# Laplace-beltrami operator simplifying to arc length derivative

I'm looking for more insight into geomatric operators on a 2d (parametric) curve. I know the laplace beltrami operator is given by $$\Delta_S u = |g|^{-1/2} \sum_{i,j}^{}\partial_i(|g|g^{ij} \partial_j u),$$

and from wikipedia : "in the usual (orthonormal) Cartesian coordinates $$x^i$$ on the Euclidean space, the metric is reduced to the Kronecker delta, giving $$|g|=1.$$ Consequently, in this case, the operator reduces to the ordinary Laplacian.

I have also seen from other posts here that the laplace beltrami operator on a 2d curve simplifies to the second derivative with respect to arc length for a curve parametrized by arc length.

I'm looking for help in piecing this all together. How would the laplace beltrami operator simplify for a 2d curve given by $$(x(r), y(r))$$ where $$r$$ is angle not arc length? I'm looking to set up a problem for a pde (say the advection-diffusion equation) set up on a parametric curve.

Even insight into how the derivation of the laplace beltrami operator $$\to$$ the derivative with respect to arc length occurs would help!

If you have a curve $$\gamma$$ parametrized by a variable $$s,$$ then this $$s$$ also forms a 1-dimensional coordinate system for the curve. In this coordinate, the metric is the $$1\times1$$ matrix with entry $$g_{ss} = g(\partial_s, \partial_s).$$ Since we're defining $$g$$ to be the induced metric from $$\mathbb R^2,$$ we have $$g(\partial_s, \partial_s) = \gamma'(s) \cdot \gamma'(s) = |\gamma'(s)|^2.$$
From this we can easily calculate the determinant and inverse as $$|g| = |\gamma'|^2$$ and $$g^{ss} = |\gamma'|^{-2}.$$
Thus the Laplace-Beltrami formula becomes $$\Delta u = \frac{1}{\sqrt {|g|}} \partial_s \left(\sqrt{|g|} g^{ss} \partial_s f \right) =\frac1{|\gamma'|} \partial_s\left( \frac 1 {|\gamma'|} \partial_sf\right).$$
When your parametrization is by arc length, you have $$|\gamma '|=1$$ and thus this just reduces to the second derivative.
• @lrs417: The metric on the curve is defined to be the metric induced by the immersion $\gamma: I \to \mathbb R^2,$ meaning that by definition we have $g(X,X) = d\gamma(X)\cdot d\gamma(X).$ Since $d\gamma(\partial_s) = \gamma'(s),$ the formula I wrote in my answer follows. (If you're not using the induced metric, the Laplace-Beltrami operator will depend on the metric you choose.) Oct 6 '20 at 0:35