Laplace-Beltrami Operator non-curvilinear coordinates

[Juan Carlos]

Homework Statement

I have to find the Laplace operator asociated to the next quasi-spherical curvilinear coordinates, for z>0.

Homework Equations

\begin{align}

x&=\rho \cos\phi\nonumber\\

y&=\rho \sin \phi\nonumber\\

z&=\sqrt{r^2-\rho^2},

\end{align}

The Attempt at a Solution

I computed the metric tensor\begin{equation}

g_{ij}=\begin{bmatrix}

\dfrac{r^2}{r^2-\rho^2} & \dfrac{r\rho}{\rho^2-r^2}& 0\\

\dfrac{r\rho}{\rho^2-r^2} & \dfrac{r^2}{r^2-\rho^2} &0 \\

0 & 0 & \rho^2 \\

\end{bmatrix},
\end{equation}
and (with help of the inverse matrix and determinant) substituting in

\begin{equation}
\nabla^2=\dfrac{1}{\sqrt{|g|}}\partial_i\left(\sqrt{|g|}g^{ij}\partial_j \,\right),
\end{equation},
explicitly

\begin{equation}

\nabla^2=\partial_\rho^2+\dfrac{1}{\rho}\partial_\rho+\dfrac{2\rho}{r}\partial_{\rho r}+\partial_{r}^2+\dfrac{2}{r}\partial_r+\dfrac{1}{\rho^2}\partial_{\phi}^2
\end{equation},
is there an easy way to check this expression?
do these coordinates have a name?

 

[mark726]

Thank you for your question. I would like to provide some feedback on your solution and answer your questions.

Firstly, your computation of the metric tensor and the substitution into the Laplace operator seems correct. However, it would be helpful to also show the calculation of the inverse matrix and determinant for completeness.

To check the expression, you can try applying it to a known function and comparing the result with the known solution. This can help you catch any errors in your derivation.

As for the name of these coordinates, they are known as oblate spheroidal coordinates or prolate spheroidal coordinates, depending on the value of the eccentricity parameter. These coordinates are commonly used in problems involving oblate or prolate spheroidal geometries.

I hope this helps. Keep up the good work in your studies.A scientist