Show HN: G=(hbar*c*2*(1+alpha/3)^2)/(m_p^2*4^64) ≈ 6.6742439706e-11 (8 ppm)

Re:Show HN: G=(hbar*c*2*(1+alpha/3)^2)/(m_p^2*4^64) ≈ 6.6742... by fatbrowndog on Tuesday December 23, 2025 @07:37AM (#46364851)

α⁻¹ = S - 1/(24α)

α⁻¹·24α = 24αS - 1

24α²S - 24α - 1 = 0

α = (24 ± √(576 + 96S))/(48S)

α = (24 + √(576 + 96·137.036...))/(48·137.036...)

α = (24 + √13,723.66...)/(6577.74...)

α = (24 + 117.12...)/(6577.74...)

α = 141.12.../6577.74...

α ≈ 0.021454...

α⁻¹ ≈ 46.61 ???

That's wrong!

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Re:Show HN: G=(hbar*c*2*(1+alpha/3)^2)/(m_p^2*4^64) ≈ 6.6742... by albert_roca on Tuesday December 23, 2025 @08:56AM (#46365338)

You transcribed the formula incorrectly.

The term is -(alpha / 24). You calculated -1 / (24 · alpha).

The correct derivation is:

  1 / alpha = S - (alpha / 24)
  1 = S · alpha - (alpha^2) / 24
  alpha^2 - 24 · S · alpha + 24 = 0

Solving this with S = 4 · π^3 + π^2 + π yields the correct value.

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