Finding the Spherical Polar Fourier Transform with Variable Change

In summary: What you want here is 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 r^2~sin(\theta)...
  • #1
ognik
643
2
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?
 
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  • #2
ognik said:
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. :)
 
  • #3
topsquark said:
\(\displaystyle \int_{-\infty}^{\infty} d^3x \to \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} r^2~sin( \theta) dr~d \theta ~d \phi\).
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?
 
  • #4
ognik said:
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
 
  • #5
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.
 

FAQ: Finding the Spherical Polar Fourier Transform with Variable Change

1. What is a Fourier Transform example?

A Fourier Transform example is a mathematical method used to decompose a complex signal into its individual frequency components. It is based on the idea that any signal can be represented as a combination of simple sine and cosine waves. This method is commonly used in fields such as signal processing, image processing, and data analysis.

2. How does a Fourier Transform work?

A Fourier Transform works by taking a signal in the time domain and converting it into the frequency domain. This is done by breaking down the signal into its individual frequency components and representing them as a spectrum. The resulting spectrum shows the amplitude and phase of each frequency component present in the original signal.

3. What are some real-world applications of Fourier Transform?

Fourier Transform has a wide range of applications in various fields such as engineering, physics, and mathematics. It is commonly used in signal and image processing to remove noise, compress data, and extract useful information. It is also used in astronomy to analyze astronomical signals, in medical imaging to analyze brain function, and in music to analyze sound waves.

4. Are there different types of Fourier Transform?

Yes, there are two main types of Fourier Transform: the continuous Fourier Transform and the discrete Fourier Transform. The continuous Fourier Transform is used for signals that are continuous in time, while the discrete Fourier Transform is used for signals that are discrete in time, such as digital signals. There are also variations of these, such as the Fast Fourier Transform (FFT), which is a more efficient algorithm for computing the discrete Fourier Transform.

5. Can Fourier Transform be used for non-periodic signals?

Yes, Fourier Transform can be used for both periodic and non-periodic signals. However, for non-periodic signals, the Fourier Transform will produce a continuous spectrum, unlike the discrete spectrum produced for periodic signals. This continuous spectrum is useful in analyzing the frequency content of non-periodic signals.

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