I've some difficulty deriving this equation from Jackson electrodynamics (The equation after 1.30)

$$\nabla^2 \Phi_a\left({\textbf{x}}\right)=-\frac{1}{\epsilon_0}\int_{0}^{R} \frac{3a^2}{\left(r^2+a^2\right)^{\frac{5}{2}}} \left[\rho\left(\textbf{x}\right)+\frac{r^2}{6}\nabla^2\rho+...\right]r^2dr+O\left(a^2\right).$$

It follows from

$$\nabla^2 \Phi_a\left({\textbf{x}}\right)=\frac{1}{4\pi\epsilon_0}\int\rho\left(\textbf{x}'\right)\left(\frac{3a^ 2}{\left(r^2+a^2\right)^{\frac{5}{2}}}\right)d^3x'$$

by Taylor expansion of the charge density $\rho(x^{\prime})$ about the point $x$ and then doing angular integration . I've difficulty working out the details of this. Here $r=x-x^{\prime}$ and $\Phi_a$ is a function from $R^{3}$ to $R^{1}$ "the electric potential".

  • 1
    $\begingroup$ As a place to start, you might look at math.stackexchange.com/q/246729 . It's a bit light on the details, but a lot of the related entries in the right column looked promising at first glance. $\endgroup$
    – David H
    Mar 26 '13 at 12:59

The Taylor expansion of $\rho(\mathbf x')$ about a point $\mathbf x$ is $$ \rho(\mathbf x') = \rho(\mathbf x) + (\mathbf x'-\mathbf x)\cdot\nabla\rho(\mathbf x)+ \frac{1}{2}\sum_{ij}(x_i'-x_i)(x_j'-x_j)\frac{\partial^2\rho}{\partial x_i'\partial x_j'}(\mathbf x) + \cdots $$ The integral we are attempting to perform is $$ \int_{\mathbb R^3} d^3\mathbf x'\,\rho(\mathbf{x})f(r), \qquad f(r) = \frac{3a^2}{(r^2+a^2)^{5/2}}, \qquad r = |\mathbf x' - \mathbf x| $$ Integration is linear, so (up to some technicalities about integration of a power series term-by-term that I'll ignore) we can Taylor expand the charge density and integrate the Taylor expansion term-by-term. The first order term in the Taylor expansion gives $$ I_1(\mathbf x) = \int d^3\mathbf x'\,f(r)(\mathbf x'-\mathbf x)\cdot\nabla\rho(\mathbf x) $$ I claim that this vanishes. To see this, note first that without loss of generality (actually the fact that there is no loss of generality here must be proven, but it's relatively straightforward, and I'll leave it to you to think about) we can choose our axes so that $\nabla\rho(\mathbf x)$ points in the positive $z$-direction, namely $\nabla\rho(\mathbf x) = |\nabla\rho(\mathbf x)|\hat{\mathbf z}$. Next, make the change of variables $\mathbf r = \mathbf x - \mathbf x'$ then $d^3 \mathbf x'=d^3\mathbf r$. Putting this together gives \begin{align} I_1(\mathbf x) &= \int d^3\mathbf r \,f(r)\,\mathbf r\cdot\nabla\rho(\mathbf x) \\ &=\int dr\,d\theta\,d\phi\,r^2\sin\theta f(r) r|\nabla\rho(\mathbf x)|\cos\theta \\ &=|\nabla\rho(\mathbf x)|\int_0^\infty dr\,r^3f(r)\int_0^{2\pi} d\phi\int_0^{\pi} d\theta\,\sin\theta\cos\theta \end{align} but notice that the $\theta$ intgeral itself vanishes; $$ \int_0^\pi d\theta\,\sin\theta\cos\theta =-\int_{-\pi/2}^{\pi/2}d\eta\,\cos\eta\sin\eta = 0 $$ The first equality came from the change of variables $\eta = \theta-\pi/2$, and the second equality comes from integrating an odd function against an even function on a symmetric interval containing zero. It follows that $$ I_1(\mathbf x) = 0 $$ It remains to show that the integral $I_2$ that comes from integrating the second order ter in the Taylor expansion gives the $\frac{r^2}{6}\nabla^2\rho$ given in Jackson. I'll leave this to you. If you're really struggling, I'll gladly give some more hints.


You can write the expansion in the form


if you consider that $r=x-x_0$, then $(x-x_0)\cdot\nabla=r\cos\theta\nabla$ and now the quadratic term can be write in the integral as

$$\frac{1}{4\pi\epsilon_0}\int_0^r \int_0^\pi \int_0^{2\pi}\frac{1}{2}r^2 \nabla^2\rho r^2\sin{\theta}\cos^2{\theta}drd\theta d\phi$$ $$=\frac{1}{4\pi\epsilon_0}\int_0^r r^4\nabla^2 \rho dr \int_0^\pi \cos^2{\theta}\sin{\theta} d{\theta}\int_0^{2\pi}d{\phi}$$$$=\frac{1}{6\ \epsilon_0}\int_0^r r^4\nabla^2 \rho dr=\frac{1}{\epsilon_0}\int_0^r (\frac{r^2}{6}\nabla^2 \rho) r^2 dr$$ This is the term showed by Jackson


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