Sphere Geodesic Problem

(a) Find the geodesic curve: the shortest path between two points on thesurface of a sphere.In spherical coordinates (r, , ), a differential element of length in three

dimensions is

ds =

q(dr)2 + (rd)2 + (r sin2 d)2, (1)

where r is the radial distance measured from the origin, is the polar anglemeasured from the positive z-axis and is the angular displacement measuredfrom the x axis in the x y plane. For displacements constrained to lie on thesurface of a sphere, r = R . The total path length between two points 1 and 2on the cylinder may, therefore, be written as

S =

Z 21

ds = R

Z 21

r1 + sin2

2

d Z 21

f(,)d. (2)

Here is chosen as the independent variable, and the dot indicates a totalderivative with respect to . The integrand of Eq.(2) does not depend on .This is one of the special cases we considered in class, and we know immediately

that f/ is a constant, say a. So we can write

f

= a =

sin2 r

1 + sin2 2. (3)

Now solve this equation for by squaring both sides and doing a bit of algebra

to find

=

a

sin psin2 a2

. (4)

Now we integrate this from some arbitrary initial point (1, 1) to a generalfinal point (, ),

1 =Z 1

a

sin psin2 a2

d. (5)

The integral can be simplified by making the substitution

sin2 = csc2 = 1 + cot2 , (6)

which results in

1 = aZ 1

csc2 p1 a2 a2 cot2

d. (7)

1

Then we letu =

a1 a2

cot , (8)

and withdu =

a1 a2

csc2 d, (9)

Eq.(7) becomes

1 = Z u()u(1)

du1 u2

=

Z u()u(1)

d cos1(u) = cos1(u()) cos1(u(1)).

(10)With the help of Eq.(8), we obtain the explicit relation between and

1 = cos1(a1 a2

cot ) cos1( a1 a2

cot 1). (11)

(b) Find a closed form expression for the path length.

Start with Eq.(2) and use Eq.(4) to eliminate. This yields the integral

expression,

S =

Z 21

R sin psin2 a2

d =

Z 21

R sin 1 a2 cos2

d. (12)

Next, introduce a new variable , defined as

cos2 = (1 a2) cos2 . (13)

This allows us to convert Eq.(12) into

S = R

Z 21

d, (14)

and, thus, to obtain the result

S = R

cos1(

cos 21 a2

) cos1( cos 11 a2

)

. (15)

The only undetermined constant in Eqs. (11) and (15) is a. The value of a canbe found by evaluating Eq.(11) at = 2 and = 2 to get

2 1 = cos1(a1 a2

cot 2) cos1(a1 a2

cot 1). (16)

It is then necessary to solve Eq.(16) for a. This actually easier than it looks. Af-ter playing around a bit with this equation, it turns into the following quadraticequation for a2, which is easily solved.

a4A(1, 2, 1, 2) + a2B(1, 2, 1, 2) + C(1, 2, 1, 2) = 0, . (17)

2

where

A = 1 + cos2 1 cos2 2 cos2(2 1) sin2 1 sin2 24 cos(2 1) sin 1 sin 2 cos 1 cos 2, (18)

B = 2 cos2(2 1) sin2 1 sin2 2 + 2 cos(2 1) sin 1 sin 2 cos 1 cos 2 sin2 1 sin2 2, (19)

andC = sin2 1 sin

2 2 cos2(2 1) sin2 1 sin2 2. (20)The value of a2 will depend on the particular end points (1, 1) and (2, 2) ofthe desired path. Quadratics have two roots, so care must be taken to select

the physical value of a2. Also, the sign of a should reflect the sign of for the

specific path (see Eq.(4)).

3