Page 139 - Intro to Tensor Calculus

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~
~
u
The direction of b in relation to T is in the same sense as the surface tangents E 1 and E 2 .Note that
b
the vector db T is perpendicular to the tangent vector T and lies in the plane which contains the vectors bn
b
ds
~
and bu. We can therefore write the curvature vector K in the component form
dT
b
~
~
~
K = = κ (n) bn + κ (g) bu = K n + K g (1.5.23)
ds
where κ (n) is called the normal curvature and κ (g) is called the geodesic curvature. The subscripts are not
indices. These curvatures can be calculated as follows. From the orthogonality condition bn·T =0 we obtain
b
dT b dbn
by differentiation with respect to arc length s the result bn · + T · =0. Consequently, the normal
b
ds ds
curvature is determined from the dot product relation
r
dbn d~ dbn
~
b
b n · K = κ (n) = −T · = − · . (1.5.24)
ds ds ds
u
By taking the dot product of b with equation (1.5.23) we find that the geodesic curvature is determined
from the triple scalar product relation
dT dT b
b
κ (g) = bu · =(bn × T) · . (1.5.25)
b
ds ds
Normal Curvature
The equation (1.5.24) can be expressed in terms of a quadratic form by writing
2
κ (n) ds = −d~r · dbn. (1.5.26)
r
The unit normal to the surface bn and position vector ~ are functions of the surface coordinates u, v with
r
∂~ ∂~ ∂bn ∂bn
r
r
d~ = du + dv and dbn = du + dv. (1.5.27)
∂u ∂v ∂u ∂v
We define the quadratic form

r
∂~ ∂~ ∂bn ∂bn
r
r
B = −d~ · dbn = − du + dv · du + dv
∂u ∂v ∂u ∂v (1.5.28)
2 2 α β
B = e(du) +2fdu dv + g(dv) = b αβ du du
where
r
r
∂~ r ∂bn ∂~ ∂bn ∂bn ∂~ ∂~ ∂bn
r
e = − · , 2f = − · + · , g = − · (1.5.29)
∂u ∂u ∂u ∂v ∂u ∂v ∂v ∂v
γ
α, β =1, 2 is called the curvature tensor and a αγ b αβ = b is an associated curvature tensor.
and b αβ
β
The quadratic form of equation (1.5.28) is called the second fundamental form of the surface. Alternative
methods for calculating the coefficients of this quadratic form result from the following considerations. The
unit surface normal is perpendicular to the tangent vectors to the coordinate curves at the point P and
therefore we have the orthogonality relationships
∂~ ∂~
r
r
· bn =0 and · bn =0. (1.5.30)
∂u ∂v