Given is this operation
$\left((\mathbb{1}-\hat{r}\hat{r})\cdot \nabla\right) \cdot v(r) = \left(\nabla-\hat{r} (\hat{r} \cdot \nabla)\right) \cdot v(r)$,
where
$r=(x,y,z) \in \mathbb{R}^3$, $\hat{r}$ is a unitvector, with $|\hat{r}|=1$,
$\nabla=(\partial_x, \partial_y, \partial_z)$ is the nabla operator, and
$v(r): \mathbb{R}^3 \rightarrow \mathbb{R}^3$, a vector field.
Surface gradient
$(\mathbb{1}-\hat{r}\hat{r})\cdot \nabla$ is also called the surface gradient, where the operator $(\mathbb{1}-\hat{r}\hat{r})\cdot a$ gives the perpendicular component of the vector $a$ with respect to the vector $\hat{r}$.
Questions
Now I have to questions:
- What does this actually calculate?
- Can this be parameterised on the unit sphere surface (for example in terms of spherical coordinates)?
Can you say, it calculates the divergence part on the surface of the sphere? And since it is projected onto the surface, I'd somehow expect, that it should be parameterisable with two angles (spherical coordinates). But at the moment, I'm not sure how to proceed.
Thanks in advance.
For the second part of the question, this can be easily solved by writing the nabla operator in terms of spherical coordinates (as JacobCheverie suggested),
$\nabla = \hat{e}_r \partial_r + \hat{e}_\theta \frac{1}{r} \partial_\theta + \hat{e}_\phi \frac{1}{r \sin(\theta)} \partial_\phi$.
Combined with the unit vector in spherical coordinates $\hat{r}=\hat{e}_r=(\sin(\theta) \cos(\phi), \sin(\theta) \sin(\phi), \cos(\theta))$, this gives the answer with respect to the parametrisation.