Why does $\nabla \to ik$ when you Fourier transform? I am reading a text that describes the scattering of light by a particle with dielectric constant $\epsilon$
After a bit of maths starting from Maxwell's equations they obtain:
$$\nabla (\nabla \cdot E(r)) - \nabla^2E(r)=\mu_0\omega^2\epsilon(r)\cdot E(r)$$
then say "Fourier transforming with respect to $r$ gives (replace $\nabla$ by $ik$):"
$$[k^2 \hat{I}-kk]\cdot E(k) = \mu_0 \omega^2 \int \epsilon(r) \cdot E(r)  \exp(-ik \cdot r) \,\mathrm{d}r$$
I don't understand why all $\nabla$ have turned into $ik$? Is there any way to visualize why this can be done?
 A: Perhaps the best way to understand this is to start simply:
Consider a function $f(x)$. Now, let's try to take the Fourier transform of its derivative $f'(x)$. Just use the definition of the Fourier transform: 
$$\mathscr{F}(f'(x))(k)=\frac{1}{\sqrt{2\pi}}\int dx\ e^{-ikx}f'(x) $$
and now use integration by parts (assuming $f(x) \to 0$ as $|x|\to \infty$, a typical assumption in this context).
A: Re: comments to the accepted answer.
There is a very natural interpretation: For a linear problem the Fourier transform is the same as a plane wave Ansatz. That is, guess that $f(x) = c e^{-ikx}$ for some $k$, and put this into your equation. What you are doing is looking for solutions that are sinusoidal oscillations with a well-defined wavevector $k$. Clearly for such an $f$, you can set $\nabla \mapsto -ik$.
Some functional analysis guarantees that general function, with reasonable boundary conditions like $\int f^2 < \infty$, is expressible as a linear combination of plane waves, $$f(x) = \int dk\, c(k) e^{-ikx} $$
where it is more common to denote the coefficients $c(k)$ by $\hat f(k)$ or just $f(k)$. This formula defines the inverse Fourier transform, since one can show that (up to some factors $2\pi$) $$\hat f(k) = \int dx \, e^{ikx} f(x)$$
which is the Fourier transform of $f$.
A: A Fourier transform can be done over spatial or temporal dimensions, but the end result is that $f(\mathbf{x},t)$ $\rightarrow$ $F(\mathbf{k},\omega)$.  The assumption is that the amplitude, $A$, of the signal is not dependent upon $\mathbf{k}$ or $\omega$, which allows one to say that $F(\mathbf{k},\omega)$ ~ $A \ e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}$.  Then one can see that $\nabla F(\mathbf{k},\omega)$ $\rightarrow$ $i \ \mathbf{k} \ F(\mathbf{k},\omega)$ and $\partial_{t} \ F(\mathbf{k},\omega)$ $\rightarrow$ -$i \ \omega \ F(\mathbf{k},\omega)$.
Side Note
This is often referred to as linearizing the equations or using a linear approximation of a fluctuating signal.  The distinction is used because a nonlinear wave can have $A$ = $A(\mathbf{k},\omega)$, though this is not strictly required for nonlinearity.
