I'm attempting to take the Fourier transform of the following function: \begin{equation} V(\mathbf{r}) = \begin{cases} V_0 & r<r_0 \\ 0 & r>r_0 \end{cases} \end{equation} My professor recommended I "choose the direction of $\mathbf{k}$ to be along the $z$-axis". In trying to understand what this means, I found this. I can solve the integral given, but I would like a hand in understanding how to perform this rotation and why it changes the $e^{-i\mathbf{k}\cdot\mathbf{r}}$ to $e^{-ik\cos{(\theta)}\mathbf{r}}$.
For reference, the answer to this FT is: \begin{equation} \tilde{V}(\mathbf{k}) = V_0 \dfrac{4\pi r_0^2}{k} j_1 (kr_0) \end{equation} Where $j_1(kr_0)$ is the first-order spherical Bessel function of the first kind.
Thanks!
You need to integrate $e^{-i\mathbf k\cdot \mathbf r}$ over the volume of the sphere with radius $r_0$. Due to the spherical symmetry of the problem, you can, in principle, rotate your reference frame (centered at the center of the sphere) any way you want, and the calculation should not change. You can write the scalar product in the exponent as $\mathbf k\cdot\mathbf r=kr\cos\alpha$, where $\alpha$ is the angle between $\mathbf k$ and $\mathbf r$. Why would you choose your $\hat z$ axis to point along $\mathbf k$? Because in this case $\alpha=\theta$. If you choose any other direction, your $\alpha$ will be a function of both $\theta$ and $\phi$. Your calculations will be more involved: you would just need to do a change of variable. But you will get the same answer in the end.