Page 145 - Intro to Tensor Calculus

140

~
where φ is the angle between the tangent T to the curve and the unit vector b e 1 .
Geodesics are curves on the surface where the geodesic curvature is zero. Since k g = 0 along a geodesic
~
surface curve, then at every point on this surface curve the normal N to the curve will be in the same
r
direction as the normal bn to the surface. In this case, we have ~ u · bn =0 and ~r v · bn = 0 which reduces to
~
~
dT dT
· ~ u =0 and · ~ v =0, (1.5.49)
r
r
ds ds
since the vectors bn and d ~ T have the same direction. In particular, we may write
ds
r
d~ r ∂~ du ∂~ dv
r
~
T = = + = ~ u u + ~r v v 0
0
r
ds ∂u ds ∂v ds
~
dT 0 2 0 2
0
= ~ uu (u ) +2~r uv u v + ~r vv (v ) + ~r u u + ~r v v 00
r
0
00
ds
Consequently, the equations (1.5.49) become
~
dT 0 2 0 2
0
· ~ u =(~r uu · ~r u )(u ) +2(~r uv · ~r u ) u v +(~r vv · ~r u )(v ) + Eu + Fv =0
00
00
0
r
ds . (1.5.50)
dT ~ 0 2 0 2
· ~ v =(~r uu · ~r v )(u ) +2(~r uv · ~r v ) u v +(~r vv · ~r v )(v ) + Fu + Gv =0.
00
00
r
0
0
ds
Utilizing the results from exercise 1.5,(See problems 4,5 and 6), we can eliminate v 00 from the equations
(1.5.50) to obtain
2
d u 1 du 2 1 du dv 1 dv 2
+ +2 + =0
ds 2 11 ds 12 ds ds 22 ds
and eliminating u 00 from the equations (1.5.50) produces the equation
2
d v 2 du 2 2 du dv 2 dv 2
+ +2 + =0.
ds 2 11 ds 12 ds ds 22 ds
In tensor form, these last two equations are written
2 α
β
d u α du du γ
+ =0, α,β,γ =1, 2 (1.5.51)
ds 2 βγ ds ds
a
2
1
where u = u and v = u . The equations (1.5.51) are the differential equations defining a geodesic curve on
a surface. We will find that these same type of equations arise in considering the shortest distance between
two points in a generalized coordinate system. See for example problem 18 in exercise 2.2.