Consider two vectors $\hat r_1$, $\hat r_2$ in a 3D Cartesian coordinate system $(O,x,y,z)$.
For $\hat r_1$, the laplacian operator could be written in spherical coordinates as \begin{equation} \hat{\nabla}_{1}^{2}=\frac{1}{r_{1}^{2}} \frac{\partial}{\partial r_{1}}\left(r_{1}^{2} \frac{\partial}{\partial r_{1}}\right)+\frac{1}{r_{1}^{2}}\left[\frac{1}{\sin \theta_{1}} \frac{\partial}{\partial \theta_{1}}\left(\sin \theta_{1} \frac{\partial}{\partial \theta_{1}}\right)+\frac{1}{\sin ^{2} \theta_{1}} \frac{\partial^{2}}{\partial \phi_{1}^{2}}\right] \end{equation} where $θ_1$ is the polar/inclination/co-latitude angle (between $\hat r_1$ and $z$-axis), $\phi_1$ is the azimuthal angle (between $\hat r_1$ and the $y$-axis). Therefore for $\hat r_2$, the Laplacian operator could be written as \begin{equation} \hat{\nabla}_{2}^{2}=\frac{1}{r_{2}^{2}} \frac{\partial}{\partial r_{2}}\left(r_{2}^{2} \frac{\partial}{\partial r_{2}}\right)+\frac{1}{r_{2}^{2}}\left[\frac{1}{\sin \theta_{2}} \frac{\partial}{\partial \theta_{2}}\left(\sin \theta_{2} \frac{\partial}{\partial \theta_{2}}\right)+\frac{1}{\sin ^{2} \theta_{2}} \frac{\partial^{2}}{\partial \phi_{2}^{2}}\right] \end{equation} My question is: how is the second Laplacian operator written as \begin{equation} \hat{\nabla}_{2}^{2}=\frac{1}{r_{2}^{2}} \frac{\partial}{\partial r_{2}}\left(r_{2}^{2} \frac{\partial}{\partial r_{2}}\right)+\frac{1}{r_{2}^{2}}\left[\frac{1}{\sin \theta_{12}} \frac{\partial}{\partial \theta_{12}}\left(\sin \theta_{12} \frac{\partial}{\partial \theta_{12}}\right)+\frac{1}{\sin ^{2} \theta_{12}} \frac{\partial^{2}}{\partial \phi_{12}^{2}}\right] \end{equation} given that $\theta_2=\theta_1+\theta_{12},\phi_2=\phi_1+\phi_{12}$?
I need a hint here.
P.S. The picture below just showed r1, but r2 is defined in the same manner, with theta12 and phi12 being the angle between r1 and r2
