Before we turn to our analytical investigations let us begin my making a few insightful geometric observations.

For a developable surface, constructing a geodesic is a straightforward task. Since a developable surface can be transformed into a region of a plane without distortion, every geodesic maps to a straight line in corresponding planar region -- provided that the surface does not require a cut that passes right through the geodesic in order to be transformed into a section of a plane.

This observation suggests a practical approach for constructing the geodesic between two points on a developable surface and that is to 1. isometrically transform the surface onto a region of a plane, 2. connect the images of the two points by a straight line, provided that line is wholly contained within the region of the plane, and 3. isometrically transform the region of a plane back into the original surface.

Of course, this method would yield an incorrect curve if the developable surface required a cut that passed through the actual geodesic. This possibility is illustrated on the surface of a cylinder in the following figure.

Let us now turn our attention to the sphere. On the surface of a sphere, the shortest path connecting two points is a segment of a great circle. Two distinct points on the surface of a sphere uniquely determine a great circle. Naturally, the shorter of the two circular segments represents the shortest distance between two points. Recall from Chapter 9 that the geodesic curvature of a great circle is zero. This is our first, albeit narrow, confirmation that geodesics are characterized by zero curvature.

The sphere also clearly shows us that geodesics are not unique: all "meridians" simultaneously represent the shortest distance between two diametrically opposed points.

Consider the practical task of finding the geodesic connecting two points $A$ and $B$ on a surface specified by analytical means. This task can be accomplished by formulating and solving a set of ordinary differential equations, known collectively as the geodesic equation, for the functions $S^{\alpha}\left( \gamma\right)$ representing the equations of the curve If our intuition regarding the relevance of curvature is correct (and it is!), then we should expect a system of second-order differential equations.

As our discussion will feature numerous instances of the derivatives of $S^{\alpha}\left( \gamma\right)$, let us borrow the "dot" notation from the Calculus of Variations and use the symbols $\dot{S}^{\alpha}$ and $\ddot {S}^{\alpha}$ to denote those derivatives, i.e. You will also find it very helpful to think of $\dot{S}^{\alpha}$ as the speed with which an image particle moves along the curve and and $\ddot{S}^{\alpha}$ as its acceleration.

We will derive the geodesic equation in two ways. In the first approach, we will expand the zero geodesic curvature condition in terms of the underlying functions $S^{\alpha}\left( \gamma\right)$. In the second approach, we will apply the techniques of the Calculus of Variations to the length functional

Regardless of the approach, however, we must recognize that such a system cannot be of the usual "Newton's second law" form i.e. one where the leading-order derivatives are expressed as explicit functions of the lower-order ones. The fundamental reason for that is this: while the solutions of such systems are -- generally speaking -- unique, the functions $S^{\alpha}\left( \gamma\right)$ cannot be unique.

The unavoidable lack of uniqueness is, of course, due to the possibility of reparameterization. After all, we have not restricted the parameterization of the geodesic in any way. As a result, there are infinitely many functions that represent the geodesic connecting the points $A$ and $B$. For example, suppose the geodesic is represented by the equations where, say, $A$ corresponds to $\gamma=0$ and $B$ corresponds to $\gamma=1$. Then it is likewise represented by the equations or, more generally, by the equations where $f\left( \gamma\right)$ is any strictly monotonically increasing function from $0$ to $1$. To use a physical analogy, such a reparameterization changes the speed of a particle moving along a trajectory but not the trajectory itself.

This a priori lack of uniqueness dictates that we must expect two features from the eventual system of differential equations. First, the system will be of the form where the leading derivatives are engaged in a set of implicit functions that cannot be solved uniquely for the leading derivatives $\ddot{S}^{\alpha }\left( \gamma\right)$. And, second, in order to achieve uniqueness, the system must require to be supplemented by a specification of the precise manner in which the parameterization evolves along the curve. For example, we may wish, as is often done, to specify that the geodesic is parameterized by arc length. Then the supplementary condition would be

In this Section, we will describe several intuitive geometric reasons why geodesics are characterized by zero geodesic curvature, i.e.

Recall that the geodesic curvature is a measure of how much the curve bends within the surface. The quintessential example of this perspective is the difference between of a "straight" road and a curved one. Surely, the shortest path between two points on the surface of the Earth is represented by a "straight" road, i.e. a curve characterized by zero geodesic curvature. Of course, a "straight" road is not actually straight but follows a great circle which, as we reminded ourselves earlier, is characterized by zero geodesic curvature.

We may also notice that a great circle has the property that its principal curvature $\mathbf{P}$ is aligned with the surface normal $\mathbf{N}$. A little bit of mental experimentation should convince us that this is a general characteristic of a geodesic. As a visual aid, take a non-spherical curved surface, such as an egg, and draw the shortest path connecting two points. If you visualize the curvature normal $\mathbf{P}$ of the resulting curve and the surface normal $\mathbf{N}$ you will observe that the two vectors are aligned. According to the equation of the three curvatures this relationship between $\mathbf{P}$ and $\mathbf{N}$ implies that the component of the geodesic normal $\mathbf{n}$ vanishes, i.e.

This argument can be augmented by a physical thought experiment. A free particle moving in a Euclidean space without the influence of any forces moves in a straight line which represents the shortest path between any two points that it visits. Suppose now that the particle is constrained to move within surface but without any additional influences, such as friction. Then two things are intuitively clear about the motion of this particle. First, it travels along the shortest path connecting any two points that it visits. Second, the force that constrains it to the motion within the surface acts in the normal direction $\mathbf{N}$. As a result, its acceleration also points along $\mathbf{N}$. This implies that the particle moves with constant velocity and therefore, as we first established in Chapter TBD of Introduction to Tensor Calculus, its acceleration, being $\mathbf{B}=\sigma\mathbf{P}$, also points along the principal normal $\mathbf{P}$. Thus, $\mathbf{N}$ and $\mathbf{P}$ are collinear and therefore and we have another confirmation of the fact that the geodesic curvature vanishes.

Note that this argument also establishes that the constraining force $\mathbf{F}$ acting on the particle is given by which gives additional meaning to the normal acceleration $B_{\beta}^{\alpha }T_{\alpha}T^{\beta}$ and therefore also to the curvature tensor $B_{\beta }^{\alpha}$.

Let us broaden the definition of a geodesic to include any surface curve characterized by zero geodesic curvature, i.e.

By definition, for a geodesic, the equation of three curvatures reduces to Thus, the absolute curvature $\sigma$ of a geodesic equals the absolute value of the normal curvature $B_{\beta}^{\alpha}T_{\alpha}T^{\beta}$, i.e. In this sense, among all possible embedded curves tangential to a given direction $T^{\alpha}$, a geodesic is the curve that is least curved or, to put it another way, is the one that is as straight as possible.

Suppose that a geodesic is described by the equations of the curve The goal of this Section is to construct a set of ordinary differential equations that determine $S^{\alpha}\left( \gamma\right)$. As we discussed earlier in Section 10.2, we expect the resulting system to be of the implicit form that cannot be uniquely solved for the leading derivatives $\ddot{S}\left( \gamma\right)$. Furthermore, we should expect to find a system that requires to be augmented by the rule for parameterizing the curve.

The task at hand can be accomplished by interpreting the analytical definition of a geodesic in terms the underlying functions $S^{\alpha}\left( \gamma\right)$. For the time being, let us switch from the symbol $\gamma$ to $S^{\Phi}$.

Recall that the geodesic curvature tensor $b_{\Phi\Psi}$ is given by and therefore However, for our purposes, we need the more general identity From this identity we see that the condition of zero geodesic curvature translates into equations

Suppose that the equations of the curve are Recall that the shift tensor $S_{\Phi}^{\alpha}$ represents the derivative of $S^{\alpha}\left( S\right)$ with respect to the curve coordinate $S^{\Phi}$, i.e. By directly appealing to the definition of the covariant derivative, we have Thus, $\nabla^{\Phi}S_{\Phi}^{\alpha}$ is given by Equating it to zero yields It is left as an exercise to show that the term $\Gamma_{\Phi\Psi}^{\Omega }S_{\Omega}^{\alpha}$ is given by

Let us now switch back to the parameter $\gamma$ and note that the term $\partial^{2}S^{\alpha}/\partial S^{\Phi}\partial S^{\Psi}$ corresponds to $\ddot{S}^{\alpha}$ while the shift tensor $S_{\Phi}^{\alpha}$ corresponds to $\dot{S}^{\alpha}$. With this in mined, note that the factor of $S^{\Phi\Psi}$ can be cancelled, and we are left with Combining the leading terms, we have which can be written in the form

With the help of the symbol it can be rewritten more compactly as We will return to the analysis of this equation after we derive it again with the help of the Calculus of Variations.

Perhaps the most straightforward analytical approach to deriving the geodesic equation is the Calculus of Variations. The classical problem of the Calculus of Variations is to minimize or maximize the functional with respect to all possible smooth functions $y\left( \gamma\right)$ for $a\leq\gamma\leq b$. The solution of the problem is represented by the celebrated Euler-Lagrange equation which reads In the case of multidimensional functions $y^{\alpha}\left( \gamma\right)$, the Euler-Lagrange equation becomes a system which reads

Thus, the problem of describing a geodesic, is precisely an example of the classical problem of the Calculus of Variations where the functions $S^{\alpha}\left( \gamma\right)$ play the role of the independent variations. The length $L$ of the segment of the embedded curve corresponding to the values of $\gamma$ between $a$ and $b$ is given by the integral where the metric tensor $S_{\alpha\beta}$ is treated as a function of $S^{\alpha}\left( \gamma\right)$.

For this functional, the Euler-Lagrange equation reads Let us now treat each of these terms one by one.

By the chain rule Recall that Substituting this identity into the previous equation, we find Notice, that owing to the symmetry of the combination $\dot{S}^{\beta}\dot {S}^{\gamma}$, we have Therefore,

The analysis of this term is certainly more involved. For the partial derivative, we have by the chain rule By the product rule, Substituting this into the preceding equation, we get Owing to the symmetry of the metric tensor, the two terms in the numerator are equal and therefore

Next comes the application of the time derivative $d/dt$. By the product rule, The first two terms are rather straightforward. Since we have For the second term, since $d\dot{S}^{\beta}/dt=\ddot{S}^{\beta}$, we have Turning our attention to the most labor-intensive third term, we have by the chain rule Then, we have by the product rule, We have already discussed how to deal with each of the terms and so we jump to the form Owing to the available symmetries, we have Thus, Finally, combining the three terms, we find

Putting the two terms together, the combined Euler-Lagrange equation reads Observe that the terms $\Gamma_{\beta,\gamma\alpha}\dot{S}^{\beta}\dot {S}^{\gamma}$ and $\Gamma_{\beta,\alpha\gamma}\dot{S}^{\beta}\dot{S}^{\gamma}$ are equivalent and therefore cancel, and therefore we are left with Next, multiply the equation by the nonzero quantity $-\sqrt{S_{\sigma\tau} \dot{S}^{\sigma}\dot{S}^{\tau}}$, which yields Next, raise the index $\alpha$, which yields Then, recall the notation We have Finally, after rearranging the terms and renaming several indices, we arrive at the exact same equation we obtained by our earlier analysis based on the zero geodesic curvature condition. Note, however, that had this been our only approach, we would have had no way of knowing the geometric interpretation of this equation.

The equation is precisely of the implicit form that we hypothesized in Section 10.2. We further hypothesized that these equations cannot be solved uniquely for the acceleration $\ddot {S}^{\beta}$. This is indeed the case for this system and to understand why this is so, let us consider the second-order system Let us ignore the placement of indices for a moment. Note that if $\dot {S}^{\alpha}$ corresponds to a vector $v$, then the object above corresponds to the symmetric matrix This matrix is necessarily nonsingular. Indeed, its null space is represented by the vector $v$ since It is left as an exercise to show that the null space of $A$ is strictly one-dimensional.

For a specific illustration, let Then and you can easily check that the null space $A$ in fact corresponds to $v$.

Taking the placement of indices into account does not change anything since it merely changes into where $M$ is the symmetric matrix corresponding to $S_{\alpha\beta}$. Then To repeat this logic in the tensor notation, observe the following chain of identities:

Recall from Linear Algebra that the general solution of a linear system where $A$ is a singular matrix with the null space spanned by a vector $v$, is given by where $x_{0}$ is any solution to the system, known as a particular solution, and $\alpha$ is an arbitrary number. Rewrite the geodesic equation in the shape of a linear system, i.e. Then the solution $\ddot{S}^{\alpha}$ of this linear system at a given moment of "time" $\gamma$ would have the form where, once again, $\ddot{S}_{0}^{\beta}$ is any solution and $c$ is an arbitrary number which we have no way of determining from the system alone.

Note, however, that the undetermined part $c\dot{S}^{\alpha}$ of the acceleration $\ddot{S}^{\alpha}$ is proportional to the velocity $\dot {S}^{\alpha}$. In other words, the particular value of $c$ determines the speed with which the imagined particle moves along the trajectory and not the trajectory itself. In other words, what is missing from the geodesic equation is the choice of parameterization and supplying one makes the system complete.

For example, let us specify that the geodesic is parameterized by arc length, i.e. It is left as an exercise to show that differentiating this identity with with respect to $\gamma$ (i.e. arc length) yields Thus, or Solving for $c$, we find Thus,

In the Euclidean plane referred to affine coordinates, the surface Christoffel symbols vanish and therefore the geodesic equations become This is undoubtedly the simplest form of the geodesic equation that still exhibits the feature associated with the arbitrariness of parameterization. Solving for $\ddot{S}^{\beta}$, we find In other words Let us stipulate that Thus So Thus and

A tremendous simplification occurs for arc-length parameterization: or This is the most familiar form of the geodesic equation

Let us recount, in broad terms, how the Euler-Lagrange equation is derived for the functional The starting point for the derivation is to consider a family of functions $y\left( t,\gamma\right)$ enumerated by a smooth parameter $t$ in which the sought-after optimal function occurs at $t=0$. Then the quantity $E$ also becomes a function of $t$, i.e. whose minimum occurs at $t=0$. Thus, the analysis proceeds by equating to $0$ the derivative of the integral with respect to $t$.

To apply this tactic to the geodesic problem, consider a family of curves given by the equations A specific family of such curves may appear as in the following figure. Correspondingly, the curve length $L$ becomes a function of $t$, i.e. where $\dot{S}^{\alpha}$ still denotes the derivative with respect to $\gamma$.