So oblate spheroidal co-ordinates are defined as:
$$x = \cosh R \cosθ \cos φ$$ $$y = \cosh R \cosθ \sin φ$$ $$z = \sinh R \sin θ .$$
To show the coordinate surfaces for $R$, $\theta,\phi$ are orthogonal, I've been trying to rewrite the coordinates in terms of $x$, $y$, $z$, and then taking the gradient, and then showing $ \nabla f\cdot\nabla g=0$ where f and g are the surfaces of constant coordinate. This is trivial for $\phi$, however, it is much harder for $R$ and $\theta$.
Is there an easier way of showing this?
Any help/hints would be greatly appreciated. Thanks.
It is not necessary to rewrite the coordinates in terms of $x,y,z$. Let's use vector notation: $$\vec r(R,\theta,\varphi)=\begin{pmatrix} \cosh R\cos \theta \cos \varphi \\ \cosh R\cos \theta \sin \varphi \\ \sinh R\sin \theta \end{pmatrix} $$ It is trivial to compute the partial derivative vectors $\vec r_R$, $\vec r_\theta$, $\vec r_\varphi$: $$\vec r_R = \begin{pmatrix} \sinh R\cos \theta \cos \varphi \\ \sinh R\cos \theta \sin \varphi \\ \cosh R\sin \theta \end{pmatrix},\quad \vec r_\theta = \begin{pmatrix} -\cosh R\sin \theta \cos \varphi \\ -\cosh R\sin \theta \sin \varphi \\ \sinh R\cos \theta \end{pmatrix}, \\ \vec r_\varphi = \begin{pmatrix} -\cosh R\cos \theta \sin \varphi \\ \cosh R\cos \theta \cos \varphi \\ 0 \end{pmatrix} $$ It is also easy to check that these three vectors are mutually orthogonal: some mental arithmetics with dot products. Therefore, the cross-product of any two of these vectors is parallel to the third one. Consequently, the three possible nontrivial cross-products $\vec r_R \times \vec r_\theta$, $\vec r_\theta\times \vec r_\varphi$, and $\vec r_\varphi\times \vec r_R$ are mutually orthogonal.
These cross-products are normal vectors to the coordinate surfaces $\varphi=c$, $R=c$, and $\theta=c$, respectively. (Taking cross-product of two tangent vectors gives a normal). This concludes the proof.