Page 149 - Intro to Tensor Calculus
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is now demonstrated. By using the results from equation (1.5.50), the tensor derivative of equation (1.5.59),
with respect to the surface coordinates, produces
2 i
∂ x i p q σ i
i
x = + x x − x (1.5.68)
α,β α β α β σ
∂u ∂u pq αβ
g a
where the subscripts on the Christoffel symbols refer to the metric from which they are calculated. Also the
tensor derivative of the equation (1.5.57) produces the result
j
i
g ij x i x + g ij x x j = a αβ,γ =0. (1.5.69)
α,γ β α β,γ
Interchanging the indices α, β, γ cyclically in the equation (1.5.69) one can verify that
g ij x i α,β γ j (1.5.70)
x =0.
The equation (1.5.70) indicates that in terms of the space coordinates, the vector x i is perpendicular to
α,β
i
i
the surface tangent vector x and so must have the same direction as the unit surface normal n . Therefore,
γ
there must exist a second order tensor b αβ such that
i
b αβ n = x i α,β . (1.5.71)
i j
By using the relation g ij n n = 1 we can transform equation (1.5.71) to the form
1 γδ i j k
j i
b αβ = g ij n x = ijk x x x . (1.5.72)
α,β α,β γ δ
2
The second order symmetric tensor b αβ is called the curvature tensor and the quadratic form
α
B = b αβ du du β (1.5.73)
is called the second fundamental form of the surface.
Consider also the tensor derivative with respect to the surface coordinates of the unit normal vector to
the surface. This derivative is
∂n i i j k
i
n ,α = + n x . (1.5.74)
α
∂u α jk
g
i j
Taking the tensor derivative of g ij n n = 1 with respect to the surface coordinates produces the result
i
i j
g ij n n = 0 which shows that the vector n j is perpendicular to n and must lie in the tangent plane to the
,α ,α
i
surface. It can therefore be expressed as a linear combination of the surface tangent vectors x and written
α
in the form
β i
n i = η x (1.5.75)
,α α β
β
where the coefficients η can be written in terms of the surface metric components a αβ and the curvature
α
i
components b αβ as follows. The unit vector n is normal to the surface so that
i j
g ij n x =0. (1.5.76)
α