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SELECCIONES MATEMATICAS´ Universidad Nacional de Trujillo ISSN: 2411-1783 (Online) Vol. 05(02): 137 - 153 (2018)

Transversal Intersection Curves of Two Surfaces in Minkowski 3-Space

Curvas dadas por la interseccion´ transversal de dos superficies en el espacio tridimesional de Minkowski Osmar Alessio´ ∗, Sayed A.-N. Badr†, Soad A. Hassan‡, Luciana A. Rodrigues §,Fabio´ N. ¶ k Silva , and M. A. Soliman Received, Aug. 21, 2018 Accepted, Dec. 03, 2018

DOI: http://dx.doi.org/10.17268/sel.mat.2018.02.02

Abstract In this paper, we study the differential geometry of the transversal intersection curve of two surfaces in Minkowski 3-space, where each pair satisfies the following types spacelike-lightlike, timelike-lightlike and lightlike-lightlike. Surfaces are generally give by their parametric or implicit equations, then the surface- surface intersection problem appear commonly as parametric-parametric, parametric-implicit and implicit- implicit. We derive the Frenet frame, Darboux frame, curvature, torsion, normal curvature and geodesic curvatures of transversal intersections for all types of intersection problems. We show the intersection curve may be spacelike (timelike, lightlike or pseudo null) curve. Finally, we show our methods by given several examples. Keywords. Minkowski 3-Space; Surface-Surface intersection; Pseudo null curve; Null curve; Null frame; Light- like surface; Darboux Frame.

Resumen En este art´ıculo, estudiamos la geometr´ıa diferencial de la curva dada por la interseccion´ transversal de dos superficies en el espacio tridimensional de Minkowski donde cada par satisface los siguientes tipos de superficies; tipo espacio - tipo luz, tipo tiempo - tipo luz y tipo luz - tipo luz. Generalmente, las superficies estan´ dadas por sus ecuaciones parametricas´ o impl´ıcitas, entonces el problema de interseccion´ superficie- superficie aparece comunmente como parametrico-param´ etrico,´ parametrico-impl´ ´ıcito e impl´ıcito-impl´ıci- to. Obtenemos el Referencial de Frenet, el Referencial de Darboux, la curvatura, la torsion,´ la curvatura normal y las curvaturas geodesicas´ de las intersecciones transversales para todos los tipos de problemas de interseccion.´ Mostramos que la curva de interseccion´ puede ser una curva similar a una curva tipo espacio ( tipo tiempo, tipo luz o pseudo nula). Finalmente, mostramos nuestros metodos´ por varios ejemplos. Palabras clave. Espacio tridimensional de Minkowski; Interseccion´ Superficie-superficie; Curva pseudo nula; Curva nula; Referencial nulo; Superficie tipo luz; Referencial de Darboux.

1. Introduction. Lorentz geometry plays an important role in the transition between modern differen- tial geometry and the mathematical physics of general relativity by giving an invariant treatment of Lorentz geometry. The fact that relativity theory is expressed in terms of Lorentz geometry is lucky for geometers, who can thus penetrate surprisingly quickly into cosmology (redshift, expanding universe, and big bang)

∗Instituto de Cienciasˆ Exatas, Naturais e Educac¸ao˜ - UFTM, Uberaba, MG, Brazil ([email protected]). †Dept. of Math. Al-Qunfudah University College - Umm Al-Qura University, Al-Qunfudah, KSA (sayed [email protected]). ‡Dept. of Math. Al-Qunfudah University College - Umm Al-Qura University, Al-Qunfudah, KSA ([email protected]). §Departamento de Matematica´ - Universidade de Bras´ılia , Bras´ılia, DF, Brazil ([email protected]). ¶ Centro das Cienciasˆ Exatas e das Tecnologias - UFOB, Barreiras, BA, Brazil ([email protected]). k Dept. of Math. Faculty of Science - Assiut University, Assiut, Egypt (m [email protected])...... This work is licensed under the Creative Commons Attribution-NoComercial-ShareAlike 4.0. 137 138 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) and, a topic no less interesting geometrically, the gravitation of a single star (perihelion precession, bending of light, and black holes) [19]. There exist four types of curves; spacelike, timelike, pseudo null, and null curves depending on their causal characters. However, the fact that the induced metric on a null curve to be degenerate leads to a much more complicated study and also different from the non degenerate case. In this geometry of null curves difficulties arise because the arc length vanishes, so it is not possible to normalize the tangent vector in the usual way. A method of proceeding is to introduce a new parameter called the pseudo-arc which normalizes the derivative of tangent vector. The primary difference between the lightlike surface and the non-degenerate is that in the first case the normal vector of surface can intersect the tangent space, in other words, a vector of the tangent space TxM can not be uniquely decomposed into ⊥ a component tangent to TxM and a component of normal space TxM . As a result, the lightlike geometry is quite different from Riemannian and semi-Riemannian geometry. From the point of view of physics, null curves and lightlike surfaces are important because they are models of different types and horizons studied in relativity theory. The intersection of two surfaces can be spacelike, timelike and null curves. The geometric properties of curves and surfaces in the classical literature on differential geometry in Euclidean 3-space R3, can be found in [11]. There is no textbook with a systematic study of curves and surfaces in Minkowski spaces such as it occurs in the Euclidean spaces. A general reference including many topics in semi-Riemannian geometry is the classical book [19]. Differential geometry of the intersection curves in R3 and R4 can be found in [14, 26, 23,1, 15,7,2,4,8,5,3]. Some results on differential geometry of the intersection curves for the transversal intersections in 3 4 Minkowski 3-space R1, and Minkowski 4-space R1, can be found in [6, 22, 16] and [13] respectively. Alessio´ and Guadalupe [6] studied the transversal intersection curve of two parametric spacelike surfaces 3 in R1 considering parametric-parametric intersection problem. Zafer and Yusuf [22] studied intersection 3 curve of two parametric timelike surfaces in R1. Karaahmetoglu and Aydemir [16] studied intersection 3 curve of two parametric spacelike and timelike surfaces in R1.Duld¨ ul¨ and C¸alıs¸kan [13] compute the Frenet vectors and the curvatures of the spacelike intersection curve of three spacelike hypersurfaces given 4 by their parametric equations in 4-dimensional Minkowski space R1. Bonnor [10] described the geometry of null curves in a Minkowski space-time and he proved the fundamental existence and congruence theorems. He introduced the Cartan frame as the most useful one and he uses this frame to study the behaviors of a null curve. Bejancu [9] gave a method for general study of the geometry to null curves in Lorentz manifolds and, more generally, in semi-Riemannian manifolds. 3 Some notations, definitions and reviews of some aspects of the differential geometry in R1, can be found in [19, 17, 18, 20, 21, 25, 12, 24]. This paper presents algorithms for computing the null frame of a given null or pseudo null curve in 3 Minkowski 3-space R1. Also, it presents the Darboux frame for a spacelike (timelike or null) curve lies on a spacelike (timelike or lightlike) surface. Moreover, it presents algorithms for computing all the differential geometry properties of the transversal intersection curves of two surfaces (spacelike, timelike and lightlike) 3 in Minkowski 3-space R1. Surfaces are generally given by their parametric or implicit equations, then the surface–surface intersection problem appears commonly as parametric-parametric, parametric-implicit and implicit-implicit. We derive the Frenet frame, Darboux frame, curvature, torsion, normal curvature and geodesic curvatures for transversal intersections for all types of the intersection problems. The intersection curve may be spacelike (timelike, null or pseudo null) curve. We obtain the transversal intersection case at a point, when the normal vector fields are linearly independent. 3 3 2. Preliminaries. Minkowski 3-space, R1 = (R , h, i1) is a three dimensional real vector space equipped with a Lorentz metric of signature (2, 1). For the two vectors u = (x1, x2, x3), v = (y1, y2, y3) ∈ 3 R1, the metric is given by

hu, vi1 = −x1y1 + x2y2 + x3y3.

A vector u 6= 0 is called spacelike, timelike or lightlike(null), if hu, ui1 > 0, hu, ui1 < 0 or hu, ui1 = 0, respectively. In particular the vector u = 0 is called spacelike. The norm of the vector u is defined by,

1 2 2 p kuk1 = |hu, ui1| = |−x1x1 + x2x2 + x3x3|.

The vector u is called a unit vector if hu, ui1 = ±1. The vectors u and v are said to be orthogonal (u⊥v) , 3 if hu, vi1 = 0. Similarly, two subsets U and W of R1 are said to be orthogonal (U⊥W ), if u⊥w for any ⊥ 3 u ∈ U and w ∈ W. Denote W = U = {w ∈ R1 : hw, ui1 = 0, ∀u ∈ U}. The vector product of u and 3 v (in that order) is the unique vector u ×1 v ∈ R1, where

−e1 e2 e3

u ×1 v = −v ×1 u = x1 x2 x3 = (x3y2 − x2y3, x3y1 − x1y3, x1y2 − x2y1),

y1 y2 y3 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) 139

3 where {e1, e2, e3} is the canonical basis of R1, which satisfy e1 ×1 e2 = −e3, e2 ×1 e3 = e1 and e3 ×1 e1 = −e2. 3 3 Proposition 1. [18] Let R1 = (R , h, i1), then • Two null (lightlike) vectors are linearly dependent if and only if they are orthogonal. • Two timelike vectors are never orthogonal. • A timelike vector is never orthogonal to a null (lightlike) vector. 3 Proposition 2. [18] For u, v, w ∈ R1 we have •h u ×1 v, ui1 = 0 and hu ×1 v, vi1 = 0. 2 •h u ×1 v, u ×1 vi1 = hu, vi1 − hu, ui1 hv, vi1 . • Let u be a spacelike vector, v be a null vector, then hu, vi1 6= 0 if and only if u ×1 v is spacelike. Also hu, vi1 = 0 if and only if u ×1 v is null. • If u and v are null vectors, then u ×1 v is a spacelike vector. • If u is a timelike vector, v is a null vector, then u ×1 v is spacelike vector. 3 Definition 1. A basis {u1, u2, u3} of R1 is orthonormal if hui, uji1 = δij, where

 ±1 if i = j, δ = ij 0 if i 6= j.

3 If u1 and u2 are non-null vectors, then B = {u1, u2, u1 ×1 u2} is a basis of R1. However, in contrast to the Euclidean space, the casual character of u1 and u2 determines if the basis is or is not positively oriented. Exactly, if u1, u2 are spacelike, then u1 ×1 u2 is timelike and B is negatively oriented because det(u1, u2, u1 ×1 u2) = hu1 ×1 u2, u1 ×1 u2i1 < 0. If u1 and u2 have different causal character, then B is positively oriented. 3 Given U ⊂ R1 a vector subspace, we consider the induced metric < u, v >U =< u, v >1 such that u, v ∈ U. If the metric in U is positive definite it is called spacelike subspace , if the metric in U has index 1 it is called timelike subspace and if the metric in U is degenerate it is called lightlike subspace . 3 Proposition 3. [18] Let P ⊂ R1 be a vector plane. Denote by N an orthogonal vector with respect to the Euclidean metric, then P is a spacelike (resp. timelike, lightlike) plane if and only if N is a timelike (resp. spacelike, lightlike) vector.

3 3 3. Differential Geometry of Curves in R1. A regular curve α(s): I ⊂ R → R1 can be locally spacelike, timelike or null (lightlike), if its velocity vector field (α0(s)) is spacelike, timelike or null (light- like), respectively. In particular, if α0(s) is spacelike and the acceleration vector field (α00(s)) is null, in this case the curve is called pseudo null curve. The notations for differentiation of the curve α(s) with respect to 0 dα 00 d2α 000 d3α (n) dnα the arc length (s) are α (s) = ds , α (s) = ds2 , α (s) = ds3 and α (s) = dsn , (n > 3). A non-null 0 0 curve α(s) is said to be parametrized by arc length parameter s, if hα (s), α (s)i1 = ±1. In this case, the curve is called a unit speed curve. A non-null curve α(s) is a straight-line, if α00(s) = 0. A null (lightlike) curve α(s) is said to be pseudo-parametrized by arc length (or distinguished) if α(s) is not a straight-line 00 00 00 00 and hα (s), α (s)i1 = 1. A null (lightlike) curve α(s), is a straight-line if hα (s), α (s)i1 = 0. 3 Definition 2. [12] The null frame F = {t, n, b} at a point of the null (or pseudo null) curve in R1 is a positively oriented 3-tuple of vectors satisfying, respectively: (3.1)  ht, ti = hb, bi = ht, ni = hn, bi = 0, hn, ni = ht, bi = 1, t ×1 b = n, n ×1 t = t,  1 1 1 1 1 1 n o 1 hu,ui1  b ×1 n = b, where b = u − t , u ∈Tα(s)M, hu, ni = 0 and hu, ti 6= 0, hu,ti1 2hu,ti1 1 1 or (3.2)  hn, ni = hb, bi = ht, ni = ht, bi = 0, ht, ti = hn, bi = 1, t ×1 n = n, n ×1 b = t,  1 1 1 1 1 1 n o 1 hu,ui1  b ×1 t = b, where b = u − n , u ∈Tα(s)M, hu, ni 6= 0 and hu, ti = 0. hu,ni1 2hu,ni1 1 1

3 3.1. Frenet Frame. The Frenet frame {t, n, b} is a basis of R1 whose assign for each point of a regular curve α(s), whose variation describes the geometry of the curve. In Euclidean space, the Frenet frame is a positive oriented orthonormal basis, with b = t ×1 n. However it has some differences in Minkowski 3-space. 140 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018)

3.1.1. Spacelike and Timelike Curves. The Frenet frame {t, n, b} of a spacelike (or a timelike) 00 00 regular curve with arc length parametrization, α(s) and hα (s), α (s)i1 6= 0 is given by

00 0 00 0 α (s) α (s)×1α (s) (3.3) t = α (s) n = 00 b = t ×1 n = 0 00 . kα (s)k1 kα (s)×1α (s)k1 The curvature κ(s) and the torsion τ(s), are given as follows

α000(s),b 00 h i1 (3.4) κ(s) = kα (s)k , τ(s) = −2 00 . 1 kα (s)k1 The Frenet formulas are given as follows

 0      t 0 2κ 0 t 0 (3.5)  n  =  −1κ 0 −3τ   n  , 0 b 0 2τ 0 b where 1 = ht, ti1, 2 = hn, ni1, 3 = hb, bi1. 3 Let β : J ⊂ R → R1 be a regular curve parametrized by parameter u, with the same trace of the curve s Z u dβ(τ) dβ(τ) ds dβ(u) dβ(u) α(s), i.e. β(u) = α(s(u)), s(u) = 1 , dτ and = 1 , . u0 du du 1 du du du 1 Thus,

2 3 dβ 2 d β d2s 3 d β ds d2s d3s dα d α 2 − 2 t d α 3 − 3 2 n + 3 t (3.6) t = = du , n = = du du , = du du du du . ds ds ds2 ds
2 ds3 ds
3 du du du 3.1.2. Pseudo Null Curves. Let α(s) = (x(s), y(s), z(s)) be a regular pseudo null with unit speed curve, such that α0(s), α00(s), α000(s) are different from zero for all s, then there exists only one Frenet 0 00 000 frame (null frame) {t, n, b} have the same orientations with {α (s), α (s), α (s)}, such that hn, ni1 = hb, bi1 = ht, ni1 = ht, bi1 = 0, ht, ti1 = hn, bi1 = 1, t ×1 n = n, n ×1 b = t, b ×1 t = b. This frame 3 is a pseudo-orthonormal basis in R1 and it can be constructed as the following, Let t = α0(s) = (x0(s), y0(s), z0(s)), n = α00(s) = (x00(s), y00(s), z00(s)) and V = (1, u, v), such that

ht, ti1 = 1, hn, ni1 = 0, ht, ni1 = 0, hV, ti1 = 0, hV, ni1 6= 0. Thus, we have

2 2 x02 = y02 + z0 − 1, x002 = y002 + z00 , x0x00 = y0y00 + z0z00, (3.7) x0 = uy0 + vz0, x00 6= uy00 + vz00.

Since n is a null vector, then x00 6= 0, then we can put

(3.8) uy00 + vz00 = 0.

Solving (3.7) and (3.8) for the u, get

x0z00 −x0y00 u = y0z00−z0y00 , v = y0z00−z0y00 . Solving (3.7) for x00, we obtain

x00 = ± (y0z00 − z0y00) .

Thus, the vector V is given by x0z00 −x0y00 (y0z00 − z0y00) V = (1, , ); δ = δx00 δx00 x00 Using (3.2), we get δ (y0z00 − z0y00) b = − ((2δx00, 2x0z00, −2x0y00) + δ x02 − 1
α00); δ = , 2x002 x00 and the null frame is given by

0 00 δ 00 0 00 0 00 02
(3.9) t = α , n = α , b = − 2x002 [(2δx , 2x z , −2x y ) + δ x − 1 n], O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) 141 where (y0z00 − z0y00) δ = x00 The Frenet formulas are given by

 t0   0 1 0   t  (3.10)  n0  =  0 −τ 0   n  . b0 −1 0 τ b

In this case the function τ is called pseudo-torsion along the curve α(s) which is given by

000 (3.11) τ(s) = − hα , bi1 .

3 Let β : J ⊂ R → R1 be a regular curve parametrized by parameter u, with the same trace of the pseudo s Z u dβ(τ) dβ(τ) ds dβ(u) dβ(u) null curve α(s), i.e. β(u) = α(s(u)), s(u) = , dτ and = , . u0 du du 1 du du du 1 Thus,

2 3 dβ 2 d β d2s 3 d β ds d2s d3s dα d α 2 − 2 t d α 3 − 3 2 n + 3 t (3.12) t = = du , n = = du du , = du du du du . ds ds ds2 ds
2 ds3 ds
3 du du du

3 0 3.1.3. Null curves. Let α(s) be a null curve in R1 parametrized with pseudo arc length s and α (s), α00(s), α000(s) are linearly independent for all s, then there exists only one Frenet frame (null frame), 0 00 000 {n, t, b} have the same orientations with {α (s), α (s), α (s)}, such that, ht, ti1 = hb, bi1 = ht, ni1 = hn, bi1 = 0, hn, ni1 = ht, bi1 = 1, t ×1 b = n, n ×1 t = t, b ×1 n = b. This frame is a pseudo- 3 orthonormal basis in R1 and it can be constructed as previous section 3.1.2, then the null frame is given by

0 00 σ 0 00 0 00 0 002
t = α , n = α , b = − 2x02 ((2σx , 2x z , −2x y ) + σ x − 1 t), where (y00z0 − z00y0) σ = . x0 The Frenet formulas are given by

 n0   0 −τ −1   n   t0  =  1 0 0   t  . b0 τ 0 0 b

In this case, the function τ is called pseudo-torsion along the curve α(s) which is given by

000 (3.13) τ(s) = − hα , bi1 .

3 Let β : J ⊂ R → R1 be a regular curve parametrized by parameter u, with the same trace of the null s Z u d2β(τ) d2β(τ) ds d2β(u) d2β(u) curve α(s), i.e. β(u) = α(s(u)), s(u) = 2 , 2 dτ and = 2 , 2 . u0 du du 1 du du du 1 Thus,

2 3 dβ 2 d β d2s 3 d β ds d2s d3s dα d α 2 − 2 t d α 3 − 3 2 n + 3 t (3.14) t = = du , n = = du du , = du du du du . ds ds ds2 ds
2 ds3 ds
3 du du du

ds Notice that, if du = 0 then the curve is a straight-line. 3 3 4. Curves on Surfaces in R1. A surface in R1 is called spacelike, timelike or lightlike, if the induced metric on the surface is positive definite, has index 1 or degenerate respectively. In other words, the surface 3 in R1 is called spacelike, timelike or lightlike, if the normal vector field on the surface is timelike, spacelike or lightlike. For more details, see [19]. 142 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018)

3 4.1. Curves on Parametric Surfaces in R1. Consider an arbitrary parametric spacelike (timelike or lightlike) surface X(u1, u2), the vector

(4.1) ζ =X1 × X2 ∂X (where Xr = (r = 1, 2) ) is the normal vector field of the surface X. It is regular if ζ 6= 0. The ∂ur coefficients of the first fundamental form are given by

gij = hXi, Xji1 ; i, j = 1, 2 .

Let ur = ur(u), r = 1, 2 be functions in the u1u2-plane which defines a regular curve parametrized by an arbitrary parameter u on the surface X, as follows

(4.2) γ(u) = X(u1(u), u2(u)) = (x1(u), x2(u), x3(u)).

Differentiating (4.2) three times with respect to the parameter u, then projecting the vectors γ0(u) and γ00(u), γ000(u) onto the normal vector field ζ of the surface X, we get

0 hγ (u), ζi1 = 0,

00 0 2 0 0 0 2 (4.3) hγ (u), ζi1 = hX11, ζi1 (u1) + 2 hX12, ζi1 u1u2 + hX22, ζi1 (u2) ,

(4.4) 000 0 3 0 3 0 2 0 0 0 2 hγ (u), ζi1 = hX111, ζi1(u1) + hX222, ζi1(u2) + 3hX112, ζi1(u1) u2 + 3hX122, ζiu1(u2)

0 0 00 0 0 00 +3(hX11, ζi1 u1 + hX12, ζi1 u2)u1 + 3(hX12, ζi1 u1 + hX22, ζi1 u2)u2 ,

2 3 ∂ X ∂ X 0 00 where Xij = , Xijk = (i, j, k = 1, 2). Projecting the vectors γ (u), γ (u) and ∂ui∂uj ∂ui∂uj∂uk 000 γ (u) onto X1 and X2, we obtain

0 0 0 0 0 0 (4.5) g11u1 + g12u2 = hγ (u), X1i1, g12u1 + g22u2 = hγ (u), X2i1,

(4.6)  g u00 + g u00 = hγ00(u), X i − hX , X i (u0 )2 − 2 hX , X i u0 u0 − hX , X i (u0 )2  11 1 12 2 1 1 11 1 1 1 12 1 1 1 2 22 1 1 2  00 00 00 0 2 0 0 0 2 g12u1 + g22u2 = hγ (u), X2i1 − hX11, X2i1 (u1) − 2 hX12, X2i1 u1u2 − hX22, X2i1 (u2) .

3 4.1.1. Curves on Parametric Spacelike and Timelike Surfaces in R1. The unit normal vector field N of a parametric spacelike (or a timelike) surface X is given by

ζ N = . kζk1 Solving (4.5) and using proposition2, we get

0 0 0 hγ (u) ×1 X2, ζi1 0 hX1 ×1 γ (u), ζi1 (4.7) u1 = . u2 = . hζ, ζi1 hζ, ζi1

Solving (4.6) and using proposition2, we get (4.8)  hγ00(u) × X , ζi − hX × X , ζi (u0 )2 − 2 hX × X , ζi u0 u0 − hX × X , ζi (u0 )2  00 1 2 1 11 1 2 1 1 12 1 2 1 1 2 22 1 2 1 2  u1 =  hζ, ζi1 .  −hγ00(u) × X , ζi + hX × X , ζi (u0 )2 + 2 hX × X , ζi u0 u0 + hX × X , ζi (u0 )2  0 1 1 1 11 1 1 1 1 12 1 1 1 1 2 22 1 1 1 2  u2 = hζ, ζi1 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) 143

3 4.1.2. Curves on Parametric Lightlike Surfaces in R1. The normal vector field ζ of a parametric lightlike surface X is given by (4.1). If Xi is lightlike, Xj is spacelike and g12 = 0, then by using the proposition2, we obtain

ζ = Xi, where {i, j} = {1, 2}. Thus the normal vector field ζ of a parametric lightlike surface X is given by  X kX × X k = kX k = 0, i = 1, 2,  i if 1 2 1 i 1 (4.9) ζ =  X1 × X2 if kX1 × X2k1 = 0. In this case (4.7) will reduced to one equation or vanish. Since the vector γ0(u) can be written as,

1 0 1 0 0 2 0 2 0 0 3 0 3 0 0 (4.10) ϑ u1 + ς u2 = x , ϑ u1 + ς u2 = y , ϑ u1 + ς u2 = z , 0 0 0 0 1 2 3 1 2 3 0 0 where γ (u)= (x , y , z ), X1 = (ϑ , ϑ , ϑ ) and X2 = (ς , ς , ς ), then we can compute u1 and u2 by 00 00 solving two suitable equations of (4.7) and (4.10). By the same way we can compute u1 and u2 . 3 4.2. Curves on Implicit Surfaces in R1. Consider an arbitrary implicit spacelike (timelike or light- like) surface

S = {(x1, x2, x3); h(x1, x2, x3) = 0}, 3 where h(x1, x2, x3): R1 −→ R is a differential function. The vector field ∂h (4.11) ζh = (−h1, h2, h3); hi = , i = 1, 2, 3, ∂xi is orthogonal to any vector of the tangent plane. We assume that the surface is regular, in other words ζh 6= 0. Let γ(u) is a regular curve on the surface S, parametrized by an arbitrary parameter u and it is defined by

(4.12) γ(u) = (x1(u), x2(u), x3(u)), h(γ(u)) = 0 Differentiating (4.12) with respect to the parameter u , then projecting the vector γ0(u) onto the normal vector field ζ of the surface S, we get

0 (4.13) hζh(u), γ (u)i1 = 0 where

  0  0 0 0  ζh(u) = −h1(γ(u)) h2(γ(u)) h3(γ(u)) , γ (u) = x1(u) x2(u) x3(u) . Differentiation (4.13) with respect to u yields,

00 D 0 h
T 0 E (4.14) hζh(u), γ (u)i1 = − γ ∗ H , (γ ) , 1 where   −h11 −h12 −h13 00  00 00 00  h γ (u) = x1 (u) x2 (u) x3 (u) ,H =  h21 h22 h23  . h31 h32 h33 and ∗ denotes to the multiplication of matrices. Differentiating (4.14) with respect to u we obtain

*  h T + 000 D 0 h
T 00E 0 dH 0 (4.15) hζh(u), γ (u)i1 = −3 γ ∗ H , γ − γ ∗ , γ , 1 du 1 where

000  000 000 000  D 00 h
T 0E D 0 h
T 00E γ (u) = x1 (u) x2 (u) x3 (u) , γ ∗ H , γ = γ ∗ H , γ , 1 1

 −h −h −h  dHh 11i 12i 13i = Hh x0 (u) + Hh x0 (u) + Hh x0 (u) and Hh = h h h . du 1 1 2 2 3 3 i  12i 22i 23i  h13i h23i h33i 144 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018)

3 4.2.1. Curves on Implicit Spacelike and Timelike Surfaces in R1. The unit normal vector field N of an implicit spacelike (or a timelike) surface S is given by

ζ N = h , kζk1

3 4.2.2. Curves on Implicit Lightlike Surfaces in R1. The normal vector field ζ of an implicit lightlike surface S is given by

ζ = (−h1, h2, h3).

4.3. Darboux Frame. 4.3.1. Darboux Frame of Spacelike, Timelike or Pseudo Null curves on a non lightlike Surfaces. In this subsection we will study the Darboux frame. Definition 3. If {t, n, b} is the Frenet frame of the curve α(s) which lies on a spacelike (timelike) 3 surface S in R1 with unit normal vector field N, then the orthonormal basis {t, U, N}, where

(4.16) U = N ×1 t, is called natural frame (Darboux frame) for the curve-surface pair (α(s),S). Definition 4. [12] The basis {t, U, N}, where  ht, ti = hU, Ui = ht, Ni = hN, Ui = 0, hN, Ni = ht, Ui = 1, t ×1 U = N, N ×1 t = t,  1 1 1 1 1 1 n o 1 hV,Vi1  U ×1 N = U, where U = V − t , V ∈ Tα(s)M, hV, Ni = 0 and hV, ti 6= 0, hV,ti1 2hV,ti1 1 1 or  hN, Ni = hU, Ui = ht, Ni = ht, Ui = 0, ht, ti = hN, Ui = 1, t ×1 N = N, N ×1 U = t,  1 1 1 1 1 1 n o 1 hV,Vi1  U ×1 t = U, where U = V − N , V ∈ Tα(s)M, hV, Ni 6= 0 and hv, ti = 0, hV,Ni1 2hV,Ni1 1 1 is called null Darboux frame at a point of the null (or pseudo null) curve in timelike surfaces (or lightlike surfaces), respectively. 3 Definition 5. [21] Let S be a non lightlike surface in R1 and α(s) be a curve on S, then the functions

0 0 0 (4.17) κg(s) = ht (s), U(s)i1 , κn(s) = ht (s), N(s)i1 , τg(s) = hN (s), U(s)i1 , are called the geodesic curvature, normal curvature and geodesic torsion of the curve α(s), respectively. 3 Theorem 1. [21, 24] Let S be a non lightlike surface in R1 and α(s) be a curve on the surface S, then the derivative formulas of the natural frame {t, U, N} , are given by

 0   ∗ ∗    t 0 2kg 3kn t 0 ∗  U  =  −1kg 0 −3τg   U  , 0 ∗ N −1kn 2τg 0 N

∗ ∗ where 1 = ht, ti1 , 2 = hU, Ui1 and 3 = hN, Ni1 . 4.3.2. Darboux Frame of Spacelike or Pseudo Null curves on Lightlike Surfaces. Let S be a 3 3 lightlike surface in R1, α(s): I → S ⊂ R1 be a unit speed spacelike curve and it lies on the surface S, ζ be the normal vector field of the surface S, and t is the unit tangent vector field of the curve α(s), then there exists only one Darboux frame (null frame) {t, ζ, U}, such that hζ, ζi1 = hU, Ui1 = ht, ζi1 = ht, Ui1 = 0, ht, ti1 = hζ, Ui1 = 1, t ×1 ζ = ζ, ζ ×1 U = t, U ×1 t = U, and it is a pseudo-orthonormal basis in 3 R1. This frame satisfies (3.2) and it can be constructed as in section 3.1.2, so it is given by

0 0 0 1 2 3 γ 1 0 3 0 2 02
(4.18) t = (x , y , z ), ζ = (ζ , ζ , ζ ), U = − 2(ζ1)2 [(2γζ , 2x ζ , −2x ζ ) + γ x − 1 ζ], where y0ζ3 − z0ζ2
γ = . ζ1 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) 145

3 Definition 6. Let S be a lightlike surface in R1 and α(s) be a curve on S, then the functions 0 0 0 (4.19) κg(s) = ht (s), U(s)i1 , κ¯n(s) = ht (s), ζ(s)i1 , τ¯g(s) = hζ (s), U(s)i1 , are called the geodesic curvature, pseudo normal curvature and pseudo geodesic torsion of the curve α(s), respectively. 3 Theorem 2. Let S be a lightlike surface in R1 and α(s) be a curve on S. Then derivative formulas of the natural frame {t, N, U} are given by

 0      t 0 κg κ¯n t 0  N  =  −κ¯n τ¯g 0   N  . 0 U −κg 0 −τ¯g U

3 4.3.3. Darboux Frame of Null Curves on Timelike Surfaces. Let S be a timelike surface in R1, 3 α : I → S ⊂ R1 be a distinguished null curve, N be the normal vector field of the surface S and t is the tangent vector field of the curve α(s), then there exists only one Darboux frame (null frame) {N, t, U}, such that ht, ti1 = hU, Ui1 = ht, Ni1 = hN, Ui1 = 0, hN, Ni1 = ht, Ui1 = 1, t ×1 U = N, N ×1 t = t, 3 U ×1 N = U, this frame is a pseudo-orthonormal basis in R1. Also, this frame satisfies (3.1) which can be constructed as in section 3.1.3, so it is given by

0 0 0 1 2 3 υ 0 1 0 1 0  1
2  t = (x , y , z ), N = (N ,N ,N ), U = − 2x02 [(2υx , 2N z , −2N y ) + υ N − 1 t], where N 2z0 − N 3y0 υ = . x0 3 Theorem 3. Let S be a timelike surface in R1 and α(s) be a curve lies on the surface S. Then derivative formulas of the natural frame {N, t, U} are given by

 0      N 0 τg −κn N 0  t  =  κn κg 0   t  , 0 U −τg 0 −κg U where

0 0 0 κg(s) = ht (s), U(s)i1 , κn(s) = ht (s), N(s)i1 , τg(s) = hN (s), U(s)i1 , which are called the geodesic curvature, normal curvature and geodesic torsion of the curve α(s), respec- tively. 4.3.4. Darboux Frame of Null Curves on Lightlike Surfaces. If {t, n, b} be the null Frenet frame 3 of a null curve α(s) which lies on a lightlike surface S in R1. In this case the tangent vector field t of the curve α(s) and the normal vector field ζ of the surface S are linearly dependent (N =ωt; ω ∈ R − {0}) and the Darboux frame is not defined ({t, N, U} where ht, Ni1 = 0 ). 3 5. Transversal Intersection Curves of Two Surfaces in R1. The intersection curve of two spacelike surfaces is a spacelike, the intersection curve of two timelike surfaces may be spacelike or timelike, the intersection curve of two lightlike surfaces may be spacelike or null, the intersection curve of spacelike and timelike surfaces is spacelike, the intersection curve of spacelike and lightlike surfaces is spacelike, and the intersection curve of timelike and lightlike surfaces spacelike or null. We obtain the transversal intersection case, when the normal vectors of the surfaces are linearly independent. For what follows we suppose α(s) be a transversal intersection curve of two arbitrary surfaces SA and SB and assume that the curve β(u) = α(s(u)) is a regular curve parametrized by a parameter u with the same trace as the curve α(s). Let ζA and ζB denote the normal vector fields of the surfaces SAand SB, respectively. Also, let NA and NB denote the unit normal vector fields of the surfaces SAand B A B A A A B B B A B ¯ S , respectively. Assume that ζ , ζ 1 = λ, ζ , ζ 1 = ε , ζ , ζ 1 = ε , N , N 1 = λ, A A A B B B N , N 1 =ε ¯ , N , N 1 =ε ¯ . A B A B Since ζ ×1 ζ is orthogonal to ζ and ζ , so by using proposition2, we obtain B A B A B λ ε 2 A B ζ ×1 ζ , ζ ×1 ζ = = λ − ε ε . 1 εA λ The surfaces are intersect transversely if λ2 − εAεB 6= 0. In particular, if λ = 0, εA = 0 and εB > 0, in this case the surfaces are intersect transversely at a null curve. 146 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018)

5.1. The Tangent Vector Field. 2 A B 0 A B • If λ − ε ε 6= 0, then The tangent vector field of the curve β(u) is given by β (u) = ζ ×1 ζ , so that the unit tangent vector field t of the transversal intersection curve is given by

A B 0 ζ ×1 ζ (5.1) α (s) = A B . kζ ×1 ζ k1 • If λ = 0, εA = 0, εB = 1, then the tangent vector field of the transversal intersection curve is null. Moreover, α0(s) and ζA are linearly dependent. From the foregoing results, we have the following preposition: Proposition 4. For the transversal intersection curve of two surfaces, we get. i) The transversal intersection curve of two spacelike or two timelike surfaces is spacelike or timelike curve if, λ¯ ∈ R − [ − 1, 1] or λ¯ ∈ ] − 1, 1[, respectively. ii) The transversal intersection curve of two lightlike surfaces is spacelike curve. iii) The transversal intersection curve of a spacelike and a timelike surfaces is spacelike curve. iv) The transversal intersection curve of a lightlike surface and a spacelike surface is spacelike curve. v) The transversal intersection curve of a lightlike surface and a timelike surface is spacelike or lightlike curve if, λ 6= 0 or λ = 0, respectively. 5.2. The Second Order Derivative Vector Field. • If λ2 − εAεB 6= 0, the vector field α00(s) must lies in the normal plane spanned by ζA and ζB, thus we can write,

00 A B (5.2) α (s) = β1ζ + β2ζ ,

00 where β1 and β2 are unknown coefficients that we need to determine. Projecting α (s) onto the normal vector fields of both surfaces and solving them for the unknown coefficients and substitu- tion into (5.2), we get εBκ¯A − λκ¯B εAκ¯B − λκ¯A (5.3) α00(s) = n n ζA + n n ζB, εAεB − λ2 εAεB − λ2 A 00 A B 00 B where κ¯n = α (s), ζ 1 and κ¯n = α (s), ζ 1 and obtained by (4.3) or (4.14). • If λ = 0, εA = 0, ε¯B = 1, then α(s) is a null curve. If α00(s) 6= 0, then α0(s) and ζA have the 0 A 0 00 00 00 same direction, hα (s), ζ i1 = 0, hα (s), α (s)i1 = 0 and hα (s), α (s)i1 = 1. In this case we can assume the null frame {NB, ζA, U}, then the vector α00(s) can write α00(s) = NB. The third vector of the null frame U can be constructed as in section 3.1.2, so it is given by

%  2  U = − ((2%ζ1A, 2N 1Bζ3A, −2N 1Bζ2A) + % N 1B
− 1 ζA), 2 (ζ1A)2 where

A 1A 2A 3A B 1B 2B 3B N 2B ζ3A−N 3B ζ2A ζ = (ζ , ζ , ζ ), N = (N ,N ,N ),% = ζ1A . 5.3. The Third Order Derivative Vector Field. • If λ2 − εAεB 6= 0 and α00(s) is not lightlike, differentiating the vector field α00(s) and using (3.5), we get

000 2 2 2 0 α (s) = −12k t + ((2 − 1)kτ + k )n − 3kτb, Since the vectors ζA and ζB lie on the normal plane of the unit tangent vector t, then we can write

000 2 2 A B (5.4) α (s) = −12κ t + β5ζ + β6ζ ,

where β5 and β6 are unknown coefficients that we need to determine. Projecting the vector field α000(s) onto the normal vector fields of both surfaces, solving for unknown coefficients and substi- tuting into (5.4) we obtain, εBp¯A − λp¯B εAp¯B − λp¯A (5.5) α000(s) = − 2κ2t + n n ζA + n n ζB, 1 3 εAεB − λ2 εAεB − λ2 A 000 A B 000 B where p¯n = α (s), ζ 1 , p¯n = α (s), ζ 1 and obtained by (4.4) or (4.15). O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) 147

• If λ2 − εAεB 6= 0 and α00(s) is lightlike, differentiating the vector field α00(s) and using (3.10), we get

000 A B (5.6) α (s) = β7ζ + β8ζ ,

where β7 and β8 are unknown coefficients that we need to determine. Projecting the vector field α000(s) onto the normal vector fields of both surfaces, solving for unknown coefficients and substi- tuting into (5.6) we obtain,

εBp¯A − λp¯B εAp¯B − λp¯A (5.7) α000(s) = n n ζA + n n ζB, εAεB − λ2 εAεB − λ2

A 000 A B 000 B where p¯n = α (s), ζ 1 , p¯n = α (s), ζ 1 and obtained by (4.4) or (4.15). • If λ = 0, εA = 0, ε¯B = 1, then the vector α000(s) can be written as

000 A (5.8) α (s) = β9ζ + β10U,

where β9 and β10 are unknown coefficients that we need to determine. Projecting the vector field α000(s) onto the vectors ζA, U and NB, solving for unknown coefficients and substituting in (5.8), we obtain

000 U A A (5.9) α (s) =p ¯n ζ +p ¯n U,

A 000 A U 000 where p¯n = α (s), ζ 1 is obtained by (4.4) or (4.15) and p¯n = hα (u), Ui1. 3 6. Characterizations of Transversal Intersection curves of two Surfaces in R1. The characteriza- tions of the transversal intersection curves of two spacelike, two timelike and spacelike-timelike surfaces can be found in [6], [22] and [16], respectively. In this section, we introduce the characterization of inter- section curves of two lightlike, lightlike-timelike and lightlike-spacelike surfaces. Theorem 4. Let α = α(s) be the transversal intersection curve of two lightlike surfaces SA and SB. The geodesic curvatures of curve are given by

κ¯B κ¯A (6.1) kA = n , kB = n . g λ g λ If α00(s) is not lightlike, then the curvature and the torsion are given by

2¯κAκ¯B  (pB hζA,bi +pB hζB ,bi ) k2 = n n , τ = − 2 n 1 n 1 . λ2 kλ

If α00(s) is lightlike, then the torsion is given by

pBhζA, bi + pBhζB, bi (6.2) τ = − n 1 n 1 . λ Proof: In this case the intersection curve is spacelike. To write the geodesic curvatures as a function of the normal curvatures, we need to write the vector UA, UB as a linear combination of ζA and ζB and applied in the equation (4.19). Thus,

A A B B A B U = x0ζ + y0ζ , U = x1ζ + y1ζ , then we have

 1 = UA, ζA = x ζA, ζA + y ζB, ζA  y = 1  1 0 1 0 1  0 λ =⇒ ,  A A 2 A A B A 2 B B  0 = U , U 1 = (x0) ζ , ζ 1 + 2x0y0 ζ , ζ 1 + (y0) ζ , ζ 1 x0 = 0

 1 = UB, ζB = x ζA, ζB + y ζB, ζB  x = 1  1 1 1 1 1  1 λ =⇒ .  B B 2 A A B A 2 B B  0 = U , U 1 = (x1) ζ , ζ 1 + 2x1y1 ζ , ζ 1 + (y1) ζ , ζ 1 y1 = 0 148 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018)

00 2 00 00 If α (s) is not lightlike the curvature κ is obtained using the equation (5.3), κ = hα , α i1 and the torsion is obtained using the equations (3.4) and (5.5). If α00(s) is lightlike the curvature is not defined and the torsion is obtained using the equations (3.11) and (5.7). Theorem 5. Let α = α(s) be the transversal intersection curve of a lightlike surface SA and a spacelike surface SB. The geodesic curvatures of intersection curve are given by

2 ζB κ¯A+2λ ζB κB ζB κ¯A+λκB (6.3) A k k n k k n B k k n n κg = 2λ2 , κg = |λ| . If α00(s) is not lightlike, then the curvature and the torsion are given by

(¯κA)2−2λκ¯AκB  −εB pA+λpB pB  (6.4) k2 = n n n , τ = − 2 n n hζA, bi + n hζB, bi . λ2 k λ2 1 λ 1 If α00(s) is lightlike, then the torsion is given by

B A B B −ε pn +λpn A pn B τ = − λ2 hζ , bi1 − λ hζ , bi1.

Proof: In this case the intersection curve is spacelike. To write the geodesic curvatures as a function of the normal curvatures, we need to write the vector UA, UB as a linear combination of ζA and ζB and applied in the equations (4.17) and (4.19). Thus, A A B U = x0ζ + y0ζ , then we have  1 = UA, ζA = x ζA, ζA + y ζB, ζA  1 0 1 0 1  A A 2 A A B A 2 B B 0 = U , U 1 = (x0) ζ , ζ 1 + 2x0y0 ζ , ζ 1 + (y0) ζ , ζ 1 ,

 y = 1  1 = UA, ζA = y λ  0 λ  1 0  =⇒ =⇒ B B 2  A A 2 B B 2  ε ζ 0 = U , U 1 = 2x0y0λ + (y0) ε ζ  x0 = − . 2λ2

2 εB ζB ζB NB =⇒ UA = − ζA + . 2λ2 λ Also, we have

B B U = N ×1 t

B A B N ×1(ζ ×1 ζ ) = A B kζ ×1 ζ k

− ζB ,NB ζA+ ζA,NB ζB h i1 h i1 = A B kζ ×1 ζ k

 B  − ζB NB ,NB ζA+ ζA, ζ ζB k kh i1 kζB k 1 = A B kζ ×1 ζ k

1 B B B A 1 A B B = A B (− ζ N , N ζ + B ζ , ζ ζ ). kζ ×1 ζ k 1 kζ k 1

1  B B A ζB  = A B − ζ ε ζ + λ B . kζ ×1 ζ k kζ k

00 2 00 00 If α (s) is not lightlike the curvature κ is obtained using the equation (5.3), κ = hα , α i1 and the torsion is obtained using the equations (3.4) and (5.5). If α00(s) is lightlike the curvature is not defined and the torsion is obtained using the equations (3.11) and (5.7). Theorem 6. Let α = α(s) be the transversal intersection curve of a lightlike surface SA and a timelike surface SB. If α0(s) is lightlike, then the torsion is given by U A A τ = −pn hζ , bi1 − pn hU, bi1. O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) 149

If α0(s) is spacelike the geodesic curvatures of intersection curve are given by

2 ζB κ¯A+2λ ζB κB − ζB κ¯A+λκB (6.5) A k k n k k n B k k n n κg = 2λ2 , κg = |λ| . If α0(s) is spacelike and α00(s) is not lightlike, then the curvature and the torsion are given by

(¯κA)2+2λκ¯AκB  −εB pA+λpB pB  k2 = n n n , τ = − 2 n n hζA, bi + n hζB, bi . λ2 k λ2 1 λ 1 If α0(s) is spacelike and α00(s) is lightlike, then the torsion of α(s) is given by

−εB pA+λpB pB (6.6) n n A n B τ = − λ2 hζ , bi1 − λ hζ , bi1.

Proof: If α0(s) is lightlike by section 4.3.4 Darboux frame is not defined to SA and torsion is obtained using the equations (3.13) and (5.9). Suppose the intersection curve is spacelike. To write the geodesic curvatures as a function of the normal curvatures, we need to write the vector UA, UB as a linear combination of ζA and ζB and applied in the equations (4.17) and (4.19). Thus,

A A B U = x0ζ + y0ζ , then we have

 1 = UA, ζA = x ζA, ζA + y ζB, ζA  1 0 1 0 1  A A 2 A A B A 2 B B 0 = U , U 1 = (x0) ζ , ζ 1 + 2x0y0 ζ , ζ 1 + (y0) ζ , ζ 1 ,

 1 y0 = λ  B B 2  A ε ζ A 1 B =⇒ B B 2 =⇒ U = − 2 ζ + ζ .  ε ζ 2λ λ  x0 = − . 2λ2 Also we have

B B U = N ×1 t

B A B N ×1(ζ ×1 ζ ) = A B kζ ×1 ζ k

− ζB ,NB ζA+ ζA,NB ζB h i1 h i1 = A B kζ ×1 ζ k

 B  − ζB NB ,NB ζA+ ζA, ζ ζB k kh i1 kζB k 1 = A B kζ ×1 ζ k

1 B B B A 1 A B B = A B (− ζ N , N ζ + B ζ , ζ ζ ). kζ ×1 ζ k 1 kζ k 1

1  B B A ζB  = A B − ζ ε ζ + λ B . kζ ×1 ζ k kζ k

00 2 00 00 If α (s) is not lightlike the curvature κ is obtained using the equation (5.3), κ = hα , α i1 and the torsion is obtained using the equations (3.4) and (5.5). If α00(s) is lightlike the curvature is not defined and the torsion is obtained using the equations (3.11) and (5.7). 7. Examples. In this section we presentation some examples of transversal intersection curves of two surfaces and we compute its Frenet frame, Darboux frame, curvature (when exists), torsion, normal curvature and geodesic curvature in a point fixed. Example 1. Consider the two implicit lightlike surfaces SA and SB as in the following √ A B 2 2 2 (7.1) S : f(x1, x2, x3) = 2x1 − x2 − x3 − 1 = 0,S : h(x1, x2, x3) = x1 − x2 − x3 = 0. 150 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018)

FIGURE 7.1. Transversal intersection of two implicit lightlike surfaces

√ A
B A B Using (7.1), we obtain ζ = − 2, −1, −1 , ζ = (−2x1, −2x2, −2x3) , therefore ε = ε = 0, for any point on the surfaces, thus SA and SB are lightlike surfaces. Considering the intersection point √ √ √  1− 3 1+ 3  p0 = 2, 2 , 2 , (see Figure 1 ), in this point we have, λ = −2 6= 0 (transversal intersection). Using (4.13), (4.15),(5.1), (5.3), (5.7) and (7.1) at the intersection point p0, we obtain √ √ √ √ √ √ 0 1 1 1 1 00 000 (7.2) t = α = (− 3, 2 2 − 2 6, − 2 2 − 2 6), α = ( 2, 1, 1), α = (0, 0, 0) . Thus, the intersection curve at this point is pseudo null. Using, (3.9), (3.11), (3.12) and (7.2), we obtain √ √ √ √ 1
(7.3) n = ( 2, 1, 1) b = − 2 2 2, 1 − 3, 1 + 3 , τ = 0. Using (4.18) and (7.2) we obtain √ √ √ √ A 1 B 1
U = 2 (2 2, 1 − 3, 1 + 3), U = 2 2, 1, 1 Using (4.19), (6.1) and (6.2), then the geodesic curvature, pseudo normal curvature and pseudo geodesic torsion are given by κ¯B κ¯A κ¯A = 0, κ¯B = 2, κA = n = −1, κ¯B = n = 0, τ¯A = 0, τ¯B = 0. n n g λ g λ g g

Example 2. Consider the parametric lightlike surface SA and the parametric spacelike surface SB as in the following

A p 2 2 (7.4) S : A(u1, u2) = ( u1 + u2, u1, u2), u1, u2 ∈ ]−5, 5[ , B S : B(v1, v2) = (−v2, − cosh v1 cos v2, − cosh v1 sin v2), v1, v2 ∈ ]−5, 5[ .

A √ −1 p 2 2 B Using (7.4), we obtain ζ = 2 2 ( u1 + u2, u1, u2) and ζ = − sinh(v1)(cosh v1, − sin v2, cos v2), u1+u2 A A B 4 B therefore ε = 0 ∀ (u1, u2) ∈ S ;(u1, u2) 6= (0, 0) and ε = − sinh (v1) ∀ (v1, v2) ∈ S ;(v1, v2) 6= A B (0, v2), thus the surfaces S and S are lightlike and spacelike, respectively. Considering the intersection π π A B π π point P0 = ( 2 , 0, 2 ) ∈ S ∩S (see Figure 2), at this point, (u1, u2) = (0, 2 ), (v1, v2) = (−1.023 2, − 2 ) 2 2 2 A B π (π −4) and λ − ε ε = 16 6= 0 (transversal intersection). Using (4.1), (4.3), (4.4), (4.7), (4.8), (5.1), (5.3), (5.5), and (7.4), we obtain t = (0.63662, −1, 0.63662), α00 = (−0.258012, −0.405285, −0.894632) , (7.5) α000 = (0.313705, 0.898052, 0.313705) Thus the intersection curve at this point is a spacelike curve. Using (3.3), (3.4), (3.5), (3.6), (6.4) and (7.5), we obtain n = (−0.272264, −0.427671, −0.944047) , κ = 0.947656, (7.6) b = (1.21631, −0.427671, 0.544527) , τ = 0.627667. O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) 151

FIGURE 7.2. Transversal intersection of parametric lightlike and spacelike surfaces

Using (4.16), (4.18), (7.5) and (7.6), we obtain

(7.7) UA=(1. 297 4, −0.636 62, 0.297 36), UB = (−0.433 84, −0.681 5, 0.636 66)

Using (4.17), (4.19), (6.3), (7.5), (7.6) and (7.7), we have the geodesic curvature, normal curvature and geodesic torsion as in the following

2 ζB = 1.4674011, ζB = 2.1532659, λ = 1.90281, B 2 A B B A A A kζ k κ¯n +2λkζ kκn κ¯n = 0.63662, τ¯g = 0, κg = 2λ2 = 0.705328, B A B B B B kζ kκ¯n +λκn κn = 0.669138, τg = 0.129093 κg = |λ| = 1.16009.

Example 3. Consider the parametric lightlike surfaces SA and the implicit timelike surfaces SB as in the following

A B (7.8) S : A(u1, u2) = (u1 cosh u2, u1 sinh u2, u1) S : f(x1, x2, x3) = −x1 + x2 + x3 = 0

FIGURE 7.3. Transversal intersection of parametric lightlike and implicit timelike surfaces

A B Considering the intersection point P0 = (1, 0, 1) ∈ S ∩ S (see Figure 3). Using (7.8) at this point, A 1 B A B B A we obtain ζ = 2 (1, 0, 1), ζ = (1, 1, 1), λ = 1 6= 0, ε = 0 and ε¯ = ε = 1, thus the surfaces S and SB are lightlike and timelike, respectively. Using (4.9), (4.10), (4.13), (4.14) and (5.3) we obtain

α0(s) = (1, 0, 1), α00(s) = (0, 0, 0) , 152 O. Alessio,´ et. al.- Selecciones Matematicas.´ 05(02): 137-153 (2018) then the intersection curve is lightlike curve and it is a straight line.

Example 4. Consider the parametric timelike surface SA and the implicit lightlike surface SB as in the following

(7.9) A u2 u2 B z 2 2 S : A(u1, u2) = ( 2 + uv + v, 2 + uv + v, u),S : f(x, y, z) = x + arctan( y ) = 0; y + z < 1

FIGURE 7.4. Transversal intersection of parametric timelile and implicit lightlike

A B z y A From (7.9), we have ζ = (u + 1, u + 1, 0) and ζ = (−1, − y2+z2 , y2+z2 ), therefore ε = 0, ∀ A B 2 2 A B (u1, u2) ∈ S and ε > 1, ∀ y + z < 1, thus the surface S is lightlike and the surface S is π π π tan(π) A B timelike. Considering the point P0 = ( 8 , 8 , − 8 8 ) ∈ S ∩ S (see Figure 4), in this point, we have λ = 1.59121, εA = 0, ε¯B = 1, then the intersection curve is transversal. Using (4.10), (4.14), (5.1), (5.3), (5.5) and (7.9) we obtain

(7.10) α0 = (−1.14379, −1.14379, 1) , α00 = (4.0765, 4.0765, 0), α000 = (43.4607, 43.4607, 0) then the intersection curve at this point is a pseudo null curve. Using (3.9),(3.11), (3.12)(6.6) and (7.10) we obtain (7.11) t = (−1.14379, −1.14379, 1), n = (4.0765, 4.0765, 0), b = (−0.283116, −0.0378078, 0.280581), τ = −10.6613.

Using (4.18), (4.19), (6.5) and (7.11) we obtain

A B A B κ¯n = 0, κn = 3.63771τ ¯g = 0, τ¯g = 3.83701,

2 − ζB κ¯A+2λ ζB κB ζB κB B A k k n k k n k k n 2.12953 κn κg = 2(λ)2 = λ = 1.59121 = 4.8684,

− ζB κ¯A+λκB B B k k n n λκn B κg = |λ| = |λ| = κn = 3.63771.

8. Conclusions. From the results obtained in this work we concluded that, we can computing the dif- ferential geometry properties of the transversal intersection curves of two surfaces, where each pair satisfies the following types spacelike-lightlike, timelike-lightlike and lightlike-lightlike using the information of surfaces involved. Moreover, the results obtained ends the study of differential geometry properties of the transversal intersection curves of two surfaces in Minkowski 3-space, because the cases spacelike-spacelike, timelike-timelike and timelike-spacelike were studied in [6], [22] and [16] respectively.

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