The Laplacian is given in any smooth local coordinates by

\begin{equation*} \Delta u = - \frac{1}{\sqrt{\det g} } \frac{\partial }{\partial x^i}\biggl(g^{ij} \sqrt{\det g} \frac{\partial u}{\partial x^j}\biggr), \end{equation*}

where \(\det g := \det (g_{ij})\) is the determinant of the component matrix of \(g\) in these coordinates and \(g^{ij}\) the components of the inverse of \((g_{ij})\).

Proof (Sketch) Link to heading

This result is simply obtained by combining the local coordinate expressions of the divergence and the gradient .

Remarks

For the euclidean metric we obtain the usual Laplacian \begin{equation*} \Delta u = - \sum_{i=1}^{n} \frac{\partial^2 u}{(\partial x^i)^2}. \end{equation*}
Using musical isomorphisms notation and alternative representation of the Laplacian, we obtain \begin{equation*} \Delta u = {u_{;i}}^i=g^{ij}u_{;ij}. \end{equation*}