The Gauss-Bonnet Formula for curves states that the integral of the curvature around a closed curve in a plane plus the sum of the turning angles at corner points is equal to 2π. The proof of this theorem in the standard sources is tedious and involved. What is given below is a marvelously simple proof that corrects an error in the usual statement of the formula.

Curvature is defined at points where the curve is smooth in terms of the rate of turning of the unit tangent vector to a curve; i.e.,

dT/ds = κ(s)N(s)
 

where T(s) is the unit tangent vector as a function of the path length parameter s, N(s) is the unit normal vector and κ(s) is the curvature.

The unit tangent vector T(s) for a plane curve is most succinctly given in terms of it direction angle α(s); i.e.,

T(s) = (cos(α(s)), sin(α(s)))
 

Since

dT/ds = (-sin(α(s)α'(s), cos(α(s)α'(s)))
= α'(s)(-sin(α(s)), cos(α(s))
it follows that
N(s) = (-sin(α(s)), cos(α(s)))
and
κ(s) = α'(s)
 

A curvature function κ*(s) could be defined as the generalized function which is the derivative of α(s) but this is usually not what is done. The important thing is that the line integral of the curvature is equal to the function α(s); i.e.,

∫₀^sκ*(z)dz = ∫₀^sκ(z)dz + Σε_i = α(s) - α(0).
 

where {ε_i} is the set of the turning angles of the curve at its corner points.

If L is the end value of the curve parameter s then

α(L) - α(0) = n(2π)
where n is an integer,
the net number of full turns
the curve makes. .
 

Obviously the correct form of the Gauss-Bonnet Formula for the path integral of curvature is

∫₀^L κ(z)dz + Σε_i = n(2π).
for some integer n.
 

A plane curve could equally well be represented by the angle function β(s) for its unit normal vectors N(s), where β(s) is at a right angle to α(s); i.e.,

β(s) = α(s)-π/2.
 

This analysis in terms of unit normals is given because the analysis of curvature for a surface has to be expressed in terms of normals rather than tangents. It is helpful to see the analysis for the simpler case of curves as a preparation for the more difficult case of surfaces.

The unit normal vector function is given in terms of β(s) as:

N(s) = (cos(β(s), sin((β(s)))
 

Since dN/ds = -κ(s)T(s) and

dN/ds = (-sin((β(s))β'(s), cos(β(s))β'(s))
= β'(s)(-sin((β(s)), cos(β(s))
= β'(s)(-sin((α(s)-π/2), cos(α(s)-π/2)
= β'(s)(-cos((α(s)), -sin(α(s))
= -β'(s)T(s)
and thus
κ = β'(s)
 

Thus the line integral of curvature could be expressed as

∫₀^s κ(z)dz + Σε_i = β(s) - β(0).
 

where {ε_i} is the set of the turning angles of the normals to the curve at its corner points, which are the same as the turning angles of the tangents to the curve at those points.

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