Thought experimentStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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RN circular null orbit example

Physical scope

A conditional calculation of equatorial circular null geodesic radii in the selected dyonic RN exterior. Under the adopted geometric-optics rule, these are ideal-light-ray orbit radii; the result makes no stability, photon-mass, observation, or measurement claim.

Statement

Let $m_{*} > 0$ and $Q_{e}, Φ_{B} \in R$ be SI mass, electric-charge and magnetic-flux inputs, and define the dyonic RN lengths by SI and geometric solution parameter identifications and Dyonic Reissner-Nordström metric and field domains. Write $Q^{2} : = q_{e}^{2} + q_{m}^{2}$ and assume $0 \leq Q^{2} \leq M^{2}$ . In the static exterior $r > r_{+} : = M + \sqrt{M^{2} - Q^{2}}$ , let $γ$ be a future affine null geodesic in the equatorial plane $θ = π / 2$ , with constants $E : = f {\dot{x}}^{0} > 0$ and $ℓ : = r^{2} \dot{ϕ} \neq 0$ . Its radial equation is ${\dot{r}}^{2} = E^{2} - ℓ^{2} V (r), V (r) : = \frac{f (r)}{r^{2}} .$ An exterior circular null orbit exists at and only at the outer radius $r_{γ}^{+} = \frac{3 M + \sqrt{9 M^{2} - 8 Q^{2}}}{2} > r_{+} .$ The second algebraic stationary radius $r_{γ}^{-} = (3 M - \sqrt{9 M^{2} - 8 Q^{2}}) / 2$ lies at or inside $r_{+}$ and is not a second exterior circular orbit. Radii are in metres. By the adopted geometric-optics assumption of Adopted clock, free-fall and light-ray assumptions of the classical gravitational framework, the exhibited null geodesic is an ideal-light-ray orbit within that model approximation. The result says nothing about orbital stability or an observed photon trajectory.

Facts & Assumptions

Given: The conditional dyonic RN model with $M > 0$ and $0 \leq Q^{2} \leq M^{2}$ , the static exterior, and the adopted ideal-light-ray geometric-optics rule.

[F1]

SI and geometric solution parameter identifications and Dyonic Reissner-Nordström metric and field domains: $M = G_{N} m_{*} / c^{2}$ , $q_{e} = \sqrt{G_{N} / (4 π ε_{0} c^{4})} Q_{e}$ , $q_{m} = α Φ_{B} / (4 π)$ , and the static metric is $g = - f (d x^{0})^{2} + f^{- 1} d r^{2} + r^{2} (d θ^{2} + \sin^{2} θ d ϕ^{2})$ with $f = 1 - 2 M / r + Q^{2} / r^{2}$ .

[F2]

Dyonic RN as adopted Einstein–Maxwell model: this dyonic RN datum on $r > 0$ , $f \neq 0$ is a conditional solution of the adopted source-free equation-based model; the field-equation proof is draft and not independently accepted.

[A1]

Adopted clock, free-fall and light-ray assumptions of the classical gravitational framework: in geometric optics an ideal light ray is assumed to be a future null geodesic of the metric-selected connection. This is an adopted model postulate, not an equation derived from the RN field solution.

[F3]

Metric compatibility, torsion freedom and the metric-selected connection class and Geodesic of an affine connection: the selected connection is torsion-free and metric-compatible, and an affine geodesic satisfies $D_{t} γ^{'} = 0$ .

[F4]

Killing fields of a Lorentzian metric and Killing fields and conserved currents, and on-shell conservation of the Hilbert stress-energy tensor: a Killing field has zero metric Lie derivative, and its metric pairing with an affine geodesic tangent is constant.

[F5]

SI unit and coordinate conventions of the classical gravitational framework: $x^{0} = c t$ and $r$ have units of metres; affine constants below are coordinate-geometric constants and are not assigned particle-energy measurements.

[F6]

Classification of future and past causal vectors by a timelike field: a nonzero causal vector whose metric pairing with the selected future timelike field is negative is future-directed.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): the inherited RN metric and geometric-flux conventions use only ${A C}_{ω}$ . The finite root and geodesic calculation makes no new selections from families and uses no full AC.

Proof

1.1F1F4F5A1

On the static exterior $f > 0$ , the fields $\partial_{x^{0}}$ and $\partial_{ϕ}$ are Killing because the metric in [F1] is independent of $x^{0}$ and $ϕ$ . Therefore [F4] makes $E = f {\dot{x}}^{0} = - g (\partial_{x^{0}}, γ^{'})$ and $ℓ = r^{2} \dot{ϕ} = g (\partial_{ϕ}, γ^{'})$ constant along the affine geodesic. Restrict to its equatorial plane, as in the Statement.

2.1F1step 1.1algebra

Substituting $θ = π / 2$ , $\dot{θ} = 0$ and these constants into $g (γ^{'}, γ^{'}) = 0$ gives $- f {\dot{x}}^{0, 2} + f^{- 1} {\dot{r}}^{2} + r^{2} {\dot{ϕ}}^{2} = 0.$ Multiplying by $f > 0$ and using $E = f {\dot{x}}^{0}$ , $ℓ = r^{2} \dot{ϕ}$ yields ${\dot{r}}^{2} = E^{2} - f ℓ^{2} / r^{2}$ .

3.1F1F3step 2.1algebra

The metric-selected Christoffel formula from [F3] gives $Γ^{r}_{00} = f f^{'} / 2$ and $Γ^{r}_{ϕ ϕ} = - f r$ on the equator. At a circular point, $\dot{r} = \ddot{r} = 0$ and the radial geodesic equation becomes $f f^{'} {\dot{x}}^{0, 2} / 2 - f r {\dot{ϕ}}^{2} = 0$ . The null condition gives $f {\dot{x}}^{0, 2} = r^{2} {\dot{ϕ}}^{2}$ , so the radial equation is equivalent to $V^{'} (r) = \frac{f^{'}}{r^{2}} - \frac{2 f}{r^{3}} = 0.$ Conversely, at an equatorial point with $f > 0$ and $V^{'} = 0$ , the same connection coefficients make the radial equation hold for the null circular tangent; the $x^{0}$ and $ϕ$ equations have only terms proportional to $\dot{r}$ , and the $θ$ equation vanishes at $θ = π / 2$ by reflection symmetry. Thus solving $V^{'} = 0$ is necessary and sufficient for an equatorial circular null geodesic in the static region.

4.1step 3.1algebra

Expanding $V = r^{- 2} - 2 M r^{- 3} + Q^{2} r^{- 4}$ gives $V^{'} (r) = - \frac{2}{r^{5}} (r^{2} - 3 M r + 2 Q^{2}) .$ The discriminant is $9 M^{2} - 8 Q^{2} \geq M^{2} > 0$ , so its real roots are $r_{γ}^{\pm} = (3 M \pm \sqrt{9 M^{2} - 8 Q^{2}}) / 2$ . Set $x = \sqrt{M^{2} - Q^{2}} \in [0, M]$ and $s = \sqrt{M^{2} + 8 x^{2}} = \sqrt{9 M^{2} - 8 Q^{2}}$ . Then $r_{γ}^{+} - r_{+} = \frac{M + s - 2 x}{2} > 0,$ because $s^{2} - (2 x)^{2} = M^{2} + 4 x^{2} > 0$ , while $r_{γ}^{-} - r_{+} = \frac{M - s - 2 x}{2} \leq 0$ because $s \geq M$ and $x \geq 0$ . Thus exactly the outer root is in $r > r_{+}$ , and there $f > 0$ .

5.1F1F6A1step 3.1step 4.1algebra

At $r = r_{γ}^{+}$ choose any $ℓ \neq 0$ and set $E = ∣ ℓ ∣ \sqrt{f (r_{γ}^{+})} / r_{γ}^{+} > 0$ . Then the null condition and $V^{'} (r_{γ}^{+}) = 0$ make the radial geodesic equation hold with constant $r$ . Taking ${\dot{x}}^{0} = E / f > 0$ and $\dot{ϕ} = ℓ / r^{2}$ makes the tangent nonzero and null; its pairing with the selected future timelike static field $ξ = \partial_{x^{0}}$ is $g (ξ, γ^{'}) = - E < 0$ , so [F6] makes it future-directed. The equatorial angular equation also holds as shown in step 3.1. This constructs the outer circular null geodesic, rather than only a formal root of an effective potential.

6.1A1F2step 5.1A2∎

The outer orbit is therefore a conditional null-geodesic prediction of the selected dyonic RN model. The geometric-optics postulate interprets such a curve as an ideal light ray within its scope; no stability or empirical claim follows.

RN circular null orbit example

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources