# Jackson Classical Electrodynamics: page 35 expansion of charge

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## Summary:

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

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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 cancelled the ## 4 \pi ## out front in the denominator. It's covered up with tape, but I think it is then ## r^2 \, dr ##.

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qnach
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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} ##.

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etotheipi, vanhees71 and 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
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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
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2019 Award
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
$$\mathrm{arsinh}(R/a)=\ln[R/a+\sqrt{1+(R/a)^2}] \simeq \ln(2 R/a)=-\ln(a/R)+\ln 2.$$