curvature data of the advanced-chart extension at the horizon r = s

Example

Let (M,g,N) be the Schwarzschild exterior datum with s>0 and f(r)=1−s/r (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation), and let (Mext,gext) with chart (v,r,ϑ,φ) and time-orientation field Next=A(r)∂v−∂r, A(r)=f(r)+f(r)2+42, be the advanced-chart extension of Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart. Write q0:=(v0,s,ϑ0,φ0) for a point on the level set {r=s} with sin⁡ϑ0≠0. Then:

  1. Metric components and nondegeneracy at the horizon. In the product chart, gext has components gvv=−f(s)=0,gvr=grv=1,gϑϑ=s2,gφφ=s2sin⁡2ϑ0,det⁡(gext)=−s4sin⁡2ϑ0<0. The two-dimensional block (0110) in the (v,r) directions is nonsingular with eigenvalues +1 and −1, and the angular directions are positive on (0,π); hence gext is nondegenerate of inertia (3,1,0) at q0 with the same signature (−,+,+,+) as in the exterior.
  2. Horizon generator and time orientation. At q0 the coordinate field ∂v satisfies gext(∂v,∂v)=−f(s)=0 and gext(∂v,∂r)=1≠0, so ∂v is a null direction tangent to the level set {r=s} (the generator direction of clause (e) of the lemma). At r=s the orientation field is A(s)=0+42=1, so Next=∂v−∂r, with gext(Next,Next)=−2<0,gext(Next,∂v)=−1<0, a timelike vector of squared norm −2 (not unit-normalized) whose pairing with the generator is negative: ∂v is future-directed at the horizon.
  3. Finite curvature at the horizon. By clause (d) of the lemma the Kretschmann scalar of gext equals 12s2/r6 on all of Mext, so at q0 its value is 12s2/s6=12/s4, finite and nonzero for every s>0.
  4. The chart component still diverges. In the exterior product chart of the datum the component g11=f(r)−1=(1−s/r)−1 diverges as r↓s, while clauses 1 to 3 show that the metric, its orientation and the Kretschmann invariant at the same value r=s are finite and nondegenerate in the extension. The divergence is therefore a property of the exterior chart's coefficient, not of the geometry at q0.

Facts & Assumptions

Given: The exterior datum with s>0, the extension (Mext,gext), its advanced chart (v,r,ϑ,φ), the function f=1−s/r, the field Next=A(r)∂v−∂r with A=(f+f2+4)/2, and the point q0=(v0,s,ϑ0,φ0) with sin⁡ϑ0≠0.

[L1]

On Mext the components of gext are gvv=−f, gvr=grv=1, gϑϑ=r2, gφφ=r2sin⁡2ϑ, all others zero; gext is smooth and nondegenerate of inertia (3,1,0) on all of r>0; Next is a time orientation with gext(Next,Next)<0 and gext(Next,∂v)<0; the level set {r=s} is a smooth null hypersurface with generator direction ∂v; and det⁡(gext)=−r4sin⁡2ϑ (Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart clauses (a), (b), (e)).

[L2]

The Kretschmann scalar of gext equals RabcdRabcd=12s2/r6 on all of Mext (Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart clause (d)).

Proof

technique · direct; evaluate the extension components and the orientation field at $r=s$, extract the two-dimensional block determinant and eigenvalues, and read the curvature value from the stated identity
1.1L1L4algebra

(Components and determinant.) Substituting r=s in the polar-chart components of [L1] gives gvv=−f(s)=0, gvr=grv=1, gϑϑ=s2, gφφ=s2sin⁡2ϑ0; in the ordered basis (∂v,∂r,∂ϑ,∂φ) the matrix has the block (0110) and the positive angular block diag⁡(s2,s2sin⁡2ϑ0). Its determinant is −s4sin⁡2ϑ0<0 by [L4]. The base block has eigenvalues ±1 and the angular block has two positive entries, so the inertia is (3,1,0) and the metric is nondegenerate at q0.

1.2L1L4algebra

(Generator and orientation.) By [L1] and [L4], gext(∂v,∂v)=−f(s)=0 and gext(∂v,∂r)=1≠0, so ∂v is a null vector whose r-component is zero in the chart: it is tangent to {r=s} and realizes the generator direction of [L1]. At r=s the coefficient of the orientation field is A(s)=f(s)+f(s)2+42=1, so Next=∂v−∂r, and gext(Next,Next)=gvv−2gvr+grr=0−2+0=−2<0, while gext(Next,∂v)=gvv−grv=0−1=−1<0. Both signs are those required of a time orientation, and the negative pairing with the null generator makes ∂v future-directed. This is clause 2.

1.3L2L4algebra

(Curvature value.) By [L2], at r=s the Kretschmann scalar equals 12s2/s6=12/s4, which is finite and positive for every s>0. This is clause 3.

2.1L1L2L3L4algebra∎

(Chart divergence.) By [L3], g11=(1−s/r)−1=rr−s for r>s; as r↓s the numerator tends to s>0 and the denominator to 0+, so g11→+∞. This divergence occurs in the exterior chart at the same value r=s at which clauses 1 to 3 exhibit finite nondegenerate metric and curvature data in the extension, so it is a coordinate artifact of the exterior chart. This is clause 4.

Remarks

  • What the example computes. The example evaluates the supplied extension formulas at r=s; it uses the lemma's identity K=12s2/r6 on Mext rather than extending the exterior-domain computation, and no value of the metric outside the extension's domain r>0 is used. No numerical value of s, c, GN or m∗ and no measurement occurs.
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Explicit causal-cone and radial-derivative checks in the Kruskal model at the horizon for s = 1 m

Example

Fix the Kruskal datum of The Kruskal model datum: domain, implicit radial function, metric and time orientation with length scale L0:=1 m and Schwarzschild radius s=L0; this sets the model parameter m∗=c2L0/(2GN) and is a normalization choice, not a measurement. Write rˉ:=r/L0, gˉ:=gK/L02, and Aˉ:=A/L02=4e−rˉ/rˉ, where A=4s3e−r/s/r is the coefficient of The causal cone, the escape horizon and decreasing areal radius in the Kruskal model clause (a). The coordinates U,V and angular coordinates are dimensionless. Fix ω0 in a spherical-polar chart with 0<ϑ0<π, and let NK=∂U+∂V be the time orientation (Kruskal model properties: smooth radial function, Lorentzian metric, four-region isometries, vacuum and Kretschmann scalar). Take the point

p0:=(U,V,ω)=(0,1,ω0)∈K.

Then:

  1. The radius and the cone coefficient at p0. U(p0)V(p0)=0, so rˉ(p0)=1 and r(p0)=L0=s (the point lies on the horizon); thus Aˉ(p0)=4e−1 and A(p0)=4L02/e. Dividing the future causal criterion of The causal cone, the escape horizon and decreasing areal radius in the Kruskal model clause (a) by L02, its normalized form at p0 reads: X is future-directed causal if and only if XU≥0, XV≥0, XU+XV>0 and 4e XUXV ≥ γp0(Xω,Xω).
  2. Four explicit tangent vectors at p0.
    • X1=∂U+∂V: X1UX1V=1≥0 and 4e⋅1≥0=γ(0,0), so X1 is future causal; indeed gˉ(X1,X1)=2gˉUV=−Aˉ(p0)=−4e<0, timelike.
    • X2=∂U: X2U=1, X2V=0, so the inequality reads 0≥0 and X2 is future causal; moreover gˉ(X2,X2)=0, so ∂U is future null at p0; it is transverse to U=0, whose horizon generator direction is ∂V.
    • X3=∂V: likewise gˉ(X3,X3)=0, gˉ(X3,NK)=−Aˉ(p0)2<0; it is future null and tangent to the horizon generator.
    • X4=∂U−∂V: X4UX4V=−1<0 violates the criterion, and indeed gˉ(X4,X4)=Aˉ(p0)=4e>0, spacelike.
  3. Angular cut-off. For X5=∂U+∂V+ω with purely angular part ω one has gˉ(X5,X5)=−4e+γp0(ω,ω), so X5 is future causal exactly when γp0(ω,ω)≤4e. In the chart with round components γ22=1, γ33=sin⁡2ϑ, the choices ω=(1,0) and ω=(2,0) give γ=1<4e (causal) and γ=4>4e (spacelike), because 1<e<4; the cut-off is the exact inequality, not a numerical approximation.
  4. Radial derivatives at p0. By clause (c) of the causal lemma, ∂Urˉ=−Ve−rˉrˉ and ∂Vrˉ=−Ue−rˉrˉ under the normalization s=L0; at p0 this gives ∂Urˉ=−e−1<0 and ∂Vrˉ=0, equivalently ∂Ur=−L0e−1 and ∂Vr=0. The normalized formula agrees with differentiating UV=−(rˉ−1)erˉ: at (0,1), V dU+U dV=−rˉerˉ drˉ, so dU=−e drˉ. Consequently the areal radius decreases along the future null direction ∂U at the horizon, in accordance with the decreasing-radius clause of the causal lemma.

Facts & Assumptions

Given: The Kruskal datum with s=L0=1 m, the point p0=(0,1,ω0), the dimensionless radius rˉ with rˉ(p0)=1, the normalized coefficients gˉ=gK/L02 and Aˉ=A/L02=4e−rˉ/rˉ, the vectors X1,…,X5, and NK=∂U+∂V.

[L1]

The Kruskal datum has r=s=L0 exactly on the axes UV=0 and rˉ=1 at p0; in a polar chart the normalized metric coefficients at the point have gˉUV=−Aˉ/2, gˉϑϑ=rˉ2, gˉφφ=rˉ2sin⁡2ϑ, with Aˉ=4e−rˉ/rˉ, and NK is future timelike (The Kruskal model datum: domain, implicit radial function, metric and time orientation, Kruskal model properties: smooth radial function, Lorentzian metric, four-region isometries, vacuum and Kretschmann scalar clauses (a)-(b), Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model).

[L2]

At every point and for every tangent vector X with chart components (XU,XV,Xω): X is causal and future-directed if and only if XU≥0, XV≥0, XU+XV>0 and A XUXV≥r2γ(Xω,Xω); and for purely angular ω, gK(∂U+∂V+ω,∂U+∂V+ω)=−A+r2γ(ω,ω) (The causal cone, the escape horizon and decreasing areal radius in the Kruskal model clause (a), Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L3]

At every point of K, ∂Ur=−s2Ve−r/sr and ∂Vr=−s2Ue−r/sr; hence for s=L0, ∂Urˉ=−Ve−rˉ/rˉ and ∂Vrˉ=−Ue−rˉ/rˉ (The causal cone, the escape horizon and decreasing areal radius in the Kruskal model clause (c)).

[L4]

The exponential constant satisfies 2<e<3 and therefore 1<e<4 (The elementary numerical bound 2<e<3); the round metric in the chosen polar chart has γ(ω,ω)=(ω2)2+sin⁡2ϑ (ω3)2, and coordinate fields form a basis there (The round metric on the sphere as an induced metric, Coordinate derivations form a basis of the tangent space).

Proof

technique · direct; evaluate the radius, coefficient, normalized metric pairings and criterion of the cited lemma at $p_0$ with the model normalization $s=L_0=1\,\mathrm m$
1.1L1L2L4algebra

(Radius and coefficient.) At p0, U(p0)V(p0)=0, so by [L1] r(p0)=s=L0 and rˉ(p0)=1; then A(p0)=4s3e−r(p0)/s/r(p0)=4L02e−1 and Aˉ(p0)=4/e. Dividing the criterion of [L2] by L02 gives exactly the displayed normalized form. This is clause 1.

1.2L1L2L4algebra

(The four tangent vectors.) Use the normalized metric gˉ=gK/L02. For X1: gˉ(X1,X1)=2gˉUV(p0)=−Aˉ(p0)=−4e<0 by [L1]; the criterion's conditions are 1≥0, 1≥0, 2>0 and 4e≥0, all true. For X2: gˉ(X2,X2)=gˉUU=0 and gˉ(X2,NK)=gˉUU+gˉUV=−Aˉ(p0)2<0, so X2 is future null; the conditions are 1≥0, 0≥0, 1>0, 4e⋅0≥0. For X3 the symmetric computation gives gˉ(X3,X3)=0, gˉ(X3,NK)=−Aˉ(p0)2<0, and X3UX3V=0. For X4: X4UX4V=(1)(−1)=−1<0, so the criterion fails; directly gˉ(X4,X4)=gˉUU−2gˉUV+gˉVV=Aˉ(p0)=4e>0, spacelike. This is clause 2.

1.3L2L4algebra

(Angular cut-off.) With X5=∂U+∂V+ω, [L2] gives gˉ(X5,X5)=−Aˉ(p0)+rˉ(p0)2γ(ω,ω)=−4e+γ(ω,ω); the criterion of [L2] with X5U=X5V=1 reads 4e≥γ(ω,ω). For ω=(1,0) in the stipulated polar chart, γ=1 and 1<4e because e<4; for ω=(2,0), γ=4 and 4>4e because e>1; both comparisons are exact by [L4]. This is clause 3.

2.1L3L4algebra∎

(Radial derivatives.) By [L3] with s=L0, rˉ(p0)=1, U(p0)=0, V(p0)=1: ∂Urˉ(p0)=−e−1 and ∂Vrˉ(p0)=0, equivalently ∂Ur(p0)=−L0e−1 and ∂Vr(p0)=0. The dimensionless implicit relation UV=−(rˉ−1)erˉ has differential V dU+U dV=−rˉerˉ drˉ; at (U,V)=(0,1) this gives dU=−e drˉ, hence the same derivatives. This proves clause 4.

Remarks

  • What the example checks. The example instantiates the cone criterion and the radial derivatives at one explicit point with the model normalization s=L0=1 m; accordingly its mass parameter is m∗=c2(1 m)/(2GN). This is a choice of units and model parameter, not a measurement, and asserts no numerical value of GN or c and no property of any observed system. The comparisons 1<4/e<4 are exact inequalities in the constant e.
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The exterior Schwarzschild charts do not cover the horizon-crossing extension: the regular horizon point (v, r) = (0, s)

Statement refuted

Statement refuted: "Let (M,g,N) be the Schwarzschild exterior datum with s>0 and let (Mext,gext) be the horizon-crossing extension of Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart. Then every point of Mext lies in the image of the exterior datum under the embedding Φ(x0,r,ω)=(x0+r∗(r),r,ω), hence in the domain of some Schwarzschild (product) chart of the datum; equivalently, every point added by the extension is reached at a bounded value of the exterior coordinate x0."

The counterexample is the point

q0:=(v,r,ϑ,φ)=(0,s,ϑ0,φ0)∈Mext,sin⁡ϑ0≠0,

on the level set {r=s}, and the following three facts refute both forms of the statement:

  1. q0∉Φ(M): the map Φ preserves the function r, so r(Φ(p))=r(p)>s for every p∈M, whereas r(q0)=s.
  2. q0 is a regular point of a smooth Lorentzian manifold, not a boundary artifact: at q0 one has det⁡(gext)=−s4sin⁡2ϑ0≠0, the inertia is (3,1,0), and the Kretschmann scalar is 12s2/r6=12/s4, finite and nonzero.
  3. Every sequence pn∈M with Φ(pn)→q0 must have r(pn)→s and exterior coordinate x0(pn)=vn−r∗(r(pn))→+∞, where vn:=v(Φ(pn))→v0=0; so the exterior coordinate x0 is unbounded along the approach to q0, and the second form of the statement ("bounded x0") fails as well.

Facts & Assumptions

Given: The exterior datum with s>0, its tortoise function r∗(r)=r+slog⁡∣r/s−1∣ on (s,∞), the extension (Mext,gext) with chart (v,r,ϑ,φ), the embedding Φ(x0,r,ω)=(x0+r∗(r),r,ω), and the point q0=(0,s,ϑ0,φ0) with sin⁡ϑ0≠0.

[L1]

Φ restricts to a diffeomorphism of M onto the open subset {r>s} of Mext and Φ∗gext=g; on this subset the chart function r of Mext pulls back to the radial coordinate r of the datum, and v=x0+r∗(r) (Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart clauses (a)-(c), The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation clauses (b)-(c)).

[L2]

gext is smooth and nondegenerate of inertia (3,1,0) on all of Mext, det⁡(gext)=−r4sin⁡2ϑ in every spherical-polar product chart, and the Kretschmann scalar of gext equals 12s2/r6 on Mext (Smooth horizon-crossing extension of the Schwarzschild exterior metric in the advanced chart clauses (a), (d), (e), Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Smooth covariant 2-tensor fields on a smooth manifold).

[L3]

On (s,∞) the tortoise function satisfies r∗(r)=r+slog⁡(r/s−1); log⁡ is the inverse of the strictly increasing bijection exp⁡:R→(0,∞), so for every M∈R the inequality log⁡t<−M is equivalent to t<e−M for t>0; hence log⁡t→−∞ as t↓0 (The natural logarithm as the inverse of the exponential function, The exponential is a continuous bijection from R onto (0,∞), The exponential function is strictly increasing, The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set, Rational powers ar of a positive base, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct; use that $\Phi$ preserves $r$ to separate $q_0$ from the image, exhibit the regular data at $q_0$, and solve $v_0=x^{0}+r_{*}(r)$ for $x^{0}$ along $r\downarrow s$
1.1L1algebra

(q0 is not in the image of the exterior datum.) Let p∈M with p=(x0,r,ω). By [L1], Φ(p)=(x0+r∗(r),r,ω), whose third-to-second chart entry equals the radial coordinate r(p) of the datum, and r(p)>s because the domain of the datum is {r>s} (The Schwarzschild exterior model datum: domain, metric, mass parameter and time orientation clause (b)). At q0 the extension chart has r(q0)=s; if q0=Φ(p) then s=r(q0)=r(p)>s, a contradiction. Hence q0∉Φ(M), and since every product chart of the datum is a chart of M, no Schwarzschild chart of the datum contains q0. This is clause 1.

1.2L2algebra

(q0 is a regular point.) By [L2], det⁡(gext)(q0)=−s4sin⁡2ϑ0, which is nonzero because s>0 and sin⁡ϑ0≠0; the inertia at q0 is (3,1,0), and the Kretschmann scalar there is 12s2/s6=12/s4, finite and positive. So q0 is an interior point of a smooth nondegenerate Lorentzian four-manifold on which all stated curvature data are finite; in particular the failure of the product charts to cover q0 is not due to q0 being a singularity or a boundary point of the extension. This is clause 2.

2.1L1L3step 1.1algebra

(The exterior coordinate is unbounded.) Let pn∈M be a sequence with Φ(pn)→q0. By [L1] the first chart coordinate of Φ(pn) is vn=x0(pn)+r∗(r(pn)) and the second is rn=r(pn); convergence to q0=(0,s,ϑ0,φ0) forces rn→s and vn→0. Then x0(pn)=vn−r∗(rn). Since rn>s for all n and rn→s, the argument rn/s−1 tends to 0+, so by [L3] log⁡(rn/s−1)→−∞ and hence r∗(rn)=rn+slog⁡(rn/s−1)→−∞; therefore x0(pn)=vn−r∗(rn)→+∞. More explicitly, for every bound B there is n0 with r∗(rn)<−B−∣vn∣ for n≥n0, whence x0(pn)>B: the exterior coordinate x0 is unbounded above along the approach. This is clause 3.

3.1L1step 1.1step 1.2step 2.1construct∎

(An approach exists.) For each integer n≥1 set rn=s+s/n and pn=(−r∗(rn),rn,ω0), with ω0 the fixed angular point of q0. Then pn∈M and Φ(pn)=(0,rn,ω0)→q0, so the limiting assertion of step 2.1 is nonvacuous. Together with steps 1.1 and 1.2, this refutes both claimed properties at the specified regular horizon point.

Remarks

  • What the counterexample says. The horizon point q0 lies in the closure of Φ(M) in Mext but not in Φ(M); it is a regular point of the extension and requires a genuinely different chart, whose existence is the content of the cited extension lemma. The refuted statement is not that the exterior chart fails somewhere — the cited lemma proves its regularity on r>s — but that the charts of the exterior datum suffice for the extension. No value of s, c, GN or m∗, no numerical integration and no measurement occurs.

Sources