Graphics Reference
In-Depth Information
so that
10
01
22
00
00
22
Ê
ˆ
Ê
ˆ
Á
Á
˜
˜
Á
Á
˜
˜
(
¢=
J
j
=
and
J
j
.
Ë
¯
Ë
¯
uv
If g(t) =j(a(t)), then
102
012
u
v
002
002
)
Ê
Ë
ˆ
¯
+¢
)
Ê
Ë
ˆ
¯
(
(
g
¢¢ =
uv
¢¢
¢¢
uv
¢
(
)
=¢¢
uv uu
,
¢¢
,
2
¢¢ +
2
vv
¢¢ + ¢ + ¢
2
u
2
v
.
It follows from equation (15.3) that the geodesics of
S
are defined by curves a(t)
satisfying
(
)
¢¢+
2
14
+
u
u
4
uvv
¢¢+
4
uu
¢+
4
uv
¢=
0
(
)
¢¢ +
2
4
uvu
¢¢ +
1
+
4
v
v
4
vu
¢ +
4
vv
¢ =
0
.
We shall not solve that system of differential equations, but will show that certain lon-
gitudinal curves (meridians) are geodesics in the kinematic sense.
Let
e
= (cos q, sin q,0) be a fixed unit vector in the plane and consider the curve
g(t) in
S
defined by
)
=
(
)
()
=+
(
2
2
g
tt t
e
001
, ,
t
cos
q
,
t
sin
q
,
t
.
The curve traces out the intersection of the vertical plane through the origin and
e
with our surface
S
. See Figure 15.2. It is easy to check that
¢
()
=
(
)
¢¢
()
=
(
)
g
t
cos
q
, sin
q
,
2
t
and
g
t
002
, ,
.
Now,
Figure 15.2.
Longitudinal curves that are
kinematic geodesics.
