1

Compact Stars – White Dwarfs, Planets, Neutron Stars

White dwarf structure dominated by P (ρ) for degenerateelectron gas.

R =

[

(n + 1)K

4πG

]1/2

ρ(1−n)/(2n)c ξ1

M = 4πR(3−n)/(1−n)[

(n + 1) K

4πG

]n/(n−1)

ξ(3−n)/(n−1)1

(

−ξ21θ′1

)

= 4π

[

(n + 1)K

4πG

]3/2

ρ(3−n)/(2n)c

(

−ξ21θ′1

)

.

Non-relativistic γ = 5/3:

K =

(

3π2)2/3

5

h̄2

me(NoYe)

5/3 = 1013Y5/3e cgs

R = 1.121 × 104ρ−1/6c,6 (2Ye)

5/6 km

M = 0.496ρ1/2c,6 (2Ye)

5/2 M� = 0.701

(

R

104 km

)−3

(2Ye)5 M�.

(1)

Relativistic γ = 4/3:

K =

(

3π2)1/3

4h̄c (NoYe)

4/3 = 1.24 × 1015Y4/3e cgs

R = 3.35 × 104ρ−1/3c,6 (2Ye)

2/3 km

M = 1.457 (2Ye)2 M�.

(2)

2

Very low density (Thomas-Fermi regime γ = 10/3):

K =31/3π3e2

10

(

4πNo

A

)10/3( 3h̄c

mec2

h̄c

e2

)6

= 1.05 × 1013(

12

A

)10/3

cgs

R = 1.18 × 105(

12

A

)5/3

ρ2/3c km

M = 0.001915

(

12

A

)5

ρ3c M�

= 2.88 × 10−8(

A

12

)5/2( R

104 km

)9/2

M�.

(3)

Physical reasoning behind the Chandrasekhar mass:

Consider N degenerate fermions in a star of radius R, sothat number density n ∝ NR−3. Momentum of a fermion is∼ h̄n1/3 and Fermi energy is EF ∼ h̄cN1/3R−1. The gravi-tational energy per fermion is ∼ −GMmBR−1 if M = NmB.The total energy is

E = EF + EG = h̄cN1/3/R − GNm2b/R.

Equilibrium is reached when this is minimized. Both termsscale as 1/R.

When E is positive, E can be decreased by increasing R.This decreases EF so that eventually the fermions become non-relativistic: then EF ∼ p2

F ∼ R−2. This then decreases fasterthan EG, so E becomes negative. However, as R → ∞, E → 0.This implies there is a minimum of E at a finite value of R.

When E is negative, E can be decreased without bound bydecreasing R so that no equilibrium state is possible and ablack hole forms.

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The maximum baryon number for equilibrium is determinedby setting E = 0:

Nmax ∼

(

h̄c

Gm2b

)3/2

∼ 2 × 1057

Mmax ∼ NmaxmB ∼ 1.5 M�.

Note that the mass is independent of the fermion’s mass.

The radius at equilibrium is set by the condition EF ≥ mc2,

R ≤h̄c

mc2

(

h̄c

Gm2B

)1/2

∼

{

5 × 103 km, m = me

3 km, m = mn .

At sufficiently high density, neutronization and pyconuclearreactions can occur. Thus, both A and N−Z will increase withdensity. The neutronization threshold for 56Fe is about 109 gcm−3. At this density, the Fermi energy of an electron is aboutmcc

2+3.695 MeV, the threshold for the inverse beta-decay 56Fe+ e− →56Mn + νe. The Mn immediately electron captures:56Mn + e− →56Cr + νe. The Cr is stable until densities above1010 g cm−3 are reached.

Lighter nuclei have other thresholds: 4He is at 20.6 MeV,12C is at 13.4 MeV, 16O is at 10.4 MeV and 20Ne is at 7.0MeV. The loss of electrons softens the EOS: the Chandrasekharmass decreases. A white dwarf at these densities will begin togravitationally collapse. Thus the maximum density <∼ 1010 gcm−3, with a minimum radius >∼ 1500 km.

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Electrostatic corrections

In a degenerate system in which the nuclei are ordered ineither a solid or liquid, there is Coulomb energy associated withthe ordering. If the nuclei are equally spaced and surroundedby a uniform density electron gas, the interaction energy perelectron is

Ec/Z = −9

10

Ze2

Rc= −

9

10

(

4π

3

)1/3

Z2/3e2n1/3e

where ne = 3Z/(4πR3c). The corresponding pressure is

Pc = n2e∂Ec/Z

∂ne= −

3

10

(

4π

3

)1/3

Z2/3e2n4/3e .

In the extreme relativistic limit this is just a constant fraction(a few percent) of the degeneracy pressure:

Pc

Pd= −

25/3

5

(

3

π

)1/3 e2

h̄cZ2/3.

In the non-relativistic limit, Pc becomes more important atlower and lower densities:

Pc

Pd= −

mee2Z2/3

(2ne)1/3 πh̄2

.

At low enough density, Pc = −Pd:

ne,c =Z2(

mee2)3

2π3h̄6

(

ρc ' 0.4Z2 g cm−3)

,

and the total pressure vanishes. For iron this is about 250 gcm−3, which is not the laboratory value of 7.86 g cm−3 becauseit is incorrect to treat the e− gas as uniform.

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The electron Fermi energy, modified by Coulomb potential

EF = −eV (r) +p2F

2me.

is constant in space, otherwise electrons would move to a lowerEF . The electron density is

ne =8π

3h3p3F =

8π

3h3[2me (EF + eV (r))]3/2 .

The potential is determined by Poisson’s equation

∇2V = 4πene + nuclear contribution

where the last term is effectively a delta function at the ori-gin. Omitting it for r > 0, we have the boundary condition:rV (r) → Ze as r → 0. The electric field should vanish atthe outer boundary, since this volume must be overall neutral:dV/dr|Rc

= 0. Poisson’s equation in spherical geometry is

d2φ

dx2=

φ3/2

x1/2, (4)

EF + eV (r) =Ze2φ (r)

r, x = r

(

128Z

9π2

)1/3 mee2

h̄2.

Eq. (4) has boundary conditions

φ (0) = 1, φ (xo)′ ≡

dφ

dx

∣

∣

xo=

φ (xo)

xo,

where xo corresponds to r = Rc. The latter condition canbe seen by evaluation of Z = 4π

∫

ner2dr over the entire vol-

ume. The equation Eq. (4) has a unique solution, when φ′(0) =−1.588071, in that as xo → ∞, φ(xo) → 144x−3

o → 0. Oth-erwise, for larger values of φ′(xo), φ doesn’t vanish anywhere

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and diverges as x → ∞. At some point, the second boundarycondition will be satisfied.

The pressure at the outer boundary is that of free particles

P =8π

15h3mep5F (Rc) =

1

10πZ2e2

(

128Z

9π2

)4/3(mee2

h̄2

)4 [φ (xo)

xo

]5/2

.

The density is the total mass divided by the volume:

ρ =3AmB

4π

128Z

9π2

(

mee2

h̄2xo

)3

=4AmBZ

3

(

2mee2

πh̄2xo

)3

.

For low densities, the solution approaches the unique solutionφ(xo) → 144x−3

o . Thus, P ∝ ρ10/3, with K given by Eq. (3).

Mass-Radius Relation for Degenerate Objects

The mass-radius diagram for cold compact objects is shownin the figure: the solid lines are the limiting expressions Eqs(1–3), the dashed line is the full result, for 12C. The maximumradius configuration has the properties, approximately, of theplanet Jupiter.

In the relativistic limit, for radii much smaller than 5000km, the equation of state will deviate from that of a γ = 4/3gas. Electron capture will reduce Ye and the value of the Chan-drasekhar mass. Therefore, a regime where dM/dρc < 0 willexist. Such a regime is dynamically unstable. At sufficientlyhigh density, where nuclear forces become important, the ef-fective value of γ will increase, the mass will reach a minimumvalue (Mmin ' 0.01 M�, where R ' 300 km), and stability isrestored. As the central density increases further, dM/dρc > 0.This is the neutron star regime. As the mass increases, and theradius shrinks, general relativity, which we have heretofore ig-nored, becomes important. The most important feature thatgeneral relativity introduces is that at densities well in excess of

7

the nuclear saturation density, ρs = 2.7 ·1014 g cm−3, the massreaches a maximum value, in the range 1.5-3 M�. Larger den-sity configurations are once again dynamically unstable. Themaximum mass is discussed in a subsequent lecture.

Cooling of white dwarfs

The interior of a white dwarf has energy transport domi-nated by conduction. The electrons are extremely degenerate,nowever, so they must have very large mean free paths. Thethermal conductivity is very high. The temperature gradientmust be rather small. The interior is roughly isothermal. Nearthe surface, isothermality breaks down as the opacity increases.The surface regions are diffusive, with a temperature gradient

dT

dr= −

3

4ac

κρ

T 3

L

4πr2.

At the high densities, Kramer’s opacity is dominant: κ =κoρT−3.5, with κo ' 4.3 × 1024Z(1 + X) cm2 g−1. With hy-drostatic equilibrium,

dP

dT=

4ac

3

4πGm (r)

κoL

T 6/5

ρ.

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The surface layer is thin, so m(r) = M . Using the nondegen-erate pressure P = NoρkT/µ, and eliminating ρ, we have

PdP =4ac

3

4πGM

κoL

kNo

µT 7.5dT.

Integrating from P = 0 at T = 0 to the interior,

ρ =

√

2

8.5

4ac

3

4πGM

κoL

µ

kNoT 3.25.

The surface approximation breaks down in the interior whenmatter becomes degenerate. This occurs when the non-degeneratepressure equals the degenerate pressure at radius r∗ where onehas ρ∗ and T∗. This results in

ρ∗ = 2/4 × 10−8T3/2∗ Y −1

e g cm−3,

L = 5.7 × 105 µY 2e

Z (1 + X)

M

M�T 3.5∗ ergs−1.

This is similar to the blackbody law L = 4πR2σT 4eff , but in-

volves the interior temperature, not the visible temperature of

the surface. It suggests a relation like Teff ∝ T7/8∗ . From L and

M , and the composition, one can deduce T∗. L of 10−2− 10−5

L� imply T∗ = 106 − 107 K, ρ∗ < 103 g cm−3 and

R − r∗R

' 4.25RNokT∗GMµ

<∼ 10−2.

The energy that is radiated as thermal energy by the whitedwarf is the residual ion thermal energy, since the electrons aredegenerate and the star can’t release any gravitational energy.For a monatomic non-degenerate ion gas, with cv = 3k/2 thetotal thermal energy is (with T∗ = T )

U =3

2

Nok

ATM.

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The cooling rate is L = −dU/dt. Using L = CMT 7/2,

t = to +3

5

kNo

AC

(

T−5/2 − T−5/2o

)

.

Taking T << To, the cooling time is

τ =3

5

NokT

A

M

L=

3

5

Nok

CA

(

CM

L

)5/7

.

For L ∼ 10−3 L�, we obtain τ ∼ 109 yr.It is interesting to compare the cooling theory with observa-

tions. Like cars on a highway, the slower they go, the more con-gested the freeway (or vice versa). The number of white dwarfsof a given luminosity should relate to their relative abundance,especially if the birth rate of white dwarfs has been roughlyconstant in time. The luminosity function is φ(L) which is thespace density of white dwarfs per unit interval of log L. Thus,with a uniform production rate,

φ (L) ∝

[

d log (L)

dt

]−1

.

If τ ∝ L−α, where our theory suggests α ' 5/7, one finds

log φ = −α log L + constant.

It turns out this is approximately matched by observations,until L <∼ 10−4 L�. Theoretical corrections to the specificheat of very cold white dwarfs imply that α → 0 below thisluminosity, but observations actually reveal that α << 0 whenL <∼ 10−4.5. This deficit of white dwarfs of low luminosity isdue to the finite age of the galactic disc. The cooling time of thewhite dwarfs where the sudden drop in α occurs then yields anestimate of the age of the galactic disc, about 10 billion years.