Chapter 16 White Dwarfs - Istituto Nazionale di Fisica Nucleare · CHAPTER 16. WHITE DWARFS 234 In...

Chapter 16

White Dwarfs

The end product of stellar evolution depends on the mass of the initial configuration. Ob-servational data and theoretical calculations indicate that stars with mass M !< 4M! afterejecting part of their mass in the form of a planetary nebula give birth to a white dwarf, withtypical mass, radius and density M ! 1M!, R ! 5000 km, and ! ! 106 gr/cm3. Whitedwarfs are composed largely of helium, carbon and oxygen, because the progenitors massesare such that the temperature never becomes high enough to burn much beyond carbon, andeven if burning may, in principle, proceed all the way to iron the time needed to completethe process would be longer than the Universe age. As we shall later show, white dwarfs ofmass exceeding the critical value MCH ! 1.4M! cannot exist.

Neutron stars or black holes are thought to be the leftover of the gravitational collapse,following a supernova explosion, of stars whose mass is greater than 4M!, but the mech-anism that may produce one or the other is still unclear. (For a review, see for instanceA. Heger, C. L. Fryer, S. E. Woosley, N. Langer, D.H. Hartmann“How massive single stars end their life”,The astrophysical Journal 59, 288-300, 2003. Also available athttp://www.journals.uchicago.edu/ApJ/journal/issues/ApJ/v591n1/57419/57419.html).Numerical simulations indicate that if the mass of the progenitor star is smaller than! [20, 30]M!, a neutron star should form, whereas bigger masses would produce a blackhole. As for white dwarfs, a critical mass exists also for neutron stars. The absolute upperlimit is in the range ! 2" 3M!; the value of the critical mass depends on the equation ofstate which is chosen to describe matter at supranuclear densities, as those prevailing in thecore of a neutron stars. Neutron stars have been observed in binary systems or as isolatedobjects. Typical parameters are M ! 1 " 3 M!, R ! 10 km, and ! ! 1012 gr/cm3.

Black holes of astrophysical origin can have very di!erent masses, ranging from a fewsolar masses of the “stellar black holes”, born in the gravitational collapse of big stars or inthe coalescence or accretion driven processes in binary systems, to supermassive black holes,with masses M ! 106 " 108M!, which sit at the center of several galaxies.

In this chapter we shall focus on the study of white dwarfs, whose structure can bedescribed using the equations of newtonian gravity; in the next chapter we shall derive theequations of stellar structure in general relativity, needed to describe neutron stars.

231

CHAPTER 16. WHITE DWARFS 232

16.1 The discovery of white dwarfs

The first white dwarf, Sirius B, was observed in 1915 by Adams. He found that the spectrumof the stellar object orbiting around Sirius, named Sirius B was that of a white star, notvery di!erent from the spectrum of Sirius. The mass of the newly discovered star was foundby applying third Kepler’s law

"2r =GMSB

r2,

and it was estimated to be in the range 0.75 " 0.95M!. Knowing the distance of thesystem from Earth, from the observed flux of radiation it was possible to estimate the e!ectivetemperature, that in this case was ! 8000 K. Since for a black-body emission L ! R2T 4

eff ,from spectral measurements it was then possible to estimate the radius of the star, whichwas, surprisingly, RSB = 18.800 km, much smaller than that of the Sun! The actual valuesof the mass and radius are MSB = 1.034 ± 0.026 M! and RSB = 0.084 ± 0.00025 R! (i.e.RSB ! 5850 km).

At that time this result was really a surprise because a star having a mass comparableto that of the Sun but a radius nearly forthy times smaller had never been observed. Inaddition, although the gravitational redshift predicted by Einstein’s theory of Relativity hadalready been measured in the famous Eddington expedition in 1919, the redshift of spectrallines of Sirius B measured by Adams in 1925 provided a much better verification of thetheory, and in fact in his book The internal constitution of stars Sir Arthur Eddington wrote“Professor Adams has killed two birds with one stone: he has carried out a new test ofEinstein’s general theory of relativity, and he has confirmed our suspicion that matter 2000times denser than platinum is not only possible, but it is actually present in our universe”.

The discovery of such an extremely dense star raised a main question: how can this“white dwarf”, as it was named, support its matter against collapse? Indeed, if the mattercomposing the star were a perfect gas its temperature would be too low to prevent thecollapse, i.e. the corresponding pressure gradient would not be su"cient to balance thegravitational attraction. About this problem Eddington wrote“It seems likely that the ordinary failure of the gas laws due to finite sizes of molecules willoccur at these high densities, and I do not suppose that the white dwarfs behave like perfectgas”.What is then that keeps white dwarfs in equilibrium? The answer to this question camea few years later, when Dirac formulated the Fermi-Dirac statistics (August 1926), R.H.Fowler identified the pressure holding up a white dwarf from collapsing with the electrondegeneracy pressure (December 1926). This was the crucial step toward the formulation ofa consistent theory of these stars that led S. Chandrasekhar to predict the existence of acritical mass above which no stable white dwarf could exist.

In order to formulate the theory, let us briefly recall some basic equations of degeneretegases.


16.1.1 Degenerate gas in quantum mechanics

A perfect gas is said to be ‘degenerate’ if its behaviour di!ers from the classical behaviour dueto the quantum properties of the system of particles. Since degenerate gases are importantin the study of the internal structure of compact stars, we shall outline some basic elementsof the theory. Consider a gas composed by particles all belonging to the same species. Ingeneral, the system will be completely described if we assign the number of particles per unitphase-space volume, i.e. the number density in the phase space

dN

d3xd3p=

g

h3f(x,p), (16.1)

where h3 is the volume of a cell in the phase-space, g = 2s+1 is the number of states ofa particle with a given value of the 3-momentum p, s is the spin, and f(x,p) is theprobability density function, i.e. the probability of finding a particle at a position betweenx and x + dx and with a 3-momentum between p and p + dp. 1 If the rest mass of aparticle is m, its total energy is E = [p2c2 + m2c4]

12 and the total energy density of the

gas is

E =!

EdN

d3xd3pd3p =

g

h3

!E f(x,p) d3p . (16.4)

The distribution function for an ideal gas of fermions or boson in equilibrium is

f =1

eEc!µ

kT ± 1, (16.5)

where the + sign holds for fermions (Fermi-Dirac statistics) and the - for bosons (Bose-Einstein statistics).

1Some useful relations:The 4-momentum of a relativistic particle is

p! = (mc!,p),

where p = m!v is the 3-momentum. Moreover, remember that the total energy of the particle is E = p0c.Since p!p! = "m2c2, it follows that

"E2

c2+ p2 = "m2c2,

where p2 is the norm of the 3-momentum, and consequently the total energy of the particle can be writtenas

E ="p2c2 + m2c4

#1/2. (16.2)

From this equation it follows that, since E = mc2!

! ="p2c2 + m2c4

#1/2

mc2

and since the norm of the particle velocity is v = p/(m!),

v =pc2

[p2c2 + m2c4]1/2. (16.3)


In eq. (16.5) Ec is the particle kinetic energy Ec = [p2c2 + m2c4]12 " mc2 and µ is

the chemical potential, which is the partial derivative of any thermodynamical potential ofthe system (the enthalpy, the internal energy, etc.) with respect to the number of moles,keeping fixed the number of moles of the other species of particles if present, and the stateparameters in terms of which the potential is expressed. For example

µi =

$#H

#ni

%

S, P, nk=const

=

$#U

#ni

%

S, V, nk=const

, (16.6)

where H is the enthalpy and U the internal energy. From eq. (16.5) we see that, since fmust be positive, the chemical potential of fermions can take any real value, either positiveor negative, whereas that of bosons is bounded to be µ < Ec.

If the temperature is high, or the energy is low (E << kT ) the Bose-Einstein and theFermi-Dirac distribution tend to the classical Maxwell-Boltzmann distribution

f ! eEc!µ

kT . (16.7)

Since f given in (16.5) only depends on Ec, i.e. it only depends on the norm of the 3-momentum p, the distribution of momenta is isotropic and we can write d3p = 4$p2dp.Thus, eq. (16.4) becomes

E =4$g

h3

! "

0

E p2 dp

eEc!µ

kT ± 1. (16.8)

The pressure can be written as

P =1

3

!pv

dN

d3xd3pd3p =

4$g

3h3

! "

0

v p3dp

eEc!µ

kT ± 1, (16.9)

where v is the particles velocity and the factor 13 comes from the hypothesis of isotropy.

This equation defines the pressure as the momentum flux.Furthermore the total number of particles and the internal energy of the system can be

written as

N =! dN

d3xd3pd3x d3p =

4$gV

h3

! "

0

p2dp

eEc!µ

kT ± 1, (16.10)

and

U =!

EcdN

d3xd3pd3x d3p =

4$gV

h3

! "

0Ec

p2dp

eEc!µ

kT ± 1.

16.1.2 A criterion for degeneracy

Let us consider the non-relativistic limit when Ec # 12mv2 = p2

2m . If we introduce thevariables

% = eµ/kT and x2 =p2

2mkT


it is easy to see that eqs. (16.10) reduce to

N =4$gV

h3(2mkT )3/2

! "

0

x2dx

%#1ex2 ± 1, (16.11)

and

U =4$gV

h3(2m)3/2 (kT )5/2

! "

0

x4dx

%#1ex2 ± 1.

In principle, these integrals can be solved and % can be found as a function of the thermo-dynamical variables. Here we shall consider explicitely the limit when % << 1, i.e., for theFermi-Dirac statistics in which we are primarily interested, when µ is negative and muchbigger than kT. In this case the integrals become

! "

0

x2dx

%#1ex2 ± 1# %

! "

0x2e#x2

dx

! "

0

x4dx

%#1ex2 ± 1# %

! "

0x4e#x2

dx;

thus, combining the expressions of N and U given in eqs. (16.11) we find

U

N= (kT )

&"0 x4e#x2

dx&"0 x2e#x2dx

and since&"0 x4e#x2

dx = 38

$$ and

&"0 x2e#x2

dx = 14

$$ we find U = 3

2NkT, which is theclassical expression of the internal energy of a perfect gas. Thus % << 1 corresponds tothe classical limit. In this limit, from the first eq. (16.11) we find

% =Nh3

gV(2$mkT )#3/2 . (16.12)

If we now put n0 = N/V , where n0 is the number of particles per cm3, and define a degeneracytemperature

Tdeg =h2

2$mk·'

n0

g

(2/3

, (16.13)

eq. (16.12) can be rewritten as

% =)

Tdeg

T

*3/2

. (16.14)

Thus, Tdeg % T , then % << 1 and the gas behaves as a classical gas; Conversely a perfectgas is said degenerate if Tdeg & T (i.e. % >> 1). When h ' 0 the degeneracy temperaturetends to zero, showing that the degeneracy of a gas is of a quantum nature. Degeneracy setsin at high densities or low temperatures.Eq. (16.13) shows that at a given density n0, Tdeg is higher for particles with smaller massm. Thus, electrons becomes degenerate earlier than heavier particles.EXAMPLES


• For a hydrogen gas in normal condition, i.e. T = 300K and n0 ! 3 · 1019cm#3

% ! 1.5 · 10#5 and the corresponding degeneracy temperature is T0 ! 0.18K, thus itbehaves as a classical perfect gas.

• For gases heavier then hydrogen % and T0 are even smaller, and consequently atordinary pressures and temperatures they are non-degenerate.

• A gas of photons is always degenerate because m = 0 and T0 = (.

• Electrons in metals are degenerate, due to their small mass (m = 9.109389 · 10#28 g)and high density (n0 ! 1023cm#3). Indeed in this case T0 ! 75.4 · 103K, and if, forexample, T = 300 K % ! 3.99 · 103.

Let us now go back to white dwarfs. As we said before, they are mainly composed ofhelium, carbon and oxygen, with heavier elements in the inner core. When the nuclearmaterial in the core has been burnt, the core contracts up to a point when the distancebetween two nuclei becomes comparable with the dimensions of the nuclei (this happenswhen ! ! 5z2g/cm3 and d ! rBhorz#

13 where z is the nuclear charge). In this situation,

there is no more space left for the external orbits of the electrons which are squeezed o!starting a pressure driven ionization process which proceeds as the density increases, pro-gressively involving the innermost orbits. As a consequence of this process a dense coreof nucleons forms, immersed in a degenerate gas of free electrons. At the same time theshells of lighter elements that surround the nucleus continue their nuclear evolution until allnuclear fuel is exhausted, and contraction and ionization processess take place also in themore exterior layers; the star then radiates its residual thermal energy and cools down. Amore accurate description of white dwarfs should take into account other e!ects, like forexample electrostatic corrections due to the fact that the positive charges are concentratedin individual nuclei rather than being uniformly distributed. 2 However, in what follows weshall neglect these e!ects. We shall consider a white dwarf at the endpoint of the evolution,assuming that the ionization process has been completed throughout the configuration andthat the star has radiated away its thermal energy, so that it is composed exclusively of adense core of nucleons, immersed in a gas of electrons that behave as a degenerate gas atzero temperature.

To describe the structure of a white dwarf we do not need General Relativity. Indeed,for a typical white dwarf the surface gravity is quite small

G M

c2R! M!(in km)

R! 1.5 km

5000= 3 · 10#4.

Thus, we shall use the newtonian equations of stellar structure, which can easily be foundas follows.

Let us consider a shell of matter of radius r and thickness dr. Be dV = dAdr the vol-ume of a fluid element belonging to the shell, where dA is its section (orthogonal to r),and be dM = ! dV its mass. The forces acting on the fluid element are the gravitational

2Electrostatic corrections have been considered by Hamada and Salpeter in 1961. (T. Hamada, E.E.Salpeter Astrophys. J. bf 134, 683, 1961).


attraction exerted by the sphere of mass M(r) and the gradient of pressure across the shell;if the fluid element is in equilibrium they balance, i.e.

"dP

drdrdA =

GM(r)

r2dM ) dP

dr= "GM(r)!(r)

r2. (16.15)

The mass contained within a sphere of radius r is

M(r) =! r

0!(r) 4$r2dr, ) dM(r)

dr= 4$r2!(r). (16.16)

Equations (16.15) and (16.16) can be solved only if we assign a further equation which relatespressure and density, i.e. and equation of state P = P (!). Finally, the equilibrium equationsto be solved are +

,,,,,,,,,-

,,,,,,,,,.

dM(r)

dr= 4$r2!(r),

dP

dr= "GM(r)

r2!(r),

P = P (!).

(16.17)

We shall now determine the equation of state (EOS) of a degenerate gas.

16.1.3 The equation of state of a degenerate gas

When T ' 0 the Fermi-Dirac distribution function becomes

f(E) =/

1 for E * EF (or p * pF , )0 for E > EF ,

(16.18)

where EF and pF are the Fermi energy and momentum. Since the temperature is zero,the particles have zero kinetic energy. If they were bosons they would occupy the lowestenergy level E = 0, as it happens in Bose condensation. But fermions cannot do this, sincePauli’s exclusion principle states that in each energy level there can be at most two electrons,one with spin up and one with spin down. Thus, electrons will fill all states with energylower than EF .

An expression of pF as a function of the density can be found as follows. The numberof levels with momenta between p and p + dp per unit volume is

d& =number of levels

unit volume=

4$p2dp

h3. (16.19)

Since Pauli’s principle establishes that two spin states are available, there are two electronsin each level; thus the number of electrons per unit volume is

n = 2! pF

0d& =

! pF

0

8$p2dp

h3=

8$

3h3p3

F . (16.20)


If there are ' nucleons for each electron (' ! 2 for stars that have used their hydrogen fuel)the mass density is

! = 'nmN , (16.21)

where mN = 1.67 · 10#24g is the mass of the nucleons. The electrons contibution to themass density is negligible since me << mN . From eqs. (16.20) and (16.21) we can find pF

as a function of the density

pF = h)

3

8$'mN!* 1

3

. (16.22)

Knowing pF , we can determine the kinetic energy-density ( and the pressure P of the gasas follows

( =U

V=

8$

h3

! pF

0{[p2c2 + m2

ec4]

12 " mec

2}p2dp, (16.23)

where Ec = [p2c2 + m2ec

4]12 " mec2 is the kinetic energy of each electron, and using eq.

(16.9) and (16.3)

P =8$

3h3

! pF

0

p4c2

[p2c2 + m2ec

4]12

dp. (16.24)

These equations can be easily integrated in two regimes: 1) the non-relativistic and 2) theultrarelativistic regime. To this purpose, it is useful to define a critical density, !crit, asthe density at which the Fermi momentum becomes equal to mec; using eq. (16.22)

!crit = ' · 8$

3mN

)me c

h

*3

= 0.98 · 106 · ' g/cm3. (16.25)

• 1) If ! << !crit, cpF << mec2 and the electrons are non relativistic. In this caseeq. (16.24) gives

P ! 8$

3h3

! pF

0

p4

medp =

8$

15h3· p5

F

me, (16.26)

and using eq. (16.22)

P =

0

1 h2

5me

)3

8$

*2/3 )1

'mN

* 53

2

3 !53 . (16.27)

Thus, the gas of degenerate electrons behaves as a perfect gas with a polytropic equa-tion of state

P = K!! , where (16.28)

K =

0

1 h2

5me

)3

8$

*2/3 )1

'mN

* 53

2

3 , and ) =5

3.

Moreover, from eq. (16.23) the kinetic energy-density is

( ! 8$

h3

! pF

0{mec

2(1 +1

2

p2c2

m2ec

4) " mec

2}p2dp =4$

h3

! pF

0

p4

medp, (16.29)



( =4$

5h3

p5F

me=

3

2P. (16.30)

• 2) If ! >> !crit, cpF >> mec2 and the electrons are ultra-relativistic. In this casefrom eq. (16.24) we find

P =8$

3h3

! pF

0p3cdp =

2$c

3h3p4

F , (16.31)


P =

0

1c h

8

)3

$

*1/3 )1

mN'

* 43

2

3 !43 . (16.32)

Again, the degenerate gas of electrons behaves as a perfect gas with a polytropicequation of state

P = K!! , where (16.33)

K =

0

1c h

8

)3

$

*1/3 )1

mN'

* 43

2

3 , and ) =4

3.

Moreover

( =8$

h3

! pF

0p3cdp, (16.34)

i.e.( = 3P. (16.35)

SUMMARY: We have shown that a degenerate gas of electrons can be described by apolytropic equation of state

P = K!!

in two di!erent regimes:

• non relativistic regime ! << !crit,

K =

0

1 h2

5me

)3

8$

*2/3 )1

'mN

* 53

2

3 = 9.9156 · 1012 · '#5/3 erg2/g8/3, and ) =5

3,

(16.36)

• ultra-relativistic regime ! >> !crit,

K =

0

1ch

8

)3

$

*1/3 )1

mN'

* 43

2

3 = 1.2316 · 1015 · '#4/3 erg2/g7/3, and ) =4

3,

(16.37)where

!crit = ' · 8$

3mN

)me c

h

*3

= 0.98 · 106 · ' g/cm3.


From these expressions we see that, in a completely degenerate gas, pressure depends onlyon density. As the density increases, degeneracy pressure increases as well, and the pressuregradients which develops inside the star is su"cient to support the equilibrium againstgravitational contraction. This is true, as we shall later see, if the mass does not exceed acritical value.It should also be noted that, either in the non releativistic and in the highly relativisticregime, a degenerate gas behaves as a perfect gas with a polytropic equation of state. Thisclearly contradicts Eddinghton’s idea that in the high density regime typical of the interiorof a white dwarf, stellar matter should not behave as a perfect fluid.

16.1.4 The structure of a White Dwarf

We shall now find the equilibrium configuration of a white dwarf solving the newtonianequations of hydrostatic equilibrium (16.17) and using the results obtained in the previoussection.

As mentioned in section 16.1.2, in order to solve eqs. (16.17) we need to know theequations of state of matter, i.e. an equation which relates pressure to density; since we areinterested in the two regimes described in section 16.1.3, i.e. the non relativistic (! << !crit),and the relativistic regimes (! >> !crit), we shall assume that the EOS has a polytropic form;thus the complete set of equations to solve by imposing appropriate boundary conditions is

+,,,,,,,,,-

,,,,,,,,,.

dM(r)

dr= 4$r2!

dP

dr= "GM(r)

r2!

P = K!! .

(16.38)

It is easy to see that the first two equations can be combined into the following second orderequation (hint: di!erentiate the second equations and replace the expression of dM(r)

dr givenby the first)

1

r2

d

dr

'r2

!

dP

dr

(

= "4$G!. (16.39)

Be !0 = !(r = 0), the central density; by putting+,,,,,,,,-

,,,,,,,,.

) = 1 +1

n, where n is called polytropic index

! = !0 #n(r)

P = K !1+ 1

n0 #(n+1)(r),

(16.40)

eq. (16.39) becomes

(n + 1)K !( 1

n#1)0

1

r2

d

dr

'

r2d#

dr

(

= "4$G #n. (16.41)


If we now introduce the following dimensionless radial coordinate

% =r

*, where * =

0

1(n + 1)K !( 1

n#1)0

4$G

2

3

12

, (16.42)

eq. (16.41) becomes1

%2

d

d%

'

%2d#

d%

(

= "#n, (16.43)

known as the Lane-Emden equation. It should be noted that this is a dimensionless equation,which depends only on the polytropic index n.

The physical boundary conditions that have to be imposed to solve the structure equa-tions are that at r = 0 the density has some assigned value !0 and that at the surface of thestar, r = R, the pressure vanishes, i.e.:

!(0) = !0, p(R) = 0. (16.44)

Since ! = !0 #n, the first condition implies that #(0) = 1; moreover, since the mass goes tozero as M(r) ! 4"

3 !0r3, from eq. (16.38) it follows that

dP

dr! "4$G

3r !2

0,

i.e. it goes to zero as ! r. From the EOS P = K!! we find

dP

dr= K) !!#1 d!

dr

from which it follows that if dPdr tends to zero d#

dr must tend to zero as well. Thus, a furthercondition to impose on # is theta #$(r = 0) = 0. In conclusion the Lane-Emden equation(16.43) must be integrated by imposing that at the center of the star

$#(0) = 1,#$(0) = 0.

(16.45)

It can be shown that if ) > 65 , #(%) vanishes for some % = %1. When # = 0 both

the density and the pressure vanish, therefore %1 is the boundary of the star, which can bedetermined numerically.

The procedure to find the stellar structure can be summarized as follows.

• Choose a value of ) (for instance ) = 53 or ) = 4

3), find the corresponding polytropicindex n = 1

!#1 , and integrate numerically eq. (16.43) with the initial conditions (16.45)

up to the value % = %1 where # = 0. For instance, for ) = 53 and ) = 4

3 we wouldfind

) = 53 n = 3

2 %1 = 3.65375 %21 #$(%1) = "2.71406 (16.46)

) = 43 n = 3 %1 = 6.89685 %2

1 #$(%1) = "2.01824 (16.47)

It should be noted that # is a monotonically decreasing function of %, that is why itsfirst derivative at the boundary is negative.


• Assign a value to ', i.e. the number of nucleons per free electrons, then find K fromeq. (16.36) or (16.37). Choose a central density !0. Knowing K and !0 the radius ofthe star can be found using the definition of % given in eqs. (16.42)

R = *%1 ' R = %1

4(n + 1)K

4$G

5 12

· !1!n2n

0 . (16.48)

• The mass of the star can now be determined as follows

M =! R

04$r2!(r)dr = 4$*3 !0

! $1

0%2 #n d%

= "4$*3 !0

! $1

0

d

d%

'

%2d#

d%

(

d%

= "4$*3 !0 %21 #$(%1)

where use has been made of eq. (16.43). Finally, the value of M as function of K and!0 can be found by using the expression of * given in (16.42)

M = 4$ %21 |#$(%1)|

4(n + 1)K

4$G

5 32

· !3!n2n

0 . (16.49)

Let us define

A =

4(n + 1)K

4$G

5

, B = 4$ %21 |#$(%1)|, (16.50)

so thatR = %1 · A1/2 · !

1!n2n

0 , (16.51)

andM = B · A3/2 · !

3!n2n

0 . (16.52)

Combinig eqs. (16.51) and (16.52), a relation between M and R can easily bederived

M =)B · A

nn!1 · %

n!31!n1

*· R

3!n1!n . (16.53)

From the procedure outlined above we understand that, having fixed the number of nucleonsper free electrons, ', and the polytropic index n, once we have found %1 and #$(%1) by numer-ical integration of the Lane-Emden equation we obtain a family of solutions parametrizedwith di!erent values of the central density !0, the radii and masses of which are given by(16.48) and (16.49).

Conversely, if we change the number of nucleons per free electrons, the new configurationcan easily be obtained by rescaling the various quantities in the following way

!$ = %"

% !, P $ = P, (16.54)

M(r)$ =6%%"

72M(r), r$ =

'

'$ r.


16.1.5 A note on the numerical integration of eq. (16.43)

Although the initial conditions (16.45) are correct, it would be impossible to integrate eq.(16.43) numerically starting from % = 0 with these conditions. Indeed, since % = 0 isa singular point, running the code we would get immediately an overflow. However thisproblem can be overcome if we start the numerical integration at some small, but finite,value of % = %start and use as initial values for the function #(%) a suitable Taylor expansion.Let us do it step by step.

Since we know from (16.45) that #(0) = 1 and #$(0) = 0, we can write the approximatesolution near % = 0 as a power series

#(%) ! 1 + #2 %2 + #3 %3 + #4 %4 + O(%5), (16.55)

(we can keep as many terms we want, but let us stop here). #1, #2, #3 .. are the constantswe need to find using eq. (16.43), therefore we also need to Taylor-expand the function #n

on the right hand side, i.e.#n ! 1 + n#2%

2 + O(%3); (16.56)

by substituting in eq. (16.43) the expansions (16.55) and (16.56) we find

6#2 + 12#3% + 20#4%2 + ... = "[1 + n#2 %2] + ... (16.57)

and this equation is satisfied only if the coe"cients of the same power of % vanish, i.e.

1 = "6#2 ' #2 = "1

6#3 = 0

20 #4 = "n#2 ' #4 =n

120;

the expansion has only even powers of % (this is true also at higher order). Thus the approx-imate solution for # and #$ near the origin is

#(%) ! 1 " 1

6%2 +

n

120%4 + O(%6) (16.58)

#$(%) ! "1

3% +

n

30%3 + O(%5).

We now have all we need to numerically integrate the Lane-Emden equation, because we canstart at, say, %start = 10#4 using as initial values the functions (16.59) computed at %start.

16.2 The Chandrasekhar limit

In section 16.1.3 we have shown that if the density is much smaller than the critical density,electrons behave as a polytropic gas with ) = 5

3 . In this regime eqs. (16.50) give

A = 2.9562 · 10#19 '#5/3, B = 34.1059


and using eq.(16.25) and (16.52) we can write the mass of the star in this form

M = 2.73 '#5/2

'!0

!c

(1/2

M!, (16.59)

where M! = 1.989 · 1033 g is the mass of the Sun. This equation shows that the mass of thestar increases with the central density. As the central density increases above the criticaldensity, the electrons start to behave as a relativistic gas with a polytropic equation of statewith ) = 4

3 . Equation (16.49) shows that in this limit the mass becomes independent ofthe central density !0 and takes the value

M = MCH = 5.74 '#2M!. (16.60)

This is a critical mass above which no stable configuration for a white dwarf can exist, andit is called the Chandrasekhar limit, as it was derived by Subrahmanyan Chandrasekhar in1931 3. It should be noted that the information on the internal composition is containedentirely in the parameter '. For instance, if we set ' = 2 we find

MCH = 1.435 M!. (16.61)

The fact that a critical mass should exist can also be understood from the followingqualitative considerations. A given configuration of matter will be in equilibrium if thegradient of pressure is balanced by the gravitational attraction.In the non relativistic case

a) P ! !53 ' P ! M

53

R5' dP

dr! M

53

R6. (16.62)

In the ultra-relativistic case

b) P ! !43 ' P ! M

43

R4' dP

dr! M

43

R5. (16.63)

The gravitational force per unit volume behaves like

Gm(r)!

r2! M2

R5. (16.64)

If the star is in equilibriumdP

dr= "Gm(r)!

r2;

in the non-relativistic case the gradient of pressure (16.62) and the gravitational force (16.64)depend on the radius to a di!erent power thus, for a given value of the mass, the star can

3The concept of a limiting mass for white dwarfs was first introduced by Chandrasekhar in a paperpublished in 1931: “The Maximum Mass of Ideal White Dwarfs”, in The Astrophysical Journal, 74 n.1, 81.The problem was subsequently investigated in a series of papers, and a complete account can be found inthe book Chandrasekhar wrote on the subject in 1939: “An introduction to the study of stellar structure”,University of Chicago Press, Chicago Illinois


‘adjust’ the radius until the two forces are equal. Conversely, in the ultra-relativistic case thegradient of pressure (16.63) and the gravitational force (16.64) have the same dependenceon the radius, and therefore the equilibrium is possible only for one value of the mass, i.e.for the critical mass. If M > MCH the gravitational attraction exceeds the gradient ofpressure and stable configurations are no longer possible.

Although the existence of a critical mass for white dwarfs seems an obvious consequenceof the theory today, it was not accepted when Chandrasekhar found it. The prejudice atthat time was that white dwarfs do represent the final state of a star, and that they couldhave any mass (neutron stars were discovered much later in 1965). The famous astronomerSir Arthur Eddington was the strongest opponent to the new theory, and called it “a stellarbu!onery”. Nobody at that time gave to Chandrasekhar any public support, although afew, as for example Rosenfeld, told him in private that they thought his result was correct4.

It should be stressed that this limit is a static limit, i.e. it refers only to the equilibriumconfiguration. It says that stars with a mass exceeding the critical mass cannot exist. How-ever, even if a star is in equilibrium it may become unstable against small perturbations. Inthis case we would call it a dynamical instability.

A second point which should be noted is that in the derivation of the critical mass generalrelativity plays no role. The basic ingredients are special relativity and the Fermi-Diracstatistics.

4An interesting account of the controversy between Eddington and Chandrasekhar on white dwarfs max-imum mass can be found in the book ”Chandra: a biography of S. Chandrasekhar”, University of ChicagoPress 1991

Chapter 17

Neutron Stars

When a star reaches the end of its thermonuclear evolution, which terminates with theformation of the heavier element that its progenitor mass allows to form, the internal pressurecan no longer sustain the gravitational attraction, and the star collapses. Current theoriesof stellar evolution show that if the progenitor mass is !< 4 M! the collapse proceeds untilit is halted by the pressure arising from electron degeneracy, as explained in Chapter 12;then, if the progenitor mass is su"ciently high to ignite at least hydrogen burning, a whitedwarf can form. White dwarfs are necessarily less massive than the Chandrasekhar limit,MCH ! 1.4 M! : the outer layers of matter are expelled by the progenitor star near theend of the nuclear burning process, and create a planetary nebula surrounding the hot corewhich collapses to form the white dwarf.

If the mass of the progenitor is in the range ! 4M! < M < 20"30M! the evolution pathis di!erent. Nuclear processes are able to burn elements heavier than carbon and oxygen,and exothermic nuclear reactions can proceed all the way to 56Fe, which is the most stableelement in nature; indeed, no element heavier than 56Fe can be generated by fusion of lighterelements through exothermic reactions. It should be noted that while iron is formed in thecore, neutrinos are produced through the reaction

56Ni '56 Fe + 2e+ + 2+e.

In addition, as the core density increases, the inverse ,-decay process, through which elec-trons are captured by protons forming neutrons and neutrinos

e# + p ' n + +e, (17.1)

becomes e"cient and nuclei richer of neutrons than 56Fe can form, like 118Kr. Since neu-trinos interact with matter very weakly, they di!use from the core to the surface and leavethe star, subtracting energy from the core. At the same time, the iron photodisintegrationprocess

) +56 Fe ' 13 4He + 4n,

which is an endothermic process, subtracts further energy to the core. Thus, all these pro-cesses tend to destabilize the stellar core, so that when the core mass becomes bigger thanthe Chandrasekhar limit, the internal pressure gradient becomes smaller than the gravita-tional attraction and the core collapses reaching, in a fraction of a second, densities typical

246

CHAPTER 17. NEUTRON STARS 247

of atomic nuclei, ! 1014 g/cm3. The core is now composed mainly of neutrons, and reacts toa further compression producing a violent shock wave that ejects, in a spectacular explosion,most of the material external to the core in the outer space. This phenomenon is calledsupernova explosion: the luminosity of the star suddenly increases to values of the order of! 109 L!, where L! is the Sun luminosity, and it is in this phase that elements heavier than56Fe are created. The remnant of this explosion is a nebula, in the middle of which sits whatremains of the core, i.e. a neutron star.

Neutron stars, whose structure and composition will be described shortly, are often ob-served as pulsars, i.e. radio sources whose emission exhibits a very sharp periodicity; pulsarsare rapidly rotating neutron stars with strong magnetic fields (B ! 1011 " 1013 Gauss),which emit beams of radio waves from the magnetic poles; the observed periodicity is dueto the fact that since the star rotates and the magnetic field is in general not aligned withthe rotation axis, the beam is visible only when it points in the direction of the detector.However, not all neutron stars are detectable as pulsars, since their beams may not pointtoward the Earth, or their magnetic field may not be su"ciently strong. Moreover, pulsarsslow down during their life because electromagnetic and gravitational emission processessubtract a substantial fraction of its rotational energy. The estimated number of neutronstars in our Galaxy is ! 109.

That neutron star could exist was first suggested by Landau (L. Landau, Zeits. Sowje-tunion 1, 285, 1932) and by Baade and Zwicky (W. Baade, F. Zwicky, Proc. Nat. Acad.Sci. 20, 255, 1934, Phys. Rev. 46, 76, 1934), who introduced the idea that neutron starsshould be the leftover of a supernova explosion. It is interesting to follow the hystorical paththat lead to the discovery of pulsars and that allowed to establish the connection betweensupernova explosions and neutron stars. In 1942 Baade identified the Crab Nebula and thestar in its center as the remnant of the supernova explosion occurred in 1054 and observedby the chinese astronomer Yang Wei-te (W. Baade, Astrophysical Journal, 96, 188, 1942).Twenty years later Hewish discovered a radio source in the Crab Nebula, whose positioncorresponded to that of the star observed by Baade (A. Hewish, S. E. Okoye, Nature 207,59, 1965) and three years later Bell and Hewish discovered four pulsars; however, none ofthem was surrounded by a nebula left by a supernova explosion (A. Hewish, S. J. Bell, J. D.H. Pilkington, P.F. Scott, R. A. Collins, Nature 217, 709, 1968). The correlation betweenpulsars and supernovae was firmly established only when a pulsar was discovered in theremnant of the Vela supernova (M. I. Large, A. F. Vaughan, B. Y. Mills, Nature 220, 340,1968) and a very fast pulsar (period=0.033 seconds) was discovered in the Crab Nebula (D.H. Staelin, E. C. Reifenstein, Science 162, 1481, 1968).

17.1 The internal structure of a neutron star

The interior of a neutron star can be modeled, as shown in figure 17.1, as a sequence of layersof di!erent composition and thickness surrounding an innermost, secret core. Proceedingfrom the exterior, we first encounter an outer crust, ! 0.3km thick, an inner crust, ! 0.5kmthick, and a core extending over about 10 km. We shall assume that the temperature ofmatter in the neutron star interior is T = 0 and that matter is transparent to neutrinos. Thefirst assumption is justified because at the densities typical of neutron stars, neutrons have a


Fermi energy EF = kTF which corresponds to temperatures TF ! 3 · 1011 " 1012 K, whereasshortly after their birth (! a year after) neutron stars reach temperatures T * 109 K << TF .The second assumption is based on the fact that the mean free path of neutrinos in nuclearmatter at T !< 109 K is much larger than the typical radius of a neutron star.

• The outer crust.The matter density ranges within ! 107 g/cm3 to the neutron drip density, !d = 4 ·1011

g/cm3. It is composed of a heavy nuclei lattice immersed in an electron gas. Proceedingfrom the external boundary to the internal one, as the density increases the inverse,-decay process becomes more and more e"cient and neutrons are produced in largenumber according to eq. 17.1. The produced neutrinos, as usual, leave the star. Inthis region pressure is mainly due to the degenerate electron gas. At ! = 4 · 1011g/cm3

all bound states available in the nuclei for neutrons are filled, neutrons can no longerlive bound to nuclei and start leaking out (neutron drip).

• The inner crust.Density ranges between !d and the nuclear density !0 = 2.67 · 1014g/cm3 and thedominant contribution to pressure is due to the neutron gas. Matter is composed of amixture of two phases: one, with density comparable to !0, is rich of protons and isindicated as PRM (Proton Rich Matter); the second phase is a neutron gas (NG). Inaddition, the electron gas is present to ensure charge neutrality. In order to determinethe fundamental state of matter in this region one has to specify the density of the twophases, !PRM and !NG, which determines the fraction of volume each phase occupies,the proton fraction in the PRM and the geometrical properties of the structures that areformed by the two phases and which strongly depend on surface e!ects at the interfacebewteen di!erent phases. For !d < ! < 0.35 !0 the minimum energy configuration isformed by spherical drops of nuclei, surrounded by a gas of electrons and neutrons.For higher densities the separation between spheres decrease up to the touching limit.Thus, for 0.35 !0 < ! < 0.5 !0 the spheres merge, forming bar-type structures, called“spaghetti” and for 0.5 !0 < ! < 0.56 !0 bars merge to form slab-type structures,called “lasagne”. When the density reaches the nuclear density !0 the two phases areno longer separated and form a homogenoeus fluid of protons, neutrons and electrons.

There is a quite general consensus on the equation of state (EOS) of matter in the outerand inner crust, because at these densities the properties of matter can be obtainedfrom experimental data on neutron rich nuclei. Conversely, densities as those prevailingin the core are presently unreachable in a laboratory, and consequently the availablemodels of EOS at supranuclear densities are based on theoretical models only partiallyconstrained by empirical data. To describe in detail the equations of state proposed todescribe matter in the neutron star core is beyond the scopes of this lectures. In thefollowing we shall just mention a number of phenomena that may occur in the core.

• The coreFor ! > 2.67 · 1014g/cm3 matter is composed of a homogenoeus fluid of p, n, e#, in


?

Uniform nuclear matter

_

Inner crust: nuclei + n + e

_

n + p + e + µ

!

!

Outer crust: nuclei + e !

~ 0.5 km~ 0.3 km

~ 10 km

3

34 x 10 g/cm11

10 g/cm~

~

~ 2 x 10 g/cm 3147

Figure 17.1: Neutron stars internal structure (not in scale!).

,-equilibrium. By minimizing the free energy one finds that matter in the core is stableonly if protons are about 10% of the total.

Several processes may develop at higher density. For instance, electrons become moreenergetic and their kinetic energy increases. So does the chemical potential; indeed thechemical potential is the energy needed to insert in a gas in equilibrium a new particlein the same state, and this energy increases if the state is more energetic. Thus, atsome density the electrons chemical potential becomes larger than the rest mass of themuon mµ! = 105MeV . At this point it is energetically more convenient to have onemuon at rest rather than such a very energetic electron, and consequently a neutroncan decay into a proton, a muon and a µ-neutrino, according to the transition

n" > p + µ# + +̄µ.

As usual neutrinos escape, and n, p, e#, µ# remain trapped in the core. It should bestressed that the main contribution to pressure in the core comes from neutrons; sincethey are more massive than electrons, the total energy also will be now provided by theneutrons themselves. Moreover, since the density extremely high (no stable neutronstar can form until the central density exceeds ! ! 1013 g/cm3), pF >> mNc2, andneutrons must be treated as ultrarelativistic particles.

Neutrons, as electrons, are fermions. However, the pressure they generate cannot beassociated only to Pauli exclusion principle, because at the core densities we can nolonger treat neutrons as non interacting particles, as we did for the degenerate electrongas in white dwarfs. Indeed, if we would assume that the neutron star is composedof non interacting neutrons, we would find a maximum mass of 0.7 M!, which is


exceedingly too low: lower than the Chandrasekhar limit, and lower than the mass ofany observed neutron star.

Depending on the particular way we choose to model neutrons interaction, we shallhave a di!erent composition. For instance in some models heavy barions may formthrough the transition

n + e# ' $# + +e.

Or, when ! ! 2 " 3 !0, $ or K mesons may form, which are bosons and therefore notsubjected to Pauli’s exclusion principle; in this case a Bose-Einstein condensate mayform in the innermost regions. Or further, since nucleons are known to be compositeobjects of size ! 0.5 " 1.0 fm, corresponding to a density ! 1015 g/cm3, if the densityin the core reaches this value matter undergoes a transition to a new phase, in whichquarks are no longer confined into nucleons or hadrons.

Thus, at the densities that are expected to occur in the inner core of a neutron star theEOS of matter at supranuclear densities depends on the modeling of neutron interactions,and the typical mass and radius these model predict range within M ! [1 " 3] M! andR ! [9 " 15] km; thus, the surface gravity is of the order of GM

c2R ! 10#1, (we remind thatGM#

c2 ! 1.47 km) and therefore general relativity must be used to determine the structureof such stars. In what follows we shall first introduce the tools that are needed to describepefect fluids in general relativity, then we shall derive the equilibrium equations of a compactstar, on the assumption that the fluid in the interior behaves as a perfect fluid.

17.2 Thermodynamics of perfect fluids in General Rel-ativity

Let us consider a perfect fluid with fixed chemical composition and in thermodynamicalequilibrium. The motion of the fluid is described by a vector field, the four-velocity u&.

Let us consider a small fluid volume and be P0 a point in the volume, for instancecoincident with its center of mass. Since the volume moves in spacetime it will describe aworldtube, defined as the congruence of geodesics described by the particles in the volume.The worldtube is plotted in figure 17.2 in 2+1 dimensions (time+2 spacelike dimensions).

At some time t = t0, let us consider a frame with origin in P0 which is locally inertialand such that P0 is at rest, i.e. the inertial frame is also comoving with P0. In the followingwe shall indicate this frame as LICF. Since P0 is at rest, its four velocity is -u = (1, 0, 0, 0),and consequently the coordinate time t of the LICF coincides with the proper time of P0.Note that:

• It is always possible to define such a locally inertial, comoving frame; indeed, givena locally inertial frame, we can make a boost and transform to a new locally inertialframe with respect to which P0 is at rest at a given time.

• Since this frame is locally inertial, the fluid in the neighborhood of P0 is freely fallingand it does not feel gravity, provided we restrict to a su"ciently short time interval.


0

worldvolume

t

y

x

LICF

worldline of P

Figure 17.2: A worldtube and a worldvolume associated to the fluid element with center ofmass in P0 is plotted in 2+1 dimensions (2 for space, 1 for time)

We shall define a worldvolume, as a portion of the 4-dimensional worldtube (see figure17.2) which is small enough to be covered by the LICF, but is large with respect to thetypical scales of the dynamics of microphysical interactions. This requires that gravitydoes not a!ect the dynamics of microscopical interactions, an hypothesis which is generallyaccepted, because the typical length scales of microscopical interactions (for instance nuclearinteractions) are much smaller than typical gravitational lengthscales (for instance curvatureradius).

Under these hypotheses a fluid element, i.e. the portion of fluid enclosed in a worldvolumeis described by the following thermodynamical variables 1:

• the particle number density n

• the energy density (

• the pressure p

• the temperature T

• the entropy per particle s,

which are scalar fields.The equation of state (EOS) of the fluid is a relation which gives one of the thermody-

namical variables in terms of two of the others, for instance

( = ((p, s) ; (17.2)

it encodes the information on the microphysics of the system. Given the values of twovariables and the EOS, the values of all other thermodynamical variables can be determined,

1we also assume, as usual in fluid mechanics, that the fluid element is large enough to contain a su!cientlylarge number of particles so that thermodynamical variables can properly be defined


as we will later show with some example; thus, in our approach a thermodynamical statedepends on two variables. Of course, this is not true for non-perfect fluids or for fluids whosechemical composition is allowed to change.

17.2.1 Baryon number conservation law

A fundamental equation in the study of stellar structure is the conservation of particlesnumber. Let us consider a fluid element of volume V and let us choose a LICF as describedbefore. The volume V contains nV particles, therefore, if . is the proper time associated tothe LICF origin, P0, the conservation of particles number can be written as

d

d.(nV ) = 0 . (17.3)

Note that this equation is not covariant, because V is not a scalar quantity. We shall nowshow that the generalization of this law, valid in any reference frame, is

(nu&);& = 0 . (17.4)

In the case of a star, n is the baryon number density; indeed, the baryon number is conservedby all interactions. If we assume that the star does not contain antimatter and that themesons content is negligible, the baryon number 2 coincides with the number of baryons.Since baryons are much heavier than electrons and neutrinos, the star “rest mass” is con-sidered as due to baryons only. Therefore, in the following we will refer to n as the baryonnumber density.

Proof of equation (17.4)To hereafter we shall use geometric units G = c = 1.Let us assume for simplicity that V is a cube of edges %x = %y = %z = L and that theLICF origin, P0, is chosen as in figure 17.3. In P0 the fluid is at rest, but within the volumeit has a small velocity

vi =dxi

dt, (17.5)

where t is the coordinate time of the chosen LICF. In order to show that the covariant formof the barion number conservation law is (17.4), we firstly expand eq. (17.3) valid in a LICF

d

d.(nV ) = 0 ' u& (nV ),& = 0 ' n,&V u& + nV,&u

& = 0

+n,&V u& + n[V,0u

0 + V,iui] , i = 1, 3 . (17.6)

Let us first evaluate the order of the various terms. In a LICF the metric reduces to /µ' andits first derivatives vanish, therefore in the small volume V gµ' = /µ'+O(|x&|2). Furthemore,

d. =8"g&(dx&dx( = dt

81 + O(|x&|2) = dt + O(|x&|2) . (17.7)

2The baryon number is B =nq " nq̄

3where where nq is the number of quarks, and nq̄ is the number of

antiquarks, and is a conserved quantum number.


Le

ee

3

12

P0

V

x

y

z

Figure 17.3: The fluid volume element V in a LICF at some fixed instant of time.

The components of u& consequently are

u0 =dt

d.= 1 + O(|x&|2)

ui =dxi

d.= vi + O(|x&|2) . (17.8)

Since the LICF is a comoving frame, vi(P0) = 0 and if we assume that #vi/#xj is finite, wefind

limvi

xj= lim

vi " vi(P0)

xj " xj(P0)=

#vi

#xj' vi =

#vi

#xjxj , (17.9)

i.e.vi = O(|x&|) and ui = O(|x&|) . (17.10)

Using eqs. (17.8) and (17.10), the term [V,0u0 + V,iui] in eq. (17.6) becomes

V,0u0 + V,iu

i = V,0[1 + O(|x&|2)] + O(|x&|) = V,0 + O(|x&|) . (17.11)

Let us evaluate how the volume V changes in a time interval 0t. The edges of each face ofthe cube change as follows

0(%x) =

4dx

dt0t

5

front face

"4dx

dt0t

5

back face

=#vx

#xL0t


0(%y) =#vy

#yL0t

0(%z) =#vz

#zL0t . (17.12)

The corresponding volume change is

0(%x%y%z) = 0(%x)%y%z+0(%y)%z%x+0(%z)%x%y =39

i=1

#vi

#xiL30t , #vi

#xiL30t (17.13)

so that#V

#t= V

#vi

#xi. (17.14)

Thus eq. (17.11) becomes

V,0u0 + V,iu

i = V#vi

#xi+ O(|x&|) ,

and since ui = vi + O(|x&|2) and u0 = 1 + O(|x&|2), we find

V,0u0 + V,iu

i = V#u&

#x&+ O(|x&|) . (17.15)

By replacing this term in eq. (17.6) we finally find

d

d.(nV ) = 0 ' n,&V u& + nV u&,& = 0 ' (nu&),& ' (nu&);& = 0 (17.16)

where we have used the property that in locally inertial frames ordinary and covariantderivative coincide. Since this is a tensorial equation, it must be valid in any referenceframe, Q.E.D.

17.2.2 The first law of Thermodynamics

Given a fluid with energy density ( and entropy per baryon s, and given a fluid element ofvolume V, formed by a given number of baryons A = nV , it will have an energy E = V (,where ( is the energy density, and an entropy S = As. The Ist law of thermodynamics

dE = "pdV + TdS (17.17)

can then be written as

d)

A

n(*

= "pd)

A

n

*+ Td(As) . (17.18)

Multiplying by n/A,

d( =( + p

ndn + nTds . (17.19)

Given the EOS( = ((n, s), (17.20)


we find that '#(

#n

(

s

=( + p

nand

'#(

#s

(

n

= nT (17.21)

and then the pressure and density of the fluid are

p(n, s) = n

'#(

#n

(

s

" ( (17.22)

T (n, s) =1

n

'#(

#s

(

n

. (17.23)

Thus, given an equation of state, namely a relation between one thermodynamical variable(( in the previous example) and two of the remaining variables (n and s), the remainingvariables (p and T in the example) can be determined in terms of them.

Another important function in the description of a fluid is the chemical potential µ, whichis the energy per baryon required to create a small extra quantity of fluid, and to insert it inthe volume V = A/n of fluid, in the same thermodynamical state. The extra fluid, composedby 0A baryons, has mass-energy 0M = )

n0A ((/n being the energy per baryon), and the workneeded to include it in the fluid is 0W = p0V = p *An , therefore

µ =0M + 0W

0A=

( + p

n=

'#(

#n

(

s

. (17.24)

17.2.3 Barotropic equation of state

If the EOS does not depend on temperature or entropy, it is named barotropic, and can bewritten as an equation relating the pressure and the energy density

p = p(() or ( = ((p). (17.25)

For instance, the EOS of matter in neutron stars a few years after birth can be consideredas barotropic, because, as explained in section 17.1, one can assume that matter behaves asa degenerate gas at zero temperature.

For a barotropic EOS, the first law of thermodynamics becomes

d( =( + p

ndn . (17.26)

Notice that di!erentiating the definition of the chemical potential µ , )+pn , using (17.26) we

find

dµ = "( + p

n2dn +

d( + dp

n=

dp

n; (17.27)

therefore for a barotropic EOS we also have

dp = ndµ . (17.28)


17.2.4 The Stress-Energy tensor of a perfect fluid

In relativity a fluid is said “perfect” if both viscosity and heat flow are absent. We shall nowshow that the stress-energy tensor of a perfect fluid is

T µ' = (( + p)uµu' + pgµ' . (17.29)

As explained in chapter 7, T 00 is the energy density; T 0i (i = 1, 2, 3) is the energy whichflows per unit time across the unit surface orthogonal to the axis xi; T ij is the amount ofith-component of momentum which flows per unit time across the unit surface orthogonal tothe axis xj .

Furthermore, since the momentum which flows per unit time is a force, T ij can also beinterpreted as the ith-component of the force per unit surface orthogonal to the axis xj .

Let us consider a fluid element and the associated LICF. In this frame the fluid is at rest(the velocity of the fluid inside the element is of order O(|x&|)) and the components of thestress-energy tensor are the following

• T 00 = (, the energy density.

• T 0i = 0, indeed the fluid element does not exchange energy with its surroundings,because the fluid is at rest and there is no heat flow.

• T ij = p0ij. Indeed in a perfect fluid no tangential stresses are allowed, which meansthat the force exerted on the surface orthogonal to the axis xj must be parallel to theaxis xj , and the force per unit surface is, by definition, the pressure;

Thus, in the chosen rest frame the fluid stress-energy tensor is

T µ' =

:

;;;<

( 0 0 00 p 0 00 0 p 00 0 0 p

=

>>>? . (17.30)

and since in this frame uµ = (1, 0, 0, 0), it can also be written as

T µ' = (( + p)uµu' + p/µ' . (17.31)

Since in a LICF gµ' , /µ' , we can also write

T µ' = (( + p)uµu' + pgµ' . (17.32)

This is a tensorial expression, and the principle of general covariance establishes that it mustbe valid in any other reference frame. Thus, eq. (17.32) is the covariant form for the stress–energy tensor of a perfect fluid in general relativity. Note that according to eq. (17.30), whichfollows from the assumption that viscosity and heat flow are absent, a comoving observersees the fluid around him as isotropic.


17.2.5 Conservation laws for the stress-energy tensor

The stress-energy tensor (17.32) satisfies the conservation law (see chapter 7)

T µ';' = 0 . (17.33)

Its contraction with uµ gives

uµTµ'

;' = uµuµu'(( + p),' + (( + p)(uµu

µu';' + uµuµ;'u

') + u'p,'

= "u'(,' " (( + p)u';' = 0, (17.34)

where we have used the relation

uµ(uµ);' =

1

2(uµuµ);' = 0 . (17.35)

Using the baryon number conservation (17.4), eq. (17.34) gives

u'(,' = "(( + p)u';' =( + p

nu'n,' (17.36)

i.e.d(

d.=

( + p

n

dn

d.; (17.37)

on the other hand, from the first law of thermodynamics (17.19),

d(

d.=

( + p

n

dn

d.+ nT

ds

d., (17.38)

and the two equations are compatible only if

ds

d.= 0 , (17.39)

which means that a fluid element does not exchange heat with its surroundings, as it mustbe for a perfect fluid.

Thus, the contraction of the stress-energy tensor conservation law with the fluid four-velocity and the baryon number conservation, implies that a perfect fluid is isentropic.

To study the space components of (17.33) we define the projector onto the subspaceorthogonal to uµ:

Pµ' = gµ' + uµu' . (17.40)

It is a projector because P 2 = P , and it projects onto the subspace orthogonal to uµ becausePµ'u' = 0.

By applying Pµ' to eq. (17.33) we find

P!&T&(

;( = P!&

@(( + p),(u

&u( + (( + p)(u(u&;( + u&u(;() + g&(p,(

A

= (g!& + u!u&)(( + p)u(u&;( + P (! p,(

= (( + p)u(u!;( + P (! p,( = 0 (17.41)

where we have used eq. (17.35). This equation gives

P (! p,( = "(( + p)u(u!;( , (17.42)

and says that the pressure gradient projected on the subspace orthogonal to uµ (that is, thespace gradient of the pressure) is equal to the fluid acceleration, u(u!;(, times the energydensity (plus the pressure); this is the relativistic generalization of one of Euler’s equation.


17.3 The equations of stellar structure in general rel-ativity

In this section we shall derive the equations which describe the structure of a non rotatingstar in static equilibrium according to general relativity. Since the spacetime generated bysuch star is static and spherically symmetric, the appropriate form of the metric is

ds2 = "e2'(r)dt2 + e2+(r)dr2 + r2(d12 + sin2 1d22). (17.43)

In this expression and in the following we shall use geometric units, setting G = c = 1. Weshall assume that the star is composed by a perfect fluid of stress-energy tensor

T &( = (( + p)u&u( + pg&(, (17.44)

where u& = dx!

d, is the four-velocity of an element of fluid, and p and ( are the pressureand the energy-density measured by an observer in a locally inertial frame locally at restwith respect to the fluid as discussed in previous sections.

It should be stressed that ( is the relativistic energy density, which reduces to the restenergy density !c2 (where ! is the mass density) in the non relativistic limit.

At this point some considerations are needed about the dimensions of the physical quan-tities we are dealing with. Since we are working in geometrical units G = c = 1, Tµ' has thesame dimensions as Gµ' , i.e.

[Tµ' ] = [l#2].

Consequently, both ( and p are [l#2] quantities. This means that

( =G

c4(phys, and p =

G

c4pphys, (17.45)

where (phys and pphys are the energy density and the pressure in physical units, i.e. [(phys] =[pphys] = [ml#1t#2].

Since by assumption the fluid is at rest, the only non vanishing component of the velocityof the generic fluid element is given by

gµ'uµu' = "1 ' u0 = e#' , u0 = "e' , (17.46)

hence the non vanishing components of the stress-energy tensor are$

T00 = (e2' T-- = r2pTrr = pe2+ T.. = sin2 1T--.

(17.47)

The pressure and energy-density are related by an assigned equation of state.The equations to solve are

$a) Gµ' = 8$Tµ'

b) T µ';' = 1%

#g//x" (

$"gT µ') + &µ

+'T'+ = 0

, (17.48)


where $T 00 = (e#2' T -- = p

r2

T rr = pe#2+ T .. = 1sin2 -

T --.(17.49)

It should be noted that eqs. (17.48) a) and b) are not independent. Indeed, as discussed inchapter 8, the divergenceless equation satisfied by the stress-energy tensor is a consequence ofthe Bianchi identities satisfied by the Riemann tensor. We write explicitly the two equationsto make the calculations easier.

In order to write explicitely eq. (17.48b) we need the expression of Christo!el’s symbols

&r00 = e2('#+)+,r &r

-- = "e#2+r (17.50)

&00r = +,r &.

-. =cos 1

sin 1&r

rr = 3,r &r.. = "e#2+r sin2 1

&-r- = 1

r &-.. = " cos 1 sin 1

&.r. = 1

r

$"g = r2e'++ sin 1.

The only non trivial component of eq. (17.48b) is µ = r which gives

1$"g

#

#x'

6$"gT r'

7+ &r

+'T'+ = 0, (17.51)

i.e.

1$"g

#

#r

6$"gT rr

7+ &r

00T00 + +&r

rrTrr + &r

--T-- + &r

..T.. = (17.52)

e#('++)

r2

6r2e('++)pe#2+

7

,r+ e2('#+)+,r(e

#2' + e#2+3,rp " 2e#2+p

r= 0,

which becomes+,r = " p,r

( + p. (17.53)

Einstein’s equations give

a) G00 = 8$T00,1

r2e2' d

dr

Br61 " e#2+

7C= 8$(e2' (17.54)

b) Grr = 8$Trr, " 1

r2e2+

61 " e#2+

7+

2

r+,r = 8$pe2+

c) G-- = 8$T--, r2e#2+

4

+,rr+2,r +

+,r

r" +,r3,r "

3,r

r

5

= 8$r2p.

If we put

m(r) =1

2r61 " e#2+

7, ' e#2+ = 1 " 2m(r)

r, (17.55)

eq. (17.54a) becomesdm(r)

dr= 4$r2(, (17.56)


which is the generalization of the newtonian equation (16.16).Eq. (17.54b) can be rewritten as

61 " e#2+

7

r2" 2e#2++,r

r= "8$p, (17.57)

and by using eq. (17.55) it becomes

+,r =m(r) + 4$r3p

[r (r " 2m(r))]. (17.58)

In the newtonian limit the pressure in geometric units is small compared to the energy-density. For example in the case of the Sun, the ratio between the central pressure anddensity is ! 10#6. In addition m(r) << r and eq. (17.58) reduces to

+,r =m(r)

r2. (17.59)

Remembering that in this limit e2' ' 1 + 2!c2 where ' is the newtonian potential and

',r = m(r)r2 , eq. (17.59) simply says that the gravitational force is that of the mass enclosed

within a sphere of radius r. From eq. (17.58) we see that in general relativity there is theadditional contribution, 4$r3p, which is due to the pressure. This should not be surprising,because dimensionally p is an energy density, thus the term 4$r3p acts as an e!ectivemass. This means that the active mass which attracts the mass shell between r and r+dris due to both contributions, and the pressure which should contrast gravity, to some extentenhance its e!ects. This phenomenon is called regeneration of the pressure.

Eqs. (17.53) and (17.58) can be combined, and the final set of equations, known as theOppenheimer-Volko! equations (TOV equations), is

+,,,,,-

,,,,,.

dm(r)

dr= 4$r2(

dp

dr= "(( + p)[m(r) + 4$r3p]

r[r " 2m(r)].

(17.60)

17.3.1 The boundary conditions

To integrate this system we need

1. to choose an equation of state connecting the pressure and the energy density

2. to impose that m(r = 0) = 0.

The reason for the condition (2) is the following. Take a tiny sphere of radius x. Thecircumference will be 2$x and the proper radius

r =! x

0e+dr # e+x, (17.61)


hence their ratio is 2$e#+. Since the spacetime is locally flat the ratio between the circum-ference of an infinitesimal sphere and the radius must be 2$. This implies that as r ' 0e+ must tend to 1. Since

e2+ =1

1 " 2m(r)r

, (17.62)

it follows that m(r) must tend to zero faster then r.For any assigned equation of state p = p((), we have a one-parameter family of solutions

identified by the value of the energy density at r = 0, i.e. ((r = 0) = (0. Once m(r), p(r),and ((r) have been determined by numerical integration, e2+ follows from eq. (17.62)and +(r) can be found by integrating eq. (17.53)

+ =!

" p,r

(( + p)dr + +0, (17.63)

where +0 is a constant to be determined. The solution of eqs. (17.60), together with (17.62)and (17.63), describes the gravitational field and the distribution of pressure and energydensity inside the star.

Outside the star p = ( = 0 and Einstein’s equations reduce to those for a vacuum,static, spherically symmetric spacetime whose unique solution is, by Birkho!’s theorem, theSchwarzschild metric. Thus, the metric computed in the interior of the star must reduce tothe Schwarzschild metric when r = R, and by imposing this condition we find the constant+0 of eq. (17.63):

e2'(r=R) , e2'0 · e2& R

0# p,r

(#+p)dr = 1 " 2M(R)

R' e2'0 =

1 " 2M(R)R

e2& R

0# p,r

(#+p)dr, (17.64)

and the constant appearing in eq. (17.63) can determined. The quantity

M(R) = 4$! R

0r2( dr, (17.65)

has the same form as in newtonian theory and one may think that it does represent the totalmass-energy of the field. However, this interpretation is incorrect, and the reason is thatsince 4$r2dr is not the element of proper volume between r and r +dr, m(r) cannotbe interpreted as the sum of all proper energies, and does not include the contribution ofthe gravitational potential energy. The true total mass-energy of the system is

M tot(R) =!

V((r) d3Vprop =

!

V((r)

8g(3) dr d1 d2 (17.66)

=! R

0

((r)r2dr8

1 " 2m(r)r

! "

0sin(1) d1

! 2"

0d2 = 4$

! R

0

((r)r2

81 " 2m(r)

r

.

The quantity M tot(R) " M(R) is the binding energy of the star.


A note on the chemical potential

Let us consider a spherical star with a barotropic equation of state p = p((); combining eqs.(17.24) and (17.28) we find ZZZ

dp = ndµ =( + p

µdµ . (17.67)

By integrating this equation, and using eq. (17.63) we find

! µ(r")

µ(r)

dµ

µ=

! p(r")

p(r)

dp

( + p= "

! '(r")

'(r)d+ = +(r) " +(r$), (17.68)

i.e.µ(r)e'(r) = µ(r$)e'(r

") = constant . (17.69)

This equation says that the chemical potential, corrected by the redshift factor e' , at anydepth in the star is a constant. In particular, for any r < R (R stellar radius)

µ(r)e'(r) =)1 " 2M

R

*1/2

µ(R) . (17.70)


17.4 The Schwarzschild solution for a homogeneousstar

An analytic solution of the equations of stellar structure (17.60) can be obtained by consid-ering the very simple equation of state:

( = const.

This solution was found by K. Schwarzschild in 1916 and this is the only exact solution ofeqs. (17.60) found up to the present time.

Although homogeneous stars are unrealistic (the speed of sound v =6

dpd)

7' (,) they

can be used as a good approximation for the core of very dense stars, and the interiorSchwarzschild solution has been used as a simplified model in a variety of situations to studythe e!ects of gravity in a regime as strong as it can ever become under the condition ofhydrostatic equilibrium.

If ( = const

m(r) =4

3$r3 (, (17.71)

and from eq. (17.55) one of the metric functions is immediately found

e2+ =

'

1 " 2m(r)

r

(#1

' e2+ =)1 " 8

3$(r2

*#1

. (17.72)

The Oppenheimer-Volko! equations reduce to

dp

dr= "4

3$r

(( + p)(( + 3p)

1 " 8"3 r2(

, (17.73)

that can be integrated to find the pressure; the integration is performed between r and theradius of the star r = R, where the pressure vanishes:

log

'( + 3p

( + p

(DDDDD

0

p(r)

=1

2log

)1 " 8$

3r2(

*DDDDD

R

r

, (17.74)

which gives

p = ((y " y1)

(3y1 " y), (17.75)

where

y2 = 1 " 8$

3r2( = 1 " 2m(r)

r, and y2

1 = y2(R) = 1 " 2M

R, (17.76)

and M , M(R).It is interesting to note that if we put r = 0 in eq. (17.75) we find

p(r = 0) = p0 = (1 "

81 " 2M

R

38

1 " 2MR " 1

. (17.77)


If the denominator of this expression is zero the central pressure becomes infinite, and neg-ative if it is smaller than zero. Thus, homogeneous stars can exist only if

3

E

1 " 2M

R" 1 > 0 ' M

R<

4

9, (17.78)

or, equivalently,

R >9

4M . (17.79)

This equation sets a lower limit on the radius that a star of a given mass can have, provided( = const. However, in the next section we will show that this result holds for a genericequation of state.

The radius of the star can be found by integrating eq. (17.73) from r = 0 (where p = p0)and r

" log

'( + 3p

( + p

(

=1

2log

)1 " 8$

3R2(

*' log

'( + 3p

( + p

(#1

= log

E)1 " 2M

R

*

(17.80)from which we find

R =2M

B1 " ()+p0)2

()+3p0)2

C . (17.81)

Thus for any assigned value of ( and of p0 we have a configuration of radius R given by(17.81).

To complete the solution we need to find the metric function +(r), which can be deter-mined from eq. (17.63)

+ " +0 = "! r

0

p,r

[( + p(r)]dr = "

! y

1

p,y

[( + p(y)]dy; (17.82)

since

p,y = (

42y1

(3y1 " y)2

5

, (( + p) = (

42y1

(3y1 " y)

5

,

we find

+ = +0 "! y

1

dy

(3y1 " y)' e2' = e2'0

'3y1 " y

3y1 " 1

(2

. (17.83)

At the boundary of the star y(R) = y1 and

e2'(R) = e2'0

'4y2

1

(3y1 " 1)2

(

; (17.84)

on the other hand we know that the metric must reduce to the Schwarzschild metric, thereforeit must also be

e2'(R) = 1 " 2M

R, y2

1, (17.85)

and by equating eq. (17.84) and (17.85) we find the value of the integration constant +0

e2'0 =(3y1 " 1)2

4,


and the solution for +(r) is completely determined

e2'(r) =(3y1 " y)2

4. (17.86)

17.5 Relativistic polytropes

In this section we shall generalize the Lane-Emden equation in general relativity. Followingwhat we did in section 12.4 for the newtonian case (see eqs. 16.38), we shall solve therelativistic equations of stellar structure (17.60) assuming that the equation of state of thematter inside the star is of a polytropic form, i.e.

+,,,,,,,,,,-

,,,,,,,,,,.

dm(r)

dr= 4$r2(

dp

dr= "(( + p)[m(r) + 4$r3p]

r[r " 2m(r)]

p = K(! ,

(17.87)

where we remind that ( is the relativistic energy density, and we shall also assume that (and p are expressible in the manner (cfr. eq. (16.40))

+,,,,,,,,-

,,,,,,,,.

) = 1 +1

n,

( = (0 #n(r)

p = K !1+ 1

n0 #(n+1)(r) = p0 #(n+1)(r), p0 = K !

1+ 1n

0 ,

(17.88)

With these substitutions eqs. (17.87) become+,,,,,-

,,,,,.

dm(r)

dr= 4$(0r

2#n(r)

d#(r)

dr= "

4(0 + p0#

p0(n + 1)

5m(r) + 4$r3p0 #(n+1)

r[r " 2m(r)].

(17.89)

By putting

*0 =(0p0

, (17.90)

these equations become+,,,,,,-

,,,,,,.

dm(r)

dr= 4$(0r

2#n(r)

d#(r)

dr= "

4*0 + #

(n + 1)

5m(r) + 4$r3 )0

&0#(n+1)]

r[r " 2m(r)].

(17.91)


As explained in section 12.7, both ( and p have dimensions [l#2], therefore the quantity$(0 has dimension [l#1] and we can use it to rescale the radial coordinate as follows. We put

% = r$(0, and M =

$(0 m (17.92)

and rewrite eqs. (17.91) in terms of the new variables (note that % and M are dimensionlessquantities) +

,,,,,,-

,,,,,,.

dM(%)

d%= 4$%2#n(%)

d#(%)

d%= "

4*0 + #

(n + 1)

5 BM(%) + 4$%3 1

&0#(n+1)

C

% [% " 2M(%)].

(17.93)

We may, at this point, multiply the second equation by %2, di!erentiate with respect to %and substituting dM($)

d$ into the resulting equation find a second order di!erential equationfor # in the Lane-Emden form (cfr. 16.43); however, the equation we would get is muchmore complicated than eq. (16.43), and it is much better to work with the system of twofirst order eqs. (17.93).

Another important di!erence with the equation of newtonian polytropes should be stressed.In the newtonian case if we assign the value of the polytropic index n and integrate the Lane-Emden equation finding #(%) up to the stellar radius %1, from this solution we can constructa family of solutions by assigning the value of K and of the central density !0; no furtherintegrations are needed and, for instance, the radius and the mass of the star can be foundby eqs. (16.48) and (16.49). The situation changes in the relativistic equations, because tosolve eqs. (17.93) we need to assign both n and *0, i.e. the ratio between the energy densityand the pressure at % = 0.

In order to integrate the structure equations numerically, we need to Taylor expand thefunctions #(%) and M(%) near the origin as in (16.55); from the newtonian expansion weknow that only even powers of # are needed and therefore M(%) will be expanded in oddpowers of % (cfr. the first eq. 17.93), therefore we shall put

#(%) ! 1 + #2 %2 + #4 %4 + O(%6),

M(%) ! m3%3 + m5%

5 + O(%7).

By inserting this expansions in eqs. (17.93) we find

3m3%2 + 5m5%

4 = 4$%2 + 4$n #2 %4

2 #2 % + 4 #4 %3 = " 1

n + 1

/)m3 +

4$

*0

*[( 1 + *0)% + #2 %3]

F

and by equating the coe"cients of the same power of % we find

m3 " 4"3 = 0 m5 "

4$n#2

5= 0 (17.94)

#2 + 1+&02(n+1)

6m3 + 4"

&0

7= 0 #4 +

#2

2(n + 1)

)m3 +

4$

*0

*= 0


from which we find

m3 = 4"3 , #2 = "2$

(1 + *0)(3 + *0)

3*0(n + 1)

m5 = 4"n"25 #4 = " #2

2(n + 1)

)m3 +

4$

*0

*.

With these initial conditions we can integrate the structure equations and find the value of %where the function # vanishes so that the pressure vanishes and we are sure we have reachedthe boundary of the star. Be %1 such value and #$

1 = #$(%1). From the second eq. (17.93)we find

#$1 =

4*0

(n + 1)

5M(%1)

%1[%1 " 2M(%1)],

from which we find the mass of the star

M(%1) =(n + 1)%2

1 |#$1|

*0 + 2%1(n + 1)|#$1|

. (17.95)

Once we know the function #(%), from eqs. (17.88) we know how the energy density andthe pressure are distributed inside the star and we can compute the function M(%)

M(%) =! $

04$(0%

"2#n(%$) d%$,

and the metric function e2+

e2+ =1

1 " 2M($)$

.

The remaining metric function e2' can be found from eq. (17.63) which now becomes

+ =! $

0" p,$

(( + p)d% + +0 =

! $

0"

(n + 1)d"d$

*0 + #d% + +0 = ln

4*0 + 1

*0 + #(%)

5(n+1)

+ +0, (17.96)

At the surface of the star the metric must reduce to the Schwarzschild metric and therefore

e2'0 ·4

(*0 + 1)

(*0 + #)

52(n+1)

= 1 " 2M(%)

%

which, using eq. (17.95), gives

e2'0 =G

*0

*0 + 1

H2(n+1)

· *0

*0 + 2%1(n + 1)|#$1|

(17.97)

Thus,

e2'($) =G

*0

*0 + 1

H2(n+1) *0

*0 + 2%1(n + 1)|#$1|·4

*0 + 1

*0 + #(%)

52(n+1)

(17.98)

and the solution is finally complete.


17.6 Buchdal’s theorem

A theorem proved by Buchdal in 1959 establishes that the result obtained in the section17.4 about the maximum value that the ratio M/R in a star of constant energy-density canreach, i.e. M/R < 4/9, is much more general. The theorem is based on the only assumptionthat the star is static, and that the energy density is positive, and monotonically decreasingfunction of the radial coordinate, i.e.

( - 0,d(

dr* 0.

No assumption is made on the equation of state that relates ( and the pressure p. Therelevant equations are

/Grr = 8$Trr

G-- = 8$T--

+-

.(1) " e2$

r2

B1 " e#2+

C+ 2

r+,r = 8$pe2+,

(2) r2e#2+B+,rr + +2

,r + ',r

r " +,r3,r " +,r

r

C= 8$r2 p,

(17.99)

By taking the following combination of eqs. (17.99)

r2e#2+EQ.(1) " EQ.(2) = 0, (17.100)

we findd

dr

4e#+

r

d

dr(e')

5

= e'++d

dr

4m(r)

r3

5

, (17.101)

where, as usual, m(r) = 4$& r0 (r$2dr$. For any r we can always define a density (r such

that

m(r) =4

3$(rr

3, (17.102)

and since ( is a monotonically decreasing function of r, (r, and consequently m(r)r3 ,

are also monotonically decreasing. Hence

d

dr

4e#+

r

d (e')

dr

5

* 0. (17.103)

It should be noted that the minimum value of (r is attained at the boundary, where

M =4

3$(RR3, (17.104)

where (R = (min. Inside the star (r - (min, and consequently

4

3$(rR

3 - 4

3$(minR3, ' m(r) - M

R3r3, (17.105)

i.e. m(r) is always bigger than what it would be if the density would be constant andequal to (min. From eq. (17.103) it follows that

e#+

r

d (e')

dr- e#+

r

d (e')

dr

DDDr=R

. (17.106)


When r = R the metric reduces to the Schwarzschild metric, therefore e2'DDDr=R

=

e#2+DDDr=R

= 1 " 2MR and eq. (17.106) gives

e#+

r

d (e')

dr- M

R3' de'

dr- re+

M

R3. (17.107)

By integrating eq. (17.107) between 0 and R we find

e'(R) " e'(0) - M

R3

! R

0re+dr, (17.108)

and since e#2+ = 1 " 2m(r)r

e'(0) *E

1 " 2M

R" M

R3

! R

0

rdr8

1 " 2m(r)r

. (17.109)

We want to establish an upper boundary for e'(0), and therefore we need to determinewhen the right hand side of eq. (17.109) attains its maximum value. From eq. (17.105) weknow that m(r) - M

R3 r3, and consequentlyE

1 " 2m(r)

r*

E

1 " 2M

R3r2, '

! R

0

rdr8

1 " 2m(r)r

-! R

0

rdr8

1 " 2MR3 r2

. (17.110)

Thus the maximum value of the right hand side of eq. (17.109) is

R.H.S.max =

E

1 " 2M

R" M

R3

! R

0

rdr8

1 " 2MR3 r2

=3

2

E

1 " 2M

R" 1

2, (17.111)

and consequently

e'(0) * 3

2

E

1 " 2M

R" 1

2. (17.112)

We shall now use the second hypothesis of the theorem, i.e. the condition that the metricis static. A static spacetime admits a timelike Killing vector, which must remain timelike inthe interior of the star, i.e.

-% · -% = g00(%0)2 < 0 ' g00 = "e2' < 0 ' e2' > 0. (17.113)

It follows from eq. (17.112) that

3

2

E

1 " 2M

R" 1

2> 0, (17.114)

and finallyM

R<

4

9. (17.115)

Q.E.D.It should be noted that, since M

R < 49 , it follows that R > 9

4M, and since theSchwarzschild radius is RS = 2M this means that a star cannot have radius smaller thanthe Schwarzschild radius.


17.7 A necessary condition for the stability of a com-pact star

A solution of the TOV equations (17.60) which satisfies the appropriate boundary conditionsdiscussed in section 17.3.1, describes a stellar configuration in hydrostatic equilibrium. Thisequilibrium can, in principle, be either stable or unstable. In this section we will study theconditions for stability.

Let us consider a sequence of equilibrium configurations obtained by integrating the TOVequations for an assigned EOS, with di!erent values of the central energy density (0. Thegravitational mass will thus be a function of (0: M = M((0).

Let us consider the profile M((0) given in figure 17.4. Each point of this curve identifiesan equilibrium configuration. Given a star in the equilibrium configuration A if a small radial

M

!"

! !" "11

22

C

## #210 0 0

Figure 17.4: The mass of equilibrium stellar configurations is plotted versus the centralenergy density, near a relative maximum.

perturbation reduces its central energy density to a value, say, (01, the new (non-equilibrium)configuration will be represented by point A1 (because the mass of the star does not change).Point A1 is above the curve, therefore the perturbed star has a mass which is larger than themass that the equilibrium configuration corresponding to (01 would have. Consequently, thestar is o! equilibrium because gravity exceeds pressure, and the star will contract, so thatits central density increases and it can return to the equilibrium configuration A.

In a similar way, if a perturbation increases the central energy density to (02, the newconfiguration is represented by a point A2 below the curve. The star in A2 has mass smallerthan that of the equilibrium configuration corresponding to (02. In this case gravity is weakerthan pressure, and the star will expand to return to the equilibrium configuration. Thus,the equilibrium in A is stable.

We can conclude that if A is a stable equilibrium configuration, in A

dM

d(0> 0 , (17.116)


Conversely, a similar discussion about the point B where

dM

d(0< 0 , (17.117)

shows that a displacement to B1, brings the star to a configuration where gravity is weakerthan pressure, so that the star expands further reducing the central density. Similarly, adisplacement to B2 brings the star to a configuration where gravity exceeds pressure, so thatthe star contracts, and the central density further increases: the equilibrium in B is unstable.

In figure 17.4, the branch on the left of the maximum C corresponds to stable config-urations, that on the right to unstable configurations. The point C is the configuration ofmaximum mass.

An example is the case of Newtonian polytropes: the function M((0), given in eq.(16.49),which we rewrite here for simplicity

M = 4$ %21 |#$(%1)|

4(n + 1)K

4$G

5 32

· !3!n2n

0 (17.118)

shows that M is an increasing function of the central density !0 for n < 3, it is stationaryfor n = 3, and decreasing for n > 3; therefore the star is stable only if n < 3.

For a realistic equation of state, the curve M((0) has a profile similar to that shown infigure 17.5. The stable branch on the left of point A represents white dwarf configurations,

M

WD

NSA

B

C

#0

Figure 17.5: Masses of equilibrium stellar configurations vs. central densities. The stablebranches corresponding to white dwarfs and neutron stars are explicitly shown.

while the stable branch BC represents neutron star configurations.

If we consider the stellar mass as a function of the radius, we find that since

dM

dR=

dM

d(0· d(0dR

, (17.119)


the stability criterion (17.116) is satisfied if

a) dR/d(0 > 0 and dM/dR > 0or ifb) dR/d(0 < 0 and dM/dR < 0.

In general, both for white dwarfs and for neutron stars the radius of the star decreasesas the central density increases, therefore the stable branches of the function M(R) arethose for which

dM

dR< 0 . (17.120)

17.7.1 Is the condition dMd"0

> 0 su!cient to say that a star is stable?

The question in the heading of this subsection can be rephrased as follows:if dM

d)0> 0, can we say that the star is stable?

The answer is No, and the reason can be understood by considering the theory of radialperturbations of stars. Since this interesting development is outside the scopes of this book,we shall just sketch the main results of the theory and give the basic notions to understandthem.

A star has an infinite set of radial proper oscillation modes, labelled by an index n =0, 1, 2, . . .; when the star oscillates in a mode, each fluid element is displaced from the equi-librium position by a radial displacement %(t, r). For the nth- mode % has the form

%n(r, t) = un(r)ei0nt (17.121)

where "n is the mode frequency and u(r) its amplitude. The mode number n correspondsto the number of nodes that u(r) has inside the star: n = 0 for zero nodes, n = 1 for 1 nodeetc. The mode frequencies are ordered:

"20 < "2

1 < "22 < . . . , (17.122)

and the mode corresponding to "0 is said the fundamental mode.If "2

n > 0, the fluid element oscillates about the equilibrium position and the mode isstable; conversely, if "2

n < 0 the radial displacement grows exponentially and the mode isunstable.

A stable fundamental mode corresponds to a global oscillation of the star, which isexpanding or contracting all at the same time; indeed, this is also called the “breathingmode”. When the star contracts the central density increases and %(t, r) < 0 throughout thestar, when it expands the central density decreases and %(t, r) < 0. The previous discussionabout stability clearly applies to this case.

However the star may oscillate in a di!erent mode with n > 0. Since in this case u(r)has one node inside the star, we may have a situation in which near the origin %(t, r) < 0and the central density increases, but in some other region of the star %(t, r) > 0 and in thatregion the density would decrease. Thus, the previous discussion about stability would notbe applicable, and we need to appeal to the theory of radial pulsation to understand what is


going on. The theory states the following: suppose we compute a sequence of stellar models(with the same EOS) di!ering for the value of (0, and for each model we compute the massand the frequency of the various radial modes. Knowing M((0) along the sequence, we cancompute dM

d)0. If for some value of (0 we find that there is an extremal point, i.e.

dM

d(0= 0 , (17.123)

then for that same (c the square of the frequency of one of the modes must cross the realaxis and change sign, therefore in that point

"2i,)c = 0. (17.124)

This means the the ith-mode becomes unstable.In general, the n = 0 mode (which is the one with lowest frequency) is the first to become

unstable.Now suppose that we have the curve shown in figure 17.6 and suppose that for (0 < (A

the fundamental mode has frequency such that "20 > 0, i.e. it is stable. A is an extremal

M

A

B

C

##A 0

Figure 17.6: Masses of equilibrium stellar configurations vs. central densities. Though inthe BC branch dM

d)0> 0, that branch may correspond to unstable configuration, as explained

in the text.

point, therefore in A "20 = 0, and all configurations belonging to the branch AB will be

unstable because their fundamental mode will have "20 < 0.

If we increase the density we reach the second extremal point B. Here two things may happen:

1) "20 changes sign again becoming positive. In this case the star corresponding to B and all

configurations of the branch BC would be stable. This is the situation we have described inthe previous section.or2) "2

0 remains negative (i.e. the fundamental mode remains unstable) and the frequency ofthe n = 1 radial mode changes sign, so that also the n = 1 mode become unstable. In thiscase all configurations on the branch BC would be unstable.


This example clearly illustrates that the fact that dMd)0

> 0 does not provide a su"cientcondition for stability.

Chapter 16 White Dwarfs - Istituto Nazionale di Fisica Nucleare · CHAPTER 16. WHITE DWARFS 234 In...

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