Addendum: A Classification of Plane
Symmetric Kinematic Self-similar
Solutions
M. Sharif ∗and Sehar Aziz †
Department of Mathematics, University of the Punjab,
Quaid-e-Azam Campus, Lahore-54590, Pakistan.
In our recent paper, we classified plane symmetric kinematic self-similar
perfect fluid and dust solutions of the second, zeroth and infinite kinds. How-
ever, we have missed some solutions during the process. In this short com-
munication, we add up those missing solutions. We have found a total of
seven solutions, out of which five turn out to be independent and cannot be
found in the earlier paper.
Keywords: Plane symmetry, Self-similar variable
Recently, we presented a classification of kinematic self-similar plane sym-
metric spacetimes [1]. We have discussed the plane symmetric solutions that
admit kinematic self-similar vectors of the second, zeroth, and infinite kinds
when the perfect fluid is tilted to the fluid flow, parallel or orthogonal. How-
ever, we missed some cases that could provide more solutions. In this ad-
dendum, we present those missing solutions, which turn out to be five in
number. Further, for the the self-similarity of the first kind (tilted), the
two-fluid formalism does not work as the self-similar variable is ξ = x
shall investigate a different approach to obtain the solution in this case. The
tilted perfect fluid yields four more solutions (one first-kind solution, two
2nd-kind solutions and one zeroth-kind solution), the parallel perfect fluid
gives one infinite kind solution, and the orthogonal perfect fluid provides two
∗msharif@math.pu.edu.pk
†sehar aziz@yahoo.com
http://arxiv.org/abs/0704.1930v1
solutions (one first-kind solution and one 2nd-kind solution). Thus, we ob-
tain total seven solutions out of which five solutions are independent. The
plane symmetric metric considered in the paper [1] is the following:
ds2 = e2ν(t,x)dt2 − dx2 − e2λ(t,x)(dy2 + dz2). (1)
We are skipping the details as the procedure can be seen elsewhere [1].
The tilted perfect fluid of the first kind implies that the energy density ρ
and pressure p must take the following forms:
ρ(ξ), (2)
p(ξ), (3)
where the self-similar variable is ξ = x/t. When the Einstein field equations
(EFEs) and the equations of motion for the matter field are satisfied, a set
of ordinary differential equations (ODEs) is obtained, hence, the EFEs and
equations of motion [1] reduce to
ρ̇ = −2λ̇(ρ+ p), (4)
2p− ṗ = ν̇(ρ+ p), (5)
ρ = −4λ̇− 3λ̇2 − 2λ̈− 1, (6)
0 = λ̇2, (7)
0 = λ̈+ λ̇2 + λ̇− λ̇ν̇, (8)
p = 1 + 2λ̇+ λ̇2 + 2ν̇ + 2λ̇ν̇, (9)
0 = 2λ̇ν̇ − 2λ̈− 3λ̇2 − 2λ̇, (10)
p = λ̈+ λ̇2 + λ̇+ λ̇ν̇ + ν̈ + ν̇2, (11)
0 = −λ̈− λ̇2 − λ̇+ λ̇ν̇. (12)
Only the EOS(3) is compatible with this kind. Equations (2) and (3) yield
p = kρ while Eqs. (7) and (4) imply that λ and ρ, respectively, are arbitrary
constants. Also, Eq. (5) gives ν̇ = 2k
. Using this value in the remaining
equations, we obtain the following solution:
ν = ln (c0ξ
2)), λ = c2,
ρ = constant, k = −3 ±
2. (13)
The corresponding metric is
ds2 = (
)(2∓2
2)dt2 − dx2 − x2(dy2 + dz2). (14)
For the self-similarity of the second kind, we obtain solutions only with
the EOS(3), and these solutions are missing in Ref. 1. The energy density ρ
and pressure p can be written as
[ρ1(ξ) +
ρ2(ξ)], (15)
[p1(ξ) +
p2(ξ)], (16)
where the self-similar variable is ξ = x/(αt)
α . A set of ODEs yield
ρ̇1 = −2λ̇(ρ1 + p1), (17)
ρ̇2 + 2αρ2 = −2λ̇(ρ2 + p2), (18)
−ṗ1 + 2p1 = ν̇(ρ1 + p1), (19)
−ṗ2 = ν̇(ρ2 + p2), (20)
ρ1 = −4λ̇− 3λ̇2 − 2λ̈− 1, (21)
α2e2νρ2 = λ̇
2, (22)
0 = λ̈+ λ̇2 + λ̇− λ̇ν̇, (23)
p1 = 1 + 2λ̇+ λ̇
2 + 2ν̇ + 2λ̇ν̇, (24)
α2e2νp2 = −2λ̈− 3λ̇2 − 2αλ̇+ 2λ̇ν̇, (25)
p1 = λ̈+ λ̇
2 + λ̇+ λ̇ν̇ + ν̈ + ν̇2, (26)
α2e2νp2 = −λ̈− λ̇2 − αλ̇+ λ̇ν̇. (27)
Proceeding along the same lines with the EOS(3), as given in Ref. 1, for
k 6= −1 we assume that ρ1 = 0 and that ρ2 is arbitrary. Thus, Eqs. (21),
(23), and (24) show that ν̇ = 0 and λ̇ = −1, and Eq. (18) implies that
α = k + 1. Equations (25) and (27) give α = 2, and we obtain the following
spacetime:
ν = c1, λ = − ln ξ + c2, ρ1 = 0 = p1, ρ2 = constant = p2,
α = 2, k = 1. (28)
The resulting plane symmetric metric becomes
ds2 = dt2 − dx2 − 2t(dy2 + dz2). (29)
When k 6= −1, we take ρ2 = 0, and ρ1 is arbitrary; hence, Eq. (22) implies
that λ̇ = 0. For the first possibility, it follows that
k + 1
ln ξ + c1, λ = c2, p1 = kρ1, ρ1 = constant,
p2 = 0 = ρ2, k = −3± 2
2; (30)
hence, the plane symmetric spacetime will take the following form:
ds2 = (
(αt)1/α
k+1dt2 − dx2 − x2(dy2 + dz2). (31)
For the self-similarity of the zeroth kind, the EFEs show that the quan-
tities ρ and p must be of the form
[ρ1(ξ) + x
2ρ2(ξ)], (32)
[p1(ξ) + x
2p2(ξ)], (33)
where the self-similar variable is ξ = x
. A set of ODEs follows such that
ρ̇1 = −2λ̇(ρ1 + p1), (34)
ρ̇2 = −2λ̇(ρ2 + p2), (35)
−ṗ1 + 2p1 = ν̇(ρ1 + p1), (36)
−ṗ2 = ν̇(ρ2 + p2), (37)
ρ1 = −4λ̇− 3λ̇2 − 2λ̈− 1, (38)
e2νρ2 = λ̇
2, (39)
0 = λ̈+ λ̇2 + λ̇− λ̇ν̇, (40)
p1 = 1 + 2λ̇+ λ̇
2 + 2ν̇ + 2λ̇ν̇, (41)
e2νp2 = 2λ̇ν̇ − 2λ̈− 3λ̇2, (42)
p1 = λ̈+ λ̇
2 + λ̇+ λ̇ν̇ + ν̈ + ν̇2, (43)
e2νp2 = −λ̈− λ̇2 + λ̇ν̇. (44)
For the EOS(3) when k 6= −1, ρ2 = 0, and ρ1 is arbitrary, Eq. (39) yields
λ̇ = 0 while Eqs. (34) and (36) show that ν̇ = 2k
. Finally, we obtain the
same solution as in the case of the second kind with the EOS(3) given by
Eq. (30) with α = 0. The corresponding metric is
ds2 = (xe−t)
k+1dt2 − dx2 − e2t(dy2 + dz2). (45)
For the self-similarity of the first kind in the orthogonal perfect fluid case,
the EFEs and the equations of motion give
e2ν(1 + ρ) = λ′
, (46)
e2ν(3− p) = 3λ′2 + 2λ′′ − 2λ′ν ′, (47)
e2ν(1− p) = λ′′ + λ′2 − λ′ν ′, (48)
2λ′(ρ+ p) = −ρ′, (49)
ρ = p. (50)
Clearly, Eq. (50) shows that this is a system with a stiff fluid. If these
equations are solved simultaneously, Eq. (49) provides the value of λ′, and
Eq. (46) gives the value of ν in terms of p. Equations (47) and (48) impose a
constraint on p, p′
p− 2(1+ p)(p′′p− p′2) = 0, and we arrive at the following
solution:
ν = ln (
(1 + p)
), λ = −
ln (p) + ln (c1), ρ = p. (51)
For the self-similarity of the second kind in the orthogonal perfect fluid
case, the quantities ρ and p must be of the forms
κρ = x−2ρ1(ξ) + x
−2αρ2(ξ), (52)
κp = x−2p1(ξ) + x
−2αp2(ξ). (53)
A set of ODEs gives
ρ′1 = −2λ
′(ρ1 + p1), (54)
ρ′2 = −2λ
′(ρ2 + p2), (55)
2p1 = α(ρ1 + p1), (56)
ρ2 = p2 (57)
ρ1 = −1, (58)
e2νρ2 = λ
′2, (59)
0 = (1− α)λ′, (60)
p1 = 1 + 2α, (61)
e2νp2 = −2λ′′ + 2λ′ν ′ − 3λ′
, (62)
p1 = α
2, (63)
e2νp2 = −λ′′ − λ′
+ λ′ν ′. (64)
Equation (57) represents a stiff fluid, and Eq. (60) gives λ′ = 0; hence, we
obtain the following solution:
ν = arbitrary, λ = c4, p2 = 0 = ρ2, ρ1 = −1,
p1 = 3± 2
2, α = 1±
2. (65)
The corresponding metric is
ds2 = x2(1±
2)dt2 − dx2 − x2(dy2 + dz2). (66)
For the self-similarity of the infinite kind in the parallel perfect fluid, a
set of ODEs is given as
− ρ = 3λ′2 + 2λ′′, (67)
p = λ′
+ 2λ′ν ′, (68)
p = λ′′ + λ′
+ λ′ν ′ + ν ′′ + ν ′
, (69)
−p′ = ν ′(ρ+ p). (70)
Solving Eqs. (67)-(70) with the assumption that p is constant, we find that
λ is a linear function of ξ. Finally, we arrive at the following solution:
ν = c1, λ = c2ξ + c3,
ρ = −3p = constant. (71)
The metric is given by
ds2 = dt2 − dx2 − e2x(dy2 + dz2). (72)
We notice that the solutions given by Eqs.(14), (31) and (66) turn out to
be dependent and the solutions given by Eqs. (29), (45), (51), and (72) are
independent. Thus, we have a total of five independent solutions. It is worth
mentioning here that the self-similar solutions in Eqs. (14), (31), and (66)
correspond to the already classified solutions [3] under particular coordinate
transformations. The metrics given by Eqs. (14), (31), and (66) correspond
to the class of metrics
ds2 = e2ν(x)dt2 − dx2 − e2λ(x)(dy2 + dz2), (73)
which has four Killing vectors admitting G3 ⊗ℜ with a timelike ℜ. We also
note that the density is either zero or positive in all the solutions, except for
the solution given by Eq. (66). The physical properties of all these solutions
can be seen in the Ref. 4.
We would like to mention here that the paper in Ref. 1 focussed on a clas-
sification of plane symmetric kinematic self-similar solutions under certain
assumptions. A complete classification for the most general plane symmetric
kinematic self-similar solutions appears elsewhere [2].
ACKNOWLEDGMENT
One of us (SA) would like to acknowledge Higher Education Commission
(HEC) for the financial support.
References
[1] M. Sharif and S. Aziz: J. Korean Physical Society 49, 21 (2006).
[2] M. Sharif and S. Aziz: Class. Quantum Grav. 24, 605 (2007).
[3] T. Feroze, A. Qadir and M. Ziad: J. Math. Phys. 42, 4947 (2001).
[4] M. Sharif and S. Aziz: J. Korean Physical Society 47, 757 (2005).

  In our recent paper, we classified plane symmetric kinematic self-similar
perfect fluid and dust solutions of the second, zeroth and infinite kinds.
However, we have missed some solutions during the process. In this short
communication, we add up those missing solutions. We have found a total of
seven solutions, out of which five turn out to be independent and cannot be
found in the earlier paper

Addendum: A Classification of Plane
Symmetric Kinematic Self-similar
Solutions
M. Sharif ∗and Sehar Aziz †
Department of Mathematics, University of the Punjab,
Quaid-e-Azam Campus, Lahore-54590, Pakistan.
In our recent paper, we classified plane symmetric kinematic self-similar
perfect fluid and dust solutions of the second, zeroth and infinite kinds. How-
ever, we have missed some solutions during the process. In this short com-
munication, we add up those missing solutions. We have found a total of
seven solutions, out of which five turn out to be independent and cannot be
found in the earlier paper.
Keywords: Plane symmetry, Self-similar variable
Recently, we presented a classification of kinematic self-similar plane sym-
metric spacetimes [1]. We have discussed the plane symmetric solutions that
admit kinematic self-similar vectors of the second, zeroth, and infinite kinds
when the perfect fluid is tilted to the fluid flow, parallel or orthogonal. How-
ever, we missed some cases that could provide more solutions. In this ad-
dendum, we present those missing solutions, which turn out to be five in
number. Further, for the the self-similarity of the first kind (tilted), the
two-fluid formalism does not work as the self-similar variable is ξ = x
shall investigate a different approach to obtain the solution in this case. The
tilted perfect fluid yields four more solutions (one first-kind solution, two
2nd-kind solutions and one zeroth-kind solution), the parallel perfect fluid
gives one infinite kind solution, and the orthogonal perfect fluid provides two
∗msharif@math.pu.edu.pk
†sehar aziz@yahoo.com
http://arxiv.org/abs/0704.1930v1
solutions (one first-kind solution and one 2nd-kind solution). Thus, we ob-
tain total seven solutions out of which five solutions are independent. The
plane symmetric metric considered in the paper [1] is the following:
ds2 = e2ν(t,x)dt2 − dx2 − e2λ(t,x)(dy2 + dz2). (1)
We are skipping the details as the procedure can be seen elsewhere [1].
The tilted perfect fluid of the first kind implies that the energy density ρ
and pressure p must take the following forms:
ρ(ξ), (2)
p(ξ), (3)
where the self-similar variable is ξ = x/t. When the Einstein field equations
(EFEs) and the equations of motion for the matter field are satisfied, a set
of ordinary differential equations (ODEs) is obtained, hence, the EFEs and
equations of motion [1] reduce to
ρ̇ = −2λ̇(ρ+ p), (4)
2p− ṗ = ν̇(ρ+ p), (5)
ρ = −4λ̇− 3λ̇2 − 2λ̈− 1, (6)
0 = λ̇2, (7)
0 = λ̈+ λ̇2 + λ̇− λ̇ν̇, (8)
p = 1 + 2λ̇+ λ̇2 + 2ν̇ + 2λ̇ν̇, (9)
0 = 2λ̇ν̇ − 2λ̈− 3λ̇2 − 2λ̇, (10)
p = λ̈+ λ̇2 + λ̇+ λ̇ν̇ + ν̈ + ν̇2, (11)
0 = −λ̈− λ̇2 − λ̇+ λ̇ν̇. (12)
Only the EOS(3) is compatible with this kind. Equations (2) and (3) yield
p = kρ while Eqs. (7) and (4) imply that λ and ρ, respectively, are arbitrary
constants. Also, Eq. (5) gives ν̇ = 2k
. Using this value in the remaining
equations, we obtain the following solution:
ν = ln (c0ξ
2)), λ = c2,
ρ = constant, k = −3 ±
2. (13)
The corresponding metric is
ds2 = (
)(2∓2
2)dt2 − dx2 − x2(dy2 + dz2). (14)
For the self-similarity of the second kind, we obtain solutions only with
the EOS(3), and these solutions are missing in Ref. 1. The energy density ρ
and pressure p can be written as
[ρ1(ξ) +
ρ2(ξ)], (15)
[p1(ξ) +
p2(ξ)], (16)
where the self-similar variable is ξ = x/(αt)
α . A set of ODEs yield
ρ̇1 = −2λ̇(ρ1 + p1), (17)
ρ̇2 + 2αρ2 = −2λ̇(ρ2 + p2), (18)
−ṗ1 + 2p1 = ν̇(ρ1 + p1), (19)
−ṗ2 = ν̇(ρ2 + p2), (20)
ρ1 = −4λ̇− 3λ̇2 − 2λ̈− 1, (21)
α2e2νρ2 = λ̇
2, (22)
0 = λ̈+ λ̇2 + λ̇− λ̇ν̇, (23)
p1 = 1 + 2λ̇+ λ̇
2 + 2ν̇ + 2λ̇ν̇, (24)
α2e2νp2 = −2λ̈− 3λ̇2 − 2αλ̇+ 2λ̇ν̇, (25)
p1 = λ̈+ λ̇
2 + λ̇+ λ̇ν̇ + ν̈ + ν̇2, (26)
α2e2νp2 = −λ̈− λ̇2 − αλ̇+ λ̇ν̇. (27)
Proceeding along the same lines with the EOS(3), as given in Ref. 1, for
k 6= −1 we assume that ρ1 = 0 and that ρ2 is arbitrary. Thus, Eqs. (21),
(23), and (24) show that ν̇ = 0 and λ̇ = −1, and Eq. (18) implies that
α = k + 1. Equations (25) and (27) give α = 2, and we obtain the following
spacetime:
ν = c1, λ = − ln ξ + c2, ρ1 = 0 = p1, ρ2 = constant = p2,
α = 2, k = 1. (28)
The resulting plane symmetric metric becomes
ds2 = dt2 − dx2 − 2t(dy2 + dz2). (29)
When k 6= −1, we take ρ2 = 0, and ρ1 is arbitrary; hence, Eq. (22) implies
that λ̇ = 0. For the first possibility, it follows that
k + 1
ln ξ + c1, λ = c2, p1 = kρ1, ρ1 = constant,
p2 = 0 = ρ2, k = −3± 2
2; (30)
hence, the plane symmetric spacetime will take the following form:
ds2 = (
(αt)1/α
k+1dt2 − dx2 − x2(dy2 + dz2). (31)
For the self-similarity of the zeroth kind, the EFEs show that the quan-
tities ρ and p must be of the form
[ρ1(ξ) + x
2ρ2(ξ)], (32)
[p1(ξ) + x
2p2(ξ)], (33)
where the self-similar variable is ξ = x
. A set of ODEs follows such that
ρ̇1 = −2λ̇(ρ1 + p1), (34)
ρ̇2 = −2λ̇(ρ2 + p2), (35)
−ṗ1 + 2p1 = ν̇(ρ1 + p1), (36)
−ṗ2 = ν̇(ρ2 + p2), (37)
ρ1 = −4λ̇− 3λ̇2 − 2λ̈− 1, (38)
e2νρ2 = λ̇
2, (39)
0 = λ̈+ λ̇2 + λ̇− λ̇ν̇, (40)
p1 = 1 + 2λ̇+ λ̇
2 + 2ν̇ + 2λ̇ν̇, (41)
e2νp2 = 2λ̇ν̇ − 2λ̈− 3λ̇2, (42)
p1 = λ̈+ λ̇
2 + λ̇+ λ̇ν̇ + ν̈ + ν̇2, (43)
e2νp2 = −λ̈− λ̇2 + λ̇ν̇. (44)
For the EOS(3) when k 6= −1, ρ2 = 0, and ρ1 is arbitrary, Eq. (39) yields
λ̇ = 0 while Eqs. (34) and (36) show that ν̇ = 2k
. Finally, we obtain the
same solution as in the case of the second kind with the EOS(3) given by
Eq. (30) with α = 0. The corresponding metric is
ds2 = (xe−t)
k+1dt2 − dx2 − e2t(dy2 + dz2). (45)
For the self-similarity of the first kind in the orthogonal perfect fluid case,
the EFEs and the equations of motion give
e2ν(1 + ρ) = λ′
, (46)
e2ν(3− p) = 3λ′2 + 2λ′′ − 2λ′ν ′, (47)
e2ν(1− p) = λ′′ + λ′2 − λ′ν ′, (48)
2λ′(ρ+ p) = −ρ′, (49)
ρ = p. (50)
Clearly, Eq. (50) shows that this is a system with a stiff fluid. If these
equations are solved simultaneously, Eq. (49) provides the value of λ′, and
Eq. (46) gives the value of ν in terms of p. Equations (47) and (48) impose a
constraint on p, p′
p− 2(1+ p)(p′′p− p′2) = 0, and we arrive at the following
solution:
ν = ln (
(1 + p)
), λ = −
ln (p) + ln (c1), ρ = p. (51)
For the self-similarity of the second kind in the orthogonal perfect fluid
case, the quantities ρ and p must be of the forms
κρ = x−2ρ1(ξ) + x
−2αρ2(ξ), (52)
κp = x−2p1(ξ) + x
−2αp2(ξ). (53)
A set of ODEs gives
ρ′1 = −2λ
′(ρ1 + p1), (54)
ρ′2 = −2λ
′(ρ2 + p2), (55)
2p1 = α(ρ1 + p1), (56)
ρ2 = p2 (57)
ρ1 = −1, (58)
e2νρ2 = λ
′2, (59)
0 = (1− α)λ′, (60)
p1 = 1 + 2α, (61)
e2νp2 = −2λ′′ + 2λ′ν ′ − 3λ′
, (62)
p1 = α
2, (63)
e2νp2 = −λ′′ − λ′
+ λ′ν ′. (64)
Equation (57) represents a stiff fluid, and Eq. (60) gives λ′ = 0; hence, we
obtain the following solution:
ν = arbitrary, λ = c4, p2 = 0 = ρ2, ρ1 = −1,
p1 = 3± 2
2, α = 1±
2. (65)
The corresponding metric is
ds2 = x2(1±
2)dt2 − dx2 − x2(dy2 + dz2). (66)
For the self-similarity of the infinite kind in the parallel perfect fluid, a
set of ODEs is given as
− ρ = 3λ′2 + 2λ′′, (67)
p = λ′
+ 2λ′ν ′, (68)
p = λ′′ + λ′
+ λ′ν ′ + ν ′′ + ν ′
, (69)
−p′ = ν ′(ρ+ p). (70)
Solving Eqs. (67)-(70) with the assumption that p is constant, we find that
λ is a linear function of ξ. Finally, we arrive at the following solution:
ν = c1, λ = c2ξ + c3,
ρ = −3p = constant. (71)
The metric is given by
ds2 = dt2 − dx2 − e2x(dy2 + dz2). (72)
We notice that the solutions given by Eqs.(14), (31) and (66) turn out to
be dependent and the solutions given by Eqs. (29), (45), (51), and (72) are
independent. Thus, we have a total of five independent solutions. It is worth
mentioning here that the self-similar solutions in Eqs. (14), (31), and (66)
correspond to the already classified solutions [3] under particular coordinate
transformations. The metrics given by Eqs. (14), (31), and (66) correspond
to the class of metrics
ds2 = e2ν(x)dt2 − dx2 − e2λ(x)(dy2 + dz2), (73)
which has four Killing vectors admitting G3 ⊗ℜ with a timelike ℜ. We also
note that the density is either zero or positive in all the solutions, except for
the solution given by Eq. (66). The physical properties of all these solutions
can be seen in the Ref. 4.
We would like to mention here that the paper in Ref. 1 focussed on a clas-
sification of plane symmetric kinematic self-similar solutions under certain
assumptions. A complete classification for the most general plane symmetric
kinematic self-similar solutions appears elsewhere [2].
ACKNOWLEDGMENT
One of us (SA) would like to acknowledge Higher Education Commission
(HEC) for the financial support.
References
[1] M. Sharif and S. Aziz: J. Korean Physical Society 49, 21 (2006).
[2] M. Sharif and S. Aziz: Class. Quantum Grav. 24, 605 (2007).
[3] T. Feroze, A. Qadir and M. Ziad: J. Math. Phys. 42, 4947 (2001).
[4] M. Sharif and S. Aziz: J. Korean Physical Society 47, 757 (2005).
