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É

2 From Boltzmann to Navier-Stokes to Euler

Reading: Ryden ch. 1, Shu chs. 2 and 3

2.1 The distribution function and the Boltzmann equation

Define the distribution function f(~x,~v, t) such that f(~x,~v, t)d3xd3v = probability of finding aparticle in phase space volume d3xd3v centered on ~x,~v at time t.The normalization is

∫ ∫

f(~x,~v, t)d3xd3v = N (# particles in system)

and[f ] = cm−3( cm s−1)−3.

Particles are not created or destroyed, so continuity implies

∂f

∂t+∑

i=1,3

(

xi∂f

∂xi+ vi

∂f

∂vi

)

=df

dt|c, (7)

where dfdt |c represents discontinuous motion of particles through phase space because of collisions.

(Collisions cannot instantaneously change particle positions, but they can instantaneously changeparticle velocities.)Substituting xi = vi and vi = gi leads to the Boltzmann equation

∂f

∂t+∑

i

vi∂f

∂xi+∑

i

gi∂f

∂vi=

df

dt|c. (8)

Stellar dynamics is based on the collisionless Boltzmann equation, with the RHS=0.In the fluid (λ ≪ L) limit, on the other hand, the collision term makes f(~v) approximatelyMaxwellian while locally conserving mass, momentum, and energy.The Boltzmann equation is hard to manage because it is 6-d, but it tells us more than we reallywant to know in most cases.We are usually happy with the density, mean velocity, and velocity dispersion as a function of ~x,since the velocity distribution function is close to Maxwellian.We can therefore get more useful equations by taking moments of the Boltzmann equation.A similar procedure is used in stellar dynamics to obtain the Jeans equations. However, these areless powerful than the hydrodynamic equations because in the absence of collisions the velocitydistribution function may be far from Maxwellian (in particular, it may be anisotropic).

The mass density is

ρ(~x, t) =

∫

mf(~x,~v, t)d3v, (9)

5

where m is the particle mass. If necessary, one can sum over different f ’s for different particletypes, but we will assume a single particle species here.The mass-weighted average of a quantity Q at position ~x is

〈Q〉 =1

ρ

∫

Qmf(~x,~v, t)d3v. (10)

2.2 The continuity equation

Multiply both sides of the Boltzmann equation by m and integrate over d3v:

∂

∂t

∫

mfd3v +∑

i

∂

∂xi

∫

mfvid3v + m

∫

∑

i

∂

∂vi(gif)d3v =

∫

mdf

dt|cd

3v.

First term is∂ρ

∂t.

Second term is∑

i

∂

∂xi(ρ〈vi〉) ≡ ~∇ · (ρ~u) ,

where

~u ≡ 〈~v〉 = mean fluid velocity at (~x, t) [ cm s−1]

~w ≡ ~v − ~u = particle random velocity [ cm s−1].

Third term is

m

∫

V

~∇v · (~gf)d3v = m

∫

Sn · (~gf)dA = 0.

The first equality follows from the divergence theorem and the second from the assumption that fvanishes as v −→ ∞.

RHS vanishes because of local mass conservation: collisions do not create or destroy particles at afixed position, only shift them in velocity space.

Result:∂ρ

∂t+ ~∇ · (ρ~u) = 0.

More transparently∂ρ

∂t+ ~u · ~∇ρ + ρ~∇ · ~u = 0,

implying a continuous change of mass density.

6

2.3 The momentum equation

Multiply Boltzmann equation by m~v and integrate.

∂

∂t

∫

mvjfd3v +∑

i

∂

∂xi

∫

mfvjvid3v + m

∫

∑

i

givj∂f

∂vid3v =

∫

mvjdf

dt|cd

3v.

First term is∂

∂t(ρuj).

Second term is∑

i

∂

∂xi(ρ〈vjvi〉) =

∑

i

∂

∂xi(ρuiuj + ρ〈wiwj〉) .

For the third term, note that∂

∂vi(vjf) = vj

∂f

∂vi+ δijf

to obtain∑

i

mgi

∫[

∂

∂vi(vjf) − δijf

]

d3v = −∑

i

giδij

∫

mfd3v = −ρgj,

where δij ≡ 1 for i = j, 0 for i 6= j, and we have used the divergence theorem to get rid of the firstterm in the brackets.

RHS vanishes because collisions conserve momentum. (Note: Ryden sets the collision term to zerobefore integrating on the grounds that it will be zero in equilibrium, but I think this is not justifiedin all circumstances.)

Result:∂

∂t(ρuj) +

∑

i

∂

∂xi(ρuiuj + ρ〈wiwj〉) = ρgj .

The diagonal terms of 〈wiwj〉 are generally much larger than the off-diagonal terms, since randomvelocities in different directions are usually almost uncorrelated.It therefore makes sense to divide the ρ〈wiwj〉 term into a contribution from pressure and a con-tribution from viscosity:

P ≡ pressure =1

3ρ〈|~w|2〉 [ dyne cm−2 or erg cm−3] (11)

πij ≡ viscous stress tensor = Pδij − ρ〈wiwj〉 [ dyne cm−2 or erg cm−3] (12)

to obtain∂

∂t(ρuj) +

∑

i

∂

∂xi(ρuiuj + Pδij − πij) = ρgj ,

or, in tensor form∂

∂t(ρ~u) + ~∇ ·

(

ρ~u~u + P↔

I −↔

π

)

= ρ~g.

7

Here (~u~u)ij = uiuj,↔

I ij= δij .

All three tensors are symmetric, and↔

π is traceless.

One can combine this form of the momentum equation with the continuity equation to get the form

∂uj

∂t+∑

i

ui∂uj

∂xi= gj −

1

ρ

∑

i

∂

∂xi(Pδij − πij)

or∂~u

∂t+(

~u · ~∇)

~u = ~g −1

ρ~∇P +

1

ρ~∇·

↔

π .

Viscosity acts to oppose shearing motion and interpenetration.

2.4 The energy equation

One can multiply by mv2 and go through a similar procedure (see Ryden and/or Shu) to derivethe internal energy equation

∂ǫ

∂t+ ~u · ~∇ǫ = −

P

ρ~∇ · ~u −

1

ρ~∇ · ~F +

1

ρΨ

ǫ ≡ specific internal energy =1

2〈|~w|2〉 [ erg g−1] (13)

~F ≡ conduction heat flux =1

2ρ〈~w|~w|2〉 [ erg cm−3 s−1 cm] (14)

Ψ ≡ viscous dissipation rate =∑

i,j

πij∂ui

∂xj[ erg cm−3 s−1]. (15)

One again makes use of the divergence theorem and of the fact that collisions conserve energy (aswell as mass and momentum).

Note that if the distribution of ~w is symmetric about zero, then ~F vanishes. If the distributionis skewed, then hot particles have a drift relative to cold particles, producing a heat flux in thedirection of the drift.In most cases, a temperature gradient produces a conductive flux ~F ∝ ~∇T .However, if ~F is uniform, heat flowing out is replaced by heat flowing in. The local thermal energychanges only if ~∇ · ~F 6= 0.

Ψ represents conversion of bulk motion of the fluid into internal energy via viscous dissipation. Itis the viscous analog of heating by PdV work.

2.5 Eulerian and Lagrangian derivatives

An Eulerian reference frame is one that stays fixed in space. (This has nothing to do with “Eulerequations,” except that it is the same Euler.)A Lagrangian reference frame is one that moves with the fluid.

8

∂Q∂t is the time derivative of Q at a fixed Eulerian position.The time derivative along an element moving with the mean flow ~u,

DQ

Dt=

∂Q

∂t+(

~u · ~∇)

Q, (16)

is known as the Lagrangian derivative.

The physical interpretation of hydrodynamics equations may be more transparent in Lagrangianformulation.Numerical calculations can be performed in either Eulerian or Lagrangian frames. Often one ismore powerful than the other for a particular application.Lagrangian codes tend to be powerful when there is a large dynamic range in density and one needshigher resolution in high density regions than in low density regions.Smoothed particle hydrodynamics (SPH) is an example of a Lagrangian hydrodynamics algorithm,where one follows fluid properties at the locations of particles, which move with the flow.

2.6 The Navier-Stokes equations

Collecting our results, we have the Navier-Stokes equations:

Dρ

Dt= ∂ρ

∂t + ~u · ~∇ρ = −ρ~∇ · ~u, (17)

D~u

Dt= ∂~u

∂t +(

~u · ~∇)

~u = ~g −1

ρ~∇P +

1

ρ~∇·

↔

π, (18)

Dǫ

Dt= ∂ǫ

∂t + ~u · ~∇ǫ = −P

ρ~∇ · ~u −

1

ρ~∇ · ~F +

1

ρΨ +

Γ − Λ

ρ(19)

Γ ≡ volumetric radiative heating rate [ erg cm−3 s−1]

Λ ≡ volumetric radiative cooling rate [ erg cm−3 s−1].

(17) represents local mass conservation. The change in density of a Lagrangian fluid element isproduced by a change in the specific volume.

(18) represents local momentum conservation. Accelerations are produced by gravity, pressuregradients, and viscous forces.

(19) represents local energy conservation. Changes in internal energy are produced by PdV work,by conduction, by viscous heating (conversion of bulk kinetic energy), and by radiative heating orcooling, which we have added to our previous results.

~g may be specified externally or, if self-gravity is important, computed from Poisson’s equation~∇ · ~g = −4πGρ.Ψ is determined once ~u and

↔

π are specified.Γ − Λ can be computed from properties of the gas and the radiation field.

9

This leaves us with 5 equations and 14 unknowns: ρ, ~u, P ,↔

π (five independent elements), ǫ, ~F .To get a closed set of equations, we must find relations among ρ, P , ǫ,

↔

π , and ~F using constitutiverelations for the gas.We have already discussed the simplest example, the equation of state for a monatomic gas, ǫ = 3

2

Pρ .

We will return to a more general discussion of the equation of state and to viscosity and heatconduction soon.

2.7 The Euler equations

Formally, the Navier-Stokes equations can be derived from the first-order expansion of the Boltz-mann equation in the parameter λ/L.In many cases, it is adequate to use the zero’th-order approximation λ/L = 0, in which case onegets the eight “constraint” equations

↔

π= 0, ~F = 0.Viscosity and heat conduction are diffusive effects (arising because particles diffuse out of their fluidelements), and these are often small compared to dynamical effects, when flows are a significantfraction of (or greater than) the sound speed, or when diffusion times are long compared to systemlifetimes.Some important exceptions in astrophysical situations are shocks, viscosity in accretion disks, con-ductive evaporation of cold clouds in a hot medium, and conduction in some kinds of stars andplanets.

If we neglect diffusive terms in the Navier-Stokes equations, we get the Euler equations

Dρ

Dt= ∂ρ

∂t + ~u · ~∇ρ = −ρ~∇ · ~u, (20)

D~u

Dt= ∂~u

∂t +(

~u · ~∇)

~u = ~g −1

ρ~∇P (21)

Dǫ

Dt= ∂ǫ

∂t + ~u · ~∇ǫ = −P

ρ~∇ · ~u +

Γ − Λ

ρ(22)

Together with the equation of state ǫ = 3

2

Pρ , the Euler equations describe the dynamics of a perfect

monatomic gas.In a perfect gas, collisions ensure that the local distribution of random velocities acquires themaximum entropy form, in which the velocity distribution in each dimension is independentlyGaussian with variance 〈w2

i 〉 = kTm .

This is a Maxwellian velocity distribution,

fMax(~w)d3w =

(

m

2πkT

)3/2

exp

(

−mw2

2kT

)

d3w. (23)

The Maxwellian distribution has 〈wiwj〉 = 0 for i 6= j and 〈w2wi〉 = 0, implying↔

π= 0 and ~F = 0.

10

2.8 The Jeans Equations

It is worth briefly noting the analogy between the Euler equations of fluid dynamics and the Jeansequations of stellar dynamics. These are also derived by taking moments of the Boltzmann equation,in this case the collisionless Boltzmann equation (Binney & Tremaine, section 4.2.)

The first Jeans equation is just the continuity equation for stellar number density, exactly equivalentto the fluid continuity equation for mass density.

The second Jeans equation, in Binney & Tremaine’s notation, is

ν∂vj

∂t+∑

i

νvi∂vj

∂xi= −ν

∂Φ

∂xj−∑

i

∂(νσ2ij)

∂xi, (24)

where v is the mean fluid velocity, ν is the stellar number density, Φ is the gravitational potential,and σ2

ij = 〈wiwj〉 where ~w is the random velocity.This is very similar to the second Euler equation, but with an anisotropic stress tensor νσ2

ij in place

of ~∇P .

The limitation of the Jeans equations is that there is no equation of state relation between den-sity and σ2

ij , no equivalent of the thermal energy equation, and no strong reason for assuming aMaxwellian random velocity distribution.Thus, stellar dynamical models that use the Jeans equations must usually assume some form ofσ2

ij, and the resulting models are only as accurate as the assumptions.

11

3 Hydrostatic Equilibrium

Reading: Shu, ch. 5, ch. 8

3.1 Timescales and Quasi-Hydrostatic Equilibrium

Consider a gas obeying the Euler equations:

Dρ

Dt= −ρ~∇ · ~u,

D~u

Dt= ~g − 1

ρ~∇P,

Dε

Dt= −P

ρ~∇ · ~u +

Γ − Λ

ρ.

Suppose that there is a substantial mismatch between the gravitational acceleration and the pressuregradient

|~g − 1

ρ~∇P | ∼ |~g|.

How long does it take for this mismatch to produce an order unity change in the density?In time ∆t

∆ρ

ρ∼ −(~∇ · ~u)∆t ∼ (−~∇ · (~g∆t))∆t.

If the gas is self-gravitating, then

~∇ · ~g = −∇2φ = −4πGρ, so

∆ρ

ρ∼ 4πGρ(∆t)2.

The density therefore changes on a dynamical timescale

∆t ∼ td ∼ (Gρ)−1/2.

Binney & Tremaine (eq. 2-30) define

td ≡(

3π

16Gρ

)1/2

≈ 3

4(Gρ)−1/2.

The relevant ρ is that of the gravitationally dominant component.Since pressure is an increasing function of density, these changes generally go in the direction ofrestoring balance between gravity and pressure gradients.

If a gas dynamical system is many dynamical times old, we generally expect it to be in hydrostaticequilibrium, with

~g =1

ρ~∇P.

Suppose that the gas is cooling, so that the pressure drops over time. On what timescale does thesystem evolve?Slow cooling: cooling time t−1

c ∼ 1

εDεDt .

12

Rapid cooling: dynamical time.

The timescale for cooling, or for other processes that change the pressure, is often much longerthan the dynamical timescale. If a system has balance between gravity and pressure gradients butevolves on a timescale td, we say it is in quasi-hydrostatic equilibrium. For such a system, weignore time derivatives in the momentum equation but not in the other equations.

3.2 Isothermal plane-parallel atmospheres

For a plane-parallel atmosphere in a uniform gravitational field, ~g = g z, hydrostatic equilibriumimplies

1

ρ

dP

dz= −g.

With P = (ρ/m)kT we obtaind lnP

dz=

1

P

dP

dz=

−gm

kT,

with solution

P = P0e−z/zh , zh ≡ kT

gm

ρ = ρ0e−z/zh , ρ0 ≡ P0

(

m

kT

)

.

For the earth, g = 980 cm s−2, m ≈ 4.8 × 10−23 g, T ≈ 290K, P0 ≈ 106 dyne cm−2, ρ0 ≈ 1.2 ×10−3 g cm−3, zh ≈ 8.5 km.Implication: it’s hard to breathe on top of Mount Everest.

We can get an order-of-magnitude estimate of the scale height from a very simple argument basedon the analogous collisionless case.The one-dimensional velocity dispersion of particles is σ2 = kT/m, since the mean kinetic energyper translational degree of freedom, 1

2mσ2, should equal 1

2kT .

A particle up from z = 0 with velocity σ comes to rest after time t ∼ σ/g, having traveled adistance h ∼ σt ∼ σ2/g = kT/(gm).

If we had assumed an adiabatic atmosphere instead of an isothermal atmosphere, with P =P0(ρ/ρ0)

γ , we would have obtained a very different result:

P (z) = P0

(

1 − z

za

)

γ

γ−1

, ρ(z) = ρ0

(

1 − z

za

)1

γ−1

, za = 30 km.

What do we make of the negative pressure that comes from this solution at z > za?Underlying equations break down as ρ −→ 0, corresponding to λ −→ ∞.

13

3.3 Scale-height of a thin disk

Consider a disk of gas orbiting at radius r around a central mass providing gravitational accelerationg = v2

φ/r.For gas at height z r above the midplane, the z-component of gravitational acceleration isgz = −gz/r.Thus

1

ρ

dP

dz= −g

rz,

andd lnP

dz=

1

P

dP

dz= − g

rσ2z,

with σ2 = kT/m.The solution is P = P0e

−z2/z2

h with

zh = σ

(

2r

g

)1/2

.

Substituting g = v2φ/r yields

zh

r=

√2

σ

vφ.

This is again what one would expect in order-of-magnitude from the collisionless case, where aparticle goes a vertical distance zh at vertical speed σ in about the same time that it goes anazimuthal distance r at azimuthal speed vφ.

3.4 Isothermal sphere in hydrostatic equilibrium

Good discussions in Binney & Tremaine §4.4(b); Shapiro, Iliev, & Raga 1999 (MNRAS, 307, 203)

Consider a self-gravitating spherical distribution of gas with temperature T .Hydrostatic equilibrium with P = nkT = (ρ/m)kT requires

~∇P = ρ~g =⇒ dP

dr=

kT

m

dρ

dr= −ρ

GM(r)

r2,

and

M(r) =

∫ r

0

4πr2ρdr =⇒ dM

dr= 4πr2ρ.

Multiply by r2mρkT to get

r2 1

ρ

dρ

dr= −Gm

kTM(r)

and differentiate to getd

dr

(

r2 d lnρ

dr

)

= −4πGm

kTr2ρ, (25)

which can be integrated to obtain ρ(r).

Singular isothermal sphere (SIS) solution:

14

Assume ρ = Cr−b, implying

lnρ = lnC − b ln r =⇒ d lnρ

dr= − b

r

d

dr(−br) = −b = −4π

Gm

kTCr2−b

b = 2, C =kT

2πGm

ρ(r) =kT

2πGmr−2 =

σ2

2πGr2

where, as usual, σ =(

kTm

)1/2

.

The mass interior to r is

M(r) =2σ2

Gr

and the circular velocity is

v2c =

GM(r)

r= 2σ2.

An isothermal (constant σ2) stellar dynamical (collisionless) sphere obeys the same equation andhas the same structure.

The solution to the differential equation (25) depends on the boundary condition as well as theequation itself.The SIS solution has a cusp at r = 0.What about the solution for finite central density ρ0?

First, we would like to identify a characteristic lengthscale so that we can obtain a general solutionin terms of dimensionless variables.The dimensional quantities in equation (25) are G, ρ, and the combination σ2 = kT/m.The combination of these quantities that yields a lengthscale is σ/(Gρ0)

1/2.Physically, this represents the typical distance a particle travels in a central dynamical time.Define the King radius (a.k.a. core radius)

r0 ≡(

9σ2

4πGρ0

)1/2

(26)

and dimensionless variablesρ =

ρ

ρ0

, r =r

r0

.

The hydrostatic equilibrium equation becomes

d

dr

(

r2 d ln ρ

dr

)

= −9r2ρ. (27)

15

The (numerical) solution is obtained by integrating outwards from r = 0 with the central boundaryconditions

ρ(0) = 1,dρ

dr= 0

(the second condition is required for hydrostatic equilibrium, since the interior mass vanishes asr −→ 0).

The solution bends from a constant ρ at r ∼< 1 to the singular isothermal sphere solution ρ = 2

9r−2

at r ∼> 3. For r ∼< 2, the solution is well described by the simple approximation

ρ = (1 + r2)−3/2,

which is accurate to better than 5%, but this approximation fails at r ∼> 3 (it has the wrongasymptotic slope).

The projected surface density of an isothermal sphere drops to 0.5013 ≈ 0.5 of its central value atr = 1.

Note that by defining characteristic dimensionless variables we have reduced the family of solutionswith different central densities and temperatures to a single solution for the appropriately scaledvariables.

What is the total mass of an isothermal sphere of central density ρ0 and temperature T ? Infinite.In order to get a system of finite total mass, we must truncate it at radius rt by confining it withan external pressure.

16

3.5 Polytropes

A spherical, self-gravitating object in hydrostatic equilibrium with a polytropic equation of stateP = Kργ is called a polytrope. The structure of a polytrope is determined by the adiabatic indexγ or by the so-called polytropic index n ≡ (γ − 1)−1.

Consider a polytrope with equation of state

P = Kργ =

(

Pc

ργc

)

ργ = Pc

(

ρ

ρc

)1+1/n

,

where Pc and ρc are the central pressure and temperature.The quantity

α ≡[

γ

γ − 1

Pc

ρc

1

4πGρc

]1/2

has units of length. We will soon learn that the sound speed is a = (γP/ρ)1/2, so α is proportionalto the speed of sound times the dynamical time in the center of the polytrope.

In terms of the dimensionless variables θ and ξ defined by

ρ = ρcθn, P = Pcθ

n+1, r = αξ,

the equation of hydrostatic equilibrium can be written

1

ξ2

d

dξ

(

ξ2 dθ

dξ

)

= −θn,

called the Lane-Emden equation for a polytropic star.The structure of a non-singular polytrope can be found by integrating this equation with the centralboundary conditions

θ = 1,dθ

dξ= 0 at ξ = 0.

The isothermal sphere corresponds to n = ∞. For n ≥ 5 the total mass is infinite. For n < 5, thedensity drops to zero at a finite radius.The cases most interesting for stars are n = 3/2 (γ = 5/3) and n = 3 (γ = 4/3), but these do nothave analytic solutions.The only analytic solutions are for n = 1 and n = 5. For n = 1 the solution is

θ =sin ξ

ξ.

Generically, polytropes with n < 5 have a roughly constant density core whose radius is ∼ αfollowed by a falling density profile that eventually drops to zero.For higher n (lower γ), the polytrope is more centrally concentrated (smaller core) because the“squishier” (more compressible) equation of state means that a higher central density is requiredto support the weight of the layers above.

17

3.6 Instability considerations

Two of the instabilities that can affect a fluid in hydrostatic equilibrium are Rayleigh-Taylor insta-bility and convective instability.

Rayleigh-Taylor instability occurs when a dense fluid sits on top of a light fluid.It is energetically favorable for dense fluid elements to sink and exchange places with light fluidelements.Although a true hydrostatic equilibrium situation with Rayleigh-Taylor instability is unlikely toarise in astrophysical situations, the instability can be important in, e.g., outflows or supernovaexplosions when a light medium tries to accelerate a dense medium.

Convective instability arises if a fluid element that moves upward (against gravity) and expandsadiabatically (because it doesn’t have time to exchange heat conductively with its surroundings)becomes less dense than its new surroundings, so that it continues to rise.The Schwarzschild criterion that determines whether a fluid is convectively unstable is

ds > 0 in the direction of gravity.

High-entropy material is buoyant and tends to move past low-entropy material.

18

4 Pressure and Viscosity

Reading: Ryden, chapter 2; Shu, chapter 4

4.1 Specific heats and the adiabatic index

First law of thermodynamics (energy conservation):

dε = −PdV + dq =⇒ dq = dε + PdV, (28)

V ≡ ρ−1 = specific volume [ cm3 g−1]

dq ≡ Tds = heat change per unit mass [ erg g−1]

s ≡ specific entropy [ erg g−1 K−1].

The specific heat at constant volume,

cV ≡(

∂q

∂T

)

V[ erg g−1 K−1] (29)

is the amount of heat that must be added to raise temperature of 1 g of gas by 1K.

At constant volume, dq = dε, and if ε depends only on temperature (not density), ε(V, T ) = ε(T ),then

cV ≡(

∂q

∂T

)

V=

(

∂ε

∂T

)

V=

dε

dT.

implyingdq = cV dT + PdV.

If a gas has temperature T , then each degree of freedom that can be excited has energy 1

2kT . (This

is the equipartition theorem of classical statistical mechanics.)The pressure

P =1

3ρ〈|~w|2〉 =

ρ

mkT

since

〈12mw2

i 〉 =1

2kT =⇒ 〈|~w|2〉 =

3kT

m.

Therefore

PV =kT

m=⇒ PdV =

k

mdT.

Using dq = cV dT + PdV , the specific heat at constant pressure is

cP ≡(

∂q

∂T

)

P= cV + P

dV

dT= cV +

k

m.

Changing the temperature at constant pressure requires more heat than at constant volume becausesome of the energy goes into PdV work.

19

For reasons that will soon become evident, the quantity γ ≡ cP /cV is called the adiabatic index.A monatomic gas has 3 degrees of freedom (translation), so

ε =3

2

kT

m=⇒ cV =

3

2

k

m=⇒ cP =

5

2

k

m=⇒ γ =

5

3.

A diatomic gas has 2 additional degrees of freedom (rotation), so cV = 5k/2m, γ = 7/5.More generally

ε =1

γ − 1

kT

m=

1

γ − 1

P

ρ.

4.2 Adiabatic evolution

As discussed in §2.7, a perfect gas has a Maxwellian velocity distribution and therefore no viscosityso it obeys the Euler equations.In the absence of radiative heating and cooling, one can combine the continuity and energy equa-tions,

Dρ

Dt= −ρ~∇ · ~u,

Dε

Dt= −P

ρ~∇ · ~u,

to findDε

Dt=

P

ρ2

Dρ

Dt= −P

DV

Dt,

and since dε = −PdV + Tds we conclude that

TDs

Dt= 0.

In the absence of radiative heating and cooling, a perfect gas undergoes only adiabatic (constantentropy) changes.

From the equation of state

ε =1

γ − 1

P

ρ

we have

dε =1

γ − 1

(

dP

ρ− P

ρ2dρ

)

.

For adiabatic (ds = 0) changes, we can combine this with the first law of thermodynamics

dε = −PdV =P

ρ2dρ

to find (after multiplying by ρ/P )

1

γ − 1

(

dP

P− dρ

ρ

)

=dρ

ρ

implyingdP

P= γ

dρ

ρ=⇒ P = P0(ρ/ρ0)

γ ,

a polytropic equation of state (P ∝ ργ).

20

4.3 Summary: single particle species equation of state

The thermal pressure is

P =1

3ρ〈|~w|2〉 =

ρ

mkT = nkT. (30)

The specific internal energy is

ε =1

γ − 1

kT

m=

1

γ − 1

P

ρ, (31)

where γ = cP /cV is the adiabatic index.A gas undergoing only adiabatic changes has a polytropic equation of state P = P0(ρ/ρ0)

γ .A change in entropy changes the “adiabat” of the gas, i.e., the relation between P0 and ρ0.

4.4 Gas entropy

Start withdε = −PdV + Tds

and consider adding or removing heat at constant ρ (dV = 0)

Tds = dε = cV dT

implying

ds = cVdT

T=⇒ s = cV lnT + const..

Since P ∝ T at constant ρ, this implies that s = cV lnP + const..

What about changes of density? We know that adiabatic changes keep Pρ−γ constant, so thesemust be lines of constant entropy (in the plane of pressure and density).

Therefore, for a single particle species gas,

s = cV ln(

Pρ−γ)

+ const. (32)

4.5 Cluster scaling relations and the “entropy floor”

Consider a simple model of galaxy clusters in which the density profile of the hot intracluster gasis ρ(r) = ρV (R/r)2 for r > r0 and ρ(r) = const. for r <= r0. Define the cluster concentrationparameter c = R/r0, implying ρ0 = c2ρV .

For Bremmstrahlung (free-free) emission, the emissivity per unit volume is proportional to ρ2T 1/2.The cluster’s X-ray luminosity will be dominated by the high density core, and we therefore expect

Lx ∼ ρ20T

1/2r30 ∼ ρ2

V c4T 1/2c−3R3 ∼ ρ2V cR3T 1/2.

Now make two fairly general assumptions about the cluster population, that they follow a virialrelation T ∼ GM/R and that they all have roughly the same average density M/R3 ∼ ρV = const.(indeed, one can define the cluster radius to be the radius where the density falls to ρV .)

21

Under these assumptions, we find

R ∼ M1/3 =⇒ T ∼ M2/3 ∼ R2

andLx ∼ cR3T 1/2 ∼ cT 2 ∼ cM4/3.

If all clusters have similar profile shapes, so that c is the same for all clusters, we therefore expectLx ∝ T 2 ∝ M4/3.

However, observed clusters show an Lx−T relation that is more like Lx ∝ T 3, perhaps even steeperin the regime of low mass groups. One popular explanation of this discrepancy is that gas was “pre-heated” by supernova winds or some other feedback mechanism before falling into clusters, givingit an “entropy floor” — a minimum level smin below which it cannot fall.

The cluster’s central entropy is s0 ∼ Tρ−2/3

0 ∼ Tc−4/3. If all clusters have s0 = smin = const., then

c ∼ T 3/4s−3/4

min=⇒ L ∼ T 2.75.

Thus, an entropy floor leads to larger cores (relative to R) in cooler clusters and thus to a steeperLx − T relation.

Alternative possibilities are that the gas is heated after falling into the clusters (though then moreenergy must be injected to “puff it up” by a significant amount) or that an approximate entropy“floor” arises because the low entropy gas cools and settles into galaxies, with higher entropy gasflowing in to replace it.

4.6 Multiple particle species and mean molecular weight

Now suppose that we have particle species j = 1, N , each with mass mj.Define the mean molecular weight

µ ≡ ρ

nmp=

∑

njmj

nmp, (33)

the ratio of the number-weighted mean particle mass to the proton mass, n =∑

nj.

Since 1

2mj〈|~w|2〉j = 3

2kT , the pressure and specific internal energy of each species is

Pj =1

3ρj〈|~w|2〉j =

ρj

mjkT = njkT,

εj =1

2〈|~w|2〉j =

3

2

kT

mj,

assuming that all species are characterized by the same temperature T .

The total pressure is

P =∑

j

Pj =∑

j

njkT = nkT =ρ

µmpkT.

22

For the specific internal energy, we must weight according to the mass density in each species,

ε =

∑

εjmjnj∑

mjnj=

∑

3

2kTnj

ρ=

3

2kT

n

ρ=

3

2

kT

µmp=

3

2

P

ρ.

So the single species formulas apply, with the substitution m −→ µmp.

For primordial composition, helium = 7% by number (23% by mass).

neutral: µ =0.93mp+4×0.07mp

(0.93+0.07)mp≈ 1.2

ionized: µ =0.93mp+4×0.07mp

(0.93+0.93+0.07+2×0.07)mp≈ 0.6 (me mp).

4.7 Molecular viscosity

Returning to the Navier-Stokes equation, the momentum conservation equation is

∂~u

∂t+

(

~u · ~∇)

~u = ~g − 1

ρ~∇P +

1

ρ~∇· ↔π,

where~∇· ↔π=

∑

i

∂

∂xiπij πij = Pδij − ρ〈wiwj〉.

In a “Newtonian fluid,”↔

π is linearly proportional to the velocity gradient ∂ui

∂xj(this was essentially

a guess on the part of Newton and Hooke).The most general symmetric tensor linear in ∂ui

∂xjis

πij = µDij + βδij

(

~∇ · ~u)

, (34)

where the deformation tensor

Dij ≡∂ui

∂xj+

∂uj

∂xi− 2

3δij

(

~∇ · ~u)

(35)

vanishes for uniform expansion or contraction, and

µ ≡ coefficient of shear viscosity = [ g cm−1 s−1]

β ≡ coefficient of bulk viscosity = [ g cm−1 s−1].

(Note that this µ has nothing to do with mean molecular weight.)

µDij represents resistance to shearing motion and βδij

(

~∇ · ~u)

represents resistance to changes in

volume.

The value of µ can be estimated at an order-of-magnitude level as described in Shu (pp. 30-32),similar to Ryden (pp. 16-17). I have not come across a similar estimate of β.

Consider a plane-parallel flow with shear, uy = uz = 0, ∂ux

∂y 6= 0, and focus on a volume ∆A∆ybounded by surfaces of area ∆A separated by ∆y.

23

If we further assume constant density and pressure and ignore gravitational accelerations, themomentum equation becomes (since ∂

∂x = ∂∂z = 0)

∂

∂t(ρux) =

∂

∂yπxy =

∂

∂y

(

µ∂ux

∂y

)

,

so ∂∂y

(

µ∂ux

∂y

)

is the viscous force per unit volume acting on the fluid element.

The rate at which particles cross the upper surface is

∼ nvT

2∆A, vT ∼

(

kT

m

)1/2

= thermal velocity.

The particles travel a distance ∼ λ before colliding and exchanging momentum with another par-ticle, so although equal numbers of particles cross the boundary in the upward and downwarddirection, there is a systematic difference in momentum

∆px ∼ 2mλ∂ux

∂y

per particle.The rate of change of momentum of the element due to particles crossing its boundaries is therefore

dpx

dt∼

[

mλ∂ux

∂ynvT∆A

]

y+∆y

−[

mλ∂ux

∂ynvT ∆A

]

y

,

since we must subtract the momentum being taken out by particles crossing the bottom surface.Dividing by the volume ∆A∆y gives the force per unit volume, so

∂

∂y

(

µ∂ux

∂y

)

∼ ∂

∂y

(

nmλvT∂ux

∂y

)

and using λ = (nσ)−1 implies

µ ∼ mvT

σ. (36)

The coefficient of shear viscosity is independent of density because if the density increases then thehigher flux of particles across the boundaries is countered by the shorter distance λ each particlegoes, and hence the smaller amount of momentum mλ ∂ux

∂y that it transfers.

Note also that a uniform shear ( ∂ux

∂y =const.) produces no net force because the drag from aboveis cancelled by the drag from below.

For neutral atomic hydrogen

µ = 6 × 10−3

(

T

104 K

)1/2

g cm−1 s−1.

Another frequently used quantity is the kinematic viscosity,

ν ≡ µ

ρ∼ vT λ ∼

(

kT

m

)1/2 1

nσ[ cm2 s−1]. (37)

24

Note from the momentum equation that ∂u∂t ∼ ν ∂2u

∂x2 .

Hydrodynamics literature is also replete with references to the Reynolds number

Re ≡ ρuL

µ=

uL

ν∼ u

vT

L

λ. (38)

For small Re, viscous forces have an important effect in altering a flow with large velocity gradients,while for Re 1 viscous forces can usually be ignored.A fluid can usually remain turbulent on scales where Re is large, while viscosity damps out turbu-lence on scales where Re∼< 1.

4.8 Heat conduction

In typical cases, the conduction heat flux ~F is proportional to the temperature gradient

~F = −K ~∇T,

where K is the coefficient of thermal conductivity. This is called Fourier’s law.

For neutral gas, the coefficient is

K =5

2cV µ ∼ k

σ

(

kT

m

)1/2

.

In the case of neutral atomic hydrogen

K = 2 × 106

(

T

104 K

)1/2

g cm s−3 K−1.

Note that the change in internal energy is proportional to ~∇ · ~F , so conduction only changes theinternal energy locally when the temperature gradient is not constant.

It is useful to note that the units of K can also be written erg s−1 cm−1 K−1. Multiplying by atemperature gradient with units K cm−1 and taking a spatial derivative (~∇ · ~F ) therefore yieldssomething with units erg s−1 cm−3.Therefore ~∇ · ~F/ρ has units of erg s−1 g−1, the same as Dε

Dt .

25

5 Accretion Flows

Reading: Pringle 1981, Ann Rev Astron Astrophys 19, sections 1-4Optional reading: Shu, chapter 7 ; Ryden chapters 8-10 (especially 9)Introduction to advection-dominated accretion flows: Narayan et al., astro-ph/9803141

5.1 General Introduction

Accretion is important because(a) it is a way for objects to grow(b) it is a way for gravitational energy to be released

If accreting gas can dissipate energy so that it is not pressure supported, then it will settle into adisk, the minimum energy configuration for fixed angular momentum.

In order for gas to accrete, some of it must lose angular momentum and move inward. Since totalangular momentum must be conserved, other gas must gain angular momentum and move outward.

Viscous dissipation, necessary to transport angular momentum, will also heat the gas. This heatcan either be radiated away near where it is generated or “advected” inward with the accretinggas.

5.2 Equations for a thin accretion disk

The equations governing a thin accretion disk with surface density profile Σ(R) can be obtainedfrom the Navier-Stokes equations in cylindrical symmetry, integrating over the height of the disk.They are:

∂Σ

∂t+

1

R

∂

∂R(RΣVR) = 0,

and∂

∂t

(

ΣR2Ω)

+1

R

∂

∂R

(

R · ΣR2Ω · VR

)

=1

R

∂

∂R

(

R · ΣR · νR∂Ω

∂R

)

,

representing mass conservation and angular momentum conservation, respectively.

Here VR = radial velocity [cm s−1]Ω = angular velocity [rad s−1] =⇒ R ∂Ω

∂R =velocity shearν = kinematic viscosity [cm2 s−1].

The left hand side of the second equation is the convective derivative of the angular momentumper unit area. The product of ν and the velocity shear gives the torque of one annulus on the next,and it is the derivative of this torque that drives changes in angular momentum.

Recall that for “molecular viscosity” ν ∼ vT λ, where vT is the thermal velocity and λ is the meanfree path.

For a point mass potential, Ω =(

GMR3

)1/2

, these equations can be combined to yield

26

∂Σ

∂t=

3

R

∂

∂R

[

R1/2 ∂

∂R

(

νΣR1/2)

]

,

a diffusion equation for the surface density.A ring that is initially thin (in ∆R as well as ∆z) spreads over time, with the inner part givingangular momentum to the outer part.

The timescale for this evolution is tacc ∼ R2/ν, as one can roughly see from the diffusion equation

by taking ∂2

∂R2 −→ 1

R2 and t−1acc = 1

Σ

∂Σ

∂t .

For molecular viscosity, ν ∼ vT λ, this timescale is usually orders of magnitude too long to ac-count for the observed phenomena, leading to the “anomalous viscosity” problem: what physicalmechanism provides the viscosity that drives the evolution of real astronomical accretion disks?

The vertical thickness H of the disk follows from hydrostatic equilibrium:

1

ρ

∂P

∂z=

∂

∂z

[

GM

(R2 + z2)1/2

]

≈ −GMz

R3for z R,

implying

−1

ρ

a2ρ

H∼ −GMH

R3=⇒ H ∼ a

(

GM

R3

)

−1/2

∼ aΩ−1 ∼ a

VφR.

The disk is thin if and only if the sound speed is rotational speed, which is the condition forpressure gradients to be small compared to centrifugal forces.

5.3 Steady-state accretion

Important simplifications occur if we require a steady state, ∂∂t = 0.

Integrating the mass conservation equation yields

M ≡ −2πRΣVR = constant,

i.e., constant mass flow at every radius.The angular momentum equation can be manipulated to yield, in the case of a Keplerian disk,

νΣ =M

3π

[

1 −(

R∗

R

)1/2]

,

where R∗

is the inner radius of the disk.The rate of viscous dissipation per unit area is

D(R) = νΣ

(

R∂Ω

∂R

)2

=3

4π

GM

R3M

[

1 −(

R∗

R

)1/2]

.

(See Pringle or Ryden for details.)

27

Magically, the viscosity vanishes from this expression. This is an important advantage of steady-state accretion, since we don’t know what the viscosity is. This result emerges because we haveimplicitly assumed that the viscosity will adjust itself to accommodate a constant mass accretionrate, and in this case many properties of the disk follow from the requirements of mass, angularmomentum, and energy conservation alone.

The total disk luminosity is

Ldisk =

∫

∞

R∗

D(R)2πRdR =1

2

GMM

R∗

,

i.e., half the gravitational energy released in accreting the gas to radius R∗. The remaining gravi-

tational energy goes into rotational energy, which may be either dissipated in a boundary layer orsucked into a black hole.

At large R, D(R) ∝ R−3, so if this energy is radiated away with emissivity σSBT 4 then T ∝ R−3/4.Integrating the blackbody spectrum over radius gives the predicted spectrum of an optically thick,geometrically thin, steady-state accretion disk

Sν ∝∫ Rout

Rin

Bν [T (R)]2πRdR,

which rises steadily from a frequency kTout/h to kTin/h, cutting off fairly sharply at higher frequencyand gently at lower frequency.

5.4 Viscosity

Even in steady state, a complete solution of disk structure (e.g., surface density profile, scale-heightprofile, vertical structure) requires specification of the viscosity.

Shakura & Sunyaev (1973) introduced the “α-disk” parameterization

ν = α aH, (39)

where a is the sound speed, H is the disk scale height (a function of radius), and α is a dimensionlessconstant.

This at least has the right units, and if one imagines that the viscosity arises from collisions of“clouds” in a turbulent medium moving in eddies of lengthscale l ∼< H at velocity v ∼< a, then oneexpects ν ∼ l · v ∼ αaH with α ∼< 1.

Typical models of disks have α ∼ 0.01 − 0.1.

For specified α, one can completely solve for the structure of a (steady-state or time-dependent)accretion disk. However, the assumption that α is constant with radius, with time, or from oneaccretion disk to another is nothing more than an assumption.

28

While the notion of “turbulent viscosity” is intuitively appealing, detailed studies suggest thathydrodynamic mechanisms alone will not produce sustained turbulence in differentially rotatingdisks (see, e.g., Hawley, Balbus, & Winters, astro-ph/9811057).

In recent years, a clear “leading contender” has emerged as the viscosity mechanism in most as-trophysical accretion disks: magneto-rotational instability. First proposed by Chandrasekhar, thecase for this mechanism has been made mainly by Balbus & Hawley, with a combination of analyticand numerical work.

Rough idea: MHD instabilities in a differentially rotating, magnetized disk drive turbulence (tap-ping the rotational energy), which in turn produces viscosity.

Three-dimensional MHD simulations that follow magneto-rotational instability in realistic situa-tions are just now becoming possible. They support the basic picture, but the details are not yetfully understood. Among other things, it is not yet clear how similar a disk with this viscositymechanism is to a steady-state α-disk.

A different mechanism may be required in proto-stellar disks, which might have very small ionizationfractions (and thus insufficient anchoring of magnetic fields to the disk).

5.5 Some open issues in accretion physics

For black hole accretion, the big questions are:

If gas falls onto a black hole of mass M at a rate M ,

What is the radiative efficiency ε ≡ L/Mc2, where L is the luminosity?What is the spectral energy distribution of the emerging radiation?

Some of the relevant questions that must be addressed along the way are:

5.5.1 What is the inner boundary condition?

The energy radiated in reaching the innermost stable orbit around a Schwarzschild black holecorresponds to efficiency ε ∼ 0.08.However, the gas at the innermost stable orbit still has a lot of rotational energy.To calculate the overall radiative efficiency at the factor of two level, we need to know what happensto this energy.The conventional assumption is that gas decouples after going inside the innermost stable orbit, soit takes its rotational energy into the black hole with it.However, it is possible that gas inside Rin remains viscously or magnetically coupled to gas in theaccretion disk, in which case the efficiency could be higher (since there is more gravitational energyto be extracted).

5.5.2 What generates the hard X-ray luminosity?

While a thin accretion disk can account for the bolometrically dominant portion of a typical quasarSED, the hard X-rays (and the radio emission, where present) must come from something else.

29

Typical models invoke a “corona” above the accretion disk that produces the hard X-rays.However, models of the corona and the mechanism that produces it are only slightly above thecartoon level.

5.5.3 What happens when the accretion rate is far below Eddington?

The Eddington accretion rate is the accretion rate for which the black hole radiates at the Eddingtonluminosity, MEdd = LEdd/εc

2.It is generally thought that when the accretion rate is ∼ 0.01 − 1MEdd, thin disk accretion is areasonable approximation.With a high accretion rate, the gas density is high, so the gas is able to radiate efficiently and staygeometrically thin.

However, if the gas density is low, the gas may be unable to radiate energy at a rate that balancesviscous heating.In this case, the heat generated by viscosity will be “advected” inwards with the flow instead ofbeing radiated. The disk becomes hot, hence geometrically thick (though perhaps optically thin),hence low density, and radiatively inefficient.Such “Advection Dominated Accretion Flows” (ADAFs) were studied by Lightman & Eardley,Rees, and others in the 1970s. They were revived in the 1990s by the work of Narayan & Yi andothers.

Relative to thin accretion disks, advection dominated flows have different structure (quasi-spherical,though angular momentum remains important), lower bolometric efficiency, and very differentspectral energy distributions, with emission over a wider range of wavelengths.ADAFs are also expected to have a 2-temperature structure, with hot ions holding most of theenergy and transferring it inefficiently to the electrons that produce most of the radiation.

Narayan and collaborators have argued, based on a combination of theoretical models and obser-vational studies of stellar mass black holes, that the inner regions of accretion disks are replacedby ADAFs when the accretion rate falls below ∼ 0.01MEdd.In stellar mass black holes (X-ray binaries), which sometimes cycle between different states, the“high soft state” (high luminosity, soft spectrum) may correspond to thin disk accretion and the“low hard state” to something like an ADAF. AGN might go through similar kinds of cycles on alonger timescale.

5.5.4 What is the physics of radiatively inefficient flows?

A follow-on from the above.

In a true ADAF, most of the energy gained by the gas ultimately disappears down the black hole,being advected all the way into the event horizon.

Blandford & Begelman have argued that most of the gas may be driven out after accreting to smallradius. They call this solution ADIOS (Advection Dominated Inflow-Outflow Solution). A smallfraction of the gas actually accretes onto the black hole, providing the energy to drive the outflow.

30

Quataert & Gruzinov have argued instead that advection dominated flows are convectively unstable.They propose CDAFs (Convection Dominated Accretion Flows), in which a small fraction of thegas accretes onto the black hole, providing energy to drive convection that returns most of the gasto large radius (but not to infinity).A CDAF cannot be indefinitely steady-state, since gas will build up at large radius.

Note that an ADAF is radiatively inefficient because most of the energy goes “down the hole,”while ADIOS and CDAF flows are inefficient because the rate at which the black hole actuallygains mass is much smaller than the “large scale M .”

Which, if any, of these solutions is the most physically realistic is a hotly debated theoretical andobservational question.On the theoretical side, one can ask what actually happens if one allows gas to accrete at a lowrate onto a black hole (but a complete solution is tough).On the observational side, one can ask which of the models makes predictions for SEDs and outflowrates that are in good agreement with observations.

5.5.5 What happens when the accretion rate is super-Eddington?

Presumably, a galaxy doesn’t “know” that it shouldn’t dump gas onto its central black hole abovethe Eddington rate. In understanding black hole growth and the quasar luminosity function,therefore, it is important to know what happens when the accretion rate is super-Eddington.

Note that the Eddington accretion rate is usually (though not always) defined for a radiativeefficiency ε ∼ 0.1.

One possibility is that the gas accretes and radiates with high efficiency, and that the black holetherefore shines at a super-Eddington luminosity. This would be impossible in spherical symmetryand steady state, but Begelman has recently argued that luminosities L ∼ 10−100LEdd are possible,with radiation escaping in “photon bubbles” that percolate out through the infalling medium.

A second possibility is that the gas accretes at a super-Eddington rate but radiates with lowefficiency, so that the luminosity is ≤ LEdd. This would be an advection dominated accretion flow,and there is indeed a class of such high accretion rate ADAF models. These are quite different fromthe low accretion rate ADAFs mentioned earlier – in these ADAFs, radiation pressure is dominant,and the radiation produced is trapped by the optically thick flow and pulled into the black holewith the gas. What is not clear is whether these high accretion rate ADAF solutions are stable.

A third possibility is that the black hole refuses to eat: if gas is dumped on at a super-Eddingtonrate, the resulting flow may drive most of the gas out, so that the actual accretion rate is sub-Eddington.

At present, any of these seems like a live possibility.

31

5.5.6 Can the accretion tap the spin energy of the black hole?

A spinning black hole has a lot of rotational energy. It may be possible for accreting gas to extractsome of this energy, in addition to its own gravitational energy.The leading model for such extraction is called the Blandford-Znajek mechanism, a messy (or, ifit’s to your taste, elegant) piece of relativistic MHD.It is generally thought that this mechanism may be important in the production of relativistic jets,and that these represent the transformation of black hole spin energy into kinetic energy of theoutflowing gas.

32

6 Sound Waves and Gravitational Instability

Reading: Ryden pp. 20-25, Shu pp. 110-112

6.1 Perturbation equations for a uniform medium

Consider the plane-parallel case =⇒ no y or z dependence

Continuity:∂ρ

∂t+ u

∂ρ

∂x+ ρ

∂u

∂x= 0

Momentum:∂u

∂t+ u

∂u

∂x= −1

ρ

∂P

∂x+ g

Poisson:∂g

∂x= −4πGρ.

Now assume a uniform static medium with ρ = ρ0, P = P0, u = 0.Symmetry =⇒ g = 0, but Poisson’s equation allows this only if ρ = 0.We will accept the “Jeans swindle” and ignore this problem with our zeroth-order solution, jumpingimmediately to the perturbed equations.

Introduce small perturbations ρ1, u1, P1, g1

ρ = ρ0 + ρ1(x, t) P = P0 + P1(x, t) u = u1(x, t) g = g1(x, t)

where |ρ1/ρ1| 1, |P1/P0| 1.

Substitute these expressions and keep only the terms that are first-order in the perturbations:

Continuity:∂ρ1

∂t+ ρ0

∂u1

∂x= 0

Momentum:∂u1

∂t= − 1

ρ0

∂P1

∂x+ g1

Poisson:∂g1

∂x= −4πGρ1.

Assume that P = P (ρ), allowing the second equation to be written

ρ0

∂u1

∂t+

dP

dρ

∣

∣

∣

0

∂ρ1

∂x= g1ρ0.

Note that this is standard first-order perturbation theory: introduce small perturbations, approxi-mate full equations by keeping only terms first-order in the perturbations, subtracting off or oth-erwise using the zeroth-order equations as necessary.

Take the time derivative of the continuity equation, subtract the spatial derivative of the momentumequation and substitute from the Poisson equation

∂2ρ1

∂t2+ ρ0

∂2u1

∂x∂t− ρ0

∂2u1

∂x∂t− dP

dρ

∣

∣

∣

0

∂2ρ1

∂x2= −ρ0

∂g1

∂x= 4πGρ0ρ1.

33

Hence∂2ρ1

∂t2− dP

dρ

∣

∣

∣

0

∂2ρ1

∂x2= 4πGρ0ρ1. (40)

6.2 Negligible self-gravity: sound waves

∂2ρ1

∂t2− a2 ∂2ρ1

∂x2= 0, a =

(

dP

dρ

∣

∣

∣

0

)1/2

is a wave equation for waves propagating at the sound speed a.If ρ1(x, t) = f(x − at) where f is an arbitrary function and f ′ and f ′′ are its first two derivativeswith respect to its argument:

∂ρ1

∂t= −af ′,

∂2ρ1

∂t2= a2f ′′,

∂2ρ1

∂x2= f ′′

=⇒ ∂2ρ1

∂t2− a2 ∂2ρ1

∂x2= 0.

For P = P0(ρ/ρ0)γ , a0 =

(

γP0

ρ0

)1/2

=(

γkT0

µmp

)1/2

.

Note that the sound speed is ∼ particle thermal speed; it is not surprising to learn that this is thespeed with which a disturbance can propagate. In physical units,

a = 1.63 × 106

(

γ

5/3

)1/2 (

µ

0.6

)

−1/2 (

kT0

1 eV

)1/2

cm s−1 ∼ 16 km s−1.

A memorable number is (kT/mp)1/2 ≈ 10 km s−1 at kT = 1 eV, T = 11, 600K.

A sound wave cannot be created with wavelength less than the mean free path λ, which is ∼ 0.2AU even in a dense molecular cloud. The frequency of such a wave is ∼ a/λ ∼ 10−8 Hz.

Ryden (pp. 22-24) shows that the effects of viscosity and heat conduction are to damp sound waves(converting acoustic energy into thermal energy) with an attenuation length

L1 ∼ Λ2

λ, Λ = sound wavelength, λ = mean free path.

=⇒ if Λ λ, sound propagates many wavelengths.For sound waves in air, Λ ≈ 300m s−1/300Hz = 1m, λ ∼ 10−4 cm = 10−6 m, L1 ∼ 106 m.

34

6.3 Non-negligible self-gravity

Go back to equation (40). Consider a perturbation of the form

ρ1(x, t) = exp[i(ωt − kx)].

Because sine waves are a complete basis set, we can represent any perturbation at time t0 as asuperposition of such perturbations.This is an especially useful approach in linear perturbation theory because the different modes donot interact to first order, so we can solve for the evolution of each one separately and add up theresults.

∂2ρ1

∂t2= −ω2ρ1

∂2ρ1

∂x2= −k2ρ1,

so equation (40) impliesω2 = k2a2

0 − 4πGρ0.

For k > kJ ≡ (4πGρ0)1/2/a0, ω is real, implying sound waves that oscillate and propagate without

growing.For k < kJ , ω is imaginary: long wavelength perturbations grow exponentially because of self-gravity.For k kJ , the growth timescale is ∼ (Gρ0)

−1/2 ∼ tdyn.

The Jeans wavelength is

λJ =2π

kJ∼ a0(Gρ0)

−1/2.

Thus, a perturbation is gravitationally unstable if a sound wave cannot cross it in a dynamicaltime. Shorter wavelength perturbations are stable because pressure gradients build up fast enoughto counter gravity.A general perturbation may decompose into short wavelength components, which do not grow, andlong wavelength perturbations, which do.In the non-linear regime, perturbations of different wavelengths influence each other.

The Jeans criterion can also be derived approximately on energetic grounds, by considering aspherical region whose density is increased by a factor (1 + δ), with δ 1.The perturbed gravitational potential energy is

∆W ∼ −GM∆M

R∼ −GM(R3ρ0δ)

R.

The perturbed thermal energy is∆U ∼ δMa2

0.

Thus ∆W + ∆U < 0 if

GM(R3ρ0δ)

R> δMa2

0 =⇒ R > a0(Gρ0)−1/2.

35

Large scale perturbations decrease their energy by growing in amplitude, but small scale perturba-tions increase their energy by growing in amplitude.

Numerical values:

λJ ≡ 2πkJ

= π1/2a0

(Gρ0)1/2= 3.6 kpc ×

(

γ

5/3

)1/2 (

0.6

µ

) (

kT0

1 eV

)1/2 (

n

1 cm−3

)

−1/2

MJ ≡ 4π3

(

λJ

2

)3

ρ0 = π6λ3

Jρ0 = 3.7 × 108M×

(

γ

5/3

)3/2 (

0.6

µ

)2 (

kT0

1 eV

)3/2 (

n

1 cm−3

)

−1/2

in a regime relevant to cosmology, or

λJ = 0.025 pc × γ1/2(

2

µ

) (

T0

10 K

)1/2 (

n106 cm−3

)

−1/2

MJ = 0.4M× γ3/2

(

2

µ

)2 (

T0

10 K

)3/2 (

n106 cm−3

)

−1/2

in a regime relevant for star formation.

36

7 Shocks

Reading: Ryden chs. 3 & 4, Shu chs. 15 & 16. For the enthusiasts, Shu chs. 13 & 14.

A good article for further reading: Shull & Draine, The physics of interstellar shock waves, inInterstellar processes; Proceedings of the Symposium, Grand Teton National Park, WY, July 1-7,1986 (A88-14501 03-90). Dordrecht, D. Reidel Publishing Co., 1987, p. 283-319.

Consider the propagation of a finite amplitude sound wave.The speed of propagation is higher at higher temperature, so the crest of the wave graduallyovertakes the trough (T ∝ ργ−1).When faster moving gas overtakes slower moving gas, we get a discontinuous change of density andvelocity, a shock.(NB: In fact, this steepening happens even for γ = 1 because of non-linearity of the equations.)

Shocks can also be produced by any supersonic compressive disturbance. This is the more commonsource of shocks in astrophysical situations.What are places where astrophysical shocks occur?Cloud-cloud collisionsHII regions expanding into neutral mediumStellar wind encountering mediumSupernova or GRB blast wave (internal and external shocks)Accretion onto compact objects: spherical or diskAccretion onto hydrostatic intracluster medium

In general a shock wave is• a pressure driven compressive disturbance• propagating faster than the “signal speed” for compressive waves• producing irreversible change in fluid state (increase of entropy)

In most cases a shock involves a “discontinuous” change of fluid properties over a scale ∼ λ.

7.1 Shock jump conditions

Consider a propagating shock wave in the rest frame of the shock.Unshocked gas approaches from the +x direction moving faster than its sound speed and passesthrough the shock.Pre-shock conditions: ρ1, u1, T1.Post-shock conditions: ρ2 > ρ1, u2 < u1, T2 > T1.

We would like to derive the relations (a.k.a. “jump conditions”) between ρ1, u1, T1 (or, equivalently,P1) and ρ2, u2, T2 (or P2), for a steady-state, plane-parallel shock (~u perpendicular to shock frontand fluid properties depend only on distance to front).

37

Within the shock front (a.k.a. “transition layer”), viscous effects are important – they cause theshock in the first place. However, outside this layer, viscous effects are small on scales larger thanthe mean free path. We will derive conservation equations of the form

d

dxQ(ρ, u, P ) = 0 =⇒ Q(ρ, u, P ) = constant,

and although the quantities Q involve viscous terms, we can ignore these outside the shock zoneand can therefore derive the jump conditions from equations that don’t involve viscosity terms.

We start from the continuity equation and the momentum equation

∂ρ

∂t+ ~∇ · (ρ~u) = 0,

∂~u

∂t+(

~u · ~∇)

~u = ~g − 1

ρ~∇P +

1

ρ~∇· ↔π,

and from the thermal energy equation, which it is helpful to write in the form (Ryden 1-43)

∂

∂t(ρε) + ~∇ · (ρε~u) = −P ~∇ · ~u − ~∇ · ~F + Ψ,

analogous to the continuity equation, and to supplement with a separate equation (Ryden 1-44)for conservation of kinetic energy

∂

∂t(1

2ρu2) + ~∇ · (1

2ρu2~u) = ρ~u · ~g − ~u · ~∇P + ~u · (~∇· ↔π ).

We now assume steady-state, ∂∂t = 0, plane-parallel, ∂

∂y = 0, ∂∂z = 0, ∂

∂x = d

dx , and ignore gravityand viscosity. These equations then become

d

dx(ρu) = 0 (41)

udu

dx= −1

ρ

dP

dx(42)

d

dx(ρεu) = −P

du

dx(43)

d

dx(1

2ρu2u) = −u

dP

dx. (44)

(45)

Equation (41) immediately gives

ρu = constant =⇒ ρ1u1 = ρ2u2.

Using

d

dx(ρu2) = 2ρu

du

dx+ u2 dρ

dx= ρu

du

dx+ u

(

ρdu

dx+ u

dρ

dx

)

= ρudu

dx+ u

d

dx(ρu) = ρu

du

dx

38

allows the equation (42) to be written

ρudu

dx+

dP

dx=

d

dx

(

ρu2 + P)

= 0 (46)

soρu2 + P = constant =⇒ ρ1u

21 + P1 = ρ2u

22 + P2. (47)

If we had kept the viscosity terms in the derivation, equation (46) would instead have been

d

dx

(

ρu2 + P − 4

3µ

du

dx

)

= 0. (48)

Within the transition zone, where µ and dudx are non-zero, ρu2 + P is not constant.

However, in the pre-shock and post-shock zones, µ and dudx are negligible, so equation (47) holds.

We could use equation (48) (together with our other equations and the constitutive relation forviscosity) to follow what happens within the transition zone. However, the fluid approximationitself breaks down within this region.

Adding together equations (43) and (44) gives

0 =d

dx

(

u(

1

2ρu2 + ρε

)

+ Pu)

=d

dx

(

ρu

(

1

2u2 + ε +

P

ρ

))

=

(

1

2u2 + ε +

P

ρ

)

d

dx(ρu) + ρu

d

dx

(

1

2u2 + ε +

P

ρ

)

.

Since d

dx(ρu) = 0 and ρu 6= 0, this equation implies that

d

dx

(

1

2u2 + ε +

P

ρ

)

= 0 =⇒ 1

2u2 + ε +

P

ρ= constant,

so1

2u2

1 + ε1 +P1

ρ1

= 1

2u2

2 + ε2 +P2

ρ2

.

If we had been more complete, the conserved quantity would also include viscosity and heat con-duction terms, but once again, these are unimportant outside of the transition zone.

In summary, we have the Rankine-Hugoniot jump conditions for a plane-parallel shock:

ρ1u1 = ρ2u2 (49)

ρ1u21 + P1 = ρ2u

22 + P2 (50)

1

2u2

1 + ε1 +P1

ρ1

= 1

2u2

2 + ε2 +P2

ρ2

. (51)

Even though the physics of the shock region may be complicated and varied, these conditionsfollow from conservation of mass, momentum, and energy alone. More precisely, the first follows

39

from mass conservation, the second from mass and momentum conservation, and the third frommass and energy conservation. If ρ1, u1, and P1 are known, we have three equations for the threeunknowns ρ2, u2, and P2.Using εi = 1

γi−1

Pi

ρi, the last of the jump conditions can be written

1

2u2

1 +γ1

γ1 − 1

P1

ρ1

= 1

2u2

2 +γ2

γ2 − 1

P2

ρ2

,

for a gas that has a polytropic equation of state.Note that γ2 may be different from γ1 if, for example, the shock dissociates molecules, or raises thetemperature so that previously inaccessible degrees of freedom become accessible.

7.2 The Mach Number

The dimensionless number that characterizes the strength of a shock is the Mach number, the ratioof the shock speed to the upstream sound speed:

M1 ≡ u1

a1

=

(

ρ1u21

γP1

)1/2

. (52)

The factor in () can be viewed as a ratio of “ram pressure” to thermal pressure in the pre-shockgas, or as a ratio of kinetic energy density to thermal energy density.

In terms of the Mach number, the shock jump conditions are (Ryden eqs. 3-51)

ρ2

ρ1

=u1

u2

=(γ + 1)M 2

1

(γ − 1)M 21 + 2

P2

P1

=ρ2kT2/m

ρ1kT1/m=

2γM21 − (γ − 1)

γ + 1.

Together these conditions imply

T2

T1

=[(γ − 1)M 2

1 + 2][2γM 21 − (γ − 1)]

(γ + 1)2M21

.

A strong shock is one with M1 1, yielding

ρ2

ρ1

=u1

u2

≈ γ + 1

γ − 1= 4 (53)

P2 ≈ 2γ

γ + 1M2

1 P1 =2

γ + 1ρ1u

21 =

3

4ρ1u

21 (54)

T2 ≈ 2γ(γ − 1)

(γ + 1)2T1M

21 =

2(γ − 1)

(γ + 1)2m

ku2

1 =3

16

m

ku2

1, (55)

where the last equalities are for γ = 5/3.

In the rest frame of a strong shock, with γ = 5/3, the post-shock kinetic energy is

1

2u2

2 ≈ 1

32u2

1,

40

and the post-shock thermal energy is3

2

kT2

m≈ 9

32u2

1.

So roughly half of the pre-shock kinetic energy is converted to thermal energy. The total energyof the post-shock gas is lower (in the shock rest frame) because of the work done on the gas byviscosity and pressure in the shock.

A weak shock has M1 − 1 = ε 1. In this limit

ρ2

ρ1

≈ 1 +4

γ + 1ε = 1 +

3

2ε (56)

P2

P1

≈ 1 +4γ

γ + 1ε = 1 +

5

2ε (57)

T2

T1

≈ 1 +4(γ − 1)

γ + 1ε = 1 + ε, (58)

where the last equalities are again for γ = 5/3.

The post-shock Mach number is

M2 ≡ u2

a2

=u1

a1

u2

u1

a1

a2

= M1

u2

u1

(

T1

T2

)1/2

.

In the weak shock limit, M2 ≈ 1 − ε.In the strong shock limit

M2 ≈ M1

(γ − 1)

(γ + 1)

[

(γ + 1)2

2γ(γ − 1)M 21

]1/2

=

(

γ − 1

2γ

)1/2

≈ 0.45.

A shock converts supersonic gas into denser, slower moving, higher pressure, subsonic gas. Itincreases the specific entropy of the gas by an amount

s2 − s1 = cV ln

(

P2

ργ2

)

− cV ln

(

P1

ργ1

)

= cV ln

(

P2

P1

)

− cV γ ln

(

ρ2

ρ1

)

.

One can also write this in the form

s2 − s1 = cP ln

(

T2

T1

)

− k

mln

(

P2

P1

)

,

which is Ryden’s eq. (3-55), using cP = cV + km = γcV .

In another terminology, a shock shifts gas to a higher adiabat.An adiabat is a locus of constant entropy (T ∝ ργ−1) in the density-temperature plane. Gas canmove adiabatically along an adiabat, while changes in entropy move it from one adiabat to another.

41

7.3 Numerical hydrodynamics and artificial viscosity

The physical scale of a shock transition layer is ∼ λ.In most systems of astrophysical interest, the scale of the system is L λ.In numerical hydrodynamics, it is usually impractical to make an individual grid cell comparableto or smaller than λ (especially in more than one dimension), so the physical scale of a shock isgenerally much smaller than one grid cell.Fortunately, we usually don’t care about the region right around the shock, just about the influenceof the shock on the physical state of the gas.The relation between the physical properties of the upstream and downstream gas are determinedby the shock jump conditions, and they follow from conservation laws, independent of the detailsof the shock.

“Shock capturing codes” work by searching for places where shocks are “likely” to occur (e.g.,where the velocity field is developing a discontinuity) and imposing shock jump conditions over thescale of one or two grid cells.

An alternative method, introduced by von Neumann and Richtmeyer in the 1950s, is to use an“artificial viscosity” that is much larger than the true viscosity one would calculate based on themicroscopic properties of the gas.High viscosity makes shocks broader, so that they can be resolved over several grid cells.Since the jump conditions follow from conservation laws, the pre- and post-shock gas should stillhave the correct relations even if the viscosity is not the true viscosity.There is a fair amount of experimentation and black magic in artificial viscosity — tuning thingsto get the desired physical behavior in as many situations as possible.

7.4 Radiative shocks

Now suppose that the gas is able to radiate energy in a “radiative relaxation layer” (rrl) after theshock.Immediately after the shock front the fluid properties are still denoted ρ2, u2, P2, T2, but theychange in the radiative relaxation layer, settling to new values ρ3, u3, P3, T3 further downstream.The rrl is always large compared to the shock front itself because many collisions are required tocool the gas (so that the size of the rrl is λ).

Within the rrl, the temperature drops, and the gas is squeezed to higher density. Since ρu =constant, the velocity drops.We can demonstrate this by returning to the steady-state equations (41)-(43) but including thecooling term

L(ρ, T ) ≡ − (Γ − Λ)

ρ> 0, [L] = erg g−1 s−1.

The equations are

d

dx(ρu) = 0

42

udu

dx= −1

ρ

dP

dx

d

dx(ρεu) = ρu

dε

dx= −P

du

dx− ρL.

Combining the last equation with

ε =1

γ − 1

P

ρ=⇒ dε

dx=

1

γ − 1

[

1

ρ

dP

dx− P

ρ2

dρ

dx

]

and1

ρ

dρ

dx= −1

u

du

dx

(since ρ ∝ u−1) impliesu

γ − 1

[

dP

dx+

P

u

du

dx

]

+ Pdu

dx= −ρL,

and substituting dPdx = −ρudu

dx implies

− ρu2

γ − 1

du

dx+

γ

γ − 1P

du

dx= −ρL

and thusa2 − u2

γ − 1

du

dx= −L

with a2 = γP/ρ. Since the post-shock flow is sub-sonic (u < a) and L > 0, we conclude thatdudx < 0, and mass conservation therefore implies that dρ

dx > 0.

The mass conservation and momentum conservation jump conditions apply as before, since theirderivation did not involve ε (which is the quantity affected by radiative cooling). Therefore

ρ3u3 = ρ2u2 = ρ1u1

ρ3u23 + P3 = ρ2u

22 + P2 = ρ1u

21 + P1.

As u drops in the rrl, ρu2 = (ρu)u drops and the pressure must rise.For a strong shock with γ = 5/3,

M2 = u2

(

5

3

P2

ρ2

)

−1/2

≈ 0.45,

so P2 ≈ 3ρ2u22 and only a small rise in pressure is required to satisfy the second jump condition

even if u3 drops to zero.The important point is that the pressure does not go down, and therefore a drop in temperaturemust be accompanied by a proportional rise in density to maintain the pressure.If there is a lot of post-shock cooling, then the density ratio ρ3/ρ1 can be very high.

43

This is probably the most important difference between a non-radiative shock and a radiative shock:a non-radiative shock can only increase the density by a factor of ∼ 4, but a radiative shock canincrease the density by a very large factor.

A specific interesting case is that of an “isothermal shock,” where T3 = T1. (This becomes thethird jump condition.)This can arise if, for instance, T1 = T3 is a temperature where the cooling time becomes very long,or a temperature where heating and cooling processes balance.Combined with the equation of state P = ρ

mkT = ρa2T , where aT = (kT/m)1/2 is the “isothermal

sound speed,” the solution to the shock jump conditions is,

ρ3

ρ1

=u1

u3

=

(

u1

aT

)2

≡ M2T

T3 = T1.

Since

ρ3u23 + P3 = ρ1u

21

(

u1

aT

)2 (aT

u1

)4

+ ρ1a2T

(

u1

aT

)2

= ρ1a2T + ρ1u

21 = P1 + ρ1u

21,

this solution satisfies the momentum jump condition.

The fact that ρ3/ρ1 = M2T means that the compression factor in an isothermal shock can be

arbitrarily high.

7.5 Oblique shocks

If the fluid enters the shock at an angle other than 90, the shock jump conditions apply only tothe perpendicular component u

⊥of u1.

The parallel component of u1 is unchanged, so the fluid is “refracted” by the shock, changing itsdirection so that it is closer to parallel to the front.

7.6 Shocks with magnetic fields: a few comments

Hydromagnetic shocks are often important in the ISM.For a shock with B parallel to the shock front:

B1u1 = B2u2,

i.e., the magnetic field lines are compressed by the same factor as the density.

The momentum jump condition becomes

ρ1u21 + P1 +

B21

8π= ρ2u

22 + P2 +

B22

8π,

i.e., there is an additional “magnetic pressure” term B2/8π.(The units of B2 are erg cm−3 = dyne cm−2.)

44

The energy jump condition becomes

1

2u2

1 +γ

γ − 1

P1

ρ1

+B2

1

4πρ1

= 1

2u2

2 +γ

γ − 1

P2

ρ2

+B2

2

4πρ2

.

Coupling of ions to magnetic fields can allow a shock front to be much narrower than the meanfree path λ for particle-particle interactions.

The situation is more complicated when B is not parallel to the shock front, and it can be muchmore complicated in multi-fluid shocks when electrons, ions, and neutrals are coupled but notperfectly coupled.One important effect is that magnetosonic waves (a.k.a. Alfven waves) can “warn” upstream ionsand electrons about an approaching density discontinuity.There are many additional classes of solutions, some of them discussed in the Shull & Draine article.

45

8 Radiative Cooling and Heating

Reading:Katz et al. 1996, ApJ Supp, 105, 19, section 3Thoul & Weinberg, 1995, ApJ, 442, 480Optional reading:Thoul & Weinberg, 1996, ApJ, 465, 608Weinberg et al., 1997, ApJ, 477, 8The latter two address the influence of photoionization on galaxy formation and the interplaybetween physics and numerical resolution in cosmological simulations.

The thermal energy equation is

Dε

Dt= −P

ρ~∇ · ~u − 1

ρ~∇ · ~F +

1

ρΨ +

Γ − Λ

ρ.

We now turn our attention to the last term, in which

Γ ≡ volumetric radiative heating rate [ erg cm−3 s−1]

Λ ≡ volumetric radiative cooling rate [ erg cm−3 s−1].

Unfortunately, due to a shortage of Greek letters, Γ is also used to represent ionization rates. Inthis section of the course and notes, therefore, I will denote the radiative heating rate by H, andwill reserve Γ for ionization rates.

I will focus on cooling and heating of a primordial composition, H/He plasma because (a) it’s simpleenough to treat fairly comprehensively, (b) you’ll encounter other cases in the ISM class, and (c)it’s the case I know the most about.

8.1 Cooling processes in a primordial plasma

The cooling processes that affect a H/He plasma are:

Collisional excitation: free electron impact knocks a bound electron to an excited state; it decays,emitting a photon.

Collisional ionization: free electron impact ionizes a formerly bound electron, taking energy fromthe free electron.

Recombination: free electron recombines with an ion; the binding energy and the free electron’skinetic energy are radiated away (only the latter counts as a “loss” here — the binding energy was“charged” to the collisional ionization process).

Free-free emission: free electron is accelerated by an ion, emitting a photon. (A.k.a. Bremsstrahlung.)

All of these processes are proportional to a function of temperature (different for each process)times the electron number density times the number density of the relevant ionic species.

46

At very high densities, formation of H2 molecules is an additional source of cooling.

8.2 Ionic abundances and ionization equilibrium

In order to calculate cooling rates using, say, the formulas for the above processes given in Table1 of Katz et al. (1996), one needs to know the density of the various ionic species: ne, nH0

, nH+,

nHe0 , nHe+ , nHe++.

At fixed total gas density, the evolution of these densities is governed by equations like

dnH0

dt= αH+

(T )nH+ne − ΓeH0

(T )nenH0− ΓγH0

nH0, (59)

where

αH+(T ) = hydrogen recombination coefficient [ cm3 s−1] (60)

ΓeH0(T ) = collisional ionization rate [ cm3 s−1] (61)

ΓγH0≡

∫

∞

νT

4πJ(ν)

hνσ(ν)dν = photoionization rate [ s−1], (62)

with

νT = ionization threshold frequency (e.g., 13.6 eV/h for H0)

σ(ν) = ionization cross-section [ cm−2]

J(ν) = radiation background intensity [erg s−1 cm−2 sr−1 Hz−1].

Suppose that the medium is ∼ 50% ionized (nH0∼ nH+

∼ ne ∼ n) and the right hand sideequation (59) is far from balance.The ionic density will evolve on a timescale

t ∼∣

∣

∣n

(

dn

dt

)

−1∣∣

∣ ∼∣

∣

∣ne(αH+(T ) − ΓeH0

(T )) − ΓγH0

∣

∣

∣

−1

.

For a typical estimate of the radiation background at z ∼ 2, J(ν) ∼ few×10−22erg s−1 cm−2 sr−1 Hz−1,

1

ΓγH0

∼ 1012 s ∼ 3 × 104 yr,

while for T ∼ 105 K

∣

∣

∣ne(αH+(T ) − ΓeH0

(T ))∣

∣

∣

−1

≈ 1

neΓeH0(T )

∼ 106

(

ne

10−5 cm−3

)

−1

yr.

If photoionization (ΓγH0) is significant, the timescale is almost always short compared to the dynam-

ical timescale of cosmological systems, and even without photoionization it is usually still shorterthan the dynamical timescale.

47

In many circumstances, therefore, one can safely assume that creation and destruction rates will bebalanced for each species, the condition known as ionization equilibrium. In this case, both sidesof equations like (59) are equal to zero.In ionization equilibrium, the abundances of ionized and neutral hydrogen are determined by alge-

braic equations (much easier than differential equations),

ΓeH0nenH0

+ ΓγH0nH0

= αH+nenH+

,

nH++ nH0

= nH

ne = nH++ nHe+ + 2nHe++

together with similar equations

destruction rate = creation rate

for nHe0 , nHe+ , nHe++(Katz et al. eqs. 25-28).

In a numerical hydrodynamics code, one must decide whether to integrate the ionic abundanceequations like (59) or compute abundances assuming ionization equilibrium.The appropriate choice depends on the physical situation. In some circumstances, departures fromionization equilibrium are physically important. However, one must be sure to integrate the time-dependent equations on a timscale short compared to the abundance evolution timescale, whichcan be very short compared to other timescales of interest.When the ionization equilibrium assumption is appropriate, it is much more efficient to use theequilibrium equations.

8.3 The collisional equilibrium cooling curve

The above equilibrium equations can be recast into the form

nH0

nH

=αH+

αH++ ΓeH0

+ ΓγH0/ne

,

with analogous equations for the fractions of neutral helium and singly ionized helium.If there is no photoionization, ΓγH0

= 0, then, since ΓeH0and αH+

depend only on temperature(not density), the relative abundances of ionic species depend only on temperature. All coolingprocesses are therefore a function of temperature times the square of the total gas density ρ (or,equivalently, the total hydrogen number density nH = Xρ/mp, where X ≈ 0.76 is the hydrogenmass fraction).

Since ionic abundances are determined by equilibrium of collisional processes, this situation is calledcollisional equilibrium (a.k.a. coronal equilibrium). Cooling rates in collisional equilibrium are fullydescribed by the function Λ(T )/n2

H, known as the cooling curve (e.g., Katz et al. Figure 1).

At high temperatures the gas is fully ionized, so the only cooling process is free-free emission, with

Λ

n2H

∼ 2.5 × 10−23

(

T

108 K

)1/2

.

48

The dominant processes at low temperatures are collisional line excitation of hydrogen (T ∼104.2 K) and collisional line excitation of He+(T ∼ 105 K).

Below 104 K, collisions are not energetic enough to ionize atoms or even excite them out of theground state, so the cooling rate drops to zero.

The timescale on which gas can radiate away its thermal energy is

tcool ∼∣

∣

∣ε

(

Dε

Dt

)

−1∣∣

∣ ∼∣

∣

∣

ερ

Λ

∣

∣

∣ ∝ ερ−1, (63)

since Λ ∝ ρ2.If the gas is confined by gravity or by ambient pressure, loss of thermal energy is usually accom-panied by an increase in density (to restore the pressure gradient or equilibrium with the ambientmedium).The increase in density usually decreases the cooling timescale, so cooling tends to be a runawayprocess.For a primordial composition cooling curve, gas that starts to cool (i.e., has tcool less than the ageof the system) usually cools fairly rapidly to 104 K, at which point it can cool no further (at leastby atomic processes).

8.4 Cooling and galaxy formation

In an expanding universe, the gravitational collapse of a homogeneous spherical perturbation pro-duces a virialized object whose average density is roughly 200 times the background critical densityat the time of collapse.For a given virial mass, one can use this density to compute a characteristic radius R, velocitydispersion σ2 ∼ GM/R, and virial temperature T ∼ GMmp/(kR), given a collapse redshift (whichdetermines the cosmic background density).A plausible assumption is that any gas that participates in this collapse is heated to this virialtemperature by shocks, and one can then calculate a cooling time from the temperature and density.

Some of the early analytic papers on galaxy formation (Binney 1977, ApJ, 215, 483; Silk 1977, ApJ,211, 638; Rees & Ostriker 1977, MNRAS, 179, 541) argued that the characteristic mass scale ofgalaxies was determined by the requirement that this cooling time be shorter than the dynamicaltime (or, perhaps, than the age of the universe).

More massive objects have higher virial temperatures and tend to collapse later (at lower density),so they tend to have longer cooling times, and the gas in them would therefore be unable to cooland form stars. The cooling argument therefore implies an upper limit on galaxy masses, at a scalethat is not too far from what is observed.

In my view, this argument is fatally flawed by the assumption that all of the collapsed gas is atthe same density. More realistically, the gas will have a density profile, and the dense gas near thecenter will be able to cool even if the more diffuse gas at larger radii cannot.

49

White & Rees (1978, MNRAS, 183, 341) introduced a more realistic picture, in which the amountof gas that cools in a collapsed dark matter halo is determined by the cooling radius, the radius atwhich the cooling time is equal to the time that the object is around before merging with anotherobject of comparable mass (which reheats the gas and “resets the clock.”) — or the age of theuniverse if that is longer.

In this case, the fraction of the mass that cools decreases as the mass of the halo increases, but thetotal cooled mass still increases. This model therefore does not yield a sharp upper cutoff in galaxymasses.

The cooling radius approach, combined with a more firmly based cosmological model and a moresophisticated framework for calculating halo merger rates, is the basis of “semi-analytic” methodsfor modeling galaxy formation. Gas cooling must be supplemented with a model for how the gasdensity determines the star formation rate and, crucially, how the star formation influences thesurrounding gas (“feedback”).

Apart from the uncertainties associated with star formation and feedback, the key uncertainty inthis approach is whether the density profiles and temperatures inferred from the spherical collapsepicture are realistic at the desired level of accuracy. In particular, there are suggestions from 3-dsimulations of the hierarchical build-up of galaxies that much of the gas that ends up in the coldcomponent is never heated to the virial temperature of a typical galaxy halo. It may yet turn outthat this suggestion is a flaw in the numerical simulations themselves, but if it is correct then itcalls into question one of the key assumptions of the semi-analytic methods.

8.5 Photoionization

At redshift z < 6 (and perhaps higher, but that is still a matter of debate), the universe is pervadedby an ultraviolet radiation background, produced by quasars and (perhaps) star-forming galaxies.(There is no question that high mass stars produce some UV photons energetic enough to ionizehydrogen and even helium. The unknown factor is what fraction of these photons escape from theISM to become part of the intergalactic UV background.)

At z ∼ 2 − 3, when the quasar population is at its peak, the hydrogen photoionization rate isestimated to be

ΓγH0∼ 10−12 s−1,

which corresponds to a background intensity at the Lyman limit frequency νT of J(ν) ∼ few ×10−22erg s−1 cm−2 sr−1 Hz−1 (see eq. 62).The helium photoionization rates ΓγHe0 and ΓγHe+ are more uncertain, probably lower by a factorof a few (ΓγHe0

) and by a factor of 10-100 (ΓγHe+).

For gas near the cosmic mean density, the photoionization rate is much higher than the recombi-nation rate, so the hydrogen is nearly all ionized and the helium is nearly all doubly ionized.

Photoionization has two important effects on gas cooling.

50

(1) It eliminates line excitation and ionization as cooling processes at low densities, by eliminatingH0, He0, and He+. Recombination cooling increases, but the net effect is to severely reduce coolingrates at T ∼ 104 − 105 K for low density gas. At high densities (where the recombination ratebecomes comparable to or larger than the photoionization rate), the cooling rates move back towardstheir collisional equilibrium values.

(2) It heats the gas because photoelectrons carry off residual energy. The heating rate is

H = nH0εH0

+ nHe0εHe0 + nHe+εHe+ , (64)

where, for example

εH0=

∫

∞

νT

4πJ(ν)

hνσ(ν)(hν − hνT )dν [ erg s−1].

The heating rate decreases with increasing temperature because the recombination rates (and hencethe fraction of neutral “targets” for the photons) decline.

At low densities, H ∝ n2H because, e.g.,

nH0∝ nH × recombination rate per atom

photoionization rate per atom∝ n2

H.

The net rate of radiative energy change per unit volume is H − Λ. The leading dependence ondensity is H − Λ ∝ n2

H, but there is a slow change with nH0because of the competition between

photoionization and recombination rates. Thus there is a density dependent “cooling curve”

H− Λ

n2H

(T ),

(e.g., Katz et al. 1996, Figure 2).

8.6 The equilibrium temperature

In the presence of photoionization, primordial (H/He) gas gets heated at low temperatures andcooled at high temperatures. There is thus a value of T at which

H(Teq) − Λ(Teq) = 0.

If other processes (e.g., shock heating, compression, expansion) can be ignored, the gas will even-tually settle to this equilibrium temperature. It is then said to be in thermal equilibrium. (Notethat this has nothing to do with local thermodynamic equilibrium.)

For photoionized H/He, Teq is always in the range 104 − 105 K. It is higher for a harder (bluer)ionizing background spectrum because the residual photoelectron energy is higher if the typicalionizing photons are more energetic.Teq decreases with increasing density because at higher densities ionization and line cooling pro-cesses become more important.

The timescale for achieving thermal equilibrium is the cooling time, equation (63), but now withΛ replaced by H− Λ.At low densities or high temperatures, tcool can be very long, so thermal equilibrium is a less robustassumption than ionization equilibrium. Systems often change on a timescale much shorter thanthat required to reach thermal equilibrium.

51

8.7 Some general lessons

There are many other radiative heating and cooling processes that are important in astrophysicalsituations, e.g.

metal-line cooling (the CII fine-structure line is especially important)Compton cooling (fast electrons lose energy by upscattering photons; at z ∼> 7, Compton coolingoff microwave background photons can be the dominant cooling process)synchrotron coolingmolecular line coolingmolecular formation coolingCompton heating by X-rays or γ-raysCosmic ray heating

However, the H/He plasma illustrates a number of features that appear in many situations.(1) For a general treatment, one must integrate time-dependent equations to compute the abun-dances of each species, then use these abundances to compute cooling rates.(2) If ionization and recombination timescales are short, one can compute abundances using ion-ization equilibrium.(3) If ionization equilibrium applies, cooling/heating rates are still usually strongly dependent ontemperature. In simple cases they are ∝ ρ2, though in general the density is more complex.(4) If there are both heating and cooling processes, then there is usually an equilibrium temperaturewhere H−Λ = 0. The gas will relax to this equilibrium temperature if it has sufficient time, thoughat low densities it often does not.

In some cases, the dependence of equilibrium temperature on density leads to thermal instability.A small amplitude perturbation can cause the gas to spontaneously separate into two phases thatare in pressure equilibrium at different values of Teq, a cool dense phase and a warm diffuse phasewith equal pressure. Thermal instability is probably important in determining the structure of theISM. (See discussion in Shu, chapter 8.)

52