STATIC SPHERICALLY SYMMETRIC INTERIOR SOLUTIONSWITH NONUNIFORM DENSITY

International Journal of Applied Mathematics & Statistical Sciences(IJAMSS) ISSN: 2319-3972 Vol.1, Issue 2 Nov 2012 17-26 © IASET

STATIC SPHERICALLY SYMMETRIC INTERIOR SOLUTIONS WITH NONUNIFORM DENSITY 1

M. A. KAUSER & 2 Q. ISLAM

1

Department of Mathematics, Chittagong University of Engineering & Technology, Chittagong-4349, Bangladesh

2

Research Centre for Mathematical and Physical Sciences, University of Chittagong, Chittagong-4331, Bangladesh

ABSTRACT For the highly nonlinearity conditions, it is so much difficult to obtain exact solution of Einstein’s field equations.

Many authors have been working on the investigation of exact solution of Einstein’s equation. One of these solutions, Schwarzschild uniform density solution is unphysical. In this paper, we have applied the Adler method to obtain Einstein’s equations solution in use of different equation of state. We got a new solution although it is Tolman’s solution no IV. But our solution is different and original. We have described the obtain solution in terms of two new variables. Also we have shown that maximum mass of 1/3 the fluid radius (in geometric units) which is less than Adler’s 2/5 or Schwarzschild 4/9.

KEYWORDS: Non-Uniform Density, Exact Solution, Equation of State, Uniform Density INTRODUCTION Uniform density solution [1] is unphysical because velocity of sound becomes infinite if

ρ  constant. A small

number of solutions by authors [2, 3, 4, 5] with non-uniform density distribution which have been obtained are valuable and interesting because they provide some insight into the effects of relativity on the interior structures of stars. In this paper, one such solution is studied. In section-2, the solution found by Adler [2] is reviewed. There is a misprint in equation (3.3d) of Adler’s paper which we have corrected in this section. Some interesting properties of Adler’s solution are studied in section 3. In section 4 we have shown our new solution by using the method found by Adler. This is a new solution but no different from Tolman’s [3] solution no IV. Finally some interesting properties of the solution are studied and defined the results in two new variables x and y in section 5. For this solution we have shown that for a finite radius ( r0 ) of the fluid sphere

m0 1 m 4  , where m 0 is the mass of the fluid sphere. The star collapses if 0 exceeds . r0 3 r0 9

Buchdahl [4] has shown that the condition

m0 4  is true in general. r0 9

ADLER’S STATIC SPHERICALLY SYMMETRIC INTERIOR SOLUTION Let us take the static spherically symmetric line element in the following form, ds 2 = -N 2 (r)dt 2 +

dr 2 G 2 (r)

+ r 2 (d θ 2  sinθ 2 d 2 )

(1)

Einstein equation without cosmological constant is

R μν 

1 g μν R  8π T μν 2

(2)

18

M. A. Kauser & Q. Islam

For a perfect fluid energy momentum tensor T is given by

Tμν  (p  ρ)uμ u ν  pgμν

(3)

For the metric (1), equation (2) gives

8π ρ  

1 dG 2 (r  G 2  1) 2 dr r

(4)

8π p 

1 dN (2rG 2  N  NG 2 ) 2 dr Nr

(5)

8π p 

1 dN 1 dG 2 d 2 N r dN dG 2 (G 2  N  rG 2 2  ) Nr dr 2 dr 2 dr dr dr

(6)

From (5) and (6) we obtain

dG 2 2(N  rN  - r 2 N )G 2 2N   dr r(N  r N ) r(N  r N ) or,

dT  P(r)T  Q(r) dr

(7)

where

T(r)  G 2 (r) , 2(N  rN  - r 2 N ) P(r)  r(N  r N ) and

Q(r) 

2N r(N  r N )

(8)

(9)

Solution of (7) becomes simple if we put P(r) = Q(r).

Now the P(r) and Q(r) are the equal if

rN - r 2 N  = 0 or,

N  1  N r

Integrating this we obtain,

log( N )  log r  log(2B) or, N   2Br

(10)

Again integrating it we get

N  A  Br 2

(11)

Static Spherically Symmetric Interior Solutions with Nonuniform Density

19

where A and B are constants.  P(r) dr I.F.of (7)  e 

Therefore, the solution of (7) is  P(r) dr  P(r)dr Te    Q(r) e  dr

or, T

Now, P(r)

 1  Ce 



P(r) dr

dr

(12)

2(A  Br 2 ) [putting the value of N from (11)] r(A  3Br 2 )

2 4Br)   P(r)dr   (  )dr r A  3Br 2

 2logr  e

P(r)dr

2 log(A  3Br 2 ) 3

r2



2

(A  3Br )

2 3

So from (12) we get

r2

T 1 C

2

(A  3Br 2 ) 3

Cr

i.e, G 2  1 

2

(13)

(A  3Br 2 )

2 3

Sometimes it is convenient to replace G(r) by another function m(r) defined by

G 2 (r)  1 

2m(r) r

So from (13) we get

1

Cr 2 2 2 3

 1

(A  3Br ) or, m(r)  1 

Cr

2m(r) r 3

2

2(A  3Br ) If

2 3

r0 is the radius of the fluid sphere and m 0 its mass, then we have

(14)

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M. A. Kauser & Q. Islam

m0   r0

ε 

Cr 0

2

2(A  3Br

(15) 2 0

)

2 3

From (5) we get

8π p 

2 2Br 1   G 2  2 (1  G 2 ) 2 r A  Br r

On the surface of the sphere p = 0,

r  r0 .

This implies

Br 0

2

A  Br 0 or,

2

(1  2ε )  2ε 2

ε A  r0 (5ε  2)B  0

(16)

Also we have by matching with the Schwarzschild exterior solution 2

A  B r0  1  2ε or,

(17)

2

ε A  ε B r0  ε 1  2ε

(18)

Subtracting (16) from (18) we get

ε

B 

2r 0

2

(19)

(1 - 2ε )

Putting this value of B in (17) we get A 

2 - 5ε

(20)

2 (1 - 2ε )

From (15) we get 2ε C  - 2 r0

(1 - ε )

2 3

(1 - 2 ε )

(21) 1 3

Now from (11) N 

2 - 5ε 2 (1 - 2 ε )



ε r ( )2 2 (1 - 2ε ) r 0 1

 or, N  (1 - 5 ε  1 ε y 2 )(1 - 2ε ) 2 2 2

Where

y

r . r0

From (13) we get

(22)

21

Static Spherically Symmetric Interior Solutions with Nonuniform Density

2

2ε (1 - ε ) 3 y 2 G2 12 5 1 (1 - ε  ε y 2 ) 3 2 2

(23)

Also from (14) we get 2

m 0 (1 - ε ) 3 y 3

m(r) 

(1 -

5 1 ε  ε y2) 2 2

(24) 2 3

From (5) pressure p(r) is given by 2

p(r) 

2

ε 5 1 5 3 [G 2 (1 - ε  ε y 2 ) -1  (1 - ε ) 3 (1 - ε  ε y 2 ) 3 ] 2 2 2 2 2 4π r0

(25)

There is a misprint in equation (3.3d) of Adler’s paper which has been correct in equation (25).

Similarly from (4), density distribution is given by 2

ε ρ(r)  { 4π r 02

(1 - ε ) 3 2

5 3 (1 - ε  ε y 2 ) 3 2 2

}{3 -

2ε y 2 } 5 3 1- ε  ε y2 2 2

(26)

PROPERTIES OF THE SOLUTION Central density

ρc is obtained by putting y = 0 in (26). We get 2

ρc 

3ε 4 π r 02

  1- ε  1- 5 ε   2

3     

We see from (27) that the central density

function of

(27)

ρc is a function of ε 

ρ0 . m0 is maximum if

dm0 0 dρ c m0 ) r0  0 [Since by assumption dρ c

d( 



dε 0 dρ c

r0 = constant]

m0 . Clearly for a fixed r0 , the mass m0 is a r0

22

M. A. Kauser & Q. Islam

dρ c  dε



(28)

From (28) we get

ε

2 5

 m0 

Thus for a fixed

m0 

r0

2 r0 5 maximum is given by m 0 

2 r0 . This may be compared with the maximum mass 5

1 r0 obtained in the Schwarzschild interior solution by demanding that the Schwarzschild radius not be exterior to 2

be fluid sphere. It should be noted that

2 1 2 r0 < r0 . In the solution found by Adler we see that m 0 cannot exceed r0 5 2 5

whereas in the Schwarzschild interior solution

m0

is bounded from above by m 0 

1 r0 .

2

p c is obtained by putting y = 0 in (25). We get

Central pressure

2

pc 

Now

pc

ε 4 π r 02

  1- ε 1 [   5 5 1 - ε  1 - ε 2  2

becomes infinite if ε 

Hence m 0 

3   ]   

(29)

2 2  m 0  r0 5 5

2 r0 5

Schwarzschild interior solution has the similar property that

pc

becomes infinite when ε 

4 4 which implies m 0 < r0 .

9 9

A NEW SOLUTION [TOLMAN’S SOLUTION NO. IV] In this section we have shown that a new solution derived by using the Adler’s method. It is nothing but Tolman’s solution no IV.

In (7) let us put

3 Q P(r)   r Q

. Then we get

2N  2rN  - 2r 2 N  3 N  N  3rN   r 2 N     r(N  r N ) r N r(N  r N ) 2

2

2

or, r N   r NN  - rNN   0 2

Put α  N in (30), then

α  2NN  , α   2N  2  2NN 

(30)

Static Spherically Symmetric Interior Solutions with Nonuniform Density

23

Thus (30) becomes

r 2 α  - rα   0 or,

α  1  α r

 α  ar 2  b , where a and b are integrating constants.  N 2  ar 2  b

(31)

 N  ar 2  b Now integrating factor of (7) is given by I.F =

 P dr e  3

=

e  ln(r Q)

=

1 rQ 3

Therefore, the solution of (7) is

T

1 1   Q  3 dr rQ rQ 3

-3

=  r dr



T 

=

=

1  d , where d is a constant of integration.

2r 2

=

2dr 2  1 2r 2

2dr 2  1 3 r Q 2r 2

Qr(2dr 2  1) 2

From (9) we get

Q(r) 

2 ar 2  b 2ar 2 r( ar 2  b  ) 2 ar 2  b

or, rQ(r)  2(ar (2ar

2 2

 b)  b)

24

M. A. Kauser & Q. Islam

T  ( 

a 2 r  1 )(1 - cr 2 ) , where –c = 2d b a 2 2 r 1 b (

or,

2

G (r) 

1  2  G (r)

For this solution

a 8π ρ   b

a 2 r  1)(1 - cr 2 ) b a 2 r2  1 b

2 (

(32)

a 2 r 1 b

a 2 r  1)(1 - cr 2 ) b

ρ and p are given by

3bc  3cr a a 1 2 r2 b

1

a 8π p   b

2(ar2  b) 2dr2  1  2 2ar2  b

2



2a  b

1 - cr 2 a (1  2 r 2 ) 2 b

bc  3cr 2 a a 1 2 r2 b

(33)

1

(34)

CONCLUDING PROPERTIES OF THE SOLUTION Central density

ρc and central pressure p c ρc 

can be obtained by putting r = 0 in (33) and (34) respectively. We obtain

1 a 3bc 2a { (1  ) } 8π b a b

or, ρ c 

3 a (  c) 8π b

(35)

Similarly

pc  Let

1 a (  c) 8π b

(36)

r0 be the radius of the fluid sphere. Then p = 0 when r = r0 . This implies 2

r0 

1 bc (1  ) 3c a

From (35) and (36) we get

(37)

25

Static Spherically Symmetric Interior Solutions with Nonuniform Density

a 4π  (ρ c  3p c ) b 3

(38)

4π (ρ c  3p c ) 3

(39)

c

Putting the values of

a and c in (37) we get b

3

2

r0 

r0 depends only on

From (40) we see that

Let

m0

r0

0

ρc ρ . It is also clear that for a finite radius we must have c > 3. pc pc

be the mass of the fluid sphere. Matching with the Schwarzschild solution, then gives

2m 1r0

or, m

(40)

ρ ρ 2 π( c  3)( c  3) pc pc

1  [1  2

( 0



(

a 2 2 r0  1)(1 - cr 0 ) b a 2 2 r0  1 b

a 2 2 r0  1)(1 - cr 0 ) b ] a 2 2 r0  1 b

Putting the values of

(41)

a 2 , c and r0 from (38), (39) and (40) in (41) we get b (42)

mo 2  ρc ro  3 pc

From (42) we see that

mo ρc mo 1 mo 1  if ρc  pc ,  if depends only on . Clearly ro pc ro 2 ro 2

m 1 mo 2   if ρc  pc . Let us put y  o and x  1 2 ro 3 ro ρc pc y = 2x The graph of (43) is a straight line passing through the origin.

ρc  pc

and

in (42). Then we get the equation

3 (43)

26

M. A. Kauser & Q. Islam

If we assume that

This implies x 

0  pc 

ρc ρ 1 is positive then the range of x is given by 0  x  . But from (40) we have c < 3. pc 3 pc

1 1 1 . Hence the range of x is further limited to 0  x  . In 0  x  for a fixed ρc we have 6 6 6

ρc .

3

Thus we have the following results:

pc 

m ρc 1 and o  .

3 ro 3

ACKNOWLEDGEMENTS We would like to thank Prof. Emeritus Jamal Nazrul Islam, Centre for Mathematical and Physical Sciences, University of Chittagong, for his encouragement, support and innumerable valuable interactions throughout this work.

REFERENCES 1.

Schwarzschild, M., et al, Astrophys. J. Vol. 129, Page-637(1959)

2.

Adler, R. J., J. Math. Phys., Vol.15, No.6, Page-727(1974)

3.

Tolman, R. C., Phys. Rev., Vol.55, Page-364(1939)

4.

Buchdahl, H. A., Phys. Rev., Vol. 116, Page-1027(1949)

5.

Islam, Q., The Chittagong Univ. J. Sc. Vol. 22(2) Page-91(1998)

6.

Kauser, M. A., M. Phil. Thesis, University of Chittagong, Bangladesh (2010).