
#1
Jan313, 06:15 PM

P: 718

I'm reading some course notes for a physics class that contain the following step in a derivation:
[itex]\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x}}}{\vec{x}^2 + m^2}[/itex] Changing to polar coordinates, and writing [itex]\vec{k}\cdot\vec{x}=kr \cos\theta[/itex], we have: [itex]\phi(\vec{x}) = \frac{1}{(2\pi)^2} \int_0^\infty dk \frac{k^2}{k^2 + m^2}\frac{2\sin kr}{kr}[/itex] I'm having a bit of difficulty seeing this step. Could someone please show some of the intermediate steps between these two and explain what's happening? I understand that k^{2} in the numerator comes from converting to spherical coordinates, but that's about all I follow. 



#2
Jan413, 06:04 AM

P: 304





#3
Jan413, 09:51 AM

P: 718

Oh dear, you're right of course. I made a transcription error, it should be:
[itex]\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x} }}{\vec{k}^2 + m^2}[/itex] It's a Fourier transform. Now that it's fixed, can you see how to proceed? 



#4
Jan413, 11:28 AM

P: 304

Stuck on a change of variables 



#5
Jan413, 04:06 PM

Sci Advisor
P: 5,942





#6
Jan413, 05:32 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi LastOneStanding!
(d^{3}k = ksinθ dkdθdφ, and so you have to ∫ ksinθ e^{ikrcosθ} dθ) 



#7
Jan413, 06:29 PM

P: 304

First, ##\phi(\vec{x})## depends only upon the radial coordinate ##r## of ##\vec{x}##. (Why?) We can therefore assume that ##\vec{x}## is ##(0,0,r)##. Then ## \theta## is the colatitude of ##\vec{k}## and in spherical coordinates ##d^3 k## becomes ##k^2 \sin \theta \,dk\,d\theta\,d\phi##. Now, it is easy to integrate ##\sin\theta\,e^{ikrcos\theta}## wrt ##\theta##. This will give the desired the result. 



#8
Jan413, 09:33 PM

P: 718





#9
Jan413, 09:39 PM

P: 718





#10
Jan413, 09:46 PM

P: 718

Oh, bother. Undone by a simple usubsitution, sigh. Thank you, Erland, I see how the rest of the integration goes.



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