Editing Helmholtz decomposition (section)

=== Derivation ===

{{math proof| proof =
Suppose we have a vector function <math>\mathbf{F}(\mathbf{r})</math> of which we know the curl, <math>\nabla\times\mathbf{F}</math>, and the divergence, <math>\nabla\cdot\mathbf{F}</math>, in the domain and the fields on the boundary. Writing the function using the [[delta function]] in the form
<math display="block">\delta^3(\mathbf{r}-\mathbf{r}')=-\frac 1 {4\pi} \nabla^2 \frac{1}{|\mathbf{r}-\mathbf{r}'|}\, ,</math>
where <math>\nabla^2</math> is the [[Laplacian]] operator, we have

<math display="block">\begin{align}
\mathbf{F}(\mathbf{r}) &= \int_V \mathbf{F}\left(\mathbf{r}'\right)\delta^3 (\mathbf{r}-\mathbf{r}') \mathrm{d}V' \\
&=\int_V\mathbf{F}(\mathbf{r}')\left(-\frac{1}{4\pi}\nabla^2\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\mathrm{d}V' \end{align}</math>

Now, changing the meaning of <math>\nabla^2</math> to the [[vector Laplacian]] operator (we have the right to do so because this laplacian is with respect to <math>\mathbf{r}</math> therefore it sees the vector field <math>\mathbf{F}(\mathbf{r'})</math> as a constant), we can move <math>\mathbf{F}(\mathbf{r'})</math> to the right of the<math>\nabla^2</math>operator.

<math display="block">\begin{align}\mathbf{F}(\mathbf{r})&=\int_V-\frac{1}{4\pi}\nabla^2\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V' \\
&=-\frac{1}{4\pi}\nabla^2 \int_V \frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V' \\
&=-\frac{1}{4\pi}\left[\nabla\left(\nabla\cdot\int_V\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)-\nabla\times\left(\nabla\times\int_V\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right] \\
&= -\frac{1}{4\pi} \left[\nabla\left(\int_V\mathbf{F}(\mathbf{r}')\cdot\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)+\nabla\times\left(\int_V\mathbf{F}(\mathbf{r}')\times\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right] \\
&=-\frac{1}{4\pi}\left[-\nabla\left(\int_V\mathbf{F}(\mathbf{r}')\cdot\nabla'\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)-\nabla\times\left(\int_V\mathbf{F} (\mathbf{r}')\times\nabla'\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right]
\end{align}</math>

where we have used the vector Laplacian identity:
<math display="block">\nabla^{2}\mathbf{a}=\nabla (\nabla\cdot\mathbf{a})-\nabla\times (\nabla\times\mathbf{a}) \ ,</math>

differentiation/integration with respect to <math>\mathbf r'</math>by <math>\nabla'/\mathrm dV',</math> and in the last line, linearity of function arguments:
<math display="block"> \nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}=-\nabla'\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\ .</math>

Then using the vectorial identities

<math display="block">\begin{align}
\mathbf{a}\cdot\nabla\psi &=-\psi(\nabla\cdot\mathbf{a})+\nabla\cdot (\psi\mathbf{a}) \\
\mathbf{a}\times\nabla\psi &=\psi(\nabla\times\mathbf{a})-\nabla \times (\psi\mathbf{a})
\end{align}</math>

we get
<math display="block">\begin{align}
\mathbf{F}(\mathbf{r})=-\frac{1}{4\pi}\bigg[
&-\nabla\left(-\int_{V}\frac{\nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'+\int_{V}\nabla'\cdot\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)
\\&
-\nabla\times\left(\int_{V}\frac{\nabla'\times\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'
- \int_{V}\nabla'\times\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\bigg].
\end{align}</math>

Thanks to the [[divergence theorem]] the equation can be rewritten as

<math display="block">\begin{align}
\mathbf{F} (\mathbf{r}) 
&=
-\frac{1}{4\pi}
\bigg[
    -\nabla\left(
        -\int_{V}
        \frac{
            \nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)
        }{
            \left|\mathbf{r}-\mathbf{r}'\right|
        } \mathrm{d}V'
        +
        \oint_{S}\mathbf{\hat{n}}'\cdot
        \frac{
            \mathbf{F}\left(\mathbf{r}'\right)
        }{
            \left|\mathbf{r}-\mathbf{r}'\right|
        }\mathrm{d}S'
    \right)
\\
&\qquad\qquad
-\nabla\times\left(\int_{V}\frac{\nabla'\times\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'
-\oint_{S}\mathbf{\hat{n}}'\times\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}S'\right)
\bigg]
\\
&= 
-\nabla\left[
  \frac{1}{4\pi}\int_{V}
  \frac{
    \nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)
  }{\left|
    \mathbf{r}-\mathbf{r}'
  \right|}
  \mathrm{d}V' 
  - 
  \frac{1}{4\pi} 
  \oint_{S}\mathbf{\hat{n}}' \cdot
  \frac{
    \mathbf{F}\left(\mathbf{r}'\right)
  }{
     \left|
       \mathbf{r}-\mathbf{r}'
     \right|
   }
 \mathrm{d}S'
\right]
\\
&\quad
+
\nabla\times
\left[
  \frac{1}{4\pi}\int_{V}
  \frac{
    \nabla '\times\mathbf{F}\left(\mathbf{r}'\right)
  }{
     \left|
       \mathbf{r}-\mathbf{r}'
     \right|
  }
  \mathrm{d}V' 
  -
  \frac{1}{4\pi}\oint_{S}
  \mathbf{\hat{n}}'
  \times
  \frac{
    \mathbf{F}\left(\mathbf{r}'\right)
  }{
    \left|
      \mathbf{r}-\mathbf{r}'
    \right|
   }
  \mathrm{d}S'
\right]
\end{align}</math>

with outward surface normal <math> \mathbf{\hat{n}}' </math>.

Defining

<math display="block">\Phi(\mathbf{r})\equiv\frac{1}{4\pi}\int_{V}\frac{\nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'-\frac{1}{4\pi}\oint_{S}\mathbf{\hat{n}}'\cdot\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}S'</math>
<math display="block">\mathbf{A}(\mathbf{r})\equiv\frac{1}{4\pi}\int_{V}\frac{\nabla'\times\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'-\frac{1}{4\pi}\oint_{S}\mathbf{\hat{n}}'\times\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}S'</math>

we finally obtain
<math display="block">\mathbf{F}=-\nabla\Phi+\nabla\times\mathbf{A}.</math>
}}