diff options
Diffstat (limited to 'mindmap')
| -rw-r--r-- | mindmap/index.org | 2 | ||||
| -rw-r--r-- | mindmap/inverse square.org | 4 | ||||
| -rw-r--r-- | mindmap/magnetostatics.org | 2 |
3 files changed, 6 insertions, 2 deletions
diff --git a/mindmap/index.org b/mindmap/index.org index 350a564..109ae12 100644 --- a/mindmap/index.org +++ b/mindmap/index.org @@ -30,6 +30,7 @@ you to make a web of notes, something close to a wiki. * I want to Break the Rules No you don't. That being said, if you really want the list of all articles, here you go: +@@html: <div class="links-page">@@ #+begin_src shell :results output raw :exports both set -f IFS=' @@ -81,3 +82,4 @@ set +f - [[file:Legendre Transformation.org][Legendre Transformation.org]] - [[file:Lagrangian mechanics.org][Lagrangian mechanics.org]] - [[file:Fourier Transform.org][Fourier Transform.org]] +@@html: </div>@@ 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 diff --git a/mindmap/magnetostatics.org b/mindmap/magnetostatics.org index ea763e7..dd9d172 100644 --- a/mindmap/magnetostatics.org +++ b/mindmap/magnetostatics.org @@ -56,7 +56,7 @@ Due to the [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law, we k \begin{align*} \vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) + (\frac{\hat{r}}{r^{2}} \cdot \vec{\nabla})\vec{J} - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}} \end{align*} -The first directional derivative is zero because $\vec{J}$ does not depend on the same coordinates as $\vec{\nabla}}$ +The first directional derivative is zero because $\vec{J}$ does not depend on the same coordinates as $\vec{\nabla}$ with the same reasoning as for the divergence, so we have: \begin{align*} \vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}} |