An introduction to geodesics: the shortest distance between two points
aa r X i v : . [ m a t h . F A ] J u l AN INTRODUCTION TO GEODESICS: THE SHORTEST DISTANCEBETWEEN TWO POINTS
ANDREW TAWFEEK
Abstract.
We give an accessible introduction and elaboration on the methods used inobtaining a geodesic, which is the curve of shortest length connecting two points lying onthe surface of a function. This is found through computing what’s known as the variationof a functional, a “function of functions” of sorts. Geodesics are of great importance withwide applications, e.g. dictating the path followed by aircraft (great-circles), how lighttravels through space, assist in the process of mapping a 2D image to a 3D surface, androbot motion planning.
Contents
1. Introduction 12. Function Spaces 23. Functionals 34. Variation of a Functional 55. The Euler-Lagrange Equation 86. Geodesics 9References 131.
Introduction
Let S be a surface defined by a vector equation ~r ( u, v ). A geodesic is the shortest curvelying on S connecting two points on the surface of S . This is done by minimizing thefunctional J [ u, v ] = Z t t p Eu ′ + 2 F u ′ v ′ + Gv ′ dt, where E , F , and G are the coefficients of the first fundamental (quadratic) form of thesurface .In this paper we provide the necessary background information needed to understand theconcept of a geodesic, rigorously proving each step along the way. In Section 4, multipleLemmas are put forth concerning function spaces which later greatly assist us in Section 5in proving that a necessary condition for a function to be an extremum for a functional isfor it to satisfy the Euler-Lagrange Differential Equation. Section 6 dives heavily into theconcept of a geodesic and provides two different methods of obtaining it; followed by a fewexamples of both methods in use. Acknowledgments.
I wish to thank the Bristol Community College Honors Programfor providing this unique opportunity to spend the semester working alongside a professor.
Additionally, I am indebted to Zachary Wolfson, my advisor, for his guidance and supportat every step of the way and enduring my endless barrage of questions curiosities every day.2.
Function Spaces
Analogous to the study of n -variable functions, where ( x , x , . . . , x n ) ∈ R n , functionspaces employ similar concepts. Each function y ( x ) is regarded to belong to a certainfunction space containing other functions with similar properties. Spaces whose elementsare functions are called function spaces .In order to properly discuss function spaces, we must first introduce the concepts of a norm and a normed linear space . Definition 1. A linear space is a set ℜ of elements x of any kind (numbers, vectors,matrices, functions, etc.), for which the operations of addition and multiplication by realnumbers a, b, . . . are defined and adhere to the following axioms:(1) x i + x j = x j + x i ;(2) ( x i + x j ) + x k = x i + ( x j + x k );(3) There exits an element 0 such that x i + 0 = x i ;(4) There exists an element − x i such that x i + ( − x i ) = 0;(5) 1 · x y = x i ;(6) a ( bx i ) = ( ab ) x i ;(7) ( a + b ) x i = ax i + bx i ;(8) a ( x i + x j ) = ax i + ax j . Definition 2.
A linear space ℜ is said to be normed if each element x ∈ ℜ is assigned anon-negative number || x || , called the norm of x , such that:(1) || x || = 0 if and only if x = 0;(2) || ax || = | a | · || x || ;(3) || x + y || ≤ || x || + || y || .Notice that by defining || x − y || as the distance between x and y , the norm induces ametric, that is, it satisfies the conditions:(1) || x + y || = 0 if, and only if, x = y ;(2) || x + y || = || x + y || ;(3) || x + y || + || y + z || ≥ || x + z || .The following normed linear spaces are important for definitions and theorems to come:(1) The space C , or more precisely C ( a, b ), consisting of all continuous functions y ( x )defined on an interval a ≤ x ≤ b . By addition and multiplication of elements of C ,we mean ordinary addition of functions and multiplication of functions by numbers.The norm of this space is defined as the maximum of the absolute value, i.e. || y || = max a ≤ x ≤ b | y ( x ) | . Which satisfies the propeties of a norm due to the properties of absolute value.Consequentially, within the space C , the distance || y ( x ) − y ( x ) || between thefunctions y ( x ) and y ( x ) does not exceed ǫ if the graph of y ( x ) lies within a stripof width 2 ǫ centered at y ( x ). Observe, for example, that for a linear space of vectors, the norm is their magnitude, and a for a linearspace of numbers, it would just be their absolute value.
N INTRODUCTION TO GEODESICS: THE SHORTEST DISTANCE BETWEEN TWO POINTS 3 (2) The space D , or D ( a, b ), consisting of all continuous functions y ( x ) defined on aninterval a ≤ x ≤ b with continuous first derivatives. The operations of addition andmultiplication are the same as C , but the norm is defined as || y || = max a ≤ x ≤ b | y ( x ) | + max a ≤ x ≤ b | y ′ ( x ) | . Hence, two functions in are regarded as close if both the functions and their deriva-tives are close together, that is, || y − z || < ǫ implies that for all x ∈ [ a, b ] that | y ( x ) − z ( x ) | < ǫ that | y ( x ) − z ( x ) | < ǫ and | y ′ ( x ) − z ′ ( x ) | < ǫ .(3) The space D n , or D n ( a, b ), consisting of all continuous functions y ( x ) defined onan interval a ≤ x ≤ b which have continuous derivatives up to n , where n ∈ N . Addition and multiplication of elements of D n are defined just as previous cases,but the norm is defined as || y || n = n X i =0 max a ≤ x ≤ b | y ( i ) ( x ) | . Thus, two functions in D n are said to be close together if the values of the functionsand all their derivatives up to n are close together.3. Functionals
Functionals are the main study of Calculus of Variations as well as our tool for obtainingthe geodesic on the surface of a function. An interpretation of this is that a functional is akind of function, where the independent variable itself is a function (or curve).
Definition 3.
A functional is a correspondence which assigns a definite (real) number toeach function (or curve) belonging to some class, that is, J : D n → R , where we chose areasonable integer n such that J is continuous. Definition 4.
The functional J [ y ] is said to be continuous at the point ˆ y ∈ ℜ if for any ǫ >
0, there is a δ > | J [ y ] − J [ˆ y ] | < ǫ provided that || y − ˆ y || < δ .As a general example, the expression J [ y ] = Z ba F ( x, y, y ′ ) dx where y ( x ) ranges over the set of all continuously differentiable functions on [ a, b ], defines afunctional.The integrand is known as the Lagrangian, and we usually assume L ( x, u, p ) to be areasonably smooth function of all three of its (scalar) variables x, u, and p representing dudx = u ′ . By choosing different Lagrangians we can generate different functions. Here aresome examples, the last of which is the two-dimensional analogue of our three-dimensionalgeodesic problem. Note that D n ⊂ D n − ⊂ · · · ⊂ D ⊂ C . ANDREW TAWFEEK
Example 1.
Let L ( x, u, p ) = p and y ( x ) be an arbitrary continuously differentiable func-tion defined on [ a, b ]. We then have Z ba L ( x, u, p ) dx = Z ba (cid:18) dydx (cid:19) dx. Example 2.
Let L ( x, u, p ) = p p dx . The corresponding functional is the length ofthe plane curve y = y ( x ) from a to b Z ba s (cid:18) dydx (cid:19) dx. In order to establish the proper connection between functionals of Calculus of Variationsand the functions treated in Analysis, we may proceed as follows:Consider the aforementioned functional J [ y ] = R ba F ( x, y, y ′ ) dx with y ( a ) = A and y ( b ) = B . Using the points a = x , x , . . . , x i , . . . , x n , x n +1 = b we divide the interval [ a, b ] into n + 1 equal parts. We then replace the curve y = y ( x ) bythe polygonal line with vertices( x , A ) , ( x , y ( x )) , . . . , ( x n , y ( x n )) , ( x n +1 , B )and the approximation of the functional J [ y ] would be given by the sum J ( y , . . . , y n ) = n +1 X i =1 F ( x i , y i , y i − y i − h )∆ x where y i = y ( x i ) and ∆ x = x − x i − .The above sum is therefore a function of the n variables y , . . . , y n and the exact valueof the functional can be found by taking lim n →∞ . In this sense, functionals can be regarded as“functions of infinitely many variables.”The reason the above analogy works so well is because functionals of the type J [ y ] = R ba F ( x, y, y ′ ) dx have a “localization property” consisting of the fact that if we divide thecurve y = y ( x ) into parts and calculate the value of the functional of each part, the sumof the values of the functional for seperate parts equals the value of the functional for thewhole curve . Example 3.
The following functional does not uphold the localization property. Assumingthe curve y ( x ), where a ≤ x ≤ b , is made of some homogeneous material, then the x -coordinate of the center of mass is provided by the functional R ba x r (cid:16) dydx (cid:17) dx R ba r (cid:16) dydx (cid:17) dx . Only functionals with the localization property are usually considered in Calculus of Variations.
N INTRODUCTION TO GEODESICS: THE SHORTEST DISTANCE BETWEEN TWO POINTS 5 Variation of a Functional
The concept of the variation is analogous to the concept of the differential of a functionof n -variables, that is, just as setting dydx = 0 assists in finding the extrema of y = y ( x ),as does setting “ δJ = 0” help locate the extrema of J = J [ y ]. This concept of δJ will bediscussed at the end of the section.We begin by introducing continuous linear functionals, then state a handful of importantlemmas the result from our definition of function spaces. Definition 5.
Given a normed linear space ℜ , let each element h ∈ ℜ be assigned a number φ [ h ], i.e., let φ [ h ] be a functional defined on ℜ . Then φ [ h ] is said to be a (continuous) linearfunctional if(1) φ [ ah ] = aφ [ h ];(2) φ [ h + h ] = φ [ h ] + φ [ h ];(3) φ [ h ] is continuous for all h ∈ ℜ .We shall make use of the following lemmas. Lemma 1. If α ( x ) is continuous in [ a, b ], and if Z ba α ( x ) h ( x ) dx = 0for every function h ( x ) ∈ C ( a, b ) such that h ( a ) = h ( b ) = 0, then, for all x ∈ [ a, b ], α ( x ) = 0. Proof.
Suppose the function α ( x ) is positive and nonzero in some point within [ a, b ]. Then α ( x ) is also positive in some interval [ x , x ] ⊆ [ a, b ]. If we set h ( x ) = ( x − x )( x − x )for x ∈ [ x , x ] and h ( x ) = 0 otherwise, then h ( x ) would clearly satisfy the conditions of thelemma. However, since Z ba α ( x ) h ( x ) dx = Z x x α ( x )( x − x )( x − x )( x ) dx > x and x ), this is a contradiction. (cid:3) Lemma 2. If α ( x ) is continuous in [ a, b ], and if Z ba α ( x ) h ′ ( x ) dx = 0for every function h ( x ) ∈ D ( a, b ) such that h ( a ) = h ( b ) = 0, then, for all x ∈ [ a, b ], α ( x ) = c , where c is a constant. Proof.
Let c be the constant defined by the condition Z ba [ α ( x ) − c ] dx = 0 , and let h ( x ) = Z xa [ ζ ( x ) − c ] dζ, ANDREW TAWFEEK so that h ( x ) automatically belongs to D ( a, b ) and satisfies the conditions h ( a ) = h ( b ) = 0.Then on one hand, Z ba [ α ( x ) − c ] h ′ ( x ) dx = Z ba α ( x ) h ′ ( x ) dx − c [ h ( b ) − h ( a )] = 0 , while on the other hand, Z ba [ α ( x ) − c ] h ′ ( x ) dx = Z ba [ α ( x ) − c ] dx. It follows that α ( x ) = c for all x ∈ [ a, b ]. (cid:3) Lemma 3. If α ( x ) is continuous in [ a, b ], and if Z ba α ( x ) h ′′ ( x ) dx = 0for every function h ( x ) ∈ D ( a, b ) such that h ( a ) = h ( b ) = 0 and h ′ ( a ) = h ′ ( b ) = 0, then,for all x ∈ [ a, b ], α ( x ) = c + c x , where c and c are constants. Proof.
Let c and c be defined by the conditions Z ba [ α ( x ) − c − c x ] dx = 0 , Z ba dx Z xa [ α ( ζ ) − c − c ζ ] dx = 0 , and let h ( x ) = Z xa dζ Z ζa [ a ( t ) − c − c ] dt, so that h ( x ) automatically belongs to D ( a, b ) and satisfies the conditions h ( a ) = h ( b ) = 0and h ′ ( a ) = h ′ ( b ) = 0. We have, on one hand, Z ba [ α ( x ) − c − c x ] h ′′ ( x ) dx = Z ba α ( x ) h ′′ ( x ) dx − c [ h ′ ( b ) − h ′ ( a )] − c Z ba xh ′′ ( x ) dx = − c [ bh ′ ( b ) − ah ′ ( a )] − c [ h ( b ) − h ( a )] = 0 , while on the other hand, Z ba [ α ( x ) − c − c x ] h ′′ ( x ) dx = Z ba [ α ( x ) − c − c x ] dx = 0 . It follows that α ( x ) − c − c x = 0, i.e., α ( x ) = c + c x , for all x in [ a, b ]. (cid:3) Lemma 4. If α ( x ) and β ( x ) are continuous in [ a, b ], and if Z ba [ α ( x ) h ( x ) + β ( x ) h ′ ( x )] dx = 0for every function h ( x ) ∈ D ( a, b ) such that h ( a ) = h ( b ) = 0, then β ( x ) is differentiable and β ′ ( x ) = α ( x ) for all x ∈ [ a, b ]. N INTRODUCTION TO GEODESICS: THE SHORTEST DISTANCE BETWEEN TWO POINTS 7
Proof.
Setting A ( x ) = Z xa α ( ζ ) dζ, and integrating by parts, we find that Z ba α ( x ) h ( x ) dx = − Z ba A ( x ) h ′ ( x ) dx, i.e., the integral within the lemma may be rewritten as Z ba [ − A ( x ) + β ( x )] h ′ ( x ) dx = 0 . According to Lemma 4.2, this implies that B ′ ( x ) = α ( x ) , for all x in [ a, b ], as asserted. Note that β ( x ) was not assumed to be differentiable but resultsfrom Lemma 4.2. (cid:3) Let J [ y ] be a functional defined on some normed linear space, and let∆ J [ h ] = J [ y + h ] − J [ y ]Be its increment , corresponding to the increment of h = h ( x ) of the ”independent variable” y = y ( x ). If y is fixed, ∆ J [ h ] is a functional of h , that is, it is a nonlinear functional, similarto that of the term f ( x + h ) − f ( x ) within Newton’s difference quotient.Now suppose that ∆ J [ h ] = φ [ h ] + ǫ || h || , where φ [ h ] is a linear functional and ǫ → || h || → J [ y ] is said to be differentiable and the principle linear part of the increment ∆ J [ h ] (i.e. the linear functional φ [ h ]) and is called the variation (or differential) of J [ y ] and is denoted by δJ [ y ]. Theorem 5.
A necessary condition for the differentiable function J [ y ] to have an extremeumfor y = ˆ y is that its variation to vanish for y = ˆ y , i.e. that δJ [ h ] = 0 for y = ˆ y and all admissable h . To analogize with Analysis, recall the following:Let F ( x , . . . , x n ) be a differentiable function of n -variables. Then F ( x , . . . , x n ) issaid to have a relative extremum at the point ( ˆ x , . . . , ˆ x n ) if∆ F = F ( x , . . . , x n ) − F ( ˆ x , . . . , ˆ x n )has the same sign for all points ( x , . . . , x n ) belonging to some neighborhood of ( ˆ x , . . . , ˆ x n ),where the extrema F ( ˆ x , . . . , ˆ x n ) is minimum if ∆ F ≥ F ≤ J [ y ] has a (relative) extremum for y = ˆ y if J [ y ] − J [ˆ y ] does not change sign in some neighborhood of the curve y = ˆ y ( x ). Taking ǫ → || h || → h → f ( x + h ) − f ( x ) h . ANDREW TAWFEEK The Euler-Lagrange Equation
We now outline a method for finding the extremum of the simplest type of functional,developed by Euler and Lagrange. This method is our main tool for obtaining geodesics.Lets assume we must face the following problem:Let F ( x, y, z ) be a function with continuous first and second (partial) derivatives withrespect to all its arguments. Then among all functions y ( x ) which are continuously differ-entiable for a ≤ x ≤ b and satisfy the boundary conditions y ( a ) = A and y ( b ) = B , find thefunction for which the functional J [ y ] = Z ba F ( x, y, y ′ ) dx has a relative minimum.Let us begin our solution. Suppose we give y ( x ) an increment h ( x ), where, in order forthe function y ( x ) + h ( x )to continue to satisfy the boundary conditions, we must have h ( a ) = h ( b ) = 0. Then, sincethe corresponding increment of the functional equals∆ J = J [ y + h ] − J [ y ]= Z ba F ( x, y + h, y ′ + h ′ ) dx − Z ba F ( x, y, y ′ ) dx = Z ba [ F ( x, y + h, y ′ + h ′ ) − F ( x, y, y ′ )] dx it follows from Taylor’s Theorem that∆ J = Z ba (cid:20) h ∂∂y F ( x, y, y ′ ) + h ′ ∂∂y ′ F ( x, y, y ′ ) (cid:21) dx + . . . where the “ . . . ” denote terms of order higher than 1 relative to h and h ′ . The integrand onthe right hand side represents the principle linear part of ∆ J , hence the variation of J [ y ] is δJ = Z ba (cid:20) h ∂∂y F ( x, y, y ′ ) + h ′ ∂∂y ′ F ( x, y, y ′ ) (cid:21) dx. According to Theorem 4.5, a necessary condition for J [ y ] to have an extremum at y = y ( x )is that δJ = Z ba [ hF y + h ′ F y ′ ] dx = 0for all admissible h . But, according to Lemma 4.4, this implies that F y − ddx F y ′ = 0a result known as the Euler-Lagrange Equation.Thus, we have proved N INTRODUCTION TO GEODESICS: THE SHORTEST DISTANCE BETWEEN TWO POINTS 9
Theorem 6.
Let J[y] be a functional of the form Z ba F ( x, y, y ′ ) dx, defined on the set of functions y ( x ) ∈ D ( a, b ) , i.e., which have continuous first derivatives in [ a, b ] , and satisfy the boundary conditions y ( a ) = A , y ( b ) = B . Then a necessary conditionfor J [ y ] to have an extremum for a given function y ( x ) is that y ( x ) satisfy the Euler-Lagrangeequation F y − ddx F y ′ = 0 . The curves satisfying the Euler-Lagrange equation are called extremals . Since it is asecond-order differential equation, its solution will in general depend on two arbitrary con-stants, which are determined from the boundary conditions y ( a ) = A and y ( b ) = B .6. Geodesics
Suppose we have a surface S defined by the vector equation(1) ~r ( u, v ) = x ( u, v )ˆ i + y ( u, v )ˆ j + z ( u, v )ˆ k. The shortest curve lying on S and connecting two points of S is called the geodesic connectingthe two points. We are now well prepared to begin our discussion of obtaining this curve.The geodesic curve lying on surface S can be specified by the equations u = u ( t ) v = v ( t )and can be found by minimizing the arc length integral L = Z t t dsdt dt = Z t t s(cid:18) dxdt (cid:19) + (cid:18) dydt (cid:19) + (cid:18) dzdt (cid:19) dt But, since x , y , and z are functions of more than one variable, chain rule states dxdt = ∂x∂u dudt + ∂x∂v dvdt (cid:18) dxdt (cid:19) = (cid:18) ∂x∂u (cid:19) (cid:18) dudt (cid:19) + 2 ∂x∂u ∂x∂v dudt dvdt + (cid:18) ∂x∂v (cid:19) (cid:18) dvdt (cid:19) and similarly for (cid:18) dydt (cid:19) and (cid:18) dzdt (cid:19) . This results in L = Z t t (cid:26)(cid:20) (cid:18) ∂x∂u (cid:19) + (cid:18) ∂y∂u (cid:19) + (cid:18) ∂z∂u (cid:19) (cid:21) (cid:18) dudt (cid:19) + 2 (cid:20) ∂x∂u ∂x∂v + ∂y∂u ∂y∂v + ∂z∂u ∂z∂v (cid:21) dudt dvdt + (cid:20) (cid:18) ∂x∂v (cid:19) + (cid:18) ∂y∂v (cid:19) + (cid:18) ∂z∂v (cid:19) (cid:21) (cid:18) dvdt (cid:19) (cid:27) dt. Which can be rewritten as J [ u, v ] = Z t t p Eu ′ + 2 F u ′ v ′ + Gv ′ dt, where E , F , and G are the coefficients of the first fundamental (quadradic) form of thesurface , i.e., E = ~r u · ~r u = (cid:18) ∂x∂u (cid:19) + (cid:18) ∂y∂u (cid:19) + (cid:18) ∂z∂u (cid:19) F = ~r u · ~r v = ∂x∂u ∂x∂v + ∂y∂u ∂y∂v + ∂z∂u ∂z∂vG = ~r v · ~r v = (cid:18) ∂x∂v (cid:19) + (cid:18) ∂y∂v (cid:19) + (cid:18) ∂z∂v (cid:19) . The Euler-Lagrange equation in this case corresponds to the two different equations F u − ddt F u ′ = 0 , F v − ddt F v ′ = 0 , hence, we obtain(2) E u u ′ + 2 F u u ′ v ′ + G u v ′ √ Eu ′ + 2 F u ′ v ′ + Gv ′ − ddt Eu ′ + F v ′ ) √ Eu ′ + 2 F u ′ v ′ + Gv ′ = 0 , (3) E v u ′ + 2 F v u ′ v ′ + G v v ′ √ Eu ′ + 2 F u ′ v ′ + Gv ′ − ddt F u ′ + Gv ′ ) √ Eu ′ + 2 F u ′ v ′ + Gv ′ = 0 . Which are the two differential equations whose solutions provide the geodesic on surface S . The above case has no restrictions due to the parametrization u = u ( t ) and v = v ( t ).Let us quickly observe a different approach with a single restriction, although resulting inhelpful special cases.Let once more the surface S be given by the vector equation 1. In terms of the differentialsof u and v , the square of the differential of arc length may be wrriten( ds ) = ( dx ) + ( dy ) + ( dz ) = Eu ′ + 2 F u ′ v ′ + Gv ′ If the given fixed points on the surface are ( u , v ) and ( u , v ), with u > u , and we limitour consideration to arcs whos equations are expressible in the form(4) v = v ( u ) v ( u ) ≤ v ≤ v ( u ) , the functional would be instead given by(5) J [ v ( u )] = Z u u ds = Z u u p E + 2 F v ′ + Gv ′ du where v ′ = v ′ ( u ) designates the derivative dvdu (hence u ′ = dudu = 1). With u and v playingthe roles of x and y , the Euler-Lagrange equation now becomes F v − ddu F v ′ = 0(6) E v + 2 v ′ F v + v ′ G v √ E + 2 F v ′ + Gv ′ − ddu (cid:18) F + Gv ′ √ E + 2 F v ′ + Gv ′ (cid:19) = 0 . Now we can observe some special cases resulting from this different approach.
N INTRODUCTION TO GEODESICS: THE SHORTEST DISTANCE BETWEEN TWO POINTS 11
Remark . In the case when E , F , and G are explicit functions of u only, we have(7) F + Gv ′ √ E + 2 F v ′ + Gv ′ = c , which can be solved for v ′ . Doing this gives us v ′ = 12 G ( G − c ) (cid:20) F ( c − G ) ± p F ( G − c ) − G ( G − c )( F − Ec ) (cid:21) . Remark . In the case where E and G are explicit functions of u only and F = 0, that is,the grid lines of u and v are perpendicular, we have v ′ = p G ( G − c ) Ec G ( G − c ) = c s EG ( G − c ) , so we get(8) v = c Z u u c s EG ( G − c ) du. Remark . Similarly, in the case where E and G are explicit functions of v only and F = 0,then E v + v ′ G v √ E + Gv ′ − ddu (cid:18) Gv ′ √ E + Gv ′ (cid:19) = 0 , so E v + v ′ G v − G p E + Gv ′ (cid:20) v ′′ √ E + Gv ′ + (cid:18) (cid:19) v ′ (2 Gv ′ v ′′ )( E + Gv ′ ) (cid:21) = 0 E v + v ′ G v − Gv ′′ + 2 G v ′ v ′′ E + Gv ′ = 0(9) Gv ′ √ E + Gv ′ − p E + Gv ′ = c which can be made even more helpful by noting v ′ = dvdu , giving us Gv ′ − ( E + Gv ′ ) = c p E + Gv ′ (cid:18) − Ec (cid:19) = E + Gv ′ E − c EGc = v ′ , finally providing(10) u = c Z v = v ( u ) v = v ( u ) r GE − c E dv
Example 4.
For a surface obtained through revolving y = f ( x ) about the x -axis over theinterval a ≤ x ≤ b parametrized by x = u y = f ( u ) cos ( v ) z = f ( u ) sin ( v )where a ≤ u ≤ b and 0 ≤ v ≤ π , the first fundamental form would be E = 1 + ( f ′ ( u )) G = ( f ( u )) F = 0 , and by applying the result of Remark 6.2, the geodesic would be given by(11) v = c Z u u p f ′ ( u )) f ( u ) p ( f ( u )) − c du. Example 5.
Consider the circular cylinder ~r ( ϕ, z ) = ( a cosϕ )ˆ i + ( a sinϕ )ˆ j + z ˆ k. Note the coeffecients of the first fundamental form of the cylinder are E = a F = 0 G = 1 , therefore the geodesics of the cylinder have the equations, according to the first method, ddt a ϕ ′ p a ϕ ′ + z ′ = 0 , ddt z ′ p a ϕ ′ + z ′ = 0 , i.e., a ϕ ′ p a ϕ ′ + z ′ = C , z ′ p a ϕ ′ + z ′ = C . Dividing the second equation by the first, we obtain dzdϕ = c , which has the solution z = c ϕ + c , representing a two-parameter family of helical lines lying on the cylinder. Example 6.
Now, consider a sphere of radius a given by the vector equation(12) ~r ( u, v ) = ( a sin ( v ) cos ( u ))ˆ i + ( a sin ( v ) sin ( u ))ˆ j + ( a cos ( v ))ˆ k where it should be simple to notice that u and v form orthogonal grid lines, due to themrepresenting the latitude and longitude, respectively. The first fundamental form of thissphere is E = a sin ( v ) G = a F = 0 . Using the result of Remark 6.3, we have u = c Z dv p a sin ( v ) − c sin ( v )= Z csc ( v ) dv r(cid:16) ( a/c ) − (cid:17) − cot ( v )= − sin − cot ( v ) p ( a/c ) − ! + c
2N INTRODUCTION TO GEODESICS: THE SHORTEST DISTANCE BETWEEN TWO POINTS 13 when rearranged provides( sin ( c )) a sin ( v ) cos ( u ) − ( cos ( c )) a sin ( v ) sin ( u ) − a cos ( v ) p ( a/c ) − . Lastly, upon which noting the parametric equations in equation 12, we may change theabove to see that the geodesic of a sphere lies on x sin ( c ) − y cos ( c ) − z p ( a/c ) − , which is a plane passing through the center of the sphere. Hence, the shortest arc connect-ing two points on the surface of a sphere is the intersection of the sphere with the planecontaining the two points and the center of the sphere, known as the great-circle arc .The concept of a geodesic can be defined not only for surface, by also for higher-dimensional manifolds. Find the geodesics of an n -dimensional manifold reduces to solvinga variational problem for a functional depending on n functions. References [1] Weinstock, R.,
Calculus of Variations, with Applications to Physics and Engineering , New York: Dover,(1974), 32–46.[2] Dacorogna, B.,
Introduction to the Calculus of Variations , Imperial College Press, (2004).[3] Gelfand, I. M., Fomin, S. V.,
Calculus of Variations , Prentice-Hall, (1963), 1–38.[4] Raussen, M.,
Elementary Differential Geometry: Curves and Surfaces , (2008), 108–111.
Bristol Community College, 777 Elsbree St, Fall River, MA
E-mail address ::