Differential-Geometry of Curves and Surfaces Normal Curvature and Geodesic Curvature, guaranteed and verified 100% Pass

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Normal Curvature and Geodesic Curvature


The shape of a surface will clearly impact the curvature of the curves on the
surface. For example, it’s possible for a curve in a plane or on a cylinder to have
zero curvature everywhere (i.e. it’s a line or a portion of a line). However, it’s not
possible for a curve on a sphere to have zero curvature everywhere. So one way
to measure how much a surface curves is by examining the curvature of the
curves on the surface, this will lead us to the second fundamental form.

Let 𝛾 be a unit speed curve on an oriented surface, 𝑆. Then, 𝛾 ′ (𝑠) is a unit
vector that is tangent to the surface. Thus, 𝛾 ′ (𝑠) is perpendicular to the unit
⃗ , of 𝑆. So 𝛾 ′ (𝑠), 𝑁
normal vector, 𝑁 ⃗ , and 𝑁
⃗ × 𝛾 ′ (𝑠) are mutually
perpendicular unit vectors.

Since 𝛾 ′ ∙ 𝛾 ′ = 1, by differentiating this equation we get:

𝛾 ′′ (𝑠) ∙ 𝛾 ′ (𝑠) = 0.
Thus, 𝛾 ′′ (𝑠) is perpendicular to 𝛾 ′ (𝑠) and must lie in the plane spanned
⃗ and 𝑁
by 𝑁 ⃗ × 𝛾 ′ (𝑠). So we can write:

⃗ + 𝑏 (𝑁
𝛾 ′′ (𝑠) = 𝑎𝑁 ⃗ × 𝛾 ′ (𝑠)).
⃗𝑝
𝑁

𝜅𝑛 𝑆
Def. We define
𝛾′′(𝑠) 𝜓 𝑝 𝛾(𝑠)
𝑎 = 𝜅𝑛 = the normal curvature of 𝛾
𝑏 = 𝜅𝑔 = the geodesic curvature of 𝛾 𝛾′(𝑠)
𝜅𝑔
so:
⃗ 𝑝 × 𝛾′(𝑠)
𝑁
𝛾 ′′ ( ⃗ + 𝜅𝑔 (𝑁
𝑠 ) = 𝜅𝑛 𝑁 ⃗ × 𝛾 ′ (𝑠)).

, 2


⃗ with −𝑁
Notice that if we replace 𝑁 ⃗ (the other unit normal of 𝑆) the
normal and geodesic curvature also change signs.



⃗
Proposition: κ𝑛 = 𝛾 ′′ (𝑠) ∙ 𝑁

⃗ × 𝛾 ′ (𝑠))
𝜅𝑔 = 𝛾 ′′ (𝑠) ∙ (𝑁

𝜅 2 = 𝜅n2 + 𝜅g2 ; where 𝜅 = curvature of 𝛾
and

𝜅𝑛 = κ cos 𝜓 , 𝜅𝑔 = ±𝜅 sin 𝜓
⃗ and the principal normal 𝑛⃗.
where 𝜓 is the angle between 𝑁
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Recall that the principal normal, 𝑛
⃗ , is defined by 𝑛⃗ = 𝛾 ′′ (𝑠).
κ(s)



Proof:

⃗ + κ𝑔 (𝑁
𝛾 ′′ (𝑠) = κ𝑛 𝑁 ⃗ × 𝛾 ′ (𝑠))



⃗ = (κ𝑛 𝑁
𝛾 ′′ (𝑠) ∙ 𝑁 ⃗ + κ𝑔 (𝑁
⃗ × 𝛾 ′ (𝑠))) ∙ 𝑁
⃗ = κ𝑛



⃗ × 𝛾 ′ (𝑠)) = (κ𝑛 𝑁
𝛾 ′′ (𝑠) ∙ (𝑁 ⃗ + κ𝑔 (𝑁
⃗ × 𝛾 ′ (𝑠))) ∙ (𝑁
⃗ × 𝛾 ′ (𝑠)) = κ𝑔



⃗ + κ𝑔 (𝑁
κ2 = ‖𝛾′′ (𝑠)‖2 = (κ𝑛 𝑁 ⃗ × 𝛾 ′ (𝑠))) ∙ (κ𝑛 𝑁
⃗ + κ𝑔 (𝑁
⃗ × 𝛾 ′ (𝑠)))

= κ2n + κ2g .