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theorem proved tactic proof high

unitSphereSurface_D3

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Unit sphere surface area in three dimensions equals four pi. Derivations of the fine structure constant from isotropic recognition coupling cite this to fix the geometric seed. The tactic proof unfolds the general hypersurface formula, evaluates the gamma function at three halves via its recurrence relation, and reduces the resulting algebraic expression to four pi with field simplification and ring tactics.

claimThe surface area of the unit sphere in three-dimensional Euclidean space satisfies $S_2 = 4π$.

background

The SolidAngleExclusivity module proves that the geometric seed in the fine structure constant derivation must be four pi because it equals the surface area of the unit sphere in three dimensions. The general definition computes this area as two pi to the D over two divided by gamma of D over two. For D set to three this yields exactly four pi, as required by isotropy in the recognition ledger.

proof idea

The proof first unfolds unitSphereSurface. It then proves the gamma identity at three halves by applying the recurrence Gamma(s+1) equals s times Gamma(s) at s equals one half, together with the known value Gamma one half equals square root of pi. A separate calculation shows pi to the three halves equals pi times square root of pi. The main simplification applies these equalities, coerces the natural number three, clears the denominator with field_simp, and finishes with ring.

why it matters in Recognition Science

This theorem supplies the concrete value needed by four_pi_unique_for_D3 to conclude that four pi is the unique factor for D equals three. It likewise discharges the rewrite in solidAngle_is_sphere_area. Within the framework it confirms the solid angle term in the alpha seed expression alpha inverse equals four pi times eleven minus gap terms, aligning with the D equals three forcing from the unified chain and the requirement of isotropic coupling.

scope and limits

formal statement (Lean)

  51theorem unitSphereSurface_D3 : unitSphereSurface 3 = 4 * Real.pi := by

proof body

Tactic-mode proof.

  52  unfold unitSphereSurface
  53  -- Γ(3/2) = (1/2)·Γ(1/2) = √π/2
  54  have hGamma : Real.Gamma ((3 : ℝ) / 2) = (Real.sqrt Real.pi) / 2 := by
  55    -- Use Γ(s+1) = s·Γ(s) with s = 1/2.
  56    have : Real.Gamma ((1 / 2 : ℝ) + 1) = (1 / 2 : ℝ) * Real.Gamma (1 / 2 : ℝ) := by
  57      simpa using (Real.Gamma_add_one (s := (1 / 2 : ℝ)) (by norm_num))
  58    -- Rewrite (1/2)+1 = 3/2 and Γ(1/2) = √π.
  59    simpa [Real.Gamma_one_half_eq, add_comm, add_left_comm, add_assoc, div_eq_mul_inv] using this
  60
  61  -- Simplify π^(3/2) = π^(1 + 1/2) = π * π^(1/2) = π * √π.
  62  have hPiPow : Real.pi ^ ((3 : ℝ) / 2) = Real.pi * Real.sqrt Real.pi := by
  63    have hpi_pos : (0 : ℝ) < Real.pi := Real.pi_pos
  64    -- 3/2 = 1 + 1/2
  65    have : (Real.pi : ℝ) ^ ((3 : ℝ) / 2) = Real.pi ^ ((1 : ℝ) + (1 / 2 : ℝ)) := by
  66      norm_num
  67    -- Expand with rpow_add.
  68    calc
  69      Real.pi ^ ((3 : ℝ) / 2)
  70          = Real.pi ^ ((1 : ℝ) + (1 / 2 : ℝ)) := this
  71      _ = Real.pi ^ (1 : ℝ) * Real.pi ^ (1 / 2 : ℝ) := by
  72            simpa [add_comm, add_left_comm, add_assoc] using (Real.rpow_add hpi_pos (1 : ℝ) (1 / 2 : ℝ))
  73      _ = Real.pi * Real.sqrt Real.pi := by
  74            -- π^1 = π, and √π = π^(1/2) (via `sqrt_eq_rpow`).
  75            simp [Real.rpow_one, Real.sqrt_eq_rpow]
  76
  77  -- Now compute the surface area.
  78  -- 2 * π^(3/2) / (√π/2) = 4π
  79  -- First normalize the coercion: (↑3 / 2) = (3 / 2 : ℝ)
  80  have hcoerce : ((3 : ℕ) : ℝ) / 2 = (3 / 2 : ℝ) := by norm_num
  81  simp only [hcoerce] at *
  82  rw [hPiPow, hGamma]
  83  -- Clear denominators carefully.
  84  have hsqrt_pos : (0 : ℝ) < Real.sqrt Real.pi := by
  85    have : (0 : ℝ) < Real.pi := Real.pi_pos
  86    exact Real.sqrt_pos.mpr this
  87  field_simp [hsqrt_pos.ne']
  88  ring
  89
  90/-- For D = 2, the "unit sphere" is a circle with circumference 2π. -/

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