A note on proper affine vector fields in non-static plane symmetric space-times Ghulam Shabbir Faculty of Engineering Sciences, GIK Institute of Engineering Sciences and Technology, Topi, Swabi, NWFP, Pakistan. Email:
[email protected] Abstract The most general form of the non-static plane symmetric space-times is considered to study proper affine vector fields by using holonomy and decomposability, the rank of the 6× 6 Riemann matrix and direct integration techniques. Studying proper
affine vector fields in each non static case of the above space-times it is shown that the very special classes of the above space-times admit proper affine vector fields. Here we also discuss the Lie algebra in each case. 1. INTRODUCTION
The most general form of non-static plane symmetric space-times is considered to study the existence of proper affine vector fields in non-static plane symmetric spacetimes by using holonomy and decomposability, the rank of the 6× 6 Rieman matrix and direct integration techinques. Affine vector fields which preserve the geodesic structure and affine parameter of a space-time carries significant information and interest in the Einstein’s theory of general relativity. It is therefor important to study this symmetry. Different approaches [2,4,5,8-13] were adopted in studying affine vector fields. It is important to mention here that we will only considering the non static cases. The cases when the above space-time becomes static are studies in [2]. Let ( M , g ) be a space-time with M a smooth connected Hausdorff four dimensional manifold and g a smooth metric of Lorentz signature (-, +, +, +) on M . The curvature tensor associated with g, through Levi-Civita connection Γ, is denoted in component form by R a bcd . The usual covariant, partial and Lie derivatives are denoted by a semicolon, a comma and the
1
symbol L , respectively. Round and square brackets denote the usual symmetrization and skew-symmetrization, respectively. The space-time M will be assumed nonflat in the sense that the Riemann tensor does not vanish over any non empty open subset of M . A vector field X on M is called an affine vector field if it satisfies X a ;bc = Rabcd X d ,
(1)
where R abcd = g af R f bcd = g af ( Γbdf ,c − Γbcf , d + Γcef Γbde − Γedf Γbce ). If one decomposes X a ;b on M into its symmetric and skew-symmetric parts
X a ;b =
1 H ab + Gab , 2
( H ab (≡ X a;b + X b;a ) = H ba , Gab = −Gba )
(2)
then equation (1) is equivalent to (i ) H ab;c = 0 (ii ) Gab;c = Rabcd X d
(iii ) Gab;c X c = 0.
(3)
The proof of the above equation (1) implies (3) or equations (3) implies (1) can be found in [3,4]. If H ab = 2cg ab , c ∈ R, then the vector field X is called homothetic (and Killing if c = 0 ). The vector field X is said to be proper affine if it is not homothetic vector field and also X is said to be proper homothetic vector field if it is not Killing vector field on M [5]. Define the subspace Z p of the tangent space T p M to M at p as those k ∈ T p M
satisfying Rabcd k d = 0.
(4)
The Lie algebra of a set of vector fields on a manifold is completely characterized by the structure constants C bca given in term of the Lie brackets by [ X b , X c ] = C bca X a ,
C bca = −C cba ,
where X a are the generators and a, b, c = 1,2,3,....n. In each case we discuss the Lie algebra for the affine vector fields.
2. Affine Vector Fields Suppose that M is a simple connected space-time. Then the holonomy group of
M is a connected Lie subgroup of the idenity component of the Lorentz group and is thus characterized by its subalgebra in the Lorentz algebra. These have been labeled into
2
fifteen types R1 − R15 [1]. It follows from [5] that the only such space-times which could admit proper affine vector fields are those which admit nowhere zero covariantly constant second order symmetric tensor field H ab . This forces the holonomy type to be either R2 , R3 , R4 , R6 , R7 , R8 , R10 , R11 or R13 [5]. A study of the affine vector fields for the
above holonomy type can be found in [5]. It follows from [6] that the rank of the 6 × 6 Riemann matrix of the above space-times which have holonomy type R2 , R3 , R4 , R6 , R7 , R8 , R10 , R11 or R13 is at most three. Hence for studying affine vector fields we are
interested in those cases when the rank of the 6 × 6 Riemann matrix is less than or equal to three. 3. MAIN RESULTS Consider a non static plane symmetric space-time in the usual coordinate system
(t , x, y, z ) (labeled by ( x 0 , x 1 , x 2 , x 3 ), respectively) with line element [7] ds 2 = −e A( t , x ) dt 2 + e B ( t , x ) dx 2 + e C ( t , x ) ( dy 2 + dz 2 ).
(5)
The Ricci tensor Segre type of the above space-time is {1,1(11)} or {2(11)} or one of its degeneracies. The above space-time admits three linearly independent Killing vector fields which are ∂ ∂ ∂ ∂ , , y −z . ∂y ∂z ∂z ∂y
(6)
The non-zero independent components of the Riemann tensor are R0101 =
A( t , x ) ( Ax2 (t , x) + 2 Axx (t , x) − Ax (t , x) B x (t , x)) − e B (t , x ) 1 e ≡ α1 , 4 ( Bt2 (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x))
e B (t , x ) (C t2 (t , x) + 2 C tt (t , x) − At (t , x) C t (t , x)) − 1 R0202 = R0303 = − e C ( t , x ) − B (t , x ) A(t , x ) ≡α2, 4 Ax (t , x) C x (t , x)) e e A( t , x ) (C x2 (t , x) + 2 C xx (t , x) − B x (t , x) C x (t , x)) − 1 R1212 = R1313 = − e C (t , x ) − A( t , x ) B ( t , x ) ≡ α3, 4 Bt (t , x) C t (t , x) e
[
]
1 R 2323 = − e A( t , x ) + B (t , x ) + 2C ( t , x ) e A( t , x ) C x2 (t , x) − e B (t , x ) C t2 (t , x) ≡ α 4 , 4
R0212 = R0313 =
1 C ( t , x ) (C t (t , x)C x (t , x) + 2 C tx (t , x) − Ax (t , x) C t (t , x) − e B (t , x) C (t , x)) ≡ α5. 4 x t
Writing the curvature tensor with components Rabcd at p as a 6 × 6 symmetric matrix
3
Rabcd
α1 0 0 = 0 0 0
0
0
0
0
α2
0
α5
0
0
α2
0
α5
α5
0
α3
0
0
α5
0
α3
0
0
0
0
0 0 0 . 0 0 α 4
(7)
As mentioned in section 2, the space-times which can admit proper affine vector fields have holonomy type R2 , R3 , R4 , R6 , R7 , R8 , R10 , R11 or R13 and the rank of the 6 × 6 Riemann matrix is at most three. Therefore we are only interested in those cases
when the rank of the 6 × 6 Riemann matrix is less than or equal to three. It is important to remind the reader that we are only considering the non static cases. The cases when the above space-time becomes static are discussed in [2]. Thus there exist the following possibilities: (1) Rank=3 Ax (t , x) = 0, Bt (t , x) = 0, C x (t , x) = 0, At (t , x) ≠ 0, B x (t , x) ≠ 0, Ct (t , x) ≠ 0 and 2
Ct (t , x) + 2 Ctt (t , x) ≠ 0.
(2) Rank=3 Ax (t , x) = 0, Bt (t , x) = 0, B x (t , x ) = 0, C x (t , x) = 0, At (t , x) ≠ 0, C t (t , x) ≠ 0 and 2
C t (t , x) + 2 C tt (t , x) ≠ 0.
(3) Rank=3,
Ax (t , x ) = 0, Bt (t , x ) = 0, C x (t , x ) = 0, At (t , x ) ≠ 0, B x (t , x ) ≠ 0, C t (t , x ) ≠ 0
and
2
C t (t , x) + 2 C tt (t , x) = 0.
(4) Rank=3,
Ax (t , x) = 0, At (t , x) = 0, Bt (t , x) = 0, C x (t , x) = 0, B x (t , x) ≠ 0, Ct (t , x) ≠ 0
and
2
C t (t , x) + 2 C tt (t , x) ≠ 0. Ax (t , x) = 0, At (t , x) = 0, Bt (t , x) = 0, Bx (t , x) = 0, C x (t , x) = 0, Ct (t , x) ≠ 0
and
C t (t , x) + 2 C tt (t , x) ≠ 0. (6) Rank=3, Ax (t , x) = 0, At (t , x) = 0, C x (t , x) = 0, Bt (t , x) ≠ 0, Bx (t , x) ≠ 0, Ct (t , x) ≠ 0
and
(5) Rank=3, 2
2
2
C t (t , x) + 2 C tt (t , x) = 0 and Bt (t , x) + 2 Btt (t , x) = 0.
(7) 2
Rank=3,
Ax (t , x) = 0, At (t , x) = 0, B x (t , x) = 0, C x (t , x) = 0, Bt (t , x) ≠ 0, C t (t , x) ≠ 0, 2
C t (t , x) + 2 C tt (t , x) = 0 and Bt (t , x) + 2 Btt (t , x) = 0. (8) Rank=3, A x (t , x ) = 0, At (t , x ) = 0, B t (t , x ) = 0, B x (t , x ) ≠ 0, C x (t , x ) ≠ 0, C t (t , x ) ≠ 0, 2
2 Ct (t , x) + 2 Ctt (t , x) = 0, Ct (t , x) C x (t , x) + 2 Ctx (t , x) = 0 and C x (t , x) + 2 C xx (t , x) ≠ 0.
4
(9)
Rank=3,
Ax (t , x ) = 0, At (t , x ) = 0, Bt (t , x ) = 0, B x (t , x ) ≠ 0, C x (t , x ) ≠ 0, C t (t , x ) ≠ 0,
2 Ct (t , x) + 2 Ctt (t , x) = 0, Ct (t , x) C x (t , x) + 2 Ctx (t , x) = 0 and C x 2 (t , x) + 2 C xx (t , x) = 0.
(10)
Rank=3,
Ax (t , x) = 0, At (t , x) = 0, Bt (t , x) = 0, B x (t , x) = 0, C x (t , x) ≠ 0, C t (t , x) ≠ 0, 2
2 Ct (t , x) + 2 Ctt (t , x) = 0, Ct (t , x) C x (t , x) + 2 Ctx (t , x) = 0 and C x (t , x) + 2 C xx (t , x) ≠ 0.
(11)
Rank=3,
Ax (t , x ) = 0, Bt (t , x ) = 0, B x (t , x ) = 0, At (t , x ) ≠ 0, C x (t , x ) ≠ 0, C t (t , x ) ≠ 0,
2
C x (t , x) + 2 C xx (t , x) = 0, Ct (t , x) C x (t , x) + 2 Ctx (t , x) = 0 and C t 2 (t , x) + 2 C tt (t , x) = 0. (12) Rank=3, Ax (t , x) = 0, Bt (t , x) = 0, B x (t , x) = 0, At (t , x) ≠ 0, C x (t , x) ≠ 0, C t (t , x) ≠ 0, 2
2 Ct (t , x) + 2 Ctt (t , x) ≠ 0, Ct (t , x) C x (t , x) + 2 Ctx (t , x) = 0 and C x (t , x) + 2 C xx (t , x) = 0.
(13)
Rank=3,
Ax (t , x) = 0, At (t , x) = 0, B x (t , x) = 0, Bt (t , x) = 0, C x (t , x) ≠ 0, C t (t , x) ≠ 0, 2
2 Ct (t , x) + 2 Ctt (t , x) ≠ 0, Ct (t , x) C x (t , x) + 2 Ctx (t , x) = 0 and C x (t , x) + 2 C xx (t , x) = 0.
(14)
Rank=1, C t (t , x ) = 0, C x (t , x ) = 0, Ax (t , x) ≠ 0, At (t , x) ≠ 0, Bt (t , x ) ≠ 0, B x (t , x) ≠ 0 2 2 Ax (t , x) + 2 Axx (t , x) − Ax (t , x) Bx (t , x) ≠ 0 and Bt (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x) ≠ 0. (15) Rank=1, Ax (t , x) = 0, C t (t , x) = 0, C x (t , x) = 0, At (t , x) ≠ 0, Bt (t , x) ≠ 0, B x (t , x) ≠ 0 and 2
Bt (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x) ≠ 0. (16) Rank=1, Ax (t , x) = 0, B x (t , x) = 0, C t (t , x) = 0, C x (t , x) = 0, At (t , x) ≠ 0, Bt (t , x) ≠ 0 and 2
Bt (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x) ≠ 0. (17) Rank=1, C t (t , x ) = 0, C x (t , x ) = 0, A x (t , x ) ≠ 0, At (t , x ) ≠ 0, B x (t , x ) ≠ 0, B t (t , x ) ≠ 0, 2
Bt (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x) ≠ 0 and Ax 2 (t , x) + 2 Axx (t , x) − Ax (t , x) B x (t , x) = 0. (18) Rank=1, C t (t , x ) = 0, C x (t , x ) = 0, Ax (t , x ) ≠ 0, At (t , x ) ≠ 0, Bt (t , x ) ≠ 0, B x (t , x ) ≠ 0 2 2 Ax (t , x) + 2 Axx (t , x) − Ax (t , x) Bx (t , x) ≠ 0 and Bt (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x) = 0.
(19)
Rank=1,
Ct (t , x) = 0, C x (t , x) = 0, Ax (t , x) ≠ 0, At (t , x) ≠ 0, B x (t , x) ≠ 0, Bt (t , x) ≠ 0,
2
2
Ax (t , x) + 2 Axx (t , x) = 0 and Bt (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x) ≠ 0. (20) Rank=1, C t (t , x) = 0, C x (t , x) = 0, Ax (t , x) ≠ 0, At (t , x) ≠ 0, B x (t , x) ≠ 0, Bt (t , x) ≠ 0, 2
2
Ax (t , x) + 2 Axx (t , x) − Ax (t , x) B x (t , x) ≠ 0 and Bt (t , x) + 2 Btt (t , x) = 0. (21) Rank=1, At (t , x ) = 0, B x (t , x ) = 0, C x (t , x ) = 0, C t (t , x ) = 0, Ax (t , x ) ≠ 0, Bt (t , x ) ≠ 0, 2
2
Ax (t , x) + 2 Axx (t , x) ≠ 0 and Bt (t , x) + 2 Btt (t , x) ≠ 0. (22) Rank=1, At (t , x ) = 0, B x (t , x ) = 0, C x (t , x ) = 0, C t (t , x ) = 0, Ax (t , x ) ≠ 0, Bt (t , x ) ≠ 0, 2
2
Bt (t , x) + 2 Btt (t , x) ≠ 0 and Ax (t , x) + 2 Axx (t , x) = 0. (23) Rank=1, Ax (t , x) = 0, At (t , x) = 0, B x (t , x) = 0, C t (t , x) = 0, C x (t , x) = 0, Bt (t , x) ≠ 0 and 2
Bt (t , x) + 2 Btt (t , x) ≠ 0. (24) Rank=1, Ax (t , x) = 0, B x (t , x) = 0, C t (t , x) = 0, C x (t , x) = 0, At (t , x) ≠ 0, Bt (t , x) ≠ 0 2
and
Bt (t , x) + 2 Btt (t , x) = 0. (25) Rank=1, Ax (t , x) = 0, B x (t , x) ≠ 0, C t (t , x) = 0, C x (t , x) = 0, At (t , x) ≠ 0, Bt (t , x) ≠ 0 and 2
Bt (t , x) + 2 Btt (t , x) = 0.
5
(26)
Rank=1,
Ct (t , x) = 0, C x (t , x) = 0, Ax (t , x) ≠ 0, At (t , x) ≠ 0, B x (t , x) ≠ 0, Bt (t , x) ≠ 0,
2
2
Ax (t , x) + 2 Axx (t , x) = 0 and Bt (t , x) + 2 Btt (t , x) = 0. (27) Rank=1, Ax (t , x ) = 0, At (t , x ) = 0, Bt (t , x ) = 0, B x (t , x) = 0, C x (t , x) = 0, C t (t , x ) ≠ 0 and 2
Ct (t , x) + 2 Ctt (t , x) = 0. (28) Rank=1, A x ( t , x ) = 0 , At ( t , x ) = 0 , B t ( t , x ) = 0 , B x ( t , x ) = 0 , C t ( t , x ) ≠ 0 , C x ( t , x ) ≠ 0 , 2 C x (t , x) + 2 C xx (t , x) = 0 and C t C x (t , x) + 2 C tx (t , x) = 0. We will consider each case in turn. Case 1
In this case we have B x (t , x) ≠ 0,
Ax (t , x) = 0, Bt (t , x) = 0,
C x (t , x) = 0
At (t , x) ≠ 0,
C t (t , x) ≠ 0, C t 2 (t , x) + 2 C tt (t , x) ≠ 0 and the rank of the 6 × 6 Riemann
matrix is three. Here, there exists a unique (up to a multiple) no where zero spacelike vector field x a = x,a satisfying x a ;b = 0. From the Ricci identity R a bcd x a = 0. From the above constraints we have Ax (t , x) = 0, Bt (t , x) = 0 and C x (t , x) = 0 ⇒ A(t , x) = α (t ), B(t , x) = β ( x) and C (t , x) = η (t ). Substituting the information of A(t , x), B(t , x) and
C (t , x) in (7) and after a rescaling of x, the line element can be written in the form ds 2 = −e α ( t ) dt 2 + dx 2 + eη ( t ) ( dy 2 + dz 2 ).
(8)
The above space-time (8) is 1+3 decomposable. Affine vector fields in this case [5] are X = (c 4 x + c 5 )
∂ + X ′, ∂x
(9)
where c 4 , c5 ∈ R and X ' is a homothetic vector field in the induced geometry on each of the three dimensional submanifolds of constant x. The completion of case 1 requires finding the homothetic vector fields in the induced geometry on the submanifolds of constant x. The induced metric g λω (where λ , ω = 0,2,3 ) has non zero components given by g 00 = −e α (t ) ,
g 22 = g 33 = eη ( t )
(10)
A vector field X ' is called a homothetic vector field if it satisfies L g λω = cg λω , c ∈ R.
(11)
X'
One can expand (11) using (10) to get
α& X 0 + 2 X 0 ,0 = c,
(12)
6
− e α X 0 , 2 + eη X 2 ,0 = 0,
(13)
− e α X 0 ,3 + eη X 3 ,0 = 0,
(14)
η& X 0 + 2 X 2 , 2 = c,
(15)
X 2 ,3 + X 3 , 2 = 0,
(16)
η& X 0 + 2 X 3 ,3 = c.
(17)
Equations (12), (13) and (14) give α
X0 =
α
α
α
− 1 −2 2 c e ∫ e dt + e 2 N 1 ( y, z ), 2
α
−η
X 2 = N 1y ( y, z ) ∫ e 2 dt + N 2 ( y, z ),
−η
X 3 = N 1z ( y, z ) ∫ e 2 dt + N 3 ( y, z ),
where N 1 ( y, z ), N 2 ( y, z ), and N 3 ( y, z ) are functions of integrartion. If one proceeds further, after a straightforward calculation one can find that the proper homothetic vector fields exist if and only if η (t ) = ln(a t + b ) and α (t ) = d , where a, b, d ∈ R. Substituting 2
this information into (7), one finds that the rank of the Riemann matrix reduces to one, thus giving a contradiction. So the homothetic vector fields in the induced geometry are the Killing vector fields which are
X 0 = 0,
X 2 = z c1 + c 2 ,
X 3 = − y c1 + c3 ,
(18)
where c1 , c 2 , c3 ∈ R. Affine vector fields in this case are given by use of equation (9) and (18)
X 0 = 0,
X 1 = c 4 x + c5 ,
X 2 = z c1 + c 2 ,
X 3 = − y c1 + c3 .
(19)
In this case the generators are X1 = z
∂ ∂ ∂ ∂ ∂ ∂ −y , X2 = , X3 = , X4 = x , X5 = . ∂y ∂z ∂y ∂z ∂x ∂x
Here the non zero components for the Lie brackets are [ X 1 , X 2 ] = X 3 , [ X 1 , X 3 ] = − X 2 , and [ X 4 , X 5 ] = − X 5 . In this case Lie algebra is closed. One can write the above equation (19) after subtracting Killing vector fields as X = (0, c 4 x + c5 ,0,0).
(20)
Clearly, in this case the above space-time (8) admits proper affine vector fields. It is important to note that in cases 2 to 5 the affine vector fields are precisely the same.
7
Case 6
In this case we have Ax (t , x) = 0, At (t , x) = 0, C x (t , x) = 0, B x (t , x) ≠ 0, Bt (t , x) ≠ 0, 2
C t (t , x) ≠ 0 C t (t , x) + 2 C tt (t , x) = 0,
2
Bt (t , x) + 2 Btt (t , x) = 0 and rank of the 6 × 6
Riemann matrix is three. Here, there exists a unique (up to a multiple) no where zero timelike vector field t a = t ,a which is the solution of equation (4). The vector field t a is not covariantly constant. From the above constraints we have Ax (t , x) = 0, At (t , x) = 0 and C x (t , x) = 0 ⇒
A(t , x) = b and C (t , x) = D(t ), where b ∈ R and D(t ) is a function of
integration.
Equations
2
Dt (t ) + 2 Dtt (t ) = 0
and
2
Bt (t , x) + 2 Btt (t , x) = 0 ⇒
D(t ) = ln(a t + d ) 2 and B(t , x) = ln( P( x) t + Q( x)), where P( x) is no where zero functions of
integration and Q(x) is function of integration and a, d ∈ R(a ≠ 0). The line element in this case can, after a rescaling of t , be written in the form ds 2 = −dt 2 + ( P( x) t + Q( x)) 2 dx 2 + (a t + d ) 2 ( dy 2 + dz 2 ).
(21)
Substituting the above information into the affine equations and after some calculation one find affine vector fields in this case are X 0 = 0,
X 1 = 0,
X 2 = z c1 + c 2 ,
X 3 = − y c1 + c3 ,
(22)
where c1 , c 2 , c3 ∈ R. Affine vector fields in this case are Killing vector fields. The generators in this case are X1 = z
∂ ∂ ∂ ∂ −y , X2 = , X3 = . ∂y ∂z ∂y ∂z
Here the non zero components for the Lie brackets are [ X 1 , X 2 ] = X 3
and
[ X 1 , X 3 ] = − X 2 . In this case Lie algebra is closed. It is important to note that the affine vector fields in the cases 7 to 13 are precisely the same. Case 14
In this case one has At (t , x ) ≠ 0, Bt (t , x) ≠ 0, C x (t , x) = 0, Ax (t , x ) ≠ 0,
C t (t , x ) = 0,
2
Ax (t , x) + 2 Axx (t , x) − Ax (t , x) B x (t , x) ≠ 0,
B x (t , x ) ≠ 0,
2 Bt (t , x) + 2 Btt (t , x) − At (t , x) Bt (t , x) ≠ 0 and the rank of the 6 × 6 Riemann matrix is one.
From the above constraints we have A = A (t , x), B = B (t , x) and C (t , x) = e, where
8
e ∈ R. Here there exist two linear independent solutions y a = y ,a and z a = z ,a of
equation (4) and satisfying y a ;b = 0 and z a;b = 0. The line element can, after rescaling of
y and z, be written in the form ds 2 = −e A( t , x ) dt 2 + e B ( t , x ) dx 2 + ( dy 2 + dz 2 ).
(23)
The above space-time (23) is 1+1+2 decomposable. Affine vector fields in this case are [5] X = (c1 y + c 2 z + c3 )
∂ ∂ + (c 4 y + c5 z + c 6 ) + X ′, ∂y ∂z
(24)
where c1 , c 2 , c3 , c 4 , c5 , c6 ∈ R and X ′ is a homothetic vector field on each of two dimensional submanifolds of constant y and z. The next step is to find the homothetic vector field in the induced geometry of the submanifolds of constant y and z. The non zero components of the induced metric on each of the two dimensional submanifolds of constant y and z are given by g 00 = −e A( t , x ) g11 = e B (t , x ) .
(25)
Expand the homothetic equation (11) in two dimension and using (25) to get At (t , x) X 0 + Ax (t , x) X 1 + 2 X ,00 = 2c,
(26)
e B ( t , x ) X ,10 − e A( t , x ) X ,01 = 0,
(27)
Bt (t , x) X 0 + B x (t , x) X 1 + 2 X ,00 = 2c.
(28)
If one proceeds further, after a straightforward calculation one can find that c = 0 ⇒ no proper homothetic vector fields exist in this case. Here, the above system of equations gives trivial solution which is X 0 = X 1 = 0. Proper affine vector fields in this case are X = (0, 0, c1 y + c 2 z + c3 , c 4 y + c5 z + c 6 ).
(29)
In this case the generators are X1 =
∂ , ∂y
X7 = y
X2 = z
∂ , ∂y
X3 =
∂ , ∂y
X4 = y
∂ , ∂z
∂ ∂ ∂ ∂ ∂ ∂ −z , X8 = y +z and X 9 = z −y . ∂z ∂z ∂z ∂y ∂y ∂y
9
X5 = z
∂ , ∂z
X6 =
∂ , ∂z
Here the non zero components for the Lie brackets are [ X 1 , X 2 ] = − X 2 , [ X 1 , X 3 ] = − X 3 , [X1, X 4 ] = X 4 ,
[X1, X 7 ] = X 8 ,
[X1, X 8 ] = X 7 ,
[X 2, X 4 ] = X 9,
[X 2, X 5 ] = −X 2,
[ X 2 , X 6 ] = − X 3 , [ X 2 , X 7 ] = X 9 , [ X 2 , X 8 ] = X 9 , [ X 2 , X 9 ] = −2 X 2 , [ X 3 , X 4 ] = X 6 , [X 3, X 7 ] = X 6,
[X 3, X8 ] = X 6,
[X 3, X 9 ] = −X 3,
[X 4, X 5 ] = X 4,
[X 4, X 7 ] = X 9,
[X 4 , X 8 ] = −X 9 , [X 4 , X 9 ] = 2X 4 , [X 5 , X 6 ] = −X 6 , [X 5, X 7 ] = −X 8, [X 5, X 8 ] = −X 7 , [ X 6 , X 7 ] = − X 3 , [ X 6 , X 8 ] = X 3 , [ X 6 , X 9 ] = X 6 , [ X 7 , X 8 ] = −2 X 9 , [ X 7 , X 9 ] = 2 X 8 and [ X 8 , X 9 ] = 2 X 7 . In this case Lie algebra is closed. Clearly, in this case the above space-time (23) admits proper affine vector fields. Affine vector fields in the cases 15 to 26 are precisely the same. Case 27
In this case we have Ax (t , x) = 0, At (t , x) = 0, Bt (t , x) = 0, B x (t , x) = 0, C x (t , x ) = 0,
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C t (t , x ) ≠ 0, C t (t , x) + 2 C tt (t , x) = 0 and the rank of the 6 × 6 Riemann
matrix is one. From the above constraints we have A(t , x) = a, B(t , x) = b and C (t ) = ln(e t + d ) 2 , where a, b, e, d ∈ R(e ≠ 0). Here there exist two linear independent solutions t a = t ,a and x a = x,a of equation (4). The vector field t a is not a covariantly constant whereas x a is covariantly constant. Substituting the information of A(t , x),
B(t , x) and C (t , x) in (7) and after rescaling of t and x, the line element can be written in the form ds 2 = −dt 2 + dx 2 + (et + d ) 2 ( dy 2 + dz 2 ).
(30)
The above space-time (30) is 1+3 decomposable but the rank of the 6 × 6 Riemann matrix is one. It follows from [8] affine vector fields in this are X 0 = (c 4 t + c5 x + c6 ), X 1 = (c7 t + c8 x + c9 ), X 2 = −c1 z + c3 , X 3 = c1 y + c 2 ,
(31)
where c1 , c 2 , c3 , c 4 , c5 , c 6 , c 7 , c8 , c9 ∈ R. In this case the generators are X1 = y
∂ ∂ −z , ∂z ∂y
X7 = t
∂ ∂ ∂ ∂ ∂ , X8 = x , X9 = and X 10 = x −t . ∂x ∂x ∂x ∂x ∂t
X2 =
∂ , ∂z
X3 =
∂ , ∂y
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X4 = t
∂ , ∂t
X5 = x
∂ , ∂t
X6 =
∂ , ∂t
Here the non zero components for the Lie brackets are [ X 1 , X 2 ] = X 3 , [ X 1 , X 3 ] = X 2 , [ X 4 , X 5 ] = − X 5 , [ X 4 , X 6 ] = − X 6 , [ X 4 , X 7 ] = X 7 , [ X 5 , X 7 ] = X 10 , [ X 5 , X 8 ] = − X 5 , [ X 5 , X 9 ] = − X 6 , [ X 5 , X 10 ] = −2 X 5 , [ X 6 , X 7 ] = X 9 , [ X 6 , X 10 ] = − X 6 , [ X 7 , X 8 ] = X 7 , [ X 7 , X 10 ] = 2 X 7 , [ X 8 , X 9 ] = − X 9 and [ X 9 , X 10 ] = X 9 . In this case Lie algebra is closed. One can write the above equation (31) after subtracting Killing vector fields as X = (c 4 t + c5 x + c 6 , c 7 t + c8 x + c9 ,0,0).
(32)
Clearly, in this case the above space-time (30) admits proper affine vector fields. Case 28
In this case we have A x (t , x ) = 0, At (t , x ) = 0, B t (t , x ) = 0, B x (t , x ) = 0, C t (t , x ) ≠ 0,
C x (t , x ) ≠ 0 ,
2
C x (t , x) + 2 C xx (t , x) = 0,
2
C x (t , x) + 2 C xx (t , x) = 0,
C t C x (t , x) + 2 C tx (t , x) = 0 and the rank of the 6 × 6 Riemann matrix is one. From the
above constraints we have A(t , x) = a, B (t , x) = b and C (t , x) = ln(c t + e x + d ) 2 , where a, b, e, d ∈ R(c, e ≠ 0). Here, there exist two linear independent solutions t a = t ,a and x a = x,a of equation (4). The vector fields t a and x a are not covariantly constant. The line element after rescaling of t and x, can be written in the form ds 2 = −dt 2 + dx 2 + (c t + e x + d ) 2 ( dy 2 + dz 2 ).
(33)
Substituting the above information into the affine equations and after some calculation one find affine vector fields in this case are Killing vector fields which are given in equation (22). SUMMARY
In this paper a study of non-static plane symmetric space-times according to their proper affine vector fields is given. An approach is adopted to study the above spacetimes by using the rank of the 6 × 6 Riemann matrix, holonomy and decomposability and direct integration techniques. From the above study we obtain the following results: (i)
We get the space-time (8) that admits proper affine vector fields (see case 1) when
the rank of the 6 × 6 Riemann matrix is three and there exists a nowhere zero
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independent covariantly constant spacelike vector field, which is the solution of equation (4). (ii)
We obtain the space-time (21) that admits affine vector fields when the rank of
the 6 × 6 Riemann matrix is three and there exists a unique nowhere zero independent timelike vector field, which is the solution of equation (4) and is not covariantly constant. In this case the affine vector fields are Killing vector fields (for details see case 6). (iii)
The space-time (23) is obtained, which admits proper affine vector fields (see
case 14) when the rank of the 6 × 6 Riemann matrix is one and there exist two independent covariantly constant spacelike vector fields being the solutions of equation (4). (iv)
The space-time (30) is obtained, which admits proper affine vector fields (see
case 27) when the rank of the 6 × 6 Riemann matrix is one and there exist two nowhere zero independent solutions of equation (4) of which only one is covariantly constant vector field. (v)
The space-time (33) is achieved, which admits affine vector fields (see case 28)
when the rank of the 6 × 6 Riemann matrix is one and there exist two nowhere zero independent timelike and spacelike vector fields being the solutions of equation (4) and are not covariantly constant. In this case affine vector fields are Killing vector fields.
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[8] G. Shabbir, International journal of Modern Mathematics, 4 (2009) 201. [9] A. V. Aminova, Sbornik Mathematics, 186 (1995) 1711. [10] A. V. Aminova, Russian Mathematical Surveys, 48 (1993) 105. [10] A. V. Aminova, Russian Mathematical Surveys, 50 (1995) 69. [11] A. V. Aminova, Journal of Mathematical Sciences, 113 (2003) 367. [12] R. Maartens, Journal of Mathematical Physics, 28 (1987) 2051. [13] A. V. Aminova and N. A. Aminov, Sbornik Mathematics, 197 (2006) 951.
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