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I am having trouble getting from one line to the next from this wiki page. I am referring to the text line

Green's function in $r$ is therefore given by the inverse Fourier transform,


$$G(r) ~=~ \frac{1}{(2\pi)^3} \iiint d^3k \frac{e^{i{\bf k}\cdot{\bf r}}}{k^2+\lambda^2}$$

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 triple integral? Divergence theorem? Stokes? Detailed steps would be much appreciated.

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Hint: Change from rectangular to spherical coordinates in $k$-space. –  Qmechanic Feb 5 '13 at 2:44
Hopefully by now it is clear that $k_r$ is the radial coordinate in $\mathbf{k}$-space, i.e. $k_r = |\mathbf{k}|.$ –  Vibert Feb 5 '13 at 23:13

2 Answers 2

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:

$k_1=k \sin{\theta} \cos{\phi}, k_2=k \sin{\theta} \sin{\phi}, k_3=k \cos{\theta}$, and $d^3 k =k^2 \sin{\theta} dkd \theta d \phi$ and $r_1=r \sin{\theta}' \cos{\phi}', r_2=r \sin{\theta}' \sin{\phi}', r_3=r \cos{\theta}'.$

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:

$k_1=k_r\sin{\theta}\cos{\phi}$, $k_2=k_r\sin{\theta}\sin{\phi}$, $k_3=k_r\cos{\theta}$, and $d^3k =k_r^2\sin{\theta}dkd{\theta}d{\phi}$ and $r_1=0, r_2=0, r_3=r$.

and the dot product is simply $k_r\cdot\ r\cdot \cos{\theta}$, with $\theta$ the angle between k and r in these "r-axes".

Now define $\gamma = k_r\cdot\ r\cdot \cos{\theta}$ and dγ = $k_r\cdot\ r\cdot \sin{\theta}d{\theta}$ and substitute in the integrals. This will give you the $k_r/r$ term.

When you integrate the $e^{i\gamma}$ from $\gamma = -k_rr$ to +$k_rr$, this will wive you the $\sin{k_rr}$.

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Hi DoM, thanks for the tex effort. I fixed up some of it but I'm worried about changing some of the meaning and spacing. You should make used of \sin and \theta and \phi rather than unicode and "sin" which is s * i * n –  Brandon Enright Jun 6 '13 at 6:03

You convert the integral to spherical coordinates and the $k \cdot r$ term becomes $kr\cos\theta$. Integrating over $\theta$ and $\phi$ gets you the single dimensional integral expression.

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