Curl Curl Curl

Many of us have struggled with the vector differential operators, grad, div and curl. There are several ways to represent vectors and several expressions for these operators, not always easy to remember. We take another look at some of their properties here.

We consider a vector ${\mathbf{V} = (u, v, w)^{\mathrm{T}}}$ which may be the velocity or something more general. For simplicity, we deal in this article only with 3-dimensional cartesian coordinates.

We define the formal vector ${\mathbf{D} = (\partial_x, \partial_y, \partial_z)^{\mathrm{T}}}$ , with which we can represent various differential operators. The gradient of a scalar variable is defined as

$\displaystyle \mathbf{grad}\ \varphi = \mathbf{D}\varphi = \left[ \begin{matrix} \partial_x\varphi \\ \partial_y\varphi \\ \partial_z\varphi \end{matrix} \right] \,.$

The divergence of ${\mathbf{V}}$ is:

$\displaystyle \mathrm{div}\ \mathbf{V} = \mathbf{D}^{\mathrm{T}}\mathbf{V} = (\partial_x, \partial_y, \partial_z) \left[ \begin{matrix} u \\ v \\ w \end{matrix} \right] = \partial_x u + \partial_y v + \partial_z w \,.$

If we define the matrix operator

$\displaystyle \mathsf{C} = \left[ \begin{matrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{matrix} \right] \,,$

we can represent the curl of ${\mathbf{V}}$ , known as the vorticity, as

$\displaystyle \mathbf{curl}\ \mathbf{V} = \mathbf{D} \boldsymbol{\times} \mathbf{V} = \mathsf{C} \mathbf{V} = \left[ \begin{matrix} w_y - v_z \\ u_z - w_x \\ v_x - u_y \end{matrix} \right]$

The scalar laplacian operator is easily computed:

$\displaystyle \nabla^2 := \mathbf{D}^{\mathrm{T}} \mathbf{D} = (\partial_x, \partial_y, \partial_z) \left[ \begin{matrix} \partial_x \\ \partial_y \\ \partial_z \end{matrix} \right] = \partial_x^2 + \partial_y^2 + \partial_z^2 \,.$

The vector laplacian is simply ${\boldsymbol{\nabla}^2 = \mathsf{I}\, \nabla^2}$ where ${\mathsf{I}}$ is the identity matrix.

Now we can demonstrate some vector identities:

$\displaystyle \mathbf{curl}\ \mathbf{grad}\ \varphi = \mathsf{C} \mathbf{D}\, \varphi = \left[ \begin{matrix} (\partial_z)_y - (\partial_y)_z \\ (\partial_x)_z - (\partial_z)_x \\ (\partial_y)_x - (\partial_x)_y \end{matrix} \right]\,\varphi = \mathbf{0}$

and

$\displaystyle \mathrm{div}\ \mathbf{curl}\ \mathbf{V} = \mathbf{D}^{\mathrm{T}} \mathsf{C} \ \mathbf{V} = (\partial_x, \partial_y, \partial_z) \left[ \begin{matrix} w_y - v_z \\ u_z - w_x \\ v_x - u_y \end{matrix} \right] = 0 \,.$

Curl Curl Curl

We easily calculate the matrix product

$\displaystyle \mathsf{C\,C} = \left[ \begin{matrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{matrix} \right] \left[ \begin{matrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{matrix} \right] = \left[ \begin{matrix} -\partial_y^2-\partial_z^2 & \partial_x \partial_y & \partial_x\partial_z \\ \partial_x \partial_y & -\partial_z^2-\partial_x^2 & \partial_y \partial_z \\ \partial_x \partial_z & \partial_y \partial_z & -\partial_x^2-\partial_y^2 \end{matrix} \right]$

We also show easily that

$\displaystyle \mathbf{D} \mathbf{D}^{\mathrm{T}} = \left[ \begin{matrix} \partial_x \\ \partial_y \\ \partial_z \end{matrix} \right] (\partial_x, \partial_y, \partial_z) = \left[\begin{matrix} \partial_x^2 & \partial_x \partial_y & \partial_x\partial_z \\ \partial_x \partial_y & \partial_y^2 & \partial_y \partial_z \\ \partial_x \partial_z & \partial_y \partial_z & \partial_z^2 \end{matrix} \right]$

Thus, we can deduce that

$\displaystyle \mathsf{C\,C} = \mathbf{D} \mathbf{D}^{\mathrm{T}} - \boldsymbol{\nabla^2} \,.$

In more conventional notation this is just the vector identity

$\displaystyle \mathbf{curl}\ \mathbf{curl} = \mathbf{grad}\ \mathrm{div} - \boldsymbol{\nabla}^2 \,.$

But now we can show that ${\mathbf{curl}}$ and ${ \boldsymbol{\nabla}^2}$ commute. First note that

$\displaystyle (\mathbf{curl}\ \mathbf{curl})\ \mathbf{curl} = \mathbf{curl}\ (\mathbf{curl}\ \mathbf{curl}) \qquad\mbox{or}\qquad (\mathsf{C}\,\mathsf{C})\,\mathsf{C} = \mathsf{C}\,(\mathsf{C}\,\mathsf{C}) \,.$

Now using the expression for ${\mathsf{C}\,\mathsf{C}}$ and the vector identities ${\mathbf{curl}\ \mathbf{grad} = \mathbf{0}}$ and ${\mathrm{div}\ \mathbf{curl} = 0}$ , we find that

$\displaystyle \boldsymbol{\nabla}^2 \mathbf{curl} = \mathbf{curl}\,\boldsymbol{\nabla}^2 \qquad\mbox{or}\qquad [ \boldsymbol{\nabla}^2,\, \mathbf{curl}] = \mathbf{0} \,,$

so that the curl and laplacian operators commute.

In a following article, I hope to consider how to express and interpret the operation of the gradient operator on a vector, and the expression for advection, ${\mathbf{V\cdot\nabla V}}$ , which is so important in fluid dynamics.

Note: A UCD course on recreational mathematics, AweSums: The Wonder, Utility and Fun of Mathematics, will be presented this Autumn by Prof Peter Lynch. Registration is now open at www.ucd.ie/lifelonglearning.

$\star \qquad \star \qquad \star$

New Collection of ThatsMaths Articles

GREATLY REDUCED PRICE from Logic Press.

Now available also in hardback form

ThatsMaths

Curl Curl Curl

Tags

Last 50 Posts

Categories

Archives

ThatsMaths

Curl Curl Curl

Related

Tags

Last 50 Posts

Categories

Archives