Why Does the Laplacian of 1/Vector r Equal Zero?

[NewtonApple]

Homework Statement

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

Homework Equations

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?

 

[TSny]

$\frac{1}{\vec{r}}$ 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?

 

[NewtonApple]

$\frac{1}{\vec{r}}$ 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]

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.

 

[HalfLostAndSearching]

Yes, you are on the right track. To show that the Laplacian of a vector is equal to zero, you need to take the divergence of the gradient of the vector. In this case, the vector is \frac{1}{\overrightarrow{r}}, so the Laplacian would be \nabla^2\left(\frac{1}{\overrightarrow{r}}\right) = \nabla \cdot \nabla \left(\frac{1}{\overrightarrow{r}}\right). Using the product rule and applying the chain rule, you correctly showed that the gradient of \frac{1}{\overrightarrow{r}} is \nabla\left(\frac{1}{\overrightarrow{r}}\right)=\hat{r}\left(\frac{-1}{\mid r\mid^{2}}\right). Taking the divergence of this, you get

\nabla \cdot \nabla \left(\frac{1}{\overrightarrow{r}}\right) = \frac{\partial}{\partial x} \left(\hat{r}\left(\frac{-1}{\mid r\mid^{2}}\right)\right) + \frac{\partial}{\partial y} \left(\hat{r}\left(\frac{-1}{\mid r\mid^{2}}\right)\right) + \frac{\partial}{\partial z} \left(\hat{r}\left(\frac{-1}{\mid r\mid^{2}}\right)\right)

= \frac{\partial}{\partial x} \left(\frac{-x}{(x^2+y^2+z^2)^{3/2}}\right) + \frac{\partial}{\partial y} \left(\frac{-y}{(x^2+y^2+z^2)^{3/2}}\right) + \frac{\partial}{\partial z} \left(\frac{-z}{(x^2+y^2+z^2)^{3/2}}\right)

= \frac{-x(x^2+y^2+z^2)^{-3/2}}{(x^2+y^2+z^2)} + \frac{-y(x^2+y^2+z^2)^{-3/2}}{(x^2+y^2+z^2)} + \frac{-z