# 4.1: Vector calculus operations

In multidimensional calculus, the role of the derivative is taken by the vector differential operator $$\vec{\nabla}$$, pronounced “del’’, or sometimes “nabla’’:

$\nabla_{i}=\frac{\partial}{\partial x_{i}}.$

This operator can be written in several different but equivalent ways:

$\vec{\nabla}=\left\{\frac{\partial}{\partial x_{1}}, \frac{\partial}{\partial x_{2}}, \frac{\partial}{\partial x_{3}}\right\}=\hat{e}^{(i)} \frac{\partial}{\partial x_{i}}=\hat{e}^{(i)} \nabla_{i}.$

## 4.1.1 Gradient of a scalar

Let $$\phi(\vec{x}$$ be a scalar field and $$\vec{x}$$ the position vector in a Cartesian coordinate system. Application of $$\vec{\nabla}$$ yields the gradient:

$\vec{G}=\vec{\nabla} \phi=\left\{\frac{\partial \phi}{\partial x_{1}}, \frac{\partial \phi}{\partial x_{2}}, \frac{\partial \phi}{\partial x_{3}}\right\}=\hat{e}^{(i)} \frac{\partial \phi}{\partial x_{i}}$

The gradient has three components and appears to be a vector, but we should check. Identifying a vector is more complicated when spatial derivatives are involved.

\begin{aligned} &G_{i}^{\prime}=\frac{\partial \phi^{\prime}}{\partial x_{i}^{\prime}}\\ &\begin{array}{ll} = & \frac{\partial \phi}{\partial x_{i}^{\prime}} \quad(\phi \text { is a scalar}) \\ = & \frac{\partial \phi}{\partial x_{j}} \frac{\partial x_{j}}{\partial x_{i}^{\prime}} \quad \left(\text {applying the chain rule and summing over } j\right) \end{array}\\ &=G_{j} \frac{\partial x_{j}}{\partial x_{i}^{\prime}}. \end{aligned} \label{eqn:1}

The partial derivative in the final term is the Jacobian matrix for the transformation. Using the reverse transformation rule Equation 3.2.7,

$\frac{\partial x_{j}}{\partial x_{i}^{\prime}}=\frac{\partial x_{k}^{\prime} C_{j k}}{\partial x_{i}^{\prime}}=C_{j k} \frac{\partial x_{k}^{\prime}}{\partial x_{i}^{\prime}}=C_{j k} \delta_{k i}=C_{j i}.\label{eqn:2}$

This useful result pertains to every orthogonal coordinate system, so we’ll highlight for later reference:

$\frac{\partial x_{j}}{\partial x_{i}^{\prime}}=C_{j i}.\label{eqn:3}$

Now, combining Equation $$\ref{eqn:1}$$ and $$\ref{eqn:2}$$, we have $$G^\prime_i=G_iG_{ij}$$, i.e., the gradient of a scalar transforms as a vector.

## 4.1.2 Divergence

The divergence normally results from applying $$\vec{\nabla}$$ to a vector:

$\vec{\nabla} \cdot \vec{u}=\frac{\partial u_{i}}{\partial x_{i}}.$

The result is a scalar. A vector whose divergence is zero is called solenoidal. The divergence may also be applied to either dimension of a matrix or a 2nd order tensor. For example,

$\frac{\partial A_{i j}}{\partial x_{j}} \hat{e}^{(i)}$

If $$\underset{\sim}{A}$$ is a tensor, the result is a vector.

## 4.1.3 Curl

The curl is applied to a vector field $$\vec{u}$$ by taking the cross product with $$\vec{\nabla}$$:

$\vec{\nabla} \times \vec{u}=\varepsilon_{i j k} \nabla_{i} u_{j} \hat{e}^{(k)}.$

The result is another vector field. It can be expanded as

$\vec{u} \times \vec{v}=\hat{e}^{(1)}\left(u_{2} v_{3}-u_{3} v_{2}\right)-\hat{e}^{(2)}\left(u_{1} v_{3}-u_{3} v_{1}\right)+\hat{e}^{(3)}\left(u_{1} v_{2}-u_{2} v_{1}\right).\label{eqn:4}$

See Appendix D, section D.3.3 for details.

## 4.1.4 Laplacian

The Laplacian results from successive applications of $$\vec{\nabla}$$:

$\vec{\nabla} \cdot \vec{\nabla}=\nabla^{2}=\nabla_{i} \nabla_{i}=\frac{\partial^{2}}{\partial x_{i}^{2}}$

The Laplacian may be applied to either a scalar or a vector:

$\nabla^{2} \phi=\vec{\nabla} \cdot \vec{\nabla} \phi=\frac{\partial^{2} \phi}{\partial x_{i}^{2}},$

or

$\nabla^{2} \vec{u}=\left\{\nabla^{2} u_{1}, \nabla^{2} u_{2}, \nabla^{2} u_{3}\right\}=\hat{e}^{(i)} \nabla^{2} u_{i}.$

The advective derivative is an operation unique to fluid mechanics. Besides $$\vec{\nabla}$$, it requires a vector field $$\vec{u}\left(\vec{x}\right)$$, which is often (though not always) chosen to be the velocity of the flow. When this choice is made, we use the more common term material derivative (discussed in section 5.1.1). The operation is

$[\vec{u} \cdot \vec{\nabla}]=u_{i} \frac{\partial}{\partial x_{i}}.$

The advective derivative can be applied to a scalar, resulting in another scalar:

$[\vec{u} \cdot \vec{\nabla}] \phi=\vec{u} \cdot \vec{\nabla} \phi=u_{i} \frac{\partial \phi}{\partial x_{i}},$

or to a vector, resulting in another vector:

$[\vec{u} \cdot \vec{\nabla}] \vec{v}=u_{i} \frac{\partial}{\partial x_{i}}\left(v_{j} \hat{e}^{(j)}\right)=\hat{e}^{(j)} u_{i} \frac{\partial v_{j}}{\partial x_{i}}.$

## 4.1.6 Vector identities

Many equations hold for only certain values of a variable, and one may want to solve the equation to find those values. In contrast, identities are equations that hold for all values of a certain class of variables, e.g., all vectors that vary continuously in space. The identity

$\vec{\nabla} \cdot(\vec{\nabla} \times \vec{u})=0$

tells us that the divergence of the curl of a vector is zero, and this is true for every continuously-varying vector $$\vec{u}$$. Such identities are tremendously useful in vector calculus. For example, if a term includes the divergence of the curl of a vector, you can throw it out regardless of what the vector is.

Appendix E lists 21 of the most useful vector identities. All of these can (and should) be proved using the methods we have covered so far. For example:

• Proof of identity #15: $$\vec{\nabla}\times\vec{\nabla}\phi=0$$. We start with the kth component of $$\vec{\nabla}\times\vec{\nabla}\phi$$:

\begin{aligned} &\begin{aligned} [\vec{\nabla} \times \vec{\nabla} \phi]_{k}=\varepsilon_{i j k} \frac{\partial}{\partial x_{i}} \frac{\partial \phi}{\partial x_{j}} &=\varepsilon_{i j k} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \\ &=\varepsilon_{i j k} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{i}} \end{aligned} \quad \text { (reverse order of differentiation) }\\ &=\varepsilon_{j i k} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \quad \left(\text { relabel } i \text { and } j \text { as each other }\right)\\ &\begin{array}{ll} = & -\varepsilon_{i j k} \frac{\partial^{2} \phi}{\partial x_{i} \partial x_{j}} \\ = & -[\vec{\nabla} \times \vec{\nabla} \phi]_{k} \end{array} \quad \left( \text { use antisymmetry of } \varepsilon\right) \end{aligned}

We have shown that $[ \vec{\nabla}\times\vec{\nabla}\phi]_k$ is equal to its own additive inverse, and therefore can have no other value but zero.

• Proof of identity #21. It’s easier if we rearrange the identity like this: $$\left(\vec{\nabla}\times\vec{u}\right)\times\vec{u}\equiv \left[\vec{u}\cdot\vec{\nabla}\right]\vec{u}-\frac{1}{2}\vec{\nabla}\left(\vec{u}\cdot\vec{u}\right)$$. Now define

$\vec{\omega}=\vec{\nabla} \times \vec{u}, \quad \text { or } \quad \omega_{k}=\varepsilon_{i j k} \frac{\partial}{\partial x_{i}} u_{j}\label{eqn:5}$

and

$\vec{F}=(\vec{\nabla} \times \vec{u}) \times \vec{u}, \quad \text { or } \quad F_{m}=\varepsilon_{k l m} \omega_{k} u_{l}\label{eqn:6}$

We now substitute Equation $$\ref{eqn:5}$$ into Equation $$\ref{eqn:6}$$ and use the $$\varepsilon - \delta$$ relation:

\begin{aligned} F_{m} &=\varepsilon_{k l m}\left(\varepsilon_{i j k} \frac{\partial}{\partial x_{i}} u_{j}\right) u_{l} \\ &=\varepsilon_{i j k} \varepsilon_{k l m} \frac{\partial u_{j}}{\partial x_{i}} u_{l} \\ &=\left(\delta_{i l} \delta_{j m}-\delta_{i m} \delta_{j l}\right) \frac{\partial u_{j}}{\partial x_{i}} u_{l} \\ &=\delta_{i l} \delta_{j m} \frac{\partial u_{j}}{\partial x_{i}} u_{l}-\delta_{i m} \delta_{j l} \frac{\partial u_{j}}{\partial x_{i}} u_{l} \\ &=\frac{\partial u_{m}}{\partial x_{l}} u_{l}-\frac{\partial u_{l}}{\partial x_{m}} u_{l}. \end{aligned}

(In that last step we had two choices: we summed over the dummy indices $$i$$ and $$j$$, but we could just as well have chosen to sum over $$i$$ and $$l$$ or $$j$$ and $$l$$.) Now it is just a matter of recognizing the remaining two terms on the right-hand side as the terms in the identity we want to prove.

\begin{aligned} F_{m} &=u_{l} \frac{\partial}{\partial x_{l}} u_{m}-\frac{1}{2} \frac{\partial}{\partial x_{m}}\left(u_{l} u_{l}\right) \\ &=[\vec{u} \cdot \vec{\nabla}] u_{m}-\frac{1}{2} \frac{\partial}{\partial x_{m}}(\vec{u} \cdot \vec{u}), \end{aligned}

or $$\vec{F}=\left[\vec{u}\cdot\vec{\nabla}\right]\vec{u}-\frac{1}{2}\vec{\nabla}\left(\vec{u}\cdot\vec{u}\right)$$, and the identity is proven.

Identity-proving is not only an excellent diversion for a rainy day; it will also, like pushups for a quarterback or scales for a musician, prepare you for great things. You are therefore encouraged to see the list of vector identities in appendix E as a fine pile of puzzles awaiting your attention. Exercise 20 lists a few that you should definitely try.