Page 153 - Intro to Tensor Calculus

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Another form of equation (1.5.104) is obtained by using the equation (1.5.83) together with the relation
σγ
a αβ = −a σα βγ . It is left as an exercise to verify the resulting form

−Ka αβ = c αβ − a σγ b σγ b αβ . (1.5.106)

Define the quantity
1 σγ
H = a b σγ (1.5.107)
2
as the mean curvature of the surface, then the equation (1.5.106) can be written in the form

c αβ − 2Hb αβ + Ka αβ =0. (1.5.108)

β
α
By multiplying the equation (1.5.108) by du du and summing, we find
C − 2HB + KA =0 (1.5.109)

is a relation connecting the first, second and third fundamental forms.

EXAMPLE 1.5-2
In a two dimensional space the Riemann Christoffel tensor has only one nonzero independent component
R 1212 . ( See Exercise 1.5, problem number 21.) Consequently, the equation (1.5.104) can be written in the
√ √
form K ae 12 ae 12 = b 22b 11 − b 21 b 12 and solving for the Gaussian curvature K we find

b 22 b 11 − b 12 b 21 b R 1212
K = = = . (1.5.110)
a 11 a 22 − a 12 a 21 a a

Surface Curvature
α
α
2
i
i
1
For a surface curve u = u (s),α =1, 2 lying upon a surface x = x (u ,u ),i =1, 2, 3, we have a two
du α
α
dimensional space embedded in a three dimensional space. Thus, if t = is a unit tangent vector to
ds
α
du du β α β
the surface curve then a αβ = a αβ t t =1. This same vector can be represented as the unit tangent
ds ds
i
dx i dx dx j
1
2
j
i
i
i
i
vector to the space curve x = x (u (s),u (s)) with T = . That is we will have g ij = g ij T T =1.
ds ds ds
i
α
The surface vector t and the space vector T are related by
i
∂x du α i α
i
T = = x t . (1.5.111)
α
∂u α ds
α
α β
The surface vector t is a unit vector so that a αβ t t =1. If we differentiate this equation intrinsically with
respect to the parameter s, we find that a αβ t α δt β =0. This shows that the surface vector δt α is perpendicular
δs δs
α
α
to the surface vector t . Let u denote a unit normal vector in the surface plane which is orthogonal to the
α α α β
tangent vector t . The direction of u is selected such that αβ t u =1. Therefore, there exists a scalar κ (g)
such that
δt α α
= κ (g) u (1.5.112)
δs