Editing Helmholtz decomposition (section)

=== Longitudinal and transverse fields ===
A terminology often used in physics refers to the curl-free component of a vector field as the '''longitudinal component''' and the divergence-free component as the '''transverse component'''.<ref name="stewart2011"/> This terminology comes from the following construction: Compute the three-dimensional [[Fourier transform]] <math>\hat\mathbf{F}</math> of the vector field <math>\mathbf{F}</math>. Then decompose this field, at each point '''k''', into two components, one of which points longitudinally, i.e. parallel to '''k''', the other of which points in the transverse direction, i.e. perpendicular to '''k'''. So far, we have

<math display="block">\hat\mathbf{F} (\mathbf{k}) = \hat\mathbf{F}_t (\mathbf{k}) + \hat\mathbf{F}_l (\mathbf{k})</math>
<math display="block">\mathbf{k} \cdot \hat\mathbf{F}_t(\mathbf{k}) = 0.</math>
<math display="block">\mathbf{k} \times \hat\mathbf{F}_l(\mathbf{k}) = \mathbf{0}.</math>

Now we apply an inverse Fourier transform to each of these components. Using properties of Fourier transforms, we derive:

<math display="block">\mathbf{F}(\mathbf{r}) = \mathbf{F}_t(\mathbf{r})+\mathbf{F}_l(\mathbf{r})</math>
<math display="block">\nabla \cdot \mathbf{F}_t (\mathbf{r}) = 0</math>
<math display="block">\nabla \times \mathbf{F}_l (\mathbf{r}) = \mathbf{0}</math>

Since <math>\nabla\times(\nabla\Phi)=0</math> and <math>\nabla\cdot(\nabla\times\mathbf{A})=0</math>,

we can get

<math display="block">\mathbf{F}_t=\nabla\times\mathbf{A}=\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times\mathbf{F}}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'</math>
<math display="block">\mathbf{F}_l=-\nabla\Phi=-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot\mathbf{F}}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'</math>

so this is indeed the Helmholtz decomposition.<ref name="littlejohn"/>