Page 143 - Intro to Tensor Calculus

138

Forming the difference
3
3
∂ ~r ∂ ~r
− =0
α
β
δ
α
∂u ∂u ∂u δ ∂u ∂u ∂u β
we find that the coefficients of the independent vectors bn and ∂~ r ω must be zero. Setting the coefficient of bn
∂u
equal to zero produces the Codazzi equations

γ γ ∂b αβ ∂b αδ
b γδ − b γβ + − =0. (1.5.45)
αβ αδ ∂u δ ∂u β
These equations are sometimes referred to as the Mainardi-Codazzi equations. Equating to zero the coefficient
δ
δ
of ∂~ r ω we find that R δ = b αβ b − b αγ b or changing indices we have the covariant form
∂u αγβ γ β
a ωδ R δ = R ωαβγ = b ωβ b αγ − b ωγ b αβ , (1.5.46)
αβγ
where
∂ δ ∂ δ ω δ ω δ
δ
R αγβ = − + − (1.5.47)
∂u γ αβ ∂u β αγ αβ ωγ αγ ωβ
is the mixed Riemann curvature tensor.
EXAMPLE 1.5-1
Show that the Gaussian or total curvature K = κ (1) κ (2) depends only upon the metric a αβ and is
R 1212
K = where a = det(a αβ ).
a
Solution:
Utilizing the two-dimensional alternating tensor e αβ and the property of determinants we can write
γ
γδ
e K = e αβ γ δ where from page 137, K = |b | = |a αγ b αβ |. Now multiply by e γζ and then contract on
b b
α β
β
ζ and δ to obtain
γδ
b b =2K
e γδ e K = e γδ e αβ γ δ
α β
δν
2K = e γδ e αβ (a γµ b αu ) a b βν
√ µν µν µν αβ
αβ
µν
µν
γµ δν
But e γδ a a = ae so that 2K = e ae b αµ b βν . Using ae = we have 2K = b αµ b βν .
Interchanging indices we can write

2K = βγ ωα b ωβ b αγ and 2K = γβ ωα b ωγ b αβ .
Adding these last two results we find that 4K = βγ ωγ (b ωβ b αγ − b ωγ b αβ )= βγ ωγ R ωαβγ . Now multiply

δ
both sides by στ λν to obtain 4K στ λν = δ βγ ωα R ωαβγ . From exercise 1.5, problem 16, the Riemann
στ λν
curvature tensor R ijkl is skew symmetric in the (i, j), (k, l) as well as being symmetric in the (ij), (kl)pair
δ
of indices. Consequently, δ βγ ωα R ωαβγ =4R λνστ and hence R λνστ = K στ λν and we have the special case
στ λν
√ √ R 1212 b
where K ae 12 ae 12 = R 1212 or K = . A much simpler way to obtain this result is to observe K =
a a
(bottom of page 137) and note from equation (1.5.46) that R 1212 = b 11 b 22 − b 12 b 21 = b.
2
β
α
Note that on a surface ds = a αβ du du where a αβ are the metrices for the surface. This metric is a
α
∂u ∂u β
tensor and satisfies ¯a γδ = a αβ and by taking determinants we find
∂¯u ∂¯u δ
γ
α β
∂u ∂u 2
¯ a = ¯a γδ γ δ = aJ
∂¯u ∂¯u