# Fourier transform of function which has only radial dependence

1. Oct 31, 2015

### LagrangeEuler

3d Fourier transform of function which has only radial dependence $f(r)$. Many authors in that case define
$$\vec{k} \cdot \vec{r}=|\vec{k}||\vec{r}|\cos\theta$$
where $\theta$ is angle in spherical polar coordinates.
So
$$\frac{1}{(2\pi)^3}\int\int_{V}\int e^{-i \vec{k} \cdot \vec{r}}f(r)=\frac{1}{(2\pi)^3}\int^{\infty}_0r^2f(r)dr\int^{\pi}_0\sin \theta d\theta \int^{2 \pi}_0 d\varphi e^{-ikr\cos \theta}$$
Ok function $f(r)$ does not depend on angles, but why here we have specially angle $\theta$?

Last edited: Oct 31, 2015
2. Oct 31, 2015

### Geofleur

You are missing a factor of $f(r)$ in your radial integral, and the exponential factor belongs inside the $\theta$ integral, which is important to note (see below). Also, the $\theta$ integral itself goes inside the radial integral.

The reason the angle $\theta$ is chosen is that we can always choose the coordinate system so that the wave vector $\mathbf{k}$ is aligned with the $z$ axis. That is what enables us to write $\mathbf{k}\cdot\mathbf{r} = kr\cos\theta$. Then the integral $\int_0^\pi \sin\theta e^{-ikr\cos\theta} d\theta$ is easy to evaluate via substitution.

If we had instead chosen for $\mathbf{k}$ to be along, say, the $x$ axis, then we would have had $\mathbf{k}\cdot\mathbf{r} = kr\cos\phi$. We would now have to evaluate the integral $\int_0^{2\pi} e^{-ikr\cos\phi} d\phi$, which is not so easy!

3. Oct 31, 2015

### LagrangeEuler

Thank you very match. And when I may use that is not so important which coordinate system I will use? Only in the case when $f(\vec{r})=f(r)$?

4. Oct 31, 2015

### Geofleur

If $f$ dependent on, say, both $r$ and $\theta$, that trick might still help. The $\theta$ integral would become $\int_0^{\pi} f(r,\theta) e^{-ikr\cos\theta}d\theta$. Depending on the form of $f(r,\theta)$, you might still be able to either evaluate the integral exactly or, failing that, do an asymptotic approximation via, e.g., the method of steepest descent or the method of stationary phase.

5. Oct 31, 2015

### LagrangeEuler

Yes but my real question here about this is

if $F[f(\vec{r})]=\tilde{f}(\vec{k})$ why should I put in general case that direction of $\vec{k}$ is z axis?

6. Oct 31, 2015

### Geofleur

Because, as a rule of thumb, it makes the resulting integrals easier to evaluate. I can't see any reason why it would have to, though (in the general case).

7. Oct 31, 2015

### LagrangeEuler

Yes but it is just when function is isotropic I think. I think that this relation is wrong for example

$$\frac{1}{(2\pi)^3}\int\int_{V}\int e^{-i \vec{k} \cdot \vec{r}}f(\vec{r})=\frac{1}{(2\pi)^3}\int^{\infty}_0r^2dr\int^{\pi}_0\sin \theta e^{-ikr\cos \theta} d\theta \int^{2 \pi}_0 d\varphi f(\vec{r})$$
is wrong because if argument of function $f(\vec{r})$ depends on two angles, then

$$\frac{1}{(2\pi)^3}\int^{\infty}_0r^2dr\int^{\pi}_0\sin \theta e^{-ikr\cos \theta} d\theta \int^{2 \pi}_0 d\varphi f(\vec{r}) \neq \frac{1}{(2\pi)^3}\int^{\infty}_0r^2dr\int^{\pi}_0\sin \theta d\theta \int^{2 \pi}_0 d\varphi e^{-ikr\cos \varphi}f(\vec{r})$$
will be different (if I calculate both somehow) for some general form of $f(\vec{r})$. This will be always the same to my mind only in case

$$\frac{1}{(2\pi)^3}\int^{\infty}_0r^2dr\int^{\pi}_0\sin \theta e^{-ikr\cos \theta} d\theta \int^{2 \pi}_0 d\varphi f(r) = \frac{1}{(2\pi)^3}\int^{\infty}_0r^2dr\int^{\pi}_0\sin \theta d\theta \int^{2 \pi}_0 d\varphi e^{-ikr\cos \varphi}f(r)$$

Last edited: Oct 31, 2015
8. Oct 31, 2015

### Geofleur

I'm not sure I understand your placement of the integrands. For example, if $f$ depends only on $r$, I would write something like

$\frac{1}{(2\pi)^3} \int_0^{2\pi} d\phi \int_0^{\infty} f(r) r^2 \left[ \int_0^{\pi} \sin\theta e^{-ikr\cos\theta} d\theta \right] dr$

because the integral in square brackets will end up depending on $r$. Also, note that if $\theta$ is the polar angle, then it's $\theta$ that appears in the exponential, not $\phi$.

I know that sometimes in physics we just pile everything in front of the integral signs; at the same time, that can be confusing if what variables to integrate over are at issue. If $f$ depends on both $r$ and $\theta$, then we would instead have, e.g.,

$\frac{1}{(2\pi)^3} \int_0^{2\pi} d\phi \int_0^{\infty} r^2 \left[ \int_0^{\pi} f(r,\theta) \sin\theta e^{-ikr\cos\theta} \right] dr$

Depending on the specific form that $f(r,\theta)$ takes, the integral may or may not be tractable using this "$\mathbf{k}$ along the $z$ axis" technique.

(sorry for the profuse editing)

Last edited: Oct 31, 2015
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