# Divergence of a radial vector field

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

$$$$\nabla \cdot \textbf{g}(r)=\textbf{g}^{\prime}\cdot \mathbf{\hat{r}},$$$$ where the vector field $$\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?

• Hint : $$\boldsymbol{\nabla\cdot}\mathbf g(r)=\dfrac{\mathrm dg_1(r)}{\mathrm dr}\dfrac{\partial r}{\partial x_1}+ \dfrac{\mathrm dg_2(r)}{\mathrm dr}\dfrac{\partial r}{\partial x_2}+\dfrac{\mathrm dg_3(r)}{\mathrm dr}\dfrac{\partial r}{\partial x_3}=\texttt{etc}$$ This is not a Physics question. It's Mathematics. Sep 13, 2021 at 21:29
• I still don't get it. I mean, shouldn't my second equation just work fine? after all, it's derived for usage in spherical coordinates. thanks
– M91
Sep 15, 2021 at 15:04
• @M91 Your equation works fine until you assume that the derivatives of $g_\theta$ and $g_\phi$ vanish. Consider e.g. the vector field $\mathbf g(r) = r\hat z$. The vector attached to the north pole of the unit sphere is $(1,0,0)$, whose radial component is $1$; the vector attached to the south pole is also $(1,0,0)$, but since it's on the south pole that corresponds to a radial component of $-1$. In other words, the vector field doesn't depend on the angular variables, but the $(r,\theta,\phi)$ components do because the $(r,\theta,\phi)$ unit vectors change with position. Sep 15, 2021 at 16:45
• ...the vector field $\mathbf g(r)$ is only nonzero in the radial direction. This is wrong. In the textbook the vector fielf $\mathbf g(r)$ is considered a function of $r$ not that it is has component along the radial direction only. So in the title the term radial is wrong. This is the reason of your wrong results. Sep 15, 2021 at 19:22
• @M91 : Yes. See equation \eqref{06} in my answer. Sep 16, 2021 at 15:50

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.

• +1 surely. But I'd write better $\mathbf{0}$ instead of $\vec{0}$. Sep 13, 2021 at 22:37

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 $$$$\mathbf g\left(r\right)\boldsymbol= \mathrm g_x\left(r\right)\mathbf{\hat{x}}\boldsymbol+\mathrm g_y\left(r\right)\mathbf{\hat{y}}\boldsymbol+\mathrm g_z\left(r\right)\mathbf{\hat{z}} \boldsymbol= \begin{bmatrix} \mathrm g_x\left(r\right) & \mathrm g_y\left(r\right) & \mathrm g_z\left(r\right) \end{bmatrix} \begin{bmatrix} \mathbf{\hat{x}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{y}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{z}}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{01}\label{01}$$$$ where $$\,\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 $$$$\begin{bmatrix} \mathbf{\hat{x}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{y}}\vphantom{\dfrac{a}{b}}\\ \mathbf{\hat{z}}\vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol= \begin{bmatrix} \sin\theta\cos\phi & \cos\theta\cos\phi & \boldsymbol-\sin\phi \vphantom{\dfrac{a}{b}}\\ \sin\theta\sin\phi & \cos\theta\sin\phi & \hphantom{\boldsymbol-}\cos\phi \vphantom{\dfrac{a}{b}}\\ \cos\theta & \boldsymbol-\sin\theta & \hphantom{\boldsymbol-}0 \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} \mathbf{\hat{r}}\vphantom{\dfrac{a}{b}}\\ \boldsymbol{\hat{\theta}}\vphantom{\dfrac{a}{b}}\\ \boldsymbol{\hat{\phi}}\vphantom{\dfrac{a}{b}} \end{bmatrix} \tag{03}\label{03}$$$$

Inserting this expression in equation \eqref{01} we have $$$$\mathbf g\left(r\right)\boldsymbol= \mathrm g_r\left(r,\theta,\phi\right)\mathbf{\hat{r}}\boldsymbol+ \mathrm g_\theta\left(r,\theta,\phi\right)\boldsymbol{\hat{\theta}}\boldsymbol+\mathrm g_\phi\left(r,\theta,\phi\right)\boldsymbol{\hat{\phi}} \tag{04}\label{04}$$$$ where \begin{align} \mathrm g_r\left(r,\theta,\phi\right) & \boldsymbol=\sin\theta\cos\phi\cdot\mathrm g_x\left(r\right)\boldsymbol+\sin\theta\sin\phi\cdot\mathrm g_y\left(r\right)\boldsymbol+\cos\theta\cdot\mathrm g_z\left(r\right) \tag{05a}\label{05a}\\ \mathrm g_\theta\left(r,\theta,\phi\right) & \boldsymbol=\cos\theta\cos\phi\cdot\mathrm g_x\left(r\right)\boldsymbol+\cos\theta\sin\phi\cdot\mathrm g_y\left(r\right)\boldsymbol-\sin\theta\cdot\mathrm g_z\left(r\right) \tag{05b}\label{05b}\\ \mathrm g_\phi\left(r,\theta,\phi\right) & \boldsymbol=\boldsymbol-\sin\phi\cdot\mathrm g_x\left(r\right)\boldsymbol+\cos\phi\cdot\mathrm g_y\left(r\right) \tag{05c}\label{05c} \end{align} So all components $$\,\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 $$$$\begin{split} \boldsymbol\nabla\boldsymbol\cdot\mathbf g\left(r\right) & \boldsymbol= \dfrac{\partial\mathrm g_x\left(r\right)}{\partial x}\boldsymbol+\dfrac{\partial\mathrm g_y\left(r\right)}{\partial y}\boldsymbol+\dfrac{\partial\mathrm g_z\left(r\right)}{\partial z}\\ &\boldsymbol= \dfrac{\mathrm d\mathrm g_x\left(r\right)}{\mathrm dr}\dfrac{\partial r}{\partial x}\boldsymbol+\dfrac{\mathrm d\mathrm g_y\left(r\right)}{\mathrm dr}\dfrac{\partial r}{\partial y}\boldsymbol+\dfrac{\mathrm d\mathrm g_z\left(r\right)}{\mathrm dr}\dfrac{\partial r}{\partial z}\\ &\boldsymbol= \dfrac{\mathrm d\mathrm g_x\left(r\right)}{\mathrm dr}\dfrac{x}{r}\boldsymbol+\dfrac{\mathrm d\mathrm g_y\left(r\right)}{\mathrm dr}\dfrac{y}{r}\boldsymbol+\dfrac{\mathrm d\mathrm g_z\left(r\right)}{\mathrm dr}\dfrac{z}{r}\\ &\boldsymbol= \dfrac{\mathrm d\mathbf g\left(r\right)}{\mathrm dr}\boldsymbol\cdot\dfrac{\mathbf r}{r}\boldsymbol=\mathbf g'\boldsymbol\cdot\mathbf{\hat{r}}\\ \end{split} \tag{06}\label{06}$$$$

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

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

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

• Thanks for this clear explanation. This is more than I could have asked for!
– M91
Sep 16, 2021 at 15:22

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.

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

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}