From 9d15d37bed3f2d96017ffa373b0d52df26f7903a Mon Sep 17 00:00:00 2001 From: Preston Pan Date: Thu, 25 Jan 2024 16:46:22 -0800 Subject: add code block in website --- mindmap/inverse square.org | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) (limited to 'mindmap/inverse square.org') diff --git a/mindmap/inverse square.org b/mindmap/inverse square.org index 132b322..3f980bf 100644 --- a/mindmap/inverse square.org +++ b/mindmap/inverse square.org @@ -122,7 +122,9 @@ k\int_{space}\sigma(\vec{r'})((x^{2} + y^{2} + z^{2})^{-3/2} - 3x^{2}(x^{2} + y^ \end{align*} If we factor out the \((x^{2} + y^{2} + z^{2})^{-\frac{5}{2}}\) term and collecting the like terms: \begin{align*} -k\int_{space}\sigma(\vec{r'})(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}})d\tau = k\int_{space}(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}})d\tau = 0. +k\int_{space}\sigma(\vec{r'})(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}})d\tau \\ += k\int_{space}(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}})d\tau \\ += 0. \end{align*} So is the divergence of this field zero? Well, not exactly. In order to understand why, we must ask: What happens to the divergence at \(\vec{0}\)? On first glance, it seems clearly undefined. After all, we're dividing by zero. However, after some amount of inspection, the assumption that our field's divergence -- cgit