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Why am I getting zero divergence of function $\vec{f} = \frac{1}{r^2} \hat{r}$, where $r$ is the distance from the origin and $\hat{r}$ is the unit vector in the radial direction. The divergence of this function over a sphere of radius $R$, which includes the origin.

$$\nabla \cdot f = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 f_r) = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{1}{r^2}) = 0$$

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  • $\begingroup$ Yes you have done it correctly.. Just look at the behaviour of the function by plotting it if you are too curious.. $\endgroup$ Feb 9, 2017 at 16:14
  • $\begingroup$ @SchrodingersCat But the answer is $4\pi $ $\endgroup$
    – Vedanshu
    Feb 9, 2017 at 16:16
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    $\begingroup$ Well, that is because the function has a singularity at $r=0$. And the divergence is not defined at that point. Better use the divergence theorem and you will get your result. $\endgroup$ Feb 9, 2017 at 16:25
  • $\begingroup$ Divergence should be $4 \pi$ but I'm getting 0. $\endgroup$
    – Vedanshu
    Feb 9, 2017 at 16:31
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    $\begingroup$ But you mean the integral over the surface of the sphere of the divergence ? $\endgroup$
    – Zubzub
    Feb 9, 2017 at 16:41

3 Answers 3

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The divergence of $\vec F(\vec r)=\frac{\vec r}{r^3}$ is not $4\pi$. By direct computation, we have

$$\nabla \cdot \vec F(\vec r)=\begin{cases}0&,r\ne 0\\\\\text{undefined}&,r=0\end{cases}$$

However, we can assign meaning to $\nabla \cdot \vec F(\vec r)$ at the origin and write

$$\bbox[5px,border:2px solid #C0A000]{\nabla \cdot \vec F(\vec r)\sim 4\pi \delta(\vec r)}$$

where $\delta(\vec r)$ is the Dirac Delta distribution. To do this, we introduce the regularization function

$$\psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}}\tag 1$$

Note that for $r\ne 0$, $\lim_{a\to 0}\psi(\vec r;a)=\vec F(\vec r)$. Taking the divergence of $\psi(\vec r;a)$ yields

$$\nabla \cdot \psi(\vec r;a)=\frac{3a^2}{(r^2+a^2)^{5/2}}$$

Now, in THIS ANSWER, I showed that for any smooth test function $\phi(\vec r)$

$$\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)\,dV&=\begin{cases}4\pi \phi(0)&,0\in V\\\\0&,0\notin V\end{cases} \end{align}$$

And it is in this sense that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$

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  • $\begingroup$ Can you explain why $\nabla \cdot \psi(\vec r;a)=\frac{3a^2}{(r^2+a^2)^{5/2}}$? I've computed divergence and it leads to a completely different answer. $\endgroup$
    – S.H.W
    Oct 23, 2019 at 22:38
  • $\begingroup$ @S.H.W. Use the quotient rule for each partial $\frac{\partial }{\partial x}$, $\frac{\partial }{\partial y}$, and $\frac{\partial }{\partial z}$. And then sum the three terms. Remember that $r^2=x^2+y^2+z^2$ and $\vec r=\hat xx+\hat yy+\hat zz$. So, you will need to take the three partial derivatives $\frac{\partial }{\partial x_i}\left(\frac{x_i}{(x^2+y^2+z^2+a^2)^{3/2}}\right)$ $\endgroup$
    – Mark Viola
    Oct 24, 2019 at 14:41
  • $\begingroup$ Okay, according to Griffiths, we have $\nabla \cdot (\frac{\hat{r}}{r^2}) = 4\pi \delta^3(\vec{r})$. Is it same as your answer? $\endgroup$
    – S.H.W
    Oct 24, 2019 at 14:55
  • $\begingroup$ @S.H.W Yes, those answers are identical. Only the notations differ. $\endgroup$
    – Mark Viola
    Oct 24, 2019 at 16:25
  • $\begingroup$ I'm little confused. I think $\delta^3(\vec{r})$ is three dimensional extension of $\delta(\vec{r})$ and different with each other. Can you explain please? $\endgroup$
    – S.H.W
    Oct 25, 2019 at 18:01
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The calculation result is:

$$\nabla \cdot \frac{1}{r^2} \hat{r} = \frac{1}{r^2} \frac{\partial}{\partial r} \left[r^2 \left( \frac{1}{r^2} \right) \right] = \frac 0 {r^2} = \begin{cases}0&,r\ne 0\\\\ \text{undefined}&,r=0\end{cases}$$

It correctly shows that the divergence is zero everywhere except the origin.

However, unfortunately, it only says that the divergence is not defined at the origin and cannot provide more information, that is, $\nabla \cdot \frac{1}{r^2} \hat{r}$ is actually positive infinity at the origin.

Nevertheless, we can use the following simple mathematical way to obtain the situation at the origin point.

Suppose there is a sphere centered on the origin, then the total flux on the surface of the sphere is : -

$$ \text {Total flux} = 4 \pi r^2 \left( \frac {1} {r^2} \right)= 4 \pi $$

Suppose the volume of the sphere be $ \mathbf {v(r)}$, so by the definition, the divergence is : -

$$ \lim_{\text {volume} \to zero} \frac {\text {Total Flux}} {\text {Volume}} = \lim_{\text {r} \to 0} \left(\frac {4 \pi} {v(r)}\right) $$

So obviously,

$$ \lim_{\text {r} \to 0} \left[ \nabla \cdot \left( \frac{1} {r^2} \hat r \right) \right] = \lim_{\text {r} \to 0} \left(\frac {4 \pi} {v(r)} \right) = \text {positive infinite} \qquad (1)$$

$$ \lim_{\text r\to 0} \int \nabla \cdot \left( \frac 1 {r^2} \hat r \right) dv(r) = \lim_{\text r\to 0}\int \frac {4 \pi} {v(r)} dv(r) = 4\pi \qquad (2)$$

Since the laplacian is zero everywhere except $r \to 0$, and according to equation (1) and (2), it is true and real that :-

$$ \nabla \cdot \left(\frac{1}{r^2} \hat r\right)=4\pi\delta^3({\bf r}) $$

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$\hat{r} = (x,y,z)/\sqrt{x^2 + y^2 + z^2}$ $$ F(x,y,z) = \frac{1}{(x^2 + y^2 + z^2)^{3/2}} (x,y,z) \\ \frac{\partial}{\partial x} F = \frac{\partial}{\partial x} \frac{x}{(x^2 + y^2 + z^2)^{3/2}} = \frac{-2x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\ \frac{\partial}{\partial y} F = \frac{\partial}{\partial y} \frac{y}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 -2y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\ \frac{\partial}{\partial z} F = \frac{\partial}{\partial z} \frac{z}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 + y^2 -2z^2}{(x^2 + y^2 + z^2)^{5/2}} \\ $$ Putting together : $$ \nabla\cdot F = \frac{\partial}{\partial x} F + \frac{\partial}{\partial y} F + \frac{\partial}{\partial z} F = \frac{0}{(x^2 + y^2 + z^2)^{5/2}} = 0 $$

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    $\begingroup$ Hopefully corrected now :| $\endgroup$
    – Zubzub
    Feb 9, 2017 at 16:22
  • $\begingroup$ answer is $4 \pi$ $\endgroup$
    – Vedanshu
    Feb 9, 2017 at 16:28
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    $\begingroup$ This doesn't interpret the divergence of $\vec F$ in the sense of distributions, which is, I believe, the area that the OP is questioning. $\endgroup$
    – Mark Viola
    Feb 9, 2017 at 17:12
  • $\begingroup$ Suppose vector F has a value F1 at (1,1,1). Now if x changes by dx, then the change in F1(x) will be 0 since partial derivative of F w.r.t to x evaluates to 0. This seems confusing. Can you explain ? $\endgroup$ Nov 12, 2021 at 18:35

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