Page 70 - Basic Structured Grid Generation

Differential geometry of surfaces in E 3 59

We obtain
dλ α
ν α ν β γ
κ g = ε να λ + ε να βγ λ λ λ
ds
ν 2 α ν β γ
√ du d u du du du
α
= a e να + e να βγ
ds ds 2 ds ds ds
2 2 β γ β γ
√ du d v dv d u du du du dv du du
1
2
= a − + βγ − βγ .
ds ds 2 ds ds 2 ds ds ds ds ds ds
Hence 3 3
2
2
κ g du d v dv d u 2 du 1 dv
√ = − + 11 − 22
a ds ds 2 ds ds 2 ds ds
2
2

du dv 1 2 dv du
1
2
+(2 12 − ) − (2 12 − ) . (3.88)
11
22
ds ds ds ds
Example: We calculate the geodesic curvature of a ‘circle of latitude’ (a parallel)
θ = θ 0 of a sphere of radius a with surface co-ordinates given by spherical polars
θ, φ.
1
2
Take u = θ, u = φ, with the usual background cartesian co-ordinates defining the
spherical surface in the parametric form y 1 = x = a sin θ cos φ, y 2 = y = a sin θ sin φ,
y 3 = z = a cos θ. The circle of latitude has radius r = a sin θ 0 , and arc-length s can
α
α
be measured from some given meridian as s = rφ. Hence this curve is u = u (s),
where
1 2
u = θ 0 ,u = s/r = s/(a sin θ 0 ).
α
Now we have the unit tangent λ , with
du 1 du 2 1
1 2
λ = = 0, λ = = .
ds ds a sin θ 0

2 2
2
∂x ∂z 2 2 2
∂y
By eqn (3.17), a 11 = + + = a , a 12 = a 21 = 0, a 22 = a sin θ,
∂θ ∂θ ∂θ
similarly, and the only non-vanishing Christoffel symbols are, from eqn (3.49), [22, 1]=
2
[12, 2]=[21, 2]= a sin θ cos θ,and 1 = 2 = 2 = sin θ cos θ.Hence the first
22 12 21
of eqns (3.87) becomes, with α = 1, 2,
1 1 2
2
0 + (sin θ 0 cos θ 0 ) = κ g ν and 0 + 0 = κ g ν .
a sin θ 0
1
2
2
1
α β
Thus ν = 0, and since ν is a unit vector, we have a αβ ν ν = a (ν )(ν ) = 1.
1
Hence ν = a −1 ,and
κ g = a −1 cot θ 0 .
3
Of course, viewed from E , a parallel is just a circle with curvature κ = r −1 =
a −1 cosecθ 0 . Note that the ‘equator’, the parallel with θ = π/2, is a geodesic with
geodesic curvature zero.