Differential Geometry of Surfaces

Lecture 3 Differential geometry of surfaces 3.1 Definition of surfaces • Implicit surfaces F (x, y, z) = 0 2 2 2 Example: xa2 + yb2 + zc2 = 1 Ellipsoid, see Figure 3.1. y z x Figure 3.1: Ellipsoid. • Explicit surfaces If the implicit equation F (x, y, z) = 0 can be solved for one of the variables as a function of the other two, we obtain an explicit surface, as shown in Figure 3.2. Example: z = 1 2 2 2 (αx + βy ) • Parametric surfaces x = x(u, v), y = y(u, v), z = z(u, v) Here functions x(u, v), y(u, v), z(u, v) have continuous partial derivatives of the r th order, and the parameters u and v are restricted to some intervals (i.e., u 1 ≤ u ≤ u2 , v1 ≤ v ≤ v2 ) leading to parametric surface patches. This rectangular domain D of u, v is called parametric space and it is frequently the unit square, see Figure 3.3. If derivatives of the surface are continuous up to the r th order, the surface is said to be of class r, denoted C r. 2 Figure 3.2: Explicit quadratic surfaces z = 12 (αx2 + βy 2 ). (a) Left: Hyperbolic paraboloid (α = −3, β = 1). (b) Right: Elliptic paraboloid (α = 1, β = 3). In vector notation: r = r(u, v) where r = (x, y, z), r(u, v) = (x(u, v), y(u, v), z(u, v)) Example: r = (u + v, u − v, u2 + v 2 ) x=u+v y =u−v z = u2 + v 2 3.2      ⇒ eliminate u, v ⇒ z = 1 2 (x + y 2 ) paraboloid 2 Curves on a surface Let r = r(u, v) be the equation of a surface, defined on a domain D (i.e., u 1 ≤ u ≤ u2 , v1 ≤ v ≤ v2 ). Let β(t) = (u(t), v(t)) be a curve in the parameter plane. Then r = r(u(t), v(t)) is a curve lying on the surface, see Figure 3.3. A tangent vector of curve β(t) is given by β̇(t) = (u̇(t), v̇(t)) A tangent vector of a curve on a surface is given by: dr(u(t), v(t)) dt (3.1) By using the chain rule: dr(u(t), v(t)) ∂r du ∂r dv = + = ru u̇(t) + rv v̇(t) dt ∂u dt ∂v dt 3 (3.2) v z r(u,v) D y u r(u(t),v(t)) β(t)=(u(t),v(t)) x 3D Space Parametric Space D Figure 3.3: The mapping of a curve in 2D parametric space onto a 3D biparametric surface . 3.3 First fundamental form (arc length) Consider a curve on a surface r = r(u(t), v(t)). The arc length of the curve on a surface is given by ds = | where du dv dr |dt = |ru + rv |dt dt dt dt = q = q = p (ru u̇ + rv v̇) · (ru u̇ + rv v̇)dt (ru · ru )du2 + 2ru rv dudv + (rv · rv )dv 2 Edu2 + 2F dudv + Gdv 2 E = r u · ru , F = r u · rv , (3.3) G = r v · rv (3.4) The first fundamental form is defined as I = dr · dr = (ru du + rv dv) · (ru du + rv dv) = Edu2 + 2F dudv + Gdv 2 (3.5) E, F , G are called first fundamental form coefficients Note that E = r u · ru > 0 and G = rv · rv > 0 if ru 6= 0 and rv 6= 0. The first fundamental form I is positive definite. That is I ≥ 0 and I = 0 if and only if du = 0 and dv = 0 since I= EG − F 2 2 1 (E du + F dv)2 + dv and EG − F 2 = |ru × rv |2 > 0. E E I depends only on the surface and not on the parametrization. The area of the surface can be derived as follows: 4 δA r(u0,v0+δv) r(u0+δu,v0) r(u0+δu,v0)−r(u0,v0) r(u0,v0+δv)−r(u0,v0) r(u0,v0) Figure 3.4: Area of an infinitessimal surface patch. ∂r δv ∂v ∂r δu r(u0 + δu, v0 ) − r(u0 , v0 ) ' ∂u r(u0 , v0 + δv) − r(u0 , v0 ) ' δA = |ru δu × rv δv| = |ru × rv |δuδv |ru × rv |2 = (ru × rv ) · (ru × rv ) Using the vector identity (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c), we get |ru × rv |2 = (ru · ru )(rv · rv ) − (ru · rv )2 = EG − F δA = p EG − F2 A= δuδv, 2 Z Z p EG − F 2 dudv (3.6) (3.7) (3.8) Example: For the hyperbolic paraboloid r(u, v) = (u, v, u 2 −v 2 ), let us derive an expression for the area of a region of its surface corresponding to a the circle u 2 + v 2 ≤ 1 in the parametric domain D. We begin by forming expressions for the derivatives of the position vector r and the first fundamental form coeffients. ru = (1, 0, 2u) rv = (0, 1, −2v) E = ru · ru = 1 + 4u2 = ru · rv = −4uv F G = rv · rv = 1 + 4v 2 Using Equation (3.8), we find EG − F 2 = (1 + 4u2 )(1 + 4v 2 ) − 16u2 v 2 = 1 + 4u2 + 4v 2 > 0 A = Z Z p 1 + 4u2 + 4v 2 dudv D 5 To compute the area, we need to evaluate the double integral over the unit disk u 2 + v 2 ≤ 1 in the parametric domain D; A= Z Z u2 +v 2 ≤1 p 1 + 4u2 + 4v 2 du dv. To perform the integration, let us change variables. u = r cos(θ), v = r sin(θ), and du dv = r dr dθ A = = = 3.4 Z Z Z 0 p 1 + 4r 2 r dr dθ r≤1 2π Z 1 0 p 1 + 4r 2 r dr dθ π √ (5 5 − 1) 6 Tangent plane Tangent plane at a point r(uo , vo ) is the union of tangent vectors of all curves on the surface pass through r(uo , vo ), as shown in Figure 3.5. Since the tangent vector of a curve on a parametric du dv surface is given by dr dt = ru dt + rv dt , the tangent plane lies on the plane of the vectors r u and rv . The equation of the tangent plane is Tp (u, v) = r(u, v) + λru (u, v) + µrv (u, v) (3.9) where λ and µ are real variables parameterizing the plane. z r=ruu+rvv Tp r(u0,v0) y x Figure 3.5: The tangent plane at a point on a surface. 3.5 Normal vector The surface normal is the vector at point r(u o , vo ) perpendicular to the tangent plane, see Figure 3.6. And therefore N= ru × r v |ru × rv | Note that ru and rv are not necessarily perpendicular. 6 (3.10) z N rv ru Tp y x Figure 3.6: The normal to the point on a surface. A regular (ordinary) point P on the surface is defined as one for which r u × rv 6= 0. A point where ru × rv = 0 is called a singular point. The condition r u × rv 6= 0 requires that at that point P the vectors ru and rv do not vanish and have different directions. Example: Elliptic Paraboloid r(u, v) = (u + v, u − v, u 2 + v 2 ) ru = (1, 1, 2u) rv = (1, −1, 2v) ru × r v = ex ey ez 1 1 2u 1 −1 2v = 2(u + v)ex + 2(u − v)ey − 2ez 6= 0 q |ru × rv | = 2 (u + v)2 + (u − v)2 + 1 p = 2 2u2 + 2v 2 + 1 > 0 ⇒ Regular ! N = = (2(u + v), 2(u − v), −2) √ 2 2u2 + 2v 2 + 1 (u + v, u − v, −1) √ 2u2 + 2v 2 + 1 at (u, v) = (0, 0), N = (0, 0, −1) Example: Circular Cone r(u, v) = (u sin α cos v, u sin α sin v, u cos α), see Figure 3.7 ru = (sin α cos v, sin α sin v, cos α) rv = (−u sin αsinv, u sin αsinv, 0) ru × r v = ex ey ez sin α cos v sin α sin v cos α −u sin α sin v u sin α cos v 0 7 z usinα p α u u α y usinαsinv singular v usinαcosv x Figure 3.7: Circular cone. = −u sin α cos α cos vex − u sin α cos α sin vey + u sin2 αez At the origin n = 0, ru × r v = 0 Therefore, the apex of the cone is a singular point. 3.6 Second fundamental form II (curvature) N S P t n kn C kg k Figure 3.8: Definition of normal curvature In order to quantify the curvatures of a surface S, we consider a curve C on S which passes through point P as shown in Figure 3.8. t is the unit tangent vector and n is the unit normal vector of the curve C at point P . dt = κn = kn + kg ds kn = κ n N (3.11) (3.12) where kn is the normal curvature vector normal to the surface, k g is the geodesic curvature vector tangent to the surface, and k = κn is the curvature vector of the curve C at point P. κn is called the normal curvature of the surface at P in the direction t. 8 Meusnier’s Theorem : All curves lying on a surface S passing through a given point p ∈ S with the same tangent line have the same normal curvature at this point. Since N · t = 0, differentiate w.r.t. s d (N · t) = N0 · t + N · t0 ds dt dN dr dN · N = −t · =− · ds ds ds ds (3.13) Recoginizing that ds · ds = dx2 + dy 2 + dz 2 = dr · dr, we can rewrite Equation 3.13 as: dt dr · dN ·N = =− ds dr · dr dt while · N = κn · N ≡ κn ds center of curvature N P N P center of curvature (a) (b) Figure 3.9: Definition of positive normal: (a) κn · N = κ n ; (b) κn · N = −κn . II = −dr · dN = −(ru du + rv dv) · (Nu du + Nv dv) = Ldu2 + 2M dudv + N dv 2 (3.14) where L = N · ruu , M = N · ruv , N = N · rvv (3.15) Therefore the normal curvature is given by κn = L + 2M λ + N λ2 II = I E + 2F λ + Gλ2 (3.16) dv where λ = du . Suppose P is a point on a surface and Q is a point in the neighborhood of P , as in Figure 3.10. Taylor’s expansion gives 1 r(u + du, v + dv) = r(u, v) + ru du + rv dv + (ruu du2 + 2ruv dudv + rvv dv 2 ) + H.O.T. (3.17) 2 9 r=r(u,v) N Q d P Tp Figure 3.10: Geometrical illustration of the second fundamental form. Therefore 1 PQ = r(u + du, v + dv) − r(u, v) = ru du + rv dv + (ruu du2 + 2ruv dudv + rvv dv 2 ) + H.O.T. 2 Thus, the projection of PQ onto N 1 d = PQ · N = (ru du + rv dv) · N + II 2 and since ru · N = rv · N = 0, we get 1 1 d = II = (Ldu2 + 2M dudv + N dv 2 ) 2 2 We want to observe in which situation d is positive and negative. When d = 0 Ldu2 + 2M dudv + N dv 2 = 0 Solve for du du = −M ± p √ −M ± M 2 − LN (M dv)2 − LN dv 2 = dv L L N N Tp P P (3.18) N Tp P Tp Figure 3.11: (a) Elliptic point; (b) Parabolic point; (c) Hyperbolic point. • If M 2 −LN < 0, there is no real root. That means there is no intersection between the surface and its tangent plane except at point P . P is called elliptic point (Figure 3.11(a)). • If M 2 − LN = 0, there is a double root. The surface intersects its tangent plane with one line du = − M L dv, which passes through point P . P is called parabolic point (Figure 3.11(b)). • If M 2 − LN > 0, there are two roots. The surface intersects its tangent plane with two √ 2 lines du = −M ± LM −LN dv, which intersect at point P . P is called hyperbolic point (Figure 3.11(c)). 10 3.7 Principal curvatures The extreme values of κn can be obtained by evaluating dκn dλ = 0 of Equation 3.16, which gives: (E + 2F λ + Gλ2 )(N λ + M ) − (L + 2M λ + N λ2 )(Gλ + F ) = 0 (3.19) Since E + 2F λ + Gλ2 = (E + F λ) + λ(F + Gλ), L + 2M λ + N λ2 = (L + M λ) + λ(M + N λ) equation (3.19) can be reduced to (E + F λ)(M + N λ) = (L + M λ)(F + Gλ) (3.20) Thus κn = L + 2M λ + N λ2 M + Nλ L + Mλ = = 2 E + 2F λ + Gλ F + Gλ E + Fλ (3.21) Therefore κn satisfies the two simultaneous equations (L − κn E)du + (M − κn F )dv = 0 (M − κn F )du + (N − κn G)dv = 0 (3.22) These equations can be simultaneously satisfied if and only if L − κn E M − κn F M − κn F N − κn G =0 (3.23) where | | denotes the determinant of a matrix. Expanding and defining K and H as K = H = LN − M 2 EG − F 2 EN + GL − 2F M 2(EG − F 2 ) (3.24) (3.25) we obtain a quadratic equation for κ n as follows: κ2n − 2Hκn + K = 0 (3.26) The values K and H are called Gauss (Gaussian) and mean curvature respectively. The discriminant D can be expressed as follows: D = H2 − K (EN + GL − 2F M )2 − 4(EG − F 2 )(LN − M 2 ) = 4(EG − F 2 )2 The denominator is always positive, so we only need to investigate the numerator. The numerator can be written as: =4 ! (EN + GL − 2F M )2 − 4(EG − F 2 )(LN − M 2 ) EG − F 2 2F (EM − F L)2 + [EN − GL − (EM − F L)]2 ≥ 0 2 E E 11 Thus, D ≥ 0. Upon solving Equation (3.26) for the extreme values of curvature, we have: κmax = H + κmin = H − p H2 − K p H2 (3.27) −K (3.28) From Equations (3.27), (3.28), it is readily seen that K = κmax κmin κmax + κmin H = 2 (3.29) (3.30) From Equation (3.24) (since EG − F 2 > 0, see Equation 3.6). K > 0 ⇒ LN > M 2 ⇒ Elliptic point K = 0 ⇒ LN = M 2 ⇒ Parabolic point K < 0 ⇒ LN < M 2 ⇒ Hyperbolic point K<0 K=0 K>0 Figure 3.12: Curvature map of a torus showing elliptic, parabolic, and hyperbolic regions. 12 Bibliography [1] P. M. do Carmo. Differential Geometry of Curves and Surfaces. Prentice-Hall, Inc., Englewood Cliffs, NJ, 1976. [2] E. Kreyszig. Differential Geometry. University of Toronto Press, Toronto, 1959. [3] M. M. Lipschutz. Theory and Problems of Differential Geometry. Schaum’s Outline Series: McGraw-Hill, 1969. [4] D. J. Struik. Lectures on Classical Differential Geometry. Addison-Wesley, Cambridge, MA, 1950. [5] T. J. Willmore. An Introduction to Differential Geometry. Clarendon Press, Oxford, 1959. 13