chapter 18 white dwarfs...chapter 18. white dwarfs 240 in eq. (18.5) e c is the particle kinetic...

Chapter 18

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 ⇠< 8M� 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 8M�, 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� 2 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.

237

CHAPTER 18. WHITE DWARFS 238

18.1 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.


18.1.1 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), (18.1)

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 =Z

EdN

d3xd3pd3p =

g

h3

ZE f(x,p) d3p . (18.4)

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

f =1

eEc�µkT ± 1

, (18.5)

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↵ = �m

2c2, it follows that

�E2

c2+ p

2 = �m2c2,

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

asE =

⇥p2c2 +m

2c4⇤1/2

. (18.2)

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

� =

⇥p2c2 +m

2c4⇤1/2

mc2

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

v =pc

2

[p2c2 +m2c4]1/2. (18.3)


In eq. (18.5) 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

, (18.6)

where H is the enthalpy and U the internal energy. From eq. (18.5) 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 (Ec << kT ) the Bose-Einstein and theFermi-Dirac distribution tend to the classical Maxwell-Boltzmann distribution

f ⇠ eEc�µkT . (18.7)

Since f given in (18.5) 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. (18.4) becomes

E =4⇡g

h3

Z 1

0

E p2 dp

eEc�µkT ± 1

. (18.8)

The pressure can be written as

P =1

3

Zpv

dN

d3xd3pd3p =

4⇡g

3h3

Z 1

0

v p3dp

eEc�µkT ± 1

, (18.9)

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 =Z dN

d3xd3pd3x d3p =

4⇡gV

h3

Z 1

0

p2dp

eEc�µkT ± 1

, (18.10)

and

U =ZEc

dN

d3xd3pd3x d3p =

4⇡gV

h3

Z 1

0Ec

p2dp

eEc�µkT ± 1

.

18.1.2 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. (18.10) reduce to

N =4⇡gV

h3(2mkT )3/2

Z 1

0

x2dx

⇠�1ex2 ± 1, (18.11)

and

U =4⇡gV

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

Z 1

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

Z 1

0

x2dx

⇠�1ex2 ± 1' ⇠

Z 1

0x2e�x2

dx

Z 1

0

x4dx

⇠�1ex2 ± 1' ⇠

Z 1

0x4e�x2

dx;

thus, combining the expressions of N and U given in eqs. (18.11) we find

U

N= (kT )

R10 x4e�x2

dxR10 x2e�x2dx

and sinceR10 x4e�x2

dx = 38

p⇡ and

R10 x2e�x2

dx = 14

p⇡ 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. (18.11) we find

⇠ =Nh3

gV(2⇡mkT )�3/2 . (18.12)

If we now put n0 = N/V , where n0 is the number of particles per unit volume, and define adegeneracy temperature

Tdeg =h2

2⇡mk· n0

g

!2/3

, (18.13)

eq. (18.12) can be rewritten as

⇠ =✓Tdeg

T

◆3/2

. (18.14)

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. (18.13) shows that at a given density n0, Tdeg is higher for particles with smaller massm. Thus, electrons becomes degenerate earlier than heavier particles.EXAMPLES


• A hydrogen gas in normal condition, i.e. T = 300K, has number density n0 ⇠ 3 ·1019cm�3 and the degeneracy temperature is Tdeg ⇠ 0.18K; therefore ⇠ ⇠ 1.5 · 10�5

and the gas behaves as a classical perfect gas.

• For gases heavier than hydrogen ⇠ and Tdeg are even smaller, and consequently atordinary pressures and temperatures they are non-degenerate.

• A gas of photons is always degenerate because m = 0 and Tdeg = 1.

• 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 Tdeg ⇠ 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. (18.15)

The mass contained within a sphere of radius r is

M(r) =Z r

0⇢(r) 4⇡r2dr, ) dM(r)

dr= 4⇡r2⇢(r). (18.16)

Equations (18.15) and (18.16) 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 8

>>>>>>>>><

>>>>>>>>>:

dM(r)

dr= 4⇡r2⇢(r),

dP

dr= �GM(r)

r2⇢(r),

P = P (⇢).

(18.17)

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

18.1.3 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 ,

(18.18)

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. (18.19)

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 = 2Z pF

0d� =

Z pF

0

8⇡p2dp

h3=

8⇡

3h3p3F . (18.20)

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

⇢ = nmN , (18.21)


where mN = 1.67 · 10�24g is the mass of the nucleons. The electrons contribution to themass density is negligible since me << mN . From eqs. (18.20) and (18.21) we can find pFas a function of the density

pF = h✓

3

8⇡mN⇢◆ 1

3

. (18.22)

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

✏ =U

V=

8⇡

h3

Z pF

0{[p2c2 +m2

ec4]

12 �mec

2}p2dp, (18.23)

where Ec = [p2c2 + m2ec

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

(18.9) and (18.3)

P =8⇡

3h3

Z pF

0

p4c2

[p2c2 +m2ec

4]12

dp. (18.24)

These equations can be easily integrated in two regimes: 1) the non-relativistic and 2) theultra-relativistic 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. (18.22)

⇢crit = · 8⇡3mN

✓me c

h

◆3

= 0.98 · 106 · g/cm3. (18.25)

• 1) If ⇢ << ⇢crit, cpF << mec2 and the electrons are non relativistic. In this caseeq. (18.24) gives

P ⇠ 8⇡

3h3

Z pF

0

p4

medp =

8⇡

15h3· p

5F

me, (18.26)

and using eq. (18.22)

P =

2

4 h2

5me

✓3

8⇡

◆2/3 ✓ 1

mN

◆ 53

3

5 ⇢53 . (18.27)

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

P = K⇢�, where (18.28)

K =

2

4 h2

5me

✓3

8⇡

◆2/3 ✓ 1

mN

◆ 53

3

5 , and � =5

3.

Moreover, from eq. (18.23) the kinetic energy-density is

✏ ⇠ 8⇡

h3

Z pF

0{mec

2(1 +1

2

p2c2

m2ec

4)�mec

2}p2dp =4⇡

h3

Z pF

0

p4

medp, (18.29)


✏ =4⇡

5h3

p5Fme

=3

2P. (18.30)


• 2) If ⇢ >> ⇢crit, cpF >> mec2 and the electrons are ultra-relativistic. In this casefrom eq. (18.24) we find

P =8⇡

3h3

Z pF

0p3cdp =

2⇡c

3h3p4F , (18.31)


P =

2

4c h8

✓3

⇡

◆1/3 ✓ 1

mN

◆ 43

3

5 ⇢43 . (18.32)

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

P = K⇢�, where (18.33)

K =

2

4c h8

✓3

⇡

◆1/3 ✓ 1

mN

◆ 43

3

5 , and � =4

3.

Moreover

✏ =8⇡

h3

Z pF

0p3cdp, (18.34)

i.e.✏ = 3P. (18.35)

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 =

2

4 h2

5me

✓3

8⇡

◆2/3 ✓ 1

mN

◆ 53

3

5 = 9.9156 ·1012 ·�5/3 erg2 s2/g8/3, and � =5

3,

(18.36)

• ultra-relativistic regime ⇢ >> ⇢crit,

K =

2

4ch8

✓3

⇡

◆1/3 ✓ 1

mN

◆ 43

3

5 = 1.2316 · 1015 · �4/3 erg cm/g4/3, and � =4

3,

(18.37)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 a


critical 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.

18.1.4 The structure of a White Dwarf

We shall now find the equilibrium configuration of a white dwarf solving the newtonianequations of hydrostatic equilibrium (18.17) and using the results obtained in the previoussection.

As mentioned in section 18.1.2, in order to solve eqs. (18.17) 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 18.1.3, 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

8>>>>>>>>><

>>>>>>>>>:

dM(r)

dr= 4⇡r2⇢

dP

dr= �GM(r)

r2⇢

P = K⇢�.

(18.38)

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

r2d

dr

r2

⇢

dP

dr

!

= �4⇡G⇢. (18.39)

Be ⇢0 = ⇢(r = 0), the central density; by putting8>>>>>>>><

>>>>>>>>:

� = 1 +1

n, where n is called polytropic index

⇢ = ⇢0 ⇥n(r)

P = K ⇢1+ 1

n0 ⇥(n+1)(r),

(18.40)

eq. (18.39) becomes

(n+ 1)K ⇢( 1n�1)

0

1

r2d

dr

r2d⇥

dr

!

= �4⇡G ⇥n. (18.41)

If we now introduce the following dimensionless radial coordinate

⇠ =r

↵, where ↵ =

2

4(n+ 1)K ⇢( 1n�1)

0

4⇡G

3

5

12

, (18.42)


eq. (18.41) becomes1

⇠2d

d⇠

⇠2d⇥

d⇠

!

= �⇥n, (18.43)

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. (18.44)

Since ⇢ = ⇢0 ⇥n, the first condition implies that ⇥(0) = 1; moreover, since the mass goes tozero as M(r) ⇠ 4⇡

3 ⇢0r3, from eq. (18.38) it follows that

dP

dr⇠ �4⇡G

3r ⇢20,

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 ⇥0(r = 0) = 0. In conclusion the Lane-Emden equation(18.43) must be integrated by imposing that at the center of the star

(⇥(0) = 1,⇥0(0) = 0.

(18.45)

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. (18.43) with the initial conditions (18.45)

up to the value ⇠ = ⇠1 where ⇥ = 0. For instance, for � = 53 and � = 4

3 we wouldfind

� = 53 n = 3

2 ⇠1 = 3.65375 ⇠21 ⇥0(⇠1) = �2.71406 (18.46)

� = 43 n = 3 ⇠1 = 6.89685 ⇠21 ⇥0(⇠1) = �2.01824 (18.47)

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. (18.36) or (18.37). Choose a central density ⇢0. Knowing K and ⇢0 the radius ofthe star can be found using the definition of ⇠ given in eqs. (18.42)

R = ↵⇠1 ! R = ⇠1

"(n+ 1)K

4⇡G

# 12

· ⇢1�n2n

0 . (18.48)

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

M =Z R

04⇡r2⇢(r)dr = 4⇡↵3 ⇢0

Z ⇠1

0⇠2 ⇥n d⇠

= �4⇡↵3 ⇢0Z ⇠1

0

d

d⇠

⇠2d⇥

d⇠

!

d⇠

= �4⇡↵3 ⇢0 ⇠21 ⇥0(⇠1)

where use has been made of eq. (18.43). Finally, the value of M as function of K and⇢0 can be found by using the expression of ↵ given in (18.42)

M = 4⇡ ⇠21 |⇥0(⇠1)|"(n+ 1)K

4⇡G

# 32

· ⇢3�n2n

0 . (18.49)

Let us define

A =

"(n+ 1)K

4⇡G

#

, B = 4⇡ ⇠21 |⇥0(⇠1)|, (18.50)

so thatR = ⇠1 · A1/2 · ⇢

1�n2n

0 , (18.51)

andM = B · A3/2 · ⇢

3�n2n

0 . (18.52)

Combinig eqs. (18.51) and (18.52), a relation between M and R can easily bederived

M =✓B · A

nn�1 · ⇠

n�31�n1

◆·R

3�n1�n . (18.53)

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 ⇥0(⇠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(18.48) and (18.49).

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

⇢0 = 0

⇢, P 0 = P, (18.54)

M(r)0 =⇣

0

⌘2M(r), r0 =

0 r.


18.1.5 A note on the numerical integration of eq. (18.43)

Although the initial conditions (18.45) are correct, it would be impossible to integrate eq.(18.43) 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 (18.45) that ⇥(0) = 1 and ⇥0(0) = 0, we can write the approximatesolution near ⇠ = 0 as a power series

⇥(⇠) ⇠ 1 +⇥2 ⇠2 +⇥3 ⇠3 +⇥4 ⇠4 +O(⇠5), (18.55)

(we can keep as many terms we want, but let us stop here). ⇥1,⇥2,⇥3 .. are the constantswe need to find using eq. (18.43), therefore we also need to Taylor-expand the function ⇥n

on the right hand side, i.e.⇥n ⇠ 1 + n⇥2⇠

2 +O(⇠3); (18.56)

by substituting in eq. (18.43) the expansions (18.55) and (18.56) we find

6⇥2 + 12⇥3⇠ + 20⇥4⇠2 + ... = �[1 + n⇥2 ⇠2] + ... (18.57)

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 ⇥0 near the origin is

⇥(⇠) ⇠ 1� 1

6⇠2 +

n

120⇠4 +O(⇠6) (18.58)

⇥0(⇠) ⇠ �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 (18.59) computed at ⇠start.

18.2 The Chandrasekhar limit

In section 18.1.3 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. (18.50) give

A = 2.9562 · 10�19 �5/3 cm g1/3, B = 34.1059


and using eq.(18.25) and (18.52) we can write the mass of the star in this form

M = 2.73 �5/2

⇢0⇢c

!1/2

M�, (18.59)

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 (18.49) shows that in this limit the mass becomes independent ofthe central density ⇢0 and takes the value

M = MCH = 5.74 �2M�. (18.60)

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�. (18.61)

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. (18.62)

In the ultra-relativistic case

b) P ⇠ ⇢43 ! P ⇠ M

43

R4! dP

dr⇠ M

43

R5. (18.63)

The gravitational force per unit volume behaves like

Gm(r)⇢

r2⇠ M2

R5. (18.64)

If the star is in equilibriumdP

dr= �Gm(r)⇢

r2;

in the non-relativistic case the gradient of pressure (18.62) and the gravitational force (18.64)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 (18.63) and the gravitational force (18.64) 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 18 white dwarfs...chapter 18. white dwarfs 240 in eq. (18.5) e c is the particle kinetic...

Documents