I got this identity from a physics textbook, but figure the Math Stack Exchange could help.
Allegedly:
$$
\left(\vec{a}\cdot\nabla\right)\nabla\frac{1}{r}=\vec{a}\frac{4\pi}{3}\delta\left(\vec{r}\right)
$$
Where $r=\left(x^2+y^2+z^2\right)^{1/2}$ ans $\vec{a}$ is a constant vector.
And I must have misunderstanding since when I carry out the computation,
$$
\nabla\frac{1}{r}=-\frac{\vec{r}}{r^3}=-\hat{x}\frac{x}{r^3}-\hat{y}\frac{y}{r^3}-\hat{z}\frac{z}{r^3}
$$
$$
\left(\vec{a}\cdot\nabla\right)\nabla\frac{1}{r}=\left(
a_x\partial_x+a_y\partial_y+a_z\partial_z
\right)\left(
-\hat{x}\frac{x}{r^3}-\hat{y}\frac{y}{r^3}-\hat{z}\frac{z}{r^3}
\right)
$$
Using:
$$
\partial_x\frac{1}{r^3}=-3\frac{x}{r^5}
$$
I get, for the x-component of $\left(\vec{a}\cdot\nabla\right)\nabla\frac{1}{r}$:
\begin{equation}
\tag{1}\label{xComponent}
a_x\frac{-1}{r^3}+3a_x\frac{x^2}{r^5}+3a_y\frac{xy}{r^5}+3a_z\frac{xz}{r^5}
\end{equation}
I see how the x-component goes to $\infty$ when $\vec{r}=0$, but do not see how it is zero for arbitrary $\vec{r}\neq 0$ and the factors of $a_y$ and $a_z$ make it seem that the x-component of $\left(\vec{a}\cdot\nabla\right)\nabla\frac{1}{r}$ is not proportional to $a_x$.
Which brings me to my question:
Why does $\left(\vec{a}\cdot\nabla\right)\nabla\frac{1}{r}=\vec{a}\frac{4\pi}{3}\delta\left(\vec{r}\right)$?
By the way, the identity can be found in 2.8 of appendix F in Blundell's Magnetism in Condensed Matter .
Thanks all in advance
Best Answer
First of all, the identity in the appendix of the book (i.e. the one from the question title) is definitely not correct, but there is more to it (and it is not just the sign error; as you have correctly shown, the LHS does not even vanish for $\mathbf{r}\neq 0$). A closer look at the problem reveals that what is really being asked is the remarkable identity $$(\mathbf{a}\cdot\nabla)(\mathbf{b}\cdot\nabla)\frac{1}{r} = -\frac{\mathbf{a}\cdot \mathbf{b}}{r^3} + 3 \frac{(\mathbf{a}\cdot\mathbf{r})(\mathbf{b}\cdot\mathbf{r})}{r^5} - \frac{4\pi}{3} \mathbf{a} \cdot \mathbf{b} \delta(\mathbf{r}).$$ The first part is just what you get from straight-forward vector calculus, so the interesting part is the $\delta(\mathbf{r})$. The intuition behind the result comes from the integral $$\int_{\mathbb{R}^3} \nabla\times\left(\nabla\times\frac{\mathbf{a}}{r}\right)\,\text{d}^3r = 4\pi \mathbf{a}\frac{2}{3}$$ that I will prove later. First, note that given this integral, we find $$\int_{\mathbb{R}^3} (\mathbf{a}\cdot\nabla) \nabla\frac{1}{r}\,\text{d}^3r = \int_{\mathbb{R}^3} \left[\nabla\times\left(\nabla\times\frac{\mathbf{a}}{r}\right) + \mathbf{a}\nabla^2 \frac{1}{r} \right]\,\text{d}^3r = 4\pi\mathbf{a} \frac{2}{3} - 4\pi \mathbf{a} = -\frac{4\pi}{3}\mathbf{a}.$$ Due to $\int_{\mathbb{R}^3} \frac{\mathbf{a}r^2 - 3\mathbf{r}(\mathbf{a}\cdot\mathbf{r})}{r^5}\,\text{d}^3r = 0$ (be careful with the $r$-integration and transform to a frame where $\mathbf{a}=a\mathbf{e}_z$ to show this), we thus find that the object $$g(\mathbf{r}) = (\mathbf{a}\cdot\nabla)(\mathbf{b}\cdot\nabla)\frac{1}{r} + \frac{\mathbf{a}\cdot \mathbf{b}}{r^3} - 3 \frac{(\mathbf{a}\cdot\mathbf{r})(\mathbf{b}\cdot\mathbf{r})}{r^5}$$ has the two properties \begin{align}(1): &&g(\mathbf{r})&=0 \qquad \text{for }\mathbf{r}\neq 0, \\ (2): &&\int_{\mathbb{R}^3} g(\mathbf{r})\,\text{d}^3r &= -\frac{4\pi}{3}\mathbf{a}\cdot\mathbf{b}.\end{align} Combined, this means that $g(\mathbf{r}) \equiv -\frac{4\pi}{3}\mathbf{a}\cdot\mathbf{b} \delta(\mathbf{r})$ in the sense of $$\int_{\mathbb{R}^3} \phi(\mathbf{r}) g(\mathbf{r})\,\text{d}^3r = -\frac{4\pi}{3}\mathbf{a}\cdot\mathbf{b} \phi(0)$$ for sufficiently "nice" functions $\phi$.
To solve the integral from above, use the usual vector identity combined with Gauss's law to get \begin{align}\int_{\mathbb{R}^3} \nabla\times\left(\nabla\times\frac{\mathbf{a}}{r}\right)\,\text{d}^3r &= \int_{\mathbb{R}^3} \nabla\left(\nabla\cdot\frac{\mathbf{a}}{r}\right)\,\text{d}^3r - \mathbf{a}\int_{\mathbb{R}^3} \nabla^2\frac{1}{r}\,\text{d}^3r \\ &= \oint_{|\mathbf{r}|=R} \left( \nabla\cdot\frac{\mathbf{a}}{r} - \mathbf{a} \nabla\frac{1}{r} \right)\,\mathbf{e}_rr^2\text{d}\Omega \\ &= \oint_{|\mathbf{r}|=R} \left( 2\mathbf{a}\cdot\frac{\mathbf{r}}{r^3} \right)\,\mathbf{e}_rr^2\text{d}\Omega\\ &= 2\oint_{|\mathbf{r}|=R} \left( \mathbf{a}\cdot\mathbf{e}_r \right)\,\mathbf{e}_r\text{d}\Omega\\ &= 2a\int_0^{2\pi}\int_0^\pi \cos\theta\,\mathbf{e}_r\sin\theta\text{d}\theta\text{d}\varphi = 4\pi\mathbf{a}\frac{2}{3}. \end{align} Similar to the other integral, justifying the simplifaction $\mathbf{a}=a\mathbf{e}_z$ requires a transform with the proper rotation matrix.