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Conformal symmetries of the energy-momentum tensor of spherically symmetric static spacetimes

Ugur Camci Department of Chemistry and Physics, Roger Williams University, One Old Ferry Road, Bristol, Rhode Island 02809, USA [email protected], [email protected]    Khalid Saifullah Department of Mathematics, Quaid-i-Azam University, Islamabad, Pakistan [email protected]
Abstract

Conformal matter collineations of the energy-momentum tensor for a general spherically symmetric static spacetime are studied. The general form of these collineations is found when the energy-momentum tensor is non-degenerate, and the maximum number of independent conformal matter collineations is fifteen. In the degenerate case of the energy-momentum tensor it is found that these collineations have infinite degrees of freedom. In some subcases of degenerate energy-momentum, the Ricci tensor is non-degenerate, that is, there exist non-degenerate Ricci inheritance collineations.

pacs:
04.20-q, 04.20.Jb, 04.70.-s, 11.30.-j

I Introduction

Recent observations indicate that the universe contains black holes whose horizons are rotating at a speed close to that of light. General relativity (GR) suggests that the dynamics near the horizon of such black holes is governed by a strong infinite-dimensional conformal symmetry - similar to the one seen near the critical points of different condensed matter systems. Researchers have explored possible observational consequences of such a symmetry (AS, ).

Symmetries and conformal symmetries play a very important role in mathematical physics. One of the fundamental symmetries on a Riemannian manifold is that of the metric tensor 𝐠\mathbf{g} written mathematically as katzin

£ξ𝐠=2σ𝐠.\pounds_{\xi}{\bf g}=2\sigma{\bf g}. (1)

Here σ\sigma is the conformal factor and £ξ\pounds_{\xi} represents the Lie derivative operator relative to the vector field ξ{\xi}, which gives isometries or Killing vectors (KVs) if the σ\sigma is zero, homothetic motions (HMs) if it is a constant, and conformal Killing vectors (CKVs) if it is a function of the coordinates xax^{a}. In component form we can write the above equation as

gab,cξc+gacξ,bc+gcbξ,ac=2σgab.g_{ab,c}\xi^{c}+g_{ac}\xi_{,b}^{c}+g_{cb}\xi^{c}_{,a}=2\sigma g_{ab}. (2)

Apart from the metric tensor the other quantities fundamental to the Einstein field equations (EFEs)

Rab12Rgab=κTab,R_{ab}-\frac{1}{2}Rg_{ab}=\kappa T_{ab}, (3)

are the stress-energy tensor, 𝐓{\bf T}, which describes the matter field in the manifold, the Ricci tensor, 𝐑\bf R, which is a contraction of the curvature tensor, and the Ricci scalar RR. In Eq. (3) κ\kappa is the coupling constant defined by κ=8πG/c4\kappa=8\pi G/c^{4}, GG and cc are Newton’s gravitational constant and the speed of light. Thus, the symmetries of both the stress-energy tensor and the Ricci tensor, play a significant role. These symmetries known as matter collineations (MCs) and Ricci collineations (RCs) hall-book , respectively, satisfy the equations

£ξ𝐓=0,and£ξ𝐑=0.\pounds_{\xi}{\bf T}=0,\,\,\,\ \text{and}\,\,\,\ \pounds_{\xi}{\bf R}=0. (4)

Similarly, one can define collineations for the curvature and Weyl tensors hall-book .

Solutions of EFEs can be classified by requiring these symmetries and thus a complete list of metrics having certain symmetry can be obtained ESEFEs . Spacetimes have been classified on the basis, for example, of KVs bq1987 ; qz1988 ; qz1995 , HMs dz1997 , CKVs maartens1995 ; keane2004 , RCs tm1990 ; bk1993 ; fqz1995 ; hall1996 ; qz1998 ; ziad2003 ; saif2003 ; ugur1 ; cb2002 ; ugur2 ; ugur3 ; tsamparlis1 and MCs ugur4 ; sharif2003 ; sharif2004 ; ugur5 ; ugur6 . This also provides a way to find new solutions of EFEs which are otherwise very difficult to solve. These collineations have been generalized to define what are called conformal collineations (or inheritance collineations duggal1 ; duggal11 ; duggal2 ). Thus we obtain conformal matter collineations (CMCs)

£𝐗𝐓=2ψ(xa)𝐓,\pounds_{\bf X}{\bf T}=2\psi(x^{a}){\bf T}, (5)

or conformal Ricci collineations (CRCs) defined by

£𝐘𝐑=2ϕ(xa)𝐑.\pounds_{\bf Y}{\bf R}=2\phi(x^{a}){\bf R}. (6)

Conformal symmetry is physically significant as CKVs, for example, generate constants of motion along the null geodesics for massless particles which are conserved quantities. On the other hand, it is of mathematical interest to obtain classification by conformal collineations and to investigate their relation with collineations. Though there has been a good amount of literature on the study of CKVs, the interest in conformal collineations is relatively recent. The complete classification of spherically symmetric static space-times by their CRCs when the conformal factor ϕ\phi is a constant has been carried out in Ref. bkk2003 . The CRCs with a non-constant ϕ\phi have been studied for the Friedmann-Robertson-Walker spacetimes cb2002 , the general static spherically symmetric spacetimes crcsph , the non-static spherically symmetric spacetimes bokhari2020 and Kantowski-Sachs spacetimes hkb2018 . For pp-waves relationship between CRCs and CKVs has been studied in Ref. keane2004 . Recently, Akhtar et al. bokhari2018 have classified static plane symmetric space-times according to CRCs. Further, the CRCs for the Einstein-Maxwell field equations in the case of non-null electromagnetic fields have been investigated as well faridi .

In this paper we classify spherically symmetric static spacetimes by their CMCs. The equation (5) for CMCs in component form can be written as

Tab,cXc+TacX,bc+TcbX,ac=2ψTab,T_{ab,c}X^{c}+T_{ac}X^{c}_{,b}+T_{cb}X^{c}_{,a}=2\psi T_{ab}, (7)

where ψ\psi is the conformal factor which is a function of all the spacetime coordinates xa=(x0,x1,x2,x3)x^{a}=(x^{0},x^{1},x^{2},x^{3}). In this paper we use the usual component notation in local charts and a partial derivative will be denoted by a comma. Note that the above equation gives MCs if ψ=0\psi=0, thus the classification of spherically symmetric static spacetimes by MCs sharif2003 becomes a special case of the classification obtained in this paper. We call a CMC as proper if it is neither a KV nor an MC. The set of all CMCs on the manifold is a vector space, but it may be infinite dimensional. If TabT_{ab} is non-degenerate, i.e. det(Tab)0det(T_{ab})\neq 0, then the Lie algebra of CMCs is finite dimensional but if TabT_{ab} is degenerate, it may be infinite. Thus, in the case of a non-degenerate energy-momentum tensor, i.e. det(Tab)0det(T_{ab})\neq 0, we use the standard results on conformal symmetries to obtain the maximal dimensions of the algebra of CMCs as 15. Since TabT_{ab} describes the distribution and motion of matter contents of a manifold, and mathematically it is very similar to the the Ricci tensor, the study of MCs and CMCs has a natural geometrical as well as physical significance.

In the next section, we setup the CMC equations for static spherically symmetric spacetimes. In Section III these equations are solved when the energy-momentum tensor is degenerate, while in Section IV we obtain results when the tensor is non-degenerate. We find that the degenerate case always gives infinite dimensional Lie algebras of CMCs. We conclude with a brief summary and discussion in Section V. Throughout the paper, we will consider four-dimensional space-times, and space-time indices will be denoted by small Latin letters (e.g., aa, bb, cc,…) and the metric has signature (+,,,)(+,-,-,-).

II Equations for Conformal Matter Collineations

We consider a general spherically symmetric static spacetime in the usual spherical coordinates

ds2=eν(r)dt2eλ(r)dr2r2(dθ2+sin2θdϕ2).ds^{2}=e^{\nu(r)}dt^{2}-e^{\lambda(r)}dr^{2}-r^{2}\left(d\theta^{2}+\sin^{2}\theta d\phi^{2}\right). (8)

The non-vanishing components of the Ricci tensor Rab=RacbcR_{ab}=R^{c}_{\,acb} for this metric are given by

R00R0(r)=14eνλ(2ν′′+ν2νλ+4rν),\displaystyle R_{00}\equiv R_{0}(r)=\frac{1}{4}e^{\nu-\lambda}\left(2\nu^{\prime\prime}+{\nu^{\prime}}^{2}-\nu^{\prime}\lambda^{\prime}+\frac{4}{r}\nu^{\prime}\right), (9)
R11R1(r)=14(2ν′′+ν2νλ4rλ),\displaystyle R_{11}\equiv R_{1}(r)=-\frac{1}{4}\left(2\nu^{\prime\prime}+{\nu^{\prime}}^{2}-\nu^{\prime}\lambda^{\prime}-\frac{4}{r}\lambda^{\prime}\right), (10)
R22R2(r)=12eλ[r(λν)2]+1,\displaystyle R_{22}\equiv R_{2}(r)=\frac{1}{2}e^{-\lambda}\left[r(\lambda^{\prime}-\nu^{\prime})-2\right]+1, (11)
R33=sin2θR2,\displaystyle R_{33}=\sin^{2}\theta R_{2}, (12)

and the Ricci scalar R=RaaR=R^{a}_{\,\,a} is

R=eλ2[2ν′′+ν2νλ+4r(νλ)+4r2(1eλ)],R=\frac{e^{-\lambda}}{2}\left[2\nu^{\prime\prime}+\nu^{\prime 2}-\nu^{\prime}\lambda^{\prime}+\frac{4}{r}(\nu^{\prime}-\lambda^{\prime})+\frac{4}{r^{2}}(1-e^{\lambda})\right], (13)

where the prime represents derivative with respect to the radial coordinate rr. Thus we can write the Ricci tensor form as

dsRic2Rabdxadxb=R0(r)dt2+R1(r)dr2+R2(r)(dθ2+sin2θdϕ2).ds^{2}_{Ric}\equiv R_{ab}dx^{a}dx^{b}=R_{0}(r)dt^{2}+R_{1}(r)dr^{2}+R_{2}(r)\left(d\theta^{2}+\sin^{2}\theta d\phi^{2}\right). (14)

The metric (8) has time-independent coefficients, and using the field equations (3) with c=1c=1 and G=18πG=\frac{1}{8\pi}, i.e., κ=1\kappa=1, the components of the energy-momentum tensor TabT_{ab} become

T00T0(r)=1r2eνλ(rλ+eλ1),\displaystyle T_{00}\equiv T_{0}(r)=\frac{1}{r^{2}}e^{\nu-\lambda}\left(r\lambda^{\prime}+e^{\lambda}-1\right), (15)
T11T1(r)=1r2(rνeλ+1),\displaystyle T_{11}\equiv T_{1}(r)=\frac{1}{r^{2}}\left(r\nu^{\prime}-e^{\lambda}+1\right), (16)
T22T2(r)=r24eλ[2ν′′+ν2νλ+2r(νλ)],\displaystyle T_{22}\equiv T_{2}(r)=\frac{r^{2}}{4}e^{-\lambda}\left[2\nu^{\prime\prime}+{\nu^{\prime}}^{2}-\nu^{\prime}\lambda^{\prime}+\frac{2}{r}(\nu^{\prime}-\lambda^{\prime})\right], (17)
T33T3(r)=sin2θT2.\displaystyle T_{33}\equiv T_{3}(r)=\sin^{2}\theta T_{2}. (18)

Similarly, the matter tensor form can be written as

dsMatter2Tabdxadxb=T0(r)dt2+T1(r)dr2+T2(r)(dθ2+sin2θdϕ2).ds^{2}_{Matter}\equiv T_{ab}dx^{a}dx^{b}=T_{0}(r)dt^{2}+T_{1}(r)dr^{2}+T_{2}(r)\left(d\theta^{2}+\sin^{2}\theta d\phi^{2}\right). (19)

Using the above energy-momentum tensor components, the Ricci curvature scalar given in (13) can be cast into the form

R=eνT0+eλT1+2r2T2.R=-e^{-\nu}T_{0}+e^{-\lambda}T_{1}+\frac{2}{r^{2}}T_{2}\,. (20)

Further, when we state the energy-momentum tensor components given above in terms of the Ricci tensor components (9)-(11) we find that

T0=12(R0+eνλR1)+eνr2R2,\displaystyle T_{0}=\frac{1}{2}\left(R_{0}+e^{\nu-\lambda}\,R_{1}\right)+\frac{e^{\nu}}{r^{2}}R_{2}, (21)
T1=12(eλνR0+R1)eλr2R2,\displaystyle T_{1}=\frac{1}{2}\left(e^{\lambda-\nu}R_{0}+R_{1}\right)-\frac{e^{\lambda}}{r^{2}}R_{2}, (22)
T2=r22(eνR0eλR1).\displaystyle T_{2}=\frac{r^{2}}{2}\left(e^{-\nu}R_{0}-e^{-\lambda}R_{1}\right). (23)

In GR, physical fields are described by the symmetric tensor TabT_{ab} which is the energy-momentum tensor of the field. We have the covariant decomposition/identity for TabT_{ab} as follows tp2019 :

Tab=ρuaubphab+2q(aub)+πab,T_{ab}=\rho u_{a}u_{b}-ph_{ab}+2q_{(a}u_{b)}+\pi_{ab}\,, (24)

where hab=gabuaubh_{ab}=g_{ab}-u_{a}u_{b} is the projection tensor, and the quantities ρ,p,qa\rho,p,q_{a} and πab\pi_{ab} are the physical variables representing the mass density, the isotropic pressure, the heat flux and the traceless stress tensor, respectively, as measured by the observers uau^{a}. In the above decomposition, TabT_{ab} is described by two scalar fields (ρ,p\rho,p), one spacelike vector (qa,qaua=0q^{a},\,q_{a}u^{a}=0), and a traceless symmetric 2-tensor (πab,gabπab=0\pi_{ab},\,g_{ab}\pi^{ab}=0). The irreducible parts of TabT_{ab} are defined as

ρ=uaubTab,\displaystyle\rho=u^{a}u^{b}T_{ab}\,, (25)
p=13habTab,\displaystyle p=-\frac{1}{3}h^{ab}T_{ab}\,, (26)
qa=habTbcuc,\displaystyle q^{a}=h^{ab}T_{bc}u^{c}\,, (27)
πab=(hachbd13habhcd)Tcd,\displaystyle\pi_{ab}=\left(h_{a}^{\,c}h_{b}^{\,d}-\frac{1}{3}h_{ab}h^{cd}\right)T_{cd}\,, (28)

where uau^{a} is a timelike unit four vector field normalized by uaua=1u^{a}u_{a}=1. The energy-momentum tensor TabT_{ab} given in (24) represents the general anisotropic fluid, and reduces to an anisotropic fluid without heat flux if qa=0q^{a}=0, an isotropic non-perfect fluid if πab=0\pi^{ab}=0, a perfect fluid if qa=0q^{a}=0 and πab=0\pi^{ab}=0, and a dust if p=0,qa=0p=0,q^{a}=0 and πab=0\pi^{ab}=0. GR is a classical theory of relativity, however, in the field equations (3), the classical spacetime geometry is also related to the stress-energy tensor of quantum matter. To overcome this inconsistency, we need to embed GR (or its generalizations) within some quantum mechanical framework, i.e., quantum gravity. For the metric ansatz (8), it is customary to choose the fluid to be at rest because the spacetime is static, i.e., ua=u0δ0au^{a}=u^{0}\delta^{a}_{0}. Then, using the normalization condition of the four-velocity, that is, uaua=1u^{a}u_{a}=1, one can find u0=eν/2u^{0}=e^{-\nu/2}. Thus, we find from (27) that for this choice of observers the heat flux vanishes (qa=0q^{a}=0), which is expected from the symmetries of the metric. Also, under the latter choice of observers, the remaining physical variables ρ,p\rho,p and πab\pi_{ab} that follow from (25), (26) and (28) are

ρ=eνT0,p=13(eλT1+2r2T2),\displaystyle\rho=e^{-\nu}T_{0}\,,\qquad p=\frac{1}{3}\left(e^{-\lambda}T_{1}+\frac{2}{r^{2}}T_{2}\right)\,, (29)
π00=0,π11=23(T1eλr2T2),π22=r22eλπ11,π33=sin2θπ22.\displaystyle\pi_{00}=0,\quad\pi_{11}=\frac{2}{3}\left(T_{1}-\frac{e^{\lambda}}{r^{2}}T_{2}\right)\,,\quad\pi_{22}=-\frac{r^{2}}{2}e^{-\lambda}\pi_{11}\,,\quad\pi_{33}=\sin^{2}\theta\,\pi_{22}\,. (30)

If the choice of observers as a timelike four-vector field ua=u0δ0au^{a}=u^{0}\delta^{a}_{0} is not appropriate for any reason, then we must apply the normalization condition of the four-velocity with uaua=1u^{a}u_{a}=-1 by choosing the four-velocity as ua=u1δ1au^{a}=u^{1}\delta^{a}_{1} which is a spacelike four-vector, since one can always normalize the four-velocity to ±1\pm 1. Then we find for the metric (8) that u1=eλ/2u^{1}=e^{-\lambda/2}. For the spacelike four-vector, the projection tensor has the form hab=gab+uaubh_{ab}=g_{ab}+u_{a}u_{b}. Then this choice effects the mass density, the isotropic pressure pp and the traceless stress tensor πab\pi_{ab} due to Eqs. (25), (26) and (28) such that

ρ=eλT1,p=13(eνT02r2T2),\displaystyle\rho=e^{-\lambda}T_{1}\,,\qquad p=-\frac{1}{3}\left(e^{-\nu}T_{0}-\frac{2}{r^{2}}T_{2}\right)\,, (31)
π11=0,π00=23(T0+eνr2T2),π22=r22eνπ00,π33=sin2θπ22.\displaystyle\pi_{11}=0,\quad\pi_{00}=\frac{2}{3}\left(T_{0}+\frac{e^{\nu}}{r^{2}}T_{2}\right)\,,\quad\pi_{22}=\frac{r^{2}}{2}e^{-\nu}\pi_{00}\,,\quad\pi_{33}=\sin^{2}\theta\,\pi_{22}\,. (32)

In GR, it is conventional to restrict the possible energy-momentum tensors by imposing energy conditions. The energy conditions for the energy-momentum tensor TabT_{ab} to represent some known matter fields are the conditions that are coordinate-independent restrictions on TabT_{ab}. In literature, there are five categories of the energy conditions. These are the trace energy conditions (TEC), null energy conditions (NEC), weak energy conditions (WEC), strong energy conditions (SEC) and dominant energy conditions (DEC). The TEC means that the trace of the energy-momentum tensor T=gabTabT=g^{ab}T_{ab} should always be positive (or negative depending on the signature of metric). The NEC mathematically states that Tabkakb0T_{ab}k^{a}k^{b}\geq 0 for any null vector kak^{a}, i.e., kaka=k^{a}k_{a}=. On the other hand, the WEC requires that Tabtatb0T_{ab}t^{a}t^{b}\geq 0 for all timelike vectors tat^{a}. The SEC states that Tabtatb12TtctcT_{ab}t^{a}t^{b}\geq\frac{1}{2}Tt^{c}t_{c} for all timelike vectors tat^{a}. The DEC includes the WEC, as well as the additional requirement that TabtbT^{ab}t_{b} is a non-spacelike vector, i.e., TabTcbtatc0T_{ab}T^{b}_{\,\,c}t^{a}t^{c}\leq 0 carrol ; capo2015 . For a perfect fluid, the energy conditions are described as

TEC:ρ3p0;NEC:ρ+p0;WEC:ρ0,ρ+p0;\displaystyle{\rm TEC:}\quad\rho-3p\geq 0\,;\qquad{\rm NEC:}\quad\rho+p\geq 0\,;\qquad{\rm WEC:}\quad\rho\geq 0\,,\,\,\rho+p\geq 0\,; (33)
SEC:ρ+3p0,ρ+p0;DEC:ρ0,pρ.\displaystyle{\rm SEC:}\quad\rho+3p\geq 0\,,\,\,\rho+p\geq 0;\quad{\rm DEC:}\quad\rho\geq 0\,,\,\,\mid p\mid\leq\rho\,. (34)

Thus it is seen that the energy conditions are simple constraints on various linear combinations of the energy density ρ\rho and the pressure pp. The matter including both positive energy density and positive pressure, which is called ”normal” matter, satisfies all the standard energy conditions. On the contrary, ”exotic” matter violates any one of the energy conditions. For example, the SEC is satisfied by ”all known forms of energy”, but not by the dark energy where p=ρp=-\rho. We note that for the static and spherically symmetric metric (8) one can find the null vector as ka=eν/2δ0a+eλ/2δ1ak^{a}=e^{-\nu/2}\delta^{a}_{0}+e^{-\lambda/2}\delta^{a}_{1} by using the condition kaka=0k^{a}k_{a}=0. Then the NEC for the general anisotropic fluid (24) becomes

ρ+3p2r2T20,\rho+3p-\frac{2}{r^{2}}T_{2}\geq 0\,, (35)

which yields the perfect fluid energy condition taking T2=pr2T_{2}=pr^{2}.

For the perfect fluid, it is easily seen that T0=ρeν,T1=peλ,T2=pr2T_{0}=\rho e^{\nu},\,\,T_{1}=pe^{\lambda},\,\,T_{2}=pr^{2}, i.e., T2=r2eλT1T_{2}=r^{2}e^{-\lambda}T_{1}, and T3=sin2θT2T_{3}=sin^{2}\theta T_{2}, which yields πab=0\pi_{ab}=0 as it should be, and these give rise to the following relations

R0=eν2(ρ+3p),R1=eλ2(ρp),R2=r22(ρp),R_{0}=\frac{e^{\nu}}{2}(\rho+3p),\quad R_{1}=\frac{e^{\lambda}}{2}(\rho-p),\quad R_{2}=\frac{r^{2}}{2}(\rho-p), (36)

where ρ\rho and pp are, respectively, density and pressure of the fluid, which are

ρ=eλr2(rλ+eλ1),p=eλr2(rνeλ+1).\rho=\frac{e^{-\lambda}}{r^{2}}\left(r\lambda^{\prime}+e^{\lambda}-1\right),\qquad p=\frac{e^{-\lambda}}{r^{2}}\left(r\nu^{\prime}-e^{\lambda}+1\right). (37)

Further, the condition of isotropy of the pressure for the perfect fluid matter yields that

2ν′′+ν2νλ2r(ν+λ)+4r2(eλ1)=0,2\nu^{\prime\prime}+\nu^{\prime 2}-\nu^{\prime}\lambda^{\prime}-\frac{2}{r}\left(\nu^{\prime}+\lambda^{\prime}\right)+\frac{4}{r^{2}}\left(e^{\lambda}-1\right)=0, (38)

which is equivalent with

r2R1eλR2=0.r^{2}R_{1}-e^{\lambda}R_{2}=0\,. (39)

In this case the energy conditions for a barotropic equation of state p=p(ρ)p=p(\rho) are given by

ρ>0,0pρ,0dpdρ1.\rho>0,\qquad 0\leq p\leq\rho,\qquad 0\leq\frac{dp}{d\rho}\leq 1\,. (40)

The linear form of a barotropic equation of state is given by p=wρp=w\rho where ww is the equation of state parameter. For w=0,1/3w=0,1/3 and 11, we get dust, incoherent radiation and stiff matter, respectively.

For the spherically symmetric static spacetimes (8), Eq. (7) takes the form:

TiX1+2TiX,ii=2Tiψ,i=0,1,2\displaystyle T^{\prime}_{i}X^{1}+2T_{i}X^{i}_{,i}=2T_{i}\psi,\,\,\ i=0,1,2 (41)
T2(X,22cotθX2X,33)=0,\displaystyle T_{2}\left(X^{2}_{,2}-\cot\theta X^{2}-X^{3}_{,3}\right)=0, (42)
T0X,j0+TjX,0j=0,j=1,2,3\displaystyle T_{0}X^{0}_{,j}+T_{j}X^{j}_{,0}=0,\,\,\,\ j=1,2,3 (43)
TjX,kj+TkX,jk=0,j,k=1,2,3(jk)\displaystyle T_{j}X^{j}_{,k}+T_{k}X^{k}_{,j}=0,\,\,\,j,k=1,2,3\,\,\,(j\neq k) (44)

In the above equations the summation convention is not assumed. For the non-degenerate energy-momentum tensor TabT_{ab}, after some tedious calculations similar to those done in Ref. crcsph we see that the general solution of equations (41)-(44) can be written as

X0\displaystyle X^{0} =\displaystyle= T2T0[sinθ(A1sinϕA2cosϕ)+A3cosθ]+A4(t,r),\displaystyle\frac{T_{2}}{T_{0}}\left[\sin\theta\left(A^{\prime}_{1}\sin\phi-A^{\prime}_{2}\cos\phi\right)+A^{\prime}_{3}\cos\theta\right]+A_{4}(t,r), (45)
X1\displaystyle X^{1} =\displaystyle= T2T1[sinθ(A1sinϕA2cosϕ)+A3cosθ]+A5(t,r),\displaystyle\frac{T_{2}}{T_{1}}\left[\sin\theta\left(A^{\prime}_{1}\sin\phi-A^{\prime}_{2}\cos\phi\right)+A^{\prime}_{3}\cos\theta\right]+A_{5}(t,r), (46)
X2\displaystyle X^{2} =\displaystyle= cosθ[A1sinϕA2cosϕ]+A3sinθ+a1sinϕa2cosϕ,\displaystyle-\cos\theta\left[A_{1}\sin\phi-A_{2}\cos\phi\right]+A_{3}\sin\theta+a_{1}\sin\phi-a_{2}\cos\phi, (47)
X3\displaystyle X^{3} =\displaystyle= cscθ[A1cosϕ+A2sinϕ]+cotθ(a1cosϕ+a2sinϕ)+a3,\displaystyle-csc\theta\left[A_{1}\cos\phi+A_{2}\sin\phi\right]+\cot\theta(a_{1}\cos\phi+a_{2}\sin\phi)+a_{3}, (48)

with the conformal function given by

ψ=(T22T2A1+A1)sinθsinϕ(T22T2A2+A2)sinθcosϕ\displaystyle\psi=\left(\frac{T^{\prime}_{2}}{2T_{2}}A^{\prime}_{1}+A_{1}\right)\sin\theta\sin\phi-\left(\frac{T^{\prime}_{2}}{2T_{2}}A^{\prime}_{2}+A_{2}\right)\sin\theta\cos\phi
+(T22T2A3+A3)cosθ+T22T2A5(t,r),\displaystyle\qquad+\left(\frac{T^{\prime}_{2}}{2T_{2}}A^{\prime}_{3}+A_{3}\right)\cos\theta+\frac{T^{\prime}_{2}}{2T_{2}}A_{5}(t,r), (49)

where A(t,r),(=1,2,3,4,5)A_{\ell}(t,r),\,(\ell=1,2,3,4,5), are integration functions and aj(j=1,2,3)a_{j}\,(j=1,2,3) are constant parameters, which give the three KVs of spherically symmetric spacetimes

𝐗𝟏=sinϕθ+cosϕcotθϕ,𝐗𝟐=cosϕθsinϕcotθϕ,𝐗𝟑=ϕ.\displaystyle\mathbf{X_{1}}=\sin\phi\,\partial_{\theta}+\cos\phi\cot\theta\,\partial_{\phi},\quad\mathbf{X_{2}}=\cos\phi\,\partial_{\theta}-\sin\phi\cot\theta\,\partial_{\phi},\quad\mathbf{X_{3}}=\partial_{\phi}. (50)

Further, the functions A(t,r)A_{\ell}(t,r) in the above equations (45)-(49) are subject to the following constraint equations

(T1T2A5)=0,\displaystyle\left(\sqrt{\frac{T_{1}}{T_{2}}}A_{5}\right)^{\prime}=0, (51)
T0A4+T1A5˙=0,\displaystyle T_{0}A_{4}^{\prime}+T_{1}\dot{A_{5}}=0, (52)
2A4˙+(T0T0T2T2)A5=0,\displaystyle 2\dot{A_{4}}+\left(\frac{T^{\prime}_{0}}{T_{0}}-\frac{T^{\prime}_{2}}{T_{2}}\right)A_{5}=0, (53)
(T2T0A˙j)=0,\displaystyle\left(\sqrt{\frac{T_{2}}{T_{0}}}\dot{A}_{j}\right)^{\prime}=0, (54)
A¨j+T02T1(T0T0T2T2)AjT0T2Aj=0,\displaystyle\ddot{A}_{j}+\frac{T_{0}}{2T_{1}}\left(\frac{T^{\prime}_{0}}{T_{0}}-\frac{T^{\prime}_{2}}{T_{2}}\right)A^{\prime}_{j}-\frac{T_{0}}{T_{2}}A_{j}=0, (55)
Aj′′+12(T2T2T1T1)AjT1T2Aj=0,\displaystyle A^{\prime\prime}_{j}+\frac{1}{2}\left(\frac{T^{\prime}_{2}}{T_{2}}-\frac{T^{\prime}_{1}}{T_{1}}\right)A^{\prime}_{j}-\frac{T_{1}}{T_{2}}A_{j}=0, (56)

where the dot represents the derivative with respect to time tt, and j=1,2,3j=1,2,3. When we solve the above constraint equations for possible cases of non-degenerate energy-momentum tensor TabT_{ab}, we obtain the corresponding CMCs for the spherically symmetric static spacetimes (see Section IV). In the following section we find CMCs for the degenerate energy-momentum tensor of the spherically symmetric static spacetimes.

III Conformal Matter Collineations for the Degenerate Matter Tensor

If the energy-momentum tensor is degenerate, that is, det(Tab)=0det(T_{ab})=0, then we have the following four possibilities: (D-A1) all of the Ta(a=0,1,2,3)T_{a}\,(a=0,1,2,3) are zero; (D-A2) one of the TaT_{a} is nonzero; (D-A3) two of the TaT_{a} are nonzero; (D-A4) three of the TaT_{a} are nonzero.


Case (D-A1). This case corresponds to the vacuum (such as the Schwarzschild) spacetime in which every vector is a CMC.


Case (D-A2). In this case, we have the subcases such that: (D-A2-i) T00,Tj=0,(j=1,2,3)T_{0}\neq 0,\,\,T_{j}=0,\,\,(j=1,2,3);   (D-A2-ii) T10,Tk=0,(k=0,2,3)T_{1}\neq 0,\,\,T_{k}=0,\,\,(k=0,2,3).

Subcase (D-A2-i). In this subcase we find

X0=X0(t),X1=(ψX˙0)2T0T0,Xα=Xα(xa),\displaystyle X^{0}=X^{0}(t),\quad X^{1}=(\psi-\dot{X}^{0})\frac{2T_{0}}{T^{\prime}_{0}}\,,\quad X^{\alpha}=X^{\alpha}(x^{a})\,, (57)

where T00T^{\prime}_{0}\neq 0 and α=2,3\alpha=2,3. If T0=0T^{\prime}_{0}=0, i.e., T0=cT_{0}=c (a constant), then the CMCs takes the following form

X0=ψ𝑑t+a,Xj=Xj(xa),\displaystyle X^{0}=\int{\psi dt}+a,\quad X^{j}=X^{j}(x^{a}), (58)

where aa is an integration constant. The corresponding Lie algebra of the vector fields in this subcase is infinite dimensional because the vector fields given in (57) and (58) have arbitrary components. For this subcase, using (21)-(23) we have

T0=2R0,R1=eλνR0,R2=r2eνR0,T_{0}=-2R_{0},\quad R_{1}=e^{\lambda-\nu}R_{0},\quad R_{2}=r^{2}e^{-\nu}R_{0}, (59)

which shows that all Ricci tensor components are nonzero, e.g. the Ricci tensor in this subcase is non-degenerate. Furthermore, in this subcase we have the equation of state p=0p=0 (dust) for the perfect fluid.

Subcase (D-A2-ii). Considering the constraints of this subcase, we obtain that

X1=1T1[ψT1𝑑r+a1],Xk=Xk(xa),\displaystyle X^{1}=\frac{1}{\sqrt{\mid T_{1}\mid}}\left[\int{\psi\,\sqrt{\mid T_{1}\mid}dr}+a_{1}\right],\qquad X^{k}=X^{k}(x^{a}), (60)

where a1a_{1} is a constant of integration. Here, we again have infinite dimensional Lie algebra of vector fields. Using (21)-(23), we find

R0=eνλR1,R2=r2eλR1,T1=2R1,R_{0}=e^{\nu-\lambda}R_{1}\,,\qquad R_{2}=-r^{2}e^{-\lambda}R_{1}\,,\qquad T_{1}=2R_{1}\,, (61)

which are independent relations for any form of the energy-momentum tensor. It is obvious that for this case the choice of observers as a timelike four-vector field ua=eν/2δ0au^{a}=e^{-\nu/2}\delta^{a}_{0} is not appropriate, since it gives ρ=0\rho=0 due to the condition T0=0T_{0}=0. Thus we use the spacelike four-velocity ua=eλ/2δ1au^{a}=e^{-\lambda/2}\delta^{a}_{1} of the observers for the metric (8), which implies that Eqs. (31) and (32) give

ρ=eλT1,p=0,πab=0.\rho=e^{-\lambda}T_{1}\,,\quad p=0\,,\quad\pi_{ab}=0\,. (62)

Thus, the dust fluid is also allowed in this subcase. Furthermore, all Ricci tensor components for this subcase are non-zero. This means that the Ricci tensor is non-degenerate even though the matter tensor is degenerate.


Case (D-A3). For this case, the possible subcases are given by (D-A3-i) Tp0,Tq=0,(p=0,1andq=2,3)T_{p}\neq 0,\,\,T_{q}=0,\,\,(p=0,1\,\,{\rm and}\,\,q=2,3) and (D-A3-ii) Tp=0,Tq0T_{p}=0,\,\,T_{q}\neq 0.

Subcase (D-A3-i). Here, by choosing a timelike four-velocity, the conditions T00T1T_{0}\neq 0\neq T_{1} and T2=0=T3T_{2}=0=T_{3} mean that the fluid represents an anisotropic fluid without heat flux, and then the physical quantities become

ρ=eνT0,p=13eλT1,\displaystyle\rho=e^{-\nu}T_{0}\,,\qquad p=\frac{1}{3}e^{-\lambda}T_{1}\,, (63)
π11=2eλp,π22=pr2,π33=pr2sin2θ.\displaystyle\pi_{11}=2e^{\lambda}p\,,\quad\pi_{22}=-pr^{2}\,,\quad\pi_{33}=-pr^{2}\sin^{2}\theta\,. (64)

Using the transformations dr¯=ψT1drd\bar{r}=\psi\sqrt{\mid T_{1}\mid}dr, where r¯=r¯(t,r)\bar{r}=\bar{r}(t,r), one finds

X0=(r¯˙+f˙)ψT0g(t)𝑑r¯,X1=r¯+f(t)T1,Xq=Xq(xa),X^{0}=-\int{\frac{(\dot{\bar{r}}+\dot{f})}{\psi T_{0}}g(t)d\bar{r}},\qquad X^{1}=\frac{\bar{r}+f(t)}{\sqrt{\mid T_{1}\mid}},\qquad X^{q}=X^{q}(x^{a}), (65)

where f(t)f(t) and g(t)g(t) are functions of integration, and the conformal factor ψ\psi has the form

ψ=2T0(r¯+f2T0)[(r¯˙+f˙ψ).dr¯T0g˙(t)].\psi=\frac{2T_{0}}{(\bar{r}+f-2T_{0})}\left[\int{\left(\frac{\dot{\bar{r}}+\dot{f}}{\psi}\right)^{.}\frac{d\bar{r}}{T_{0}}}-\dot{g}(t)\right]. (66)

For this case, it follows from the condition Tq=0T_{q}=0 that

R0=eνλ2r(ν+λ),R1=12r(ν+λ),\displaystyle R_{0}=\frac{e^{\nu-\lambda}}{2r}(\nu+\lambda)^{\prime}\,,\qquad R_{1}=\frac{1}{2r}(\nu+\lambda)^{\prime}\,, (67)
T0+eνλT1=2R0,eνT0eλT1=2r2R2.\displaystyle T_{0}+e^{\nu-\lambda}T_{1}=2R_{0}\,,\qquad e^{-\nu}T_{0}-e^{-\lambda}T_{1}=\frac{2}{r^{2}}R_{2}\,. (68)

We find from Eqs. (63) and (68) that

ρ+3p=2eνR0,ρ3p=2r2R2.\rho+3p=2e^{-\nu}R_{0}\,,\qquad\rho-3p=\frac{2}{r^{2}}R_{2}\,. (69)

These equations show that it should be satisfied the conditions R0eν/2R_{0}\geq e^{\nu}/2 and R20R_{2}\geq 0 in order to be valid the SEC and TEC, respectively. There is only one constraint equation 2ν′′+ν2νλ+2r(νλ)=02\nu^{\prime\prime}+{\nu^{\prime}}^{2}-\nu^{\prime}\lambda^{\prime}+\frac{2}{r}(\nu^{\prime}-\lambda^{\prime})=0 following from the condition T2=0T_{2}=0 for this subcase. Also, all the Ricci tensor components are again non-zero, i.e. det(Rab)0\det(R_{ab})\neq 0, even if the matter tensor is degenerate.

Subcase (D-A3-ii). For this subcase, considering the constraint T0=0=T1T_{0}=0=T_{1}, we find

λ=1r(1eλ)=ν,\lambda^{\prime}=\frac{1}{r}\left(1-e^{\lambda}\right)=-\nu^{\prime}\,, (70)

which yields the following solution

ν(r)=ln(1λ0r),λ(r)=ln(11λ0r),\nu(r)=\ln\left(1-\frac{\lambda_{0}}{r}\right)\,,\qquad\lambda(r)=\ln\left(\frac{1}{1-\frac{\lambda_{0}}{r}}\right)\,, (71)

where λ0\lambda_{0} is an integration constant. The above solution is just the Schwarzschild metric which gives Rab=0R_{ab}=0, i.e., all RiR_{i}’s and TiT_{i}’s vanish identically. Therefore there is a contradiction with the condition T20T3T_{2}\neq 0\neq T_{3} as an assumption that is not possible in this subcase.


Case (D-A4). In this case, the possible subcases are (D-A4-i) T0=0,Tj0,(j=1,2,3)T_{0}=0,\,\,T_{j}\neq 0,\,\,(j=1,2,3) and (D-A4-ii) T1=0,Tk0,(k=0,2,3)T_{1}=0,\,\,T_{k}\neq 0,\,\,(k=0,2,3).

Subcase (D-A4-i). In this subcase, the constraint T0=0T_{0}=0 gives

λ=1r(1eλ),\lambda^{\prime}=\frac{1}{r}\left(1-e^{\lambda}\right)\,, (72)

and

R0=2eν(eλR1+1r2R2),T1=12(R1+4r2eλR2),T2=32r2eλR1R2,\displaystyle R_{0}=-2e^{\nu}\left(e^{-\lambda}R_{1}+\frac{1}{r^{2}}R_{2}\right),\quad T_{1}=-\frac{1}{2}\left(R_{1}+\frac{4}{r^{2}}e^{\lambda}R_{2}\right),\quad T_{2}=-\frac{3}{2}r^{2}e^{-\lambda}R_{1}-R_{2}, (73)

where T1=1r(ν+λ)T_{1}=\frac{1}{r}\left(\nu+\lambda\right)^{\prime}. For this case the choice of a timelike four velocity of the observers is not allowed since T0=0T_{0}=0 gives ρ=0\rho=0. So we need to choose a spacelike four-velocity such as ua=eλ/2δrau^{a}=e^{-\lambda/2}\delta^{a}_{r}. Using Eqs. (31) and (32), this choice gives rise to an anisotropic fluid without heat flux as follows:

ρ=eλT1,p=23r2T2,π00=peλ,π11=0,π22=12pr2,π33=sin2θπ22.\displaystyle\rho=e^{-\lambda}T_{1}\,,\qquad p=\frac{2}{3r^{2}}T_{2}\,,\quad\pi_{00}=pe^{\lambda}\,,\quad\pi_{11}=0\,,\quad\pi_{22}=\frac{1}{2}pr^{2}\,,\quad\pi_{33}=\sin^{2}\theta\,\pi_{22}\,. (74)

Here Eq. (72) has the following solution

λ(r)=ln(11λ1r),\lambda(r)=\ln\left(\frac{1}{1-\frac{\lambda_{1}}{r}}\right)\,, (75)

where λ1\lambda_{1} is a constant of integration. The equations given in (73) are second order ordinary differential equations in terms of ν\nu. Then, using the λ(r)\lambda(r) given by (75) in any of the three equations of (73), one can solve the obtained second order differential equation to find ν(r)\nu(r) as

ν(r)=2ln{ν041λ1r+ν12[r3λ1+32λ11λ1rln(r+r(rλ1)λ12)]},\nu(r)=2\ln\left\{-\frac{\nu_{0}}{4}\sqrt{1-\frac{\lambda_{1}}{r}}+\frac{\nu_{1}}{2}\left[r-3\lambda_{1}+\frac{3}{2}\lambda_{1}\sqrt{1-\frac{\lambda_{1}}{r}}\ln\left(r+\sqrt{r(r-\lambda_{1})}-\frac{\lambda_{1}}{2}\right)\right]\right\}\,, (76)

where ν0\nu_{0} and ν1\nu_{1} are constants of integration. The Ricci scalar of the obtained metric given by (75) and (76) is

R=8ν1ν0r(rλ1)+ν1[2r(r3λ1)+3λ1r(rλ1)ln(r+r(rλ1)λ12)],R=\frac{8\nu_{1}}{-\nu_{0}\sqrt{r(r-\lambda_{1})}+\nu_{1}\left[2r(r-3\lambda_{1})+3\lambda_{1}\sqrt{r(r-\lambda_{1})}\ln\left(r+\sqrt{r(r-\lambda_{1})}-\frac{\lambda_{1}}{2}\right)\right]}\,, (77)

and the RiR_{i}’s and TiT_{i}’s for this solution are

R0=14r2{ν0r(rλ1)+ν1[2r(r3λ1)+3λ1r(rλ1)ln(r+r(rλ1)λ12)]},R_{0}=\frac{1}{4r^{2}}\left\{-\nu_{0}\sqrt{r(r-\lambda_{1})}+\nu_{1}\left[2r(r-3\lambda_{1})+3\lambda_{1}\sqrt{r(r-\lambda_{1})}\ln\left(r+\sqrt{r(r-\lambda_{1})}-\frac{\lambda_{1}}{2}\right)\right]\right\}\,, (78)
R1=0,R2=2ν1r2ν0r(rλ1)ν1[2r(r3λ1)+3λ1r(rλ1)ln(r+r(rλ1)λ12)],R_{1}=0\,,\,R_{2}=\frac{2\nu_{1}r^{2}}{\nu_{0}\sqrt{r(r-\lambda_{1})}-\nu_{1}\left[2r(r-3\lambda_{1})+3\lambda_{1}\sqrt{r(r-\lambda_{1})}\ln\left(r+\sqrt{r(r-\lambda_{1})}-\frac{\lambda_{1}}{2}\right)\right]}\,, (79)
T1=2r(λ1r)R2,T2=R2.T_{1}=\frac{2}{r(\lambda_{1}-r)}R_{2}\,,\qquad T_{2}=-R_{2}\,. (80)

Further, the matter density and pressure that come from Eqs. (74), (75) and (80) are

ρ=2r2T2,p=13ρ,\rho=\frac{2}{r^{2}}T_{2}\,,\qquad p=\frac{1}{3}\rho\,, (81)

which implies that the equation of state parameter ww yields an incohorent radiation, i.e., w=1/3w=1/3. Finally, we conclude that the Ricci tensor is also degenerate in this case.

For this subcase, we obtain that X0=X0(xa)X^{0}=X^{0}(x^{a}) and Xj=Xj(r,θ,ϕ)X^{j}=X^{j}(r,\theta,\phi) where the form of XjX^{j} is the same as in equations (46) - (48), and the constraint equations (51) - (56) yield the following solution

Aj=bjcoshr¯+djsinhr¯,A5=T2T1,A_{j}=b_{j}\cosh\bar{r}+d_{j}\sinh\bar{r},\qquad A_{5}=\ell\sqrt{\frac{T_{2}}{T_{1}}}, (82)

where bj,djb_{j},d_{j} and \ell are constants. In this case, the conformal factor is given by

ψ=(T2,r¯2T2A1,r¯+A1)sinθsinϕ(T2,r¯2T2A2,r¯+A2)sinθcosϕ\displaystyle\psi=\left(\frac{T_{{2},\bar{r}}}{2T_{2}}A_{1,\bar{r}}+A_{1}\right)\sin\theta\sin\phi-\left(\frac{T_{{2},\bar{r}}}{2T_{2}}A_{2,\bar{r}}+A_{2}\right)\sin\theta\cos\phi
+(T2,r¯2T2A3,r¯+A3)cosθ+T2,r¯2T2,\displaystyle\qquad+\left(\frac{T_{2,\bar{r}}}{2T_{2}}A_{3,\bar{r}}+A_{3}\right)\cos\theta+\ell\frac{T_{2,\bar{r}}}{2T_{2}}, (83)

where we have used the transformation dr=(T2/T1)1/2dr¯dr=(T_{2}/T_{1})^{1/2}d\bar{r}. By considering (79) and (80), the latter transformation yields r¯=2ln(r+r(rλ1)λ12)\bar{r}=-\sqrt{2}\ln\left(r+\sqrt{r(r-\lambda_{1})}-\frac{\lambda_{1}}{2}\right). Thus, in addition to the three KVs given in (50), it follows that the remaining CMCs and corresponding conformal factors are

𝐗4=sinhr¯cosθr¯+coshr¯sinθθ,ψ4=cosθ(T2,r¯2T2sinhr¯+coshr¯),\displaystyle\mathbf{X}_{4}=\sinh\bar{r}\cos\,\theta\partial_{\bar{r}}+\cosh\bar{r}\sin\theta\,\partial_{\theta}\,,\quad\psi_{4}=\cos\theta\left(\frac{T_{{2},\bar{r}}}{2T_{2}}\sinh\bar{r}+\cosh\bar{r}\right)\,,
𝐗5=coshr¯cosθr¯+sinhr¯sinθθ,ψ5=cosθ(T2,r¯2T2coshr¯+sinhr¯),\displaystyle\mathbf{X}_{5}=\cosh\bar{r}\cos\theta\,\partial_{\bar{r}}+\sinh\bar{r}\sin\theta\,\partial_{\theta}\,,\quad\psi_{5}=\cos\theta\left(\frac{T_{{2},\bar{r}}}{2T_{2}}\cosh\bar{r}+\sinh\bar{r}\right)\,,
𝐗6=sinhr¯sinθsinϕr¯coshr¯ξ1,ψ6=sinθsinϕ(T2,r¯2T2sinhr¯+coshr¯),\displaystyle\mathbf{X}_{6}=\sinh\bar{r}\sin\theta\sin\phi\,\partial_{\bar{r}}-\cosh\bar{r}\,{\xi_{1}}\,,\quad\psi_{6}=\sin\theta\sin\phi\left(\frac{T_{{2},\bar{r}}}{2T_{2}}\sinh\bar{r}+\cosh\bar{r}\right)\,,
𝐗7=sinhr¯sinθcosϕr¯coshr¯ξ2,ψ7=sinθcosϕ(T2,r¯2T2sinhr¯+coshr¯),\displaystyle\mathbf{X}_{7}=\sinh\bar{r}\sin\theta\cos\phi\,\partial_{\bar{r}}-\cosh\bar{r}\,{\xi_{2}}\,,\quad\psi_{7}=\sin\theta\cos\phi\left(\frac{T_{{2},\bar{r}}}{2T_{2}}\sinh\bar{r}+\cosh\bar{r}\right)\,, (84)
𝐗8=coshr¯sinθsinϕr¯sinhr¯ξ1,ψ8=sinθsinϕ(T2,r¯2T2coshr¯+sinhr¯),\displaystyle\mathbf{X}_{8}=\cosh\bar{r}\sin\theta\sin\phi\,\partial_{\bar{r}}-\sinh\bar{r}\,{\xi_{1}}\,,\quad\psi_{8}=\sin\theta\sin\phi\left(\frac{T_{{2},\bar{r}}}{2T_{2}}\cosh\bar{r}+\sinh\bar{r}\right)\,,
𝐗9=coshr¯sinθcosϕr¯sinhr¯ξ2,ψ9=sinθcosϕ(T2,r¯2T2coshr¯+sinhr¯),\displaystyle\mathbf{X}_{9}=\cosh\bar{r}\sin\theta\cos\phi\,\partial_{\bar{r}}-\sinh\bar{r}\,{\xi_{2}}\,,\quad\psi_{9}=\sin\theta\cos\phi\left(\frac{T_{{2},\bar{r}}}{2T_{2}}\cosh\bar{r}+\sinh\bar{r}\right)\,,
𝐗10=r¯,ψ10=T2,r¯2T2,\displaystyle\mathbf{X}_{10}=\partial_{\bar{r}}\,,\quad\qquad\qquad\qquad\qquad\qquad\quad\,\,\,\psi_{10}=\frac{T_{{2},\bar{r}}}{2T_{2}}\,,
𝐗11=F(t,r,θ,ϕ)t,ψ11=0,\displaystyle\mathbf{X}_{11}=F(t,r,\theta,\phi)\partial_{t}\,,\quad\qquad\qquad\qquad\quad\psi_{11}=0\,,

where F(t,r,θ,ϕ)F(t,r,\theta,\phi) is an arbitrary function, and we have defined ξ1{\xi}_{1} and ξ2{\xi_{2}} as follows

ξ1=cosθsinϕθ+cscθcosϕϕ,ξ2=cosθcosϕθcscθsinϕϕ.\displaystyle{\xi_{1}}=\cos\theta\sin\phi\,\partial_{\theta}+\csc\theta\cos\phi\,\partial_{\phi}\,,\qquad{\xi_{2}}=\cos\theta\cos\phi\,\partial_{\theta}-\csc\theta\sin\phi\,\partial_{\phi}\,. (85)

In order to construct a closed algebra for vector fields (84), we find that F=F(t)F=F(t). Hence we have finite dimensional Lie algebra of CMCs which has the following non-vanishing commutators

[𝐗𝟏,𝐗𝟐]=𝐗𝟑,[𝐗𝟏,𝐗𝟑]=𝐗𝟐,[𝐗𝟏,𝐗𝟒]=𝐗𝟔,[𝐗𝟏,𝐗𝟓]=𝐗𝟖,[𝐗𝟏,𝐗𝟔]=𝐗𝟒,\displaystyle\left[\bf{X}_{1},\bf{X}_{2}\right]=\bf{X}_{3},\qquad\left[\bf{X}_{1},\bf{X}_{3}\right]=-\bf{X}_{2},\quad\left[\bf{X}_{1},\bf{X}_{4}\right]=-\bf{X}_{6},\quad\left[\bf{X}_{1},\bf{X}_{5}\right]=\bf{X}_{8},\qquad\left[\bf{X}_{1},\bf{X}_{6}\right]=\bf{X}_{4},
[𝐗𝟏,𝐗𝟖]=𝐗𝟓,[𝐗𝟐,𝐗𝟑]=𝐗𝟏,[𝐗𝟐,𝐗𝟒]=𝐗𝟕,[𝐗𝟐,𝐗𝟓]=𝐗𝟗,[𝐗𝟐,𝐗𝟕]=𝐗𝟒,\displaystyle\left[\bf{X}_{1},\bf{X}_{8}\right]=\bf{X}_{5},\qquad\left[\bf{X}_{2},\bf{X}_{3}\right]=\bf{X}_{1},\qquad\left[\bf{X}_{2},\bf{X}_{4}\right]=-\bf{X}_{7},\quad\left[\bf{X}_{2},\bf{X}_{5}\right]=-\bf{X}_{9},\quad\left[\bf{X}_{2},\bf{X}_{7}\right]=\bf{X}_{4},
[𝐗𝟐,𝐗𝟗]=𝐗𝟓,[𝐗𝟑,𝐗𝟔]=𝐗𝟕,[𝐗𝟑,𝐗𝟕]=𝐗𝟔,[𝐗𝟑,𝐗𝟖]=𝐗𝟗,[𝐗𝟑,𝐗𝟗]=𝐗𝟖,\displaystyle\left[\bf{X}_{2},\bf{X}_{9}\right]=\bf{X}_{5},\qquad\left[\bf{X}_{3},\bf{X}_{6}\right]=\bf{X}_{7},\qquad\left[\bf{X}_{3},\bf{X}_{7}\right]=-\bf{X}_{6},\quad\left[\bf{X}_{3},\bf{X}_{8}\right]=\bf{X}_{9},\quad\,\,\,\,\left[\bf{X}_{3},\bf{X}_{9}\right]=-\bf{X}_{8},
[𝐗𝟒,𝐗𝟓]=𝐗𝟏𝟎,[𝐗𝟒,𝐗𝟔]=𝐗𝟏,[𝐗𝟒,𝐗𝟕]=𝐗𝟐,[𝐗𝟒,𝐗𝟏𝟎]=𝐗𝟓,[𝐗𝟓,𝐗𝟖]=𝐗𝟏,\displaystyle\left[\bf{X}_{4},\bf{X}_{5}\right]=-\bf{X}_{10},\quad\left[\bf{X}_{4},\bf{X}_{6}\right]=\bf{X}_{1},\qquad\left[\bf{X}_{4},\bf{X}_{7}\right]=\bf{X}_{2},\quad\,\,\,\left[\bf{X}_{4},\bf{X}_{10}\right]=-\bf{X}_{5},\,\,\,\,\left[\bf{X}_{5},\bf{X}_{8}\right]=-\bf{X}_{1}, (86)
[𝐗𝟓,𝐗𝟗]=𝐗𝟐,[𝐗𝟓,𝐗𝟏𝟎]=𝐗𝟒,[𝐗𝟔,𝐗𝟕]=𝐗𝟑,[𝐗𝟔,𝐗𝟖]=𝐗𝟏𝟎,[𝐗𝟔,𝐗𝟏𝟎]=𝐗𝟖,\displaystyle\left[\bf{X}_{5},\bf{X}_{9}\right]=-\bf{X}_{2},\,\,\quad\left[\bf{X}_{5},\bf{X}_{10}\right]=-\bf{X}_{4},\quad\left[\bf{X}_{6},\bf{X}_{7}\right]=-\bf{X}_{3},\,\,\,\left[\bf{X}_{6},\bf{X}_{8}\right]=-\bf{X}_{10},\,\,\,\,\left[\bf{X}_{6},\bf{X}_{10}\right]=-\bf{X}_{8},
[𝐗𝟕,𝐗𝟗]=𝐗𝟏𝟎,[𝐗𝟕,𝐗𝟏𝟎]=𝐗𝟗,[𝐗𝟖,𝐗𝟗]=𝐗𝟑,[𝐗𝟖,𝐗𝟏𝟎]=𝐗𝟔,[𝐗𝟗,𝐗𝟏𝟎]=𝐗𝟕.\displaystyle\left[\bf{X}_{7},\bf{X}_{9}\right]=-\bf{X}_{10},\quad\left[\bf{X}_{7},\bf{X}_{10}\right]=-\bf{X}_{9},\quad\left[\bf{X}_{8},\bf{X}_{9}\right]=\bf{X}_{3},\quad\,\,\left[\bf{X}_{8},\bf{X}_{10}\right]=-\bf{X}_{6},\,\,\,\,\left[\bf{X}_{9},\bf{X}_{10}\right]=-\bf{X}_{7}.

Subcase (D-A4-ii). For this subcase, where T1=0T_{1}=0, we have ν=1r(eλ1)\nu^{\prime}=\frac{1}{r}\left(e^{\lambda}-1\right) and thus T0=1reνλ(λ+ν)T_{0}=\frac{1}{r}e^{\nu-\lambda}\left(\lambda+\nu\right)^{\prime}. Also, T0T_{0} and T2T_{2} in terms of RiR_{i}’s (i=0,1,2i=0,1,2) become

T0=R0+eνλR1,T2=r22(eνR0eλR1),T_{0}=R_{0}+e^{\nu-\lambda}R_{1}\,,\qquad T_{2}=\frac{r^{2}}{2}\left(e^{-\nu}R_{0}-e^{-\lambda}R_{1}\right)\,, (87)

and

R2=r22(eνR0+eλR1).R_{2}=\frac{r^{2}}{2}\left(e^{-\nu}R_{0}+e^{-\lambda}R_{1}\right)\,. (88)

In this subcase one can choose a timelike four-velocity of the observers such that ua=eνδtau^{a}=e^{-\nu}\delta^{a}_{t}, which yields an anisotropic fluid without heat flux as

ρ=eνT0,p=23r2T2,π11=eλp,π22r22p,π33=sin2θπ22.\rho=e^{-\nu}T_{0}\,,\quad p=\frac{2}{3r^{2}}T_{2}\,,\quad\pi_{11}=-e^{\lambda}p\,,\quad\pi_{22}\frac{r^{2}}{2}p\,,\quad\pi_{33}=\sin^{2}\theta\,\pi_{22}\,. (89)

It is interesting to point out that we have a variable equation of state parameter w=23r2eνw=\frac{2}{3\,r^{2}}e^{\nu}, i.e., p=23r2eνρp=\frac{2}{3\,r^{2}}e^{\nu}\rho when T0=T2T_{0}=T_{2}.


If T0=T2T_{0}=T_{2}, it follows from the constraint equations (51)-(56) that AjA_{j} and A4A_{4} have the following solutions

Aj=bjcoshr¯+djsinhr¯,A4=,A_{j}=b_{j}\cosh\bar{r}+d_{j}\sinh\bar{r}\,,\qquad A_{4}=\ell\,, (90)

where bj,djb_{j},d_{j} and \ell are integration constants. Then, the components of the CMC vector field are obtained as

X0=sinθ[A1sinϕA2cosϕ]+A3cosθ+,\displaystyle X^{0}=\sin\theta\left[A^{\prime}_{1}\sin\phi-A^{\prime}_{2}\cos\phi\right]+A^{\prime}_{3}\cos\theta+\ell\,,
X1=2T0T0[ψsinθ(A1sinϕA2cosϕ)A3cosθ],\displaystyle X^{1}=\frac{2T_{0}}{T^{\prime}_{0}}\left[\psi-\sin\theta(A_{1}\sin\phi-A_{2}\cos\phi)-A_{3}\cos\theta\right]\,, (91)
X2=cosθ(A1sinϕA2cosϕ)+A3sinθ+a1sinϕa2cosϕ,\displaystyle X^{2}=-\cos\theta(A_{1}\sin\phi-A_{2}\cos\phi)+A_{3}\sin\theta+a_{1}\sin\phi-a_{2}\cos\phi\,,
X3=cscθ(A1cosϕ+A2sinϕ)+cotθ(a1cosϕ+a2sinϕ)+a3,\displaystyle X^{3}=-\csc\theta(A_{1}\cos\phi+A_{2}\sin\phi)+\cot\theta(a_{1}\cos\phi+a_{2}\sin\phi)+a_{3}\,,

where T00T^{\prime}_{0}\neq 0, and ψ\psi is an arbitrary conformal factor, that is, the component X1X^{1} is an arbitrary function of the coordinates and so we have infinite dimensional algebra of CMCs. If T0T2T_{0}\neq T_{2}, then, in addition to the three KVs given in (50), we have the following CMCs

𝐗𝟒=sinθsinϕG(r)rξ1,ψ4=sinθsinϕG(r),\displaystyle\mathbf{X_{4}}=\sin\theta\sin\phi\,G(r)\,\partial_{r}-{\xi}_{1}\,,\quad\,\,\,\psi_{4}=\sin\theta\sin\phi\,G(r)\,,
𝐗𝟓=sinθcosϕG(r)r¯+ξ2,ψ5=sinθcosϕG(r),\displaystyle\mathbf{X_{5}}=\sin\theta\cos\phi\,G(r)\,\partial_{\bar{r}}+{\xi}_{2}\,,\quad\,\,\,\psi_{5}=\sin\theta\cos\phi\,G(r)\,, (92)
𝐗𝟔=cosθG(r)r¯+sinθθ,ψ6=cosθG(r),\displaystyle\mathbf{X_{6}}=\cos\theta\,G(r)\,\partial_{\bar{r}}+\sin\theta\,\partial_{\theta}\,,\quad\,\,\,\,\psi_{6}=\cos\theta\,G(r)\,,
𝐗𝟕=f˙(t)G(r)r¯+f(t)t,ψ7=f˙(t)[1G(r)],\displaystyle\mathbf{X_{7}}=\dot{f}(t)\,G(r)\,\partial_{\bar{r}}+f(t)\,\partial_{t}\,,\qquad\,\,\psi_{7}=\dot{f}(t)\left[1-G(r)\right]\,,

where G(r)2T0/T2(T0/T2)G(r)\equiv\frac{2T_{0}/T_{2}}{(T_{0}/T_{2})^{\prime}}, (T0/T2)0(T_{0}/T_{2})^{\prime}\neq 0, and f(t)f(t) is an integration function. Thus, we again have an finite dimensional Lie algebra of CMCs, and non-zero commutators of the Lie algebra have the following form:

[𝐗𝟏,𝐗𝟐]=𝐗𝟑,[𝐗𝟏,𝐗𝟑]=𝐗𝟐,[𝐗𝟏,𝐗𝟒]=𝐗𝟔,[𝐗𝟏,𝐗𝟔]=𝐗𝟒,\displaystyle\left[\bf{X}_{1},\bf{X}_{2}\right]=\bf{X}_{3}\,,\qquad\left[\bf{X}_{1},\bf{X}_{3}\right]=-\bf{X}_{2}\,,\,\,\quad\left[\bf{X}_{1},\bf{X}_{4}\right]=\bf{X}_{6},\quad\,\,\,\,\,\left[\bf{X}_{1},\bf{X}_{6}\right]=-\bf{X}_{4}\,,
[𝐗𝟐,𝐗𝟑]=𝐗𝟏,[𝐗𝟐,𝐗𝟓]=𝐗𝟔,[𝐗𝟐,𝐗𝟔]=𝐗𝟓,[𝐗𝟑,𝐗𝟒]=𝐗𝟓,\displaystyle\left[\bf{X}_{2},\bf{X}_{3}\right]=\bf{X}_{1}\,,\qquad\left[\bf{X}_{2},\bf{X}_{5}\right]=\bf{X}_{6}\,,\qquad\left[\bf{X}_{2},\bf{X}_{6}\right]=-\bf{X}_{5}\,,\quad\left[\bf{X}_{3},\bf{X}_{4}\right]=\bf{X}_{5}\,, (93)
[𝐗𝟑,𝐗𝟓]=𝐗𝟒,[𝐗𝟒,𝐗𝟓]=𝐗𝟑,[𝐗𝟒,𝐗𝟔]=𝐗𝟏,[𝐗𝟓,𝐗𝟔]=𝐗𝟐.\displaystyle\left[\bf{X}_{3},\bf{X}_{5}\right]=-\bf{X}_{4}\,,\quad\left[\bf{X}_{4},\bf{X}_{5}\right]=-\bf{X}_{3}\,,\quad\,\,\left[\bf{X}_{4},\bf{X}_{6}\right]=-\bf{X}_{1}\,,\quad\left[\bf{X}_{5},\bf{X}_{6}\right]=-\bf{X}_{2}\,.

IV Conformal Matter Collineations for the Non-degenerate Matter Tensor

In this section, we consider the CMCs in non-degenerate case, i.e. det(Tab)0\det(T_{ab})\neq 0, admitted by the static spherically symmetric spacetimes. Here we consider the following five possibilities of the non-degenerate matter tensor.


Case (ND-A). For this case, where none of the Ta(a=0,1,2,3)T^{\prime}_{a}\,(a=0,1,2,3) is zero, applying the transformation dr=T2/T1dr¯dr=\sqrt{T_{2}/T_{1}}d\bar{r}, we find that the the number of CMCs is fifteen such that there are three minimal KVs given in (50), and the remaining ones are

𝐗4=f1(t)[sinθsinϕh1(r¯)𝐘h2(r¯)ξ1],ψ4=f1(t)H1(r¯)sinθsinϕ,\displaystyle{\bf X}_{4}=f_{1}(t)\left[\sin\theta\sin\phi\,h_{1}(\bar{r}){\bf Y}-h_{2}(\bar{r})\xi_{1}\right]\,,\qquad\quad\,\,\,\psi_{4}=f_{1}(t)H_{1}(\bar{r})\sin\theta\sin\phi\,,
𝐗5=f2(t)[sinθsinϕh1(r¯)𝐘h2(r¯)ξ1],ψ5=f2(t)H1(r¯)sinθsinϕ,\displaystyle{\bf X}_{5}=f_{2}(t)\left[\sin\theta\sin\phi\,h_{1}(\bar{r}){\bf Y}-h_{2}(\bar{r})\xi_{1}\right]\,,\qquad\quad\,\,\,\psi_{5}=f_{2}(t)H_{1}(\bar{r})\sin\theta\sin\phi\,,
𝐗6=1b[sinθsinϕh2(r¯)𝐘h1(r¯)ξ1],ψ6=H2(r¯)bsinθsinϕ,\displaystyle{\bf X}_{6}=\frac{1}{b}\left[\sin\theta\sin\phi\,h_{2}(\bar{r}){\bf Y}-h_{1}(\bar{r})\xi_{1}\right]\,,\qquad\qquad\quad\psi_{6}=\frac{H_{2}(\bar{r})}{b}\sin\theta\sin\phi\,,
𝐗7=f1(t)[sinθcosϕh1(r¯)𝐘+h2(r¯)ξ2],ψ7=f1(t)H1(r¯)sinθcosϕ,\displaystyle{\bf X}_{7}=f_{1}(t)\left[-\sin\theta\cos\phi\,h_{1}(\bar{r}){\bf Y}+h_{2}(\bar{r})\xi_{2}\right]\,,\qquad\,\,\,\psi_{7}=-f_{1}(t)H_{1}(\bar{r})\sin\theta\cos\phi\,,
𝐗8=f2(t)[sinθcosϕh1(r¯)𝐘+h2(r¯)ξ2],ψ8=f2(t)H1(r¯)sinθcosϕ,\displaystyle{\bf X}_{8}=f_{2}(t)\left[-\sin\theta\cos\phi\,h_{1}(\bar{r}){\bf Y}+h_{2}(\bar{r})\xi_{2}\right]\,,\qquad\,\,\,\psi_{8}=-f_{2}(t)H_{1}(\bar{r})\sin\theta\cos\phi\,,
𝐗9=1b[sinθcosϕh2(r¯)𝐘+h1(r¯)ξ2],ψ9=H2(r¯)bsinθcosϕ,\displaystyle{\bf X}_{9}=\frac{1}{b}\left[-\sin\theta\cos\phi\,h_{2}(\bar{r}){\bf Y}+h_{1}(\bar{r})\xi_{2}\right]\,,\qquad\qquad\psi_{9}=-\frac{H_{2}(\bar{r})}{b}\sin\theta\cos\phi\,, (94)
𝐗10=f1(t)[cosθh1(r¯)𝐘+h2(r¯)sinθθ],ψ10=f1(t)H1(r¯)cosθ,\displaystyle{\bf X}_{10}=f_{1}(t)\left[\cos\theta\,h_{1}(\bar{r}){\bf Y}+h_{2}(\bar{r})\sin\theta\,\partial_{\theta}\right]\,,\qquad\quad\psi_{10}=f_{1}(t)H_{1}(\bar{r})\cos\theta\,,
𝐗11=f2(t)[cosθh1(r¯)𝐘+h2(r¯)sinθθ],ψ11=f2(t)H1(r¯)cosθ,\displaystyle{\bf X}_{11}=f_{2}(t)\left[\cos\theta\,h_{1}(\bar{r}){\bf Y}+h_{2}(\bar{r})\sin\theta\,\partial_{\theta}\right]\,,\qquad\quad\psi_{11}=f_{2}(t)H_{1}(\bar{r})\cos\theta\,,
𝐗12=1b[cosθh2(r¯)𝐘+h1(r¯)sinθθ],ψ12=H2(r¯)bcosθ,\displaystyle{\bf X}_{12}=\frac{1}{b}\left[\cos\theta\,h_{2}(\bar{r}){\bf Y}+h_{1}(\bar{r})\sin\theta\,\partial_{\theta}\right]\,,\qquad\qquad\,\,\,\psi_{12}=\frac{H_{2}(\bar{r})}{b}\cos\theta\,,
𝐗13=f¨1sinhr¯ah2(r¯)t+f˙1r¯,ψ13=f˙1T2,r¯2T1T2,\displaystyle{\bf X}_{13}=-\ddot{f}_{1}\frac{\sinh\bar{r}}{a\,h_{2}(\bar{r})}\partial_{t}+\dot{f}_{1}\,\partial_{\bar{r}}\,,\qquad\qquad\qquad\qquad\quad\,\psi_{13}=\dot{f}_{1}\frac{T_{2,\bar{r}}}{2\sqrt{T_{1}T_{2}}}\,,
𝐗14=f¨2sinhr¯ah2(r¯)t+f˙2r¯,ψ14=f˙2T2,r¯2T1T2,\displaystyle{\bf X}_{14}=-\ddot{f}_{2}\frac{\sinh\bar{r}}{a\,h_{2}(\bar{r})}\partial_{t}+\dot{f}_{2}\,\partial_{\bar{r}}\,,\qquad\qquad\qquad\qquad\quad\,\psi_{14}=\dot{f}_{2}\frac{T_{2,\bar{r}}}{2\sqrt{T_{1}T_{2}}}\,,
𝐗15=t,ψ15=0,\displaystyle{\bf X}_{15}=\partial_{t}\,,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,\psi_{15}=0\,,

where H1(r¯),H2(r¯),h1(r¯),h2(r¯),f1(t),f2(t)H_{1}(\bar{r}),H_{2}(\bar{r}),h_{1}(\bar{r}),h_{2}(\bar{r}),f_{1}(t),f_{2}(t) and 𝐘{\bf Y} are defined as follows:

H1(r¯)=h1(r¯)T1T2,r¯2T22+h2(r¯),H2(r¯)=h2(r¯)T1T2,r¯2T22+h1(r¯),\displaystyle H_{1}(\bar{r})=h_{1}(\bar{r})\frac{T_{1}T_{2,\bar{r}}}{2T_{2}^{2}}+h_{2}(\bar{r})\,,\quad H_{2}(\bar{r})=h_{2}(\bar{r})\frac{T_{1}T_{2,\bar{r}}}{2T_{2}^{2}}+h_{1}(\bar{r})\,, (95)
h1(r¯)=asinhr¯+bcoshr¯,h2(r¯)=acoshr¯+bsinhr¯,\displaystyle h_{1}(\bar{r})=a\sinh\bar{r}+b\cosh\bar{r}\,,\qquad h_{2}(\bar{r})=a\cosh\bar{r}+b\sinh\bar{r}\,, (96)
f1(t)={1αsinh(αt),forα2>0,1|α|sin(|α|t),forα2<0,\displaystyle f_{1}(t)=\left\{\begin{array}[]{l}\frac{1}{\alpha}\sinh(\alpha t)\,,\quad\,\,\,{\rm for}\,\,\alpha^{2}>0,\\ \frac{1}{|\alpha|}\sin(|\alpha|t)\,,\quad{\rm for}\,\,\alpha^{2}<0,\end{array}\right. (99)
f2(t)={1αcosh(αt),forα2>0,1|α|cos(|α|t),forα2<0,\displaystyle f_{2}(t)=\left\{\begin{array}[]{l}\frac{1}{\alpha}\cosh(\alpha t)\,,\qquad\,\,{\rm for}\,\,\alpha^{2}>0,\\ -\frac{1}{|\alpha|}\cos(|\alpha|t)\,,\quad{\rm for}\,\,\alpha^{2}<0,\end{array}\right. (102)
𝐘=T1T2T0t+T2T1r=T1T2T0t+r¯,\displaystyle{\bf Y}=\frac{\sqrt{T_{1}T_{2}}}{T_{0}}\partial_{t}+\sqrt{\frac{T_{2}}{T_{1}}}\partial_{r}=\frac{\sqrt{T_{1}T_{2}}}{T_{0}}\partial_{t}+\partial_{\bar{r}}\,, (103)

and T0=h2(r¯)2T2T_{0}=h_{2}(\bar{r})^{2}T_{2}  , aa and bb are integration constants, and α\alpha is a constant of separation such that

α2={a2b2,forα2>0,b2a2,forα2<0.\alpha^{2}=\left\{\begin{array}[]{l}a^{2}-b^{2}\,,\quad{\rm for}\,\,\alpha^{2}>0\,,\\ b^{2}-a^{2}\,,\quad{\rm for}\,\,\alpha^{2}<0\,.\end{array}\right. (104)

For α2=0\alpha^{2}=0, after solving the constraint equations (51)-(56), we find twelve CMCs as follows: the KVs 𝐗1,𝐗2,𝐗3{\bf X}_{1},{\bf X}_{2},{\bf X}_{3} given in (50), and

𝐗4=eβr¯[βsinθsinϕ𝐘ξ1],ψ4=H(r¯)eβr¯sinθsinϕ;𝐗5=c0t𝐗4,ψ5=c0tψ4;\displaystyle{\bf X}_{4}=e^{\beta\bar{r}}\left[\beta\,\sin\theta\sin\phi\,{\bf Y}-\xi_{1}\right]\,,\qquad\psi_{4}=H(\bar{r})e^{\beta\bar{r}}\sin\theta\sin\phi\,;\qquad{\bf X}_{5}=c_{0}\,t\,{\bf X}_{4}\,,\quad\psi_{5}=c_{0}\,t\,\psi_{4}\,;
𝐗6=eβr¯[βsinθcosϕ𝐘+ξ2],ψ6=eβr¯H(r¯)sinθcosϕ;𝐗7=c0t𝐗6,ψ7=c0tψ6;\displaystyle{\bf X}_{6}=e^{\beta\bar{r}}\left[-\beta\,\sin\theta\cos\phi\,{\bf Y}+\xi_{2}\right]\,,\quad\psi_{6}=-e^{\beta\bar{r}}H(\bar{r})\sin\theta\cos\phi\,;\quad{\bf X}_{7}=c_{0}\,t\,{\bf X}_{6}\,,\quad\psi_{7}=c_{0}\,t\,\psi_{6}\,;
𝐗8=eβr¯[βcosθ𝐘+sinθθ],ψ8=eβr¯H(r¯)cosθ;𝐗9=c0t𝐗8,ψ9=c0tψ8;\displaystyle{\bf X}_{8}=e^{\beta\bar{r}}\left[\beta\,\cos\theta\,{\bf Y}+\sin\theta\,\partial_{\theta}\right]\,,\qquad\psi_{8}=e^{\beta\bar{r}}H(\bar{r})\cos\theta\,;\qquad{\bf X}_{9}=c_{0}\,t\,{\bf X}_{8}\,,\quad\psi_{9}=c_{0}\,t\,\psi_{8}\,; (105)
𝐗10=12(e2βr¯βc02βt2)t+tr¯,ψ10=tT2,r¯2T1T2,\displaystyle{\bf X}_{10}=\frac{1}{2}\left(\frac{e^{-2\beta\bar{r}}}{\beta\,c_{0}^{2}}-\beta\,t^{2}\right)\partial_{t}+t\,\partial_{\bar{r}}\,,\qquad\psi_{10}=\frac{t\,T_{2,\bar{r}}}{2\sqrt{T_{1}T_{2}}}\,,
𝐗11=βtt+r¯,ψ11=T2,r¯2T1T2;𝐗12=t,ψ12=0,\displaystyle{\bf X}_{11}=-\beta t\,\partial_{t}+\partial_{\bar{r}}\,,\quad\psi_{11}=\frac{T_{2,\bar{r}}}{2\sqrt{T_{1}T_{2}}}\,;\quad{\bf X}_{12}=\partial_{t}\,,\quad\ \psi_{12}=0\,,

where H(r¯)H(\bar{r}) is defined as

H(r¯)=βT1T2,r¯2T22+1,H(\bar{r})=\beta\frac{T_{1}T_{2,\bar{r}}}{2T_{2}^{2}}+1\,, (106)

and T0=c02e2βr¯T2T_{0}=c^{2}_{0}e^{2\beta\bar{r}}T_{2}, T1=(c1e2βr¯+β)T2T_{1}=(c_{1}e^{-2\beta\bar{r}}+\beta)T_{2}, β\beta is a separation constant such that β=±1\beta=\pm 1, c0c_{0} and c1c_{1} are integration constants.


Case (ND-B). Three of the TaT^{\prime}_{a} are zero. In this case we have the possibilities: (ND-B-i) T00,Tj=0,(j=1,2,3)T^{\prime}_{0}\neq 0,\,\,T^{\prime}_{j}=0,\,\,(j=1,2,3) and (ND-B-ii) T10,Tk=0,(k=0,2,3)T^{\prime}_{1}\neq 0,\,\,T^{\prime}_{k}=0,\,\,(k=0,2,3).

Subcase (ND-B-i). For this subcase we have 15 CMCs which are the same form as (94) together with the KVs 𝐗1,𝐗2,𝐗3{\bf X}_{1},{\bf X}_{2},{\bf X}_{3} given in (50), under the transformations r¯kr\bar{r}\rightarrow k\,r, h1(r¯)h1(r)=asinh(kr)+bcosh(kr)h_{1}(\bar{r})\rightarrow h_{1}(r)=a\sinh(kr)+b\cosh(kr), h2(r¯)h2(r)=acosh(kr)+bsinh(kr)h_{2}(\bar{r})\rightarrow h_{2}(r)=a\cosh(kr)+b\sinh(kr), H1(r¯)h2(r)H_{1}(\bar{r})\rightarrow h_{2}(r) and H2(r¯)h1(r)H_{2}(\bar{r})\rightarrow h_{1}(r), where k=c1/c2k=c_{1}/c_{2}, and a,b,c1,c2a,b,c_{1},c_{2} are non-zero constants, and

T1=±c12,T2=±c22,α2=±(a2b2)/c22,T0=[acosh(kr)+bsinh(kr)]2.\displaystyle T_{1}=\pm c_{1}^{2},\quad\,\,T_{2}=\pm c_{2}^{2},\quad\,\,\alpha^{2}=\pm(a^{2}-b^{2})/c_{2}^{2}\,,\qquad T_{0}=\left[a\cosh(k\,r)+b\sinh(k\,r)\right]^{2}\,. (107)

Here we note that the vector fields 𝐗13,𝐗14{\bf X}_{13},{\bf X}_{14} and 𝐗15{\bf X}_{15} are MCs since the scale factors for those are zero, i.e., ψ13=0,ψ14=0\psi_{13}=0,\psi_{14}=0 and ψ15=0\psi_{15}=0.

When α2=0\alpha^{2}=0, one finds that there are 15 CMCs which are the KVs 𝐗1,𝐗2,𝐗3{\bf X}_{1},{\bf X}_{2},{\bf X}_{3} given in (50) and

𝐗4=βK(t,r)sinθsinϕ𝐘K+(t,r)ξ1,ψ4=K+(t,r)sinθsinϕ,\displaystyle{\bf X}_{4}=\beta\,K_{{}_{-}}(t,r)\,\sin\theta\sin\phi\,{\bf Y}-K_{+}(t,r)\,\xi_{1}\,,\qquad\quad\,\,\,\psi_{4}=K_{+}(t,r)\sin\theta\sin\phi\,,
𝐗5=c0βc1eβrt[sinθsinϕ𝐘ξ1],ψ5=c0βc1eβrtsinθsinϕ,\displaystyle{\bf X}_{5}=\frac{c_{0}\,\beta}{c_{1}}e^{\beta\,r}t\left[\sin\theta\sin\phi\,{\bf Y}-\xi_{1}\right]\,,\qquad\qquad\qquad\quad\psi_{5}=\frac{c_{0}\,\beta}{c_{1}}e^{\beta\,r}t\,\sin\theta\sin\phi\,,
𝐗6=eβr[sinθsinϕ𝐘ξ1],ψ6=eβrsinθsinϕ,\displaystyle{\bf X}_{6}=e^{\beta\,r}\left[\sin\theta\sin\phi\,{\bf Y}-\xi_{1}\right]\,,\qquad\qquad\qquad\qquad\quad\,\psi_{6}=e^{\beta\,r}\sin\theta\sin\phi\,,
𝐗7=βK(t,r)sinθcosϕ𝐘+K+(t,r)ξ2,ψ7=K+(t,r)sinθcosϕ,\displaystyle{\bf X}_{7}=-\beta\,K_{{}_{-}}(t,r)\sin\theta\cos\phi\,{\bf Y}+K_{+}(t,r)\xi_{2}\,,\qquad\,\,\,\,\,\psi_{7}=-K_{+}(t,r)\sin\theta\cos\phi\,,
𝐗8=c0βc1eβrt[sinθcosϕ𝐘+ξ2],ψ8=c0βc1eβrtsinθcosϕ,\displaystyle{\bf X}_{8}=\frac{c_{0}\,\beta}{c_{1}}e^{\beta\,r}t\left[-\sin\theta\cos\phi\,{\bf Y}+\xi_{2}\right]\,,\qquad\qquad\quad\,\,\,\,\,\psi_{8}=-\frac{c_{0}\,\beta}{c_{1}}e^{\beta\,r}t\sin\theta\cos\phi\,,
𝐗9=eβr[sinθcosϕ𝐘+ξ2],ψ9=eβrsinθcosϕ,\displaystyle{\bf X}_{9}=e^{\beta\,r}\left[-\sin\theta\cos\phi\,{\bf Y}+\xi_{2}\right]\,,\qquad\qquad\qquad\quad\,\,\,\,\,\,\psi_{9}=-e^{\beta\,r}\sin\theta\cos\phi\,, (108)
𝐗10=βK(t,r)cosθ𝐘+K+(t,r)sinθθ,ψ10=K+(t,r)cosθ,\displaystyle{\bf X}_{10}=\beta\,K_{{}_{-}}(t,r)\cos\theta\,{\bf Y}+K_{+}(t,r)\sin\theta\,\partial_{\theta}\,,\qquad\quad\psi_{10}=K_{+}(t,r)\cos\theta\,,
𝐗11=c0βc1eβrt[cosθ𝐘+sinθθ],ψ11=c0βc1eβrtcosθ,\displaystyle{\bf X}_{11}=\frac{c_{0}\,\beta}{c_{1}}e^{\beta\,r}t\left[\cos\theta\,{\bf Y}+\sin\theta\,\partial_{\theta}\right]\,,\qquad\qquad\qquad\,\,\psi_{11}=\frac{c_{0}\,\beta}{c_{1}}e^{\beta\,r}t\,\cos\theta\,,
𝐗12=eβr[cosθ𝐘+sinθθ],ψ12=eβrcosθ,\displaystyle{\bf X}_{12}=e^{\beta\,r}\left[\cos\theta\,{\bf Y}+\sin\theta\,\partial_{\theta}\right]\,,\qquad\qquad\qquad\qquad\,\,\,\psi_{12}=e^{\beta\,r}\cos\theta\,,
𝐗13=12(T1e2βrβc02βt2)t+tr,ψ13=0,\displaystyle{\bf X}_{13}=\frac{1}{2}\left(\frac{T_{1}\,e^{-2\beta\,r}}{\beta\,c_{0}^{2}}-\beta\,t^{2}\right)\partial_{t}+t\,\partial_{r}\,,\qquad\qquad\quad\,\,\psi_{13}=0\,,
𝐗14=βtt+r,ψ14=0,\displaystyle{\bf X}_{14}=-\beta\,t\,\partial_{t}+\partial_{r}\,,\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\psi_{14}=0\,,
𝐗15=t,ψ15=0,\displaystyle{\bf X}_{15}=\partial_{t}\,,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,\psi_{15}=0\,,

where β\beta is an integration constant, T0=c02e2βr,T1=±c12,β2=T1/T2T_{0}=c_{0}^{2}e^{2\beta r},T_{1}=\pm c_{1}^{2},\beta^{2}=T_{1}/T_{2}, 𝐘{\bf Y} is given in (103), and K±(t,r)K_{\pm}(t,r) is defined as

K±(t,r)=12(c0βc1eβrt2±c1c0βeβr).K_{\pm}(t,r)=\frac{1}{2}\left(\frac{c_{0}\,\beta}{c_{1}}e^{\beta\,r}t^{2}\pm\frac{c_{1}}{c_{0}\,\beta}e^{-\beta\,r}\right)\,. (109)

It is explicitly seen from (108) that in addition to the KVs 𝐗1,𝐗2,𝐗3{\bf X}_{1},{\bf X}_{2},{\bf X}_{3} given in (50), the vector fields 𝐗13,𝐗14{\bf X}_{13},{\bf X}_{14} and 𝐗15{\bf X}_{15} are MCs.

Subcase (ND-B-ii). In this case, where T0,T2T_{0},T_{2} and T3T_{3} are constants, one can easily find that the number of CMCs is six, and these reduce to the MCs which are given by three KVs 𝐗1,𝐗2,𝐗3{\bf X}_{1},{\bf X}_{2},{\bf X}_{3} given in (50), and the remaining ones

𝐗4=t,𝐗5=r¯,𝐗6=r¯ttr¯,\displaystyle{\bf X}_{4}=\partial_{t}\,,\qquad{\bf X}_{5}=\partial_{\bar{r}}\,,\qquad{\bf X}_{6}=\bar{r}\partial_{t}-t\partial_{\bar{r}}\,, (110)

where we have used the rescaling dr¯=T1drd\bar{r}=\sqrt{T_{1}}dr.


Case (ND-C). Two of the TaT^{\prime}_{a} are zero. In this case, the possible subcases are (ND-C-i) Tp0,Tq=0,(p=0,1andq=2,3)\ T^{\prime}_{p}\neq 0,\,\,T^{\prime}_{q}=0,\,\,(p=0,1\,\,{\rm and}\,\,q=2,3) and (ND-C-ii) Tp=0,Tq0T^{\prime}_{p}=0,\,\,T^{\prime}_{q}\neq 0.

Subcase (ND-C-i). For this subcase, where T2,T3T_{2},T_{3} are constants, we find 15 CMCs that are similar to the ones given in (94). Here, the functions f1(t)f_{1}(t) and f2(t)f_{2}(t) are respectively the same form given in Eqs. (99) and (102), and the vector field 𝐘{\bf Y} has the form 𝐘=T1T0t+r¯{\bf Y}=\frac{\sqrt{T_{1}}}{T_{0}}\partial_{t}+\partial_{\bar{r}}. Also, the functions h1(r¯),h2(r¯),H1(r¯)h_{1}(\bar{r}),h_{2}(\bar{r}),H_{1}(\bar{r}) and H2(r¯)H_{2}(\bar{r}) in this subcase have the following form

h1(r¯)=asinh(r¯c2)+bcosh(r¯c2),h2(r¯)=1c2[acosh(r¯c2)+bsinh(r¯c2)],\displaystyle h_{1}(\bar{r})=a\sinh\left(\frac{\bar{r}}{c_{2}}\right)+b\cosh\left(\frac{\bar{r}}{c_{2}}\right)\,,\quad h_{2}(\bar{r})=\frac{1}{c_{2}}\left[a\cosh\left(\frac{\bar{r}}{c_{2}}\right)+b\sinh\left(\frac{\bar{r}}{c_{2}}\right)\right]\,, (111)
H1(r¯)=h2(r¯),H2(r¯)=h1(r¯),\displaystyle H_{1}(\bar{r})=h_{2}(\bar{r})\,,\quad H_{2}(\bar{r})=h_{1}(\bar{r})\,, (112)

with dr=dr¯/|T1|,T0=T2h2(r¯)2,T2=c22dr=d\bar{r}/\sqrt{{|T_{1}|}},T_{0}=T_{2}\,h_{2}(\bar{r})^{2}\,,T_{2}=c_{2}^{2} and α2=±(a2b2)/T2\alpha^{2}=\pm(a^{2}-b^{2})/T_{2}. The CMCs 𝐗13,𝐗14{\bf X}_{13},{\bf X}_{14} and 𝐗15{\bf X}_{15} of this subcase reduce to MCs since the scale factors of these vector fields are zero.

When α2=0\alpha^{2}=0, there are twelve CMCs which are in the form (105), by replacing β𝐘𝐘=T1T0t+r¯\beta{\bf Y}\rightarrow{\bf Y}=\frac{\sqrt{T_{1}}}{T_{0}}\partial_{t}+\partial_{\bar{r}} and H(r¯)1H(\bar{r})\rightarrow 1. For this subcase, the CMCs 𝐗10,𝐗11{\bf X}_{10},{\bf X}_{11} and 𝐗12{\bf X}_{12} become MCs due to ψ10=0=ψ11\psi_{10}=0=\psi_{11} and ψ12=0\psi_{12}=0.

Subcase (ND-C-ii). In this subcase, where T0,T1T_{0},T_{1} are constants, there appears a constant of separation α\alpha that is given by the following constraint equation

T02T1T2(T2T2)=α2.\frac{T_{0}}{2T_{1}\sqrt{T_{2}}}\left(\frac{T_{2}^{\prime}}{\sqrt{T_{2}}}\right)^{\prime}=\alpha^{2}\,. (113)

Then, it follows for α20\alpha^{2}\neq 0 that there are 15 CMCs such that the KVs 𝐗1,𝐗2,𝐗3{\bf X}_{1},{\bf X}_{2},{\bf X}_{3} given in (50) and

𝐗4=T0f3(t)(T2,r¯2sinθsinϕ𝐙+1T2ξ1),ψ4=f3(t)H3(r¯)sinθsinϕ,\displaystyle{\bf X}_{4}=-\sqrt{T_{0}}\,f_{3}(t)\left(\frac{T_{2,\bar{r}}}{2}\sin\theta\sin\phi\,{\bf Z}+\frac{1}{\sqrt{T_{2}}}\xi_{1}\right)\,,\quad\,\,\,\psi_{4}=f_{3}(t)H_{3}(\bar{r})\sin\theta\sin\phi\,,
𝐗5=T0f4(t)(T2,r¯2sinθsinϕ𝐙+1T2ξ1),ψ5=f4(t)H3(r¯)sinθsinϕ,\displaystyle{\bf X}_{5}=-\sqrt{T_{0}}\,f_{4}(t)\left(\frac{T_{2,\bar{r}}}{2}\sin\theta\sin\phi\,{\bf Z}+\frac{1}{\sqrt{T_{2}}}\xi_{1}\right)\,,\quad\,\,\,\psi_{5}=f_{4}(t)H_{3}(\bar{r})\sin\theta\sin\phi\,,
𝐗6=T1bsinθsinϕ𝐙bT2,r¯T2ξ1,ψ6=H4(r¯)sinθsinϕ,\displaystyle{\bf X}_{6}=-T_{1}\sqrt{b}\sin\theta\sin\phi\,{\bf Z}-\frac{bT_{2,\bar{r}}}{\sqrt{T_{2}}}\xi_{1}\,,\qquad\qquad\qquad\quad\psi_{6}=H_{4}(\bar{r})\sin\theta\sin\phi\,,
𝐗7=T0f3(t)(T2,r¯2sinθcosϕ𝐙+1T2ξ2),ψ7=f3(t)H3(r¯)sinθcosϕ,\displaystyle{\bf X}_{7}=\sqrt{T_{0}}\,f_{3}(t)\left(\frac{T_{2,\bar{r}}}{2}\sin\theta\cos\phi\,{\bf Z}+\frac{1}{\sqrt{T_{2}}}\xi_{2}\right)\,,\qquad\,\,\,\psi_{7}=-f_{3}(t)H_{3}(\bar{r})\sin\theta\cos\phi\,,
𝐗8=T0f4(t)(T2,r¯2sinθcosϕ𝐙+1T2ξ2),ψ8=f4(t)H3(r¯)sinθcosϕ,\displaystyle{\bf X}_{8}=\sqrt{T_{0}}\,f_{4}(t)\left(\frac{T_{2,\bar{r}}}{2}\sin\theta\cos\phi\,{\bf Z}+\frac{1}{\sqrt{T_{2}}}\xi_{2}\right)\,,\qquad\,\,\psi_{8}=-f_{4}(t)H_{3}(\bar{r})\sin\theta\cos\phi\,,
𝐗9=T1bsinθcosϕ𝐙+bT2,r¯T2ξ2,ψ9=H4(r¯)sinθcosϕ,\displaystyle{\bf X}_{9}=T_{1}\sqrt{b}\sin\theta\cos\phi\,{\bf Z}+\frac{bT_{2,\bar{r}}}{\sqrt{T_{2}}}\xi_{2}\,,\qquad\qquad\qquad\quad\,\,\,\,\psi_{9}=-H_{4}(\bar{r})\sin\theta\cos\phi\,, (114)
𝐗10=T0f3(t)(T2,r¯2cosθ𝐙+sinθT2θ),ψ10=f3(t)H3(r¯)cosθ,\displaystyle{\bf X}_{10}=\sqrt{T_{0}}\,f_{3}(t)\left(-\frac{T_{2,\bar{r}}}{2}\cos\theta\,{\bf Z}+\frac{\sin\theta}{\sqrt{T_{2}}}\,\partial_{\theta}\right)\,,\qquad\,\,\,\,\psi_{10}=f_{3}(t)H_{3}(\bar{r})\cos\theta\,,
𝐗11=T0f4(t)(T2,r¯2cosθ𝐙+sinθT2θ),ψ11=f4(t)H3(r¯)cosθ,\displaystyle{\bf X}_{11}=\sqrt{T_{0}}\,f_{4}(t)\left(-\frac{T_{2,\bar{r}}}{2}\cos\theta\,{\bf Z}+\frac{\sin\theta}{\sqrt{T_{2}}}\,\partial_{\theta}\right)\,,\qquad\,\,\,\,\psi_{11}=f_{4}(t)H_{3}(\bar{r})\cos\theta\,,
𝐗12=T1bcosθ𝐙+bT2,r¯T2sinθθ,ψ12=H4(r¯)cosθ,\displaystyle{\bf X}_{12}=-T_{1}\sqrt{b}\cos\theta\,{\bf Z}+\frac{bT_{2,\bar{r}}}{\sqrt{T_{2}}}\sin\theta\,\partial_{\theta}\,,\qquad\qquad\qquad\psi_{12}=H_{4}(\bar{r})\cos\theta\,,
𝐗13=(a2f1T1T0f¨1r¯)t+T2f˙1r¯,ψ13=f˙1T2,r¯2,\displaystyle{\bf X}_{13}=\left(\frac{a}{2}f_{1}-\frac{T_{1}}{T_{0}}\ddot{f}_{1}\bar{r}\right)\partial_{t}+T_{2}\dot{f}_{1}\,\partial_{\bar{r}}\,,\qquad\qquad\qquad\,\,\,\,\,\psi_{13}=\dot{f}_{1}\frac{T_{2,\bar{r}}}{2}\,,
𝐗14=(a2f2T1T0f¨2r¯)t+T2f˙2r¯,ψ14=f˙2T2,r¯2,\displaystyle{\bf X}_{14}=\left(\frac{a}{2}f_{2}-\frac{T_{1}}{T_{0}}\ddot{f}_{2}\bar{r}\right)\partial_{t}+T_{2}\dot{f}_{2}\,\partial_{\bar{r}}\,,\qquad\qquad\qquad\,\,\,\,\,\psi_{14}=\dot{f}_{2}\frac{T_{2,\bar{r}}}{2}\,,
𝐗15=t,ψ15=0,\displaystyle{\bf X}_{15}=\partial_{t}\,,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\psi_{15}=0\,,

where H3(r¯),H4(r¯),f3(t),f4(t)H_{3}(\bar{r}),H_{4}(\bar{r}),f_{3}(t),f_{4}(t) and 𝐙{\bf Z} are defined as follows:

H3(r¯)=T0T2(1T2,r¯24T2),H4(r¯)=(bT1bT2)T2,r¯T2,\displaystyle H_{3}(\bar{r})=\sqrt{\frac{T_{0}}{T_{2}}}\left(1-\frac{T_{2,\bar{r}}^{2}}{4T_{2}}\right)\,,\quad H_{4}(\bar{r})=\left(b-\frac{T_{1}\sqrt{b}}{T_{2}}\right)\frac{T_{2,\bar{r}}}{\sqrt{T_{2}}}\,, (115)
f3(t)={1αsin(αt),forα2>0,1|α|sinh(|α|t),forα2<0,\displaystyle f_{3}(t)=\left\{\begin{array}[]{l}\frac{1}{\alpha}\sin(\alpha t)\,,\qquad\,\,\,\,{\rm for}\,\,\alpha^{2}>0,\\ \frac{1}{|\alpha|}\sinh(|\alpha|t)\,,\quad{\rm for}\,\,\alpha^{2}<0,\end{array}\right. (118)
f4(t)={1αcos(αt),forα2>0,1|α|cosh(|α|t),forα2<0,\displaystyle f_{4}(t)=\left\{\begin{array}[]{l}\frac{1}{\alpha}\cos(\alpha t)\,,\qquad\quad\,\,\,{\rm for}\,\,\alpha^{2}>0,\\ -\frac{1}{|\alpha|}\cosh(|\alpha|t)\,,\quad{\rm for}\,\,\alpha^{2}<0,\end{array}\right. (121)
𝐙=1T0t+T2T1r¯.\displaystyle{\bf Z}=\frac{1}{T_{0}}\partial_{t}+\frac{\sqrt{T_{2}}}{T_{1}}\partial_{\bar{r}}\,. (122)

Here, Eq. (113) becomes

T0T2,r¯r¯2T1=α2,\frac{T_{0}T_{2,\bar{r}\bar{r}}}{2T_{1}}=\alpha^{2}\,, (123)

by using the transformation dr=dr¯/|T2|dr=d\bar{r}/\sqrt{|T_{2}|}, and it has a solution T2=α2T1T0r¯2+ar¯+bT_{2}=\frac{\alpha^{2}T_{1}}{T_{0}}\bar{r}^{2}+a\,\bar{r}+b, where aa and bb are integration constants. Further, we have a relation T0=4bα2T1/(a24T1)T_{0}=4b\,\alpha^{2}T_{1}/(a^{2}-4T_{1}) from the constraint equation (56).

If we consider the possibility α2=0\alpha^{2}=0 in (123), it yields T2=ar¯+bT_{2}=a\,\bar{r}+b. Then one can find 15 CMCs which are three KVs 𝐗1,𝐗2,𝐗3{\bf X}_{1},{\bf X}_{2},{\bf X}_{3} given in (50), and the following proper CMCs

𝐗4=T0t(a2sinθsinϕ𝐙+1T2ξ1),ψ4=tH3(r¯)sinθsinϕ;𝐗5=t2𝐗4,ψ5=t2ψ4\displaystyle{\bf X}_{4}=-\sqrt{T_{0}}\,t\left(\frac{a}{2}\sin\theta\sin\phi\,{\bf Z}+\frac{1}{\sqrt{T_{2}}}\xi_{1}\right)\,,\,\,\,\psi_{4}=tH_{3}(\bar{r})\sin\theta\sin\phi\,;\,\,{\bf X}_{5}=\frac{t}{2}{\bf X}_{4}\,,\,\,\psi_{5}=\frac{t}{2}\psi_{4}\,
𝐗6=12asinθsinϕ𝐙1T2ξ1,ψ6=H5(r¯)sinθsinϕ,\displaystyle{\bf X}_{6}=-\frac{1}{2a}\sin\theta\sin\phi\,{\bf Z}-\frac{1}{\sqrt{T_{2}}}\xi_{1}\,,\qquad\qquad\psi_{6}=H_{5}(\bar{r})\sin\theta\sin\phi\,,
𝐗7=T0t(a2sinθcosϕ𝐙+1T2ξ2),ψ7=tH3(r¯)sinθcosϕ;𝐗8=t2𝐗7,ψ8=t2ψ7,\displaystyle{\bf X}_{7}=\sqrt{T_{0}}\,t\left(\frac{a}{2}\sin\theta\cos\phi\,{\bf Z}+\frac{1}{\sqrt{T_{2}}}\xi_{2}\right)\,,\,\,\,\psi_{7}=-tH_{3}(\bar{r})\sin\theta\cos\phi\,;\,\,{\bf X}_{8}=\frac{t}{2}{\bf X}_{7}\,,\,\,\psi_{8}=\frac{t}{2}\psi_{7}\,,
𝐗9=12asinθcosϕ𝐙+1T2ξ2,ψ9=H5(r¯)sinθcosϕ,\displaystyle{\bf X}_{9}=\frac{1}{2a}\sin\theta\cos\phi\,{\bf Z}+\frac{1}{\sqrt{T_{2}}}\xi_{2}\,,\qquad\qquad\psi_{9}=-H_{5}(\bar{r})\sin\theta\cos\phi\,, (124)
𝐗10=T0t(a2cosθ𝐙+sinθT2θ),ψ10=tH3(r¯)cosθ;𝐗11=t2𝐗10,ψ11=t2ψ10,\displaystyle{\bf X}_{10}=\sqrt{T_{0}}\,t\left(-\frac{a}{2}\cos\theta\,{\bf Z}+\frac{\sin\theta}{\sqrt{T_{2}}}\,\partial_{\theta}\right)\,,\,\,\,\psi_{10}=tH_{3}(\bar{r})\cos\theta\,;\,\,{\bf X}_{11}=\frac{t}{2}{\bf X}_{10}\,,\,\,\psi_{11}=\frac{t}{2}\psi_{10}\,,
𝐗12=12acosθ𝐙+sinθT2θ,ψ12=H5(r¯)cosθ,\displaystyle{\bf X}_{12}=-\frac{1}{2a}\cos\theta\,{\bf Z}+\frac{\sin\theta}{\sqrt{T_{2}}}\,\partial_{\theta}\,,\qquad\qquad\psi_{12}=H_{5}(\bar{r})\cos\theta\,,
𝐗13=a2(t22r¯T0)t+2T2atr¯,ψ13=t,\displaystyle{\bf X}_{13}=\frac{a}{2}\left(\frac{t^{2}}{2}-\frac{\bar{r}}{\sqrt{T_{0}}}\right)\partial_{t}+\frac{2T_{2}}{a}t\,\partial_{\bar{r}}\,,\quad\psi_{13}=t\,,
𝐗14=tt+2T2ar¯,ψ14=1;𝐗15=t,ψ15=0,\displaystyle{\bf X}_{14}=t\,\partial_{t}+\frac{2T_{2}}{a}\,\partial_{\bar{r}}\,,\quad\psi_{14}=1\,;\,\,{\bf X}_{15}=\partial_{t}\,,\,\,\,\psi_{15}=0\,,

where r=2aar¯+b,T1=a2/4r=\frac{2}{a}\sqrt{a\,\bar{r}+b},T_{1}=a^{2}/4, and H5(r¯)H_{5}({\bar{r}}) is defined by

H5(r¯)=1T2(1a34T2).\displaystyle H_{5}(\bar{r})=\frac{1}{\sqrt{T_{2}}}\left(1-\frac{a^{3}}{4T_{2}}\right)\ . (125)

Case (ND-D). One of the TaT^{\prime}_{a} is zero. In this case, the possibilities are (ND-D-i) T0=0,Tj0T^{\prime}_{0}=0,\,\,T^{\prime}_{j}\neq 0, (j=1,2,3)(j=1,2,3) and (ND-D-ii) T1=0,Tk0,(k=0,2,3)\ T^{\prime}_{1}=0,\,\,T^{\prime}_{k}\neq 0,\,\,(k=0,2,3).

In the subcase (ND-D-i), if α20\alpha^{2}\neq 0, then one obtains 15 CMCs as given by (50) and (94) in the case (ND-A), in which the differences are the conditions T0=c02T_{0}=c_{0}^{2}, T2=[acoshr¯+bsinhr¯]2T_{2}=[a\cosh\bar{r}+b\sinh\bar{r}]^{-2}, and α2=±T0(a2b2)\alpha^{2}=\pm T_{0}(a^{2}-b^{2}) which is a separation constant. Also, when α2=0\alpha^{2}=0, we find 12 CMCs which are the same form as (105) in the case (ND-A), where the differences are T0=c02T_{0}=c_{0}^{2} and T2=c22e2βr¯T_{2}=c_{2}^{2}e^{-2\beta\bar{r}}.

For the subcase (ND-D-ii), we again find 15 CMCs for α20\alpha^{2}\neq 0, which are the same form as given in (50) and (94), and the differences come from the constraints as T1=c12,T0=T2[acoshr¯+bsinhr¯]2T_{1}=c_{1}^{2},T_{0}=T_{2}\left[a\cosh\bar{r}+b\sinh\bar{r}\right]^{2}, and α2=±(a2b2)/T1\alpha^{2}=\pm(a^{2}-b^{2})/T_{1}. Further, if α2=0\alpha^{2}=0, then we find 12 CMCs which are the same form as in (50) and (105) together with the constraints T1=c12,T0=c02T2e2βr¯T_{1}=c_{1}^{2},T_{0}=c_{0}^{2}T_{2}e^{2\beta\bar{r}} and β2=1\beta^{2}=1.


Case (ND-E). All TaT^{\prime}_{a} are zero. In this case, the constraints are T0=c0,T1=c1,T2=c2T_{0}=c_{0},\,T_{1}=c_{1},\,T_{2}=c_{2}, where c0,c1c_{0},c_{1} and c2c_{2} are non-zero constants. Using these constraints, it follows that in addition to the three KVs given in (50) there are three additional CMCs such as

𝐗4=trc1c0rt,𝐗5=r,𝐗6=t,{\bf X}_{4}=t\,\partial_{r}-\frac{c_{1}}{c_{0}}r\,\partial_{t}\,,\qquad{\bf X}_{5}=\partial_{r}\,,\qquad{\bf X}_{6}=\partial_{t}\,, (126)

with ψ4,ψ5,ψ6=0\psi_{4},\psi_{5},\psi_{6}=0, which means that these CMCs reduce to MCs.

Here, using the constraints T0=c0,T1=c1,T2=c2T_{0}=c_{0},\,T_{1}=c_{1},\,T_{2}=c_{2} in Eqs. (15), (16) and (17), we have the following ordinary differential equations:

λ=1r(1eλ)+c0reλν,\displaystyle\lambda^{\prime}=\frac{1}{r}\left(1-e^{\lambda}\right)+c_{0}\,r\,e^{\lambda-\nu}\,, (127)
ν=1r(eλ1)+c1r,\displaystyle\nu^{\prime}=\frac{1}{r}\left(e^{\lambda}-1\right)+c_{1}\,r\,, (128)
2ν′′+ν2νλ+2r(νλ)4r2c2eλ=0.\displaystyle 2\nu^{\prime\prime}+\nu^{\prime 2}-\nu^{\prime}\lambda^{\prime}+\frac{2}{r}(\nu^{\prime}-\lambda^{\prime})-\frac{4}{r^{2}}c_{2}e^{\lambda}=0\,. (129)

Then, putting λ\lambda^{\prime} and ν\nu^{\prime} given above into (129), yields

e2λν(c1r2+1)eλν+(3c1r24c2)eλc0r2+c1c0(c1r2+1)=0,e^{2\lambda-\nu}-(c_{1}\,r^{2}+1)e^{\lambda-\nu}+(3c_{1}\,r^{2}-4c_{2})\frac{e^{\lambda}}{c_{0}\,r^{2}}+\frac{c_{1}}{c_{0}}(c_{1}r^{2}+1)=0\,, (130)

which gives

eν=c0eλ[eλ(c1r2+1)](4c2r23c1)eλc1(c1r2+1).e^{\nu}=\frac{c_{0}\,e^{\lambda}\left[e^{\lambda}-(c_{1}\,r^{2}+1)\right]}{\left(\frac{4c_{2}}{r^{2}}-3c_{1}\right)e^{\lambda}-c_{1}(c_{1}\,r^{2}+1)}\,. (131)

Thus, the physical variables ρ,p\rho,p and πab\pi_{ab} for this case are

ρ=(4c2r23c1)c1(c1r2+1)eλeλc1r21,p=13(c1eλ+2c2r2),\rho=\frac{\left(\frac{4c_{2}}{r^{2}}-3c_{1}\right)-c_{1}(c_{1}\,r^{2}+1)\,e^{-\lambda}}{e^{\lambda}-c_{1}\,r^{2}-1}\,,\qquad p=\frac{1}{3}\left(c_{1}\,e^{-\lambda}+\frac{2c_{2}}{r^{2}}\right)\,, (132)
π11=23(c1c2r2eλ),π22=r22eλπ11,π33=sin2θπ22,\pi_{11}=\frac{2}{3}\left(c_{1}-\frac{c_{2}}{r^{2}}e^{\lambda}\right)\,,\qquad\pi_{22}=-\frac{r^{2}}{2}e^{-\lambda}\pi_{11}\,,\qquad\pi_{33}=\sin^{2}\theta\pi_{22}\,, (133)

with the choice of timelike observers. Note that the prefect fluid which requires πab=0\pi_{ab}=0 is not allowed in this case since Eqs. (127) - (129) are not equivalently satisfied for eλ(r)=c1r2/c2e^{\lambda(r)}=c_{1}\,r^{2}/c_{2} and ν(r)\nu(r) that comes from (131).

The vector fields for cases (ND-A)-(ND-D) are CMCs of the original metric (8) which are fifteen for α20\alpha^{2}\neq 0 and twelve for α2=0\alpha^{2}=0 in almost all cases. Also, we can employ an anisotropic fluid without heat flux for all cases in this section.

V Conclusions

Symmetries of the metric tensor on a manifold, like KVs, HMs and CKVs have finite dimensional Lie algebras as the metric tensor is always non-degenerate. The maximum dimension for the Killing algebras (in four dimensional space) is 10, for HMs it is 11 and for CKVs 15. But there is no such guarantee for other tensors which can be degenerate as well as non-degenerate. Thus the Lie algebras for RCs, MCs etc. can be finite as well as infinite. When the tensor is non-degenerate the algebra of RCs and MCs is finite and maximum dimension is 10, but for the degenerate tensor it can be finite as well as infinite. For conformal collineations the maximum dimension is 15 for the non-degenerate tensor. For the degenerate case finite dimensionality is not guaranteed.

In this work, we have completely classified CMCs for static spherically symmetric spacetimes which are not of Bertotti-Robinson type. We have seen in Section III that if the energy-momentum tensor is degenerate, i.e., det(Tab)=0\det(T_{ab})=0, then the CMCs have infinite degrees of freedom. In Section IV in which the non-degenerate case where det(Tab)0\det(T_{ab})\neq 0 is considered, there are fifteen CMCs if the separation constant is not zero, or twelve CMCs if the separation constant vanishes. In the cases (ND-B-ii) and (ND-E) of six dimensional Lie algebras the conformal factor comes out to be zero, and thus, they are actually MCs and not CMCs.

We point out that in degenerate case (D-A4-i) we found an exact spherically symmetric solution of the form

ds2=[ν0412Mr+ν12{r6M+3M12Mrln(r+r(r2M)M)}]2dt2\displaystyle ds^{2}=\left[-\frac{\nu_{0}}{4}\sqrt{1-\frac{2M}{r}}+\frac{\nu_{1}}{2}\left\{r-6M+3M\sqrt{1-\frac{2M}{r}}\ln\left(r+\sqrt{r(r-2M)}-M\right)\right\}\right]^{2}dt^{2}
dr212Mrr2(dθ2+sin2θdϕ2),\displaystyle\qquad\qquad-\frac{dr^{2}}{1-\frac{2M}{r}}-r^{2}\left(d\theta^{2}+\sin^{2}\theta\,d\phi^{2}\right)\,, (134)

where we have taken λ1=2M\lambda_{1}=2M. This new metric is radiation-dominated fluid solution for EFEs. It is interesting to note from Eq. (35) that the NEC for the above solution reads p0p\geq 0. For this metric the Ricci scalar becomes

R=8ν1ν0r(r2M)+2ν1[r(r6M)+3Mr(r2M)ln(r+r(r2M)M)].R=\frac{8\nu_{1}}{-\nu_{0}\sqrt{r(r-2M)}+2\nu_{1}\left[r(r-6M)+3M\sqrt{r(r-2M)}\ln\left(r+\sqrt{r(r-2M)}-M\right)\right]}\,. (135)

This metric reduces to the well-known form of Schwarzschild solution if ν0=4\nu_{0}=-4 and ν1=0\nu_{1}=0. The above solution admits eleven CMCs, which are given in (50) and (84), and they have finite dimensional Lie algebra.

For the sake of completeness, we shortly touch on the possible extension of similar studies in generic dimensions. It is well-known that the spherically symmetric metrics continue to exist in space-times of dimension greater than four. However, it will be much more complicated to solve the symmetry equations even for static and spherically symmetric space-times in higher dimensions.

Acknowledgements

KS acknowledges a research grant from the Higher Education Commission of Pakistan.

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