Laplacian of a Vector

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).
  • #1

Homework Statement


Show that [itex]\nabla^{2}\left(\frac{1}{\overrightarrow{r}}\right)=0[/itex]


Homework Equations



The Attempt at a Solution


Let [itex]\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}[/itex]

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

[itex]\hat{r}=\hat{i}+\hat{j}+\hat{k}[/itex]

First calculate [itex]\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][/itex]

[itex] = \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}} [/itex]

Am I doing it the right way?
 
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  • #2
##\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?
 
  • #3
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
image024.gif

The problem is from Mathematical Methods for Physicists by Tai L. Chow.
 
  • #4
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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