Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Exterior potential sphere and zero ideal force — conditional physical prediction

Physical scope

Explicit smooth potential on exterior sphere, slip/no-through condition; exact inviscid Euler pressure force zero; not viscous/no-slip. Worked example is the actual explicit field/parameter selection in that section. B leaf cannot supply any later item. Constant density with spatially varying constraint pressure, not barotropic EOS pressure.

Statement

Let $a, ρ_{0} > 0$ , $p_{\infty} \in R$ , $U \in R^{3}$ , with lengths in metres, velocity in $m s^{- 1}$ , density in $k g m^{- 3}$ and pressure in pascals, and $Ω_{a} = {x \in R^{3} : r = ∣ x ∣ > a}$ . Define $n = x / r$ , Cartesian $div u = \partial_{x} u_{x} + \partial_{y} u_{y} + \partial_{z} u_{z}$ , $ϕ (x) = (U \cdot x) (1 + a^{3} / (2 r^{3}))$ , $u = \nabla ϕ$ , and $p = p_{\infty} + ρ_{0} (∣ U ∣^{2} - ∣ u ∣^{2}) / 2$ . In the selected unforced incompressible Euler model these smooth fields solve $(u \cdot \nabla) u + ρ_{0}^{- 1} \nabla p = 0$ , div $u = 0$ , tend uniformly in direction to $U, p_{\infty}$ at infinity, and have $u \cdot n = 0$ at $r = a$ . On the surface the velocity is $u = \frac{3}{2} (U - (U \cdot n) n)$ , generally nonzero tangent to the sphere. Thus the boundary is impermeable with slip, not no-slip. For the inviscid stress $σ_{E} = - p I$ and sphere normal $n = x / a$ pointing from center into exterior, $\int_{∣ x ∣ = a} σ_{E} n d S = - \int_{∣ x ∣ = a} p n d S = 0.$ This is an exact inviscid mathematical-model prediction, neither zero viscous drag nor an observed force.

Statement references

Stewart, MIT 8.09, Fluid Mechanics, §6.4, printed pp. 120–123

Fitzpatrick, Fluid Mechanics, UT Austin, potential flow past a sphere and zero resultant

Mandre, Fluid Dynamics Lecture Notes, Chapter 6, potential flow

Facts & Assumptions

Given: $a, ρ_{0} > 0$ , $U, p_{\infty}$ , exterior sphere, exact potential and pressure, far field and slip boundary.

[A1]

Single-component nonpolar continuum and material balances: continuum balances are adopted physical inputs, not measured force laws.

[A2]

Specified wall, thermal, periodic and initial data: far-field and impermeability data are separate; no no-slip condition is adopted.

[F1]

Incompressible inviscid Euler model with pressure multiplier: the selected local equations are div $u = 0$ and $ρ_{0} D_{t} u + \nabla p = 0$ , with inviscid stress $σ_{E} = - p I$ .

[F2]

Exterior potential sphere and zero ideal force: the displayed potential sphere field solves steady Euler, has no-through slip, and has zero pressure-only resultant with the stated orientation.

[F3]

Formal parameterized fluid systems and observables: for $u = (u_{x}, u_{y}, u_{z})$ in Cartesian coordinates, $curl u = (\partial_{y} u_{z} - \partial_{z} u_{y}, \partial_{z} u_{x} - \partial_{x} u_{z}, \partial_{x} u_{y} - \partial_{y} u_{x})$ ; in the smooth potential field the mixed second derivatives commute, giving $curl (\nabla ϕ) = 0$ .

Proof

1.1givenA1A2F1F2F3

The potential field in [F2] is divergence-free, solves the local steady Euler equation in [F1], tends to $U$ , and has zero radial velocity at $r = a$ . For $u = \nabla ϕ$ , the Cartesian curl convention in [F3] also gives $curl u = 0$ by equality of the mixed second derivatives of the smooth potential. Its tangential surface value is the stated projection; this is slip, not no-slip.

2.1F2step 1.1constructalgebra

On $r = a$ , [F2] gives $u = \frac{3}{2} (U - (U \cdot n) n)$ , so $∣ u ∣^{2} = \frac{9}{4} (∣ U ∣^{2} - (U \cdot n)^{2})$ and pressure depends on $n$ only through $(U \cdot n)^{2}$ . Parameterize $S_{a}$ by $Φ (θ, φ) = a (\sin θ \cos φ, \sin θ \sin φ, \cos θ)$ , with $0 \leq θ \leq π$ and $0 \leq φ \leq 2 π$ ; then $d S = a^{2} \sin θ d θ d φ$ . The change $(θ, φ) \mapsto (π - θ, φ + π m o d 2 π)$ sends $Φ$ to $- Φ$ and preserves this area element because its absolute Jacobian is one and $\sin (π - θ) = \sin θ$ . Thus it preserves $p$ and reverses $n$ , so the two antipodal contributions cancel and $\int_{S_{a}} p n d S = 0$ .

3.1A2F1step 2.1∎

The resultant uses only $σ_{E} = - p I$ and contains no viscous shear term, so it makes no no-slip or viscous-drag claim.

Exterior potential sphere and zero ideal force — conditional physical prediction

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources