Coordinate systems. 2.1 Cartesian coordinates

Transcription

1 2 Coordinate systems The practical description of dynamical systems involves a variety of coordinates systems. While the Cartesian coordinates discussed in section 2. are probably the most commonly used, many problems are more easily treated with special coordinate systems. The differential geometry of curves is studied in section 2.2 and leads to the concept of path coordinates, treated in section 2.3. Similarly, the differential geometry of surfaces is investigated in section 2.4 and leads to the concept of surface coordinates, treated in section 2.5. Finally, the differential geometry of three-dimensional maps is studied in section 2.6 and leads to orthogonal curvilinear coordinates developedinsection Cartesian coordinates The simplest way to represent the location of a point in three-dimensional space is to make use of a reference frame, F =[O, I =(ī, ī 2, ī 3 )], consisting of an orthonormal basis I with its origin and point O, as described in section.2.2. The time-dependent position vector of point P is represented by its Cartesian coordinates, x (t), x 2 (t), andx 3 (t), resolved along unit vectors, ī, ī 2,andī 3, respectively, i i 3 O x 2 r P x 3 x i 2 r(t) =x (t)ī + x 2 (t)ī 2 + x 3 (t)ī 3, (2.) Fig. 2.. Cartesian coordinate system. where t denotes time. Figure 2. depicts the situation: Cartesian coordinate x =ī T r is the projection of the position vector of point P along unit vector ī. Similarly, Cartesian coordinates x and x 2 are the projections of the same position vector along unit vectors ī 2 and ī 3, respectively. The components of the velocity vector are readily obtained by differentiating the expression for the position vector, eq. (2.), to nd O. A. Bauchau, Flexible Multibody Dynamics, DOI 0.007/ _2 Springer Science+Business Media B.V. 20

2 32 2 Coordinate systems v(t) =ẋ (t)ī +ẋ 2 (t)ī 2 +ẋ 3 (t)ī 3 = v (t)ī + v 2 (t)ī 2 + v 3 (t)ī 3. (2.2) The Cartesian components of the velocity vector are simply the time derivatives of the corresponding Cartesian components of the position vector: v (t) =ẋ (t), v 2 (t) =ẋ 2 (t), andv 3 (t) =ẋ 3 (t). Finally, the acceleration vector is obtained by taking a time derivative of the velocity vector to nd a(t) =ẍ (t)ī +ẍ 2 (t)ī 2 +ẍ 3 (t)ī 3 = a (t)ī + a 2 (t)ī 2 + a 3 (t)ī 3. (2.3) Here again, the Cartesian components of the acceleration vector are simply the derivatives of the corresponding Cartesian components of the velocity vector, or the second derivatives of the position components: a (t) = v (t) = ẍ (t), a 2 (t) = v 2 (t) =ẍ 2 (t), anda 3 (t) = v 3 (t) =ẍ 3 (t). Cartesian coordinates are simple to manipulate and are the most commonly used coordinate system in computational applications that deal with problems presenting arbitrary topologies. On the other hand, several other coordinate systems, such as those discussed in the rest of this chapter, are often used because they can ease the solution process for specic problems. In such cases, a specic coordinate system is used solve a specic problem. For instance, polar coordinates are very efcient to describe the behavior of a particle constrained to move along a circular path. 2.2 Differential geometry of a curve This section investigates the differential geometry of a curve, leading to the concept of path coordinates. Both intrinsic and arbitrary parameterizations will be considered. Frenet s triad is dened and its derivatives evaluated Intrinsic parameterization i 3 s b Figure 2.2 depicts a curve, denoted C, in three- dimensional space. A curve is the locus of the points generated by a single parameter, such that the position vector,, of such points can be written as i i 2 n Fig Conguration of a curve in space. t = (s), (2.4) where s is the parameter that generates the curve. If parameter s is the curvilinear coordinate that measures length along the curve, it is said to dene the intrinsic parameterization or natural parameterization of the curve. Frenet s triad A differential element of length, ds, along the curve is written as ds 2 =dp T 0 d, and in follows that (d /ds) T (d /ds) =.Theunit tangent vector to the curve is dened as

3 2.2 Differential geometry of a curve 33 t = d ds. (2.5) By construction, this is a unit vector because t T t =. Taking a derivative of this relationship with respect to the curvilinear coordinate leads to t T d t/ds =0. Vector d t/ds is normal to the tangent vector. The unit normal vector to the curve is dened as n = ρ d t ds, (2.6) where ρ is the radius of curvature of the curve, such that ρ = d t. (2.7) ds The quantity /ρ is the curvature of the curve, and ρ its radius of curvature. The two unit vector, t and n, are said to form the osculating plane of the curve. An orthonormal triad is now constructed by dening the binormal vector, b, as the cross product of the tangent by the normal vectors, b = t n. (2.8) The unit tangent, normal, and binormal vectors form an orthonormal triad, called Frenet s triad, depicted in g Derivatives of Frenet s triad First, the derivative of the normal vector is resolved in Frenet s triad as d n/ds = α t + β n+ γ b,whereα, β,andγ are unknown coefcients. Pre-multiplying this relationship by n T yields β = n T d n/ds =0, because n is a unit vector. Pre-multiplying by t T yields α = t T d n/ds = n T d t/ds = /ρ, where eq. (2.6) was used. Finally, pre-multiplying by b T yields γ = b T d n/ds =/τ. Combining all these results yields d n ds = ρ t + τ b, (2.9) where τ is the radius of twist of the curve, dened as τ = b T d n ds. (2.0) Next, the derivative of the binormal vector is resolved in Frenet s triad as d b/ds = α t+β n+γ b,whereα, β,andγ are unknown coefcients. Pre-multiplying this relationship by b T yields γ = b T d b/ds =0, because b is a unit vector. Premultiplying by t T yields α = t T d b/ds = b T d t/ds = b T n/ρ =0. Finally, premultiplying by n T yields β = n T d b/ds = b T d n/ds = /τ, where eq. (2.0) was used. Combining all these results yields d b ds = n. (2.) τ

4 34 2 Coordinate systems It follows that the twist of the curve can also be written as τ = d b. (2.2) ds If the binormal vector has a constant direction at all points along the curve, d b/ds =0, and the curve entirely lies in the plane dened by vectors t and n, i.e., the osculating plane is the same at all points of the curve. The curve is then a planar curve, and eq. (2.2) implies that /τ =0, i.e., the twist of the curve vanishes. The derivatives of Frenet s triad can be expressed in a compact manner by combining eqs. (2.6), (2.9), and (2.), d t n ds b = 0 /ρ 0 t /ρ 0 /τ n 0 /τ 0 b. (2.3) Arbitrary parameterization The previous section has developed a representation of a curve based on its natural or intrinsic parameterization. In many instances, however, this parameterization is difcult to obtain; instead, the curve is dened in terms of a single parameter, η,that does not measure length along the curve, see g The position vector of a point on the curve is now = (η). The derivatives of the position vector with respect to parameter η will be denoted as p = d dη, p 2 = d2 dη 2, p 3 = d3 dη 3, p 4 = d4 dη 4. A similar notation will be used for the tangent and normal vectors, t i = di t dη i, n i = di n dη i. The differential element of length along the curve can be written as ds 2 = (d /dη) T (d /dη) dη 2. The ratio of the increment in length along the curve, ds, to the increment in parameter value, dη,isthen ds dη = p T p = p. (2.4) Notation ( ) will be used to indicate a derivative with respect to η, and hence, d/ds = ( ) /p. The unit tangent vector to the curve is evaluated with the help of eq. (2.5) as t = p (2.5) p Next, the derivative of the tangent vector is found as

5 t = p p 2 p (p T p 2 )/p p Differential geometry of a curve 35 = ( t t T )p p 2 = ] [p p 2 ( t T p 2 ) t. (2.6) From eq. (2.7), the radius of curvature now becomes ρ = d t ds = t. p It follows that t = t = p /ρ. For a straight line, the tangent vector has a xed direction in space, t =0. It follows that for a straight line /ρ =0, i.e., its radius of curvature is innite. The curve s curvature is found to be ρ = p 2 p2 2 (pt 2 p )2 Higher-order derivatives of the tangent vector are found in a similar manner and t 2 = p [ p 3 ( t T p 3 + t T p 2 ) t 2( t T p 2 ) t ], t 3 = p [ p 4 ( t T p 4 +2 t T p 3 + t T 2 p 2 ) t 3( t T p 3 + t T p 2 ) t 3( t T p 2 ) t 2 ]. Next, the normal vector denedineq.(2.6)becomes p 3 (2.7) n = t t = t t. (2.8) For a straight line, t =0, and hence, the normal vector is not dened. In fact, any vector normal to a straight line is a normal vector. The derivative of the normal vector with respect to η then follows as The second-order derivative is then n = t [ t 2 ( n T t 2 ) n ]. (2.9) n 2 = t [ t 3 ( n T t 3 + n T t 2 ) n 2( n T t 2 ) n ]. (2.20) The binormal vector is readily expressed as b = t n = t t t = ρ p 3 p p 2. (2.2) Because the normal vector is not dened for a straight line, the binormal vector is not dened in that case. In fact, any vector normal to a straight line is a binormal vector. The derivative of the binormal vector becomes

6 36 2 Coordinate systems Using eq. (2.0), the twist of the curve is found to be b =( ρ p 3 ) p p 2 + ρ p 3 p p 3. (2.22) τ = p n T b = ρ p 5 Finally, introducing eq. (2.22) leads to [ p 2 pt 2 (pt p 2 )pt ] b. τ = ρ2 p 6 p T p 2 p 3. (2.23) The twist of the curve is closely related to the volume dened by vectors p, p 2,and p 3. Note that a straight line has a vanishing twist, /τ =0. Derivatives of the binormal vector are more easily expressed as b = t n+ t n = t n,and b 2 = t n + t n 2 = ñ t 2 + t n 2, where eqs. (2.8) and (2.9) were used. Example 2.. The helix Figure 2.3 depicts a helix, which is a three-dimensional curve denedbythefollowing position vector (η) =a cos η ī + a sin η ī 2 + kη ī 3, (2.24) where a and k are two parameters dening the shape of the curve. The derivatives of the position vector are p = a sin η ī + a cos η ī 2 + k ī 3, p 2 = a cos η ī a sin η ī 2,andp 3 = a sin η ī a cos η ī 2. The curvature and twist of the helix are found with the help of eqs. (2.7) and (2.23), respectively, as ρ = a a 2 + k 2, τ = k a 2 + k 2. Note that both curvature and twist are constant along the helix. The unit tangent vector is evaluated with the help of eq. (2.5) as t = a2 + k p = 2 a2 + k ( a sin ηī 2 + a cos ηī 2 + kī 3 ). (2.25) The ratio between an increment in length along the curve and the increment in the parameter value is then ds = a 2 + k 2 dη, see eq. (2.4). Next, the derivative of the tangent vector is computed with the help of eq. (2.6) as t = p 2 /p and the normal vector then follows as n = cos η ī sin η ī 2. Finally, the binormal vector found from eq. (2.2) b = a2 + k 2 [k sin η ī k cos η ī 2 + a ī 3 ]. The derivatives of Frenet s triad are found with the help of eq. (2.3) as d t ds = a a 2 + k 2 n, d n ds = a a 2 + k t k 2 + a 2 + k b, 2 d b ds = k a 2 + k 2 n.

7 2.2 Differential geometry of a curve 37 i 3 i 2 i a i 2 r r=a i Fig Conguration of a helix in threedimensional space. Fig Conguration of a planar linear spiral. Example 2.2. The linear spiral Figure 2.4 depicts a linear spiral, which is a planar curve dened by the following position vector = aθ cos θ ī + aθ sin θ ī 2, (2.26) where a is a parameter dening the shape of the curve. The derivatives of the position vector are p = a [(cos θ θ sin θ)ī +(sinθ + θ cos θ)ī 2 ], p 2 = a [ (2 sin θ + θ cos θ)ī +(2cosθ θ sin θ)ī 2 ]. It is readily veried that p 2 = a 2 ( + θ 2 ), p 2 2 = a 2 (4 + θ 2 ) and p T p 2 = a2 θ. The curvature of the linear spiral is found with the help of eq. (2.7) a ρ = 2+θ2 ( + θ 2 ) 3/2. Note that the curvature varies along the spiral. Of course, the twist is zero since the curve is planar. The unit tangent vector is evaluated with the help of eq. (2.5) as Finally, the normal vector becomes t = (cos θ θ sin θ)ī +(sinθ + θ cos θ)ī 2. +θ 2 n = [ 2sinθ + θ cos θ(2 + θ 2 ) ] ī + [ 2cosθ θ sin θ(2 + θ 2 ) ] ī 2 4+θ2 (2 + θ 2 ) 2. Example 2.3. Using polar coordinates to represent curves Cams play an important role in numerous mechanical systems: cam-follower pairs typically transform the rotary motion of the cam into a desirable motion of the follower. Figure 2.5 depicts a typical cam whose outer shape is denedbyacurve. It is convenient to dene this curve using the polar coordinate system indicated on the gure: for each angle α, the distance from point O to point P is denoted r. The complete curve is then dened by function r = r(α); angle α provides an arbitrary parameterization of the curve. If r(α) is a periodic function of angle α, the curve will be a closed curve, as expected for a cam.

8 38 2 Coordinate systems e O e 2 t r P n e r e / Fig Conguration of a cam. Fig Curvature distribution for the cam. Vectors, p,andp 2 now become = rc α ē + rs α ē 2, p =(r C α rs α )ē +(r S α + rc α )ē 2, p 2 =(r C α 2r S α rc α )ē +(r S α +2r C α rs α )ē 2, (2.27a) (2.27b) (2.27c) where the notation ( ) indicates a derivative with respect to α, S α =sinα, and C α =cosα. It then follows that p 2 = r 2 +r 2 and p 2 2 =(r r) 2 +4r 2. The various properties of the curve can then be evaluated; for instance, eqs. (2.5) and (2.7) yield the tangent vector and curvature along the curve, respectively. The curve depicted in g. 2.5 is dened by the following equation, r(α) = cosα +0.5 cos 2α and g. 2.6 shows the curvature distribution as a function of angle α. Figure 2.5 shows the unit tangent vector, t, at point P of the curve and denes angles β =(ē, t) and γ =(ē r, t); note that γ = β α. The unit tangent vector can now be written as t = C β ē +S β ē 2 = p /p, where the second equality follows from eq. (2.5). Pre-multiplying this relationship by ē T and ē T 2 yields p C β = r C α rs α and p S β = r S α + rc α, respectively. Solving these two equations for r and r and using elementary trigonometric identities then leads to r = p sin(β α) =p S γ, r = p cos(β α) =p C γ, (2.28a) (2.28b) where S γ =sinγ, andc γ =cosγ. The quotient of these two equations then yields the following relationship dα =tanγ dr r. (2.29) The derivative of the unit tangent vector with respect to the curvilinear coordinate along the curve is d t/ds =( S γ ē + C γ ē 2 )dβ/ds, and the curvature is then /ρ = dβ/ds. If the curve is convex, which is generally the case for cams, angle β is a monotonically increasing function of s, and hence, /ρ =dβ/ds. The chain rule

9 2.3 Path coordinates 39 for derivatives implies dβ =(/ρ)(ds/dα)(dα/dr)dr and introducing eqs. (2.4), (2.28a), and (2.29) then yields dβ = dr. (2.30) ρc γ It is left to the reader to verify that eq. (2.30) yields an alternative, simplied expression for the curvature of the cam ρ = 2r 2 rr + r 2. (2.3) Finally, an increment in angle γ can be expressed as dγ =dβ dαand introducing eqs. (2.30) and (2.29) yields ( dγ = tan γ ) dr. (2.32) ρc γ r p Path coordinates Consider a particle moving along a curve such that its position, s(t), is a given function of time. The velocity vector, v, of the particle is then v = d dt = d ds ds dt = v t, (2.33) where v =ds/dt is the speed of the particle, Clearly, the velocity vector of the particle is along the tangent to the curve. Next, the particle acceleration vector, a, becomes a = dv dt = dv dt t + v d t ds ds dt = v t + v2 ρ n. (2.34) The acceleration vector is contained in the osculating plane, and can be written as a = a t t + a n n, wherea t and a n are the tangential and normal components of acceleration, respectively. The tangential component of acceleration, a t = v, simply measures the change in particle speed. The normal component, a n = v 2 /ρ,isalways directed towards the center of curvature since v 2 /ρ is a positive number. This normal acceleration is clearly related to the curvature of the path; in fact, when the path is a straight line, /ρ =0, and the normal acceleration vanishes Problems Problem 2.. Prove identity Prove that /ρ = p 2/p 2 sin α, wherep 2 = p 2 and α is the angle between vectors p and p 2.

10 40 2 Coordinate systems Problem 2.2. Study of a curve Consider the following spatial curve: = a(η +sinη)ī + a( + cos η)ī 2 + a( cos η)ī 3, where a>0 is a given parameter. () Find the tangent, normal, and binormal vectors for this curve. (2) Determine the curvature, radius of curvature, and twist of the curve. Is this a planar curve? Is the tangent vector dened at all points of the curve? Problem 2.3. Study of a curve Consider the following spatial curve: = ρ(cos αη)(cos η)ī + ρ(cos αη)(sin η)ī 2 + ρ(sin αη)ī 3,whereρ > 0 and α are given parameters. () Find the tangent, normal, and binormal vectors for this curve. (2) Determine the curvature, radius of curvature, and twist of the curve. Problem 2.4. Short questions () A particle of mass m is sliding along a planar curve. Find the component of the particle s acceleration vector along the binormal vector of Frenet s triad. (2) A particle of mass m is sliding along a three-dimensional curve. Find the component of the particle s acceleration vector along the binormal vector of Frenet s triad. (3) State the criterion used to ascertain whether a curve is planar or three-dimensional. Problem 2.5. Study of a curve dened in polar coordinates The outer surface of a cam is specied by the following curve dened in polar coordinates, r(α) =.0 0.5cosα +0.8 cos 2α. () Plot the curve. (2) Plot the curvature distribution for α [0, 2π]. 2.4 Differential geometry of a surface This section investigates the differential geometry of surfaces, leading to the concept of surface coordinates. The differential geometry of surfaces is more complex than that of curves. The rst and second metric tensors of surfaces are introduced rst, and the analysis of the curvature of surfaces leads to the concept of lines of curvatures and associated principal radii of curvature. Finally, the base vectors and their derivatives are evaluated, leading to Gauss and Weingarten s formulæ The rst metric tensor of a surface Figure 2.7 depicts a surface, denoted S, in three-dimensional space. A surface is the locus of the points generated by two parameters, η and η 2, such that the position vector,, of such points can be written as = (η,η 2 ). (2.35) If η 2 is kept constant, η 2 = c 2, = (η,c 2 ) denes a curve embedded into the surface; such curve is called an η curve. Figure 2.7 shows a grid of such curves for various values of c 2. Similarly, η 2 curves can be dened, corresponding to = (c,η 2 ); a grid of η 2 curves obtained for different constant c is also shown on the gure. In general, parameters η and η 2 do not measure length along these

11 2.4 Differential geometry of a surface 4 embedded curves, and hence, they do not dene intrinsic parameterizations of the curves. The surface base vectors are dened as follows a = η, a 2 =, (2.36) gent to the surface at this point. Since η and η 2 do not form an intrinsic parameterization, vectors a and a 2 are not unit tangent vectors. Furthermore, these two vectors are not, in general, orthogonal to each other. The rst metric tensor of the surface, A, is dened as [ a T A = a a T a ] 2 a T 2 a at 2 a = 2 [ ] a a 2, (2.37) a 2 a 22 i 3 n P (, 2 ) and are shown in g Clearly, vectors a and a 2 are tangent to the η and η 2 curves that intersect at point P, respectively. Consequently, they lie in the plane tani i 2 a 2 a 2 Tangent plane Fig The base vectors of a surface. and its determinant is denoted a =det(a). A differential element of length on the surface is found as ds 2 =dp T 0 d = ( a T dη + a T 2 dη 2 ) (a dη + a 2 dη 2 )=dη T A dη. (2.38) where dη T = { dη, dη 2 }. Clearly, the rst metric tensor is closely related to length measurements on the surface. Because the base vectors dene the plane tangent to the surface, the unit vector, n, normal to the surface is readily found as n = ãa 2 ã a 2 = ãa a 2. (2.39) The area of a differential element of the surface then becomes da = ã a 2 dη dη 2 = ã a 2 dη dη 2 = a dη dη 2. (2.40) Curve on a surface Figure 2.8 depicts a curve, C, entirely contained within surface S. Letthecurve be dened by its intrinsic parameter, s, the curvilinear variable along curve C. The tangent vector, t, tocurvec is dened by eq. (2.5). This unit tangent vector clearly lies in the plane tangent to S, and hence, it can be resolved along the base vectors, t = λ a + λ 2 a 2. Because t is a unit vector, it follows that

12 42 2 Coordinate systems t T t = λ T A λ =, (2.4) where λ T = { λ,λ 2 }. On the other hand, eq. (2.38) can be recast as i 3 P (, 2 ) i i 2 a 2 a 2 Fig A curve, C, entirely contained within surface, S dη T ds Adη ds =. (2.42) Because eqs. (2.4) and (2.42) must be identical for all curves on the surface, λ = dη ds. (2.43) This result is expected since ds is an increment of length along C, and t is tangent to C. Angles θ =( t, a ) and θ 2 =( t, a 2 ) can be obtained by expanding the dot products t T a and t T a 2, respectively, to nd { a cos θ a22 cos θ 2 } = A λ. (2.44) The second metric tensor of a surface Consider once again a curve, C, entirely contained within surface S, as depicted in g The unit tangent vector clearly lies in the plane tangent to the surface, but the curvature vector d t/ds will have components in and out of this tangent plane, d t ds = κ n n + κ g ρ, (2.45) where κ n is the normal curvature, κ g the geodesic curvature, and ρ a unit vector belonging to the plane tangent to S. The normal curvature can be evaluated as κ n = n T d t T d n = t ds ds = dpt d n 0 ds 2, (2.46) where the normality condition, t T n =0, was used. The numerator can be written as dp T d n = ( a T 0 dη + a T 2 dη ) ( n 2 dη + n ) dη 2 η [ ( ) ] = a T n dη 2 η + n at 2 dη2 2 η + a T n + a T n 2 dη dη 2, 2 η [ = n T a dη 2 + n T a ( 2 dη2 2 + n T a + n T a ) ] 2 dη dη 2, η η where the orthogonality conditions, n T a =0and n T a 2 =0, were used to obtain the last equality.

13 2.4 Differential geometry of a surface 43 The second metric tensor of the surface is dened as n T a n T a B = η n T a 2 n T a n T 2 = n T 2 [ ] η 2 η η n T 2 n T 2 p = b b 2, (2.47) 0 b 2 b 22 2 η η2 2 and its determinant is denoted b = det(b). The second equality shows that the second metric tensor is a symmetric tensor. It follows that dp T 0 d n =dηt Bdη,and the normal curvature, eq. (2.46), becomes κ n = dηt Bdη ds 2 = dηt ds B dη ds = λt B λ. (2.48) Analysis of curvatures Figure 2.9 shows a plane, P, containing the normal, n, to surface S. Let curve C n be at the intersection of plane P and surface S. Because curve C n is a planar curve, its curvature vector is in plane P. Next, let plane P rotate about n. For each new orientation of the plane, a new curve, C n, is generated with its own normal curvature κ n. The following problem will be investigated: what is the orientation of plane P that maximizes the normal curvature κ n? In mathematical terms, the maximum value of κ n = λ T B λ is sought, under the normality constraint, λ T A λ =. This constrained maximization problem will be solved with the help of Lagrange s multiplier technique [ ] max λ T B λ μ(λ T A λ ), λ,μ i 3 n P (, 2 ) i i 2 a 2 2 a n Fig Intersection of surface, S, with plane, P, that contains the normal to the surface. where μ is the Lagrange multiplier used to enforce the constraint. The solution of this problem implies (B μa)λ =0, and the normality condition λ T A λ =. Pre-multiplying this equation by λ T yields the physical interpretation of the Lagrange multiplier: λ T B λ μλ T A λ =0or, in view of the normality constraint, μ = λ T B λ = κ n, Hence, Lagrange s multiplier can be interpreted as the normal curvature itself. The condition for maximum normal curvature can now be written as (B κ n A)λ =0. This set of homogeneous algebraic equations admits the trivial solution λ =0, but this solution violates the normality constraint. Non-trivial solutions correspond to the eigenpairs of the generalized eigenproblem B λ = κ n A λ. Because A and B are symmetric and A is positive-denite, the eigenvalues are always real, and mutually orthogonal eigenvectors can be constructed.

14 44 2 Coordinate systems or The eigenvalues are the solution of the quadratic equation det(b κ n A)=0, κ 2 n 2κ mκ n + b =0, (2.49) a where κ m =(a b 22 + a 22 b 2a 2 b 2 )/2a. The solutions of this quadratic equation are called the principal curvatures The mean curvature is dened as κ m = κi n + κii n 2 and the Gaussian curvature as κ I n,κii n = κ m ± κ 2 m b/a. (2.50) = a b 22 + a 22 b 2a 2 b 2, (2.5) 2a κ I n κii n = b a. (2.52) When b/a > 0, the principal curvatures have the same sign, corresponding to a convex shape; when b/a < 0, the principal curvatures are of opposite sign, corresponding to a saddle shape; nally, when b/a = 0, one of the principal curvatures is zero, the surface S has zero curvature in one of the principal curvature directions Lines of curvature A line of curvature of a surface is dened as a curve whose tangent vector always points along the principal curvature directions of the surface. Consider now a set of coordinates, η and η 2, such that a 2 = b 2 =0. It follows that a = a a 22, b = b b 22 and κ m =(b /a + b 22 /a 22 )/2. The principal curvatures then simply become κ I n = b a, κ II n = b 22 a 22. (2.53) On the other hand, in view of eq. (2.4), η or η 2 curves are characterized by λ T = { / a, 0 } or λ T = { 0, / } a 22, respectively. Their normal curvature then follows from eq. (2.48) as κ n = b /a and κ n = b 22 /a 22, respectively. It is now clear that when a 2 = b 2 =0,theη and η 2 curves are indeed the lines of curvatures. It is customary to introduce the principal radii of curvature, R and R 2, dened as κ I n = b =, κ II n a R = b 22 =. (2.54) a 22 R Derivatives of the base vectors At this point, the discussion will focus exclusively on surface parameterizations dening lines of curvatures. In this case, vectors a, a 2 and n form a set of mutually orthogonal vectors, although the rst two are not necessarily unit vectors. An orthonormal triad can be constructed as follows

15 2.4 Differential geometry of a surface 45 ē = a a, ē 2 = a 2 a 2, ē 3 = n. (2.55) To interpret the meaning of these unit vectors, the chain rule for derivatives is used to write a = η = s ds dη = ds dη ē, where s is the arc length measured along the η curve. Because / s =ē is the unit tangent vector to the η curve, see eq. (2.5), it follows that a = h = ds dη, a 2 = h 2 = ds 2 dη 2. (2.56) Notation h = a was introduced to simplify the writing. Clearly, h is a scale factor, the ratio of the innitesimal increment in length, ds,totheinnitesimal increment in parameter η, dη, along the curve. It is interesting to compute the derivatives of the base vectors. To that effect, the following expression is considered Expanding the derivatives leads to 2 η = a = a 2 η = (h ē ) = (h 2ē 2 ) η. h ē + h ē = h 2 η ē 2 + h 2 ē 2 η. (2.57) Pre-multiplying this relationship by ē T yields the following identity ē T ē 2 η = h 2 h. To obtain this result, the orthogonality of the base vectors, ē T ē2 =0, was used; furthermore, ē T ē / =0,sinceē is a unit vector. In terms of intrinsic parameterization, this expression becomes ē T where T is the rst radius of twist of the surface. Next, eq. (2.57) is pre-multiplied ē T 2 to yield ē T 2 ē 2 = ē T ē 2 = h =, (2.58) s s h s 2 T ē = ē T ē 2 = h 2 =, (2.59) s 2 s 2 h 2 s T 2 where T 2 is the second radius of twist of the surface. Since the parameterization denes lines of curvatures, b 2 =0, and eq. (2.47) then implies ē T 2 n = n T ē 2 =0, ē T n = n T ē =0. s s s 2 s 2

16 46 2 Coordinate systems The denitions of the diagonal terms, b and b 22, of the second metric tensor, eq. (2.47), lead to ē T n = n T ē =, ē T n 2 = n T ē 2 =, s s R s 2 s 2 R 2 where the principal radii of curvature, R and R 2,weredened in eq. (2.54). The derivatives of the surface base vector ē can be resolved in the following manner ē s = c e + c 2 e 2 + c 3 n, (2.60) where the unknown coefcients c, c 2,andc 3 are readily found by pre-multiplying the above relationship by ē T, ē T 2,and n T to nd A similar development leads to ē s = T ē 2 + R n. (2.6) ē s 2 = T 2 ē 2. (2.62) The derivatives of the surface base vector ē 2 are found in a similar manner ē 2 s = T ē, ē 2 s 2 = T 2 e + R 2 n. (2.63) These results are known as Gauss formulæ. Proceeding in a similar fashion, the derivatives of the normal vector are resolved in the following manner n s = R ē, n s 2 = R 2 ē 2. (2.64) These results are known as Weingarten s formulæ. Gauss and Weingarten s formulæ can be combined to yield the derivatives of the base vectors in a compact manner as ē 0 /T /R ē ē 2 s = /T 0 0 ē 2, (2.65a) n /R 0 0 n ē ē 2 s 2 = 0 /T 2 0 ē /T 2 0 /R 2 ē 2. (2.65b) n 0 /R 2 0 n These equations should be compared to the derivatives of Frenet s triad, eq. (2.3).

17 2.4 Differential geometry of a surface 47 Example 2.4. The spherical surface The spherical surface in three-dimensional space depicted in g. 2.0 is dened by following position vector = R (sin η cos η 2 ī +sinη sin η 2 ī 2 +cosη ī 3 ), where R is the radius of the sphere. The surface base vectors are readily evaluated as a = / η = R (cos η cos η 2 ī +cosη sin η 2 ī 2 sin η ī 3 ),and a 2 = / = R ( sin η sin η 2 ī +sinη cos η 2 ī 2 ). The rst metric tensor of the sphere now becomes [ ] R 2 0 A = 0 R 2 sin 2. η Clearly, h = R, h 2 = R sin η,and a = R 2 sin η. The normal vector is then evaluated with the help of eq. (2.39), to nd n = ãa 2 ã a 2 =sinη cos η 2 ī +sinη sin η 2 ī 2 +cosη ī 3. i 3 n i 3 P i 2 i 2 i r i 2 Fig Spherical surface conguration. Fig. 2.. Parabolic surface of revolution. The second metric tensor of the spherical surface now follows from eq. (2.47) [ ] R 0 B = 0 R sin 2. η Note that since a 2 =0and b 2 =0, the coordinates used here are lines of curvature for the spherical surface. The orthonormal triad to the surface is ē =cosη cos η 2 ī +cosη sin η 2 ī 2 sin η ī 3, ē 2 = sin η 2 ī +cosη 2 ī 2, n =sinη cos η 2 ī +sinη sin η 2 ī 2 +cosη ī 3. These expressions are readily inverted to nd ī =cosη cos η 2 ē sin η 2 ē 2 +sinη cos η 2 n, ī 2 =cosη sin η 2 ē +cosη 2 ē 2 +sinη sin η 2 n, ī 3 = sin η ē +cosη n.

18 48 2 Coordinate systems The mean curvature, eq. (2.5), and Gaussian curvature, eq. (2.52), are κ m = ( R 2 R 2 R ) sin2 η R 2 sin 2 = η R, κi n κii n = R2 sin 2 η R 4 sin 2 = η R 2. Finally, the principal curvatures, eq. (2.53), become κ I n = R, κii n = R. As expected, the principal radii of curvature R = R 2 = R are equal to the radius of sphere. The twists of the surface now follow from eqs. (2.58) and (2.59) T = h h 2 h =0, = h 2 = cos η. (2.66) T 2 h h 2 η R sin η Problems Problem 2.6. The parabola of revolution Figure 2. depicts a parabolic surface of revolution. It is dened by the following position vector = r cos φ ī + r sin φ ī 2 + ar 2 ī 3,wherer 0 and 0 φ 2π. The following notation was used η = r and η 2 = φ. () Find the rst and second metric tensors of the surface. (2) Find the orthonormal triad ē, ē 2,and n. (3) Find the mean curvature, the Gaussian curvature, and the principal radii of curvature of the surface. (4) Find the twists of the surface. Problem 2.7. Jacobian of the transformation Consider two parameterizations of a surface dened by coordinates (η,η 2)and(ˆη, ˆη 2). Show that the base vectors in the two parameterizations are related as follows â = J a + J 2a 2 and â 2 = J 2a + J 22a 2,whereJ is the Jacobian of the coordinate transformation [ ] η J J 2 J = = ˆη ˆη J 2 J 22 η. ˆη 2 ˆη 2 If A and B are the rst and second metric tensors in coordinate system (η,η 2)and and ˆB the corresponding quantities in coordinate system (ˆη, ˆη 2), show that  = J A J T and ˆB = J B J T. Problem 2.8. Finding the line of curvature system Using the notations dened in problem 2.7, let (η,η 2) be a known coordinate system and (ˆη, ˆη 2) the unknown line of curvature system. Find the Jacobian of the coordinate transformation that will bring (η,η 2) to the desired line of curvature system (ˆη, ˆη 2). Show that the principal radii of curvature are R = Hint: write the Jacobian as b + γ(2b2 + γb22) a + γ(2a 2 + γa, = 22) R 2 J = [ ] γ, α b + α(2b2 + αb22) a + α(2a 2 + αa 22). and compute the coefcients α and γ so as to enforce â 2 = ˆb 2 =0. The solution of the problem is α = C α/[δ/(+αγ)] and γ = C γ/[δ/(+αγ)] where C α = a 22b 2 b 22a 2, C γ = a b 2 b a 2, Δ = a b 22 b a 22,andΔ/(+αγ) =Δ/2± (Δ/2) 2 + C αc γ.

19 2.5 Surface coordinates 2.6 Differential geometry of a three-dimensional mapping 49 A particle is moving on a surface and its position is given by the lines of curvature coordinates, η (t) and η 2 (t). The velocity vector is computed with the help of the chain rule for derivatives v = d dt = η η + η 2 =ṡ ē +ṡ 2 ē 2. (2.67) Note the close similarity between this expression and that obtained for path coordinates, eq. (2.33). The velocity vector is in the plane tangent to the surface, and the speed of the particle is v = ṡ 2 +ṡ2 2. Next, the acceleration vector is computed as a = s ē +ṡ ē + s 2 ē 2 +ṡ 2 ē 2 = s ē + s 2 ē 2 +ṡ ( ē s ṡ + ē s 2 ṡ 2 Introducing Gauss formulae, eq. (2.6) to (2.63), then yields ( a = s + ṡṡ ) ( 2 ṡ2 2 ē + s 2 + ṡṡ ) 2 ṡ2 ē 2 + T T 2 T 2 T ) ( ē2 +ṡ 2 ṡ + ē ) 2 ṡ 2. s s 2 ( ṡ2 R + ṡ2 2 R 2 ) n. (2.68) Note here again the similarity between this expression and that obtained for path coordinates, eq. (2.34). The acceleration component along the normal to the surface is related to the principal radii of curvatures, R and R 2. For a curve, the radius of curvature is always positive, see eq. (2.7), whereas for a surface, the radii of curvatures could be positive or negative, see eq. (2.54). Hence, the normal component of acceleration is not necessarily oriented along the normal to the surface. The components of acceleration in the plane tangent to the surface are related to the second time derivative of the intrinsic parameters, as expected. Additional terms, however, associated with the surface radii of twist also appear. Clearly, the acceleration of a particle moving on the surface is affected by the surface radii of curvature and twist; the particle feels the curvatures and twists of the surface as it moves. 2.6 Differential geometry of a three-dimensional mapping This section investigates the differential geometry of mappings of the threedimensional space onto itself. The differential geometry of such mappings is more complex than that of curves or surfaces. For simplicity, the analysis focuses on orthogonal mappings, leading to the denition of the curvatures of the coordinate system and orthogonal curvilinear coordinates. Two orthogonal curvilinear coordinate systems of great practical importance, the cylindrical and spherical coordinate systems are reviewed.

20 50 2 Coordinate systems 2.6. Arbitrary parameterization Consider the following mapping of the three-dimensional space onto itself in terms of three parameters, η, η 2,andη 3, (η,η 2,η 3 )=x (η,η 2,η 3 )ī + x 2 (η,η 2,η 3 )ī 2 + x 3 (η,η 2,η 3 )ī 3. (2.69) This relationship denes a mapping between the parameters and the Cartesian coordinates x = x (η,η 2,η 3 ), x 2 = x 2 (η,η 2,η 3 ), x 3 = x 3 (η,η 2,η 3 ). (2.70) Let η 2 and η 3 be constants whereas η only is allowed to vary: a general curve in three-dimensional space is generated. The analysis of section 2.2 would readily apply to this curve, called an η curve. Similarly, η 2 and η 3 curves could be dened. Next, let η be a constant, whereas η 2 and η 3 are allowed to vary: a general surface in three-dimensional space is generated. The analysis of section 2.4 would readily apply to this surface, called an η surface. Here again, η 2 and η 3 surfaces could be similarly dened. A point in space with parameters (η,η 2,η 3 ) is at the intersection of three η, η 2, and η 3 curves, or at the intersection of three η, η 2,andη 3 surfaces. Furthermore, an η curve forms the intersection of η 2 and η 3 surfaces. The inverse mapping denes the parameters as functions of the Cartesian coordinates η = η (x,x 2,x 3 ), η 2 = η 2 (x,x 2,x 3 ), η 3 = η 3 (x,x 2,x 3 ). (2.7) It is assumed here that eqs. (2.70) and (2.7) dene a one to one mapping, which implies that the Jacobian of the transformation, x x x η η 3 x 2 x 2 x 2 J =, (2.72) η η 3 x 3 x 3 x 3 η η 3 has a non vanishing determinant at all points in space. Next, the base vectors associated with the parameters are dened as g = η, g 2 =, g 3 = η 3. (2.73) For an arbitrary parameterization, the base vectors will not be unit vectors, nor will they be mutually orthogonal. Consider the example of the cylindrical coordinate system dened by the following parameterization

21 2.6 Differential geometry of a three-dimensional mapping 5 x = r cos θ, x 2 = r sin θ, x 3 = z, where r 0 and 0 θ<2π. The following notation was used: η = r, η 2 = θ and η 3 = z. The inverse mapping is readily found as r = x 2 + x x2 2,θ=tan 2,z= x 3. x Note that det J = r, and hence, vanishes at r =0. Indeed, cylindrical coordinates are not dened at the origin since when r =0, any angle θ maps to the same point, the origin. i 3 P g 3 Figure 2.2 depicts this mapping; clearly, the fa- miliar polar coordinates are used in the (ī, ī 2 ) plane and z is the distance point P is above this plane. The Jacobian of the transformation becomes cos θ r sin θ 0 J = sin θ rcos θ r z i i 2 r g 2 g Fig The cylindrical coordinate system. The base vectors of this coordinate system are g =cosθī +sinθ ī 2, g 2 = r sin θ ī + r cos θ ī 2,andg 3 =ī 3. Note that g is a unit vector, since g =, but g 2 is not, g 2 = r. Also note that for cylindrical coordinates, g T g = 2 3 gt g = 3 g T g =0, the base vectors are mutually orthogonal, as shown in g Orthogonal parameterization When the base vectors associated with the parameterization are mutually orthogonal, the parameters dene an orthogonal parameterization of the three-dimensional space. The rest of this section will be restricted to such parameterization. In this case, it is advantageous to dene a set of orthonormal vectors ē = g g, ē 2 = g 2 g 2, ē 3 = g 3 g 3. (2.74) To interpret the meaning of these unit vectors, the chain rule for derivatives is used to write g = = ds ds =ē, (2.75) η s dη dη where s is the arc length measured along the η curve. Because / s =ē is the unit tangent to the η curve, see eq. (2.5), it follows that g = h = ds dη, g 2 = h 2 = ds 2 dη 2, g 3 = h 3 = ds 3 dη 3. (2.76) Notation h = g is introduced to simplify the notation. Clearly, h is a scale factor, the ratio of the innitesimal increment in length, ds,totheinnitesimal increment in parameter η, dη, along the curve.

22 52 2 Coordinate systems Derivatives of the base vectors Here again, the derivatives of the base vectors will be evaluated. To that effect, the following expression is considered Expanding the derivatives leads to 2 η η 2 = g = g 2 η = (h ē ) = (h 2ē 2 ) η. (2.77) h ē + h ē = h 2 η ē 2 + h 2 ē 2 η. (2.78) Pre-multiplying this relationship by ē T yields the following identity ē T ē 2 η = h 2 h. (2.79) To obtain this result, the orthogonality of the base vectors, ē T ē 2 =0, was used; furthermore, ē T ē / =0,sinceē is a unit vector. Next, eq. (2.78) is pre-multiplied ē T 2 to yield ē T ē 2 = ē T ē 2 = h 2. (2.80) h η Finally, pre-multiplication by ē T 3 leads to h ē T 3 ē = h 2 ē T ē 2 3. (2.8) η Since ē T 3 ē 2 / η = ē T 2 ē 3 / η, this result can be manipulated as follows h ē T 3 ē = h 2 ē T ē 3 2 = h h 2 ē T ē 2, (2.82) η h 3 η 3 where identity (2.8) was used with a permutation of the indices. Using the same identities once again leads to h ē T 3 This result clearly implies ē = h h 2 ē T ē 2 = h ē T ē 3 = h ē T ē 3. h 3 η 3 The derivatives of the base vector can be resolved as ē T 3 ē =0. (2.83) ē η = c ē + c 2 ē 2 + c 3 ē 3, where the unknown coefcients c, c 2,andc 3 are found by pre-multiplying this expression by ē, ē 2,andē 3, respectively, and using identities (2.79), (2.80) and (2.83) to nd

23 2.7 Orthogonal curvilinear coordinates 53 ē η = h 2 h ē 2 h 3 h η 3 ē 3. Proceeding in a similar manner, the derivatives of base vector ē with respect to η 2 and η 3 are found as ē = h h 2 η ē 2, ē η 3 = h h 3 η ē 3. Similar expression are readily found for the derivatives of the unit base vectors ē 2 and ē 3 through index permutations and are summarized as ē 0 /R 3 /R 2 ē ē 2 s = /R ē 2, (2.84a) ē 3 /R ē 3 ē 0 /R 23 0 ē ē 2 s 2 = /R 23 0 /R 2 ē 2, (2.84b) ē 3 0 /R 2 0 ē 3 ē 0 0 /R 32 ē ē 2 s 3 = 0 0 /R 3 ē 2, (2.84c) ē 3 /R 32 /R 3 ē 3 where the curvatures of the system were dened as = h, R 2 h s 3 = h 2, R 2 h 2 s 3 = h 3, R 3 h 3 s 2 = h, R 3 h s 2 = h 2, R 23 h 2 s = h 3. R 32 h 3 s (2.85a) (2.85b) (2.85c) 2.7 Orthogonal curvilinear coordinates Consider a particle moving in three-dimension space. The position of this particle can be dened by eq. (2.69) in terms of an orthogonal parameterization of space. These parameter dene a set of orthogonal curvilinear coordinates for the particle. The velocity vector is computed with the help of the chain rule for derivatives v = d dt = s ṡ + s 2 ṡ 2 + s 3 ṡ 3 =ṡ ē +ṡ 2 ē 2 +ṡ 3 ē 3. (2.86) The expression for the acceleration vector will involve term in s ē and ṡ ē, and similar terms for the other two indices. The latter term is further expanded using the chain rule for derivatives, and expressing the derivatives of the base vectors using eqs. (2.84) then yields

24 54 2 Coordinate systems a = [ s ṡ 2 2 /R 23 +ṡ 2 3 /R ] 32 ṡ ṡ 2 /R 3 +ṡ ṡ 3 /R 2 ē + [ s 2 +ṡ 2 /R 3 ṡ 2 3/R 3 +ṡ ṡ 2 /R 23 ṡ 2 ṡ 3 /R 2 ] ē2 + [ s 3 ṡ 2 /R 2 +ṡ 2 2/R 2 ṡ ṡ 3 /R 32 +ṡ 2 ṡ 3 /R 3 ] ē3. (2.87) Note here again the similarity between this expression and that obtained for path or surface coordinates, eqs. (2.34) or (2.68), respectively. The acceleration components in each direction involve the second time derivative of the intrinsic parameters, as expected. Additional terms, however, associated with the radii of curvature of the curvilinear coordinate system also appear Cylindrical coordinates The cylindrical coordinate system, depicted in g. 2.3, is an orthogonal curvilinear coordinate system dened as follows = r cos θ ī + r sin θ ī 2 + z ī 3, (2.88) where r 0 and 0 θ<2π. The following notation was used: η = r, η 2 = θ,and η 3 = z. Note that if z =0, the cylindrical coordinate system reduces to coordinates r and θ in plane (ī, ī 2 ) and are then often called polar coordinates. i 3 i 3 e e 3 e 2 e r z i 2 i Fig The cylindrical coordinate system. i r e 3 i 2 e 2 Fig The spherical coordinate system. The following summarizes important formulæ in cylindrical coordinates. The scale factors are h =, h 2 = r, andh 3 =. The curvatures of the cylindrical coordinate system all vanish, except that R 23 = r. The base vectors expressed in terms of the Cartesian system are ē = cosθ ī +sinθ ī 2, (2.89a) ē 2 = sin θ ī +cosθ ī 2, (2.89b) ē 3 = ī 3. (2.89c) The time derivatives of the based vectors resolved along this triad are ē = θ ē 2, (2.90a) ē 2 = θ ē, (2.90b) ē 3 = 0. (2.90c)

25 2.7 Orthogonal curvilinear coordinates 55 Finally, the position, velocity, and acceleration vectors, resolved along the base vectors of the cylindrical coordinate system are = r ē + z ē 3, v =ṙ ē + r θ ē 2 +ż ē 3, a =( r r θ 2 )ē +(r θ +2ṙ θ)ē 2 + z ē 3. (2.9a) (2.9b) (2.9c) respectively Spherical coordinates The spherical coordinate system, depicted in g. 2.4, is an orthogonal curvilinear coordinate system dened as follows = r sin φ cos θ ī + r sin φ sin θ ī 2 + r cos φ ī 3, (2.92) where r 0, 0 φ π,and0 θ<2π. The following notation was used: η = r, η 2 = φ, andη 3 = θ. The following summarizes important formulæ in spherical coordinates. The scale factors are h =, h 2 = r, andh 3 = r sin φ. The curvatures of the spherical coordinate system all vanish, except that R 23 = r, R 3 = r tan φ and R 32 = r. The base vectors expressed in terms of the Cartesian system are ē = sinφ cos θ ī +sinφsin θ ī 2 +cosφ ī 3, (2.93a) ē 2 = cosφ cos θ ī +cosφsin θ ī 2 sin φ ī 3, (2.93b) ē 3 = sin θ ī + cosθ ī 2. (2.93c) The time derivatives of the based vectors resolved along this triad are ē = φ ē 2 + θ sin φ ē 3, (2.94a) ē 2 = φ ē + θ cos φ ē 3, ē 3 = θ(sin φ ē +cosφ ē 2 ). (2.94b) (2.94c) Finally, the position, velocity, and acceleration vectors, resolved along the base vectors of the spherical coordinate system are = r ē, v =ṙ ē + r φ ē 2 + r θ sin φ ē 3, (2.95a) (2.95b) a =( r r φ 2 r θ 2 sin 2 φ)ē +(r φ +2ṙ φ r θ 2 sin φ cos φ) ē 2 +(r θ sin φ +2ṙ θ sin φ +2r φ θ cos φ) ē 3. (2.95c)

26