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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 $g_{11} = 1 / f$ (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

$R (ρ) : = ρ (1 + \frac{s}{4 ρ})^{2}, ρ > \frac{s}{4},$

be the isotropic-to-areal radius map of The isotropic coordinate form of the Schwarzschild exterior metric and the isothermal radius shift and let $ρ_{0} : (s, \infty) \to (s / 4, \infty)$ be its inverse, so that the isotropic coefficient of the spatial metric is the function $ρ \mapsto (1 + s / (4 ρ))^{4}$ and the areal coefficient is $r \mapsto (1 - s / r)^{- 1}$ . 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.

Exact form and positivity. $R (ρ) = ρ + \frac{s}{2} + \frac{s^{2}}{16 ρ}$ and $R^{'} (ρ) = 1 - \frac{s^{2}}{16 ρ^{2}} > 0$ on $(s / 4, \infty)$ ; the inverse is $ρ_{0} (r) = \frac{(r - \frac{s}{2}) + \sqrt{r (r - s)}}{2}$ .
Conversion and areal expansion. With $ρ = ρ_{0} (r)$ , $ρ_{0} (r) = r - \frac{s}{2} - \frac{s^{2}}{16 r} + O ⁣ (\frac{s^{3}}{r^{2}}), \frac{1}{ρ_{0} (r)} = \frac{1}{r} + \frac{s}{2 r^{2}} + O ⁣ (\frac{s^{2}}{r^{3}}) .$ These are expansions as $r / s \to \infty$ with fixed $s > 0$ . The areal coefficient satisfies, exactly for $r > s$ , $(1 - s / r)^{- 1} = 1 + \frac{s}{r} + \frac{(s / r)^{2}}{1 - s / r},$ so its expansion is $1 + \frac{s}{r} + \frac{s^{2}}{r^{2}} + O ⁣ (\frac{s^{3}}{r^{3}})$ .
Isotropic coefficient after argument substitution. With $ρ = ρ_{0} (r)$ one has $\frac{s}{4 ρ_{0} (r)} = \frac{s}{4 r} + \frac{s^{2}}{8 r^{2}} + O (s^{3} / r^{3})$ and $(1 + \frac{s}{4 ρ_{0} (r)})^{4} = 1 + \frac{s}{r} + \frac{7}{8} \frac{s^{2}}{r^{2}} + O ⁣ (\frac{s^{3}}{r^{3}}) .$ Hence these scalar functions agree through first order and differ at second order by $\frac{1}{8} \frac{s^{2}}{r^{2}}$ .
Failure of the unconverted comparison. Evaluating the isotropic coefficient at $ρ = r$ instead of at $ρ = ρ_{0} (r)$ gives $(1 + \frac{s}{4 r})^{4} = 1 + \frac{s}{r} + \frac{3}{8} \frac{s^{2}}{r^{2}} + O ⁣ (\frac{s^{3}}{r^{3}}),$ which differs from the areal coefficient of clause 2 by $\frac{5}{8} \frac{s^{2}}{r^{2}}$ 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 $(s / r)^{2}$ .
Radial tensor transformation. Put $a (r) = s / (4 ρ_{0} (r))$ . Differentiating $R (ρ_{0} (r)) = r$ gives $ρ_{0}^{'} (r) = (1 - a (r)^{2})^{- 1}$ . Consequently the radial coefficient in the areal basis is exactly $g_{r r} = (1 + a (r))^{4} (ρ_{0}^{'} (r))^{2} = (1 - s / r)^{- 1} .$ Moreover $(ρ_{0}^{'})^{2} = 1 + s^{2} / (8 r^{2}) + O (s^{3} / r^{3})$ , so this coefficient is $1 + s / r + s^{2} / r^{2} + O (s^{3} / r^{3})$ : 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 $R$ , $ρ_{0}$ , the areal coefficient $1 / f$ and the isotropic coefficient $(1 + s / (4 ρ))^{4}$ .

[L1]

$R (ρ) = ρ (1 + s / (4 ρ))^{2}$ is a smooth bijection of $(s / 4, \infty)$ onto $(s, \infty)$ with smooth inverse $ρ_{0} (r) = \frac{(r - s / 2) + \sqrt{r (r - s)}}{2}$ , $R^{'} (ρ) = 1 - s^{2} / (16 ρ^{2}) > 0$ , the exact algebraic identity $R (ρ) = ρ + \frac{s}{2} + \frac{s^{2}}{16 ρ}$ , and the asymptotic expansion $ρ_{0} (r) = r - \frac{s}{2} - \frac{s^{2}}{16 r} + O (s^{3} / r^{2})$ for fixed $s > 0$ as $r / s \to \infty$ (The isotropic coordinate form of the Schwarzschild exterior metric and the isothermal radius shift clauses (a) and (c), 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, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Rational powers $a^{r}$ of a positive base).

[L2]

For $0 < t < 1$ , $(1 - t)^{- 1} = 1 + t + t^{2} (1 - t)^{- 1}$ by multiplication by $1 - t$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Rational powers $a^{r}$ of a positive base).

[L3]

In the isotropic chart of the datum the spatial coefficient is $g_{ρ ρ} = (1 + s / (4 ρ))^{4}$ and in a spherical-polar product chart $g_{11} = (1 - s / r)^{- 1}$ (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)).

[L4]

The identity $(1 + x)^{4} = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}$ holds for every real $x$ ; on $(s, \infty)$ one has $r > s$ so $s / r \in (0, 1)$ and $r \geq s$ , and the interval $(s, \infty)$ is nonempty (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Rational powers $a^{r}$ of a positive base).

[L5]

The chain rule for differentiable functions on open real intervals gives $R^{'} (ρ_{0} (r)) ρ_{0}^{'} (r) = 1$ from $R \circ ρ_{0} = i d$ (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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ); the radial term of the metric in [L3] is $(1 + s / (4 ρ))^{4} d ρ^{2}$ , so $d ρ = ρ_{0}^{'} (r) d r$ 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], $R (ρ) = ρ + \frac{s}{2} + \frac{s^{2}}{16 ρ}$ , so $R^{'} (ρ) = 1 - \frac{s^{2}}{16 ρ^{2}}$ , which is positive exactly because $ρ^{2} > s^{2} / 16$ , that is $ρ > s / 4$ ; [L1] also gives the displayed inverse $ρ_{0}$ , and the identity $R (ρ_{0} (r)) = r$ holds on $(s, \infty)$ . This is clause 1.

1.2L1algebra

(Expansion of the inverse.) By [L1], for fixed $s > 0$ and $r / s \to \infty$ , $ρ_{0} (r) = r - \frac{s}{2} - \frac{s^{2}}{16 r} + O ⁣ (\frac{s^{3}}{r^{2}}) .$ Write $ρ_{0} (r) = r (1 + δ (r))$ , where $δ (r) = - s / (2 r) - s^{2} / (16 r^{2}) + O (s^{3} / r^{3}) = O (s / r)$ . For all sufficiently large $r / s$ , $∣ δ (r) ∣ \leq 1 / 2$ , and the exact identity $(1 + δ)^{- 1} = 1 - δ + δ^{2} / (1 + δ)$ gives $\frac{1}{ρ_{0} (r)} = \frac{1}{r} + \frac{s}{2 r^{2}} + O ⁣ (\frac{s^{2}}{r^{3}}) .$ All remainder estimates here are for fixed $s > 0$ and $r / s \to \infty$ . This proves the two expansions of clause 2.

1.3L2L4algebra

(Areal expansion.) For $r > s$ the identity $(1 - s / r)^{- 1} = 1 + \frac{s / r}{1 - s / r} = 1 + \frac{s}{r} + \frac{(s / r)^{2}}{1 - s / r}$ is exact by [L2]; since $0 < s / r < 1$ by [L4], the last term is positive and satisfies $\frac{(s / r)^{2}}{1 - s / r} = (s / r)^{2} (1 + O (s / r))$ , so $(1 - s / r)^{- 1} = 1 + \frac{s}{r} + \frac{s^{2}}{r^{2}} + O (s^{3} / r^{3})$ with remainder at most $\frac{s^{3} / r^{3}}{1 - s / r}$ , which is $O (s^{3} / r^{3})$ on $r \geq 2 s$ . This is the second display of clause 2.

2.1step 1.2L3L4algebra

(Isotropic coefficient after conversion.) By step 1.2, $\frac{s}{4 ρ_{0} (r)} = \frac{s}{4 r} + \frac{s^{2}}{8 r^{2}} + O (s^{3} / r^{3})$ , so with $x : = \frac{s}{4 ρ_{0} (r)}$ , [L4] gives $(1 + x)^{4} = 1 + 4 x + 6 x^{2} + O (x^{3})$ , hence $(1 + x)^{4} = 1 + \frac{s}{r} + \frac{s^{2}}{2 r^{2}} + \frac{6 s^{2}}{16 r^{2}} + O (s^{3} / r^{3}) = 1 + \frac{s}{r} + \frac{7}{8} \frac{s^{2}}{r^{2}} + O (s^{3} / r^{3})$ . Combining with step 1.3, the two coefficients agree in their constant and $1 / r$ terms and differ in their $1 / r^{2}$ terms by $\frac{1}{8} \frac{s^{2}}{r^{2}}$ ; this is clause 3.

2.2step 1.3L3L4algebra

(Unconverted comparison.) Evaluating at $ρ = r$ gives $x = \frac{s}{4 r}$ and by [L4] $(1 + \frac{s}{4 r})^{4} = 1 + \frac{s}{r} + \frac{6 s^{2}}{16 r^{2}} + O (s^{3} / r^{3}) = 1 + \frac{s}{r} + \frac{3}{8} \frac{s^{2}}{r^{2}} + O (s^{3} / r^{3})$ ; subtracting the areal expansion of step 1.3 termwise gives the mismatch $(1 - \frac{3}{8}) \frac{s^{2}}{r^{2}} = \frac{5}{8} \frac{s^{2}}{r^{2}}$ 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], $ρ_{0}^{'} = 1 / (1 - a^{2})$ with $0 < a < 1$ . Since $f (r) = (1 - a)^{2} / (1 + a)^{2}$ for $r = R (ρ_{0} (r))$ , the transformed radial coefficient is $(1 + a)^{4} / (1 - a^{2})^{2} = (1 + a)^{2} / (1 - a)^{2} = f (r)^{- 1}$ . Step 1.2 gives $a^{2} = s^{2} / (16 r^{2}) + O (s^{3} / r^{3})$ ; for $u = a^{2}$ , the exact remainder identity $(1 - u)^{- 2} - 1 - 2 u = u^{2} (3 - 2 u) / (1 - u)^{2}$ gives $(ρ_{0}^{'})^{2} = 1 + s^{2} / (8 r^{2}) + O (s^{3} / r^{3})$ . 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 $r / s \to \infty$ : 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 $1 / ρ_{0}$ and $3 / 8$ from the fourth-power expansion. The areal coefficient has second-order coefficient $1$ , so their difference is $1 / 8$ ; the $O (s^{2} / r^{3})$ remainder of $1 / ρ_{0}$ 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.

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

Example

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources