9.7: Geodesic Curvature in Orthogonal Coordinates

For a curve X(t) in the horizontal plane in 3-space, we have defined the curvature κg(t) by the condition T'(t) = κgs'(t)U(t), where U(t) = (0,0,1) × T(t). We may then write

T(t) = cosθ(t)E1 + sinθ(t)E2,        so
U(t) = -sinθ(t)E1 + cosθ(t)E2            

We then have

T'(t) = -sinθ(t)θ'(t)E1 + cosθ(t)θ'(t)E2 = θ'(t)U(t)

So, κg(t)s'(t) = θ'(t) in the case of a plane curve. For a curve X(t) = X(u(t),v(t)) on a surface in 3-space with unit normal vector N(t), we define U(t) = N(t) × T(t) and we define the geodesic curvature, κg(t), as

κg(t)s'(t) = T'(t) · U(t)

Let E1(t) and E2(t) form an orthonormal frame in the tangent space to X at X(t), so that

N(t) × E1(t) = E2(t)        and         N(t) × E2(t) = -E1(t)

Then we may write

T(t) = cosθ(t)E1(t) + sinθ(t)E2(t)        and
U(t) = -sinθ(t)E1(t) + cosθ(t)E2(t)             

It follows that

T'(t) = -sinθ(t)θ'(t)E1(t) + cosθ(t)θ'(t)E2(t) + cosθ(t)E1'(t) + sinθ(t)E2'(t)

Note that E1(t) · E2(t) = 0 implies that -E1'(t) · E2(t) = E2'(t) · E1(t). Using this observation, we can derive an expression for κg(t)s'(t) for a curve on a surface in 3-space as we did for a curve in the horizontal plane:
       κg(t)s'(t) = T'(t) · U(t)
                     = θ'(t) + cosθ(t)E1(t) · (-sinθ(t)E1(t) + cosθ(t)E2(t)) + sinθ(t)E2'(t)(-sinθ(t)E1(t) + cosθ(t)E2(t))
                     = θ'(t) + cos2θ(t)E1'(t) · E2(t) - sin2θ(T)E2'(t) · E1(t)
                     = θ'(t) + E1'(t) · E2(t)

In the case of a plane curve, E1(t) = E1 and E2(t) = E2 are constant so this result reduces to the previous calculation. Furthermore, for plane curves we have:

abκg(t)s'(t)dt = abθ'(t)dt = θ(b) - θ(a).

If the curve is closed, continuous, and non-self-intersecting, then θ(b) - θ(a) = 2π. For closed curves that contain vertices, we have to add in the external angles, so that θ(b) - θ(a) + ∑ext.angles = 2π. Therefore,

abκgds + ∑ext.angles = 2π

In the case of a general non-self-intersecting curve bounding a domain D in a coordinate patch of surface in 3-space, we have

∂Dκgds + ∑ext.angles = ∂D(θ'(t) + E1'(t) · E2(t))dt = 2π + ∂DE1'(t) · E2(t))dt

We will show that ∂DE1'(t) · E2(t))dt = -ΚdA, and that will complete the intrinsic form of the Gauss-Bonnet Theorem. In particular, we will show that E1'(t) · E2(t) is an intrinsic quantity.

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