Tilted Cylindrically Symmetric Self-Similar Solutions
aa r X i v : . [ g r- q c ] M a y Tilted Cylindrically SymmetricSelf-Similar Solutions
M. Sharif ∗ and Sajid Sultan † Department of Mathematics, University of the Punjab,Quaid-e-Azam Campus Lahore-54590, Pakistan.
Abstract
This paper is devoted to explore tilted kinematic self-similar so-lutions of the the general cylindrical symmetric spacetimes. Thesesolutions are of the first, zeroth, second and infinite kinds for the per-fect fluid and dust cases. Three different equations of state are used toobtain these solutions. We obtain a total of five independent solutions.The correspondence of these solutions with those already available inthe literature is also given.
Keywords:
Cylindrical symmetry, Self-similar solution.
Einstein’s theory of General Relativity (GR) relates the geometry (curva-ture) to the physical content of spacetime (matter) through the Einsteinfield equations (EFEs) given by R ab − Rg ab = κT ab , (1)where R ab is the Ricci tensor, R is the Ricci scalar, κ is the coupling constantand T ab is the energy-momentum tensor. These equations are coupled, second ∗ [email protected] † [email protected] p = kρ . This work was extended[4] to discuss the physical aspects of the solutions.Sintes et al. [5] investigated KSS solutions of the infinite kind in thecases of plane, spherically or hyperbolically symmetric spacetime. Bicknelland Henriksen [6] considered the self-similar growth of black holes for a classof equation of state p = c s ρ , where c s is the constant sound speed. Theyconcluded that this growth is possible in the sufficiently bounded regionssurrounding the black hole. Mitsuda and Tomimatsu [7] investigated thestability of self-similar solutions of gravitational collapse from the perspec-tive of their nature as an attractor. They studied the critical phenomenaand stability of a naked singularity. Harada and Maeda [8] explored spher-ical collapse of a perfect fluid with equation of state p = kρ by full generalrelativistic numerical simulations. Ori and Piran [9] described a family ofgeneral relativistic solutions for self-similar spherical collapse of an adia-batic perfect fluid including naked singularities. A counterexample to thecosmic-censorship hypothesis was also provided. In another paper, they [10]examined the structure of general relativistic spherical collapse solution fora perfect fluid with a barotropic equation of state.Sharif and Aziz [11]-[12] discussed perfect fluid and dust solutions for thespecial and the most general plane symmetric spacetimes. They explored thefirst, second, zeroth and infinite kinds with different equations of state whenthe KSS vector is tilted, orthogonal and parallel to the fluid flow. The same2uthors [13]-[15] discussed the properties of the self-similar solutions of thefirst kind for spherically, cylindrically and plane symmetric spacetimes. Theyhave also studied the self-similar solutions for a special cylindrically symmet-ric spacetime [16]. Recently, self-similar solutions of the general cylindricallysymmetric spacetime have been investigated [17] for the parallel and orthog-onal cases. In this paper, we take the most general cylindrically symmetricspacetimes and find the KSS solutions of the first, zeroth, second and infinitekinds when the KSS vector is tilted to the fluid flow both for perfect fluidand dust cases.The paper is organized as follows. Section is devoted in formulationof the KSS vector for different kinds of self-similarity. In section and ,we investigate self-similar solutions for the tilted perfect fluid and dust casesrespectively. Finally, Section is furnished with summary of the resultsobtained. The line element for the general cylindrically symmetric spacetimes is givenas [18] ds = e ν ( t,r ) dt − e φ ( t,r ) dr − e µ ( t,r ) dθ − e λ ( t,r ) dz , (2)where ν, φ, µ and λ are functions of t and r only. Notice that the fourstructure functions are used since we are using a co-moving frame. Theenergy-momentum tensor for a perfect fluid is given by T ab = [ ρ ( t, r ) + p ( t, r )] u a u b − p ( t, r ) g ab , ( a, b = 0 , , , , (3)where ρ and p are the density and pressure respectively and u a is the 4-velocity of the fluid. In co-moving coordinate system, the 4-velocity can bewritten as u a = ( e ν , , , πGρ = e − ν ( µ t φ t + λ t φ t + λ t µ t ) + e − φ ( − µ rr + µ r φ r − µ r − λ rr + λ r φ r − λ r − µ r λ r ) , (4)0 = − µ tr − λ tr + µ t ν r + λ t ν r + φ t µ r + λ r φ t − λ r λ t − µ t µ r , (5)8 πGp = e − ν ( − µ tt − λ tt + µ t ν t + λ t ν t − λ t µ t − µ t − λ t )3 e − φ ( µ r ν r + λ r ν r + λ r µ r ) , (6)8 πGp = e − ν ( − φ tt − λ tt + ν t φ t + λ t ν t − λ t φ t − φ t − λ t )+ e − φ ( ν rr + ν r − ν r φ r + λ r ν r + λ rr − λ r φ r + λ r ) , (7)8 πGp = e − ν ( − φ tt − µ tt + ν t φ t + µ t ν t − µ t φ t − φ t − µ t )+ e − φ ( ν rr + ν r − ν r φ r + µ r ν r + µ rr − µ r φ r + µ r ) . (8)The conservation of energy-momentum tensor, T ab ; b = 0 , yields the followingequations φ t = − ρ t ρ + p − µ t − λ t , ν r = − p r ρ + p . (9)For a cylindrically symmetric spacetime, the vector field ξ can have the fol-lowing form ξ a ∂∂x a = h ( t, r ) ∂∂t + h ( t, r ) ∂∂r , (10)where h and h are arbitrary functions of t and r . The tilted perfect fluidhas both h and h non-zero while h = 0 gives orthogonal case and h = 0the parallel case. This paper is devoted to investigate the KSS solutions forthe tilted perfect fluid and dust cases.A kinematic self-similar vector ξ is defined by £ ξ h ab = 2 δh ab , £ ξ u a = αu a , (11)where h ab = g ab − u a u b is the projection tensor and α, δ are dimensionlessconstants. We can have different kinds of self-similarity according as δ = 0or δ = 0 given below:( ∗ ) δ = 0 : Here the KSS vector for the tilted perfect fluid case takes theform ξ a ∂∂x a = ( αt + β ) ∂∂t + r ∂∂r . (12)The similarity index, αδ , yields the following three possibilities (i) α = 1 ( β can be taken to zero) - first kind, (ii) α = 0 ( β can be taken to unity) - zeroth kind, (iii) α = 0 , β can be taken to zero) - second kind,4here δ can be taken as unity. The self-similar variable for self-similarity ofthe first kind turns out to be ξ = rt . In the zeroth kind, i.e., α = 0, theself-similar variable is ξ = re t . For the second kind, the self similar variablebecomes ξ = r/ ( αt ) α . For all these kinds, the metric functions are ν ( t, r ) = ν ( ξ ) , φ ( t, r ) = φ ( ξ ) , e µ ( t,r ) = re µ ( ξ ) , e λ ( t,r ) = re λ ( ξ ) . (13)( ∗∗ ) δ = 0 : Here the KSS vector can take the following form (when α = 0) ξ a ∂∂x a = t ∂∂t + r ∂∂r (14)and the corresponding self-similar variable is ξ = rt . The metric function willbecome ν ( t, r ) = ν ( ξ ) , φ ( t, r ) = − ln r + φ ( ξ ) , µ ( t, r ) = µ ( ξ ) , λ ( t, r ) = λ ( ξ ) . (15)The following equations of state (EOS) are used.EOS(1): p = kρ γ , where k and γ are constants.EOS(2): p = kn γ , ρ = m b n + pγ − , where k = 0 and γ = 0 , p = kρ, − ≤ k ≤ , k = 0. Using Eq.(13) in Eqs.(4) and (6)-(8), the mass density and pressure musttake the following forms [19] κρ ( t, r ) = 1 r ρ ( ξ ) , (16) κp ( t, r ) = 1 r p ( ξ ) , (17)where the self-similar variable is ξ = r/t . If the EFEs and the equations ofmotion for the matter field are satisfied for O [( r ) − ], we obtain a set of ODEsgiven by ˙ ρ = − ( ˙ φ + ˙ µ + ˙ λ )( ρ + p ) , (18)2 p − ˙ p = ˙ ν ( ρ + p ) , (19)5 = ˙ µ ˙ φ + ˙ λ ˙ φ + ˙ λ ˙ µ, (20) ρe φ = − ¨ µ − ¨ λ − ˙ µ − ¨ λ − µ − λ + 2 ˙ φ + ˙ µ ˙ φ + ˙ λ ˙ φ − ˙ λ ˙ µ − , (21)0 = ¨ µ + ¨ λ + ˙ µ + ¨ λ + ˙ µ + ˙ λ − ˙ µ ˙ ν − ˙ λ ˙ ν − φ − ˙ φ ˙ µ − ˙ φ ˙ λ, (22)0 = − ¨ µ − ¨ λ − ˙ µ − ¨ λ − ˙ µ − ˙ λ − ˙ λ ˙ µ + ˙ µ ˙ ν + ˙ λ ˙ ν, (23) pe φ = 1 + ˙ µ + ˙ λ + 2 ˙ ν + ˙ µ ˙ ν + ˙ λ ˙ ν + ˙ λ ˙ µ, (24)0 = − ¨ φ − ¨ λ − ˙ φ − ¨ λ − ˙ φ − ˙ λ − ˙ φ ˙ λ + ˙ φ ˙ ν + ˙ λ ˙ ν, (25) pe φ = ¨ ν + ¨ λ + ˙ ν + ˙ λ + ˙ λ + ˙ λ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ λ − ˙ φ, (26)0 = − ¨ φ − ¨ µ − ˙ φ − ¨ µ − ˙ φ − ˙ µ − ˙ φ ˙ µ + ˙ φ ˙ ν + ˙ µ ˙ ν, (27) pe φ = ¨ ν + ¨ µ + ˙ ν + ˙ µ + ˙ µ + ˙ µ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ µ − ˙ φ. (28)where dot represents derivative with respect to ξ . Sine the first kind is notcompatible with EOS(1) and EOS(2), hence no solution exists. If a perfect fluid satisfies EOS(3), then we have the following solution ν = ln( c ξ ±√ ) , φ = c , µ = c , λ = c ,p = ρ = constant (29)corresponding to the metric ds = ( r ) ± √ dt − dr − r ( dθ + dz ) . (30) Here the quantities ρ and p take the form κρ = 1 r { ρ ( ξ ) + r ρ ( ξ ) } , (31) κp = 1 r { p ( ξ ) + r p ( ξ ) } , (32)where the self-similar variable is ξ = re − t and the set of ODEs become˙ ρ = − ( ˙ φ + ˙ µ + ˙ λ )( ρ + p ) , (33)6 ρ = − ( ˙ φ + ˙ µ + ˙ λ )( ρ + p ) , (34)2 p − ˙ p = ˙ ν ( ρ + p ) , (35) − ˙ p = ˙ ν ( ρ + p ) , (36) ρ e φ = − ¨ µ − ¨ λ − ˙ µ − ¨ λ − µ − λ + 2 ˙ φ + ˙ µ ˙ φ + ˙ λ ˙ φ − ˙ λ ˙ µ − , (37) ρ e ν = ˙ µ ˙ φ + ˙ λ ˙ φ + ˙ λ ˙ µ, (38)0 = ¨ µ + ¨ λ + ˙ µ + ¨ λ + ˙ µ + ˙ λ − ˙ µ ˙ ν − ˙ λ ˙ ν − φ − ˙ φ ˙ µ − ˙ φ ˙ λ, (39) p e φ = 1 + ˙ µ + ˙ λ + 2 ˙ ν + ˙ µ ˙ ν + ˙ λ ˙ ν + ˙ λ ˙ µ, (40) p e ν = − ¨ µ − ¨ λ − ˙ µ − ¨ λ − ˙ λ ˙ µ + ˙ µ ˙ ν + ˙ λ ˙ ν, (41) p e φ = ¨ ν + ¨ λ + ˙ ν + ˙ λ + ˙ λ + ˙ λ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ λ − ˙ φ, (42) p e ν = − ¨ φ − ¨ λ − ˙ φ − ¨ λ − ˙ φ ˙ λ + ˙ φ ˙ ν + ˙ λ ˙ ν, (43) p e φ = ¨ ν + ¨ µ + ˙ ν + ˙ µ + ˙ µ + ˙ µ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ µ − ˙ φ, (44) p e ν = − ¨ φ − ¨ µ − ˙ φ − ¨ µ − ˙ φ ˙ µ + ˙ φ ˙ ν + ˙ µ ˙ ν. (45) For EOS(1) with k = 0 and γ = 0 ,
1, Eqs.(31) and (32) become ρ = p = 0 , p = kκ γ − ρ γ . [ CaseI ] (46)For EOS(2) with k = 0 and γ = 0 ,
1, Eqs.(31) and (32) take the form ρ = 0 = p , p = kκ γ − m γb ( ρ − γ − p ) γ . [ CaseII ] (47)EOS(3) yields two more cases for k = − III ] and for k = − IV ]. Case I:
Solving the equations simultaneously, we obtain ν = c , φ = − ln ξ + ln( ξ − c ) + c , µ = − ln ξ + c , λ = − ln ξ + c ,p = 0 = ρ , p = constant, ρ = − ξ + c ) e c ( ξ − c ) . (48)The corresponding metric is ds = dt − ( r − c e t re t ) dr − e t ( dθ + dz ) . (49)7he second solution is ν = c , φ = 12 ln ξ + c , µ = 12 ln ξ + c , λ = − ln ξ + c ,ρ = 0 = p , ρ = p = constant (50)and the corresponding metric is ds = dt − re t ( dr + r dθ ) − e t dz . (51)The third solution is ν = c , φ = 12 ln ξ + c , µ = − ln ξ + c , λ = 12 ln ξ + c ,ρ = 0 = p , ρ = p = constant (52)and the metric takes the form ds = dt − re t dr − e t dθ − r e t dz . (53)The Case II gives the same solutions as the Case I . Case III:
Here the solution becomes ν = c , φ = − ln ξ + c , µ = − ln ξ + c , λ = − ln ξ + c ,ρ = 0 = p , ρ = p = constant (54)and the corresponding metric is ds = dt − e t r ( dr + r ( dθ + dz )) . (55) Case IV : This case gives the two solutions out of which the first is ν = c , φ = 2 ln ξ + c , µ = − ln ξ + c , λ = − ln ξ + c ,ρ = 0 = p , ρ = p = constant (56)and the corresponding metric is ds = dt − r e t dr − e t ( dθ + dz ) . (57)The second solution is ν = (1 ± √
2) ln ξ + c , φ = c , µ = c , λ = c ,ρ = p = constant, ρ = 0 = p (58)and the corresponding spacetime is ds = ( re t ) ± √ dt − dr − r ( dθ + dz ) . (59)8 .3 Self-similarity of the Second Kind Here the EFEs imply that that κρ = 1 r { ρ ( ξ ) + r t ρ ( ξ ) } , (60) κp = 1 r { p ( ξ ) + r t p ( ξ ) } , (61)where the self-similar variable is ξ = r/ ( αt ) /α . The set of ODEs become˙ ρ = − ( ˙ φ + ˙ µ + ˙ λ )( ρ + p ) , (62)˙ ρ + 2 αρ = − ( ˙ φ + ˙ µ + ˙ λ )( ρ + p ) , (63)2 p − ˙ p = ˙ ν ( ρ + p ) , (64) − ˙ p = ˙ ν ( ρ + p ) , (65) ρ e φ = − ¨ µ − ¨ λ − ˙ µ − ¨ λ − µ − λ + 2 ˙ φ + ˙ µ ˙ φ + ˙ λ ˙ φ − ˙ λ ˙ µ − , (66) α ρ e ν = ˙ µ ˙ φ + ˙ λ ˙ φ + ˙ λ ˙ µ, (67)0 = ¨ µ + ¨ λ + ˙ µ + ¨ λ + ˙ µ + ˙ λ − ˙ µ ˙ ν − ˙ λ ˙ ν − φ − ˙ φ ˙ µ − ˙ φ ˙ λ, (68) p e φ = 1 + ˙ µ + ˙ λ + 2 ˙ ν + ˙ µ ˙ ν + ˙ λ ˙ ν + ˙ λ ˙ µ, (69) α p e ν = − ¨ µ − ¨ λ − ˙ µ − ¨ λ − α ˙ µ − α ˙ λ − ˙ λ ˙ µ + ˙ µ ˙ ν + ˙ λ ˙ ν, (70) p e φ = ¨ ν + ¨ λ + ˙ ν + ˙ λ + ˙ λ + ˙ λ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ λ − ˙ φ, (71) α p e ν = − ¨ φ − ¨ λ − ˙ φ − ¨ λ − α ˙ φ − α ˙ λ − ˙ φ ˙ λ + ˙ φ ˙ ν + ˙ λ ˙ ν, (72) p e φ = ¨ ν + ¨ µ + ˙ ν + ˙ µ + ˙ µ + ˙ µ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ µ − ˙ φ, (73) α p e ν = − ¨ φ − ¨ µ − ˙ φ − ¨ µ − α ˙ φ − α ˙ µ − ˙ φ ˙ µ + ˙ φ ˙ ν + ˙ µ ˙ ν. (74) For EOS(1) ( k = 0 and γ = 0 , p = 0 = ρ , α = γ, p = kκ γ − γ ξ − γ ρ γ , [ CaseI ] (75) p = 0 = ρ , α = 1 γ , p = kκ γ − γ γ ξ ρ γ . [ CaseII ] (76)9hen a perfect fluid satisfies EOS(2) for k = 0 and γ = 0 ,
1, it follows fromEqs.(60) and (61) that p = km γb γ κ γ − ξ − γ ρ γ = ( γ − ρ ,p = 0 , α = γ, [ CaseIII ] (77) p = km γb γ γ κ γ − ξ ρ γ = ( γ − ρ ,p = 0 , α = 1 γ . [ CaseIV ] (78)EOS(3) yields two more cases (Case V and Case VI) for k = − k = − Case I:
Solving the ODEs simultaneously, we obtain the solutions as ν = c , φ = 12 ln ξ + c , µ = − ln ξ + c , λ = − ln ξ + c ,ρ = ρ = 0 = p = p , α = 32 (79)and the corresponding metric is ds = dt − ( 23 t ) rdr − ( 3 t ( dθ + dz ) . (80)The second solution is ν = c , φ = 2 ln ξ + c , µ = − ln ξ + c , λ = 2 ln ξ + c ,ρ = ρ = 0 = p = p , α = − . (81)Its metric form is ds = dt − r ( − t ) dr − − t ) dθ − r ( − t ) dz . (82)The third solution is ν = c , φ = 2 ln ξ + c , µ = 2 ln ξ + c , λ = − ln ξ + c ,ρ = ρ = 0 = p = p , α = − ds = dt − r ( − t ) ( dr + r dθ ) − − t ) dz . (84)10t is mentioned here that the Cases II, III and IV have the same solutions asgiven by Eqs.(80), (82) and (84). It is also noted that the Case V has onlyone solution given by Eq.(80). Case VI:
For this case, we obtain the following solutions. The first solutionis ν = c , φ = − ln ξ + c , µ = − ln ξ + c , λ = − ln ξ + c ,ρ = 0 = p , ρ = p = constant, k = 2 α −
33 (85)and the corresponding metric is ds = dt − ( αt ) α r dr − ( αt ) α ( dθ + dz ) . (86)The second solution is ν = c , φ = (2 − α ) ln ξ = c , µ = − ln ξ + c , λ = − ln ξ + c ,ρ = 0 = p , ρ = p = constant, k = 1 (87)and the corresponding spacetime is ds = dt − r − α ) ( αt ) − α ) α dr − ( αt ) α ( dθ + dz ) . (88)The third solution is ν = c , φ = − ln ξ = c , µ = − ln ξ + c , λ = − ln ξ + c ,ρ = 0 = p , ρ = p = constant, k = 1 , α = 3 (89)and its metric is ds = dt − (3 t ) r dr − (3 t ) ( dθ + dz ) . (90)The fourth solution is ν = (1 ± √
2) ln ξ + c , φ = c , µ = c , λ = c ,ρ = p = constant, ρ = 0 = p , k = − ± √ ds = ( r ( αt ) /α ) ± √ dt − dr − r ( dθ + dz ) . (92)11 .4 Self-similarity of the Infinite Kind Here the quantities ρ and p can be written in terms of ξ as κρ = ρ ( ξ ) + 1 t ρ ( ξ ) , (93) κp = p ( ξ ) + 1 t p ( ξ ) , (94)where the self-similar variable is ξ = r/t and the set of ODEs become˙ ρ = − ( ˙ φ + ˙ µ + ˙ λ )( ρ + p ) , (95)˙ ρ + 2 ρ = − ( ˙ φ + ˙ µ + ˙ λ )( ρ + p ) , (96) − ˙ p = ˙ ν ( ρ + p ) , (97) − ˙ p = ˙ ν ( ρ + p ) , (98) ρ e φ = − ¨ µ − ¨ λ − ˙ µ − ¨ λ + ˙ µ ˙ φ + ˙ λ ˙ φ − ˙ λ ˙ µ, (99) ρ e ν = ˙ µ ˙ φ + ˙ λ ˙ φ + ˙ λ ˙ µ, (100)0 = ¨ µ + ¨ λ + ˙ µ + ¨ λ − ˙ µ ˙ ν − ˙ λ ˙ ν − ˙ φ ˙ µ − ˙ φ ˙ λ, (101) p e φ = ˙ µ ˙ ν + ˙ λ ˙ ν + ˙ λ ˙ µ, (102) p e ν = − ¨ µ − ¨ λ − ˙ µ − ¨ λ − ˙ µ − ˙ λ − ˙ λ ˙ µ + ˙ µ ˙ ν + ˙ λ ˙ ν, (103) p e φ = ¨ ν + ¨ λ + ˙ ν + ˙ λ + ˙ λ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ λ, (104) p e ν = − ¨ φ − ¨ λ − ˙ φ − ¨ λ − ˙ φ − ˙ λ − ˙ φ ˙ λ + ˙ φ ˙ ν + ˙ λ ˙ ν, (105) p e φ = ¨ ν + ¨ µ + ˙ ν + ˙ µ + ˙ µ ˙ ν − ˙ φ ˙ ν − ˙ φ ˙ µ, (106) p e ν = − ¨ φ − ¨ µ − ˙ φ − ¨ µ − ˙ φ − ˙ µ − ˙ φ ˙ µ + ˙ φ ˙ ν + ˙ µ ˙ ν. (107) For EOS(1) with k = 0 and γ = 0 ,
1, Eqs.(95) and (96) become ρ = 0 = p , p = kκ γ − ρ γ . [ Case I ] (108)For EOS(2) with k = 0 and γ = 0 ,
1, Eqs.(95) and (96) take the form ρ = 0 = p , p = km γb κ γ − ( ρ − p γ − γ . [ Case II ] (109)EOS(3) yields two more cases for k = − III ] and for k = − IV ]. 12 ase I : Solving the ODEs simultaneously for this case, we have ν = c , φ = c , µ = c , λ = c , ρ = ρ = 0 = p = p (110)and the corresponding metric is ds = dt − r dr − dθ − dz (111)which corresponds to Minkowski spacetime. The second solution is ν = ln(ln ξ − ln c ) + c , φ = c , µ = c , λ = c ,ρ = ρ = 0 = p = p . (112)The corresponding metric is ds = [ln( rc t )] dt − r dr − ( dθ + dz ) . (113)The third solution becomes ν = c , φ = ln( ξ − c ξ ) + c , ˙ µ = 0 , ˙ λ = 0 , ρ = ρ = 0 = p = p . (114)The corresponding metric is ds = dt − ( r − c t ) r dr − ( dθ + dz ) . (115)It is mentioned here that the Cases II, III and IV have the same solutions asthe Case I. It is well-known that a perfect fluid is characterized by the pressure and itreduces to the dust fluid when p = 0. When we substitute p = 0 in Eqs.(18)-(28), it follows from Eq.(19)˙ νρ = 0 (116)13hich gives the following two solutions. The first solution turns out to be ν = c , φ = c , µ = c , λ = − ln ξ + c , ρ = 0 (117)with metric ds = dt − dr − r dθ − t dz . (118)The second solution is ν = c , φ = c , µ = − ln ξ + c , λ = c , ρ = 0 (119)and the corresponding metric is ds = dt − dr − t dθ − r dz . (120) When we take p = 0 and p = 0 in Eqs.(33)-(45), we obtain contradictionand consequently there is no self-similar solution of the zeroth kind. When we substitute p = 0 and p = 0 in Eqs.(62)-(74), we obtain thefollowing solution ν = c , φ = ln( 2 ξ − c ξ ) , µ = − ln ξ + c , λ = − ln ξ + c ,ρ = 0 , ρ = 49 e c ( 3 c c − ξ ) . (121)The corresponding metric is ds = dt − ( 3 t ( 4 r − tc rt ) dr − ( 3 t ( dθ + dz ) . (122) Taking p = 0 = p in Eqs.(97)-(109), it yields the same solutions as givenby Eqs.(111) and (115). 14 Summary and Discussion
Recently, Sharif and Aziz [16] have found the KSS solutions for a specialcylindrically symmetric spacetimes. This work has been extended to investi-gate the KSS solutions of the general cylindrically symmetric spacetime whenthe fluid flow is parallel as well as orthogonal [17]. However, we have left thetilted case to deal it separately. In this paper, we have studied the case whenKSS vector is tilted to the fluid flow. The solutions are found for the first,zeroth, second and the infinite kinds both for perfect fluid and dust cases.The summary of the independent solutions is given in the form of table . Table 1.
Tilted KSS Solutions.
S. NO. Metric Equation of state Physical I ds = ( r ) ± √ dt − dr − r ( dθ + dz ) ρ = − πr < ,p = − (3 ± √ ρ > ds = dt − re t ( dr + r dθ ) − e t dz ρ = p = − π < ds = dT − e T ( dR + R d Θ + dZ ) ,R > , − ∞ < T < + ∞ , < Θ < π, − ∞ < Z < + ∞ ρ = − p = π > ds = dt − r (3 t ) ( dr + r dθ ) − t ) dz ρ = 0 = p PhysicalV ds = dT − dR − R d Θ + dZ ,R > , − ∞ < T < + ∞ , < Θ < π, − ∞ < Z < + ∞ ρ = 0 = p PhysicalThe results can be discussed as follows:I. ρ = − πr < , p = − (3 ± √ ρ >
0. This solution is unphysical asenergy density is negative.II. For this solution, we have ρ = p = − π < > p < ρ = 0 = p . This is an empty but curved spacetime as R = 0 , R ab =0, but R abcd R abcd = t = 0 singular at t = 0 (extendible).15. This gives ρ = 0 = p, R abcd R abcd = 0, i.e. empty flat, Minkowski spacetimein cylindrical coordinates.We have seen that the number of independent solutions are very few.Also, most of these solutions coincide with those already available in theliterature. Thus we can conclude that one may obtain already known space-times by specializing a map adapted to some symmetries. However, the lineelement may look more complicated in such a map. Moreover, such a mapmay cover only a fragment of the spacetime. Acknowledgment
We would like to appreciate the referee’s useful comments.
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