3D Fourier Transform - Angle between $\mathbf{k}$ and $\mathbf{r}$

Question

The definition of the Fourier transform for three dimensions is

$$\mathcal{F}[f(\mathbf{r})](\mathbf{k})=\int e^{-i\mathbf{k}\cdot \mathbf{r}}f(\mathbf{r})\,d^3 r$$

If the function $f(\mathbf{r})$ is spherically symmetric (i.e. f($\mathbf{r}$) can be written as $f(r)$, where $r=|\mathbf{r}|$), the transform can be written as

$$\mathcal{F}[f(\mathbf{r})](\mathbf{k})=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}e^{-i|\mathbf{k}|r\cos (\theta)}f(\mathbf{r}) r^2\sin(\theta) dr d\theta d\phi$$

This is what is given in my book, but I have no idea where this $\cos(\theta)$ term came from. I read this math.SE question and answer, but I still don't get why the $\cos(\theta)$ term is there.

In the triple integral, $\theta$ is designated with respect to the $z$-axis and the $\mathbf{r}$ vector. I don't see why it is also claimed that it is the angle between the $\mathbf{r}$ vector and the arbitrary $\mathbf{k}$ vector.

Could you help clear up my confusion?

Answer 1

Wow, I'm surprised nobody has answered this question. It's so simple.

In my question, $\mathbf{r}$ corresponds to the vector whose underlying space we are integrating over - emphasis on "integrating over". It doesn't matter how you choose to "orient the space" that you're integrating over, because, in a sense, you'll "get all of it in the end". Since we are free to choose any coordinate system or axes to do that integral, we can conveniently choose spherical polar coordinates with the polar axis lined up with the $\mathbf{k}$-vector. This will make $\mathbf{k}\cdot\mathbf{r}=k r \cos( \theta )$, where $\theta$ really is the angle between the integration vector $\mathbf{r}$ and our judiciously chosen polar axis.