Another way in the solution of the problem is the relationship between the Frenet vectors of the curves (see [6

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http://dx.doi.org/10.5831/HMJ.2014.36.3.475

CURVE COUPLES AND SPACELIKE FRENET PLANES IN MINKOWSKI 3-SPACE

Ali Uc¸um, Kazim ˙Ilarslan^∗, and Siddika ¨Ozkaldi Karakus¸

Abstract. In this study, we have investigated the possibility of whether any spacelike Frenet plane of a given space curve in Minkowski 3-space E³1also is any spacelike Frenet plane of another space curve in the same space. We have obtained some characterizations of a given space curve by considering nine possible case.

1. Introduction

In the theory of curves in Euclidean space, one of the important and interesting problem is characterization of a regular curve. In the solution of the problem, the curvature functions k₁(or κ) and k2(or τ ) of a regular curve have an effective role. For example: if k1 = 0 = k2, then the curve is a geodesic or if k₁ = constant 6= 0 and k₂ = 0, then the curve is a circle with radius (1/k1), etc. Thus we can determine the shape and size of a regular curve by using its curvatures.

Another way in the solution of the problem is the relationship between the Frenet vectors of the curves (see [6]). For instance Bertrand curves:

In 1845, Saint Venant (see [6] and [8]) proposed the question whether upon the surface generated by the principal normal of a curve, a second curve can exist which has for its principal normal the principal normal of the given curve. This question was answered by Bertrand in 1850 in a paper which he showed that a necessary and sufficient condition for the existence of such a second curve is that a linear relationship with constant coefficients shall exist between the first and second curvatures of the given original curve. In other word, if we denote first and second

Received April 16, 2014. Accepted June 18, 2014.

2010 Mathematics Subject Classification. 53C40.

Key words and phrases. Frenet planes, curvatures, circular helix, generalized helix, rectifying curve, Mannheim curve, Salkowski and anti-Salkowski curve, Minkowski 3-space.

∗Corresponding author

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curvatures of a given curve by k1 and k2 respectively, then for λ, µ ∈ R we have λk₁ + µk₂ = 1. Since the time of Bertrand’s paper, pairs of curves of this kind have been called Conjugate Bertrand Curves, or more commonly Bertrand Curves.

Another interesting example is Mannheim curves: If there is exist a corresponding relationship between the space curves α and β such that, at the corresponding points of the curves, the principal normal lines of α coincides with the binormal lines of β, then α is called a Mannheim curve, β is called Mannheim partner curve of α. Mannheim partner curves was studied by Liu and Wang (see [7]) in Euclidean 3-space and Minkowski 3-space.

The other way in the solution of the problem is the relationship between the Frenet planes of the curves. In ([5]), the authors asked the following question and investigated the possible answers of the question:

Is it possible that one of the Frenet planes of a given curve in E³ be a Frenet plane of another space curve in the same space? Then they give many interesting results.

In this paper, we consider the same question for only spacelike Frenet planes of given curves in Minkowski 3-space E³1. We study the nine possible cases and we get the following results. The timelike and lightlike cases for Frenet planes of given curves in Minkowski 3-space E³1 will be considered in our next papers.

2. Preliminaries

The Minkowski space E³₁ is the Euclidean 3-space E³ equipped with indefinite flat metric given by

g = − dx²₁+ dx²₂+ dx²₃,

where (x₁, x₂, x₃) is a rectangular coordinate system of E³₁. Recall that a vector v ∈ E³1\{0} can be spacelike if g(v, v) > 0, timelike if g(v, v) < 0 and null (lightlike) if g(v, v) = 0 and v 6= 0. In particular, the vector v = 0 is a spacelike. The norm of a vector v is given by ||v|| =p|g(v, v)|, and two vectors v and w are said to be orthogonal, if g(v, w) = 0. An arbitrary curve α(s) in E³1, can locally be spacelike, timelike or null (lightlike), if all its velocity vectors α⁰(s) are respectively spacelike, timelike or null. A spacelike or a timelike curve α(s) has unit speed, if g(α⁰(s), α⁰(s)) = ±1 ([4]).

Let {T, N, B} be the moving Frenet frame along a curve α in E³1, consisting of the tangent, the principal normal and the binormal vector

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fields respectively. If α is a non-null curve, the Frenet equations are given by ([6]):

(2.1)



 T⁰ N⁰ B⁰



=





0 2k1 0

−₁k₁ 0 ₃k₂ 0 −₂k2 0







 T N B





where k1 and k2 are the first and the second curvature of the curve respectively. Moreover, the following conditions hold:

g(T, T ) = 1 = ±1, g(N, N ) = 2 = ±1, g(B, B) = 3= ±1, and g(T, N ) = g(T, B) = g(N, B) = 0.

3. On The Curves And Spacelike Frenet Planes In Minkowski 3-Space

Let us consider the given two space curves C and C, defined on the same open interval I ⊂ R. Let us attach moving triads {C, T, N, B} and {C, T , N , B} to C and C at the corresponding points of C and C. We denote the arcs, curvatures and torsions of C and C by s, k1, k2and s, k1, k₂respectively. At each point C(s) of the curve C, the planes spanned by {T, N }, {N, B} , {T, B} are known respectively as the osculating plane, the normal plane and the rectifying plane. We denote these planes by OP, N P and RP , respectively. Now, we assume that C be a arbitrary unit speed space curve with curvatures k1, k2 and Frenet vectors T , N , B. At each point C(s) of the curve C, the planes spanned by T , N ,

N , B , T , B are known respectively as the osculating plane, the normal plane and the rectifying plane. We denote these planes by OP , N P and RP , respectively. Let

ds ds = f⁰.

In this section, we ask the following question:

“Is it possible that one of the spacelike Frenet planes of a given curve be a spacelike Frenet plane of another space curve? ” and we investigate the answer of the question. For this, we consider the following possible cases:

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Case Frenet plane of C Frenet plane of C Condition 1 sp {T, N } = OP spT , N = OP OP = OP 2 sp {T, N } = OP spN , B = N P OP = N P 3 sp {T, N } = OP spT , B = RP OP = RP 4 sp {N, B} = N P spT , N = OP N P = OP 5 sp {N, B} = N P spN , B = N P N P = N P 6 sp {N, B} = N P spT , B = RP N P = RP 7 sp {T, B} = RP spT , N = OP RP = OP 8 sp {T, B} = RP spN , B = N P RP = N P 9 sp {T, B} = RP spT , B = RP RP = RP Now, we investigate these possible cases step by step.

If the OP = sp {T, N } is spacelike, then T spacelike, N spacelike and B timelike. We have the following Frenet formulae:

(3.1)



 T⁰ N⁰ B⁰



=





0 k1 0

−k₁ 0 −k₂ 0 −k₂ 0







 T N B



,

g(T, T ) = 1 g(N, N ) = 1 g(B, B) = −1

Case 1. OP = OP

In this case, we investigate the answer of the following question:

“Is it possible that the spacelike osculating plane of a given space curve be the spacelike osculating plane of another space curve in E³1?”.

Now, we investigate the answer of the question. We assume that the spacelike osculating plane of the given curve C is the spacelike osculating plane of another space curve C. Since the osculating plane of C is a spacelike plane and spanned by spacelike vector T and spacelike vector N , its binormal vector field B is a timelike vector. Then, C is spacelike curve satisfying the Frenet formulae (2.1) with ₃ = −1 and ₁ = ₂ = 1.

Thus, we have the following relation

(3.2) X = X + aT + bN, a 6= 0, b 6= 0

where X and X are the position vectors of the curves C and C respectively, and a and b are the non-zero functions of the parameter s. By taking the derivative of (3.2) with respect to s and applying the Frenet formulas given in (3.1), we get

(3.3) T f⁰ = (1 + a⁰− bk₁)T + (ak₁+ b⁰)N − bk₂B.

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Since T ∈ Sp {T, N } , we can write T = λT + µN, for some constant λ and µ. From (3.3) we have,

(3.4) (λT + µN ) f⁰ = (1 + a⁰− bk₁)T + (ak1+ b⁰)N − bk2B.

Multiplying Eq.(3.4) by B, we obtain bk₂ = 0.

Thus, b or k₂ must be zero which is contradiction with our assumption.

Therefore, we give the following theorem:

Theorem 1. There is no exist a pair of space curves C, C for which spacelike osculating plane of C is spacelike osculating plane of C.

Case 2. OP = N P

In this case, we investigate the answer of the following question: “Can the spacelike osculating plane of a given space curve be the spacelike normal plane of another space curve?”.

Now, we investigate the answer of the question. We assume that the spacelike osculating plane of the given curve C is the spacelike normal plane of another space curve C. Since the normal plane of C is a spacelike plane and spanned by spacelike vector N and spacelike vector B, its tangent vector field T is a timelike vector. Then, C is a timelike curve satisfying the Frenet formulae (2.1) with ₁ = −1 and ₂ = ₃= 1. Thus, we have the following relation

(3.5) X = X + aT + bN, a 6= 0, b 6= 0,

where X and X are the position vectors of the curves C and C respectively, and a and b are the non-zero functions of the parameter s. By taking the derivative of (3.5) with respect to s and applying the Frenet formulae given in (3.1), we get

(3.6) T f⁰ = (1 + a⁰− bk₁)T + (ak₁+ b⁰)N − bk₂B.

First, multiplying Eq.(3.6) by B, we obtain

(3.7) b = −f⁰

k2

. Next, multiplying Eq.(3.6) by N , we have

a = 1 k₁

f⁰ k₂

0

. Thus, we prove the following theorem:

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Theorem 2. Let C be a given unit speed curve with non-zero curvatures k₁, k₂ and Frenet vectors T, N , B. If the spacelike osculating plane of the curve C is the spacelike normal plane of another space curve C, then C has the following form

(3.8) C = C + 1

k1

1 k2

ds ds

0

T − 1 k2

ds dsN.

Conclusion 1. Without loss of generality, we assume that the curves C and C have the same parameter s and k₁= k₂ = constant 6= 0.Then, the curve C is a spacelike pseudo spherical helix lying in S₁². In this case, C is a timelike pseudo spherical helix lying in H₂⁰ with curvature k1= ±k2 and k1 = k2 = k1.

Case 3. OP = RP

In this case, we investigate the answer of the following question: “Can the spacelike osculating plane of a given space curve be the spacelike rectifying plane of another space curve?”.

Now, we investigate the answer of the question. We assume that the spacelike osculating plane of the given curve C is the spacelike rectifying plane of another space curve C. Since the rectifying plane of C is a spacelike plane and spanned by spacelike vector T and spacelike vector B, its normal vector field N is a timelike vector. Then, C is a spacelike curve satisfying the Frenet formulae (2.1) with 2 = −1 and 1 = 3 = 1.

(3.9) X = X + aT + bN, a 6= 0, b 6= 0,

(3.10) T f⁰ = (1 + a⁰− bk₁)T + (ak1+ b⁰)N − bk2B.

Since T ∈ Sp {T, N } , we can write T = λT + µN, for some constant λ and µ. From (3.10) we have,

(3.11) (λT + µN ) f⁰ = (1 + a⁰− bk₁)T + (ak₁+ b⁰)N − bk₂B.

Multiplying Eq.(3.11) by B, we obtain bk₂ = 0.

Thus, b or k₂ must be zero which is contradiction with our assumption.

Therefore, we give the following theorem:

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Theorem 3. There is no exist a pair of space curves C, C for which spacelike osculating plane of C is spacelike rectifying plane of C.

If the normal plane of C, that is N P = sp {N, B}, is a spacelike plane, then N spacelike, B spacelike and T timelike. Thus, the curve C is a timelike curve satisfying the following Frenet formulae:

(3.12)



 T⁰ N⁰ B⁰



=





0 k₁ 0

k1 0 k2

0 −k₂ 0







 T N B



,

g(T, T ) = −1 g(N, N ) = 1 g(B, B) = 1

Case 4. N P = OP

In this case, we investigate the answer of the following question: “Can the spacelike normal plane of a given space curve be the spacelike osculating plane of another space curve?”.

Now, we investigate the answer of the question. Let assume that the spacelike normal plane of the given curve C is the spacelike osculating plane of another space curve C. Since the osculating plane of C is a spacelike plane and spanned by spacelike vector T and spacelike vector N , its binormal vector field B is a timelike vector. Then, C is a spacelike curve satisfying the Frenet formulae (2.1) with ₃ = −1 and ₁ = ₂ = 1.

(3.13) X = X + aN + bB, a 6= 0, b 6= 0,

where X and X are the position vectors of the curves C and C respectively, and a and b are the non-zero functions of the parameter s. By taking the derivative of (3.13) with respect to s and applying the Frenet formulae given in (3.12), we have

(3.14) T f⁰ = (1 + ak1)T + (a⁰− bk₂)N + (b⁰+ ak2)B.

Multiplying Eq.(3.14) by T , we obtain

(3.15) a = − 1

k1

. Substituting (3.15) in (3.14), we get

(3.16) T f⁰= (a⁰− bk₂)N + (b⁰+ ak₂)B.

If we take derivative of (3.16) with respect to s, we obtain

T f⁰⁰+ (f⁰)²k₁N = (a⁰k₁− bk₂k₁)T + (a⁰⁰− 2b⁰k₂− bk⁰₂− ak₂²)N + (2a⁰k2+ ak⁰₂+ b⁰⁰− bk₂²)B.

(3.17)

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Multiplying Eq.(3.17) by T , we get

(3.18) b = k₁⁰

k₂k²₁. If we put (3.15) and (3.18) in (3.13), we have

(3.19) X = X − 1

k₁N + k₁⁰ k2k²₁B.

Thus, we prove the following theorem:

Theorem 4. Let C be a given unit speed curve with non-zero curvatures k1, k2 and Frenet vectors T, N , B. If the spacelike normal plane of the curve C is the spacelike osculating plane of another space curve C, then C has the following form

(3.20) C = C − 1

k1

N + k₁⁰ k2k²₁B.

Conclusion 2. Let C and C be space curves with non-zero curvatures in E³1. If the normal plane of the space curve C is osculating plane of the curve C, then C is the locus of the centers of the osculating spheres of the curve C.

Case 5. N P = N P

In this case, we investigate the answer of the following question: “Can the spacelike normal plane of a given space curve be the spacelike normal plane of another space curve?”.

Now, we investigate the answer of the question. We assume that the spacelike normal plane of the given curve C is the spacelike normal plane of another space curve C. Since the normal plane of C is a spacelike plane and spanned by spacelike vector N and spacelike vector B, its tangent vector field T is a timelike vector. Then, C is a timelike curve satisfying the Frenet formulae (2.1) with ₁ = −1 and ₂= ₃ = 1. Thus, we have the following relation

(3.21) X = X + aN + bB, a 6= 0, b 6= 0,

(3.22) T f⁰ = (1 + ak₁)T + (a⁰− bk₂)N + (b⁰+ ak₂)B.

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Multiplying Eq.(3.22) by T , we have

(3.23) a = f⁰− 1

k1

. Next, multiplying Eq.(3.22) by N , we get

b = 1 k2

f⁰− 1 k1

0

. Thus, we give the following theorem:

Theorem 5. Let C be a given unit speed curve with non-zero curvatures k₁, k₂ and Frenet vectors T, N , B. If the spacelike normal plane of the curve C is the spacelike normal plane of another space curve C, then C has the following form

(3.24) C = C + 1 k₁

−1 + ds ds

N + 1 k₂

1 k₁

−1 +ds ds

0

B.

Conclusion 3. Without loss of generality, we assume that the curves C and C have the same parameter, then C = C.

Case 6. N P = RP

In this case, we investigate the answer of the following question: “Can the spacelike normal plane of a given space curve be the spacelike rectifying plane of another space curve?”.

Now, we investigate the answer of the question. We assume that the spacelike normal plane of the given curve C is the spacelike rectifying plane of another space curve C. Since the rectifying plane of C is a spacelike plane and spanned by spacelike vector T and spacelike vector B, its normal vector field N is a timelike vector. Then, C is a spacelike curve satisfying the Frenet formulae (2.1) with ₂ = −1 and ₁ = ₃ = 1.

(3.25) X = X + aN + bB, a 6= 0, b 6= 0,

(3.26) T f⁰ = (1 + ak1)T + (a⁰− bk₂)N + (b⁰+ ak2)B.

Multiplying Eq.(3.26) by T , we obtain

(3.27) a = − 1

k₁.

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Substituting (3.27) in (3.26), we get

(3.28) T f⁰= (a⁰− bk₂)N + (b⁰+ ak2)B.

If we denote

(3.29) λ = a⁰− bk₂

f⁰ and µ = b⁰+ ak₂ f⁰ , we have

(3.30) T = λN + µB.

Differentiating the equation (3.30), we find

(3.31) −(f⁰)k₁N = λk₁T + (λ⁰− k₂µ)N + (λk₂+ µ⁰)B.

Multiplying Eq.(3.31) by N , we obtain

(3.32) λ⁰− k₂µ = 0.

Substituting Eq.(3.29) in (3.32), we get

(3.33) b⁰+ δb = γ

and

b(s) = e⁻^{R δds}

Z

e^{R δds}γ ds + t

, where t ∈ R.

where

(3.34) δ = f⁰k₂⁰ − f⁰⁰k₂ 2k₂f⁰ and

(3.35) γ = −(_k¹

1)⁰⁰f⁰+ (_k¹

1)⁰f⁰⁰+ ^f_k⁰^k²²

1

2k₂f⁰ . Thus, we give the following theorem:

Theorem 6. Let C be a given unit speed curve with non-zero curvatures k1, k2 and Frenet vectors T, N , B. If the spacelike normal plane of the curve C is the spacelike rectifying plane of another space curve C, then C has the following form

(3.36) C = C − 1 k1

N + e⁻^{R δds}

Z

e^{R δds}γ ds + t

B where

(3.37) δ = f⁰k₂⁰ − f⁰⁰k2

2k₂f⁰

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and

(3.38) γ = −(_k¹

1)⁰⁰f⁰+ (_k¹

1)⁰f⁰⁰+ ^f

0k²₂ k1

2k2f⁰ .

Conclusion 4. Without loss of generality, we assume that the curves C and C have the same parameter s (s = s) and if k₁ = constant 6= 0 and k2= constant 6= 0, then we get

b = k₂ 2k1

s + t,

for some t ∈ R. In this case, it is clear that the curve C is a circular helix.

Conclusion 5. Without loss of generality, we assume that the curves C and C have the same parameter s (s = s) and if k1 = constant 6= 0 and k₂6= 0, k₂ nonconstant function, then we get

b = 1

2k₁√ k₂

Z

(k2)³²ds + 1

√k₂t,

for some t ∈ R. In this case, the curve C is a timelike Salkowski curve in E³1.

Conclusion 6. Without loss of generality, we assume that the curves C and C have the same parameter s (s = s) and if k2 = constant 6= 0 and k₁6= 0, k₁ nonconstant function , then we get

b = − 1 2k₂

1 k₁

0

+k₂ 2

Z 1

k₁ds + c,

for some c ∈ R. In this case, the curve C is a timelike anti-Salkowski curve in E³1.

If the RP = sp {T, B} is spacelike then T spacelike, B spacelike and N timelike. We have the following Frenet formulae:

(3.39)



 T⁰ N⁰ B⁰



=





0 −k₁ 0

−k₁ 0 k₂ 0 −k₂ 0







 T N B



,

g(T, T ) = 1 g(N, N ) = −1

g(B, B) = 1

Case 7. RP = OP

In this case, we investigate the answer of the following question: “Can the spacelike rectifying plane of a given space curve be the spacelike osculating plane of another space curve?”.

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Now, we investigate the answer of the question. Let assume that the spacelike rectifying plane of the given curve C is the spacelike osculating plane of another space curve C. Since the osculating plane of C is a spacelike plane and spanned by spacelike vector T and spacelike vector N , its binormal vector field B is a timelike vector. Then, C is a spacelike curve satisfying the Frenet formulae (2.1) with 3 = −1 and 1 = 2 = 1.

(3.40) X = X + aT + bB, a 6= 0, b 6= 0,

where X and X are the position vectors of the curves C and C respectively, and a and b are the non-zero functions of the parameter s. By taking the derivative of (3.40) with respect to s and applying the Frenet formulae (3.39), we have

(3.41) T f⁰= (1 + a⁰)T + (bk₂− ak₁)N + b⁰B.

Then, multiplying Eq.(3.41) by N , we obtain

(3.42) bk2− ak₁= 0.

Substituting (3.42) in (3.41), we have

(3.43) T f⁰ = (1 + a⁰)T + b⁰B.

Differentiating equation (3.43) with respect to s, we have

(3.44) T f⁰⁰+ (f⁰)²k1N = a⁰⁰T + (b⁰k2− a⁰k1− k₁)N + b⁰⁰B.

Multiplying Eq.(3.44) by N, we obtain

(3.45) b⁰k₂− a⁰k₁− k₁= 0.

Differentiating (3.42) and substituting (3.45), we have a = k₁k₂

k₂k₁⁰ − k₁k⁰₂ = k₁ k₂

1 (^k_k¹

2)⁰, b = k²₁

k₂k₁⁰ − k₁k⁰₂ = − 1 (^k_k²

1)⁰. (3.46)

Substituting (3.46) in (3.40), we have (3.47) X = X + k1k2

k₂k₁⁰ − k₁k₂⁰T + k²₁

k₂k₁⁰ − k₁k₂⁰B.

Thus, we give the following theorem:

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Theorem 7. Let C be a given unit speed curve with non-zero curvatures k₁, k₂ and Frenet vectors T, N , B. If the spacelike rectifying plane of the curve C is the spacelike osculating plane of another space curve C, then C has the following form

(3.48) C = C + k₁k₂

k2k₁⁰ − k₁k⁰₂T + k²₁

k2k₁⁰ − k₁k₂⁰B.

Thus, we give the following conclusions:

Conclusion 7. Let be M and M be any points on the curves C and C respectively. If we denote the position vectors −−→

OM = X and

−−→

OM = X, by using (3.40) and (3.41), we have

(3.49)







−−−→

M M = aT + bB, T =

1 + a⁰

T + b⁰B

1 r. From (3.42) and (3.45) we find that −−−→

M M is parallel to T . This means that the vector −−−→

M M is tangent to the curve C at the point of the M . Conclusion 8. Since−−−→

M M ∈ Sp {T, B} , we can write

−−−→

M M = sin θT + cos θB, where θ is denote the angle between the vector

−−−→

M M and B. So, a

b = tan θ.

Also from (3.46) we get

tan θ = k2

k₁. Then the vector−−−→

M M is parallel to Darboux vector of C at the point M.

Definition 1. The vector

−−−→

M M is called rectifying line of the curve C at the point M ∈ C.

Conclusion 9. The θ in Conclusion 7 is constant if and only if the curve C is a general helix.

If we take the derivative of (3.43) and using (3.46) we obtain (3.50) ak₁⁰ − bk⁰₂= k₁.

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From (3.43) and (3.50) we get

(3.51) a = k₁

k2

1

k1

k2

0, and

(3.52) b = − 1

k2

k1

0.

Also, from (3.51) and (3.52), we can easily obtain that

(3.53)

M M

2 = k²₁ k²₁+ k₂² k2k₁⁰ − k₁k₂⁰ . Conclusion 10. If a = ^k_k¹

2

1

k1 k2

0 = constant, then we get k1

k₂ = c2e^c¹^s. Conclusion 11. If b = − ¹

k2 k1

⁰ = constant, then we get k₂

k1 = cs + d, c ∈ R, d ∈ R, which means that the curve C is a rectifying curve.

We note that such curves also called canonical geodesics by Izumiya and Takeuchi in [3].

Conclusion 12. If M M

2 = ^k

2 1(^k1²+k₂²)

k2k⁰₁−k₁k⁰₂ = constant, then we get k₂

k1

= sinh (c₁s + c₂) ,

which means that the curve C is a Mannheim curve (see [7]).

Case 8. RP = N P

In this case, we investigate the answer of the following question: “Can the spacelike rectifying plane of a given space curve be the spacelike normal plane of another space curve?”.

Now, we investigate the answer of the question. We assume that the spacelike rectifying plane of the given curve C is the spacelike normal plane of another space curve C. Since the normal plane of C is a spacelike plane and spanned by spacelike vector N and spacelike vector B, its

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tangent vector field T is a timelike vector. Then, C is a timelike curve satisfying the Frenet formulae (2.1) with ₁ = −1 and ₂ = ₃= 1. Thus, we have the following relation

(3.54) X = X + aT + bB, a 6= 0, b 6= 0,

where X and X are the position vectors of the curves C and C respectively, and a and b are the non-zero functions of the parameter s. By taking the derivative of (3.54) with respect to s and applying the Frenet formulae (3.39), we get

(3.55) T f⁰= (1 + a⁰)T + (bk₂− ak₁)N + b⁰B.

Multiplying Eq.(3.55) by N , T and B respectively, we obtain

(3.56) bk2− ak₁= f⁰,

(3.57) a = −s + c1 and b⁰= 0,

for some constant c1. Substituting (3.57) in (3.56), we find (3.58) b = f⁰+ (−s + c₁)k₁

k₂ = constant

for some constant c₁. Substituting (3.57) and (3.58) in (3.54), we have (3.59) X = X + (−s + c1)T + f⁰+ (−s + c1)k1

k2

B.

Thus, we give the following theorem:

Theorem 8. Let C be a given unit speed curve with non-zero curvatures k₁, k₂ and Frenet vectors T, N , B. If the spacelike rectifying plane of the curve C is the spacelike normal plane of another space curve C, then C has the following form

(3.60) C = C + (−s + c1)T + 1 k₂

ds

ds + (−s + c1)k1

B for some constant c1.

Case 9. RP = RP

In this case, we investigate the answer of the following question: “Can the spacelike rectifying plane of a given space curve be the spacelike rectifying plane of another space curve?”.

Now, we investigate the answer of the question. We assume that the spacelike rectifying plane of the given curve C is the spacelike rectifying plane of another space curve C. Since the rectifying plane of C is a spacelike plane and spanned by spacelike vector T and spacelike vector

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B, its normal vector field N is a timelike vector. Then, C is a spacelike curve satisfying the Frenet formulae (2.1) with ₂ = −1 and ₁ = ₃ = 1.

(3.61) X = X + aT + bB, a 6= 0, b 6= 0,

where X and X are the position vectors of the curves C and C respectively, and a and b are the non-zero functions of the parameter s. By taking the derivative of (3.61) with respect to s and applying the Frenet formulae (3.39), we get

(3.62) T f⁰= (1 + a⁰)T + (bk2− ak₁)N + b⁰B.

Multiplying Eq.(3.62) by N, we obtain

(3.63) bk₂− ak₁= 0.

(3.64) T f⁰ = (1 + a⁰)T + b⁰B.

If we denote,

(3.65) λ = 1 + a⁰

f⁰ and µ = b⁰ f⁰ we get,

(3.66) T = λT + µB.

Differentiating equation (3.66) with respect to s, we have (3.67) −f⁰k1N = λ⁰T + (µk2− λk₁)N + µ⁰B.

Multiplying Eq.(3.67) by T and B respectively, we obtain

(3.68) µ⁰ = 0 and λ⁰= 0.

(3.69) a = −s + c₁

Z

f⁰ ds + c₂ and

b = d₁ Z

f⁰ ds + d₂

for some constant c1, c2, d1, and d2. Thus we give the following theorem:

Theorem 9. Without loss of generality, we assume that the curves C and C have the same parameter s (s = s). Let C be a given unit speed curve in E³₁ with non-zero curvatures k₁, k₂ and Frenet vectors T,

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N , B. If the rectifying plane of the curve C is the rectifying plane of another space curve C, then C has the following form

(3.70) C = C + (c₁s + c₂) T + (d₁s + d₂) B.

where c1, c2, d1, d2∈ R.

From (3.63), we get

(3.71) k2

k1

= c1s + c2

d1s + d2

. Thus, we give the following conclusions:

Conclusion 13. If c₁d₂− c₂d₁= 0, then ^k_k¹

2 = constant. In this case, the curve of C is a circular helix or a generalized helix in E³1.

Example 1. Let us consider the circular helix C(s) =

√1

2cosh s ,

√1

2sinh s,^√¹

2s

in E³₁. The Frenet vectors and curvatures of the curve are easily obtained as follows:

T (s) =

sinh s√

2 ,^{cosh s}^√₂ ,^√¹₂

N (s) = (− cosh s, − sinh s, 0) B(s) =

−^{sinh s}^√

2 , −^{cosh s}^√

2 ,^√¹

2

k₁= ^√¹

2, k₂ = ^√¹

2.

Now, we assume that there exist a space curve C(s) which have the same rectifying plane with C. Then, by using the equations (3.70) and (3.71) and taking c1 = d1 = 0, c2 = d2 = √

2, we obtain that C(s) =

√1

2cosh s,^√¹

2sinh s,^√¹

2s + 2

. It is easily shown that C(s) and C(s) are congruent curves according to the Fundamental theorem of curves in E³₁.

Conclusion 14. If we take d₂ = 0 in (3.71) then we get k₁ k2

= c₁ d1

+ c₂

d₁ 1

s. In this case the curve of C is a rectifying curve up to parametriza- tion of the curve C.

Conclusion 15. If the rectifying plane of C is the rectifying plane of C, then (C, C) are Bertrand Mates.

Acknowledgement 1. The authors express thank to the referees for their valuable suggestions.

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References

[1] Chen, B. Y., When does the position vector of a space curve always lie in its rectifying plane?, Amer. Math. Monthly 110 (2003), 147-152.

[2] Chen, B. Y. and Dillen, F., Rectifying curves as centrodes and extremal curves, Bull. Inst. Math. Academia Sinica 33(2) (2005), 77-90.

[3] Izumiya, S. and Takeuchi, N., New special curves and developable surfaces, Turk- ish J. Math. 28(2) (2004), 153-163.

[4] O’Neill, B., Semi-Riemannian Differential Geometry, Academic Press, New York, 1983.

[5] ¨Ozkaldı Karaku¸s S., ˙Ilarslan K. and Yaylı, Y., A new approach for characterization of curve couples in Euclidean 3-space, Honam Mathematical J. 36(1) (2014), 113-129.

[6] Kuhnel, W., Differential geometry: curves-surfaces-manifolds, Braunschweig, Wiesbaden, 1999.

[7] Liu, H. and Wang, F., Mannheim partner curves in 3-space, Journal of Geometry, 88 (2008), 120-126.

[8] Millman, R. S. and Parker, G. D., Elements of differential geometry, Prentice- Hall, New Jersey, 1977.

Ali U¸cum

Department of Mathematics, Faculty of Science and Arts, University of Kırıkkale, Kırıkkale-TURKEY.

E-mail: aliucum05@gmail.com Kazim ˙Ilarslan

Department of Mathematics, Faculty of Science and Arts, University of Kırıkkale, Kırıkkale-TURKEY.

E-mail: kilarslan@yahoo.com Siddika ¨Ozkaldi Karaku¸s

Department of Mathematics, Faculty of Sciences and Arts, Bilecik S¸eyh Edebali University, Bilecik-TURKEY.

E-mail: sozkaldi98@yahoo.com