Laplacian of a vector $\Delta\cdot b$, where $b=(-\sin\phi , \cos\phi , 0)$

Question

I have a very quick question. How do I calculate the Laplacian of this vector:

$$\Delta\cdot b$$

where

$$b=(-\sin\phi , \cos\phi , 0)$$

Thanks.

Answer 1

So the vector is obviously dependent on the polar coordinate $\phi \in [0,2\pi)$. that is if you simple take the zylinder coordinates as: \begin{align} x = r cos\phi,\quad y = rsin\phi,\quad z = z\quad (r,\phi,z) \in [0,\infty)\times [0,2\pi)\times \mathbb{R} \end{align} where $r = \sqrt{x^2+y^2+z^2}$ and use the chain rule in the first step and the coordinate $x$ and get: \begin{align} \frac{\partial v(r,\phi,z) }{\partial x} = \frac{\partial v(r,\phi,z) }{\partial r}\frac{\partial r(x,y,z) }{\partial x} + \frac{\partial v(r,\phi,z) }{\partial z}\frac{\partial z }{\partial x} + \frac{\partial v(r,\phi,z) }{\partial \phi}\frac{\partial \phi(x,y,z) }{\partial x} \end{align} Applying this again should give you the solution, which you can compute with some motivation.