– 1 –

Order of Magnitude Astrophysics - a.k.a. Astronomy 111

Stellar Remnants

When the nuclear fuel in the star is exhausted, the gravitational force will start contracting the matter again and the density will increase. Eventually, the density will be high enough that the quantum degeneracy pressure will dominate over the thermal pressure. The equilibrium condition for such a system will require the degeneracy pressure of matter to be large enough to balance

gravitational pressure. Equivalently, the Fermi energy εF(n) must be larger than the gravitational

2 2/3 1/3 potential energy εg ≈ GmpN n . When the particles are non-relativistic they obey the equation: 2 2 2/3 2/3 εF(n) = (~ /2me)(3π ) n and the condition εF ≥ εg can be satisfied at equality if  2  2 Gmpme  n1/3 =   N2/3. (1) (3π2)2/3  ~2 

3 With n = 3N/(4πR ) and N = (M/mp), this reduces to the following mass-radius relation:

1/3 −1 1/3 −3 1/3 RM ≈ αG oemp ≈ 8.7 × 10 R M . (2)

When the density is still higher, the Fermi energy has to be supplied by relativistic

1/3 1/3 particles, and εF now becomes εF ≈ ~cn , which sclaes as n just like εg. Therefore, the

equilibrium condition is invariant to the number-density; i.e. εF > εg can be satisfied only if

2 2/3 −3/2 ~c ≥ GmpN or N ≤ αG . The corresponding mass bound (called the Chandrasekhar limit) is −3/2 Mch ≤ mpαG ≈ 1M . Such structures are called white dwarfs. A with M ≈ M −2 6 will have R ≈ 10 R and ρ ≈ 10 ρ (Figure 1). As the density increases, electrons combine with protons through inverse beta decay (also called “electron-capture”) to form neutrons, which

can also provide degeneracy pressure. Equation [1] is still applicable with me replaced with mn;

−3 correspondingly the right-hand side of relation [2] is reduced by (on/oe) = (me/mn) ≈ 10 . Such

−5 15 objects - called neutron stars – will have a radius R ≈ 10 R and ρ ≈ 10 ρ if M = M . If

−3/2 the mass of the stellar remnant is higher than αG mp, no physical process can provide support against the gravitational collapse. In such a case, the star will form a black hole. Dead Stars and Black Holes a.k.a. Astronomy 15 Winter Quarter 2008 Gravity Triumphant: Key Concepts

white dwarf structure Pe = P (ρ, T, C) (1) – 2 –

from Schatzman (1958) -1.5 1+ 1 Pe = Keρ n (2)

Van Maanen 2 o2 Eri B AC 70 8247 n = 3/2 -2.0 polytrope B W 219 5/3 )

. P =RossK 6271ρ (3) (R/R

10

log -2.5 L 930-80

n = 34 polytrope/3 P = K"Chandrasekhar2ρ (4) Mass" -3.0

(r + R)3 G(M + m) -3.5 = (5) -0.6 -0.4⇒ -0.2 T 02 0.2 0.4 0.64π2

log 10 (M/M . )

Fig. 1.— White dwarfs in the log R - log M plane. 1 GMm mv2 + − = 0 + 0 (6) 2 esc r 2GM v = c = (7) esc R ! S

0.5M 5 1010 yr (8) ! → × 1.0M 1010 yr (9) ! → 10M 2 107 yr (10) ! → × > 30M 3 106 yr (11) ! → × (12) – 3 –

Compact Object Accretion

The process of accretion works along the following lines. When a mass m falls from infinite distance to radius R, in the gravitational field of a massive object with mass M, it gains the kinetic energy E = (GMm/R). If this kinetic energy is converted into radiation with efficiency , then the of the accreting system will be ! ! dE GM  dm L =  =  (3) dt R dt

The photons that are emitted by this process will be continuously interacting with the in-falling particles and will be exerting a force on the ionized gas. When this force is comparable with the gravitational force attracting the gas towards the central object, the accretion will effectively stop. The number density n(r) of photons crossing a sphere of radius r, centered at the accreting object of luminosity L , is (L/4πr2)(~ω)−1, where ω is some average frequency. The rate of collisions

between photons and electrons in the ionized matter will be n(r)σT and each collision will transfer momentum ~ω/c. Because electrons and ions are strongly coupled in a plasma, this force will be transferred to the protons. Hence the outward force on an in-falling proton at a distance r will be ! ~ω  Lσ  f ≈ nσ = T . (4) rad T c 4πcr2

2 This force will exceed the gravitational force attracting the proton, fg = (GMmp/r ), if L > LE, where ! 4πGmpc 38 M −1 LE = M ≈ 1.3 × 10 erg s (5) σT M

is called the Eddington luminosity. the temperature of a system of size R radiating at LE will be

2 4 determined by (4πR )σT = LE, that is

!1/4  −1/2 8 M R TE ≈ 1.8 × 10 . (6) M 1 km

For a this radiation will peak in the x-ray band.