How Does Changing to Polar Coordinates Simplify This Physics Integral?

[VantagePoint72]

I'm reading some course notes for a physics class that contain the following step in a derivation:

\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x}}}{\vec{x}^2 + m^2}
Changing to polar coordinates, and writing \vec{k}\cdot\vec{x}=kr \cos\theta, we have:
\phi(\vec{x}) = \frac{1}{(2\pi)^2} \int_0^\infty dk \frac{k^2}{k^2 + m^2}\frac{2\sin kr}{kr}

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.

 

[Erland]

\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x}}}{\vec{x}^2 + m^2}

This seems strange. How can it be meaningful to integrate w.r.t $d^3k$ ?

 

[VantagePoint72]

Oh dear, you're right of course. I made a transcription error, it should be:
\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x} }}{\vec{k}^2 + m^2}

It's a Fourier transform. Now that it's fixed, can you see how to proceed?

 

[Erland]

Oh dear, you're right of course. I made a transcription error, it should be:
\phi(\vec{x}) = \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k}\cdot\vec{x} }}{\vec{k}^2 + m^2}

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]

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]

Hi LastOneStanding!

… 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 e^ikrcosθ dθ)

 

[Erland]

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.

 

[VantagePoint72]

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 e^ikrcosθ dθ)

I don't follow...the spherical volume element is d^3x = r^2 \sin\theta dr d\theta d\phi. 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"?

 

[VantagePoint72]

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.

 

[VantagePoint72]

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