# Show that $\nabla\cdot\left(\dfrac{\mathbf{e}_r}{r^2}\right)=4\pi\delta(\mathbf{r})$ using the divergence theorem.

The book answer goes as follows:

By the divergence theorem, in spherical coordinates we find $$\color{red}{\iiint_\limits{\large\text{volume}\,\tau}\nabla\cdot\left(\dfrac{\mathbf{e}_r}{r^2}\right)\mathrm{d}\tau}=\color{blue}{\iint_\limits{\large\text{surface enclosing}\, \tau}\dfrac{\mathbf{e}_r}{r^2}\cdot\mathbf{e}_r\,\mathrm{d}\sigma}=\color{#180}{\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\pi}\frac{1}{r^2}r^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi}=4\pi$$ Thus $\nabla\cdot\left(\dfrac{\mathbf{e}_r}{r^2}\right)$ has the properties that it is zero $\forall\,r\gt 0$ but its integral over any volume including the origin is $4\pi$; this suggests that it is equal to $4\pi\delta(\mathbf{r})$.

As mentioned in a comment below; $\mathbf{e}_r$ is a unit radial vector.

I know that the $\color{red}{\mathrm{red}}$ and $\color{blue}{\mathrm{blue}}$ integrals are a statement of the divergence theorem. The only thing I can't understand is how the $\color{#180}{\mathrm{green}}$ integral was obtained from the $\color{blue}{\mathrm{blue}}$ integral.

I know that $$\mathrm{d}\sigma=\left|\frac{\partial r}{\partial \theta}\times\frac{\partial r}{\partial \phi}\right|\,\mathrm{d}\theta\,\mathrm{d}\phi\tag{1}$$ I think that equation $(1)$ has been used but I'm not sure how to use it. Could someone please explain how the $\color{#180}{\mathrm{green}}$ integral was reached?

• $\boldsymbol{e}_r$ is the radial unit vector.
– vnd
Commented Mar 7, 2016 at 18:36
• Additionally, $d\tau$ is the volume element. If the origin is taken to be surrounded by a sphere of radius $r$, then the surface element is $r^2do \boldsymbol{e}_r$, where do is the solid angle subtended at the origin by the surface element. It is simple to take $do = \sin\theta d\theta d\phi$.
– vnd
Commented Mar 7, 2016 at 18:45
• The two conditions (the first provided by the divergence theorem and the second that the integrand is zero for $r\neq 0$ are satisfied if we take the integrand to be $4\pi\delta (\vec r)$. That's all the author is saying and probably why he uses the word "suggests". Commented Mar 7, 2016 at 19:42
• Using the parameterization given by spherical coordinates, when you compute $\vert \vec r_{\theta}\times \vec r_{\phi}\vert$ and substitute, you will get the green integrand. The limits are obtained by realizing you are integrating over a sphere of radius r. Commented Mar 7, 2016 at 21:10
• Not much more I can give, that can aid you further than you already are on this - but I will say that this question in my (trivial) opinion anyway, is an example of a brilliantly constructed question. Well done BLAZE..
– user284001
Commented Mar 8, 2016 at 10:46

The surface integral can be evaluated in spherical coordinates:

1). Set

\begin{align} x&=r\sin\theta\cos\phi\;,\\ y&=r\sin\theta\sin\phi\;,\\ z&=r\cos\theta\;, \end{align}

2). Write

$\vec r(\theta,\phi)=r\sin\theta\cos\phi \vec i+r\sin\theta\sin\phi\vec j+r\cos\theta \vec k$

3). Then

$\vec r_{\theta}(\theta, \phi)=r\cos\theta\cos\phi \vec i+r\cos\theta\sin\phi\vec j-r\sin\theta \vec k$

and

$\vec r_{\phi}(\theta, \phi)=-r\sin\theta\sin\phi \vec i+r\sin\theta\cos\phi\vec j$

4). Find the magnitude of the cross product:

$\color{red}{\vert \vec r_{\phi}\times \vec r_{\theta}\vert =r^2\sin \theta.}$ $(\sin \theta \geq 0$ since $0\leq \theta <\pi)$,

so that

$d\sigma =r^2\sin \theta d\theta d\phi$.

5). Find the limits of integration:

$0\leq \phi <2\pi;\ 0\leq \theta <\pi$ because the surface is a sphere.

6). Substitute into the integral:

$\color{blue}{\iint_\limits{\large\text{surface enclosing}\, \tau}\dfrac{\mathbf{e}_r}{r^2}\cdot\mathbf{e}_r\,\mathrm{d}\sigma }=\int \int _{\sigma }\frac{1}{r^{2}}d\sigma =\int^{2\pi }_0 \int^{\pi }_0 \frac{1}{r^{2}}(r^{2}\sin \theta )d\theta d\phi=$

• How did you deduce the part marked red in your answer? Commented Mar 8, 2016 at 20:57
• I took the magnitude of the cross porduct. In general, fora parameterization $X(u,v)=(x(u,v),y(u,v),z(u,v))$ it is easy to show that$dS=\sqrt{|X_u|^2 |X_v|^2 - (X_u \cdot X_v)^2} du dv$ Commented Mar 8, 2016 at 23:36
• Thank you for your excellent answer. I just needed to see some justification of the part marked $\color{red}{\mathrm{red}}$ in your answer. I think that I understand now and have given an answer of my own. Kindest Regards. Commented Mar 11, 2016 at 0:24

The vector $\vec r$ is a function of $\theta$ and $\phi$; which is given by $$\vec r=r\sin\theta\cos\phi \widehat i+r\sin\theta\sin\phi\widehat j+r\cos\theta \widehat k$$ and the unit vector $\widehat r$ is given by $$\color{teal}{\widehat r=\sin\theta\cos\phi \widehat i+\sin\theta\sin\phi\widehat j+\cos\theta \widehat k}\tag{2}$$

Therefore,

$$\frac{\partial r}{\partial \theta}=r\cos\theta\cos\phi \widehat{i}+r\cos\theta\sin\phi\widehat{j}-r\sin\theta \widehat{k}$$ and $$\frac{\partial r}{\partial \phi}=-r\sin\theta\sin\phi \widehat i+r\sin\theta\cos\phi\widehat j$$

So \begin{align}\frac{\partial r}{\partial \theta}\times\frac{\partial r}{\partial \phi}&=\begin{vmatrix} \widehat i & \widehat j & \widehat k \\ r\cos\theta\cos\phi & r\cos\theta\sin\phi & -r\sin\theta \\ -r\sin\theta\sin\phi & r\sin\theta\cos\phi & 0 \end{vmatrix}\\&=r^2\sin^2\theta\cos\phi\widehat i +r^2\sin^2\theta\sin\phi\widehat j+ r^2\cos\theta\sin\theta\cos^2\phi\widehat k+r^2\cos\theta\sin\theta\sin^2\phi\widehat k\\&=r^2\sin^2\theta\cos\phi\widehat i +r^2\sin^2\theta\sin\phi\widehat j+ r^2\cos\theta\sin\theta\left(\cos^2\phi+\sin^2\phi\right)\widehat k\\&=r^2{\left(\sin^2\theta\cos\phi\widehat i+\sin^2\theta\sin\phi\widehat j+\cos\theta\sin\theta\widehat k\right)}\\&=r^2\sin\theta\color{purple}{\underbrace{\left(\sin\theta\cos\phi\widehat i+\sin\theta\sin\phi\widehat j+\cos\theta\widehat k\right)}_{\color{teal}{\Large\text{From (2)}\,=\,\widehat r}}}\\&=r^2\sin\theta\,\widehat r\end{align}

Taking the magnitude gives $$\left|\frac{\partial r}{\partial \theta}\times\frac{\partial r}{\partial \phi}\right|=\sqrt{\left(r^2\sin\theta\,\widehat r\right)^2}={\sqrt{r^4\sin^2\theta \,\widehat r^2}}=\underbrace{r^2\sin\theta}_{\Large\text{Since}\,\color{teal}{\widehat r}^2=1}$$ Therefore $$\color{#F80}{\mathrm{d}\sigma=\left|\frac{\partial r}{\partial \theta}\times\frac{\partial r}{\partial \phi}\right|\,\mathrm{d}\theta\,\mathrm{d}\phi}\tag{1}$$ $$=r^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi$$

As helpfully pointed out in a comment given by @Chilango we are integrating over a sphere of radius $r$. So the azimuthal angle $\phi$ has a limit range given by $0\leq \phi \lt 2\pi$ and the polar angle $\theta$ has a limit range given by $\ 0\leq \theta \lt\pi$.

Finally, substitution of $(1)$ and the limits into the $\color{blue}{\mathrm{blue}}$ integral gives the $\color{#180}{\mathrm{green}}$ integral as required.

A special thanks goes to @Chilango for his/her answer that gave me the ideas necessary to give this answer with full justification that $$\left|\frac{\partial r}{\partial \theta}\times\frac{\partial r}{\partial \phi}\right|=r^2\sin\theta$$

A note on what I think to be a misleading trend in some texts, where I have recently stumbled.

As a proper Riemann integral, $$\iiint_{\tau}\nabla\cdot\left(\frac{\mathbf{e}_r}{r^2}\right)d\tau$$ where $\frac{\mathbf{e}_r}{r^2}=\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}$ (it seems that $\mathbf{y}=\mathbf{0}$ in your case), is $0$ if $\mathbf{y}\notin\bar{\tau}$ and does not exist if $\mathbf{y}\in\bar{\tau}$, because the integrand of a Riemann integral, in the usual calculus definitions of it, has to be defined on all the domain.

As the limit $$\lim_{\varepsilon\to 0}\iiint_{\tau\setminus B(\mathbf{y},\varepsilon)}\nabla\cdot\left(\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\right)dx_1dx_2dx_3$$ it is $0$ because $\lim_{\varepsilon\to 0}0=0$ and the same holds for the Lebesgue integral $$\int_{\tau}\nabla\cdot\left(\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\right)d\mu_{\mathbf{x}}$$which is calculated as the preceding limit.

This shows that, under these definitions of the integral and the usual definition of the derivative, the divergence theorem, certainly valid if, instead of $\frac{\mathbf{e}_r}{r^2}$ you had a vector field $\mathbf{F}\in C^1(\mathring{A})$ with $\tau\subset\mathring{A}$ satisfying opportune assumptions, cannot be applied.

Since $\forall\mathbf{x}\in\mathbb{R}^3\setminus\{\mathbf{y}\}\quad\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}=-\nabla\left(\frac{1}{\|\mathbf{x}-\mathbf{y}\|}\right)$ and the divergence of the gradient is the Laplacian $\nabla\cdot\nabla=\nabla^2$, we see that $\nabla\cdot\left(\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\right)=-\nabla^2\left(\frac{1}{\|\mathbf{x}-\mathbf{y}\|}\right)$. Then by reading the integral $$\int_{\tau}-\nabla^2\left(\frac{1}{r}\right)\varphi \,d\tau$$ where $\varphi\in C^2(\mathbb{R}^3)$ (typically $\varphi\in C^\infty(\mathbb{R}^3)$) is such that $\forall\mathbf{x}\notin\tau\quad\varphi(\mathbf{x})=0$, in the symbolic way of the Laplacian of the distribution defined by $-\frac{1}{r}$, it can be shown, as it is here, that $-\int_{\tau}\nabla^2\left(\frac{1}{r}\right)\varphi \,d\tau=4\pi\int_{\tau}\delta_{\mathbf{y}}\varphi \,d\tau:=4\pi\varphi(\mathbf{y})$ (where $\delta_{\mathbf{y}}(\mathbf{x}):=\delta(\mathbf{x}-\mathbf{y})$), but that is$$\lim_{\varepsilon\to 0}\iiint_{\tau\setminus B(\mathbf{y},\varepsilon)}\frac{-\nabla^2\varphi(\mathbf{x})}{\|\mathbf{x}-\mathbf{y}\|}dx_1dx_2dx_3=\int_{\tau}\frac{-\nabla^2\varphi(\mathbf{x})}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{x}}$$if we use the usual Riemann (one th left) or Lebesgue (on the right) integrals, while $$\lim_{\varepsilon\to 0}\iiint_{\tau\setminus B(\mathbf{y},\varepsilon)}\nabla\cdot\left(\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\right)\varphi(\mathbf{x})dx_1dx_2dx_3=\int_{\tau}\nabla\cdot\left(\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\right)\varphi(\mathbf{x})d\mu_{\mathbf{x}}\equiv 0$$for all $\mathbf{y}$ and all functions $\varphi$.

• Thank you for taking the time to explain this to me; it is appreciated (+1) Commented Mar 11, 2016 at 20:37
• It's a pleasure to share one's (little, in my case) knowledge! Commented Mar 12, 2016 at 9:42
• Now that I have your attention maybe you could shed some light on this simple question, best regards. Commented Mar 12, 2016 at 12:47
• I see an answer has been given there. By the way, I have no idea what "change" means... if it means $\lim_{x\to x'^{+}}\frac{\mathrm{d}}{\mathrm{d}x} G-\lim_{x\to x'^{-}}\frac{\mathrm{d}}{\mathrm{d}x} G$, then your book's answer is quite trivial, but I'm not sure what it means... Commented Mar 12, 2016 at 16:15
• Honoured by your comment. I would be glad to help, but I only know (as you can see in my profile, I'm no expert) that as the very definition of the Legendre polynomials $P_n$: $$P_n(x):=\frac{1}{n!2^n}\frac{d^n}{dx^n}(x^2-1)^n$$because the textbook I've followed, Kolmogorov-Fomin's, defines them like that. Commented Mar 15, 2016 at 9:40