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So my textbook says that $\nabla \cdot \big(\frac{\hat{r}}{r^{2}}\big) = 4\pi\delta^{3}(r)$ and I was wondering if anyone could help me. I understand how the delta function is used and that it is necessary to have in this situation, but I am confused with multiplying a function by a delta function. Wouldn't $f(x)\delta^{3}(r)$ equal zero everywhere and then $\infty$ at the center? So with $\nabla \cdot \big(\frac{\hat{r}}{r^{2}}\big)$, It is zero everywhere and then $4\pi$ at the origin. Doesn't $4\pi\delta^{3}(r)$, because of the delta function, equal infinity at the center?

Thanks and I hope my question was clear.

JohnColtraneisJC
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Space20
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    The delta function is a distribution, not a function. It does not really make full sense to say it "equals infinity" at the origin, instead it makes more sense to say it "represents a point mass of size 1" at the origin. Thus $4\pi \delta(x)$ represents a point mass of size $4\pi$ at the origin (which I guess is what was confusing you). As a practical matter, what this is saying is that $\int_{\mathbb{R}^3} \nabla \cdot \left ( \frac{\hat{r}}{r^2} \right ) f(\mathbf{x}) d \mathbf{x}=4 \pi f(0)$, if $f$ is a "good enough" function. – Ian Aug 28 '17 at 17:29
  • Are you using Electrodynamics by Griffiths ? –  Aug 28 '17 at 18:00
  • I am using Electrodynamics by Griffiths. But, because the delta function is not really a function, and instead a distribution, $f(x)\delta(x)$ is not really two functions being multiplied together, but rather a function and a distribution that describes the function? – Space20 Aug 28 '17 at 18:09
  • $f(x) \delta(x)$ is intuitively a thing you can integrate over the whole space. When you do, you get $f(0)$ back. We write it that way because there are functions $\delta_n$ such that $\int f(x) \delta_n(x) dx \to f(0)$ for all "good enough" $f$ (where this is just ordinary multiplication and ordinary integration). – Ian Aug 28 '17 at 18:18
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    Does this answer your question? Divergence of $\vec{F} = \frac{\hat{\mathrm{r}}}{r^{2}}$ – Mark Viola Sep 18 '20 at 17:02

1 Answers1

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Take $\phi \in C_c^\infty(\mathbb R^3)$ and consider the integral $$I := \int_{\mathbb R^3} \bigg( \nabla \cdot \frac{\mathbf x}{|\mathbf x|^3} \bigg) \phi(\mathbf x) \, d\mathbf x$$

We can rewrite this using $\nabla \cdot (\mathbf F \phi) = (\nabla \cdot \mathbf F) \phi + \mathbf F \cdot \nabla \phi$: $$I = \int_{\mathbb R^3} \nabla \cdot \bigg( \frac{\mathbf x}{|\mathbf x|^3} \phi(\mathbf x) \bigg) \, d\mathbf x - \int_{\mathbb R^3} \frac{\mathbf x}{|\mathbf x|^3} \cdot \nabla \phi(\mathbf x) \, d\mathbf x$$

Since $\phi(\mathbf x) = 0$ for big $|\mathbf x|$ the first term vanishes. Therefore, $$I = - \int_{\mathbb R^3} \frac{\mathbf x}{|\mathbf x|^3} \cdot \nabla \phi(\mathbf x) \, d\mathbf x$$

Now we change to polar coordinates, $\mathbf x = r \mathbf n$: $$I = - \int_0^\infty \int_{S^2} \frac{\mathbf n}{r^2} \cdot \nabla \phi(\mathbf x) \, r^2 d\Omega(\mathbf n) \, dr = - \int_0^\infty \int_{S^2} \mathbf n \cdot \nabla \phi(\mathbf x) \, d\Omega(\mathbf n) \, dr$$ where $d\Omega$ is the area measure on $S^2 = \{ \mathbf x \in \mathbb R^3 : |\mathbf x| = 1 \}$.

In polar coordinates, $\nabla \phi = \frac{\partial\phi}{\partial r} \mathbf n + (\text{terms orthogonal to $\mathbf n$})$ so $$I = - \int_0^\infty \int_{S^2} \frac{\partial\phi}{\partial r}(\mathbf x) \, d\Omega(\mathbf n) \, dr$$

Let us do the $r$ integral first: $$I = - \int_{S^2} \int_0^\infty \frac{\partial\phi}{\partial r}(\mathbf x) \, dr \, d\Omega(\mathbf n) = - \int_{S^2} \big[ \phi(r \mathbf n) \big]_{r=0}^{\infty} \, d\Omega(\mathbf n) = \int_{S^2} \phi(\mathbf 0) \, d\Omega(\mathbf n)$$

The left integral is trivial since we now have no dependency on $\mathbf n$ and the area of the unit sphere is $4\pi$: $$I = 4\pi \, \phi(\mathbf 0)$$ which can be written as $$I = 4\pi \int_{\mathbb R^3} \delta(\mathbf x) \phi(\mathbf x) \, d\mathbf x$$

Thus, $$\int_{\mathbb R^3} \bigg( \nabla \cdot \frac{\mathbf x}{|\mathbf x|^3} \bigg) \phi(\mathbf x) \, d\mathbf x = \int_{\mathbb R^3} 4\pi \, \delta(\mathbf x) \phi(\mathbf x) \, d\mathbf x$$

Since this is valid for all $\phi \in C_c^\infty(\mathbb R^3)$ we have $$\nabla \cdot \frac{\mathbf x}{|\mathbf x|^3} = 4\pi \, \delta(\mathbf x)$$

Edit

I got a question why $\int_{\mathbb R^3} \nabla \cdot \left( \frac{\mathbf x}{|\mathbf x|^3} \phi(\mathbf x) \right) \, d\mathbf x$ vanishes. It does so by divergence theorem. Since $\phi$ has compact support, for big enough $R$ we have $$ \int_{|\mathbf{x}|<R} \nabla \cdot \bigg( \frac{\mathbf x}{|\mathbf x|^3} \phi(\mathbf x) \bigg) \, d\mathbf x = \oint_{|\mathbf{x}|=R} \frac{\mathbf x}{|\mathbf x|^3} \phi(\mathbf x) \, \mathbf n \cdot dS(\mathbf x) = 0. $$ Therefore, $$ \int_{\mathbb R^3} \nabla \cdot \bigg( \frac{\mathbf x}{|\mathbf x|^3} \phi(\mathbf x) \bigg) \, d\mathbf x = \lim_{R\to\infty} \int_{|\mathbf{x}|<R} \nabla \cdot \bigg( \frac{\mathbf x}{|\mathbf x|^3} \phi(\mathbf x) \bigg) \, d\mathbf x = \oint_{|\mathbf{x}|=R} \frac{\mathbf x}{|\mathbf x|^3} \phi(\mathbf x) \, \mathbf n \cdot dS(\mathbf x) = 0. $$

md2perpe
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