Table Of Content9 Slant helices in Euclidean 4-space E4
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Ahmad T. Ali
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Mathematics Department
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2 Faculty of Science, Al-Azhar University
Nasr City, 11448, Cairo, Egypt
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email: [email protected]
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[ Rafael L´opez∗
1 Departamento de Geometr´ıa y Topolog´ıa
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Universidad de Granada
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18071 Granada, Spain
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3 email: [email protected]
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Abstract
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r We consider a unit speed curve α in Euclidean four-dimensional space E4
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anddenote theFrenet frameby{T,N,B1,B2}. We say that αis a slanthelix
if its principal normal vector N makes a constant angle with a fixed direction
U. In this work we give different characterizations of such curves in terms of
their curvatures.
MSC: 53C40, 53C50
Keywords: Euclidean 4-space; Frenet equations; slant helices.
∗Partially supported by MEC-FEDER grant no. MTM2007-61775.
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1 Introduction and statement of results
A helix in Euclidean 3-space E3 is a curve whose tangent lines make a constant angle
with a fixed direction. A helix curve is characterized by the fact that the ratio τ/κ
is constant along the curve, where τ and κ denote the torsion and the curvature,
respectively. Helices are well known curves in classical differential geometry of space
curves [10] and we refer to the reader for recent works on this type of curves [5,
14]. Recently, Izumiya and Takeuchi have introduced the concept of slant helix by
saying that the normal lines make a constant angle with a fixed direction [6]. They
characterize a slant helix if and only if the function
κ2 τ ′
(κ2 +τ2)3/2 κ
(cid:16) (cid:17)
is constant. This article motivated generalizations in a twofold sense: first, by
increasing thedimension ofEuclidean space[8,12]; second, byconsidering analogous
problems in other ambient spaces, mainly, in Minkowski space En [1, 2, 3, 4, 7, 13].
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In this work we consider the generalization of the concept of slant helix in Euclidean
4-space E4. Let α : I ⊂ R → E4 be an arbitrary curve in E4. Recall that the
curve α is said to be of unit speed (or parameterized by arclength function s) if
′ ′
hα(s),α(s)i = 1, where h,i is the standard scalar product in the Euclidean space
E4 given by
hX,Yi = x1y1 +x2y2 +x3y3 +x4y4,
for each X = (x1,x2,x3,x4), Y = (y1,y2,y3,y4) ∈ E4.
Let {T(s),N(s),B1(s),B2(s)} be the moving frame along α, where T,N,B1 and
B2 denote the tangent, the principal normal, the first binormal and second binor-
mal vector fields, respectively. Here T(s), N(s), B1(s) and B2(s) are mutually
orthogonal vectors satisfying
hT,Ti = hN,Ni = hB1,B1i = hB2,B2i = 1.
The Frenet equations for α are given by
T′ 0 κ1 0 0 T
N′ −κ1 0 κ2 0 N
B′1 = 0 −κ2 0 κ3 B1 . (1)
B′2 0 0 −κ3 0 B2
Recall the functions κ1(s), κ2(s) and κ3(s) are called respectively, the first, the
second and the third curvatures of α. If κ3(s) = 0 for any s ∈ I, then B2(s) is
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a constant vector B and the curve α lies in a three-dimensional affine subespace
orthogonal to B, which is isometric to the Euclidean 3-space E3.
We will assume throughout this work that all the three curvatures satisfy κ (s) 6= 0
i
for any s ∈ I, 1 ≤ i ≤ 3.
Definition 1.1. A unit speed curve α : I → E4 is said to be a generalized helix if
there exists a non-zero constant vector field U and a vector field X ∈ {T,N,B1,B2}
such that the function
s 7−→ hX(s),Ui, s ∈ I
is constant.
Among the possibilities to choose the vector field X we have:
1. If X is the unit tangent vector field T, α is called a cylindrical helix. It is
known that α(s) is a cylindrical helix if and only if the function
κ21 1 κ1 ′ 2
+
κ22 κ3 κ2
h (cid:16) (cid:17)i
is constant. See [9, 11].
2. If X is the vector field B2, then the curve is called a B2-slant curve. Moreover
α is a such curve if and only if the function
κ23 1 κ3 ′ 2
+
κ22 κ1 κ2
h (cid:16) (cid:17)i
is constant. See [12].
Definition 1.2. A unit speed curve α : I → E4 is called slant helix if its unit
principal normal vector N makes a constant angle with a fixed direction U.
Our main result in this work is the following characterization of slant helices.
Theorem 1.3. Let α : I → E4 be a unit speed curve in E4. Then α is a slant helix
if and only if the function
2 1 κ1 ′ κ2 2 κ1 2
κ1(s)ds + κ1ds + + κ1ds
κ3 κ2 κ3 κ2
(cid:16)Z (cid:17) h (cid:16) Z (cid:17) i (cid:16) Z (cid:17)
is constant. Moreover, this constant agrees with tan2θ, being θ the angle that makes
N with the fixed direction U that determines α.
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2 Proof of Theorem 1.3
Let α be a unit speed curve in E4. Assume that α is a slant curve. Let U be
the direction with which N makes a constant angle θ (suppose that hU,Ui = 1).
Consider the differentiable functions a , 1 ≤ i ≤ 4,
i
U = a1(s)T(s)+a2(s)N(s)+a3(s)B1(s)+a4(s)B2(s), s ∈ I, (2)
that is,
a1 = hT,Ui, a2 = hN,Ui, a3 = hB1,Ui, a4 = hB2,Ui.
Because the vector field U is constant, a differentiation in (2) together (1) gives the
following ordinary differential equation system
′
a1 −κ1a2 = 0
′
a2 +κ1a1 −κ2a3 = 0
(3)
′
a3 +κ2a2 −κ3a4 = 0
′
a4 +κ3a3 = 0
Then the function a2(s) = hN(s),Ui is constant, andit agrees with cosθ. Then (3)
gives
′
a1 −κ1a2 = 0
κ1a1 −κ2a3 = 0
(4)
′
a3 +κ2a2 −κ3a4 = 0
a′4 +κ3a3 = 0
The first third equations in (4) lead to
a1 = a2 κ1ds
κ1
a3 = a2κR2 κ1ds (5)
a4 = a2 1R κ1 κ1ds ′ + κ2
κ3 κ2 κ3
h (cid:16) R (cid:17) i
We do the change of variables:
s dt
t(s) = κ3(u)du, = κ3(s).
ds
Z
In particular, and from (4), we have
′ κ2
a3(t) = a4 −a2 .
κ3
4
As a consequence, if α is a slant helix, the last equation of (4) yields
′′ κ2(t)
a4(t)+a4(t)−a2 = 0. (6)
κ3(t)
The general solution of this equation is
κ2(t) κ2(t)
a4(t) = a2 A− sintdt cost+ B + costdt sint , (7)
" κ3(t) κ3(t) #
(cid:16) Z (cid:17) (cid:16) Z (cid:17)
where A and B are arbitrary constants. Then (7) takes the following form
a4(s) = a2 A− κ2(s)sin κ3(s)ds ds cos κ3(s)ds
(8)
+hB(cid:16) + R κh2(s)cos Rκ3(s)ds ids (cid:17)sin Rκ3(s)ds .
(cid:16) h i (cid:17) i
R R R
From (4), the function a3 is given by
a3(s) = a2 A− κ2(s)sin κ3(s)ds ds sin κ3(s)ds
(9)
−hB(cid:16) + R κh2(s)cos Rκ3(s)ds ids (cid:17)cos Rκ3(s)ds .
(cid:16) h i (cid:17) i
R R R
From (8), (9) and (5) we have the following two conditions:
1 κ1 ′ κ2
κ1ds + = A− κ2(s)sin κ3(s)ds ds cos κ3(s)ds
κ3 κ2 κ3 (10)
(cid:16) (cid:17) (cid:16) h i (cid:17)
R + B +R κ2(s)coRs κ3(s)ds ds siRn κ3(s)ds.
(cid:16) h i (cid:17)
R R R
and
κ1
κ1ds = A− κ2(s)sin κ3(s)ds ds sin κ3(s)ds
κ2 (11)
(cid:16) h i (cid:17)
R − B +R κ2(s)coRs κ3(s)ds ds coRs κ3(s)ds.
(cid:16) h i (cid:17)
The condition (11) can be writteRn as follows:R R
κ1(s) κ1(s)ds = A− κ2(s)sin κ3(s)ds ds κ2(s)sin κ3(s)ds
R (cid:16)− B +R h κ2(s)coRs κ3(s)dis d(cid:17)s κ2(s)cRos κ3(s)ds.
(cid:16) h i (cid:17)
R R R
If we integrate the above equation we have
2 2
κ1(s)ds = C − A− κ2(s)sin κ3(s)ds ds
(12)
2
(cid:16) (cid:17) (cid:16) h i (cid:17)
R − B + Rκ2(s)cos κR3(s)ds ds ,
(cid:16) h i (cid:17)
R R
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where C is a constant of integration. From Equations (10) and (11), we get
1 κ1 ′ κ2 2 κ1 2 2
κ1ds + + κ1ds = A− κ2(s)sin κ3(s)ds ds
κ3 κ2 κ3 κ2
h (cid:16) (cid:17) i (cid:16) (cid:17) (cid:16) h i (cid:17)2
R R + B + R κ2(s)cos Rκ3(s)ds ds .
(cid:16) h i (cid:17)(13)
R R
Now Equations (12) and (13) give
2 1 κ1 ′ κ2 2 κ1 2
κ1(s)ds + κ1ds + + κ1ds = C. (14)
κ3 κ2 κ3 κ2
(cid:16)Z (cid:17) h (cid:16) Z (cid:17) i (cid:16) Z (cid:17)
Moreover this constant C calculates as follows. From (14), together the three equa-
tions (5) we have
2 2 2 2
a +a +a 1−a
1 3 4 2 2
C = = = tan θ,
a2 a2
2 2
where we have used (2) and the fact that U is a unit vector field.
We do the converse of the proof. Assume that the condition (14) is satisfied for a
curve α. Let θ ∈ R be so that C = tan2θ. Define the unit vector U by
U = cosθ κ1dsT+N+ κ1 κ1dsB1 + 1 κ1 κ1ds ′ + κ2 B2 . (15)
" κ2 κ3 κ2 κ3 #
Z Z h (cid:16) Z (cid:17) i
dU
By taking account (14), a differentiation of U gives that = 0, which it means
ds
that U is a constant vector. On the other hand, the scalar product between the unit
principal normal vector field N with U is
hN(s),Ui = cosθ.
Thus α is a slant curve. This finishes with the proof of Theorem 1.3.
3 Further characterizations of slant helices
In this section we present two new characterizations of slant helices. The first one
is a consequence of Theorem 1.3.
Theorem 3.1. Let α : I ⊂ R → E4 be a unit speed curve in Euclidean space E4.
Then α is a slant helix if and only if there exists a C2-function f such that
κ1 ′ d κ3κ1
κ3f(s) = κ1ds +κ2, f(s) = − κ1ds. (16)
κ2 ds κ2
(cid:16) Z (cid:17) Z
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Proof. Let now assume that α is a slant helix. A differentiation of (14) gives
′ κ1 κ1 ′
κ1(s)ds κ1(s)ds + κ1ds κ1ds
(cid:16)R 1 κ1 (cid:17)(cid:16)R ′ κ2(cid:17) 1(cid:16)κκ21R (cid:17)(cid:16)′ κ2κR2 ′ (cid:17) (17)
+ κ1ds + κ1ds + = 0.
κ3 κ2 κ3 κ3 κ2 κ3
h (cid:16) (cid:17) ih (cid:16) (cid:17) i
R R
After some manipulations, the equation (17) takes the following form
κ1κ3 1 κ1 ′ κ2 ′
κ1ds+ κ1ds + = 0. (18)
κ2 κ3 κ2 κ3
Z h (cid:16) Z (cid:17) i
If we define f = f(s) by
κ1 ′
κ3f(s) = κ1ds +κ2.
κ2
(cid:16) Z (cid:17)
Then Equation (18) writes as
d κ3κ1
f(s) = − κ1ds.
ds κ2
Z
Conversely, if (16) holds, we define a unit constant vector U by
U = cosθ κ1dsT+N+ κ1 κ1dsB1 +f(s)B2 .
" κ2 #
Z Z
We have that hN(s),Ui = cosθ is constant, that is, α is a slant helix.
We end giving an integral characterization of a slant helix.
Theorem 3.2. Let α : I ⊂ R → E4 be a unit speed curve in Euclidean space E4.
Then α is a slant helix if and only if the following condition is satisfied
κ1
κ1ds = A− κ2(s)sin κ3(s)ds ds sin κ3(s)ds
κ2 (19)
(cid:16) h i (cid:17)
R − B +R κ2(s)coRs κ3(s)ds ds coRs κ3(s)ds,
(cid:16) h i (cid:17)
R R R
for some constants A and B.
Proof. Suppose that α is a slant helix. By using Theorem 3.1, let define m(s) and
n(s) by
s
φ = φ(s) = κ3(u)du, (20)
Z
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κ1
m(s) = f(s)cosφ+ κ1ds sinφ+ κ2sinφ ds,
(cid:16)κκ12 (cid:17) h i (21)
n(s) = f(s)sinφ− R κ1ds cosφ−R κ2cosφ ds.
κ2
(cid:16) (cid:17) h i
R R
If we differentiate Equations (21) with respect to s and taking into account of (20)
dm dn
and (16), we obtain = 0 and = 0. Therefore, there exist constants A and
ds ds
B such that m(s) = A and n(s) = B. By substituting into (21) and solving the
κ1
resulting equations for κ1ds, we get
κ2
R
κ1
κ1ds = A− κ2(s)sinφ ds sinφ− B + κ2(s)cosφ ds cosφ.
κ2
Z (cid:16) Z h i (cid:17) (cid:16) Z h i (cid:17)
Conversely, suppose that (19) holds. In order to apply Theorem 3.1, we define
f = f(s) by
f(s) = A− κ2(s)sinφ ds cosφ+ B + κ2(s)cosφ ds sinφ, (22)
(cid:16) Z h i (cid:17) (cid:16) Z h i (cid:17)
with φ(s) = κ3(u)du. A direct differentiation of (19) gives
R ′
κ1
κ1ds = κ3f(s)−κ2.
κ2 !
Z
This shows the left condition in (16). Moreover, a straightforward computation
′ κ3κ1
leads to f (s) = − κ1ds, which finishes the proof.
κ2
R
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