# Laplacian of a Vector

• NewtonApple
In summary, the problem is from the book Mathematical Methods for Physicists by Tai L. Chow and involves showing that the Laplacian of ##1/|\vec{r}|## is equal to 0, except at ##r = 0##. There may be some misprints in the problem, specifically in part (c).

## 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?

##\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?

TSny said:
##\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.

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.

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