Suppose $M$ is a two-dimensional manifold with metric $\bar{g}$, and $r: M \to \mathbb{R}^3$ is a (not necessarily isometric) embedding of $M$ into $\mathbb{R}^3$ with first fundamental form $g$ and second fundamental form $b$.
I know that $$\operatorname{tr}(g^{-1}b)N = \Delta_g r,$$ where $N$ is the (Euclidean) surface normal of $r(M)$, and $\Delta$ is the Laplace-Beltrami operator on $M$ with respect to the pullback metric $g$.
Does a similar relationship hold for the original metric $\bar{g}$, i.e. is it true that $$\operatorname{tr}(\bar{g}^{-1}b)N = \Delta r,$$ possibly for a different interpretation of $N$?
