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I am having trouble getting from one line to the next from this wiki page. http://en.wikipedia.org/wiki/Screened_Poisson_equation. I am referring to " Green's function in r is therefore given by the inverse Fourier transform" where

$$G(r) = \frac{1}{(2\pi)^3} \int\int\int d^3k \frac{e^{ik*r}}{k^2+\lambda^2}$$

goes to

$$G(r) = \frac{1}{2\pi^2r} \int^{+\infty}_0 dk_r \frac{k_r \sin(k_r r)}{k_r^2+\lambda^2}$$

where does the $\frac{1}{r}$ term come from and what is $k_r$. How did they simplify the tripple integral? Divergence theorem? Stokes? Detailed steps would be much appreciated.

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2 Answers 2

They used spherical coordinates. $k_1 = k \sin{\theta} \cos{\phi}$, $k_2 = k \sin{\theta} \sin{\phi}$, $k_3 = k \cos{\theta}$, and

$$\mathrm{d^3}\vec{k} = k^2 \sin{\theta} \, dk \, d \theta \, d \phi $$


$$\begin{align}G(r) &= \frac{1}{(2 \pi)^3} \int_{\mathbb{R^3}} d^3 \vec{k} \frac{e^{i \vec{k} \cdot \vec{r}}}{k^2 + \lambda^2} \\ &= \frac{1}{(2 \pi)^3} \int_0^{\infty} dk \: k^2 \int_0^{\pi} d \theta \sin{\theta} \: \int_0^{2 \pi} d \phi \frac{e^{i k r_{\perp} \sin{\theta} \cos{(\phi - \phi')}} e^{i k z \cos{\theta}} }{k^2 + \lambda^2} \\ &= \frac{1}{(2 \pi)^2} \int_0^{\infty} dk \: \frac{k^2}{k^2 + \lambda^2} \int_0^{\pi} d \theta \sin{\theta} \: e^{i k z \cos{\theta}} J_0(k r_{\perp} \sin{\theta}) \end{align}$$

where $J_0$ is the Bessel function (zeroth order, 1st kind), and $r_{\perp} = \sqrt{x^2+y^2}$. Now, it seems to me that your integral above was derived using $z=0$. In this case, the integral over $\theta$ is

$$\int_0^{\pi} d \theta \sin{\theta} \: J_0(k r_{\perp} \sin{\theta}) = \frac{\sin{k r_{\perp}}}{k r_{\perp}}$$

(Go here and here for details.) Therefore, we get

$$G(r_{\perp}) = \frac{1}{(2 \pi)^2 r_{\perp}} \int_0^{\infty} dk \: \frac{k \sin{(k r_{\perp})}}{k^2 + \lambda^2} $$

which is what you had above. This in turn simplifies to

$$G(r_{\perp}) = \frac{e^{- |\lambda| r_{\perp}}}{8 \pi r_{\perp}}$$

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I know, that this answer is more than 2 years old, but do you happen to know the steps taken from the second last equation to the last equation? Wikipedia states that this can be done by Contour Integratiion, but I fail to see how. –  sonystarmap Jul 7 at 8:30
@sonystarmap: yes, that last integral may be done via contour integration. If I have time, I will outline the steps. –  Ron Gordon Jul 7 at 11:48
That would be perfekt, my problem however is actually related to the fractional situation where I have $k^\alpha+\lambda^2$, but I would like to understand the integer case first. Thanks –  sonystarmap Jul 7 at 11:53

I ran into the same problem today and didn't like the use of the Bessel functions. So here's my simpler approach to it: no scary stuff.

Put everything into spherical coordinates:

$k_1=ksinθcosϕ, k_2=ksinθsinϕ, k_3=kcosθ, and d^3k =k^2sinθdkdθdϕ$ and $r_1=rsinθ'cosϕ', r_2=rsinθ'sinϕ', r_3=rcosθ'.$

Then you make the dot product and it gives $k\cdot r \cdot [...]$, where $[...]$ is an ugly term full of θ,θ',ϕ and ϕ' for the angle between k and r.

However, you can choose a different orthogonal axis: say one where z' in parallel to r. In these "r-ortogonal" axes: $\hat x_r$' = $\hat x_r$ ^ $\hat z_r$' and $\hat y_r$' = $\hat z_r$' ^ $\hat x_r$' and $\hat z_r$' = r/r.

So now your spherical coordinates are:

$k_1=k_rsinθcosϕ$, $k_2=k_rsinθsinϕ$, $k_3=k_rcosθ$, and $d^3k =k_r^2sinθdkdθdϕ$ and $r_1=0, r_2=0, r_3=r$.

and the dot product is simply $k_r\cdot\ r\cdot cosθ$, with θ the angle between k and r in these "r-axes".

Now define γ = $k_r\cdot\ r\cdot cosθ$ and dγ = $k_r\cdot\ r\cdot sinθdθ$ and substitute in the integrals. This will give you the $k_r/r$ term.

When you integrate the $e^{iγ}$ from γ = -$k_rr$ to +$k_rr$, this will wive you the sin($k_rr$).

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