# Curvature of contour lines of a scalar field

How can I compute the curvature of the contour lines (equipotential lines) $\phi (\vec{r})=c$ for the scalar field $\phi (\vec{r})$ ? I expect the direction of the curvature vector to be along the gradient of the field, in analogy to the electric field vectors which are orthogonal to the equipotential lines. Wwhat about the magnitude?

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@Tarek: Sure. However, it's not the coordinates matter, it's the metric and its Levi-Civita connection. For an oriented Riemannian surface $(S,g)$ with Levi-Civita connection $\nabla$ and a function $f:S\to \mathbb{R}$ with $df\not=0$, then the geodesic curvature of the level sets of $f$ is the function $$\kappa = \frac{\nabla^2f(\ J\nabla f,\ J\nabla f\ )}{|\nabla f|^3},$$ where $J:TS\to TS$ is the usual 'rotation by $90$ degrees' operator on the tangent spaces. – Robert Bryant Apr 21 '13 at 23:51
@Tarek: I realized, after I made my comment, that you had asked about the 'general $N$-dimensional curvilinear system of coordinates'. In that case, the answer is essentially the same, but now you are asking about the second fundamental form of the level sets. The formula in that case is just $$II(v,w) = \frac{\nabla^2f(v,w)}{|\nabla f|}$$ when $v,w\in T_xS$ satisfy $df(v)=df(w)=0$. – Robert Bryant Apr 22 '13 at 0:00