Geodesic curvature and boundary orientation

Try

A latitude circle on a sphere has two kinds of bending. Part of its acceleration points normal to the sphere, and part turns within the tangent plane. At the equator the tangent-plane part vanishes, so the curve is geodesic. Away from the equator it does not.

For a unit-speed curve on an oriented surface, write T=γ′ and use the orthonormal directions T, N×T and N. The signed geodesic curvature is k_g=⟨γ″,N×T⟩; normal curvature is k_n=⟨γ″,N⟩.

Predict

Predict why unit speed matters. Differentiating T·T=1 eliminates acceleration in the T direction. With a nonunit parameter, tangential acceleration also records changes of speed.

Predict the effect of reversing traversal while keeping N fixed. The acceleration at the same geometric point stays the same under arc-length reversal, but N×T changes sign. Thus signed geodesic curvature reverses.

Experiment

On the unit sphere, consider the positively oriented boundary of a north polar cap at colatitude θ.

θ	k_g=cot θ	Circle length	Integrated k_g
π/2	0	2π	0
π/3	1/√3	π√3	π
π/6	√3	π	π√3

Multiply the middle columns to check the integral. The cap boundary is traversed with increasing longitude, putting the cap on the left relative to the outward sphere orientation.

Understand

Because γ″·T=0, the acceleration decomposes as γ″=k_g(N×T)+k_nN. The two components are orthogonal, so the squared space curvature is κ²=k_g²+k_n². This explains how a plane curve can have nonzero κ while k_n=0, or a sphere geodesic can have nonzero κ while k_g=0.

For the unit sphere latitude, T=eφ=(−sin φ,cos φ,0), and arc length satisfies ds=sin θ dφ. Thus dT/ds=(−cos φ,−sin φ,0)/sin θ. The oriented tangent normal is N×T=−eθ. Their dot product is cos θ/sin θ, proving k_g=cot θ with the chosen boundary orientation.

For a piecewise smooth boundary, tangent direction can jump at corners. The signed exterior turning angle records that jump. Straight polygon edges in the plane have zero k_g, yet the corner turns sum to 2π for a positively oriented simple polygon. Ignoring corners would incorrectly produce zero total turning.

Apply

Repair “Geodesic curvature is always nonnegative.” Its sign depends on both surface orientation and traversal. The magnitude describes bending, while the sign is essential in boundary-integral formulas. Inner boundary components of an oriented annulus run in the opposite direction from the outer boundary.

For optional depth, a curve with k_g=0 and constant speed is geodesic because its acceleration has no tangent component. This gives an intrinsic characterization even when its ambient path is curved. Conversely, normal curvature is controlled by II and its tangent direction, while k_g depends on how the curve turns within the surface.

The separation of these two curvatures is the bridge between surface geometry and boundary turning.

Continue

Before integrating k_g, draw or state the orientation and list all corner contributions. Continue to local Gauss–Bonnet to prove the exact balance between these boundary quantities and interior Gaussian curvature.