Mathematics

# 3D Curl

[latexpage]

## Introduction

Consider a perfect sphere rotating in some direction at some velocity

• Q: what is a numerical way to perfectly describe this rotation?

Let’s first consider a 2D circle with radius $R$ rotating at a certain rate

• We want to define this rate based on the number of rotations that occur over some unit time
• e.g., a 2D circle that rotates every 5 sec has a rotation rate of $1/5 \mathrm{sec} = 0.2 s^{-1}$ ⇒ $0.2$ rotations every second
• By convention, we say a single negative rotation occurs in the counterclockwise direction and a single negative roation occurs in the clockwise direction
• Thus, we can use a single number to perfectly describe 2D rotation
• Note: in physics and math, we actually write this number with units of $\mathrm{radians/sec}$ rather than rotations per second
• Recall: there are exactly $2\pi$ radians for a single rotation

• Conversion between rotations/sec and radians/sec can be found by multiplying $2\pi$ with our rotation/sec value: $\frac{\mathrm{rotations}}{\mathrm{sec}} = 2\pi \frac{\mathrm{radians}}{\mathrm{sec}}$
• Take-home-message: we can use a single number to perfectly describe rotation in 2D

In the 3D case, a single number is not sufficient to describe rotation

• There are two things we need to account for:
1. The axis around which our sphere is rotating (i.e., the line you can draw such that all rotation occurs around that line) ⇒ direction component
2. The rate at which rotation occurs ⇒ magnitude component
• We can use a 3D vector to acccount for these two components!
• The direction or this vector indicates the the axis around which the sphere rotates
• The magnitude/length of this vector indicates the rate at which the sphere is rotating (usually units are given in $\mathrm{rad} / \mathrm{sec}$ – i.e., angular velocity!!!)

### Introduction to the Right-Hand-Rule

Question: how do we interpret rotation for two otherwise identical vectors that point in opposite directions?

• Answer: we use a convention called the right hand rule

Take-home-message: we can pack a lot of information into a 3D vector

• We only need three numbers to describe a rotating sphere
• This is leading up to how 3D curl relates to fluid flow in 3D AND how that induces a rotation at every point in the vector field
• Can use a 3D vector field to determine what rotation is a specific input point is induces by a certain fluid flow

## Intuition behind 3D curl

### Prototype Example

Consider the following vector field:

$\overrightarrow{v}(x,y) = \begin{bmatrix} P(x,y) \\ Q(x,y) \\ R(x,y) \end{bmatrix} = \begin{bmatrix} y^3 - 9y \\ x^3 - 9x \\ 0 \end{bmatrix}$

• Curling your right-hand fingers in the direction of the trajectories surrounding the two points marked with colored-in dots should cause your thumb to point directly away from the screen (i.e., the positive $z$ direction)
• Likewise, curling your right-hand fingers in the direction of the trajectories surrounding the two points marked with crosses should cause your thumb to point directly towards the screen (i.e., the negative $z$ direction)

Now, we want a function that tells us what our “$z$  direction vectors” are for each point in the vector field

• Note: the reason these vectors point purely in the $z$-direction is because our actual vector field lives purely in the $xy$-plane
• It turns out that the actual expression for the curl of this vector field is…

$\mathrm{curl} \overrightarrow{v}(x,y) = \begin{bmatrix} 0 \\ 0 \\ 3(x^2 - y^2) \end{bmatrix}$

• Since our original vector field only lies in $xy$-planes, the curl of this vector field is going to generate a new vector field whose output only gives vectors that point purely in the $z$ direction
• Note: so far, we haven’t actually derived the general form for 3D curl, it was given to us here as a precursor in order to gain better intuition when we do actually derive the general form
• We can think of this example as a prototype for 3D curl since our vector field only exists in the $xy$-plane (i.e., it only takes in $x$ and $y$ as inputs

We want a general expression that associates a “curl vector” with each input point in a 3D input → 3D output vector field

$\overrightarrow{v}(x,y,z) = \begin{bmatrix} P(x,y,z) \\ Q(x,y,z) \\ R(x,y,z) \end{bmatrix} = \begin{bmatrix} y^3 - 9y \\ x^3 - 9x \\ 0 \end{bmatrix}$

If we plot our prototype example vector field in a 3D input space, we wend up with a 3D vector field where each “slice” of input space that lies parallel to the $xy$-plane and perpendicular to the $z$-axis depicts a copy of the vector field plotted in the first diagram:

• Since the $z$-component of the output of $\overrightarrow{v}(x,y,z)$ is zero, all “$xy$-slices” are identical as you go up/down the $z$-axis
• Now, let’s imagine the 3D fluid flow corresponding with $\overrightarrow{v}(x,y,z)$
• In Figure 3, the regions near $x$ and $y$ input values that correspond with positive/negative curl trajectories in Figure 1 will have “tornado-like” trajectories along the $z$-axis
• This is because any given $z$ value in the input of $\overrightarrow{v}(x,y,z)$ has corresponding output vectors that all point in the same direction with the same length as $x$ and $y$ input values vary
•  Note: the tricky part about 3D curl is, for any rotating “tornado-like”  trajectory…
• Looking down at the trajectory from above will depict trajectories rotating in one direction
• Looking up at the trajectory from below will depict trajectories rotating in the opposite direction
• This is why the right-hand-rule is a useful convention!
• Thumb will always point in the direction corresponding to the rotation of the vector field

Usually, a 3D vector field will be much more complicated than our example in Figure 3 (e.g., curl usually varies along the $z$ axis)

• This can make it difficult to visually identify positive/zero/negative curl at any given point in the input space
• Recall: imagine a perfect sphere exists at each point in an arbitrary 3D vector field
• Question: what is the rotation induced on the sphere due to the “flow of fluid” in the vector field?
• Once we know this rotation, we can perfectly describe it using a 3D vector
• This is exactly what the curl of a vector field tells us!!

## 3D Curl Formula (Finally)

Definition: curl is a measure of how much rotation is induced around a given point by an arbitrary vector field

Consider an arbitrary 3D vector field

$\overrightarrow{v}(x,y,z) = \begin{bmatrix} P(x,y,z) \\ Q(x,y,z) \\ R(x,y,z) \end{bmatrix}$

• Importantly, every point in 3D space has a 3D output vector associated with it (i.e., this is a MIMO system with a 3D input and a 3D output):
• We end up with a total of 3 + 3 = 6 dimensions

It turns out, the curl of this is the cross product between the gradient and the vector field

• Recall the gradient is an analog of the partial derivative in 1D that can be extended to $n$-dimension:

$\nabla = \begin{bmatrix} \frac { \partial }{ \partial x_{ 1 } } \\ \vdots \\ \frac { \partial }{ \partial x_{ n } } \end{bmatrix}$

$\nabla \times \overrightarrow{v}(x,y,z) = \begin{bmatrix} \frac { \partial }{ \partial x} \\ \frac { \partial }{ \partial y} \\ \frac { \partial }{ \partial z} \end{bmatrix} \times \begin{bmatrix} P(x,y,z) \\ Q(x,y,z) \\ R(x,y,z) \end{bmatrix}$

• Also, recall in 2D,

$\nabla \times \overrightarrow{v}(x,y) = \begin{bmatrix} \frac { \partial }{ \partial x} \\ \frac { \partial }{ \partial y} \end{bmatrix} \times \begin{bmatrix} P(x,y) \\ Q(x,y) \end{bmatrix} = \frac { \partial }{ \partial x} Q(x,y) - \frac { \partial }{ \partial y} P(x,y)$

Proof:

$\nabla \times \overrightarrow{v}(x,y,z) = \begin{bmatrix} \frac { \partial }{ \partial x} \\ \frac { \partial }{ \partial y} \\ \frac { \partial }{ \partial z} \end{bmatrix} \times \begin{bmatrix} P(x,y,z) \\ Q(x,y,z) \\ R(x,y,z) \end{bmatrix} = \begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ \frac { \partial }{ \partial x } & \frac { \partial }{ \partial y } & \frac { \partial }{ \partial z } \\ P & Q & R \end{vmatrix}$

•  Note: the matrix on the RHS of the equation is a determinant
• The first row is composed of unit vectors
• The second row is composed of partial derivative operators
• Specifically, these components correspond to the components in the transpose of the (3D) gradient operator
• The third row is comprised of the components that correspond those contained in the transpose of $\overrightarrow{v}(x,y,z)$

$\nabla \times \overrightarrow{v}(x,y,z) = \hat { i } \begin{vmatrix} \frac { \partial }{ \partial y } & \frac { \partial }{ \partial z } \\ Q & R \end{vmatrix}-\hat { j } \begin{vmatrix} \frac { \partial }{ \partial x } & \frac { \partial }{ \partial z } \\ P & R \end{vmatrix}+\hat { k } \begin{vmatrix} \frac { \partial }{ \partial x } & \frac { \partial }{ \partial y } \\ P & Q \end{vmatrix}$

$\nabla \times \overrightarrow{v}(x,y,z) = \hat{i}(\frac{ \partial }{ \partial y } R - \frac{ \partial }{ \partial z } Q ) - \hat{j}(\frac{ \partial }{ \partial x } R - \frac{ \partial }{ \partial z } P ) + \hat{k}(\frac{ \partial }{ \partial x } Q - \frac{ \partial }{ \partial y } P)$

$\nabla \times \overrightarrow{v}(x,y,z) = \hat{i}(\frac{ \partial R }{ \partial y } - \frac{ \partial Q }{ \partial z }) + \hat{j}(\frac{ \partial P}{ \partial z } - \frac{ \partial R}{ \partial x}) + \hat{k}(\frac{ \partial Q}{ \partial x } - \frac{ \partial P}{ \partial y })$

⇒ Formula for 3D curl of a 3D vector field:

$\nabla \times \overrightarrow{v}(x,y,z) = \begin{bmatrix} \frac { \partial R }{ \partial y } -\frac { \partial Q }{ \partial z } \\ \frac { \partial P }{ \partial z } -\frac { \partial R }{ \partial x } \\ \frac { \partial Q }{ \partial x } -\frac { \partial P }{ \partial y } \end{bmatrix}$

• Note: our $z$ component in the curl vector is the formula for our 2D curl!!
• Gives the curl for the part of the vector field that lies purely in the $xy$-plane
• Similarly, our $x$ and $y$ components of the curl vector indicate the curl for the part of the vector field lying purely in the $yz$ and $xz$ planes (respectively)

## 3D Curl Computation Example

Consider the following vector field:

$\overrightarrow{v}(x,y,z) = \begin{bmatrix} xy \\ cos(z) \\ z^2 + y \end{bmatrix}$

Find the curl for the following vector field:

$\nabla \times \overrightarrow{v}(x,y,z) = \begin{bmatrix} \frac { \partial }{ \partial x} \\ \frac { \partial }{ \partial y} \\ \frac { \partial }{ \partial z} \end{bmatrix} \times \begin{bmatrix} xy \\ cos(z) \\ z^2 + y \end{bmatrix}$

$\nabla \times \overrightarrow{v}(x,y,z) = \begin{bmatrix} \frac { \partial }{ \partial y } (z^{ 2 }+y)\quad -\quad \frac { \partial }{ \partial z } \cos (z) \\ \frac { \partial }{ \partial z } (xy)\quad -\quad \frac { \partial }{ \partial x } (z^{ 2 }+y) \\ \frac { \partial }{ \partial x } \cos (z)\quad -\quad \frac { \partial }{ \partial y } (xy) \end{bmatrix} = \begin{bmatrix} 1\quad -\quad (-\sin (z)) \\ 0\quad -\quad 0 \\ 0\quad -\quad x \end{bmatrix}$

$\nabla \times \overrightarrow{v}(x,y,z) = \begin{bmatrix} 1 + \sin (z) \\ 0 \\- x \end{bmatrix} = (1 + \sin(z)) \hat{i} - x \hat{k}$