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Diffstat (limited to 'mindmap/inverse square.org')
| -rw-r--r-- | mindmap/inverse square.org | 4 |
1 files changed, 3 insertions, 1 deletions
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 |