The magnetic field strength, $\textbf{B}$, is related to the vector potential, $\textbf{A}$, by $ \textbf{B} = \nabla \times \textbf{A}$.
With $\textbf{A} = \frac{\mu_0}{4\pi} \frac{ \pmb{\mu} \times \pmb{\hat{r}}}{r^2}$, I need to calculate $\nabla \times \biggl ( \frac{ \pmb{\mu} \times \pmb{\hat{r}}}{r^2} \biggl )$, and one of the terms I have to deal with in the calculation is $(\pmb{\mu} \cdot \nabla) \pmb{\hat{r}}$. After some research, I've found a vector identity that says $$(\pmb{\mu} \cdot \nabla) \pmb{\hat{r}} = \frac{1}{r}[\pmb{\mu} – \pmb{\hat{r}}(\pmb{\mu} \cdot \pmb{\hat{r}}],$$
where $\pmb{\mu}$ is a constant vector.
Please can someone show me how this vector identity can be derived.
Best Answer
You can do this with components fairly simple. (Ab)using Einstein's notation we have
\begin{eqnarray} (\pmb{\mu} \cdot \nabla)\hat{\pmb{r}} &=& (\mu_i \partial_i)\left(\frac{x_j}{r} \hat{\pmb{e}}_j\right) \\ &=& \mu_i \hat{\pmb{e}}_j \left(\partial_i \frac{x_j}{r}\right) \\ &=& \mu_i \hat{\pmb{e}}_j \left(\frac{\partial_ix_j}{r}-\frac{x_j\partial_i r}{r^2}\right) \\ &=& \mu_i \hat{\pmb{e}}_j \left(\frac{\delta_{ij}}{r}-\frac{x_jx_i}{r^3}\right) \\ &=& \frac{1}{r}\mu_i \hat{\pmb{e}}_i -\frac{1}{r}\left(\mu_i \frac{x_i}{r}\right)\left(\frac{x_j \hat{\pmb{e}}_j}{r}\right) \\ &=& \frac{1}{r}\left[\pmb{\mu} - \frac{(\pmb{\mu} \cdot \hat{\pmb{r}})\hat{\pmb{r}}}{r} \right] \end{eqnarray}