����������� ������ 53 (2001), 21–27

UDK 514.752.2

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SOME CHARACTERIZATIONS OF THE LORENTZIAN SPHERICAL TIMELIKE AND NULL CURVES

Miroslava Petrović-Torgašev and Emilija Šućurović

Abstract. In [5] and [6] the authors have characterized the Lorentzian spherical spacelike curves in the Minkowski 3-space E_{1}^{3}. In this paper, we shall characterize the Lorentzian spherical timelike and null curves in the same space.

1. Introduction

In the Euclidean space E^{3} a spherical unit speed curves and their characterizations are given in [3], [9] and [10]. In [5] and [6] the authors have characterized the Lorentzian spherical spacelike curves in the Minkowski 3-space E_{1}^{3}. In this paper, we shall characterize the Lorentzian spherical timelike and null curves in the same space.

2. Preliminaries

The Minkowski 3-space E_{1}^{3} is the Euclidean 3-space E^{3} provided with the Lorentzian inner product

g(a, b) = −a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3},

where a = (a_{1}, a_{2}, a_{3}) and b = (b_{1}, b_{2}, b_{3}).

An arbitrary vector a = (a_{1}, a_{2}, a_{3}) in E_{1}^{3} can have one of three Lorentzian causal characters: it is spacelike if g(a, a) > 0 or a = 0, timelike if g(a, a) < 0 and null (lightlike) if g(a, a) = 0 and a ≠ 0. Similarly, an arbitrary curve α = α(s) in E_{1}^{3} is locally spacelike, timelike or null (lightlike), if all of its velocity vectors α′(s) are respectively spacelike, timelike or null, for each s ∈ I ⊂ R. Recall that the pseudo-norm of an arbitrary vector a ∈ E_{1}^{3} is given by

∥ a ∥ = \sqrt{| g(a, a) |},

AMS Subject Classification: 53 C 50, 53 C 40.

Keywords and phrases: Lorentzian 3-space, Lorentzian sphere, causal character, curvature, torsion.

and that the velocity v of the curve α is given by v = ∥α′(s)∥. Therefore, α is a unit speed curve if and only if g(α′(s), α′(s)) = ±1.

The Lorentzian sphere of center m = (m_{1}, m_{2}, m_{3}) and radius r ∈ R^{+} in the space E_{1}^{3} is defined by

S_{1}^{2} = {a = (a_{1}, a_{2}, a_{3}) ∈ E_{1}^{3} | g(a − m, a − m) = r^{2}}.

The vectors a, b ∈ E_{1}^{3} are orthogonal if and only if g(a, b) = 0.

Denote by {T (s), N(s), B(s)} the moving Frenet frame along the curve α = α(s) parameterized by a pseudo-arclength parameter s, i.e. g(α′(s), α′(s)) = ±1. In particular, null curve α(s) in E_{1}^{3} is parameterized by a pseudo-arclength s if g(α′′(s), α′′(s)) = 1. Let T (s) = α′(s), N(s) = α′′(s)/∥α′′(s)∥ and B(s) be the tangent, the principal normal and the binormal vector of the curve α(s) respectively. If α is a timelike curve, i.e. if T is a timelike vector, then the Frenet formulae read:

\begin{aligned}T′ = κN, N′ & = κT + τB, B′ = −τN, \\ g(T, T) = −1, g(N, N) & = g(B, B) = 1, g(T, N) = g(T, B) = g(N, B) = 0. \end{aligned}

On the other hand, if α is a null curve, i.e. if T is a null vector, then the Frenet formulae read:

\begin{aligned}T′ = κN, N′ & = τT − κB, B′ = −τN, \\ g(T, T) = g(B, B) = 0, g(N, N) & = 1, g(T, N) = g(N, B) = 0, g(T, B) = 1 \end{aligned}

where κ takes only two values: κ = 0 when α is a straight null line or κ = 1 in all other cases. The functions κ = κ(s) and τ = τ(s) are called the curvature and the torsion of α respectively [8].

3. The Lorentzian spherical timelike curves

Theorem 3.1. Let α(s) be a plane unit speed timelike curve with a curvature κ = κ(s). Then α lies on the Lorentzian sphere of center m and radius r ∈ R^{+} in E_{1}^{3} if and only if κ = constant ≠ 0 and

α − m = (1/κ) N ± \sqrt{r^{2} − (1/κ)^{2}}B.

Proof. Let us first suppose that α lies on the Lorentzian sphere of center m and radius r ∈ R^{+}. Then g(α − m, α − m) = r^{2}, for each s ∈ I ⊂ R. By differentiation with respect to s of the previous relation, we find that

g(T, α − m) = 0.  (3.1)

Further, the differentiation with respect to s of (3.1) gives

\begin{matrix}g(T′, α − m) + g(T, T) = 0, \\ κ g(N, α − m) = 1, \end{matrix}

where we have used the corresponding Frenet formula. It follows that κ ≠ 0 for each s ∈ I ⊂ R and that

g(N, α − m) = 1/κ.  (3.2)

Next, decompose the vector α − m as

α − m = aT + bN + cB,  (3.3)

where a = a(s), b = b(s) and c = c(s) are arbitrary functions. Then the relations (3.1) and (3.2) imply that

g(T, α − m) = −a = 0, g(N, α − m) = b = 1/κ, g(B, α − m) = c.

Further, the differentiation of (3.2) with respect to s gives

g(N′, α − m) + g(N, α′) = (1/κ)′.

By assumption α is a plane curve. Hence τ = 0 and using the corresponding Frenet formula we get that κg(T, α − m) = (1/κ)′. Then the relation (3.1) implies (1/κ)′ = 0 and thus 1/κ = constant ∈ R, i.e. κ = constant ∈ R. Since κ ≠ 0 for each s, it follows that κ = constant ≠ 0. Further, the substitution of the coefficients a, b and c in (3.3) gives

α − m = (1/κ) N + cB.

Now it is easy to see that g(α − m, α − m) = (1/κ)^{2} + c^{2} = r^{2}, so it follows that c = ±\sqrt{r^{2} − (1/κ)^{2}}. Consequently,

α − m = (1/κ) N ± \sqrt{r^{2} − (1/κ)^{2}}B.

Conversely, if κ = constant ≠ 0 and

α − m = (1/κ) N ± \sqrt{r^{2} − (1/κ)^{2}}B,

m ∈ E_{1}^{3} is an arbitrary vector and r ∈ R^{+}, we shall prove that m = constant. Since

m = α − (1/κ) N ± \sqrt{r^{2} − (1/κ)^{2}}B,

by differentiation with respect to s of the previous equation and using the corresponding Frenet formulae we get m′ = 0. It follows that m = constant and that g(α − m, α − m) = r^{2}. Therefore, α lies on the Lorentzian sphere of center m and radius r.

Remark. In [8] a classification of all W -curves (i.e. a curves for which a curvature and a torsion are constants) in space E_{1}^{3} is given. Since α is a curve with κ = constant ≠ 0 and τ = 0, by that classification it is a part of an orthogonal hyperbola.

Theorem 3.2. Let α(s) be a unit speed timelike curve in E_{1}^{3} with a curvature κ(s) ≠ 0 and a torsion τ(s) ≠ 0 for each s ∈ I ⊂ R. Then α lies on the Lorentzian sphere of radius r ∈ R^{+} if and only if

(1/κ)^{2} + ((1/τ)(1/κ)′)^{2} = r^{2}.

Proof. Let us first suppose that α lies on the Lorentzian sphere of center m and radius r. Then g(α − m, α − m) = r^{2}. By three differentiations with respect to s of the previous equation and using the corresponding Frenet formulae, we get

g(B, α − m) = (1/τ)(1/κ)′.

Next, decompose the vector α − m as

α − m = aT + bN + cB,  (3.4)

where a = a(s), b = b(s) and c = c(s) are arbitrary functions. Then

g(T, α − m) = −a = 0, g(N, α − m) = b = 1/κ, g(B, α − m) = c = (1/τ)(1/κ)′.

Therefore, substitution of the coefficients a, b and c in (3.4) gives

α − m = (1/κ) N + (1/τ)(1/κ)′B.

Thus

g(α − m, α − m) = r^{2} = (1/κ)^{2} + ((1/τ)(1/κ)′)^{2}.

Conversely, if

(1/κ)^{2} + ((1/τ)(1/κ)′)^{2} = r^{2},  (3.5)

where r ∈ R^{+}, we may consider the vector m ∈ E_{1}^{3} of the form

m = α − (1/κ) N − (1/τ)(1/κ)′B.  (3.6)

We shall prove that m = constant. By differentiation with respect to s of the previous equation, we have that

\begin{aligned}m′ & = T − (1/κ)′N − (1/κ)(κT + τB) − ((1/τ)(1/κ)′)′B + (1/τ)(1/κ)′(τN) \\ & = (−τ/κ − ((1/τ)(1/κ)′)′) B. \end{aligned}

(3.7)

By differentiation with respect to s of the assumption (3.5), we have

(2/κ)(1/κ)′ + (2/τ)(1/κ)′((1/τ)(1/κ)′)′ = 0

and thus

(τ/κ) + ((1/τ)(1/κ)′)′ = 0.  (3.8)

Substituting the last relation in (3.7), we find that m′ = 0 for each s ∈ I ⊂ R and thus m = constant. The relation (3.6) implies that

g(α − m, α − m) = (1/κ)^{2} + ((1/τ)(1/κ)′)^{2} = r^{2}.

Hence α lies on the Lorentzian sphere of center m and radius r.

Theorem 3.3. Let α(s) be a unit speed timelike curve, with a curvature κ(s) ≠ 0 and a torsion τ(s) ≠ 0 for each s ∈ I ⊂ R. Then α lies on a Lorentzian sphere in E_{1}^{3} if and only if

(τ/κ) = −((1/τ)(1/κ)′)′.

Proof. Let us first assume that α is a curve lying on the Lorentzian sphere of radius r ∈ R^{+}. Then by the Theorem 3.2 it follows that the relation (3.5) holds, so differentiation with respect s of the relation (3.5) implies the relation (3.8).

Conversely, suppose that the equation (3.8) holds for each s ∈ I ⊂ R. Since (3.8) is the differential of the equation

(1/κ)^{2} + ((1/τ)(1/κ)′)^{2} = c = constant > 0,

we may take c = r^{2}, r ∈ R^{+}. Finally, by Theorem 3.2 it follows that image of the curve α lies on a Lorentzian sphere of radius r.

Theorem 3.4. A unit speed timelike curve α(s) with κ(s) ≠ 0 and τ(s) ≠ 0 for each s ∈ I ⊂ R lies on a Lorentzian sphere in E_{1}^{3} if and only if κ(s) > 0 and there is a differentiable function f(s) such that fτ = (1/κ)′ and f′ + τ/κ = 0.

Proof. Let us first assume that α(s) is a curve lying on the Lorentzian sphere. Then by the Theorem 3.3 we have that τ/κ = −((1/τ)(1/κ)′)′. Next, define the differentiable function f = f(s) by

f = (1/τ)(1/κ)′.

Consequently, f′ = −τ/κ. Since κ(s) = ∥T′∥ ≥ 0 and κ(s) ≠ 0 for each s ∈ I ⊂ R, it follows that κ(s) > 0.

Conversely, assume that α is a curve for which κ > 0 for each s ∈ I ⊂ R and that there is a differentiable function f(s) such that fτ = (1/κ)′ and f′ = −τ/κ. Next, since f = (1/τ)(1/κ)′, we have that

((1/τ)(1/κ)′)′ = −τ/κ.

Hence by the Theorem 3.3 it follows that α lies on a Lorentzian sphere.

Theorem 3.5. A unit speed timelike curve α(s) with κ(s) ≠ 0 and τ(s) ≠ 0 lies on a Lorentzian sphere in E_{1}^{3} if and only if there are constants A, B ∈ R such that the equation

κ(A cos ( ∫_{0}^{s} τ(s) ds) + B sin ( ∫_{0}^{s} τ(s) ds)) = 1.

holds for each s ∈ I ⊂ R.

Proof. Let us first suppose that α(s) is a curve lying on a Lorentzian sphere. Then by the Theorem 3.4 there is a differentiable function f(s) such that fτ = (1/κ)′ and f′ = −τ/κ. Next, define the C^{2} function θ(s) and the C^{1} functions g(s) and h(s) by θ(s) = ∫_{0}^{s} τ(s) ds,

g(s) = (1/κ) cos θ − f(s) sin θ, h(s) = (1/κ) sin θ + f(s) cos θ.  (3.9)

Differentiation with respect to s of the functions θ, g and h easily gives θ′(s) = τ(s), g′(s) = h′(s) = 0 and therefore g(s) = A, h(s) = B, so the relation (3.9) becomes

(1/κ) cos θ − f(s) sin θ = A, (1/κ) sin θ + f(s) cos θ = B.

Multiplying the first of the previous equations with cos θ and the second with sin θ and adding, we find that 1/κ = A cos θ + B sin θ. Thus the equation

κ(A cos ( ∫_{0}^{s} τ(s) ds) + B sin ( ∫_{0}^{s} τ(s) ds)) = 1,

is satisfied.

Conversely, let A and B be the real constants, such that the equation

κ(A cos ( ∫_{0}^{s} τ(s) ds) + B sin ( ∫_{0}^{s} τ(s) ds)) = 1  (3.10)

holds for each s ∈ I ⊂ R. Then obviously κ(s) ≠ 0 and therefore κ(s) = ∥T′∥ > 0 for each s. The differentiation with respect to s of the relation (3.10) gives

τ( − A sin ( ∫_{0}^{s} τ(s) ds) + B cos ( ∫_{0}^{s} τ(s) ds)) = (1/κ)′.  (3.11)

Next, define the differentiable function f(s) by

f(s) = −A sin ( ∫_{0}^{s} τ(s) ds) + B cos ( ∫_{0}^{s} τ(s) ds).  (3.12)

Then the relations (3.11) and (3.12) give (1/κ)′ = τf, that is f = (1/τ)(1/κ)′. By differentiation with respect to s of (3.12) and using (3.10), we find that

f′ = −τ(A cos ( ∫_{0}^{s} τ(s) ds) + B sin ( ∫_{0}^{s} τ(s) ds)) = −τ/κ.

Therefore, by the Theorem 3.4 it follows that α(s) lies on a Lorentizan sphere.

4. The Lorentzian spherical null curves

Theorem 4.1. There are no null curves α(s) lying on the Lorentzian sphere in E_{1}^{3}.

Proof. Assume that α(s) is a null curve lying on the Lorentzian sphere of center m ∈ E_{1}^{3} and radius r ∈ R^{+}. Then we have

g(α − m, α − m) = r^{2},  (4.1)

for each s ∈ I ⊂ R. If α is a straight null line with the equation α(s) = p + sq, p, q ∈ E_{1}^{3}, then by differentiation with respect to s of the relation (4.1) we get g(p + sq − m, q) = 0 and therefore g(q, p) = g(q, m) = constant. It follows that p = m and consequently α − m = sq. But then g(α − m, α − m) = 0, which is a contradiction. On the other hand, if α is not a straight null line, by differentiation with respect to s of the relation (4.1), we find that

g(T, α − m) = 0.  (4.2)

By differentiation with respect to s of the relation (4.2), we get

g(T′, α − m) + g(T, T) = 0, κg(N, α − m) = 0,

and since in this case we have κ = 1 for each s ∈ I ⊂ R, it follows that

g(N, α − m) = 0.  (4.3)

By differentiation of (4.3) and using the corresponding Frenet formula, we find that

τg(T, α − m) − κg(B, α − m) = 0,

which together with the relation (4.2) gives −κg(B, α − m) = 0, and consequently

g(B, α − m) = 0.  (4.4)

Next, decompose the vector α − m as

α − m = aT + bN + cB,  (4.5)

where a = a(s), b = b(s) and c = c(s) are arbitrary functions. Then by the relations (4.2), (4.3) and (4.4), we have that

g(T, α − m) = c = 0, g(N, α − m) = b = 0, g(B, α − m) = a = 0.

Therefore, the equation (4.5) implies that α = m, which is a contradiction.

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(received 28.07.2000, in revised form 01.12.2001)

Faculty of Science, Radoja Domanovića 12, 34000 Kragujevac, Yugoslavia E-mail: mirapt@uis0.uis.kg.ac.yu, emilija@knez.uis.kg.ac.yu