CCFU Proof 15 — One-Step Lorentz Factor γ₁ = √5/2

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CCFU Proof 15 — One-Step Lorentz Factor γ₁ = √5/2

05/17/2026

Given.
Let φ = (1+√5)/2, so φ² = φ + 1.
C₂ one-step rapidity: η = ln φ.

Part A — Lorentz factor.

    γ₁ = cosh(ln φ)
       = (φ + 1/φ) / 2

Since 1/φ = φ − 1:

    φ + 1/φ = 2φ − 1

Squaring:

    (2φ − 1)² = 4φ² − 4φ + 1
              = 4(φ+1) − 4φ + 1    [since φ² = φ+1]
              = 5

Since 2φ − 1 > 0:  2φ − 1 = √5.

    γ₁ = √5/2  ∎

Part B — Reciprocal of 2φ.

    1/(2φ) = 1/(1 + √5)
           = (√5 − 1) / ((√5 + 1)(√5 − 1))
           = (√5 − 1) / 4
           = 0.30902...  ∎