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Stuck on a change of variables

by LastOneStanding
Tags: stuck, variables
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LastOneStanding
#1
Jan3-13, 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 k2 in the numerator comes from converting to spherical coordinates, but that's about all I follow.
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Erland
#2
Jan4-13, 06:04 AM
P: 345
Quote Quote by LastOneStanding View Post
[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]
This seems strange. How can it be meaningful to integrate w.r.t ##d^3k## ???
LastOneStanding
#3
Jan4-13, 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?

Erland
#4
Jan4-13, 11:28 AM
P: 345
Stuck on a change of variables

Quote Quote by LastOneStanding View Post
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?
I still don't understand what ##d^3k## is. Shouldn't it be ##dx## or something like that?
mathman
#5
Jan4-13, 04:06 PM
Sci Advisor
P: 6,071
Quote Quote by Erland View Post
I still don't understand what ##d^3k## is. Shouldn't it be ##dx## or something like that?
It is a short hand notation to indicate that the integral is 3 dimensional, dx would be 1 dimensional.
tiny-tim
#6
Jan4-13, 05:32 PM
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Hi LastOneStanding!
Quote Quote by LastOneStanding View Post
I understand that k2 in the numerator comes from converting to spherical coordinates
no, only one k comes from the conversion, the other k cancels with the k in the denominator

(d3k = ksinθ dkdθdφ, and so you have to ∫ ksinθ eikrcosθ dθ)
Erland
#7
Jan4-13, 06:29 PM
P: 345
Quote Quote by mathman View Post
It is a short hand notation to indicate that the integral is 3 dimensional, dx would be 1 dimensional.
Ok, I did never see this notation before. Now, I got it:

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.
LastOneStanding
#8
Jan4-13, 09:33 PM
P: 718
Quote Quote by tiny-tim View Post
Hi LastOneStanding!


no, only one k comes from the conversion, the other k cancels with the k in the denominator

(d3k = ksinθ dkdθdφ, and so you have to ∫ ksinθ eikrcosθ dθ)
I don't follow...the spherical volume element is [itex]d^3x = r^2 \sin\theta dr d\theta d\phi[/itex]. For an integral in k-space, this means a factor of k2. What do you mean "the other k cancels with the k in the denominator"?
LastOneStanding
#9
Jan4-13, 09:39 PM
P: 718
Quote Quote by Erland View Post
Ok, I did never see this notation before. Now, I got it:

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.
As I said in the original question, I understand the part of the step that comes from converting to spherical coordinates. So, unfortunately it is precisely the part of the calculation you suppressed by saying "it is easy to integrate..." which is what I'm stuck on.
LastOneStanding
#10
Jan4-13, 09:46 PM
P: 718
Oh, bother. Undone by a simple u-subsitution, sigh. Thank you, Erland, I see how the rest of the integration goes.


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