From 16b94d8f622f5270689a71a91fd05596ce99e76b Mon Sep 17 00:00:00 2001
From: Alexa Amundson <118287761+blackboxprogramming@users.noreply.github.com>
Date: Fri, 27 Feb 2026 03:48:56 -0600
Subject: [PATCH] Update README.md

Co-authored-by: Copilot <175728472+Copilot@users.noreply.github.com>
---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index c6c5eba..8da2524 100644
--- a/README.md
+++ b/README.md
@@ -2830,7 +2830,7 @@ The presence of π does not indicate simulation. It indicates that the system su
 $$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$$
 
 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.