Divergence of a Vector Field - Surprising Result [duplicate]

This question already has an answer here:

I'm following the text Introduction to Electrodynamics by Griffiths, and I came across the following in an in-text problem:

Sketch the vector function v = $$\frac{\boldsymbol{\hat{\mathbf{r}}}}{r^2}$$, and compute its divergence. The answer may surprise you... can you explain it?

Well, the answer did surprise me, for the sketch of the function is indeed indicating a diverging field (like field lines from a point positive charge), yet the math claims the divergence to be zero. What's going wrong?

This is the solution I have, from a manual, which also doesn't make sense to me:

The answer is that $$\nabla·v = 0$$ everywhere except at the origin, but at the origin our calculation is no good, since $$r = 0$$, and the expression for $$v$$ blows up. In fact, $$\nabla·v$$ is infinite at that one point, and zero elsewhere.

Could someone please help me understand the situation? Any help would be appreciated, thanks!

P.S. I understand that this has been asked earlier on Physics SE, but I didn't understand the answers. The one with most upvotes said:

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. (Answer by @genneth)

What's the delta distribution in conversation?

marked as duplicate by Aaron Stevens, John Rennie homework-and-exercises StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 17 at 9:21

• "the math claims the divergence to be zero", how can you take the derivative of a function at zero if it diverges at zero? – Wolphram jonny Aug 17 at 6:56
• Except at the origin, the divergence is zero everywhere. – arya_stark Aug 17 at 6:58
• I thought you were claiming it was zero at the origin – Wolphram jonny Aug 17 at 7:00
• Just keep reading there book. Also, not understanding an answer doesn't make the question not a duplicate, in my opinion. – Aaron Stevens Aug 17 at 8:25

The "problem" is there because one assumes that the charge is a point at $$r=0$$.

To see how one might round the "problem" in a non-rigerous way suppose instead one assumes a uniform charge density $$\rho$$.

In a sphere of volume $$V$$, the total charge is $$\displaystyle\int_{\rm V} \rho\,dV$$.

The electric field due to a point charge $$q$$ is $$\vec E = - \dfrac {1}{4\pi \epsilon_0}\dfrac {q}{r^2} \hat r = k \dfrac {q}{r^2} \hat r$$

In our case the charge is not a point charge but distributed over a volume $$V$$ and so $$\vec E = k \dfrac {\int_{\rm V} \rho\,dV}{r^2} \hat r$$.

The electric flux through the surface of a sphere of radius $$R$$ is $$\displaystyle \int_{\rm S}\vec E \cdot d\vec s = k \dfrac {\int_{\rm V} \rho\,dV}{R^2}\,4\pi R^2=\int _{\rm V}4 \pi k \rho\, dV= \int _{\rm V}\nabla\cdot \vec E \,dV$$.

So the divergence of the electric field is $$4 \pi k\rho = \dfrac {\rho}{\epsilon _0}$$ in a more familiar form.

Now consider what a point charge implies.
As the radius of the sphere decreases and tends towards zero then the charge density must tend towards infinity.

To get around this "problem" a function is "loosely" defined as having the property that at $$r=0$$ the area under the function is $$1$$ and for all other values of $$r$$ the area under the function is zero.
It is called a delta function $$\delta (r)$$ and has infinite height and zero width but with a finite area of $$1$$ at $$r=0$$..

So now the divergence of the electric field from a point charge $$q$$ is given by $$\nabla \cdot \vec E= \dfrac{q}{\epsilon_0} \,\delta (r)$$.

At $$r=0$$ the divergence of the electric field is $$\dfrac{q}{\epsilon_0}$$ and the divergence is zero everywhere else as you have found.

• Why the downvote(s) as I was trying to answer the question in simple terms which are not necessarily the most rigorous?. – Farcher Aug 17 at 8:31
• This problem isn't about electric fields or point charges yet. Really the OP just needs to understand why mathematically $r=0$ is a problem. Not how to get around that problem. The question doesn't ask how to get around this problem, it's explained later in the text. Also, these answers are already given in duplicate questions. – Aaron Stevens Aug 17 at 8:34
• The divergence of the E-field isn't $q/\epsilon_0$ at $r=0$, it is "infinite" because of the delta function. – Puk Aug 17 at 14:59

We want to find $$\nabla \cdot \vec{v}$$, where $$\vec{v} = \hat{r}/r^2$$

First, let's write $$\hat{r}$$ in terms of cartesian coordinates:

$$\hat{r} = \vec{r}/|r|$$

$$= \frac{x\hat{x}+y\hat{y}+z\hat{z}}{\sqrt{x^2+y^2+z^2}}$$

Sub this value back into $$\vec{v}$$:

$$\vec{v} = \frac{x\hat{x}+y\hat{y}+z\hat{z}}{(x^2+y^2+z^2)^{3/2}}$$

$$\nabla \cdot \vec{v}$$ is the sum of the $$\hat{x}$$ component differentiated by x, the $$\hat{y}$$ component differentiated by y and the $$\hat{z}$$ component differentiated by z. This is computationally intensive, but easily found using software like Mathematica.

$$\nabla \cdot \vec{v} = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}} + \frac{3}{(x^2+y^2+z^2)^{3/2}} = 0$$

Of course, we made this calculation with the assumption that $$(x,y,z) > 0$$, otherwise we'll have zero on the denominator, and thus an infinite divergence.

Edit: To answer your question about what the dirac-delta distribution is: It's a function that's zero everywhere except for the origin where it's infinite.

• What's the element of surprise, then, as stated by the author? – arya_stark Aug 17 at 7:46
• Note that this problem is basically the same thing as taking the divergence of the electric field of a point particle. We expect such a divergence to be non-zero since the electric field lines radiate outward from the centre, but in fact the divergence at any point is zero as we've just shown. – Visipi Aug 17 at 8:17
• The OP claims they have already done this work. The key is the end of your answer. Explaining why $r=0$ is a problem. – Aaron Stevens Aug 17 at 8:37