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Regular smooth counterpoised sphere

Example

Assume ${A C}_{ω}$ . Let $b > 0$ have length units, let $x = (x^{1}, x^{2}, x^{3}) \in R^{3}$ , and define its Euclidean radius by $r : = \sqrt{(x^{1})^{2} + (x^{2})^{2} + (x^{3})^{2}}$ . Take $a = b$ and $s = + 1$ in the exact family of Exact smooth superharmonic counterpoised dust family, and define $H (x) = 1 + b / \sqrt{r^{2} + b^{2}}$ on all of $R^{3}$ . For the family field strength $F$ , set $n : = H \partial_{0}$ , $e_{i} : = H^{- 1} \partial_{i}$ , and define the static-frame electric components by $E_{i} : = c F (e_{i}, n)$ for $i = 1, 2, 3$ . This is an everywhere regular smooth solution with center values $H (0) = 2, \nabla H (0) = 0, Δ H (0) = - \frac{3}{b^{2}}, ρ (0) = \frac{3 c^{2}}{32 π G_{N} b^{2}}, S_{g} (0) = \frac{3}{4 b^{2}}, E_{i} (0) = 0.$ Its proper mass and signed charge are $c^{2} b / G_{N}$ and $Q_{s} c^{2} b / G_{N}$ , respectively. The model density is positive everywhere, so its topological support, defined as the closure in $R^{3}$ of ${x \in R^{3} : ρ (x) \neq 0}$ , is all of $R^{3}$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , $a = b > 0$ , $s = + 1$ , the smooth family, the arbitrary-positive- $H$ curvature table, and the proper-integral result.

[F1]

Exact smooth superharmonic counterpoised dust family supplies the exact coupled metric and fields, including $A = s (c / Q_{s}) H^{- 1} d x^{0}$ , $U = c H \partial_{0}$ , and $ρ = - c^{2} Δ H / (4 π G_{N} H^{3})$ , with $Q_{s} > 0$ and the stated geometric conventions. The observer components used below are computed locally from this potential.

[F2]

Arbitrary-positive-H conformastatic Ricci and Einstein components gives $S_{g} = - 2 Δ H / H^{3}$ .

[F3]

Globally regular sphere, proper integrals and dilute bounds gives finite proper mass $c^{2} a / G_{N}$ and charge $s Q_{s} c^{2} a / G_{N}$ for the stated profile, under ${A C}_{ω}$ .

[F4]

The positive-root function is smooth with derivative $s' (q) = 1 / (2 s (q))$ for $q > 0$ (The positive square-root branch is smooth). The chain, product and quotient rules hold (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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , 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 $g (c) \neq 0$ ).

[F5]

In a smooth coordinate chart, $d (A_{a} d x^{a}) = \sum_{a} d A_{a} \land d x^{a}$ (The local coordinate formula for the exterior derivative), so $(d A)_{i 0} = \partial_{i} A_{0} - \partial_{0} A_{i}$ .

[A1]

Assume the indexed Axiom ${A C}_{ω}$ of The Axiom of Countable Choice ( ${A C}_{ω}$ ): every family $(X_{n})_{n \in N}$ of nonempty sets has a function $f$ on $N$ with $f (n) \in X_{n}$ for every $n$ . This is the assumption required by [F3]; no full Axiom of Choice is assumed.

Verification

1.1givenF3F4algebra

Put $Ψ = (r^{2} + b^{2})^{1 / 2}$ . Since $r^{2} + b^{2} \geq b^{2} > 0$ , [F4] and the chain rule give $\partial_{i} Ψ = x^{i} / Ψ$ . Differentiating $H = 1 + b / Ψ$ with the quotient rule in [F4] gives $H_{i} = - b x^{i} / Ψ^{3}$ , hence $H_{i} (0) = 0$ . At $r = 0$ , $Ψ = b$ and $H (0) = 2$ . The Laplacian in [F3] gives $Δ H (0) = - 3 b^{3} / b^{5} = - 3 / b^{2}$ .

1.2givenA1F1F3algebra

Applying [F3] with $a = b$ and $s = + 1$ yields $M_{p r o p} = c^{2} b / G_{N}$ and $Q_{p r o p} = Q_{s} c^{2} b / G_{N}$ . The exact-solution conclusion is [F1].

2.1step 1.1F1F2F4F5algebra

Substituting these values into [F1] gives $ρ (0) = 3 c^{2} / (4 π G_{N} b^{2} \cdot 2^{3}) = 3 c^{2} / (32 π G_{N} b^{2})$ , and [F2] gives $S_{g} (0) = - 2 (- 3 / b^{2}) / 2^{3} = 3 / (4 b^{2})$ . For the electric components, the metric in [F1] gives the future unit vector $n = H \partial_{0}$ and positively oriented orthonormal spatial vectors $e_{i} = H^{- 1} \partial_{i}$ by direct contraction. Since $A_{0} = s (c / Q_{s}) H^{- 1}$ and $A_{i} = 0$ , [F4]–[F5] give $F_{i 0} = - s (c / Q_{s}) H_{i} / H^{2}$ . Therefore $E_{i} : = c F (e_{i}, n) = c F_{i 0} = - s c^{2} H_{i} / (Q_{s} H^{2})$ , which vanishes at the center by step 1.1.

3.1givenF1step 1.1algebra∎

The formula in [F1] has $Δ H = - 3 b^{3} / (r^{2} + b^{2})^{5 / 2} < 0$ and $H > 0$ at every $x \in R^{3}$ , hence $ρ > 0$ throughout $R^{3}$ . The denominator never vanishes because $b > 0$ , so the metric, potential, and fields extend smoothly through $r = 0$ . The nonzero set of $ρ$ is therefore $R^{3}$ , whose closure is $R^{3}$ ; by the Statement's definition, this is its topological support.

Remarks

This is a regular ideal mathematical solution, not a reported object. The dust is held static by its electromagnetic force; no equilibrium stability, formation history, or real-particle composition is inferred.

Regular smooth counterpoised sphere

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources