How to calculate spherical coordinate components of dipole field? I understand well enough how to calculate the radial and tangential components in spherical coordinates at a point due to a magnetic dipole field using the magnetic potential gradient ($\overrightarrow{B}=\nabla\times \overrightarrow{W}$). Is it possible to calculate radial ($B_r=\frac{2\mu_om\cos\theta}{4\pi r^3}$) and tangential ($B_\theta=\frac{\mu_om\sin\theta}{4\pi r^3}$) components in spherical coordinates without using the magnetic potential? I.e. I would like to know if it is possible to demonstrate these components in spherical coordinates starting out from the magnetic field due to a monopole ($B=\frac{\mu_oQ}{4\pi r^2}$).
 A: $\def\m{\mu}
\def\p{\pi}
\def\th{\theta}
\def\vm{\mathbf{m}}
\def\vr{\mathbf{r}}
\def\vs{\mathbf{s}}
\def\VB{\mathbf{B}}
\def\ur{\mathbf{\hat{r}}}
\def\uz{\mathbf{\hat{z}}}
\def\uth{\boldsymbol{\hat{\th}}}
\def\rd{\mathrm{d}}
\def\o{\cdot}$If we assume 
$$\VB = \frac{\m_0 g}{4\p r^3}\vr$$
for a monopole then the dipole field will be given by 
$$\VB_\rd = \frac{\m_0 g}{4\p r_+^3}\vr_+
- \frac{\m_0 g}{4\p r_-^3}\vr_-,$$
where $\vr_\pm = \vr\mp\vs/2$, $\vr$ points from the center of the dipole to the point of interest, $\vs$ points from the negative to the positive charge, and where $s/r\ll 1$.
It is a straightforward exercise to expand in small $s$ with the result
$$\VB_\rd = \frac{\m_0}{4\p r^3}(3(\vm\cdot\ur)\ur-\vm),$$
where $\vm = g\vs$ is the dipole moment. 
Letting $\vm = m\uz$ and noting that 
\begin{align*}
(3(\vm\cdot\ur)\ur-\vm)\cdot\ur &= 2m\cos\th\\
(3(\vm\cdot\ur)\ur-\vm)\cdot\uth &= m\sin\th
\end{align*}
we find
$$\VB_\rd = \frac{\m_0 m\cos\th}{2\p r^3}\ur + \frac{\m_0 m\sin\th}{4\p r^3}\uth$$
as claimed. 
Addendum
Some important steps for the expansion referred to above:
\begin{align*}
\frac{1}{r_+^3} &= (\vr_+\o\vr_+)^{-3/2} \\
&= ((\vr-\vs/2)\o(\vr-\vs/2))^{-3/2} \\
&= (r^2-\vr\o\vs+s^4/4)^{-3/2} \\
&= \frac{1}{r^3}(1-\ur\o\vs/r+s^4/(4r^2))^{-3/2} \\
&= \frac{1}{r^3}(1-(\ur\o\vs/r-s^4/(4r^2))^{-3/2} \\
&\approx \frac{1}{r^3}(1+3\ur\o\vs/(2r)). 
\end{align*}
Likewise 
$$\frac{1}{r_-^3} \approx \frac{1}{r^3}(1-3\ur\o\vs/(2r)).$$
Here we use the generalized binomial theorem, 
$(1+x)^s \approx 1+s x$ (also known as the Taylor series of $(1+x)^s$ about $x=0$). 
Use these approximations to derive the final result. 
