Fourier Transform of the vector $\vec{E}(\vec{x}, t)$ and its direction

Question

In chapter 13.3 - Density Effect in Collisional Energy Loss of "Classical Eletrodynamics" by Jackson, it says that:

The problem of finding the electric field in the medium due to the incident fast particle moving with constant velocity can be solved most readily by Fourier Transforms.

This approach is easier because we assume that macroscopic dielectric field is a function of $\omega$, $\epsilon (\omega)$. But I wonder what physically means $\vec{E}(\omega, \vec{k})$, the Fourier Transform of $\vec{E}(\vec{x}, t)$.

Furthermore, if, initially, I have that $\vec{E}(\vec{x}, t)= (E_x, E_y, E_z)$ and I take the Fourier Transform, I get $\vec{E}(\omega, \vec{k}) = (E_1, E_2, E_3)$. Could anyone explain me what happened to the direction of the old vector $\vec{E}(\vec{x}, t)$?

Moreover, If I calculate $\vec{S}(\omega, \vec{k})= \vec{E}(\omega, \vec{k})\times \vec{B}(\omega, \vec{k})$ and apply the Inverse Fourier Transform, do I have that $\vec{S}(\vec{x}, t) = \vec{E}(\vec{x}, t) \times \vec{B}(\vec{x}, t)$?

Answer 1

The Fourier transform is a linear process. Therefore, it commutes with any other linear operation. What I mean by that is that we can interchange the order in which we apply these operations. Since Maxwell's equations are linear operations (provided that the medium is linear), we can apply the Fourier transform on these equations in arbitrary order.

The role of the Fourier transform is to represent the electric field for example in terms of basis functions - plane waves. These plane waves are solutions of the Helmholtz equation that follows from Maxwell's equations. So, the spectrum that we get from the Fourier transform represents the coefficient function telling us how to recover the original electric field by recombining the plane waves. So now we only need to work with the spectrum.

A vector field can be represented in terms of the individual components. When the Fourier transform is applied to a vector field, one can see it as a Fourier transform on each of the components individually. Therefore, we end up with a spectrum that is also a vector quantity.

The Poynting vector requires a product operation (the curl), which is nonlinear. When we compute the Fourier transform of the product of two functions we end up with the convolution of the spectra of the two functions. So the expression in the Fourier domain does not look the same as in space and time.