Ch1 lecture notes and labs

book/1_gradient_divergence_curl/curl.md

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+# Curl
+
+The curl is a peculiar operator because it turns a vector field into another vector field. Often a determinant expression is used, but we can give it in terms of even and odd permutations of the triplet $(x,y,z)$, which is easy to remember. Let $\boldsymbol v(\boldsymbol r,t)$ be a continuously differentiable space-time function. Then
+
+$$
+\nabla\times\boldsymbol v = \hat{\boldsymbol x}(\partial_y v_z - \partial_z v_y) + \hat{\boldsymbol y}(\partial_z v_x - \partial_x v_z) + \hat{\boldsymbol z}(\partial_x v_y - \partial_y v_x).
+$$ (eq:curl)
+
+Notice that the three terms have the triplet pairs $(xyz)-(xzy)$, $(yzx)-(yxz)$, and $(zxy)-(zyx)$ in (1) direction, (2) differentiation, and (3) vector component. The first in each pair is an even permutation of $(xyz)$ and the second is an odd permutation. It can be seen that in each term the direction of the resulting vector $\boldsymbol w = \nabla\times\boldsymbol v$ is perpendicular both to the direction of differentiation and to the component of $\boldsymbol v$ that is being differentiated. The curl is interpreted as the rate of circulation of the vector field $\boldsymbol v$.
+
+## The circulation integral
+
+Let us first look at the situation of an elementary area with centre point $\boldsymbol r_p = \left(\tfrac{1}{2}\mathrm{d}x,\ \tfrac{1}{2}\mathrm{d}y,\ \tfrac{1}{2}\mathrm{d}z\right)$. The circulation integral for the path from $y=0$ to $y=\mathrm{d}y$ at $x=\tfrac{1}{2}\mathrm{d}x$ and $z=0$ is given by
+
+$$
+\int_{y=0}^{\mathrm{d}y} v_y\!\left(\tfrac{1}{2}\mathrm{d}x,\,y,\,0\right)\mathrm{d}l ,
+$$
+
+and completing the circulation for all four sides of the elementary area as shown in {numref}`fig-circulation`, we find
+
+```{figure} figures/circulation.png
+:name: fig-circulation
+:width: 55%
+
+Net circulation integral on a rectangle around the point $\boldsymbol r_p$.
+```
+
+$$
+\begin{aligned}
+\oint_{\boldsymbol r}\boldsymbol\tau\cdot\boldsymbol v\,\mathrm{d}l
+&= \int_{y=0}^{\mathrm{d}y} v_y\!\left(\tfrac{1}{2}\mathrm{d}x,\,y,\,0\right)\mathrm{d}l
+ + \int_{z=0}^{\mathrm{d}z} v_z\!\left(\tfrac{1}{2}\mathrm{d}x,\,\mathrm{d}y,\,z\right)\mathrm{d}l \\
+&\quad - \int_{y=0}^{\mathrm{d}y} v_y\!\left(\tfrac{1}{2}\mathrm{d}x,\,y,\,\mathrm{d}z\right)\mathrm{d}l
+ - \int_{z=0}^{\mathrm{d}z} v_z\!\left(\tfrac{1}{2}\mathrm{d}x,\,0,\,z\right)\mathrm{d}l ,
+\end{aligned}
+$$ (eq:circint)
+
+where the symbol $\oint$ stands for a closed loop integral, $\boldsymbol\tau$ is the local unit tangent vector along the edges of $\mathbb{S}$, and $l$ is the arclength along the path.
+
+Earlier we introduced the Taylor series expansion for a function in one variable, and we give it here for each vector component of $\boldsymbol v$ around the point $\boldsymbol r_p$,
+
+$$
+\begin{aligned}
+v_x(\boldsymbol r) &= v_x(\boldsymbol r_p) + \partial_x v_x\left(x-\tfrac{1}{2}\mathrm{d}x\right) + \partial_y v_x\left(y-\tfrac{1}{2}\mathrm{d}y\right) + \partial_z v_x\left(z-\tfrac{1}{2}\mathrm{d}z\right) + \text{higher order terms}, \\
+v_y(\boldsymbol r) &= v_y(\boldsymbol r_p) + \partial_x v_y\left(x-\tfrac{1}{2}\mathrm{d}x\right) + \partial_y v_y\left(y-\tfrac{1}{2}\mathrm{d}y\right) + \partial_z v_y\left(z-\tfrac{1}{2}\mathrm{d}z\right) + \text{higher order terms}, \\
+v_z(\boldsymbol r) &= v_z(\boldsymbol r_p) + \partial_x v_z\left(x-\tfrac{1}{2}\mathrm{d}x\right) + \partial_y v_z\left(y-\tfrac{1}{2}\mathrm{d}y\right) + \partial_z v_z\left(z-\tfrac{1}{2}\mathrm{d}z\right) + \text{higher order terms},
+\end{aligned}
+$$
+
+which is the Taylor series expansion to first order. Substituting these results in {eq}`eq:circint` leads to
+
+$$
+\oint_{\boldsymbol r}\boldsymbol\tau\cdot\boldsymbol v\,\mathrm{d}l = (\partial_y v_z - \partial_z v_y)\,\mathrm{d}y\,\mathrm{d}z + \text{higher order terms}.
+$$
+
+Similarly, the net circulation integral for the elementary surface area perpendicular to the $\hat{\boldsymbol y}$-axis is $(\partial_z v_x - \partial_x v_z)\,\mathrm{d}x\,\mathrm{d}z$, and the net circulation integral for the elementary surface area perpendicular to the $\hat{\boldsymbol z}$-axis is $(\partial_x v_y - \partial_y v_x)\,\mathrm{d}x\,\mathrm{d}y$.
+
+For this reason, the physical interpretation of the curl is the net circulation per unit of surface area at the point $\boldsymbol r_p$, and it can be expressed as
+
+$$
+\hat{\boldsymbol n}\cdot(\nabla\times\boldsymbol v) = \lim_{S\rightarrow 0}\frac{\oint_{\boldsymbol r}\boldsymbol\tau\cdot\boldsymbol v\,\mathrm{d}l}{A},
+$$ (eq:circ1)
+
+in which
+
+$$
+A = \int_{\boldsymbol r\in\mathbb{S}}\mathrm{d}S
+$$
+
+is the surface area of the surface $\mathbb{S}$.
+
+:::{admonition} The curl in words
+:class: tip
+The curl of a vector field is the net circulation of that field per unit of surface area, and it points along the normal of the surface for which that circulation is largest.
+:::
+
+## Stokes' integral theorem
+
+The unit normal $\hat{\boldsymbol n}$ in {eq}`eq:circ1` is the normal vector to the surface area $\mathbb{S}$ and is oriented to the side of advance of a right-hand screw as it is turned in the direction of $\boldsymbol\tau$ around the boundary of $\mathbb{S}$. This leads to Stokes' integral theorem, given by
+
+$$
+\oint_{\boldsymbol r}\boldsymbol\tau\cdot\boldsymbol v\,\mathrm{d}l = \int_{\boldsymbol r\in\mathbb{S}}\hat{\boldsymbol n}\cdot(\nabla\times\boldsymbol v)\,\mathrm{d}S .
+$$ (eq:stokes)
+
+## Exercises
+
+1. Use Stokes' theorem to evaluate $\oint_{\boldsymbol r}\boldsymbol\tau\,\mathrm{d}l$, where $\boldsymbol\tau$ is the unit tangent along the closed boundary of the area $\mathbb{S}$. The integration runs in the direction of circulation that forms a right-handed system with the unit normal vector on $\mathbb{S}$.
+2. Show that when $\boldsymbol v(\boldsymbol r) = \boldsymbol a\,p(\boldsymbol r)$, where $\boldsymbol a$ is an arbitrary constant vector and $p(\boldsymbol r)$ is a continuously differentiable scalar function, Stokes' integral theorem gives
+
+    $$
+    \oint_{\boldsymbol r}\boldsymbol\tau\,p\,\mathrm{d}l = \int_{\boldsymbol r\in\mathbb{S}}(\hat{\boldsymbol n}\times\nabla)p\,\mathrm{d}S,
+    $$
+
+    which is Stokes' theorem for the gradient.
+3. Show that when $\boldsymbol v(\boldsymbol r) = \boldsymbol a\times\boldsymbol w(\boldsymbol r)$, where $\boldsymbol a$ is an arbitrary constant vector and $\boldsymbol w(\boldsymbol r)$ is a continuously differentiable vector function, Stokes' integral theorem gives
+
+    $$
+    \oint_{\boldsymbol r}\boldsymbol\tau\times\boldsymbol w\,\mathrm{d}l = \int_{\boldsymbol r\in\mathbb{S}}(\hat{\boldsymbol n}\times\nabla)\times\boldsymbol w\,\mathrm{d}S .
+    $$
+4. Ampère's law is given by $\nabla\times\boldsymbol H = \boldsymbol J$, which states that the electric current is equal to the curl of the magnetic field. Convert this to integral form using Stokes' theorem, by using a flat surface $\mathbb{S}$ and choosing a unit normal vector $\hat{\boldsymbol n}$ on $\mathbb{S}$. You should find
+
+    $$
+    \oint_{\boldsymbol r}\boldsymbol H\cdot\boldsymbol\tau\,\mathrm{d}l = \int_{\boldsymbol r\in\mathbb{S}}\hat{\boldsymbol n}\cdot\boldsymbol J\,\mathrm{d}S .
+    $$
+
+    Show in which direction the unit tangent vector $\boldsymbol\tau$ is circulating at the boundary of $\mathbb{S}$, and in which direction through the surface a positive current should run.