0
$\begingroup$

In the context of a cosmology text, I have found the following function, called a spatial top-hat filter:

$$W_{TH,R}(r)=\dfrac{3\theta(R-r)}{4\pi R^3}$$

It is claimed that this leads to:

$$W_{TH,R}(r)=3\dfrac{\sin(kR)-(kR)\cos(kR)}{(kR)^3}$$

which I assume is the Fourier transform of the expression given above. I want to be able to reproduce this result, so I attempted the following:

$$W_{TH,R}(k)=\int d^3r\ e^{-i\vec{k}\cdot\vec{r}}\cdot W_{TH,R}(r)=\int d^3r\ e^{-i\vec{k}\cdot\vec{r}}\cdot\dfrac{3\theta(R-r)}{4\pi R^3}=$$

$$=\int_0^{2\pi}d\phi\int_0^\pi d\theta\int_0^\infty dr\ r^2\sin\theta\cdot e^{-ikr\cos\theta}\cdot\dfrac{3\theta(R-r)}{4\pi R^3}=$$

$$=2\pi\dfrac{3}{4\pi R^3}\int_0^\pi d\theta\int_0^\infty dr\ r^2\theta(R-r)\sin\theta\cdot e^{-ikr\cos\theta}=$$

$$\dfrac{3}{2R^3}\int_0^\infty dr \int_0^\pi d\theta\ r^2\theta(R-r)\sin\theta\cdot e^{-ikr\cos\theta}$$

But I cannot imagine how to proceed any further. I would like to be able to change variables as $u=\cos\theta$, but the extra $\theta$ in front of the exponential makes it seem like this would not solve anything.

Any help would be greatly appreciated!

$\endgroup$
4
  • 3
    $\begingroup$ It is not the same $\theta$, one is the Heavyside function $\endgroup$ Commented Jul 12, 2023 at 14:25
  • 2
    $\begingroup$ I may suppose that $\theta(R-r)=1$ for $r\in[0;R]$, and zero otherwise. If this supposition is correct, this theta has nothing to do with the polar angle in the spherical system of coordinates. $\endgroup$ Commented Jul 12, 2023 at 14:27
  • 1
    $\begingroup$ Assuming $r \ge 0$, you might start with the substitution $\Theta(R-r) = 1 - \Theta(r-R)$. $\endgroup$ Commented Jul 13, 2023 at 22:45
  • 1
    $\begingroup$ And you might want to use this formula for the Fourier Transform when there is n-dimensional symmetry: math.stackexchange.com/a/3029986/441161 $\endgroup$ Commented Jul 13, 2023 at 22:51

2 Answers 2

Reset to default
1
$\begingroup$

Presumably $\Theta(R-r)$ signifies the indicator function of the 3D ball of radius $R$ ( using Heaviside function notation). If so, the Fourier transform is described here:Fourier transform of the indicator of the unit ball

Then the fractional-order Bessel function that so arises can be represented using this technique: Bessel function $J_{3\over 2}(x)$

$\endgroup$
0
$\begingroup$

The Heaviside step function $\theta(R-r)$ is defined here. It is not to be confused with the integration variable $\theta$ (which you should rename in this particular calculation, e.g. $\alpha$). The Heaviside step function basically allows you to change the integration bounds of your $r$ integral if you think about it:

\begin{equation} \int_{0}^{\infty}dr\, f(r)\, \theta(R-r) = \int_{0}^{R}dr\, f(r). \end{equation}

So your integral actually turns out to actually simplify a lot:

\begin{align} \tilde{W}_{TH,R}(k) &= \,...\\ & = \frac{3}{2R^3} \int_{0}^{R}dr\, \int_{0}^{\pi}d\alpha\, r^2 \sin(\alpha)e^{-ikr\cos(\alpha)} \\ & = \frac{3}{kR^3} \int_{0}^{R}dr\, r\sin(kr) \\ & = 3 \frac{\sin(kR) - kR\cos(kR)}{(kR)^3}, \end{align}

using the properties of the Heaviside step function in the first step, your proposed substitution in the second step and integration by parts in the third step.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.