# Fourier transform of a function in spherical coordinates

• I
• redtree
In summary, the conversation discusses the relationship between Fourier conjugates in the spherical basis for two functions, where one is the Fourier transform of the other. The conversation also includes a coordinate transformation from Cartesian to spherical coordinates and addresses mistakes made in the transformation. Finally, it is mentioned that if a function's Fourier transform is independent of certain variables, then the function itself is also independent of those variables.

#### 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?

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

##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##

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

\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##

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 said:

\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?

These equations

redtree said:

\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
jasonRF said:
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!

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}

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)]##?