# Laplacian of a Vector

## Homework Statement

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

## The Attempt at a Solution

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?

## Answers and Replies

TSny
Homework Helper
Gold Member
##\frac{1}{\vec{r}}## doesn't make sense because it involves dividing by a vector.

Are you sure you aren't supposed to take the Laplacian of ##\frac{1}{|\vec{r}|}## instead?

##\frac{1}{\vec{r}}## doesn't make sense because it involves dividing by a vector.

Are you sure you aren't supposed to take the Laplacian of ##\frac{1}{|\vec{r}|}## instead?

It looks like a vector

The problem is from Mathematical Methods for Physicists by Tai L. Chow.

TSny
Homework Helper
Gold Member
You're right, it does look like ##1/\vec{r}##. I think it must be a misprint. It does turn out that ##\nabla^2(1/r)=0## (except at ##r = 0##).

So, I'm thinking that's what was meant.

Part (c) also seems to have some misprints, unless the author is purposely trying to test whether you can tell if an expression is meaningless.

Last edited: