Page 138 - Intro to Tensor Calculus

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Here and throughout the remainder of this section, we adopt the convention that Greek letters have the
range 1,2, while Latin letters have the range 1,2,3.
Construct at a general point P on the surface the unit normal vector bn at this point. Also construct a
plane which contains this unit surface normal vector bn. Observe that there are an infinite number of planes
which contain this unit surface normal. For now, select one of these planes, then later on we will consider
r
all such planes. Let ~ = ~(s) denote the position vector defining a curve C which is the intersection of the
r
selected plane with the surface, where s is the arc length along the curve, which is measured from some fixed
point on the curve. Let us find the curvature of this curve of intersection. The vector T = d~ r , evaluated
b
ds
at the point P, is a unit tangent vector to the curve C and lies in the tangent plane to the surface at the
point P. Here we are using ordinary differentiation rather than intrinsic differentiation because we are in
a Cartesian system of coordinates. Differentiating the relation T · T =1, with respect to arc length s we
b b
db T db T
b
b
find that T · = 0 which implies that the vector is perpendicular to the tangent vector T. Since the
ds ds
coordinate system is Cartesian we can treat the curve of intersection C as a space curve, then the vector
~
~
K = db T , evaluated at point P, is defined as the curvature vector with curvature |K| = κ and radius of
ds
curvature R =1/κ. A unit normal N to the space curve is taken in the same direction as db T so that the
b
ds
dT b
~
curvature will always be positive. We can then write K = κN = . Consider the geometry of figure 1.5-1
b
ds
and define on the surface a unit vector b = bn × T which is perpendicular to both the surface tangent vector
u
b
i
i
i
T and the surface normal vector bn, such that the vectors T ,u and n forms a right-handed system.
b

Figure 1.5-1 Surface curve with tangent plane and a normal plane.