Gradient of a Scalar Field

Let $\phi (x, y, z)$ is a scalar field, which is a function of space variables $x, y, z$ . Then the gradient of scalar field is defined as operation of $\overrightarrow{\nabla}$ on the scalar field.
That is:
$Grad\phi (x, y, z)$ = $\overrightarrow{\nabla} \phi (x, y, z)$ $\equiv \overrightarrow\nabla} \phi$
Here the operator $\overrightarrow{\nabla}$ is called Del or Nabla vector. It is given by the following expression:

(1) $\begin{equation*} \overrightarrow{\nabla}= \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \end{equation*}$

Please note that $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along X, Y and Z axes respectively in cartesian system of cordinates.
Thus using the description given in Eq: (1), the expression for gradient of a scalar field can also be written as:
$Grad\phi (x, y, z)$ = $\overrightarrow{\nabla} \phi (x, y, z)$ $\equiv \overrightarrow\nabla} \phi$ = $\hat{i}\frac{\partial\phi}{\partial x} + \hat{j}\frac{\partial\phi}{\partial y} + \hat{k}\frac{\partial\phi}{\partial z}$

$\implies \boxed{{\overrightarrow\nabla} \phi $ = $\hat{i}\frac{\partial\phi}{\partial x} + \hat{j}\frac{\partial\phi}{\partial y} + \hat{k}\frac{\partial\phi}{\partial z}}$

Example: 1 If $\overrightarrow{r}$ is the position vector of a point (x, y, z) in space, then find the gradient of $\frac{1}{r}$ .
Solution: Since $\overrightarrow{r}$ is the position vector of a point (x, y, z) in space, therefore, it is given as follows:

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

The magnitude of this vector is given as follows:

(2) $\begin{equation*} r =\sqrt{x^{2}+y^{2}+z^{2}} \end{equation*}$

Thus

(3) $\begin{equation*} \frac{1}{r}=(x^{2}+y^{2}+z^{2})^{-\frac{1}{2}}$ \end{equation*}$

The gradient of $\frac{1}{r}$ is given by:

(4) $\begin{equation*} \overrightarrow\nabla}\left(\frac{1}{r}\right) = \hat{i}\frac{\partial}{\partial x}\left(\frac{1}{r}\right) + \hat{j}\frac{\partial}{\partial y}\left(\frac{1}{r}\right) + \hat{k}\frac{\partial}{\partial z}}\left(\frac{1}{r}\right) \end{equation*}$

Let us now evaluate each term on the right hand side of equation (4) one by one and then solve the problem.
The first term can be calculated using equation (3) as follows:

(5) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial x}\left(\frac{1}{r}\right) &=-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}(2x)\\ &=-\frac{x}{(\sqrt{x^{2}+y^{2}+z^{2}})^{3}}\\ &=\frac{-x}{r^{3}} \end{array} \end{equation*}$

(using equation (2))
Similarly, we can find that

(6) $\begin{equation*} \frac{\partial}{\partial y}\left(\frac{1}{r}\right) = \frac{-y}{r^{3}} \end{equation*}$

and

(7) $\begin{equation*} \frac{\partial}{\partial z}\left(\frac{1}{r}\right) = \frac{-z}{r^{3}} \end{equation*}$

Put the values from equations (5), (6) and (7) in equation (4), we get

(8) $\begin{equation*} \overrightarrow{\nabla}\left(\frac{1}{r}\right) = \frac{-(x\hat{i}+y\hat{j}+z\hat{k})}{r^{3}} \nonumber \end{equation*}$

$\implies\boxed{\overrightarrow\nabla\left(\frac{1}{r}\right) = -\frac{\overrightarrow{r}}{r^{3}}}$

Example:2 If $\overrightarrow{r~}$ and $\overrightarrow{r~}'$ be the position vectors of points $(x, y, z)$ and $(x', y', z')$ in space, then find the gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ .

Solution: Since $\overrightarrow{r~}$ is the position vector of a point (x, y, z) in space, therefore, it is given as follows:

$\overrightarrow{r~}=x\hat{i}+y\hat{j}+z\hat{k}$

Similarly

$\overrightarrow{r~}'={x'}\hat{i}+{y'}\hat{j}+{z'}\hat{k}$

Therefore

$\overrightarrow{r~}-\overrightarrow{r~}'={(x-x')}\hat{i}+{(y-y')}\hat{j}+{(z-z')}\hat{k}$

The magnitude of this difference or displacement vector is given by:

(9) $\begin{equation*} |\overrightarrow{r~}-\overrightarrow{r~}'|=\sqrt{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2} \end{equation*}$

(10) $\begin{equation*} \implies \frac{1}{|\overrightarrow{r~}-\overrightarrow{r~}'|}=[{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2}]^{-\tfrac{1}{2}} \end{equation*}$

The gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ is given by:

(11) $\begin{equation*} \overrightarrow{\nabla}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \hat{i}\frac{\partial}{\partial x}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{j}\frac{\partial}{\partial y}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{k}\frac{\partial}{\partial z}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) \end{equation*}$

Let us now evaluate each term on the right hand side of equation (11) one by one and then solve the problem.
The first term can be calculated using equation (10) as follows:

(12) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial x}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=-\frac{1}{2}\left({(x-x')}^2+{(y-y')}^2+{(z-z')}^2\right)^{-\frac{3}{2}}[2(x-x')]\\ &=\frac{-(x-x')}{(\sqrt{(x-x')^2+(y-y')^2+(z-z')^2})^{3}}\\ &=\frac{-(x-x')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

(using equation (9))
Similarly, we can find that

(13) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial y}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{-(y-y')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

and

(14) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial z}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{-(z-z')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

Put the values from equations (12), (13) and (14) in equation (11), we get

(15) $\begin{equation*} \overrightarrow{\nabla}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{-[(x-x')\hat{i}+(y-y')\hat{j}+(z-z')\hat{k}]}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \nonumber \end{equation*}$

(16) $\begin{equation*} \implies\boxed{\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = -\frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}}} \end{equation*}$

Example: 3 If $\overrightarrow{r~}$ and $\overrightarrow{r~}'$ be the position vectors of points $(x, y, z)$ and $(x', y', z')$ in space, then find the gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ with respect to the point $(x', y', z')$ . That is, find expression for $\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$

Solution: Since $\overrightarrow{r~}$ is the position vector of a point (x, y, z) in space, therefore, it is given as follows:

$\overrightarrow{r~}=x\hat{i}+y\hat{j}+z\hat{k}$

Similarly

$\overrightarrow{r~}'={x'}\hat{i}+{y'}\hat{j}+{z'}\hat{k}$

Therefore

$\overrightarrow{r~}-\overrightarrow{r~}'={(x-x')}\hat{i}+{(y-y')}\hat{j}+{(z-z')}\hat{k}$

The magnitude of this difference or displacement vector is given by:

(17) $\begin{equation*} |\overrightarrow{r~}-\overrightarrow{r~}'|=\sqrt{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2} \end{equation*}$

(18) $\begin{equation*} \implies \frac{1}{|\overrightarrow{r~}-\overrightarrow{r~}'|}=[{{(x-x')}^2+{(y-y')}^2+{(z-z')}^2}]^{-\tfrac{1}{2}} \end{equation*}$

The gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ with respect to the point $(x', y', z')$ is given by:

(19) $\begin{equation*} \overrightarrow{\nabla'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \hat{i}\frac{\partial}{\partial x'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{j}\frac{\partial}{\partial y'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) + \hat{k}\frac{\partial}{\partial z'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) \end{equation*}$

Let us now evaluate each term on the right hand side of equation (19) one by one and then solve the problem.
The first term can be calculated using equation (18) as follows:

(20) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial x'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=-\frac{1}{2}\left({(x-x')}^2+{(y-y')}^2+{(z-z')}^2\right)^{-\frac{3}{2}}[2(x-x')(-1)]\\ &=\frac{(x-x')}{(\sqrt{(x-x')^2+(y-y')^2+(z-z')^2})^{3}}\\ &=\frac{(x-x')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

(using equation (17))
Similarly, we can find that

(21) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial y'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{(y-y')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

and

(22) $\begin{equation*} \begin{array}{ll} \frac{\partial}{\partial z'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) &=\frac{(z-z')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{array} \end{equation*}$

Put the values from equations (20), (21) and (22) in equation (19), we get

(23) $\begin{equation*} \overrightarrow{\nabla'}\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{[(x-x')\hat{i}+(y-y')\hat{j}+(z-z')\hat{k}]}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \nonumber \end{equation*}$

(24) $\begin{equation*} \implies\boxed{\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}}} \end{equation*}$

Example: 4 If $\overrightarrow{r~}$ and $\overrightarrow{r~}'$ be the position vectors of points $(x, y, z)$ and $(x', y', z')$ in space, then find the gradient of $\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}$ with respect to the points $(x, y, z)$ and $(x', y', z')$ . That is, find expression for $\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$ and $\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$ . Hence show that these two gradients are equal and opposite. That is $\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)=-\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)$
Solution: We have already proved in Example 2 and Example 3 that

(25) $\begin{equation*} \overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = -\frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{equation*}$

and

(26) $\begin{equation*} \overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right) = \frac{(\overrightarrow{r~}-\overrightarrow{r~}')}{|\overrightarrow{r~}-\overrightarrow {r~}'|^{3}} \end{equation*}$

Comparing equations (25) and (26), we get the desired result. That is

$\boxed{\overrightarrow\nabla'\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)=-\overrightarrow\nabla\left(\frac{1}{|\overrightarrow{r~}-\overrightarrow {r~}'|}\right)}$

Note: This relation is very useful in the derivation and solution of many complex problems of electromagnetism.

If you are interested to know more about scalar and vector fields, then watch video lecture by me on this topic at the following link: https://youtu.be/H4eWdwaGWYI.

Another video, that may be of interest to the reader of this post is ‘Transverse nature of electromagnetic waves’. The link to this video lecture by me is given here: https://youtu.be/k7399IiFEzo

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