Section 1.19
Solid Mechanics Part III
Kelly
177
3
,
3
3
,
3
a
a
a
a
⋅
−
=
⋅
α
(1.19.10)
so that, from Eqn 1.18.6,
0
33
3
3
=
Γ
=
Γ
(1.19.11)
Further, since
0
/
3
3
=
Θ
∂
∂
a
,
0
,
0
333
33
=
Γ
=
Γ
(1.19.12)
These last two equations imply that the
ijk
Γ
vanish whenever two or more of the
subscripts are 3.
Next, differentiate 1.19.1 to get
β
a
a
a
a
⋅
−
=
⋅
,
3
3
,
,
a
a
a
a
⋅
−
=
⋅
,
3
3
,
(1.19.13)
and Eqns. 1.18.6 now lead to
βα
αβ
3
3
3
3
Γ
−
=
Γ
−
=
Γ
=
Γ
(1.19.14)
From 1.18.8, using 1.19.11,
0
3
3
33
3
3
3
3
3
3
3
33
3
3
3
=
Γ
=
Γ
+
Γ
=
Γ
Γ
=
Γ
+
Γ
=
Γ
γ
αβγ
g
g
g
g
(1.19.15)
and, similarly {
▲
Problem 1}
0
3
33
33
3
3
=
Γ
=
Γ
=
Γ
(1.19.16)
1.19.2
The Curvature Tensor
In this section is introduced a tensor which, with the metric coefficients, completely
describes the surface.
First, although the base vector
3
a
maintains unit length, its direction changes as a
function of the coordinates
2
1
,
Θ
Θ
, and its derivative is, from 1.18.2 or 1.18.5 (and using
1.19.15)
a
a
a
3
3
3
Γ
=
Γ
=
Θ
∂
∂
k
k
,
a
a
a
3
3
3
Γ
−
=
Γ
−
=
Θ
∂
∂
k
k
(1.19.17)
Define now the
curvature tensor
K
to have the covariant components
K
, through