When dealing with Maxwell's equations, we often need to switch between \((\mathbf{r},t)\)- and \((\mathbf{k},\omega)\)-space, i.e. between space-time and Fourier domain. This is advantageous, since derivatives in space and time are simple products in Fourier domain.

The Fourier transformation, or often short Fourier transform, of the electric field is given by:

\begin{eqnarray*}\mathbf{E}\left(\mathbf{r},t\right)&=&\intop_{-\infty}^{\infty}\mathbf{\overline{E}}\left(\mathbf{r},\omega\right)e^{-i\omega t}d\omega\\\mathbf{\overline{E}}\left(\mathbf{r},\omega\right)&=&\frac{1}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{i\omega t}dt \end{eqnarray*} and \begin{eqnarray*} \mathbf{E}\left(\mathbf{r},t\right)&=&\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{k},t\right)e^{i\mathbf{kr}}d\mathbf{r}\\\mathbf{E}\left(\mathbf{k},t\right)&=&\frac{1}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}\end{eqnarray*}

Some Properties Of Fourier Transformation

Fourier Transformation of the Heaviside Function

The Heaviside step function is very important in physics. It often models a sudden switch-on phenomenon and is therefore present in a lot of integrals. For example, the derivation of the Kramers-Kronig Relations can be significantly simplified once we know the Fourier-Transform \(\bar{\theta}(\omega)\) of the Heaviside function \(\theta(t)\). Although the function appears to be quite simple, the calculation of its Fourier transform can be quite challenging. Let's find out!

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