Fourier transform of a function in spherical coordinates

[redtree]

TL;DR Summary

I am trying to understand the relationship between Fourier conjugate bases in spherical coordinates

I am trying to understand the relationship between Fourier conjugates in the spherical basis. Thus for two functions ##f(\vec{x}_3)## and ##\hat{f}(\vec{k}_3)##, where

\begin{equation}

\begin{split}

\hat{f}(\vec{k}_3) &= \int_{\mathbb{R}^3} e^{-2 \pi i \vec{k}_3 \cdot \vec{x}_3} f(\vec{x}_3 d\vec{k}_3

\end{split}

\end{equation}

where ##\vec{x}_3 = [x_1,x_2,x_3]## and ##\vec{k}_3 = [k_1,k_2,k_3]##
In spherical 3-space coordinates,

\begin{equation}

\begin{split}

\hat{f}(\varrho, \xi_1, \xi_2) &= \int_{0}^{\infty} \int_{0}^{1} \int_{0}^{1/2} e^{-2 \pi i (\varrho r + \xi_1 \theta_2 + \xi_2 \theta_2)} f(r,\theta_1,\theta_2) dr d\theta_1 d\theta_2

\end{split}

\end{equation}

where ##\vec{x}_3 = [r,\theta_1,\theta_2]## and ##\vec{k}_3 = [\varrho,\xi_1,\xi_2]##
Thus, for a function ##\hat{f}\left( \big(\vec{k}_3\big)^2 \right)##, where in spherical coordinates ##\big(\vec{k}_3\big)^2 = \big( \varrho \big)^2##,

\begin{equation}

\begin{split}

\hat{f}\left( \big(\vec{k}_3\big)^2 \right) &= \hat{f}\left(\big( \varrho \big)^2 \right)

\\

&= \int_{0}^{\infty} e^{-2 \pi i \varrho r} f(r^2) dr

\end{split}

\end{equation}

such that ##\hat{f}\left( \big(\vec{k}_3\big)^2 \right)## is independent of ##\theta_1## and ##\theta_2##. Is that correct? Am I missing something?

 

[jasonRF]

How are ##\theta_1## and ##\theta_2## defined?

 

[redtree]

##0 \leq \theta_1 \leq 1/2 ##, such that ## 0 \leq 2 \pi \theta_1 \leq \pi##
## 0 \leq \theta_2 \leq 1##, such that ## 0 \leq 2 \pi \theta_2 \leq 2 \pi##

 

[jasonRF]

I’m not sure I really understand, but in any case you definitely did your coordinate transformation wrong. Show us how you did it.

 

[redtree]

\begin{equation}
\begin{split}
x_1 &= r \sin{2 \pi \theta_1} \cos{2 \pi \theta_2}
\\
x_2 &= r \sin{2 \pi \theta_1} \sin{2 \pi \theta_2}
\\
x_3 &= r \cos{2 \pi \theta_1}
\end{split}
\end{equation}
where ##r \geq 0##, ##0 \leq \theta_1, \leq \frac{1}{2}## and ##0 \leq \theta_2 \leq 1##

 

[jasonRF]

Of course your first integral should be over physical space, not wave-vector space. That integral should include (using your notation I think) ##d\vec{x}_3## which has dimensions of volume. However, when you transformed the coordinates you somehow have ##dr d\theta_1 d\theta_2## which has dimensions of length. That should clue you in that it cannot possibly be correct. Have you transformed integrals from Cartesian to spherical coordinates before?

 

[redtree]

\begin{equation}
\begin{split}
x_1 &= r \sin{2 \pi \theta_1} \cos{2 \pi \theta_2}
\\
x_2 &= r \sin{2 \pi \theta_1} \sin{2 \pi \theta_2}
\\
x_3 &= r \cos{2 \pi \theta_1}
\end{split}
\end{equation}
where ##r \geq 0##, ##0 \leq \theta_1, \leq \frac{1}{2}## and ##0 \leq \theta_2 \leq 1##

Where is the mistake in the coordinate transformation?

 

[jasonRF]

These equations

\begin{equation}
\begin{split}
x_1 &= r \sin{2 \pi \theta_1} \cos{2 \pi \theta_2}
\\
x_2 &= r \sin{2 \pi \theta_1} \sin{2 \pi \theta_2}
\\
x_3 &= r \cos{2 \pi \theta_1}
\end{split}
\end{equation}
where ##r \geq 0##, ##0 \leq \theta_1, \leq \frac{1}{2}## and ##0 \leq \theta_2 \leq 1##

Are fine, but when you used them to change the variables of integration you did most of it wrong. For example ##d\vec{x}_3## should transform (if I did the math right) to ##4\pi^2 \, r^2 \, \sin(2\pi\theta_1) \, dr \, d\theta_1 \, d\theta_2##. Also
$$
\vec{k}_3\cdot \vec{x}_3 = k_1\, r\, \sin{2 \pi \theta_1} \cos{2 \pi \theta_2} + k_2\, r\, \sin{2 \pi \theta_1} \sin{2 \pi \theta_2} + k_3\, r\, \cos{2 \pi \theta_1}
$$

This is standard stuff for changing coordinates in multiple integrals, as learned in a standard calculus sequence. Have you learned multivariable calculus?

jason

 

[redtree]

These equationsAre fine, but when you used them to change the variables of integration you did most of it wrong. For example ##d\vec{x}_3## should transform (if I did the math right) to ##4\pi^2 \, r^2 \, \sin(2\pi\theta_1) \, dr \, d\theta_1 \, d\theta_2##. Also
$$
\vec{k}_3\cdot \vec{x}_3 = k_1\, r\, \sin{2 \pi \theta_1} \cos{2 \pi \theta_2} + k_2\, r\, \sin{2 \pi \theta_1} \sin{2 \pi \theta_2} + k_3\, r\, \cos{2 \pi \theta_1}
$$

This is standard stuff for changing coordinates in multiple integrals, as learned in a standard calculus sequence. Have you learned multivariable calculus?

jason

Got it. Thanks!

 

[redtree]

such that,
\begin{equation}
\begin{split}
\hat{f}(\varrho,\xi_1,\xi_2) &= \int_{0}^{1} \int_{0}^{1/2} \int_{0}^{\infty} \text{Exp}\left[-2 \pi i \varrho r \big(\cos{2 \pi \theta_1 } \cos{2 \pi \xi_1 } + \cos{2 \pi (\theta_2 - \xi_2) } \sin{2 \pi \theta_1 } \sin{2 \pi \xi_1 }\big) \right]
\\
&4\pi^2 \, r^2 \, \sin(2\pi\theta_1) \, dr \, d\theta_1 \, d\theta_2
\end{split}
\end{equation}

 

[redtree]

Does it remain true that if ##f\left( (\vec{x}_3)^2 \right) = r^2##, then ##f(\vec{x}_3) = f(r)##, where ##f(r) = \mathscr{F}^{-1}[\hat{f}(\varrho)]##?