Palestine Journal of Mathematics Vol. 8(1)(2019) , 397–411 © Palestine Polytechnic University-PPU 2019 A new approach to find a generalized evolute and involute curve in 4-dimensional Minkowski space-time Muhammad Hanif and Zhong Hua Hou Communicated by Zafar Ahsan MSC 2010 Classifications: : Primary 53-XX, 53Axx; Secondary 53A35. Keywords and phrases: Evolute, Involute, Null Cartan curve, Minkowski space. This work is supported by the National Natural Science Foundation of China (No.61473059). Abstract. In this study, we introduce a new approach to find a special kind of generalized 4 evolute and involute curve in Minkowski space-time E1 . Necessary and sufficient condition for the curve possessing generalized evolute as well as involute curve were obtained. Furthermore, Cartan null curve is also discussed in detail. 1 Introduction Many mathematicians did work about the general theory of the curves in Euclidean space. Now, we have much understanding on their local geometry as well as their global geometry. Identi- fication of a regular curve is one of the important and interesting complication in the theory of curves in Euclidean space. Using two different approaches this complication can be solved, the relation between the frenet vectors of the curve [1], and determining the shape and size of the curve by curvatures κ1 and κ2. In differential geometry an evolute is the envelope of the normals of the specific curve. An evolute and its involute are defined in mutual pairs. The evolute of any curve is defined as the locus of the centers of curvatures of the curve. The original curve is then described as the involute of the evolute. Evolutes and involutes (also known as evolvents) were studied by C. Huygens [2]. Later, in [3] the author explore that if evolute occur then the evolute of parallel arc also occur. In [4] the author fixed that evolute Frenet apparatus can be establish by involute apparatus in four dimensional Euclidean space so by this approach another orthonormal of the same space is acquired. In [5] author resolved that the iteration of involutes create a pair of sequences of curves with respect to Minkowski metric and its dual. In (1845), Saint Venant [6] suggested a question either the principle normal of a curve is the principle normal of another on the surface produced by the principle normal of the specific curve. Bertrand answered this question in [7]. He proved that a necessary and sufficient condition for the existence of such a second curve is required. Using this method we define a kind of generalized evolute-involute curve in Minkowski space time. We acquire the necessary and sufficient conditions for the curves with spacelike (1,3)- normal plane to be (1,3)-Evolute curves and we also prove its converse by using the condition of Evolute curve that is spanned by principle normal and the second binormal. In the end we give some examples for these curves. Evolute curves and their identification were studied by some researchers in Minkowski space [8], [9], [10], [11], [12], [13], [14], [15], [16] as well as in Euclidean space. We see that mainly evolute-involute curves have been studied but not so much research has been carried out to find the mate curves of Cartan null curves. In this paper, a kind of generalized evolute and involute curve is considered for Cartan null curve in Minkowski space-time. The necessary and sufficient conditions for a curve possessing generalized evolute as well as involute mate curves is obtained. 2 Preliminaries 4 4 Consider the Minkowski space-time (E1 ;H) where E1 = fz = (z1; z2; z3; z4)jzi 2 Rg and 2 2 2 2 H = −dz1 + dz2 + dz3 + dz4. For any U = (x1; x2; x3; x4) and V = (y1; y2; y3; y4) 2 TzE, we 398 Muhammad Hanif and Zhong Hua Hou denote U · V = H(U; V ) = −x1y1 + x2y2 + x3y3 + x4y4. 4 4 Let I be an open interval in R and G : I ! E1 be a regular curve in E1 parameterized by arc length parameter s and fT;N1;N2;N3g be a moving Frenet frame along G, consisting of tangent vector T , principal normal vector N1, the first binormal vector N2 and the second binormal vector 4 N3 respectively, so that T ^ N1 ^ N2 ^ N3 coincides with the standard orientation of E1 . From [17] Frenet seret formula. 2 0 3 2 3 2 3 T 0 2κ1 0 0 T 6N 07 6− κ 0 κ 0 7 6N 7 6 17 = 6 1 1 3 2 7 6 17 ; (2.1) 6 07 6 7 6 7 4N25 4 0 −2κ2 0 −123κ35 4N25 0 N3 0 0 −3κ3 0 N3 where H(T;T ) = 1;H(N1;N1) = 2;H(N2;N2) = 3; H(N3;N3) = 4; 1234 = −1; i 2 f1; −1g; i 2 f1; 2; 3; 4g: In specific, the succeeding conditions exist: H(T;N1) = H(T;N2) = H(T;N3) = H(N1;N2) = H(N1;N3) = H(N2;N3) = 0: 4 0 A curve G(s) in E1 can be spacelike, timelike, or null if its velocity vectors G (s) are com- monly spacelike, timelike, or null [18]. A null curve G is parametrized by pseudo-arc s if H(G00(s); G00(s)) = 1 [19]. Further more nonnull curve G, we have this condition H(G0(s); G0(s)) = ±1. From [19, 20] if G is null Cartan curve, the Cartan Frenet frame is given by 2 0 3 2 3 2 3 T 0 κ1 0 0 T 6N 07 6 κ 0 −κ 0 7 6N17 6 17 = 6 2 1 7 6 7 ; (2.2) 6 07 6 7 6 7 4N25 4 0 −κ2 0 κ35 4N2 5 0 N3 −κ3 0 0 0 N3 where κ1(s) = 0 if G(s) is a null straight line or κ1(s) = 1 in all other cases. In this case T · T = N2 · N2 = 0;N1 · N1 = N3 · N3 = 1, T · N1 = T · N3 = N1 · N2 = N1 · N3 = N2 · N3 = 0, T · N2 = 1. We established some terminologies in this study. At any point of G, the plane spanned by fT;N2g is called the (0; 2)-tangent plane of G. The plane spanned by fN1;N3g is called the (1; 3)-normal plane of G. 4 ∗ 4 4 Let G : I ! E and G : I ! E1 be two regular curves in E1 where s is the arc-length parameter of G. Denote s∗ = f(s) to be the arc-length parameters of G∗. For any s 2 I, if the (0; 2)-tangent plane of G at G(s) of coincides with the (1; 3)-normal plane at G∗(s) of G∗, then ∗ 4 ∗ G is called the (0; 2)-involute curve of G in E1 and G is called the (1; 3)-evolute curve of G in 4 E1 . 4 2.1 (1,3)-involute curve of a given curve in E1 In this section, we proceed to study the existence and expression of the (1; 3)-evolute curve of a 4 given curve in E1 . 4 Let G : I ! E1 be a regular curve with arc-length parameter s so thatκ1, κ2 and κ3 are not ∗ 4 ∗ ∗ ∗ ∗ zero. Let G : I ! E1 be the (1; 3)-evolute curve of G. Denote fT ;N1 ;N2 ;N3 g to be the ∗ ∗ ∗ ∗ ∗ Frenet frame along G and κ1 , κ2 and κ3 to be the curvatures of G . Then ∗ ∗ ∗ ∗ spanfT;N2g = spanfN1 ;N3 g; spanfN1;N3g = spanfT ;N2 g: 4 Generalized involute-evolute curves in E1 399 3 Theorem 4 Let G : I ! E1 be a regular curve with arc-length parameter s so that κ1, κ2 and κ3 are not zero. ∗ 4 ∗ ∗ ∗ ∗ Let G : I ! E1 be the (1; 3)-evolute curve of G. Denote fT ;N1 ;N2 ;N3 g to be the Frenet ∗ ∗ ∗ ∗ ∗ frame along G and κ1 , κ2 and κ3 to be the curvatures of G if and only if there there exists scalar functions F, Y of arc-length parameter s and real constant numbers α =6 ±1, β satisfying 0 F0(s) =6 0; Y (s) =6 0; (3.1) 0 0 F (s) = αY (s); (3.2) βαk1(s) = αk2(s) − k3(s); (3.3) 2 2 2 2 [−κ2κ3(α − 1) + α(κ3 − κ1 − κ2)] =6 0 (3.4) for all s 2 I. 4 Proof. Let G : I ! E1 be a regular curve with arc-length parameter s so that κ1, κ2 and κ3 are ∗ 4 ∗ ∗ ∗ ∗ not zero. Let G : I ! E1 be the (1; 3)-evolute curve of G. Denote fT ;N1 ;N2 ;N3 g to be the ∗ ∗ ∗ ∗ ∗ Frenet frame along G and κ1 , κ2 and κ3 to be the curvatures of G . Then ∗ ∗ ∗ ∗ spanfT;N2g = spanfN1 ;N3 g; spanfN1;N3g = spanfT ;N2 g: Moreover, we can write the curve G∗ as follows ∗ ∗ G (s ) = G(s) + F(s)N1(s) + Y(s)N3; (3.5) for all s∗ 2 I∗, s 2 I where F(s) and Y(s) are C1 functions on I. Taking derivative of (3:5), using the equation (2:1), we get ∗ 0 0 0 T f = (1 − F1κ1)T (s) + F (s)N1(s) + Y (s)N3 + 3(F(s)κ2 − Y(s)κ3)N2: (3.6) ∗ ∗ 1 fT ;N g?fT;N2g so 1 − a1κ1 = 0 and 3(aκ2 − bκ3) = 0 from these we have F = ,Y = 2 1κ1 1κ2 . κ1κ3 Equation(3.6) gets the form 0 ∗ 0 0 f T = F (s)N1 + Y (s)N3: (3.7) So (3:6) gets the form 0 0 F Y T ∗ = N + N : (3.8) f 0 1 f 0 3 Multiplying (3:7) by itself, we get ∗ 0 2 0 2 0 2 1 (f ) = 2(F ) + 4(Y ) : (3.9) If we denote 0 0 F Y η = ; ζ = : (3.10) f 0 f 0 Using equation (3.10) in (3.7), we get ∗ T = ηN1 + ζN3: (3.11) Taking derivative of equation (3.11) using (2.1), we get 0 ∗ ∗ 0 0 f κ1 N1 = η N1 − 1ηκ1T + ζ N3 + 3(ηκ2 − ζκ3)N2: (3.12) Taking inner product on both side of (3.12) by N1 and N3 respectively, we get η0 = 0; ζ0 = 0: (3.13) 400 Muhammad Hanif and Zhong Hua Hou 0 ∗ ∗ f κ1 N1 = −1ηκ1T + 3(ηκ2 − ζκ3)N2: (3.14) Multiplying (3.14) by itself, we get 0 2 ∗ 2 2 2 2 (f ) (κ1 ) = η κ1 + (ηκ2 − ζκ3) : (3.15) ∗ ∗ ∗ We know that curvatures κ1 , κ2 , and κ3 =6 0 , so we obtained result (3.1) 0 F0 =6 0; Y =6 0: From (3:10), we get the result (3.2) 0 0 F = αY : (3.16) F−η Integrating (3.16), we get F = αY + η and Y = α .