On Special Kinds of Involute and Evolute Curves in 4-Dimensional Minkowski Space

Recently, extensive research has been done on evolute curves in Minkowski space-time. However, the special characteristics of curves demand advanced level observations that are lacking in existing well-known literature. In this study, a special kind of generalized evolute and involute curve is considered in four-dimensional Minkowski space. We consider (1,3)-evolute curves with respect to the casual characteristics of the (1,3)-normal plane that are spanned by the principal normal and the second binormal of the vector fields and the (0,2)-evolute curve that is spanned by the tangent and first binormal of the given curve. We restrict our investigation of (1,3)-evolute curves to the (1,3)-normal plane in four-dimensional Minkowski space. This research contribution obtains a necessary and sufficient condition for the curve possessing the generalized evolute as well as the involute curve. Furthermore, the Cartan null curve is also discussed in detail.


Introduction
In the theory of curves, one of the important and interesting problems is the characterization of regular curves, in particular, the involute-evolute of a given curve.Evolutes and involutes (also known as evolvents) were studied by C. Huygens [1].According to D. Fuchs [2], an involute of a given curve is a curve to which all tangents of the given curve are normal.He also defined the equation for an enveloping curve of the family of normal planes for a space curve.Suleyman and Seyda [3] determined the concept of parallel curves, which means that if the evolute exists, then the evolute of the parallel arc will also exist and the involute will coincide with the evolute.Brewster and David [4] stated that a curve is composed of two arcs with a common evolute, and the common evolute of two arcs must be a curve with only one tangent in each direction.In general, the evolute of a regular curve has singularities, and these points correspond to vertices.Emin and Suha [5] determined that an evolute Frenet apparatus can be formed by an involute apparatus in four dimensional Euclidean space, so, in this way, another orthonormal of the same space can be obtained.Shyuichi Izumiya [6] defined evolutes as the loci of singularities of space-like parallels and geometric properties of non-singular space-like hyper surfaces corresponding to the singularities of space-like parallels or evolutes.Takami Sato [7] investigated the singularities and geometric properties of pseudo-spherical evolutes of curves on a space-like surface in three-dimensional Minkowski-space.Marcos Craizer [8] stated that the iteration of involutes generates a pair of sequences of curves with respect to the Minkowski metric and its dual.
According to Boaventura Nolasco and Rui Pacheco [9], correspondence between plane curves and null curves in Minkowski three-space exists.He also described the geometry of null curves in terms of the curvature of the corresponding plane curves.M. Turgut and S. Yilmaz [10] obtained the Frenet apparatus of a given curve by defining the space-like involute-evolute curve couple in Minkowski space-time.Some researchers have investigated evolute curves and their characterization in Minkowski space [11][12][13][14][15][16] as well as in Euclidean space.Many researchers have dealt with evolute-involute curves, but no research has been carried out on the Cartan null curve.In this study, a special kind of generalized evolute and involute curve is considered in four-dimensional Minkowski space.We obtained necessary and sufficient conditions for the curve possessing a generalized evolute as well as an involute.

Preliminaries
Consider the Minkowski space-time, (E 1 be a regular curve in E 4 1 that is parameterized by the arc length parameter, s, and {T, N, B 1 , B 2 } is the moving Frenet frame along α, consisting of the tangent vector, T; the principal normal vector, N; the first binormal vector, B 1 , and the second binormal vector, B 2 , respectively, so that T ∧ N ∧ B 1 ∧ B 2 coincides with the standard orientation of E 4  1 .Then, In particular, the following conditions hold: In accordance with reference [17], the Frenet-Serret formula for α in E 4 1 is given by We introduce some methodologies in this paper.At any point of α, the plane spanned by {T, B 1 } is called the (0,2)-tangent plane of α.The plane spanned by {N, B 2 } is called the (1,3)-normal plane of α.

The (0,2)-Involute Curve of a Given Curve in E 4 1
In this section, we proceed to study the existence and expression of the (0,2)-involute curve of a given curve in E 4  1 .
By differentiating (3) with respect to s and using the Frenet formula (1), we get Taking the inner product on both sides of (4) with T and B 1 , respectively, we get 1 + φ = 0 and ϕ = 0, which implies that ϕ is constant and φ = ϕ 0 − s, where φ 0 is the integration constant.So, (4) turns into (5) If we denote then (5) turns into Case 1: By differentiating ( 7) with respect to s and using the Frenet formula (1), we get * By taking the inner product from both sides of (9) with N and B 2 , respectively, we get µ = 0 and ν = 0, which implies that µ and ν are constants.So, (9) turns into * then (10) turns into g f = t 2 implies that g = t 2 f and From Equations ( 8) and ( 13), we have By differentiating ( 12) with respect to s using the Frenet formula (1), we get By taking inner product on both side of ( 16) by T and B 1 respectively, we get f = 0 and g = 0, which implies that f and g are constants.In this case, (16) By substituting (7) and ( 15) into (17), we get From ( 18), we may choose that By differentiating ( 19) about s and using the Frenet formula (1), we get from which we obtain * From (21), we may choose that From Equations ( 14), ( 15), ( 18) and ( 22), we can easily acquire our theorem.
Case 2: If ϕ = 0, we have the following theorem.
Moreover, the three curvatures of α * are given by The associated Frenet frames are given by In this case, (4) turns into By differentiating (24) with respect to s and using the Frenet Formula (1), we get from which we may assume that By differentiating the second equation of (26) about s and using the Frenet Formula (1), we get * Suppose that By differentiating (27) about s, we obtain that f and g are constants: * Suppose that By differentiating (30) about s, we obtain * Suppose that From Equations ( 27), ( 30) and (33), we have achieved the desired theorem.

The (1,3)-Evolute Curve of a Given Curve in E 4 1
In this section, we want to study the (1,3)-evolute curve of a given curve in E 4 1 .
Theorem 3. Let α : I → E 4 1 be a regular curve with arc length parameter s so that k 1 , k 2 and k 3 are not zero, If α possesses the (1,3)-evolute mate curve, α * (s) = α(s) + 1 ik 1 (s) [iN(s) + jB 2 (s)] − 1 k 3 (s) B 2 (s), then k 1 , k 2 and k 3 satisfy 1 ik 1 + 3 (jk 2 − jk 3 ) = 0, where i and j are given constants.Three curvatures of α * are given by The associated Frenet frames are given by Proof.Let α : I → E 4 1 be a regular curve with arc-length parameter s so that k 1 , k 2 and k 3 are not zero.Let α * : I → E Moreover, α * can be expressed as where p(s) and q(s) are C ∞ functions on I. Differentiating (35) with respect to s using Frenet Formula (1), we get By taking the inner product from both sides of (36) with T and B 1 , respectively, we get then (37) turns into By differentiating (39) with respect to s and using the Frenet formula (1), we get * By taking inner product on both sides of (40) with N and B 2 respectively, we get i = o and j = 0, which implies that i and j are constants.
From (38), we obtain Moreover, (40) turns into * then (42) turns into Moreover, we have Case 1: t = 0.By differentiating (44) about s and using the Frenet Formula (1), we get By taking inner product on both sides of (46) with T and B 1 respectively, we get r = 0 and t = 0, which implies that r and t are constants.In this case, (46) turns into then (47) turns into Since T * ⊥ B * 1 , it follows from (40) and (50) that σ ς = − j i , which implies that From ( 45) and (50), we can see that Since t = 0, it follows from (51) that t = r.Hence, (49) turns into By differentiating (52) about s using (1), we get from which we obtain It follows from (54) that From ( 43), ( 52) and (55), we can easily acquire our desired theorem.
Case 2: If t = 0, we have the following theorem.
Theorem 4. Let α : I → E 4 1 be a regular curve parameterized by arc-length s so that k 1 , k 2 and k 3 are not zero.If α possesses the (1,3)-evolute mate curve, α * (s) = α(s) + 1 ik 1 (s) [iN(s) + jB 2 (s)], then k 2 and k 3 satisfy ik 2 − jk 3 = 0, where i and j are given constants.Moreover, the three curvatures of α * are given by The associated Frenet frames are given by Proof.For this case, we may suppose that From ( 41) and the third equation of (58), we acquire By differentiating (58) about s and using (1), we get It follows that we may choose By differentiating (61) about s using the Frenet Formula (1) and third equation of (58), we get From ( 58), ( 61) and (62), we can easily acquire our desired theorem.

The (1,3)-Evolute Curve of a Cartan Null Curve in E 4 1
In this section, we proceed to study the existence and expression of the (1,3)-evolute curve of a given Cartan null curve in E 4  1 .At any point of α, the plane spanned by {N, B 2 } is called the (1,3)-normal plane of α.
Theorem 5. Let α : I → E 4 1 be a null Cartan curve with arc length parameter s so that k 1 = 1, and k 2 k 3 are not zero, if α possesses the (1,3)-evolute mate curve, α * (s) = α(s) k 2 and k 3 satisfy i + ik 2 − jk 3 = 0, where i and j are given constants.Three curvatures of α * are given by Moreover, the associated Frenet frames are given by Proof.Let α : I → E 4 1 be a Cartan null curve parameterized by the pseudo-arc parameter s with curvatures k 1 = 1, and k 2 and k 3 are not zero.Let α * : I → E 4  1 be the (1,3)-evolute curve of α.Denote {T * , N * , B * 1 , B * 2 } as the Frenet frame along α * and k * 1 ,k * 2 and k * 3 as the curvatures of α * .Then Moreover, α * can be expressed as where p(s) and q(s) are C ∞ functions on I.By differentiating (66) with respect to s using the Frenet Formula (2), we get By taking the inner product on both sides of (67) with T and B 1 , respectively, we get By differentiating (70) with respect to s and using the Frenet formula (2), we get By taking the inner product on both sides of (71) with N and B 2 respectively, we get i = o and j = 0 which implies that i and j are constants.From (69), we get Moreover, (71) turns into then (73) turns into Case 1: t = 0.By differentiating (75) about s and using the Frenet Formula (2), we get By taking the inner product from both sides of (77) with T and B 1 respectively, we get r = 0 and t = 0, which implies that r and t are constants.In this case, (77) turns into then (78) turns into Since T * ⊥ B * 1 , it follows from (70) and (80) that σ ς = − j i , which implies that From ( 76) and (81), we can see that Since t = 0, it follows from (82) that t = r.Hence (80) turns into By differentiating (83) about s using (2), we get From which we have It follows from (85) that From ( 74), ( 83) and (86), we easily acquire our desired theorem.
Case 2: For t = 0, we have the following theorem.
Theorem 6.Let α : I → E 4 1 be a null Cartan curve with arc-length parameter s so that k 1 = 1, k 2 and k 3 are not zero.If α possesses the (1,3)-evolute mate curve, α * (s) = α(s) + 1 ik 1 (s) [iN(s) + jB 2 (s)], then k 2 and k 3 satisfy ik 2 − jk 3 = 0, where i and j are given constants.Moreover, the three curvatures of α * are given by k The associated Frenet frames are given by Proof.For this case, we may suppose that then (99) turns into By differentiating (101) with respect to s and using the Frenet formula (2), we get By taking the inner product from both sides of (102) with N and B 2 , respectively, we get i = o and j = 0 which implies that i and j are constants.From (100), we get Moreover, (102) turns into then (104) turns into Case 1: t = 0.By differentiating (106) about s and using the Frenet Formula (2), we get By taking the inner product on both sides of (108) with T and B 1 , respectively, we get r = 0 and t = 0, which implies that r and t are constants.In this case, (108) turns into then (109) turns into Since T * ⊥ B * 1 , it follows from (101) and (111) that σ ς = − j i , which implies that From ( 107) and (112), we get Since t = 0, it follows from (113) that t = r.Hence, (111) turns into By differentiating (114) about s using (2), we get From which we have It follows from (116) that From ( 106), ( 114) and (117), we can easily acquire our desired theorem.
Case 2: For t = 0, we have the following theorem.
Theorem 8. Let α : I → E 4 1 be a null Cartan curve with arc-length parameter s so that k 1 = 1, k 2 and k 3 are not zero.If α possesses the (1,3)-evolute mate curve α * (s) = α(s) + 1 ik 1 (s) [iN(s) + jB 2 (s)], then k 2 and k 3 satisfy ik 2 − jk 3 = 0, where i and j are given constants.Moreover, the three curvatures of α * are given by ). (118) The associated Frenet frames are given by Proof.In this case, we may suppose that Moreover, from (112) and the third equation of (120), we get By differentiating (120) about s and using (2), we get It follows that we can choose By differentiating (123) about s using the Frenet Formula (2) and using the third equation of (120), we get From ( 120), ( 123) and (124), we can easily acquire our desired theorem.

Conclusions
This paper established new kinds of generalized evolute and involute curves in four-dimensional Minkowski space by providing the necessary and sufficient conditions for the curves possessing generalized evolute and involute curves.Furthermore, the study invoked a new type of (1,3)-evolute and (0,2)-evolute curve in four-dimensional Minkowski space.The study also provided a new kind of generalized null Cartan curve in four-dimensional Minkowski space.For this new type of curve, the study provided several theorems with necessary and sufficient conditions and obtained significant 4 1 be the (1,3)-evolute curve of α. {T * , N * , B * 1 , B * 2 } is the Frenet frame along α * and k * 1 , k * 2 and k * 3 are the curvatures of α * .Then, 4 1 , G), where E 4 1 = {y = (y 1 , y 2 , y 3 , y 4 )|y i ∈ R} and G = −dy 2 1 + dy 2 2 + dy 2 3 + dy 2 4 .For any M