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Generalized Timelike Mannheim Curves in 1 Minkowski space-time E4 1 1 0 2 M. AKYI˙GˇI˙T, S. ERSOY, I˙. O¨ZGU¨R, M. TOSUN n a J 9 Department of Mathematics, Faculty of Arts and Sciences 1 Sakarya University, 54187 Sakarya/TURKEY ] G Abstract D WegiveadefinitionofgeneralizedtimelikeMannheimcurveinMinkowski . space-time E14. The necessary and sufficient conditions for the general- h izedtimelikeMannheimcurveobtain. Weshowsomecharacterizationsof t a generalized Mannheim curve. m MathematicsSubjectClassification(2010): 53B30,53A35,53A04. Keywords: Mannheim curve, Minkowski space-time. [ 1 v 1 Introduction 6 8 6 The geometry of curves has long captivated the interests of mathematicians, 3 from the ancient Greeks through to the era of Isaac Newton (1647-1727) and . 1 the invention of the calculus. It is branch of geometry that deals with smooth 0 curves in the plane and in the space by methods of differential and integral 1 calculus. The theory of curves is the simpler and narrower in scope because 1 : a regular curve in a Euclidean space has no intrinsic geometry. One of the v mostimportanttoolsusedtoanalyzecurveistheFrenetframe,amovingframe i X thatprovidesacoordinatesystemateachpoint ofcurvethatis ”bestadopted” r to the curve near that point. Every person of classical differential geometry a meets early in his course the subject of Bertrand curves, discovered in 1850 by J. Bertrand. A Bertrand curve is a curve such that its principal normals are the principal normals of a second curve. There are many works related with Bertrand curves in the Euclidean space and Minkowski space, [1]-[3]. AnotherkindofassociatedcurveiscalledMannheimcurveandMannheimpart- ner curve. The notion of Mannheim curves was discovered by A. Mannheim in 1878. These curves in Euclidean 3-space are characterized in terms of the curvature and torsionas follows: A space curveis a Mannheim curve if andonly if its curvature κ and torsion τ satisfy the relation k =β k2+k2 1 1 2 (cid:0) (cid:1) 1 for some constant β. The articles concerning Mannheim curves are rather few. In [4], a remarkable class of Mannheim curves is studied. General Mannheim curves in the Euclidean 3-space are obtained in [5]-[7] Recently, Mannheim curves are generalized and some characterizations and examples of generalized Mannheim curves are given in Euclidean 4-space E4 by [8]. In this paper, we study the generalized spacelike Mannheim partner curves in 4 dimensionalMinkowskispace-time. We willgive the necessaryandsufficient − conditions for the generalized spacelike Mannheim partner curves. 2 Preliminaries To meet the requirements in the next sections,the basic elements of the theory of curves in Minkowski space-time E4 are briefly presented in this section. A 1 more complete elementary treatment can be found in [9]. Minkowskispace-timeE4 isanEuclideanspaceprovidedwiththe standardflat 1 metric given by , = dx2+dx2+dx2+dx2 h i − 1 2 3 4 where (x , x , x , x ) is a rectangular coordinate system in E4. 1 2 3 4 1 Since , is anindefinite metric, recallthat a v E4 canhaveone ofthe three causalhchiaracters;itcanbespacelikeif v,v >0∈orv1 =0,timelikeif v,v <0 andnull(ligthlike)if v,v =0andv=h0. Siimilarly,anarbitrarycurvehc=ic(t) h i 6 in E4 can locally be spacelike, timelike or null (lightlike) if all of its velocity 1 vectorsc′(t)are,respectively,spacelike,timelikeornull. Thenormofv E4 is givenby v = v,v . If c′(t) = c′(t),c′(t) =0forallt I,∈then1C k k |h i| k k |h i|6 ∈ isaregularcurveinE4. Atimelike(spacelike)regularcurveC isparameterized p 1 p byarc-lengthparametertwhichisgivenbyc:I E4,thenthetangentvector c′(t)alongC hasunit length,thatis, c(t),c(t)→=11, ( c(t),c(t) = 1)for h i h i − all t I. ∈ Hereafter, curves considered are timelike and regular C∞ curves in E4. Let 1 T (t)= c′(t) for all t I, then the vector field T (t) is timelike and it is called ∈ timelike unit tangent vector field on C. The timelike curve C is called special timelike Frenet curve if there exist three smoothfunctionsk , k , k onCandsmoothnon-nullframefield T, N, B , B 1 2 3 1 2 { } along the curve C. Also, the functions k , k and k are called the first, the 1 2 3 secondand the third curvature function on C, respectively. For the C∞ special timelike Frenet curve C, the following Frenet formula is T′ 0 k 0 0 T 1 N′ k 0 k 0 N  B ′ =  01 k 02 k  B  1 2 3 1  B2′   0 −0 −k3 0  B2       [9]. 2 Here, due to characters of Frenet vectors of the timelike curve, T, N, B and 1 B are mutually orthogonalvector fields satisfying equations 2 T , T = 1 , N, N = B , B = B , B =1 . 1 1 2 2 h i − h i h i h i For t I, the non-null frame field T, N, B , B and curvature functions 1 2 ∈ { } k , k and k are determined as follows 1 2 3 1st step T(t)=c′(t) 2nd step k (t)= T′(t) >0 1 k k N(t)= 1 T′(t) k1(t) 3rd step k (t)= N′(t) k (t)T (t) >0 2 1 k − k B (t)= 1 (N′(t) k (t)T (t)) 1 k2(t) − 1 4th step B (t)=ε 1 B ′(t)+k (t)N(t) 2 kB1′(t)+k2(t)N(t)k 1 2 where ε is determined by the fact that orthono(cid:0)rmal frame field T(cid:1)(t), N(t), { B (t), B (t) , is of positive orientation. The function k is determined by 1 2 3 } k (t)= B ′(t) , B (t) =0. 3 1 2 6 So the function k3 never vanishes(cid:10). (cid:11) Inorderto makesurethat the curve C is a specialtimelike Frenetcurve,above steps must be checked, from 1st step to 4th step, for t I. ∈ Let T, N, B , B be the moving Frenet frame along a unit speed timelike 1 2 { } curveC inE4,consistingofthetangent,theprincipalnormal,thefirstbinormal 1 and the second binormal vector field, respectively. Since C is a timelike curve, its Frenet frame contains only non-null vector fields. 3 Generalized timelike Mannheim curves in E4 1 Mannheim curves are generalized in by [8]. In this paper, we have investigated generalizationof timelike Mannheim curves in Minkowski space E4. 1 Definition 3.1 AspecialtimelikecurveC inE4 isageneralizedtimelikeMannheim 1 curve if there exists a special timelike Frenet curve C∗ in E4 such that the first 1 normal line at each point of C is included in the plane generated by the second normal line and the third normal line of C∗ at the corresponding point under φ. Here φ is a bijection from C to C∗. The curve C∗ is called the generalized timelike Mannheim mate curve of C. By the definition, a generalized Mannheim mate curve C∗ is given by the map c∗ :I∗ E4 such that → 1 c∗(t)=c(t)+β(t)N(t), t I. (3.1) ∈ Hereβ isasmoothfunctiononI.Generally,the parametertisn’tanarc-length of C∗. Let t∗ be the arc-length of C∗ defined by 3 t dc∗(t) t∗ = dt. dt Z (cid:13) (cid:13) 0 (cid:13) (cid:13) If a smooth function f :I I∗ is giv(cid:13)(cid:13)en by f(cid:13)(cid:13)(t)=t∗, then for t I, we have → ∀ ∈ dt∗ dc∗(t) f′(t)= = = (1+β(t)k (t))2+(β′(t))2+(β(t)k (t))2 . dt dt − 1 2 (cid:13) (cid:13) r(cid:12) (cid:12) The representati(cid:13)(cid:13)on of tim(cid:13)(cid:13)elike(cid:12)(cid:12)curve C∗ with arc-length parameter t∗ is (cid:12)(cid:12) (cid:13) (cid:13) c∗ : I∗ E4 t∗ → c∗1(t∗). → For a bijection φ : C C∗ defined by φ(c(t)) = c∗(f(t)), the reparameteri- → zation of C∗ is c∗(f(t))=c(t)+β(t)N(t) where β is a smooth function on I. Thus, we have dc∗(f(t)) dc∗(t∗) = f′(t)=f′(t)T∗(f(t)), t I. dt dt ∈ (cid:12)t∗=f(t) (cid:12) Theorem 3.1 If a special time(cid:12)(cid:12)like Frenet curve C in E14 is a generalized timelike Mannheim curve, then the following relation between the first curvature function k and the second curvature function k holds: 1 2 k (t)= β k2(t) k2(t) , t I (3.2) 1 − 1 − 2 ∈ where β is a constant number. (cid:0) (cid:1) Proof Let C be a generalized timelike Mannheim curve and C∗ be the generalized timelike Mannheim mate curve of C, as following diagram c c∗ ·· ·· f : I I∗ → ↓ ↓ φ : E4 E4 1 → 1 A smooth function h is defined by f(t) = dc∗(t) dt = t∗ and t∗ is the arc- dt length parameter of C∗. Also φ is a bijectio(cid:13)n whic(cid:13)h is defined by φ(c(t)) = R (cid:13) (cid:13) c∗(f(t)). Thus, the timelike curve C∗ is repa(cid:13)rametr(cid:13)izedas follows c∗(f(t))=c(t)+β(t)N(t) (3.3) where β : I R R is a smooth function. By differentiating both sides of ⊂ → equation (3.3) with respect to t, we have f′(t)T∗(f(t))=(1+β(t)k (t))T +β′(t)N(t)+β(t)k (t)B (t). (3.4) 1 2 1 4 On the other hand, since the first normalline at the each point of C is lying in the plane generated by the second normal line and the third normal line of C∗ at the corresponding points under bijection φ, the vector field N(t) is given by N(t)=g(t)B ∗(f(t))+h(t)B ∗(f(t)) 1 2 wheregandharesomesmoothfunctionsonI R. Ifwetakeintoconsideration ⊂ T∗(f(t)), g(t)B ∗(f(t))+h(t)B ∗(f(t)) =0 1 2 h i andthe equation(3.4),thenwe haveβ′(t)=0Sowerewritethe equation(3.4) as f′(t)T∗(f(t))=(1+βk (t))T (t)+βk (t)B (t), (3.5) 1 2 1 that is, (1+βk (t)) βk (t) T∗(f(t))= 1 T (t)+ 2 B (t) f′(t) f′(t) 1 where f′(t)= (1+βk (t))2+(βk (t))2 . 1 2 − r (cid:12) (cid:12) Bytaking differentiationboth(cid:12)sides ofthe equations(3.5)w(cid:12) ithrespecttot I, (cid:12) (cid:12) ∈ we get ′ f′(t)k∗(f(t))N∗(f(t))= 1+βk1(t) T(t) 1 f′(t) + (1+βk1(cid:16)(t))k1(t)−β(cid:17)(k2(t))2 N(t) (3.6) f′(t) ′ +(cid:16)βk2(t) B (t)+ βk2(t)(cid:17)k3(t) B (t). f′(t) 1 f′(t) 2 (cid:16) (cid:17) (cid:16) (cid:17) Since N∗(f(t)), g(t)B ∗(f(t))+h(t)B ∗(f(t)) =0. 1 2 h i The coefficient of N(t) in equation (3.6) vanishes, that is, (1+βk (t))k (t) β(k (t))2 =0. 1 1 2 − Thus, this completes the proof. Theorem 3.2 In E4, let C be a special timelike Frenet curve such that its 1 non-constant first and second curvature functions satisfy the equality k (s) = 1 β k2(t) k2(t) for all t I R. If the timelike curve C∗ given by − 1 − 2 ∈ ⊂ (cid:0) (cid:1) c∗(t)=c(t)+βN(t) is a special timelike Frenet curve, then C∗ is a generalized timelike Mannheim mate curve of C. 5 Proof The arc-length parameter of C∗ is given by t dc∗(t) t∗ = dt , t I. dt ∈ Z (cid:13) (cid:13) 0 (cid:13) (cid:13) (cid:13) (cid:13) Under the assumption of (cid:13) (cid:13) k (t)= β k2(t) k2(t) , 1 − 1 − 2 we obtain f′(t)= 1+βk (t) , t (cid:0)I. (cid:1) 1 Differentiating the eq|uation c∗(|f(t)∈)= c(t)+βN(t) with respect to t the we p reach f′(t)T∗(f(t))=(1+βk (t))T (t)+βk (t)B (t). 1 2 1 Thus, it is seen that 1+βk (t) βk (t) T∗(f(t))= 1 T (t)+ 2 B (t) , t I. (3.7) 1 1+βk1(t) 1+βk1(t) ! ∈ | | | | The differentiationpof the last equation wpith respect to t is ′ f′(t)k∗(f(t))N∗(f(t))= 1+βk (t) T(t) 1 | 1 | + (1+βk√1((cid:16)|t1)p+)kβ1k(1t)(−t)β|k22(t) N(cid:17) (t) (3.8) (cid:18) ′ (cid:19) + βk2(t) B (t)+ βk2(t)k3(t) B (t). √|1+βk1(t)| 1 √|1+βk1(t)| 2 (cid:18) (cid:19) (cid:18) (cid:19) From our assumption, we have k (t)+βk2(t) βk2(t) 1 1 − 2 =0. 1+βk (t) 1 | | Thus, the coefficient of N(t)pin the equation (3.8) is zero. It is seen from the equation (3.7), T∗(f(t)) is a linear combination of T (t) and B (t). Ad- 1 ditionally, from equation (3.8), N∗(f(t)) is given by linear combination of T (t), B (t) and B (t). On the otherhand, C∗ is a special timelike Frenet 1 2 curve that the vector N(t) is given by linear combination of T∗(f(t)) and N∗(f(t)). Therefore, the first normal line C lies in the plane generated by the second normal line and third normal line of C∗ at the corresponding points under a bijection φ which is defined by φ(c(t))=c∗(f(t)). This,completes the proof. Remark 3.1 In 4-diemsional Minkowski space E4, a special timelike Frenet 1 curveC withcurvaturefunctionsk andk satisfyingk (t)= β k2(t) k2(t) , 1 2 1 − 1 − 2 it is not clear that a smooth timelike curve C∗ given by (3.1) is a special Frenet (cid:0) (cid:1) curve. Thus, it is unknown whether the reverse of Theorem 3.1 is true or false. 6 Theorem 3.3 Let C be a special timelike curve in E4 with non-zero third cur- 1 vature function k . If there exists a timelike special Frenet curve C∗ in E4 such 3 1 that the first normal line of C is linearly dependent with the third normal line of C∗ at the corresponding points c(t) and c∗(t), respectively, under a bijection φ:C C∗, iff the curvatures k and k of C are constant functions. 1 2 → Proof Let C be a timelike Frenet curve in E4 with the Frenet frame field 1 T, N,B , B andcurvaturefunctionsk , k andk . Also,weassumethatC∗ 1 2 1 2 3 b{eatimelikesp}ecialFrenetcurveinE4withtheFrenetframefield T∗, N∗, B ∗, B ∗ and curvature functions k∗, k∗1and k∗. Let the first norma{l line of C 1be 2 } 1 2 3 linearlydependentwiththe thirdnormallineofC∗ atthe correspondingpoints C and C∗, respectively. Then the parameterization of C∗ is c∗(f(t))=c(t)+β(t)N(t), t I. (3.9) ∈ If the arc-length parameter of C∗ is given t∗, then t t∗ = (1+β(t)k (t))2+(β′(t))+(β(t)k (t))2 dt (3.10) 1 2 − Z0 r(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) and f : I I∗ → t f(t)=t∗. → Moreover, φ:C C∗ is a bijection given by φ(c(t))=c∗(f(t)). → Differentiating the equation (3.9) with respect to t and using Frenet formulas, we get f′(t)T∗(f(t))=(1+β(t)k (t))T(t)+β′(t)N(t) 1 (3.11) +β(t)k (t)B (t). 2 1 Since B ∗(f(t))= N(t), then 2 ∓ (1+β(t)k (t))T (t)+β′(t)N(t) f′(t)T∗(f(t)), B ∗(f(t)) = 1 , h 2 i +β(t)k2(t)B1(t), N(t) (cid:28) ∓ (cid:29) that is, 0= β′(t). ∓ From last equation, it is easily seen that β is a constant. Hereafter, we can denote β(t)=β, for all t I. ∈ From the equation (3.10), we have f′(t)= (1+βk (t))2+(βk (t))2 >0. 1 2 − r (cid:12) (cid:12) (cid:12) (cid:12) Thus, we rewrite the equatio(cid:12)n (3.11) as follows; (cid:12) 1+βk (t) βk (t) T∗(f(t))= 1 T (t)+ 2 B (t). f′(t) f′(t) 1 (cid:18) (cid:19) (cid:18) (cid:19) 7 The differentiation of the last equation with respect to t is ′ f′(t)k∗(f(t))N∗(f(t))= 1+βk1(t) T (t) 1 f′(t) + (1+(cid:16)βk1(t))k1(t(cid:17))−βk22(t) N(t) (3.12) f′(t) ′ +(cid:16)βk2(t) B (t)+ βk2(cid:17)(t)k3(t) B (t). f′(t) 1 f′(t) 2 (cid:16) (cid:17) (cid:16) (cid:17) Since f′(t)k∗(f(t))N∗(f(t)), B ∗(f(t)) = 0 and B ∗(f(t)) = N(t) for h 1 2 i 2 ∓ all t I, we obtain ∈ k (t)+βk2(t) βk2(t)=0 1 1 − 2 is satisfied. Then k (t) β = 1 (3.13) −k2(t) k2(t) 1 − 2 is a non-zero constant number. Thus, from the equation (3.12), we reach ′ ′ N∗(f(t))= 1 1+βk1(t) T (t)+ 1 βk2(t) B (t) f′(t)K(t) f′(t) f′(t)K(t) f′(t) 1 + 1 β(cid:16)k2(t)k3(t) (cid:17)B (t) (cid:16) (cid:17) f′(t)K(t) f′(t) 2 (cid:16) (cid:17) where K(t) = k∗(f(t)) for all t I. Differentiating the last equation with 1 ∈ respect to t, then we have ′ ′ f′(t)[k∗(f(t))T∗(f(t))+k∗(f(t))B (f(t))]= 1 1+βk1(t) T(t) 1 2 1 f′(t)K(t) f′(t) ′ (cid:18) (cid:16) ′ (cid:17)(cid:19) + k1(t) 1+βk1(t) k2(t) βk2(t) N(t) f′(t)K(t) f′(t) − f′(t)K(t) f′(t) (cid:18) (cid:16) (cid:17)′ ′ (cid:16) (cid:17)(cid:19) + 1 βk2(t) k3(t) βk2(t)k3(t) B (t) (cid:18)f′(t)K(t)(cid:16) f′(t) (cid:17)(cid:19) − f′(t)K(t)(cid:16) f′(t) (cid:17)! 1 ′ ′ + 1 βk2(t)k3(t) + k3(t) βk2(t) B (t) f′(t)K(t) f′(t) f′(t)K(t) f′(t) 2 (cid:18)(cid:16) (cid:16) (cid:17)(cid:17) (cid:16) (cid:17)(cid:19) for all t I. Considering ∈ f′(t)(k∗(f(t))T∗(f(t))+k∗(f(t))B ∗(f(t))) , B ∗(f(t)) =0 h 1 2 1 2 i and B ∗(f(t))= N(t), 2 ∓ then we get 1+βk (t) ′ βk (t) ′ k (t) 1 k (t) 2 =0. 1 f′(t) − 2 f′(t) (cid:18) (cid:19) (cid:18) (cid:19) Arranging the last equation, we find 8 β k (t)k ′(t) k (t)k ′(t) f′(t) k (t)+βk2(t) βk2(t) f′′(t)=0. 1 1 − 2 2 − 1 1 − 2 (3.14) (cid:2) (cid:3) (cid:2) (cid:3) Moreover,the differentiation of the equation (3.13) with respect to t is k′ (t)+2β(k (t)k′ (t) k (t)k′ (t))=0. 1 1 1 − 2 2 From the above equation, it is seen that k′ (t) 1 =β(k (t)k′ (t) k (t)k′ (t)). (3.15) 1 1 2 2 − 2 − Substituting the equations(3.13)and(3.15)intothe equation(3.14),weobtain k′ (t) 1 =0. − 2 This means that the first curvature function is constant (that is, positive constant). Additionally, from the equation (3.15) it is seen that the second curvature function k is positive constant, too. 2 Conversely, suppose that C is a timelike Frenet curve E4 in with the Frenet 1 frame field T, N, B , B and curvature functions k , k and k . The first 1 2 1 2 3 { } curvature function k and the second curvature function k of C are of positive 1 2 constant. Thus, k1 is a positive constant number, say β. k22−k12 The representation of timelike curve C∗ with arc-length parameter t is c∗ : I E4 t → c1∗(t)=c(t)+β(t)N(t). (3.16) → Let t∗ denote the arc-length parameter of C∗, we have f : I I∗ → t t∗ =f(t)= 1+βk t. 1 → | | Then, we obtain f′(t)= 1+βk and p 1 | | f′(t)T∗p(f(t))=T (t)+βN′(t) =(1+βk )T (t)+βk B (t), 1 2 1 that is βk T∗(f(t))= 1+βk T (t)+ 2 B (t). (3.17) 1 1 | | 1+βk 1 p | | By differentiating both sides of the above equaplity with respect to t we find dT∗(t∗) βk f′(t) = 1+βk T′(t)+ 2 B ′(t) dt∗ (cid:12)(cid:12)t∗=f(t) p| 1| |1+βk1| 1 (cid:12) p (cid:12) 9 = k1(1+βk1)−βk22 N(t)+ βk2k3(t) B (t) √|1+βk1| √|1+βk1| 2 (cid:20) (cid:21) (cid:20) (cid:21) = βk2k3(t) B (t). √|1+βk1| 2 (cid:20) (cid:21) Hence, since k doesn’t vanish, we get 3 dT∗(t∗) βk k (t) k∗(f(t))= =ε 2 3 >0 where ε=sign(k13) denotes(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)thedsti∗gn o(cid:12)(cid:12)(cid:12)(cid:12)tf∗=fufn(tc)t(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)ion k31. T+hβakt1is, ε is −1 or +1. We can put 1 dT∗(t∗) N∗(t∗)= , t I. k∗(t∗) dt∗ ∈ 1 Then, we get N∗(f(t))= B (t). 2 ∓ Differentiating of the last equation with respect to t, we reach dN∗(t∗) k f′(t) = ε 3 B (t) dt∗ (cid:12)t∗=f(t) − |1+βk1| 1 (cid:12) and we have (cid:12) p (cid:12) dN∗(t∗) βk k (t) f′(t) k∗(f(t))T∗(f(t))= ε 2 3 T (t) ε 1+βk B (t). dt∗ (cid:12)(cid:12)t∗=f(t)− 1 − |1+βk1| − p| 1| 1 Since εk3(t) is(cid:12)(cid:12)positive for t I, we have p ∈ k∗(f(t))= dN∗(t∗) k∗(f(t))T∗(f(t)) 2 dt∗ t∗=f(t)− 1 = (cid:13)(cid:13)(cid:13)(cid:13)−β2k122+(kβ(cid:12)(cid:12)(cid:12)3k(1t))2 +(1+βk1)(k3(t))2 (cid:13)(cid:13)(cid:13)(cid:13) = q(k (t))2 =εk (t)>0. 3 3 Thus, we can put q B ∗(f(t))= 1 dN∗(t∗) k∗(f(t))T∗(f(t)) 1 k2∗(f(t)) dt∗ t∗=f(t)− 1 = −√|1β+k2βk1(cid:18)|T (t)−(cid:12)(cid:12)(cid:12) |1+βk1|B1(t) , t∈I(cid:19). Differentiation of the above with respecpt to t, we get dB ∗(t∗) k f′(t) 1 = 2 N(t) k (t) 1+βk B (t). dt∗ (cid:12)(cid:12)t∗=f(t) |1+βk1| − 3 p| 1| 2 (cid:12) p (cid:12) 10