- #1

Cole A.

- 12

- 0

## Homework Statement

(From Plonsey, R. and R. C. Barr, "Bioelectricity: A Quantitative Approach")

Show

[tex] \nabla^2 r = 0 [/tex]

given

[tex] r = \sqrt{x^2 + y^2 + z^2}. [/tex]

## Homework Equations

[tex] \nabla = \frac{\partial}{\partial x}{\bf i} + \frac{\partial}{\partial y}{\bf j} + \frac{\partial}{\partial z}{\bf k} [/tex] (the "abuse of notation")

## The Attempt at a Solution

This is my first attempt at a post on this forum, and I'm embarrassed to say I'm really struggling with this question. (Not a homework problem per se, just an in-text side note). It has been a good while since I've studied vector calculus, so if somebody could point out my error, I would appreciate it.

Beginning with

[tex]

\begin{equation}

r = (x^2 + y^2 + z^2)^{1/2}

\end{equation}

[/tex]

taking the gradient results in

[tex]

\begin{equation}

\begin{aligned}

\nabla r &= \frac{\partial r}{\partial x} {\bf i} + \frac{\partial r}{\partial y} {\bf j} + \frac{\partial r}{\partial z} {\bf k} \\

&= \left[ \frac{x}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf i} + \left[ \frac{y}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf j} + \left[ \frac{z}{(x^2 + y^2 + z^2)^{1/2}} \right]{\bf k}.

\end{aligned}

\end{equation}

[/tex]

Now taking the divergence of this gradient [itex] - [/itex] which amounts cumulatively to taking the Laplacian of [itex] r - [/itex] gives

[tex]

\begin{equation}

\begin{aligned}

\nabla \cdot \nabla r &= \nabla^2 r \\

&= \frac{\partial}{\partial x}\left[ \frac{x}{(x^2 + y^2 + z^2)^{1/2}} \right] + \frac{\partial}{\partial y}\left[ \frac{y}{(x^2 + y^2 + z^2)^{1/2}} \right] + \frac{\partial}{\partial z}\left[ \frac{z}{(x^2 + y^2 + z^2)^{1/2}} \right] \\

&= \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{x^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{y^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{1}{(x^2 + y^2 + z^2)^{1/2}} - \frac{z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] \\

&= \left[ \frac{y^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{x^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \right] + \left[ \frac{x^2 + y^2}{(x^2 + y^2 + z^2)^{3/2}} \right] \\

&= \frac{2}{(x^2 + y^2 + z^2)^{1/2}}

\end{aligned}

\end{equation}

[/tex]

which is not zero, although I very much wish it was. (Sorry for the long-winded calculations.)