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260

Spherical solutions for stars

is therefore

z = e(r1) 1.

(10.13)

This important equation attaches physical significance to e . (Compare this calculation with the one in Ch. 2.)

The Einstein tensor

We can show that for the metric given by Eq. (10.7), the Einstein tensor has components

1

e2

d

e2 )],

 

G00 =

 

 

 

[r(1

(10.14)

r2

dr

1

 

 

 

2

 

Grr = −

 

e2 (1 e2 ) +

 

,

(10.15)

r2

r

Gθ θ = r2e2 [ + ( )2 + /r − − /r],

(10.16)

Gφφ = sin2 θ Gθ θ ,

 

 

 

(10.17)

where := d/dr, etc. All other components vanish.

10.3 S t a t i c p e r f e c t fl u i d E i n s t e i n e q u a t i o n s

Stress–energy te nsor

We are interested in static stars, in which the fluid has no motion. The only nonzero

0

component of U is therefore U . What is more, the normalization condition

U · U = −1

(10.18)

implies, as we have seen before,

 

 

U0 = e,

U0 = −e .

(10.19)

Then T has components given by Eq. (4.38):

 

 

T00 = ρ e2 ,

(10.20)

Trr = p e2 ,

(10.21)

Tθ θ = r2p,

 

(10.22)

Tφφ = sin2 θ Tθ θ .

(10.23)

All other components vanish.

261

10.3 Static perfect fluid Einstein equations

Equation of state

The stress–energy tensor involves both p and ρ, but these may be related by an equation of state. For a simple fluid in local thermodynamic equilibrium, there always exists a relation of the form

p = p(ρ, S),

(10.24)

which gives the pressure p in terms of the energy density ρ and specific entropy S. We often deal with situations in which the entropy can be considered to be a constant (in particular, negligibly small), so that we have a relation

p = p(ρ).

(10.25)

These relations will of course have different functional forms for different fluids. We will suppose that some such relation always exists.

Equations of motion

The conservation laws are (Eq. (7.6))

Tαβ ;β = 0.

(10.26)

These are four equations, one for each value of the free index α. Because of the symmetries, only one of these does not vanish identically: the one for which α = r. It implies

d

dp

 

(ρ + p) dr

= − dr .

(10.27)

This is the equation that tells us what pressure gradient is needed to keep the fluid static in the gravitational field, the effect of which depends on d /dr.

Einstein equations

The (0, 0) component of Einstein’s equations can be found from Eqs. (10.14) and (10.20). It is convenient at this point to replace (r) with a different unknown function m(r) defined as

m(r) := 21 r(1 e2 ),

(10.28)

262

 

Spherical solutions

for

stars

 

 

 

 

 

 

 

 

 

 

 

 

 

 

or

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

grr = e2 = 1 2

 

r

.

(10.29)

 

 

 

 

 

 

 

m(r)

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Then the (0,0) equation implies

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dm(r)

= 4π r2ρ.

 

 

(10.30)

 

 

 

dr

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This has the same form as the Newtonian equation, which calls m(r) the mass inside the sphere of radius r. Therefore, in relativity we call m(r) the mass function, but it cannot be interpreted as the mass energy inside r since total energy is not localizable in GR. We shall explore the Newtonian analogy in § 10.5 below.

The (r, r) equation, from Eqs. (10.15) and (10.21), can be cast in the form

 

d

=

 

m(r) + 4π r3 p

.

(10.31)

 

 

 

 

dr

 

r[r

2m(r)]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

If we have an equation of state of the form Eq. (10.25), then Eqs. (10.25), (10.27), (10.30), and (10.31) are four equations for the four unknowns , m, ρ, p. If the more general equation of state, Eq. (10.24), is needed, then S is a completely arbitrary function. There is no additional information contributed by the (θ , θ ) and (φ, φ) Einstein equation, because (i) it is clear from Eqs. (10.16), (10.17), (10.22), and (10.23) that the two equations are essentially the same, and (ii) the Bianchi identities ensure that this equation is a consequence of Eqs. (10.26), (10.30), and (10.31).

10.4 T h e ex t e r i o r g e o m e t r y

Schwarzschild metric

In the region outside the star, we have ρ = p = 0, and we get the two equations

dm

= 0,

 

 

 

(10.32)

dr

 

 

 

d

=

 

m

 

.

(10.33)

 

 

 

 

dr

r(r

2m)

 

 

 

 

 

 

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