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 Total Downloads : 2118
 Authors : Purushottam, R.B.S. Yadav
 Paper ID : IJERTV1IS9214
 Volume & Issue : Volume 01, Issue 09 (November 2012)
 Published (First Online): 29112012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Spherically Symmetric SelfGravitating Fluid with Specified Equation of State
by 

Purushottam 
R.B.S. Yadav 

Nalanda College of Engineering 
P.G. Deptt. of Mathematics 

Govt. of Bihar 
Magadh University 

Chandi, Nalanda. 
ABSTRACT 
BodhGaya, Bihar. 
In this paper 
some exact, 
static spherically 
symmetric solution of Einsteins field equations for the perfect fluid with equation of state p=a, where a[0,1] has been obtained taking suitable choice of g11 or g44
(e.g.
e k or
e Arn ). Many previously known
solutions are contained here in as a particular case. Various physical and geometrical properties have been also studied.
Key words: Exact solution, Perfect fluid, Equation of state. Homogenous, Homaloidal.

INTRODUCTION
Perfect fluid spheres with homogeneous density and isotopic pressure in general relativity were firstly considered by Schwarzschild
[10] and the solutions of relativistic field equations were obtained.Tolman [16] developed a mathematical method for solving Einsteins field equations applied to a static fluid sphere in such a manner as to provide explicit solutions in terms of known functions. A number of new
solutions were thus obtained and the properties of three of them were examined in detail.
No stationary inhomogeneous solutions to Einsteins equation for an irrotational perfect fluid have featured equations of state. p= (Letelier [14], Letelier and Tabensky [15] and Singh and Yadav [23]). Solutions to Einsteins equation with a simple equation of state have been found in various cases, e.g. for +3p=constant (Whittaker [7]) for =3p (Klein [12], Singh and Abdussattar [11], Feinstein and Senovilla [1], Kramer [2]); for p=+constant (Buchdahl and Land [6], Alluntt [9]) and for
(1 a) p ap (Buchdahl [4]). But if one takes, e.g. polytropic fluid
sphere
ap
1 1
n (Klein [12], Tooper [18], Buchdahl [5]) or a mixture of
ideal gas and radiation (Suhonen [3]), one soon has to use numerical methods. Yadav and Saini [20] have also studied the static fluid sphere with equation of state p= (i.e. stiff matter). Davidson [25] has presented a solution that provides a non stationary analog to the static case when
p 1 .
3
1
In the present paper, we have obtained some exact, static spherically symmetric solutions of Einsteins field equations for the perfect fluid with equation of state p=a, where a [0,1]. We have also
taken
e k
in one case while
e Arn
in second case. For different
values of a and n we get many previously known solutions. To overcome
the difficulty of infinite density at the centre, it is assumed that distribution has a core of radius ro and constant density o which is surrounded by the fluid with the specified equation of state.

THE FIELD EQUATIONS AND THEIR SOLUTIONS
We take the line element in the form
(2.1) ds2 e dr2 r2 (d2 sin2 d2 ) edt 2
where and are functions of r only.
The field equations
(2.2)
R i 1 Ri 8Ti
j 2 j j
for (2.1) are [1]
(2.3)
8T1 e '
1 1
1 r 2 r 2
r
2 3 "
''
'2
1 '
(2.4)
8T2
8T3 e
2 4
4 2r
(2.5)
8T4 e '
1 1
4 r 2 r 2
r
j
j
j
where a prime denotes differentiation with respect to r. The energy momentum tensor for perfect fluid is given by
(2.6)
Ti ( b)ui u

i p.
We choose the equation of state as (2.7) =ap
where a is positive constant a[0,1] In this case we find that
j
Ti 0(i j)
We use commoving coordinate so that
u1 u2 u3 0
and
u 4 e 2
The nonvanishing components of the energy momentum tensor are
T1 T2 T3 p and T4
1 2 3 4
We can then write
(2.8)
8p e '
r
1 1 ,
r 2
r 2
" '' '2
1 '
(2.9)
8p e
2 4 4
,
2r
r
(2.10) 8 e '
1 1
r 2
r 2
Using equations (2.7), (2.8) and (2.10) we get
r
(2.11) ae '
1 a
r 2
r 2
e '
r
1 1 .
r 2
r 2
Case I
1
We choose e k
(2.12) ' 1 (1 k
(a constant) which reduces (2.11) to the form
a
1
) 1 0
1
r
Integrating w.r.t. r, we get
(k 1)1 1
1
(2.13) e
k2r
a
where k2 is a constant. Now (2.8) and (2.9) lead to k1=2, so that
1 1
a
(2.14) e
k 2 r
Hence the metric (2.1) can be cast into the form
1 1
(2.15)
ds2
2dr 2

r 2
(d2

sin 2
d2
) k 2 r
a dt 2
Absorbing the constant k2 is the coordinate differential dt the metric (2.15) is reduced to the form.
1 1
(2.16) ds2
2dr 2


r 2
(d2

sin 2
d2
) r
a dt 2
The non zero components of ReimannChristoffel curvature tensor Rhijk for the metric (2.16) are
R 1
1 1
1
1 1 1
(2.17) R
sin 2 2323 1
r a
R
1
r
a sin 2
2424
2 a
3434
4 a
Choosing the orthonormal tetrad j as
i
i 1
1 , 0, 0, 0
2
i 1
2 0, , 0, 0
r
(2.18) i 1
3 0, 0,
, 0
r sin
i
0,
1
0, 0,
4 a 1
r 2a
The physical components R(abcd) of the curvature tensor defined by
R(abcd) =
h (a)
i
(b)
j (c)
k
(d) R
hijk
are
(2.19) R
a 1R
R (a 1) .
2424
2323
2
3434
4ar 2
Since a is finite +ve constant, we see that
R (abcd) 0 as
r .
Hence it follows that the space time is asymptotically homaloidal.
For the metric (2.16) the fluid velocity ui is given by
1
a1
1
(2.20) u1 u 2 u3 0 u
u 2
u3
and
u 4
a1
, u 4 r 2a .
;i
The scalar of expansion = u i
r 2a
is identically zero. The non
vanishing components of the tensor of rotation ij is defined by (2.1) ij = ui;juj;i
are
a 1
1a
2a
(2.22) 14=41=
r
2a
The components of the shear tensor ij defined by
(2.23) ij =
1 u 2
i; j

u j;1
1 h
3 ij
with the projection tensor (2.24) hij = gij uiuj
are
1 a 1
1a
2a
(2.25) 14 41
.
2
2a
r .
while other components are zero.
For the particular values of constant a, several previously known solutions are contained here in. When a=1, results of this case reduce to that of Singh and Yadav [23]. Also in this case the relative mass m of a particle in the gravitational field of (2.16) is related to its proper mass mo (Narlikar [24]) through
(2.26)
m k 2
o

r 2
k being a constant. As the particle moves towards the origin, m increases and r, m 0 i.e. the relative mass goes on decreasing continuously.
The case when a=3 gives the distribution of disordered radiation already obtained by Singh and Abdussattar [11].
Case II
From (2.11) we see that if is known, can be obtained. So we choose
(2.27) e Arn
where A is constant
Use of (2.27) reduces the equation (2.11) to the form (2.28) e (an a 'r 1) a 1.
We put y e so that equation (2.28) is transformed to
(2.29)
dy (an a 1) y a 1
dr r r
which is a linear differential equations whose solution is
(2.30) y
a 1
an a 1
E
ran a 1
where E is integration constant.
Therefore we get
(2.31) e
a 1
an a 1
E .
rana1
Consequently the metric (2.1) can be put into that form
(2.32)
ds2 Arn dt 2 a 1
E 1
dt 2 r 2 (d2 sin 2 d2 )
an a 1
rana1
Absorbing the constant A in coordinate differential dt, the metric (2.32) goes to the form
(2.33)
ds2 rn dt2 a 1
E 1
dr2 r2 (d2 sin2 d2 )
an a 1
rana1
The non vanishing components of ReimannChristoffel curvature tensor Rhijk for the metric (2.33) are
R
E(an a 1) ,
1212
a 1
2
E an a 1

.r
an a 1
r an a 1
R1313
E(an a 1) sin 2
,
2r an a 1 a 1

E
an a 1
n 2 2n n 2
r an a 1 .
nE(an a 1)
(2.34)
R1414 r
4
4r{n (a 1)(a 3)} a 1
an a 1
,
E r an a 1
R r 2 sin 2
a 1



E
2323
nr n
an a 1 a 1
r an a 1 , E
R 2424
2 an a 1 r an a 1 ,
nr n sin 2
a 1 E
R 3434
2 an a 1 r an a 1 .
Choosing the orthonormal tetrad
i
j as
i
1
a 1
1
r
E 2 , 0,
0, 0
an a 1
an a 1
i 1
2 0,
, 0, 0
r
(2.35)
3
i
0, 0,
0
,
1
r sin
i 1
4 0, 0, 0, n
r 2
the physical components R(abcd) of the curvature tensor are
R
1212
R
1313
(an a 1)E , 2r an a 3
(an a 1)E ,
2r an a 3 .
R
1414
2n n 2
a 1 E
nE(an a 1) ,
4r 2
an a 1
r an a 1
4r a (n1)3
(2.36)
1 a 1 E
R 2323 r 2 an a 1 r an a 1 ,
n
a 1 E
R 2424
2r 2 an a 1
an a 1 , r
n a 1 E
R 3434
.
2r 2 an a 1
r an a 1
We see that R(abcd)0 as r. It follows that the spacetime is asymptotically homaloidal.
4
Also the metric (2.33) the fluid velocity ui is given by
1
(2.37) u1 u 2 u3 0 u
u 2
u3
and
u r n / 2 , u 4
1 .
r n / 2
;i
The scalar of expansion = u i is identically zero. The non
vanishing components of the tensor of rotation ij are
(2.38) 14 =41 =

n 1
r 2
2
The non zero components of the shear tensor ij are
(2.39) 14 =41 = n
2
n 1
r 2 .


SOLUTION FOR THE PERFECT FLUID CORE
Pressure and density for the metric (2.33) are
(3.1)
8p 8 n 1 a 1
E 1
a r 2 an a 1
rana1 r 2
It follows from (3.1) that the density of the distribution tends to infinity as r tends to zero. In order to get rid of the singularity at r=0 in the density we visualize that the distribution has a core of radius ro and constant density o. The field inside the core is given by Schwarzschild internal solution.
r 2
r 2
2
e 1
, e A B1
R 2
R
1
(3.2)
2
r 2
3B1
A
r
R
8p
.
R 2
A B1
2 1/ 2
R 2
Where A and B are constants and
R 2
3 .
8
The continuity conditions for the metric (2.33) and (3.2) at the boundary
gives
r2
R
2 o ,
E
an
an a 1
rana1
n
A r 2
nR 2
o
r
2
1 o
o 2r2n
R 2
(3.3)
o 2
1
2 r2 2
B nR 1 o ,
2r2n R 2
o 2
R 2
ana1 an
r2
E ro
an a 1
o .
and the density of the core
3 an E
.
(3.4)
o
8r
2 an a 1
r
ana 1
which complete the solution for the perfect fluid core of radius ro surrounded by considered fluid. The energy condition Tij Ui Uj >0 and the Hawking and Penrose condition (Hawking and Penrose, 1970).
T 1 i
ij
2 gijT u u j 0,
both reduces to >0 which is obviously satisfied.
For different values of a and n, solutions obtained above in case II provide many previously known solutions. For a=1, n=1, we get the results due to Yadav and Saini [20]. For n=2 and by suitable adjustment of constant we get the solution due to Singh and Yadav [23]. Also for a=3
and n=2 we get the solution due to Yadav and Purushottom [21] and Yadav et al [22] by suitable adjustment of constants.

DISCUSSION
In this paper the equation of state for the fluid has been taken as p=a which (for a=1) describes several important cases, e.g. relativistic degenerate Fermi gas and probably very dense baryon matter (Zeldovich andNovikov [26]; Walecka [8]). The casual limit for ideal gas has also form =p (Zeldovich and Novkove [26]).
Furthermore, if the fluid satisfies the equation of state p= and if in addition its motion is irrotational, then such a source has the same stress energy tensor as that of a m assless scalar field (Tabensky and Taub [19]). Also the solution in this case can be transformed to the solution of BransDicke Theory in vacuum. (Dicke [17]).
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