Divergence of a radial vector field

Question

I am reading Modern Electrodynamics by Zangwill and cannot verify equation (1.61) [page 7]:

\begin{equation} \nabla \cdot \textbf{g}(r)=\textbf{g}^{\prime}\cdot \mathbf{\hat{r}}, \end{equation} where the vector field $\textbf{g}(r)$ is only nonzero in the radial direction.

By using the divergence formula in Spherical coordinates, I get:

\begin{align} \nabla \cdot \textbf{g}(r)&=\frac{1}{r^2} \partial_r (r^2 g_r) + \frac{1}{r \sin \theta} \partial_{\theta} (g_{\theta} \sin \theta) + \frac{1}{r \sin \theta} \partial_{\phi} g_{\phi}\\ &=\frac{2}{r}g_r + \frac{d}{dr}g_r\\ &= \frac{2}{r}\textbf{g}\cdot \mathbf{\hat{r}}+\textbf{g}^{\prime}\cdot \mathbf{\hat{r}} \end{align}

What is going wrong?

Answer 1

The function $\mathbf g(r)$ is not a radial vector field; it is a vector field which depends only on $r\equiv\sqrt{x^2+y^2+z^2}$. Indeed unless the vector field is the trivial $\mathbf g(r) = \vec 0$, then it cannot be radial; you can see this by noting that if $\mathbf g$ is radial, then the vectors at the north and south poles must point in opposite directions, but if $\mathbf g=\mathbf g(r)$ then the vectors at the north and south poles must be the same.

Answer 2

In $\S\texttt{1.3 Derivatives }$ of the textbook $''$Modern Electrodynamics$''$ by A.Zangwill the function $\,\mathbf g\left(r\right)\,$ is considered as a vector function of the position radius $\,r$. Following the notation therein this means that \begin{equation} \mathbf g\left(r\right)\boldsymbol= \mathrm g_x\left(r\right)\mathbf{\hat{x}}\boldsymbol+\mathrm g_y\left(r\right)\mathbf{\hat{y}}\boldsymbol+\mathrm g_z\left(r\right)\mathbf{\hat{z}} \boldsymbol= \begin{bmatrix} \mathrm g_x\left(r\right) & \mathrm g_y\left(r\right) & \mathrm g_z\left(r\right) \end{bmatrix} \begin{bmatrix} \mathbf{\hat{x}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{y}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{z}}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{01}\label{01} \end{equation} where $\,\mathbf{\hat{x}},\mathbf{\hat{y}}\,$ and $\,\mathbf{\hat{z}}\,$ the unit vectors along the cartesian coordinates axes $\,x,y\,$ and $\,z\,$ respectively.

Now, for the analysis of this vector function by components in the spherical coordinate system we make use of the relation between the local spherical coordinate system $\,\{\mathbf{\hat{r}},\boldsymbol{\hat{\theta}},\boldsymbol{\hat{\phi}}\}\,$ and the global cartesian $\,\{\mathbf{\hat{x}},\mathbf{\hat{y}},\mathbf{\hat{z}}\}\,$ \begin{align} \mathbf{\hat{x}} & \boldsymbol=\left(\sin\theta\cos\phi\right)\mathbf{\hat{r}}\boldsymbol+\left(\cos\theta\cos\phi\right)\boldsymbol{\hat{\theta}}\boldsymbol+\left(\boldsymbol-\sin\phi\right)\boldsymbol{\hat{\phi}} \tag{02a}\label{02a}\\ \mathbf{\hat{y}} & \boldsymbol=\left(\sin\theta\sin\phi\right)\mathbf{\hat{r}}\boldsymbol+\left(\cos\theta\sin\phi\right)\boldsymbol{\hat{\theta}}\boldsymbol+\left(\cos\phi\right)\boldsymbol{\hat{\phi}} \tag{02b}\label{02b}\\ \mathbf{\hat{z}} & \boldsymbol=\left(\cos\theta\right)\mathbf{\hat{r}}\boldsymbol+\left(\boldsymbol-\sin\theta\right)\boldsymbol{\hat{\theta}} \tag{02c}\label{02c} \end{align} or in matrix form \begin{equation} \begin{bmatrix} \mathbf{\hat{x}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{y}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{z}}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol= \begin{bmatrix} \sin\theta\cos\phi & \cos\theta\cos\phi & \boldsymbol-\sin\phi \vphantom{\dfrac{a}{b}}\\ \sin\theta\sin\phi & \cos\theta\sin\phi & \hphantom{\boldsymbol-}\cos\phi \vphantom{\dfrac{a}{b}}\\ \cos\theta & \boldsymbol-\sin\theta & \hphantom{\boldsymbol-}0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \mathbf{\hat{r}}\vphantom{\dfrac{a}{b}}\\ \boldsymbol{\hat{\theta}}\vphantom{\dfrac{a}{b}}\\ \boldsymbol{\hat{\phi}}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{03}\label{03} \end{equation}

Inserting this expression in equation \eqref{01} we have \begin{equation} \mathbf g\left(r\right)\boldsymbol= \mathrm g_r\left(r,\theta,\phi\right)\mathbf{\hat{r}}\boldsymbol+ \mathrm g_\theta\left(r,\theta,\phi\right)\boldsymbol{\hat{\theta}}\boldsymbol+\mathrm g_\phi\left(r,\theta,\phi\right)\boldsymbol{\hat{\phi}} \tag{04}\label{04} \end{equation} where \begin{align} \mathrm g_r\left(r,\theta,\phi\right) & \boldsymbol=\sin\theta\cos\phi\cdot\mathrm g_x\left(r\right)\boldsymbol+\sin\theta\sin\phi\cdot\mathrm g_y\left(r\right)\boldsymbol+\cos\theta\cdot\mathrm g_z\left(r\right) \tag{05a}\label{05a}\\ \mathrm g_\theta\left(r,\theta,\phi\right) & \boldsymbol=\cos\theta\cos\phi\cdot\mathrm g_x\left(r\right)\boldsymbol+\cos\theta\sin\phi\cdot\mathrm g_y\left(r\right)\boldsymbol-\sin\theta\cdot\mathrm g_z\left(r\right) \tag{05b}\label{05b}\\ \mathrm g_\phi\left(r,\theta,\phi\right) & \boldsymbol=\boldsymbol-\sin\phi\cdot\mathrm g_x\left(r\right)\boldsymbol+\cos\phi\cdot\mathrm g_y\left(r\right) \tag{05c}\label{05c} \end{align} So all components $\,\mathrm g_r,\mathrm g_\theta,\mathrm g_\phi\,$ are functions of $\,r,\theta,\phi\,$ and especially $\,\mathrm g_\theta,\mathrm g_\phi\,$ are not zero in general.

It's not convenient to find the divergence of such a vector field using spherical coordinates. To the contrary, by cartesian coordinates \begin{equation} \begin{split} \boldsymbol\nabla\boldsymbol\cdot\mathbf g\left(r\right) & \boldsymbol= \dfrac{\partial\mathrm g_x\left(r\right)}{\partial x}\boldsymbol+\dfrac{\partial\mathrm g_y\left(r\right)}{\partial y}\boldsymbol+\dfrac{\partial\mathrm g_z\left(r\right)}{\partial z}\\ &\boldsymbol= \dfrac{\mathrm d\mathrm g_x\left(r\right)}{\mathrm dr}\dfrac{\partial r}{\partial x}\boldsymbol+\dfrac{\mathrm d\mathrm g_y\left(r\right)}{\mathrm dr}\dfrac{\partial r}{\partial y}\boldsymbol+\dfrac{\mathrm d\mathrm g_z\left(r\right)}{\mathrm dr}\dfrac{\partial r}{\partial z}\\ &\boldsymbol= \dfrac{\mathrm d\mathrm g_x\left(r\right)}{\mathrm dr}\dfrac{x}{r}\boldsymbol+\dfrac{\mathrm d\mathrm g_y\left(r\right)}{\mathrm dr}\dfrac{y}{r}\boldsymbol+\dfrac{\mathrm d\mathrm g_z\left(r\right)}{\mathrm dr}\dfrac{z}{r}\\ &\boldsymbol= \dfrac{\mathrm d\mathbf g\left(r\right)}{\mathrm dr}\boldsymbol\cdot\dfrac{\mathbf r}{r}\boldsymbol=\mathbf g'\boldsymbol\cdot\mathbf{\hat{r}}\\ \end{split} \tag{06}\label{06} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

ADDENDUM : $\texttt{ Gymnastics with Spherical Coordinates}$

If we insist to follow the difficult path of finding the divergence using spherical coordinates then \begin{equation} \boldsymbol\nabla\boldsymbol\cdot\mathbf g\left(r\right) \boldsymbol=\underbrace{\dfrac{1}{r^2}\dfrac{\partial\left(r^2\mathrm g_r\right)}{\partial r}}_{\boxed{\boldsymbol 1}}\boldsymbol+\underbrace{\dfrac{1}{r\sin\theta}\dfrac{\partial\left(\sin\theta\,\mathrm g_\theta\right)}{\partial \theta}}_{\boxed{\boldsymbol 2}}\boldsymbol+\underbrace{\dfrac{1}{r\sin\theta}\dfrac{\partial\mathrm g_\phi}{\partial \phi}}_{\boxed{\boldsymbol 3}} \tag{07}\label{07} \end{equation} Using the expressions \eqref{05a},\eqref{05b} and \eqref{05c} for the components $\,\mathrm g_r,\mathrm g_\theta\,$ and $\,\mathrm g_\phi\,$ respectively we have in sequence \begin{equation} \begin{split} \boxed{\boldsymbol 1} & \boldsymbol= \dfrac{1}{r^2}\dfrac{\partial\left(r^2\mathrm g_r\right)}{\partial r}\boldsymbol=\dfrac{2}{\,r\,}\mathrm g_r\boldsymbol+\dfrac{\partial\mathrm g_r}{\partial r}\quad \boldsymbol\implies\\ \boxed{\boldsymbol 1} & \boldsymbol= \dfrac{2\sin\theta\cos\phi}{r}\mathrm g_x\boldsymbol+\dfrac{2\sin\theta\sin\phi}{r}\mathrm g_y\boldsymbol+\dfrac{2\cos\theta}{r}\mathrm g_z\\ & \hphantom{\boldsymbol=}\boldsymbol+\sin\theta\cos\phi\dfrac{\mathrm d\mathrm g_x}{\mathrm dr}\boldsymbol+\sin\theta\sin\phi\dfrac{\mathrm d\mathrm g_y}{\mathrm dr}\boldsymbol+\cos\theta\dfrac{\mathrm d\mathrm g_z}{\mathrm dr}\\ \end{split} \tag{08}\label{08} \end{equation} and \begin{equation} \begin{split} \boxed{\boldsymbol 2} & \boldsymbol= \dfrac{1}{r\sin\theta}\dfrac{\partial\left(\sin\theta\,\mathrm g_\theta\right)}{\partial \theta}\boldsymbol=\dfrac{\cot\theta }{r}\mathrm g_\theta\boldsymbol+\dfrac{1}{r}\dfrac{\partial\mathrm g_\theta}{\partial \theta}\quad \boldsymbol\implies\\ \boxed{\boldsymbol 2} & \boldsymbol=\dfrac{\cos^2\theta\cos\phi }{r\sin\theta}\mathrm g_x\boldsymbol+\dfrac{\cos^2\theta\sin\phi }{r\sin\theta}\mathrm g_y\boldsymbol-\dfrac{\cos\theta }{r}\mathrm g_z\\ & \hphantom{\boldsymbol=}\boldsymbol-\dfrac{\sin\theta\cos\phi}{r}\mathrm g_x\boldsymbol-\dfrac{\sin\theta\sin\phi}{r}\mathrm g_y\boldsymbol-\dfrac{\cos\theta}{r}\mathrm g_z\\ \end{split} \tag{09}\label{09} \end{equation} and \begin{equation} \begin{split} \boxed{\boldsymbol 3} & \boldsymbol= \dfrac{1}{r\sin\theta}\dfrac{\partial\mathrm g_\phi}{\partial \phi}\quad \boldsymbol\implies\\ \boxed{\boldsymbol 3} & \boldsymbol=\boldsymbol-\dfrac{\cos\phi}{r\sin\theta}\mathrm g_x\boldsymbol-\dfrac{\sin\phi}{r\sin\theta}\mathrm g_y\\ \end{split} \tag{10}\label{10} \end{equation} So \begin{equation} \begin{split} \boldsymbol\nabla\boldsymbol\cdot\mathbf g\left(r\right) &\boldsymbol=\boxed{\boldsymbol 1} \boldsymbol+\boxed{\boldsymbol 2} \boldsymbol+\boxed{\boldsymbol 3}\\ &\boldsymbol=\overbrace{\left(\dfrac{2\sin\theta\cos\phi}{r}\boldsymbol+\dfrac{\cos^2\theta\cos\phi }{r\sin\theta}\boldsymbol-\dfrac{\sin\theta\cos\phi}{r}\boldsymbol-\dfrac{\cos\phi}{r\sin\theta}\right)}^{0}\mathrm g_x\\ & \hphantom{\boldsymbol=}\boldsymbol+\overbrace{\left(\dfrac{2\sin\theta\sin\phi}{r}\boldsymbol+\dfrac{\cos^2\theta\sin\phi }{r\sin\theta}\boldsymbol-\dfrac{\sin\theta\sin\phi}{r}\boldsymbol-\dfrac{\sin\phi}{r\sin\theta}\right)}^{0}\mathrm g_y\\ & \hphantom{\boldsymbol=}\boldsymbol+\overbrace{\left(\dfrac{2\cos\theta}{r}\boldsymbol-\dfrac{\cos\theta}{r}\boldsymbol-\dfrac{\cos\theta}{r}\right)}^{0}\mathrm g_z\\ & \hphantom{\boldsymbol=}\boldsymbol+\underbrace{\sin\theta\cos\phi}_{x/r}\dfrac{\mathrm d\mathrm g_x}{\mathrm dr}\boldsymbol+\underbrace{\sin\theta\sin\phi}_{y/r}\dfrac{\mathrm d\mathrm g_y}{\mathrm dr}\boldsymbol+\underbrace{\cos\theta}_{z/r}\dfrac{\mathrm d\mathrm g_z}{\mathrm dr}\\ &\boldsymbol=\dfrac{\mathrm d\mathbf g\left(r\right)}{\mathrm dr}\boldsymbol\cdot\dfrac{\mathbf r}{r}\boldsymbol=\mathbf g'\boldsymbol\cdot\mathbf{\hat{r}}\\ \end{split} \tag{11}\label{11} \end{equation} qed.

Answer 3

The first equation is presumptive written without rigorous thinking, and it is not right. For a radial vector field, $\vec g(r) = g_r(r) \hat r $, its divergence is: \begin{align} \nabla \cdot \vec g(r) &= \vec \nabla \cdot g_r(r) \hat r \\ &= \vec \nabla g_r(r) \cdot \hat r + g_r(r) \vec \nabla \cdot \hat r \\ &= g'_r(r) + g_r(r) \vec \nabla \cdot \frac{\vec r}{r}\\ &= g'_r(r) + g_r(r) \left(\frac{\vec \nabla \cdot \vec r}{r} - \frac{\vec r \cdot \vec \nabla r}{r^2} \right)\\ &= g'_r(r) + g_r(r) \left(\frac{3}{r} - \frac{r}{r^2} \right) \\ &= g'_r(r) + g_r(r) \frac{2}{r}\\ & = \vec g'(r) \cdot \hat r + \frac{2}{r} \vec g \cdot \vec r. \end{align} The result is same as your second equation.

Answer 4

$ \renewcommand\vec\mathbf \newcommand\R{\mathbb R} \newcommand\adj\overline \newcommand\diff\underline $

I want to add that the desired result is a straightforward application of an extremely general chain rule for $\nabla$. For any $f : \R^n \to R$ where $R = \R$ or $R = \R^n$ $$ \nabla = \adj f_{\vec x}(\partial_f). $$ This requires some unpacking.

• This rewriting of $\nabla$ can be applied in any expression where $\nabla$ appears in a linear slot, i.e. where if you replaced $\nabla$ with a vector $\vec v$ then the resulting expression is linear in $\vec v$.
• $\nabla = \nabla_{\vec x}$ is implicitly differentiating with respect to $\vec x$.
• $\partial_f = \partial/\partial f$ when $R = \R$ and $\partial_f = \nabla_f$ when $R = \R^n$; in either case, what this notation means is that for an appropriate function $g$ $$ \partial_fg(f(x)) = [\partial_yg(y)]_{y=f(x)}, $$ or similar expressions for divergence, curl, etc.
• $\adj f_{\vec x} : R \to \R^n$ is the adjoint of the total differential $\diff{f_{\vec x}} : \R^n \to \R$ at $\vec x$ under a suitable inner product, either multiplication for $R = \R$ or the standard inner product for $R = \R^n$. The total differential $\diff{f_{\vec x}}$ at $\vec x$ is the best linear approximation of $f$ at that point; the matrix of $v \mapsto \diff{f_{\vec x}}(v)$ in the standard basis is the Jacobian. When $R = \R$, we can write $$ \adj f_{\vec x}(v) = \nabla(vf(\vec x)), $$ and when $R = \R^n$ we can write $$ \adj f_{\vec x}(\vec v) = \nabla(\vec v\cdot f(\vec x)). $$

In your specific case, $f(\vec x) = r = |\vec x|$, and the chain rule becomes $$\begin{aligned} \nabla\cdot\vec g(r) &= \nabla\cdot\vec g(f(\vec r)) = \adj f_{\vec x}\left(\frac\partial{\partial f}\right)\cdot\vec g(f) = (\nabla|\vec x|\frac\partial{\partial f})\cdot\vec g(f) \\ &= (\nabla|\vec x|)\cdot\frac{\partial\vec g}{\partial f} = \frac{\vec x}{|\vec x|}\cdot\vec g'(f(\vec x)) \\ &= \hat{\vec r}\cdot\vec g'(r). \end{aligned}$$