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

        radians
        Source link to image
      • 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 axis of rotation sphere.jpg
    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

    Right-hand_grip_rule
    Source link

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}

2D curl example_d_black
Figure 1: Vector field diagram on the xy-plane for \overrightarrow{v}(x,y) with 9 marked sample points.  Sample points with positive curl are marked with flled-in orange dots, points with zero curl are makred with empty orange circles, and points with negative curl are marked with orange crosses.
  • 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

2D curl example_e1
Figure 2
  • 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:

2D curl example_f
Figure 3
  • 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): 3D-curl system diagram.jpg.png
    • 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}

 

 

3D-curl example3
Figure 4

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}

Final Answer:

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

References

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