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Generalized Null Bertrand Curves In Minkowski Space-Time

Generalized Null Bertrand Curves In Minkowski Space-Time Coken and Ciftci proved that a null Cartan curve in Minkowski space¸¨ ¸ time E4 is a null Bertrand curve if and only if k2 is nonzero constant and k3 is zero. 1 That is, the null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . 1 So, in this paper we defined a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the similar idea of generalized 1 Bertrand curve given by Matsuda and Yorozu and we called it a null (1, 3)-Bertrand curve. Also, we proved that if a null curve with non-zero curvatures in Minkowski spacetime E4 is a null (1, 3)-Bertrand curve then it is a null helix. We give an example of such 1 curves. Mathematics Subject Classification 2010: 53C50, 53B30. Key words: Minkowski space-time, null curve, Frenet vectors, Bertrand curves. 1. Introduction In the classical differential geometry of curves in Euclidean space, Saint Venant (see [15]) proposed the question whether upon the surface generated by the principal normal of a curve, a second curve can exist which has for its principal normal the principal normal of the given curve. This question was answered by Bertrand in 1850 in a paper (see [3]) in which he showed that a necessary and sufficient condition for the existence of such a second curve is that a linear relationship with constant coefficients exists between the first and second curvatures of the given original curve. In other ¨ 490 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN words, if we denote first and second curvatures of a given curve by k1 and k2 , respectively then we have k1 + µk2 = 1 where , µ R. Since Bertrand published his paper, curve pairs of this type have been called Conjugate Bertrand Curves or more commonly Bertrand Curves (see [11]). In 1888, Bioche [4] gave a new theorem to obtain Bertrand curves by using the given two curves C1 and C2 in Euclidean 3-space. Later, Burke [6] gave a theorem related with Bioche's theorem on Bertrand curves. The following properties of Bertrand curves are well known: If two curves have the same principal normals: (i) corresponding points are a fixed distance apart; (ii) the tangents at corresponding points are at a fixed angle. These well known properties of Bertrand curves in Euclidean 3space were extended by Pears in [14] to Riemannian n-space and found general results for Bertrand curves. When we apply these general results to Euclidean n-space, it is easily found that either k2 or k3 is zero. In other words, Bertrand curves in En (n > 3) are degenerate curves. This result was restated in [12] by Matsuda and Yorozu. They proved that there is no special Bertrand curve in En (n > 3) and they defined new type which is called (1, 3)-type Bertrand curve in 4-dimensional Euclidean space. In differential geometry of curves in Minkowski space, there are three different kinds of curves called spacelike, timelike and null ( lightlike) depending on their causal characters. Many of the classical results from Riemannian geometry have Lorentz counterparts. In fact, spacelike curves or timelike curves can be studied by a similar approach to that in positive definite Riemannian geometry. However, since the induced metric of a null curve is degenerate, this case is much more complicated and also different from a non-degenerate case. The presence of null curves often causes important and interesting differences as will be the case in the present study. In Minkowski 3-space (also in a Lorentzian manifold), spacelike and timelike Bertrand curves and their characterizations were studied in [2, 7, 9, 10]. Null Bertrand curves in Minkowski 3- space were studied by Balgetir, Bektas and Inoguchi in [1] and they proved the following theorem for a ¸ null Cartan curve to be a Bertrand curve: Theorem A. Let be a Cartan framed null curve. Then it is a Bertrand curve if and only if is a null geodesic or a Cartan framed null curve with constant second curvature k2 . Null Bertrand curves were studied in a Lorentzian manifold by Jin [10]. Coken and Ciftci [5] proved the following theorem for Bertrand curves ¸ ¸ in Minkowski space-time Theorem B. A Cartan curve in Minkowski space-time E4 is a null 1 Bertrand curve if and only if the curvature k2 is a non-zero constant and k3 is zero. That is, the null curve with non-zero curvature k3 is not a Bertrand ¨¸ curve in Minkowski space-time E4 . Also, Gocmen and Keles [8] obtained ¸ 1 some new results for Cartan framed null Bertrand curves in R4 . 1 In this paper, we define a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the 1 similar idea of generalized Bertrand curve given by Matsuda and Yorozu [12] and we called it a null (1, 3)-Bertrand curve. Also, we prove that if a null curve with non-zero curvatures in Minkowski space-time E4 is a null 1 (1, 3)-Bertrand curve then it is a null helix. Also, some properties of null (1, 3)-Bertrand curves in Minkowski space-time are given. We complete the paper with an example of such curves. 2. Preliminaries The Lorentzian 4 -space E4 is the Euclidean 4-space E4 equipped with 1 indefinite flat metric given by g= -dx2 1 i=2 dx2 , i where (x1 , x2 , x3 , x4 ) is a rectangular coordinate system of E4 . Recall that 1 a vector v E4 \{0} is spacelike if g(v, v) > 0, timelike if g(v, v) < 0 and 1 null (lightlike) if g(v, v) = 0 and v = 0. In particular, the vector v = 0 is a spacelike vector. The norm of a vector v is given by ||v||L = |g(v, v)| and two vectors v and w are said to be orthogonal if g(v, w) = 0. An arbitrary curve in E4 can locally be spacelike, timelike or null (lightlike) 1 if all of its velocity vectors (s) are spacelike, timelike or null, respectively. Recall that a spacelike curve in E4 is called pseudo-null curve or partially1 null curve if its principal normal vector is null and its first binormal vector is null, respectively. A spacelike or a timelike curve has unit speed, if g( (s), (s)) = ±1 ([13]). A null curve is parametrized by arclength function s if g( (s), (s)) = 1. In particular, a pseudo-null or a partiallynull curve has unit speed if g( (s), (s)) = 1. ¨ 492 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Let {T, N1 , N2 , N3 } be the moving Frenet frame along a null curve in E4 . Then the Frenet frame field of the curve satisfies the following Frenet 1 equations: T N1 N2 N3 = = = = k1 N1 , k2 T - k1 N2 , -k2 N1 + k3 N3 , -k3 T (2.1) where the first curvature k1 (s) = 0 if is a straight line or k1 (s) = 1 in all other cases [16]. Such curve has two non zero curvatures k2 (s) and k3 (s) . Moreover, this moving Frenet frame {T, N1 , N2 , N3 } satisfies the following conditions: g(T, T ) = g(N2 , N2 ) = 0, g(N1 , N1 ) = g(N3 , N3 ) = 1, g(T, N1 ) = g(T, N3 ) = g(N1 , N3 ) = g(N1 , N2 ) = g(N2 , N3 )=0, g(T, N2 )=1. In this study we consider the curve is not a straight line, that is, the first curvature of the curve is equal to one. 3. Null (1, 3)-bertrand curves in minkowski space-time From [5], we know that a null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . In this section, we give the 1 definition of null (1, 3)- Bertrand curve for a null curve with non-zero curvature k3 in E4 . Also we give some characterizations of such curves. 1 Definition 3.1. Let : I R E4 and : I R E4 be null curves 1 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and k1 ((s)), k 2 ((s)), k3 ((s)), respectively, where : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. If the Frenet (1, 3)-normal plane at each point (s) of the curve coincides with the Frenet (1, 3)-normal plane at corresponding point (s) = ((s)) of the curve for all s I then is called a null (1, 3)-Bertrand curve in E4 and is called a null (1, 3)-Bertrand mate of 1 the curve . Theorem 3.1. Let : I R E4 be a null curve with curvature 1 functions k1 (s) = 1, k2 (s) and k3 (s) = 0. Then is a null (1, 3)-Bertrand curve if and only if there exist constant real numbers , , and µ = 0 satisfying (3.1-a) (3.1-b) (3.1-c) (3.1-d) for all s I. Proof. We assume that is a null (1, 3)-Bertrand curve parametrized by arc-length s and is the null (1, 3)-Bertrand mate of the curve with arc-length s. Then we can write the curve as (3.2) (s) = ( (s)) = (s) + (s)N1 (s) + µ(s)N3 (s) = 0, 1 + k2 (s) - µk3 (s) = 0, (k2 (s))2 + (k3 (s))2 = - k2 (s) = , k3 (s) 2 , 4 for all s I where (s) and µ(s) are C -functions on I. Differentiating (3.2) with respect to s and by using the Frenet equations, we have (3.3) T ( (s)) (s) = [1 + (s) k2 (s) - µ (s) k3 (s)] T (s) + (s)N1 (s) -(s)k1 (s)N2 (s) + µ (s)N3 (s) for all s I. Since the plane spanned by N1 (s) and N3 (s) coincides with the plane spanned by N 1 ( (s)) and N 3 ( (s)), we can write (3.4) (3.5) N 1 ( (s)) = cos (s) N1 (s) + sin (s) N3 (s), N 3 ( (s)) = - sin (s) N1 (s) + cos (s) N3 (s). And then by using (3.4) and (3.5), we have g(N 1 ( (s)), T ( (s)) (s)) = (s) cos (s) + µ (s) sin (s) = 0, g(N 3 ( (s)), T ( (s)) (s)) = - (s) sin (s) + µ (s) cos (s) = 0. Thus we get (s) = 0, µ (s) = 0. That is, and µ are constant functions on I. So, we can rewrite (3.2) and (3.3) for all s I as follows: (3.6) (s) = ( (s)) = (s) + N1 (s) + µN3 (s) ¨ 494 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN and (3.7) T ( (s)) (s) = [1 + k2 (s) - µk3 (s)] T (s) - k1 (s)N2 (s). (s) = 1 + k2 (s) - µk3 (s) k1 (s) , (s) = - (s) (s) If we denote (3.8) for all s I. We can easily obtain that (3.9) T ( (s)) = (s)T (s) + (s)N2 (s), where (s) and (s) are C -functions on I. Since T ( (s)) , T (s) and N2 (s) are lightlike vectors, we get (3.10) (s)(s) = 0. That is, 1 + k2 (s) - µk3 (s) = 0 or = 0. We assume that = 0 and 1 + k2 (s) - µk3 (s) = 0. In that case, we can write T ( (s)) = (s)T (s) and if we differentiate the last equation with respect to s, we get dT ((s)) (s) = ds (s)T (s) + (s)T (s). By using the Frenet equations of and null curves, we have k1 ((s)))N 1 ( (s)) (s) = (s)T (s)+ (s)k1 (s)N1 (s). From (3.4), it holds (s) = 0. That is, (s) is non-zero constant function on I. So we get k1 ((s))N 1 ( (s)) (s) = k1 (s)N1 (s), where (s) = || . Since the null curves and are not straight lines, the principal curvature functions of the null curves and are equal to one. That is, k1 ((s)) = 1 and k1 (s) = 1. So we have N 1 ( (s)) = ±N1 (s), for all s I. This implies that is a null Bertrand curve. But by Theorem B, this fact is a contradiction. Thus we must consider only the case of = 0 and 1 + k2 (s) - µk3 (s) = 0. Then we obtain the relations (3.1-a) and (3.1-b). Hence we can write (3.11) T ( (s)) = (s)N2 (s). Differentiating (3.11) and by using the Frenet equations, we obtain (3.12) k1 ((s)))N 1 ( (s)) (s) = -(s)k2 (s)N1 (s) + (s)N2 (s) + (s)k3 (s)N3 (s). Since N 1 ( (s)) is expressed by linear combination of N1 (s) and N3 (s), it holds that (s) = 0. That is, (s) is a non-zero constant function. Also from (3.8), we can write (3.13) (s) = - . Since = 0, it follows (s) = 0. Hence there exists a regular map : I I defined by s = (s) = - s + , where is a real constant. We can rewrite (3.12) as k1 ((s)))N 1 ( (s)) (s) = -k2 (s)N1 (s) + k3 (s)N3 (s). and we can easily see that (3.14) (s) k1 ((s)) = 2 (k2 (s))2 + (k3 (s))2 . By substituting (3.13) into (3.14) and using k1 ((s)) = 1, we obtain the relation (3.1-c). From (3.12), we have N 1 ( (s)) = - where (3.15) cos (s) = - k2 (s) , (s) sin (s) = k3 (s) . (s) k2 (s) k3 (s) N1 (s) + N3 (s), (s) (s) Differentiating (3.4) with respect to s and using the Frenet equations, we obtain (3.16) (s) k2 ((s))T ((s)) - (s) k1 ((s))N 2 ( (s)) = (cos (s)) N1 (s) + (sin (s)) N3 (s) + (cos (s) k2 (s) - sin (s) k3 (s)) T (s) - cos (s) k1 (s)N2 (s) for all s I. From the above fact, it holds (3.17) (cos (s)) = 0, (sin (s)) = 0. That is, is a constant function on L with value 0 . Let = (cos 0 ) (sin 0 )-1 k2 (s) be a constant number. So from (3.15) , we get k3 (s) = -. Thus we obtain the relation (3.1-d). Conversely, we assume that : I R E4 is a null curve with cur1 vature functions k1 (s) = 1, k2 (s) and k3 (s) = 0 satisfying the relation (3.1 - a) , (3.1 - b) , (3.1 - c) and (3.1 - d) for constant numbers , , and µ. Then we define a null curve : I R E4 such as 1 (3.18) (s) = (s) + N1 (s) + µN3 (s) for all s I. Differentiating (3.18) with respect to s and by using the Frenet equations, we have d (s) = (1 + k2 (s) - µk3 (s)) T (s) + (-k1 (s)) N2 (s). ds ¨ 496 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN By using (3.1-b), we obtain d (s) = -k1 (s)N2 (s) ds for all s I. From (3.1-a), it follows d(s) = 0. We consider that the curve ds is not a straight line, that is, the first curvature of the curve is equal to one. So we can write (3.19) (s) = -N2 (s). (s) = (k2 (s)N1 (s) - k3 (s)N3 (s)) . There exists a regular map : I I defined by Differentiating (3.19) with respect to s, we get s = (s) = g (t), (t) dt, (s I) where s denotes the pseudo-arc length parameter of the curve . From (3.1-c), we obtain (3.20) (s) = for all s I, where : I I is a regular C -function and = 1, -1, >0 . <0 Thus the curve is rewritten as follows: (3.21) (s) = ((s)) = (s) + N1 (s) + µN3 (s). Differentiating (3.21) with respect to s, we obtain (s) From (3.1-b) (3.22) T ((s)) = - N2 (s), (s) d (s) ds = (1 + k2 (s) - µk3 (s)) T (s) - N2 (s). s=(s) 9 where T ((s)) = (3.23) d(s) ds . By substituting (3.20) into (3.22), we get T ((s)) = -N2 (s) for all s I. Differentiating (3.23) with respect to s and by using the Frenet equations, we have dT ((s)) k2 (s) k3 (s) = N1 (s) - N3 (s) ds (s) (s) and (3.24) 2 (k2 (s))2 + (k3 (s))2 dT ((s)) = . ds ( (s))2 Since the curve is a null (1, 3)-Bertrand curve, the curvatures of satisfy (3.1-c). If we substitute (3.1-c) into (3.24), we get (3.25) k1 ((s)) = dT ((s)) = 1. ds Then we can define a unit vector field N 1 ((s)) along the curve by N 1 ((s)) = = 1 dT ((s)) ds k1 ((s)) k2 (s) k3 (s) N1 (s) - N3 (s) . (s) (s) Since N 1 ((s)) is expressed by linear combination of N1 (s) and N3 (s), we can put N 1 ((s)) = cos (s) N1 (s) + sin (s) N3 (s) , where (3.26) cos (s) = k2 (s) (s) sin (s) = - k3 (s) (s) for all s I and (s) is a C -function. (3.1-c) and (3.1-d) imply that the curvatures of the curve k2 (s) and k3 (s) are constants. On the other hand, (s) = =constant. Thus cos (s) and sin (s) are constants, that is, (s) = 0 =constant. We can rewrite as (3.27) N 1 ((s)) = cos 0 N1 (s) + sin 0 N3 (s) . ¨ 498 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Differentiating (3.27) with respect to s and by using the Frenet equations, we have dN 1 ((s)) (s) = (k2 cos 0 - k3 sin 0 ) T (s) - cos 0 N2 (s) . ds By using (3.26) in the above equality, we get k2 dN 1 ((s)) = T (s) - N2 (s) ds ( (s))2 for all s I. From the Frenet equations (3.28) 1 k2 ((s)) = - g 2 dN 1 ((s)) dN 1 ((s)) , ds ds = k2 . ( (s))2 Thus we can define a unit vector field N 2 ((s)) along the curve by N 2 ((s)) = k2 ((s))T ((s)) - that is, N 2 ((s)) = - T (s) dN 1 ((s)) , ds Next we can define a unit vector N 3 ( (s)) along the curve by N 3 ((s)) = - sin 0 N1 (s) + cos 0 N3 (s) , that is, N 3 ((s)) = Thus we obtain (3.29) k3 ((s)) = -g dN 3 (s) , N 2 (s) ds =- k3 (s) . ( (s))2 k2 (s) k3 (s) N1 (s) + N3 (s) . (s) (s) Notice that g T , T = g N 2 , N 2 = 0, g N 1 , N 1 = g N 3 , N 3 = 1 and g T , N 1 = g T , N 3 = g N 1 , N 3 = g N 1 , N 2 = g N 2 , N 3 = 0, g T , N 2 = 1 for all s I, where T , N 1 , N 2 , N 3 is moving Frenet 4 frame along null curve in E1 . And it is trivial that the Frenet (1,3)normal plane at each point (s) of the curve coincides with the Frenet (1,3)-normal plane at corresponding point (s) of the curve . Hence is a null (1,3)- Bertrand curve in E4 . This completes the proof. 1 Corollary 3.1. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and : I R E4 be 1 a null (1, 3)-Bertrand mate of the curve with curvature functions k1 (s), k2 (s), k3 (s) where s and s denote the arc-length parameter of the curves and , respectively. Then the relations between these curvature functions are k2 (s) k3 (s) k1 ((s)) = 1, k 2 ((s)) = , 2 , k 3 ((s)) = - ( (s)) ( (s))2 where : I I, s = (s) is a regular C -function for all s I. Proof. It is obvious from the Theorem (3.1). Corollary 3.2. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve and : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. Then the distance between the points (s) and (s) is constant for all s I. Proof. Let : I R E4 be a null (1, 3)-Bertrand curve with 1 curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve . We assume that is distinct from . Let the pairs of (s) and (s) = ((s)) (where : I I, s = (s) is a regular C -function) be corresponding points of the curves and . Then we can write (s) = ( (s)) = (s) + N1 (s) + µN3 (s), where and µ are non-zero constants. Thus we can write (s) - (s) = N1 (s) + µN3 (s) and (s) - (s) = 2 + µ2 . So, d ( (s) , (s)) = constant. This completes the proof. Corollary 3.3. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)Bertrand mate of the curve with curvature functions k1 ((s)), k2 ((s)), k3 ((s)). Then the curvatures of the curves and are constants. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s) = 0, k3 (s) = 0 and is a null (1, 3)Bertrand mate of the curve . In that case, the relations (3.1-c) and (3.1-d) imply that k2 (s), k3 (s) are constant. From the Corollary 3.1, it can be easily seen that the curvatures of the curve are constant curvatures. ¨ 500 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Corollary 3.4. If : I R E4 is a null (1, 3)-Bertrand curve with 1 non-zero curvatures and is a null (1, 3)-Bertrand mate of then and are null helices. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and is a null (1, 3)-Bertrand mate of the curve . Then both the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k2 ((s)), k3 ((s)) belong to the curve are nonzero constants. Hence they are null helices. Corollary 3.5. Let : I R E4 be a null (1, 3)-Bertrand curve 1 and be a null (1, 3)-Bertrand mate of . Then the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k 2 ((s)), k3 ((s)) belong to the curve satisfy k 2 k3 + k2 k3 = 0. Proof. It is obvious from (3.28) and (3.29). 4 Example 3.1. Let be a null curve in E1 given by 1 (s) = (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 The Frenet frame of the curve is given by T (s) = N1 (s) = N2 (s) = N3 (s) = 1 (cosh (s) , sinh (s) , cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , - sin (s) , - cos (s)) , 2 1 (- cosh s, - sinh s, cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 Then we get the curvatures of as follows k1 (s) = 1, k2 (s) = 0, k3 (s) = -1. By using the Theorem B, the curve is not a null Bertrand curve. But if we take constants , , and µ as = 1, = 1, = 0, µ = -1, then it is trivial that the relations (3.1-a), (3.1-b), (3.1-c) and (3.1-d) hold. Therefore, the curve is a null (1, 3)-Bertrand curve in E4 . In this case, the 1 null (1, 3)-Bertrand mate of the curve is given by 1 (s) = (sinh (s) , cosh (s) , - sin (s) , - cos (s)) . 2 By using (3.25), (3.28), (3.29), we obtain the curvatures of the curve as k1 = 1, k2 = 0 and k 3 = 1. Acknowledgement. The authors would like to express their sincere gratitude to the referee for valuable suggestions to improve the paper. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annals of the Alexandru Ioan Cuza University - Mathematics de Gruyter

Generalized Null Bertrand Curves In Minkowski Space-Time

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Abstract

Coken and Ciftci proved that a null Cartan curve in Minkowski space¸¨ ¸ time E4 is a null Bertrand curve if and only if k2 is nonzero constant and k3 is zero. 1 That is, the null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . 1 So, in this paper we defined a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the similar idea of generalized 1 Bertrand curve given by Matsuda and Yorozu and we called it a null (1, 3)-Bertrand curve. Also, we proved that if a null curve with non-zero curvatures in Minkowski spacetime E4 is a null (1, 3)-Bertrand curve then it is a null helix. We give an example of such 1 curves. Mathematics Subject Classification 2010: 53C50, 53B30. Key words: Minkowski space-time, null curve, Frenet vectors, Bertrand curves. 1. Introduction In the classical differential geometry of curves in Euclidean space, Saint Venant (see [15]) proposed the question whether upon the surface generated by the principal normal of a curve, a second curve can exist which has for its principal normal the principal normal of the given curve. This question was answered by Bertrand in 1850 in a paper (see [3]) in which he showed that a necessary and sufficient condition for the existence of such a second curve is that a linear relationship with constant coefficients exists between the first and second curvatures of the given original curve. In other ¨ 490 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN words, if we denote first and second curvatures of a given curve by k1 and k2 , respectively then we have k1 + µk2 = 1 where , µ R. Since Bertrand published his paper, curve pairs of this type have been called Conjugate Bertrand Curves or more commonly Bertrand Curves (see [11]). In 1888, Bioche [4] gave a new theorem to obtain Bertrand curves by using the given two curves C1 and C2 in Euclidean 3-space. Later, Burke [6] gave a theorem related with Bioche's theorem on Bertrand curves. The following properties of Bertrand curves are well known: If two curves have the same principal normals: (i) corresponding points are a fixed distance apart; (ii) the tangents at corresponding points are at a fixed angle. These well known properties of Bertrand curves in Euclidean 3space were extended by Pears in [14] to Riemannian n-space and found general results for Bertrand curves. When we apply these general results to Euclidean n-space, it is easily found that either k2 or k3 is zero. In other words, Bertrand curves in En (n > 3) are degenerate curves. This result was restated in [12] by Matsuda and Yorozu. They proved that there is no special Bertrand curve in En (n > 3) and they defined new type which is called (1, 3)-type Bertrand curve in 4-dimensional Euclidean space. In differential geometry of curves in Minkowski space, there are three different kinds of curves called spacelike, timelike and null ( lightlike) depending on their causal characters. Many of the classical results from Riemannian geometry have Lorentz counterparts. In fact, spacelike curves or timelike curves can be studied by a similar approach to that in positive definite Riemannian geometry. However, since the induced metric of a null curve is degenerate, this case is much more complicated and also different from a non-degenerate case. The presence of null curves often causes important and interesting differences as will be the case in the present study. In Minkowski 3-space (also in a Lorentzian manifold), spacelike and timelike Bertrand curves and their characterizations were studied in [2, 7, 9, 10]. Null Bertrand curves in Minkowski 3- space were studied by Balgetir, Bektas and Inoguchi in [1] and they proved the following theorem for a ¸ null Cartan curve to be a Bertrand curve: Theorem A. Let be a Cartan framed null curve. Then it is a Bertrand curve if and only if is a null geodesic or a Cartan framed null curve with constant second curvature k2 . Null Bertrand curves were studied in a Lorentzian manifold by Jin [10]. Coken and Ciftci [5] proved the following theorem for Bertrand curves ¸ ¸ in Minkowski space-time Theorem B. A Cartan curve in Minkowski space-time E4 is a null 1 Bertrand curve if and only if the curvature k2 is a non-zero constant and k3 is zero. That is, the null curve with non-zero curvature k3 is not a Bertrand ¨¸ curve in Minkowski space-time E4 . Also, Gocmen and Keles [8] obtained ¸ 1 some new results for Cartan framed null Bertrand curves in R4 . 1 In this paper, we define a new type of Bertrand curve in Minkowski space-time E4 for a null curve with non-zero curvature k3 by using the 1 similar idea of generalized Bertrand curve given by Matsuda and Yorozu [12] and we called it a null (1, 3)-Bertrand curve. Also, we prove that if a null curve with non-zero curvatures in Minkowski space-time E4 is a null 1 (1, 3)-Bertrand curve then it is a null helix. Also, some properties of null (1, 3)-Bertrand curves in Minkowski space-time are given. We complete the paper with an example of such curves. 2. Preliminaries The Lorentzian 4 -space E4 is the Euclidean 4-space E4 equipped with 1 indefinite flat metric given by g= -dx2 1 i=2 dx2 , i where (x1 , x2 , x3 , x4 ) is a rectangular coordinate system of E4 . Recall that 1 a vector v E4 \{0} is spacelike if g(v, v) > 0, timelike if g(v, v) < 0 and 1 null (lightlike) if g(v, v) = 0 and v = 0. In particular, the vector v = 0 is a spacelike vector. The norm of a vector v is given by ||v||L = |g(v, v)| and two vectors v and w are said to be orthogonal if g(v, w) = 0. An arbitrary curve in E4 can locally be spacelike, timelike or null (lightlike) 1 if all of its velocity vectors (s) are spacelike, timelike or null, respectively. Recall that a spacelike curve in E4 is called pseudo-null curve or partially1 null curve if its principal normal vector is null and its first binormal vector is null, respectively. A spacelike or a timelike curve has unit speed, if g( (s), (s)) = ±1 ([13]). A null curve is parametrized by arclength function s if g( (s), (s)) = 1. In particular, a pseudo-null or a partiallynull curve has unit speed if g( (s), (s)) = 1. ¨ 492 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Let {T, N1 , N2 , N3 } be the moving Frenet frame along a null curve in E4 . Then the Frenet frame field of the curve satisfies the following Frenet 1 equations: T N1 N2 N3 = = = = k1 N1 , k2 T - k1 N2 , -k2 N1 + k3 N3 , -k3 T (2.1) where the first curvature k1 (s) = 0 if is a straight line or k1 (s) = 1 in all other cases [16]. Such curve has two non zero curvatures k2 (s) and k3 (s) . Moreover, this moving Frenet frame {T, N1 , N2 , N3 } satisfies the following conditions: g(T, T ) = g(N2 , N2 ) = 0, g(N1 , N1 ) = g(N3 , N3 ) = 1, g(T, N1 ) = g(T, N3 ) = g(N1 , N3 ) = g(N1 , N2 ) = g(N2 , N3 )=0, g(T, N2 )=1. In this study we consider the curve is not a straight line, that is, the first curvature of the curve is equal to one. 3. Null (1, 3)-bertrand curves in minkowski space-time From [5], we know that a null curve with non-zero curvature k3 is not a Bertrand curve in Minkowski space-time E4 . In this section, we give the 1 definition of null (1, 3)- Bertrand curve for a null curve with non-zero curvature k3 in E4 . Also we give some characterizations of such curves. 1 Definition 3.1. Let : I R E4 and : I R E4 be null curves 1 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and k1 ((s)), k 2 ((s)), k3 ((s)), respectively, where : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. If the Frenet (1, 3)-normal plane at each point (s) of the curve coincides with the Frenet (1, 3)-normal plane at corresponding point (s) = ((s)) of the curve for all s I then is called a null (1, 3)-Bertrand curve in E4 and is called a null (1, 3)-Bertrand mate of 1 the curve . Theorem 3.1. Let : I R E4 be a null curve with curvature 1 functions k1 (s) = 1, k2 (s) and k3 (s) = 0. Then is a null (1, 3)-Bertrand curve if and only if there exist constant real numbers , , and µ = 0 satisfying (3.1-a) (3.1-b) (3.1-c) (3.1-d) for all s I. Proof. We assume that is a null (1, 3)-Bertrand curve parametrized by arc-length s and is the null (1, 3)-Bertrand mate of the curve with arc-length s. Then we can write the curve as (3.2) (s) = ( (s)) = (s) + (s)N1 (s) + µ(s)N3 (s) = 0, 1 + k2 (s) - µk3 (s) = 0, (k2 (s))2 + (k3 (s))2 = - k2 (s) = , k3 (s) 2 , 4 for all s I where (s) and µ(s) are C -functions on I. Differentiating (3.2) with respect to s and by using the Frenet equations, we have (3.3) T ( (s)) (s) = [1 + (s) k2 (s) - µ (s) k3 (s)] T (s) + (s)N1 (s) -(s)k1 (s)N2 (s) + µ (s)N3 (s) for all s I. Since the plane spanned by N1 (s) and N3 (s) coincides with the plane spanned by N 1 ( (s)) and N 3 ( (s)), we can write (3.4) (3.5) N 1 ( (s)) = cos (s) N1 (s) + sin (s) N3 (s), N 3 ( (s)) = - sin (s) N1 (s) + cos (s) N3 (s). And then by using (3.4) and (3.5), we have g(N 1 ( (s)), T ( (s)) (s)) = (s) cos (s) + µ (s) sin (s) = 0, g(N 3 ( (s)), T ( (s)) (s)) = - (s) sin (s) + µ (s) cos (s) = 0. Thus we get (s) = 0, µ (s) = 0. That is, and µ are constant functions on I. So, we can rewrite (3.2) and (3.3) for all s I as follows: (3.6) (s) = ( (s)) = (s) + N1 (s) + µN3 (s) ¨ 494 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN and (3.7) T ( (s)) (s) = [1 + k2 (s) - µk3 (s)] T (s) - k1 (s)N2 (s). (s) = 1 + k2 (s) - µk3 (s) k1 (s) , (s) = - (s) (s) If we denote (3.8) for all s I. We can easily obtain that (3.9) T ( (s)) = (s)T (s) + (s)N2 (s), where (s) and (s) are C -functions on I. Since T ( (s)) , T (s) and N2 (s) are lightlike vectors, we get (3.10) (s)(s) = 0. That is, 1 + k2 (s) - µk3 (s) = 0 or = 0. We assume that = 0 and 1 + k2 (s) - µk3 (s) = 0. In that case, we can write T ( (s)) = (s)T (s) and if we differentiate the last equation with respect to s, we get dT ((s)) (s) = ds (s)T (s) + (s)T (s). By using the Frenet equations of and null curves, we have k1 ((s)))N 1 ( (s)) (s) = (s)T (s)+ (s)k1 (s)N1 (s). From (3.4), it holds (s) = 0. That is, (s) is non-zero constant function on I. So we get k1 ((s))N 1 ( (s)) (s) = k1 (s)N1 (s), where (s) = || . Since the null curves and are not straight lines, the principal curvature functions of the null curves and are equal to one. That is, k1 ((s)) = 1 and k1 (s) = 1. So we have N 1 ( (s)) = ±N1 (s), for all s I. This implies that is a null Bertrand curve. But by Theorem B, this fact is a contradiction. Thus we must consider only the case of = 0 and 1 + k2 (s) - µk3 (s) = 0. Then we obtain the relations (3.1-a) and (3.1-b). Hence we can write (3.11) T ( (s)) = (s)N2 (s). Differentiating (3.11) and by using the Frenet equations, we obtain (3.12) k1 ((s)))N 1 ( (s)) (s) = -(s)k2 (s)N1 (s) + (s)N2 (s) + (s)k3 (s)N3 (s). Since N 1 ( (s)) is expressed by linear combination of N1 (s) and N3 (s), it holds that (s) = 0. That is, (s) is a non-zero constant function. Also from (3.8), we can write (3.13) (s) = - . Since = 0, it follows (s) = 0. Hence there exists a regular map : I I defined by s = (s) = - s + , where is a real constant. We can rewrite (3.12) as k1 ((s)))N 1 ( (s)) (s) = -k2 (s)N1 (s) + k3 (s)N3 (s). and we can easily see that (3.14) (s) k1 ((s)) = 2 (k2 (s))2 + (k3 (s))2 . By substituting (3.13) into (3.14) and using k1 ((s)) = 1, we obtain the relation (3.1-c). From (3.12), we have N 1 ( (s)) = - where (3.15) cos (s) = - k2 (s) , (s) sin (s) = k3 (s) . (s) k2 (s) k3 (s) N1 (s) + N3 (s), (s) (s) Differentiating (3.4) with respect to s and using the Frenet equations, we obtain (3.16) (s) k2 ((s))T ((s)) - (s) k1 ((s))N 2 ( (s)) = (cos (s)) N1 (s) + (sin (s)) N3 (s) + (cos (s) k2 (s) - sin (s) k3 (s)) T (s) - cos (s) k1 (s)N2 (s) for all s I. From the above fact, it holds (3.17) (cos (s)) = 0, (sin (s)) = 0. That is, is a constant function on L with value 0 . Let = (cos 0 ) (sin 0 )-1 k2 (s) be a constant number. So from (3.15) , we get k3 (s) = -. Thus we obtain the relation (3.1-d). Conversely, we assume that : I R E4 is a null curve with cur1 vature functions k1 (s) = 1, k2 (s) and k3 (s) = 0 satisfying the relation (3.1 - a) , (3.1 - b) , (3.1 - c) and (3.1 - d) for constant numbers , , and µ. Then we define a null curve : I R E4 such as 1 (3.18) (s) = (s) + N1 (s) + µN3 (s) for all s I. Differentiating (3.18) with respect to s and by using the Frenet equations, we have d (s) = (1 + k2 (s) - µk3 (s)) T (s) + (-k1 (s)) N2 (s). ds ¨ 496 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN By using (3.1-b), we obtain d (s) = -k1 (s)N2 (s) ds for all s I. From (3.1-a), it follows d(s) = 0. We consider that the curve ds is not a straight line, that is, the first curvature of the curve is equal to one. So we can write (3.19) (s) = -N2 (s). (s) = (k2 (s)N1 (s) - k3 (s)N3 (s)) . There exists a regular map : I I defined by Differentiating (3.19) with respect to s, we get s = (s) = g (t), (t) dt, (s I) where s denotes the pseudo-arc length parameter of the curve . From (3.1-c), we obtain (3.20) (s) = for all s I, where : I I is a regular C -function and = 1, -1, >0 . <0 Thus the curve is rewritten as follows: (3.21) (s) = ((s)) = (s) + N1 (s) + µN3 (s). Differentiating (3.21) with respect to s, we obtain (s) From (3.1-b) (3.22) T ((s)) = - N2 (s), (s) d (s) ds = (1 + k2 (s) - µk3 (s)) T (s) - N2 (s). s=(s) 9 where T ((s)) = (3.23) d(s) ds . By substituting (3.20) into (3.22), we get T ((s)) = -N2 (s) for all s I. Differentiating (3.23) with respect to s and by using the Frenet equations, we have dT ((s)) k2 (s) k3 (s) = N1 (s) - N3 (s) ds (s) (s) and (3.24) 2 (k2 (s))2 + (k3 (s))2 dT ((s)) = . ds ( (s))2 Since the curve is a null (1, 3)-Bertrand curve, the curvatures of satisfy (3.1-c). If we substitute (3.1-c) into (3.24), we get (3.25) k1 ((s)) = dT ((s)) = 1. ds Then we can define a unit vector field N 1 ((s)) along the curve by N 1 ((s)) = = 1 dT ((s)) ds k1 ((s)) k2 (s) k3 (s) N1 (s) - N3 (s) . (s) (s) Since N 1 ((s)) is expressed by linear combination of N1 (s) and N3 (s), we can put N 1 ((s)) = cos (s) N1 (s) + sin (s) N3 (s) , where (3.26) cos (s) = k2 (s) (s) sin (s) = - k3 (s) (s) for all s I and (s) is a C -function. (3.1-c) and (3.1-d) imply that the curvatures of the curve k2 (s) and k3 (s) are constants. On the other hand, (s) = =constant. Thus cos (s) and sin (s) are constants, that is, (s) = 0 =constant. We can rewrite as (3.27) N 1 ((s)) = cos 0 N1 (s) + sin 0 N3 (s) . ¨ 498 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Differentiating (3.27) with respect to s and by using the Frenet equations, we have dN 1 ((s)) (s) = (k2 cos 0 - k3 sin 0 ) T (s) - cos 0 N2 (s) . ds By using (3.26) in the above equality, we get k2 dN 1 ((s)) = T (s) - N2 (s) ds ( (s))2 for all s I. From the Frenet equations (3.28) 1 k2 ((s)) = - g 2 dN 1 ((s)) dN 1 ((s)) , ds ds = k2 . ( (s))2 Thus we can define a unit vector field N 2 ((s)) along the curve by N 2 ((s)) = k2 ((s))T ((s)) - that is, N 2 ((s)) = - T (s) dN 1 ((s)) , ds Next we can define a unit vector N 3 ( (s)) along the curve by N 3 ((s)) = - sin 0 N1 (s) + cos 0 N3 (s) , that is, N 3 ((s)) = Thus we obtain (3.29) k3 ((s)) = -g dN 3 (s) , N 2 (s) ds =- k3 (s) . ( (s))2 k2 (s) k3 (s) N1 (s) + N3 (s) . (s) (s) Notice that g T , T = g N 2 , N 2 = 0, g N 1 , N 1 = g N 3 , N 3 = 1 and g T , N 1 = g T , N 3 = g N 1 , N 3 = g N 1 , N 2 = g N 2 , N 3 = 0, g T , N 2 = 1 for all s I, where T , N 1 , N 2 , N 3 is moving Frenet 4 frame along null curve in E1 . And it is trivial that the Frenet (1,3)normal plane at each point (s) of the curve coincides with the Frenet (1,3)-normal plane at corresponding point (s) of the curve . Hence is a null (1,3)- Bertrand curve in E4 . This completes the proof. 1 Corollary 3.1. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and : I R E4 be 1 a null (1, 3)-Bertrand mate of the curve with curvature functions k1 (s), k2 (s), k3 (s) where s and s denote the arc-length parameter of the curves and , respectively. Then the relations between these curvature functions are k2 (s) k3 (s) k1 ((s)) = 1, k 2 ((s)) = , 2 , k 3 ((s)) = - ( (s)) ( (s))2 where : I I, s = (s) is a regular C -function for all s I. Proof. It is obvious from the Theorem (3.1). Corollary 3.2. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve and : I I, s = (s) is a regular C -function such that each point (s) of the curve corresponds to the point (s) = ((s)) of the curve for all s I. Then the distance between the points (s) and (s) is constant for all s I. Proof. Let : I R E4 be a null (1, 3)-Bertrand curve with 1 curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)-Bertrand mate of the curve . We assume that is distinct from . Let the pairs of (s) and (s) = ((s)) (where : I I, s = (s) is a regular C -function) be corresponding points of the curves and . Then we can write (s) = ( (s)) = (s) + N1 (s) + µN3 (s), where and µ are non-zero constants. Thus we can write (s) - (s) = N1 (s) + µN3 (s) and (s) - (s) = 2 + µ2 . So, d ( (s) , (s)) = constant. This completes the proof. Corollary 3.3. Let : I R E4 be a null (1, 3)-Bertrand curve 1 with curvature functions k1 (s) = 1, k2 (s), k3 (s) = 0 and be a null (1, 3)Bertrand mate of the curve with curvature functions k1 ((s)), k2 ((s)), k3 ((s)). Then the curvatures of the curves and are constants. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s) = 0, k3 (s) = 0 and is a null (1, 3)Bertrand mate of the curve . In that case, the relations (3.1-c) and (3.1-d) imply that k2 (s), k3 (s) are constant. From the Corollary 3.1, it can be easily seen that the curvatures of the curve are constant curvatures. ¨ 500 FERDAG KAHRAMAN AKSOYAK, ISMAIL GOK and KAZIM ILARSLAN Corollary 3.4. If : I R E4 is a null (1, 3)-Bertrand curve with 1 non-zero curvatures and is a null (1, 3)-Bertrand mate of then and are null helices. Proof. We assume that : I R E4 is a null (1, 3)-Bertrand curve 1 with curvatures k1 (s) = 1, k2 (s), k3 (s) = 0 and is a null (1, 3)-Bertrand mate of the curve . Then both the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k2 ((s)), k3 ((s)) belong to the curve are nonzero constants. Hence they are null helices. Corollary 3.5. Let : I R E4 be a null (1, 3)-Bertrand curve 1 and be a null (1, 3)-Bertrand mate of . Then the curvatures k2 (s), k3 (s) belong to the curve and the curvatures k 2 ((s)), k3 ((s)) belong to the curve satisfy k 2 k3 + k2 k3 = 0. Proof. It is obvious from (3.28) and (3.29). 4 Example 3.1. Let be a null curve in E1 given by 1 (s) = (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 The Frenet frame of the curve is given by T (s) = N1 (s) = N2 (s) = N3 (s) = 1 (cosh (s) , sinh (s) , cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , - sin (s) , - cos (s)) , 2 1 (- cosh s, - sinh s, cos (s) , - sin (s)) , 2 1 (sinh (s) , cosh (s) , sin (s) , cos (s)) . 2 Then we get the curvatures of as follows k1 (s) = 1, k2 (s) = 0, k3 (s) = -1. By using the Theorem B, the curve is not a null Bertrand curve. But if we take constants , , and µ as = 1, = 1, = 0, µ = -1, then it is trivial that the relations (3.1-a), (3.1-b), (3.1-c) and (3.1-d) hold. Therefore, the curve is a null (1, 3)-Bertrand curve in E4 . In this case, the 1 null (1, 3)-Bertrand mate of the curve is given by 1 (s) = (sinh (s) , cosh (s) , - sin (s) , - cos (s)) . 2 By using (3.25), (3.28), (3.29), we obtain the curvatures of the curve as k1 = 1, k2 = 0 and k 3 = 1. Acknowledgement. The authors would like to express their sincere gratitude to the referee for valuable suggestions to improve the paper.

Journal

Annals of the Alexandru Ioan Cuza University - Mathematics – de Gruyter

Published: Nov 24, 2014

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