### Divergence

• Calculate divergence.

• Understand physical interpretation and applications of divergence.

Divergence is calculated for a vector, but since it is a dot product between the del ($\nabla$) operator (which is a vector) and another vector, the final result is a scalar.

$\nabla . \mathbf{V} = \left ( \frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k} \right ) .(v_{1} i+ v_{2} j+ v_{3}k)$

Let us take an example of a variable vector which keeps on changing depending upon the location (coordinates).
e.g. if $\bar{a} = x^3zi + xy^2j + yzk$ $\nabla . \mathbf{V} = \left ( \frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k} \right ) .(x^3z i+ xy^2 j+ yz k)$ $\nabla . \mathbf{V} =(3x^2z + 2xy + z )$ If you want to calculate at some location, let’s say, at (1,2-1) then simply substitute the coordinates, $\nabla . \mathbf{V} \mathrm{\ at\ } (1,2-1) =(3(1)^2(-1) + 2(1)(2) + (-1) ) = 0$

## Physical Interpretation of the Divergence:

• Consider a vector field $\mathbf{F}$ that represents a fluid velocity: The divergence of $\mathbf{F}$ at a point in a fluid is a measure of the rate at which the fluid is flowing away from or towards that point.

• A positive divergence is indicating a flow away from the point.

• Physically divergence means that either the fluid is expanding or that fluid is being supplied by a source external to the field.

• The lines of flow diverge from a source and converge to a sink.

• If there is no gain or loss of fluid anywhere then div $\mathbf{F}$ = 0. Such a vector field is said to be solenoidal.

### Solved Example:

#### 4-10-01

The divergence of vector $\textbf{i} = x\textbf{i} + y\textbf{j} + z\textbf{k}$ is: (ISRO SDSC SA MPC Physics Feb 2017)

Solution:

Divergence is defined as $\nabla$.r, where:

$r = x\overline {i} + y\overline {j} + z\overline {k}$ \begin{align*} \nabla \cdot r&=\left( \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right). \left( x\overline {i}+y\overline {j}+z\overline {k}\right)\\ &= 1 + 1 + 1\\ &= 3 \end{align*}

### Solved Example:

#### 4-10-02

The divergence of the vector field $x^2y \mathbf{i}+ xy \mathbf{j} + z^2 \mathbf{k}$ at P (1, 1, 1) is:

Solution:
$\nabla . \mathbf{V} = \left ( \dfrac{\partial}{\partial x} \mathbf{i} + \dfrac{\partial}{\partial y} \mathbf{j} + \dfrac{\partial}{\partial z} \mathbf{k} \right ) .(v_{1} i+ v_{2} j+ v_{3} k)$ $\nabla . \mathbf{V} = \left ( \dfrac{\partial}{\partial x} \mathbf{i} + \dfrac{\partial}{\partial y} \mathbf{j} + \dfrac{\partial}{\partial z} \mathbf{k} \right ) .(x^2y\ i+ xy\ j+ z^2\ k)$ $\nabla . \mathbf{V} = \dfrac{\partial}{\partial x} x^2y + \dfrac{\partial}{\partial y} xy + \dfrac{\partial}{\partial z} z^2 = 2xy + x + 2z$ Substitute x = 1, y = 1 and z = 1 (Given coordinates) $\nabla . \mathbf{V} = 5$

### Solved Example:

#### 4-10-03

Determine the divergence of the vector: $\overline{V}$ = $(x^2) \overline{i} + (-xy) \overline{j} + (xyz) \overline{k}$ at the point (3,2,1).

Solution:
$\nabla =\dfrac {\partial}{\partial x} \overline{i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline{k}, \quad \overline {V}=\left(x^{2}\right) \overline {i}+\left( -xy\right) \overline {j}+ \left(xyz\right) \overline {k}$ \begin{align*} \nabla \cdot \overline {V} &= \left( \dfrac {\partial }{\partial x}\right) \left( x^{2}\right) +\dfrac {\partial }{\partial y}\left( -xy\right) +\dfrac {\partial }{\partial z}\left( xyz\right)\\ &= 2x-x+xy\\ &=x+xy \end{align*}

At x = 3, y = 2 and z = 1,

$\nabla \cdot \overline {V}$ = 3+6 = 9

### Solved Example:

#### 4-10-04

The divergence of the vector field $(x - y) \textbf{i} + (y - x)\textbf{j} + (x + y + z)\textbf{k}$ is:

Solution:
\begin{align*} \overline {V}&=\left( x-y\right) \overline {i}+\left( y-x\right) \overline {j}+\left( x+y+z\right) \overline {k},\\ \nabla &=\left( \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right) \end{align*} \begin{align*} \nabla \cdot \overline {V} &=\dfrac {\partial }{\partial x}\left( x-y\right) +\dfrac {\partial }{\partial y}\left( y-x\right) +\dfrac {\partial }{\partial z}\left( x+y+z\right)\\ &=1+1+1\\ &=3 \end{align*}

### Solved Example:

#### 4-10-05

The divergence of the vector field, $3xzi + 2xyj - yz^2k$ at a point (1,1,1) is equal to:

Solution:

Let, $V = 3xzi + 2xyj - yz^2k$. We know divergence vector field of V is given by ($\nabla$ $\cdot$V)

\begin{align*} \nabla \cdot V &=\left( \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right) \cdot \left(3xz\overline {i}+2y\overline {j}-yz^{2}\overline {k}\right)\\ &=3z+2x-2yz \end{align*} At point P(1,1,1),

$\left( \nabla \cdot V\right) =3\times 1+2\times 1-2\times 1\times 1 =3$

### Solved Example:

#### 4-10-06

Let f be a scalar field, and let $\bar{F}$ be a vector field. Which of the following expressions is meaningful:

Solution:

$\nabla$ is a vector operator. When it is applied to a scalar, you get a vector result. Hence $\nabla$f is a vector.

When it is applied to a vector as a dot product, you get a scalar result. Hence $\nabla$.$\bar{F}$ is a scalar.

When it is applied to a vector as a cross product, you get a vector result. Hence $\nabla$ $\times$ $\bar{F}$ is a vector.

You CANNOT add scalar with a vector.