# Divergence of $\frac{\hat{r}}{r^2}$

In David J. Griffiths's Introduction to Electrodynamics, the author gave the following problem in an exercise.

Sketch the vector function $$\vec{v} ~=~ \frac{\hat{r}}{r^2},$$ and compute its divergence, where $$\hat{r}~:=~ \frac{\vec{r}}{r} , \qquad r~:=~|\vec{r}|.$$ The answer may surprise you. Can you explain it?

I found the divergence of this function as $$\frac{1}{x^2+y^2+z^2}$$ Please tell me what is the surprising thing here.

• convert your expression for $r$ into Cartesian coordinates, and then compute the divergence in these coordinates. You definitely have the wrong answer. – Jerry Schirmer Feb 6 '12 at 14:38
• Sorry, the numerator 'r' is a vector. I do not know how to put a hat over the 'r' here in this website. vecor v = vector r/ r^2. – Inquisitive Feb 6 '12 at 14:43
• Yes, but still, your answer should be half of what you've written. – Manishearth Feb 6 '12 at 14:46
• Wait, r hat or r vector? R hat means that it is a unit vector, whereas r vector means that it is a full r vector. $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, $\hat{r}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}$. Mouseover the above two formula and right-click, show source to get an idea of how to make vectors in TeX. – Manishearth Feb 6 '12 at 14:50
• ya, now I made it right. The denominator is the equation of the sphere , is that the surprising thing? or anything else important here? – Inquisitive Feb 6 '12 at 14:50

Pretty sure the question is about $\frac{\hat{r}}{r^2}$, i.e. the electric field around a point charge. Naively the divergence is zero, but properly taking into account the singularity at the origin gives a delta-distribution.

I have the same book, so I take it you are referring to Problem 1.16, which wants to find the divergence of $\frac{\hat{r}}{r^2}$.

If you look at the front of the book. There is an equation chart, following spherical coordinates, you get $\nabla\cdot\vec{v} = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left (r^2 v_r\right) + \text{ extra terms}$. Since the function $\vec{v}$ here has no $v_\theta$ and $v_\phi$ terms the extra terms are zero. Hence $\nabla\cdot\vec{v} = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(r^2 \frac{1}{r^2}\right) = \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(1\right) = 0$.

At least this is how I interpret the surprising element of the question.

For me another surprising thing about this question was that the divergence was not negative, seeing as the flow decreases as we move radially outwards. I found an excellent explanation of this here:

http://mathinsight.org/divergence_subtleties

• This looks more like a comment rather than answer – freude May 30 '13 at 12:38
• @freude The question is 'Please tell me what is the surprising thing here'. How is this not an answer to that question? – user25210 Jun 1 '13 at 14:54

You may wish to check if the divergence is finite everywhere.

• But what is the physical meaning of infinite divergence? – Inquisitive Feb 6 '12 at 16:38
• @sree: Infinite charge density (if the vector field is the electric field). – akhmeteli Feb 6 '12 at 16:46
• The electron is a single point here. Countably many charge in an infinitely small point -> infinite charge density. – Martin Ueding Feb 6 '12 at 19:36

## protected by Qmechanic♦Oct 4 '15 at 9:29

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