The Curvature Tensor

This entire Chapter is devoted to the exploration of the curvature tensor BαβB_{\alpha\beta}. We will begin by recalling the key relationships involving the curvature tensor BαβB_{\alpha\beta} that we have established in the earlier Chapters.
Our analytical exploration of curvature started with the realization that the spatial variability of the surface covariant basis Sα\mathbf{S}_{\alpha} comes from two sources: the potential non-uniformity of the surface coordinates SαS^{\alpha} and the non-flatness -- in other words, curvature -- of the surface. While the non-uniformity of the surface coordinates can be effectively managed by switching from partial to covariant differentiation, the curvature of the surface is a matter of actual geometry. We therefore embrace it rather than seek to eliminate it from consideration by some analytical device.
The opportunity to quantify curvature presented itself when we discovered that each of the vectors in the tensor αSβ\nabla_{\alpha}\mathbf{S}_{\beta} is orthogonal to the surface and therefore collinear with the unit normal N\mathbf{N}. We then assembled corresponding coefficients of proportionality into a second-order system called the curvature tensor BαβB_{\alpha \beta}, i.e.
αSβ=NBαβ.(2.79)\nabla_{\alpha}\mathbf{S}_{\beta}=\mathbf{N}B_{\alpha\beta}. \tag{2.79}
Dotting both sides of the above equation with N\mathbf{N} yields an explicit expression for BαβB_{\alpha\beta} captured by the equation
Bαβ=NαSβ.(2.80)B_{\alpha\beta}=\mathbf{N}\cdot\nabla_{\alpha}\mathbf{S}_{\beta}. \tag{2.80}
Since the covariant basis Sβ\mathbf{S}_{\beta} is the derivative of the position vector R\mathbf{R}, i.e.
Sβ=βR,(6.1)\mathbf{S}_{\beta}=\nabla_{\beta}\mathbf{R,}\tag{6.1}
the above identities may also be written in forms
αβR=NBαβ.(6.2)\nabla_{\alpha}\nabla_{\beta}\mathbf{R}=\mathbf{N}B_{\alpha\beta}.\tag{6.2}
and
Bαβ=NαβR(6.3)B_{\alpha\beta}=\mathbf{N}\cdot\nabla_{\alpha}\nabla_{\beta}\mathbf{R}\tag{6.3}
in terms of the second derivatives of R\mathbf{R}. Note that while, in general, the covariant derivatives α\nabla_{\alpha} and β\nabla_{\beta} do not commute, they do commute when applied to a variant of order zero. Indeed, for a zeroth-order variant UU, we have
αβU=2USαSβΓαβωUSω.(6.4)\nabla_{\alpha}\nabla_{\beta}U=\frac{\partial^{2}U}{\partial S^{\alpha }\partial S^{\beta}}-\Gamma_{\alpha\beta}^{\omega}\frac{\partial U}{\partial S^{\omega}}.\tag{6.4}
Since the Christoffel symbol is symmetric in its subscripts, i.e.
Γαβγ=Γβαγ,(6.5)\Gamma_{\alpha\beta}^{\gamma}=\Gamma_{\beta\alpha}^{\gamma},\tag{6.5}
both of the terms in the expression for αβU\nabla_{\alpha}\nabla_{\beta}U are symmetric in α\alpha and β\beta and therefore
αβU=βαU.(6.6)\nabla_{\alpha}\nabla_{\beta}U=\nabla_{\beta}\nabla_{\alpha}U.\tag{6.6}
Therefore αβR\nabla_{\alpha}\nabla_{\beta}\mathbf{R} is symmetric, i.e.
αβR=βαR.(6.7)\nabla_{\alpha}\nabla_{\beta}\mathbf{R}=\nabla_{\beta}\nabla_{\alpha }\mathbf{R.}\tag{6.7}
In other words,
αSβ=βSα,(2.78)\nabla_{\alpha}\mathbf{S}_{\beta}=\nabla_{\beta}\mathbf{S}_{\alpha}, \tag{2.78}
from which it follows that the curvature tensor BαβB_{\alpha\beta} is symmetric, i.e.
Bαβ=Bβα(2.81)B_{\alpha\beta}=B_{\beta\alpha} \tag{2.81}
or, equivalently,
Bβα=Bβα.(2.82)B_{\cdot\beta}^{\alpha}=B_{\beta}^{\cdot\alpha}. \tag{2.82}
Because of the symmetry of the curvature tensor, its mixed form
Bβα(6.8)B_{\beta}^{\alpha}\tag{6.8}
can be written in this simplified form without the use of the dot placeholder. This subtle point was discussed in Chapter TBD of Introduction to Tensor Calculus. Finally, recall that, as we discussed in Chapter 2, the covariant derivative in the identity
Bαβ=NαSβ(2.80)B_{\alpha\beta}=\mathbf{N}\cdot\nabla_{\alpha}\mathbf{S}_{\beta} \tag{2.80}
may be replaced with the partial derivative, i.e.
Bαβ=NSβSα.(2.83)B_{\alpha\beta}=\mathbf{N}\cdot\frac{\partial\mathbf{S}_{\beta}}{\partial S^{\alpha}}. \tag{2.83}
In Chapter 5, we observed that BαβB_{\alpha\beta} can be computed in the component space by the equation
Bαβ=NiαZβi(5.54)B_{\alpha\beta}=N_{i}\nabla_{\alpha}Z_{\beta}^{i} \tag{5.54}
and the Weingarten equation
Bαβ=ZβiαNi,(5.55)B_{\alpha\beta}=-Z_{\beta}^{i}\nabla_{\alpha}N_{i}, \tag{5.55}
where ZβiZ_{\beta}^{i} is the shift tensor obtained by differentiating the equations of the surface Zi(S)Z^{i}\left( S\right) , i.e.
Zαi=Zi(S)Sα.(3.68)Z_{\alpha}^{i}=\frac{\partial Z^{i}\left( S\right) }{\partial S^{\alpha}}. \tag{3.68}
Importantly, the "sign" of the curvature tensor depends on the choice of normal. Recall that there are two unit normals pointing in opposite directions and, as we agreed in Chapter 2, the symbol N\mathbf{N} represents either one of them. Thus, depending on the choice of normal, the resulting elements in the tensor BαβB_{\alpha\beta} may summarily change sign. As a result, the curvature tensor inherits the same sign ambiguity of the unit normal. By extension, the same is true of the mean curvature BααB_{\alpha }^{\alpha}. The Gaussian curvature KK, on the other hand, being the determinant of the 2×22\times2 matrix corresponding to BβαB_{\beta}^{\alpha}, is unchanged when the sign of the matrix is flipped and is therefore independent of the choice of normal. The product NBαβ\mathbf{N}B_{\alpha\beta}, known as the vector curvature tensor, is also independent of the choice of normal since both elements in the product change sign when the normal is reversed.
This Section calls for a brief Linear Algebra preamble. As we described in Chapter TBD of Introduction to Tensor Calculus, a matrix AA representing a linear transformation corresponds to a mixed second-order tensor AβαA_{\cdot\beta}^{\alpha}. Such a matrix AA is characterized by nn fundamental invariants -- its eigenvalues λ1\lambda_{1}, λ2\lambda_{2}, \cdots, λn\lambda_{n}. From these invariants, numerous others can be constructed. Two of the most noteworthy such invariants are the sum of the eigenvalues, which equals the trace of AA, i.e.
trace(A)=λ1+λ2++λn,(6.9)\operatorname{trace}\left( A\right) =\lambda_{1}+\lambda_{2}+\cdots +\lambda_{n},\tag{6.9}
and their product, which equals the determinant, i.e.
det(A)=λ1λ2λn.(6.10)\det\left( A\right) =\lambda_{1}\lambda_{2}\cdots\lambda_{n}.\tag{6.10}
While the eigenvalues of a general n×nn\times n matrix may be difficult to calculate (in fact, no finite algorithm for calculating eigenvalues is possible for a matrix of dimension greater than 44), the trace and the determinant are readily available.
For a self-adjoint transformation characterized by a symmetric tensor AβαA_{\cdot\beta}^{\alpha}, i.e.
Aβα=Aβα,(6.11)A_{\cdot\beta}^{\alpha}=A_{\beta}^{\cdot\alpha},\tag{6.11}
all eigenvalues are real and the corresponding eigenvectors are orthogonal -- or, in the case of repeated eigenvalues, can be chosen to be orthogonal. The eigenvectors represent the critical points of the quadratic form
F(U)=AαβUαUβ,(6.12)F\left( U\right) =A_{\alpha\beta}U^{\alpha}U^{\beta},\tag{6.12}
subject to the unit-length constraint
SαβUαUβ=1.(6.13)S_{\alpha\beta}U^{\alpha}U^{\beta}=1.\tag{6.13}
When we transfer these facts onto the curvature tensor BβαB_{\beta}^{\alpha}, we find that each of the objects mentioned above are of great geometric importance. The eigenvalues κ1\kappa_{1} and κ2\kappa_{2} are known as the principal curvatures while the corresponding eigenvectors are known as the principal directions. The sum κ1+κ2\kappa_{1}+\kappa_{2} is, of course, the mean curvature BααB_{\alpha}^{\alpha} while their product κ1κ2\kappa_{1}\kappa_{2} is, of course, the Gaussian curvature. We will now take a detailed look into each of these elements, starting with mean curvature.

6.2.1The mean curvature

The invariant
Bαα(6.14)B_{\alpha}^{\alpha}\tag{6.14}
is known as the mean curvature of the surface. The mean curvature is often denoted by the letter κ\kappa but we will almost always prefer the symbol BααB_{\alpha}^{\alpha}.
The mean curvature is a particularly important invariant that appears in many fundamental identities. It is, by virtue of its ubiquity, the star of our entire narrative. Many physical phenomena associated with capillary pressure (i.e. the pressure differential across an interface between two fluids) are described by equations that invariably feature the mean curvature. Thus, not surprisingly, it plays an important role in the dynamics of fluid films. Furthermore, mean curvature plays an important role in the study of minimal surfaces, i.e. surfaces with a given boundary that have the least possible area. As we will demonstrate in a future book on the Calculus of Moving Surfaces, minimal surfaces are characterized by zero mean curvature.
The vector NBαα\mathbf{N}B_{\alpha}^{\alpha} can be referred to as the vector mean curvature or, by analogy with curves that we will discover in Chapter 8, as the curvature normal. As we learned in Chapter 2, it equals the surface divergence of the covariant basis Sα\mathbf{S}^{\alpha} or the surface Laplacian of the position vector R\mathbf{R}, i.e.
NBαα=αSα=ααR.(6.15)\mathbf{N}B_{\alpha}^{\alpha}=\nabla_{\alpha}\mathbf{S}^{\alpha} =\nabla_{\alpha}\nabla^{\alpha}\mathbf{R.}\tag{6.15}
Thus, an alternative expression for the mean curvature is
Bαα=NααR.(6.16)B_{\alpha}^{\alpha}=\mathbf{N}\cdot\nabla_{\alpha}\nabla^{\alpha}\mathbf{R.}\tag{6.16}
The advantage of these identities is, of course, their pure geometric nature. For example, the first of the two equations will readily explain why the integral of the curvature normal over a closed surface is zero.

6.2.2The Gaussian curvature

The determinant BB of the mixed curvature tensor BβαB_{\beta}^{\alpha} of a two-dimensional surface is the Gaussian curvature KK. However, this is not the definition of Gaussian curvature but rather a consequence of the Gauss equations which we mentioned in Chapter 2 and will describe in detail in Chapter 7. As we stated in Chapter 2, by definition, the Gaussian curvature is the invariant KK in the two-dimensional identity
Rαβγδ=Kεαβεγδ.(7.52)R_{\alpha\beta\gamma\delta}=K\varepsilon_{\alpha\beta}\varepsilon _{\gamma\delta}. \tag{7.52}
where RαβγδR_{\alpha\beta\gamma\delta} is the Riemann-Christoffel tensor. Meanwhile, as we showed in Chapter TBD of Introduction to Tensor Calculus, any second-order system BβαB_{\beta}^{\alpha} in a two-dimensional space satisfies the equation
BαγBβδBαδBβγ=Bεαβεγδ.(6.17)B_{\alpha\gamma}B_{\beta\delta}-B_{\alpha\delta}B_{\beta\gamma}=B\varepsilon _{\alpha\beta}\varepsilon_{\gamma\delta}.\tag{6.17}
Now, this is where the Gauss equations enter. From the Gauss equations
BαγBβδBαδBβγ=Rαβγδ(7.38)B_{\alpha\gamma}B_{\beta\delta}-B_{\alpha\delta}B_{\beta\gamma}=R_{\alpha \beta\gamma\delta} \tag{7.38}
we conclude that
Bεαβεγδ=Kεαβεγδ(6.18)B\varepsilon_{\alpha\beta}\varepsilon_{\gamma\delta}=K\varepsilon_{\alpha \beta}\varepsilon_{\gamma\delta}\tag{6.18}
and therefore
K=B.(7.61)K=B. \tag{7.61}
We will save the rest of our discussion of Gaussian curvature for Chapter 7.

6.2.3The principal curvatures

The eigenvalues of BβαB_{\beta}^{\alpha} are known as the principal curvatures. Thus, two-dimensional surfaces are characterized by two eigenvalues, κ1\kappa_{1} and κ2\kappa_{2}. As mentioned earlier, the principal curvatures κ1\kappa_{1} and κ2\kappa_{2} correspond to the minimum and the maximum values of the quadratic form
BαβXαXβ(6.19)B_{\alpha\beta}X^{\alpha}X^{\beta}\tag{6.19}
subject to the unit normalization condition
SαβXαXβ=1.(6.20)S_{\alpha\beta}X^{\alpha}X^{\beta}=1.\tag{6.20}
Furthermore their sum is the mean curvature, i.e.
Bαα=κ1+κ2.(6.21)B_{\alpha}^{\alpha}=\kappa_{1}+\kappa_{2}.\tag{6.21}
and their product is the determinant BB of BβαB_{\beta}^{\alpha} which, by the Gauss equations, equals the Gaussian curvature KK, i.e.
K=B=κ1κ2.(7.62)K=B=\kappa_{1}\kappa_{2}. \tag{7.62}
The geometric interpretation of the principal curvatures κ1\kappa_{1} and κ2\kappa_{2} and the corresponding eigenvectors will be described below in a Section devoted entirely to that topic.
Note that some texts define the mean curvature as the average of the principal curvatures, i.e.
κ1+κ22,(6.22)\frac{\kappa_{1}+\kappa_{2}}{2},\tag{6.22}
which explains the use of the word mean in mean curvature. Such a definition has some advantages over our definition. For example, as we demonstrate below, the mean curvature BααB_{\alpha}^{\alpha} of a sphere of radius RR is 2/R-2/R which, in nn dimensions, generalizes to n/R-n/R. The definition based on the average, on the other hand, yields the value of 1/R-1/R in all dimensions. Nevertheless, we will stick with our definition since, in numerous applications, including the Calculus of Moving Surfaces, it is BααB_{\alpha}^{\alpha}, rather than Bαα/2B_{\alpha}^{\alpha}/2, that is most frequently encountered.

6.2.4The third fundamental tensor of a surface

The metric tensor SαβS_{\alpha\beta} is sometimes referred to as the first fundamental tensor of the surface while the curvature tensor BαβB_{\alpha\beta} is referred to as the second fundamental tensor. The third fundamental tensor CβαC_{\beta}^{\alpha}, which arises frequently in applications, is the tensor CβαC_{\beta}^{\alpha} defined by the equation
Cβα=BγαBβγ.(6.23)C_{\beta}^{\alpha}=B_{\gamma}^{\alpha}B_{\beta}^{\gamma}.\tag{6.23}
In other words, CβαC_{\beta}^{\alpha} corresponds to the matrix square of the curvature tensor. Note that the fundamental tensors often figure as binlinear forms, such as SαβUαVβS_{\alpha\beta}U^{\alpha}V^{\beta}. For this reason, they can also be referred to as fundamental forms or groundforms.
It is left as an exercise to show that
αNβN=Cαβ(6.24)\nabla_{\alpha}\mathbf{N}\cdot\nabla_{\beta}\mathbf{N}=C_{\alpha\beta}\tag{6.24}
Note the similarity among the equations that can act as definitions of the first groundform SαβS_{\alpha\beta}, the second groundform BαβB_{\alpha\beta}, and the third groundform CαβC_{\alpha\beta}, especially if we denote Sα=αR\mathbf{S}_{\alpha}=\nabla_{\alpha}\mathbf{R} by Rα\mathbf{R}_{\alpha} and αN\nabla_{\alpha}\mathbf{N} by Nα\mathbf{N}_{\alpha}. Then we have
Sαβ=RαRβ          (6.25)Bαβ=NαRβ          (6.26)Cαβ=NαNβ.          (6.27)\begin{aligned}S_{\alpha\beta} & =\mathbf{R}_{\alpha}\cdot\mathbf{R}_{\beta}\ \ \ \ \ \ \ \ \ \ \left(6.25\right)\\-B_{\alpha\beta} & =\mathbf{N}_{\alpha}\cdot\mathbf{R}_{\beta}\ \ \ \ \ \ \ \ \ \ \left(6.26\right)\\C_{\alpha\beta} & =\mathbf{N}_{\alpha}\cdot\mathbf{N}_{\beta}.\ \ \ \ \ \ \ \ \ \ \left(6.27\right)\end{aligned}
As we know from Linear Algebra, the eigenvalues of CβαC_{\beta}^{\alpha} are κ12,κ22,,κn2\kappa_{1}^{2},\kappa_{2}^{2},\cdots,\kappa_{n}^{2}. Therefore, the invariant CααC_{\alpha}^{\alpha} is given by
Cαα=κ12+κ22++κn2,(6.28)C_{\alpha}^{\alpha}=\kappa_{1}^{2}+\kappa_{2}^{2}+\cdots+\kappa_{n}^{2},\tag{6.28}
while the determinant CC of CβαC_{\beta}^{\alpha} is given by
C=κ12κ22κn2.(6.29)C=\kappa_{1}^{2}\kappa_{2}^{2}\cdots\kappa_{n}^{2}.\tag{6.29}

6.2.5The special case of planar curves

We have already studied planar curves in Chapter TBD of Introduction to Tensor Caclulus, where our analysis was based on parameterizing the curve by the arc length ss. We will now approach curves from a different angle -- that is, as a special one-dimensional case of an embedded surface. In other words, we will, to the extent possible, ignore their one-dimensional nature and focus on what the general theory of surfaces can tell us about them.
A planar curve is a curve embedded in a Euclidean space. Therefore, the dimension of a planar curve trails that of its ambient space by 11, which makes it a hypersurface. As a hypersurface, it is characterized by a unique (within sign) unit normal N\mathbf{N}. Thus, the concept of the curvature tensor applies in full force, as is the concept of mean curvature. The curvature tensor BαβB_{\alpha\beta} is still defined by the equation
αSβ=NBαβ(2.79)\nabla_{\alpha}\mathbf{S}_{\beta}=\mathbf{N}B_{\alpha\beta} \tag{2.79}
or, equivalently,
Bαβ=NαSβ,(2.80)B_{\alpha\beta}=\mathbf{N}\cdot\nabla_{\alpha}\mathbf{S}_{\beta}, \tag{2.80}
while the mean curvature BααB_{\alpha}^{\alpha} is given by
αSα=NBαα(6.30)\nabla_{\alpha}\mathbf{S}^{\alpha}=\mathbf{N}B_{\alpha}^{\alpha}\tag{6.30}
or, equivalently,
Bαα=NαSα.(2.80)B_{\alpha}^{\alpha}=\mathbf{N}\cdot\nabla_{\alpha}\mathbf{S}^{\alpha}. \tag{2.80}
Compare this equation to the identity
κ=NT(s)(6.31)\kappa=\mathbf{N}\cdot\mathbf{T}^{\prime}\left( s\right)\tag{6.31}
from Chapter TBD of Introduction to Tensor Caclulus, where κ\kappa is signed curvature and T(s)\mathbf{T}\left( s\right) is the unit tangent as a function of the arc length ss. Since, as it is left as an exercise to show, αSα\nabla_{\alpha}\mathbf{S}^{\alpha} is equivalent to T(s)\mathbf{T}^{\prime}\left( s\right) , we observe that for planar curves, the concept of mean curvature is equivalent to that of signed curvature.
Below, we will calculate the curvature tensor as well as the mean curvature for general planar curves represented in Cartesian and polar coordinates.
The geometric interpretation of principal curvatures involves curves embedded in two-dimensional surfaces. This configuration creates a fascinating interplay among three spaces: the one-dimensional curve, the two-dimensional surface, and the three-dimensional ambient Euclidean space. This intricate arrangement will yield a great number of intriguing relationships and insights. Some of these will appear in this book, but a great many more are beyond our scope -- and many more still are yet to be discovered.
Note that we have not yet developed a theory of surfaces embedded in non-Euclidean spaces, which is what a curve embedded in a general two-dimensional surface represents. Such a theory will be developed in Chapter 9. Fortunately, in order to understand the geometric interpretation of principal curvatures, we will only need to consider planar curves which, as we saw in the previous Section, we already have under our belt.
A planar curve arises when a surface is cut by a plane, as illustrated in the following figure.
(6.32)
Our analysis will focus on a single point PP which lies on the curve at the intersection of the surface and the cutting plane. Furthermore, we will assume that the cutting plane is orthogonal to the surface at the point PP. In other words, the cutting plane contains within in it the surface normal N\mathbf{N} at PP.
(6.33)
This condition does not specify the plane uniquely as there is an entire family of such planes obtained by rotation about the straight line containing N\mathbf{N}.
The curve at the intersection of the surface and the plane can be analyzed from three distinct points of view. First, it can be viewed as a curve embedded in the two-dimensional surface. In this context, the surface can be described as the ambient space with respect to the curve. Second, it can be viewed as a curve embedded in the overall three-dimensional Euclidean space. And finally, it can be viewed as a curve embedded in the two-dimensional Euclidean cutting plane. Naturally, it is the last embedding with which we will concern ourselves here.
The curve represents a hypersurface with the respect to the plane. It is therefore characterized by its signed curvature κ\kappa which, as we observed in the previous section, is equivalent to its mean curvature in the context of its embedding in the plane. Naturally, κ\kappa depends on two characteristics: the curvature characteristics of the ambient surface and the orientation of the cutting plane. In this Section, we will succeed in capturing these dependencies by the elegant equation
κ=BαβTαTβ,(6.34)\kappa=B_{\alpha\beta}T^{\alpha}T^{\beta},\tag{6.34}
where TαT^{\alpha} are the surface components of the unit tangent T\mathbf{T} to the curve. Clearly, the curvature tensor BαβB_{\alpha\beta} captures the curvature characteristics of the surface while the tangent T\mathbf{T} captures the orientation of the surface. The reader is invited to attempt deriving this identity on their own before we lay out the details below, as it relies on the techniques that we have developed earlier.
Before we turn to the derivation of the above identity, let us make one important geometric observation that will prove relevant later. Consider the unit normal N~\mathbf{\tilde{N}} to the curve when viewed as a hypersurface within the cutting plane. At the point PP, and likely only at the point PP, it is colinear with the surface normal N\mathbf{N}. This is by construction: the surface normal N\mathbf{N} at the point PP is orthogonal to all curves in the surface that pass through PP, including the one formed by the intersection of the surface and the plane. The same cannot be said for any other point on the curve since there is no guarantee that the surface normal is contained within the cutting plane.
Despite our present affinity for treating curves as a special case of surfaces, we will, for the remainder of this Section, return to the style of analysis based on parameterizing the curve by its arc length ss, which we employed in Chapter TBD of Introduction to Tensor Calculus. As before, denote the function representing the position vector R\mathbf{R} at the points on the curve by R(s)\mathbf{R}\left( s\right) , i.e.
R=R(s).(6.35)\mathbf{R}=\mathbf{R}\left( s\right) .\tag{6.35}
Recall that the first derivative of R(s)\mathbf{R}\left( s\right) is the unit tangent T\mathbf{T}, i.e.
T=R(s),(6.36)\mathbf{T}=\mathbf{R}^{\prime}\left( s\right) ,\tag{6.36}
while the second derivative is the curvature normal B\mathbf{B}, i.e.
B=R(s).(6.37)\mathbf{B}=\mathbf{R}^{\prime\prime}\left( s\right) .\tag{6.37}
(Note that the determinant BB of the surface curvature tensor BβαB_{\beta }^{\alpha} is unrelated to the magnitude of the vector B\mathbf{B}.) Importantly, for all ss, the vectors T(s)\mathbf{T}\left( s\right) and B(s)\mathbf{B}\left( s\right) both lie within the cutting plane. Meanwhile, the curvature normal B\mathbf{B} is collinear with the in-plane normal N~\mathbf{\tilde{N}} to the curve which coincides with the surface normal N\mathbf{N} at the point PP. Thus, at the point PP, the signed curvature κ\kappa, given by the identity
B=κN~,(6.38)\mathbf{B}=\kappa\mathbf{\tilde{N},}\tag{6.38}
is also given by the identity
B=κN.(6.39)\mathbf{B}=\kappa\mathbf{N.}\tag{6.39}
All but the last equation in the previous paragraph are characteristic of the curve with respect to its embedding in the cutting plane. Thus, in order to derive the equation
κ=BαβTαTβ,(6.34)\kappa=B_{\alpha\beta}T^{\alpha}T^{\beta}, \tag{6.34}
we must find a meaningful way to engage the differential characteristic of the surface.
Suppose that the curve within the surface is given by the equations
Sα=Sα(s).(6.40)S^{\alpha}=S^{\alpha}\left( s\right) .\tag{6.40}
Then the function R(s)\mathbf{R}\left( s\right) can be constructed by composing the function R(S)\mathbf{R}\left( S\right) , i.e. R\mathbf{R} as a function of the surface coordinates SαS^{\alpha}, with the equations of the curve Sα(s)S^{\alpha}\left( s\right) , i.e.
R(s)=R(S(s)).(6.41)\mathbf{R}\left( s\right) =\mathbf{R}\left( S\left( s\right) \right) .\tag{6.41}
As we always do after establishing a relationship such as above, we will differentiate both sides with respect to the independent variable, i.e. the arc length ss. By the chain rule, we have
R(s)=R(S)SαdSα(s)ds.(6.42)\mathbf{R}^{\prime}\left( s\right) =\frac{\partial\mathbf{R}\left( S\right) }{\partial S^{\alpha}}\frac{dS^{\alpha}\left( s\right) }{ds}.\tag{6.42}
On the left, we recognize R(s)\mathbf{R}^{\prime}\left( s\right) as the unit tangent T\mathbf{T}. On the right, the partial derivative R(S)/Sα\partial \mathbf{R}\left( S\right) /\partial S^{\alpha} is, of course, the surface covariant basis Sα\mathbf{S}_{\alpha}. In other words, we have obtained the identity
T=dSα(s)dsSα(6.43)\mathbf{T}=\frac{dS^{\alpha}\left( s\right) }{ds}\mathbf{S}_{\alpha}\tag{6.43}
which tells us that the derivatives
dSα(s)ds(6.44)\frac{dS^{\alpha}\left( s\right) }{ds}\tag{6.44}
of the equations of the curve Sα(s)S^{\alpha}\left( s\right) can be interpreted as the surface coordinates TαT^{\alpha} of T\mathbf{T}. In this respect, the equation
T=dSα(s)dsSα(6.43)\mathbf{T}=\frac{dS^{\alpha}\left( s\right) }{ds}\mathbf{S}_{\alpha} \tag{6.43}
is similar to a number of identities we have derived in the past, such as
V=dZi(t)dtZi(6.45)\mathbf{V}=\frac{dZ^{i}\left( t\right) }{dt}\mathbf{Z}_{i}\tag{6.45}
in Chapter TBD of Introduction to Tensor Calculus.
In order to get to curvature, we must differentiate the identity
T(s)=dSα(s)dsSα(s)(6.43)\mathbf{T}\left( s\right) =\frac{dS^{\alpha}\left( s\right) } {ds}\mathbf{S}_{\alpha}\left( s\right) \tag{6.43}
a second time. By the product rule,
T(s)=d2Sα(s)ds2Sα+dSα(s)dsdSα(s)ds.(6.46)\mathbf{T}^{\prime}\left( s\right) =\frac{d^{2}S^{\alpha}\left( s\right) }{ds^{2}}\mathbf{S}_{\alpha}+\frac{dS^{\alpha}\left( s\right) }{ds} \frac{d\mathbf{S}_{\alpha}\left( s\right) }{ds}.\tag{6.46}
As we recalled above, the derivative T(s)\mathbf{T}^{\prime}\left( s\right) is the curvature normal B\mathbf{B}. Meanwhile, on the right, the object of utmost interest is the derivative
dSα(s)ds(6.47)\frac{d\mathbf{S}_{\alpha}\left( s\right) }{ds}\tag{6.47}
which captures the rate of change of the covariant surface basis Sα\mathbf{S}_{\alpha} along the curve. Similarly to R(s)\mathbf{R}\left( s\right) , the function Sα(s)\mathbf{S}_{\alpha}\left( s\right) can be constructed by substituting the equations of the curve Sα=Sα(s)S^{\alpha}=S^{\alpha }\left( s\right) into Sα(S)\mathbf{S}_{\alpha}\left( S\right) , i.e.
Sα(s)=Sα(S(s)).(6.48)\mathbf{S}_{\alpha}\left( s\right) =\mathbf{S}_{\alpha}\left( S\left( s\right) \right) .\tag{6.48}
Then, by the chain rule, we have
dSα(s)ds=Sα(S)SβdSβ(s)ds.(6.49)\frac{d\mathbf{S}_{\alpha}\left( s\right) }{ds}=\frac{\partial \mathbf{S}_{\alpha}\left( S\right) }{\partial S^{\beta}}\frac{dS^{\beta }\left( s\right) }{ds}.\tag{6.49}
The derivative Sα(S)/dSβ\partial\mathbf{S}_{\alpha}\left( S\right) /dS^{\beta} is where the curvature tensor makes its appearance. Since
βSα=NBαβ,(2.79)\nabla_{\beta}\mathbf{S}_{\alpha}=\mathbf{N}B_{\alpha\beta}, \tag{2.79}
and by definition of the covariant derivative
βSα=Sα(S)SβΓαβωSω,(6.50)\nabla_{\beta}\mathbf{S}_{\alpha}=\frac{\partial\mathbf{S}_{\alpha}\left( S\right) }{\partial S_{\beta}}-\Gamma_{\alpha\beta}^{\omega}\mathbf{S} _{\omega},\tag{6.50}
we find
Sα(S)Sβ=NBαβ+ΓαβωSω.(6.51)\frac{\partial\mathbf{S}_{\alpha}\left( S\right) }{\partial S^{\beta} }=\mathbf{N}B_{\alpha\beta}+\Gamma_{\alpha\beta}^{\omega}\mathbf{S}_{\omega}.\tag{6.51}
Thus,
dSα(s)ds=Sα(S)SβdSβ(s)ds(6.52)\frac{d\mathbf{S}_{\alpha}\left( s\right) }{ds}=\frac{\partial \mathbf{S}_{\alpha}\left( S\right) }{\partial S^{\beta}}\frac{dS^{\beta }\left( s\right) }{ds}\tag{6.52}
becomes
dSα(s)ds=(NBαβ+ΓαβωSω)dSβds(6.53)\frac{d\mathbf{S}_{\alpha}\left( s\right) }{ds}=\left( \mathbf{N} B_{\alpha\beta}+\Gamma_{\alpha\beta}^{\omega}\mathbf{S}_{\omega}\right) \frac{dS^{\beta}}{ds}\tag{6.53}
and therefore
T(s)=d2Sα(s)ds2Sα+dSα(s)dsdSα(s)ds(6.46)\mathbf{T}^{\prime}\left( s\right) =\frac{d^{2}S^{\alpha}\left( s\right) }{ds^{2}}\mathbf{S}_{\alpha}+\frac{dS^{\alpha}\left( s\right) }{ds} \frac{d\mathbf{S}_{\alpha}\left( s\right) }{ds} \tag{6.46}
becomes
B=(NBαβ+ΓαβωSω)dSαdsdSβds+Sαd2Sαds2(6.54)\mathbf{B}=\left( \mathbf{N}B_{\alpha\beta}+\Gamma_{\alpha\beta}^{\omega }\mathbf{S}_{\omega}\right) \frac{dS^{\alpha}}{ds}\frac{dS^{\beta}} {ds}+\mathbf{S}_{\alpha}\frac{d^{2}S^{\alpha}}{ds^{2}}\tag{6.54}
where we have dropped the arguments of the functions for the sake of conciseness. Combining normal and tangential terms, we find
B=NBαβdSαdsdSβds+(d2Sαds2+ΓβωαdSβdsdSωds)Sα.(6.55)\mathbf{B}=\mathbf{N}B_{\alpha\beta}\frac{dS^{\alpha}}{ds}\frac{dS^{\beta} }{ds}+\left( \frac{d^{2}S^{\alpha}}{ds^{2}}+\Gamma_{\beta\omega}^{\alpha }\frac{dS^{\beta}}{ds}\frac{dS^{\omega}}{ds}\right) \mathbf{S}_{\alpha}.\tag{6.55}
On the one hand, the combination
d2Sαds2+ΓβωαdSβdsdSωds,(6.56)\frac{d^{2}S^{\alpha}}{ds^{2}}+\Gamma_{\beta\omega}^{\alpha}\frac{dS^{\beta} }{ds}\frac{dS^{\omega}}{ds},\tag{6.56}
which represents the tangential component of the curvature normal B\mathbf{B} , is quite intriguing as it reminds us of the δ/δt\delta/\delta t-derivative introduced in Section TBD of Introduction to Tensor Calculus. It will also reemerge in the future as we study the concept of the geodesic curvature for curves embedded in surfaces. On the other hand, since the curvature normal B\mathbf{B} is colinear with the surface normal N\mathbf{N} at the point PP as we established above, we can conclude that the tangential component of B\mathbf{B} vanishes, i.e.
d2Sαds2+ΓβωαdSβdsdSωds=0.(6.57)\frac{d^{2}S^{\alpha}}{ds^{2}}+\Gamma_{\beta\omega}^{\alpha}\frac{dS^{\beta} }{ds}\frac{dS^{\omega}}{ds}=0.\tag{6.57}
Therefore, the curvature normal B\mathbf{B} is given by
B=NBαβdSαdsdSβds.(6.58)\mathbf{B}=\mathbf{N}B_{\alpha\beta}\frac{dS^{\alpha}}{ds}\frac{dS^{\beta} }{ds}.\tag{6.58}
Finally, since
B=κN,(6.39)\mathbf{B}=\kappa\mathbf{N,} \tag{6.39}
we arrive at the identity
κ=BαβdSαdsdSβds.(6.59)\kappa=B_{\alpha\beta}\frac{dS^{\alpha}}{ds}\frac{dS^{\beta}}{ds}.\tag{6.59}
In other words,
κ=BαβTαTβ,(6.34)\kappa=B_{\alpha\beta}T^{\alpha}T^{\beta}, \tag{6.34}
as we set out to show.
The geometric interpretation of the principal curvatures κ1\kappa_{1} and κ2\kappa_{2} follows immediately from the identity
κ=BαβTαTβ(6.34)\kappa=B_{\alpha\beta}T^{\alpha}T^{\beta} \tag{6.34}
Recall that the principal curvatures κ1\kappa_{1} and κ2\kappa_{2} of a surface are defined as the eigenvalues of BβαB_{\beta}^{\alpha}. From Linear Algebra, we know that these values are precisely the least and greatest values of the quadratic form
BαβTαTβ(6.60)B_{\alpha\beta}T^{\alpha}T^{\beta}\tag{6.60}
subject to the constraint
SαβTαTβ=1.(6.61)S_{\alpha\beta}T^{\alpha}T^{\beta}=1.\tag{6.61}
Therefore, the principal curvatures correspond to the least and the greatest signed curvatures of curves formed by intersecting the surface with orthogonal planes. Such values always exist and the corresponding directions, characterized by the unit tangents T1\mathbf{T}_{1} and T2\mathbf{T}_{2} are orthogonal. The following figure shows the two cutting planes corresponding to the principal directions.
(6.62)
The following figure shows a point where the interplay between the principal curvatures is more interesting -- one is positive and the other is negative.
(6.63)
This interpretation of the principal curvatures immediately tells us the curvature invariants on the sphere of radius RR.
(6.64)
A normal plane cuts the sphere in a great circle of radius RR. Thus, with respect to the outward normal, for both principal curvatures, we have
κ1=κ2=1R.(6.65)\kappa_{1}=\kappa_{2}=-\frac{1}{R}.\tag{6.65}
Thus, the mean curvature of the sphere with radius RR with respect to the outward normal is given by
Bαα=κ1+κ2=2R.(6.66)B_{\alpha}^{\alpha}=\kappa_{1}+\kappa_{2}=-\frac{2}{R}.\tag{6.66}
The Gaussian curvature KK, which is independent of the choice of normal, is given by
K=κ1κ2=1R2.(6.67)K=\kappa_{1}\kappa_{2}=\frac{1}{R^{2}}.\tag{6.67}
In this Section, we will expand the analyses presented in Chapter 4 to include the curvature tensor. One of the more convenient ways to calculate the curvature tensor by hand is represented by the form
Bαβ=ZβiαNi,(5.55)B_{\alpha\beta}=-Z_{\beta}^{i}\nabla_{\alpha}N_{i}, \tag{5.55}
of Weingarten's equation. When the ambient space is referred to Cartesian coordinates, αNi\nabla_{\alpha}N_{i} coincides with the partial derivative Ni/Sα\partial N_{i}/\partial S^{\alpha} and the expression for the curvature tensor BαβB_{\alpha\beta} becomes
Bαβ=ZβiNiSα.(6.68)B_{\alpha\beta}=-Z_{\beta}^{i}\frac{\partial N_{i}}{\partial S^{\alpha}}.\tag{6.68}
This identity corresponds to a product of two matrices which, leveraging the notation introduced in Chapter 4, reads
[Bαβ]=[Zβi]T[NiSα].(6.69)\left[ B_{\alpha\beta}\right] =-\left[ Z_{\beta}^{i}\right] ^{T}\left[ \frac{\partial N_{i}}{\partial S^{\alpha}}\right] .\tag{6.69}

6.4.1A sphere of radius RR

For a sphere of radius RR referred to coordinates θ\theta and φ\varphi embedded in a Euclidean space referred to Cartesian coordinates, recall the following values:
[Zαi]=[Ziα]=[RcosθcosφRsinθsinφRcosθsinφRsinθcosφRsinθ0]          (4.2)[Sαβ]=[1R2001R2sin2θ]          (4.5)[Ni]=[Ni]=[sinθcosφsinθsinφcosθ].          (4.18)\begin{aligned}\left[ Z_{\alpha}^{i}\right] & =\left[ Z_{i\alpha}\right] =\left[ \begin{array} {rr} R\cos\theta\cos\varphi & -R\sin\theta\sin\varphi\\ R\cos\theta\sin\varphi & R\sin\theta\cos\varphi\\ -R\sin\theta & 0 \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.2\right)\\\left[ S^{\alpha\beta}\right] & =\left[ \begin{array} {cc} \frac{1}{R^{2}} & 0\\ 0 & \frac{1}{R^{2}\sin^{2}\theta} \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.5\right)\\\left[ N^{i}\right] & =\left[ N_{i}\right] =\left[ \begin{array} {l} \sin\theta\cos\varphi\\ \sin\theta\sin\varphi\\ \cos\theta \end{array} \right] . \ \ \ \ \ \ \ \ \ \ \left(4.18\right)\end{aligned}
Note that the components NiN^{i} in the above table correspond to the outward normal.
Differentiating the components of the normal with respect to the surface coordinates, we find
[NiSα]=[cosθcosφsinθsinφcosθsinφsinθcosφsinθ0].(6.70)\left[ \frac{\partial N^{i}}{\partial S^{\alpha}}\right] =\left[ \begin{array} {rr} \cos\theta\cos\varphi & -\sin\theta\sin\varphi\\ \cos\theta\sin\varphi & \sin\theta\cos\varphi\\ -\sin\theta & 0 \end{array} \right] .\tag{6.70}
Therefore, for the curvature tensor BαβB_{\alpha\beta}, we have
[Bαβ]=[Zαi]T[NiSα]=[R00Rsin2θ].(6.71)\left[ B_{\alpha\beta}\right] =-\left[ Z_{\alpha}^{i}\right] ^{T}\left[ \frac{\partial N_{i}}{\partial S^{\alpha}}\right] =\left[ \begin{array} {cc} -R & 0\\ 0 & -R\sin^{2}\theta \end{array} \right] .\tag{6.71}
Raising the index α\alpha and, subsequently, β\beta, we find
[Bβα]=[Sαβ][Bαβ]=[1R001R]          (6.72)[Bαβ]=[Sαβ][Bβα]=[1R3001R3sin2θ].          (6.73)\begin{aligned}\left[ B_{\beta}^{\alpha}\right] & =\left[ S^{\alpha\beta}\right] \left[ B_{\alpha\beta}\right] =\left[ \begin{array} {cc} -\frac{1}{R} & 0\\ 0 & -\frac{1}{R} \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.72\right)\\\left[ B^{\alpha\beta}\right] & =\left[ S^{\alpha\beta}\right] \left[ B_{\beta}^{\alpha}\right] =\left[ \begin{array} {cc} -\frac{1}{R^{3}} & 0\\ 0 & -\frac{1}{R^{3}\sin^{2}\theta} \end{array} \right] .\ \ \ \ \ \ \ \ \ \ \left(6.73\right)\end{aligned}
From the expression for BβαB_{\beta}^{\alpha}, we can confirm our earlier calculation of the mean curvature BααB_{\alpha}^{\alpha}, i.e.
Bαα=2R.(6.74)B_{\alpha}^{\alpha}=-\frac{2}{R}.\tag{6.74}
Finally, the principal curvatures κ1\kappa_{1} and κ2\kappa_{2} are given by
κ1=κ2=1R.(6.75)\kappa_{1}=\kappa_{2}=-\frac{1}{R}.\tag{6.75}
Note that the values of the curvature tensors, the mean curvature, and the principal curvatures correspond to the outward normal and would have the opposite sign had the other normal been selected.

6.4.2A cylinder of radius RR

Recall that
[Zαi]=[Ziα]=[Rsinθ0Rcosθ001],          (4.51)[Sαβ]=[1R2001], and          (4.53)[Ni]=[cosθsinθ0].          (4.56)\begin{aligned}\left[ Z_{\alpha}^{i}\right] & =\left[ Z_{i\alpha}\right] =\left[ \begin{array} {ll} -R\sin\theta & 0\\ \phantom{-} R\cos\theta & 0\\ \phantom{-} 0 & 1 \end{array} \right] ,\ \ \ \ \ \ \ \ \ \ \left(4.51\right)\\\left[ S^{\alpha\beta}\right] & =\left[ \begin{array} {cc} \frac{1}{R^{2}} & 0\\ 0 & 1 \end{array} \right] ,\text{ and}\ \ \ \ \ \ \ \ \ \ \left(4.53\right)\\\left[ N^{i}\right] & =\left[ \begin{array} {r} \cos\theta\\ \sin\theta\\ 0 \end{array} \right] . \ \ \ \ \ \ \ \ \ \ \left(4.56\right)\end{aligned}
Thus,
[NiSα]=[sinθ0cosθ000].(6.76)\left[ \frac{\partial N^{i}}{\partial S^{\alpha}}\right] =\left[ \begin{array} {rr} -\sin\theta & 0\\ \cos\theta & 0\\ 0 & 0 \end{array} \right] .\tag{6.76}
and therefore the various forms of the curvature tensors are as follows:
[Bαβ]=[Zαi]T[NiSα]=[R000]          (6.77)[Bβα]=[Bαβ][Sαβ]=[1R000]          (6.78)[Bαβ]=[Bβα][Sαβ]=[1R3000].          (6.79)\begin{aligned}\left[ B_{\alpha\beta}\right] & =\left[ Z_{\alpha}^{i}\right] ^{T}\left[ \frac{\partial N^{i}}{\partial S^{\alpha}}\right] =\left[ \begin{array} {rr} -R & 0\\ 0 & 0 \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.77\right)\\\left[ B_{\beta}^{\alpha}\right] & =\left[ B_{\alpha\beta}\right] \left[ S^{\alpha\beta}\right] =\left[ \begin{array} {rr} -\frac{1}{R} & 0\\ 0 & 0 \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.78\right)\\\left[ B^{\alpha\beta}\right] & =\left[ B_{\beta}^{\alpha}\right] \left[ S^{\alpha\beta}\right] =\left[ \begin{array} {rr} -\frac{1}{R^{3}} & 0\\ 0 & 0 \end{array} \right] .\ \ \ \ \ \ \ \ \ \ \left(6.79\right)\end{aligned}
From BβαB_{\beta}^{\alpha}, we see that the mean curvature BααB_{\alpha}^{\alpha } is given by
Bαα=1R.(6.80)B_{\alpha}^{\alpha}=-\frac{1}{R}.\tag{6.80}
Finally, the principal curvatures are given by
κ1=1R and          (6.81)κ2=0.          (6.82)\begin{aligned}\kappa_{1} & =-\frac{1}{R}\text{ and}\ \ \ \ \ \ \ \ \ \ \left(6.81\right)\\\kappa_{2} & =0.\ \ \ \ \ \ \ \ \ \ \left(6.82\right)\end{aligned}

6.4.3A torus with radii RR and rr

Recall that
[Zαi]=[Ziα]=[(R+rcosφ)sinθrcosθsinφ(R+rcosφ)cosθrsinθsinφ0rcosφ]          (4.71)[Sαβ]= [(R+rcosφ)200r2]          (4.73)[Ni]=[cosθcosφsinθcosφsinφ].          (4.77)\begin{aligned}\left[ Z_{\alpha}^{i}\right] & =\left[ Z_{i\alpha}\right] =\left[ \begin{array} {ll} -\left( R+r\cos\varphi\right) \sin\theta & -r\cos\theta\sin\varphi\\ \phantom{-} \left( R+r\cos\varphi\right) \cos\theta & -r\sin\theta\sin\varphi\\ \phantom{-} 0 & \phantom{-} r\cos\varphi \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.71\right)\\\left[ S^{\alpha\beta}\right] & =\ \left[ \begin{array} {cc} \left( R+r\cos\varphi\right) ^{-2} & 0\\ 0 & r^{-2} \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.73\right)\\\left[ N^{i}\right] & =\left[ \begin{array} {r} \cos\theta\cos\varphi\\ \sin\theta\cos\varphi\\ \sin\varphi \end{array} \right] . \ \ \ \ \ \ \ \ \ \ \left(4.77\right)\end{aligned}
Thus,
[NiSα]=[sinθcosφcosθsinφcosθcosφsinθsinφ0cosφ].(6.83)\left[ \frac{\partial N^{i}}{\partial S^{\alpha}}\right] =\left[ \begin{array} {rr} -\sin\theta\cos\varphi & -\cos\theta\sin\varphi\\ \cos\theta\cos\varphi & -\sin\theta\sin\varphi\\ 0 & \cos\varphi \end{array} \right] .\tag{6.83}
As a result, the various forms of the curvature tensor are
[Bαβ]=[(R+rcosφ)cosφ00r]          (6.84)[Bβα]=[cosφR+rcosφ001r]          (6.85)[Bαβ]=[cosφ(R+rcosφ)3001r3].          (6.86)\begin{aligned}\left[ B_{\alpha\beta}\right] & =\left[ \begin{array} {cc} -\left( R+r\cos\varphi\right) \cos\varphi & 0\\ 0 & -r \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.84\right)\\\left[ B_{\beta}^{\alpha}\right] & =\left[ \begin{array} {cc} -\frac{\cos\varphi}{R+r\cos\varphi} & 0\\ 0 & -\frac{1}{r} \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.85\right)\\\left[ B^{\alpha\beta}\right] & =\left[ \begin{array} {cc} -\frac{\cos\varphi}{\left( R+r\cos\varphi\right) ^{3}} & 0\\ 0 & -\frac{1}{r^{3}} \end{array} \right] .\ \ \ \ \ \ \ \ \ \ \left(6.86\right)\end{aligned}
From BβαB_{\beta}^{\alpha}, we see that the mean curvature BααB_{\alpha}^{\alpha } is given by
Bαα=R+2rcosφr(R+rcosφ).(6.87)B_{\alpha}^{\alpha}=-\frac{R+2r\cos\varphi}{r\left( R+r\cos\varphi\right) }.\tag{6.87}
Finally, the principal curvatures are given by
κ1=cosφR+rcosφ and          (6.88)κ2=1r.          (6.89)\begin{aligned}\kappa_{1} & =-\frac{\cos\varphi}{R+r\cos\varphi}\text{ and}\ \ \ \ \ \ \ \ \ \ \left(6.88\right)\\\kappa_{2} & =-\frac{1}{r}.\ \ \ \ \ \ \ \ \ \ \left(6.89\right)\end{aligned}

6.4.4A surface of revolution

Consider a surface of revolution given by the functions G(γ)G\left( \gamma\right) and H(γ)H\left( \gamma\right) as described in Sections 3.2.3 and 4.3, and let GγG_{\gamma}, GγγG_{\gamma\gamma}, HγH_{\gamma}, and HγγH_{\gamma\gamma} denote the derivatives of the functions GG and HH. Recall that
[Zαi]=[Ziα]=[GsinθGγcosθGcosθGγsinθ0Hγ]          (4.58)Sαβ=[G200(Gγ2+Hγ2)2]          (4.60)Ni=1Gγ2+Hγ2[HγcosθHγsinθGγ]          (4.63)\begin{aligned}\left[ Z_{\alpha}^{i}\right] & =\left[ Z_{i\alpha}\right] =\left[ \begin{array} {ll} -G\sin\theta & G_{\gamma}\cos\theta\\ \phantom{-} G\cos\theta & G_{\gamma}\sin\theta\\ \phantom{-} 0 & H_{\gamma} \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.58\right)\\S^{\alpha\beta} & =\left[ \begin{array} {cc} G^{-2} & 0\\ 0 & \left( G_{\gamma}^{2}+H_{\gamma}^{2}\right) ^{-2} \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.60\right)\\N^{i} & =\frac{1}{\sqrt{G_{\gamma}^{2}+H_{\gamma}^{2}}}\left[ \begin{array} {l} \phantom{-} H_{\gamma}\cos\theta\\ \phantom{-} H_{\gamma}\sin\theta\\ -G_{\gamma} \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.63\right)\end{aligned}
We have
[NiSα]=[HγGγ2+Hγ2sinθGγ(GγHγγGγγHγ)(Gγ2+Hγ2)3/2cosθHγGγ2+Hγ2cosθGγ(GγHγγGγγHγ)(Gγ2+Hγ2)3/2sinθ0Hγ(GγHγγGγγHγ)(Gγ2+Hγ2)3/2].(6.90)\left[ \frac{\partial N^{i}}{\partial S^{\alpha}}\right] =\left[ \begin{array} {ll} -\frac{H_{\gamma}}{\sqrt{G_{\gamma}^{2}+H_{\gamma}^{2}}}\sin\theta & \frac{G_{\gamma}\left( G_{\gamma}H_{\gamma\gamma}-G_{\gamma\gamma}H_{\gamma }\right) }{\left( G_{\gamma}^{2}+H_{\gamma}^{2}\right) ^{3/2}}\cos\theta\\ \phantom{-} \frac{H_{\gamma}}{\sqrt{G_{\gamma}^{2}+H_{\gamma}^{2}}}\cos\theta & \frac{G_{\gamma}\left( G_{\gamma}H_{\gamma\gamma}-G_{\gamma\gamma}H_{\gamma }\right) }{\left( G_{\gamma}^{2}+H_{\gamma}^{2}\right) ^{3/2}}\sin\theta\\ \phantom{-} 0 & \frac{H_{\gamma}\left( G_{\gamma}H_{\gamma\gamma}-G_{\gamma\gamma }H_{\gamma}\right) }{\left( G_{\gamma}^{2}+H_{\gamma}^{2}\right) ^{3/2}} \end{array} \right] .\tag{6.90}
As a result, the various forms of the curvature tensor are
[Bαβ]=[GHγGγ2+Hγ200GγγHγGγHγγGγ2+Hγ2]          (6.91)[Bβα]=[HγGGγ2+Hγ200GγγHγGγHγγ(Gγ2+Hγ2)3/2]          (6.92)[Bαβ]=[HγG3Gγ2+Hγ200GγγHγGγHγγ(Gγ2+Hγ2)5/2]          (6.93)\begin{aligned}\left[ B_{\alpha\beta}\right] = & \left[ \begin{array} {cc} -\frac{GH_{\gamma}}{\sqrt{G_{\gamma}^{2}+H_{\gamma}^{2}}} & 0\\ 0 & \frac{G_{\gamma\gamma}H_{\gamma}-G_{\gamma}H_{\gamma\gamma}} {\sqrt{G_{\gamma}^{2}+H_{\gamma}^{2}}} \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.91\right)\\\left[ B_{\beta}^{\alpha}\right] = & \left[ \begin{array} {cc} -\frac{H_{\gamma}}{G\sqrt{G_{\gamma}^{2}+H_{\gamma}^{2}}} & 0\\ 0 & \frac{G_{\gamma\gamma}H_{\gamma}-G_{\gamma}H_{\gamma\gamma}}{\left( G_{\gamma}^{2}+H_{\gamma}^{2}\right) ^{3/2}} \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.92\right)\\\left[ B^{\alpha\beta}\right] = & \left[ \begin{array} {cc} -\frac{H_{\gamma}}{G^{3}\sqrt{G_{\gamma}^{2}+H_{\gamma}^{2}}} & 0\\ 0 & \frac{G_{\gamma\gamma}H_{\gamma}-G_{\gamma}H_{\gamma\gamma}}{\left( G_{\gamma}^{2}+H_{\gamma}^{2}\right) ^{5/2}} \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.93\right)\end{aligned}
Therefore, the mean curvature BααB_{\alpha}^{\alpha} is given by
Bαα=GHγGγγHγGγ2GGγHγγHγ3G(Gγ2+Hγ2)3/2.(6.94)B_{\alpha}^{\alpha}=\frac{GH_{\gamma}G_{\gamma\gamma}-H_{\gamma}G_{\gamma} ^{2}-GG_{\gamma}H_{\gamma\gamma}-H_{\gamma}^{3}}{G\left( G_{\gamma} ^{2}+H_{\gamma}^{2}\right) ^{3/2}}.\tag{6.94}
For the special case
H(γ)=γ(6.95)H\left( \gamma\right) =\gamma\tag{6.95}
used very commonly in applications, we have
[Bαβ]=[G1+Gγ200Gγγ1+Gγ2]          (6.96)[Bβα]=[1G1+Gγ200Gγγ(1+Gγ2)3/2]          (6.97)[Bαβ]=[1G31+Gγ200Gγγ(1+Gγ2)5/2].          (6.98)\begin{aligned}\left[ B_{\alpha\beta}\right] = & \left[ \begin{array} {cc} -\frac{G}{\sqrt{1+G_{\gamma}^{2}}} & 0\\ 0 & \frac{G_{\gamma\gamma}}{\sqrt{1+G_{\gamma}^{2}}} \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.96\right)\\\left[ B_{\beta}^{\alpha}\right] = & \left[ \begin{array} {cc} -\frac{1}{G\sqrt{1+G_{\gamma}^{2}}} & 0\\ 0 & \frac{G_{\gamma\gamma}}{\left( 1+G_{\gamma}^{2}\right) ^{3/2}} \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.97\right)\\\left[ B^{\alpha\beta}\right] = & \left[ \begin{array} {cc} -\frac{1}{G^{3}\sqrt{1+G_{\gamma}^{2}}} & 0\\ 0 & \frac{G_{\gamma\gamma}}{\left( 1+G_{\gamma}^{2}\right) ^{5/2}} \end{array} \right] .\ \ \ \ \ \ \ \ \ \ \left(6.98\right)\end{aligned}
Meanwhile, the mean curvature is given by
Bαα=GGγγGγ21G(1+Gγ2)3/2(6.99)B_{\alpha}^{\alpha}=\frac{GG_{\gamma\gamma}-G_{\gamma}^{2}-1}{G\left( 1+G_{\gamma}^{2}\right) ^{3/2}}\tag{6.99}
and the principal curvatures are
κ1=1G1+Gγ2 and          (6.100)κ2=Gγγ(1+Gγ2)3/2.          (6.101)\begin{aligned}\kappa_{1} & =-\frac{1}{G\sqrt{1+G_{\gamma}^{2}}}\text{ and}\ \ \ \ \ \ \ \ \ \ \left(6.100\right)\\\kappa_{2} & =\frac{G_{\gamma\gamma}}{\left( 1+G_{\gamma}^{2}\right) ^{3/2}}.\ \ \ \ \ \ \ \ \ \ \left(6.101\right)\end{aligned}

6.5.1In Cartesian coordinates

Recall, that for a general planar curve described in Cartesian coordinates by the equations
x=x(γ)          (6.102)y=y(γ),          (6.103)\begin{aligned}x & =x\left( \gamma\right)\ \ \ \ \ \ \ \ \ \ \left(6.102\right)\\y & =y\left( \gamma\right) ,\ \ \ \ \ \ \ \ \ \ \left(6.103\right)\end{aligned}
we have
[Zαi]=[Ziα]=[xγyγ]          (4.83)[Sαβ]=[1xγ2+yγ2]          (4.85)[Ni]=1xγ2+yγ2[yγxγ]          (4.89)\begin{aligned}\left[ Z_{\alpha}^{i}\right] & =\left[ Z_{i\alpha}\right] =\left[ \begin{array} {c} x_{\gamma}\\ y_{\gamma} \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.83\right)\\\left[ S^{\alpha\beta}\right] & =\left[ \frac{1}{x_{\gamma}^{2} +y_{\gamma}^{2}}\right] \ \ \ \ \ \ \ \ \ \ \left(4.85\right)\\\left[ N^{i}\right] & =\frac{1}{\sqrt{x_{\gamma}^{2}+y_{\gamma}^{2}} }\left[ \begin{array} {r} -y_{\gamma}\\ x_{\gamma} \end{array} \right] \ \ \ \ \ \ \ \ \ \ \left(4.89\right)\end{aligned}
Thus,
[NiSα]=xγyγγxγγyγ(xγ2+yγ2)3/2[xγyγ](6.104)\left[ \frac{\partial N^{i}}{\partial S^{\alpha}}\right] =\frac{x_{\gamma }y_{\gamma\gamma}-x_{\gamma\gamma}y_{\gamma}}{\left( x_{\gamma}^{2} +y_{\gamma}^{2}\right) ^{3/2}}\left[ \begin{array} {c} x_{\gamma}\\ y_{\gamma} \end{array} \right]\tag{6.104}
As a result, the various forms of the curvature tensor are
Bαβ=[xγyγγxγγyγxγ2+yγ2]          (6.105)Bβα=[xγyγγxγγyγ(xγ2+yγ2)3/2]          (6.106)Bαβ=[xγyγγxγγyγ(xγ2+yγ2)5/2].          (6.107)\begin{aligned}B_{\alpha\beta} & =\left[ \frac{x_{\gamma}y_{\gamma\gamma}-x_{\gamma\gamma }y_{\gamma}}{\sqrt{x_{\gamma}^{2}+y_{\gamma}^{2}}}\right]\ \ \ \ \ \ \ \ \ \ \left(6.105\right)\\B_{\beta}^{\alpha} & =\left[ \frac{x_{\gamma}y_{\gamma\gamma} -x_{\gamma\gamma}y_{\gamma}}{\left( x_{\gamma}^{2}+y_{\gamma}^{2}\right) ^{3/2}}\right]\ \ \ \ \ \ \ \ \ \ \left(6.106\right)\\B^{\alpha\beta} & =\left[ \frac{x_{\gamma}y_{\gamma\gamma}-x_{\gamma\gamma }y_{\gamma}}{\left( x_{\gamma}^{2}+y_{\gamma}^{2}\right) ^{5/2}}\right] .\ \ \ \ \ \ \ \ \ \ \left(6.107\right)\end{aligned}
The mean curvature BααB_{\alpha}^{\alpha} equals the sole element of BβαB_{\beta }^{\alpha}, i.e.
Bαα=xγyγγxγγyγ(xγ2+yγ2)3/2.(6.108)B_{\alpha}^{\alpha}=\frac{x_{\gamma}y_{\gamma\gamma}-x_{\gamma\gamma} y_{\gamma}}{\left( x_{\gamma}^{2}+y_{\gamma}^{2}\right) ^{3/2}}.\tag{6.108}
Let us now document two special cases. When arc length ss is used to parameterize the curve, we have
Bαβ,Bβα,Bβα=[xsyssxssys](6.109)B_{\alpha\beta},B_{\beta}^{\alpha},B_{\beta}^{\alpha}=\left[ x_{s} y_{ss}-x_{ss}y_{s}\right]\tag{6.109}
and BααB_{\alpha}^{\alpha} is, of course, given by
Bαα=xsyssxssys.(6.110)B_{\alpha}^{\alpha}=x_{s}y_{ss}-x_{ss}y_{s}.\tag{6.110}
If the curve is given by the graph of a function y=y(x)y=y\left( x\right) , then we have
Bαβ=[yxx1+yx2]          (6.111)Bβα=[yxx(1+yx2)3/2]          (6.112)Bαβ=[yxx(1+yx2)5/2].          (6.113)\begin{aligned}B_{\alpha\beta} & =\left[ \frac{y_{xx}}{\sqrt{1+y_{x}^{2}}}\right]\ \ \ \ \ \ \ \ \ \ \left(6.111\right)\\B_{\beta}^{\alpha} & =\left[ \frac{y_{xx}}{\left( 1+y_{x}^{2}\right) ^{3/2}}\right]\ \ \ \ \ \ \ \ \ \ \left(6.112\right)\\B^{\alpha\beta} & =\left[ \frac{y_{xx}}{\left( 1+y_{x}^{2}\right) ^{5/2} }\right] .\ \ \ \ \ \ \ \ \ \ \left(6.113\right)\end{aligned}
The mean curvature BααB_{\alpha}^{\alpha} equals the sole element of BβαB_{\beta }^{\alpha}, i.e.
Bαα=yxx(1+yx2)3/2.(6.114)B_{\alpha}^{\alpha}=\frac{y_{xx}}{\left( 1+y_{x}^{2}\right) ^{3/2}}.\tag{6.114}

6.5.2In polar coordinates

For a general planar curve described in polar coordinates by the equations
r=r(γ)          (6.115)θ=θ(γ),          (6.116)\begin{aligned}r & =r\left( \gamma\right)\ \ \ \ \ \ \ \ \ \ \left(6.115\right)\\\theta & =\theta\left( \gamma\right) ,\ \ \ \ \ \ \ \ \ \ \left(6.116\right)\end{aligned}
the expressions for the mean curvature are too cumbersome to give here. We will therefore restrict ourselves to curves that represent graph of function r=r(θ)r=r\left( \theta\right) . It is left as an exercise to show that the various forms of the curvature tensor are given by
Bαβ=[rγγr2rγr2r2+rγ2]          (6.117)Bβα=[rγγr2rγr2(r2+rγ2)3/2]          (6.118)Bαβ=[rγγr2rγr2(r2+rγ2)5/2].          (6.119)\begin{aligned}B_{\alpha\beta} & =\left[ \frac{r_{\gamma\gamma}r-2r_{\gamma}-r^{2}} {\sqrt{r^{2}+r_{\gamma}^{2}}}\right]\ \ \ \ \ \ \ \ \ \ \left(6.117\right)\\B_{\beta}^{\alpha} & =\left[ \frac{r_{\gamma\gamma}r-2r_{\gamma}-r^{2} }{\left( r^{2}+r_{\gamma}^{2}\right) ^{3/2}}\right]\ \ \ \ \ \ \ \ \ \ \left(6.118\right)\\B^{\alpha\beta} & =\left[ \frac{r_{\gamma\gamma}r-2r_{\gamma}-r^{2} }{\left( r^{2}+r_{\gamma}^{2}\right) ^{5/2}}\right] .\ \ \ \ \ \ \ \ \ \ \left(6.119\right)\end{aligned}
In particular, the mean curvature BααB_{\alpha}^{\alpha} equals the sole element of BβαB_{\beta}^{\alpha}, i.e.
Bαα=rγγr2rγr2(r2+rγ2)3/2.(6.120)B_{\alpha}^{\alpha}=\frac{r_{\gamma\gamma}r-2r_{\gamma}-r^{2}}{\left( r^{2}+r_{\gamma}^{2}\right) ^{3/2}}.\tag{6.120}
In this Section, we will demonstrate how to calculate the mean curvature BααB_{\alpha}^{\alpha} of a surface specified by an implicit equation  \;
F(Z)=c,(6.121)F\left( Z\right) =c,\tag{6.121}
where F(Z)F\left( Z\right) is a function of the ambient coordinates ZiZ^{i} and cc is a constant. In other words, the surface is a level set of the function F(Z)F\left( Z\right) . The goal is to express BααB_{\alpha} ^{\alpha} in terms of the covariant derivatives of F(Z).F\left( Z\right) .
The following figure illustrates an implicit surface in an ambient space referred to Cartesian coordinates.
(6.122)
To be clear, our analysis will apply to arbitrary ambient coordinates. The above figure features Cartesian coordinates only for the sake of visual clarity.
Naturally, we are not able to calculate to full curvature tensor BβαB_{\beta }^{\alpha} since the posed problem does not allow for surface coordinates SαS^{\alpha}. On the other hand, the task of finding an expression for the mean curvature BααB_{\alpha}^{\alpha} in terms of the derivatives of F(Z)F\left( Z\right) should be feasible since BααB_{\alpha}^{\alpha} is an invariant characteristic of the surface. In other words, it is independent of the choice of the surface coordinates SαS^{\alpha}. Nevertheless, in order to find such an expression for BααB_{\alpha}^{\alpha}, we may still introduce a surface coordinate system SαS^{\alpha}, as long as we make sure that the final expression is free of any traces of it.
With the help of temporary surface coordinates SαS^{\alpha}, represent the surface by the equations
Zi=Zi(S).(6.123)Z^{i}=Z^{i}\left( S\right) .\tag{6.123}
Plugging in these equation of the surface into the implicit equation
F(Z)=c,(6.121)F\left( Z\right) =c, \tag{6.121}
we obtain the identity
F(Z(S))=c(6.124)F\left( Z\left( S\right) \right) =c\tag{6.124}
in the surface coordinates SαS^{\alpha}. Differentiating with SαS^{\alpha}, we find
FZiZiSα=0,(6.125)\frac{\partial F}{\partial Z^{i}}\frac{\partial Z^{i}}{\partial S^{\alpha}}=0,\tag{6.125}
or, equivalently,
iF Zαi=0.(6.126)\nabla_{i}F~Z_{\alpha}^{i}=0.\tag{6.126}
Comparing this identity with the definition of the unit normal
NiZαi=0,(3.134)N_{i}Z_{\alpha}^{i}=0, \tag{3.134}
we recognize that iF\nabla_{i}F must is a scalar multiple of NiN_{i}. Therefore,
Ni=iFjFjF.(6.127)N_{i}=\frac{\nabla_{i}F}{\sqrt{\nabla_{j}F\nabla^{j}F}}.\tag{6.127}
Note that by adopting this equation over the alternative
Ni=iFjFjF,(6.128)N_{i}=-\frac{\nabla_{i}F}{\sqrt{\nabla_{j}F\nabla^{j}F}},\tag{6.128}
we have selected one of the two available normal directions.
In order to go from the unit normal to the curvature tensor, recall the corollary
Bβα=ZiαβNi(6.129)B_{\beta}^{\alpha}=-Z_{i}^{\alpha}\nabla_{\beta}N^{i}\tag{6.129}
of Weingarten's equation. By a formal application of the chain rule to βNi\nabla_{\beta}N_{i}, we have
βNi=ZβjjNi(6.130)\nabla_{\beta}N_{i}=Z_{\beta}^{j}\nabla_{j}N^{i}\tag{6.130}
and therefore
Bβα=ZiαZβjjNi.(6.131)B_{\beta}^{\alpha}=-Z_{i}^{\alpha}Z_{\beta}^{j}\nabla_{j}N^{i}.\tag{6.131}
Here we must pause and note that, ordinarily, the combination jNi\nabla_{j} N^{i} is nonsensical since NiN^{i} is defined only on the surface and not in the surrounding ambient space, and is therefore not subject to the ambient covariant derivative. In the present context, however, we can extend the normal field NiN^{i} to the surrounding space by interpreting the implicit equation
F(Z)=c(6.121)F\left( Z\right) =c \tag{6.121}
not as a single surface corresponding to a single value of cc, but as a family of surfaces corresponding to a range of values of cc. Then the corresponding family of surface normals NiN^{i} fills the surrounding ambient space, as illustrated in the following figure, and we may therefore treat NiN^{i} as a spatial tensor field subject to the ambient covariant derivative.
(6.132)
In the context of this interpretation of the object NiN^{i}, the formula
Bβα=ZiαZβjjNi(6.133)B_{\beta}^{\alpha}=-Z_{i}^{\alpha}Z_{\beta}^{j}\nabla_{j}N^{i}\tag{6.133}
is justified. Contracting on α\alpha and β\beta, we find
Bαα=ZiαZαjjNi.(6.134)B_{\alpha}^{\alpha}=-Z_{i}^{\alpha}Z_{\alpha}^{j}\nabla_{j}N^{i}.\tag{6.134}
By the projection formula
ZαiZjα=δjiNiNj,(3.161)Z_{\alpha}^{i}Z_{j}^{\alpha}=\delta_{j}^{i}-N^{i}N_{j}, \tag{3.161}
we have
Bαα=(δijNiNj)jNi(6.135)B_{\alpha}^{\alpha}=-\left( \delta_{i}^{j}-N_{i}N^{j}\right) \nabla_{j}N^{i}\tag{6.135}
or, equivalently,
Bαα=iNi+NiNjjNi.(6.136)B_{\alpha}^{\alpha}=-\nabla_{i}N^{i}+N_{i}N^{j}\nabla_{j}N^{i}.\tag{6.136}
Note that with this equation, we have already accomplished the goal of expressing BααB_{\alpha}^{\alpha} in terms of the covariant derivatives of FF since we already have such an expression for NiN^{i}. However, we can greatly simplify this equation by showing that the second term term on the right vanishes.
This is so because the combination NijNiN_{i}\nabla_{j}N^{i} represents the derivative of a unit vector analogous to the blueprint example discussed in Section TBD of Introduction to Tensor Calculus. Recall that the unit-length condition for the normal NiN^{i} reads
NiNi=1.(3.130)N_{i}N^{i}=1. \tag{3.130}
An application of the covariant derivative j\nabla_{j} to both sides yields
jNi Ni+NijNi=0.(6.137)\nabla_{j}N_{i}~N^{i}+N_{i}\nabla_{j}N^{i}=0.\tag{6.137}
Since the two terms of the left are equivalent, we find
NijNi=0,(6.138)N_{i}\nabla_{j}N^{i}=0,\tag{6.138}
as we set out to show.
Thus, in terms of NiN^{i}, the mean curvature BααB_{\alpha}^{\alpha} is given by the remarkably simple expression
Bαα=iNi.(6.139)B_{\alpha}^{\alpha}=-\nabla_{i}N^{i}.\tag{6.139}
Recalling that
Ni=iFjF jF,(6.127)N^{i}=\frac{\nabla^{i}F}{\sqrt{\nabla_{j}F~\nabla^{j}F}}, \tag{6.127}
gives us the explicit expression for BααB_{\alpha}^{\alpha} in terms of the covariant derivatives of FF, i.e.
Bαα=i(iFjF jF).(6.140)B_{\alpha}^{\alpha}=-\nabla_{i}\left( \frac{\nabla^{i}F}{\sqrt{\nabla _{j}F~\nabla^{j}F}}\right) .\tag{6.140}
We have therefore accomplished our objective.
It is left as an exercise to show that the above equation is equivalent to
Bαα=ijF iF jFiiF jF jF(kF kF)3/2(6.141)B_{\alpha}^{\alpha}=\frac{\nabla_{i}\nabla^{j}F~\nabla^{i}F~\nabla_{j} F-\nabla_{i}\nabla^{i}F~\nabla_{j}F~\nabla^{j}F}{\left( \nabla_{k} F~\nabla^{k}F\right) ^{3/2}}\tag{6.141}
If we denote iF\nabla_{i}F by FiF_{i} and ijF\nabla_{i}\nabla_{j}F by FijF_{ij}, then the above equation reads
Bαα=FijFiFjFiiFjFj(FkFk)3/2.(6.142)B_{\alpha}^{\alpha}=\frac{F_{i}^{j}F^{i}F_{j}-F_{i}^{i}F_{j}F^{j}}{\left( F_{k}F^{k}\right) ^{3/2}}.\tag{6.142}
It is also left as an exercise to show that, in terms of NiN^{i}, the trace BβαBαβB_{\beta}^{\alpha}B_{\alpha}^{\beta} of the third groundform BβαBγβB_{\beta }^{\alpha}B_{\gamma}^{\beta} is given by
BβαBαβ=jNiiNj.(6.143)B_{\beta}^{\alpha}B_{\alpha}^{\beta}=\nabla_{j}N^{i}\nabla_{i}N^{j}.\tag{6.143}
Therefore, the Gaussian curvature KK of a is given by
K=12(iNijNjjNiiNj).(6.144)K=\frac{1}{2}\left( \nabla_{i}N^{i}\nabla_{j}N^{j}-\nabla_{j}N^{i}\nabla _{i}N^{j}\right) .\tag{6.144}
Exercise 6.1Show that
αNβN=Cαβ.(6.24)\nabla_{\alpha}\mathbf{N}\cdot\nabla_{\beta}\mathbf{N}=C_{\alpha\beta}. \tag{6.24}
Exercise 6.2With regard to the equations
Bαα=NαSα(2.80)B_{\alpha}^{\alpha}=\mathbf{N}\cdot\nabla_{\alpha}\mathbf{S}^{\alpha} \tag{2.80}
and
κ=NT(s)(6.145)\kappa=\mathbf{N}\cdot\mathbf{T}^{\prime}\left( s\right)\tag{6.145}
in Section 6.2.5, explain why αSα\nabla_{\alpha} \mathbf{S}^{\alpha} is equivalent to T(s)\mathbf{T}^{\prime}\left( s\right) .
Exercise 6.3Show that at points where the mean curvature is zero,
BβαBαβ=2K.(6.146)B_{\beta}^{\alpha}B_{\alpha}^{\beta}=-2K.\tag{6.146}
Exercise 6.4Show that the catenoid, which is a shape of revolution given by the equation
G(z)=acoshzba,(6.147)G\left( z\right) =a\cosh\frac{z-b}{a},\tag{6.147}
has zero mean curvature.
Exercise 6.5Show that for a sphere,
Bαβ=1RSαβ(6.148)B_{\alpha\beta}=-\frac{1}{R}S_{\alpha\beta}\tag{6.148}
in any surface coordinate system.
Exercise 6.6Show that various forms of the curvature tensor for a cone of angle Θ\Theta are given by equations
[Bαβ]=[rcosΘsinΘ000]          (6.149)[Bβα]=[cotΘr000]          (6.150)[Bαβ]=[cosΘr3sin3Θ000].          (6.151)\begin{aligned}\left[ B_{\alpha\beta}\right] & =\left[ \begin{array} {cc} -r\cos\Theta\sin\Theta & 0\\ 0 & 0 \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.149\right)\\\left[ B_{\beta}^{\alpha}\right] & =\left[ \begin{array} {cc} -\frac{\cot\Theta}{r} & 0\\ 0 & 0 \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.150\right)\\\left[ B^{\alpha\beta}\right] & =\left[ \begin{array} {cc} -\frac{\cos\Theta}{r^{3}\sin^{3}\Theta} & 0\\ 0 & 0 \end{array} \right] .\ \ \ \ \ \ \ \ \ \ \left(6.151\right)\end{aligned}
Thus, the mean curvature BααB_{\alpha}^{\alpha} is given by
Bαα=cotΘr(6.152)B_{\alpha}^{\alpha}=-\frac{\cot\Theta}{r}\tag{6.152}
while the Gaussian curvature KK vanishes, i.e.
K=B=0.(6.153)K=B=0.\tag{6.153}

6.7.1The tangent developable of a curve

The tangent developable of a curve is the surface swept out by tangent lines to the curve. If the curve is given by the vector equation
R=R(γ),(6.154)\mathbf{R}=\mathbf{R}\left( \gamma\right) ,\tag{6.154}
then the vector equation for the tangent developable reads
R(γ,t)=R(γ)+tR(γ).(6.155)\mathbf{R}\left( \gamma,t\right) =\mathbf{R}\left( \gamma\right) +t\mathbf{R}^{\prime}\left( \gamma\right) .\tag{6.155}
For the sake of simplicity, assume that the curve is parameterized by the arc length ss. Recall that the derivative R(s)\mathbf{R}^{\prime}\left( s\right) is the unit tangent T\mathbf{T}, i.e.
R(s)=T(s),(6.156)\mathbf{R}^{\prime}\left( s\right) =\mathbf{T}\left( s\right) ,\tag{6.156}
and that the derivatives of T(s)\mathbf{T}\left( s\right) are governed by the Frenet equations
[TsPsQs]=[σσττ][TPQ],(8.13)\left[ \begin{array} {c} \mathbf{T}_{s}\\ \mathbf{P}_{s}\\ \mathbf{Q}_{s} \end{array} \right] =\left[ \begin{array} {rrr} & \sigma & \\ -\sigma & & \tau\\ & -\tau & \end{array} \right] \left[ \begin{array} {c} \mathbf{T}\\ \mathbf{P}\\ \mathbf{Q} \end{array} \right] , \tag{8.13}
where P\mathbf{P} is the principal normal, Q\mathbf{Q} is the binormal, σ\sigma is the absolute curvature, τ\tau is the torsion and the subscript ss denotes differentiation with respect to the arc length ss.
Exercise 6.7Show that the covariant basis Sα\mathbf{S}_{\alpha} is given by
S1=T+tσP          (6.157)S2=T.          (6.158)\begin{aligned}\mathbf{S}_{1} & =\mathbf{T}+t\sigma\mathbf{P}\ \ \ \ \ \ \ \ \ \ \left(6.157\right)\\\mathbf{S}_{2} & =\mathbf{T.}\ \ \ \ \ \ \ \ \ \ \left(6.158\right)\end{aligned}
Exercise 6.8Show that the surface normal N\mathbf{N} is collinear with the binormal Q\mathbf{Q} an may therefore be chosen so that it coincides with Q\mathbf{Q} , i.e.
N(s,t)=Q(t).(6.159)\mathbf{N}\left( s,t\right) =\mathbf{Q}\left( t\right) .\tag{6.159}
Exercise 6.9Show that the covariant metric tensor SαβS_{\alpha\beta} is given by
[Sαβ]=[1+t2σ2111](6.160)\left[ S_{\alpha\beta}\right] =\left[ \begin{array} {cc} 1+t^{2}\sigma^{2} & 1\\ 1 & 1 \end{array} \right]\tag{6.160}
and therefore the contravariant basis SαβS^{\alpha\beta} is given by
[Sαβ]=1t2σ2[111t2σ2+1].(6.161)\left[ S^{\alpha\beta}\right] =\frac{1}{t^{2}\sigma^{2}}\left[ \begin{array} {cc} \phantom{-} 1 & -1\\ -1 & t^{2}\sigma^{2}+1 \end{array} \right] .\tag{6.161}
Meanwhile, the area element S\sqrt{S} is given by
S=σt.(6.162)\sqrt{S}=\sigma\left\vert t\right\vert .\tag{6.162}
Exercise 6.10Show that the curvature tensors BαβB_{\alpha\beta} and BβαB_{\beta}^{\alpha}, that correspond to the choice of the normal N\mathbf{N} that coincides with Q\mathbf{Q}, are given by
[Bαβ]=[tστ000]          (6.163)[Bβα]=[τtσ0τtσ0].          (6.164)\begin{aligned}\left[ B_{\alpha\beta}\right] & =\left[ \begin{array} {cc} t\sigma\tau & 0\\ 0 & 0 \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.163\right)\\\left[ B_{\beta}^{\alpha}\right] & =\left[ \begin{array} {rr} \frac{\tau}{t\sigma} & 0\\ -\frac{\tau}{t\sigma} & 0 \end{array} \right] .\ \ \ \ \ \ \ \ \ \ \left(6.164\right)\end{aligned}
Thus, the mean curvature BααB_{\alpha}^{\alpha} is given by
Bαα=τtσ.(6.165)B_{\alpha}^{\alpha}=\frac{\tau}{t\sigma}.\tag{6.165}
Recall that the surface Christoffel symbols are not needed for this calculation since BαβB_{\alpha\beta} is given by the idntity
Bαβ=NSβSα(2.83)B_{\alpha\beta}=\mathbf{N}\cdot\frac{\partial\mathbf{S}_{\beta}}{\partial S^{\alpha}} \tag{2.83}
which includes only partial derivatives.

6.7.2The polar developable

The polar developable of a curve is the surface swept out by axes of the osculating circles. Thus, if the curve is given by the vector equation
R=R(γ),(6.166)\mathbf{R}=\mathbf{R}\left( \gamma\right) ,\tag{6.166}
then the vector equation for the tangent developable reads
R(γ,t)=R(γ)+1σ(γ)P(γ)+tQ(γ).(6.167)\mathbf{R}\left( \gamma,t\right) =\mathbf{R}\left( \gamma\right) +\frac {1}{\sigma\left( \gamma\right) }\mathbf{P}\left( \gamma\right) +t\mathbf{Q}\left( \gamma\right) .\tag{6.167}
Let
Σ=σs+tσ2τ.(6.168)\Sigma=\sigma_{s}+t\sigma^{2}\tau.\tag{6.168}
begin{exercise} Show that the covariant basis vectors are given by
S1=σs+tσ2τσ2P+τσQ          (6.169)S2=Q.          (6.170)\begin{aligned}\mathbf{S}_{1} & =-\frac{\sigma_{s}+t\sigma^{2}\tau}{\sigma^{2}} \mathbf{P}+\frac{\tau}{\sigma}\mathbf{Q}\ \ \ \ \ \ \ \ \ \ \left(6.169\right)\\\mathbf{S}_{2} & =\mathbf{Q.}\ \ \ \ \ \ \ \ \ \ \left(6.170\right)\end{aligned}
end{exercise}
Exercise 6.11Show that the surface normal N\mathbf{N} corresponds to the tangent T\mathbf{T}, i.e.
N(s,t)=T(s).(6.171)\mathbf{N}\left( s,t\right) =\mathbf{T}\left( s\right) .\tag{6.171}
Exercise 6.12Show that the covariant metric tensor SαβS_{\alpha\beta} is given by
[Sαβ]=[(σs+tσ2τ)2σ4+τ2σ2τστσ1](6.172)\left[ S_{\alpha\beta}\right] =\left[ \begin{array} {cc} \frac{\left( \sigma_{s}+t\sigma^{2}\tau\right) ^{2}}{\sigma^{4}}+\frac {\tau^{2}}{\sigma^{2}} & \frac{\tau}{\sigma}\\ \frac{\tau}{\sigma} & 1 \end{array} \right]\tag{6.172}
and the contravariant metric tensor SαβS^{\alpha\beta} is given by
[Sαβ]=σ4(σs+tσ2τ)2[1τστσ(σs+tσ2τ)2σ4+τ2σ2].(6.173)\left[ S^{\alpha\beta}\right] =\frac{\sigma^{4}}{\left( \sigma_{s} +t\sigma^{2}\tau\right) ^{2}}\left[ \begin{array} {cc} 1 & -\frac{\tau}{\sigma}\\ -\frac{\tau}{\sigma} & \frac{\left( \sigma_{s}+t\sigma^{2}\tau\right) ^{2} }{\sigma^{4}}+\frac{\tau^{2}}{\sigma^{2}} \end{array} \right] .\tag{6.173}
Exercise 6.13Show that the nonzero elements of the Christoffel symbol are
Γ111=2σs2+σσss+σ2τ2+tσ3τsσ(σs+tσ2τ)          (6.174)Γ121=Γ211=σ2τσs+tσ2τ          (6.175)Γ112=σssτσsτs+t2σ3τ3+3tσsστ2+στ3σ(σs+tσ2τ)          (6.176)Γ122=Γ212=στ2σs+tσ2τ.          (6.177)\begin{aligned}\Gamma_{11}^{1} & =\frac{-2\sigma_{s}^{2}+\sigma\sigma_{ss}+\sigma^{2} \tau^{2}+t\sigma^{3}\tau_{s}}{\sigma\left( \sigma_{s}+t\sigma^{2}\tau\right) }\ \ \ \ \ \ \ \ \ \ \left(6.174\right)\\\Gamma_{12}^{1} & =\Gamma_{21}^{1}=\frac{\sigma^{2}\tau}{\sigma_{s} +t\sigma^{2}\tau}\ \ \ \ \ \ \ \ \ \ \left(6.175\right)\\\Gamma_{11}^{2} & =\frac{\sigma_{ss}\tau-\sigma_{s}\tau_{s}+t^{2}\sigma ^{3}\tau^{3}+3t\sigma_{s}\sigma\tau^{2}+\sigma\tau^{3}}{\sigma\left( \sigma_{s}+t\sigma^{2}\tau\right) }\ \ \ \ \ \ \ \ \ \ \left(6.176\right)\\\Gamma_{12}^{2} & =\Gamma_{21}^{2}=-\frac{\sigma\tau^{2}}{\sigma_{s} +t\sigma^{2}\tau}.\ \ \ \ \ \ \ \ \ \ \left(6.177\right)\end{aligned}
Exercise 6.14Show that the curvature tensors BαβB_{\alpha\beta} and BβαB_{\beta}^{\alpha}, that correspond to the choice of the normal N\mathbf{N} that coincides with Q\mathbf{Q}, are given by
[Bαβ]=[σs+tσ2τσ000]          (6.178)[Bβα]=σ3σs+tσ2τ[10τσ0].          (6.179)\begin{aligned}\left[ B_{\alpha\beta}\right] & =\left[ \begin{array} {cc} \frac{\sigma_{s}+t\sigma^{2}\tau}{\sigma} & 0\\ 0 & 0 \end{array} \right]\ \ \ \ \ \ \ \ \ \ \left(6.178\right)\\\left[ B_{\beta}^{\alpha}\right] & =\frac{\sigma^{3}}{\sigma_{s} +t\sigma^{2}\tau}\left[ \begin{array} {cc} 1 & 0\\ \frac{\tau}{\sigma} & 0 \end{array} \right] .\ \ \ \ \ \ \ \ \ \ \left(6.179\right)\end{aligned}
Thus, the mean curvature BααB_{\alpha}^{\alpha} is given by
Bαα=σ3σs+tσ2τ.(6.180)B_{\alpha}^{\alpha}=\frac{\sigma^{3}}{\sigma_{s}+t\sigma^{2}\tau}.\tag{6.180}

6.7.3The δ/δt\delta/\delta t-derivative

The next set of exercises analyzes the motion of a particle along a trajectory embedded in a surface. The trajectory is described by the equations
Sα=Sα(t)(6.181)S^{\alpha}=S^{\alpha}\left( t\right)\tag{6.181}
that specify the surface coordinates SαS^{\alpha} as a function of time tt.
Exercise 6.15Show that the surface components VαV^{\alpha} of the particle's velocity V\mathbf{V} are given by ordinary time derivatives of Sα(t)S^{\alpha}\left( t\right) , i.e.
Vα=dSα(t)dt.(6.182)V^{\alpha}=\frac{dS^{\alpha}\left( t\right) }{dt}.\tag{6.182}
Hint: R(t)=R(S(t))\mathbf{R}\left( t\right) =\mathbf{R}\left( S\left( t\right) \right) .
Exercise 6.16 Show that the acceleration A\mathbf{A} of the particle is given by
A=δVαδtSα+NBαβVαVβ,(6.183)\mathbf{A}=\frac{\delta V^{\alpha}}{\delta t}\mathbf{S}_{\alpha} +\mathbf{N}B_{\alpha\beta}V^{\alpha}V^{\beta},\tag{6.183}
where
δVαδt=dVαdt+ΓβγαVβVγ.(6.184)\frac{\delta V^{\alpha}}{\delta t}=\frac{dV^{\alpha}}{dt}+\Gamma_{\beta\gamma }^{\alpha}V^{\beta}V^{\gamma}.\tag{6.184}
Conclude that δVα/δt\delta V^{\alpha}/\delta t is a tensor with respect to coordinate transformations on the surface.
Exercise 6.17Note that the term
BαβVαVβ(6.185)B_{\alpha\beta}V^{\alpha}V^{\beta}\tag{6.185}
corresponds to centripetal acceleration. It is a generalization of the familiar expression
V2R(6.186)-\frac{V^{2}}{R}\tag{6.186}
for the centripetal acceleration of a particle moving with uniform speed around a circle of radius RR. In fact, show that for a sphere of radius RR, we have
BαβVαVβ=(V)2R(6.187)B_{\alpha\beta}V^{\alpha}V^{\beta}=-\frac{\left( V\right) ^{2}}{R}\tag{6.187}
where VV is the magnitude of the velocity vector V\mathbf{V}, i.e.
(V)2=V1V1+V2V2.(6.188)\left( V\right) ^{2}=V_{1}V^{1}+V_{2}V^{2}.\tag{6.188}
Exercise 6.18Inspired by equation
δVαδt=dVαdt+ΓβγαVβVγ,(6.184)\frac{\delta V^{\alpha}}{\delta t}=\frac{dV^{\alpha}}{dt}+\Gamma_{\beta\gamma }^{\alpha}V^{\beta}V^{\gamma}, \tag{6.184}
define the δ/δt\delta/\delta t-derivative along the trajectory according to the equation
δTβαδt=dTβαdt+VγΓγωαTβωVγΓγβωTωα.(6.189)\frac{\delta T_{\beta}^{\alpha}}{\delta t}=\frac{dT_{\beta}^{\alpha}} {dt}+V^{\gamma}\Gamma_{\gamma\omega}^{\alpha}T_{\beta}^{\omega}-V^{\gamma }\Gamma_{\gamma\beta}^{\omega}T_{\omega}^{\alpha}.\tag{6.189}
In his Application of Tensor Calculus, A.J. McConnell refers to this operator as the intrinsic derivative, but we will prefer the term δ/δt\delta/\delta t-derivative. Show that the δ/δt\delta/\delta t-derivative satisfies the tensor property.
Exercise 6.19Show that the δ/δt\delta/\delta t-derivative satisfies the product rule.
Exercise 6.20Show that the δ/δt\delta/\delta t-derivative commutes with contraction.
Exercise 6.21Show that for a time-dependent variant TβαT_{\beta}^{\alpha} defined on the broader surface, the δ/δt\delta/\delta t-derivative satisfies the chain rule
δTβαδt=Tβα(t,S)t+VγγTβα.(6.190)\frac{\delta T_{\beta}^{\alpha}}{\delta t}=\frac{\partial T_{\beta}^{\alpha }\left( t,S\right) }{\partial t}+V^{\gamma}\nabla_{\gamma}T_{\beta}^{\alpha }.\tag{6.190}
Exercise 6.22Conclude that the δ/δt\delta/\delta t-derivative is metrinilic with respect to all the surface metrics, except Sα\mathbf{S}_{\alpha} and Sα\mathbf{S}^{\alpha }. Meanwhile,
δSαδt=NVγBγα.(6.191)\frac{\delta\mathbf{S}_{\alpha}}{\delta t}=\mathbf{N}V^{\gamma}B_{\gamma \alpha}.\tag{6.191}
Exercise 6.23Use the newly developed calculus to rederive the equation
A=δVαδtSα+NBαβVαVβ.(6.183)\mathbf{A}=\frac{\delta V^{\alpha}}{\delta t}\mathbf{S}_{\alpha} +\mathbf{N}B_{\alpha\beta}V^{\alpha}V^{\beta}. \tag{6.183}
Exercise 6.24Show that the jolt J=A(t)\mathbf{J}=\mathbf{A}^{\prime}\left( t\right) is given by the formula
J=(δ2Vαδt2BβαBγδVβVγVδ)Sα          (6.192)            +(3BαβδVαδtVβ+αBβγVαVβVγ)N.          \begin{aligned}\mathbf{J} & =\left( \frac{\delta^{2}V^{\alpha}}{\delta t^{2}}-B_{\beta }^{\alpha}B_{\gamma\delta}V^{\beta}V^{\gamma}V^{\delta}\right) \mathbf{S} _{\alpha}\ \ \ \ \ \ \ \ \ \ \left(6.192\right)\\& \ \ \ \ \ \ \ \ \ \ \ \ +\left( 3B_{\alpha\beta}\frac{\delta V^{\alpha} }{\delta t}V^{\beta}+\nabla_{\alpha}B_{\beta\gamma}V^{\alpha}V^{\beta }V^{\gamma}\right) \mathbf{N}.\ \ \ \ \ \ \ \ \ \ \end{aligned}

6.7.4Implicit surfaces

The next set of problems concerns surfaces specified by the implicit equation
F(Z)=c.(6.121)F\left( Z\right) =c. \tag{6.121}
Define the tensor BjiB_{j}^{i} according to the equation
Bji=BβαZαiZjβ.(6.193)B_{j}^{i}=B_{\beta}^{\alpha}Z_{\alpha}^{i}Z_{j}^{\beta}.\tag{6.193}
The object BjiB_{j}^{i} may be referred to as the upshifted curvature tensor.
Exercise 6.25Show that BjiB_{j}^{i} is symmetric, i.e.
Bij=Bji.(6.194)B_{ij}=B_{ji}.\tag{6.194}
Exercise 6.26Show that
Bii=Bαα.(6.195)B_{i}^{i}=B_{\alpha}^{\alpha}.\tag{6.195}
Exercise 6.27Show that
Bji=jNi.(6.196)B_{j}^{i}=-\nabla_{j}N^{i}.\tag{6.196}
Exercise 6.28Show that
BβαBαβ=jNiiNj,(6.143)B_{\beta}^{\alpha}B_{\alpha}^{\beta}=\nabla_{j}N^{i}\nabla_{i}N^{j}, \tag{6.143}
or
BβαBαβ=BjiBij.(6.197)B_{\beta}^{\alpha}B_{\alpha}^{\beta}=B_{j}^{i}B_{i}^{j}.\tag{6.197}
Exercise 6.29With the help of the equation
BααBββBαβBβα=2K,(6.198)B_{\alpha}^{\alpha}B_{\beta}^{\beta}-B_{\alpha}^{\beta}B_{\beta}^{\alpha}=2K,\tag{6.198}
show that the Gaussian curvature KK is given by
K=12(iNijNjiNjjNi)(6.199)K=\frac{1}{2}\left( \nabla_{i}N^{i}\nabla_{j}N^{j}-\nabla_{i}N^{j}\nabla _{j}N^{i}\right)\tag{6.199}
or
K=12(BiiBjjBjiBij).(6.200)K=\frac{1}{2}\left( B_{i}^{i}B_{j}^{j}-B_{j}^{i}B_{i}^{j}\right) .\tag{6.200}
Exercise 6.30Test the formulas derived in Section 6.6 for a sphere of radius RR, implicitly specified in Cartesian coordinates by the equation
x2+y2+z2=R2.(6.201)x^{2}+y^{2}+z^{2}=R^{2}.\tag{6.201}
Exercise 6.31In Cartesian coordinates, an egg-shaped ellipsoid with semi-axes aa, aa, and cc is given by the implicit equation
x2a2+y2a2+z2c2=1.(6.202)\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}+\frac{z^{2}}{c^{2}}=1.\tag{6.202}
Derive the expressions in terms of xx, yy, and zz for its mean curvature BααB_{\alpha}^{\alpha} and Gaussian curvature KK.
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