The essence is that, modulo subtleties, any vector field \(\vect{P}\) can be expressed as
the sum of a pure divergence and a pure curl:
\[\begin{gather*}
\vect{P} = -\vect{\nabla}\phi + \vect{\nabla}\times \vect{A}.
\end{gather*}\]
This decomposition is not unique, and is invariant under the following transformation:
\[\begin{gather*}
\phi \rightarrow \phi + \delta \phi, \qquad
\vect{A} \rightarrow \vect{A} + \delta\vect{A} + \vect{\nabla}\varphi
\end{gather*}\]
where
\[\begin{gather*}
\vect{\nabla}\times \delta\vect{A} = \vect{\nabla}\delta\phi,
\end{gather*}\]
and \(\varphi\) is an arbitrary scalar field. Taking the divergence of this equation
gives
\[\begin{gather*}
\nabla^2\delta\phi = 0,
\end{gather*}\]
so \(\delta \phi\) is a harmonic function. To see that this works, simply effect the transformation
\[\begin{gather*}
\delta\vect{P} = -\vect{\nabla}\delta\phi + \vect{\nabla}\times(\delta\vect{A} +
\vect{\nabla} \varphi)\\
= \underbrace{-\vect{\nabla}\delta\phi + \vect{\nabla}\times\delta\vect{A}}_{0} +
\underbrace{\vect{\nabla}\times(\vect{\nabla} \varphi)}_{0}.
\end{gather*}\]
This decomposition allows one to express \(\vect{P}\) in terms of a distribution of
sources \(s\) and currents \(\vect{c}\):
\[\begin{gather*}
\vect{\nabla}\cdot\vect{P} = s = -\nabla^2 \phi, \qquad
\vect{\nabla}\times\vect{P} = \vect{c} = \vect{\nabla}\times(\vect{\nabla}\times\vect{A}).
\end{gather*}\]
See Helmholtz Decomposition for details.
Spherical Coordinates
Our goal is to get quickly to the formula for things like the gradient and Laplacian in
spherical coordinates.
\[\begin{gather*}
\vect{r} = \begin{pmatrix}
x\\y\\z
\end{pmatrix}
=
\begin{pmatrix}
r\sin\theta\cos\phi\\
r\sin\theta\sin\phi\\
r\cos\theta
\end{pmatrix},\qquad
r\in[0, \infty), \qquad
\theta \in [0, \pi], \qquad
\phi \in [0, 2\pi).
\end{gather*}\]
One easy approach is to compute the Jacobian matrix for the coordinate transformation
\[\begin{gather*}
\d{\vect{r}} = \mat{J}\cdot
\begin{pmatrix}
\d{r}\\
\d{\theta}\\
\d{\phi}
\end{pmatrix},\qquad
\mat{J} = \begin{pmatrix}
\sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi\\
\sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi\\
\cos\theta & -r\sin\theta & 0
\end{pmatrix}.
\end{gather*}\]
The volume element follows from the determinant of this matrix
\[\begin{gather*}
\d^3\vect{r} = \d{x}\d{y}\d{z} = \det(\mat{J})\;\d{r}\d{\theta}\d{\phi}
= r^2\sin\theta \; \d{r}\d{\theta}\d{\phi}
= -r^2\d{r}\;\d{(\cos\theta)}\;\d{\phi}.
\end{gather*}\]
Similarly, the metric tensor is
\[\begin{gather*}
\mat{g} = \mat{J}^T\mat{J} = \begin{pmatrix}
1 & 0 & 0\\
0 & r^2 & 0\\
0 & 0 & r^2\sin^2\theta
\end{pmatrix},
\end{gather*}\]
so that
\[\begin{gather*}
\d{\vect{r}}^T\d{\vect{r}} =
\begin{pmatrix}
\d{r} &
\d{\theta} &
\d{\phi}
\end{pmatrix}
\mat{g}
\begin{pmatrix}
\d{r}\\
\d{\theta}\\
\d{\phi}
\end{pmatrix}.
\end{gather*}\]
