# Darboux Vector in Four-Dimensional Space-Time

Authors: Na Hu; Tingting Zhang; Yang Jiang
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

As the space-time model of the theory of relativity, four-dimensional Minkowski space is the basis of the theoretical framework for the development of the theory of relativity. In this paper, we introduce Darboux vector fields in four-dimensional Minkowski space. Using these vector fields, we define some new planes and curves. We find that the new planes are the instantaneous rotation planes of rigid body moving in four-dimensional space-time. In addition, according to some characteristics of Darboux vectors in geometry, we define some new space curves in four-dimensional space-time and describe them with curvature functions. Finally, we give some examples.


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## Body

## 1. Introduction

On the basis of the principle of relativity and Lorentz transformation, in 1907, Minkowski proposed to add a time dimension on the basis of three space dimensions, thus forming a four-dimensional space-time, and this space-time is also called Minkowski 4-space. The metric tensorg in E14 is given by
(1)g=dx12+dx22+dx32−dx42,where x1,x2,x3,x4 is a standard rectangular coordinate system in E14. Minkowski space is not only closely related to physics but also provides theoretical and methodological support for the study of astrophysics and cosmology [1–4]. The study of submanifolds in Minkowski space is of interest in relativity theory; therefore, more and more geometers and physicists are committed to the study of submanifolds in Minkowski space. For example, in [5], the authors studied some local properties of slant geometry on spacelike submanifolds of codimension two in Lorentz-Minkowski space and investigate spacelike curves in Lorentz-Minkowski 3-space from different viewpoints as another special case. In [6], the authors studied null helices of 1-dimensional lightlike submanifolds and gave some characterizations of null helices in ℝ13. We refer the reader to [7–17] and the references therein for more related works.The Darboux vector is the local speed vector of the Frenet frame of space curves, which was discovered and named after Gaston Darboux [18]. If an object moves along a regular curve, we can use the Frenet frame of space curves to describe the motion of the object in terms of two vectors: the translation vector and the rotation vector, where the rotation vector is the Darboux vector. Because the Darboux vector is directly related to the angular momentum, it is also called the angular momentum vector.In the past few decades, many researchers have mainly studied Darboux vectors in 3-dimensional space [19–26] and have obtained some interesting conclusions. For example, in 2012, Ziplar introduced and studied Darboux helices in Euclidean 3-space and proved that Darboux helices coincide with slant helices [19]. In [20], Öztürk and Nešovic′ defined the pseudo null and null Cartan Darboux helices in Minkowski 3-space and obtained the relationship between pseudo null, null Cartan Darboux helices, and slant helices. In [21], the quasi Darboux vector field of null curve in Minkowski 3-space was defined, and some interesting conclusions about osculating developable of null curve which is defined by quasi Darboux vector field of null curve were obtained. Wang and Pei defined the Darboux vector of the null curve in [23] and described the direction of the rotation axis of the Cartan frame in Minkowski 3-space. Later, in 2017, Düldül [27] extended the Darboux frame field to four-dimensional Euclidean space and gave the relationship between the curvature of Frenet frame and Darboux frame. In [28], Düldül defined some new vector fields in four-dimensional Euclidean space and showed that the determined new planes play the role of the Darboux vector. İlarslan and Yildirim [29] defined the Darboux helices in four-dimensional Euclidean space as a curve whose Darboux vector makes a constant angle with some fixed direction and obtained relation between the curves Darboux helix, general helix, and V4-slant helix in a special case.Motivated by those ideas, in this paper, we construct four new vector fields along the space curve whose curvatures do not disappear in four-dimensional space-time. Based on these vector fields, we define some new planes and helices in four-dimensional space-time. The corresponding curvature functions are given when the position vectors of the curves lie on different planes. Moreover, we define Darboux helices in Minkowski 4-space and give some descriptions of their curvature functions.


## 2. Preliminaries

Four-dimensional space-timeE14 is the real four-dimensional vector space ℝ4 equipped with the standard flat metric given by
(2)x,y=x1y1+x2y2+x3y3−x4y4.For any three vectorsx=x1,x2,x3,x4, y=y1,y2,y3,y4, and z=z1,z2,z3,z4 in E14, their exterior product is given by
(3)x×y×z=e1e2e3−e4x1x2x3x4y1y2y3y4z1z2z3z4,where e1,e2,e3,e4 is an orthogonal basis in E14, that is,
(4)e1=1,0,0,0,e2=0,1,0,0,e3=0,0,1,0,e4=0,0,0,1.A vectorv in E14 is called spacelike, timelike, or null (lightlike), if v,v>0 or v,v<0, v,v=0, respectively. In particular, the vector v=0 is said to be spacelike. A curve γs:I⟶E14 is called spacelike, timelike, or null (lightlike) if all of its velocity vectors γ′s satisfy γ′,γ′>0, γ′,γ′<0, or γ′,γ′=0, respectively. The norm of a vector v in E14 is given by ∥v∥=∣v,v∣[15].Definition 1 (see [30]).
Letγs be a null curve parameterized by null arc length s (i.e., ∥γ′′s∥=1) in E14. Then, γs can be framed by a Cartan Frenet frame T,N,B1,B2 such that
(5)T′N′B1′B2′=0100k20−100−k20k3−k3000TNB1B2,where
(6)N,B2=T,B2=N,B1=T,N=T,T=B1,B1=B1,B2=0,N,N=B2,B2=T,B1=1,T×N×B1=B2,N×B1×B2=B1,B1×B2×T=N,B2×T×N=T.In sequence,T,N,B1,B2 are called the tangent, principal normal, first binormal, and second binormal vector field of γs and k2 and k3 are first curvature and second curvature of the curve γs, respectively.Definition 2 (see [16]).
Letγs be a pseudo null curve parameterized by arc length s (i.e., ∥γ′′s∥=0) in E14. Then, the Frenet equation is defined by
(7)T′N′B1′B2′=010000k100k20−k1−10−k20TNB1B2,where
(8)B2,B2=B1,B2=N,B1=T,B1=T,N=T,B2=N,N=0,N,B2=B1,B1=T,T=1,T×N×B1=N,N×B1×B2=T,B1×B2×T=B2,B2×T×N=B1,and k1 and k2 are first curvature and second curvature of the curve γs, respectively.


## 3. Darboux Helix and Planes of Null Curve

When the Frenet frameT,N,B1,B2 of a nongeodesic null curve makes an instantaneous helix motion in E14, there exists an axis of the frame’s rotation. The direction of such axis is given by the vector
(9)D1=k2T+B1,D2=B2,D3=T,D4=k3N+k2B2,and we call them the Darboux vectors for the null curves in E14. The Darboux vectors satisfy the Darboux equations
(10)T′=D1×D2×T,N′=D2×D1×N,B1′=D3×D4×B1,B2′=D4×D3×B2.From (10), we know that Frenet vectors T and N rotate around the D1D2 plane, and Frenet vectors B1 and B2 rotate around the D3D4 plane. We find that the D1D2 plane and D3D4 plane play the role of Darboux vector in three-dimensional space. We also note that D2 and D3 are Frenet vectors of the null curve, D1,D2,D3,D4 is linearly independent, and D1 is orthogonal to D2 and D4. We are going to use the subspace spanned by D1,D2 and D1,D4 to represent D1D2 plane and D1D4 plane, respectively.Inspired by [10, 28], we discuss the situation when the curve γs lies in D1D2 and D1D4 planes.Theorem 3.
Letγs:I⟶E14 be a null curve parameterized by null arc length s. k2,k3 are the curvature functions of the null curve γs. If γs lies in D1D2 plane, then the curvature functions k2,k3 satisfy
(11)1/k3′k3+k2′/k3′′=0,and in addition, the curve γs can be expressed as
(12)γs=cD1+ck2′−1k3D2,where c is nonzero constant.Proof.
We may assume that(13)γs=λsD1+μsD2,and we take the derivative of (13) according to s, and we obtain
(14)T=γ′s=λk2′−μk3T+λ′B1+λk3+μ′B2.
Hence,(15)λk2′−μk3=1,λ′=0,λk3+μ′=0.
From the second equation of (15), we get
(16)λ=c,and substituting (16) into the first equation of (15), we have
(17)μ=ck2′−1k3.
Then, the curveγs can be denoted as
(18)γs=cD1+ck2′−1k3D2.
From the third equation of (15), we get
(19)c=1/k3′k3+k2′/k3′,that is,
(20)1/k3′k3+k2′/k3′′=0.
This ends the proof.Corollary 4.
In particular, whenk3=a=constant≠0, we have k2=a2/2s2+b, and the curve γs can be expressed as
(21)γs=cD1+ca2s−1aD2,where a and b are constants.Theorem 5.
Letγs:I⟶E14 be a null curve parameterized by null arc length s. k2,k3 are the curvature functions of the pseudonull curve γs. If the γs lies in D1D4 plane, then the curvature functions k2,k3 satisfy
(22)k2=s+c3c1s+c2,k32=c1s+c32c1s+c221−s+c32,and in addition, the curve γs can be expressed as
(23)γs=c1s+c2D1+c1k3D4,where c1, c2, and c3 are constants.Proof.
Assume that(24)γs=λD1+μD4.
Differentiating equation (24) with respect to s, we have
(25)T=γ′=λk2′T+μk3′N+λ′−μk3B1+μk2′+λk3B2.
Then, we obtain the system of differential equations(26)λk2′=1,μk3′=0,λ′−μk3=0,μk2′+λk3=0.
From the second equation of (26), we get
(27)μk3=c1.
Substituting (27) into the third equation of (26), we have
(28)λ=c1s+c2.
Then, the curveγs can be denoted as
(29)γs=c1s+c2D1+c1k3D4.
Substituting (28) into the first equation of (26), we have
(30)k2=s+c3c1s+c2.
Substituting (27) and (28) into the fourth equation of (26), we can calculate that
(31)k32=c1s+c32c1s+c221−s+c32.
This ends the proof.Definition 6.
Letγs:I⟶E14 be a null curve with parameterized by null arc length s. If there exists a fixed direction V≠0 such that
(32)D1,V=a,a∈ℝ,then the null curve γs is called the null Darboux helix, and the fixed direction V is called an axis of the null Darboux helix.Theorem 7.
Letγs:I⟶E14 be a null curve with parameterized by null arc length s. If γs is a null Darboux helix in E14 whose fixed direction V satisfies
(33)D1,V=a,a∈ℝ,then V is given by
(34)V=a−bk2e∫ξdsT+bξe∫ξdsN+be∫ξdsB1−bk2′k3e∫ξdsB2,and the curvature functions k2,k3 satisfy
(35)2k2−ξ2−ξ′e∫ξds=ab,where
(36)ξ=k3−k2′/k3′k2′/k3,and b∈ℝ0,k3≠0,k2≠const.Proof.
Letγs:I⟶E14 be a null Darboux helix with parameterized by null arc length s. Then, for a fixed direction V satisfying
(37)D1,V=a,a∈ℝ,we can assume
(38)V=u1T+u2N+u3B1+u4B2.
By using (5), we can obtain
(39)D1,V=k2u3+u1=a,(40)D1′,V=k2′u3+k3u4=0.
Taking the derivative of equation (39) according to s, we obtain
(41)u1′=−k2u3′.
Differentiating equation (38) and using the Frenet equation (5), we have
(42)u1′−k3u4+k2u2=0,u1+u2′−k2u3=0,u2−u3′=0,u4′+k3u3=0.
By (40), we can obtain
(43)u4=−k2′k3u3,
Substituting (43) into the fourth equation of (42), we can obtain
(44)u3=be∫ξds,ξ=k3−k2′/k3′k2′/k3.
From (39), (44), and the third equation of (42), we have
(45)u2=u3′=bξe∫ξds,u1=a−k2u3=−bξe∫ξds+a.
Thus,(46)V=a−bk2e∫ξdsT+bξe∫ξdsN+be∫ξdsB1−bk2′k3e∫ξdsB2.
From the second equation of (42), the relationship between k2 and k3 can be expressed as
(47)2k2−ξ2−ξ′e∫ξds=ab,where b is given by the relation (44), and if b=0, the axis V=0, which is a contradiction. Hence, b≠0, which completes the proof.Corollary 8.
In particular, whena=0,b=1, we have
(48)V=−k2T+ξN+B1−k2′k3B2e∫ξds,and the curvature functions k2,k3 satisfy
(49)ξ′+ξ2−2k2=0,where
(50)ξ=k3−k2′/k3′k2′/k3.Some examples of null Darboux helix inE14 are given below.Example 1.
Letγs:I⟶E14 be a null curve with the arc length s and the curvature
(51)k2=s2+12,k3=se−s2+1,and then, γs is a Darboux helix whose fixed direction is given by
(52)V=−s2+12T+sN+B1−s2+1B2es2/2.Example 2.
Letγs:I⟶E14 be a null curve with the arc length s and the curvature
(53)k2=sec2s+tan2s2,k3=2secstans,and then, γs is a Darboux helix whose fixed direction is given by
(54)V=−1+sin2s2cos2sT+tansN+B1−secsB2secs.Example 3.
Letγs:I⟶E14 be a null curve with the arc length s and the curvature
(55)k2=1s2,k3=−2s2,and then, γs is a Darboux helix whose fixed direction is given by
(56)V=−1s2T−1sN+B1−1sB21s.


## 4. Darboux Helix and Planes of Pseudo null Curve

When the Frenet frameT,N,B1,B2 of a nongeodesic pseudonull curve makes an instantaneous helix motion in E14, there exists an axis of the frame’s rotation. The direction of such axis is given by the vector
(57)D1=N,D2=k2T,D3=T,D4=−k2N−k1B2,and we call them the Darboux vectors for the pseudo null curves in E14. The Darboux vectors satisfy the Darboux equations
(58)T′=D1×D2×T,N′=D3×D4×N,B1′=D4×D3×B1,B2′=D2×D1×B2.From (58), we know that Frenet vectors T and B2 rotate around the D1D2 plane, and Frenet vectors N and B1 rotate around the D3D4 plane. We find that the D1D2 plane and D3D4 plane play the role of Darboux vector in three-dimensional space. We also note that D1 and D3 are Frenet vectors of the pseudonull curve, D1,D2,D3,D4 is linearly independent, and D4 is orthogonal to D2 and D3. We are going to use the subspace spanned by D2,D4 and D3,D4 to represent D2D4 plane and D3D4 plane, respectively.Similar to Section3, we discuss the situation when the curve γs is in D2D4 and D3D4 planes.Theorem 9.
Letγs:I⟶E14 be a pseudonull curve with parameterized by arc length s. k1,k2 are the curvature functions of the pseudonull curve γs. If γs lies in D2D4 plane, then the curvature functions k1,k2 satisfy
(59)c2k2′′+c1k2k2′+k2=0,and in addition, the curve γs can be expressed as
(60)γs=c1D2+c2k2D4,where c1 and c2 are nonzero constants.Proof.
We may assume that(61)γs=λD2+μD4,and we take the derivative of (61) according to s, and we obtain
(62)T=γ′=μk1−λk2′T−μk2′N+λ′B1−λk1+μk1′B2.
Hence,(63)u1k1−λk2′=1,μk2′=0,λ′=0,λk1+μk1′=0.
From the second and the third equations of (63), we get
(64)λ=c1,μk2=c2,where c1 and c2 are nonzero constants.
Then, the curveγs can be denoted as
(65)γs=c1D2+c2k2D4.
Substituting (64) into the fourth equation of (63), we have
(66)c2k1=c1k2k2′+k2.
From (66) and the first equation of (63), we get
(67)c2k2′′+c1k2k2′+k2=0.
This ends the proof.Theorem 10.
Letγs:I⟶E14 be a pseudonull curve with parameterized by arc length s. k1,k2 are the curvature functions of the pseudonull curve γs. If the γs lies in D3D4 plane, then the curvature functions k1,k2 satisfy
(68)k2k1=1−c1s22c1+c2sc1+c3c1,and in addition, the curve can be expressed as
(69)γs=c2+1−c1sD3+c1k1D4,where c1,c2,c3 are constants.Proof.
Assume that(70)γs=λD3+μD4.
Differentiating equation (70) with respect to s, we have
(71)T=γ′=μk1+λ′T−λ−μk2′N−μk1′B2.
So we obtain the system of differential equations(72)μk1+λ′=1,λ−μk2′=0,μk1′=0.
From the first and the third equations of (72), we get
(73)λ=1−c1s+c2,(74)μk1=c1.
Then, the curveγs can be denoted as
(75)αs=1−c1s+c2D3+c1k1D4.
Substituting (73) and (74) into the second equation of (72), we have
(76)μk2=121−c1s2+c2s+c3.
From equations (74) and (76), we can obtain
(77)k2k1=1−c1s22c1+c2sc1+c3c1,where c1,c2,c3 are constants.Corollary 11.
In particular, whenc1=1/1,c2=c3=0, we have k2/k1=s2. Let k1=3s/s2+12 and k2=3s3/s2+12. Then, the curve γs can be expressed as
(78)γ′′=3ss2+12s2+132s,s2,1−s2−s44,1+s2+s44.Definition 12.
Letγs:I⟶E14 be a pseudo null curve with parameterized by arc length s. If there exists a fixed direction V≠0 such that
(79)D2,V=a,a∈ℝ,then the pseudo null curve γs is called the pseudo null Darboux helix, and the fixed direction V is called an axis of the pseudo null Darboux helix.Theorem 13.
Letγs:I⟶E14 be a pseudonull curve with parameterized by arc length s. If γs is a pseudonull Darboux helix in E14 whose fixed direction V satisfies
(80)D2,V=a,a∈ℝ,then V is given by
(81)V=u1T−k2′k1u1N+u1k2+aB1+u1′B2,and the curvature functions k1,k2 satisfy
(82)u1′′−k1k2u1=ak1,where
(83)ξ=1+k22−k2′/k1′k2′/k1,and a,b∈ℝ,k1≠0,k2≠const.Proof.
Letγs:I⟶E14 be a pseudo null Darboux helix with parameterized by arc length s. Then, for a fixed direction V satisfying
(84)D2,V=a,a∈ℝ,we can assume
(85)V=u1T+u2N+u3B1+u4B2.
By using (84), we can obtain
(86)D2,V=u3−u1k2=a,(87)D2′,V=−k1u2−k2′u1=0.
Taking the derivative of equation (86) according to s, we obtain
(88)u3′=k2u1′.
Differentiating equation (85) and using the Frenet equation (7), we have
(89)u1′−u4=0,u1+u2′+k2u3=0,k1u2+u3′−k2u4=0,u4′−k1u3=0.
Substituting (86) and (87) into the second equation of (89), we can obtain
(90)u1′−1+k22−k2′/k1′k2′/k1u1=ak2.
By (90), we can obtain
(91)u1=e∫ξdsak2e−∫ξds+b,where
(92)ξ=1+k22−k2′/k1′k2′/k1.
From (87), (88), and the first equation of (89), we have
(93)u2=−k2′k1u1=−k2′k1e∫ξdsak2e−∫ξds+b,u3=u1k2+a=k2e∫ξdsak2e−∫ξds+b+a,u4=u1′=c1k2+ξe∫ξdsak2e−∫ξds+b.
Thus,(94)V=u1T−k2′k1u1N+u1k2+aB1+u1′B2.
From the fourth equation of (89), the relationship between k1 and k2 can be expressed as
(95)u1′′−k1k2u1=ak1,a∈ℝ.
This ends the proof.Corollary 14.
In particular, whena=0,b=1, we have
(96)V=T−k2′k1N+k2B1+ξB2e∫ξds,and the curvature functions k2,k3 satisfy
(97)ξ′+ξ2−k1k2=0,where
(98)ξ=1+k22−k2′/k1′k2′/k1.Some examples of pseudo null Darboux helix inE14 are given below.Example 4.
Letγs:I⟶E14 be a pseudo null curve, and s is the pseudoarc length. The curvature function k1,k2 satisfies
(99)k1=1ss2+1s2+1,k2=ss2+1,and then, γs is a Darboux helix whose fixed direction is given by
(100)V=s2+1T−ss2+1N+sB1+ss2+1B2.Example 5.
Letγs:I⟶E14 be a pseudo null curve with the arc length s and the curvature
(101)k1=2sec2s2sec2s−1,k2=2sec2s−1,and then, γs is a Darboux helix whose fixed direction is given by
(102)V=tansT−tan2sN+tans2sec2s−1B1+sec2sB2.Example 6.
Letγs:I⟶E14 be a pseudo null curve with the arc length s and the curvature
(103)k1=12es−1,k2=2es−1,and then, γs is a Darboux helix whose fixed direction is given by
(104)V=esT−e2sN+es2es−1B1+esB2.


## 5. Conclusion

In this paper, we discuss some new space curves and planes in four-dimensional space-time and give characterizations of them in terms of the curvature functions. Before this study, most researchers studied Darboux vector in three-dimensional space and four-dimensional Euclidean space. In this paper, the Darboux vector fields in three-dimensional space are extended to four-dimensional space-time by mathematical method. By defining Darboux vector fields in four-dimensional space-time in the form of vector products, we find that the Frenet vectors rotate around a plane spanned by two new vector fields, and this plane plays the role that the Darboux vector plays in three-dimensional space. This paper gives a new description of Darboux vector in four-dimensional space-time, which promotes the further development of angular momentum vector in physics and geometry.



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