# Fourier Transformation

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

### Linearity

To show that the FT is linear we need to prove $$F[a\cdot f(t)+b\cdot g(t)]=a\cdot F[f(t)]+b\cdot F[g(t)]$$. Therefore we will use a simple example:

\begin{eqnarray*} FT\left[aE_{1}(t)+bE_{2}(t)\right]&=&\intop_{-\infty}^{\infty}\left[aE_{1}(t)+bE_{2}(t)\right]e^{i\omega t}dt\\&=&a\intop_{-\infty}^{\infty}E_{1}(t)e^{i\omega t}dt+b\intop_{-\infty}^{\infty}E_{2}(t)e^{i\omega t}dt\\&=&aFT\left[E_{1}(t)\right]+bFT\left[E_{2}(t)\right] \end{eqnarray*}

### Translation

Here we consider an electric field with an offset in time. The Fourier transformation leads to:

\begin{eqnarray*} FT\left[E(t-t_{0})\right](\omega)&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}dt\ E(t-t_{0})e^{i\omega t}\\&=&\frac{1}{2\pi}e^{i\omega t_{0}}\int_{-\infty}^{\infty}dt\ E(t-t_{0})e^{i\omega(t-t_{0})} \end{eqnarray*}

Now substitute $$t^{\prime}=t-t_{0}$$, which leads to:

\begin{eqnarray*}FT\left[E(t-t_{0})\right](\omega)&=&\frac{1}{2\pi}e^{i\omega t_{0}}\int_{-\infty}^{\infty}dt^{\prime}\ E(t^{\prime})e^{i\omega t^{\prime}}\\&=&\frac{1}{2\pi}e^{i\omega t_{0}}\widetilde{E}(\omega)  \end{eqnarray*}

### Modulation

How you can easy check:

$FT\left[e^{i\omega_{0}t}E(t)\right](\omega)=\widetilde{E}(\omega-\omega_{0})$

### Derivation Rules

Going into Fourier space, we just have to replace derivatives via:

$\frac{\partial}{\partial t}\longleftrightarrow-i\omega~\mathrm{and}~\frac{\partial}{\partial\mathbf{r}}\longleftrightarrow i\mathbf{k}.$

## How to show that $$\frac{\partial}{\partial t}\longrightarrow-i\omega$$

In order to show that $$\frac{\partial}{\partial t}\longrightarrow-i\omega$$, we will start with the inverse Fourier transform for $$\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]$$, obtaining:

\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt&_{P.I.}^{=}&\frac{1}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}|_{-\infty}^{\infty}-\frac{i\omega}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt\\&=&-\frac{i\omega}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt=-i\omega\mathbf{\overline{E}}\left(\mathbf{r},\omega\right) \end{eqnarray*}

To obtain a general rule for the $$n^{\textrm{th}}$$-derivative, we perform the second derivative:

\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial^{2}}{\partial t^{2}}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt&=&\frac{1}{2\pi}\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}|_{-\infty}^{\infty}-\frac{i\omega}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial}{\partial t}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt\\&=&-\frac{i\omega}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}|_{-\infty}^{\infty}+\frac{\omega^{2}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt=\omega^{2}\mathbf{\overline{E}}\left(\mathbf{r},\omega\right) \end{eqnarray*}

Hence, we get for the $$n^{\textrm{th}}$$-derivative:

$\frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\mathbf{\frac{\partial^{n}}{\partial t^{n}}E}\left(\mathbf{r},t\right)\right]e^{-i\omega t}dt=\frac{\left(-i\omega\right)^{n}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\omega t}dt=\left(-i\omega\right)^{n}\mathbf{\overline{E}}\left(\mathbf{r},\omega\right)$

with $$n\in\mathbb{N}$$

## How to show that $$\frac{\partial}{\partial\mathbf{r}}\longrightarrow i\mathbf{k}$$

To show that $$\frac{\partial}{\partial\mathbf{r}}\longrightarrow i\mathbf{k}$$ we will use the same strategy as above. Starting with the Fourier transform for $$\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]$$ we obtain:

$\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}&_{P.I.}^{=}&\frac{1}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}|_{-\infty}^{\infty}+\frac{i\mathbf{k}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}\\&=&\frac{i\mathbf{k}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}=i\mathbf{k}\mathbf{E}\left(\mathbf{k},t\right) \end{eqnarray*}$

Again, the second order derivative:

\begin{eqnarray*} \frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial^{2}}{\partial\mathbf{r}^{2}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}&_{P.I.}^{=}&\frac{1}{2\pi}\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}|_{-\infty}^{\infty}+\frac{i\mathbf{k}}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial}{\partial\mathbf{r}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}\\&=&\frac{i\mathbf{k}}{2\pi}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}|_{-\infty}^{\infty}-\frac{\mathbf{k^{2}}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}=\mathbf{-k^{2}}\mathbf{E}\left(\mathbf{k},t\right) \end{eqnarray*}

We find analogously the general form:

$\frac{1}{2\pi}\intop_{-\infty}^{\infty}\left[\frac{\partial^{n}}{\partial\mathbf{r}^{n}}\mathbf{E}\left(\mathbf{r},t\right)\right]e^{-i\mathbf{kr}}d\mathbf{k}=\frac{\left(i\mathbf{k}\right)^{n}}{2\pi}\intop_{-\infty}^{\infty}\mathbf{E}\left(\mathbf{r},t\right)e^{-i\mathbf{kr}}d\mathbf{k}=\left(i\mathbf{k}\right)^{n}\mathbf{E}\left(\mathbf{k},t\right)$

Hence, we can write in general:

$\frac{\partial^{n}}{\partial t^{n}}\longleftrightarrow(-i\omega)^{n}~\mathrm{and}~\frac{\partial^{n}}{\partial\mathbf{r}^{n}}\longleftrightarrow(i\mathbf{k})^{n}.$

Any linear partial differential equation transforms to an algebraic equation or set of equations in Fourier space, which can be solved conveniently. The general solution is time and space can be found by superposition and inverse Fourier transform.

### Symmetry

Here consider the integral:

$\left(\widetilde{E}(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}E(t)e^{i\omega t}dt\right)^{\star}$

Take the complex conjugate inside the brackets, obtaining:

$\widetilde{E}^{\star}(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}E(t)e^{i(-\omega)t}dt=\widetilde{E}(-\omega)$

### Parseval Theorem

\begin{eqnarray*}\int\limits _{-\infty}^{\infty}|f(t)|^{2}dt&=&\int\limits _{-\infty}^{\infty}|f(\omega)|^{2}d\omega\\&=&\int\limits _{-\infty}^{\infty}\left[\int\limits _{-\infty}^{\infty}f(t)e^{i\omega t}dt\int\limits _{-\infty}^{\infty}f^{*}(t')e^{-i\omega t'}dt'\right]d\omega\\&=&\int\limits _{-\infty}^{\infty}\left[\int\limits _{-\infty}^{\infty}\int\limits _{-\infty}^{\infty}f(t)f^{*}(t')e^{i\omega\left(t-t'\right)}dtdt'\right]d\omega\\&=&\int\limits _{-\infty}^{\infty}\int\limits _{-\infty}^{\infty}f(t)f^{*}(t')\delta\left(t-t'\right)dtdt'\\&=&\int\limits _{-\infty}^{\infty}f(t)f^{*}(t)dt \end{eqnarray*}.

Therefore:

$\int \limits_{-\infty}^{\infty}|f(t)|^{2}dt=\int\limits _{-\infty}^{\infty}|f(\omega)|^{2}d\omega$