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

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 ##.

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- #3

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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} ##.

## \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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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.

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## \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.

- #6

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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

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.

$$f(a)=\mathcal{O(a^2,a^2 \ln a)}$$

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

$$\mathrm{arsinh}(R/a)=\ln[R/a+\sqrt{1+(R/a)^2}] \simeq \ln(2 R/a)=-\ln(a/R)+\ln 2.$$

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