Table of Contents

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Prerequisites: vectors will be denoted as: $\mathbf{v}$, and unit vectors: $\mathbf{\hat{i}}$. Furthermore, $\phi$ will mostly be used to represent a scalar field, and $\mathbf{a}$ will be used to represent a vector field.

Parametric Surfaces

A parametric representation of a surface is given as: $$ \mathbf{r}(u,v)=x(u,v)\mathbf{\hat{i}}+y(u,v)\mathbf{\hat{j}}+z(u,v)\mathbf{\hat{k}} $$ Example: A paraboloid is given by $$ \mathbf{r}(u,v)=u\cos(v)\mathbf{\hat{i}}+u\sin(v)\mathbf{\hat{j}}+u^2\mathbf{\hat{k}},\qquad0\leq u\leq 1,\quad 0\leq v\leq 2\pi $$

Infinitesimal displacement within a surface

The infinitesimal displacement $d\mathbf{r}$ within the surface due to an infinitesimal change in $u$ and an infinitesimal change in $v$ is derived from the chain rule: $$ d\mathbf{r}=\frac{\partial\mathbf{r}}{\partial u}du+\frac{\partial\mathbf{r}}{\partial v}dv $$ Some shorthand: The vectors $\partial_u\mathbf{r}, \partial_v\mathbf{r}$, are tangent to the surface, where $\partial_u$ is shorthand for $\partial/\partial u$, and $\partial_v$ for $\partial/\partial v$.

Surface Area

The surface area is obtained by summing over the areas of a finite number of small surface elements and letting the number of elements tend to infinity. Thus $$ \textrm{surface area}=\lim_{N\to\infty}\sum_N\Delta S=\int ds $$ where $dS$ is the infinitesimal scalar area element given by $$ dS=|\partial_u\mathbf{r}\;du\times\partial_v\mathbf{r}\;dv|=|\partial_u\mathbf{r}\times\partial_v\mathbf{r}|\;du\;dv $$ hence: $$ S=\int^{u_2}_{u_1}\int^{v_2}_{u_1}|\partial_u\mathbf{r}\times\partial_v\mathbf{r}|du\;dv $$

Volume of a 3D region

You can expand this idea to volumes: $$ \mathbf{r}(u,v,w)=x(u,v,w)\mathbf{\hat{i}}+y(u,v,w)\mathbf{\hat{j}}+z(u,v,w)\mathbf{\hat{k}} $$ The infinitesimal displacement $d\mathbf{r}$ is $$ d\mathbf{r}=\frac{\partial \mathbf{r}}{\partial u}du+\frac{\partial\mathbf{r}}{\partial v}dv+\frac{\partial\mathbf{r}}{\partial w}dw $$ the volume $dV$ of the corresponding infinitesimal parallelepiped is $$ dV=|\partial_u\mathbf{r}\;du\cdot(\partial_v\mathbf{r}\;dv\times\partial_w\mathbf{r}\;dw)| $$ Hence, the volume of the 3D region is $$ V=\int^{u_2}_{u_1}\int^{v_2}_{v_1}\int^{w_2}_{w_1}|\partial_u\mathbf{r}\cdot(\partial_v\mathbf{r}\times\partial_w\mathbf{r})|dw\;dv\;du $$

Grad, Div and Curl

Three important operations on scalar / vector fields

Mathematical Definitions

So what is Nabla ($\nabla$)? Nabla is both a vector operator and a vector: $$ \nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right) $$ Now, the complicated bits: $$\begin{align} \nabla\phi&=\frac{\partial\phi}{\partial x}\mathbf{\hat{i}}+\frac{\partial\phi}{\partial y}\mathbf{\hat{j}}+\frac{\partial\phi}{\partial z}\mathbf{\hat{k}}:\;\;\textrm{The gradient of $\phi$. (becomes vector field)}\\ \nabla\cdot\mathbf{a}&=\frac{\partial a_x}{\partial x}+\frac{\partial a_y}{\partial y}+\frac{\partial a_z}{\partial z}:\;\;\textrm{The divergence of $\mathbf{a}$ (becomes scalar field)}\\ \nabla\times\mathbf{a}&=\left|\begin{matrix} \mathbf{\hat{i}}&\mathbf{\hat{j}}&\mathbf{\hat{k}}\\ \partial_x&\partial_y&\partial_z\\ a_x&a_y&a_z\\ \end{matrix}\right|:\;\;\textrm{The curl of $\mathbf{a}$ (stays vector field)}\\ \end{align}$$ ...as you can see, the divergence is basically the dot product of $\nabla$ and $\mathbf{a}$, and the curl is basically the cross product of $\nabla$ and $\mathbf{a}$

So what are they, and what do they do?

The gradient

...Creates a vector field from a scalar. Every point is a vector, whose direction and magnitude represents the gradients (of the scalar field) direction and magnitude. The direction corresponds to the MAXIMUM rate of change. For example: The gradient of this level surface of the scalar field: ($\phi=y^2-x^2$)

...looks like ($\nabla\phi=-2x\hat{i}+2y\hat{j}$):

Divergence

This tells you how much a vector field is "sourcing" or "sinking". You can imagine this like a tap or a drain in a sink. It is positive is the field is exiting a point more than it is entering, and negative if vice versa. I will only visualise the field here (in 2D) as I do not have time to do it in 3D. Example for the field $\mathbf{a}=x\mathbf{\hat{i}}+y\mathbf{\hat{j}}$:

$$ \nabla\cdot\mathbf{a}=1+1=2\gt 0 $$ hence this field has a divergence of 2 everywhere, i.e. the field is exiting more than entering at all points. Note: A field is solenoidal over a region if the divergence at all point in that region is 0.

Curl

This tells you how much a vector field is rotating around a point. It is more of a 3D idea, but it is hard to plot so I will simplify to 2D. The curl (like the cross product) in 2D exists only in the $\hat{k}$ direction. In 3D this is not true. Example for the field $\mathbf{a}=-y\mathbf{\hat{i}}+x\mathbf{\hat{j}}$:

$$ \nabla\times\mathbf{a}=1-(-1)=2\gt 0 $$ Note: A field is irrotational over a region if the curl at all point in that region is 0.

Identities Involving $\nabla$

Considering that the various operations involving $\nabla$ are just specific cases of first-order partial differentiation with respect to $x$, $y$, $z$, it is unsurprising that $\nabla$ has similar properties to its 1D counterpart $d/dx$ . $\nabla$ is distributive: $$\begin{align} \nabla(\phi+\psi)&=\nabla\phi+\nabla\psi\\ \nabla\cdot(\mathbf{a}+\mathbf{b})&=\nabla\cdot\mathbf{a}+\nabla\cdot\mathbf{a}\\ \nabla\times(\mathbf{a}+\mathbf{b})&=\nabla\times\mathbf{a}+\nabla\times\mathbf{a}\\ \end{align}$$ ...and satisfies the product rule: $$\begin{align} \nabla(\phi\psi)&=(\nabla\phi)\psi+\phi(\nabla\psi)\\ \nabla\cdot(\phi\mathbf{a})&=(\nabla\phi)\cdot\mathbf{a}+\phi(\nabla\cdot\mathbf{a})\\ \nabla\times(\phi\mathbf{a})&=(\nabla\phi)\times\mathbf{a}+\phi(\nabla\times\mathbf{a})\\ \end{align}$$ A bit of thought also reveals that $\nabla$ satisfies the chain rule when acting on a scalar field $f(\psi)$ that has been expressed as a function of the scalar field $\psi$: $$ \nabla\left[f(\psi)\right]=\frac{df}{d\psi}\nabla\psi $$ Second Order Two very important identites: $$\begin{align} \nabla\times(\nabla\phi)&=0\\ \nabla\cdot(\nabla\times\mathbf{a})&=0\\ \end{align}$$

The Laplacian ($\nabla^2$)

The Laplacian acts differently on scalar and vector fields. Its action on scalar fields is defined as: $$ \nabla^2\phi=\nabla\cdot(\nabla\phi) $$ The action of the Laplacian on vector fields is more subtle. It is defined as follows: $$ \nabla^2\mathbf{a}=\nabla(\nabla\cdot\mathbf{a})-\nabla\times(\nabla\times\mathbf{a}) $$

Level Surfaces Of a Scalar Field

A level surface is where we set a scalar field to be constant. I.e.: $$ \phi(x,y,z)=c $$ Geometrically, this surface sits within the field, and is just a collection of all of the points where the field is equal to $c$. A normal to the level surface is obtained by working out $\nabla\phi$ at the level surface $\phi(x,y,z)=c$. We write this as $$ \left.\mathbf{n}=\nabla\phi\right|_{\phi=c} $$ to remind us that $\mathbf{n}$ lives on on the level surface

Important Note

If we express this level surface parametrically, ($\mathbf{r}(u,v)$), then we can say that the normal must be $\partial_u\mathbf{r}\times\partial_v\mathbf{r}$. Hence the identity: $$ \partial_u\mathbf{r}\times\partial_v\mathbf{r}=\lambda\mathbf{n} $$ where $\lambda$ is some function of $u$ and $v$ and $\left.\mathbf{n}=\nabla\phi\right|_{\phi=c}$.

Curvilinear Coordinate Systems

When you express $x,\;y,\;z$ in terms of $u_1,\;u_2,\;u_3$ such that: $$x=x(u_1,u_2,u_3),\quad y=y(u_1,u_2,u_3),\quad z=z(u_1,u_2,u_3)... $$

There are the two known examples of cylindrical and spherical polar coordinates. For cylindrical coordinates, you have: $$\begin{align} x=&r\cos(\phi)\\ y=&r\sin(\phi)\\ z=&z \end{align}$$ and for spherical: $$\begin{align} x=&\rho\sin(\theta)\cos(\phi)\\ y=&\rho\sin(\theta)\sin(\phi)\\ z=&\rho\cos(\theta) \end{align}$$ an easy way to remember spherical coordinates is to use cylindrical coordinates and the fact that $\rho=r\sin(\theta)$

Bases for Vector Fields

There are two methods to construct a vector basis from curvilinear coordinates ($u_1,\;u_2,\;u_3$). Tangents to coordinate curves, or normals to coordinate surfaces.

1: Tangents to coordinate curves

If you imagine fixing all but one of your parameters that make up your coordinate system, and plot a curve as you vary the unfixed parameter, the tangent to this curve will form a base for the vector field. Repeat this through your parameters and you have a basis: $$ \mathbf{e}_1=\frac{\partial\mathbf{r}}{\partial u_1},\quad\mathbf{e}_2=\frac{\partial\mathbf{r}}{\partial u_2},\quad\mathbf{e}_3=\frac{\partial\mathbf{r}}{\partial u_3} $$ where $\mathbf{r}=x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z\mathbf{\hat{k}}$ It is sometimes important to normalise these, such that their lengths are $1$. To find the normalised basis ($\mathbf{\hat{e}}_1$), we simply compute: $$ \mathbf{\hat{e}}_1=\frac{\mathbf{e}_1}{|\mathbf{e}_1|} $$ Example: Cylindrical Polar Coordinates $$ \mathbf{r}(\rho,\phi,z)=\rho\cos(\phi)\mathbf{\hat{i}}+\rho\sin(\phi)\mathbf{\hat{j}}+z\mathbf{\hat{k}} $$ We can obtain: $$\begin{align} \mathbf{e}_\rho&=\partial_\rho\mathbf{r}=\cos(\phi)\mathbf{\hat{i}}+\sin(\phi)\mathbf{\hat{j}}\\ \mathbf{e}_\phi&=\partial_\phi\mathbf{r}=-\rho\sin(\phi)\mathbf{\hat{i}}+\rho\cos(\phi)\mathbf{\hat{j}}\\ \mathbf{e}_z&=\partial_z\mathbf{r}=\mathbf{\hat{k}}\\ \end{align}$$ then $$\begin{align} \mathbf{\hat{e}}_\rho&=\cos(\phi)\mathbf{\hat{i}}+\sin(\phi)\mathbf{\hat{j}}\\ \mathbf{\hat{e}}_\phi&=-\sin(\phi)\mathbf{\hat{i}}+\cos(\phi)\mathbf{\hat{j}}\\ \mathbf{\hat{e}}_z&=\mathbf{\hat{k}}\\ \end{align}$$

2: Normals To Coordinate Surfaces

First, consider the inverse transformation: $$ u_1=u_1(x,y,z),\quad u_2=u_2(x,y,z),\quad u_3=u_3(x,y,z) $$ Setting $u_1$, $u_2$ or $u_3$ to constants will yield three level surfaces. This leads to the basis $\epsilon_1, \epsilon_2, \epsilon_3$ given by: $$ \epsilon_1=\nabla u_1,\quad \epsilon_1=\nabla u_2,\quad \epsilon_1=\nabla u_3,\quad $$ It is sometimes important to normalise these, such that their lengths are 1. To find the normalised basis ($\mathbf{\hat{\epsilon}}_1$), we simply compute: $$ \mathbf{\hat{\epsilon}}_1=\frac{\epsilon_1}{|\epsilon_1|} $$ The unit bases from both methods are not always the same. They are only equivalent when the coordinate curves intersect at right angles, i.e. the basis $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$ is orthogonal. We call such coordinate systems orthogonal and Cartesian coordinates, cylindrical polar coordinates and spherical polar coordinates are examples of orthogonal coordinate systems.

Gradient In Curvilinear Coordinates

Since $$ \frac{\partial\phi}{\partial u_1}=\frac{\partial\phi}{\partial x}\frac{\partial x}{\partial u_1}+\frac{\partial\phi}{\partial y}\frac{\partial y}{\partial u_1}+\frac{\partial\phi}{\partial z}\frac{\partial z}{\partial u_1} $$ using the chain rule, it follows that $$ \frac{\partial\phi}{\partial u_1}=\frac{\partial\mathbf{r}}{\partial u_1}\cdot\nabla\phi $$ with more manipulation (and an identity) $$ \nabla\phi=\frac{1}{h_1}\frac{\partial\phi}{\partial u_1}\mathbf{\hat{e}}_1+\frac{1}{h_2}\frac{\partial\phi}{\partial u_2}\mathbf{\hat{e}}_2+\frac{1}{h_3}\frac{\partial\phi}{\partial u_3}\mathbf{\hat{e}}_3 $$ where $h_n$ is the scale factor $|\mathbf{e_n}|$

Divergence In Curvilinear Coordinates

We do not need to remember the derivation, so I shall just state them: Cylindrical Polar Coordinates: $$ \nabla\cdot\mathbf{a}=\frac{1}{\rho}\frac{\partial(\rho a_\rho)}{\partial\rho}+\frac{1}{\rho}\frac{\partial a_\phi}{\partial\phi}+\frac{\partial a_z}{\partial z} $$ Spherical Polar Coordinates: $$ \nabla\cdot\mathbf{a}=\frac{1}{\rho^2}\frac{\partial(\rho^2 a_\rho)}{\partial\rho}+\frac{1}{\rho\sin(\phi)}\frac{\partial\sin(\phi)a_\theta}{\partial \theta}+\frac{1}{\rho\sin(\phi)}\frac{\partial a_\phi}{\partial\phi} $$

Curl In Curvilinear Coordinates

We do not need to remember the derivation, so I shall just state them: Cylindrical Polar Coordinates: $$ \nabla\times\mathbf{a}=\frac{1}{\rho}\left|\begin{matrix} \mathbf{\hat{e}}_\rho&\rho\mathbf{\hat{e}}_\phi&\mathbf{\hat{e}}_z\\ \partial_\rho&\partial_\phi&\partial_z\\ a_\rho&\rho a_\phi&a_z \end{matrix}\right| $$ Spherical Polar Coordinates: $$ \nabla\times\mathbf{a}=\frac{1}{\rho^2\sin(\theta)}\left|\begin{matrix} \mathbf{\hat{e}}_\rho&\rho\mathbf{\hat{e}}_\theta&\rho\sin(\theta)\mathbf{\hat{e}}_\phi\\ \partial_\rho&\partial_\theta&\partial_\phi\\ a_\rho&\rho a_\theta&\rho\sin(\theta)a_\phi \end{matrix}\right| $$

Line Integrals

In practice, a good way to evaluate such integrals is to introduce a parametric form $r=r(u)$ for $C$: $$\begin{align} \int_C\phi\;d\mathbf{r}&=\int^{u_2}_{u_1}\phi(\mathbf{r}=\mathbf{r}(u))\frac{d\mathbf{r}}{du}du\\ \int_C\mathbf{a}\cdot d\mathbf{r}&=\int^{u_2}_{u_1}\mathbf{a}(\mathbf{r}=\mathbf{r}(u))\cdot\frac{d\mathbf{r}}{du}du\\ \int_C\mathbf{a}\times d\mathbf{r}&=\int^{u_2}_{u_1}\mathbf{a}(\mathbf{r}=\mathbf{r}(u))\times\frac{d\mathbf{r}}{du}du\\ \end{align}$$ Example:

Green's Theorem

Quick Note: we always traverse boundaries with the surface/volume/area to our left. Green's theorem is essentially a 2D stokes theorem. It tells us that the total amount of field in line with the boundary $C$, is equal to the sum of the curl of the field over the entire area. $$ \boxed{\oint_C\left(P\;dx+Q\;dy\right)=\iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\;dxdy} $$ ...or in more general form: $$ \oint_C\mathbf{a}\cdot d\mathbf{r}=\iint_R(\nabla\times\mathbf{a})\; dA $$ where $\mathbf{a}=P(x,y)\mathbf{\hat{i}}+Q(x,y)\mathbf{\hat{j}}$

Surface Integral

Pretty much the same as line integrals. Remember that: $$ d\mathbf{S}=\mathbf{\hat{n}}\;dS $$ and $$ dS=|\partial_u\mathbf{r}\times\partial_v\mathbf{r}|du\;dv $$ hence $$ d\mathbf{S}=\partial_u\mathbf{r}\times\partial_v\mathbf{r}\;du\;dv $$

Volume Integrals & Jacobian

The volume $dV$ of the parallelepiped formed by the infinitesimal displacements $\mathbf{e_1}du_1,\;\mathbf{e_2}du_2,\;\mathbf{e_3}du_3$ is $$ dV=|\mathbf{e}_1\cdot(\mathbf{e}_2\times\mathbf{e}_3)|du_1\;du_2\;du_3 $$ ...Which the elegant way of writing is: $$ dV=\left|\begin{matrix} \frac{\partial x}{\partial u_1}&\frac{\partial y}{\partial u_1}&\frac{\partial z}{\partial u_1}\\ \frac{\partial x}{\partial u_2}&\frac{\partial y}{\partial u_2}&\frac{\partial z}{\partial u_2}\\ \frac{\partial x}{\partial u_3}&\frac{\partial y}{\partial u_3}&\frac{\partial z}{\partial u_3}\\ \end{matrix}\right| du_1\;du_2\;du_3 $$ for $\mathbf{r}(u_1,u_2,u_3)=x(u_1,u_2,u_3)\mathbf{\hat{i}}+y(u_1,u_2,u_3)\mathbf{\hat{j}}+z(u_1,u_2,u_3)\mathbf{\hat{k}}$

The Jacobian

The Jacobian is part of that expression above. Specifically: $$ J=|\mathbf{e}_1\cdot(\mathbf{e}_2\times\mathbf{e}_3)|=\left|\begin{matrix} \frac{\partial x}{\partial u_1}&\frac{\partial y}{\partial u_1}&\frac{\partial z}{\partial u_1}\\ \frac{\partial x}{\partial u_2}&\frac{\partial y}{\partial u_2}&\frac{\partial z}{\partial u_2}\\ \frac{\partial x}{\partial u_3}&\frac{\partial y}{\partial u_3}&\frac{\partial z}{\partial u_3}\\ \end{matrix}\right| $$ Hence, in curvilinear coordinates: $$ \iiint_VdV=\iiint_VJ\;du_1du_2du_3 $$ The Jacobian is commonly denoted: $$ J=\frac{\partial (x,y,z)}{\partial (u_1,u_2,u_3)} $$ Note that the volume element $dV$ has a particularly simple form when expressed in orthogonal curvilinear coordinates. The scale factors $h_1,\;h_2,\;h_3$ satisfy: $$ \mathbf{e}_1=h_1\mathbf{\hat{e}}_1,\quad\mathbf{e}_2=h_2\mathbf{\hat{e}}_2,\quad\mathbf{e}_3=h_3\mathbf{\hat{e}}_3 $$ therefore for orthogonal curvilinear coordinates ONLY: $$ J=h_1h_2h_3 $$

Gauss' Divergence Theorem

This theorem basically says: The flux of a field through a closed surface, is equal to the divergence of the field within the volume bound by the surface.

More simply: Net flow out of region = Sum of sources in region. $$ \boxed{\iiint_V\nabla\cdot\mathbf{a}\;dV=\oint_S\mathbf{a}\cdot d\mathbf{S}} $$

Stokes Theorem

A more general form of Green's theorem. The total field in line with the boundary $C$, is equal to the flux of the curl of the field through surface. $$ \boxed{\oint_C\mathbf{a}\cdot d\mathbf{r}=\iint_S(\nabla\times\mathbf{a})\cdot d\mathbf{S}} $$ The direction of $C$ is that the surface is on the left. The direction of $dS$ is normal to the surface $S$. Specifically, the direction of $dS$ is the cross product of the direction of $C$ and a tangent to $S$.