W CURVES IN MINKOWSKI SPACE TIME

Transcription

1 Novi Sad J Math Vol 3, No, 00, W CURVES IN MINKOWSKI SPACE TIME Miroslava Petrović Torgašev 1, Emilija Šućurović 1 Abstract In this paper we complete a classification of W curves in Minkowski space time Namely, we classify all spacelike curves with constant curvatures in E 4 1 primary 53C50, sec- AMS Mathematics Subject Classification (000): ondary 53C40 Key words and phrases:w-curves, Minkowski space-time 1 Introduction It is well known that to each unit speed curve α : I E n in the Euclidean space E n whose successive derivatives α (s), α (s),, α (n) (s) are linearly independent vectors, one can associate the orthonormal frame {V 1, V, V 3,, V n } and n 1 functions k 1,, k n 1 : I R called the Frenet curvatures, such that the following Frenet formulas hold ([6]): V 1 V V 3 V n 1 V n = 0 k k 1 0 k k 0 k k n k n 1 0 V 1 V V 3 V n 1 V n In particular, the first curvature k 1 is also called the curvature k, and the second curvature k is also called the torsion τ Recall that a curve α is called a W curve (or a helix), if it has constant Frenet curvatures W curves in the Euclidean space E n have been studied intensively The simplest examples are circles as planar W curves and helices as non planar W curves in E 3 A parameterization of a unit speed W curve in E k+1 is given by (11) γ(s) = γ 0 + ase 0 + k ( ) r i cos(ai s)e i 1 + sin(a i s)e i, i=1 where {e 0, e 1,, e k } is an orthonormal basis of E k+1, a R, a 1 < a < < a k are positive real numbers satisfying the equation a + k i=1 (r ia i ) = 1 If 1 Institute of Mathematics, Faculty of Science, Radoja Domanovića 1, Kragujevac, Yugoslavia, e mail: miraptuis0uiskgacyu, emilija$uis0uiskgacyu

2 56 M Petrović-Torgašev, E Šućurović a 0 the curve γ lies fully in E k+1 Otherwise, γ lies fully in E k and on a hypersphere in that space We remark that a W curve is closed if and only if a = 0 and a i = p i r, p i N, r R 0 + Further, we mention that W curves in En are the examples of the finite type curves ([3]) In particular, closed W curves in E 4 are spherical type curves ([4]) All W curves in the Minkowski 3 space E 3 1 are completely classified in [10] For example, the only planar spacelike W curves are circles and hyperbolas In this paper, we classify all spacelike W curves in the Minkowski space time E 4 1 Since all three curvatures k 1, k and k 3 are constant, the classification is reduced mainly to differential equations with constant coefficients and a method well developed by B Y Chen The examples of null W curves in the Minkowski space time E 4 1 are given in [1] Timelike W curves in the same space have been studied in [8] Preliminaries Let E 4 1 denote the 4 dimensional Minkowski space time, ie the Euclidean space E 4 with the standard flat metric given by (1) g = dx 1 + dx + dx 3 + dx 4, where (x 1, x 4 ) is a rectangular coordinate system of E1 4 Since g is indefinite metric, recall that a vector v in E1 4 can have one of three causal characters: it can be spacelike if g(v, v) > 0 or v = 0, timelike if g(v, v) < 0, and null if g(v, v) = 0 and v 0 The norm of a vector v is given by v = g(v, v) Therefore, v is a unit vector if g(v, v) = ±1 Next, vectors v and w are said to be orthogonal if g(v, w) = 0 An arbitrary curve α : I E1 4 in the space E1 4 can locally be spacelike, timelike or null, if respectively all of its velocity vectors α (s) are spacelike, timelike or null Next, α is a unit speed curve if g(α (s), α (s)) = ±1 Recall that a curve α in E1 n is said to be of k type for some natural number k, if its position vector α(s) can be written as a finite sum of eigenfunctions s, cos(ps), sin(ps), cosh(qs), sinh(qs) of its Laplace operator = ± d ds which has exactly k mutually different eigenvalues {λ 1,, λ k } In particular, if one of the eigenvalues is equal to zero, α is said to be of null k type Therefore, α is a type curve in E1 4 if and only if it has one of the following forms: (i) α(s) = a 0 + (ii) α(s) = a 0 + (iii) (a i cos(p i s) + b i sin(p i s)); i=1 (a i cosh(p i s) + b i sinh(p i s)); i=1 α(s) = a 0 + a 1 cos(p 1 s) + b 1 sin(p 1 s) + a cosh(p s) + b sinh(p s); where a 0, a 1, a, b 1, b E 4 1 are constant vectors and 0 < p 1 < p, p 1, p N

3 W-curve in Minkowski space-time 57 In particular, α is a null type curve in E 4 1 if and only if it has one of the following forms: (iv) (v) α(s) = a 0 + b 0 s + a 1 cos(ps) + b 1 sin(ps); α(s) = a 0 + b 0 s + a 1 cosh(ps) + b 1 sinh(ps); where a 0, a 1, b 0, b 1 E 4 1 are constant vectors and p N Denote by {T (s), N(s), (s), B (s)} the moving Frenet frame along the spacelike curve α, where s is a pseudo arclength parameter Then T (s) is a spacelike tangent vector, so depending on the causal character of the principal normal vector N(s) and the binormal vector (s), we have the following Frenet formulas ([10]): Case (1) N and are spacelike; T 0 k Ṅ B 1 = k 1 0 k 0 0 k 0 k k 3 0 B where T, N,, B are mutually orthogonal vectors satisfying the equations T N B, g(t, T ) = g(n, N) = g(, ) = 1, g(b, B ) = 1 Case () N is spacelike, is timelike; T 0 k Ṅ B 1 = k 1 0 k 0 0 k 0 k k 3 0 B where T, N,, B are mutually orthogonal vectors satisfying the equations T N B, g(t, T ) = g(n, N) = g(b, B ) = 1, g(, ) = 1 Case (3) N is spacelike, is null; T 0 k Ṅ B 1 = k 1 0 k k k 0 k 3 B where T, N,, B satisfy the equations T N B, g(t, T ) = g(n, N) = 1, g(, ) = g(b, B ) = 0, g(t, N) = g(t, ) = g(t, B ) = g(n, ) = g(n, B ) = 0, g(, B ) = 1

4 58 M Petrović-Torgašev, E Šućurović Case (4) N is timelike, is spacelike; T Ṅ B = 0 k k 1 0 k 0 0 k 0 k k 3 0 where T, N,, B are mutually orthogonal vectors satisfying the equations T N B, g(t, T ) = g(, ) = g(b, B ) = 1, g(n, N) = 1 Case (5) N is null, is spacelike; T Ṅ B = 0 k k 0 0 k 3 0 k k 1 0 k 3 0 where the curvature k 1 can only take two values: 0 if α is a straight line, or 1 in all other cases In this case, the vectors T, N,, B satisfy the equations T N B, g(t, T ) = g(, ) = 1, g(n, N) = g(b, B ) = 0, g(t, N) = g(t, ) = g(t, B ) = g(n, ) = g(, B ) = 0, g(n, B ) = 1 The notion of causal character of vectors has a natural generalization to vector subspaces A subspace V of E1 4 can be spacelike, timelike or lightlike if respectively g V is positive definite, g V is nondegenerate of index 1 or g V is degenerate For a subspace V of the Minkowski space time E1, 4 recall that V is a subspace defined by V = {v E1 4 : v V } Then the following simple property holds: a subspace V is timelike (spacelike) if and only if V is spacelike (timelike) ([7]) Moreover, if V is a timelike (spacelike) subspace, then E1 4 = V V, where denotes the direct sum of subspaces Next, a subspace V is lightlike if and only if V is lightlike, but then V V is not all of E1 4 Recall some of the most important hyperquadrics in E 4 1 The pseudo Riemannian sphere and the pseudo hyperbolic space in E 4 1 are defined respectively by () S 3 1(c, r) = {x E 4 1 : g(x c, x c) = r }, (3) H 3 (c, r) = {x E 4 1 : g(x c, x c) = r }, where r > 0 is a radius and c E 4 1 is a center of the mentioned hyperquadrics Finally, the light cone C(c) with the vertex at a point c in E 4 1 is defined by (4) C(c) = {x E 4 1 : g(x c, x c) = 0}

5 W-curve in Minkowski space-time 59 3 A classification of spacelike W curves First we give some introductory results which characterize spacelike curves in the Minkowski space time E 4 1 In [10] it is proved that a spacelike curve α in E 3 1 with g( α, α) 0 has the second curvature k 0 if and only if α is a planar curve Therefore, it is easy to prove that in E 4 1 the following analogous theorem holds Theorem 31 Let α be a spacelike unit speed curve in E 4 1 with curvature k 1 > 0 Then α has k 0 if and only if α lies fully in a dimensional subspace of E 4 1 The following theorems characterize spacelike curves with respect to their third curvature k 3 Theorem 3 Let α be a spacelike unit speed curve in E1 4 with a spacelike principal normal N, a spacelike binormal and with curvatures k 1 > 0, k 0 Then α has k 3 0 if and only if α lies fully in a spacelike hyperplane of E1 4 Proof If α has k 3 0, then by using the Frenet equations we obtain α = T, α = k 1 N, α= k1t + k 1 N + k 1 k, α = 3k 1 k 1 T + ( k 1 k1 3 k 1 k)n + (k 1 k + k 1 k ) Next, all higher-order derivatives of α are linear combinations of vectors α, α, α, so by using the MacLaurin expansion for α given by (31) α(s) = α(0) + α(0) s + α(0) s! + α (0) s3 3! +, we conclude that α lies fully in a spacelike hyperplane of the space E 4 1, spanned by { α(0), α(0), α (0)} Conversely, assume that α satisfies the asumptions of the theorem and lies fully in a spacelike hyperplane π of E 4 1 Then there exist points p, q E1, 4 such that α satisfies the equation of π given by g(x(s) p, q) = 0, where q π is a timelike vector Differentiation of the last equation yields g( α, q) = g( α, q) = g( α, q) = 0 Therefore, α, α, α π Since T = α, N = α α, it follows that g(t, q) = g(n, q) = 0 Next, differentiation of the equation g(n, q) = 0 gives g(ṅ, q) = 0 From the Frenet equations we obtain = 1 k (Ṅ + k 1T ), so g(, q) = 0 Since B (s) is the unique timelike unit vector perpendicular to {T, N, }, it follows that B (s) = q q Thus B (s) = k 3 = 0 for each s and therefore k 3 0 Theorem 33 Let α be a spacelike unit speed curve in E1 4 with a spacelike (timelike) principal normal N, a timelike (spacelike) binormal and with curvatures k 1 > 0, k 0 Then α has k 3 0 if and only if α lies fully in a timelike hyperplane of E1 4

6 60 M Petrović-Torgašev, E Šućurović We omit the proof, as it is analogous to the proof of Theorem 3 Next, recall that a spacelike curve with a spacelike principal normal N and a null binormal is called a partially null spacelike curve Theorem 34 A partially null spacelike unit speed curve α in E 4 1 with curvatures k 1 > 0, k 0 lies fully in a lightlike hyperplane of E 4 1 and has k 3 0 Proof By using the Frenet formulas for this case, we obtain α = T, α = k 1 N, α= k 1 T + k 1 N + k 1 k, α = 3k 1 k 1 T + ( k 1 k1)n 3 + (k 1 k 1 + k 1 k + k 1 k k 3 ) Thus α, α, α are linearly independent vectors, while α, α, α, α are not linearly independent Moreover, all higher-order derivatives of α are linear combinations of α, α, α, so by using the MacLaurin expansion (31) it follows that α lies fully in a lightlike hyperplane π of E1, 4 spanned by { α(0), α(0), α (0)} Therefore, we may assume that there exist points p, q E1, 4 such that α satisfies the equation of π given by g(x(s) p, q) = 0, where q π is a null vector Since q is a null vector perpendicular to, it follows that q = λ, λ R 0 Then q = λk 3 = 0 and thus k 3 0 Remark 31 Also by making a null rotation from one null tetrad to another null tetrad, we can make k 3 0 For more details, see [] Next, recall that a spacelike curve with a null principal normal is called a pseudo null spacelike curve Such curves are characterized by the following theorem Theorem 35 A pseudo null spacelike unit speed curve α in E 4 1 with curvatures k 1 > 0, k 0 lies fully in the space E 4 1 Proof The Frenet formulas imply the equations α = T, α = N, α= k, α = k k 3 N + k kb Therefore, the vectors α, α, α, α are linearly independent On the other hand, all higher order derivatives of α may be expressed as linear combinations of vectors α, α, α, α Thus by using the MacLaurin expansion (31) for α, we conclude that α lies fully in the space E1, 4 spanned by { α(0), α(0), α (0), α (0)} Theorem 36 Let α be a spacelike unit speed curve in E1, 4 with a spacelike principal normal N and a spacelike binormal Then α has: (i) k 1 = c 1, k = c, k 3 = 0, c 1, c R 0 if and only if α can be parameterized by (3) α(s) = 1 λ (0, c λs, c 1 sin(λs), c 1 cos(λs)), λ = c 1 + c ; (ii) k 1 = c 1, k = c, k 3 = c 3, c 1, c, c 3 R 0 if and only if α can be parameterized by (33) α(s) = 1 λ 1 (V 1 sinh(λ 1 s) + V cosh(λ 1 s)) + 1 λ (V 3 sin(λ s) V 4 cos(λ s))

7 W-curve in Minkowski space-time 61 with λ 1 = K+ K +4c 1 c 3, λ = K+ K +4c 1 c 3, K = c 1 + c c 3, where V 1, V, V 3, V 4 are mutually orthogonal vectors satisfying the equations g(v 1, V 1 ) = g(v, V ) = λ c 1, g(v λ 3, V 3 ) = g(v 4, V 4 ) = λ 1 +c 1 1 +λ λ 1 +λ Proof (i) If α has constant curvatures, then by using the Frenet formulas we find T +(c 1 + c ) T = 0 Solving this equation, we easily obtain T = A + B cos( c 1 + c s) + C sin( c 1 + c s), where A, B, C E4 1 are constant vectors Next, the equation g(t, T ) = 1 implies that g(b, B) = g(c, C), g(a, B) = g(a, C) = g(b, C) = 0, g(a, A) = 1 g(b, B) On the other hand, the equation g( T, T ) = c 1 gives g(b, B) = c 1 Therefore, we may c 1 +c c take A = (0,, 0, 0), B = (0, 0, c 1, 0), C = (0, 0, 0, c 1 ) and up to c 1 +c c 1 +c c 1 +c isometries of E1 4 the curve α has the form (3) Conversely, if α has the form (3), then it lies fully in a spacelike hyperplane of E1, 4 with the equation x 1 = 0 Then the Theorem 3 implies that k 3 0 Next, from the Frenet equations we get g( T, T ) = k1, g(ṅ, Ṅ) = k 1 + k Since T = α and N = α α, we find g( T, T ) = c 1, g(ṅ, Ṅ) = c 1 + c Accordingly, k 1 = c 1 and k = c (ii) First assume that α has constant curvatures different from zero Then from the Frenet formulas we obtain the equation T +(c 1+c c 3) T c 1c 3T = 0 Solving the previous equation, we find T = V 1 cosh(λ 1 s) + V sinh(λ 1 s) + V 3 cos(λ s) + V 4 sin(λ s) where V 1, V, V 3, V 4 E1 4 are constant vectors, λ 1 = K+ K +4c 1 c 3, λ = K+ K +4c 1 c 3, K = c 1 + c c 3 Next, the equation g(t, T ) = 1 implies g(v 1, V 1 ) = g(v, V ), g(v 3, V 3 ) = g(v 4, V 4 ), g(v 1, V 1 )+g(v 3, V 3 ) = 1, g(v i, V j ) = 0 for i j (i, j {1,, 3, 4}) Finally, by using the equation g( T, T ) = c 1, we get g(v 3, V 3 ) = λ 1 +c 1 Accordingly, α has the form (33) λ 1 +λ Conversely, if α can be parameterized by (33), then it has spacelike principal normal N and spacelike binormal, so that the Frenet formulas imply g( T, T ) = k1, g(ṅ, Ṅ) = k 1 + k Since T = α and N = α α, we get g( T, T ) = c 1, g(ṅ, Ṅ) = c 1 + c Thus k 1 = c 1 and k = c Finally, by the Frenet equations we get g( B 1, B 1 ) = k k3, and on the other hand since B 1 = 1 c ( N + c 1N) we obtain g( B 1, B 1 ) = c c 3 Consequently, k 3 = c 3 Remark 3 The curve (3) lies on a circular cylinder in E1 4 with the equation x 3 + x c 4 = 1 The curve (33) lies on some hyperquadric in E (c 1 4 More 1 +c ) precisely, if c 3 > c, c 3 < c, or c 3 = c, then respectively α lies on pseudo Riemannian sphere with the equation x 1 + x + x 3 + x 4 = c 3 c c 1 c 3, pseudo hyperbolic space with the same equation or light cone with the equation x 1 +

8 6 M Petrović-Torgašev, E Šućurović x + x 3 + x 4 = 0 By the Theorem 33, all spacelike W curves with a spacelike (timelike) principal normal N and a timelike (spacelike) binormal which have k 3 0 lie fully in E 3 1, so their classification is given in [10] In the next two theorems we consider the remaining cases and omite the proofs, since they are very similar with the proof of the Theorem 36 Theorem 37 A spacelike unit speed curve α in E 4 1 with a spacelike principal normal N and a timelike binormal has k 1 = c 1, k = c, k 3 = c 3, c 1, c, c 3 R 0 if and only if α can be parameterized by (34) α(s) = 1 λ 1 (V 1 sinh(λ 1 s) + V cosh(λ 1 s)) + 1 λ (V 3 sin(λ s) V 4 cos(λ s)), with λ 1 = K+ K +4c 1 c 3, λ = K+ K +4c 1 c 3, K = c 1 c c 3, where V 1, V, V 3, V 4 are mutually orthogonal vectors satisfying the equations g(v 1, V 1 ) = g(v, V ) = λ c 1, g(v λ 3, V 3 ) = g(v 4, V 4 ) = c 1 +λ 1 1 +λ λ 1 +λ Remark 33 The curve (34) lies on pseudo Riemannian sphere with the equation x 1 + x + x 3 + x 4 = c +c 3 c 1 c 3 Theorem 38 A spacelike unit speed curve α in E 4 1 with a timelike principal normal N has k 1 = c 1, k = c, k 3 = c 3, c 1, c, c 3 R 0 if and only if α can be parameterized by (35) α(s) = 1 λ 1 (V 1 sinh(λ 1 s) + V cosh(λ 1 s)) + 1 λ (V 3 sin(λ s) V 4 cos(λ s)), with λ 1 = K+ K +4c 1 c 3, λ = K+ K +4c 1 c 3, K = c 3 c 1 c, where V 1, V, V 3, V 4 are mutually orthogonal vectors satisfying the equations g(v 1, V 1 ) = g(v, V ) = λ +c 1, g(v λ 3, V 3 ) = g(v 4, V 4 ) = λ 1 c 1 1 +λ λ 1 +λ Remark 34 The curve (35) lies on some hyperquadric in E 4 1 If c > c 3, c < c 3, c = c 3, then respectively α lies on pseudo Riemannian sphere with the equation x 1 + x + x 3 + x 4 = c c 3, pseudo hyperbolic space with the c 1 c 3 same equation or light cone with the equation x 1 + x + x 3 + x 4 = 0 By the Theorem 34, a partially null spacelike curve α has k 3 (s) = 0 for each s In the following theorem, we classify all partially null spacelike W curves in E 4 1 Theorem 39 A partially null spacelike unit speed curve α in E 4 1 has k 1 = c 1 R 0, k = constant 0 if and only if α is a part of a partially null spacelike

9 W-curve in Minkowski space-time 63 helix (36) α(s) = (as, as, 1 c 1 sin(c 1 s), 1 c 1 cos(c 1 s)), a R 0 Proof First assume that α has non zero constant curvatures Then by the Frenet equations we find T +c 1T = 0 Solving this equation, we get T = V 1 + V cos(c 1 s) + V 3 sin(c 1 s), where V 1, V, V 3 E1 4 are constant vectors Next, the equation g(t, T ) = 1 implies that g(v 1, V 1 ) + g(v, V ) = 1, g(v 1, V ) = g(v 1, V 3 ) = g(v, V 3 ) = 0, g(v, V ) = g(v 3, V 3 ) Finally, by using the equation g( T, T ) = c 1, we obtain g(v, V ) = 1 Therefore, we may take V 1 = (a, a, 0, 0, ), a R 0, V = (0, 0, 1, 0), V 3 = (0, 0, 0, 1), so up to isometries of E1 4 the curve α has the form (36) On the other hand, if α can be parameterized by (36), then we obtain that α = (0, 0, c 1 sin(c 1 s), c 1 cos(c 1 s)) and thus g( α, α) = c 1 However, from the Frenet formulas we get g( T, T ) = g( α, α) = k1 It follows that k 1 = c 1 Next, since N = α α, we obtain that α = (0, 0, c 3 1 sin(c 1 s), c 3 1 cos(c 1 s)) = c 3 1N However, by the Frenet equations we get α = k1n 3 + k 1 k It follows that k 1 k = 0 and therefore k = constant 0 Remark 35 The curve (36) lies on a circular cylinder in E1 4 with the equation x 3 + x 4 = 1 c 1 Theorem 310 Let α be a pseudo null spacelike unit speed curve in E 4 1 Then α has: (i) k 1 = 1, k = c, k 3 = 0, c R 0, if and only if α can be parameterized by (37) α(s) = 1 c (cosh( c s), sinh( c s), sin( c s), cos( c s)); (ii) k 1 = 1, k = c, k 3 = c 3, c, c 3 R 0 if and only if α can be parameterized by (38) α(s) = 1 λ 1 (V 1 sinh(λ 1 s) + V cosh(λ 1 s)) + 1 λ (V 3 sin(λ s) V 4 cos(λ s)), with λ 1 = K + K + c, λ = K + K + c, K = c c 3, where V 1, V, V 3, V 4 are mutually orthogonal vectors satisfying the equations g(v 1, V 1 ) = g(v, V ) = λ, g(v λ 3, V 3 ) = g(v 4, V 4 ) = λ 1 1 +λ λ 1 +λ Proof (i) First assume that α has k 1 = 1, k = c, k 3 = 0 Then by using the Frenet equations we find T c T = 0 Solving the previous equation, we obtain that T = V 1 cosh( c s) + V sinh( c s) + V 3 cos( c s) + V 4 sin( c s),

10 64 M Petrović-Torgašev, E Šućurović where V 1, V, V 3, V 4 E1 4 are constant vectors Further, the equation g(t, T ) = 1 implies that g(v 1, V 1 ) = g(v, V ), g(v 3, V 3 ) = g(v 4, V 4 ), g(v 1, V 1 )+g(v 3, V 3 ) = 0, g(v i, V j ) = 0 for i j, (i, j {1,, 3, 4}) Finally, by using the equation g( T, T ) = 0, we find g(v 3, V 3 ) = 1 Consequently, we may take V 1 = 1 (0,, 0, 0), V = ( 1 1, 0, 0, 0), V 3 = (0, 0, 1, 0), V 4 = (0, 0, 0, ) Accordingly, up to isometries of E1 4 the curve α has the form (37) Conversely, if α can be parameterized by (37), then we find g( α, α) = 0 and therefore k 1 = 1 Next, we find g( α, α) = g(ṅ, Ṅ) = c However, the Frenet formulas give g(ṅ, Ṅ) = k It follows that k = c Finally, the Frenet equations imply g( B 1, B 1 ) = k k 3 and on the other hand since = α, α we obtain that g( B 1, B 1 ) = 0 Therefore, k 3 = 0 (ii) Suppose that α has constant curvatures k 1 = 1, k = c, k 3 = c 3 Then by the Frenet formulas we find T c c 3 T c T = 0 Solving this equation, we obtain T = V 1 cosh(λ 1 s) + V sinh(λ 1 s) + V 3 cos(λ s) + V 4 sin(λ s), where V 1, V, V 3, V 4 E1 4 are constant vectors, λ 1 = K+ K + c, λ = K+ K + c, K = c c 3 Next, the equation g(t, T ) = 1 implies that g(v 1, V 1 ) = g(v, V ), g(v 3, V 3 ) = g(v 4, V 4 ), g(v 1, V 1 ) + g(v 3, V 3 ) = 1, g(v i, V j ) 0 for i j (i, j {1,, 3, 4}) Finally, from the equation g( T, T ) = 0, we get g(v 1, V 1 ) = λ λ 1 +λ Consequently, α has the form (38) On the other hand, if α can be parameterized by (38), then we find that g( α, α) = 0 and thus k 1 = 1 Further, we find that g( α, α) = c and from the Frenet formulas we get g(ṅ, Ṅ) = k It follows that k = c Finally, since = α, we obtain g( B α 1, B 1 ) = c c 3 However, the Frenet equations imply g( B 1, B 1 ) = k k 3 and consequently k 3 = c 3 Remark 36 The curve (37) lies on a light cone in E1 4 with the equation x 1 + x + x 3 + x 4 = 0 The curve (38) lies on some hyperquadric in E1 4 If c c 3 > 0 or c c 3 < 0, then respectively α lies on pseudo Riemannian sphere with the equation x 1 + x + x 3 + x 4 = c3 c or on pseudo hyperbolic space with the same equation Finally, note that some of a W curves are the curves of type The proof of the following theorem follows immediately from definition of type curves Theorem 311 The curves (33), (34), (35) and (38) for which λ 1, λ N are a type curves The curve (37) for which c N is a type curve The curves (3) and (36) for which respectively λ N, c 1 N are a null type curves

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