Rewriting the Laplacian on a curved manifold

Question

I guess there is a sense in which the following is true:

"The Laplacian written on a Riemannian manifold $(M,g)$ can be seen as adding a coordinate dependent mass field to the Laplacian on Euclidean space."

• Can someone kindly refer me to a place where this is exactly derived? (or feel free to write in the derivation here if its not too long!)

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Just so that we are on the same page :

For ``nice" real valued functions $f$ on $(M,g)$ we have for the square of the gradient of $f$,

$\Vert {\nabla_g f} \Vert ^2 = g(\nabla_g f,\nabla_g f) =  \sum_{j=1}^n  \sum_{i=1}^n g^{ij} \partial_i f  \partial_j  f$

and the Laplacian of $f$ being,

$\nabla_g^2 f := \frac{1}{\sqrt{\det(g)}} \sum_{i,j=1}^n \frac{\partial }{\partial x_i} \left (  \sqrt{\det(g)} g^{ij} \frac{f}{\partial x_j}\right )$

where we define the metric as $g = [g_{ij}] = g \left ( \partial_{x_i}, \partial_{x_j} \right )$ and $g^{-1} = [g^{ij}]$.

Answer 1

1. Recall that the quantum Hamiltonian$^1$ $\hat{H}=-\frac{\hbar^2}{2}\Delta$ for a point particle is (in the Schrödinger representation) associated with the Laplacian $\Delta$.

2. The corresponding classical Hamiltonian is $H=\frac{1}{2}\sum_{i,j=1}^np_i g^{ij} p_j$.

3. In flat space the Hamiltonian for a free particles is $H=\sum_{i=1}^n\frac{p_ip_i}{2m_i}$.

4. Comparing 2 & 3 we see that the inverse metric $g^{ij}$ plays the role of a position-dependent inverse mass matrix ${\rm diag}(1/m_1,\ldots, 1/m_n)$, which is essentially the statement quoted by OP.

For more details, see e.g. this related Phys.SE post.

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$^1$ We ignore for simplicity potential terms $V$.