# Spherical wave as sum of plane waves

How can we do this computation?

$\iiint_{R^3} \frac{e^{ik'r}}{r} e^{ik_1x+k_2y+k_3z}dx dy dz$ where $r=\sqrt{x^2+y^2+z^2}$ ? I think we must use distributions...

Physically, it's equivalent to find wave vectors k distribution and to write a spherical wave as sum of plane waves... I know the formula for the inverse problem: write a plane wave as sum of spherical waves...The solution in this case is a serie of spherical harmonics and shperical bessel functions...

Someone can help me please? :-)

• This is the 3D Fourier transform of $\frac{e^{ik'r}}r$? If so, the formula should be $\iiint_V \frac{e^{ik'r}}r e^{{\color{red}-}i\mathbf k\cdot\mathbf r} d^3\mathbf r$ – kennytm Dec 1 '10 at 18:43
• This question might have a physical motivation but I think it is purely mathematical in nature. You should probably ask for the answer at math.SE. – Marek Dec 1 '10 at 19:34
• @Marek: I think I agree... I mean, the title suggests a physics question, but in essence it is just about how to do an integral. I'm voting to close it (but only because it requires four other people to agree before the question actually gets closed - I wouldn't be comfortable unilaterally closing this if I had the power to do so). – David Z Dec 1 '10 at 19:42
• @David: right, I also wouldn't be comfortable to just close it on my own. @Boy: no need to be sorry, closing a question is no big deal. It's really more about setting boundaries of what should and shouldn't be asked on this site so that people know in the future. – Marek Dec 1 '10 at 20:24
• I know I'm in the minority, but I like seeing these "mathematical" questions here. Physicists have a different culture of math than mathematicians, and are more likely to get a useful answer from other physicists. – nibot Dec 2 '10 at 0:17

From your description, I believe you want to find the Fourier transform of $$f(\mathbf r) = \frac{e^{ik'r}}r,$$ and the wave can be recovered from the linear superposition of plane waves identified by k $$f(\mathbf r) = \frac1{(2\pi)^{3/2}}\iiint \mathcal F[f](\mathbf k)e^{i\mathbf k\cdot\mathbf r} d^3 \mathbf k.$$
The spherical wave have spherical symmetry, so what you should do is to perform the integration in spherical coordinates instead of Cartesian. WLOG, assume k is along the z axis, thus \begin{aligned} \mathcal F[f](k\hat{\mathbf z}) &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-i\mathbf k\cdot \mathbf r} d^3\mathbf r \\ &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-ikr\cos\theta} r^2 \sin\theta dr d\theta d\phi \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty \left(re^{ik'r} \int_0^{\pi} e^{-ikr\cos\theta} \sin\theta d\theta\right) dr \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty r e^{ik'r} \frac{2 \sin kr}{kr} dr \\ &= \frac1k\sqrt{\frac2\pi} \int_0^\infty e^{ik'r} \sin kr dr \\ &= \sqrt{\frac2\pi}\frac1{k^2 - k'^2} \end{aligned}
• thank you! And how have you do the last passage? $1/k \int_0^{+infty} e^{ik'r}sinkrdr= \frac{1}{k^2+k'^2}$ – Boy Simone Dec 1 '10 at 19:53
• I think he wants to know how your last step of the calculation is done, i.e. the integral over $e^{ik' r}\sin kr$. – Lagerbaer Dec 1 '10 at 20:06
• @Boy: Change $\sin kr$ to $(e^{ikr}-e^{-ikr})/2i$. Then notice that $\int_0^\infty e^{iKr} dr = i/K$ (if we ignore convergence stuff. Actually the answer isn't right when $k'$ doesn't have a positive imaginary part since the integral diverges.) – kennytm Dec 1 '10 at 20:06
• Yes, then I arrived at that consideration too. My doubt is just it, a k' purely real means no absorption, but I think that you can't explain a purely spherical wave with plane waves is very strange, because the inverse is possible...Maybe there's the way to do convergence of that integral in distrubution sense...like $\int e^{ikx}=2\pi \delta(k)$...I post the question on the mathematical section too, because that's really purely mathematical...Thank you :-) – Boy Simone Dec 1 '10 at 20:34