Jackson Classical Electrodynamics: page 35 expansion of charge

[qnach]

Could anyone explain how did Jackson obtain the Taylor distribution of charge distribution at the end of section 1.7 (version 3)?

 

[Charles Link]

Without verifying it in detail, I believe it may be a second order Taylor series expansion, where first order terms $\nabla \rho(x) \cdot (x-x')$ will integrate to zero because of the even integrand, and similarly for the second order mixed partials. @vanhees71 Is this assessment correct?
Edit: Note: I'm getting a factor of $\frac{1}{2}$, instead of $\frac{1}{6}$ on the $\nabla^2$ term. See Perhaps it is because of the spherical symmetry, and the $(\frac{\partial^2 \rho}{\partial x^2})$ term gets a factor of $(x-x_o)^2$, etc. With $(\frac{\partial^2 \rho}{\partial x^2}) (x-x_o)^2=(\nabla^2 \rho) r^2/3$, (when integrated, etc., along a single direction $dr$), it may then be a $\frac{1}{6}$, (edit=I'm handwaving here). Note also the $4 \pi r^2$ factor, that canceled the $4 \pi$ out front in the denominator. It's covered up with tape, but I think it is then $r^2 \, dr$.

 

[Charles Link]

The print isn't real clear, but I worked the next step. (The exponent in the denominator is $\frac{5}{2}$. It looks like $\frac{3}{2}$, but correctly it should be $\frac{5}{2}$).
$\int\limits_{0}^{+\infty} \frac{3 a^2 r^2}{(r^2+a^2)^{5/2}} \, dr$ can be readily solved by letting $r=a \tan{\theta}$.
To resolve the handwaving in post 2, $\nabla^2 \rho =(\frac{\partial^2{\rho}}{\partial{x^2}}) +$ y and z second partial terms at $\vec{x}$, so that $\nabla^2 \rho =3 (\frac{\partial^2{\rho}}{\partial{x^2}})$ at $\vec{x}$. This $\nabla^2$ term is a constant when integrating over dr. It also is useful to look at the integrals of $\int x^2 \, d^3 r$ vs. $\int r^2 \, d^3 r$, etc. Upon working through all the details, I agree with his $\frac{1}{6}$.

 

[qnach]

Could anyone please explain the meaning of the last term of the second last equation in this page.
What does O(a^2,a^2 log a) mean? The O notation contains only one term inside the bracket.
But this guy has two inside.

 

[vanhees71]

The print isn't real clear, but I worked the next step. (The exponent in the denominator is $\frac{5}{2}$. It looks like $\frac{3}{2}$, but correctly it should be $\frac{5}{2}$).

$\int\limits_{0}^{+\infty} \frac{3 a^2 r^2}{(r^2+a^2)^{5/2}} \, dr$ can be readily solved by letting $r=a \tan{\theta}$.

To resolve the handwaving in post 2, $\nabla^2 \rho =(\frac{\partial^2{\rho}}{\partial{x^2}}) +$ y and z second partial terms at $\vec{x}$, so that $\nabla^2 \rho =3 (\frac{\partial^2{\rho}}{\partial{x^2}})$ at $\vec{x}$. This $\nabla^2$ term is a constant when integrating over dr. It also is useful to look at the integrals of $\int x^2 \, d^3 r$ vs. $\int r^2 \, d^3 r$, etc. Upon working through all the details, I agree with his $\frac{1}{6}$.

I think that's correct though I find the conclusion by Jackson a bit too quick, but after some detailed analysis, I think it's correct too.

What's behind it is a formal proof of the very important formula
$$-\Delta_x G(\vec{x},\vec{x}')=-\Delta_x \frac{1}{4 \pi |\vec{x}-\vec{x}'|}=\delta^{(3)}(\vec{x}-\vec{x}').$$
The standard proof is to integrate a test function in the sense of distributions
$$\phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' G(\vec{x},\vec{x}') \rho(\vec{x}')$$
and then show that indeed
$$-\Delta \phi(\vec{x})=\rho(\vec{x}).$$
The integral is evaluated by integrating not over $\mathbb{R}^3$, because of the singularity of the Green's function at $\vec{x}'=\vec{x}$ but to leave out a little spherical ball or radius $\epsilon$ around the singular point $\vec{x}$ and then taking the limit $\epsilon \rightarrow 0$, using the fact that
$$\Delta_x G(\vec{x},\vec{x}')=\Delta_{x'} G(\vec{x},\vec{x}')=0 \quad \text{for} \quad \vec{x}' \neq \vec{x}.$$
Jackson provides the proof in a different way, i.e., he regularized the Green's function and then takes the weak limit.

The advantage is that the original integral over entire $\mathbb{R}^3$ is well defined to begin with, and
$$\Delta \phi_a(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \Delta_x G_a(\vec{x},\vec{x}') \rho(\vec{x}')$$
is well defined.

Now he wants to show that for $a \rightarrow 0$ this integral goes to $-\rho(\vec{x})$. The first observation is that the only contribution can come from an arbitrarily little region around $\vec{x}'=\vec{x}$, i.a., for taking the limit you can just evaluate the integral over a sphere of radius $R$ around $\vec{x}$, because for the rest of the integral you can exchange the limit with the integral, giving $0$.

So he sets $\vec{x}'=\vec{x}+\vec{r}$ with $\vec{r}=\vec{x}'-\vec{x}$. Now
$$\Delta G_a(\vec{x},\vec{x}')=-\frac{3 a^2}{(r^2+a^2)^{5/2}}.$$
So the integral is
$$I(\vec{r})=-3 a^2\int_{K_R} \mathrm{d}^3 r \frac{1}{(r^2+a^2)^{5/2}} \rho(\vec{x}+\vec{r}),$$
where $K_R$ is the solid sphere of radius $R$ around the origin $\vec{r}=0$. Next Jackson expands $\rho$ around $\vec{r}=0$. Formally
$$\rho(\vec{x}+\vec{r})=\exp(\vec{r} \cdot \vec{\nabla}) \rho(\vec{x}).$$
Now all you need is to evaluate the integrals
$$I_{j_1,\ldots,j_k}=-3 a^2\int_{K_R} \mathrm{d}^3 r \frac{1}{(r^2+a^2)^{5/2}} r_{j_1} \cdots r_{j_k}.$$
Of course one introduces spherical coordinates for $\vec{r}$ to do the integral. Fortunately all that's interesting here is the integral over $r=|\vec{r}|$. We don't need the complicated integration over the angles for all $k$. So let's only do the $r$ integral:
$$I_k=-3a^2 \int_0^R \mathrm{d} r \frac{r^{k+2}}{(r^2+a^2)^{5/2}}.$$
Substitution of $r=a \sinh \lambda$ leads to
$$I_k=-3a^k \int_0^{\mathrm{arsinh(R/a)}} \mathrm{d} \lambda \frac{\sinh^{k+2} \lambda}{\cosh^4 \lambda}.$$
Now the integrand is increasing for $k \geq 4$, i.e., for these $k$
$$|I_k| \leq \frac{3 a^2 R^{k+2}}{(a^2+R^2)^2}\mathrm{arsinh}(R/a) \rightarrow 0 \quad \text{for} \quad a \rightarrow 0.$$
So we need only the cases $k=0$ and $k=1$, for which
$$I_0=-\frac{R^3}{(a^2+R^2)^{3/2}} \rightarrow -1 \quad \text{for} \quad a \rightarrow 0$$
and
$$I_1=a \left (-2+\frac{a(2a^2+3R^2)}{(a^2+R^2)^{3/2}} \right) \rightarrow 0 \quad \text{for} \quad a \rightarrow 0.$$
So the only contribution is for $k=0$, and together with the integral over the angles, which gives $4 \pi$ indeed leads to
$$\lim_{a \rightarrow 0} \Delta \phi_a(\vec{x})=-\rho(\vec{x}),$$
as claimed.

 

[vanhees71]

Could anyone please explain the meaning of the last term of the second last equation in this page.
What does O(a^2,a^2 log a) mean? The O notation contains only one term inside the bracket.
But this guy has two inside.

That's the Landauer big-O symbol. It tells you how in some limit of a variable a function behaves. In this case it's the limit $a \rightarrow 0$, and
$$f(a)=\mathcal{O(a^2,a^2 \ln a)}$$
means that
$$f(a)=A a^2 + B a^2 \ln a$$
with some constants $A$ and $B$. My lengthy calculation shows that this is correct since
$$\mathrm{arsinh}(R/a)=\ln[R/a+\sqrt{1+(R/a)^2}] \simeq \ln(2 R/a)=-\ln(a/R)+\ln 2.$$