MATH 2031 Lecture Notes (Michaelmas) - 4 Differentiating vector fields

There are two commonly-used differential operators on vector fields in $\Real^3$: the “curl” and the “divergence”.

These are more useful than individual partial derivatives of the components, because they are defined in a way that is independent of a particular choice of coordinates.

4.1 Curl

The curl of $\fb(\xb)$ is the vector field $\nabla\times\fb$ whose component in the direction $\hat{\nb}$ is \[ (\nabla\times\fb)\cdot\hat{\nb} = \lim_{|A|\to 0}\frac{1}{|A|}\oint_C\fb\cdot\,d\xb, \] where $C$ is a loop in the plane with normal $\hat{\nb}$, and $|A|$ is the area enclosed. The circulation is calculated in the right-handed direction relative to $\hat{\nb}$.

Thus curl represents circulation per unit area of $\fb$ around any possible direction. By definition it points in the direction with maximum circulation around it.

To find the Cartesian components of $\nabla\times\fb$, we can use Green’s Theorem (Theorem 3.3). Let $\hat{\nb}=\eb_3$, so $C$ lies in the $xy$-plane. Then \[\begin{align*} (\nabla\times\fb)\cdot\eb_3 &= \lim_{|A|\to 0}\frac{1}{|A|}\oint_C\fb\cdot\,d\xb\\ &=\lim_{|A|\to 0}\frac{1}{|A|}\int_A\left(\ddy{f_2}{x} - \ddy{f_1}{y}\right)\,dA\\ &= \ddy{f_2}{x} - \ddy{f_1}{y}. \end{align*}\]

The last step is rigorously justified by the Mean Value Theorem for Double Integrals, which says that if $g(\xb)$ is continuous on a closed, bounded and connected subset $A\subset\Real^2$ then there exists $\xb_0\in A$ such that \[ \int_A g(\xb)\,dA = |A|g(\xb_0). \]

This is a generalisation of the 1-d version $\displaystyle\int_a^bf(x)\,dx = (b-a)f(x_0)$, as illustrated here:

The Mean Value Theorem for Integrals in 1-d.

Setting $\displaystyle g(\xb)=\left(\ddy{f_2}{x} - \ddy{f_1}{y}\right)$ gives \[ \lim_{|A|\to 0}\frac{1}{|A|}\int_Ag(\xb)\,dA = \lim_{|A|\to 0}\frac{1}{|A|}|A|g(\xb_0) = \lim_{|A|\to 0}g(\xb_0) = g(\xb). \]

Applying Green’s Theorem to curves in the $yz$-plane or $xz$-plane, and taking care with the orientations, gives the other two Cartesian components:

\[\begin{align*} \nabla\times\fb &= \left(\ddy{f_3}{y} - \ddy{f_2}{z} \right)\eb_1 + \left(\ddy{f_1}{z} - \ddy{f_3}{x} \right)\eb_2 + \left(\ddy{f_2}{x} - \ddy{f_1}{y} \right)\eb_3\\ &= \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ \displaystyle \ddy{}{x} & \displaystyle\ddy{}{y} & \displaystyle\ddy{}{z}\\ f_1 & f_2 & f_3 \end{vmatrix}. \end{align*}\]

This last expression explains the $\nabla\times\fb$ notation, if we think of $\nabla$ as the vector differential operator \[ \nabla = \eb_1\ddy{}{x} + \eb_2\ddy{}{y} + \eb_3\ddy{}{z}. \] You will also see the curl written as “$\mathrm{curl}\,\fb$” or in older books as “$\mathrm{rot}\,\fb$” (standing for “rotation”).

Example. Find the curl of $\fb(\xb)=\omega(-y\eb_1 + x\eb_2)$ where $\omega\in\Real$.

\[\begin{align*} \nabla\times\fb &= \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ \displaystyle \ddy{}{x} & \displaystyle\ddy{}{y} & \displaystyle\ddy{}{z}\\ -\omega y & \omega x & 0 \end{vmatrix}\\ &= \left(\ddy{}{y}(0) - \ddy{}{z}(\omega x)\right)\eb_1 - \left(\ddy{}{x}(0) - \ddy{}{z}(-\omega y)\right)\eb_2 + \left(\ddy{}{x}(\omega x) - \ddy{}{y}(-\omega y)\right)\eb_3\\ &= 2\omega \eb_3. \end{align*}\]

This tells us that $\fb$ has local circulation only around the $z$-direction.

If $\omega>0$, then $(\nabla\times\fb)\cdot\eb_3 > 0$, indicating anti-clockwise circulation of the vector field. It is independent of $x, y$, so $\fb$ has the same circulation around every infinitesimal loop in the $xy$-plane:

A physical way to think of curl is to imagine $\fb(\xb)$ to be the velocity field $\vb(\xb)$ of a fluid. A small “paddle wheel” will align its axis with the direction of $\nabla\times\vb$ and rotate with angular speed $\propto|\nabla\times\vb|$.

The previous example corresponds to a rigid rotation of the fluid, where each particle of fluid follows the trajectory $\xb(t) = r\cos(\omega t)\eb_1 + r\sin(\omega t)\eb_2$. The corresponding velocity field is \[ \vb(\xb) = \ddt{\xb} = -\omega r\sin(\omega t)\eb_1 + \omega r\cos(\omega t)\eb_2 = \omega(-y\eb_1 + x\eb_2). \] This velocity field has $|\nabla\times\vb| = 2|\omega|$, i.e. twice the angular speed. The non-zero curl tells us that the paddle wheel rotates as it moves around the origin:

Example. Find the curl of $\displaystyle\fb(\xb)=\frac{-y}{x^2+y^2}\eb_1 + \frac{x}{x^2+y^2}\eb_2$ for $\xb\neq\bfzero$.

Clearly the partial derivatives don’t exist at $\xb=\bfzero$ so the curl is not defined there.

For any other point, we have \[\begin{align*} \nabla\times\fb &= \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ \displaystyle \ddy{}{x} & \displaystyle\ddy{}{y} & \displaystyle\ddy{}{z}\\ \displaystyle\frac{-y}{x^2+y^2} & \displaystyle\frac{x}{x^2+y^2} & 0 \end{vmatrix}\\ &= (0)\eb_1 + (0)\eb_2 + \left[\ddy{}{x}\left(\frac{x}{x^2+y^2}\right) - \ddy{}{y}\left(\frac{-y}{x^2+y^2}\right)\right]\eb_3\\ &= \left[\frac{(x^2+y^2) - 2x^2}{(x^2+y^2)^2} + \frac{(x^2+y^2) - 2y^2}{(x^2+y^2)^2}\right]\eb_3\\ &= \bfzero. \end{align*}\]

So (at least away from the origin) this vector field has no local circulation even though it is clearly rotating globally.

Rotation of a paddle wheel in the vortex.

[This more closely approximates the motion when you pull the plug in a bathtub – in fluid mechanics it is called a line vortex.]

From the Cartesian expression, we see that we the $i$th component of $\nabla\times\fb$ may be written in index notation as \[ \big[\nabla\times\fb\big]_i = \epsilon_{ijk}\ddy{}{x_j}(f_k) = \epsilon_{ijk}\ddy{f_k}{x_j}. \]

Example. If $\fb(\xb) = r^2\ab$, where $r=|\xb|$ and $\ab$ is a constant vector, use index notation to show that $\nabla\times\big(\nabla\times\fb\big) = -4\ab$.

The $i$th component is \[\begin{align*} \Big[\nabla\times(\nabla\times\fb)\Big]_i &= \epsilon_{ijk}\ddy{}{x_j}\left(\big[\nabla\times\fb\big]_k \right)\\ &= \epsilon_{ijk}\ddy{}{x_j}\Big(\epsilon_{klm}\ddy{}{x_l}(f_m)\Big)\\ &= \epsilon_{ijk}\epsilon_{klm}\ddy{}{x_j}\Big(\ddy{}{x_l}(f_m)\Big) \quad \textrm{[since $\epsilon_{klm}$ is a constant]}\\ &= \epsilon_{ijk}\epsilon_{klm}\ddy{}{x_j}\Big(\ddy{}{x_l}(r^2a_m)\Big)\\ &= \epsilon_{ijk}\epsilon_{klm}a_m\ddy{}{x_j}\Big(\ddy{}{x_l}(r^2)\Big) \quad \textrm{[since $a_m$ is a constant]}. \end{align*}\] Note that $r^2=x^2 + y^2 + z^2$. So $\displaystyle\ddy{}{x_l}(r^2) = 2x_l$. Therefore \[\begin{align*} \Big[\nabla\times(\nabla\times\fb)\Big]_i &=\epsilon_{ijk}\epsilon_{klm}a_m\ddy{}{x_j}(2x_l)\\ &= 2\epsilon_{ijk}\epsilon_{klm}a_m\delta_{jl}\\ &= 2(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\delta_{jl}a_m\\ &= 2\delta_{il}\delta_{jm}\delta_{jl}a_m - 2\delta_{im}\delta_{jl}\delta_{jl}a_m\\ &= 2\delta_{ij}a_j - 2\delta_{jj}a_i\\ &= 2a_i - 6a_i\\ &= -4a_i. \end{align*}\]

Proposition 4.1 (Properties of curl.) Let $a,b$ be scalar constants, $\fb(\xb)$, $\gb(\xb)$ be vector fields, and $h(\xb)$ be a scalar field, all in $\Real^3$. Then

$(i)\;\;$ $\nabla\times(a\fb + b\gb) = a\nabla\times\fb + b\nabla\times\gb.$

$(ii)\;\;$ $\nabla\times(h\fb) = (\nabla h)\times\fb + h\nabla\times\fb.$

$(iii)\;\;$ $\nabla\times(\nabla h) = \bfzero$.

Proof. $(i)$ This just follows from the linearity of partial differentiation.

$(ii)$ Using index notation, we have \[ \big[\nabla\times(h\fb)\big]_i = \epsilon_{ijk}\ddy{}{x_j}(hf_k). \] Since $h$ and the $f_k$ are both scalars, we can use the normal product rule from Calculus to find \[\begin{align*} \big[\nabla\times(h\fb)\big]_i &= \epsilon_{ijk}\left(\ddy{h}{x_j}f_k + h\ddy{f_k}{x_j} \right)\\ &= \epsilon_{ijk}\big[\nabla h\big]_j f_k + h \epsilon_{ijk}\ddy{f_k}{x_j}\\ &= \Big[(\nabla h) \times\fb \Big]_i + \Big[ h\nabla\times\fb \Big]_i. \end{align*}\] Here we used the fact that the $j$th component of $\nabla h$ is $\displaystyle\ddy{h}{x_j}$.

$(iii)$ Again using index notation, we have \[\begin{align*} \Big[\nabla\times(\nabla h)\Big]_i &= \epsilon_{ijk}\ddy{}{x_j}\left(\ddy{h}{x_k}\right)\\ &= \epsilon_{ijk}\ddy{}{x_k}\left(\ddy{h}{x_j}\right) \quad \textrm{[assuming $h$ is sufficiently smooth]}\\ &= \epsilon_{ikj}\ddy{}{x_j}\left(\ddy{h}{x_k}\right) \quad \textrm{[relabelling $j\leftrightarrow k$]}\\ &= -\epsilon_{ijk}\ddy{}{x_j}\left(\ddy{h}{x_k}\right) \quad \textrm{[using antisymmetry $\epsilon_{ikj}=-\epsilon_{ijk}$]}. \end{align*}\] Since this expression is equal to minus itself, it must be zero.

By “sufficiently smooth” in $(iii)$, I mean that the second partial derivatives $\displaystyle\ddy{^2h}{x\partial y}$ and $\displaystyle\ddy{^2h}{y\partial x}$ are continuous. This is the condition required for them to be equal, by Clairaut’s Theorem.

4.2 Divergence

The divergence of $\fb(\xb)$ is the scalar field \[ \nabla\cdot\fb = \lim_{|V|\to 0}\frac{1}{|V|}\oint_S\fb\cdot\dS, \] where $S$ is a closed surface enclosing a volume $|V|$.

The divergence represents flux per unit volume of $\fb$ away from a given point. A positive divergence means an outward flux, while a negative divergence means an inward flux.

To find the Cartesian expression for $\nabla\cdot\fb$, consider an infinitesimal cube of side $\delta$:

From the definition, $\displaystyle \nabla\cdot\fb = \lim_{\delta\to 0}\frac{1}{\delta^3}\sum_{i=1}^6\int_{S_i}\fb\cdot\hat{\nb}\,dS$, where $S_i$ is the $i$th face.

The front face (say $S_1$) has normal $\hat{\nb}=\eb_1$, so \[ \int_{S_1}\fb\cdot\hat{\nb}\,dS = \int_{S_1}\fb\cdot\eb_1\,dS = \int_{S_1}f_1\,dS = \delta^2f_1(\xb_1), \] where the last step used the Mean Value Theorem for Double Integrals, so that $\xb_1$ is some (unknown) point on $S_1$.

Repeating for the other five faces gives \[\begin{align*} \nabla\cdot\fb &= \lim_{\delta\to 0}\left(\frac{f_1(\xb_1)-f_1(\xb_2)}{\delta} + \frac{ f_2(\xb_3)-f_2(\xb_4)}{\delta} + \frac{f_3(\xb_5)-f_3(\xb_6)}{\delta}\right)\\ &= \ddy{f_1}{x} + \ddy{f_2}{y} + \ddy{f_3}{z}. \end{align*}\]

Example. Find the divergence of $\fb(\xb)=-y\eb_1 + x\eb_2.$

\[ \nabla\cdot\fb = \ddy{}{x}(-y) + \ddy{}{y}(x) + \ddy{}{z}(0) = 0. \] So here every infinitesimal cube has zero net flux of $\fb$, in this case equal “inflow” and “outflow”:

A vector field with zero divergence is sometimes called solenoidal.

Example. Find the divergence of $\fb(\xb)=x\eb_1 + y\eb_2 + z\eb_3.$

\[ \nabla\cdot\fb = \ddy{}{x}(x) + \ddy{}{y}(y) + \ddy{}{z}(z) = 3. \]

We can visualise $\fb$ in the $z=0$ plane:

Since $\nabla\cdot\fb$ is constant, any infinitesimal cube has the same (net) outward flux. (If instead $\nabla\cdot\fb<0$ then it would be an inward flux.)

Think of points of positive divergence as sources and points of negative divergence as sinks.

From the Cartesian expression, we see that $\nabla\cdot\fb$ may be written in index notation as \[ \nabla\cdot\fb = \ddy{}{x_j}(f_j) = \ddy{f_j}{x_j}. \]

Proposition 4.2 (Properties of divergence.) Let $a,b$ be scalar constants, $\fb(\xb)$, $\gb(\xb)$ be vector fields, and $h(\xb)$ be a scalar field, all in $\Real^3$. Then

$(i)\;\;$ $\nabla\cdot(a\fb + b\gb) = a\nabla\cdot\fb + b\nabla\cdot\gb$.

$(ii)\;\;$ $\nabla\cdot(h\fb) = (\nabla h)\cdot\fb + h\nabla\cdot\fb$.

$(iii)\;\;$ $\nabla\cdot(\nabla\times\fb) = \bfzero$.

Proof. $(i)$ This just follows from the linearity of partial differentiation.

$(ii)$ By the product rule, $\displaystyle\nabla\cdot(h\fb) = \ddy{}{x_j}(hf_j) = \ddy{h}{x_j}f_j + h\ddy{f_j}{x_j} = (\nabla h)\cdot\fb + h\nabla\cdot\fb.$

$(iii)$ We have \[\begin{align*} \ddy{}{x_j}\left(\epsilon_{jkl}\ddy{f_l}{x_k}\right) &= \epsilon_{jkl}\ddy{^2 f_l}{x_j\partial x_k}\\ &= \epsilon_{jkl}\ddy{^2 f_l}{x_k\partial x_j} \quad\textrm{[assuming $f_l$ is sufficiently smooth]}\\ &= \epsilon_{kjl}\ddy{^2 f_l}{x_j\partial x_k} \quad\textrm{[relabelling $j\leftrightarrow k$]}\\ &= -\epsilon_{jkl}\ddy{^2 f_l}{x_j\partial x_k} \quad\textrm{[using antisymmetry $\epsilon_{kjl}=-\epsilon_{jkl}$]}. \end{align*}\] Since this expression is equal to minus itself, it must be zero.

We define the Laplacian of a scalar field $h(\xb)$ to be the second-derivative operator \[ \nabla^2 h = \nabla\cdot(\nabla h). \] (Sometimes written instead as $\Delta h$.)

The Laplacian is the differential operator appearing in the Laplace equation $\nabla^2 h = 0$ (hence the name), as well as the heat and wave equations. More on these partial differential equations next term!

Example. Compute $\nabla^2 h$ for $h(\xb)=x^3 - 3xy^2$.

In Cartesian coordinates, we have \[ \nabla\cdot(\nabla h) = \ddy{}{x_j}\left(\ddy{h}{x_j}\right) = \ddy{^2h}{x^2} + \ddy{^2h}{y^2} + \ddy{^2h}{z^2}. \] So here \[ \nabla^2 h = \ddy{}{x}(3x^2 - 3y^2) + \ddy{}{y}(-6xy) = 6x - 6x = 0. \]

Functions like this that solve $\nabla^2 h=0$ are called harmonic functions.

If you study Complex Analysis II, you will know that both the real and imaginary parts of a differentiable complex function $f(x+iy)=u(x,y) + iv(x,y)$ satisfy the Cauchy-Riemann equations \[ \ddy{u}{x} = \ddy{v}{y}, \quad \ddy{u}{y}=-\ddy{v}{x}. \] By differentiating you can show that $\nabla^2 u = \nabla^2 v = 0$.

4.3 The Divergence Theorem

This is a generalisation of Green’s Theorem to a three-dimensional volume.

Theorem 4.1 (Divergence Theorem.) Let $\fb(\xb)$ be a vector field with continuous partial derivatives on a region $V\subset\Real^3$, and let $S$ be the oriented closed surface that bounds $V$ (with outward unit normal). Then \[ \oint_S\fb\cdot\dS = \int_V(\nabla\cdot\fb)\dV. \]

This is also known as Gauss’ Theorem or sometimes Ostrogradsky’s Theorem.

An intuitive justification for Theorem 4.1 is given by approximating $V$ with infinitesimal cubes:

The volume integral is the sum of $\nabla\cdot\fb$ over each cube. By definition of divergence, this is (in the limit) the integral of $\fb\cdot\hat{\nb}$ over the cube surface, where $\hat{\nb}$ is the outward normal. These surface integrals cancel between neighbouring cubes, leaving only the contribution from the boundary $S$.

Proof. We follow a similar argument to Green’s Theorem (Theorem 3.3).

First, let $\fb(\xb)=f_3(\xb)\eb_3$ and assume that $V$ has the form $(x,y)\in A$, $z_0(x,y)\leq z\leq z_1(x,y)$.

Domain for the Divergence Theorem proof.

Then \[\begin{align*} \int_V(\nabla\cdot\fb)\dV &= \int_V\ddy{f_3}{z}\,dV\\ &= \int_A\left(\int_{z_0(x,y)}^{z_1(x,y)}\ddy{f_3}{z}\,dz\right)\,dA\\ &= \int_A\Big[f_3\big(x,y,z_1(x,y)\big) - f_3\big(x,y,z_0(x,y)\big)\Big]\,dA \quad \textrm{[Fund. Thm. of Calculus]}. \end{align*}\] On the other hand, the top surface $z=z_1(x,y)$ has parametrisation $\xb = x\eb_1 + y\eb_2 + z_1(x,y)\eb_3$ giving normal vector \[ \ddy{\xb}{x}\times\ddy{\xb}{y} = \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ 1 & 0 & \displaystyle\ddy{z_1}{x}\\ 0 & 1 & \displaystyle\ddy{z_1}{y} \end{vmatrix} = -\ddy{z_1}{x}\eb_1 - \ddy{z_1}{y}\eb_2 + \eb_3, \] and hence surface integral \[ \int_{z=z_1(x,y)}(f_3\eb_3)\cdot\dS = \int_A(f_3\eb_3)\cdot\left(\ddy{\xb}{x}\times\ddy{\xb}{y}\right)\,dA = \int_Af_3\big(x,y,z_1(x,y)\big)\,dA. \] Similarly the bottom surface has normal $-\eb_3$ giving \[ \int_{z=z_0(x,y)}(f_3\eb_3)\cdot\dS = - \int_Af_3\big(x,y,z_0(x,y)\big)\,dA. \] There is no contribution from the vertical side face, so putting this together gives \[ \int_V\ddy{f_3}{z}\,dV = \oint_S(f_3\eb_3)\cdot\dS. \] For suitable domains a similar argument gives \[ \int_V\ddy{f_1}{x}\,dV = \oint_S(f_1\eb_1)\cdot\dS \quad \textrm{and} \quad \int_V\ddy{f_2}{y}\,dV = \oint_S(f_2\eb_2)\cdot\dS, \] and summing gives Theorem 4.1.

If $V$ is a more complicated shape, it must first be broken into a sum over simpler subdomains.

The Divergence Theorem explains why the equation \[ \ddy{\phi}{t} + \nabla\cdot\Fb = 0 \] is called a conservation law. Integrating over some volume $V$ (and assuming $\displaystyle\ddy{\phi}{t}$ is continuous) gives \[ \ddt{}{\int_V\phi\,dV = -\oint_{\partial V}\Fb\cdot\dS}. \] This integral form of the conservation law says that the total amount of $\phi$ in $V$ changes only if there is a flux, $\Fb$, through the boundary.

Example. For a vector field $\fb(\xb)$ and scalar field $h(\xb)$ on a volume $V$ with boundary $S$, show that $\displaystyle\int_V(\fb\cdot\nabla h)\,dV = \oint_S(h\fb)\cdot\dS - \int_V(h\nabla\cdot\fb)\dV$.

Recall from earlier that \[ \nabla\cdot(h\fb) = (\nabla h)\cdot\fb + h\nabla\cdot\fb. \] Integrating over a volume $V$ gives \[\begin{align*} \int_V\nabla\cdot(h\fb)\dV &= \int_V(\fb\cdot\nabla h)\dV + \int_V(h\nabla\cdot\fb)\dV\\ \iff\quad \oint_S(h\fb)\cdot\dS &= \int_V(\fb\cdot\nabla h)\dV + \int_V(h\nabla\cdot\fb)\dV \quad \textrm{[by the Divergence Theorem]}. \end{align*}\] Rearranging gives the required formula.

This is analogous to integration by parts in single-variable Calculus.

Unlike curl, the divergence is defined for $\fb\in\Real^n$ for any dimension $n>0$. Moreover, the Divergence Theorem holds for any $n$, where $V$ is an $n$-dimensional volume bounded by an $(n-1)$-dimensional surface $S$.

4.4 Stokes’ Theorem

This is a second generalisation of Green’s Theorem.

Theorem 4.2 (Stokes’ Theorem.) Let $S$ be an oriented surface whose boundary is the closed curve $C$, and let $\fb(\xb)$ be a vector field on $S$ with continuous partial derivatives. Then \[ \int_S(\nabla\times\fb)\cdot\dS = \oint_{C}\fb\cdot\,d\xb, \] where the orientations of $\dS$ and the line integral must be related in a right-handed sense.

If $S$ is a region in the $xy$-plane, then $\dS = \eb_3\,dS$ and Theorem 4.2 reduces to Green’s Theorem (Theorem 3.3). So Stokes’ Theorem is a generalisation of Green’s Theorem to curved surfaces.

An intuitive justification for Theorem 4.2 is given by approximating $S$ as a polyhedron – for example by “triangulating”:

Approximating a surface by a polyhedron.

The surface integral is the sum of $(\nabla\times\fb)\cdot\hat{\nb}\,dS$ over each triangle. By definition of curl, this is (in the limit) the circulation around the triangle. Since $\hat{\nb}$ has a consistent orientation, the line integrals cancel between neighbouring triangles, leaving only the contribution from the boundary $C$.

Example. Verify Stokes’ Theorem for the vector field $\fb(\xb)=z\eb_1 + x\eb_2 - x\eb_3$ when $S$ is the hemisphere $x^2+y^2+z^2= 1$, $z\geq 0$.

Left-hand side. The curl is \[ \nabla\times\fb = \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ \displaystyle\ddy{}{x} & \displaystyle\ddy{}{y} & \displaystyle\ddy{}{z}\\ z & x & -x \end{vmatrix} = 2\eb_2 + \eb_3. \] We can parametrize the surface using spherical polar coordinates as \[ \xb(\theta,\phi) = \sin\theta\cos\phi\eb_1 + \sin\theta\sin\phi\eb_2 + \cos\theta\eb_3, \quad \theta\in[0,\pi/2], \quad \phi\in[0,2\pi]. \] As we found earlier, $\displaystyle \ddy{\xb}{\theta}\times\ddy{\xb}{\phi} = \sin\theta\xb$, which points upward ($z$-component > $0$). The surface integral is \[\begin{align*} \int_S(\nabla\times\fb)\cdot\dS &= \int_0^{2\pi}\int_0^{\pi/2}(\nabla\times\fb)\cdot\left(\ddy{\xb}{\theta}\times\ddy{\xb}{\phi}\right)\,d\theta\,d\phi\\ &= \int_0^{2\pi}\int_0^{\pi/2}\sin\theta(2\eb_2 + \eb_3)\cdot(x\eb_1 + y\eb_2 + z\eb_3)\,d\theta\,d\phi\\ &= \int_0^{2\pi}\int_0^{\pi/2}(2y+z)\sin\theta\,d\theta\,d\phi\\ &= \int_0^{2\pi}\int_0^{\pi/2}(2\sin^2\theta\sin\phi + \sin\theta\cos\theta)\,d\theta\,d\phi\\ &= \int_0^{2\pi}\int_0^{\pi/2}\frac12\sin(2\theta)\,d\theta\,d\phi\\ &= -\frac{\pi}{2}\Big[\cos(2\theta)\Big]_0^{\pi/2} = \pi. \end{align*}\]

Right-hand side. The boundary $C$ is the circle $x^2+y^2=1$, $z=0$, which has the standard polar coordinate parametrisation \[ \xb(t) = \cos t\eb_1 + \sin t\eb_2, \quad t\in[0,2\pi]. \] Notice that this traces $C$ anti-clockwise, which is consistent with the upward orientation of the surface normal. This parametrisation gives tangent vector $\displaystyle \ddt{\xb}= -\sin t\eb_1 + \cos t\eb_2,$ so the circulation is \[\begin{align*} \oint_C\fb\cdot\,d\xb &= \int_0^{2\pi}\fb\cdot\ddt{\xb}\,dt\\ &= \int_0^{2\pi}(\cos{t}\eb_2 - \cos{t}\eb_3)\cdot(-\sin t\eb_1 + \cos t\eb_2)\,dt\\ &= \int_0^{2\pi}\cos^2{t}\,dt = \pi. \end{align*}\]

Stokes’ Theorem implies that $\displaystyle\int_S(\nabla\times\fb)\cdot\dS$ will be the same for all surfaces $S$ that share the same boundary curve $C$.

Sometimes they are called capping surfaces for $C$.

Example. Previous example, but with $S$ replaced by $S'$, the unit disk $x^2+y^2\leq 1$, $z=0$.

We can parametrise the disk with polar coordinates \[ \xb(r,\theta) = r\cos\theta\eb_1 + r\sin\theta\eb_2, \quad r\in[0,1],\quad \theta\in[0,2\pi]. \] Then \[ \ddy{\xb}{r}\times\ddy{\xb}{\theta} = \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ \cos\theta & \sin\theta & 0\\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} = r\eb_3, \] so \[\begin{align*} \int_{S'}(\nabla\times\fb)\cdot\dS &= \int_0^{2\pi}\int_0^1(\nabla\times\fb)\cdot\left(\ddy{\xb}{r}\times\ddy{\xb}{\theta}\right)\,dr\,d\theta\\ &= \int_0^{2\pi}\int_0^1r(2\eb_2 + \eb_3)\cdot\eb_3\,dr\,d\theta\\ &= \int_0^{2\pi}\int_0^1r\,dr\,d\theta = \int_0^{2\pi}\frac12\,d\theta = \pi. \end{align*}\] We get the same answer as before, because $S'$ and $S$ share the same boundary. (Again, we have to choose the appropriate orientation for it to work.)

Proof (of Theorem 4.2). Suppose $S$ has parametrisation $\xb(u,v)$, so \[\begin{align*} \int_S(\nabla\times\fb)\cdot\dS &= \int_U(\nabla\times\fb)\cdot\left(\ddy{\xb}{u}\times\ddy{\xb}{v}\right)\,du\,dv. \end{align*}\] The basic idea will be to apply Green’s Theorem to the region $U$ in the $uv$-plane.

To simplify the integrand here it is neatest to use index notation. We have \[\begin{align*} (\nabla\times\fb)\cdot\left(\ddy{\xb}{u}\times\ddy{\xb}{v}\right) &= \left(\epsilon_{ijk}\ddy{f_k}{x_j}\right)\left( \epsilon_{ilm}\ddy{x_l}{u}\ddy{x_m}{v}\right)\\ &= \epsilon_{jki}\epsilon_{ilm}\ddy{f_k}{x_j}\ddy{x_l}{u}\ddy{x_m}{v}\\ &= (\delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl})\ddy{f_k}{x_j}\ddy{x_l}{u}\ddy{x_m}{v} \quad \textrm{[by Lemma 1.1]}\\ &= \ddy{f_k}{x_j}\ddy{x_j}{u}\ddy{x_k}{v} - \ddy{f_k}{x_j}\ddy{x_k}{u}\ddy{x_j}{v}\\ &= \left(\ddy{f_k}{x_j}\ddy{x_j}{u} \right)\ddy{x_k}{v} - \left( \ddy{f_k}{x_j}\ddy{x_j}{v} \right)\ddy{x_k}{u}\\ &= \ddy{f_k}{u}\ddy{x_k}{v} - \ddy{f_k}{v}\ddy{x_k}{u} \quad \textrm{[by the chain rule]}\\ &= \ddy{\xb}{v}\cdot\ddy{\fb}{u} - \ddy{\xb}{u}\cdot\ddy{\fb}{v}. \end{align*}\]

Putting this into our surface integral gives \[\begin{align*} \int_S(\nabla\times\fb)\cdot\dS &= \int_U\left(\ddy{\xb}{v}\cdot\ddy{\fb}{u} - \ddy{\xb}{u}\cdot\ddy{\fb}{v} \right)\,du\,dv\\ &= \int_U\left[\ddy{}{u}\left(\fb\cdot\ddy{\xb}{v} \right) -\fb\cdot\ddy{^2\xb}{u\partial v} - \ddy{}{v}\left(\fb\cdot\ddy{\xb}{u}\right) + \fb\cdot\ddy{^2\xb}{v\partial u} \right]\,du\,dv\\ &= \int_U\left[\ddy{}{u}\left(\fb\cdot\ddy{\xb}{v} \right) - \ddy{}{v}\left(\fb\cdot\ddy{\xb}{u}\right) \right]\,du\,dv\\ &= \oint_{\partial U}\left(\fb\cdot\ddy{\xb}{u}\,du + \fb\cdot\ddy{\xb}{v}\,dv \right) \quad \textrm{[by Green's Theorem]}\\ &= \oint_C\fb\cdot\,d\xb. \end{align*}\]

A classic application of Stokes’ Theorem is to derive Ampère’s law. Empirically, the magnetic field $\Bb(\xb)$ integrated around a closed loop is proportional to the electric current passing through the loop: \[ \oint_C\Bb\cdot\,d\xb = \mu_0\int_S\Jb\cdot\dS, \] where $\Jb$ is the current density (current per unit area) and $\mu_0$ is a constant (the “permeability of free space”). Applying Stokes’ Theorem shows that \[ \int_S(\nabla\times\Bb)\cdot\,d\xb = \mu_0\int_S\Jb\cdot\dS, \] but since the loop was arbitrary we deduce Ampère’s Law \[ \nabla\times\Bb=\mu_0\Jb. \] Note the trick of deriving the local (i.e. differential) form of a physical law from the integral form.

There is a much more general form of Stokes’ Theorem in Differential Geometry, written elegantly as \[ \int_\Omega\mathrm{d}\omega = \int_{\partial\Omega}\omega. \] Here $\Omega$ is a manifold, the operator $\mathrm{d}$ is the exterior derivative, and the object $\omega$ is a differential form. In fact, Green’s Theorem, our Stokes’ Theorem and the Divergence Theorem are all special cases of this result.

A mathematical corollary of Stokes’ Theorem is the following. Recall that a conservative vector field $\fb=\nabla\phi$ has $\nabla\times\fb=\bfzero$, because $\nabla\times\nabla\phi=\bfzero$. But does $\nabla\times\fb=\bfzero$ imply $\fb=\nabla\phi$?

Corollary 4.1 If $\fb(\xb)$ has continuous first derivatives with $\nabla\times\fb=\bfzero$ in a simply connected region $V\subset\Real^3$, then there exists a differentiable scalar potential $\phi$ in that region such that $\fb=\nabla\phi$.

A region $V\subset\Real^3$ is simply connected if any closed curve in $V$ can be continuously shrunk to a point in $V$. A ball (the interior of a sphere) is simply connected, but a toroid (interior of a torus) is not:

Proof. Recall that $\fb=\nabla\phi$ if and only if $\fb$ has path independent line integrals (Theorem 3.2), so consider two arbitrary curves $C_1, C_2$ between the same two endpoints $\ab, \bb \in V$. Then \[ \int_{C_1}\fb\cdot\,d\xb - \int_{C_2}\fb\cdot\,d\xb = \int_C\fb\cdot\,d\xb, \] where $C$ is defined as “$C_1-C_2$”:

Since $V$ is simply connected, we can always find a capping surface $S$ for $C$ that lies wholly within $V$. Then $\nabla\times\fb=\bfzero$ everywhere on that surface so Stokes’ Theorem gives \[ \int_C\fb\cdot\,d\xb = 0 \quad \implies \int_{C_1}\fb\cdot\,d\xb = \int_{C_2}\fb\cdot\,d\xb. \]

Example. Show that the “line vortex” $\displaystyle\fb(\xb)=\frac{-y}{x^2+y^2}\eb_1 + \frac{x}{x^2+y^2}\eb_2$ cannot be written as $\nabla\phi$ in the region $\xb\neq\bfzero$.

Recall that this vector field has $\nabla\times\fb=\bfzero$ for $\xb\neq\bfzero$.

But we cannot apply Stokes’ Theorem since the region $\Real^2\backslash\{\bfzero\}$ is not simply connected. (There is a “hole” at $\xb=\bfzero$.)

Indeed, if $C$ is the unit circle $x^2+y^2=1$, then Stokes’ Theorem would suggest that $\displaystyle\oint_C\fb\cdot\,d\xb=\int_S(\bfzero)\cdot\dS=0$, but in fact a direct calculation (check!) shows that $\displaystyle\oint_C\fb\cdot\,d\xb = 2\pi.$

This non-zero circulation means that $\fb$ is not a conservative vector field.

This result shows us that there is a deep relation between the topology of a domain and the types of vector fields that can exist on that domain, owing to the constraints of continuity.