the unizednatin abdus salam educational, scientiffic .a a eqinS I organhzation itraonlXA0203566 centre enrpaen, for theoretical physics
COLLINEATIONS OF TILE RICCI TENSOR FOR CYLINDRICAL SPACETIMES
Asghar Qadir
K. Saifuliah
and
M. Ziad 3 4~ Available at: http://ww.ictp. trieste. it-pub- of f IC/2002/65
United Nations Educational Scientific and Cultural Organization and International Atomic Energy Agency THE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS
COLLINEATIONS OF THE RICCI TENSOR FOR CYLINDRICAL SPACETIMES
Asghar Qadir' Department of Mathematics, Quaid-i-Azamn University, Islama bad, Pakistan and Department of Mathematics, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia,
K. Saifullah 2 Department of Mathematics, Quaid-i-Azam University, Islamabad, Pakistan and The Abdus Salam International Centre for Theoretical Physics, Tries te, Italy
and
M. Ziad' Department of Mathematics, Quaid-i-Azam University, Islamabad, Pakistan.
MIRAMARE - TRIESTE July 2002
1 aqadirs~comsats.net.pk 2 saifulahqauedupk 3 mziad~qau.edu.pk Abstract
A complete classification of cylindrically symmetric static Lorentzian manifolds according to their Picci collneations (M) is provided. It is found that the Lie algebras of the RCs of these manifolds (for non-degenerate Ricci tensor) has dimension ranging from 3 to 10 excluding 8 and 9. The comparison of the RCs with the Killing vectors (KVs) and homothetic motions (HMs) is made. Corresponding to each Lie algebra there arise highly non-linear differential constraints on the metric coefficients. These constraints are solved to construct examples which include some exact solutions admitting proper R~s. Their physical interpretation is given. The classification of plane symmetric static spacetimes (Farid et al., J. Math. Phys. 36 (1995) 5812) emerges as a special case of this classification when the cylinder is unfolded.
2 1 Introduction
Spacetime symmetries not only make it possible to obtain exact solutions of the Einstein field equations (EFEs)
Rab - - Rg~b = STab,(1 2 but also provide an invariant basis for their classification. Isometries (KVs), homothetic mo- tions (HMs), Ricci collineations (RCs) and matter collineations (MCs) are some of these symme- tries [1]. A vector field V on a Lorentzian Manifold M is called an isometry if the Lie derivative of g along V is zero, i.e.
£vgO= . (2)
For HMs the right side is merely replaced by Ag, where is a constant. Similarly, B is an RC for a Ricci tensor R, if
LER = 0, (3)
which reduces, in component form, to
BcRab. ± R..c ± RbcBe =b0. (4)
The R~s, like the KVs, ae purely geometrical in construction, but like the MCs give physical information by virtue of the EFEs [2]. Ever since the first investigations of cylindrically symmetric spacetimes by Levi-Givita [3] and Weyl [4] and, later by T. Lewis [5], these spacetimes have been studied extensively for their mathematical and physical properties. These have recently been studied particularly in the con- text of black holes [6], gravitational waves and cosmic strings [7, 8, 9. Some examples of well known cylindrically symmetric astrophysical and cosmological solutions discussed in the litera- ture include Einstein-Maxwell fields [10], the gravitational field inside a rotating hollow cylinder [11], vacuum solutions [12], dust solutions 13], perfect fluid solutions with and without rigid rotations [14], gravitational waves [15, 7], magnetic strings [16], static gravitational fields [17] and a large number of cosmic string solutions. This paper classifies cylindrically symmetric static spacetimes according to their RCs. As a result of this classification, cases of 10, 7, 6, 5, 4 and 3 RCs for non-degenerate Ricci tensor i.e. when det (Rab) $ 0 have been obtained . The corresponding metrics are implicitly given in the form of constraints on the Ricci tensor components. Concrete examples have been constructed by solving these constraints. Physical nature of the spaces thus obtained is also discussed where possible. The cases of proper (i.e. non-isometric) RCs are 10, 7, 6, 5, 4 and 3 dimensional. It may be worth mentioning here that plane symmetry may locally be thought of as a special case of cylindrical symmetry [18]. As such, the classification of plane symmetric static metrics [19] can be obtained as a special case of this classification. The RC equations ae given in Section 2.
3 The plan of the paper is as follows. The RCs for the non-degenerate Ricci tensor, are given in Section 3 while the degenerate cases, where det (Rab) = 0, ae discussed in Section 4. While the RCs for the non-degenerate cases are always finite dimensional, those for the degenerate cases are mostly, but not always, infinite dimensional. The comparison of RCs with KVs is made in Section 5 where the non-trivial RCs are discussed in detail. Concluding remarks axe given in Section 6.
2 The Ricci Collineation Equations
The most general cylindrically symmetric static metric can be written as [18]
2 2 2 2 ds = evpdt- dp - a e\(P)d9 - ,PdZ (5) where a has the dimensions of length. For this metric the surviving components of the Ricci tensor are
= ~~ + ,2 + VA' + ''
2 42 2 4 ~~~~~~~~~~~~(6) 2 = i - (A ±A + V'A' + I
Here ( 0xlx 2 X3 ) (t, p,0, z) and prime and dot will denote differentiation w.r.t. p and t respectively. For the sake of brevity &4 will be written as R, for i = 0, 1, 2, 3. The RC equations [1] then can be written as
R'B + 2RaB a = 0, (7)
RaBa±+RbB' = 0, (a, b =0,1,2,3). (8)
Eq. (7) (which drops the summation convention) gives four equations and Eq. (8) are six equations. These constitute together ten highly non-linear partial differential equations involving four components of the arbitrary RC vector B -(B 0 , B', B 2 , B 3 ), four components of the Ricci tensor, R0 , R,, R 2, R3 and their partial derivatives. Here the B's ae functions of t, po, 0 and z; and R, depend on p only. The minimal symmetry is given by 49t, a96, 9,,, which has the algebra
3 Ricci Collineations for the Non-Degenerate Ricci Tensor
In this section Eqs. (7) and (8) are solved for the RCs of the non-degenerate Ricci tensor i.e., when R, 0 0, i = 0, 1, 2,3. In the beginning the calculations are given in some detail in order to explain the procedure, but later only the R~s and the metrics corresponding to the constraints on Rjs are given. Metrics for the non-trivial RCs are discussed in Section 5.
4 Taking a = 0, 2 and b = 2,3 in Eqs. (8) and differentiating these relative to z, and t, one
obtains 3 equations. Now, solving them simultaneously yields RB1B 3 = 0. This gives rise to two cases: (A) R' #4 0 or (B) R' = 0.
3.1 Case (A) R' $~0, (.B1 3 =0)
3 Here, taking a = 1 and b = 2,3 in Eq. (8), one gets B23 = B j.UigtsinE.()wh
a = 2 and b =3 gives (2) B2 = 0. This gives rise to further two cases: A(I) (R2 #0 or A(II) (R)' =0. Case A(I) ( ) 0 (which implies that B2 = 0) Eq. (8) implies that 23= 0 and so from Eq. (7) with a = 2 and 3, R~Bl = 0= MB1 This further gives rise to three possibilities depending on whether one or both of Rand R3' is/are non-zero. Case AI(a) R = , R' 760
In this case note that the RC Eqs. (7) and (8) imply that B = B = B 2= B 2 =B2 B2= B' = 0. Thus B 2 = c, a constant. Integrating Eq. (7) for a = 1 w.r.t. p gives
v1 where A, (t, z) is an arbitrary function of t and z. Using Eq. (9) in Eq. (7) for a = 0, and Eq. (8) for a = 0 and b = 1, and differentiating w.r.t. p and t respectively, and comparing gives
A1 (t, z) KRVj A, (t,z) = 0. (10)
Similarly from Eq. (7) for a = 3 and Eq. (8) for a = 1 and b = 3 one gets
A,~ R 3 / R 3 N' A,33 (t, Z)- VRk1 2R VfR-) A,(t,z) =O. (11)
Eqs. (10) and (11) imply that and (EL. fl"adI'en separating constants. Here again aise four cases depending upon whether none, one or two of these constants is zero. Case Ala(l) ce 0, /3= 0
In hiscas l R =kian 2RIV =k 2, where k,and k2 are nonzero constants. The solution of Eqs. (10) and (11) in this case is
A, (t, z)= C2 tZ +Cat +C4 Z +C 5 . (12)
Putting this value of A, in Eq. (9) one gets from Eq. (7) for a =0 and 3 respectively
0 B = -ki C24-Z+ C3-+ C4 Zt C5 ] + A2 (P,Z) ,(13)
3 B = -k2[C2t--+C 3tZ +C 4 - + C5ZJ+ A3(A Z) .(14)
5 Using Eqs. (13) and (14) in Eqs. (7) and (8), and integrating w.r.t. p yields
A 2 ( ) = - (C2 Z +C 3 ) 1 Ro dp +B 3 (z), (15)
A 3 (t, ) = - (C2 t+ C4 ) f ~dp +B 4 (t). (16)
With these values in Eqs. (13) and (14), one gets from Eq. (8) for a =0, b =3
kR' [ki (2- + c4t) - B3 ,3 (z] +
2 ~~C2-- ~=0
L 2 L- k k 2 ](7 where R0 kR3"2, k being a constant of integration, has been used. Clearly c2 =0 and one is left with
kR k2 [c4k,t - B3,3 (z) + C3 k2 Z- h4 (t) = 0. (18)
Nowif k - 1= 0, Eq. (18) gives
B4 (t) - c4kkit = CAk2 Z- kB 3,3 (z) C6 (say) .(19)
On integrating and substituting in Eqs. (15) and (16), noting that c3 =C4 = 0 (from Eq. (8) for a = 0, b = 1; and a = 1, b = 3 ), the R~s from Eqs. (13), (9) and (14) finally take the form
BO =-C 2 kit -C3 + C5 1 B'- C B =cl,B = -C 2 k,z+ C3 t +C 4. (20) where, the constants cis have been renamed, and one has 5 dimensional RC vector. However, if
k2 1#O~ , the 4RCs are
B = -C2kit+C 4 , B'- C B =cl,BB= -C 2 Z +C3 . (21)
Case Ala(2) a$0, 8 = 0 Now, a can be greater or less than zero. When a > 0, one gets a case of minimal symmetry. When a < 0, if ( o)' : 0 one again gets the minimal symmetry, however, for()=0 writing =-y, the RCs become
0 2 3 B = cl1-Z + 3 , B' = 0, B = C2, B = 1 t + C4 .(22) -y Case Ala(3) a =0, j3 0 In this case the RCs ae obtained just by interchanging the role of t and z in the previous case, Case AMa(2). Case Aa(4) a#O~ , 0:6 Now, a and fi can be positive or negative which gives rise to four further cases depending on whether one or both of these constants are greater than or less than zero.
6 CaseAMa4(i) a >0, /3> 0 The solution for Eq. (10) in this case is
A, (t, z) = B, (z) coshs,/-at + B2 (z) sinh x/fat. (23)
Using this in Eq. (11) gives
Bl,33 (z) cosh ~/at +B 2,33 (z) sinh V't - /3[Bl (z) cosh V/at +B 2 (z) sinh,/'at] 0, from where one gets
B, (z) = 2 cosh V,-6 + c3 sinh Vpz, (24)
B 2 (Z) = C4 cosh V'10z +c 5 sinh V/lz. (25)
Now, satisfying Eq. (8) for a = 0 and b = 3 gives rise to two possibilities. Either C2 = C3 = C4=
C5 = 0 and RoB 3 ,3 ()= R3 A4 (t), which gives 2 3 B' = c 2 Z+C 3 , B' =0, B = Cl, B = C2t +c4. (26) when R0 = R 3 (when RO $~R one gets minimal symmetry); or R f R' dp - 13 =0 and
RoB 3 ,3 () = -R 3 A4 (t), which again gives minimal symmetry. Case AIa4 (ii-iv) The results in these three cases will be similar to those obtained in the previous case, Case AIa4(i), the difference being that in the first case the solution of Eq.(11), in the second case the solution of Eq. (10) and in the third case the solution of both these equations will involve trigonometric functions instead of hyperbolic functions.
Case AI(b) R'2 # 0, R =O Note that the RC Equations (7) and (8), are symmetric with respect to interchange in 9 and z (i.e. the indices 2 and 3). Thus this case is similar to the previous case, Case AI(a), except for the interchange of 9 and z in all the equations. Case AI(c) ~0 R2 R'-#0 In this case one notes that B0 BO (t, p, 0, z), B' = BI (t, p), B2 = B 2 (t, 0) and B 3 3 0 0 0 0 B (t, z). From the RC equations the further constraints are B 0 = B 1 = B 0- B -
BO3= B 2 =B33 = 0. In view of this information if one integrates Eq. (8) for a =0, b = 1, w.r.t. p, and Eq. (7) for a = 1, 2 and 3 w.r.t. p, 0 and z respectively one finds that B' take the form
0 B -] vR dp A, (t) +A 3 (t)+A 4 (0) +A 5 (Z) ,(27)
B' A, (t) (28)
2 B = ( 2 R ~A,(t) 0+A 6 (t) ,(29)
B 3 -AI R3(t) z + A7 (t) ~~~~~~~(30)
7 Now, either both the terms in the brackets in Eqs. (29) and (30) are constants or otherwise A, (t) will be zero. These two cases are discussed here.
Case Alc(l) A, (t) = 0 Here, from Eq. (27) one gets B' = 0, which on using Eqs. (27) and (7) implies that
A 3 (t)= cl. Now, using Eqs. (27) and (29) in Eqs. (7) and (8) for a = 0 and b 3, gives
RoA 4 ,2 (0)+R 2 A6 (t) = 0,
RoA5, 3 (Z) +R 3 A (t) = 0.
These imply that A 6 (t)= C2t + C3 , A 7 (t)= c4t ± C5 and therefore, one has
A4,2 () = - C2, A5, (z)= -Rc4 (31)
This gives rise to four further cases depending upon whether both, one or none of RoR2 and Ro3 are constants. Case AIcl(i) (R o , ()'o
Putting R2 = k and R3= c2 ,,Eqs. (31) imply
A 4 () k-kC 20+ 6 , A 5 (z = k 2 C4 Z+C 7 , and finally Eqs. (27) and (29) take the form
0 B =- k1 C2 0 - k2 C4 Z + cl, B' = 0, (32) 2 B = C2t+ 3 , B = C4 t+ 5 . (33)
Here, + c6 + C7 has been written as c. Case Alcl(ii) (R)=0 Ro)$ In this case 2. = ki and c4 = and therefore, Eqs. (31) give
A 4 ()=k1C 2 + C6 , A,5 (z) =C7 , (34) and one gets from Eqs. (27) and (29)
0 B = - kc 2 + cl, B'=0, (35) 2 3 B = C2t+ C3,B = 4. (36) where, the constant cl + c6 + c7 has been written as c and c5 as C4. Case AIcl(iii) (R) R0o() This is similar to the previous case; just the indices 2 and 3 are interchanged. Case Alcl(iv) (R) R0o~)#
Here, one has C2 = c4 = 0 from Eqs. (31) and therefore, A4 () = 6 and A5 () =C7, and Eqs. (27) and (29) become
2 3 B = cl ,B' =0,B = C2 ,B = 3 . (37)
8 Case AMc(2) (•..2R3=0, R =0
Let - kRan - 3 = , where kland k2 are nonzero constants. Therefore, one has
2 3 B = k1 Al(t) 0+ Ar,(t) , B =k 2 A, (t) z±+A7 (t) ,(38) and Eqs. (27) give
0 B = - 1dp A(t) +A (t) R[ki A, )9-A (t) 0]
-- [ k2 A1 (t i 7(t) l + C3, (39) RL 2 where C3 = c + C2. Now, the value of B , i = 0, 1, 2, 3, as given by Eqs. (27) , (38) and (39) identically satisfy all the RC equations except Eq. (7) and Eq. (8) which take the form:
2Ro0 rR7- IRO
R2 (t) 02=0, (40
Ro ~~~~2 ±( 2[A21 (-A (t)] 0. (41)
Case A~~c2(i) = 0, k 0 6()0 Ro~~~~~~R
2 Ro~i~~c1+ A t - A7 (t)Z = 0. (42) whichr imples tha( 'an (t)3c +sae/ qatifi Now,zr. otnt,b thesesult the quals
2R0 V~~~~~~R
3 LetR2=3.InthsBas -q k(41 elsc t) 5 ,B=cl ( =kC 2.Teeoe q 44)e Ro~~~~~~~~~~ Otherwise, one has
0 B = - k3 C10+ C2 ,B' O, (45) 2 3 B = Cit+C3, B = C4. (46)
Case Akc2(ii) Ro 0,Ro This is similar to the previous case; only indices 2 and 3 (i.e. coordinates 9Oand z) are interchanged. (R '~ ,(g Case Akc2(iii) Ro , ~
In this case if 2Rov(,/R--R-- is a constant, the result is the 4 R~s
0 B = k4 Ct+ C3 , B' cl (47)
2 3 B = kliO + c2 , B - k2 CZ + 4 , (48)
Otherwise, one gets the minimal symmetry.
Case A () (R )' = 0
This means that R 2/R 3 = constant = l/d (say) , d :$ 0. Putting R3 =dR in the RC equations and proceeding as previously, one arrives at the following possibilities:
(a) (~) 0 or (b) ( R2 V ) = 0 Case AII(a) Here if (/ 0 ,let =k andthe RCsare
0 B = -(c 5 dz+C 6 0)±Cl, B'=0,
2 3 B = C Z +C kt+ , B =-C -0+ kt +c . (49) 2 6 3 2d 5 4
So, one has six RCs. And if V R) 0, the RCs are
0 B -cl, Bl=0 B =C 2Z+C 3 , B 0C-+C (50) so that the number of RCs is 4.
Case AII(b) ('W) =0 Let -~~~~~~~~~,R-
Let R2 - a, a constant which can be zero as well as non-zero. Case AIIb(l) o Here arise further three cases:
(iii) ( 2
Case AIIbl(i) (R2)' o -~ R. '0
10 Here the RCs are
0 B = c B' =0, B = C2 Z+C 3 , C -+C9 (51)
Case AIIbl(ii) #R):o R = o
Let R =S /3. if 6 0, the RCs are
2/3 1 2 B = C3Z +C20 +C4, B C-+CZ+C9
So, it has 5 RCs. And if3 =0, one has
0 B cl,Bl - 1 (2Z - C3 0 + C4) (53)
2 B = 3( 0+2R)C9C9+5+
B = Z_- Ci -- C4- C0Z - C - + C, d~ ~~~~ 5 which is a case of 7 RCs.
Case AIIbl(iii) -
Here, writing R2 = 5R0, where 3 is a constant, the result is the 10 RCs.
Bn = -t ____ 0 _ d6Z2 + c tZ -Crd~z - Ct--c 9+ cot + C (54) k co 2Ro /d 6 9 1 (5 B' V (C5 t +C 6 Z - C8 +CIO) (56)
B=C5t +C6ZC8(--+ d2R + z+C9t +C100±C2Z±+C3 (6
3 B = C5 Zt -C 6 (t 1 920 z2\ 9zCt 1Z- 2 980 (7
Case AIM(2) a = 0
This means that R' = 0 i.e, R 2 is a constant. This gives rise to two possibilities discussed below. Case AIIb2(i) (-R,) 0
(V"&~~~~~1 Let (VRs' 7 constant. Now, if y)'$0 the 10 RCs axe 0 B = [dz (c,5 sirnyt - c6 cos -yt) +O0(C7 sin-yt - c8 cosyt) -
(Cg Sin -rt - clo cos-yt)J + cl, (58)
B' [dz (c,5 cos ^t +c 6 sin-yt) +O0(C7 COS It +C Sin-Yt)-
(Cg cos 'yt + clo Sin 7ft)], (59)
B 2 =-~R (C7 COS-t +C8 sin-yt) +c2z +cS, (60) 'yR2
3 B = -Vf-(c 5 cos -yt + csinyt) -c 2 0+ 4 , (61)
and if y =0 i.e. R0 is constant, the RCs are
0 B = (c 6 dz + C8 ) - f x/R-dp + cl, (62)
7=(c 5 dz+ C70) + 7=(ct + C1) ,(63)
2 B R2 J ~JldP + C8t + C2Z + c3 , (64)
3 B - V, dp +c 6 t - -C2 +C4. (65)
Case AIIb2(ii)(V)] Here, proceeding as before, one arrives at the following two cases.
Case AIM2(ii)a &~ (- = constant = r, (say)
If i7 = 0 one has Ro= constant =A$0, and the RCs ae
0 B = C5 J - At 2) - c 6 At +cl, B'- 1 ,Ct+ (66)
B 2 = C Z +c , B3= -C 0+C . 2 4 2 d 3
In the case tj 0 one gets
BO= ~ C5eVA1-t c~e-vt) + cl, B' 1( t + c6 e-,/"-t), (67) B2 3 0 3 B=C 2 Z +C4, B =-C2- ±C3
Case AIIb2(ii)3& 6 $0 Here the result is
0 2 B = cB'=0, B = C2 Z +C 4 , 0 C3(68)
Now, Case (B) will be taken up.
12 3.2 Case (B) 1%=
Let R0 = a, a non-zero constant. In this case the RC equations give B% = B 2 = B 3 = B 3 = 1 2 2 3 3 B' B' 00B'20 -B-B'=B'3 -=B0=- ,3 = -=Bo ,2~- =0 q. (7) fora= 1, can thus be written as
B' - (clt +A 2 (O, )) ,(69)
Using this in Eq. (8) for a = O and b =1, and integrating w.r.t. p gives
B0 c f /dpA 4 ()±+A5 . (70)
Now, Eqs. (8) for a = 0 and b = 2, 3 yield
2 at B = _-A 4 ,2 (0) +A 6 ( ,Z) ,(71) at 3 B = at-A5 ,3 (z) + A7 (, 9, Z). (72)
Putting these values in Eqs. (8) for a = 1 and b = 2,3 gives
R' A4,2 (9) = 0 , R' A 5,3 (z) 0, (73) which gives rise to four possibilities depending upon whether both, one or none of R' and R' are/is zero. CaseB1(I) R' = 0, R
Letting R 2 /3 and R 3 = y and proceeding as previously gives the final form of the R~s as 0 B =cl f X/RftdP + C20 + C4Z + C3 , (74) aJ
B' = (C~t+ C5 0+ C6 Z+ 7 ) ,(75)
2 a 1 B = -_C2 t- -C5 VRdp + C8 z+C 9 , (76) 3 B = a 1 B _ C4 t - -C 6 IRdp -- C8 0+ C0 (77) which is a 10 dimensiomal RC vector. Case B(II) R~ = , R' #~0,
Putting I?2 = / and A 5 (Z) = c2 (Eq. (73)), one gets from Eq. (7) for a =2, 3 and Eq. (8) for a= 1,2 and b=2,3
A 4 () = C30 +C4 , (78)
A 2 (0, Z) = 0A 8 ()± A 9 (z) ,(79)
A 6 (P, Z) = -jA8 (Z) f ./7Ridp+Alo (z) (80)
1 A7 (P, 9, Z) =- J R dp [9A8,3 (Z) + A9,3 (z)] + Al (, z) ,(81)
Al (, z) = 0A12 (z) + A13 (Z) ,(82)
13 A833 -[ I ( ' A (z) =0(3 8,3\J '/IR- 2R 3v/7) 8 = (3
Depending upon whether the term in the square brackets in the last equation is constant or not, one needs to discuss two further cases. Case BII(a) R7 23W Here the RC vector is
0 B = C20+ClM=O, (84) 2 B = 3 B 3~~~= C4. (5
Case BII(b) VT- (2R 3/R7 k Here there are three further possibilities: k, 0. Case BIM(1) k, > 0 In this case the solution of Eq. (83) can be written as
As8z)=C5 e vkl + c6e"~,klZ(86) and the result is the 6 RCs
B0 = C20 + Cl (87) B' = C3e k-z+ C4e- -,'), (88)
2 B = -C 2t+ 5 (89)
B 3= -1 R C3e A/-lz Ce-vklz+C6(0 3 3 B 1/k2R 3 Vij e(0
Case BIM(2) k, < 0 This case is similar to the previous one, the difference being that in Eqs. (86) the arguments of the exponential functions will be complex instead of real and one gets a similar RC vector with 6 RCs. Case BIM(3) kl = 0
This means that -- =k2 , where k2 is a non-zero constant. rom Eq.(83), one has 2 R3 V/WT
A 8 (z = c5Z + C6 . (91) giving the result as
B0 = C,,B 12 C3 4 (92)
B = -- C2 t+C5 B~ /C d-2C 4 +C6. (93) - / c3 dp 2k
Case B(MI) R~#~0, R'= 0
14 As the RC equations remain unchanged if one interchanges indices 2 and 3, the results for this case can be obtained by interchanging these indices (i.e. and z coordinates) in Case B(II). Case B(V) R~ #~0, R' #~0
In this case Eqs. (73) yields A4 () = 2 and A5 (z) = C3 and B' take the form
0 B = C2 +C 3 B'- A 2 (, Z) (94)
2 3 B =- A 6 ( ,Z) , B =A 7 ( ,Z) .(95)
Differentiating Eq. (8) for a = 1, b =2, w.r.t. and Eq.(7) for a =2 w.r.t. p and subtracting yields
A 2,22 (, Z) - 2~'-j A2 (, Z)=0. (96) Here again there are two cases depending upon whether the term in the squaxe brackets is constant or not. Case BIV(a) R W v'lR71 2R2V/R1 0 Here if R2 is a constant one has R3
B0 = Cl ,B' = 0 (97) 2 3 B =C2Z+C 3 , B =-aC2 9+C 4 , (98) otherwise, one gets the minimal symmetry.
Case BIVb [ R,-( 2R 2iflT)J =0.
Cuttie B=b1 2~-i2 3k3 , there are three possibilities for the constant k3: k3 = 0.
Eq. (96) can be solved to give
A2 (, Z) = A8 (z) e"'v ±3A 9 (z) e- v , (99) where A8 (z) and A9 (z) satisfy (from the RG equations)
A 8,33 (z - R3 2R 3v'~ A 8 (z = 0, (100)
A9 ,3 3 ( - R3 ( R13 -- , = 0. (101) ,R-K2R 3Y/R Y As before, there are two possibilities depending upon whether the term in the bracket is a constant or not. Case BIVbl(i) R (0,~)]'o Here, the RG vector is
B0 = cl ,B 0 =0, (102) 2 B = CZcRzc C, - C2 0+ 4 . (103) R3
15 Case BIM1(ii) [ R' =0 Here B' are of the form
B0 = cl,B'=O, (104)
2 3 B = C2 Z + 3 B = _R 20 .C (105)
Here, R2 is a constant otherwise c2 will be zero. R3
Case BIM(2) k3 < 0 In this case the solution (Eq. (99)) of Eq. (96) will involve complex arguments for exponential functions and the results can be obtained similarly as in Case IVb(l). Case BIM(3) 3= 0. In this case, one gets
A 2 (,z)= 0A 8 (z)±+A9 (z) .(106) where As (z) and A9 (z) again satisfy satisfy Eqs. (100) and (101). This again implies two possibilities, depending on whether the term in the square brackets is constant or not.
Case BIVb3(i) [v~ R3 W 0 Eqs. (100) and (101) imply that
A8 (z)=0=A 9 (z) ,(107) and the solution is similar to Case BIVbl(i).
Case BIVb3(ii) [v' 'R. 3 -
One can put 2RV3 = k , where k is a constant which can be greater than, equal to or less than zero.
Case BIVb3(i)a k > 0
This solution here is similar to Case BIVb1(ii)-y, where k = 0, k3 > 0. Case BIVb3(ii)8 k, < 0
This case is similar to the subcase of Case BIVb(2) where k = 0, k3 < 0. Case BIM3(ii)y kI = 0. Here if (R3)' $0 the RC vector is
0 B = clB'= C2, (108)
B = 20 3B =- k 2 C2 Z +C4 , (109)
16 and if (3)' = 0 one has the case of 7 R~s given by
B0 cl, (110)
B' = (C2 + C3 Z+ 4) ,(111)
B 2 -C2I dp+ k- -k 2 - - k 2 C3 0Z -k 2 C4 0 -C5Z +C6, (112) B= -C2 R22 2/
3 B = -k 2 C2 Z - C3 (f -fdp +k 2 -- k 2 - - k 2 c4 Z +C5 0 + C7 (113) R 2 2 2 4 Ricci Collineations for the Degenerate Ricci Tensor
In the degenerate case there are 15 sub-cases depending upon which one (or more) of the four components of the Ricci tensor is zero. In Cases I-IV one of R, , i = 0,1,2,3 respectively is zero, in Gases V-X two and in Cases XI-XIV three of these are zero. In Case XV Ri- = 0, for all i. As the RG equations remain unchanged if any two of the three indices 0, 2 and 3 ae interchanged, note that one needs to discuss only one case each from Gases 1,111I and IV; Gases V, VIII and IX; Gases VI, VII and X; and Cases XI, XII and XIV. The results for the similar cases can be obtained by interchanging the role of any two of the coordinates t, 9 and z. Cases II, XIII and XV ae independent. So, here only 7 cases are discussed instead of 15. It is found that only Case II (RI = 0, R,. #, 0 , i = 0,2,3) admits finite dimensional RCs and for all other cases the RCs are infinite dimensional. Therefore, this case is discussed first and in some detail; for other cases only the results are given.
Case II R =0, Ro$0, R 2 :h0, R 3 #:0 In this case one notes that BO -B 2 -B 3 -B 0 -0(114) ,1-1 -,j- ,230
and therefore, B' = B' (t, 9, z) , i = 0, 2, 3 and one can assume B0 of the form
0 B = A, (t, 0) + A 2 (t, Z) (115)
Substituting this in RC equations following the previous procedure yields
B' = -2R [A1l(t,o) ±A 2 (t, Z)] (116) 0 2 B = __fA, 2 (t,) dt +A 3 (, Z) ,(117)
B 3= -- 2A 2 ,3 (t,z) dt +A 4 (,Z) (118) R 3 j In view of Eq. (114) these give the following equations.
RO)fA 1 ,2 (t,0) dt = 0, (119)
17 (R)'f A, (t, z)dt = 0, (120)
which suggest four possibilities of taking both, one or none of (O)' and () 'equal to zero.
Putting RO = a and RO /one has 10 R~s of the form 0~2 aZ2 t 0 B = clI- a +C2tG+c30+c4tZ+C Z+C6t+CIO , (121) 2 /2 2/ 5
B'= (-aClt + C20 + C4Z + C6) 12
B= -aCIt9 +C 2 ( -a- + - -~~ 13
aC3 t+ C40Z +C 6 0 - aC7Z + C,
3 B -aCtZ +C 2 0Z +C 4 (- + - +~ - (124)
,6C.5t + C6Z + C70 - C .
In this case Eq. (120) implies A2 (t, Z) = A2 (t) . With this in Eqs. (1 15) - (118), one gets from Eqs. (7) for a = 3
(R3RO)' [A, (t,O0) +A 2 (t)] =0, (125) giving rise to further two cases depending upon whether isa ostnto nt
Case IIb(l) (Ro ' 0' ( R3R'0 , Taking ('&)3 - the RC vector has the form
0 B =Cit+ C2t +C 3 , (126) B' =_2Ro (127)
2 B - _aclt +C20+ 4 , (128)
3 B -kC 2 Z +C 5 . (129)
Case IM(2) In this case the R~s are
0 B =C20 + C , (130) B' =0, (131)
2 *B -aC 2 t +c 3 , (132)
3 *B =C4.- (133)
18 Case II(c) (Ro)' o , R As the RC equations are invariant under the interchange of indices 2 and 3, if one change these indices (i.e. coordinates 9 and z ) in Case II(b), one gets the results for this case.
Case II(d) (R)'#0, R 0 Now, Eqs. (119) and (120) imply
A1,2 (t, 9) A2 ,3 (t, Z) 0 ,(134) or
A, (t,90) = A, (t) , A2 (t,90) = A2 (t) .(135)
In this case Eqs. (115) - (118) have the following form
B' = A, (t,0)±+A2 (t) ,(136)
B'- 2R0 A, (t, 9) + A 2 (t),(17
2 B = A 3 (9, Z) ,(138) 3 B = A 4 (0, Z) .(139)
Now, Eqs. (7) for a =2,3 become
A, (t)+ A2 (t) - R(OR) A3,2 (, Z) = 0, (140)
A I(t)+A,,(t) - R(RiT A 4 ,3 (9,Z) = 0. (141)
This gives rise to four cases depending upon whether the terms in the brackets in the above two equations are constants or not. However, it turns out that if both the terms are constant, the R~s are
B0 = 0 , (142)
B' - 0 , (143)
B 2 =C1Z +C2 , (144)
B 3 = _21 3,(145)
where R2is a constant, otherwise c will be zero. If this is not the case one has
B0 = c , (146) B' = 0, (147)
2 B = C2Z +C3 , (148) B3 = _RC2 4(149)
19 Here again one notes that 2. is a constant, otherwise C2 will be zero.
Case Ro =O, R1 #l0, R 2 #:A0, R 3 #O6 In this case from the RC equations one can see that B' isan arbitrary function of t,p and z, and B', i = 1,2,3 are independent of t and, therefore, Eq. (7) for a 1 on integration gives
1 B=A,B'= IR1(10(, z)(10 Using this in Eq. (7) for a = 2 and 3 and then substituting in Eq. (8) for a = 1, b = 2 yields
A1,2(0 Z)+ R R I)A, (0,z) =0. (151) ,./RPJ 2R 2 1V&
This gives rise to two possibilities: either -~-- R~ ''is a constant or not. If not, A (, z) = 0 and the result is
B0 is totally arbitrary, 2 3 B' 0, B =CIZ + C2 ,B =C 3 09+c 4 . (152)
Otherwise, putting R ( R~ ) = ki, a constant gives rise to further three cases: k,>0. Case (a) k, > 0 The solution of Eq. (151) in this case can be written as
A, (, z) = B 3 (z) ei8 + B 4 (z) e-Vi .(153)
Following the previous procedure, one finds that R3 (~~ =" k2 is a constant (in order to have a non-trivial solution) and there are three further possibilities, k2 > 0.
Case la(1) k2 > 0
2 3 B' =O,B = CIZ C2 , B =C10±C3 . (154)
Case Ia(2) k2= 0 In this case the RCs are
B' 1 Pv[ek"0 (C1z + c2) + e-"1k' (C3Z + C4)j (155)
2 B -= [ke CZ+ 2 (C3Ze'~l + cC4, (156) ir!,K L 2 eVB(1 + +4)]
3 2 2 B = -k3 [eiv'k--1 (Cl z + C2Z) + e - ,/k6 (C3 Z + C4Z)] (157)
- [cle'k1 + C3e- lkj] IR dp +c6 (158)
where - - k3, is a constant. R3 vPJ Case Ia(3) k2 < 0 The argument of exponential function which was complex in the previous case will be real in this case similar results are obtained on parallel lines.
20 Case (b) I= 0
There are two possibilities: either R!2 is zero or not. Case Ib(l) R = 0 Here, if -b-- (-- 'is not a constant, the RCs are
2 3 B'=0 ,B = Z+ C2 ,B =C 3 +C 4 . (159)
However, if =' 4 a constant, there are three possibilities k Ž0. RJ R 4 < Case Ibl(i) k4 > 0 Here, the solution is
B' - 1 [clei,/k- Z+ Cei'/-Z) (160)
B 2 -O, (161)
B = -______ce c2 e 4Z) (162)
Case Ibl(ii) k4= 0
Here one can write (R' k,5, a constant. Now, if R' :h 0 or k5 0, the RCs are
2 2 B' - 1(CIZ +C2 ) B = O,3=-k 5 (Cl Z ±+C2 Z) - f dpi (163)
and if R' = 0 iLe. k5 =0 the R~s are 3~~~~
Case Ibl(iii) k4 < 0 This case is similar to the Case Ibl(i) except for the difference that the argument of the exponential functions will be real and not complex. Case Ib(2) R' #, 0
Now, if the quantity -R- R--- k4 is not a constant, then the solution is
2 3 B' =0, B = c1z + C2, B R210+c3 ,(165) R3 where, R' is constant. On the other hand if k4 is a constant, one has the possibilities k4 > 0.
Case b2(i) k4 > 0
2 B' =0, B = cZ+ C2 , B =-_Cl - +C 3 (166)
Case b2(ii) =4 0
This means that -a-- k5 is a constant. In this case the solution becomes 2R3 Y/R1
2 3 B'-c B = 3 + 2 B ckzc.(167)
21 Case b2(iii) k4 < 0. In this case the results can be obtained similarly as in the Case 1b2(i).
Case V RO =0, R 1 =0, R2#O-, R 3 0 In this case one notes that B 2 = B 2 (9,z), B 3 = B 3 (9,z) and B0 (t,P, 9,z) is a completely arbitrary function. Now, differentiating Eq. (7) for a = 1, w.r.t. p gives (3)'B 3 = 0, which gives rise to two cases.
Case V(a) R
Let R3= k #40. Now, subtracting Eq. (8) for a =1, b =2 from Eq. (8) for a =0,b =3 , one gets
B,2- B3= 0. (168)
Differentiating Eq. (8) for a =2, b =3 w.r.t. and z respectively and using Eq. (168) gives the following two equations.
3213 = 0, (169) B k+B 22 B = 0. (170)
For k > 0, the solution of Eqs. (169) and (170) can be written as
2 B =f+(9±i )+k(- i) ,(171)
3 B -k +( + ) g (0 j ). (172)
Therefore, from Eq. (7) for a = 2 one gets
B' 2R [f+,2 (f-,k2(90+i - )] ,(173)
0 provided that R!2 # 0. B is an arbitrary function of t,p, 9 and z. If = 0, one sees that B = B3 = 0, and, therefore
2 B = ClZ+ C2 , (174) 3 B = C10 +C3, (175) and, B0 and B become arbitrary. Case V(b) (R3) , 0 In this case one notes from Eq. (7) for a = 1, that B3 = 0 = B 2. Therefore, B 2 B 2 (), B 3 = B 3 () , and one is only left with Eq. (8) for a = 0, b = 3 and a = 1, b = 2, for which there are the following three subcases. Case Vb(1) R = 0, R' #40
2 3 B' = R3 A4,3 (z), B =cl,= B (Z) =A 4(z)+ 2 , (176)
22 and B' (t, p, 9, z) is arbitrary. Case Vb(2) R~ ~60, R'= 0 Interchanging the role of the indices 2 and 3 in the previous case gives the result for this case.
Case Vb(3) R~ #~0, RS #~0
3 B' = c R,- B =C10+C 2 , B -~~ 4 (177)
and B' (t, p, 0, z) is arbitrary.
Case VI Ro =0, R 2 =0, R1 #0, R 3 0 In this case from RC equations one notes that B0 is an arbitrary function of t, p, and z;
and B' = B'(p, z) ,i = 1, 2, 3. Eq. (7) yields
B' = AR(-) (178)
Using this in Eq. (8) for a = 0, 2 and a = 1, 2 gives
A1,3 3 (zWR~?i Al (z) =0, (179)
which gives rise to the following two cases.
Case VI(a) [., 3(R VR'i)] #0
In this case A, (z) =0 so that one has
B' =0, B 3 ==Cl (180)
with B0 (t, p, 9, z) and B2 (p, z) arbitrary. Case VI(b) [ 3 R ' 0
One may write 3 (R) =k , where k is a constant with the following three possibilities. Case VIb(1) k > 0
In this case the solution of Eq. (179) can be written as AI (z) = clevk~ + C2e- .k-z and one has
3 B _- k ce f-C 2 e-'/ vIdp + Yx/- cl e/z ~C2 e-v-k,)J\/R1dp
R______3 clvk Ce- ') +C (182) 2R 3 V-R7V/rk' with B' (t, p, 0, z) and B 2 (p, z) arbitrary. (Note that f dpanarcosnt) Case VIb(2) k= 0
23 Here the solution of Eq. (179) can be written as A, (z) =c 1 z + C2 and the result is
B' = cl z +c2, (183)
3 2 B = -Cf IR--,dp R3§ (ClZ + C2 Z) + C3, (184)
2 with B' (t, p, 9, z) and B (p, z) arbitrary. (Note that f 1Rdp and le -/ax ontnt. Case VIb(3) k < 0
Here the solution from Eq. (179) can be written as A, (z) = ce i,/1z + c2 e-ivk~z and the results can be obtained similarly as in Case VIb(l).
Case XI Ro =0, R 1 =O, R 2 =, R 3 ~0 3 rom RC equations one has B - B' (t) and
B' 2R3-ffl (t) (185)
0 2 provided that R3 #00, and, B and B are completely arbitrary functions of t, P, and z.
Case XIII R, #00, Ro =0, R 2 =0, R 3 =O In this case one has
B 1 cl (186)
and B0 , B 2 and B 3 are arbitrary functions of t, p, and z.
Case XV Ro =O, R 1 =0, R 2 =O, R 3 =0 In this case B is totally arbitrary and one has infinite dimensional RCs, which means that every direction is an RC.
5 Metrics with their KVs and RCs
Solving the constraints on the components of the Ricci tensor, R,-, obtained in the previous sections, gives explicit form of metrics corresponding to different cases, discussed earlier. In some cases particular spacetimes are found while in others a class or family of metrics is obtained. The physical nature of some of these spacetimes is also discussed. These metrics for the non- degenerate and the degenerate cases are listed here, and the RCs of these metrics are compared with their KVs [20] and HMs [21]. It is well known that every KV is an HM and every HM is an RC but the converse is not true in general [1]. This gives rise to the interesting cases where one obtains RCs which are not KVs. When the results obtained in the previous sections are compared with the classification of cylindrically symmetric spacetimes according to KVs [20], numerous cases of these proper (i.e. non-isometric) RCs have been found.
24 5.1 The Non-Degenerate Case
It is well known that the spaces of non-zero constant curvature have RCs coinciding with their KVs, for example, the anti-de Sitter universe with 10 KVs and 10 RCs (Case AIIbl(iii)). These are not listed below. However, there are some non-trivial cases of coinciding RCs and KVs which do not have constant curvature, and they are included in the list. The solution of constraints on Ricci tensor to obtain the metrics is explained in the first example; for other examples, only the metrics are given.
Ni. The constraints on the components of Rlicci tensor, R,-, for Case BIVb3(ii)-y2 are (R 3 /R 2 )'
0,1R' = 0, R~$760, and =0. These suggest taking R 3 = R2 and A (in Eq.
(5)), so that R0 becomes RO= -"+ e2v,! (v' +2A')
Now, R' = 0 suggests a possibility, ev" 2k 2 (a constant) and v' + 2A' = 0, which can be solved to yield v =ln cosh 2 kp and A p = n (cos hkp)' so that all the constraints for this case ae satisfied and the line element is
2 2 2 2 2 1 2 1 2 ds = cos h kpdt _ dp - a (coshkp) - dO - (coshkp)- dZ ,
where k is a constant. It admits 7 RCs and minimal (3) KVs and, therefore, is a case of proper RCs. Now, solving the EFEs for the energy momentum tensor gives anisotropic fluid with energy density oscillating between negative and positive values. However, with the cosmological constant put in, the energy density becomes positive definite.
2 2 2 2 2 N2. ds= - Ap+B (dt - a dO - dZ ) -dp ,
A and B are constants. It admits 10 RCs (Case AlIbl(iii)) and 3 KVs. This is a case of non-trivial RCs. It represents a spacetime with non-zero cosmological constant and represents an anisotropic inhomogeneous perfect fluid.
2 2 2 2 2 N3. ds = (p/p0) adt - d p _ (p/p0) b & dO2 _-pP)2d'
a, b, c, a and po are constants. If a, b, c $~0, 1, it admits 4 RCs (Case AIc(l). If one takes b c a with a, b $40, 1 one gets 3 KVs, 4 HMs and 5 RCs (Case AIIbl(ii)). It is a perfect fluid spacetime when b = a(a - 1)/(a ± 1) and a non-null electromagnetic field for b = a 1.
N4. Taking a = b = c 0, 1, in metric (N3) gives a metric of Petrov type D admitting 6 KVs, 7 HMs and 10 RCs (Case AIIbl(iii)). It represents a tachyonic fluid but with a suitable cosmological constant an anisotropic fluid.
N5. Setting p = v (in Eq. (5)), and solving the constraints on R,., for Case AIa(2) gives the metric
25 2 2 2 2 2 2 1 2 ds = (coshf cit - dp - a cosh kpd8 - (coshkp)- dZ ,
where k is a constant. It admits 5 R~s and minimal isometry group and hence is a case of proper RCs.
N6. Taking a = 1, b c o 1, in metric (N3) gives a metric with 7 RCs (Case BIVb3(ii)-y2) and minimal isometries.
N7. Taking b = c = 1; a :, 0, 1, in metric (N3) gives a metric with 4 KVs and 6 RCs (Case AMb2(ii)a).
N8. ds2 = el (t 2 - dZ2 ) - dp2 - ea2 dO2 , (A` :#~0, ," #0 0).- It admits 4 KVs and 4 RCs given by Case AIcl(iii).
2 2 2 2 2 2 N9. ds = e" (t - a d8 ) - dp - e~dZ , (,, 7 0, el" $, 0).
It admits 4 KVs and 4 RCs given by Case AIcl(ii).
N10. Taking a = b 4 c; a, c 0, 1, in metric (N3) gives a metric with 4 KVs and 4 RCs (Case AIc2(i)).
Nil. Taking a = c 4 b; a, b ~0, 1, in metric (N3) gives a metric with 4 KVs and 5 RCs (Case AIc(2)).
N12. ds2 = eL (t 2 - a2 dO2 - dZ2 ) - dp2 , (,, : 0).
It has 6 KVs and 6 RCs given in the Case AII(a).
2 2 2 2 2 N13. ds = eA (t - a d ) - eBPdz - dp , A and B are constants (A #7B). It has 5 KVs and 5 RCs (Case AIc2(i)a). It does not seem to have physical significance.
2 2 2 2 2 N14. ds = Apdt - dp - eBP (a dO -dZ)
A and B are constants. It has 5 KVs and 5 RCs (Case AIIbl(ii)). It does not seem to have physical significance.
N15. ds2 =-A (cit2 & ) -dp 2 eBPa2 dO2
A and B are constants. It has 5 KVs and 5 RCs (Case AIc2(ii)). It does not seem to have physical significance.
2 2 2 2 2 2 N16. is -=eApct _ dp - eBPa dO - eCPdZ ,
A, B and C are constants. Ithas 4 KVs and 4 RCs (Case AIc2(iii)a). Itappears to be non-physical unless a cosmological constant is introduced.
2 2 2 1 2 2 2 N17. is = eL~dt _ dp - & (a ci - dZ ), (,, $ 0, i, =~0).
It admits 4 KVs and 4 RCs given by Case AII(a).
26 5.2 The Degenerate Case
As seen before the RCs for the degenerate Ricci tensor are mostly arbitrary functions of coordi- nates t, p, 0 and z. However, in Case II where RI = 0 and Rj are non-zero for i = 0, 2,3 RCs of dimensions 10, 5, 4 and 3 have been obtained. Here, the metrics which admit degenerate Ricci tensor are given.
D1. If a = b = c = 1, in metric (N3), one gets R = 0 and Rj are non-zero constants for i = 0, 2,3. It admits 10 RCs given in Case II(a). This is the only case of degenerate Ricci tensor which admits finite dimensional RC vector. It is an exact solution of EFEs with non-zero cosmological constant and represents an anisotropic inhomogeneous perfect fluid.
D2. The conditions of Case IV of the degenerate Ricci tensor are satisfied if one takes Ricci 2 2 tensor of the form R0 cccosh kp, R, = constant, R2 = csinh kp and R3 = 0. Setting p= 0 and choosing v, Aand the constants appropriately gives the form of the metric as
2 2 2 2 2 2 2 2 ds = (b + b sinh ap) dt _ dp _ 2 sinh apd0 - dZ ,
where a and b are constants. It admits minimal KVs and infinite RCs. If the coordinate p is transformed according to the relation p = sinh 1 and one takes b2 -=1-8E>O0, the above metric corresponds to the string solution
2 2 1 2 2 2 ds = (1 -8e+a r )t + ( -S 8+a2r2) dr2 ± r dO - dz ,
discussed in [8]. The string is naked for < 1/8. If one sets p= cosh' one~ gets the corresponding solution for > 1/8, in which case there is a trapping horizon at 1 = (E - 1)1/2 . The transformation p = Inr corresponds to the solution for E 1/8.
2 2 2 2 2 D3. ds = dt - d p - a d - dZ ,
It has 10 KVs and infinite RCs (Case XV). It is wrapped Minkowski space.
2 2 2 2 2 D4. ds = - dp - a d - dZ , A 54 0. It has 10 KVs and infinite RCs (Case XV). It is wrapped Minkowski.
2 2 2 D5. ds - _t dp _- ra2dO2 - dZ ,
It has 10 KVs and infinite RCs (Case XV). It gives a string solution.
2 2 2 2 D.d - t - p 0O rZ2,2
* ~~Ithas 10 KVs and infinite RCs (Case XV). It is Bertotti-Robinson like metric.
2 2 2 2 2 2 D7. ds - _tdp _ a dO _ (p/pu) dZ ,
It admits 11 HMs and infinite RCs (Case VI).
2 2 2 2 2 D8. ds = dt _ dp _ a&d0 - dZ ,
It has 11 HMs and infinite RCs (Case XV).
27 2 2 2 2 2 2 D9. ds - (p/Pu) dt _ dp _ cr dO - dZ ,
It has 11 HMs and infinite R~s (Case X).
2 2 2 2 2 2 D10. ds = eAp (dt - dz ) - dP - a d0 ,
A is a non-zero constant. It has 7 KVs and infinite RCs (Case III). It is anti-Einstein and anisotropic with negative energy.
2 2 2 2 2 Dll. ds = e Ap (dt - a d9 ) - dp - dZ ,
A is a non-zero constant. It has 7 KVs and infinite RCs (Case IV). It is anti-Einstein and anisotropic with negative energy.
2 2 2 2 2 D12. ds = dt - dp - e APa dO - eBpdZ ,
A and B are non-zero constants. It admits infinite RCs (Case Ib(ii)). It admits 7 KVs if A = B, and it is anti-Einstein. When A #~B, it admits 4 KVs and is non-physical unless a cosmological constant is introduced.
D13. ds = dt _ dp 2 - a2 dO2 - eApdZ 2 ,
A :y 0. It has 6 KVs and infinite dimensional RCs of the form given in Case VI. It is anisotropic with negative energy.
2 2 2 2 2 D14. ds = dt - dp - eAPa dO - dZ ,
A #40. It has 6 KVs and infinite RCs (Case VII). It is anisotropic with negative energy.
2 2 2 2 2 2 D15. ds - eAPdt - dp - a d - dZ ,
A :, 0. It has 6 KVs and infinite RCs (Case X). It has zero energy, zero radial pressure and is isotropic along the cylindrical direction.
2 2 2 2 2 2 2 D16. ds = cosh (A + Bp)dt - dp - a dO - dZ ,
A and a ae constants. It has 6 KVs and infinite dimensional R~s (Case X). It is a Bertotti-Robinson like metric.
2 2 2 2 2 2 D17. ds = dt _ dp - a dO - cosh (A + Bp)dZ ,
A and a are constants. It has 6 KVs and infinite dimensional RCs of the form given in Case VI. It is a Bertotti-Robinson like metric.
2 2 2 2 2 2 D18. ds = dt _ dp - cosh (A + Bp)a dO - dZ ,1
A and a are constants. It has 6 KVs and infinite dimensional RCs (Case VII). It is Bertotti-Robinson with traceless stress-energy tensor.
2 2 2 2 2 2 D19. ds = (/P0) dt _ dp _ p d - dZ ,
It is of Petrov type D and admits 5 HMs and infinite RCs (Case IV).
28 2 2 2 2 2 2 2 D20. ds = (P/Pu) dt _ dp - a dO - (/Po) dZ ,
It is of Petrov type D. It has 5 HMs and infinite RCs (Case III).
2 2 2 2 2 2 2 D21. d8 t - dp - a (p/p0) dO _ (p/Pu) dZ ,
a is a constant. It has 5 HMs and infinite R~s (Case I). It represents a non-null electro- magnetic field.
2 2 2 D22. If R 1 = 0 is taken for metric (N3), one gets a + b + c = a + b + C . Now, if the Kasner conditions, a +b +c = a2 ++b 2 +c 2 = 1, are applied, one gets R,. 0, i = 0,2,3 , corresponding to Case XV for infinite dimensional RCs. These conditions are satisfied by metrics representing different physical situations [22]. For example, the standard metric for a conical spacetime with an angle deficit
2 2 2 2 2 ds = dt _ dp - (1 - k) p dO - dz ,
which is a fat metric, or the solution representing gravitationally collapsed cylindrical matter distribution which is totally disconnected from the external space [17], given by
2 2 2 2 ds = (1 ± 4GM In (Pp)) (dt - dZ ) - (1 + 4GM In(p/Po)Y 1 /2 (a/p) 2 (dp2 + p2 dO2 ),
where G and M represent the gravitational constant and mass respectively. Another solution is
2 2 2 2 2 2 2 ds = (1 + 2GM n (p/pa)) dt _ (Pa/P) (dp + p d0 ) - dz ,
which is again a fiat spacetime with cylindrical topology.
=2m 1+2m b - 2m+1 C -2m the Kasner conditions are If one sets a 4m +2m+1' 4m+2m+1'm 4m +2m+1' satisfied and with an appropriately chosen Pa, one gets the cosmic string solution [8]
2 2 2 2 ds - (1 - 8m - 8c) 2(4m2+2~m+1) (4M + 2m + 1)ap adt
2 02 + (1 2(44m+2 1 422 2 2 ±dp ±( - 8m - 8e) 2(4m2+2mn1 (4 2m ± ) bp2bd0 2
-4m 212 2 (1 - 8m - 8E) 2(4m2+2+) (m+2m+1cpcdz which after a simple coordinate transformation can be cast into the form given in [8]. Here m is the mass per specific length and - is the energy per specific length. If m is replaced by -a, in the above definition of the so-called Kasner parameters a, b, c and, the constants are redefined, this metric takes the form
2 2 2 2 2 ds = a (4cr - 2 + 1) c p2cdt2 + dp + (4cr - 2 + ) bp2bdO2+
2 2 (4cr - 2cr ±1) a p2dZ2,
which is the Levi-Civita spacetime discussed in [23], written in different coordinates. When oa = 0, the spacetime is fat. When ua 1/2, it is again fat but for or = -1/2, it becomes 2 3 2 2 413 2 4 3 2 ds = a (3) 2/3 p- / dt + dp + (3 )4/3 p d0 + (3 )4/3 p / dZ ,
29 which is not flat and admits an extra KV. This type of metric can also be obtained by putting a = -1/3, b = c 2/3, in metric (N3), which corresponds to Einstein-Maxwell field of Petrov type D. It admits 4 HMs and infinite RCs.
2 2 2 2 2 D23. ds = (/,0) a dt _ dp - p dO _ dZ ,
a $6 0, 1 and po are constants. It is of Petrov type I and admits 4 HMs and infinite RCs (Case IV).
2 2 2 2 2 2 D24. ds (lpo) dt _ dp - a dO _ (P/Pa) dZ ,
a, a and po are constants. It is of Petrov type I and admits 4 HMs and infinite RCs (Case III).
2 2 2 2 2 D25. ds = e Apdt - dp - a dO - eBpdZ ,
A 7~B. It has 4 KVs and infinite RCs (Case Ib(ii)). It appears to be non-physical unless a cosmological constant is introduced.
2 2 2 2 2 2 D26. ds - eAPdt - dp - Bpa dO - dZ ,
A #~ B. It has 4 KVs and infinite RCs (Case IV). It appears to be non-physical unless a cosmological constant is introduced.
6 Conclusion
Cylindrically symmetric static spacetimes have been classified according to their RCs. Cases of 10, 7, 6, 5, 4 and 3 RCs for non-degenerate Ricci tensor were obtained. The cases of non-trivial (i.e. non-isometric) RCs ae 10, 7, 6, 5, 4 and 3 dimensional. Taking plane symmetry (locally) as a special case of cylindrical symmetry, the classification of plane symmetric static metrics [19] can be obtained as a special case from this classification. It is found that the RCs for the non-degenerate Ricci tensor are always finite, while for the degenerate cases they are mostly, but not always, infinite; Case II of the degenerate case has finite dimensional RCs. The comparison of present classification with the classification of cylindrically symmetric spacetimes according to KVs has given rise to the interesting cases of proper RCs given in Section 5. The corresponding metrics are implicitly given in the form of constraints on the Ricci tensor components. Concrete examples have been constructed by solving these constraints and their physics (where possible) is discussed. Acknowledgments: This work was supported in part by Pakistan Science Foundation under Project No. C- QU/MATHS (21). One of the authors (KS) gratefully acknowledges the hospitality at the Abdus Salam International Centre for Theoretical Physics, Trieste, Italy, where a part of this work was done, and a grant for this work from the Quaid-i-Azam University Research Fund.
30 Another (AQ) would like to thank KFUPM for their excellent research facilities and support to attend GR16, where this work was pursued and presented.
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