Spherically Symmetric Self-Gravitating Fluid with Specified Equation of State

Spherically Symmetric Self-Gravitating 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	Bodh-Gaya, 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.

1. 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.

2. 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 co-ordinate so that

     u1 u2 u3 0

     and

     u 4 e 2

     The non-vanishing 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 co-ordinate 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 Reimann-Christoffel 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

     1. 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 co-ordinate 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 Reimann-Christoffel 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 space-time 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 =

   1. 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 .

3. 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.

4. 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 Brans-Dicke Theory in vacuum. (Dicke [17]).

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