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Co-authored-by: Copilot <175728472+Copilot@users.noreply.github.com>
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Alexa Amundson
2026-02-27 03:48:56 -06:00
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@@ -2830,7 +2830,7 @@ The presence of π does not indicate simulation. It indicates that the system su
$$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$$ $$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$$
pulls π from the geometry of the two-dimensional case, not from any circular shape in the one-dimensional distribution. pulls π from the geometry of the two-dimensional case, not from any circular shape in the one-dimensional distribution.
**Field theory:** 4π appears in Coulomb's law and gravitational flux because the flux spreads over a sphere. The surface area of a unit sphere is 4π — the solid angle subtended by the full sphere in steradians. **Field theory:** 4π appears in Coulomb's law and gravitational flux because the flux spreads over a sphere. For a sphere of radius \(r\), the surface area is \(4\pi r^2\), so a unit sphere (\(r = 1\)) has area \(4\pi\). A full sphere also subtends a total solid angle of \(4\pi\) steradians, but in Coulomb's law the 4π specifically comes from the \(1/r^2\) field spreading over the spherical surface area \(4\pi r^2\).
**Shannon entropy:** The continuous version of H involves ln(2π) in the entropy of a Gaussian distribution. Again: the circle appears because a Gaussian is the maximum-entropy distribution for given variance, and that extremization connects to the rotational symmetry of the two-dimensional problem. **Shannon entropy:** The continuous version of H involves ln(2π) in the entropy of a Gaussian distribution. Again: the circle appears because a Gaussian is the maximum-entropy distribution for given variance, and that extremization connects to the rotational symmetry of the two-dimensional problem.