<<

GEODESICS ON SURFACES BY VARIATIONAL CALCULUS

J Villanueva Florida Memorial University 15800 NW 42nd Ave Miami, FL 33054 [email protected]

1. Introduction 1.1 The problem by variational calculus 1.2 The Euler-Lagrange

2. The geodesic problem: general formulation

3. Examples 3.1 Plane 3.2 3.3 Right circular cylinder 3.4 Right circular cone 3.5 Hyperbolic paraboloid

4. Applications 4.1 between any two cities on the

References:

1. Livio, M, 2005. The Equation that Couldn’t be Solved. NY: Simon & Schuster. 2. Oprea, J, 1997. Differential . NJ: Prentice Hall. 3. Stewart, J, 2003. Multivariable Calculus. CA: Brooks/Cole Thomson. 4. Thomas, G, 1972. Calculus. NY: Addison Wesley. 5. Weinstock, R, 1974. . NY: Dover. 6. Wolfram Mathworld, Internet. 7. Wylie, CR, & LC Barnett, 1995. Advanced . NY: McGraw-Hill.

Geodesics are of shortest distance on a given . Apart from their intrinsic interest, they are of practical importance in the transport of goods and passengers at minimal expense of time and energy. They are also of paramount importance as escape routes during flights. Finding geodesics can be accomplished using the methods of . We will use instead the calculus of variation, which we have used before in solving the brachistochrone problem.

The fundamental equation in the calculus of variations is the Euler-Lagrange equation:

휕푓 푑 휕푓 (1) − ( ) = 0, 푥 ≤ 푥 ≤ 푥 . 휕푦 푑푥 휕푦′ 1 2

In , we are looking for those values of 푥 which give some function 푓(푥) its maximum or minimum values. In the calculus of variation, we are seeking the function 푓 itself that makes some integral of 푓, satisfying certain conditions, a maximum or minimum. In the geodesic problem, we wish to find that 푓 = 푓(푥, 푦, 푦′) that joins two points on a given surface such that the distance between them is minimized. Thus, the problem is to find that integrand 푓 which minimizes the integral of the arcclength:

푥 (2) 퐿 = ∫ 푑푠 = ∫ 2 √1 + (푦′)2푑푥, 푓 = √1 + (푦′)2. 푥1

Here, 푓 = 푓(푥, 푦, 푦′),

푑 휕푓 휕 휕푓 휕 휕푓 휕 휕푓 and, ( ) = ( ) + ( ) + ( ). 푑푥 휕푦′ 휕푥 휕푦′ 휕푦 휕푦′ 휕푦′ 휕푦′

Hence, (1) is equivalent to:

′ ′′ (3) 푓푦 − 푓푦′푥푓푦′푦푦 − 푓푦′푦푦 = 0.

There are first integrals of (1) based on two simplifying assumptions: (i) If 푓 is independent of 푦, 휕푓 i.e., = 0. Then 휕푦

푑 휕푓 휕푓 (4) ( ) = 0 ⇒ = 퐶 푑푥 휕푦′ 휕푦′ a first-order degree 1 differential equation.

(ii) If 푓 is independent of 푥:

푑 휕푓 푑 휕푓 휕푓 휕푓 휕푓 푑 휕푓 휕푓 (5) (푦′ − 푓) = 푦′ ( ) − − 푦′ = −푦′ [ − ( )] − , 푑푥 휕푦′ 푑푥 휕푦′ 휕푥 휕푦 휕푦 푑푥 휕푦′ 휕푥

The last term = 0, ⇒

휕푓 (5’) 푦′ − 푓 = 퐶. 휕푦′

The geodesic problem may be recast in curvilinear coordinates. The surface 푔(푥, 푦, 푧) = 0 in parametric form: (6) 푥 = 푥(푢, 푣), 푦 = 푦(푢, 푣), 푧 = 푧(푢, 푣).

The arclength 푑푠 is:

(7) (푑푠)2 = (푑푥)2 + (푑푦)2 + (푑푧)2 = 푃푑푢2 + 2푄푑푢푑푣 + 푅푑푣2

휕푥 2 휕푦 2 휕푧 2 휕푥 2 휕푦 2 휕푧 2 with 푃 = ( ) + ( ) + ( ) , 푅 = ( ) + ( ) + ( ) , 휕푢 휕푢 휕푢 휕푣 휕푣 휕푣

휕푥 휕푥 휕푦 휕푦 휕푧 휕푧 푄 = + + . 휕푢 휕푣 휕푢 휕푣 휕푢 휕푣

And the integral (2) to minimize is:

푢 (8) 퐿 = ∫ 2[푃 + 2푄푣′ + 푅푣′2]1/2푑푢. 푢1

The first integrals (4) and (5’) become (9) and (10):

푑푣 with 푥 → 푢, 푦 → 푣, 푣′ = : 푑푢

휕푃 휕푄 2휕푅 +2푣′ +푣′ 푑 푄+푅푣′ 휕푣 휕푣 휕푣 − ( ) = 0. 2√푃+2푄푣′+푅푣′2 푑푢 2 √푃+2푄푣′+푅푣′

If 푃, 푄, 푅 are explicit functions of 푢 alone, the first term is zero, and further if 푄 = 0, then

푅푣′ 2 = 퐶 √푃+푅푣′

And the solution is:

√푃 (9) 푣 = 퐶 ∫ 푑푢. √푅2−퐶2푅

If 푄 = 0 and 푃, 푅 are explicit functions of 푣 alone:

휕푓 푣′ − 푓 = 퐶, 휕푣′

2 푅푣′ √ ′2 or, 2 − 푃 + 푅푣 = 퐶, √푃+푅푣′ with the solution:

√푅 (10) 푢 = 퐶 ∫ 푑푣. √푃2−퐶2푃

푥 Examples of geodesics. Surface 1: A plane. Our problem is: 퐿 = ∫ 2 푓(푥, 푦, 푦′)푑푥, 푥1 with 푓 = √1 + 푦′2, independent of 푦, thus, we have solution (4):

휕푓 푦′ = = 퐶 휕푦′ √1+푦′2

퐶2 푑푦 ⇒ 푦′2 = 퐶2(1 + 푦′2) ⇒ 푦′2 = = 퐶′2 ⇒ = 퐶′ 1−퐶2 푑푥

(11) ∴ 푦 = 퐶′푥 + 퐶′′.

Thus, 푦 is a linear function of 푥, with the constants C’ and C’’ determined by joining 푥1 to 푥2. The geodesic on a plane is a straight between the two points (Figure 1).

Figure 1. Geodesic on a plane

Surface 2: Sphere

푥 = 푎 sin 푣 cos 푢, 푦 = 푎 sin 푣 sin 푢, 푧 = 푎 cos 푣, with 푢 = , 푣 = colatitude, a = radius.

Then, 푃 = 푎2푠𝑖푛2푣, 푄 = 0, 푅 = 푎2, and 푓 = √푃 + 2푄푣′ + 푅푣′2.

The solution is given by (10):

√푅 푎푑푣 푐푠푐2푣푑푣 푢 = 퐶 ∫ 2 2 푑푣 = 퐶 ∫ 4 4 2 2 2 = ∫ √푃 −퐶 푃 √푎 푠푖푛 푣−퐶 푎 푠푖푛 푣 푎 2 2 √[( ⁄퐶) −1]−푐표푡 푣

cot 푣 = −푠𝑖푛−1 + 퐶′. 푎 2 √( ⁄퐶) −1

The last equation may be rewritten: 푎 cos 푣 (sin 퐶′)푎 sin 푣 cos 푢 − (cos 퐶′)푎 sin 푣 sin 푢 − = 0 푎 2 √( ⁄퐶) −1

푧 (12) ∴ 푥 sin 퐶′ − 푦 cos 퐶′ − = 0. 푎 2 √( ⁄퐶) −1

The geodesic is the intersection of the sphere with a plane through its center connecting the two points on its surface – a great circle.

Figure 2. Spherical coordinates (휃 → 푢, 휙 → 푣)

Figure 3. Geodesic on a sphere: a great circle

Surface 3: Right circular cylinder

푥 = 푎 cos 휃, 푦 = 푎 sin 휃, 푧 = 푢, 푎 = radius of the cylinder;

푑푠2 = 푑푥2 + 푑푦2 + 푑푧2 = (−푎 sin 휃)2 + (푎 cos 휃)2 + 푑푧2 = (푎푑휃)2 + 푑푧2. With 푧 → 푢, 휃 → 푣, 휃′ → 푣′: 퐿 = ∫ 푑푠 = ∫ √푎2푣′2 + 1푑푢 ⇒ 푓 = √푎2푣′2 + 1.

Here, 푓 is independent of 푢 and 푣, and we have solution (9): 푣′ = 퐶′ ⇒

(13) 휃 = 퐶′푧 + 퐶′′.

The geodesic on a right circular cylinder is a cylindrical spiral – a helix.

Figure 4. Cylindrical coordinates (푧 → 푢, 휃 → 푣)

Figure 5. Cylindrical spiral

Surface 4: Right circular cone. Here,

푥 = 푢 sin 훼 cos 푣, 푦 = 푢 sin 훼 sin 푣, 푧 = 푢 cos 푣

푑푠2 = 푑푢2 + (푢 sin 훼)2 푑푣2, 푟 → 푢, 휙 → 푣, 훼 = apex angle, 퐿 = ∫ 푑푠 = ∫ √1 + (푢 sin 훼)2 푣′2푑푢, 푓 = √1 + (푢 sin 훼)2 푣′2, independent of 푣.

휕푓 The solution is given by (9): = 퐶 휕푣′

퐶 ⇒ (푢 sin 훼)푣′ = √(푢 sin 훼)2−퐶2

푑푟 1 sin 훼 (14) ⇒ 휙 = ∫ = 푠푒푐−1 [( ) 푟 + 퐶′]. 2 sin 훼 퐶 √ 푟 sin 훼 푟 sin 훼 ( ⁄퐶) −1

Thus, the geodesics are spirals on the surface of the cone.

Figure 6. Right circular conical coordinates

Figure 7. Cone geodesic

Surface 5: Hyperbolic paraboloid. This is the saddle-point surface: 푥 = sinh 푢, 푦 = cosh 푢, 푧 = 푣, with parabolas for cross-sections in the 푥푧 – plane and 푦푧 – plane, and hyperbolas in the 푥푦 – plane. We have:

푑푠2 = (푐표푠ℎ2푢 + 푠𝑖푛ℎ2푢)푑푢2 + 푑푣2 = cosh 2푢푑푢2 + 푑푣2

퐿 = ∫ 푑푠 = ∫ √cosh 2푢 + 푣′2 푑푢, 푓 = √cosh 2푢 + 푣′2, independent of 푣.

휕푓 The solution is given by (9): = 퐶 휕푣′

(15) ⇒ 푣 = 퐶′ ∫ √cosh 2푢 푑푢.

Alas, there is no closed form to this integral. It turns out that this integral can be put in the form of an incomplete elliptic integral of the second kind, with solution:

푣 = −𝑖퐸(𝑖푢|2) + 퐶′′.

Figure 8. Hyperbolic paraboloid: 푧 = 푦2 − 푥2

For this less familiar surface, it is known that the sum of the interior angles of a triangle on the surface is less than 1800. It would have been interesting to see what kind of geodesics it has, but they could only be displayed parametrically.

Figure 9. Triangles on a plane, a sphere, and a saddle-point surface As an application, consider the calculation of great circle routes on a sphere, in particular, for any two cities on the Earth. The airlines have long discovered this, with flight paths between cities as sections of planes through the Earth’s center.

Figure 10. Approximate great circle routes for airlines

Figure 11. Great circle path between two cities on the Earth Let P and Q represent any two cities on the Earth’s surface. The of the vectors 푂푃⃗⃗⃗⃗⃗ and 푂푄⃗⃗⃗⃗⃗⃗ from the origin O gives the central angle 훼 between 푂푃⃗⃗⃗⃗⃗ and 푂푄⃗⃗⃗⃗⃗⃗ , in radians, and the formula 푠 = 푅훼 gives the arclength 푠 along a great circle between points P and Q. 푅 = 3960 푚𝑖, the Earth’s radius.

푂푃⃗⃗⃗⃗⃗ = (푥1, 푦1, 푧1), 푂푄⃗⃗⃗⃗⃗⃗ = (푥2, 푦2, 푧2). Convert to spherical coordinates:

푥 = 푅 sin 휙 cos 휃, 푦 = 푅 sin 휙 sin 휃, 푧 = 푅,

휃 = longitude 휙 = colatitude

→ 휃, if East of Greenwich → 900 − 휙, if North Hemisphere

→ 3600 − 휃, if West of Greenwich → 900 + 휙, if South Hemisphere

푂푃⃗⃗⃗⃗⃗ ∙ 푂푄⃗⃗⃗⃗⃗⃗ = 푅2 cos 푎, and ∴ 푠 = 푅훼, miles.

Figure 12. Great circle distance s between P and Q

Example 1. Miami (25047′푁, 80013′푊), Manila (14035′푁, 120058′푁) → 푠 = 9,308 mi

Example 2. Chicago (41053′푁, 87038′푊), Tokyo (35041′푁, 139042′퐸) → 푠 = 6,306 mi

Example 3. NP (900푁, 00퐸), Quito (0015′푆, 78035′푊) → 6,238 mi.

This last example serves as a check: between the North Pole and the Equator is a quarter of the Earth’s circumference, 푠 = 2휋푅⁄4 = 6,220 mi, the difference between the values to account for the slight deviation of Quito into the South Hemisphere.

Geodesics are minimal arcs between two points on a surface. There are several ways to calculate geodesics. Here we found them directly by the calculus of variations. The main process is solving the Euler-Lagrange partial differential equation for the function that minimizes the arclength between two points on the surface. The integral is the parametric equation of the geodesic. For familiar surfaces, like the plane, sphere, cylinder, and cone, the results were also familiar because the integrals of the Euler-Lagrange equation could be put in standard forms and worked out nicely. For a less familiar surface, such as the hyperbolic paraboloid, the integration could get very complicated and foils our attempt at getting a simple classical standard form. For application, the calculation of great circle between any two cities on the Earth was very straightforward.