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250

6 Stellar Equilibrium

and simpler way of determining the radial behavior of the pressure is provided by writing such a conservation law. In flat anholonomic indices we have:

cTabηcb = 0

 

 

(6.3.29)

cTabηca + ωc|af Tgbηca ηf g + ωc|bf Tga ηca ηf g = 0

 

Using the particular structure of Tab and choosing for instance b = 1 we get:

1T11 ωc|a1T11ηca + ωc|1f Tag ηca ηf g = 0

 

 

(6.3.30)

1T11 0|01 ω2|21 ω3|31)T11 = 0

Inserting the explicit form (5.9.5) of the spin connection for a spherically symmetric static metric, the non-vanishing components with an index 1 are:

 

 

ω0|01 = −a eb

 

 

 

ω2|21

= −

eb

(6.3.31)

 

 

 

r

 

 

ω3|31

= −

eb

 

 

 

 

r

 

while the intrinsic derivative 1 is defined by

 

 

 

 

 

1 eb

1

 

(6.3.32)

 

 

 

 

∂r

Inserting (6.3.32) and (6.3.31) into 6.3.30) we find:

 

 

d

 

 

 

 

 

 

p(r) a (p + ρ) = 0

(6.3.33)

dr

which combined with (6.3.21) yields the Tolman-Oppenheimer-Volkoff relativistic equation of stellar equilibrium:

d

p(r)

(p

ρ)

M(r) +

4π r3p(r)

(6.3.34)

 

[ −

]

dr

= − +

 

r r

2M(r)

 

6.3.1Integration of the Pressure Equation in the Case of Uniform Density

A very simple and idealized model of a star corresponds to choosing a uniform density:

ρ(r) = ρ0 = const

(6.3.35)

M(r) r

6.3 Interior Solutions and the Stellar Equilibrium Equation

251

Fig. 6.5 The total force exerted by pressure on the spherical stratum of matter contained between the spherical surface of radius r and the spherical surface of radius r + dr is given by [p(r + dr) p(r)] × 4π r2. On the other hand the gravitational force exerted on the same stratum of matter by the matter contained in the sphere of radius r is, by Newton’s law, ρ0 × 4πρ0r2 × . Hence the equilibrium equation is obtained by balancing these two forces

In this case the function M(r) is immediately determined and we obtain:

M(r) =

4

πρ0r3

(6.3.36)

3

6.3.1.1 Solution in the Newtonian Case

If we consider the stellar equilibrium problem in the contest of Newtonian Physics, what we have to write is simply the following equation:

dp

= −ρ0

M(r)

= −

4

πρ0r

(6.3.37)

dr

r2

3

which expresses the balancing of the pressure repulsive force with the gravitational attractive force (see Fig. 6.5). Equation (6.3.37) is immediately integrated to give:

p(r) = −

2

πρ02r2 + const

(6.3.38)

3

The integration constant is fixed by imposing the obvious boundary condition that the pressure should vanish where the star ends, namely p(R) = 0 if R is the radius of the spherical star. The solution of the differential equation with this boundary condition becomes:

p(r) = 2 πρ2 R2 r2

3 0

If we denote the total mass of the star by

M 4 πρ0R3

3

then the Newtonian solution (6.3.39) can be rewritten as follows:

 

 

 

3

 

 

2

1

 

 

r

 

p(r)

=

 

 

M

 

 

 

 

 

 

8π R4

 

R

(6.3.39)

(6.3.40)

(6.3.41)

252

6 Stellar Equilibrium

Equation (6.3.41) is written in natural units G = c = 1. It is worth to reinstall the physical units and correspondingly the fundamental physical constants. Dimensionwise we have:

[G] = 3t2m1

(6.3.42)

[c] = t1

In natural units we have:

M(r) = [ ];

r3p = [ ]

so that [p(r)] = [ 2]. The dimension of the physical pressure is:

P (r) = Force = m 1t2

Area

Hence we conclude that:

G p = P c4

while we already know that:

M = MG c2

(6.3.43)

(6.3.44)

(6.3.45)

(6.3.46)

where M denotes the mass of the star in physical units. Hence (6.3.41) translates into:

P (r)

=

 

3

 

GM2

1

 

r

 

(6.3.47)

8π

 

 

 

 

R4

 

R

 

In particular from (6.3.47) we estimate the central pressure of a star of mass M and radius R:

Pc = P (0) =

3 GM2

8π R4

Let us feed into (6.3.48) the relevant parameters for the Sun:

R, = 6.96 × 1010 cm

M, = 1.99 × 1033 g

G = 6.670 × 108dyn × cm2 × g2

We get the following value for the central pressure:

PcSun = 1.343 × 1015

dyn

cm2

(6.3.48)

(6.3.49)

(6.3.50)

Let us compare this pressure with the pressure of a weight positioned on the surface of the earth. The force experienced by somebody holding a kilogram is

6.3 Interior Solutions and the Stellar Equilibrium Equation

253

9.8 × 105 dyn 106 dyn. Hence the central pressure in a uniform density star with a stellar mass and a size of the order of the sun size is of the order of 109 kilograms per square centimeter. It is a very large but perfectly finite pressure. In the next section we will see the qualitative difference provided by the integration of the relativistic pressure equation. In General Relativity the central pressure can become infinite if either the mass is too large or the star radius is too small. In other words General Relativity implies that there are critical densities beyond which gravitational attraction is so strong that cannot be balanced by pressure. For the sun the average density is:

ρ, =

3 1.99

10331030 g cm3 = 1.41

g

(6.3.51)

4π

 

(6.96)2

cm3

which, as we will see, is much below the critical density.

6.3.1.2 Integration of the Relativistic Pressure Equation

We consider (6.3.34) and we substitute M(r) = 43 πρ0r3. Then using the definition (6.3.40) of the total mass in natural units we obtain:

 

 

 

r

3

3

R3p

dp

 

M

+ 4π

r

= −(p + ρ0)

R

R3

dr

r r 2M

r

3

R

which we can rewrite as follows:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r

3

 

 

 

 

3p

 

 

 

 

 

 

 

 

 

dp

 

 

 

 

 

 

 

 

M M

 

 

1

+ ρ0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

R

 

 

 

 

 

 

 

 

 

 

= −(p

+ ρ0)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dr

R

 

r

r 2M

r

3

 

 

 

 

 

 

 

 

 

 

R

R

 

 

 

 

Dividing by ρ0 we get:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

p

 

 

 

 

p

 

 

 

 

 

M

 

 

 

r

 

3

1 +

3p

 

 

 

 

 

 

 

 

 

 

 

 

R

 

ρ0

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= −

 

 

 

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ρ

 

 

 

 

ρ

 

 

 

 

R r

 

 

r

 

 

 

2M

r

3

 

 

 

0

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

R

R

R

R

 

 

introducing rescaled variables

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ξ =

r

;

 

 

 

 

 

 

h =

 

 

p

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

R

 

 

 

 

 

 

ρ0

 

 

 

 

 

 

 

 

 

 

(6.3.54) is rewritten as follows:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

= (1 + h)(1

+ 3h)M ξ(Rξ

ξ 3

 

 

 

 

 

 

 

 

 

h

23)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

which is immediately reduced to quadratures in the form:

(6.3.52)

(6.3.53)

(6.3.54)

(6.3.55)

(6.3.56)

 

 

 

dh

 

 

= −

M

 

ξ dξ

(6.3.57)

 

+

 

+

 

 

22

(1

h)(1

3h)

 

R

 

 

 

 

 

 

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