Mannheim partner curves in 3-space

Transcription

1 J. Geom. 88 (2008) /08/ Birkhäuser Verlag, Basel, 2008 DOI /s Mannheim partner curves in 3-space Huili Liu and Fan Wang Abstract. In this paper, we study Mannheim partner curves in three dimensional space. We obtain the necessary and sufficient conditions for the Mannheim partner curves in Euclidean space E 3 and Minkowski space E 3 1, respectively. Some examples are also given. Mathematics Subject Classification (2000): 53A04, 53B30. Key wor: Mannheim partner curve, curvature, torsion, Minkowski space. 1. Introduction In the study of the fundamental theory and the characterizations of space curves, the corresponding relations between the curves are the very interesting and important problem. The well-known Bertrand curve is characterized as a kind of such corresponding relation between the two curves. For the Bertrand curve Ɣ, it shares the normal lines with another curve Ɣ 1, called Bertrand mate or Bertrand partner curve of Ɣ. In this paper, we are concerned with another kind of associated curves, called Mannheim curve and Mannheim mate (partner curve). In this work, we call them simply as Mannheim pair. DEFINITION 1. Let E 3 be the 3-dimensional Euclidean space with the standard inner product,. If there exists a corresponding relationship between the space curves Ɣ and Ɣ 1 such that, at the corresponding points of the curves, the principal normal lines of Ɣ coincides with the binormal lines of Ɣ 1, then Ɣ is called a Mannheim curve, and Ɣ 1 a Mannheim partner curve of Ɣ. The pair {Ɣ, Ɣ 1 } is said to be a Mannheim pair. From the elementary differential geometry we have the well-known characterizations of Bertrand pair. But there are rather few works on Mannheim pair. It is just known that a space curve in E 3 is a Mannheim curve if and only if its curvature κ and torsion satisfy the formula κ = λ(κ ), where λ is a nonzero constant. In this paper, we study the Mannheim partner curves in three dimensional Euclidean space E 3 and three dimensional Minkowski space E 3 1. We will give the necessary and sufficient Supported by NSFC, No ; Joint Research of NSFC and KOSEF, NEU. 120

2 Vol. 88, 2008 Mannheim partner curves in 3-space 121 conditions for the Mannheim partner curves in Euclidean space E 3 and Minkowski space E 3 1, respectively. In [2], Prof. B. Y. Chen characterizes the curve which satisfies κ = as+b, a = 0. Here, our examples provide curves which satisfy κ = sinh s. 2. Mannheim partner curves in E 3 Let Ɣ : x(s)be a Mannheim curve in E 3 parameterized by its arc length s and Ɣ 1 : x 1 (s 1 ) the Mannheim partner curve of Ɣ with an arc length parameter s 1. Denote by {α(s), β(s), γ (s)} the Frenet frame field along Ɣ : x(s), that is, α(s) is the tangent vector field, β(s)the normal vector field and γ(s) the binormal vector field of the curve Ɣ, respectively. The famous Frenet formulas are given by α(s) = κ(s)β(s) β(s) = κ(s)α(s) + (s)γ (s) γ(s) = (s)β(s). Here and in the following, we use dot to denote the derivative with respect to the arc length parameter of a curve. THEOREM 1. Let Ɣ : x(s) be a Mannheim curve in E 3 with the arc length parameter s. Then Ɣ 1 : x 1 (s 1 ) is the Mannheim partner curve of Ɣ if and only if the curvature κ 1 and the torsion 1 of Ɣ 1 satisfy the following equation for some nonzero constant λ. 1 = d 1 = κ 1 1 λ (1 + λ2 1 2 ) Proof. Suppose that Ɣ : x(s) is a Mannheim curve. Then by the definition we can assume that x(s 1 ) = x 1 (s 1 ) + λ(s 1 )γ 1 (s 1 ) (2.1) for some function λ(s 1 ). By taking the derivative of (2.1) with respect to s 1 and applying the Frenet formulas, we have α = α 1 + λγ 1 λ 1 β 1. (2.2) 1 Since γ 1 is coincident with β in direction, we get λ(s 1 ) = 0. This means that λ is a nonzero constant. Thus we have α 1 = α 1 λ 1 β 1. (2.3)

3 122 Huili Liu and Fan Wang J. Geom. On the other hand, we have α = α 1 cos θ + β 1 sin θ, (2.4) where θ is the angle between α and α 1 at the corresponding points of Ɣ and Ɣ 1. By taking the derivative of this equation with respect to s 1, we obtain κβ = (κ 1 + θ)sin θα 1 + (κ 1 + θ)cos θβ sin θγ 1. 1 From this equation and the fact that the direction of β is coincident with γ 1,weget { (κ1 + θ)sin θ = 0 (κ 1 + θ)cos θ = 0. Therefore we have θ = κ 1. (2.5) From (2.3), (2.4) and notice that α 1 is orthogonal to β 1, we find that Then we have = 1 1 cos θ = λ 1 sin θ. λ 1 = tan θ. By taking the derivative of this equation and applying (2.5), we get λ 1 = κ 1 (1 + λ ), that is 1 = κ 1 λ (1 + λ2 1 2 ). Conversely, if the curvature κ 1 and torsion 1 of the curve Ɣ 1 satisfy 1 = κ 1 λ (1 + λ2 1 2 ) for some nonzero constant λ, then define a curve Ɣ by x(s 1 ) = x 1 (s 1 ) + λγ 1 (s 1 ) (2.6) and we will prove that Ɣ is a Mannheim curve and Ɣ 1 is the partner curve of Ɣ. By taking the derivative of (2.6) with respect to s 1 twice, we get α 1 = α 1 λ 1 β 1, (2.7)

4 Vol. 88, 2008 Mannheim partner curves in 3-space 123 ( ) 2 κβ + α d2 s = λκ 1 1 α 1 + (κ 1 λ 1 )β 1 λ1 2 γ 1, (2.8) respectively. Taking the cross product of (2.7) with (2.8) and noticing that κ 1 λ 1 + λ 2 κ = 0, we have ( ) 3 κγ = λ α 1 + λ1 2 β 1. (2.9) 1 By taking the cross product of (2.9) with (2.7), we obtain also ( ) 4 κβ = λ1 2 (1 + λ2 1 2 )γ 1. 1 This means that the principal normal direction β of Ɣ : x(s) coincides with the binormal direction γ 1 of Ɣ 1 : x 1 (s 1 ). Hence Ɣ : x(s) is a Mannheim curve and Ɣ 1 : x 1 (s 1 ) is its Mannheim partner curve. REMARK 1. By a simple parameter transformation, the condition 1 = κ 1 λ (1 + λ2 2 1 ) can be written as 1 = 1 λ tan( κ c 0 ). Therefore, for each Mannheim curve, there is an unique Mannheim partner curve. We have the following Examples (Helices as Mannheim partner curves). PROPOSITION 1. Let Ɣ : x(s) be a Mannheim curve in E 3 with the arc length parameter s and Ɣ 1 : x 1 (s 1 ) the Mannheim partner curve of Ɣ with the arc length parameter s 1. If Ɣ : x(s) is a generalized helix, then Ɣ 1 : x 1 (s 1 ) is a straight line. Proof. Let α, β, γ be the tangent, principal normal and binormal vector field of the curve Ɣ : x(s), respectively. From the properties of generalized helices and the definition of Mannheim curves, we have γ 1 p = β p = 0 for some constant vector p. Then it is easy to obtain that 1 = κ 1 0.

5 124 Huili Liu and Fan Wang J. Geom. PROPOSITION 2. If a generalized helix is the Mannheim partner curve of some curve Ɣ : x(s) in E 3, then the ratio of torsion and curvature of the curve Ɣ : x(s) is κ = c 2 2 ec 1s 1 2c 2 e c 1s for some nonzero constant c 1 and c 2 and s is the arc length parameter of Ɣ. In particular, if we put c 1 = c 2 = 1, we have κ = es e s = sinh s. 2 Proof. Let α, β, γ be the tangent, principal normal and binormal vector field of the curve Ɣ : x(s), respectively. From the properties of generalized helices and the definition of Mannheim curves, we have β p = cos θ 0 for some constant vector p and some constant angle θ 0. From Proposition 1 we know that cos q 0 = 0 and = constant. By taking the derivative of this equation with respect to s κ twice, we get κα p + γ p = 0, κα p + γ p = (κ ) cos θ 0. By a direct calculation and using κ = λ(κ ), we obtain α p = λκ d(/κ) cos θ 0, 1 γ p = λ d(/κ) cos θ 0. Taking the derivative, we have d 2 (/κ) κ = 1 λ 1 ( 2 ) d(/κ) 2, κ d 2 (/κ) = ( 2 ) d(/κ) 2, λ

6 Vol. 88, 2008 Mannheim partner curves in 3-space 125 respectively. From these equations, we find that κ = d 2 (/κ) ( ) 2 d(/κ) 2 κ d 2. (/κ) 2 Let /κ = y(s), then we get the following differential equation Solving this equation, we obtain that (1 + y 2 ) d2 y 2 y ( ) dy 2 = 0. y(s)= c 0 or y(s)= c 2 2 ec 1s 1 2c 2 e c 1s for some nonzero constants c 0, c 1 and c 2. Thus, the proposition is proved. REMARK 2. It is well known that a twisted curve in E 3 is a generalized helix if and only if the ratio /κ is a nonzero constant (see [1]). It is also known that a twisted curve is congruent to a rectifying curve if and only if the ratio /κ is a nonconstant linear function of the arc length parameter (see [2]). Proposition 2 provides some characterizations of the curves whose slope /κ is hyperbolic sine function in arc length s, i.e., /κ = sinh s. 3. Mannheim partner curves in E 3 1 Let E 3 1 be the 3-dimensional Minkowski space with the indefinite inner product, = dx dx2 2 dx2 3 in terms of natural coordinates (x 1,x 2,x 3 ). A vector α = 0inE 3 1 is called spacelike, timelike or lightlike, if α, α > 0, α, α < 0or α, α =0, respectively. In this section, we extend the main result of Mannheim partner curves in E 3 to the Minkowski 3-space E 3 1. By a similar calculation, we obtain the following theorem. THEOREM 3. Let Ɣ : x(s) be a curve in E 3 1 and α, β, γ be tangent, principal normal and binormal vector field of Ɣ : x(s), respectively. Then

7 126 Huili Liu and Fan Wang J. Geom. (i) in case that α and β are spacelike vectors, γ is timelike vector, we have the following Frenet formulas α = κβ; β = κα + γ; γ = β. The necessary and sufficient condition for Mannheim partner curves is = κ λ (1 + λ2 2 ). (ii) in case that α and γ are spacelike vectors, β is timelike vector, the corresponding Frenet formulas are α = κβ; β = κα + γ; γ = β. The necessary and sufficient condition for Mannheim partner curves is = κ λ (λ2 2 1). (iii) in case that α is timelike vector, β and γ are spacelike vectors, the corresponding Frenet formulas are α = κβ; β = κα + γ; γ = β. The necessary and sufficient condition for Mannheim partner curves is Where λ is a nonzero constant. = κ λ (1 λ2 2 ). References [1] M.P. do Carmo, Differential Geometry of Curves and Surfaces, Pearson Education, [2] B.Y. Chen, When does the position vector of a space curve always lie in its rectifying plane?, Amer. Math. Monthly 110 (2003) [3] B.Y. Chen and F. Dillen, Rectifying curves as centrodes and extremal curves, Bull. Inst. Math. Acad. Sinica 33 (2005) [4] B. O Neill, Semi-Riemannian Geometry, Academic Press, Orland, [5] F. Wang and H.L. Liu, Mannheim partner curves in 3-Euclidean space, Math. Practice Theory 37 (2007) Huili Liu Fan Wang Department of Mathematics Department of Mathematics Northeastern University Nanjing AgriculturalUniversity Shenyang Nanjing P. R. China P. R. China liuhl@mail.neu.edu.cn wangfan84@eyou.com Received 1 September 2005; revised 23 January 2007.