APM3713
Assignment 6
Due Year 2025
, Solution
Question 1
Consider the surface parametrized by
√ √
r(u, v) = u2 + 1 cos v, u2 + 1 sin v, u .
We take x1 = u, x2 = v.
(a) Line Element
Tangent vectors:
√ √
ru = √ u cos v, √ u sin v, 1 , rv = − u2 + 1 sin v, u2 + 1 cos v, 0 .
u2 +1 u2 +1
Metric coefficients:
2u2 + 1
guu = , guv = 0, gvv = u2 + 1.
u2 + 1
Thus the line element is
2u2 + 1 2
ds2 = du + (u2 + 1) dv 2 .
u2 + 1
(b) Metric and Inverse Metric
u2 + 1
2
2u + 1
0 0
2
g ij = 2u + 1
2
gij = u + 1 , .
1
0 u2 + 1 0 2
u +1
(c) Christoffel Symbols
Using
Γkij = 21 g kℓ (∂i gℓj + ∂j gℓi − ∂ℓ gij ),
we compute
u u(u2 + 1) u
Γ111 = 2 , Γ122 =− , Γ212 = Γ221 = .
(u + 1)(2u2 + 1) 2u2 + 1 u2 + 1
1
Assignment 6
Due Year 2025
, Solution
Question 1
Consider the surface parametrized by
√ √
r(u, v) = u2 + 1 cos v, u2 + 1 sin v, u .
We take x1 = u, x2 = v.
(a) Line Element
Tangent vectors:
√ √
ru = √ u cos v, √ u sin v, 1 , rv = − u2 + 1 sin v, u2 + 1 cos v, 0 .
u2 +1 u2 +1
Metric coefficients:
2u2 + 1
guu = , guv = 0, gvv = u2 + 1.
u2 + 1
Thus the line element is
2u2 + 1 2
ds2 = du + (u2 + 1) dv 2 .
u2 + 1
(b) Metric and Inverse Metric
u2 + 1
2
2u + 1
0 0
2
g ij = 2u + 1
2
gij = u + 1 , .
1
0 u2 + 1 0 2
u +1
(c) Christoffel Symbols
Using
Γkij = 21 g kℓ (∂i gℓj + ∂j gℓi − ∂ℓ gij ),
we compute
u u(u2 + 1) u
Γ111 = 2 , Γ122 =− , Γ212 = Γ221 = .
(u + 1)(2u2 + 1) 2u2 + 1 u2 + 1
1
