Fourier transform of a function in spherical coordinates

[redtree]

TL;DR

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 $f(r) = \mathscr{F}^{-1}[\hat{f}(\varrho)]$?