# Fourier transform of a function in spherical coordinates

• I
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{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}

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{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}

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{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}

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?

Gold Member
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##

Gold Member
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{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}

where ##r \geq 0##, ##0 \leq \theta_1, \leq \frac{1}{2}## and ##0 \leq \theta_2 \leq 1##

Gold Member
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{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}

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?

Gold Member
These equations

\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}

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
redtree
These equations

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
Got it. Thanks!

redtree
such that,

\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}

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)]##?