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(k,m)-type slant helices for partially null and pseudo null curves in Minkowski space 𝔼14{\rm{\mathbb E}}_1^4


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Introduction

The curve theory has been one of the most studied research area because of having many application area from geometry to the various branch of science. Especially the characterizations on the curvature and torsion play important role to define special curve types such as so-called helices. The curves of this type have drawn great attention in science. Helices appear naturally in structures of DNA, nanosprings. They are also widely used in engineering and architecture. The concept of slant helix defined by Izumiya and Takeuchi [6] based works have been studied in various spaces. For instance in [1] authors extended slant helix concept to En and conclude that there are no slant helices with nonzero constant curvatures in the space E4. The subject is also considered in 3–, 4–, and n–dimensional Eucliedan spaces, respectively in [7, 10, 12]. Moreover different properties of helices are also considered in [8, 9, 10, 11, 13, 18]. On the other hand in A.T. Ali, R. Lopez and M. Turgut extended this study to the k-type slant helix in E14E_1^4 . In this study they called α curve as k-type slant helix if there exists on (non-zero) constant vector field UE14U \in E_1^4 such that 〈 Vk + d, U〉 = const, for 0 ≤ k ≤ 3. Here Vk+1 shows the Frenet vectors of this curve [2].

One may easily conclude that O-type slant helices are general helices and 1-type slant helices correspond just slant helices. k-type slant helices for partially null and pseudo null curves are also studied. In accordance with above studies, the authors introduced (k,m)-type slant helices in E4 and we show that there do not exist (1, m) type slant helices in E4 [15].

In the present work, we define the notion of (k,m)-type slant helices in Minkowski 4-space and express some characterizations for partially and pseudo null curves in E14E_1^4 .

Preliminaries

Because of the indefiniteness of the Lorentzian metric g in 𝔼14{\rm{\mathbb E}}_1^4 vector v in this space can have one of three causal characters called spacelike (g(u,u) > O or u = O), timelike (g(u,u) < O) and lightlike (null) (g(u,u) = O, uO), respectively. In accordance with the metric, a curve in 𝔼14{\rm{\mathbb E}}_1^4 is called spacelike, timelike or lightlike if its velocity vectors al α(s) are spacelike, timelike and lightlike, respectively. In 𝔼14{\rm{\mathbb E}}_1^4 , if u is a unit vector, we know that g(u,u) = ±1 and the norm of a vector u is given by ||u||=|g(u,u)|||u|| = \sqrt {|g(u,u)|} . In addition a spacelike or timelike curve is said to be arclength parametrized if a α(s) is a unit vector for any s [2].

Suppose that α = α(s) is a spacelike curve with its arclength parameter. The Frenet frame along the curve a can be denoted by {T (s), N(s), B1(s), B2(s)}. Here T,N,B1, B2 are called tangent, principal normal, the first binormal and the second binormal vector fileds of the curve α, respectively. Because of the indefiniteness of the metric g, the vectors N,B1 and B2 have different causal characters. In this study we will assume that T is spacelike, since α is a spacelike curve.

Definition 2.1

A spacelike curve called partially null curve if N is spacelike and B1 is lightlike [16, 17].

For partially null curves the second binormal B2 is the only lightlike vector orthogonal to T and N such that g(B1, B2) = 1. The Frenet equations are given as follows [TNB1B2]=[0κ00κ0τ000σ00τ0σ][TNB1B2][16,17]\left[ {\matrix{{T'} \cr {N'} \cr {B'_1} \cr {B'_2} \cr}} \right] = \left[ {\matrix{0 & \kappa & 0 & 0 \cr {- \kappa} & 0 & \tau & 0 \cr 0 & 0 & \sigma & 0 \cr 0 & {- \tau} & 0 & {- \sigma} \cr}} \right]\left[ {\matrix{T \cr N \cr {{B_1}} \cr {{B_2}} \cr}} \right]\;\;\;[16,17] where κ, τ and σ are first, second and third curvature of the curve α, respectively. Note that after a null rotation of the ambient space the third curvature σ can be chosen as zero, and τ is determined up to a constant which means that any partially null curve lies in a three dimensional lightlike subspace orthogonal to B1.

Definition 2.2

A spacelike curve called pesudo null curve if α(s) is a lightlike vector for all s where the normal vector is N = T [16, 17].

For the case N is lightlike the curve a lies in the lightlike plane which we omit this trivial case. For the other cases, B1 is a unit spacelike vector orthogonal to {T,N} and B2 is the only lightlike vector orthogonal to T and B1 such that g(N,B2) = 1. The Frenet equations are given as follows [TNB1'B2']=[0κ0000τ00σ0τκ0σ0][TNB1B2][16,17]\left[ {\matrix{{T'} \cr {N'} \cr {B'_1} \cr {B'_2} \cr}} \right] = \left[ {\matrix{0 & \kappa & 0 & 0 \cr {- 0} & 0 & \tau & 0 \cr 0 & \sigma & 0 & {- \tau} \cr {- \kappa} & {- 0} & {- \sigma} & 0 \cr}} \right]\left[ {\matrix{T \cr N \cr {{B_1}} \cr {{B_2}} \cr}} \right]\;\;\;[16,17]

In this case the first curvature κ, can take only 0 and 1 values. As is well known the curve is a straight line if curvature vanishes. We will focus on the cases κ = 1 and στO [2].

(k,m)-type slant helices for partially null curves in 𝔼14{\rm{\mathbb E}}_1^4

In this section, we will define (k,m) type partially null slant helices in 𝔼14{\rm{\mathbb E}}_1^4 .

From (2.1) we have that σ = O. We also suppose that κ, τO.

Definition 3.1

Let α be a regular unit speed curve in 𝔼14{\rm{\mathbb E}}_1^4 with Frenet frame {V1, V2, V3, V4}. We call α is a (k,m) type slant helix if there exists a non-zero constant vector field U𝔼14U \in {\rm{\mathbb E}}_1^4 satisfies 〈Vk, U〉 = α and 〈Vm, U〉 = b(a,b constant) for 1 ≤ k,m ≤ 4, km. The constant vector U is on axis of (k,m)-type slant helix. We decompose U with respect to Frenet frame {T,N,B1, B2} as U = u1T + u2N + u3B1 + u4B2, where ui = ui(s) are differentiable functions of s. Here we denote V1 = T,V2 = N,V3 = B1, V4 = B2. From now on, in sake of easinesss, we will use these notations and assume that kiO, (1 ≤ i ≤ 3).

Theorem 3.1

There are no (1, 2) type partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (1, 2) type slant helix. Then for a constant vector field U. g(T,U) = a is constant and g(N,U) = b is constant. Differentiating this equation and using Frenet equations, we obtain κg(N,U) = O means that U is orthogonal to N. Hence there are no (1, 2) type partially null slant helices in 𝔼14{\rm{\mathbb E}}_1^4 .

Theorem 3.2

Let α be a (1, 3) partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then α is a general helix.

Proof

Assume that α is a (1, 3) partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we may write 〈T, U〉 = const = aB, U〉 = const = b. Also taking account Theorem 3.1 we decompose U as follows U=αT+bB1+u1B2U = \alpha T + b{B_1} + {u_1}{B_2}0=U=a(κN)+b(σB1)+u1'B2+u1(τNσB2)0 = U' = a(\kappa N) + b(\sigma {B_1}) + u_1'{B_2} + {u_1}(- \tau N - \sigma {B_2})0=(aκu1τ)N+(bσ)B1+(u1'u1σ)B20 = (a\kappa - {u_1}\tau)N + (b\sigma){B_1} + (u_1' - {u_1}\sigma){B_2}aκu1τ=0a\kappa - {u_1}\tau = 0(u1'u1σ)=0- (u_1' - {u_1}\sigma) = 0

From Definition 2.1 we have chosen σ = O which means that u1=0{u'_1} = 0 hence u1 = constant. From (3.3) we get κτ=constant{\kappa \over \tau} = constant . Then this completes the proof.

Theorem 3.3

Let α be a (1, 4) partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then α is a general helix.

Proof

Assume that α is a (1, 4) type slant helix. Then we may write U=aT+u1B1+bB2U = aT + {u_1}{B_1} + b{B_2}0=U=a(κN)+u1'B1+u1(σB1)+b(τNσb2)0 = U' = a(\kappa N) + u_1'{B_1} + {u_1}(\sigma {B_1}) + b(- \tau N - \sigma {b_2})0=(aκbτ)N+(u1'+u1σ)B1+bσb20 = (a\kappa - b\tau)N + (u_1' + {u_1}\sigma){B_1} + b\sigma {b_2}

From Definition 2.1 we have chosen σ = O which means that u1=0{u'_1} = 0 hence u1 = constant. From (3.3) we get κτ=constant{\kappa \over \tau} = constant . Then this completes the proof.

Theorem 3.4

There are no (2, 3) partially null type slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (2, 3) partially null type slant helix in type slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we may write U=u1T+aN+bB1+u2B2U = {u_1}T + aN + b{B_1} + {u_2}{B_2}0=U=ua(κN)+u1'T+a(κT+τB1)+b(σB1)+u2'B2+u2(τNσB2)0 = U' = {u_a}(\kappa N) + u_1'T + a(- \kappa T + \tau {B_1}) + b(\sigma {B_1}) + u_2'{B_2} + {u_2}(- \tau N - \sigma {B_2})0=U=(u1'aκ)T+(u1κu2τ)N+(aτ+bσ)B1+(u2'u2σ)B20 = U' = (u_1' - a\kappa)T + ({u_1}\kappa - {u_2}\tau)N + (a\tau + b\sigma){B_1} + (u_2' - {u_2}\sigma){B_2}

Taking into account of σ = O we get a = O which is a contradiction. Hence there are no (2, 3) type slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Theorem 3.5

There are no (2, 4) type partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (2, 4) partially null type slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we may write U=u1T+aN+u2B1+bB2U = {u_1}T + aN + {u_2}{B_1} + b{B_2}0=U=u1'T+u1(κN)+a(κT+τB1)+u2'B1+u2(σB1)+b(τNσB2)=(u1'0aκ)T+(u1κbτ)N+(aτ+u2'+u2σ)B1+(bσ)B2\matrix{{0 = U' = u_1'T + {u_1}(\kappa N) + a(- \kappa T + \tau {B_1}) + u_2'{B_1} + {u_2}(\sigma {B_1}) + b(- \tau N - \sigma {B_2})} \hfill \cr {\,\,\,\,\, = (u_1'0a\kappa)T + ({u_1}\kappa - b\tau)N + (a\tau + u_2' + {u_2}\sigma){B_1} + (- b\sigma){B_2}} \hfill \cr}

Noting that σ = O we get b = O which means that there are no (2, 4) type partially null type slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Theorem 3.6

There are no (3, 4) type partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (3, 4) type partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we may write U=u1T+u2N+aB1+bB2U = {u_1}T + {u_2}N + a{B_1} + b{B_2}0=U=u1'T+u1(κN)+u2'N+u2(κT+τB1)+a(σB1)+b(τNσB2)0 = U' = u_1'T + {u_1}(\kappa N) + u_2'N + {u_2}(- \kappa T + \tau {B_1}) + a(\sigma {B_1}) + b(- \tau N - \sigma {B_2})0=(u1'u2κ)T+(u1κ+u2'bτ)N+(u2τ+aσ)B1+(u2τ+aσ)B1+(bσ)B20 = (u_1' - {u_2}\kappa)T + ({u_1}\kappa + u_2' - b\tau)N + ({u_2}\tau + a\sigma){B_1} + ({u_2}\tau + a\sigma){B_1} + (- b\sigma){B_2}

In virtue of σ = O we get b = O which means that there are no (3, 4) type partially null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

(k,m)-type slant helices for pseudo null curves in 𝔼14{\rm{\mathbb E}}_1^4

In this part, we will focus on the (k,m)-type pesudo null slant helices. Recall that we assume κ = 1 and σ, τO.

Theorem 4.1

There are no (1, 2) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (1, 2) type pseudo null slant helix. Then for a constant vector field U. g(T,U) = a is constant and g(N,U) = b is constant. Differentiating this equation and using Frenet equations, we obtain κg(N,U) = O means that U is orthogonal to N. Hence there are no (1, 2) type pseudo null slant helices in 𝔼14{\rm{\mathbb E}}_1^4 .

Theorem 4.2

There are no (1, 3) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (1, 3) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we may write 〈T,U〉 = const = aB,U〉 = const = b. Also taking account Theorem 3.1 we decompose U as follows U=aT+bB1+u1B2U = aT + b{B_1} + {u_1}{B_2}0=U=a(N)+b(σNτB2)+u1'B2+u1(TσB1)0 = U' = a(N) + b(\sigma N - \tau {B_2}) + u_1'{B_2} + {u_1}(- T - \sigma {B_1})0=(u1)T+(a+bσ)N+(u1σ)B1+(u1'bτ)B20 = (- {u_1})T + (a + b\sigma)N + (- {u_1}\sigma){B_1} + (u_1' - b\tau){B_2}u1=0{u_1} = 0(a+bσ)=0(a + b\sigma) = 0u1'+bτ=0u_1' + b\tau = 0

From (3.25) and (3.27) we conclude b = O which is a contradiction. Hence there do not exist (1, 3) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Theorem 4.3

There are no (1, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that a is a (1, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we decompose U as follows U=aT+u1B1+bB2U = aT + {u_1}{B_1} + b{B_2}0=U=a(N)+b(TσB1)+u1'B1+u1(σNτB2)0 = U' = a(N) + b(- T - \sigma {B_1}) + u_1'{B_1} + {u_1}(\sigma N - \tau {B_2})0=(b)T+(a+u1σ)N+(u1'bσ)B1+(u1τ)B20 = (- b)T + (a + {u_1}\sigma)N + (u_1' - b\sigma){B_1} + (- {u_1}\tau){B_2}

Using (3.25) we get b = O which is a contradiction. Hence there do not exist (1, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

From the theorem proved above we conclude the following corollary.

Corollary 4.1

There do not exist (1, k) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 for 2 ≤ k ≤ 4.

Theorem 4.4

Let α be a (2, 3) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 if and only (aστdsτ)bτds=0- \left({a\sigma {{\int \tau ds} \over \tau}} \right) - b\int \tau ds = 0

Proof

Assume that α is a (2, 3) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we decompose U as follows U=u1T+aN+bB1+u2B2U = {u_1}T + aN + b{B_1} + {u_2}{B_2}0=U=u1'T+u1(N)+a(τB1)+b(σNτB2)+u2'B2+u2(TσB1)0 = U' = u_1'T + {u_1}(N) + a(\tau {B_1}) + b(\sigma N - \tau {B_2}) + u_2'{B_2} + {u_2}(- T - \sigma {B_1})0=(u1'u2)T+(u1+bσ)N+(aτu2σ)B1+(u2'bτ)B20 = (u_1' - {u_2})T + ({u_1} + b\sigma)N + (a\tau - {u_2}\sigma){B_1} + (u_2' - b\tau){B_2}

Hence we get u1'u2=0u_1' - {u_2} = 0u1+bσ=0{u_1} + b\sigma = 0aτu2σ=0a\tau - {u_2}\sigma = 0(u2'bτ)=0(u_2' - b\tau) = 0 Using (4.17) we get u2 = bτds and taking into account of (4.15) with (4.14) we get u1=aστdsτ{u_1} = - a\sigma {{\int \tau ds} \over \tau} . Considering these facts in (4.13) we conclude the desired proof.

Corollary 4.2

An axis of a (2, 3) type pseudo null slant helix is the vector given by D=[στdsτ]T+N+τdsτB1+[bτds]B2D = \left[ {\sigma {{\int \tau ds} \over \tau}} \right]T + N + {{\int \tau ds} \over \tau}{B_1} + \left[ {b\int \tau ds} \right]{B_2}

Theorem 4.5

There are no (2, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (2, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we decompose U as follows U=u1T+aN+u2B1+bB2U = {u_1}T + aN + {u_2}{B_1} + b{B_2}0=U=u1T+u1(N)+a(τB1)+b(TσB1)+u2'B+u2(σNτB2)0 = U' = {u'_1}T + {u_1}(N) + a(\tau {B_1}) + b(- T - \sigma {B_1}) + u_2'B + {u_2}(\sigma N - \tau {B_2})0=(u1'b)T+(u1+u2σ)N+(u2'+aτbσ)B1+(u2τ)B20 = (u_1' - b)T + ({u_1} + {u_2}\sigma)N + (u_2' + a\tau - b\sigma){B_1} + (- {u_2}\tau){B_2}u1'b=0u_1' - b = 0u1+u2σ=0{u_1} + {u_2}\sigma = 0u2'+aτbσ=0u_2' + a\tau - b\sigma = 0u2τ=0- {u_2}\tau = 0

From (4.23) we conclude that u2 = O and considering this fact in (4.20) we also see that u1 = O which means a contradiction. Hence there do not exist (2, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Theorem 4.6

There are no (3, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 .

Proof

Assume that α is a (3, 4) type pseudo null slant helix in 𝔼14{\rm{\mathbb E}}_1^4 . Then we decompose U as follows U=u1T+u2N+aB1+bB2U = {u_1}T + {u_2}N + a{B_1} + b{B_2}0=U=u1'T+u1(N)+u2'N+u2(τB1)+a(σNτB2)+b(TσB1)0 = U' = u_1'T + {u_1}(N) + u_2'N + {u_2}(\tau {B_1}) + a(\sigma N - \tau {B_2}) + b(- T - \sigma {B_1})9=(u1'b)T+(u1+u2'+aσ)N+(u2τbσ)B1+(aτ)B29 = (u_1' - b)T + ({u_1} + u_2' + a\sigma)N + ({u_2}\tau - b\sigma){B_1} + (- a\tau){B_2}u1'b=0u_1' - b = 0u1+u2'+aσ=0{u_1} + u_2' + a\sigma = 0u2τbσ=0{u_2}\tau - b\sigma = 0aτ=0- a\tau = 0

Taking into account of (3.55) we see that a = O and this is a contradiction.

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