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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.

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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. –  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

4 Answers 4

up vote 6 down vote accepted

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.

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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.

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

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Thanks, I got it. –  Inquisitive Feb 6 '12 at 16:33
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

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:

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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

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