Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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,

where

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

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.

share|improve this question
1  
Hint: Change from rectangular to spherical coordinates in $k$-space. – Qmechanic Feb 5 at 2:44
1  
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 at 23:13

1 Answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.