$\nabla \cdot \mathbf{\delta u_{perp}} = 0$ where $\mathbf{\delta u_{perp}}$ is a function of both x and y coordinates and perpendicular to z axis. Moreover, $\delta u_{perp}$ along z axis is $0$.
I need to prove that in spherical coordinates $$\mathbf{\delta u_{perp}} = \delta u (-\sin \phi, \cos \phi)$$
I was trying : Let's consider the vector field $\mathbf{\delta u_{perp}}$ in Cartesian coordinates, where $\mathbf{\delta u_{perp}} = (\delta u_x, \delta u_y, \delta u_z)$. Given that $(\nabla \cdot \mathbf{\delta u_{perp}} = 0)$, we can express the divergence in Cartesian coordinates as:
$$\nabla \cdot \mathbf{\delta u_{perp}} = \frac{\partial \delta u_x}{\partial x} + \frac{\partial \delta u_y}{\partial y} + \frac{\partial \delta u_z}{\partial z} = 0$$
Since $(\mathbf{\delta u_{perp}})$ is only a function of $(x)$ and $(y)$ and is zero along the $(z)$-axis, the partial derivative with respect to $(z)$ is zero, i.e., $(\frac{\partial \delta u_z}{\partial z} = 0)$.
Now, expressing the vector field $(\mathbf{\delta u_{perp}})$ in spherical coordinates $((r, \theta, \phi))$. The spherical coordinates are related to Cartesian coordinates as follows:
$$x = r \sin \theta \cos \phi$$ $$y = r \sin \theta \sin \phi$$ $$z = r \cos \theta$$
Expressing the Cartesian components of $(\mathbf{\delta u_{perp}})$ in terms of spherical coordinates as $(\delta u_x = \delta u_r \sin \theta \cos \phi)$, $(\delta u_y = \delta u_r \sin \theta \sin \phi)$, and $(\delta u_z = \delta u_r \cos \theta)$, where $(\delta u_r)$ is the radial component.
Now, the divergence in spherical coordinates:
\begin{align} \nabla \cdot \mathbf{\delta u_{perp}} &= \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \delta u_r \sin \theta \cos \phi) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}(\delta u_r \sin \theta \cos \phi) \\ &\quad + \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi}(\delta u_r \sin \theta \sin \phi) \end{align}
Now, we need to evaluate these partial derivatives and set the result equal to zero to satisfy $(\nabla \cdot \mathbf{\delta u_{perp}} = 0)$. Since the original problem statement provides a specific solution, we can check if the given solution is consistent with the requirement $(\nabla \cdot \mathbf{\delta u_{perp}} = 0)$.
Given $(\nabla \cdot \mathbf{\delta u_{perp}} = 0)$, we find that:
$$\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \delta u_r \sin \theta \cos \phi) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}(\delta u_r \sin \theta \cos \phi) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi}(\delta u_r \sin \theta \sin \phi) = 0 $$
Now, how can I prove $(\mathbf{\delta u_{perp}} = \delta u (-\sin \phi, \cos \phi))$?