In Chapter 3, we discussed the possibility of choosing a signed arc length parameterization $s$ on a curve. Denote by $\mathbf{R}\left( s\right)$ the corresponding vector equation of the curve. In selecting an arc length parameterization, we have two arbitrary choices to make: the location of the origin, i.e. the point where $s=0$ (which is distinct from the origin $O$), and the orientation of the parameterization, i.e. the direction in which $s$ increases.

An important intuitive characteristic of an arc-length parameterization is the regular spacing in the sense of the distance along the curve between points corresponding to coordinates separated by constant values. This is illustrated in the figure above. Note that while the figure shows a planar curve, arc-length parameterization works perfectly well for curves in a three-dimensional space and throughout the discussion, the reader should try to visualize three-dimensional configurations.

As we established in Chapter 4 for general vector equations of the curve, the derivative $\mathbf{R}^{\prime}\left( s\right)$ is tangential to the curve at each point. However, in the special case of an arc length parameterization, $\mathbf{R}^{\prime}\left( s\right)$ satisfies the additional property that it is unit length, i.e. As a result, the vector $\mathbf{R}^{\prime}\left( s\right)$ is referred to as the unit tangent. We will denote it by the symbol $\mathbf{T}$, i.e.

To see why $\mathbf{R}^{\prime}\left( s\right)$ is unit length, let us examine its construction according to the limiting procedure described in Chapter 4. Consider two nearby points on the curve corresponding to the values of the arc length $s$ and $s+h$. For any value of $h$, the distance $\Delta s$ along the curve between the two points is, by definition, $h$. When we consider a small $h$, and zoom in on the section of the curve between the two points, it will appear to be essentially straight. Therefore, the length of the vector will essentially equal $h$, although, in actuality, it will be slightly shorter. The following figure this insight at $s=1$ for $h=0.4$, where the length of the vector $\Delta\mathbf{R}$ is $0.3998$. We chose this value of $h$ because it is small enough to illustrate the near-equality of $\Delta s$ and the length of $\Delta\mathbf{R}$, yet still large enough for us to see that the curve is not actually straight. Had we chosen $h=0.04$, the corresponding section of the curve would be visually indistinguishable from $\Delta\mathbf{R}$, whose length would be approximately $0.039998$, i.e. $100$ times closer to $\Delta s$ compared to $h=0.4$.

Since for small $h$, the length of the vector $\Delta\mathbf{R}$ is essentially $h$, the length of the vector is essentially $1$ and, in the limit as $h\rightarrow0$, it is $1$, i.e. In other words, as we set out to prove.

Thus, referring to $\mathbf{R}^{\prime}\left( s\right)$ as the unit tangent is justified. Note, however, that at each point there are two unit tangents since the vector $-\mathbf{T}$ is also one. Nevertheless, having arbitrarily selected one unit tangent out of the two available options, we can refer to it as the unit tangent to indicate that the choice has already been made. Between these two legitimate unit tangents, the derivative $\mathbf{R}^{\prime}\left( s\right)$ corresponds to the one that points in the direction of increasing arc-length $s$. In the future, we will observe this interesting lack of uniqueness for other important geometric objects, including the unit normal $\mathbf{N}$ characterizing planar curves.

Let us now return to a curve parameterized by an arbitrary parameter $\gamma$. Our present goal is to demonstrate that the integral where $\gamma_{0} \lt \gamma_{1}$, yields the arc length of the section of the curve between the points $A$ and $B$ corresponding to $\gamma_{0}$ and $\gamma_{1}$. Note that the integrand is simply the length of $\mathbf{R}^{\prime}\left( \gamma\right)$. However we prefer the expression $\sqrt{\mathbf{R}^{\prime }\left( \gamma\right) \cdot\mathbf{R}^{\prime}\left( \gamma\right) }$ since it yields itself more readily to analytical manipulations.

The integral is our first example of a central concept in Tensor Calculus -- that of an invariant. It is an invariant because it yields the same value for any parameterization $\gamma$, provided that the limits of integration correspond to the section of the curve between the points $A$ and $B$ and that $\gamma_{0} \lt \gamma_{1}$. The fact that the above integral is an invariant means that its value is a characteristic of the curve itself rather than that of a particular parameterization. And more often than not, invariants, and especially those given by simple expressions, have a clear geometric meaning, although that meaning may not always be apparent. In the particular case of the above integral, it is the length of the curve between $A$ and $B.$

Before we prove this, we would like to point out two special parameterizations for which it is intuitively so. First, if $\gamma$ represents time and the curve, therefore, corresponds to the trajectory of a material particle, then the vector $\mathbf{R}^{\prime}\left( \gamma\right)$ represents the velocity of the particle, while its magnitude $\sqrt {\mathbf{R}^{\prime}\left( \gamma\right) \cdot\mathbf{R}^{\prime}\left( \gamma\right) }$ represents speed. Thus, is an integral of speed over the period of time from $\gamma_{0}$ to $\gamma_{1}$, and thus represents the total distance traveled between those times. In other words, it is the length of the curve between the points $A$ and $B$.

Second, suppose that $\gamma$ corresponds to arc length $s$, and let the points $A$ and $B$ correspond to the values $s_{0}$ and $s_{1}$, where $s_{0} \lt s_{1}$. The presumed length-of-curve integral for this parameterization reads In the previous Section, we established that the length of $\mathbf{R} ^{\prime}\left( s\right)$ is $1$. Therefore, and we have By definition, the difference $s_{1}-s_{0}$ is precisely to the length of the curve between the points $A$ and $B$, as we set out to show.

The arc length parameterization example can now serve as a starting point for the general demonstration. Having shown that the integral yields the length of the curve between the points $A$ and $B$ for one particular parameterization, we only need to prove that the above integral is independent of parameterization. In other words, we must show that if the curve were alternatively parameterized with the help of another variable $\xi$, then where the values $\xi_{0}$ and $\xi_{1}$ also correspond to the endpoints $A$ and $B$, and $\xi_{0} \lt \xi_{1}$. If the values of both variables $\gamma$ and $\xi$ increase in the same direction along the curve, then $\xi_{0}$ must correspond to $A$ and $\xi_{1}$ to $B$. Otherwise, $\xi_{0}$ must correspond to $B$ and $\xi_{1}$ to $A$.

An important remark regarding our notation is in order. Even though the symbols $\mathbf{R}\left( \gamma\right)$ and $\mathbf{R}\left( \xi\right)$ use the same letter $\mathbf{R}$, they represent different, albeit related, functions. For example, the symbol $\mathbf{R}\left( 1\right)$ may represents both $\mathbf{R}\left( \gamma\right)$ evaluated at $\gamma=1$ and $\mathbf{R}\left( \xi\right)$ evaluated at $\xi=1$. Naturally, those are generally completely different vectors. Nevertheless, the two functions $\mathbf{R}\left( \gamma\right)$ and $\mathbf{R}\left( \xi\right)$ represent the same geometric quantity, i.e. the position vector $\mathbf{R}$, and are therefore closely related enough to justify using the same letter. We can still easily distinguish the two functions by the letter representing the argument, and doing so is a common and necessary practice in Tensor Calculus that will be used extensively throughout our narrative.

Before we give the analytical argument that proves the identity we would like to present a geometric argument that will give you an intuitive feel for it. The essence of the argument is that if the alternative parameterization $\mathbf{R}\left( \xi\right)$ is such that the vector $\mathbf{R}^{\prime}\left( \xi\right)$ is longer (or shorter) compared to $\mathbf{R}^{\prime}\left( \gamma\right)$, then the interval of integration shrinks (or expands) correspondingly, leaving the value of the integral unchanged.

For an illustration, consider the two alternative parameterizations for the same curve in the following figure. Notice that the relationship between $\gamma$ and $\xi$ is Whether we describe this change of variables as stretching or shrinking the parameterization, let us show that the old tangent vector $\mathbf{R}^{\prime}\left( \gamma\right)$ is twice the length of the new tangent vector $\mathbf{R}^{\prime}\left( \xi\right)$.

Consider the point on the curve that corresponds to $\gamma=6$ and $\xi=12$. Next, identify the points that correspond to increasing each parameter by $h=0.5$. The point that corresponds to $\gamma+h=6.5$ is located roughly half way between the points corresponding to $\gamma=6$ and $\gamma=7$. The point corresponding to $\xi+h=12.5$ is located roughly only a quarter of the way between the same two points. Thus, for the same value of $h$, the vector is roughly twice as great as and, in the limit as $h$ approaches zero, we find Thus, the integrand in is twice the integrand in for corresponding values of $\gamma$ and $\xi$. However, the interval of integration in the first integral is half that in the second integral for any given section of the curve. Thus, the two effects balance each other and, as a result, the two integrals yield the same value, as we set out to show.

Let us now turn to the formal proof of the fact that the integral is independent of the parameterization. Consider an alternative parameterization of the curve by the variable $\xi$, where $\xi_{0}$ corresponds to $A$, $\xi_{1}$ corresponds to $B$. Furthermore, assume that $\xi_{0} \lt \xi_{1}$, i.e. the two parameterizations have the same orientation, meaning that the values of $\gamma$ and $\xi$ increase in the same direction along the curve. The proof for the case of opposite orientations is left as an exercise.

Suppose that $\gamma$ and $\xi$ are related by the function $\xi\left( \gamma\right)$, i.e. where For reasons that will become apparent shortly, it is essential to use the same letter $\xi$ to denote both the variable $\xi$ itself and the function that translates the values of $\gamma$ to $\xi$. Since the two parameterizations have the same orientation, $\xi\left( \gamma\right)$ is monotonically increasing and therefore has a positive derivative, i.e.

The functions $\mathbf{R}\left( \gamma\right)$ and $\mathbf{R}\left( \xi\right)$ are related by substituting the function $\xi\left( \gamma\right)$ into $\mathbf{R}\left( \xi\right)$, i.e. This identity makes it apparent why it was necessary to reuse the letter $\xi$ for the function relating the variables $\xi$ and $\gamma$. Had we denoted this function by a different symbol, say $h$, then, in the resulting expression $\mathbf{R}\left( h\left( \gamma\right) \right)$, it would be unclear whether $\mathbf{R}$ refers to $\mathbf{R}\left( \gamma\right)$ or $\mathbf{R}\left( \xi\right)$. In the expression $\mathbf{R}\left( \xi\left( \gamma\right) \right)$, on the other hand, the appearance of the letter $\xi$ makes it clear that $\mathbf{R}$ stands for $\mathbf{R}\left( \xi\right)$.

The equation represents an identity in the independent variable $\gamma$ and can therefore be differentiated with respect to $\gamma$. By the chain rule, we have Substituting this result into the integral we find where, in bringing $\xi^{\prime}\left( \gamma\right)$ from under the square root, we used the fact that $\xi^{\prime}\left( \gamma\right)$ is positive. The resulting integral is tailor-made for an application of the change-of-variables formula from ordinary Calculus, which yields as we set out to show.

For a given parameterization $\gamma$, denote by $s\left( \gamma\right)$ the signed arc length with origin at $\gamma=\gamma_{0}$, i.e. and assume that $s$ and $\gamma$ have the same orientation. Then, thanks to the result we have just established, $s\left( \gamma\right)$ can be expressed by the integral By the Fundamental Theorem of Calculus, the derivative $s^{\prime}\left( \gamma\right)$ is given by

Since $s\left( \gamma\right)$ is monotonically increasing, there also exists an inverse function $\gamma\left( s\right)$ that expresses the parameter $\gamma$ in terms of the signed arc length $s$. Its derivative is, of course, given by It is a common occurrence when working with inverse functions that the derivative of $\gamma\left( s\right)$ is expressed in terms of the values $\gamma$ of the function $\gamma\left( s\right)$ rather than its independent variable $s$. This may seem like an inconvenience, but it will soon prove to our advantage.

If the arc length $s$ has the opposite orientation relative to the parameter $\gamma$, then the function $s\left( \gamma\right)$ is given by the equation while its derivative is given by Then the derivative of the inverse function $\gamma^{\prime}\left( s\right)$ is given by

This completes our discussion of the arc length integral and we will now turn our attention to curvature.

Curvature is one of the central themes in Differential Geometry and we are about to take our initial steps towards its analytical description. Our analysis will be based exclusively on parameterizing the curve by its arc length $s$.

Recall from earlier that the derivative of the position vector $\mathbf{R} \left( s\right)$ is a tangent vector of unit length known as the unit tangent $\mathbf{T}$, i.e. When we want to call attention to the fact that $\mathbf{T}$ is a function of $s$, we will use the symbol $\mathbf{T}\left( s\right)$. Generally, the argument of a function may be included for several purposes. First, when the represented quantity is being differentiated, the argument indicates the independent variable with respect to which the differentiation is taking place. Second, the argument can be used to distinguish between two different functions denoted by the same later -- for example, $\mathbf{T}\left( s\right)$ and $\mathbf{T}\left( \gamma\right)$. Finally, it may be included simply to emphasize the fact that the symbol denotes a function rather than an isolated object.

Let us now consider the derivative $\mathbf{T}^{\prime}\left( s\right)$ of the unit tangent vector $\mathbf{T}\left( s\right)$ and denote it by the symbol $\mathbf{B}$, i.e. In Chapter 4, we demonstrated that the derivative of a vector of constant length is orthogonal to it. Since $\mathbf{T}\left( s\right)$ is a vector of unit (and therefore constant) length, its derivative $\mathbf{B}$, is orthogonal to it, and since $\mathbf{T}$ represents the instantaneous direction of the curve, $\mathbf{B}$ is said to be orthogonal, or normal, to the curve.

Furthermore, since $\mathbf{T}\left( s\right)$ is of constant length, its derivative measures solely the rate at which $\mathbf{T}\left( s\right)$ changes direction. What is the underlying phenomenon responsible for $\mathbf{T}\left( s\right)$ changing its direction? Of course, it is what we intuitively understand to be curvature. The greater the curvature, the greater the rate of change in the direction of $\mathbf{T}\left( s\right)$. The vector $\mathbf{B}$, therefore, quantifies the concept of curvature. Thanks to its two signature properties -- being normal to the curve and being characteristic of curvature -- the vector $\mathbf{B}$ is known as the curvature normal.

Note that $\mathbf{B}$ can be expressed as the second derivative of the position vector $\mathbf{R}\left( s\right)$, i.e. Thus, curvature is a second-derivative phenomenon.

Simply by imagining the way the unit tangent changes its direction as you travel along the curve, you should be able to convince yourself that the curvature normal $\mathbf{B}$ points in the "inward" direction, i.e. the direction towards which the curve is bending. The figure above shows $\mathbf{B}$ for a planar curve. The following figure shows the curvature normal $\mathbf{B}$ for a three-dimensional helical funnel. Note that the displayed length of $\mathbf{B}$ is affected by our angle of view.

Another way to get a sense for the direction of $\mathbf{B}$ is to imagine yourself traveling by car with unit speed along the curve. The velocity of the car corresponds to the tangent vector $\mathbf{T} =\mathbf{R}^{\prime}\left( s\right)$, while the acceleration corresponds to the curvature normal $\mathbf{B}=\mathbf{T}^{\prime}\left( s\right)$. As you go around a bend, you will feel yourself being pulled in the outward direction by the apparent centrifugal force. This is so because your actual acceleration, i.e. $\mathbf{B}$, points inward. Note that this effect is independent of the direction in which the car is travelling and we will now take a closer look at this phenomenon.

Unlike the unit normal $\mathbf{T}$, whose direction depends on the orientation of the parameterization, the curvature normal $\mathbf{B}$ is independent of it. This can be seen in a number of insightful ways. The first way is to consider the finite difference for two parameterizations with opposite orientations, as illustrated in the following figure. Note that in both scenarios, the tangent vector turns inward as $s$ increases and therefore the vector $\Delta\mathbf{T}$ also points in the inward direction. Thus, in the limit as $h\rightarrow0$, the before $\mathbf{B=T} ^{\prime}\left( s\right)$ also points inward.

A related intuitive way to justify the independence of $\mathbf{B}$ from the orientation of the parameterization is to interpret the parameter $s$ as time and to imagine a hammer thrower in the act of spinning the "hammer", i.e. a metal ball attached by a wire to a handle. Prior to its release, the ball is kept in circular motion by the string's tension. The acceleration of the ball points in the direction of the tension force, i.e. inward, which is the case regardless of the direction in which the hammer is is spun.

Finally, the independence of $\mathbf{B}$ from the orientation of the parameterization can be demonstrated by a formal analytical calculation. Let $s_{1}$ be an arc length parameterization that has the opposite orientation with respect to $s$. Assuming $s$ and $s_{1}$ share the same origin, we have If we treat $s$ as a function $s\left( s_{1}\right)$ of $s_{1}$, then Consider the two functions $\mathbf{R}\left( s\right)$ and $\mathbf{R} \left( s_{1}\right)$ that represent the position vector with respect to $s$ and $s_{1}$. Then $\mathbf{R}\left( s_{1}\right)$ can be expressed as a composition of $\mathbf{R}\left( s\right)$ and $s\left( s_{1}\right)$, i.e. Differentiating the above identity with respect to $s_{1}$, we find In other words, at corresponding values of $s$ and $s_{1}$, which confirms that the unit tangents associated with each parameterization point in the opposite directions.

Next, convert the above equation into an identity with respect to $s_{1}$, i.e. Differentiating once again with respect to $s_{1}$, we find In summary, or as we set out to show. In general, odd-ordered derivatives of $\mathbf{R} \left( s_{1}\right)$ and $\mathbf{R}^{\prime}\left( s\right)$ are opposite of each other while even-ordered derivatives are identical.

The magnitude of $\mathbf{B}$ is known as the absolute curvature and is denoted by $\sigma$, i.e. The term absolute refers to the fact that $\sigma$ is always nonnegative. However, absolute is often omitted and $\sigma$ is referred to simply as curvature, although we have to exercise care since we will also work with the related concept of signed curvature $\kappa$ for planar curves which may also be casually referred to as curvature, but may differ in sign from $\sigma$.

The unit vector $\mathbf{P}$ that points in the same direction as $\mathbf{B}$ is called the principal normal. Thus, the curvature normal $\mathbf{B}$ is the product of the absolute curvature $\sigma$ and the principal normal $\mathbf{P}$, i.e.

The principal normal field for a planar curve is illustrated in the following figure. Note, importantly, that the principal normal is undefined at points of inflection where it undergoes a nonremovable discontinuity. The following figure illustrates the principal normal $\mathbf{P}$ for a three-dimensional helical funnel.

By definition, the principal normal $\mathbf{P}$ is found in the normal plane, and thus singles out a particular direction in the two-dimensional normal space. It can therefore act as an element in a natural basis for the normal plane. Later in this Chapter, we will supplement with a vector $\mathbf{Q}$, orthogonal to both $\mathbf{P}$ and $\mathbf{T}$, that will complete the basis.

Let us briefly turn our attention to planar curves. Whereas for curves in three dimensions we find a two-dimensional normal space, planar curves are characterized by a one-dimensional normal space. In other words, the normal direction is unique, and this is a common characteristic of hypersurfaces, i.e. geometric shapes whose dimension trails that of the surrounding space by $1$, such as planar curves and two-dimensional surfaces in a three-dimensional space.

Thus, we know the direction (within the sign) of the principal normal $\mathbf{P}$ as soon as we have calculated the tangent $\mathbf{T}$. In fact, for hypersurfaces, it is common to establish the unique normal direction prior to the analysis of curvature. This is usually accomplished by arbitrarily selecting one of two opposite normal directions. The unit vector $\mathbf{N}$ pointing in the selected direction is referred to as the unit normal. As in the case of the unit tangent, the article the indicates that the choice has already been made. The freedom to choose one of the two opposite unit normals is referred to as the choice of normal. In practice, it is common to choose a consistent normal direction so that the resulting normal field is globally continuous. The following figure shows two alternative choices of normal that result in continuous normal fields.

Recall that unlike the unit normal $\mathbf{N}$, the curvature normal $\mathbf{B}$ is uniquely determined by the shape of the curve. Since $\mathbf{B}$ also points in the normal direction, it is colinear with the unit normal $\mathbf{N}$. Thus, $\mathbf{B}$ is a scalar multiple $\kappa$ of $\mathbf{N}$, i.e. The number $\kappa$ is known as the signed curvature. It is a special case of the beautiful concept of mean curvature that characterizes general hypersurfaces. Mean curvature will be described in a future book and, from the moment of its introduction, will play a crucial role in most of our subsequent explorations.

Crucially, the sign of $\kappa$ depends on the choice of normal. It equals the magnitude of $\mathbf{B}$ when the latter points in the same direction as $\mathbf{N}$ and minus the magnitude of $\mathbf{B}$ otherwise. This fact can be expressed by the equation The same equation in terms of the unit tangent $\mathbf{T}\left( s\right)$ reads

Much like the absolute curvature $\sigma$, the signed curvature $\kappa$ is also often referred to simply as curvature, making the term curvature ambiguous for planar curves. To catalog the relationship between $\sigma$ and $\kappa$, recall that where $\mathbf{P}$ is the principal normal. Note that both $\kappa\mathbf{N}$ and $\sigma\mathbf{P}$ equal the curvature normal $\mathbf{B}$, i.e. Thus, $\sigma=\kappa$ when $\mathbf{B}$ (and $\mathbf{P}$) point in the same direction as $\mathbf{N}$, and $\sigma=-\kappa$ otherwise. The relationship among $\sigma$, $\kappa$, $\mathbf{B}$, and $\mathbf{N}$ is illustrated in the following figure.

To reiterate, our entire discussion of signed curvature has been predicated on the choice of normal $\mathbf{N}$ having been arbitrarily made. Had we made the opposite choice of normal, $\kappa$ would have had the opposite value. Thus, the phrase signed curvature with respect to the normal $\mathbf{N}$ is often used to describe $\kappa$. Meanwhile, the product $\kappa\mathbf{N}$ is independent of the choice of normal since both $\kappa$ and $\mathbf{N}$ change sign when the choice of normal is reversed.

Finally, note that the ambiguity in the choice of normal can be removed by coordinating the unit normal with the unit tangent. For example, we could agree to always choose $\mathbf{N}$ so that the set $\mathbf{T},\mathbf{N}$ is positively oriented. In other words, $\mathbf{N}$ is obtained from $\mathbf{T}$ by a $90^{\circ}$ counterclockwise rotation. There is an advantage to this approach: choosing $\mathbf{N}$ in this way enables us to determine which way the curve is bending simply by examining the sign of $\kappa$. Namely, the curve is bending counterclockwise when $\kappa\gt 0$ and clockwise when $\kappa \lt 0$.

Curvature characterizes the rate at which a curve deviates from being straight. Torsion characterizes the rate which a curve deviates from being planar.

The plane spanned by the tangent and the principal normal directions $\mathbf{T}$ and $\mathbf{P}$ is known as the osculating plane. The verb to osculate comes from the Latin word osculum which means to kiss. While the tangent line captures the instantaneous direction of the curve, the osculating plane captures its instantaneous plane of "motion" as it accommodates both the "velocity" $\mathbf{T}$ and the "acceleration" $\mathbf{B}=\sigma\mathbf{P}$.

The orthonormal vectors $\mathbf{T}$ and $\mathbf{P}$ form a basis for the osculating plane. As a set, $\mathbf{T}$ and $\mathbf{P}$ are one vector short of spanning the entire three-dimensional space. This shortfall can be remedied by adding a unit vector $\mathbf{Q}$ that is orthogonal to both $\mathbf{T}$ and $\mathbf{P}$. Between the two vectors that satisfy this condition, choose $\mathbf{Q}$ so that the set $\mathbf{T},\mathbf{P},\mathbf{Q}$ is positively oriented. (The concept of the orientation of a set of vectors was discussed in Chapter 3.) We wish the vectors $\mathbf{T}$, $\mathbf{P}$, and $\mathbf{Q}$ had arisen in alphabetical order, but alas! Fortunately, the set $\mathbf{P},\mathbf{Q,T}$, which features the same vectors in alphabetical order, has the same orientation as $\mathbf{T},\mathbf{P},\mathbf{Q}$.

The resulting vector $\mathbf{Q}$ is known as the binormal vector or, simply, the binormal. The binormal can be expressed in terms of the unit tangent $\mathbf{T}$ and the principal normal $\mathbf{P}$ with the help of the cross product, i.e. Collectively, the vectors $\mathbf{T}$, $\mathbf{P}$, and $\mathbf{Q}$ are known as a Frenet or Frenet-Serret frame, but can also be referred to as a local frame. The principal normal $\mathbf{P}$ and the binormal $\mathbf{Q}$ form a basis in the normal space.

For the sake of brevity, we will now begin to drop the argument $s$ from most symbols and denote the derivatives by the subscript $s$ rather than a prime, i.e.

Recall that the derivative of the position vector $\mathbf{R}$ with respect to $s$ yields the unit tangent $\mathbf{T}$. The derivative of $\mathbf{T}$, in turn, leads to the concepts of absolute curvature $\sigma$ and principal normal $\mathbf{P}$. As we might expect, the derivative of $\mathbf{P}$ gives rise to yet another differential characteristic of the curve.

Let us represent $\mathbf{P}_{s}$ as a linear combination of $\mathbf{T}$, $\mathbf{P}$, and $\mathbf{Q}$, i.e. and determine each component in turn.

The component $\alpha$ of $\mathbf{P}_{s}$ with respect to $\mathbf{T}$ is given by the dot product By the product rule, Since $\mathbf{P}$ is orthogonal to $\mathbf{T}$, the first term vanishes and we have Recall that $\mathbf{T}_{s}$ is, by definition, the curvature normal $\mathbf{B}=\sigma\mathbf{P}$, i.e. and, therefore, Finally, since $\mathbf{P}$ is unit length, its dot product with itself equals $1$ and therefore This identity makes a great deal of intuitive sense, especially for a planar curve. After all, the unit vectors $\mathbf{P}$ and $\mathbf{T}$ are locked in an orthogonal relationship. As a result, whatever the rate at which $\mathbf{T}$ is rotating -- which, of course, is proportional to $\sigma$ -- $\mathbf{P}$ is rotating at the same rate, as illustrated in the following figure. Therefore, the projection of $\mathbf{T}_{s}$ onto $\mathbf{P}$ equals the projection of $\mathbf{P}_{s}$ onto $-\mathbf{T}$, which is what the above equation is telling us.

The coefficient $\beta$ of $\mathbf{P}_{s}$ with respect to $\mathbf{P}$, given by the dot product is easily seen to be zero, i.e. since $\mathbf{P}\left( s\right)$ is of constant length and therefore $\mathbf{P}_{s}$ is orthogonal to $\mathbf{P}$.

Finally, let us turn our attention to the final coefficient $\gamma$ of $\mathbf{P}_{s}$ with respect to $\mathbf{Q}$ given by Unlike the components of $\mathbf{P}_{s}$ with respect to $\mathbf{T}$ and $\mathbf{P}$, its component with respect to the binormal $\mathbf{Q}$ cannot be expressed in terms of the quantities that have already been introduced. After all, a curve in a three-dimensional space cannot be described by its curvature alone. The absolute curvature characterizes the shape of the curve within the osculating plane spanned by $\mathbf{T}$ and $\mathbf{P}$. However, the derivative $\mathbf{P}_{s}$ is not necessarily contained in the osculating plane. Therefore, its component $\gamma$ with respect to the binormal $\mathbf{Q}$, which is orthogonal to $\mathbf{T}$ and $\mathbf{P}$, captures the rate at which the curve escapes the osculating plane and corresponds to a new geometric quantity.

The component of $\mathbf{P}_{s}$ with respect to the binormal $\mathbf{Q}$ is known as the torsion $\tau$, i.e. In particular, the full expansion of $\mathbf{P}_{s}$ with respect to $\mathbf{T}$, $\mathbf{P}$, and $\mathbf{Q}$ reads

Unlike the absolute curvature $\sigma$, the torsion $\tau$ can be both positive and negative, as will be demonstrated in the upcoming example involving two helices of opposite orientations. Also note that the value of torsion does not depend on the orientation of the parameter $s$. Indeed, recall when the parameterization is reversed, $\mathbf{T}$ changes its direction and $\mathbf{P}$ remains unchanged. Meanwhile $\mathbf{Q}$, being the cross product of $\mathbf{T}$ and $\mathbf{P}$, also changes its direction. It is left as an exercise to show that the derivative $\mathbf{P}_{s}$ also changes its direction. Thus, since both $\mathbf{P}_{s}$ and $\mathbf{Q}$ change their directions, their dot product, i.e. the torsion $\tau$, remains unchanged.

The shape of a spring is known as a helix. A helix is great for illustrating the interplay between curvature and torsion as it is characterized by constant values of both quantities. Helical shapes commonly encountered in everyday life include the aforementioned springs, threads on bolts, spiral staircases, and (with apologies to anyone reading these lines before 1940 or after 2025) slinkies.

A helix can have one of two orientations. Suppose that the axis of a helix is aligned with a ray pointing in the direction arbitrarily labeled as up. Then a right-handed helix twists in the counterclockwise direction as it goes up, while a left-handed helix twists in the clockwise direction. Threads on most bolts and screws are right-handed and are sometimes described by the mnemonic righty-tighty. Importantly, rigidly rotating a helix upside down does not change its orientation. In other words, one of the helices cannot be transformed into the other by turning it upside down. Instead, it will remain equivalent to itself. We know this from our everyday experience and will confirm it analytically by calculating the torsion of each helix.

In the above figure, the helix on the left is right-handed while the one on the right is left-handed. As we discussed above, the orientation of the parameterization does not affect the value of torsion. Therefore, assume that the parameter $s$ increases in the upward direction for both helices. Then, for both shapes, the unit tangent $\mathbf{T}$ points slightly up, while the principal normal $\mathbf{P}$ points directly towards the axis in the strictly horizontal plane. (This property of $\mathbf{P}$ is intuitively clear from the fact that the upward "motion" is uniform and, thus, the "acceleration" has no vertical component.) By the right-hand rule, the binormal $\mathbf{Q}$ points mostly up for the right-handed helix and mostly down for the left-handed helix.

Now, let us get a sense for the derivative $\mathbf{P}_{s}$ of the principal normal $\mathbf{P}$. Since $\mathbf{P}$ remains in the horizontal plane for all values of $s$, its derivative $\mathbf{P}_{s}$ is also found in that plane. Within that plane, $\mathbf{P}$ rotates in the counterclockwise direction for the right-handed helix and in the clockwise direction for the left-handed helix. Furthermore, we know that $\mathbf{P}_{s}$ is also contained in the plane spanned by $\mathbf{T}$ and $\mathbf{Q}$. Thus, let us examine how it is arranged in that plane relative to $\mathbf{T}$ and $\mathbf{Q}$. The following figure shows the plane spanned by $\mathbf{T}$ and $\mathbf{Q}$ for both helices. Since $\mathbf{P}$ rotates in the counterclockwise direction for the right-handed helix, $\mathbf{P}_{s}$ points to the left. Correspondingly, $\mathbf{P}_{s}$ points to the right for the left-handed helix. Observe that for the right-handed helix, the component $\mathbf{P}_{s}$ with respect to $\mathbf{Q}$, i.e. the torsion $\tau$, is positive. Meanwhile, for the left-handed helix, it is negative.

This also proves that one type of helix cannot be transformed into the other by a rigid rotation. Indeed, since torsion depends only on the shape of the curve, it cannot be changed by a rigid rotation. Therefore, the torsion on the helix, say, on the left will remain positive if that helix were rotated upside down. Therefore, the upside-down version of the helix on the left is still fundamentally distinct from the helix on the right whose torsion is negative.

Finally, this is an opportune moment to reiterate an important point that we made at the beginning of the Chapter. Namely, despite the great geometric insight that we have been able to achieve, our geometric approach does not enable us to calculate either the absolute curvature $\sigma$ or the torsion $\tau$ for a specific curve -- at least, not easily. After all, the helix is the simplest conceivable three-dimensional curve with nonvanishing curvature and torsion and we do not yet have an effective tool for determining the numerical value of $\sigma$ or $\tau$. This observation makes clear the need for a more robust analytical network. This will be achieved by the introduction of a coordinate system in the ambient space and the subsequent development of a tensor framework, which will enable us to work with the components of vectors rather than the vectors themselves since the components of vectors can be analyzed by the powerful methods of Calculus or -- when a reasonable analytical approach is not feasible -- numerical methods.

It is only natural to wonder whether the derivative $\mathbf{Q}_{s}$ of the binormal $\mathbf{Q}$ produces yet another differential characteristic of the curve that would join the ranks of curvature and torsion. To this end, let us decompose $\mathbf{Q}_{s}$ with respect to the local frame $\mathbf{T,P,Q}$, i.e.

Since $\mathbf{Q}\left( s\right)$ is unit length, $\mathbf{Q}_{s}$ is orthogonal to $\mathbf{Q}$ and therefore

The components $\alpha$ and $\beta$ of $\mathbf{Q}_{s}$ with respect to $\mathbf{T}$ and $\mathbf{P}$ are given by the dot products In both cases, an application of the product rule will transfer the derivative from $\mathbf{Q}_{s}$ onto the other vector. For $\alpha$, we have Since $\mathbf{Q}$ is orthogonal to $\mathbf{T}$, the first term vanishes and thus Recall that $\mathbf{T}_{s}=\sigma\mathbf{P}$ and therefore Finally, since $\mathbf{Q}$ is orthogonal to $\mathbf{P}$, we conclude that $\alpha$ vanishes, i.e.

For the component $\beta$ of $\mathbf{Q}_{s}$ with respect to the principal normal $\mathbf{P}$, the same approach yields the following chain of identities.

In summary, the complete decomposition of $\mathbf{Q}_{s}$ with respect to $\mathbf{T}$, $\mathbf{P}$, and $\mathbf{Q}$ reads Thus, the analysis of $\mathbf{Q}_{s}$ has not lead to a new differential characteristic of a curve. It would therefore appear that the curvature and the torsion capture all of the available information about the local behavior of a curve. In fact, as we demonstrate below, if the functions $\sigma\left( s\right)$ and $\tau\left( s\right)$ are stated a priori, they are sufficient to reconstruct the shape of the curve in a three-dimensional space.

Let us combine the decompositions of $\mathbf{T}_{s}$, $\mathbf{P}_{s}$, and $\mathbf{Q}_{s}$ in terms of $\mathbf{T}$, $\mathbf{P}$, and $\mathbf{Q}$ into a single set, i.e. Collectively, these equations are known as the Frenet formulas or the Frenet-Serret formulas. In matrix form, the equations read and it becomes apparent how these formulas might generalize to higher-dimensional spaces. Indeed, this generalization will be accomplished in Chapter 20 on Riemannian spaces.

One of the insights provided by the Frenet equations is that the shape of a curve is fully specified by the curvature and torsion as functions of arc length. Suppose that the functions $\sigma\left( s\right)$ and $\tau\left( s\right)$ are given, along with the ambient location of one point on the curve, say, at $s=s_{0}$, and the curve's orientation at that point. In other words, let the values $\mathbf{R}_{0}=\mathbf{R}\left( s_{0}\right)$, $\mathbf{T}_{0}=\mathbf{T}\left( s_{0}\right)$, $\mathbf{P}_{0} =\mathbf{P}\left( s_{0}\right)$, and therefore $\mathbf{Q}_{0} =\mathbf{Q}\left( s_{0}\right) =\mathbf{T}_{0}\times\mathbf{P}_{0}$, be known. Then the entire curve can be reconstructed by solving the system of ordinary differential equations subject to the initial conditions Once $\mathbf{T}$ is calculated, $\mathbf{R}$ can be reconstructed by solving the equation subject to the initial condition Of course, $\mathbf{R}$ could have been included among the unknowns, resulting in a system with $4$ equations and $4$ unknowns, but that would have diminished the elegance of the system and its closure under the functions $\mathbf{T}\left( s\right)$, $\mathbf{P}\left( s\right)$, and $\mathbf{Q}\left( s\right)$.

Because $\sigma\left( s\right)$ and $\tau\left( s\right)$, along with the initial conditions, are sufficient to describe the shape, the location, and the orientation of the curve, these functions are known as the intrinsic, or natural, equations of the curve. In the absence of initial conditions, the functions $\sigma\left( s\right)$ and $\tau\left( s\right)$ are still sufficient to determine the shape of the curve.

The foregoing analysis took fundamental advantage of the special parameterization of the curve with the help of the arc length $s$. In practice, however, an arc-length parameterization is difficult to achieve for specific curves. Meanwhile, a different parameterization may be readily available. It is of interest, then, to adapt the above analysis to a parameterization with an arbitrary variable $\gamma$. Since the only parameter-dependent operation in our analysis is the derivative, all we need to do is express $d/ds$ in terms of $d/d\gamma$.

Suppose that a quantity $T$, scalar or vector, is defined on the curve. Consider it simultaneously as functions $T\left( \gamma\right)$ and $T\left( s\right)$ of $\gamma$ and $s$. The functions $T\left( \gamma\right)$ and $T\left( s\right)$ are related by the identity where $\gamma\left( s\right)$ is dependence of the variable $\gamma$ on the arc length $s$. Then, by the chain rule, we find Isolating the operators $d/ds$ and $d/d\gamma$, we have Recall that the derivative $\gamma^{\prime}\left( s\right)$ is given by the equation This is where the fact that the derivative $\gamma^{\prime}\left( s\right)$ is naturally expressed as a function of $\gamma$ rather than $s$, is advantageous since it enables us to express $d/ds$ explicitly in terms of $\gamma$ and $d/d\gamma$, i.e. With the help of this equation, can now rewrite the Frenet equations for a curve subject to an arbitrary parameterization. Letting we have