The Maxwell equation and the Fourier Spectrum of an Electric Field

1. Mar 8, 2014

Gabriel Maia

Hi. I have an electric field E(r) which can be equivalently characterized by its Fourier spectrum $\tilde{E}$(k) through

E(r)$\propto$$\int$$\tilde{E}$(k)exp[ik$\cdot$r]dk

The Maxwell equation states that in a homogeneous and isotropic medium

∇$\cdot$E=0

So, applying this equation to my Fourier representation of the electric field I'm supposed to find

k$\cdot$$\tilde{E}$(k)=0

Now... doesn't $\tilde{E}$(k) have components in the k-space? I was under the impression its components were ($\tilde{E}$$_{kx}$,$\tilde{E}$$_{ky}$,$\tilde{E}$$_{kz}$)

So how come we can apply the divergence operator (in the coordinates space) on a function on the k-space?

Thank you

2. Mar 8, 2014

Einj

When you apply the divergence operator (as you said, with respect to the space coordinates) you are doing that on the field $\vec E(\vec x)$ which is a function of the space. However, it can be expressed in term of its Fourier transform, such that the coordinate dependence is shifted to the exponential inside the integral. Therefore (keep in mind that the derivative with respect with the coordinate can be brought inside the integrale with respect to the wave number):

$$0=\nabla\cdot \vec E(\vec x)=\sum_{i=1}^3\frac{\partial}{\partial x_i}E_i(\vec x)=\sum_{i=1}^3\int d^3k \tilde E_i(\vec k)\frac{\partial}{\partial x_i}e^{i\vec k\cdot\vec x}=\sum_{i=1}^3\int d^3k ik_i\tilde E_i(\vec k)e^{i\vec k\cdot\vec x},$$
and so $\sum_{i=1}^3k_i\tilde E_i(\vec k)=\vec k\cdot\vec{\tilde E}(\vec k)=0$.