# Equivalence between Differential Geometry and Mechanics?

Given a metric

$$ds^{2}~=~ g_{a,b}dx^{a}dx^{b}.$$

Here Einstein's summation convention is assumed for $a$ and $b$.

Then given the Laplacian over that metric, can then we find a metric $g_{a,b}$ so

$$\Delta _{g}f ~=~ - \nabla^2 f +V(x)f.$$

• the first term on the left is the Laplacian operator in curvilinear coordinates.

• the term on the right is just the ${\rm div}({\rm grad})f$ in Euclidean coordinates.

• $V(x)$ is a potential in $n$-dimensions.

The idea, if this is possible, is to turn physical problems into geometrical problems, for example solving the Schroedinger equation would be equivalent to solve a Dirichlet problem for some Laplacian.

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Unfortunately, this is not a good idea. The Beltrami-Laplacian $\Delta_{g}$ is in local coordinates given as

$$\Delta_{g}\Psi ~=~ \frac{1}{\sqrt{g}}\frac{\partial}{\partial x^i}\left(\sqrt{g}~ g^{ij} \frac{\partial \Psi}{\partial x^j} \right).$$

A nice feature of $\Delta_{g}$ is that it takes a scalar $\Psi$ into a scalar $\Delta_{g}\Psi$.

Most importantly, a decomposition

$$\Delta_{g}\Psi~=~g^{ij}\frac{\partial^2 \Psi}{\partial x^i\partial x^j} + \frac{1}{2} g^{ij} \left(\frac{\partial \ln g}{\partial x^i}\right) \frac{\partial \Psi}{\partial x^j}$$

would not contain a zero-order term that could be identified with a potential $V$, but rather a first-order term.

Finally, let me mention that for a generic metric $g_{ij}=g_{ij}(x)$ on a 3-dimensional Riemannian manifold $(M,g)$ (rather than just curvilinear coordinates for $\mathbb{R}^3$), it might not be possible to find flat coordinates in an open neighborhood, as also discussed here (in the context of GR).

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For scalar $f$s, no. Given two metrics $g_{ij}$ and $h_{ij}$, the difference in their corresponding Laplace-Beltrami operators on scalar functions is

$$\triangle_g f - \triangle_h f = \tilde{\Gamma}^i \partial_i f$$

where $\tilde{\Gamma}^i$ is the difference of the traces of the connection coefficients for the metrics $g$ and $h$

$$\tilde{\Gamma}^{i} = g^{jk}{}^{(g)}\Gamma_{jk}^i - h^{jk}{}^{(h)}\Gamma_{jk}^i$$

And so in particular it cannot be a scalar potential $V\cdot f$.

One can also quite simply see this fact by noting that for $V(x)$ not identically vanishing, if $f$ is the constant function 1, then $\triangle_g f = 0$ on your left hand side, but on your right hand side $\nabla f = 0$ while $V(x) f(x) = V(x) \neq 0$.

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