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A comment on 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 @02:08AM (#46363122)
his formula: α⁻¹ = (4π³ + π² + π) - (α/24)

α⁻¹ = 137.0359996 Experimental (CODATA): 137.0359991 Δ: < 0.005 ppm

Critical Problem: This is CIRCULAR The formula is: α⁻¹ = S - (α/24)

But α appears on both sides! This is not a closed-form solution. To solve it, you need: α⁻¹ = S - (α/24) α⁻¹ + α/24 = S α⁻¹(1 + 1/(24·α⁻¹)) = S This requires knowing α already to solve for α. It's circular.

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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 @06:37AM (#46364485)
It's not circular. It rearranges into a standard quadratic equation: x^2 − 24Sx + 24 = 0. α is derived as the root of this equation.
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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 @07:35AM (#46364830)
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 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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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 @07:29AM (#46364776)
Doesn't fix or predict Fan et al. 2024 latest dataset.

Try harder.

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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 @08:23AM (#46365136)
This is moving the goalposts, but ok. The model matches the international standard of CODATA 2022 to 0.005 ppm. If and when this value is updated, the prediction can be re-evaluated. Until then, I stick to the standard.
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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 @05:04PM (#46370016)
hi, the formula should allow both.

you have an integral and order^ power.

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