# Divergence of a Gradient

## The Attempt at a Solution

(a)[/B] Divergence of a gradient is a Laplacian.

(b) I suppose to do it in Cartesian coordinates.

Let $\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$

and $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}$,

$\hat{r}=\hat{i}+\hat{j}+\hat{k}$

First calculate $\nabla\left(\frac{1}{\overrightarrow{r}}\right)=\nabla\left(\frac{1}{\mid r\mid\hat{r}}\right)=\nabla\left(\frac{1}{\mid r\mid}\hat{r}\right)=\left[\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right]\left[\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\right]$

$= \hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}$

Am I doing it the right way?

PS. cross posted at

## Answers and Replies

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Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

View attachment 77030

## The Attempt at a Solution

(a)[/B] Divergence of a gradient is a Laplacian.

(b) I suppose to do it in Cartesian coordinates.

Let $\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$

and $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}$,

$\hat{r}=\hat{i}+\hat{j}+\hat{k}$

First calculate $\nabla\left(\frac{1}{\overrightarrow{r}}\right)=\nabla\left(\frac{1}{\mid r\mid\hat{r}}\right)=\nabla\left(\frac{1}{\mid r\mid}\hat{r}\right)=\left[\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right]\left[\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\right]$

$= \hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{-1}{2}}$

Am I doing it the right way?

PS. cross posted at
No. ##\hat{r}## is a vector, so ##\nabla(\hat{r}/r)## does not make sense: you need to take the gradient of a scalar, not of a vector. Anyway, why do you define ##\hat{r}=\hat{i}+\hat{j}+\hat{k}##? I cannot see its relation to anything in the problem. If by ##\hat{r}## you mean the unit vector in the direction of ##\vec{r}##, then that is most certainly not equal to what you wrote.

Also, never, never write something like ##\frac{1}{|r|\hat{r}}## because that is meaningless: it is a fraction of the form 1/vector, and those things do not exist in any usual form.

BvU
Homework Helper
Dear Isaac,

If you yourself write ##\vec r = \hat {\bf r} | {\bf r}|## you really should go back to ##\vec {r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}## and correct the hideous ##\hat{r}=\hat{\imath}+\hat{\jmath}+\hat{k}## to ##\hat{r}={x\over \sqrt{x^{2}+y^{2}+z^{2}}} \hat{\imath} + {y\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{\jmath} + {z\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{k}##

 sorry, pressed wrong button.

But, as Ray (And TSny) indicate, you don't need ##\hat r##

In your defence: The original problem formulation is confusing because it uses a boldface r in the 1/r. Many of us are conditioned to see that as a vector.

Dear Isaac,

If you yourself write ##\vec r = \hat {\bf r} | {\bf r}|## you really should go back to ##\vec {r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}## and correct the hideous ##\hat{r}=\hat{\imath}+\hat{\jmath}+\hat{k}## to ##\hat{r}={x\over \sqrt{x^{2}+y^{2}+z^{2}}} \hat{\imath} + {y\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{\jmath} + {z\over \sqrt{x^{2}+y^{2}+z^{2}}}\hat{k}##

 sorry, pressed wrong button.

But, as Ray (And TSny) indicate, you don't need ##\hat r##

In your defence: The original problem formulation is confusing because it uses a boldface r in the 1/r. Many of us are conditioned to see that as a vector.
Dear BvU, I think Author referred it as a vector. In the book (Mathematical Methods for Physicists by Tai L. Chow) scalars are mentioned as non bold, such as in same exercise page other problems are

BvU
Homework Helper
I count it as a misprint. What could possibly be the interpretation of ##1/\;\vec {\bf r}## ?

Ok, thx for the input.

## Homework Statement

Show that $\nabla^{2}\left(\frac{1}{r}\right)=0$

## Homework Equations

Let $\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$

and $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}, \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}$

## The Attempt at a Solution

$\nabla\left(\frac{1}{r}\right)=\nabla\left(\frac{1}{\mid r\mid}\right)=\left[\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right]\left[\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\right]$

$\nabla\left(\frac{1}{r}\right)= \hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}$

$\nabla\left(\frac{1}{r}\right)=-\hat{i}\frac{2x}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{j}\frac{2y}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{k}\frac{2z}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}$

$\nabla.\nabla\left(\frac{1}{r}\right)=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right).\left(-\hat{i}x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{j}y\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-\hat{k}z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)$

$\nabla^{2}\left(\frac{1}{r}\right)=\frac{\partial}{\partial x}\left(-x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)+\frac{\partial}{\partial y}\left(-y\left(x^{2}+y^{2}+z^{2}\right)\right)+\frac{\partial}{\partial z}-\left(z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}\right)$

$\nabla^{2}\left(\frac{1}{r}\right)=-x\left(2x\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-y\left(2y\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}$
$- \left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-z\left(2z\right)\left(-\frac{3}{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}$

$\nabla^{2}\left(\frac{1}{r}\right)=3x^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}+3y^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}$
$+3z^{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}$

$\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}$

$\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{5}{2}+1}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}$

$\nabla^{2}\left(\frac{1}{r}\right)=3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}-3\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{3}{2}}$

$\nabla^{2}\left(\frac{1}{r}\right)=0$

Hence Showed

Last edited:
BvU
Homework Helper
Yes, but we've to it in both - Cartesian and Spherical coordinates.

Yes, but we've to it in both - Cartesian and Spherical coordinates.
^ solve it

Solving part (c). As suggested above it's also a misprint. It should be
$\overrightarrow{r}.\left(\nabla.\overrightarrow{r}\right)\neq\left(r\nabla\right)r$