Editing Helmholtz decomposition (section)

=== Derivation from the Fourier transform ===
Note that in the theorem stated here, we have imposed the condition that if <math>\mathbf{F}</math> is not defined on a bounded domain, then <math>\mathbf{F}</math> shall decay faster than <math>1/r</math>. Thus, the [[Fourier transform]] of <math>\mathbf{F}</math>, denoted as <math>\mathbf{G}</math>, is guaranteed to exist. We apply the convention
<math display="block">\mathbf{F}(\mathbf{r}) = \iiint \mathbf{G}(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{r}} dV_k </math>

The Fourier transform of a scalar field is a scalar field, and the Fourier transform of a vector field is a vector field of same dimension.

Now consider the following scalar and vector fields:
<math display="block">\begin{align}
G_\Phi(\mathbf{k}) &= i \frac{\mathbf{k} \cdot \mathbf{G}(\mathbf{k})}{\|\mathbf{k}\|^2} \\
\mathbf{G}_\mathbf{A}(\mathbf{k}) &= i \frac{\mathbf{k} \times \mathbf{G}(\mathbf{k})}{\|\mathbf{k}\|^2}  \\ [8pt]
\Phi(\mathbf{r}) &= \iiint G_\Phi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k \\
\mathbf{A}(\mathbf{r}) &= \iiint \mathbf{G}_\mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k
\end{align} </math>

Hence

<math display="block">\begin{align}
\mathbf{G}(\mathbf{k}) &= - i \mathbf{k} G_\Phi(\mathbf{k}) + i \mathbf{k} \times \mathbf{G}_\mathbf{A}(\mathbf{k}) \\ [6pt]
\mathbf{F}(\mathbf{r}) &= -\iiint i \mathbf{k} G_\Phi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k + \iiint i \mathbf{k} \times \mathbf{G}_\mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k \\
&= - \nabla \Phi(\mathbf{r}) +  \nabla \times \mathbf{A}(\mathbf{r})
\end{align}</math>