128
2 Variational Problems with Fixed Boundaries
Solve for θ
from the above equation and θ
is recast into differential form, we
have
dθ =
c 1 dϕ
sin ϕ
sin
2
ϕ − c
2
1
(6)
Let η = cot ϕ, then there is dϕ = − sin
2
ϕdη = −
1
1+η 2 dη, substituting it into
Eq. (6), we obtain
dθ =
−dη
1
c
2
1
− 1 − η 2
(7)
Let
1
c 2 =
1
c
2
1
− 1, substituting it into Eq. (7) and integrating, we obtain
θ = arccos(cη) + θ 0 = arccos(c cot ϕ) + θ 0
(8)
or
sin ϕ cos(θ − θ 0 ) = c cos ϕ
(9)
where, θ 0 is an integral constant. Using the angle sum formula of the trigonometric
function, there is
cos(θ − θ 0 ) = cos θ cos θ 0 + sin θ sin θ 0
(10)
Substituting Eqs. (1) and (10) into Eq. (9), we obtain
x cos θ 0 + y sin θ 0 − cz = 0
(11)
This is an equation of a plane through the origin. The circle that the sphere
intercepted by a plane through the spherical center is called the great circle. The arc
greater than the semicircle is called the major arc or superior arc, the arc less than
the semicircle is called the minor arc or inferior arc. This example shows that on
a sphere the shortest curve connecting two fixed points is the minor arc on the great
circle through the two points. Take notice that cos θ 0 is even function, sin θ 0 is odd
function, if θ 0 of the Eq. (11) is changed to −θ 0 , then there is
x cos θ 0 − y sin θ 0 − cz = 0
( 1 2 )