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Derivation

In section 7 of lab 8, we provided an explicit formula for geodesic curvature in terms of extrinsic quantities.

κg = (X''(t) · U(t))/s'(t)2

Here we derive an intrinsic formula for geodesic curvature. We start with the fact that U(t) = N(t) × T(t) = (N(t) × X'(t))/s'(t).

κg = [X''(t) · (N(t) × X'(t))]/s'(t)3

In order to simplify the cross product term N(t) × X'(t), let us assume that the partial derivative vectors are perpendicular, so that we can choose E1(t) = Xu(t)/√g11 and E2(t) = Xv(t)/√g22 as an orthogonal basis for the tangent space, such that

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

With these conditions in place, we can expand X'(t) and find the cross product:

N(t) × X'(t) = N(t) × (Xu(t)u'(t) + Xv(t)v'(t))
                              = (N(t) × Xu(t))u'(t) + (N(t) × Xv(t)v'(t))
                                           = (√g11/√g22)Xv(t)u'(t) - (√g22/√g11)Xu(t)v'(t)

To calculate the dot product X''(t) · (N(t) × X'(t)), we need to expand X''(t) as well:

X''(t) = Xuu(t)u'(t)2 + Xuv(t)u'(t)v'(t) + Xu(t)u''(t) + Xvv(t)v'(t)2 + Xvu(t)u'(t)v'(t) + Xv(t)v''(t)

Dotting X''(t) with N(t) × X'(t) gives us the following expression in which the Christoffel symbols are used to simplify things.

X''(t) · (N(t) × X'(t)) = u'(t)√(g11/g22)(Γ112u'(t)2 + Γ222v'(t)2 + 2Γ122u'(t)v'(t) + v''(t))g22
                                     - v'(t)√(g22/g11)(Γ111u'(t)2 + Γ221v'(t)2 + 2Γ121u'(t)v'(t) + u''(t))g11

Finally, we can substitute the intrinisic formulas for each of the Christoffel symbols to get an equation in terms of the metric coefficients:

κgs'(t)3 = u'(t)√(g11/g22)[(-1/2)(∂g11/∂v)u'(t)2 + (∂g22/∂u)u'(t)v'(t) + (1/2)(∂g22/∂v)v'(t)2]
                - v'(t)√(g22/g11)[(1/2)(∂g11/∂u)u'(t)2 + (∂g11/∂v)u'(t)v'(t) - (1/2)(∂g22/∂u)v'(t)2]
+ √(g11g22)[u'(t)v''(t) - v'(t)u''(t)]