Extended Darboux Frame Field in Minkowski Space-Time E

Malaya Journal of Matematik, Vol. 6, No. 3, 473-477, 2018

https://doi.org/10.26637/MJM0603/0002

Extended Darboux frame ﬁeld in Minkowski 4 space-time E1 Bahar Uyar Duld¨ ul¨ 1*

Abstract In this paper, we extend the Darboux frame field along a non-null curve lying on an orientable non-null hyper- 4 surface into Minkowski space-time E1 in two cases which the curvature vector and the normal vector of the hypersurface are linearly independent or dependent. Then the normal curvature, the geodesic curvature(s), and the geodesic torsion(s) of the hypersurface are given when the curve lying on the hypersurface is an asymptotic or geodesic curve. Keywords Curves on hypersurface, Darboux frame field, curvatures, Minkowski space-time. AMS Subject Classification 53A04, 53A07.

1Yildiz Technical University, Education Faculty, Department of Mathematics and Science Education, Istanbul, Turkey. *Corresponding author: 1 [email protected] Article History: Received 24 March 2018; Accepted 09 May 2018 c 2018 MJM.

Contents orientable non-null hypersurface in Minkowski space-time 4 E1 and call as ”extended Darboux frame field” or shortly 1 Introduction...... 473 ”ED-frame field”. After, we obtain the derivative equations of 2 Preliminaries...... 473 this frame field and give the normal curvature, the geodesic curvature(s) and the geodesic torsion(s) of the hypersurface. 4 3 Extended Darboux frame field in E1 ...... 474 We hope that this new frame field will provide the basis References...... 477 for future works to be done in this area.

1. Introduction 2. Preliminaries The most known frame fields of differential geometry are the 4 4 The Minkowski space-time E1 is the Euclidean space E pro- Frenet frame field and the Darboux frame field, and these vided with the indefinite flat metric given by frame fields occupy an important place in the study of curves and surfaces. There are many studies about generalization of 2 2 2 2 h,i = −dx1 + dx2 + dx3 + dx4, the Frenet frame in higher dimensional spaces in the literature, 4 but there is no study about generalization of Darboux frame where (x1,x2,x3,x4) is a rectangular coordinate system of E1. to higher dimensional spaces, except for the article given by Since the above metric is an indefinite metric, we know that 4 Duld¨ ul¨ et al., [2]. a vector in E1 can have one of the three causal characters: As we know, in differential geometry the Darboux frame The arbitrary vector v is called a spacelike, a timelike, and a field along a curve lying on a surface denoted by {T,V,N}, null or lightlike vector if hv,vi > 0 or v = 0, hv,vi < 0, and where T is the unit tangent vector of the curve, N is the surface hv,vi = 0 for v 6= 0, respectively. The norm of a vector v is normal restricted to the curve, and V = T × N. The normal defined by kvk = p|hv,vi| and two vectors v and w are called curvature, the geodesic curvature and the geodesic torsion orthogonal if hv,wi = 0. A vector v satisfying hv,vi = ±1 is 4 of the surface can be calculated by means of the derivative called a unit vector. For an arbitrary curve α(s) in E1, the equations of this frame field, [1,5,7,8,10]. curve is called a spacelike, a timelike and a null or lightlike In this paper, similar to given in Euclidean 4-space we curve, if all of its velocity vectors α0(s) are spacelike, timelike construct a frame field along a non-null curve lying on an and null or lightlike, respectively, [6]. 4 Extended Darboux frame field in Minkowski space-time E1 — 474/477

4 A hypersurface in the Minkowski space-time E1 is called is called n×n L-identity matrix according to L-multiplication. m a spacelike or a timelike hypersurface if the induced metric For every A ∈ Ln , Im.LA = A .LIn = A . on the hypersurface is a positive definite Riemannian metric or a Lorentzian metric, respectively. The normal vector on iii) An n × n matrix A is called L-invertible if there exists the spacelike or the timelike hypersurface is a timelike or a an n × n matrix B such that A .LB = B.LA = In. Then B spacelike vector, respectively. is called the L-inversible of A and is denoted by A −1. 4 The ternary (or vector) product of the vectors u = ∑ uiei, T n i=1 iv) The matrix A = [a ji] ∈ Lm is called the transpose of 4 4 m 4 the matrix A = [ai j] ∈ Ln . v = ∑ viei, and w = ∑ wiei in E1 is defined by i=1 i=1 n v) The matrix A ∈ Ln is called L-orthogonal matrix if −e1 e2 e3 e4 −1 T A = A ,[3]. u1 u2 u3 u4 u ⊗ v ⊗ w = − , v1 v2 v3 v4 4 3. Extended Darboux frame field in E w1 w2 w3 w4 1 Let M be an orientable non-null hypersurface, N be its non- where {e ,e ,e ,e } is the standard basis of 4. The equa- 1 2 3 4 E1 null unit normal vector field in 4, and α(s) be a non-null tions; E1 Frenet curve parametrized by arc-length parameter s lying on e1 ⊗ e2 ⊗ e3 = e4, e2 ⊗ e3 ⊗ e4 = e1, M. If the non-null unit tangent vector field of α is denoted by T, and the non-null unit normal vector field of M restricted to e3 ⊗ e4 ⊗ e1 = e2, e4 ⊗ e1 ⊗ e2 = −e3 α is denoted by N, we have α0(s) = T(s) and N(α(s)) = N(s). 4 are satisfied for the vectors ei, 1 ≤ i ≤ 4, [11]. As in Euclidean 4-space E [2], the extended Darboux Let M be an orientable non-null hypersurface and α : I ⊂ frame can be constructed in two different cases in Minkowski 4 4 {N,T, 00} R → M be a unit speed non-null curve in E1. Let {t,n,b1,b2} space-time E1 according to whether the set α is lin- be the moving Frenet frame along α. Then t, n, b1, and b2 are early independent or linearly dependent. Let us denote the the unit tangent, the principal normal, the first binormal, and ED-frame field is the first kind and the second kind if the 00 the second binormal vector fields, respectively. If k1, k2, and set {N,T,α } is linearly independent and linearly dependent, k3 are the curvature functions of the unit speed non-null curve respectively. α, then for the non-null frame vectors we have the following Now, let us construct the ED-frame field of the first kind in Frenet equations: Case 1 and the second kind in Case 2 along the non-null Frenet 4  0 curve α in E1. As explained in [2], using the Gram-Schmidt  t = εnk1n,  0 orthonormalization method, we have n = −εtk1t + εb1 k2b1, b0 = −ε k n − ε ε ε k b , α00 − hα00,NiN  1 n 2 t n b1 3 2 E =  0 00 00 b2 = −εb1 k3b1, ||α − hα ,NiN|| where εt = ht,ti, εn = hn,ni, εb1 = hb1,b1i, εb2 = hb2,b2i for Case 1 and 000 000 000 whereby εt,εn,εb1 ,εb2 ∈ {−1,1}, 1 ≤ i ≤ 4 and εtεnεb1 εb2 = α − hα ,NiN − hα ,TiT E = −1, [4]. ||α000 − hα000,NiN − hα000,TiT|| m n Definition 2.1. i) Let A = [ai j] ∈ Rn and B = [b jk] ∈ Rp be for Case 2. If we define −D = N ⊗ T ⊗ E for both cases, we m n two matrices, where Rn and Rp are the real vector spaces obtain the orthonormal frame field {T,E,D,N} another from with the matrix addition and the scalar-matrix multiplica- Frenet frame field {T,n,b1,b2} along the curve α. With re- tion. Then, the Lorentzian matrix multiplication or shortly spect to the orthonormal frame {T,E,D,N}, the vector fields L-multiplication of the matrices A and B is defined by T0,E0,D0,N0 have the following decompositions: n 0 0 0 0 . = − a b + a b . T = ε1 hT ,TiT + ε2 hT ,EiE + ε3 hT ,DiD A LB i1 1k ∑ i j jk 0 j=1 +ε4 hT ,NiN, 0 0 0 0 m E = ε1 hE ,TiT + ε2 hE ,EiE + ε3 hE ,DiD The real vector space Rn with L-multiplication is denoted by 0 m +ε4 hE ,NiN, n . 0 0 0 0 L D = ε1 hD ,TiT + ε2 hD ,EiE + ε3 hD ,DiD 0 +ε4 hD ,NiN, ii) The matrix 0 0 0 0 N = ε1 hN ,TiT + ε2 hN ,EiE + ε3 hN ,DiD 0  −1 0 ... 0  +ε4 hN ,NiN,  0 1 ... 0  =   where ε1 = hT,Ti, ε2 = hE,Ei, ε3 = hD,Di, ε4 = hN,Ni In  . . .   . . .  whereby ε1,ε2,ε3,ε4 ∈ {−1,1}. Besides, when εi = −1, then 0 0 ... 0 ε j = 1 for all j 6= i, 1 ≤ i, j ≤ 4 and ε1ε2ε3ε4 = −1.

474 4 Extended Darboux frame ﬁeld in Minkowski space-time E1 — 475/477

Similar operations are performed in [2], we obtain hT0,Di = Proof. Since β is the orthogonal projection curve onto the 0 0 0 0 for Case 1 and hT ,Ei = hT ,Di = 0 and hN ,Di = 0 for tangent hyperplane at α(s0) of α, we can write 0 0 1 Case 2. If we use hT ,Ni = κn and denote hE ,Ni = τg , 0 2 0 1 0 2 i i β(s) = α(s) − hα(s) − α(s0),N(s0)iN(s0). hD ,Ni = τg , hT ,Ei = κg , hE ,Di = κg , where κg and τg are the geodesic curvature and the geodesic torsion of order If we differentiate both sides of this equation three times i, (i = 1,2), respectively, then the differential equations for according to s, we find ED-frame field have the form for Case 1: 0 0  0   1   β (s0) = α (s0) = T(s0), T 0 ε2κg 0 ε4κn T 0 1 2 1  E   −ε1κg 0 ε3κg ε4τg  E  β 00(s ) = T0(s ) = ε κ1(s )E(s ),  0  =  2 2  , 0 0 2 g 0 0  D   0 −ε2κg 0 ε4τg  D  0 1 2 β 000(s ) = −ε ε (κ1)2(s ) − ε ε (κ )2(s ) T(s ) N −ε1κn −ε2τg −ε3τg 0 N 0 1 2 g 0 1 4 n 0 0 1 0 1 + ε2(κg ) (s0) − ε2ε4κn(s0)τg (s0) E(s0) and for Case 2: 1 2 2 + ε2ε3κg (s0)κg (s0) − ε3ε4κn(s0)τg (s0) D(s0)  0     T 0 0 0 ε4κn T (s ) = (s ) 0 2 1 at the point β 0 α 0 . Then, we obtain  E   0 0 ε3κg ε4τg  E   0  =  2  . 0 00 000  D   0 −ε2κg 0 0  D  β (s0) ⊗ β (s0) ⊗ β (s0) 0 1 b2(s0) = εb1 0 00 000 = (0,0,0,εb1 ε2), N −ε1κn −ε2τg 0 0 N ||β (s0) ⊗ β (s0) ⊗ β (s0)|| 0 00 Now, let us consider the normal curvature, the geodesic b2(s0) ⊗ β (s0) ⊗ β (s0) curvatures, and the geodesic torsions of the curve α and let us b1(s0) = −εn 0 00 = (0,0,εnεb1 ,0), ||b2(s0) ⊗ β (s0) ⊗ β (s0)|| give the geometrical results of these real valued functions. In 0 0 b (s ) ⊗ b (s ) ⊗ β (s ) both cases, we know that κn = hT ,Ni is the normal curvature 1 0 2 0 0 n(s0) = 0 = (0,εnε2,0,0), of the hypersurface in the direction of the tangent vector T. ||b1(s0) ⊗ b2(s0) ⊗ β (s0)|| Therefore, α is an asymptotic curve if and only if κn = 0 and along α. hn(s ),β 00(s )i kβ (s ) = 0 0 = 1(s ). Theorem 3.1. Let us consider a unit speed non-null curve 1 0 0 2 εnκg 0 ||β (s0)|| α on an orientable non-null hypersurface M in Minkowski 4 space-time E1. Let M1 and M2 be the non-null hyperplanes at α(s ) ∈ M determined by {T(s ),D(s ),N(s )} and {T(s ), 0 0 0 0 0 Theorem 3.3. Let α be a unit speed non-null asymptotic E(s0),N(s0)}, respectively. Denoting the transversal intersec- 4 curve on an orientable non-null hypersurface M in E1. Let us tion curve of M1, M2, and M with β, then the first curvature β β denote the non-null orthogonal projection curve of α onto the k (s ) of β at the point β(s ) is given by k (s ) = |κn(s )|, 1 0 0 1 0 0 non-null hyperplane determined by {T(s0),E(s0),N(s0)} at where κn is the normal curvature of the hypersurface M in the γ α(s0) with γ. Then the first curvature k1(s0) of the projection direction of T. γ 1 curve γ at the point γ(s0) is given by k1(s0) = εnκg (s0). Proof. According to the transversal intersection of three hy- Proof. If we write persurfaces, T is the tangent vector of the intersection curve β. Using the similar calculations in [9], since the normal vectors γ(s) = α(s) − hα(s) − α(s0),D(s0)iD(s0) of the hypersurfaces are orthogonal at the point β(s0), for the first curvature of β at β(s0) we find and do the similar calculations at the proof of Theorem 3.2, q we find the desired result. kβ (s ) = |ε (κ1)2(s ) + ε (κ2)2(s ) + ε (κ3)2(s )|, 1 0 2 n 0 3 n 0 4 n 0 Now, let us take into consideration the non-null Frenet 1 0 2 0 3 0 frame {T,n,b1,b2} along the non-null curve α. Since n, b1, where κn = hT ,Ei, κn = hT ,Di, κn = hT ,Ni. Then, we obtain b2, E, D, N are orthogonal to T, we can write q β 2 Y = A .LX, (3.1) k1 (s0) = |ε4(κn) (s0)|.

β where Since ε4 = ±1, we get k1 (s0) = |κn(s0)|.  n   E  Theorem 3.2. Let α be a unit speed non-null curve on an Y =  b1 , X =  D , orientable non-null hypersurface M in 4. Let us denote the E1 b2 N non-null orthogonal projection curve of α onto the non-null (3.2) tangent hyperplane at α(s ) with β. Then the ﬁrst curvature   0 sinhφ1 sinhφ2 sinhφ3 β k1 (s0) of the projection curve β at the point β(s0) is equal to A =  sinhψ1 sinhψ2 sinhψ3  1 1 εnκg (s0), where κg is the geodesic curvature of order 1 of M. sinhθ1 sinhθ2 sinhθ3

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Since the matrix A is an L − orthogonal matrix, we may write we obtain

T 2 0 0 I3.LX = A .LY, κg = hE ,Di = φ1 coshφ1 sinhφ2εn 0 0 +ψ coshψ1 sinhψ2εb + θ coshθ1 sinhθ2εb 1 1 1 2 (3.8) where +k2 sinhψ1 sinhφ2 − sinhφ1 sinhψ2  −1 0 0  +k3 sinhψ1 sinhθ2 − sinhθ1 sinhψ2 I3 =  0 1 0  . and 0 0 1 3×3 1 0 0 0 τg = hE ,Ni = φ1 coshφ1 sinhφ3εn + ψ1 coshψ1 sinhψ3εb1 Then we have 0 +θ coshθ1 sinhθ3εb + k2 sinhψ1 sinhφ3 − sinhφ1 sinhψ3 1 2 E = −sinhφ1n + sinhψ1b1 + sinhθ1b2, +k3 sinhψ1 sinhθ3 − sinhθ1 sinhψ3 . D = −sinhφ2n + sinhψ2b1 + sinhθ2b2, (3.3) (3.9) N = −sinhφ3n + sinhψ3b1 + sinhθ3b2. Substituting (3.6) and (3.7) into (3.8) and (3.9), we get 0 Therefore, if we use Frenet formula T = εnk1n and (3.3), we 2 get κg = −k3 1 0 κg = T ,E = −k1 sinhφ1 (3.4) and and 1 τg = −k2. 0 κn = T ,N = −k1 sinhφ3, (3.5) where k1 is the ﬁrst curvature of α. Theorem 3.5. Let α be a unit speed non-null curve with arc- length parameter s on an orientable non-null hypersurface M Theorem 3.4. Let α be a unit speed non-null curve with arc- in 4. If is an asymptotic curve on M, then length parameter s on an orientable non-null hypersurface M E1 α 4 in E1. If α is a geodesic curve on M, then 1 2 1 κg = k1, κg = k2 sinhψ2, τg = k2 sinhψ3, 2 1 κn = −k1, κg = −k3, τg = −k2, τ2 = ψ0 coshψ sinhψ ε + θ 0 coshθ sinhθ ε + where k (i = 1,2,3) denotes the i-th curvature functions of α. g 2 2 3 b1 2 2 3 b2 i k3 sinhψ2 sinhθ3 − sinhθ2 sinhψ3 , Proof. Since α is a geodesic curve, by the proper orientation of the hypersurface with N(s) = n(s), Case 2 is valid. In this case, E and D coincide with b1 and b2, respectively. So, where ki (i = 1,2,3) denotes the i-th curvature functions of α. the frame {T,E,D,N} coincides with the frame {T,b1,b2,n}. Using (3.1) and (3.2), since Proof. Since α is an asymptotic curve, then κn = 0. In this 0 1 case, T = εnk1n = ε2κg E, i.e. n and E are linearly dependent. hn,Ei = 0, hn,Di = 0, hn,Ni = ε4, So Case 1 is valid. Using (3.1) and (3.2), since hb1,Ei = ε2, hb1,Di = 0, hb1,Ni = 0, hn,Ei = −ε sinhφ , hn,Di = 0, hb2,Ei = 0, hb2,Di = ε3, hb2,Ni = 0 2 1 hb1,Ei = 0, hb1,Di = ε3 sinhψ2, we obtain hb2,Ei = 0, hb2,Di = ε3 sinhθ2,

φ1(s) = φ2(s) = ψ2(s) = ψ3(s) = θ1(s) = θ3(s) = 0 (3.6) hn,Ni = 0, and hb1,Ni = ε4 sinhψ3, hb2,Ni = ε4 sinhθ3 sinhφ3(s) = −sinhψ1(s) = sinhθ2(s) = 1 (3.7) we have along α. Substituting (3.7) into (3.5), we ﬁnd

sinhφ1(s) = −1, φ2(s) = φ3(s) = ψ1(s) = θ1(s) = 0 (3.10) κn = −k1.

On the other hand, since along α. Substituting (3.10) into (3.4), (3.8) and (3.9) yield 0 0 1 = k , E = k1εT sinhφ1T + − φ1 coshφ1 − k2εn sinhψ1 n κg 1 0 + ψ1 coshψ1 − k2εb1 sinhφ1 − k3εb1 sinhθ1 b1 0 κ2 = k sinhψ , + θ1 coshθ1 − k3εTεnεb1 sinhψ1 b2 g 2 2

476 4 Extended Darboux frame ﬁeld in Minkowski space-time E1 — 477/477 and ????????? ISSN(P):2319 − 3786 1 τg = k2 sinhψ3. Malaya Journal of Matematik ISSN(O):2321 − 5666 Besides, since ????????? 0 0 D = k1εT sinhφ2T + − φ2 coshφ2 − k2εn sinhψ2 n 0 + ψ2 coshψ2 − k2εb1 sinhφ2 − k3εb1 sinhθ2 b1 0 + θ2 coshθ2 − k3εTεnεb1 sinhψ2 b2 we get

2 0 0 τg = hD ,Ni = φ2 coshφ2 sinhφ3εn 0 +ψ2 coshψ2 sinhψ3εb1 0 +θ coshθ2 sinhθ3εb 2 2 +k2 sinhψ2 sinhφ3 − sinhφ2 sinhψ3 +k3 sinhψ2 sinhθ3 − sinhθ2 sinhψ3 . Using (3.10) yields 2 0 τg = ψ2 coshψ2 sinhψ3εb1 0 +θ coshθ2 sinhθ3εb 2 2 +k3 sinhψ2 sinhθ3 − sinhθ2 sinhψ3 .

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