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author | Preston Pan <preston@nullring.xyz> | 2023-07-23 09:12:03 -0700 |
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committer | Preston Pan <preston@nullring.xyz> | 2023-07-23 09:12:03 -0700 |
commit | 80da24887ac760a9d18936634d8d46c0643521ee (patch) | |
tree | 20c7846353ca983a10c724d965631e28d3fe0587 /mindmap/del operator.org | |
parent | c335c05f511a373681d8644500d7750a519f58fa (diff) |
add a lot of mindmap articles
Diffstat (limited to 'mindmap/del operator.org')
-rw-r--r-- | mindmap/del operator.org | 103 |
1 files changed, 103 insertions, 0 deletions
diff --git a/mindmap/del operator.org b/mindmap/del operator.org new file mode 100644 index 0000000..7410e30 --- /dev/null +++ b/mindmap/del operator.org @@ -0,0 +1,103 @@ +:PROPERTIES: +:ID: 4bfd6585-1305-4cf2-afc0-c0ba7de71896 +:END: +#+title: del operator +#+author: Preston Pan +#+html_head: <link rel="stylesheet" type="text/css" href="../style.css" /> +#+html_head: <script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script> +#+html_head: <script id="MathJax-script" async src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js"></script> +#+options: broken-links:t + +* Definition +The operator /del/ in \( n \) dimensional euclidean space is defined as follows: +\begin{align*} +\vec{\nabla} := \sum_{i = 1}^{n} \hat{e_{i}}\frac{\partial}{\partial e_{i}} +\end{align*} +Where \( \frac{\partial}{\partial e_{k}}\) is the [[id:3993a45d-699b-4512-93f9-ba61f498f77f][partial derivative]] with respect to the \(k^{th}\) orthogonal axis, and \( \hat{e}_{k} \) is the +orthogonal basis vector pointing in that direction. In three dimensional euclidean +space using Cartesian coordinates, the del operator would look like: + +\begin{align*} +\vec{\nabla} = \begin{bmatrix} +\frac{\partial}{\partial x} \\ +\frac{\partial}{\partial y} \\ +\frac{\partial}{\partial z} +\end{bmatrix} += \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} +\end{align*} + +The del operator is what is called a /linear operator/ because it is consistent with operations +pertaining to linear algebra. +* Usage +The del operator is useful for representing the gradient, divergence, and curl of a given +scalar or vector field. + +** Gradient +:PROPERTIES: +:ID: 3587c3b4-c3d8-4ff1-b0ba-8eecb1ef0e4c +:END: +Multiplying the del operator by a scalar field yields a vector that is called the *gradient* +of a function: +\begin{align*} +\vec{\nabla}f = \begin{bmatrix} +\frac{\partial f}{\partial x} \\ +\frac{\partial f}{\partial y} \\ +\frac{\partial f}{\partial z} +\end{bmatrix} += \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k} +\end{align*} +Where this vector points in the direction of the greatest rate of change, and has a magnitude corresponding +with the slope. The reason why is somewhat intuitive, if you think about it a little. + +** Divergence +:PROPERTIES: +:ID: 12a2d5b3-f98c-45e5-9107-5560288b5aa8 +:END: +Taking the dot product of the del operator with a vector field yields a scalar function, which is called the divergence: +\begin{align*} +\vec{\nabla} \cdot \vec{f} = \frac{\partial f_{x}}{\partial x} + \frac{\partial f_{y}}{\partial y} + \frac{\partial f_{z}}{\partial z} +\end{align*} +Where \( f_{n} \) is the \( n \) component of \( \vec{f} \). + +You can think of it as measuring the rate of change of the outwards or inwards direction of a vector field. +In order to think about this more clearly, we can think about the two dimensional case with just x and y. +Given a two-dimensional vector field, a two-dimensional divergence would look like this: +\begin{align*} +\vec{\nabla} \cdot \vec{f} = \frac{\partial f_{x}}{\partial x} + \frac{\partial f_{y}}{\partial y} +\end{align*} +and to explain this further, let's take a vector \( \vec{v} \), as well as two other vectors to compare it with, +\( \vec{v_{up}}\), and \( \vec{v_{right}} \). Then, we take \( \vec{r} = \vec{f}(\vec{v}) \) and compare it to +\( \vec{r_{up}} \) and \( \vec{r_{right}}\). We then compare the x component of the right vector with the original one, +and we compare the y component of the up vector with the original one, by taking the difference. We then sum these +differences, and what we are left with is a measurement of how spread apart the directions and magnitudes of these vectors +are in this local area. If these \( \vec{r} \) vectors are infinitely close to each other, we can consider this comparison to be analogous to the divergence at that point. +This argument naturally extends to three dimensions. + +** Curl +:PROPERTIES: +:ID: b25e0e44-c764-4f0a-a5ad-7f9d79c7660d +:END: +The curl of a vector field is defined as follows: +\begin{align*} +\vec{\nabla} \times \vec{f} = \hat{i}(\frac{\partial f_{z}}{\partial y} - \frac{\partial f_{y}}{\partial z}) - \hat{j}(\frac{\partial f_{z}}{\partial x} - \frac{\partial f_{x}}{\partial z}) + \hat{k}(\frac{\partial f_{y}}{\partial x} - \frac{\partial f_{x}}{\partial y}). +\end{align*} +Where the equation above is derived from the definition of the cross product. It represents the rate of change of a +vector field "perpendicular" to the divergence of the field. In fact, if you have any field \( \vec{f} \), +you can represent this field as an addition of a curl-less field and a divergence-less field. +Another way to think of it is that you are measuring the strength of rotational component of the vector field about a certain axis. + +** Laplacian +:PROPERTIES: +:ID: 65004429-a6b7-41f2-8489-07605841da3d +:END: +The Laplacian is defined as follows: +\begin{align*} +\nabla^{2}\vec{f} = \nabla \cdot \nabla\vec{f} +\end{align*} +It returns a scalar field and is the multivariable analogue to the second derivative. Because both the divergence +and gradient have been described, I feel it is trivial to understand the Laplacian. + +** Product Rules +The product rules pertaining to the del operator are consistent with that of linear algebra. +For example, \( \vec{\nabla} \times \vec{\nabla}f = 0\). You can show this yourself quite easily, so I find no need to go over it here. +When in doubt, just assume the del works the same way as any old vector, and you will usually be correct. |