Finding the Spherical Polar Fourier Transform with Variable Change

[ognik]

Show that the 3-D FT of a radially symmetric function may be rewritten as a Fourier sin transform

i.e. $ \frac{1}{({2\pi})^{{3}_{2}}} \int_{-\infty}^{\infty}f(r)e^{ik \cdot r} \,d^3x = \frac{1}{k} \sqrt{\frac{2}{\pi}} \int_{-\infty}^{\infty} \left[ rf(r) \right] sin(kr) \,dr $

The example then changes $d^3x$ to $d^3r$ and 'write the integral of this exercise in spherical polar coordinates, make the change of variable $Cos \theta = t$

They then write : $ \frac{1}{({2\pi})^{{3}_{2}}} \int_{0}^{\infty} r^2 f(r) dr \int_{0}^{2\pi} \,d\phi \int_{-1}^{1} e^{ikrt} \,dt $

I don't know this coordinate transform, it doesn't look like the $x = r sin \theta cos \phi$ etc. that I am familiar with, could someone tell me what this transform is please? Also how does the 't' get into the exponent?

 

[topsquark]

Show that the 3-D FT of a radially symmetric function may be rewritten as a Fourier sin transform

i.e. $ \frac{1}{({2\pi})^{{3}_{2}}} \int_{-\infty}^{\infty}f(r)e^{ik \cdot r} \,d^3x = \frac{1}{k} \sqrt{\frac{2}{\pi}} \int_{-\infty}^{\infty} \left[ rf(r) \right] sin(kr) \,dr $

The example then changes $d^3x$ to $d^3r$ and 'write the integral of this exercise in spherical polar coordinates, make the change of variable $Cos \theta = t$

They then write : $ \frac{1}{({2\pi})^{{3}_{2}}} \int_{0}^{\infty} r^2 f(r) dr \int_{0}^{2\pi} \,d\phi \int_{-1}^{1} e^{ikrt} \,dt $

I don't know this coordinate transform, it doesn't look like the $x = r sin \theta cos \phi$ etc. that I am familiar with, could someone tell me what this transform is please? Also how does the 't' get into the exponent?

Your notation has some serious issues. You have r's where you should only have x's. Let's say that we have a function \(\displaystyle F( \vec{x} )\) that is radially symmetric. Then its Fourier sine transform is
\(\displaystyle \frac{1}{(2 \pi )^{3/2}} \int _{-\infty}^{\infty} F( \vec{x} ) sin(\vec{k} \cdot \vec{x})~d^3x\)

Putting this into a more radial form means we can define a \(\displaystyle f(r) = F( \vec{x} )\), \(\displaystyle \vec{k} \cdot \vec{x} = kr~cos(\theta)\), \(\displaystyle \int_{-\infty}^{\infty} d^3x \to \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} r^2~sin( \theta) dr~d \theta ~d \phi\).

So noting that the cosine part of the transform integrates out (as suggested by the problem statement) we get, for the Fourier transform:
\(\displaystyle \frac{1}{(2 \pi )^{3/2}} \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} r^2 f(r) sin(\theta)~e^{ikr~cos(\theta)}~dr~d \theta ~d \phi\)

Now, letting \(\displaystyle t = cos(\theta)\) provides a considerable simplification. First \(\displaystyle \int_0^{\pi} d \theta \to \int_{-1}^1 \frac{dt}{\sqrt{1 - t^2}}\) and \(\displaystyle sin(\theta) = \sqrt{1 - t^2}\) so your Fourier transform becomes
\(\displaystyle \frac{1}{(2 \pi )^{3/2}} \int_0^{\infty} \int_{-1}^1 \int_0^{2 \pi} r^2 f(r) \sqrt{1 - t^2} ~e^{ikrt}~dr~\frac{dt}{\sqrt{1 - t^2}} ~d \phi\)

-Dan

Addendum: If memory serves there is another mode of attack instead of using \(\displaystyle t = cos(\theta)\) and that is to expand the integrand in term of Bessel functions. In this case the function f is independent of \(\displaystyle \theta\) so we wouldn't want to do that here.

Final edit, then I'm done with it! Did I miss a 1/k factor somewhere? I'll let you sort that out. I'm feeling lazy now. :)

 

[ognik]

\(\displaystyle \int_{-\infty}^{\infty} d^3x \to \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} r^2~sin( \theta) dr~d \theta ~d \phi\).

Thanks Dan - it seems as though the book is aware of the notation issue, but deliberately left it that way ... anyway that is secondary...

But its the step above that is causing me difficulty. From $x=r sin \theta cos \phi$ I would do this:-

$dx = sin \theta cos \phi \,dr, dx = r cos \theta cos \phi \,d\theta, dx = -r sin \theta sin \phi \,d\phi $
But that's not going to give me $r^2 sin \theta \,rd \,d\theta \,d\phi $ as above?

 

[topsquark]

Thanks Dan - it seems as though the book is aware of the notation issue, but deliberately left it that way ... anyway that is secondary...

But its the step above that is causing me difficulty. From $x=r sin \theta cos \phi$ I would do this:-

$dx = sin \theta cos \phi \,dr, dx = r cos \theta cos \phi \,d\theta, dx = -r sin \theta sin \phi \,d\phi $
But that's not going to give me $r^2 sin \theta \,rd \,d\theta \,d\phi $ as above?

I would have thought you'd've seen this by now at your level.

I'm not going to derive it because I'll get it wrong somehow. I always do. What you want is the Jacobian. (Look at Examples 2 and 3 for how to do it. The rest is a bit complicated and you may want a more primary source. You should be able to find it in a Calc III level text.)

What you want here is \(\displaystyle d^3x = dx~dy~dz = \left | \frac{\partial (x, y, z)}{\partial (r, \theta, \phi )} \right | dr~d \theta ~d \phi\). That will get you the correct \(\displaystyle r^2~sin(\theta)\) factor.

-Dan

 

[ognik]

Thanks Dan, perfect - yes I had encountered this but it just didn't come to mind ...

There is a $\frac{1}{ikr}$ factor which emerges from integrating the exponent term; incidentally I left $t = cos \theta, dt= - sin \theta d\theta $ thus removing the $sin \theta$ with a u substitution.