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Arbitrary-positive-H conformastatic Ricci and Einstein components — Examples

1 · Prerequisites

2 · Summary

Scope

These examples specialize the exact smooth family to a regular sphere and to a quantified dilute parameter range. The first checks the center values, proper mass and charge, and everywhere-positive density. The second records uniform metric and density comparison bounds and shows that, away from the origin, the static dust's nonzero electromagnetic force per unit mass equals its nonzero acceleration; at the origin both vanish and the central static worldline is geodesic. Both are exact mathematical calculations, not measurements or stability claims.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Regular smooth counterpoised sphere

Example

Assume ACω. Let b>0 have length units, let x=(x1,x2,x3)∈R3, and define its Euclidean radius by r:=(x1)2+(x2)2+(x3)2. Take a=b and s=+1 in the exact family of Exact smooth superharmonic counterpoised dust family, and define H(x)=1+b/r2+b2 on all of R3. For the family field strength F, set n:=H∂0, ei:=H−1∂i, and define the static-frame electric components by Ei:=cF(ei,n) for i=1,2,3. This is an everywhere regular smooth solution with center values H(0)=2,∇H(0)=0,ΔH(0)=−3b2,ρ(0)=3c232πGNb2,Sg(0)=34b2,Ei(0)=0. Its proper mass and signed charge are c2b/GN and Qsc2b/GN, respectively. The model density is positive everywhere, so its topological support, defined as the closure in R3 of {x∈R3:ρ(x)≠0}, is all of R3.

Facts & Assumptions

Given: ACω, 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/Qs)H−1dx0, U=cH∂0, and ρ=−c2ΔH/(4πGNH3), with Qs>0 and the stated geometric conventions. The observer components used below are computed locally from this potential.

[F3]

Globally regular sphere, proper integrals and dilute bounds gives finite proper mass c2a/GN and charge sQsc2a/GN for the stated profile, under ACω.

[F5]

In a smooth coordinate chart, d(Aa dxa)=∑adAa∧dxa (The local coordinate formula for the exterior derivative), so (dA)i0=∂iA0−∂0Ai.

[A1]

Assume the indexed Axiom ACω of The Axiom of Countable Choice (ACω): every family (Xn)n∈N of nonempty sets has a function f on N with f(n)∈Xn for every n. This is the assumption required by [F3]; no full Axiom of Choice is assumed.

Verification

1.1givenF3F4algebra

Put Ψ=(r2+b2)1/2. Since r2+b2≥b2>0, [F4] and the chain rule give ∂iΨ=xi/Ψ. Differentiating H=1+b/Ψ with the quotient rule in [F4] gives Hi=−bxi/Ψ3, hence Hi(0)=0. At r=0, Ψ=b and H(0)=2. The Laplacian in [F3] gives ΔH(0)=−3b3/b5=−3/b2.

1.2givenA1F1F3algebra

Applying [F3] with a=b and s=+1 yields Mprop=c2b/GN and Qprop=Qsc2b/GN. The exact-solution conclusion is [F1].

2.1step 1.1F1F2F4F5algebra

Substituting these values into [F1] gives ρ(0)=3c2/(4πGNb2⋅23)=3c2/(32πGNb2), and [F2] gives Sg(0)=−2(−3/b2)/23=3/(4b2). For the electric components, the metric in [F1] gives the future unit vector n=H∂0 and positively oriented orthonormal spatial vectors ei=H−1∂i by direct contraction. Since A0=s(c/Qs)H−1 and Ai=0, [F4]–[F5] give Fi0=−s(c/Qs)Hi/H2. Therefore Ei:=cF(ei,n)=cFi0=−sc2Hi/(QsH2), which vanishes at the center by step 1.1.

3.1givenF1step 1.1algebra∎

The formula in [F1] has ΔH=−3b3/(r2+b2)5/2<0 and H>0 at every x∈R3, hence ρ>0 throughout R3. 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 R3, whose closure is R3; 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.

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Quantified dilute exact coupled family

Example

Assume ACω. Let b>0 have length units and let δ be dimensionless with 0<δ≤1/10. Put a=δb and choose either charge sign s∈{−1,+1} in the exact coupled family. For x=(x1,x2,x3)∈R3, define the Euclidean radius r:=(x1)2+(x2)2+(x3)2 and h0(x):=b/r2+b2, so H=1+δh0. Then ∣g00+(1−2δh0)∣≤3δ2≤3100,∣gij−(1+2δh0)δij∣≤δ2≤1100,0≤1−ρρlin≤3δ≤310, where ρlin=3c2δb3/[4πGN(r2+b2)5/2]. These are exact ideal-family bounds, not measured errors. On M=Rx0×R3, the static dust velocity field is U=cH∂0. For each fixed x∈R3 and x∗0∈R, define its static worldline by γx,x∗0:R→M,γx,x∗0(τ):=(x∗0+cH(x)τ,x). This is a smooth future-directed timelike curve with γ˙=U and gH(γ˙,γ˙)=−c2; τ∈R, measured in seconds, is its proper-time parameter in the adopted length-coordinate convention. Here a static worldline is freely falling exactly when it is a geodesic, ∇γ˙γ˙=0, with ∇ the connection of the exact family. Define the model's electromagnetic force per unit mass by fb:=(σ/ρ)FbcUc; this is defined everywhere for this profile since ρ>0. For this profile, Hi=−δbxi/(r2+b2)3/2, so when r>0 both ai and fi are nonzero and ai=fi. At r=0 both vectors vanish, and the central static worldline is geodesic.

Facts & Assumptions

Given: ACω, b>0, 0<δ≤1/10, a=δb, s∈{−1,+1}, and the exact family and dilute comparison of the corollary.

[F1]

Globally regular sphere, proper integrals and dilute bounds supplies H=1+δh0, the exact metric and density profile, and the bounds ∣H−2−(1−2δh0)∣≤3δ2, ∣H2−(1+2δh0)∣≤δ2, and 0≤1−ρ/ρlin≤3δ.

[F2]

Exact smooth superharmonic counterpoised dust family supplies U=cH∂0, ai=−c2Hi/H3, and the identity ρai=σFicUc.

[F4]

Continuous real functions on a compact interval are Riemann integrable (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion). If a function is continuous on [u,v], differentiable inside, and its derivative has an integrable extension, Newton--Leibniz gives its endpoint difference (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative). Pointwise order and nonnegativity are preserved by integrals on u<v (If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a)).

[A1]

0<δ≤1/10, so 3δ2≤3/100, δ2≤1/100, and 3δ≤3/10.

Verification

1.1givenF1A1algebra

Put z=δh0. Since r2+b2≥b2 and both square roots are positive, 0<h0≤1 and 0<z≤δ. By the metric formula in [F1], g00=−(1+z)−2, so the time-component discrepancy is ∣(1+z)−2−1+2z∣.

2.1givenF3F4A1step 1.1algebra

Define f(u):=(1+u)−2−1+2u for u≥0. Twice applying the quotient/product rules [F3] gives f(0)=f′(0)=0 and f′′(u)=6(1+u)−4, between 0 and 6. For fixed z>0, set Gz(u):=(z−u)f′(u)+f(u) on [0,z]. It is continuous with continuous derivative Gz′(u)=(z−u)f′′(u), which is integrable by [F4]. Newton--Leibniz [F4] gives f(z)=Gz(z)−Gz(0)=∫0z(z−u)f′′(u) du. The integrand lies between 0 and 6(z−u), both continuous comparison functions. Integral monotonicity [F4], and Newton--Leibniz for the polynomial primitive 6zu−3u2, give 0≤f(z)≤3z2. Thus the discrepancy in step 1.1 is at most 3z2≤3δ2; [A1] further gives 3δ2≤3/100. At z=0 the same sharper comparison is zero by direct evaluation, without applying the nondegenerate-interval theorem.

3.1step 1.1step 2.1F1A1algebra

The spatial bound in [F1] gives ∣gij−(1+2z)δij∣≤δ2, since δij is 0 or 1. The density identity in [F1] gives 0≤1−ρ/ρlin≤3δ. Applying [A1] and combining with step 2.1 yields the three numerical bounds in the Example.

4.1givenF1F2step 3.1algebra∎

For the worldline γx,x∗0 defined in the Statement, γ˙=U, so its covariant acceleration is the supplier field ab. Here ρ>0 by [F1] and Hi=−δbxi/(r2+b2)3/2; hence ai=−c2Hi/H3 is nonzero when r>0, while ai=0 at r=0. The supplier also gives a0=0. Dividing its all-component identity ρab=σFbcUc by positive ρ gives ab=fb=(σ/ρ)FbcUc everywhere. Thus both vectors are nonzero for r>0 and vanish at the center; only the noncentral static worldlines fail the geodesic definition of free fall.

Remarks

The numerical fractions are consequences of the ideal choice δ≤1/10 and are not experimental uncertainties. No measurement, statistical estimate, or general perturbation theorem is asserted.

Sources