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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.

Qmechanic
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Inquisitive
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    convert your expression for $r$ into Cartesian coordinates, and then compute the divergence in these coordinates. You definitely have the wrong answer. – Zo the Relativist Feb 06 '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 06 '12 at 14:43
  • Yes, but still, your answer should be half of what you've written. – Manishearth Feb 06 '12 at 14:46
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    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 06 '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 06 '12 at 14:50
  • Sorry, it is a r vector not r hat. Anyway thanks for the source info. – Inquisitive Feb 06 '12 at 14:51
  • I doubt that that is the surprising thing, a symmetric situation usually gives a nice spherical answer. Ill think about it, got to go now. – Manishearth Feb 06 '12 at 14:55
  • Possible duplicates: https://physics.stackexchange.com/q/488220/2451 and links therein. – Qmechanic Feb 11 '21 at 10:28

4 Answers4

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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.

genneth
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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.

Manishearth
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Harry Song
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  • The surprising thing is the answer of zero in the context of the plot which naively has non zero divergence! – Kashmiri Sep 11 '20 at 04:27
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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

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

akhmeteli
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