Problem in finding the divergence at a point [duplicate]

Question

I am solving a problem given as

Divergence of $\frac{\textbf{r}}{r^3}$ is
a) zero at the origin
b) zero everywhere
c) zero everywhere except the origin
d) nonzero everywhere

The answer is given as (c) zero everywhere except the origin
But I get answer (b)

I am presenting my solution
$\textbf{r}=x\hat i+y\hat j+z\hat k$
$r=\sqrt{x^2+y^2+z^2}$
So, $\vec\nabla.\Big(\frac{\textbf{r}}{r^3}\Big)=\frac{\partial x/r^3}{\partial x}+\frac{\partial y/r^3}{\partial y}+\frac{\partial z/r^3}{\partial z}$
$\frac{\partial x/r^3}{\partial x}=\frac{\partial \big(x/(x^2+y^2+z^2)^{3/2}\big)}{\partial x}$
$\implies \frac{\partial x/r^3}{\partial x}= \frac{(x^2+y^2+z^2)^{3/2}-3x^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3}$
Similarly, $\frac{\partial y/r^3}{\partial y}= \frac{(x^2+y^2+z^2)^{3/2}-3y^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3}$
and $\frac{\partial z/r^3}{\partial z}= \frac{(x^2+y^2+z^2)^{3/2}-3z^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3}$

So, $\vec\nabla.\Big(\frac{\textbf{r}}{r^3}\Big)=\frac{3(x^2+y^2+z^2)^{3/2}-3(x^2+y^2+z^2)(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3}$
Thus, $\vec\nabla.\Big(\frac{\textbf{r}}{r^3}\Big)=\frac{3(x^2+y^2+z^2)^{3/2}-3(x^2+y^2+z^2)^{3/2}}{(x^2+y^2+z^2)^3}=0$

Hence, I get the answer (b) Zero everywhere

I am not able to understand how the divergence in non-zero at origin and if it is non-zero then what will be its value at origin?
Please clarify the doubt.

Answer 1

Since $\frac{\mathbf r}{r^3}$ is not defined at $0$, it does not make sense to compute its divergence there in the naive way. Your calculation shows that the divergence is $0$ everywhere but at the origin, where it is not defined.

To define the divergence over all space, you need to compute as a distribution. That is, for a function $f:\mathbb R^3 \to \mathbb R$ (smooth with compact support, if you want all the mathematical details), you compute : \begin{align} \int {\rm{d}}^3 x f(\vec x)\vec {\nabla} \cdot \left(\frac{\vec x}{|\vec x|^3}\right) &= -\int {\rm{d}}^3 x \frac{x}{|x|^3} \cdot \nabla f(x) \\ &= -\lim_{\epsilon \to 0} \int_{|x|\geqslant \epsilon} {\rm{d}}^3 x \frac{\vec x}{|x|^3} \cdot \vec \nabla f(\vec x) \\ &= \lim_{\epsilon \to 0}\int_{|x|= \epsilon} {\rm{d}}^2 \vec x \cdot \frac{\vec x}{|\vec x|^3} f(\vec x)\\ &= 4\pi f(0) \\ &=\int {\rm{d}}^3 x f(\vec x) 4\pi\delta(\vec x) \end{align}

To get the third line, we perform the standard integration by parts (with Gauss's theorem), since, away from the origin, we are dealing with ordinary functions.

Since this is true for any function $f$, we have : $$\vec {\nabla} \cdot \left(\frac{\vec x}{|\vec x|^3}\right) = 4\pi\delta(\vec x)$$ Also, notice that away from the origin, this agrees with your calculation.

Answer 2

The divergence defined as $$\text{div}\ \mathbf{F}=\lim_{\Delta V\rightarrow 0\ \text{about}\ (x,y,z)}\frac{1}{\Delta V}\iint_\mathcal{S}\mathbf{F}\cdot \hat{\mathbf{n}}\ dS$$ It's required that field have a well defined value at a point otherwise the limit would not exist.

Now $\hat{r}/r^2$ is not defined at $(0,0,0)$.

Now for $(x,y,z)\not=0$, you have computed the divergence to be zero and the other case can be found here.

Answer 3

To understand how $\vec{\partial}\cdot(\vec{x}/r^3)$ boils down to a delta function, it may be more or less helpful to consider regularization.

The vector field $\vec{x}/r^3$ ($r:=|\vec{x}|$) is singular at $r=0$. We can go around this difficulty by regularizing the singularity with a small parameter $\epsilon$. For instance, we can consider a vector field \begin{equation} \vec{G}(\vec{x}) = \frac{\vec{x}}{(r^2+\epsilon^2)^{3/2}} \end{equation} first, calculate $\vec{\partial}\cdot\vec{G}(\vec{x})$, and then take the $\epsilon\to0$ limit. \begin{align} \vec{\partial}\cdot\vec{G}(\vec{x}) &=\frac{3}{(r^2+\epsilon^2)^{3/2}} - \frac{3}{2} \frac{2\vec{x}\cdot\vec{x}}{(r^2+\epsilon^2)^{5/2}} =\frac{1}{r^2} \left[ \frac{3\epsilon^2r^2}{(r^2+\epsilon^2)^{5/2}} \right] \end{align} The bracketed term approaches to $\delta(r)$ as $\epsilon\to0$, so we have $\frac{1}{r^2}\delta(r) = \frac{1}{4\pi} \delta^{(3)}(\vec{r})$. A rigorous proof (distribution approach) is presented in Jackson's electrodynamics 2ed p.39-40. But if you want to be sloppy, you can consider \begin{equation} \int^\infty_0 \hspace{-0.35em}dr\,\frac{3\epsilon^2r^2}{(r^2+\epsilon^2)^{5/2}} = \left[ \frac{r^3}{(r^2+\epsilon^2)^{3/2}} \right]^\infty_0 = 1, \end{equation} while the peak position and value of the integrand goes like $\sim \epsilon^{-1}$ as $\epsilon\to0$.

I think one should somehow appeal to distribution theory to answer this question correctly. In this light, I think SolubleFish gave the clearest explanation. Besides, I think my answer can help you to grasp the "picture" of how $\vec{\partial}\cdot(\vec{x}/r^3)$ approaches to delta function.

(Also consider taking a look at https://math.stackexchange.com/questions/368155/laplacians-and-dirac-delta-functions?noredirect=1&lq=1!)