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RN conformal data on a punctured exterior

Statement

Let $m > 0$ be a geometric mass length, let $q \in R$ be a signed geometric electric-charge length, and define

$r_{*} : = \max {0, \frac{∣ q ∣ - m}{2}}, D_{m, q} : = (r_{*}, \infty), ψ (r) : = \sqrt{{(1 + \frac{m}{2 r})}^{2} - \frac{q^{2}}{4 r^{2}}} (r \in D_{m, q}),$

where the square root is positive. On $Σ = D_{m, q} \times S^{2}$ , use the Euclidean seed metric $\overset{ˉ}{h} = d r^{2} + r^{2} γ_{S^{2}}$ , with $γ_{S^{2}}$ the unit round metric, and set

$\overset{ˉ}{E} = q r^{- 2} \partial_{r}, \overset{ˉ}{B} = 0, \overset{ˉ}{A} = 0, τ = 0, Λ = 0.$

With $h = ψ^{4} \overset{ˉ}{h}$ , $K = 0$ , $E = ψ^{- 6} \overset{ˉ}{E}$ , and $B = 0$ , the exact transformed equations are

${\overset{ˉ}{D}}_{i} {\overset{ˉ}{E}}^{i} = 0, {\overset{ˉ}{D}}_{i} {\overset{ˉ}{B}}^{i} = 0, - 8 Δ_{\overset{ˉ}{h}} ψ - 2 ∣ \overset{ˉ}{E} ∣_{\overset{ˉ}{h}}^{2} ψ^{- 3} = 0,$

and the separate momentum equation is $0 = - 2 ψ^{- 6} (\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B})_{i}$ . They are satisfied by the displayed fields, so the corresponding SI-convention data satisfy the original constraints. The puncture $r = 0$ is omitted from $Σ$ , so the assertion concerns smooth equations on this domain and introduces no delta-function source there.

For $m > ∣ q ∣$ , $D_{m, q} = (0, \infty)$ . The centered coordinate sphere $S_{r} = {r} \times S^{2}$ has area $4 π R_{a r e a l} (r)^{2}$ , where

$R_{a r e a l} (r) : = r ψ (r)^{2} = r + m + \frac{m^{2} - q^{2}}{4 r} .$

On each $S_{r}$ , the outward unit normal is $ν : = ψ^{- 2} \partial_{r}$ . Define its mean-curvature scalar by $H : = {div}_{h} ν$ ; by convention here, $S_{r}$ is minimal exactly when $H (r) = 0$ . Its mean curvature vanishes exactly at $r_{0} = \frac{1}{2} \sqrt{m^{2} - q^{2}}$ , where $R_{a r e a l} (r_{0}) = m + \sqrt{m^{2} - q^{2}}$ . Both ends are asymptotically Euclidean in the following explicit sense. In the Euclidean seed Cartesian coordinates $x^{i} = r n^{i}$ with $n \in S^{2}$ , the $r \to \infty$ end has $h_{i j} = (1 + 2 m / r + O (r^{- 2})) δ_{i j}$ , $\partial_{k} h_{i j} = O (r^{- 2})$ , and $\partial_{ℓ} \partial_{k} h_{i j} = O (r^{- 3})$ . For the inner end define inverted Cartesian coordinates ${\tilde{x}}^{i} : = (r_{0}^{2} / r^{2}) x^{i}$ , whose radius is $\tilde{r} : = ∣ \tilde{x} ∣ = r_{0}^{2} / r$ ; the same metric and derivative falloffs hold as $r ↓ 0$ in these coordinates. On the branch $r > r_{0}$ , put $R : = R_{a r e a l} (r)$ . The comparator spatial metric is

$h_{R N}^{s p a t i a l} : = f (R)^{- 1} d R \otimes d R + R^{2} γ_{S^{2}}, f (R) : = 1 - \frac{2 m}{R} + \frac{q^{2}}{R^{2}},$

with $R > m + \sqrt{m^{2} - q^{2}}$ and hence $f (R) > 0$ ; the spatial metric $h$ agrees with this static Reissner–Nordström exterior metric under the displayed areal-radius coordinate. These statements make no global event-horizon or maximal-extension claim.

Statement references

Conformal electrovacuum seed data

Exact conformal electrovacuum constraint transformation

Dyonic Reissner-Nordström metric and field domains

Facts & Assumptions

Given: $m > 0$ , $q \in R$ , the positive domain $D_{m, q}$ , the seed metric and fields in the Statement, and the stated conformal constraint identities.

[F1]

Conformal electrovacuum seed data: $\overset{ˉ}{E}$ and $\overset{ˉ}{B}$ are geometric-normalized contravariant fields, $K = (τ / 3) h + ψ^{- 2} \overset{ˉ}{A}$ , and $h = ψ^{4} \overset{ˉ}{h}$ .

[F2]

Exact conformal electrovacuum constraint transformation: source-free seed divergences and the transformed Hamiltonian equation imply the corresponding SI constraints; the momentum equation is separately satisfied when $K = 0$ and $\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B} = 0$ .

[F3]

Dyonic Reissner-Nordström metric and field domains: with mass length $m$ , electric length $q$ , magnetic length zero, and areal coordinate $R$ , the RN metric has radial factor $f (R)^{- 1} d R^{2}$ , angular factor $R^{2} γ_{S^{2}}$ , and $f (R) = 1 - 2 m / R + q^{2} / R^{2}$ where $f (R) \neq 0$ .

Proof

1.1givenF1algebra

Put $C : = m^{2} - q^{2}$ and $w : = ψ^{2}$ . Since $2 r + m > 0$ , the condition $w > 0$ is equivalent to $2 r + m > ∣ q ∣$ , giving exactly $D_{m, q}$ . Thus $ψ$ is smooth and positive there. In Euclidean spherical coordinates, ${\overset{ˉ}{D}}_{i} {\overset{ˉ}{E}}^{i} = r^{- 2} \partial_{r} (r^{2} q r^{- 2}) = 0$ , while ${\overset{ˉ}{D}}_{i} {\overset{ˉ}{B}}^{i} = 0$ .

1.2givenF1F2algebra

On this domain $w : = ψ^{2} = 1 + m / r + C / (4 r^{2})$ , $w^{'} = - m / r^{2} - C / (2 r^{3})$ , and $w^{''} = 2 m / r^{3} + 3 C / (2 r^{4})$ . Since $Δ_{R^{3}} ψ = ψ^{''} + 2 ψ^{'} / r$ , $ψ^{'} / ψ = w^{'} / (2 w)$ and $ψ^{''} / ψ = w^{''} / (2 w) - (w^{'})^{2} / (4 w^{2})$ give $Δ_{R^{3}} ψ / ψ = (w^{''} / 2 + w^{'} / r) / w - (w^{'})^{2} / (4 w^{2}) = (C w / r^{4} - (w^{'})^{2}) / (4 w^{2}) = - q^{2} / (4 r^{4} w^{2})$ ; here $w^{''} / 2 + w^{'} / r = C / (4 r^{4})$ and direct expansion gives $C w / r^{4} - (w^{'})^{2} = - q^{2} / r^{4}$ . Thus $Δ_{R^{3}} ψ = - q^{2} / (4 r^{4} ψ^{3})$ . With $\overset{ˉ}{R} = 0$ , $\overset{ˉ}{A} = τ = Λ = \overset{ˉ}{B} = 0$ , the transformed scalar equation is $- 8 Δ ψ - 2 q^{2} r^{- 4} ψ^{- 3} = 0$ .

2.1F2step 1.1step 1.2algebra

The seed Gauss constraint holds by step 1.1, and step 1.2 verifies the Hamiltonian equation. The momentum right side vanishes because $\overset{ˉ}{B} = 0$ , and its left side vanishes because $K = τ = 0$ . The omitted point $r = 0$ is not in the manifold, so the smooth divergence calculation on $D_{m, q} \times S^{2}$ makes no distributional assertion at the puncture.

2.2givenstep 1.2algebra

Suppose $m > ∣ q ∣$ , so $C > 0$ and $r_{0} = \sqrt{C} / 2$ . Then $R_{a r e a l} = r + m + C / (4 r)$ , $R_{a r e a l}^{'} = 1 - C / (4 r^{2})$ , and $R_{a r e a l}^{''} = C / (2 r^{3}) > 0$ . Hence $R_{a r e a l}$ has its unique minimum on $(0, \infty)$ at $r_{0}$ , with value $m + \sqrt{C}$ . The area formula follows directly from the induced metric on $S_{r}$ , namely $r^{2} ψ^{4} γ_{S^{2}}$ .

3.1F1step 2.2algebra

The outward unit normal to $S_{r}$ is $ν = ψ^{- 2} \partial_{r}$ . Its mean curvature, defined as the trace of the second fundamental form, is $H = {div}_{h} ν = (\sqrt{\det h})^{- 1} \partial_{r} (\sqrt{\det h} ψ^{- 2}) = 2 R_{a r e a l}^{'} / (ψ^{2} R_{a r e a l})$ . Thus $H (r_{0}) = 0$ , and $H$ is nonzero for every other centered coordinate sphere.

3.2F3step 2.2algebra

Write $r_{0}^{2} = C / 4$ and $R = R_{a r e a l} (r)$ . On $r > r_{0}$ , $R > m + 2 r_{0}$ , so $f (R) > 0$ . Direct algebra gives $R^{2} f (R) = (R - m)^{2} - C = (r - r_{0}^{2} / r)^{2}$ and $d R / d r = 1 - r_{0}^{2} / r^{2} = (r - r_{0}^{2} / r) / r$ . Therefore $f (R)^{- 1} (d R / d r)^{2} = R^{2} / r^{2} = ψ^{4}$ , which proves that $h = ψ^{4} d r^{2} + R^{2} γ_{S^{2}}$ equals the radial and angular spatial metric in [F3] on this branch.

4.1F1step 2.2algebra∎

As $r \to \infty$ , $ψ^{4} = 1 + 2 m / r + O (r^{- 2})$ ; differentiating this explicit rational expression for $ψ^{4}$ gives first- and second-coordinate-derivative falloffs $O (r^{- 2})$ and $O (r^{- 3})$ . For the inner end put $\tilde{r} = r_{0}^{2} / r$ . Then $ψ (r)^{2} r^{2} / r_{0}^{2} = 1 + m / \tilde{r} + r_{0}^{2} / {\tilde{r}}^{2} = ψ (\tilde{r})^{2}$ and $d r^{2} + r^{2} γ_{S^{2}} = (r^{4} / r_{0}^{4}) (d {\tilde{r}}^{2} + {\tilde{r}}^{2} γ_{S^{2}})$ . Hence the inner metric is exactly the same isotropic form in $\tilde{r}$ , and it has the same falloff and derivative estimates. This proves the two stated end descriptions without extending the static areal chart through $f = 0$ .

RN conformal data on a punctured exterior

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources