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Theoretical Astrophysics

Matthias BartelmannInstitut für Theoretische Astrophysik

Universität Heidelberg

Inhaltsverzeichnis

1 Macroscopic Radiation Quantities, Emission and Absorpti-on 1

1.1 Specific Intensity . . . . . . . . . . . . . . . . . . . . 1

1.2 Relativistic Invariant . . . . . . . . . . . . . . . . . . 2

1.2.1 Lorentz Transformation of Iν . . . . . . . . . . 2

1.2.2 Example: The CMB Dipole . . . . . . . . . . 4

1.3 Einstein coefficients and the Planck spectrum . . . . . 5

1.3.1 Transition Balance . . . . . . . . . . . . . . . 5

1.3.2 Example: The CMB Spectrum . . . . . . . . . 6

1.4 Absorption and Emission . . . . . . . . . . . . . . . . 7

1.5 Radiation Transport in a Simple Case . . . . . . . . . 8

1.6 Emission and Absorption in the Continuum Case . . . 10

2 Scattering 13

2.1 Maxwell’s Equations and Units . . . . . . . . . . . . . 13

2.2 Radiation of a Moving Charge . . . . . . . . . . . . . 14

2.3 Scattering off Free Electrons . . . . . . . . . . . . . . 15

2.3.1 Polarised Thomson Cross Section . . . . . . . 15

2.3.2 Unpolarised Thomson Cross Section . . . . . . 16

2.4 Scattering off Bound Charges . . . . . . . . . . . . . . 17

2.5 Radiation Drag . . . . . . . . . . . . . . . . . . . . . 19

2.5.1 Time-Averaged Damping Force . . . . . . . . 19

2.5.2 Energy Transfer to a Radiation Field . . . . . . 20

2.6 Compton Scattering . . . . . . . . . . . . . . . . . . . 21

3

4 INHALTSVERZEICHNIS

2.6.1 Energy-Momentum Conservation . . . . . . . 21

2.6.2 Energy Balance . . . . . . . . . . . . . . . . . 22

2.7 The Kompaneets Equation . . . . . . . . . . . . . . . 24

3 Radiation Transport and Bremsstrahlung 27

3.1 Radiation Transport Equations . . . . . . . . . . . . . 27

3.2 Local Thermodynamical Equilibrium . . . . . . . . . . 29

3.3 Scattering . . . . . . . . . . . . . . . . . . . . . . . . 32

3.4 Bremsstrahlung . . . . . . . . . . . . . . . . . . . . . 34

3.4.1 Spectrum of a Moving Charge . . . . . . . . . 34

3.4.2 Hyperbolic Orbits . . . . . . . . . . . . . . . . 35

3.4.3 Integration over the Electron Distribution . . . 37

4 Synchrotron Radiation, Ionisation and Recombination 41

4.1 Synchrotron Radiation . . . . . . . . . . . . . . . . . 41

4.1.1 Electron Gyrating in a Magnetic Field . . . . . 41

4.1.2 Beaming and Retardation . . . . . . . . . . . . 42

4.1.3 Synchrotron Spectrum . . . . . . . . . . . . . 44

4.2 Photo-Ionisation . . . . . . . . . . . . . . . . . . . . . 45

4.2.1 Transition Amplitude . . . . . . . . . . . . . . 45

4.2.2 Transition Probability . . . . . . . . . . . . . 46

4.2.3 Transition Matrix Element . . . . . . . . . . . 48

4.2.4 Cross Section . . . . . . . . . . . . . . . . . . 50

5 Spectra 53

5.1 Natural Width of Spectral Lines . . . . . . . . . . . . 53

5.2 Cross Sections and Oscillator Strengths . . . . . . . . 53

5.2.1 Transition Probabilities . . . . . . . . . . . . . 54

5.3 Collisional Broadening of Spectral Lines . . . . . . . . 56

5.4 Velocity Broadening of Spectral Lines . . . . . . . . . 57

5.5 The Voigt Profile . . . . . . . . . . . . . . . . . . . . 58

INHALTSVERZEICHNIS 5

5.6 Equivalent Widths and Curves-of-Growth . . . . . . . 59

6 Energy-Momentum Tensor and Equations of Motion 63

6.1 Boltzmann Equation and Energy-Momentum Tensor . 63

6.1.1 Boltzmann Equation . . . . . . . . . . . . . . 63

6.1.2 Moments; Continuity Equation . . . . . . . . . 64

6.1.3 Energy-Momentum Tensor . . . . . . . . . . . 66

6.2 The Tensor Virial Theorem . . . . . . . . . . . . . . . 69

6.2.1 A Corollary . . . . . . . . . . . . . . . . . . . 69

6.2.2 Second Moment of the Mass Distribution . . . 70

7 Ideal and Viscous Fluids 73

7.1 Ideal Fluids . . . . . . . . . . . . . . . . . . . . . . . 73

7.1.1 Energy-Momentum Tensor . . . . . . . . . . . 73

7.1.2 Equations of Motion . . . . . . . . . . . . . . 75

7.1.3 Entropy . . . . . . . . . . . . . . . . . . . . . 77

7.2 Viscous Fluids . . . . . . . . . . . . . . . . . . . . . . 78

7.2.1 Stress-Energy Tensor; Viscosity and Heat Con-ductivity . . . . . . . . . . . . . . . . . . . . 78

7.2.2 Estimates for Heat Conductivity and Viscosity 80

7.2.3 Equations of Motion for Viscous Fluids . . . . 82

7.2.4 Entropy . . . . . . . . . . . . . . . . . . . . . 83

7.3 Generalisations . . . . . . . . . . . . . . . . . . . . . 84

7.3.1 Additional External Forces; Gravity . . . . . . 84

7.3.2 Example: Cloud in Pressure Equilibrium . . . 85

7.3.3 Example: Self-Gravitating Gas Sphere . . . . . 86

8 Flows of Ideal and Viscous Fluids 89

8.1 Flows of Ideal Fluids . . . . . . . . . . . . . . . . . . 89

8.1.1 Vorticity and Kelvin’s Circulation Theorem . . 89

8.1.2 Bernoulli’s Constant . . . . . . . . . . . . . . 91

8.1.3 Hydrostatic Equlibrium . . . . . . . . . . . . . 92


8.1.4 Curl-Free and Incompressible Flows . . . . . . 93

8.2 Flows of Viscous Fluids . . . . . . . . . . . . . . . . . 94

8.2.1 Vorticity; Incompressible Flows . . . . . . . . 94

8.2.2 The Reynolds Number . . . . . . . . . . . . . 95

8.3 Sound Waves in Ideal Fluids . . . . . . . . . . . . . . 96

8.3.1 Linear Perturbations . . . . . . . . . . . . . . 96

8.3.2 Sound Speed . . . . . . . . . . . . . . . . . . 98

8.4 Supersonic Flows . . . . . . . . . . . . . . . . . . . . 99

8.4.1 Mach’s Cone; the Laval Nozzle . . . . . . . . 99

8.4.2 Spherical Accretion . . . . . . . . . . . . . . . 100

9 Shock Waves and the Sedov Solution 105

9.1 Steepening of Sound Waves . . . . . . . . . . . . . . . 105

9.1.1 Formation of a Discontinuity . . . . . . . . . . 105

9.1.2 Specific Example . . . . . . . . . . . . . . . . 107

9.2 Shock Waves . . . . . . . . . . . . . . . . . . . . . . 110

9.2.1 The Shock Jump Conditions . . . . . . . . . . 110

9.2.2 Propagation of a One-Dimensional Shock Front 111

9.2.3 The Width of a Shock . . . . . . . . . . . . . 114

9.3 The Sedov Solution . . . . . . . . . . . . . . . . . . . 115

9.3.1 Dimensional Analysis . . . . . . . . . . . . . 115

9.3.2 Similarity Solution . . . . . . . . . . . . . . . 117

10 Instabilities, Convection, Heat Conduction, Turbulence 121

10.1 Rayleigh-Taylor Instability . . . . . . . . . . . . . . . 121

10.2 Kelvin-Helmholtz Instability . . . . . . . . . . . . . . 124

10.3 Thermal Instability . . . . . . . . . . . . . . . . . . . 126

10.4 Heat Conduction and Convection . . . . . . . . . . . . 130

10.4.1 Heat conduction . . . . . . . . . . . . . . . . 130

10.4.2 Convection . . . . . . . . . . . . . . . . . . . 132

10.5 Turbulence . . . . . . . . . . . . . . . . . . . . . . . 133


11 Collision-Less Plasmas 135

11.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . 135

11.1.1 Shielding; the Debye length . . . . . . . . . . 135

11.1.2 The plasma frequency . . . . . . . . . . . . . 136

11.2 The Dielectric Tensor . . . . . . . . . . . . . . . . . . 137

11.2.1 Polarisation and dielectric displacement . . . . 137

11.2.2 Structure of the dielectric tensor . . . . . . . . 138

11.3 Dispersion Relations . . . . . . . . . . . . . . . . . . 139

11.3.1 General form of the dispersion relations . . . . 139

11.3.2 Transversal and longitudinal waves . . . . . . 140

11.4 Longitudinal Waves . . . . . . . . . . . . . . . . . . . 141

11.4.1 The longitudinal dielectricity . . . . . . . . . . 141

11.4.2 Landau Damping . . . . . . . . . . . . . . . . 143

11.5 Waves in a Thermal Plasma . . . . . . . . . . . . . . . 144

11.5.1 Longitudinal and transversal dielectricities . . 144

11.5.2 Dispersion Measure and Damping . . . . . . . 147

12 Magneto-Hydrodynamics 149

12.1 The Magneto-Hydrodynamic Equations . . . . . . . . 149

12.1.1 Assumptions . . . . . . . . . . . . . . . . . . 149

12.1.2 The induction equation . . . . . . . . . . . . . 150

12.1.3 Euler’s equation . . . . . . . . . . . . . . . . 151

12.1.4 Energy and entropy . . . . . . . . . . . . . . . 153

12.1.5 Magnetic advection and diffusion . . . . . . . 154

12.2 Generation of Magnetic Fields . . . . . . . . . . . . . 155

12.3 Ambipolar Diffusion . . . . . . . . . . . . . . . . . . 157

12.3.1 Scattering cross section . . . . . . . . . . . . . 157

12.3.2 Friction force; diffusion coefficient . . . . . . . 159

13 Waves in Magnetised Plasmas 161

13.1 Waves in magnetised cold plasmas . . . . . . . . . . . 161


13.1.1 The dielectric tensor . . . . . . . . . . . . . . 161

13.1.2 Contribution by ions . . . . . . . . . . . . . . 163

13.1.3 General dispersion relation . . . . . . . . . . . 164

13.1.4 Wave propagation parallel to the magnetic field 165

13.1.5 Faraday rotation . . . . . . . . . . . . . . . . 166

13.1.6 Wave propagation perpendicular to the magneticfield . . . . . . . . . . . . . . . . . . . . . . . 167

13.2 Hydro-Magnetic Waves . . . . . . . . . . . . . . . . . 168

13.2.1 Linearised perturbation equations . . . . . . . 168

13.2.2 Alfvén waves . . . . . . . . . . . . . . . . . . 170

13.2.3 Slow and fast hydro-magnetic waves . . . . . . 171

14 Jeans Equations and Jeans Theorem 173

14.1 Collision-less motion in a gravitational field . . . . . . 173

14.1.1 Motion in a gravitational field . . . . . . . . . 173

14.1.2 The relaxation time scale . . . . . . . . . . . . 174

14.2 The Jeans Equations . . . . . . . . . . . . . . . . . . 177

14.2.1 Moments of Boltzmann’s equation . . . . . . . 177

14.2.2 Jeans equations in cylindrical and spherical coor-dinates . . . . . . . . . . . . . . . . . . . . . 179

14.2.3 Application: the mass of a galaxy . . . . . . . 180

14.3 The Virial Equations . . . . . . . . . . . . . . . . . . 181

14.3.1 The tensor of potential energy . . . . . . . . . 181

14.3.2 The tensor virial theorem . . . . . . . . . . . . 182

14.4 The Jeans Theorem . . . . . . . . . . . . . . . . . . . 184

15 Equilibrium, Stability and Disks 187

15.1 The Isothermal Sphere . . . . . . . . . . . . . . . . . 187

15.1.1 Phase-space distribution function . . . . . . . 187

15.1.2 Isothermality . . . . . . . . . . . . . . . . . . 188

15.1.3 Singular and non-singular solutions . . . . . . 189

15.2 Equilibrium and Relaxation . . . . . . . . . . . . . . . 190


15.3 Stability . . . . . . . . . . . . . . . . . . . . . . . . . 191

15.3.1 Linear analysis and the Jeans swindle . . . . . 191

15.3.2 Jeans length and Jeans mass . . . . . . . . . . 193

15.4 The rigidly rotating disk . . . . . . . . . . . . . . . . 194

15.4.1 Equations for the two-dimensional system . . . 194

15.4.2 Analysis of perturbations . . . . . . . . . . . . 195

15.4.3 Toomre’s criterion . . . . . . . . . . . . . . . 196

16 Dynamical Friction, Fokker-Planck Approximation 199

16.1 Dynamical Friction . . . . . . . . . . . . . . . . . . . 199

16.1.1 Deflection of point masses . . . . . . . . . . . 199

16.1.2 Velocity changes . . . . . . . . . . . . . . . . 201

16.1.3 Chandrasekhar’s formula . . . . . . . . . . . . 202

16.2 Fokker-Planck Approximation . . . . . . . . . . . . . 204

16.2.1 The master equation . . . . . . . . . . . . . . 204

16.2.2 The Fokker-Planck equation . . . . . . . . . . 205

Kapitel 1

Macroscopic RadiationQuantities, Emission andAbsorption

further reading: Shu, “The Phy-sics of Astrophysics, Vol I: Radia-tion”, chapter 1; Rybicki, Light-man, “Radiative Processes inAstrophysics”, chapter 1; Pad-manabhan, “Theoretical Astro-physics, Vol. I: AstrophysicalProcesses”, sections 6.1–6.3

1.1 Specific Intensity

• to begin with, radiation is considered as a stream of particles;energy, momentum and so on of this stream will be investigated aswell as changes of its properties;

• a screen of area dA is set up; which energy is streaming per timeinterval dt enclosing the angle θ with the direction normal to thescreen into the solid angle element dΩ and within the frequencyinterval dν?

• we begin with the occupation number: let nα~p be the number the occupation number is thenumber density of occupiedstates per phase space element

density of photons with momentum ~p and the polarisation state α(α = 1, 2);

• the energy per photon is E = hν = cp (because the photon haszero rest mass); thus

p = |~p| =hνc

; (1.1)

• the volume element in momentum space is d3 p; the number ofindependent phase-space cells is Heisenberg’s uncertainty relation

implies that points in phase spacecannot be observed; rather, obser-vable cells in phase space have afinite volume

d3 p(2π~)3 =

p2dpdΩ

h3 =ν2dνdΩ

c3 , (1.2)

where momentum has been expressed by frequency ν in the laststep;

1

2KAPITEL 1. MACROSCOPIC RADIATION QUANTITIES, EMISSION AND ABSORPTION

• in terms of these quantities, the following amount of energy isflowing through the screen: (number of phase space cells) times(photon occupation number) times (energy per photon) times (vo-lume filled by the photons); thus

dE =ν2dνdΩ

c3

2∑α=1

nα~p hν dA cos θ dt (1.3)

• the energy flowing through the screen per unit time, frequency andsolid angle is

dEdtdνdAdΩ

=

2∑α=1

nα~phν3

c2 cos θ ≡ Iν cos θ , (1.4)

where Iν is called specific intensity of the radiation;

• for unpolarised light, we obviously have

Iν =2hν3

c2 nα~p ; (1.5)

1.2 Relativistic Invariant

1.2.1 Lorentz Transformation of Iν

• switching from one reference frame to another, the transformationproperties of the physical quantities is important to be known; weshall now show by Lorentz transformation that the quantity

Iνν3 (1.6)

is relativistically invariant;

• let us assume two observers O and O′, which are moving relative-ly to each other with velocity v in x3 direction; O′ is collectingphotons on a screen dA′ in the x′1-x′2 plane which move under theangle θ′ with respect to the area normal into the solid angle dΩ′;he finds

dN′ = 2p′2dp′dΩ′

(2π~)3 n′p′ dA′ c cos θ′ dt′ (1.7)

photons on his screen;

• likewise, observer O expects the same screen to collect the photonnumber

dN = 2p2dpdΩ

(2π~)3 np dA(c cos θ − v) dt′ (1.8)

and of course the two numbers must be equal, dN′ = dN;

1.2. RELATIVISTIC INVARIANT 3

• the Lorentz transformation relating O and O′ is

Λ =

γ 0 0 βγ0 1 0 00 0 1 0βγ 0 0 γ

(1.9)

with β ≡ v/c and γ ≡ (1 − β2)−1/2;

• for the screen at rest in O′, dx′3 = 0, thus

dx0 = cdt = γ dx′0 = γ cdt′ , (1.10)

so that dt = γ dt′, which is the usual relativistic time dilation;

• energy and momentum are combined in the four-vector

pµ =

(Ec, ~p

)≡

(p0, ~p

); p0 = |~p| because |~p| =

Ec, (1.11)

and we obtain

p0 = γ(p′0 + βp′3) ; p3 = γ(βp′0 + p′3) , (1.12)

and the other components are p1 = p′1, p2 = p′2;

• since, from simple geometry, p′3 = cos θ′ |~p′| = p′0 cos θ′ andp3 = p0 cos θ, we then find

cos θ =p3

p0 =βp′0 + p′3

p′0 + βp′3=

β + cos θ′

1 + β cos θ′(1.13)

for the Lorentz transformation of the angle θ;

• this implies for the solid-angle element

d2Ω = d(cos θ)dφ = dφ′d[β + cos θ′

1 + β cos θ′

]=

d2Ω′

γ2(1 + β cos θ′)2 ;

(1.14)

• summarising, we find for the number of photons in the system O:

dN =2h3

[γ(1 + β cos θ′)

]3 p′2dp′︸︷︷︸=p2dp

d2Ω′

γ2(1 + β cos θ′)2︸︷︷︸=d2Ω

× np dA′︸︷︷︸=dA

[c(β + cos θ′

1 + β cos θ′

)− v

]︸︷︷︸

=c cos θ−v

γdt′︸︷︷︸=dt

; (1.15)

equating this to dN′ from (1.7) yields

n′p′ cos θ′ =

[γ2(1 + β cos θ′)

(β + cos θ′

1 + β cos θ′− β

)]np

= γ2(1 − β2) cos θ′np = np cos θ′ ; (1.16)


• thus, the occupation number is obviously a relativistic invariant,

np = n′p′ , (1.17)

and Iν ∝ ν3np implies the claimed invariance (1.6),

Iνν3 =

I′ν′ν′3

; (1.18)

the Lorentz transformation of the solid angle (1.14) will be usedlater in the discussion of synchrotron radiation

1.2.2 Example: The CMB Dipole

• this relativistic invariance of Iν/ν3 allows the dipole of the cosmicmicrowave background to be computed: a photon flying at an angleθ relative to the x axis of the observer will be redshifted by anamount which directly follows from Lorentz transformation;

• using pµ = (E/c, ~p) and p1 = |~p| cos θ, the Lorentz transformationyields

p′µ =

γ βγ 0 0βγ γ 0 00 0 1 00 0 0 1

E/cp1

p2

p3

=

γE/c + βγ|~p| cos θβγE/c + γ|~p| cos θ

p2

p3

, (1.19)

i.e. the energy in the primed system is

E′ = c(γ

Ec

+ βγEc

cos θ)

= γ(1 + β cos θ)E ; (1.20)

the frequency is thus increased to

ν′ = γ(1 + β cos θ)ν ; (1.21)

• with the occupation number np being a relativistic invariant,

np =1

ehν/kT − 1=

1ehν′/kT ′ − 1

= n′p′ , (1.22)

the temperature T must change exactly as the frequency ν, thus

T ′ = Tγ(1 + β cos θ) ; (1.23)

• for non-relativistic velocities v c, γ ≈ 1, and thus

T ′ ≈ T(1 +

v

ccos θ

); (1.24)

the motion of the Earth relative to the microwave backgroundthus causes a dipolar pattern in its measured temperature; withv ∼ 10−3c and T ∼ 3 K, the amplitude of the dipole is of order afew milli-Kelvins;

1.3. EINSTEIN COEFFICIENTS AND THE PLANCK SPECTRUM5

1.3 Einstein coefficients and the Planck spec-trum

1.3.1 Transition Balance

• we consider mean transition rates in an emission- and absorptionprocess between two energy levels E1 and E2; the rates of absorp-tion and of stimulated emission will be proportional to the specific stimulated emission is a conse-

quence of the Bose character ofphotons: if a quantum state is oc-cupied by photons, an increase inthe occupation number is morelikely

intensity, absorption rate ∝ IνB12 and stimulated emission rate ∝IνB21, while the rate of spontaneous emission will not depend on Iν,spontaneous emission rate ∝ A21; A and B are called the Einsteincoefficients;

• now, let N1 and N2 be the mean number of states with the energiesE1 and E2; equilibrium between transitions will require as manytransitions from E1 to E2 as there are from E2 to E1, thus

N1IνB12 = N2 [A21 + IνB21] , (1.25)

which can be satisfied if the specific intensity is

Iν =N2A21

N1B12 − N2B21=

A21N1N2

B12 − B21=

A21

B21

(N1N2− 1

) , (1.26)

where we have used that B12 = B21 (E1 and E2 are eigenstates ofthe Hamilton operator);

• according to the definition of A21 and B21, we must have [cf. Eq. (1.5)]

A21 = 2hν3

c2 B21 ; (1.27)

• if there is thermal equilibrium between the states E1 and E2, wehave the Boltzmann factor between N1 and N2,

N2

N1= exp

(−

hνkT

), (1.28)

where E2 = E1 + hν;

• under this condition, (1.28) implies

Iν =2hν3

c2

1ehν/kT − 1

≡ Bν , (1.29)

which is the Planck spectrum;

• limiting cases of the Planck spectrum for high and low frequenciesare

Bν ≈2hν3

c2 e−hν/kT for ν kTh

(Wien’s law) (1.30)


and

Bν ≈2ν2

c2 kT for ν kTh

(Rayleigh-Jeans law) (1.31)

• the spectral energy density is

dUν =dE

dνdx3 =dE

dνdA(cdt)=

Iνc

d2Ω , (1.32)

thuscUν =

∫Iνd2Ω , (1.33)

which equals 4πIν for isotropic radiation;

• a unit for the spectral energy density which is frequently used inastronomy is the Jansky, defined bynote that 1 Jy is not the unit of

specific intensity, which wouldbe Jy/sr 1 Jy = 10−26 W

m2 Hz= 10−23 erg

cm2 s Hz; (1.34)

1.3.2 Example: The CMB Spectrum

• for example, the spectral energy density of the CMB is given by

Uν =4πc

Bν =4πc

2hν3

c2

1ehν/kT − 1

= 2.4 × 10−25 ergcm2 s Hz

= 23.9 mJy (1.35)

at a frequency of ν = 30 GHz;

• the maximum of the Planck spectrum is located at

x ≡hνkT≈ 2.82 ; (1.36)

for the CMB, this corresponds to a frequency of

ν ≈ 1.60 × 1011 Hz = 160 GHz ; (1.37)

• inserting the Planck spectrum for Iν in (1.28) and integrating overall frequencies yields

U =

∫ ∞

0Uνdν =

π2

15(kT )4

(~c)3 (1.38)

for the energy density of a Planckian radiation field;

• the number density of the photons is clearly

n =

∫ ∞

0

Uν

hνdν =

2ζ(3)π2

(kT~c

)3

, (1.39)

where the Riemann ζ function takes the numerical value ζ(3) ≈1.202;

1.4. ABSORPTION AND EMISSION 7

• for the cosmic microwave background, T = 2.7 K, and thus

n ≈ 400 cm−3 , U ≈ 4.0 × 10−13 ergcm3 ; (1.40)

• the Rayleigh-Jeans law is often used to define a radiation tempera-ture Trad by requiring

2ν2

c2 kTrad!= Iν ; (1.41)

obviously, this agrees well with the thermodynamic temperature ifhν/kT 2.82 and Iν = Bν, but the deviation becomes considera-ble for higher frequencies;

1.4 Absorption and Emission

• the absorption coefficient αν is defined in terms of the energyabsorbed per unit volume, time and frequency from the solid angled2Ω,

ανIν =

(dE

d3xdtdνd2Ω

)abs

; (1.42)

• since the stimulated emission is also proportional to Iν, an analo-gous definition applies for the “induced” emission,

αindν Iν =

(dE

d3xdtdνd2Ω

)ind

; (1.43)

• for the spontaneous emission, we define the emissivity

jν =

(dE

d3xdtdνd2Ω

)spn

, (1.44)

i.e. the spontaneous energy emission per unit volume, time andfrequency into the solid-angle element d2Ω;

• the effective net absorption is

αnetν = αν − α

indν ; (1.45)

• since the unit of Iν is

energytime × area × frequency × solid angle

, (1.46)

αν must obviously have the dimension (length)−1; the “mean freepath” for a photon of frequency ν is thus approximately α−1

ν ;


• let σν be the cross section of an atom, molecule or other particlefor the absorption of light of frequency ν, then

αν = nσν ≡ ρκ , (1.47)

where n is the number density of absorbing systems and ρ is theirmass density; κ is called “opacity”, whose physical meaning is theabsorption cross section per unit mass,

[κ] =cm2

g; (1.48)

• if matter is in equilibrium with a radiation field, the emitted andabsorbed amounts of energy must equal, hence

jν + αindν Iν = ανIν ⇒ jν = αnet

ν Iν ; (1.49)

using (1.28) and (1.29), we then find

Iν =jναnetν

= 2hν3

c2

(N1

N2− 1

)−1

; (1.50)

i.e. if the occupation numbers are known, the emission and absorp-tion properties in equilibrium can be calculated, and vice versa;

• in particular, in thermal equilibrium with matter, we have

Iν = Bν ⇒ αnetν =

jνBν

; (1.51)

1.5 Radiation Transport in a Simple Case

• we consider an emitting and absorbing medium which does notscatter for now and is being irradiated by a light bundle; let themedium be characterised by an emissivity jν and a net absorptioncoefficient αnet

ν ;

• per unit of the traversed distance, the intensity of the light bundlechanges according to

dIν = jν dl︸︷︷︸emission

− ανIν dl︸︷︷︸absorption

, (1.52)

from which we obtain the equation of radiation transport in itssimplest case,

dIνdl

= jν − ανIν . (1.53)

1.5. RADIATION TRANSPORT IN A SIMPLE CASE 9

• the homogeneous equation (1.53) is easily solved:

dIνdl

= −ανIν ⇒ d ln Iν = −ανdl , (1.54)

thus

Iν = C1 exp(−

∫ανdl

); (1.55)

• for solving the inhomogeneous equation (1.53), we assume C1 =

C1(l) and find

dIνdl

=ddl

[C1(l) exp

(−

∫ανdl

)](1.56)

=[C′1(l) −C1(l)αν

]exp

(−

∫ανdl

)!= jν − ανIν = jν − ανC1(l) exp

(−

∫ανdl

);

this implies

C′1(l) exp(−

∫ανdl

)= jν , (1.57)

which has the solution

C1(l) =

∫dl

[jν exp

(∫ανdl

)]+ C2 ; (1.58)

• if αν is a constant along the light path, the integral is simply∫ανdl = ανl , (1.59)

and then we have

C1(l) =jναν

eανl , Iν(l) =jναν−C2e−ανl ; (1.60)

• for example, if the intensity satisfies the boundary condition Iν = 0at l = 0, the intensity as a function of path length becomes

Iν(l) =jναν

(1 − e−ανl

); (1.61)

• interesting limiting cases: let L be the entire path length throughthe medium; if

ανL 1 : Iν(L) =jναν

(ανL) = jνL ,

ανL 1 : Iν(L) =jναν

(1.62)

the former is the “optically thin”, the latter the “optically thick”case; this amounts to comparing the mean free path α−1

ν to the totalpath length L;


• if the radiation is in thermal equilibrium with the irradiated mate-rial, we must have

Iν = Bν =jναν

(1 − e−ανL

), (1.63)

which implies that the source emits at most the intensity of thePlanck spectrum

• we consider optically thin, thermal emission of radio waves; op-tically thin implies ανL 1 and Iν = jνL, thermal equilibriumrequires Iν = Bν, and in the radio regime we have

hνkT 1 , Bν ≈

2ν2

c2 kT ; (1.64)

combining these conditions, we find

Iν ≈ jνL = ανBνL =2ν2

c2 ανkT L =2ν2

c2 kTb , (1.65)

where Tb is the observed temperature, which is obviously relatedto the emission temperature T by

Tb ≈ ανLT ; (1.66)

this absurd conclusion shows indicates that the two assumptions,thermal equilibrium and optically-thin radiation, are in conflictwith each other;

1.6 Emission and Absorption in the Continu-um Case

• in the discrete case, the energy balance for the emitted energy was

N2A21︸︷︷︸transition number

× hν12︸︷︷︸energy per transition

= δE (1.67)

• the emissivity (per unit solid angle) is

jν =N2A21hν12

4π→

N2A21hν4π

δD(ν − ν12) , (1.68)

with the Dirac delta function modeling a sharp line transition;

• correspondingly, we generalise this expression by a “line profilefunction” φ(ν),

jν =N2A21hν

4πφ(ν) , (1.69)

where φ(ν) quantifies the transition probability as a function offrequency;

1.6. EMISSION AND ABSORPTION IN THE CONTINUUM CASE11

• by an analogous procedure for the absorption coefficient, we find

αν =N1B12

4πhν φ(ν) ; (1.70)

• we now consider an electron of energy E which emits the energy

dεdνdt

≡ P(ν, E) (1.71)

per unit time and unit frequency; let further f (~p) be the momentumdistribution of the electrons, then the number of electrons withenergies between E and E + dE is

n(E)dE = f (~p)d3 pdE

dE = 4πp2 dpdE

f (~p) dE , (1.72)

if we assume the distribution to be isotropic in momentum space;since each electron emits the energy

dε = P(ν, E) dνdt , (1.73)

we obtain for the emissivity

4π jν =

∫ ∞

0n(E)P(ν, E)dE = 4π

∫ ∞

0p2 f (p)

dpdE

P(ν, E)dE

(1.74)

• by definition, we have for a continuous transition

P(ν, E2) = hν∫ E2

0A21φ(ν)dE1 , (1.75)

i.e. electrons with the energy E2 can emit in transitions to allpossible states with E1 < E2; thus

P(ν, E2) = hν2hν3

c2

∫ E2

0B21φ(ν)dE1 ; (1.76)

• likewise, the net absorption coefficient is

αν =hν4π

∫dE1

∫dE2

n(E1)B12︸︷︷︸absorption

− n(E2)B21︸︷︷︸stim. emission

φ(ν) ;

(1.77)

• the second term in this expression can be written

hν4π

∫dE1

∫dE2 n(E2)B21φ(ν)

=hν4π

∫dE2n(E2)

∫ E2

0dE1 B21φ(ν)

=c2

8πhν3

∫dE2 n(E2)P(ν, E2) , (1.78)


while the first term reads

hν4π

∫dE1

∫dE2 n(E1)B12φ(ν)

=hν4π

∫dE2 n(E2 − hν)

∫dE1 B12φ(ν)

=c2

8πhν3

∫dE2 n(E2 − hν)P(ν, E2) ; (1.79)

• we thus obtain for the absorption coefficient

αν =c2

8πhν3

∫dE [n(E − hν) − n(E)] P(ν, E) ; (1.80)

• in thermal equilibrium and far from the Fermi edge, the electronnumber density is

n(E) ∝ exp(−

EkT

), (1.81)

thus

n(E − hν) − n(E) = n(E)[exp

(hνkT

)− 1

], (1.82)

from which we obtain

αν =c2

8πhν3

(ehν/kT − 1

) ∫dE n(E) P(ν, E)

=c2

2hν3

(ehν/kT − 1

)jν =

jνBν

, (1.83)

just as in the discrete case;

Kapitel 2

Scattering

further reading: Rybicki, Light-man, “Radiative Processes inAstrophysics”, chapter 7; Pad-manabhan, “Theoretical Astro-physics, Vol. I: AstrophysicalProcesses”, sections 6.4–6.7

2.1 Maxwell’s Equations and Units

• we use cgs units, i.e. the dielectric constant and the magneticpermeability of the vacuum are both unity, ε0 = 1 = µ0; Maxwell’sequations in vacuum then read

~∇ · ~E = 4πρ , ~∇ · ~B = 0 ,

~∇ × ~E = −1c∂~B∂t

, ~∇ × ~B =4πc~j +

1c∂~E∂t

, (2.1)

where ρ is the charge density and ~j is the current density;

• the Lorentz force per unit charge is

~fL = ~E +~v

c× ~B ; (2.2)

• the energy density of the electromagnetic field is

U =1

8π

(~E2 + ~B2

); (2.3)

• consequently, the field components ~E and ~B have dimension( ergcm3

)1/2=

(g cm2

s2 cm3

)1/2

=

( gcm s2

)1/2(2.4)

• forces have the dimensiong cm

s2 ≡ dyn ; (2.5)

thus, the Lorentz force

[ ~FL] = [q][~E] ⇒g cm

s2 = [q]( gcm s2

)1/2, (2.6)

13

14 KAPITEL 2. SCATTERING

implies that the unit of charge must be

[q] =g1/2cm3/2

s; (2.7)

• in these units, the elementary charge is

e = 4.8033 × 10−10 cm3/2g1/2

s; (2.8)

• the Poynting vector, i.e. the vector of the energy current density ofthe electromagnetic field, is

~S =c

4π

(~E × ~B

), (2.9)

with dimension[~S ] =

cms

ergcm2s

(2.10)

which is obvious because the unit of ~E2 is

[~E2] =ergcm3 ; (2.11)

2.2 Radiation of a Moving Charge

• far from its source, the electric field of an accelerated charge is, inthe non-relativistic limit |~β| 1see, for example, Jackson, Classi-

cal Electrodynamics, eq. (14.18)~E =

qcR

[~e ×

(~e × ~β

)], (2.12)

where ~e is the unit vector pointing from the radiating charge to theobserver, and R is the distance;

• since ~B = ~e × ~E and ~E = ~B × ~e in vacuum, the ~B field is

~B = −q

cR

(~e × ~β

), (2.13)

and the Poynting vector is

~S =c

4π

[(~B × ~e

)× ~B

]=

c4π

[~B2~e − (~B · ~e)~B

]=

c4π~B2~e (2.14)

because ~e · ~B = 0;

• per unit time, the energydEdt

= ~S · d~A (2.15)

is radiated through the area element d~A; since d~A is related to thesolid-angle element d~Ω as d~A = R2~edΩ, we find

dEdt

=c

4π~B2R2dΩ ; (2.16)

thus, the energy radiated per unit time into the solid-angle elementdΩ is

dEdtdΩ

=c

4π

∣∣∣∣R~B∣∣∣∣2 ; (2.17)

2.3. SCATTERING OFF FREE ELECTRONS 15

2.3 Scattering off Free Electrons

2.3.1 Polarised Thomson Cross Section

• a point charge q is accelerated by an incoming electromagneticwave with the electric field component ~E′; the equation of motionfor the charge is

m~x = ~FL = q(~E′ + ~β × ~B′

)≈ q~E′ + O(β) , (2.18)

i.e. the last approximation employs the non-relativistic limit of theLorentz force; thus, the acceleration is

~x = c~β =qm~E′ ; (2.19)

• using the dipole moment ~d ≡ q~x, we can write eq. (2.13) for themagnetic field in the form

~B = −~e × ~dc2R

; (2.20)

• according to (2.19), the second time derivative of the dipole mo-ment is

~d =q2

m~E′ , (2.21)

which, when combined with (2.20) and (2.17), implies

dEdtdΩ

=c

4π

(1c2

∣∣∣∣~e × ~d∣∣∣∣)2

=q4

4πc3m2

∣∣∣∣~e × ~E′∣∣∣∣2 =q4

4πc3m2~E′2 sin2 α , (2.22)

where α was introduced as the angle between the incoming electricfield ~E′ and the direction of the outgoing radiation, ~e;

• the incoming energy current density is

S ′ =c

4π~E′2 ; (2.23)

thus the differential scattering cross section is

dσdΩ

=1S ′

dEdtdΩ

=

(q2

mc2

)2

sin2 α ; (2.24)

• suppose the elementary charge −e is homogeneously distributedon the surface of a sphere with radius re; then, its absolute potentialenergy is

∼e2

re; (2.25)


equating this to an electron’s rest-mass energy mec2, we can solvefor re,

e2

re

!= mec2 ⇒ re =

e2

mec2 ≈ 2.8 × 10−13 cm ; (2.26)

this is the so-called “classical electron radius”;

• generally, the radius

r0 ≡q2

mc2 (2.27)

is associated with a particle of charge q and rest-mass m; usingthis radius, the differential scattering cross section reads

dσdΩ

= r20 sin2 α ; (2.28)

• the total cross section is

σ = r20

∫sin2 αdΩ = 2πr2

0

∫ π

0sin3 αdα =

8π3

r20 ; (2.29)

for electrons, we obtain the Thomson cross section,

σT =8π3

r2e =

8π3

(e2

mec2

)2

≈ 6.6 × 10−25 cm2 ; (2.30)

2.3.2 Unpolarised Thomson Cross Section

• this scattering cross section is valid for one particular polarisa-tion direction; we now average over all incoming polarisationdirections; for doing so, we introduce the angle ϕ in the plane per-pendicular to the incoming direction ~n′; the polarisation directionis then

~e′ =

cosϕsinϕ

0

, (2.31)

if ~e′ is parallel to the x3 axis; the outgoing direction of the scatteredradiation is

~e =

sin θ0

cos θ

; (2.32)

• using this, one obtains the differential scattering cross section

dσdΩ

= r20 sin2 α = r2

0(1 − cos2 α) = r20

[1 − (~e · ~e′)2

]; (2.33)

2.4. SCATTERING OFF BOUND CHARGES 17

using ~e · ~e′ = sin θ cosϕ, averaging over ϕ yields⟨dσdΩ

⟩=

r20

2π

∫ 2π

0dϕ

(1 − sin2 θ cos2 ϕ

)= r2

0

[1 −

sin2 θ

2π

∫ 2π

0dϕ cos2 ϕ

]=

r20

2(1 + cos2 θ) ; (2.34)

this is the unpolarised Thomson cross section;

2.4 Scattering off Bound Charges

• an accelerated charge radiates energy and thus damps the incoming,accelerating wave; the non-relativistic Larmor formula asserts that cf. Jackson, Classical Electrody-

namics, eq. (14.22)a non-relativistic, accelerated charge q emits the power

P =2q2

3c3

∣∣∣~v∣∣∣2 ; (2.35)

• this is interpreted as damping with a force ~FD,

− ~FD ·~v = P ⇒ ~v · ~FD = −2q2

3c3

∣∣∣~v∣∣∣2 ; (2.36)

• the temporal average over a time interval T is⟨dEdt

⟩=

1T

∫ T

0dt

2q2

3c3

∣∣∣~v∣∣∣2=

1T

2q2

3c3

[~v~v

∣∣∣T0−

∫ T

0dt~v~v

]; (2.37)

• the first term vanishes for bound charges and large T , thus

⟨~FD ·~v

⟩=

2q2

3c3

⟨...~x ·~v

⟩; (2.38)

we thus identify the expression

~FD =2q2

3c3

...~x (2.39)

with the time-averaged damping force;

• for bound orbits with an angular frequency of ω0, we have

~x = −ω20~x ⇒

...~x = −ω2

0~x ; (2.40)


thus, the equation of motion reads

~x + ω20~x =

qm~E0e−iωt − γ~x (2.41)

with the damping term

γ =2q2

3c3ω20 ; (2.42)

the first term on the right-hand side of (2.41) is the external exci-tation, the second is the damping; this equation models a driven,damped harmonic oscillator, whose solution is known to read

~x =qm

~E0e−iωt

ω20 − ω

2 − iωγ; (2.43)

• we put this back into Larmor’s equation (2.35) and obtain

dEdt

= P =2q2

3c3

∣∣∣~x∣∣∣2 =2q2

3c3 ~x · ~x∗

=2q4

3m2c3~E2

0ω4

(ω2 − ω20)2 + ω2γ2

; (2.44)

• the incoming energy current is |~S | = c~E20/(4π), and thus the scatte-

ring cross section becomes

σ =1

|~S |

dEdt

=8π3

r20

ω4

(ω2 − ω20)2 + ω2γ2

(2.45)

with the typical resonance behaviour at ω = ω0;

• interesting limiting cases are:

ω ω0 : σ ≈8π3

r20 = σT ;

(binding forces are then irrelevant)

ω ω0 : σ ≈ σT

(ω

ω0

)4

;

(Rayleigh scattering)

ω ≈ ω0 : σ ≈ σTω2

0

4(ω − ω0)2 + γ2

=2π2q2

mc

[γ/(2π)

(ω − ω0)2 + (γ/2)2

], (2.46)

where the term in square brackets defines the so-called Lorentzprofile;

2.5. RADIATION DRAG 19

2.5 Radiation Drag

2.5.1 Time-Averaged Damping Force

• in the case of Thomson scattering, the scattering charge damps themotion which is caused by the incoming electric field accordingto the damping force (2.39)

FD =2q2

3c3

...~x ; (2.47)

• an incoming electromagnetic wave exerts the Lorentz force

FL = q(~E − ~β × ~B) = m~x (2.48)

on the scattering charge; the last equality in (2.48) assumes that~FD ~FL, i.e. the back reaction of the radiation by the charge wasneglected;

• from (2.48), we find...~x =

qm

(~E + ~β × ~B + ~β × ~B

)=

qm

~E + ~β × ~B +~xc× ~B

=

qm

[~E + ~β × ~B +

qmc

(~E + ~β × ~B

)× ~B

]; (2.49)

in the non-relativistic limit, we can drop the terms proportional to~β and find

...~x =

qm

[~E +

qmc

(~E × ~B)], (2.50)

and thus~FD =

2q3

3mc3

[~E +

qmc

(~E × ~B)]

; (2.51)

• averaging over time yields

〈 ~FD〉 =2q3

3mc3 〈~E〉︸︷︷︸=0

+23

(q2

mc2

)2

〈~E × ~B〉 ; (2.52)

using

〈~E × ~B〉 = 〈~E + (~e × ~E)〉 = ~E2~e − (~E · ~e)︸︷︷︸=0

~E

= 4πU~e , (2.53)

we finally find

〈 ~FD〉 =8π3

r2qU~e = σTU~e (2.54)

for the time-averaged damping force;


2.5.2 Energy Transfer to a Radiation Field

• we now consider a charge moving with a relative velocity~v througha radiation field which is isotropic in its rest frame; in the rest frameof the radiation, we have

〈~E〉 = 0 = 〈~B〉 , 〈~E2〉 = 4πU = 〈~B2〉 ; (2.55)

• in the rest frame of the charge, the Lorentz force is

~F′L = q(~E′ + ~β′ × ~B′) = q~E′ , (2.56)

because ~β′ = 0 in the charge’s rest frame; in addition, we have

~E′ = ~E′⊥ + ~E′‖ = γ(~E⊥ + ~β × ~B

)+ ~E‖ , (2.57)

where γ is the usual Lorentz factor; for Larmor’s equation, wefurther need

dEdt

=2q2

3c3

∣∣∣~x′∣∣∣2 , ~x′ =~F′Lm

; (2.58)

• in a first step, we compute⟨∣∣∣~x′∣∣∣⟩ =q2

m2

⟨[γ(~E⊥ + ~β × ~B) + ~E‖

]2⟩

(2.59)

=q2

m2

⟨[γ(~E − ~E‖ + ~β × ~B) + ~E‖

]2⟩

=q2

m2

⟨[γ(~E + ~β × ~B) + (1 − γ)~E‖

]2⟩

=q2

m2

[γ2〈~E2〉 + γ2~β2〈~B2〉〈sin2 θ〉 + (1 − γ2)〈~E2

‖ 〉],

where we have used that

〈~E‖ · (~β × ~B)〉 = 0 (2.60)

because the direction of ~β is random with respect to the directionof ~E × ~B; thus, we obtain⟨∣∣∣~x′∣∣∣2⟩ = 4πγ2U

q2

m2

(1 +

2β2

3+

1 − γ2

3γ2

)= 4πγ2U

q2

m2

(1 +

β2

3

); (2.61)

• with that, we find the result

dEdt

= 4πγ2U2q4

3m2c3

(1 +

β2

3

)= σTUcγ2

(1 +

β2

3

)(2.62)

for the radiation which is on average radiated by the charge;

2.6. COMPTON SCATTERING 21

• according to the radiation damping, the energy which is on averageabsorbed by the charge is(

dEdt

)abs

= σTUc , (2.63)

and thus the total energy change of the radiation field per unit timeis

dEdt

= σTUc[γ2

(1 +

β2

3

)− 1

]=

43σTUcγ2β2 ; (2.64)

this amount of energy is added to the radiation field per unit timeby a single charge;

• the number of collisions between the charge and photons per unittime is

dNc

dt= σTc

Uhν

; (2.65)

combining this with (2.64), we find the energy gain of the radiationfield by scattering of the charge per scattering process,⟨

∆Eγ

⟩=

dEc

dt

(dNc

dt

)−1

=43

hνγ2β2 =43γ2β2Eγ ; (2.66)

2.6 Compton Scattering

2.6.1 Energy-Momentum Conservation

• we now consider electromagnetic radiation as being composedof photons; if an ensemble of charges is embedded into a radia-tion field, energy is transfered by scattering from the photons tothe charges and back; if the radiation temperature is higher thanthe temperature of the charge ensemble, energy flows from theradiation to the charges; this process is called Compton scattering;in astrophysics, the inverse Compton scattering is typically moreimportant, during which energy is transfered from the charges tothe radiation field;

• an incoming photon with momentum hν~e/c hits an electron withmomentum ~p; after scattering, the photon and the electron havemomenta hν′~e′/c and ~p′;

• conservation of momentum and energy imply

hν~e + c~p = hν′~e′ + c~p′ , hν + E = hν′ + E′ , (2.67)

whereE2 = c2 p2 + m2c4 (2.68)

according to the relativistic energy-momentum relation;


• solving the energy equation for E′2 and inserting (2.68) yields

c2~p′2 = c2~p2 + h2(ν − ν′)2 + 2Eh(ν − ν′) , (2.69)

while the momentum equation implies

c2~p′2 = c2~p2 + h2(ν~e − ν′~e′)2 + 2h(ν~e − ν′~e′)c~p ; (2.70)

• subtracting (2.69) from (2.70) and cancelling suitable terms gives

hνν′(1 − cos θ) = E(ν − ν′) − c~p(ν~e − ν′~e′) , (2.71)

where θ is the angle between ~e and ~e′;

• if the electron is originally at rest, ~p = 0 and E = mc2, and (2.71)simplifies to

ν − ν′ =h

mc2 νν′(1 − cos θ) , (2.72)

and in the limit of very low photon energy, hν mc2, we find forthe relative energy change of the Compton-scattered photon

∆Eγ

Eγ

=ν′ − ν

ν= −

Eγ

mc2 (1 − cos θ) , (2.73)

and if hν & mc2, quantum electrodynamics must be used anyway;

• averaging (2.73) over all scattering angles θ, we find the meanenergy loss per Compton scattering,

〈Eγ〉 = −E2γ

mc2

∫ 1

−1(1 − cos θ)d(cos θ) = −

E2γ

mc2 ; (2.74)

2.6.2 Energy Balance

• the total energy transfer to the radiation field due to the motion ofa single charge is given by the difference between the energy gain(2.66) per scattering and the energy loss per Compton scattering(2.74), ⟨

∆Eγ

⟩=

(43γ2β2 −

Eγ

mc2

)Eγ ; (2.75)

• for photons with Eγ mc2 in the relativistic limit, β ≈ 1, and⟨∆Eγ

⟩≈

43γ2 Eγ , (2.76)

which can become a very large number; in that way, for example,CMB photons can be converted to X-ray photons;

2.6. COMPTON SCATTERING 23

• in the thermal limit of (2.75), we can approximate v c, thusγ ≈ 1, and mv2 = 3kTe; then

⟨∆Eγ

⟩≈

(4v2

3c2 −Eγ

mc2

)Eγ =

(4kTe − Eγ

) Eγ

mc2 ; (2.77)

thus, the photons gain energy (on average), if

4kTe > Eγ (2.78)

(inverse Compton scattering), and lose energy otherwise (Comptonscattering)

• Compton scattering causes fast charges to lose energy; typical timescales are, according to (2.64)

tc ≡E

dE/dt=

γmc2

43σTUcγ2β2

=34

mc2

γβ2σTU; (2.79)

for non-relativistic, thermal electrons, E = 3kTe/2 and γ ≈ 1, and

tc =

32kTec2

43σTUcv2

=98

mcσTU

; (2.80)

• after Ns scatterings, the total energy transfer from thermal electronsto the photons is

E′

E=

(1 +

4kTe

mc2

)Ns

≈ exp(4kTeNs

mc2

)≡ e4y , (2.81)

where the Compton parameter

y ≡4kTeNs

mc2 (2.82)

was introduced;

• if the electron number density is ne, the number of scatterings perpath length dl is

dNs = neσTdl ⇒ Ns =

∫σTnedl , (2.83)

and thus the Compton-y parameter becomes

y =kTe

mc2

∫σTnedl ; (2.84)


2.7 The Kompaneets Equation

• we need an additional equation which specifies how the photonspectrum is changed due to the scatterings; for deriving it, weassume that a homogeneous, thermal distribution of electrons islocated in a homogeneous sea of radiation, such as, for example, agalaxy cluster in the microwave background; the collisions withthe electrons change the photon energy, but not their number, andthus their spectrum cannot remain a Planck spectrum;

• let n(ν) be the occupation number of photon states with frequencyν; then, the Boltzmann equation requires

∂n(ν)∂t

=

∫d3 p

∫dΩ

(dσdΩ

)c (2.85)

×n(ν′) [1 + n(ν)] N(E′) − n(ν)

[1 + n(ν′)

]N(E)

;

this equation has the following meaning: the occupation numberat the frequency ν changes due to scattering from ν to ν′, and fromν′ to ν; the term

n(ν)[1 + n(ν′)

]N(E) (2.86)

quantifies how many photons there are at frequency ν, correctedby the factor for stimulated emission from ν to ν′, and multiplieswith the number of collision partners N(E) at energy E; in otherwords, it quantifies the number of collisions away from frequencyν; analogously, the term

n(ν′) [1 + n(ν)] N(E′) (2.87)

quantifies the opposite scattering, i.e. scattering processes increa-sing the occupation number at frequency ν; of course, the energydifference between photon frequencies ν and ν′ must be balancedby the difference between the energies E and E′; the integral overd3 p integrates over the electron distribution, and the factor

dσdΩ

dΩ (2.88)

specifies the probability for scattering photons from frequency νto frequency ν′ or backward;

• we assume thermal photon and electron distributions, and restrictourselves to the limit of Thomson scattering, which applies if

hν mc2 ; (2.89)

moreover, we assume small changes in the photon frequency, hence

δν ≡ ν′ − ν ν ; (2.90)

2.7. THE KOMPANEETS EQUATION 25

moreover, the electron energy distribution is

N(E) ∝ exp(−

EkTe

), (2.91)

and energy conservation requires

E′ = hν − hδν ; (2.92)

• now, both n(ν) and N(E) can be expanded in Taylor series up tosecond order,

n(ν′) = n(ν) +∂n∂νδν +

12∂2n∂ν2 δν

2 + O(δν3) , (2.93)

N(E′) = N(E) −∂N∂E

hδν +12∂2N∂E2 h2δν2 + O(δν3) ,

where (2.91) allows us to use

∂N∂E

= −N(E)kTe

,∂2N∂E2 =

N(E)(kTe)2 ; (2.94)

• for simplification, we now define the dimension-less photon energy,scaled by the thermal electron energy

x ≡hνkTe

(2.95)

and find

n(x′) ≈ n(x) +∂n∂xδx +

12∂2n∂x2 δx2 ,

N(E′) ≈ N(E)[1 + δx +

δx2

2

]; (2.96)

• with these approximations, we return to the original equation (2.86)for n(ν) and obtain

∂n∂t

=

[∂n∂x

+ n(n + 1)]

I1

+12

[∂2n∂x2 + 2(1 + n)

∂n∂x

+ n(n + 1)]

I2 , (2.97)

with the abbreviations

Ii ≡

∫d3 p

∫dΩ

dσdΩ

cδxiN(E) (2.98)

• the energy change of a photon scattering off a moving electronfollows from (2.71), adopting the non-relativistic limit

E = mc2 +p2

2m(2.99)

and using (2.89) and (2.90); this yields

hδν = −hνmc

(~e − ~e′) · ~p ⇒ δx = −x

mc(~e − ~e′) · ~p ; (2.100)


• using this result, the integrals Ii can be carried out straightforward-ly; with the unpolarised Thomson cross section (2.34), we firstfind

I2 = 2σTneckTex2

mc2 ; (2.101)

• for evaluating I1, we note that I1 is the mean rate of relative energytransfer, quantified by δx from the electrons to the photons, andtherefore the mean energy transfer rate, divided by kTe; from(2.77), we know that this is⟨

∆Eγ

⟩=

x(kTe)2

mec2 (4 − x) (2.102)

per scattering, and multiplying with the collision rate neσTc gives

I1 =kTe

mec2 neσTc x(4 − x) ; (2.103)

• with these two expression for Ii, we find the time derivative of n tobe

mec2

kTe

1neσTc

∂n∂t

=1x2

∂

∂x

[x4

(∂n∂x

+ n + n2)]

; (2.104)

• we finally transform the time t to the Compton parameter, using

dy =kTe

mec2 neσTc dt (2.105)

to find the Kompaneets equation

∂n∂y

=1x2

∂

∂x

[x4

(∂n∂x

+ n + n2)]

; (2.106)

• the hot gas in galaxy clusters is much hotter than the cosmicbackground radiation; then, we can approximate the right-handside of (2.106) to lowest order in x,

∂n∂y≈ x2∂

2n∂x2 + 4x

∂n∂x

; (2.107)

• inserting here the occupation number in thermal equilibrium, n ≈(ex − 1)−1, we find

δnn

= δy

(x2ex(1 + ex)

(ex − 1)2 −4xex

ex − 1

)(2.108)

for the relative change of the occupation number, where x is nowhν/kT and no longer hν/kTe!

Kapitel 3

Radiation Transport andBremsstrahlung

further reading: Shu, “The Phy-sics of Astrophysics, Vol I: Ra-diation”, chapters 2, 3, and 15;Rybicki, Lightman, “RadiativeProcesses in Astrophysics”, chap-ter 5; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophysi-cal Processes”, sections 6.8–6.9

3.1 Radiation Transport Equations

• we start with the collision-less Boltzmann equation for describingthe temporal change of the photon distribution function in phasespace,

∂n∂t

+ ~∇ ·∂n∂~x

+ ~p ·∂n∂~p

= 0 , (3.1)

which is valid in absence of collisions;

• for photons, we have ~v = c~e, where ~e is the unit vector in thedirection of light propagation; moreover, ~p = 0 in absence ofsystematic external forces (such as gravitational lensing); sincethe intensity Iν is proportional to n, the Boltzmann equation forphotons can also be written as

1c∂Iν∂t

+ ~e ·∂Iν∂~x

= 0 ; (3.2)

• we now define the following quantities:

~Fν ≡

∫dΩ~e · Iν , Pν,i j ≡

1c

∫dΩ eie jIν (3.3)

and recall the spectral energy density

Uν =1c

∫dΩ Iν ; (3.4)

• integrating the Boltzmann equation (3.2) first over dΩ, we obtainthe equation

∂Uν

∂t+ ~∇ · ~Fν = 0 ; (3.5)

27

28KAPITEL 3. RADIATION TRANSPORT AND BREMSSTRAHLUNG

which has the form of a continuity equation and identifies ~Fν asthe spectral radiation current density (spectral because it retainsthe dependence on frequency ν); this equation expresses energyconservation in the radiation field;

• if we multiply (3.2) with ei first before integrating over dΩ, wefind

1c

∫dΩ ei

∂Iν∂t

+

∫dΩ eie j

∂Iν∂x j

= 0 , (3.6)

and hence1c∂Fν,i

∂t+ c

∂Pν,i j

∂x j= 0 ; (3.7)

• this equation describes the change of the momentum current den-sity, because (Uν

c

)(c~e) (3.8)

is the momentum density of the radiation field, and thus

1c∂~F∂t

=1c2

∂

∂t

∫dΩ Iν~e (3.9)

is c times the temporal change of the momentum current density;Eq. (3.7) expresses momentum conservation;

• in presence of emission, stimulated emission and absorption, weknow from the first chapter that the energy equation must beaugmented by source and sink terms on its right-hand side; we had

dIνdl

= jν − ανIν =1c

dIνdt

; (3.10)

integrating over dΩ, and assuming that jν and αν are isotropic, wefind

dUν

dt= 4π jν − ανUνc = 4π jν − ρκνcUν ; (3.11)

we now re-define the emissivity,

4π jν → ρ j′ν ≡ ρ jν , (3.12)

i.e. we refer it to the mass density, and write

dUν

dt= ρ( jν − κνcUν) ; (3.13)

• likewise, the momentum-conservation equation

1c

dIνdt

= jν − ανIν (3.14)

becomes after multiplication with ~e and integration over dΩ

1c

ddt

∫dΩ~e · Iν =

∫dΩ jν~e −

∫dΩανIν~e , (3.15)

3.2. LOCAL THERMODYNAMICAL EQUILIBRIUM 29

and thus1c

d ~Fdt

= −αν ~Fν = −ρκν ~Fν , (3.16)

where we have assumed again that jν and κν are isotropic

• including the emission and absorption terms, the transport equati-ons are modified to read

∂Uν

∂t+ ~∇ · ~Fν = ρ( jν − κνcUν)

1c∂Fν,i

∂t+ c

∂Pν,i j

∂x j= −ρκνFν,i ; (3.17)

these equations do not contain scattering terms yet!

• since the change in the momentum current density corresponds toa force density, and this force is caused by the interaction betweenradiation and matter, an oppositely directed and equally strongforce must act on the matter as radiation pressure force; thus

~frad =ρ

c

∫ ∞

0κν ~Fν dν (3.18)

is the density of the radiation pressure force;

3.2 Local Thermodynamical Equilibrium

• the moment equations for Uν and ~Fν are by no means easier tohandle than the Boltzmann equation whose moments they are;we obviously need an additional approximation, or condition, inorder to “close” the moment equations; the “closure” means thatthey can then be solved without progressing indefinitely to higherorders of moments;

• often, the mean free path of the photons is much smaller thanthe dimensions of the system under consideration; then, we canassume that thermodynamical equilibrium is locally establishedbetween the radiation field and the matter; under this condition,

Iν ≈ Bν(T ) , (3.19)

i.e. the specific intensity of the radiation field is the Planckianintensity of a black body, and

Uν =1c

∫dΩ Iν =

4πc

Bν(T ) ; (3.20)

• under such circumstances, there is obviously no radiation flux anymore because the radiation field is isotropic; in order to estimatethe flux nonetheless, we study the orders of magnitude of thedifferent terms in the moment equations;


• time derivatives can typically be neglected because temporal chan-ges of the quantities Uν, ~Fν and Pν,i j occur on an evolutionary timescale, while the other terms change according to the streaming ofthe photons, thus approximately on time scales of order (mean freepath)/c;

• if we first ignore ∂~Fν/∂t, we obtain

Fν,i ≈ −cρκν

∂Pν,i j

∂x j(3.21)

in the approximation of Local Thermodynamical Equilibrium(LTE), we further have

Pν,i j ≈ Pνδi j =Uν

3δi j (3.22)

because of the (local) isotropy of the radiation field, and thus


∂Uν

∂xi≈ −

cρκν

Uν

R, (3.23)

where R is a typical dimension of the system; the mean free pathλν is determined by

λνnσν = λνρκν ≈ 1 , (3.24)

and (3.23) can thus be approximated by

|Fν,i| ≈ cUν

(λνR

), (3.25)

which is smaller by a factor λν/R compared to the transparent case(in which κ → 0 and λν → R;

• using this estimate for ~Fν, we return to the Eq. (3.17) for the partialtime derivative of Uν; as before, we ignore the time derivative, suchthat the only term remaining on the left-hand side is

~∇ · ~Fν ≈cUν

RλνR

; (3.26)

the second term on the right-hand side is

ρκνcUν ≈cλν

Uν ≈

(Rλν

)2~∇ · ~Fν ~∇ · ~Fν ; (3.27)

thus, because of the assumption of local thermodynamical equili-brium, the divergence of ~Fν is negligibly small; consequently, wemust require

ρ jν ≈ ρκνcUν ⇒ Uν ≈ρκνcαν

=4πc

Bν(T ) , (3.28)

as anticipated;

3.2. LOCAL THERMODYNAMICAL EQUILIBRIUM 31

• accordingly, if λν R and tevol λν/c, the solutions of themoment equations are


∂Pν,i j

∂x j, Uν ≈

4πc

Bν(T ) ; (3.29)

• because of (local) isotropy, we had

Pν,i j ≈Uν

3δi j ≈

4π3c

Bν(T ) δi j , (3.30)

and thus

Fν,i ≈ −4π

3ρκν

(∂Bν

∂T

)∂T∂xi

, (3.31)

i.e. the flux will become proportional to the temperature, which ischaracteristic for diffusion processes;

• for convenience, we now introduce the Rosseland mean opacity,

κ−1R ≡

∫ ∞0

dν(κ−1ν

∂Bν(T )∂T

)∫ ∞

0dν

(∂Bν(T )∂T

) ; (3.32)

here, we can use the fact that∫ ∞

0dν

(∂Bν(T )∂T

)=

∂

∂T

∫ ∞

0dν Bν(T ) =

∂

∂T

(caT 4

4π

), (3.33)

where

a ≡π2

15k4

(~c)3 = 7.6 × 10−15 ergK4cm3 (3.34)

is the so-called Stefan-Boltzmann constant;

• using this, we obtain the expression

~F =

∫ ∞

0dν ~Fν = −

4π3ρ∂T∂~x

∫ ∞

0

dνκν

∂Bν

∂T

= −4π

3ρκR

~∇Td

dT

(caT 4

4π

)= −

c3ρκR

~∇(aT 4

)(3.35)

for the radiative energy flux;

• the energy which is streaming away interacts with the absorbingmatter and thus exerts a force on it, which is determined by theright-hand side of the momentum-conservation equation, as des-cribed above:

~frad =ρ

c

∫ ∞

0dν κν ~Fν = −

ρ

c4π3ρ

∫ ∞

0dν∂Bν

∂T∂T∂~x

= −13~∇(aT 4) = −~∇P , (3.36)

which equals just the negative pressure gradient;


• a remark on units: the unit of Uν is

[Uν] =erg

cm3 Hz, (3.37)

the unit of κν is

[κν] =cm2

g, (3.38)

and thus the unit of ~Fν is

[ ~Fν] = [c][Uν] =erg

cm2 t Hz, (3.39)

and the unit of ~frad is

[ ~frad] =g

cm3

scm

cm2

gerg

cm2 s HzHz =

ergcm4 (3.40)

=g cm2

s4

1cm4 =

g cms2

1cm3 =

dyncm3 =

forcevolume

,

as it should be;

3.3 Scattering

• so far, we have only considered emission and absorption, butneglected scattering; scattering changes the distribution functionof the photons by exchanging photons with different momenta; ifwe assume for simplicity that the scattering process changes thephoton’s momentum, but not its energy, we can write the scatteringcross section in the form

dσ(~e→ ~e′)dΩ

= σφ(~e, ~e′) , (3.41)

where ~e and ~e′ are unit vectors in the propagation directions ofthe incoming and the outgoing photon; the function φ(~e, ~e′) isnormalised, symmetric in its arguments and dimension-less anddescribes the directional distribution of the scattered photons;

• scattering increases the distribution function n(~e) according to[dn(~e)

dt

]+

=

∫dΩ′

[Necσφ(~e, ~e′)

]︸︷︷︸# of scatterings ~e→ ~e′

n(~e′) [1 + n(~e)] ,

(3.42)where the factor [1 + n(~e)] is included for describing stimulatedemission of photons with momentum direction ~e;

3.3. SCATTERING 33

• analogously, losses due to scattering are given by[dn(~e)

dt

]−

=

∫dΩ′

[Necσφ(~e, ~e′)

]n(~e) [1 + n(~e′)] , (3.43)

and thus the total change of n(~e) due to scatterings becomes

dn(~e)dt

=

∫dΩ′

[Necσφ(~e, ~e′)

]×

n(~e′) [1 + n(~e)] − n(~e) [1 + n(~e′)]

=

∫dΩ′

[Necσφ(~e, ~e′)

] [n(~e′) − n(~e)

], (3.44)

in which the terms from stimulated emission cancel exactly;

• since the integral over the solid angle only concerns the directionof ~e′, we obtain from (3.44)

n(~e)dt

= −Necσn(~e) + Necσ∫

dΩ′ φ(~e, ~e′)n(~e′) , (3.45)

and thus

1c

dn(~e)dt

= −ρκscaν n(~e) + ρκsca

ν

∫dΩ′ φ(~e, ~e′)n(~e′) , (3.46)

where we have introduced the scattering opacity through κscaν =

Neσ;

• since the intensity at fixed frequency is proportional to the occupa-tion number, the same equation (3.46) also holds for Iν; therefore,the transport equation for the specific intensity is changed in pre-sence of scattering to

1c

dIνdt

=1c∂Iν∂t

+ ~e ·∂Iν∂~x

(3.47)

=ρ jν4π− ρκabs

ν Iν − ρκscaν

[Iν −

∫dΩ′φ(~e, ~e′)Iν(~e′)

];

• again, we now take the moments of the transport equation in orderto see how the moment equations are changed by scattering; thefirst moment is obtained by integrating (3.48) over dΩ,

∂Uν

∂t+ ~∇ · ~Fν = ρ jν − ρκabs

ν cUν − ρκscaν cUν + ρκsca

ν cUν (3.48)

due to the normalisation of φ(~e, ~e′); therefore, the scattering termscancel, and the equation for Uν remains unchanged;

• the next moment equation simplifies if we further assume that∫dΩφ(~e, ~e′)~e = 0 =

∫dΩ′φ(~e, ~e′)~e′ , (3.49)


which holds for many scattering processes (e.g. Thomson scatte-ring); then, the second moment equation reads

1c∂Fν,i

∂t+ c

∂Pν,i j

∂x j= −ρκabs

ν Fν,i − ρκscaν Fν,i

= −ρ(κabsν + κsca

ν )Fν,i , (3.50)

i.e. the scattering opacity is simply added to the absorption opacityhere; with a suitable modification of the Rosseland mean opacity,the diffusion approximation remains valid which we have obtainedabove;

3.4 Bremsstrahlung

3.4.1 Spectrum of a Moving Charge

• a radiation process which is very important in astrophysics is dueto electrons which are scattered off ions and radiate due to theacceleration they experience; in order to describe it, we start againfrom Larmor’s equation, which says

dEdt

=2e2

3c3

∣∣∣~x∣∣∣2 , (3.51)

where e is the charge of the accelerated particle (the electron, inmost cases), ~x is its acceleration, and dE/dt is the power radiatedaway;

• as a function of frequency, this equation can be written as follows:

dE =2e2

3c3

∣∣∣~x∣∣∣2 ⇒ E =2e2

3c3

∫ ∞

−∞

dt∣∣∣~x∣∣∣2 ; (3.52)

if we Fourier-transform the particle’s trajectory,

~x(ω) =

∫ ∞

−∞

dt ~x(t)eiωt , ~x(t) =

∫ ∞

−∞

dω2π

~x(ω)e−iωt , (3.53)

we first have

~x =

∫ ∞

−∞

dω2π

[−ω2~x(ω)

]e−iωt ⇒

ˆ~x = −ω2~x(ω) , (3.54)

and we can use Parseval’s equation,∫ ∞

−∞

dt f 2(t) =

∫ ∞

−∞

dω2π

∣∣∣ f (ω)∣∣∣2 ; (3.55)

combining these results yields

E =2e2

3c3

∫ ∞

−∞

dω2π

∣∣∣∣ ˆ~x(ω)∣∣∣∣2 =

4e2

3c3

∫ ∞

0

dω2πω4

∣∣∣∣~x(ω)∣∣∣∣2 , (3.56)

3.4. BREMSSTRAHLUNG 35

and, by differentiation,

dEdω

=2e2

3πc3ω4∣∣∣∣~x(ω)

∣∣∣∣2 ; (3.57)

this is a general expression valid for all radiation processes; inorder to make progress, we need the Fourier transform of thespecific particle trajectory;

3.4.2 Hyperbolic Orbits

• classically, the electron follows a hyperbola around the ion in theorbital plane perpendicular to the (conserved) angular momentum;the focal point of the hyperbola is the centre of mass, which weassume to coincide with the centre of the scattering ion, i.e. weneglect the mass of the electron; in polar coordinates, the trajectoryis given by

r(ϕ) =p

1 + ε cosϕ, (3.58)

with the parameters

p =L2

z

αm= a(ε2 − 1) and ε ≡

(1 +

2L2z E

α2m

)1/2

, (3.59)

where Lz = bmv∞ is the angular momentum in z direction, v∞ isthe initial velocity at infinity, E = mv2

∞/2 is the kinetic energy atinfinity and thus the total energy, and α quantifies the couplingstrength; for electrons orbiting nuclei at rest with charge Ze,

α = Ze2 (3.60)

• as for solving the Kepler problem, we introduce a parameter ψ (theeccentric anomaly), of which we require that it satisfy

r = a(ε coshψ − 1) ; (3.61)

we find the relation between ϕ and ψ by inserting (3.61) into (3.58),using (3.59)

a(ε2 − 1)1 + ε cosϕ

= a(ε coshψ − 1) ⇒ cosϕ =ε − coshψε coshψ − 1

,

(3.62)if we want ψ = 0 where ϕ = 0;

• energy conservation implies that the time t when the electronreaches the distance r from the scattering ion is

t =

∫ r

r0

dx[2m

(E + α

x −L2

z2mx2

)]1/2 =

∫ r

r0

xdx[2m

(Ex2 + αx − L2

z2m

)]1/2 ,

(3.63)


whereL2

z = mαa(ε2 − 1) (3.64)

was used from (3.59); furthermore, we have

a =p

ε2 − 1=

L2z

αmα2m2L2

z E=

α

2E⇒ E =

α

2a; (3.65)

combining, we first find

t =

√m2α

∫ r

r0

xdx[x2

2a + x − a2 (ε2 − 1)

]1/2 , (3.66)

which can be transformed with (3.61) to obtain

t =

√ma3

α

∫ ψ

0(ε coshψ− 1)dψ =

√ma3

α(ε sinhψ−ψ) ; (3.67)

• the coordinates x and y in the orbital plane satisfy

x = r cosϕ = a(ε coshψ − 1)ε − coshψε coshψ − 1

= a(ε − coshψ)

y = r sinϕ = a√ε2 − 1 sinhψ , (3.68)

where we have used (3.61) and (3.62); with these expressions, wereturn to the Fourier transform of x and y

• sinceˆx = −iωx(ω) , (3.69)

we have

x(ω) =iω

ˆx(ω) =iω

∫ ∞

−∞

dt x(t)eiωt

=iω

∫ ∞

−∞

dtdxdψ

dψdt

eiωt(ψ)

= −iaω

∫ ∞

−∞

dψ sinhψeiωt(ψ) ; (3.70)

• using (3.67), we can write

eiωt(ψ) = exp

iω√

ma3

α(ε sinhψ − ψ)

; (3.71)

from (3.65), we finda =

α

mv2∞

, (3.72)

moreover,

ω

√ma3

α= ω

√mα2

m3v6∞

=αω

mv3∞

≡ µ , (3.73)

and thuseiωt(ψ) = eiµ(ε sinhψ−ψ) (3.74)


• putting this result into (3.70), we write

x(ω) = −iaω

∫ ∞

−∞

dψ sinhψeiµ(ε sinhψ−ψ)

y(ω) =ia√ε2 − 1ω

∫ ∞

−∞

dψ coshψeiµ(ε sinhψ−ψ) ; (3.75)

these expressions can be analytically integrated and lead to first-order Hankel functions, which will not be discussed in detail here;

• forming|~x(ω)|2 = x(ω)x∗(ω) + y(ω)y∗(ω) (3.76)

and inserting the result into (3.57) yields the desired bremsstrah-lung spectrum;

3.4.3 Integration over the Electron Distribution

• having obtained the spectrum dE/dω for a single charge, we nowhave to integrate over a distribution of charges; we do this by inte-grating over all impact parameters b from 0 to∞ after multiplyingdE/dω with

nine v · 2πbdb , (3.77)

which is the number of scatterings per unit volume and unit timebetween ions and electrons with the number densities ni and ne,respectively; for a fully ionised pure hydrogen gas, ni = ne ≡ n;

• preparing the integration, we note from (3.59) and (3.72) that

ε2 = 1 +b2m2v4

∞

α2 = 1 +b2

a2 , (3.78)

such that the integration over b can be transformed into an integra-tion over ε,

εdε = εbdbεa2 =

bdba2 ⇒ bdb = a2εdε , (3.79)

where ε ∈ [1,∞) while b ∈ [0,∞);

• inserting suitable approximations for the first-order Hankel func-tions, we find after carrying out the b integration and inserting(3.60)

dEdVdtdω

≈16Z2e6n2

3m2c3v

ln(

2γ

mv3

Ze2ω

) (ω mv3

Ze2

)π√

3

(ω mv3

Ze2

) , (3.80)

where γ ≈ 1.78 is Euler’s constant;


• we write this result as

dEdVdtdω

=16πZ2n2e6

3√

3m2c3

1vgff(v, ω) , (3.81)

introducing the so-called gaunt factor gff, which usually dependsat most weakly on v;

• in a dilute thermal plasma, the electrons have a Maxwellian veloci-ty distribution, but for emitting a photon of energy ~ω, an electronneeds at least an energy of

m2v2

min = ~ω ⇒ vmin =

√2~ωm

; (3.82)

the thermal average of the inverse velocity is then⟨1v

⟩=

( m2πkT

)3/2 ∫ ∞

vmin

4πv2dv1v

e−mv2/2kT

=

√2mπkT

e−~ω/kT ; (3.83)

• replacing 1/v in (3.81) by the average (3.83) finally yields thespecific emissivity of a thermal plasma due to bremsstrahlung,

dEdVdtdω

≡ j(ω) =16πZ2n2e6

3√

3m2c3

√2mπkT

e−~ω/kT gff(v, ω) ; (3.84)

• the volume emssivity is the integral of j(ω) over frequency ω,

j =dE

dVdt=

16Z2n2e6

3~m2c3

√2mkTπ

gff(v, ω) ∝ n2√

T ; (3.85)

• numerically, the volume emissivity in cgs units is

j = Z2 gff

( ncm−3

)2

6.69 × 10−20(

kTerg

)1/2

2.68 × 10−24(

kTkeV

)1/2

7.86 × 10−28(TK

)1/2

(3.86)

• as an example, we consider the X-ray emission of a massive galaxycluster with kT = 10 keV; typical clusters reach electron densitiesof n ≈ 10−3 cm−3 in their cores; let us assume for simplicity thatthe X-ray emitting gas with that electron density fills a volume of1 Mpc3;


• assuming fully ionised pure hydrogen, we put Z = 1 and gff = 1for simplicity; then, (3.86) yields

LX ≈ V j ≈ (3.1 × 1024)3 · 10−6 · 2.68 × 10−24 ·√

10≈ 2.5 × 1044 erg s−1 , (3.87)

which makes galaxy clusters the most luminous X-ray sources onthe sky;

• with an average energy of ∼ 10 keV = 1.6e−8 erg per photon, thisluminosity corresponds to

NX ≈ 1.6 × 1052 s−1 (3.88)

photons emitted by the cluster per second; if the cluster is at adistance of, say, 100 Mpc ≈ 3.1 × 1026 cm, these photons aredistributed over an area of ≈ 1.2 × 1054 cm2, such that an X-raydetector with a typical collecting area of a few hundred cm2 sees

≈ 1001.6 × 1052

1.2 × 1054 s−1 ≈ 1 s−1 (3.89)

i.e. this enormous X-ray luminosity produces a flux of approxima-tely one photon per second in a typical X-ray detector;

Kapitel 4

Synchrotron Radiation,Ionisation and Recombination

further reading: Shu, “The Phy-sics of Astrophysics, Vol I: Ra-diation”, chapters 18, 19, 21–23; Rybicki, Lightman, “Radia-tive Processes in Astrophysics”,chapters 6 and 10; Padmanabhan,“Theoretical Astrophysics, Vol. I:Astrophysical Processes”, secti-ons 6.10–6.12

4.1 Synchrotron Radiation

4.1.1 Electron Gyrating in a Magnetic Field

• a further very important radiation process is the emission of ra-diation by electrons moving in a magnetic field ~B; in such a field,electrons spiral around field lines, with their angular frequencygiven by

ωB =ceBE

=eBγmc

, (4.1)

where E is the electron energy, and γ is the usual Lorentz factor;numerically, we have

ωB ≈ 17.6γ−1 MHz( B1 G

), (4.2)

i.e. synchrotron radiation is typically emitted at radio frequencies;

• the radius of the projection of the spiral orbit perpendicular to themagnetic field is

rB =v

ωB=γmcv

eB, (4.3)

and the complete motion is the circular motion around ~B, superpo-sed by a drift along ~B;

• we employ Larmor’s equation

dEdt

=2e2

3c3

∣∣∣~x∣∣∣2 (4.4)

again to calculate the emission; assuming ~E = 0, the Lorentz forceis

~FL =γec

(~v × ~B) ⇒ ~x =γemc

(~v × ~B) ; (4.5)

41

42KAPITEL 4. SYNCHROTRON RADIATION, IONISATION AND RECOMBINATION

• let now α be the angle between~v and ~B, then |~v× ~B| = vB sinα and

|~x|2 =

( emc

)2(γvB sinα)2 , (4.6)

and, from Larmor’s equation (4.4),

dEdt

=2e4γ2v2B2 sin2 α

3m2c5 (4.7)

=23

(e2

mc2

)cγ2β2B2 sin2 α = 2cσT

B2

8πγ2β2 sin2 α ,

where we have identified the Thomson scattering cross section;

• averaging over all pitch angles α, finding

〈sin2 α〉 =12

∫ π

0sin3 αdα =

23, (4.8)

we obtain the radiation energy emitted per unit time by an isotropicelectron distribution in a magnetic field,

dEdt

=43

B2

8πcσTγ

2β2 =43

UB cσTγ2β2 , (4.9)

where we have inserted the energy density UB = B2/(8π) of themagnetic field;

4.1.2 Beaming and Retardation

• we had seen in Sect. 2 of Chap. 1 that solid angles are deformedby Lorentz transformations according to

dΩ′ =dΩ

γ2(1 − β cos θ)2 ; (4.10)

relativistically moving charges thus focus their emission into theirforward direction; for estimating the opening angle ∆θ of theresulting cone, we require that

11 − β cos ∆θ

≈1

2(1 − β), (4.11)

i.e. we want to estimate the half-width of the cone; approximatingthe cosine in (4.11) to second order,

1 − β +∆θ2

2≈ 2(1 − β) , (4.12)

thus∆θ2 ≈ 2(1 − β) ≈ (1 + β)(1 − β) = γ−2 (4.13)

where we have approximated 2 ≈ (1 + β) for relativistic particles,and

∆θ ≈1γ

(4.14)

4.1. SYNCHROTRON RADIATION 43

• due to the narrow emission cone with the opening angle ∆θ ≈ γ−1,an observer sees an electron spiralling in a magnetic field onlyduring the short moment while the cone is moving past him;

• moreover, this means that the radiated electric field can depend onθ only through the combination ∆θγ, because a change in γ causesan immediate change in ∆θ;

• finally, the arrival time of the radiation at the observer dependsalso on the angle θ

• we now consider two radiation signals which leave the electron attimes t1 and t2 = t1 + t; if the velocity of the electron is v and itsorbital radius is rB, we have

rBθ = vt ⇒ t =rBθ

v; (4.15)

the signal leaving at t2 needs less time to get to the observer,namely by the amount

rBθ

c; (4.16)

this is of course the usual retardation due to the finite light speed;the observing time for the second signal is thus

tobs = t −rBθ

c=

rBθ

v

(1 −

v

c

); (4.17)

• now we obtain for the angle, where the electron was when itemitted the radiation which arrives at the observer at time tobs

θobs =vtobs

rB

(1 −

v

c

)−1; (4.18)

approximating the squared Lorentz factor by

γ2 =

(1 −

v2

c2

)−1

=

(1 −

v

c

)−1 (1 +

v

c

)−1≈

12

(1 −

v

c

)−1(4.19)

for v ≈ c, we finally find

θobs =2vtobs

rBγ2 , (4.20)


γθobs ≈2vtobs

rBγ3 = tobs

(2ωBγ

3 sinα)≡

43ωctobs , (4.21)

because v = rBωB sinα; moreover, the last equation defines theangular frequency

ωc ≡32ωBγ

3 sinα =32

( eBmc

sinα)γ2

≈ 100 MHz(

BµG

) ( EGeV

)2

, (4.22)

where the factor 3/2 was introduced for later convenience;


4.1.3 Synchrotron Spectrum

• since the electric field can depend on ∆θ only through the factorγ∆θ and γ∆θ depends only through the factor ωctobs on the ob-serving time, a Fourier transform must find that the spectrum ofsynchrotron radiation can depend on frequency ω only through theratio

ω

ωc; (4.23)

• qualitatively, synchrotron radiation is determined by the followingproperties:

– the basic frequency for γ ≈ 1 is the cyclotron frequency

ωcyc =eBmc

; (4.24)

– for higher velocities, overtones of the cyclotron frequencywill be added whose amplitudes will decrease by powers ofv/c;

– since ωB ∝ γ−1, the electron’s angular orbital frequency will

decrease as v/c increases;– finally, the radiation is limited to a cone which allows the

observer to see the radiation only during time intervals

∆t ∝ γ−3ω−1B ; (4.25)

• therefore, we expect a spectrum which consists of a sequence ofsharp maxima at ωB and its overtones and cuts off at

ωc ∝1∆t∝ ωBγ

3 ; (4.26)

• while γ is increasing, the contribution due to the overtones willincrease, and the maxima will be broadened due to the electrondistribution in the Lorentz factor γ and the pitch angle α; thus, acontinuum is formed, which ends near ωc;

• most of the energy will be emitted near ωc; since

ωc ∝ ωBγ3 ∝ Bγ2 ∝ BE2 (4.27)

the electron energy E is proportional to

E ∝√ωc

B, (4.28)

more preciselyE ≈ 4ν1/2

c B−1/2 erg , (4.29)

where νc = ωc/(2π) is and all quantities are expressed in cgs units;

• synchrotron radiation is strongly linearly polarised because isresults from the orbital motion of electrons perpendicular to themagnetic field; a circular polarisation parallel to the magnetic fieldis effectively averaged out by the pitch-angle distribution;

4.2. PHOTO-IONISATION 45

4.2 Photo-Ionisation

4.2.1 Transition Amplitude

• many astrophysical radiation processes are accompanied by io-nisation and recombination; in ionising processes, a sufficientlyenergetic photon removes an electron from a bound state and pla-ces it into an unbound state, and the reverse process happens inrecombinations;

• as an example, we consider the cross section for ionising a hy-drogen atom from its ground state; for doing so, we first need tocompute the transition rate between bound and unbound stateswhich is caused by a “perturbing” photon;

• according to the quantum-mechanical perturbation theory, thetransition probability within a time interval ∆t is given by

P(∆t) = |a2(∆t)| , (4.30)

where a is the probability amplitude

a(∆t) =

∫ ∆t

0

dt~〈E′|H|E〉 , (4.31)

and |E〉 and |E′〉 are the initial and final states of the electron,respectively, which are the bound and unbound states here;H isthe Hamiltonian of the “perturbing” interaction; the states |E〉 and|E′〉 are written as eigenstates of the Hamiltonian;

• the canonical momentum of a particle in an electromagnetic fieldcan be written as

~P = ~p −ec~A , (4.32)

where ~A is the vector potential; accordingly, the total Hamiltonianis

H =~P2

2m=

12m

(~p −

ec~A)2

=~p2

2m−

emc

~A · ~p +

(ec

)2 ~A2

2m; (4.33)

the Hamiltonian of the interaction between the electron and thefield is thus

H = −e

mc~A · ~p ; (4.34)

• we have assume here that ~A satisfies the transversal (or Coulomb)gauge, in which the scalar potential Φ = 0 and ~∇ · ~A = 0; then, ~Asatisfies the wave equation(

1c2

∂2

∂t2 −~∇2

)~A = 0 ; (4.35)


• if ~A(~x, t) is decomposed into plane waves,

~A(~x, t) =

∫d3k

(2π)3~Ak(t)e−i~k·~x , (4.36)

the wave equation implies for the Fourier amplitudes

~Ak(t) + c2k2 ~Ak(t) = 0 (4.37)

because the individual amplitudes ~Ak(t) are independent of eachother; therefore, ~Ak(t) satisfies an oscillator equation and can bewritten in the form

~Ak(t) = A~e eiωt ; (4.38)

the dispersion relation ω = kc holds here, ~e is the unit vector inpolarisation direction, and Ak is the amplitude of the given modeof the vector potential;

• the interaction Hamiltonian between field and electron is thus

H = −eAmc

ei(ωt−~k·~x) ~e · ~p ; (4.39)

• the transition matrix element 〈E′|H|E〉 can now be calculated asfollows: first, the state are represented by the wave functions

|E〉 = φ(~x)eiEt/~ , |E′〉 = φ′(~x)eiE′t/~ , (4.40)

in which φ(~x) and φ′(~x) are the spatial amplitudes, and the phasefactors describe the time evolution; using these, the probabilityamplitude (4.31) turns into

a(∆t) =A~c

∫ ∆t

0dt

∫ d3x φ′∗(~x)e ei~k·~x

m~e · ~p φ(~x)

ei(E′/~−E/~−ω)t ;

(4.41)

4.2.2 Transition Probability

• we abbreviate the notation by introducing the matrix elementbetween the initial and the final state,

~Mfi ≡e ei~k·~x

m

∫d3x φ′∗(~x)

em~p φ(~x) , (4.42)

in terms of which we can write the probability amplitude (4.41) as

a(∆t) =A~c

(~e · ~Mfi)∫ ∆t

0dt ei(E′/~−E/~−ω)t ; (4.43)


• the time integral can be evaluated as follows:∫ ∆t

0dt eiεt =

∫ ∆t

0dt (cos εt+i sin εt) , ε ≡

E′

~−

E~−ω ; (4.44)

taking the absolute square of this expression, which we shall laterneed, we find ∣∣∣∣∣∣

∫ ∆t

0dt eiεt

∣∣∣∣∣∣ =

(sin(ε∆t/2)

ε/2

)2

; (4.45)

in the limit of ∆t → ∞ this expression approaches a δ function,

lim∆t→∞

sin2(ε∆t/2)(ε/2)2 → 2π∆tδD(ε) ; (4.46)

• combining these results, the transition probability (4.31) turns outto be

P(∆t) =2πA2∆t~2c2 δD(ε)

∣∣∣∣~e · ~Mfi

∣∣∣∣2 , (4.47)

and thus the transition rate is

R =P(∆t)

∆t=

2πA2

~2c2 δD

(E′

~−

E~− ω

) ∣∣∣∣~e · ~Mfi

∣∣∣∣2 ; (4.48)

we further introduce the abbreviation

ωfi ≡E′ − E~

, (4.49)

i.e. we assign an angular frequency to the energy difference E′ − Ebetween the final and initial states;

• what remains is the determination of the field amplitude Ak andthe matrix element ~Mfi; we first consider the amplitude Ak;

• the electric field belonging to the vector potential

~Ak(t) = Ak ~e ei(ωt−~k·~x) (4.50)

is

~Ek = −1c∂~Ak(t)∂t

= −iωAk~e

cei(ωt−~k·~x) , (4.51)

and the energy density in an electromagnetic field in vacuum canbe written as

U =~E2 + ~B2

8π=~E2

4π; (4.52)

• similarly, the energy density in photons of angular frequency ω is

12

N~ωV

, (4.53)


where the factor 1/2 appears because the two independent pola-risation directions need to be distinguished; we set N = 1 so thatwe shall only have to multiply with the number of photons in thevolume V later; thus, we find by comparing (4.51), (4.52) and(4.53)

N~ω2V

=A2

kω2

4πc2 ⇒ Ak =

√2π~c2

Vω(4.54)

for a single photon (N = 1); the unit of Ak is

[Ak] =

(erg s cm2 s

cm3 s2

)1/2

=

(ergcm

)1/2=

g1/2cm1/2

s, (4.55)

from which we find for the unit of the electric field

[~Ek] = [Ak][ω

c

]= [Ak] cm−1 =

g1/2

cm1/2 s, (4.56)

as it must be;

• for the transition rate (4.48), we now have

R =4π2

~Vω

∣∣∣∣~e · ~Mfi

∣∣∣∣2 δD(ωfi − ω) ; (4.57)

• according to our choice of the amplitude Ak, this is the transi-tion rate caused by a single photon; we must now multiply theexpression (4.57) with the number of available photons, which is

n(ω)Vd3k(2π)3 , (4.58)

in which n(ω) is the occupation number of photons with frequencyω and polarisation state ~e, i.e. the photon number density perphase-space cell;

• since ω = ck, the volume element in k-space is

d3k = k2dkdΩ =ω2dω

c3 dΩ , (4.59)

and we obtain the number of ionisation transitions per unit time,unit frequency and unit solid angle as

dPdtdωdΩ

=ωn(ω)2π~c3

∣∣∣∣~e · ~Mfi

∣∣∣∣2 δD(ωfi − ω) ; (4.60)

4.2.3 Transition Matrix Element

• finally, we have to calculate the matrix element ~Mfi as defined in(4.42); if the wave length of the incoming light, λ = 2π/k, is much


larger than the extent of the wave function of the bound electron,we can approximate

ei~k·~x ≈ 1 (4.61)

and are left with

~Mfi ≈

∫d3x φ′∗(~x)

em~p φ(~x) ; (4.62)

this approximation is called “dipole approximation” for the follo-wing reason: the momentum operator ~p can be expressed by thecommutator

~p =im~

[H , ~x] =im~

(H~x − ~xH) ; (4.63)

therefore, the matrix element ~Mfi (4.62) can be transformed to

~Mfi ≈em

im~

(E′ − E)∫

d3x φ′∗(~x) ~x φ(~x) = iωfi e~x , (4.64)

i.e. it turns into a dipole matrix element;

• during the transition from the bound to the free state, the Hamil-tonian changes, and thus it is preferable in this context to use themomentum operator instead; we now insert the wave function ofthe ground state in the hydrogen atom as the wave function for theinitial state,

φ(~x) =e−r/a0√πa3

0

, (4.65)

where a0 is Bohr’s radius

a0 ≡~2

me2 = 4.7 × 10−8 cm , (4.66)

while the free electron is described by the plane wave

φ′(~x) =ei~ke·~x

√V, (4.67)

where ~ke is the wave vector of the free electron, which is related tothe momentum by ~pe = ~~ke;

• we first confirm that the dipole approximation can be applied here;for short-wave light, λ ≈ 1000 Å = 10−5 cm, which is almost threeorders of magnitude larger than Bohr’s radius a0, λ a0;

• now, the transition probability between the initial and the finalstate equals the reverse transition rate,∣∣∣⟨final|~e · ~p|initial

⟩∣∣∣2 =∣∣∣⟨initial|~e · ~p|final

⟩∣∣∣2 , (4.68)


and inserting the momentum operator ~p = i~~∇, we find

⟨initial|~e · ~p|final

⟩=~~e · ~ke√πVa3

0

∫d3x e−x/a0e−i~ke·~x ; (4.69)

• denoting the angle between ~ke and ~x by θ, we can write∫d3x e−x/a0e−i~ke·~x = 2π

∫ ∞

0x2dx

∫ π

0sin θdθ e−x/a0e−ike x cos θ

= 2π∫ ∞

0x2dx

∫ 1

−1d(cos θ) e−x/a0e−ike x cos θ

= 4π∫ ∞

0x2dxe−x/a0

sin kexkex

=8πa3

0

(1 + k2ea2

0)2, (4.70)

which turns the square of the transition matrix element into

∣∣∣∣~e · ~Mfi

∣∣∣∣2 =64π~2e2a3

0

Vm2

|~e · ~ke|2

(1 + ~k2ea2

0)4, (4.71)

and the number of transitions per unit time, unit frequency andunit solid angle (4.60) becomes

dPdtdωdΩ

=ωn(ω)2π~c3

∣∣∣∣~e · ~Mfi

∣∣∣∣2 δD(ωfi − ω) (4.72)

=32~e2ωn(ω)a3

0

Vm2c3

|~e · ~ke|2

(1 + ~k2ea2

0)4δD(ωfi − ω) ;

4.2.4 Cross Section

• the cross section for photo-ionisation is determined by the relation

dPdt

dNe =dPdt

Vd3 pe

(2π~)3 =dPdt

Vd3ke

(2π)3

!= σ(ω) cn(ω)

d3 pγ(2π~)3 ; (4.73)

this means that the flux of incoming photons, cn(ω) times thenumber density of states in phase space, multiplied with the crosssection, is the number of transitions per unit time; this must equalthe transition rate times the number of available final states for theelectron;


• in order to obtain the cross section, we first need to integrate theexpression

dPdtdωdΩ

Vd3ke

(2π)3 =4~e2ωn(ω)a3

0

π3m2c3

|~e · ~ke|2 d3ke

(1 + ~k2ea2

0)4δD(ωfi − ω) (4.74)

over all solid angles; we had assumed that the electron is photo-ionised into a free final state, and thus we must have k2

e a−20 ;

using that, we can approximate

(1 + k2ea2

0)4 ≈ a80k8

e , (4.75)

and we introduce a coordinate system in which the photon wavevector ~k and the polarisation vector ~e are parallel to the z- andx-axes, respectively; then, ~ke is described by the two angles (θ, ϕ),

~ke = ke

sin θ cosϕsin θ sinϕ

cos θ

, (4.76)

and~e · ~ke = ke sin θ cosϕ ; (4.77)

• integrating over the solid angle then yields∫ 2π

0dϕ

∫ π

0dθ sin3 θ cos2 φ = π

∫ π

0dθ sin3 θ =

4π3, (4.78)

which leads us to write the number of transitions per unit time andunit photon frequency

dPdtdω

Vd3ke

(2π)3 =16~e2ωn(ω)3π2m2c3a5

0

dke

k4eδD(ωfi − ω) ; (4.79)

• finally, if the electron is to become unbound,

~ω =pe

2m=~2k2

e

2m, (4.80)

thus

ke =

(2mω~

)1/2

and dke =mdω~ke

, (4.81)

anddke

k4e

=

(~

2mω

)5/2 mdω~

; (4.82)

• usingd3 pγ

(2π~)3 =ω2dω(2πc)3 , (4.83)


we finally obtain from (4.73)

dPdtdω

Vd3ke

(2π)3 =16e2ωn(ω)3π2mc3a5

0

(~

2mω

)5/2

dω

!= σ(ω) cn(ω)

ω2dω(2πc)3 (4.84)

and thus, for the cross section,

σ =128πe2

3mca50ω

7/2

(~

2m

)5/2

; (4.85)

identifying the (dimension-less) fine-structure constant,

α ≡e2

~c, (4.86)

and Bohr’s radius (4.66), we finally write this in the intuitive form

σ(ω) = πa20

(2α)9/2

3

(ω0

ω

)7/2, (4.87)

with ω0 ≡ c/a0;

Kapitel 5

Spectra

further reading: Shu, “The Phy-sics of Astrophysics, Vol I: Ra-diation”, chapters 22–23; Ry-bicki, Lightman, “Radiative Pro-cesses in Astrophysics”, chapters10; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophysi-cal Processes”, sections 7.1–7.3

5.1 Natural Width of Spectral Lines

5.2 Cross Sections and Oscillator Strengths

• the analysis of photo-ionisation is equally applicable to systems inwhich transitions occur between two bound levels; as before, wehave from (4.60) and (4.64)

dPdt

=4e2ω3

fin(ωfi)~c3 |~xfi|

2 (5.1)

for the dipole transition probability, where |~xfi| is the matrix ele-ment of the position operator ~x between the two bound states;

• since the photon flux per unit frequency can be expressed by theoccupation number n(ω) times the number of states in phase space,

n(ω)cd3k

(2π)3dω=

n(ω)c 4π k2dk(2π)3dω

=n(ω)ω2

2π2c2 , (5.2)

we can identify the cross section

σ(ω) =

(2π2c2

n(ωfi)ω2fi

) (4e2ω3

fin(ωfi)~c3 |~xfi|

2)δD(ω − ωfi)

=4πe2ωfi

~c|~xfi|

2 2πδD(ω − ωfi)

≡πe2

mcffi φ(ω) , (5.3)

in whichφ(ω) = 2πδD(ω − ωfi) (5.4)

is the line profile function, and ffi is called the “oscillator strength”;the factor

e2

mc(5.5)

53

54 KAPITEL 5. SPECTRA

has the dimension length2 time−1, the profile function has the di-mension time, i.e. the cross section (5.3) does have the dimensionof an area, as it needs be; the profile function is normalised suchthat its integral over frequency is unity,

2π∫ ∞

0δD(ω − ωfi)dν =

∫ ∞

0δD(ω − ωfi)dω = 1 ; (5.6)

accordingly, the oscillator strength is

ffi =4πe2ωfi

3~c|~xfi|

2 mcπe2 =

4mωfi

3~|~xfi|

2 ; (5.7)

it is dimension-less because its unit is

g s−1 cm2

erg s=

g cm2 s2

s2 g cm2 = 1 ; (5.8)

for electric dipole transitions, f ≈ 1 typically;

5.2.1 Transition Probabilities

• transitions between bound states occur spontaneously only with acertain transition probability, i.e. there is an uncertainty in energycorresponding to their uncertainty in time; if Γ−1 is the mean lifetime of the initial state, the uncertainty in energy is approximately

∆E1Γ≈ ~ ⇒ ∆E ≈ Γ~ ; (5.9)

• therefore, spectral lines are not infinitely sharp but somewhatwidened; the profile function φ(ω) will thus not typically be a δD

function; we shall now calculate the shape of this function if themean life time Γ−1 is finite (rather than infinite);

• doing so, we consider a radiative transition between two boundstates 1 and 2 with energy eigenvalues E1 and E2; generally, theSchrödinger equation requires for the amplitude an of the energyeigenstate n

i~∂an

∂t=

∑m

〈n|H|m〉 amei(En−Em)t/~ , (5.10)

whereH is again the interaction Hamiltonian between the electronand the electromagnetic field;

• modeling the transition between the states |2〉 and |1〉, we can seta2 = 1 and am = 0 for m , 2; due to the finite life time of the state|2〉, we can use the ansatz

a2(t) = e−Γt/2 (5.11)

5.2. CROSS SECTIONS AND OSCILLATOR STRENGTHS 55

and obtain

i~∂a1

∂t= 〈1|H|2〉 a2(t) ei[(E1−E2)/~−ω]t

= 〈1|H|2〉 exp

i[E1 − E2 + ~ω

~t]−

Γt2

= 〈1|H|2〉 exp

[i(ω − ω12)t −

Γt2

](5.12)

whereω12 ≡

E1 − E2

~(5.13)

is the angular frequency corresponding to the transition energy;upon integrating (5.12) with the boundary condition a1(t = 0) = 0,we find

a1(t) =i〈1|H|2〉~

1 − exp [i(ω − ω12)t − Γt/2]ω − ω12 + iΓ/2

; (5.14)

• in the limit of very long times, t Γ−1, thus Γt 1, part of theexponential factor can be approximated as

e−Γt/2 → 0 , (5.15)

and the transition rate becomes

|a1(t)|2 =|〈1|H|2〉|2

~2

∣∣∣∣∣ 1ω − ω12 + iΓ/2

∣∣∣∣∣2=|〈1|H|2〉|2

~2

1(ω − ω12)2 + Γ2/4

; (5.16)

• this changes the profile function to read

φ(ω) =Γ

(ω − ω12)2 + (Γ/2)2 , (5.17)

which is again defined such that its integral over frequency ν (ratherthan angular frequency ω) is unity;

• we now need to determine Γ; we had seen above that the transitionrate is given by (5.1), which can be written as

dPdt

=4mωfin(ωfi)

3~|~xfi|

2 e2ω2fi

mc3 =e2ω2

12n(ω12)mc3 f12 , (5.18)

which corresponds to the decay rate; since f12 ≈ 1 for the transiti-ons which are of interest here, a good estimate for Γ is

Γ ≈e2ω2

12

mc3 ; (5.19)


• the cross section for the line transition is

σ12(ω) =πe2

mcf12 φ(ω) =

πe2

mcf12

Γ

(ω − ω12)2 + (Γ/2)2 ; (5.20)

in the centre of the line, i.e. at ω = ω12, we have

σ12 =πe2

mcf12

4Γ

Γ2 =4πe2

mcΓf12 ; (5.21)

using Γ from (5.18), we find

σ12 =4πe2

mcmc3

e2ω212

= 4π(

cω12

)2

∝ λ212 , (5.22)

i.e. the cross section in the centre of the line is proportional to thesquare of the absorbed wave length;

5.3 Collisional Broadening of Spectral Lines

• collisions between atoms can change the occupation numbers andthe life times of states and modify the line profile of emission orabsorption in this way; the effect of collisions can be described byrandom changes of the phase of a1,

i~∂a1

∂t= 〈1|H|2〉 exp [i(ω − ω12)t − Γt/2] eiφ(t) , (5.23)

where φ(t) is a random function such that

⟨eiφ(t)

⟩=

1 if there was no collision

until time t0 else

, (5.24)

which means that the average phase factor (5.24) expresses theprobability that the individual system under consideration experi-enced no collision until time t; in this way, (5.23) formalises theexpectation that (sufficiently energetic) collisions can change thephase of a1 completely;

• extending this consideration from a single system to an ensembleof systems and averaging over them, the ensemble average willturn into an exponential if we assume that the number of collisionsin the ensemble until time t follows a Poisson distribution,⟨

eiφ(t)⟩→ e−Γct/2 , (5.25)

in which Γ−1c is the mean time between collisions; using this, we

find the change in a1 after time t to be proportional to the integralof (5.23),

∆a1 ∝

∫dt exp

[i(ω − ω12)t −

Γt2

+ iφ(t)]

; (5.26)

5.4. VELOCITY BROADENING OF SPECTRAL LINES 57

the averages over time and over all systems in the ensemble thenyield, using (5.25),

〈∆a1〉 ∝

∫dt exp

[i(ω − ω12)t −

Γt2

] ⟨eiφ(t)

⟩=

∫dt exp

[i(ω − ω12)t −

Γt2−

Γct2

]; (5.27)

• obviously, therefore, the sum of the mean decay and collision ratesΓ and Γc now takes the role that Γ had before, i.e. the collisionsshorten the mean life time to

1Γ→

1Γ + Γc

=1Γ

11 + Γc/Γ

(5.28)

and thus broadens the line profile;

5.4 Velocity Broadening of Spectral Lines

• a further broadening mechanism is caused by the Doppler effect;if the emitting atoms (or molecules) move along the line-of-sight,we observe the frequency

ν = ν0

(1 +

v‖

c

)(5.29)

instead of the frequency ν0, if v‖ is the velocity component alongthe line-of-sight;

• it is often appropriate the assume a Gaussian velocity distribution;the observed line profile is then given by∫ ∞

−∞

dv‖√2πσ2

v

δD

[ν − ν0

(1 +

v‖

c

)]exp

[−

(v‖ − v)2

2σ2v

], (5.30)

where v is the mean velocity of the emitting system, and σv is thevelocity dispersion of its particles; using the identity

δD(ax) =1aδD(x) , (5.31)

the Gaussian line profile

cν0

1√2πσ2

v

exp− 1

2σ2v

(ν − ν0

ν0c − v

)2=

cν0

1√2πσ2

v

exp− c2

2σ2v

(ν − ν

ν0

)2 (5.32)


follows, where we have defined

ν ≡ ν0

(1 +

v

c

), (5.33)

i.e. ν is the central line frequency, shifted by the Doppler effectdue to the mean motion of the emitting or absorbing medium;

• according to the definition (5.3) of the scattering cross section, wehave for the Doppler-broadened line

σ =πe2

mcf12

cν0

1√2πσ2

v

exp− c2

2σ2v

(ν − ν

ν0

)2 ; (5.34)

• if the motion of the particles in the medium is thermal, we obtainfor the velocity dispersion

m2σ2v =

kT2

⇒ σ2v =

kTm

, (5.35)

which gives the cross section

σ =πe2

mcf12

1ν0

√mc2

2πkTexp

−mc2

2kT

(ν − ν

ν0

)2 ; (5.36)

the cross section in the centre of the line, i.e. at ν = ν, is thus

σ =πe2

mcf12

ν0

√mc2

2πkT(5.37)

and thus proportional to the inverse root of the temperature;

5.5 The Voigt Profile

• natural line width and collisional broadening result in the Lorentzprofile, while the (typically thermal) Doppler broadening resultsin a Gaussian profile; if all effects need to be taken into account,the resulting profile is a convolution of a Lorentz and a Gaussianprofile,

φ(ω) =

∫ ∞

−∞

Γ

(ω − ω12)2 + (Γ/2)2

1√2πσ2

v

exp

− v2‖

2σ2v

dv‖ ,

(5.38)where we have neglected for simplicity that the line may be shiftedas a whole due to the mean motion with velocity v;

• in (5.38), we need to replace ω12 by

ω12

(1 +

v‖

c

)(5.39)

5.6. EQUIVALENT WIDTHS AND CURVES-OF-GROWTH 59

to take the Doppler shift of the emission frequency into considera-tion; this yields

φ(ω) =

∫ ∞

−∞

Γ

(ω − ω12 − ω12v‖/c)2 + (Γ/2)2

×1√

2πσ2v

exp

− v2‖

2σ2v

dv‖ ; (5.40)

• we now set

v0 ≡√

2σv , u ≡ω − ω12

ω12

cv0, a ≡

Γ

2ω12

cv0

(5.41)

andq ≡

v‖√

2σv

=v‖

v0(5.42)

and obtain

φ(ω) =2ac√πv0ω12

∫ ∞

−∞

e−q2dq

(u − q)2 + a2 , (5.43)

which is the so-called Voigt profile;

• near its centre, this line profile has a Gaussian shape, while itswings retain the Lorentzian shape;

5.6 Equivalent Widths and Curves-of-Growth

• two concepts have been introduced for describing the informationcontained in observed spectral lines, namely the equivalent widthand the curve-of-growth;

• the equivalent width quantifies the area under a spectral line; if I0

is the specific intensity of the spectral continuum, the equivalentwith is defined as

W ≡∫

I0 − I(ν)I0

dν , (5.44)

where I(ν) is the spectral (specific) intensity within the line; thus,the equivalent width of an absorption line is a measure for theintensity removed from the spectrum, or added to the spectrum byan emission line;

• the optical depth within the line is

τ = N Lσ(ν) , (5.45)


where N is the number of absorbers and L is the extent of theabsorbing medium; the specific intensity is then

I(ν) = I0 e−τ , (5.46)

and thus the equivalent width is

W =

∫dν

[1 − e−τ(ν)

]; (5.47)

• since the cross section is proportional to the profile function φ(ω),(5.47) can equally be written as

W =

∫dν

[1 − e−Cφ(ω)

], (5.48)

with a constant (frequency-independent) C;

• for small optical depths, τ 1, the exponential function in (5.46)or (5.47) can be expanded into a Taylor series; this results in

W =

∫dνN Lσ(ν) = N L

πe2

mcf12 (5.49)

because the profile function had been normalised such that its inte-gral over frequency ν yields unity; thus, for small optical depths,we have

W ∝ N , (5.50)

i.e. the equivalent width is simply growing linearly with the num-ber of absorbers;

• if τ 1, the function1 − e−τ (5.51)

behaves like a step function across the absorption line whose stepwidth is determined by how τ approaches unity; let ∆ be this stepwidth in frequency space, then

W ≈ 2∆ ; (5.52)

• for Doppler-broadened lines, we have

τ = N Lσ = N Lπe2

mcf12

ν0

√mc2

2πkTexp

[−

mc2

2kT

(∆

ν0

)]; (5.53)

the condition τ ≈ 1 then requires

exp[−

mc2

2kT∆2

ν20

]!=

1NL

(πe2 f12

mcν0

)−1 (mc2

2πkT

)−1/2

≡C1

C2

1NL

(5.54)

with the abbreviations

C1 ≡

√2πkTmc2 ν0 , C2 ≡

πe2

mcf12 ; (5.55)

5.6. EQUIVALENT WIDTHS AND CURVES-OF-GROWTH 61

• from this and (5.54), we obtain

∆2 =2kTmc2 ν

20 ln

(NLC2

C1

)⇒ ∆ =

C1√π

ln1/2(

NLC2

C1

),

(5.56)i.e. W(N) grows approximately as

√ln N;

• for a Lorentz profile,

φ(ω) =Γ

(ω − ω12)2 + (Γ/2)2 , (5.57)

we can approximate the cross section in the limit |ω−ω12| Γ by

σ =πe2

mcf12

Γ

(ω − ω12)2 =C2Γ

(ω − ω12)2 ; (5.58)

again, we conclude from

τ = N Lσ != ⇒ NLC2Γ = ∆2 ⇒ ∆ =

√NLC2Γ ,

(5.59)i.e. the equivalent width grows in this case as

√N;

• summarising, the curve-of-growth W(N) behaves as

W(N) ∝

N small Nln1/2 N intermediate N√

N large N(5.60)

• for determining N, lines with different oscillator strengths f areused because then the spectral lines fall into different sections ofthe curve-of-growth W(N) for the same number N of absorbers;this may prove difficult when some lines fall into the flat sectionof W(N) where W(N) ∝

√ln N;

Kapitel 6

Energy-Momentum Tensor andEquations of Motion

further reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapters 1–3; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophy-sical Processes”, sections 8.1–8.4; Landau, Lifshitz, “Theoreti-cal Physics, Vol VI: Hydrodyna-mics”, chapter I

6.1 Boltzmann Equation and Energy-MomentumTensor

6.1.1 Boltzmann Equation

• there are many possible approaches to hydrodynamics; one of themost direct ones starts by considering the particles of a fluid asthey interact by collisions; the distribution of the particles in phasespace (~x, ~p) then becomes the fundamental physical quantity;

• let f (~x, ~p, t) be the distribution function in phase space, i.e. thequantity

dN = f (~x, ~p, t) d3xd3 p (6.1)

is the number of particles in the six-dimensional phase spaceelement d3xd3 p;

• forces acting on the particles can be distinguished according towhether they are “smooth” or “rough” on microscopic lengthscales; forces which are “rough” on a microscopic scale are due tothe direct interactions between the particles and are summarisedas “collisions”; forces which are smooth on a microscopic scaleare described by a potential U,

~F = −~∇U ; (6.2)

• without collisions, the distribution function would satisfy thecollision-less Boltzmann equation,

d f (~x, ~p, t)dt

= 0 , (6.3)

63

64KAPITEL 6. ENERGY-MOMENTUM TENSOR AND EQUATIONS OF MOTION

in which the time derivative is to be taken along the particle tra-jectories; this equation says that in absence of collisions, particletrajectories are not lost in phase space, which is a consequence ofLiouville’s theorem for Hamiltonian systems;

• writing the time derivative in (6.3) explicitly, the collision-lessBoltzmann equation reads

∂ f∂t

+ ~x∂ f∂~x

+ ~p∂ f∂~p

= 0 (6.4)

or, using (6.2) and ~p = ~F,

∂ f∂t

+ ~x · ~∇ f − ~∇U∂ f∂~p

= 0 (6.5)

• in presence of collisions, the right-hand side of the Boltzmannequation is changed from zero to

0→ C[ f (~x, ~p, t)] (6.6)

where C[ f ] is a functional of the distribution function f whichdescribes how it is changed by collisions;

• collisions happen between particles with momenta ~p1 and ~p2 andlead to momenta ~p′1 and ~p′2; the scattering cross section as a func-tion of the solid angle be σ(Ω); if the distribution functions areabbreviated as follows

f (~x, ~p1, t) ≡ f1 , f (~x, ~p2, t) ≡ f2

f (~x, ~p′1, t) ≡ f ′1 , f (~x, ~p′2, t) ≡ f ′2 , (6.7)

the collision term can be written as

C[ f ] =

∫dΩd3 p2

[σ(Ω) |~v1 −~v2|

(f ′1 f ′2 − f1 f2

)]; (6.8)

this is a result of kinetic theory; we shall later see that this termdrops out when moments of the Boltzmann equation are formed;

6.1.2 Moments; Continuity Equation

• hydrodynamics builds upon the central assumption that the meanfree path λ of the particles making up a fluid is very much smallerthan the extent L of the system under consideration, λ L; thismeans that integrations over phase space cells typically averageover many collisions;

6.1. BOLTZMANN EQUATION AND ENERGY-MOMENTUM TENSOR65

• this suggests to solve Boltzmann’s equation by taking its moments(as we did before when studying radiation transport); we assumefor now that there are no external forces, ~F = −~∇U = 0, and formthe first and second moments of the equation

∂ f∂t

+ ~x · ~∇ f = C[ f ] ; (6.9)

• the moments are formed by multiplying Eq. (6.9) with 1 or the four-momentum pµ and integrating over momentum space, weightingwith a factor E−1(~p), in short, by applying the operators∫

d3 pE(~p)

and∫

d3 pE(~p)

pµ (6.10)

to (6.9);

• because of momentum conservation during collisions, the collisionterms on the right-hand side of (6.9) drop out: no net momentum isexchanged on average because under the basic underlying assump-tion λ L, collisions are very frequent and it can be assumed thatcollisions “away from ~p” and “to ~p” are in equilibrium; note thatthis is different from starting with a collision-less equation in thefirst place; here, the collision terms are present, but have no neteffect because of the very short mean free path;

• the first moment yields∫d3 pE(~p)

∂ f∂t

+

∫d3 pE(~p)

~x · ~∇ f = 0 ; (6.11)

• using the definition

Jµ ≡ c∫

d3 pE(~p)

pµ f (xµ, ~p) , (6.12)

we findJ0 =

∫d3 p f (xµ, ~p) (6.13)

because p0 = E/c; quite obviously, J0 is the particle density n(xµ)at the position ~x and the time t = x0/c;

• similarly, the spatial components of Jµ are

Ji = c∫

d3 pE(~p)

pi f (xµ, ~p) , (6.14)

or, using the expressions E = γmc2 and ~p = γm~x which are validfor relativistic and non-relativistic particles alike,

~J =

∫d3 p

~xc

f (xµ, ~p) ; (6.15)


since the average particle velocity is

~v =

∫d3 p ~x f∫d3 p f

=1n

∫d3 p ~x f , (6.16)

the spatial vector ~J turns out to be the average particle current n~vdivided by the light speed c; therefore, the first moment equationcan be written in the form

∂J0

x0 +∂Ji

∂xi =∂n∂t

+ ~∇ · (n~v) = 0 , (6.17)

upon multiplying this equation for the evolution of the number den-sity n with the particle mass m, we obtain the continuity equationfor the mass,

∂ρ

∂t+ ~∇ · (ρ~v) = 0 , (6.18)

which can also be expressed by the vanishing four-divergence

∂Jµ

∂xµ= 0 (6.19)

of the current Jµ, which is the relativistic generalisation of thecontinuity equation (6.18);

6.1.3 Energy-Momentum Tensor

• we now consider the second moments, which are obtained byapplying the second integral operator from (6.10) to the Boltzmannequation (6.9); we first study the tensor

c2∫

d3 pE(~p)

pµpν f ≡ T µν ; (6.20)

• in the non-relativistic limit, we have

p0 =Ec

=γmc2

c= mc

1 +~x2

2c2

, (6.21)

and thus the time-time component of the tensor T µν is

T 00 = mc2∫

d3 p1 +

~x2

2c2

f

= ρc2 +

∫d3 p

m~x2

2f = ρc2 + nε , (6.22)

where ε is the mean kinetic energy of the fluid particles

ε ≡

∫d3 p m

2 ~x2 f∫

d3 p f=

1n

∫d3 p

m2~x2 f ; (6.23)

6.1. BOLTZMANN EQUATION AND ENERGY-MOMENTUM TENSOR67

• the space-time part of the tensor T µν reads

T 0i = c∫

d3 p pi f =

∫d3 p γmcxi f

≈

∫d3 p

1 +~x2

2c2

mcxi f =

(ρc~v +

~qc

)i

, (6.24)

where ~q is the current of the kinetic energy

~q ≡∫

d3 pm~x2

2

~x f ; (6.25)

the first term on the right-hand side of (6.24), ρc~v, is the masscurrent times the light speed;

• finally, the space-space components of T µν are

T i j = c2∫

d3 pE(~p)

pi p j f = c2∫

d3 pmc2 γm2 xi x j f

≈

∫d3 p mxi x j f , (6.26)

which can be interpreted as the (three-dimensional) stress-energytensor;

• we now return to the (simplified) Boltzmann equation (6.9) andrewrite it somewhat, using

E = γmc2 , ~p = γm~x ⇒~pE

=~xc2 (6.27)

as well as p0 = E/c and x0 = ct; with that, we first obtain

c∂ f∂x0 + c2 ~p

E~∇ f = C[ f ] , (6.28)

and from that,

c2 p0

E∂ f∂x0 + c2 ~p

E~∇ f = C[ f ] ; (6.29)

• this can obviously be written in the form of a four-divergence,

∂

∂x0

(f c2 p0

E

)+∂

∂~x·

(f c2 ~p

E

)= C[ f ] , (6.30)

and an integration over∫

d3 p pν yields

∂T 0ν

∂xν= 0 , (6.31)

i.e. this moment of the Boltzmann equation is equivalent to thevanishing four-divergence of T 0ν;


• we can proceed in much the same way with the spatial componentsT iν of T µν; doing so, we first write

cpi ∂ f∂x0 + pi~x · ~∇ f = cpi ∂ f

∂x0 + pi x j ∂ f∂x j = C[ f ] pi (6.32)

and insert

x j = c2 p j

E(6.33)

to find

c2 pi EcE

∂ f∂x0 + c2 pi p j

E∂ f∂x j = C[ f ] pi , (6.34)

which equals

c2

Ep0 pi ∂ f

∂x0 +c2

Epi p j ∂ f

∂x j = C[ f ] pi ; (6.35)

after integration over d3 p, this yields

∂

∂x0 T i0 +∂

∂x j Ti j =

∂T iν

∂xν= 0 ; (6.36)

• thus, the moment equations of the Boltzmann equation can be ex-pressed by the vanishing four-divergence of the energy-momentumtensor T µν,

∂T µν

∂xν= 0 ; (6.37)

this indicates how the hydrodynamical equations can be relativisti-cally generalised; we now return to the non-relativistic expressionsfor T µν;

• the time-component of the divergence equation implies

1c∂

∂t

(ρc2 + nε

)+ ~∇ ·

(ρc~v +

~qc

)= 0 ; (6.38)

using the continuity equation (6.18) to eliminate the partial timederivative of the density ρ from (6.38), we find

− c~∇ · (ρ~v) +∂

∂t

(nεc

)+ c~∇ · (ρ~v) +

1c~∇ · ~q = 0 (6.39)

and thus∂(nε)∂t

+ ~∇ · ~q = 0 ; (6.40)

this is the continuity equation for the energy, i.e. the expression ofenergy conservation;

• the spatial part of the divergence equation implies

1c∂

∂t

(cρvi +

qi

c

)+

∂

∂x j Ti j = 0 ; (6.41)

6.2. THE TENSOR VIRIAL THEOREM 69

here, we can neglect the term ~q c−2 because the energy flow willbe much slower than the speed of light, and obtain

∂(ρvi)∂t

+∂T i j

∂x j = 0 , (6.42)

which is the equation of momentum conservation in the fluidapproximation;

• summarising, we obtain the equations of motion

∂(nε)∂t

+ ~∇ · ~q = 0 , with ~q ≡∫

d3 pm~x2

2

~x f

∂(ρvi)∂t

+∂T i j

∂x j = 0 , with T i j ≡

∫d3 p mxi x j f (6.43)

• we now split the velocities ~x of the particles into the mean velocity~v and a (usually thermal) velocity ~u about the mean,

~x = ~v + ~u ; (6.44)

the kinetic energy then reads

nε =

∫d3 p

m2

(~v + ~u

)2 f =

∫d3 p

m2~v2 f +

∫d3 p

m2~u2 f ,

(6.45)because the (thermal) velocities ~u vanish on average, by definition;we thus obtain

nε =ρ

2~v2 +

nm2

⟨~u2

⟩, (6.46)

where⟨~u2

⟩is the mean-squared thermal velocity,

⟨~u2

⟩≡

∫d3 p~u2 f∫

d3 p f; (6.47)

the second term in (6.46) is the internal energy ε, the first is thekinetic energy of the mean fluid motion; if the internal energy isthermal,

ε =32

nkT ; (6.48)

6.2 The Tensor Virial Theorem

6.2.1 A Corollary

• an important theorem for all systems satisfying the equations wehave derived is the tensor virial theorem, which holds independent


of the particular form of the energy-momentum tensor T µν; inorder to prove it, we first demonstrate that the relation

ddt

∫V

d3x ρF =

∫V

d3x ρdFdt

(6.49)

holds for arbitrary functions F(~x, t) and integration volumina V;

• the proof proceeds as follows: first, the total time derivative of theintegral is

ddt

∫V

d3x ρF =

(∂

∂t+~v · ~∇

) ∫V

d3x ρF =

∫V

d3x∂(ρF)∂t

(6.50)

because the integration over d3x removes the dependence on ~x andmakes the gradient vanish; then, the remaining integral over thepartial time derivative is∫

Vd3x

∂(ρF)∂t

=

∫V

d3x(ρ∂F∂t− F~∇ · (ρ~v)

)=

∫V

d3x(ρ∂F∂t− ~∇ · (Fρ~v) + ρ~v · ~∇F

)=

∫V

d3x(ρ∂F∂t

+ ρ~v · ~∇F)

=

∫V

d3x ρdFdt

, (6.51)

where Gauss’ theorem was employed again to remove the diver-gence term; this completes the proof;

6.2.2 Second Moment of the Mass Distribution

• we now consider the second spatial moment of the mass distributi-on,

Ii j ≡

∫V

d3x ρxix j (6.52)

and its second time derivative,

d2Ii j

dt2 =ddt

[ddt

∫V

d3x ρxix j

]=

ddt

∫V

d3x ρd(xix j)

dt, (6.53)

where the theorem (6.49) was used with F = xix j; notice that thevolume V is fixed, so that the coordinates xi introduced in (6.52)do not explicitly depend on time;

6.2. THE TENSOR VIRIAL THEOREM 71

• the integral on the right-hand side of (6.53) can be transformed toread ∫

Vd3x ρ

d(xix j)dt

=

∫V

d3x ρ~v · ~∇(xix j)

=

∫V

d3x ρvk∂(xix j)∂xk

=

∫V

d3x ρ(vkδi

kx j + vkxiδjk

)=

∫V

d3x ρ(vix j + v jxi

); (6.54)

we now take the second total time derivative remaining from (6.53),which can again be replaced by a partial time derivative as in (6.50)before,

d2Ii j

dt2 =

∫V

d3x[∂(ρvix j)∂t

+∂(ρv jxi)∂t

]; (6.55)

• now we establish a connection with the energy-momentum tensorby noting that

∂(xiT jk)∂xk = T jkδi

k + xi∂T jk

∂xk = T ji − xi∂(ρv j)∂t

(6.56)

holds because of momentum conservation (6.42); we can use againthat the coordinates xi do not explicitly depend on time to writethis last result in the form

∂(ρxiv j)∂t

= T ji − xi∂T jk

∂xk (6.57)

• inserting this into (6.55), we can bring the time derivative of Ii j

into the following form:

d2Ii j

dt2 =

∫V

d3x[(

T i j + T ji)−

∂

∂xl

(T ilx j + T jlxi

)]; (6.58)

the second term in this last equation is a divergence, which may betransformed by Gauss’ theorem into a vanishing boundary term;the symmetry of T i j then implies

d2Ii j

dt2 = 2∫

Vd3x T i j , (6.59)

which is the tensor virial theorem;

• although the described prorcedure of taking moments is mathema-tically meaningful and correct, it has not brought us closer to asolution of Boltzmann’s equation: we do not know the form of thestress-energy tensor yet; only when this is defined can we solvethe equations of motion;


• here, too, there are numerous possibilities for “closing” the mo-ment equations; they are similar to the procedure which we haveapplied to describe radiation transport in the local thermodynami-cal equilibrium: the fluid approximation asserts that the mean freepath, λ, of the fluid particles is much smaller than the dimensionof the system; accordingly, we can define a small parameter

ε ≡λ

L, ε 1 (6.60)

and expand the distribution function in powers of ε, such as

f = f0 + ε f1 + ε2 f2 + . . . , (6.61)

where f0 is the distribution function in the limit ε→ 0;

Kapitel 7

Ideal and Viscous Fluidsfurther reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapters 3–4; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophy-sical Processes”, sections 8.5–8.7; Landau, Lifshitz, “Theoreti-cal Physics, Vol VI: Hydrodyna-mics”, chapters I and II

7.1 Ideal Fluids

7.1.1 Energy-Momentum Tensor

• we first start with a distribution function f0 which describes aninfinitely extended medium in thermal equilibrium; then, f0 is theMaxwellian distribution

f0 =4n

√π(2kTm)3/2

exp(−

m~u2

2kT

), (7.1)

which contains the velocity ~u relative to the mean velocity of thefluid, according to our previous notation

~u = ~x −~v ; (7.2)

• the stress-energy tensor is then

T i j =

∫d3 p m(vi + ui)(v j + u j) f0 =

= mviv j∫

d3 p f0 + m∫

d3 uiu j f0

+ mvi∫

d3 p u j f0 + mv j∫

d3 p ui f0 ; (7.3)

the two latter terms vanish because f0 is isotropic, the first term is

mviv j∫

d3 p f0 = ρviv j , (7.4)

and the second term is

m∫

d3 p uiu j f0 = 0 for i , j (7.5)

73

74 KAPITEL 7. IDEAL AND VISCOUS FLUIDS

and nm3

⟨~u2

⟩for i = j (7.6)

which is the gas pressure P,

P =ρ

3

⟨~u2

⟩; (7.7)

• this interpretation also follows from our earlier considerations; wehad

ε =32

nkT =ρ

2〈~u2〉 ⇒ ρ〈~u2〉 = 3nkT = 3P (7.8)

in case of thermal motion;

• in this way, we obtain the complete stress-energy tensor,

T i j = ρviv j + Pδi j ; (7.9)

• in addition, we require the flux ~q of the kinetic energy,

~q =

∫d3 p

m2

(~v + ~u

)2 (~u +~v

)f0 ; (7.10)

we write it in components,

qi =

∫d3 p

m2

(~v2 + ~u2 + 2~u ·~v

)(vi + ui) f0

=ρ

2~v2vi +

ρ

2~v2〈ui〉 +

ρ

2vi〈~u2〉 +

ρ

2〈ui~u2〉

+ ρviv j〈u j〉 + ρv j〈uiu j〉 , (7.11)

where we have used again the averages of arbitrary quantities Q,

〈Q〉 ≡1n

∫d3 p (Q f ) ; (7.12)

due to the isotropy of f0, the second, fourth and fifth term on theright-hand side of (7.13) vanish, the third term equals εvi, and forthe last term we use again

m∫

d3 p (uiu j f0) = Pδi j = ρ〈uiu j〉 , (7.13)

thusρv j〈uiu j〉 = v jPδi j = Pvi ; (7.14)

• then, the flux of kinetic energy becomes

~q =

(~v2

2+ε

ρ+

Pρ

)ρ~v ≡

(~v2

2+ w

)ρ~v , (7.15)

wherew ≡

ε + Pρ

(7.16)

is the “heat function” (enthalpy) per mass; the enthalpy occurshere instead of the energy because the the pressure work exertedby the fluid needs to be taken into account;

7.1. IDEAL FLUIDS 75

7.1.2 Equations of Motion

• substituting now the expressions (7.9) for the stress-energy tensorT i j and (7.15) for the flux of kinetic energy into the equations ofmotion, we obtain the equations of motion for an “ideal” fluid:first, the equation of continuity,

∂ρ

∂t+ ~∇ · (ρ~v) =

∂ρ

∂t+ (~v · ~∇)ρ + ρ~∇ ·~v =

dρdt

+ ρ~∇ ·~v = 0 ; (7.17)

next, the equation of energy transport,

∂

∂t

(ρ~v2

2+ ε

)+ ~∇ ·

[(~v2

2+ w

)ρ~v

]= 0 , (7.18)

and finally the equation for the momentum transport,

∂(ρvi)∂t

+∂

∂x j

(ρviv j + Pδi j

)= 0 ; (7.19)

• we rewrite the last two equations in order to bring them into aparticularly manageable form; from energy conservation (7.18),we have

0 =~v2

2∂ρ

∂t+ρ

2∂~v2

∂t+∂ε

∂t

+ ρ~v · ~∇~v2

2+~v2

2~∇ · (ρ~v) + ~∇(ε~v) + ~∇ · (P~v) , (7.20)

which we can simplify with the aid of the continuity equation,

ρ

2∂~v2

∂t+∂ε

∂t+ ρ~v · ~∇

~v2

2+ ~∇(ε~v) + ~∇ · (P~v) = 0 ; (7.21)

momentum conservation requires

0 = vi∂ρ

∂t+ ρ

∂vi

∂t

+ viv j ∂ρ

∂x j + ρv j ∂vi

∂x j + ρv j ∂vj

∂x j +∂P∂x j δ

i j , (7.22)

which can be written in vector form as

0 = ~v∂ρ

∂t+ ρ

∂~v

∂t+ ~v(~v · ~∇)ρ + ρ(~v · ~∇)~v + ρ~v(~∇ ·~v) + ~∇P ; (7.23)

for simplifying that, we first use the continuity equation again inits form

∂ρ

∂t+~v · ~∇ρ + ρ~∇ ·~v = 0 (7.24)


to get rid of the first, third, and fifth terms on the right-hand sideof (7.23); we thus obtain

ρ∂~v

∂t+ ρ(~v · ~∇)~v + ~∇P = 0 ; (7.25)

moreover, we can use the identity

(~v · ~∇)~v = ~∇

(~v2

2

)+ (~∇ ×~v) ×~v (7.26)

to find∂~v

∂t+ ~∇

(~v

2

)+ (~∇ ×~v) ×~v = −

~∇Pρ

; (7.27)

• multiplying (7.25) with ~v yields

ρ

2∂~v2

∂t+ρ

2(~v · ~∇)~v2 +~v · ~P = 0 ; (7.28)

• inserting this into the equation of energy conservation, the first,second, and next-to-last terms cancel to yield

∂ε

∂t+ ~∇ · (ε~v) = −P~∇ ·~v ; (7.29)

using here the energy per mass, ε ≡ ε/ρ, this equation reads

(ερ)∂t

+ ~∇ · (ρε~v) = ρ∂ε

∂t+ ε

∂ρ

∂t+ ε~∇ · (ρ~v) + ρ~v · ~∇ε

= ρ

(∂ε

∂t+~v · ~∇ε

)= −P~∇ ·~v ; (7.30)

• we have now arrived at Euler’s equations,

∂ρ

∂t+ ~∇ · (ρ~v) = 0 ,

∂ε

∂t+ ~∇ · (ε~v) = −P~∇ ·~v ,

∂~v

∂t+ (~v · ~∇)~v = −

~∇Pρ

, (7.31)

which described the conservation laws for mass, energy, and mo-mentum in the approximation of an ideal fluid;

• had we allowed (conservative) external forces, with

~Fext = m~v = −m~∇Φ (7.32)

with the potential Φ of the force, the right-hand side of the momentum-conservation equation had acquired an additional potential gradi-ent,

∂~v

∂t+ (~v · ~∇)~v = −

~∇Pρ− ~∇Φ ; (7.33)

7.1. IDEAL FLUIDS 77

7.1.3 Entropy

• the entropy of an ideal, monatomic gas is

s =3k2m

ln(

Pργ

)(7.34)

per unit mass, where we have omitted an additive constant; γ = 5/3is the adiabatic index; the total time derivative of this specificentropy is

∂s∂t

+ (~v · ~∇)s =ργ

P

(ρ−γ

∂P∂t− Pγρ−γ−1∂ρ

∂t

)(7.35)

+

(vi ∂

∂xi

)ln

(Pργ

)=

1P∂P∂t−γ

ρ

∂ρ

∂t

+ viργ

P

(ρ−γ

∂P∂xi − γρ

−γ−1P∂ρ

∂xi

)=

1P∂P∂t−γ

ρ

∂ρ

∂t+vi

P∂P∂xi −

γvi

ρ

∂ρ

∂xi ;

• according to the continuity equation, we can simplify

γ

ρ

[∂ρ

∂t+ (~v · ~∇)ρ

]=γ

ρ

[∂ρ

∂t+ ~∇ · (ρ~v) − ρ~∇ ·~v

]= −γ~∇·~v , (7.36)

and we further hadP = nkT =

23ε , (7.37)

and therefore1P∂P∂t

=1ε

∂ε

∂tand

1P∂P∂xi =

1ε

∂ε

∂xi ; (7.38)

• thus, the entropy equation (7.36) reads

∂s∂t

+ (~v · ~∇)s =1ε

∂ε

∂t+

1ε

(~v · ~∇)ε + γ~∇ ·~v

=1ε

[∂ε

∂t+ (~v · ~∇)ε + γε~∇ ·~v

]; (7.39)

finally, we use

γε =53·

32

nkT =52

nkT =

(32

+ 1)

nkT = ε + P , (7.40)

which allows us to conclude

∂s∂t

+ (~v · ~∇)s =1ε

[∂ε

∂t+ ~∇ · (ε~v) + P~∇ ·~v

]= 0 , (7.41)

because the expression in square brackets vanishes due to energyconservation;


• we thus obtain the conservation of entropy,

∂s∂t

+ (~v · ~∇)s =dsdt

= 0 , (7.42)

which is intuitively expected in the absence of dissipation;

7.2 Viscous Fluids

7.2.1 Stress-Energy Tensor; Viscosity and Heat Con-ductivity

• so far, we have neglected gradients in the temperature because wehave assumed a Maxwellian velocity distribution belonging to a fi-xed temperature for the particles; likewise, the stress-energy tensorof the ideal fluid does not contain velocity- or density gradients;such terms will now be included;

• the form of the corresponding expressions in the stress-energytensor and in the energy current ~q can be computed by expandingthe phase-space distribution function to the next order beyond theideal-fluid term f0; we abbreviate here and justify the form of theappearing terms by physical arguments;

• differential velocity terms can be expressed by the tensor

vij ≡

∂vi

∂x j , (7.43)

whose trace is the divergence of ~v,

trvij =

∂vi

∂xi = ~∇ ·~v ; (7.44)

we subtract this trace from vij in order to obtain a trace-free residu-

al,

vij −

13δi

j~∇ ·~v ; (7.45)

this expression describes pure shear flows which deform the medi-um, while the part proportional to ~∇ ·~v describes the compressionof the medium; we thus obtain the shear tensor

σij ≡ 2η

(vi

j −13δi

j~∇ ·~v

)− ζδi

j~∇ ·~v , (7.46)

in which η and ζ are constants which remain to be determined anddescribe the strength of shear flows and compression, respectively;

7.2. VISCOUS FLUIDS 79

• the tensor vij from (7.43) can be split into a symmetric and an

antisymmetric part:

∂vi

∂x j =12

(∂vi

∂x j +∂v j

∂xi

)+

12

(∂vi

∂x j −∂v j

∂xi

); (7.47)

• if the velocity field is caused by rigid rotation,

~v = ~ω × ~x , vi = ε ijkω

jxk , (7.48)

the antisymmetric part turns into

12

(∂vi

∂x j −∂v j

∂xi

)=

12

∂(ε iklω

kxl)∂x j −

∂(ε jklω

kxl)∂xi

=

12

(ε i

k j − εjki

)ωk

= −ε ijkω

k , 0 , (7.49)

while the symmetric part vanishes;

• in order to prevent rigid rotation from causing dissipation, we useonly the symmetric part and write from now on

vij ≡

12

(∂vi

∂x j +∂v j

∂xi

); (7.50)

• we augment the energy-momentum tensor of the ideal fluid nowby this shear tensor and obtain

T ij = ρviv j + Pδi

j − σij , (7.51)

in which the minus sign is conventional;

• accordingly, we can modify the energy current; first, a temperaturegradient will cause an energy current against the gradient whichwill be quantified by a heat conductivity κ,

− κ∂T∂xi ; (7.52)

moreover, we need to add a contribution to the energy transportwhich is due to the flow of the velocity gradient,

− v jσij ; (7.53)

and therefore the energy current then reads

qi =

(ρ

2~v2 + w

)vi − κ

∂T∂xi− v jσi

j ; (7.54)


• the additional contributions are thus characterised by three coef-ficients, i.e. the heat conductivity κ, and the two viscosity coeffi-cients η and ζ;

• the form of these coefficients can be computed by inserting theansatz f = f0 + f1 for the phase-space distribution function intothe Boltzmann equation and iteratively searching for solutions,evaluating the collision terms;

• κmust have the dimension (energy density× velocity)/(temperaturegradient), hence

[κ] =ergcm3

cms

cmK

=erg

cm s K; (7.55)

• similarly, one finds that the dimension of η is (energy densi-ty)/(velocity gradient), or

[η] =ergcm3

scm

=erg scm3 ; (7.56)

7.2.2 Estimates for Heat Conductivity and Viscosity

• we consider a gaseous system in thermal equilibrium with a tem-perature T whose particles are moving randomly in all directions;let ∆A be the area of a screen perpendicular to the y axis;

• per unit time,n v∆A

6(7.57)

particles will fly through the screen, either from left to right or theother way round; the factor of 1/6 is owed to the fact that typicallyonly 1/3 of the particles is flying along the y axis, and of those,only 1/2 in either direction;

• the mean free path is λ = (nσ)−1, if σ is the collisional crosssection of the particles; particles coming from the left transportproperties of the gas from y−λ to y, and particles coming from theopposite direction transport properties from y + λ to y; interestingeffects occur if these properties have gradients;

• if the particle number density, n, changes along y, ∂n/∂y , 0, thenet number of flowing particles is

n(y + λ)v∆A6

−n(y − λ)v∆A

6≈v∆A

62∂n∂y

λ , (7.58)

where we have implicitly assumed that the mean free path λ is veryshort compared to the typical length scale of the number-densitygradient;


• the diffusion coefficient D, which relates the particle current perunit area to the number-density gradient,

∆N∆A

= D∂n∂y

, (7.59)

thus isD =

vλ

3; (7.60)

• if the temperature changes along y, ∂T/∂y , 0, the particlestransport energy,

∆ε

∆A=

nv∆A6∆A

[ε(y + λ) − ε(y − λ)

](7.61)

=nvλ3

(∂ε

∂T∂T∂y

)=

nv cvλ

3∂T∂y

,

where cv is the heat capacity at constant volume; using∆ε

∆A!= κ

∂T∂y

, (7.62)

we findκ =

nv cvλ

3=v cv

3σ; (7.63)

since ε is the energy per particle, we have

cv =3k2

(7.64)

and thusκ =

vk2σ

, (7.65)

which has the physical unit

[κ] =cms

ergK

1cm2 =

ergcm s K

, (7.66)

as expected from (7.55);

• in complete analogy, the transport of momentum is∆px

∆t=

nv∆A6

2mλ∂vx

∂y, (7.67)

i.e. momentum in the x direction is transported in this way alongthe y axis;

• the change of momentum per unit time is a force; the force perunit area is

Fx =∆px

∆t ∆A=

nvmλ3

∂vx

∂y

!= η

∂vx

∂y, (7.68)

where η is the viscosity coefficient; from this, we obtain

η =nvmλ

3=

mv3σ

=2m3kκ , (7.69)

which clarifies where the form of the expressions for κ and ηoriginate from;


7.2.3 Equations of Motion for Viscous Fluids

• again, we can find the equations of motion by inserting ~q and T i j

into the general results (6.18) and (6.43), where

nε =ρ

2~v2 +

32

nkT ≡ρ

2~v2 + ε ; (7.70)

obviously, the continuity equation remains valid without change;

• the force equation now reads, with the stress-energy tensor (7.51)and the shear tensor (7.46)

∂(ρvi)∂t

+∂(ρviv j)∂x j +

∂P∂xi

= 2η

∂vi j

∂x j −13∂(~∇ ·~v)∂xi

+ ζ∂(~∇ ·~v)∂xi ; (7.71)

• the right-rand side of this equation can be simplified to read

η∂

∂x j

(∂vi

∂x j +∂v j

∂xi

)−

2η3

∂

∂xi

∂v j

∂x j + ζ∂

∂xi

∂v j

∂x j

= η~∇2vi +

(η

3+ ζ

)∂(~∇ ·~v)∂xi ; (7.72)

• the left-hand side of (7.71) can be transformed by means of thecontinuity equation,

ρ∂~v

∂t+~v

∂ρ

∂t+~v · (~v · ~∇)ρ + ρ(~v · ~∇)~v + ρ~v · (~∇ ·~v) =

= ρ∂~v

∂t+~v

(∂ρ

∂t+ (~v · ~∇)ρ + ρ~∇ ·~v

)+ ρ(~v · ~∇)~v =

= ρd~vdt

; (7.73)

• this leads us to the Navier-Stokes equation,

ρd~vdt

= −~∇P + η~∇2vi +

(η

3+ ζ

)∂(~∇ ·~v)∂xi , (7.74)

which simplifies to Euler’s equation if the viscosity parametersvanish, η = 0 = ζ;

• the energy-conservation equation is

∂

∂t

(ρ~v2

2+ ε

)+

∂

∂xi

[ρ

(~v2

2+ w

)vi − κ

∂T∂xi− v jσ

i j

]= 0 ; (7.75)

this expression can be simplified as follows:


• the equation of momentum conservation can be written as

∂(ρvi)∂t

+∂

∂x j

(ρviv j + Pδi j − σi j

)= 0 ; (7.76)

multiplying this with vi and using the continuity equation enablesus to write

∂

∂t

(ρ

2~v2

)+

∂

∂x j

(12ρ~v2v j

)= −vi

∂P∂xi

+ vi∂σi j

∂x j ; (7.77)

subtracting this from the energy conservation equation yields

∂ε

∂t+∂(ρwvi)∂xi −

∂

∂xi

(κ∂T∂xi

)− (~v · ~∇)P − σi j∂v j

∂xi = 0 ; (7.78)

• using the definition of the enthalpy (7.16), we can transform

∂(ρwvi)∂xi =

∂[(ε + P)vi]∂xi = (~v · ~∇)(ε + P) + (ε + P)~∇ ·~v , (7.79)

and the energy conservation equation can be cast into the form

∂ε

∂t+ ~∇ · (ε~v) + P~∇ ·~v = ~∇ · (κ~∇T ) + σi j ∂vi

∂x j ; (7.80)

7.2.4 Entropy

• we now introduce again the energy per unit mass, ε ≡ ε/ρ, to write

∂(ρε)∂t

+ ~∇ · (ρε~v) = ε

[∂ρ

∂t+ ~∇ · (ρ~v)

]+ ρ

(∂ε

∂t+~v · ~∇ε

)= ρ

dεdt

; (7.81)

this first implies the equation

ρdεdt

+ P~∇ ·~v = ~∇ · (κ~∇T ) + σi j ∂vi

∂x j ; (7.82)

• the term P~∇ · ~v can also be rewritten; because of the continuityequation, we first obtain

dρdt

=∂ρ

∂t+~v · ~∇ρ =

∂ρ

∂t+ ~∇ · (ρ~v) − ρ~∇ ·~v = −ρ~∇ ·~v , (7.83)

which allows us to write

P~∇ ·~v = P(−

1ρ

dρdt

)= −

Pρ

dρdt

= Pρdρ−1

dt= Pρ

dVdt

, (7.84)

where we have used the specific volume V ≡ ρ−1;


• thus, the left-hand side of the energy equation (7.82) can be castinto the form

ρdεdt

+ P~∇ ·~v = ρ

(dεdt

+ PdVdt

)= ρT

dsdt

= ρT[∂s∂t

+ (~v · ~∇)s], (7.85)

where s is again the specific entropy; therefore, we finally obtain

ρTdsdt

= ρT[∂s∂t

+ (~v · ~∇)s]

= ~∇ · (κ~∇T ) + σi j ∂vi

∂x j ; (7.86)

this describes how the entropy is changed due to heat conductionand viscous dissipation; obviously, the entropy is conserved ifκ = 0 = σi j;

7.3 Generalisations

7.3.1 Additional External Forces; Gravity

• the equations derievd so far can be generalised in obvious ways ifexternal forces act such as gravity, −ρ~∇Φ, or a radiation pressureforce ~frad; such additional forces leave the continuity equationunchanged; the force equation acquires the additional force densityterms

− ρ~∇Φ + ~frad ; (7.87)

• in the energy-conservation equation, two additional terms appearwhich describe the work done by the external forces, thus

− ρ~v · ~∇Φ +~v · ~frad ; (7.88)

• the stress-energy tensor of the gravitational field is

T i j =1

4πG

(∂Φ

∂xi

∂Φ

∂x j−

12δi j ∂Φ

∂xk

∂Φ

∂xk

)(7.89)

whose trace is

T ii =

14πG

(∂Φ

∂xi

∂Φ

∂xi−

32∂Φ

∂xk

∂Φ

∂xk

)= −

18πG

∂Φ

∂xi

∂Φ

∂xi; (7.90)

the spatial integral of the trace,∫V

d3x T ii = −

18πG

∫V

d3x∂Φ

∂xi

∂Φ

∂xi=

= −1

8πG

∫V

d3x[~∇ · (Φ~∇Φ) − Φ~∇2Φ

]=

12

∫V

d3x Φρ , (7.91)

7.3. GENERALISATIONS 85

is the potential energy in the gravitational field; in the last equalityof (7.90), we have used the Poisson equation,

~∇2Φ = 4πGρ , (7.92)

and dropped the boundary term which results from the divergence~∇ · (Φ~∇Φ) when we use Gauss’ theorem;

• then, the total stress-energy tensor of a self-gravitating fluid is

T i jges = T i j

gas + T i jgrav = ρviv j + Pδi j +

+1

4πG

(∂Φ

∂xi

∂Φ

∂x j−

12δi j ∂Φ

∂xk

∂Φ

∂xk

)(7.93)

with the trace

T ii ges = ρ~v2 + 3P −

(~∇Φ)2

8πG, (7.94)

whose volume integral is∫V

d3x T ii ges =

∫V

d3x(ρ~v2 + 3P +

12

Φρ

)= 2T + U + 3

∫V

d3x P , (7.95)

where T and U are the kinetic and potential energies, respectively;

• the tensor virial theorem (6.59) tells us

d2Ii j

dt2 = 2∫

Vd3x T i j ⇒

d2Iii

d2t= 2

∫V

d3x T ii ; (7.96)

which means that the integral over the trace of T i j must vanish ifthe system under consideration is static;

7.3.2 Example: Cloud in Pressure Equilibrium

• now, we briefly consider two astrophysical consequences of thisresult; first, let a homogeneous, spherical cloud be given of massM and radius R which has the temperature T ; it be embedded intothe constant pressure P;

• its kinetic energy is32

Mm

kT , (7.97)

and the potential energy is

U = −αGM2

Rwith α =

310

(7.98)

for the homogeneous sphere; for other mass distributions, α willbe different but remains of order unity;


• our earlier result now asserts

3kTMm− α

GM2

R+ 3PV = 0 (7.99)

for a static configuration; with V = 4πR3/3, we find for the pressu-re

P =1

4π

(αGM2

R4 −3kT MmR3

); (7.100)

the external pressure must thus be reduced by the amount of thegravitational force compared to the thermal pressure NkT/V ofthe gas in the sphere;

• at the critical mass

Mcr =3kTmR3 ·

R4

αG=

3kTRmGα

, (7.101)

the pressure P vanishes, such that the sphere is in equilibrium withits self-gravity;

7.3.3 Example: Self-Gravitating Gas Sphere

• a further example concerns isolated systems in which the kineticenergy of the gas is

T =32

(γ − 1)Uint , (7.102)

where Uint is the internal energy of the gas and γ the adiabaticindex; for such a static system, the tensor virial theorem requires

3(γ − 1)Uint + Ugrav = 0 , (7.103)

if we denote the gravitational potential energy by Ugrav

• the total energy is E = Uint + Ugrav, and thus (7.103) implies

3(γ − 1)E −[3(γ − 1) − 1

]Ugrav

= 3(γ − 1)E − (3γ − 4)Ugrav = 0 (7.104)

and therefore

E =3γ − 4

3(γ − 1)Ugrav ; (7.105)

since Ugrav < 0 and E < 0 for a bound system, we require

γ >43

; (7.106)

7.3. GENERALISATIONS 87

• in order to see what happens in the limiting case γ → 4/3, wewrite

12

d2Idt2 = 3(γ − 1)E − (3γ − 4)Ugrav ⇒ (7.107)

E =γ − 4/3γ − 1

Ugrav +1

6(γ − 1)d2Idt2 ;

for γ → 4/3, the first term on the right-hand side vanishes, andbecause of E < 0, we must have

d2Idt2 < 0 , (7.108)

which typically implies a collapse;

Kapitel 8

Flows of Ideal and ViscousFluids

further reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapters 6 and14; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophy-sical Processes”, sections 8.6–8.9; Landau, Lifshitz, “Theoreti-cal Physics, Vol VI: Hydrodyna-mics”, chapters I, II and VIII

8.1 Flows of Ideal Fluids

8.1.1 Vorticity and Kelvin’s Circulation Theorem

• ideal fluids are such in which dissipative effects are unimportantor absent, i.e. fluids for which we can put the viscosity coefficientsand the thermal conductivity to zero, ζ = η = κ = 0; in such fluids,the entropy is conserved along flow lines,

dsdt

=∂s∂t

+~v · ~∇s = 0 ; (8.1)

• Euler’s equation for the force per unit mass is

∂~v

∂t+ (~v · ~∇)~v = −

~∇Pρ− ~∇Φ ; (8.2)

here, we employ the identity

~∇(~a · ~b) = (~a · ~∇)~b + (~b · ~∇)~a + ~a × (~∇ × ~b) + ~b × (~∇ × ~a) (8.3)

and put ~a = ~v = ~b; then, the relation

~∇(~v2) = 2(~v · ~∇)~v + 2~v × (~∇ ×~v) (8.4)

follows, and thus

(~v · ~∇)~v =12~∇(~v2) −~v × (~∇ ×~v) , (8.5)


∂~v

∂t−~v × (~∇ ×~v) = −

12~∇(~v2) −

~∇Pρ− ~∇Φ ; (8.6)

89

90 KAPITEL 8. FLOWS OF IDEAL AND VISCOUS FLUIDS

• the curl of the velocity,

~Ω ≡ ~∇ ×~v , (8.7)

is called the vorticity of the flow; if we take the curl of Euler’sequation in its form (8.6), we find the equation for the vorticity

∂~Ω

∂t= ~∇ × (~v × ~Ω) − ~∇ ×

~∇Pρ

= ~∇ × (~v × ~Ω) +

~∇ρ × ~∇Pρ2 ; (8.8)

• if the pressure P is a function only of ρ and not of other quantitiessuch as ~v, the gradients of P and ρ must align,

~∇P ‖ ~∇ρ ⇒ ~∇P × ~∇ρ = 0 ; (8.9)

for such barotropic fluids, the vorticity equation simplifies to

∂Ω

∂t= ~∇ × (~v × ~Ω) ; (8.10)

• we consider now the so-called circulation, which is the line integralover the velocity along closed curves,

Γ ≡

∮C~v · d~l ; (8.11)

we are interested in the total change with time of the circulationembedded into the flow, i.e. taking into consideration that thecontour C is deformed by the flow; we first use Stokes’ theoremto write

Γ =

∫A(~∇ ×~v) · d~A =

∫A

~Ω · d~A , (8.12)

where A is the area enclosed by the contour C; d~A is the directedarea element pointing along the local normal to the area A;

• the total time derivative of Γ now is

dΓ

dt=

∫A

∂Ω

∂t· d~A +

∫A

~Ω ·∂~A∂t

, (8.13)

and the change of the area due to the deformation of the contour is∮C

~Ω · (~v × d~l) , (8.14)

for (~vdt) × d~l gives the differential change of area per time intervaldt;

8.1. FLOWS OF IDEAL FLUIDS 91

• using (8.14) yields with (8.13)

dΓ

dt=

∫A

∂Ω

∂t· d~A +

∫C

(~Ω ×~v) · d~l

=

∫A

∂~Ω∂t+ ~∇ × (~Ω ×~v)

· d~A = 0 , (8.15)

since the term in square brackets vanishes because of the vorticityequation (8.10); we have further used that

~Ω · (~v × d~l) = (~Ω ×~v) · d~l and~∇ × (~Ω ×~v) = −~∇ × (~v × ~Ω) ; (8.16)

thus, the circulation Γ along contours comoving with the flow isconserved in barotropic flows; this is Kelvin’s circulation theorem;

8.1.2 Bernoulli’s Constant

• if the flow is stationary, all partial derivatives with respect to timevanish; in such cases, flow lines can be introduced which are theintegral curves of the velocity field; obviously,

dxvx

= dt =dyvy

=dzvz

(8.17)

for the flow lines;

• in ideal fluids, the specific entropy is constant,

dsdt

=∂s∂t

+ (~v · ~∇)s = 0 thus (~v · ~∇)s = 0 (8.18)

for a stationary flow because ∂s/∂t = 0, i.e. the entropy remainsconstant along flow lines; moreover, the enthalpy satisfes

dw = dε + Pd(ρ−1) + ρ−1dP , (8.19)

because it is, per definition,

w =ε + Pρ

= ε +Pρ, (8.20)

and withdε = Tds − PdV = Tds − Pd(ρ−1) (8.21)

we finddw = Tds + ρ−1dP = ρ−1dP , (8.22)

since ds = 0 along flow lines;


• again for stationary flows, ∂~v/∂t = 0, Euler’s equation in its form(8.6) implies

12~∇(~v2) −~v × ~Ω = −

~∇Pρ− ~∇Φ ; (8.23)

let now ~l be a unit tangent vector to a flow line; if we multiply(8.23) with ~l, the first term quantifies the change of ~v2/2 along theflow line; the second term vanishes, because it is perpendicular to~v and thus also to ~l, and this implies that

B ≡~v2

2+ w + Φ (8.24)

is constant along flow lines; this is Bernoulli’s equation,

B = constant along flow lines ; (8.25)

• we have merely used that ds = 0, i.e. such flows can be calledadiabatic because no heat is exchanged between flow lines; if theflow is furthermore isentropic, i.e. if s = const. holds everywhereand thus all flow lines have the same value of s, then

s = s(P, ρ) = const. (8.26)

means that P must be a function of ρ alone, i.e. the flow is thenalso barotropic, and the circulation Γ is conserved;

8.1.3 Hydrostatic Equlibrium

• in the hydrostatic case, ~v = 0, and Euler’s equation simplifies to

~∇P = −ρ~∇Φ , (8.27)

and Poisson’s equation

~∇2Φ = 4πGρ (8.28)

relates the gravitational potential to the density; taking the diver-gence of (8.27) yields

~∇ ·

~∇Pρ

= −4πGρ ; (8.29)

only assumptions on mechanical, but not on thermodynamicalequilibrium entered here; the curl of (8.27) shows that

0 = ~∇ × (ρ~∇Φ) = ~∇ρ × ~∇Φ , (8.30)

which shows that the gradients of ρ and Φ are then parallel to eachother, i.e. ρ and Φ have the same iso-surfaces, i.e. equipotentialsurfaces are then also iso-density contours;

8.1. FLOWS OF IDEAL FLUIDS 93

• in the spherically symmetric limiting case, we finally have

~∇P =dPdr~er , ~∇ · ~f =

1r2

d(r2 fr)dr

with fr ≡ ~f · ~er , (8.31)

and thus1r2

ddr

(r2

ρ

dPdr

)= −4πGρ ; (8.32)

if there is a barotropic relation, P = P(ρ), this is an ordinary,second-order differential equation for the density ρ which candirectly be integrated;

• Euler’s equation in its form (8.32) is often applied to self-gravitatingsystems such as galaxy clusters where the gravitational potentialis largely caused by the dark matter, which requires us to separatethe gas density ρgas from the dark-matter density ρDM,

1r2

ddr

(r2

ρgas

dPdr

)= −4πGρDM ; (8.33)

• with the equation of state for an ideal gas,

P = nkT =ρgas

mkT , (8.34)

where m is the (mean) mass of a gas particle, we find

r2

ρgas

km

d(ρgaskT )dr

= −4πG∫ r

0r′2dr′ ρDM = −GM(r) , (8.35)

where M(r) is the dark mass enclosed in a sphere of radius r; thus,

M(r) = −kr2

mGρgas

(T

dρgas

dr+ ρgas

dTdr

)=

= −r2kTmG

(d ln ρgas

dr+

d ln Tdr

)=

= −rkTmG

(d ln ρgas

d ln r+

d ln Td ln r

), (8.36)

i.e. if the two logarithmic gradients can be estimated or determinede.g. from X-ray observations, the dark mass can be found;

8.1.4 Curl-Free and Incompressible Flows

• if the velocity field is curl-free, ~∇ ×~v = 0 = ~Ω, the velocity fieldcan be written as the gradient of a velocity potential, ~v = ~∇ψ, andEuler’s equation (8.6) then reads

∂(~∇ψ)∂t

+12~∇~v2 = −

~∇Pρ− ~∇Φ , (8.37)


or, if we combine the gradients (cf. 8.22)

~∇

(∂ψ

∂t+ w +

~v2

2+ Φ

)= 0 ; (8.38)

the quantity

B′ ≡∂ψ

∂t+ w +

~v2

2+ Φ (8.39)

can then only be a function of time,

B′ = f (t) , (8.40)

which can be set to zero because it can be absorbed into the velocitypotential ψ without changing the relation ~v = ~∇ψ; thus,

B′ =∂ψ

∂t+ w +

~v2

2+ Φ = 0 , (8.41)

which generalises Bernoulli’s equation for such cases in which thevelocity field is curl-free, ~v = ~∇ψ, instead of stationary, ∂~v/∂t = 0;

• finally, if ρ = const. in addition, like for incompressible fluids, itsgradient vanishes, ~∇ρ = 0, and the vorticity equation (8.8) implies

∂(~∇ ×~v)∂t

= ~∇ ×[~v × (~∇ ×~v)

], (8.42)

i.e. the flow is then described solely by one equation for the velo-city field, because the continuity equation shrinks to ~∇ ·~v = 0;

• if the velocity field is also curl-free, ~∇ ×~v = 0, i.e. if the flow isincompressible and curl-free, ~∇ ·~v = 0 and ~v = ~∇ψ imply that thevelocity potential ψ has to satisfy the Laplace equation,

~∇2ψ = 0 ; (8.43)

8.2 Flows of Viscous Fluids

8.2.1 Vorticity; Incompressible Flows

• Euler’s equation for viscous fluids was

∂~v

∂t+ (~v · ~∇)~v = −

~∇Pρ

+η

ρ~∇2~v +

1ρ

(ζ +

η

3

)~∇(~∇ ·~v) ; (8.44)

taking the curl of this equation and using the vector identity

~∇ ×[(~v · ~∇)~v

]= −~∇ ×

[~v × (~∇ ×~v)

], (8.45)

8.2. FLOWS OF VISCOUS FLUIDS 95

we obtain the equation for the vorticity in viscous fluids,

∂~Ω

∂t− ~∇ × (~v × ~Ω) =

η

ρ~∇2~Ω (8.46)

+~∇ρ

ρ2 ×

[~∇P − η~∇2~v −

(ζ +

η

3

)~∇(~∇ ·~v)

];

here, the first term on the right-hand side describes the diffusionof vorticity caused by the viscosity η, and the second term arisesexclusively through the density gradient;

• thus, if the flow is incompressible, ρ = const., the second term onthe right-hand side of (8.47) drops out, and the equation is reducedto

∂~Ω

∂t− ~∇ × (~v × ~Ω) =

η

ρ~∇2~Ω ; (8.47)

in incompressible fluids, the divergence of ~v must also vanish,~∇ ·~v = 0, thus

~∇ × (~v × ~Ω) = (~Ω · ~∇)~v − (~v · ~∇)~Ω , (8.48)

which allows us to write the vorticity equation as

∂~Ω

∂t− (~Ω · ~∇)~v + (~v · ~∇)~Ω =

η

ρ~∇2~Ω , (8.49)

or, if we identify the total time derivative of ~Ω,

d~Ωdt− (~Ω · ~∇)~v =

η

ρ~∇2~Ω ; (8.50)

• the divergence of Euler’s equation implies, with ~∇ · ~v = 0 forincompressible fluids,

~∇ ·[(~v · ~∇)~v

]= −

~∇2Pρ

; (8.51)

together with ~∇·~v = 0, these equations (8.50) and (8.51) determinethe flow of incompressible, viscous fluids: ρ is constant, the curlof ~v is the vorticity, ~∇ ×~v = ~Ω, and the divergence of ~v vanishes;this determines the velocity field, and the pressure P follows from(8.51);

8.2.2 The Reynolds Number

• the only dimensional physical parameter in equation (8.50) for thevorticity of a viscous, incompressible flow is

ν ≡η

ρ, (8.52)


whose dimension is

erg scm3

cm3

g=

cm2

s, (8.53)

thus squared length over time; a body characterised by a geometri-cal dimension L moving with velocity u through a viscous fluidthus introduces the length scale L and the time scale L/u; fromthem and ν, we can form the following dimension-less quantity:

L2

L/u·

1ν≡ R =

uLν

=uLρη

, (8.54)

which is the so-called Reynolds number;

• since no other dimensional physical quantities occur, the velocity~v of the fluid can be scaled with the velocity u of the body and thespatial coordinate ~x with the length scale L; in this way, dimension-less quantities emerge whose change is described by quantitieswhich otherwise contain only the Reynolds number; bodies ofdifferent size but otherwise equal shape embedded into flows whichare scaled as described here thus create self-similar flows as long asthe Reynolds number remains the same; conversely, the Reynoldsnumber classifies such self-similar solutions of the flow equations;the transition to ideal fluids is characterised by R → ∞;

8.3 Sound Waves in Ideal Fluids

8.3.1 Linear Perturbations

• we now consider small perturbations of a fluid which is otherwi-se flowing according to a background solution characterised bydensity ρ0 and pressure P0; we further transform into a coordinateframe in which the unperturbed fluid is locally at rest, thus ~v0 = 0;

• let the flow be ideal for now; the perturbations be small, and wecan consequently linearise Euler’s equation around the backgroundsolution, i.e. we write~v = 0+~v′, P = P0+P′, ρ = ρ0+ρ′ and neglectterms of higher than first order in ~v′, P′ and ρ′; this transforms thecontinuity equation to

∂ρ

∂t+ ~∇ · (ρ~v) = 0 =

∂(ρ0 + ρ′)∂t

+ ~∇ · (ρ0~v′) ; (8.55)

the background density ρ0 must also satisfy the continuity equation,which reads

∂ρ0

∂t= 0 (8.56)

8.3. SOUND WAVES IN IDEAL FLUIDS 97

because we are in a frame locally co-moving with the fluid, ~v0 = 0,and thus the linearised continuity equation reads

∂ρ′

∂t+ ~∇ · (ρ0~v

′) = 0 ; (8.57)

• the linearised Euler equation reads

∂~v′

∂t+~∇P′

ρ0= 0 (8.58)

if terms of higher than first order in the perturbations are ignoredagain;

• a relation between density and pressure is established as follows:since the ideal fluid flows adiabatically, we can set

P′ =

(∂P∂ρ

)sρ′ (8.59)

for the pressure perturbations, where the subscript s denotes thatthe derivative must be taken at constant entropy; if we furtherneglect the density gradient of the background solution on thelength scale of the perturbation, ~∇ρ0 = 0, the continuity equationrequires

∂P′

∂t+ ρ0

(∂P∂ρ

)s

~∇ ·~v′ = 0 ; (8.60)

• this equation connects the curl-free part of ~v′ to the evolution ofthe pressure; if we set ~v′ = ~∇ψ, we first have

∂P′

∂t+ ρ0

(∂P∂ρ

)s

~∇2ψ = 0 , (8.61)

and the linearised Euler equation further requires

~∇∂ψ

∂t+~∇P′

ρ0= 0 , (8.62)

which means that, up to an irrelevant constant, the pressure pertur-bation is

P′ = −ρ0∂ψ

∂t; (8.63)

we thus find the wave equation

∂2ψ

∂t2 −

(∂P∂ρ

)s

~∇2ψ = 0 ; (8.64)


8.3.2 Sound Speed

• this is the usual d’Alembert equation, whose solutions are arbi-trary functions f (x) which propagate with a velocity cs in eitherdirection, f (x ± cst), where

cs ≡

[(∂P∂ρ

)s

]1/2

(8.65)

is the sound speed; obviously,

∂2 f (x ± cst)∂t2 =

∂[±cs f ′(x ± cst)]∂t

= c2s f ′′(x ± cst) (8.66)

and∂2 f (x ± cst)

∂x2 = f ′′(x ± cst) , (8.67)

if the primes denote derivatives of f (x) with respect to its argument;these solutions represent arbitrarily shaped waves propagating atthe sound speed cs into the positive or negative x direction;

• the condition imposed during the derivation of this wave equationwas that the perturbations are small; since

v′x =∂ψ

∂x= f ′(x ± cst) (8.68)

as well asP′ = −ρ0

∂ψ

∂t= ∓ρ0 f ′(x ± cst) (8.69)

andP′ = c2

sρ′ , (8.70)

we findv′x = ∓

P′

ρ0cs= ∓

csρ′

ρ0; (8.71)

thus, this condition is satisfied as long as |v| cs, i.e. for sub-sonicflows;

• sound waves describe how small perturbations propagate throughthe fluid; because of ~v′ = ~∇ψ, they oscillate in propagation directi-on, i.e. they are longitudinal waves; however, we have to take intoaccount that entropy is conserved in an ideal fluid,

dsdt

= 0 (8.72)

along flow lines, which means that entropy perturbations can on-ly propagate with the flow velocity ~v since they have to remainconstant along the flow lines; the same holds for the circulation ofbarotropic fluids;

• in viscous fluids, sound waves are dissipated, i.e. they are conver-ted to heat;

8.4. SUPERSONIC FLOWS 99

8.4 Supersonic Flows

8.4.1 Mach’s Cone; the Laval Nozzle

• perturbations propagate with the sound speed into all directionsrelative to the fluid, i.e. relative to a coordinate frame which isco-moving with the fluid; within a given time t, they reach allpoints around their origin with radius cst;

• if the fluid is flowing with velocity ~v, it depends on the modulus |~v|where perturbations may propagate to; the situation for |~v| < cs isillustrated in a diagram (to be inserted);

• sound waves can still propagate into all directions, but the situationchanges according to another diagram (to be inserted) if |~v| > cs;then, as seen from the laboratory frame, sound waves can onlyreach points within a cone with the half opening angle

α = arcsincs

v; (8.73)

this means that sound waves cannot reach an area in the directionof the flow because they are passed by the flow; from the pointof view of a body which is at rest in the laboratory frame, thisimplies that the flow is meeting with the body “blindly”, withouthaving been “informed” about its presence by sound waves; thishas far-reaching implications;

• as an example for the (steady) transition from sub- to supersonicflow, we consider a nozzle with variable circular cross section A;mass conservation requires

ρvA = const. (8.74)

ord(ρv)ρv

= −dAA

; (8.75)

from Euler’s equation, we obtain in the stationary case

~v ·d~vdt

= −~v · ~∇Pρ

= −1ρ

dPdt

, (8.76)

for ∂P/∂t = 0 due to the assumed stationarity; in the rotationally-symmetric and thus effectively one-dimensional case considered,this implies

vdvdt

= −1ρ

dPdt

⇒

vdv = −dPρ

= −1ρ

(∂P∂ρ

)sdρ = −

c2s

ρdρ , (8.77)


and thusdρρ

= −vdvc2

s; (8.78)

• with this result, we obtain

dAA

= −vdρ + ρdv

ρv= −

dρρ−

dvv

= −

(1 −

v2

c2s

)dvv

(8.79)

ordAA

= −dρρ

(1 −

c2s

v2

); (8.80)

• we first consider a flow which enters with v < cs into a nozzlewhich widens in the flow direction; in this case, dA/A > 0 and(

1 −c2

s

v2

)> 0 , (8.81)

thus dv/v < 0; therefore, the flow decelerates, but becomes denser,dρ/ρ > 0;

• the reverse case happens if the nozzle narrows in the flow direction,dA/A < 0; then, dv/v > 0 and the flow accelerates;

• exactly the opposite occurs if the flow is supersonic on entrance,v > cs, since then

dAA

> 0 ⇒dvv> 0 (8.82)

anddAA

< 0 ⇒dvv< 0 , (8.83)

i.e. a supersonic flow accelerates in a widening nozzle; we nowconsider a nozzle which is assembled as follows (graphic to beinserted); if sub-sonic gas is entering the nozzle, v < cs, it ac-celerates; if it remains subsonic up to the smallest cross section,its velocity decreases again; however, if it reaches v = cs at thenarrowest cross section of the nozzle, the gas accelerates furtherbeyond the sound speed;

8.4.2 Spherical Accretion

• now we consider the example of spherical accretion, i.e. of a bodyof mass M with radius R which is embedded in a gas cloud fromwhich it attracts mass; the continuity equation requires in thisspherically-symmetric case

1r2

∂

∂r(r2ρv) = 0 (8.84)


if the accretion flow is stationary and thus ∂ρ/∂t = 0; the quantityr2ρv is therefore spatially constant, and we set it equal to theaccretion rate,

4πr2ρv ≡ −M , (8.85)

i.e. the mass accepted by the central body per unit time, whichis defined with a negative sign because v < 0; likewise for astationary, spherically-symmetric flow, Euler’s equation says

(~v · ~∇)~v = vdvdr

= −1ρ

dPdr−

GMr

, (8.86)

because the gravitational potential is

Φ = −GM

r⇒

dΦ

dr= ~er · ~∇Φ =

GMr2 ; (8.87)

• we now further assume a polytropic equation-of-state, i.e. P ∝ ργ

with the adiabatic index 1 ≤ γ ≤ 5/3; moreover, the pressuregradient can be written as

dPdr

=dPdρ

dρdr

= c2sdρdr

; (8.88)

• we now use these three equations and rewrite them; first, we canconclude from the continuity equation

1r2

d(r2ρv)dr

= 0 =1r2

[ρ

d(r2v)dr

+ r2vdρdr

]=

ρ

r2

d(r2v)dr

+ vdρdr

, (8.89)

thus1ρ

dρdr

= −1

r2v

d(r2v)dr

; (8.90)

• with this result, Euler’s equation can be written as

vdvdr

+c2

s

ρ

dρdr

= vdvdr−

c2s

r2v

d(r2v)dr

+GMr2 = 0 , (8.91)

where we can use

vdvdr

=12

dv2

dr(8.92)

to arrive at the more convenient equation

12

(1 −

c2s

v2

)dv2

dr= −

GMr2

(1 −

2c2s r

GM

); (8.93)


• far away from the central mass, the term(1 −

2c2s r

GM

)(8.94)

is negative, thus the right-hand side of (8.93) is positive; sinced(v2)/dr < 0 in order to have inflowing gas, the factor 1 − c2

s/v2

must be negative, and hence the flow at large radii must be sub-sonic, v < cs;

• the gas flows towards the accreting object and will reach a point atsufficiently small rc where(

1 −2c2

s rGM

)= 0 , (8.95)

from which the critical radius rc can be read off to be

rc =GM2c2

s; (8.96)

inserting the sound speed from

c2s =

dPdρ

=[.P0(ρ/ρ0)γ]

dρ=γP0

ργ0

ργ−1

=γPρ

=γnkTρ

=γkTm

(8.97)

yields

rc =GMm2γkT

; (8.98)

• this critical radius is typically far beyond the radius of the centralobject, i.e. the accretion flow passes into the supersonic regimethere;

• Euler’s equation can be integrated; first,∫drρ

dPdr

=

∫drρ

ddr

[P0

(ρ

ρ0

)γ]=γP0

ργ0

∫dr ργ−2 dρ

dr

=γP0

ργ0

∫dr

ddr

(ργ−1

γ − 1

)=

γP0ργ−1

(γ − 1)ργ0, (8.99)

then, we can use the squared sound speed

c2s =

γP0

ργ0

ργ−1 (8.100)

from (8.97) to find ∫drρ

dPdr

=c2

s

γ − 1, (8.101)

where c2s is a function of radius r, of course;


• thus, Euler’s equation implies

v2

2+

c2s

γ − 1−

GMr

= const. ≡ C ; (8.102)

• since v→ 0 at r → ∞, the constant must equal

C =c2

s (∞)γ − 1

, (8.103)

but on the other hand we must satisfy at the critical radius

v = cs at r = rc =GM2c2

s, (8.104)

so that we can conclude from (8.102)

c2s (∞)γ − 1

=c2

s (rc)2

+c2

s (rc)γ − 1

− 2c2s (rc) , (8.105)

or

c2s

(1

γ − 1− 2 +

12

)=

c2s (∞)γ − 1

, (8.106)

and thus

cs(rc) = cs(∞)[

11 − 3/2(γ − 1)

]1/2

= cs

√2

5 − 3γ; (8.107)

• because of the proportionality c2s ∝ ρ

γ−1, the density follows from

ρ(rc) = ρ(∞)[

cs(rc)cs(∞)

]2/(γ−1)

; (8.108)

this also specifies the accretion rate, since

M = 4πr2ρv = const. = 4πr2cρ(rc)cs(rc) , (8.109)

and therefore

M = 4πr2cρ(∞)

(2

5 − 3γ

)1/(γ−1)

cs(∞)(

25 − 3γ

)1/2

= πG2M2 ρ(∞)c3

s (∞)

(2

5 − 3γ

)1/(γ−1)+1/2−2

= πG2M2 ρ(∞)c3

s (∞)

(2

5 − 3γ

)(5−3γ)/[2(γ−1)]

; (8.110)

• the solution for the velocity finally follows from mass conservati-on,

−4πr2ρv = M ⇒ (8.111)

v =−M

4πr2ρ(r)=

−M4πr2ρ(∞)

[cs(∞)cs(r)

]2/(γ−1)

;


• inserting this into equation (8.102) yields the solution for v(r),and mass conservation then yields ρ(r); indeed, the maximumaccretion rate at given r is reached exactly when that radius iscritical, r = rc; i.e. if the flow velocity reaches the sound velocitythere;

Kapitel 9

Shock Waves and the SedovSolution

further reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapter 15 and17; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophy-sical Processes”, sections 8.10–8.12; Landau, Lifshitz, “Theore-tical Physics, Vol VI: Hydrodyna-mics”, chapter IX

9.1 Steepening of Sound Waves

9.1.1 Formation of a Discontinuity

• as long as perturbations remain small in the sense that δρ ρ, δP P and δv v, they propagate with the sound speedand retain their initial (arbitrary) shape, as guaranteed by thed’Alembert equation governing them;

• in a polytropic gas, the pressure is P ∝ ργ and thus the sound speedcs ∝ ρ(γ−1)/2 increases if the density of the medium increases;waves thus run faster in denser regions and can even pass lessdense parts of the waves;

• the density varies along the wave; in denser regions, the wavepropagates faster, which makes the wave front steepen; however,since the density must be unique at any location, this behaviourmust lead to discontinuities, i.e. to sharp density jumps;

• we analyse this behaviour in the simple case of an isentropic flow,which means s = const. and the same everywhere in the fluid; inthis case, the pressure is a function of density alone, P = P(ρ), andlikewise the velocity along the wave can be written as a functionof density alone, v = v(ρ); for a wave propagating into the positivex direction, the continuity equation requires

∂ρ

∂t+∂(ρv)∂x

= 0 =∂ρ

∂t+

d(ρv)dρ

∂ρ

∂x; (9.1)

• similarly, Euler’s equation requires

∂v

∂t+ v

∂v

∂x+

1ρ

∂P∂x

= 0 =∂v

∂t+

(v +

1ρ

dPdv

)∂v

∂x; (9.2)

105

106 KAPITEL 9. SHOCK WAVES AND THE SEDOV SOLUTION

• the total change in density is

dρ =∂ρ

∂tdt +

∂ρ

∂xdx , (9.3)

i.e. the change in x at constant ρ is(∂x∂t

)ρ

= −

(∂ρ

∂t

) (∂ρ

∂x

)−1

=d(ρv)

dρ= v + ρ

dvdρ

, (9.4)

where (9.1) was used;

• the change of x at constant v follows similarly,

dv = 0 =∂v

∂tdt +

∂v

∂xdx ⇒(

∂x∂t

)v

= −

(∂v

∂t

) (∂v

∂x

)−1

, (9.5)

or, with the help of Euler’s equation (9.2),(∂x∂t

)v

= v +1ρ

dPdv

; (9.6)

• however, since ρ is a function of v alone, the change of x withconstant ρmust equal the change of x with constant v, and thereforethe two partial derivatives (9.4) and (9.6) must equal,

v + ρdvdρ

= v +1ρ

dPdv

= v +1ρ

dPdρ

dρdv

; (9.7)

• from this, we conclude (dvdρ

)2

=c2

s

ρ2 , (9.8)

hence the velocity can be expressed as

v = ±

∫csdρρ

= ±

∫dPρcs

; (9.9)

• for our polytropic equation of state,

P = P0

(ρ

ρ0

)γand P =

ρkTm

, (9.10)

and the squared sound speed is

c2s =

∂P∂ρ

= γP0ργ−1

ργ0

= P0γ

ρ

(ρ

ρ0

)γ=γkTm

, (9.11)

9.1. STEEPENING OF SOUND WAVES 107

thus c2s ∝ T , and according to (9.10) the temperature scales with

the density as

T ∝ ργ−1 or ρ ∝ T 1/(γ−1) , (9.12)

which can be used to express the density in terms of the soundspeed,

ρ = ρ0

(cs

c0

)2/(γ−1)

; (9.13)

• the velocity follows from (9.6), which can be combined with (9.8)to find x,(

∂x∂t

)v

= v +1ρ

dPdv

= v +1ρ

dPdρ

dρdv

= v ± cs , (9.14)

which yieldsx(v, t) = (v ± cs)t + f (v) , (9.15)

where f (v) is an arbitrary function of v to be specified by theboundary conditions;

• with

dρρ

= d ln(ρ

ρ0

)= d ln

( cs

c0

)2/(γ−1)=

2γ − 1

d lncs

c0=

2γ − 1

dcs

cs, (9.16)

the sound speed can be rewritten from (9.9)

v = ±

∫csdρρ

= ±2(cs − c0)γ − 1

⇒ cs = c0 ±γ − 1

2v ; (9.17)

• putting this into (9.15) finally yields

x =

(±c0 +

γ + 12

v

)t + f (v) ; (9.18)

9.1.2 Specific Example

• to give an example, we study an infinitely long pipe along thex direction which is closed from the left side with a piston andfilled with gas from the right side; until t = 0, the piston be atrest at x = 0, and it be accelerated into the pipe with a constantacceleration a afterwards;


• the velocity and position of the piston are thus

vp = at , xp =a2

t2 , (9.19)

and since the velocity of the gas has to equal the velocity of thepiston vp at the piston’s position xp, we find from (9.18) at theposition of the piston

a2

t2 =

(c0 +

γ + 12

at)

t + f (v) ; (9.20)

using t = v/a, we can solve for f (v),

f (v) = −

(c0 +

γv

2

)v

a= −

c0v

a−γv2

2a, (9.21)

which gives us the general solution for the relation between x andt to the right of the piston

x =

(c0 +

γ + 12

v

)t −

c0v

a−γv2

2a; (9.22)

• the velocity at position x and time t is thus determined by

γv2

2a+

(c0

a−γ − 1

2t)v − (c0t − x) = 0 , (9.23)

which can be solved for v to yield

v =1γ

(γ + 1

2at − c0

)±

√(γ + 1

2at − c0

)2

+ 2γa(c0t − x)

;

(9.24)this is the velocity of the gas for all points x to the right of thepiston, i.e. for x ≥ at2/2;

• for x = 0 and t = 0, the velocity must vanish, which selects fromthe two branches of the solution (9.24) the one with the positivesign,

v =1γ

(γ + 1

2at − c0

)±

√(γ + 1

2at − c0

)2

+ 2γa(c0t − x)

;

(9.25)

• a discontinuity is formed where the velocity changes suddenly,which is where (

∂v

∂x

)t= ∞ ⇒

(∂x∂v

)t= 0 (9.26)

9.1. STEEPENING OF SOUND WAVES 109

holds; physically, this means that parts of the wave with differentvelocities meet at the same position; differentiating (9.22) withrespect to v, we see that this happens in our example when

γ + 12

t −c0

a−γv

a= 0 (9.27)

at the timetc =

2a(γ + 1)

(c0 + γv) ; (9.28)

there, the velocity drops to zero, v = 0, and thus

tc =2c0

a(γ + 1); (9.29)

• specifically, setting γ = 5/3 and a = c0/τ with a characteristicacceleration time scale τ, the gas velocity is

v =3c0

5

(

4t3τ− 1

)+

√(4t3τ− 1

)2

+103

(tτ−

xc0τ

) , (9.30)

where x must obey

x ≥at2

2=

c0t2

2τ=

c0τ

2

( tτ

)2⇒

xc0τ≥

12

( tτ

)2; (9.31)

• now, we define

xc0τ−θ2

2≡ ξ ;

tτ≡ θ ;

v

c0≡ η , (9.32)

which enables us to write (9.22) as

η =35

(4θ3− 1

)+

√(4θ3− 1

)2

+103

(θ −

θ2

2− ξ

) , (9.33)

where ξ ≥ 0 is now the distance from the piston in units of c0τ;

• the discontinuity occurs at the time

tc =3τ4

or θc =34

; (9.34)

since the velocity must vanish there,

v = 0 = η , (9.35)

the position of the discontinuity is determined by

θc −θ2

c

2= ξc ⇒ ξc =

1532

; (9.36)

• the sound speed is

cs = c0 +γ − 1

2v = c0

(1 +

η

3

)(9.37)

and thus the density is

ρ

ρ0=

(1 +

η

3

)3; (9.38)


9.2 Shock Waves

9.2.1 The Shock Jump Conditions

• shock waves occur when a supersonic flow hits an obstacle, forinstance because it impinges on a solid, resting body, or becausea body moves at supersonic speeds through the fluid; in the im-mediate vicinity of the body, the velocity must drop tp zero, andbecause the flow is supersonic, no information on the obstacle canpropagate upstream against the flow;

• we approximate the discontinuity as a plane and consider a super-sonic fluid flow perpendicularly hitting the plane from the left; letρ1, P1 and v1 be the density, pressure and velocity of the fluid tothe left of the discontinuity, and ρ2, P2 and v2 be to its right;

• deriving Euler’s equations, we had identified the following cur-rents:

ρ~v mass current(~v2

2 + w)ρ~v energy current

ρviv j + Pδi j momentum current(9.39)

• when specialised to the situation considered here, in which thediscontinuity is perpendicular to the flow along the x axis, theconservation of these currents requires the following conditions

ρ1v1 = ρ2v2(~v2

1

2+ w1

)ρ1~v1 =

(~v2

2

2+ w2

)ρ2~v2

ρ1v21 + P1 = ρ2v

22 + P2 ; (9.40)

• the enthalpy and the sound speed for a polytropic gas are

w =γ

γ − 1Pρ, c2

s =γPρ

; (9.41)

• writing the velocity left of the discontinuity as

v1 =M1cs (9.42)

with the Mach numberM1 > 1, the equations (9.40) can be re-written in the following way

ρ2

ρ1=

(γ + 1)M21

(γ + 1) + (γ − 1)(M21 − 1)

=v1

v2

P2

P1=

(γ + 1) + 2γ(M21 − 1)

γ + 1, (9.43)

and for an ideal gas with P = ρkT/m, the temperature ratio is

T2

T1=

P2

P1

ρ1

ρ2; (9.44)

9.2. SHOCK WAVES 111

• by construction, the Mach numberM1 > 1, which implies ρ2 > ρ1

because

(γ + 1) + (γ − 1)(M21 − 1) = 2 + (γ + 1)M2

1 − 2M21

= 2(1 −M21) + (γ + 1)M2

1

< (γ + 1)M21 , (9.45)

and therefore the density ratio from (9.43) is larger than unity;correspondingly, v2 < v1, P2 > P1 and T2 > T1;

• in the limiting case of a highly supersonic flow,M1 → ∞ and

limM1→∞

ρ2

ρ1=γ + 1γ − 1

; (9.46)

for a gas with adiabatic index γ = 5/3, the maximum density ratiois therefore

γ + 1γ − 1

= 4 , (9.47)

which is called the maximum shock strength; in the same limit,

P2

P1→ ∞ ,

T2

T1→ ∞ ; (9.48)

9.2.2 Propagation of a One-Dimensional Shock Front

• now we consider a fluid pipe with a piston, which remains at restat x = 0 until t = 0 and is then instantly accelerated to a velocity uinto the positive x direction;

• a discontinuity forms at t = 0 which propagates with a velocity vs

to the right; then, there exists a region ahead of the shock where thedensity, pressure and temperature still have their original valuesρ1, P1 and T1; in the region between the shock and the piston, thegas moves with the velocity u of the piston; the difference of thevelocities between the two regions is thus u;

• in order to use the jump conditions derived before, we need totransform into a coordinate frame in which the shock is at rest; inthis (primed) frame, the gas velocities are obviously

v′1 = −vs ; v′2 = u − vs = u + v′1 , (9.49)

and the velocity difference must remain the same, of course,

v′2 − v′1 = u (9.50)


• eliminating the Mach numberM1, the jump conditions (9.43) forthe density ρ and the pressure P can be combined to read

ρ2

ρ1=

P2(γ + 1) + P1(γ − 1)P2(γ − 1) + P1(γ + 1)

; (9.51)

the matter current is j = ρ1v′1, thus

j2 = ρ21v′21 = ρ2

2v′22 , (9.52)

and therefore

j2 =12

(ρ2

1v′21 + ρ2

2v′22

)=

12

[j2 + (P2 − P1)ρ1 + ρ1ρ2v

′22

], (9.53)

where we have used the jump condition for the momentum from(9.40); we thus obtain

j2 =12

[j2 + (P2 − P1)ρ1 + j2ρ1

ρ2

](9.54)

orj2 =

P1 − P2

ρ1 − ρ2ρ1ρ2 ; (9.55)

• the velocity difference can be written as follows:

v′2 − v′1 = u =

ρ1v′2 − ρ1v

′1

ρ1=

1ρ1

(ρ1

ρ2ρ2v′2 − ρ1v

′1

)= j

ρ1 − ρ2

ρ1ρ2; (9.56)

with (9.55), this turns into

u =

[(P1 − P2)(ρ1 − ρ2)

ρ1ρ2

]1/2

(9.57)

• according to (9.51), the density ratio is

ρ1 − ρ2

ρ1ρ2=

1ρ1

(ρ1

ρ2− 1

)=

1ρ1

[P1(γ + 1) + P2(γ − 1)P1(γ − 1) + P2(γ + 1)

− 1]

=2ρ1

P1 − P2

P1(γ − 1) + P2(γ + 1), (9.58)

which allows us to write the velocity difference as

u = cs|1 − π|

√2

γ(γ − 1) + πγ(γ + 1), (9.59)

where π ≡ P2/P1 is the pressure ratio, and we have expressed theratio P1/ρ1 in terms of the sound speed cs in the unshocked gasahead of the piston from (9.41), P1/ρ1 = c2

s/γ;

9.2. SHOCK WAVES 113

• this can be re-written as a quadratic equation for the pressure ratioπ,

π2 − π

[2 +

γ(γ + 1)u2

2c2s

]+

[1 −

γ(γ − 1)u2

2c2s

]= 0 (9.60)

which has the solutions

π = 1 +γ(γ + 1)u2

4c2s

(9.61)

±

√(1 +

γ(γ + 1)u2

4c2s

)2

−

(1 −

γ(γ − 1)u2

2c2s

);

the pressure ratio needs to exceed unity, π ≥ 1, which excludesthe negative branch; the solution for the pressure ratio can thus besimplified to

π = 1 +γ(γ + 1)u2

4c2s

+γucs

√1 +

(γ + 1)2u2

16c2s

; (9.62)

note that, if the piston is at rest, u = 0 and π = 1, as expected;

• equations (9.56) and (9.58) together yield v′1,

u = 2v′11 − π

γ − 1 + π(γ + 1), (9.63)

which can be simplified by means of (9.59) to read

v′1 = −cs√2γ

√γ − 1 + π(γ + 1) ; (9.64)

this is the velocity of the unshocked gas in the rest frame of theshock front; obviously, the velocity of the shock front in the restframe of the unshocked gas is

vs = −v′1 , (9.65)

hence (9.65) also yields the (negative) velocity of the shock frontin our laboratory system;

• the physical conditions in the unshocked gas, expressed by (ρ1, P1,T1)and the velocity u of the piston, first yield π ≡ P2/P1 from (9.62),from which ρ2 and T2 immediately follow using the shock jumpconditions; the velocity of the shock relative to the velocity of thepiston is given by (9.64); if u cs, we find

π ≈γ(γ + 1)u2

2c2s

, (9.66)

and the shock velocity becomes

vs ≈γ + 1

2u & u , (9.67)

i.e. for a gas with γ = 5/3, the shock moves ∼ 33% faster than thepiston;


9.2.3 The Width of a Shock

• one of our assumptions was that the energy is conserved across thediscontinuity of a shock front; this is not necessarily so because itcan be transported away due to radiative losses, friction, diffusionof fast particles and similar processes;

• if the energy remains conserved, the shock front is called adiaba-tic; otherwise, viscous and thermal effect need to be taken intoconsideration; energy conservation then demands

const. = ρ~v

(~v2

2+ w

)− κ~∇T (9.68)

+ ζ~v(~∇ ·~v) +23η~v(~∇ ·~v) − 2η(~v · ~∇)~v ,

which can be specialised to our case of a one-dimensional shockfront,

ρv

(v2

2+ w

)− κ

∂T∂x−

(43η − ζ

)v∂v

∂x= const. ; (9.69)

• we now assume that η dominates over ζ and put in coarse ap-proximation ζ = 0; moreover, the viscous friction term must becomparable to the kinetic energy if it is to play an important role;thus

ρvv2

2≈

4η3v∂v

∂x; (9.70)

if the velocity changes by ∆v across a distance ∆x, we estimate

ηv∂v

∂x≈ ηv

∆v

∆x≈ ρvν

∆v

∆x, (9.71)

from which we can estimate ∆x,

ρv2

2≈

4η3ρν

∆v

∆x⇒ ∆x ≈

8ν3v

∆v

v; (9.72)

across a strong shock, ∆v ≈ v, which yields the shock width

∆x ≈8ν3v

; (9.73)

since the viscosity is ν ≈ λv, the shock with turns out to be ∆x ≈8λ/3, i.e. it is a factor of order unity times the mean-free pathlength;

9.3. THE SEDOV SOLUTION 115

9.3 The Sedov Solution

9.3.1 Dimensional Analysis

• one example for a shock wave is given by an explosion, i.e. by anevent in which in very short time energy is being released within avery small volume; we consider such an event under the followingsimplifying assumptions: (1) the shock is very strong, meaningthat the pressure of the surrounding medium can be neglected,P1 P2; (2) the energy E is released instantaneously; and (3) theenergy of the surrounding material is negligible compared to E,i.e. the explosion energy dominates that of the surroundings; andfinally (4) the gas be polytropic with an adiabatic index γ;

• under these conditions, our shock jump condition for the density is

ρ1

ρ2=

P1(γ + 1) + P2(γ − 1)P1(γ − 1) + P2(γ + 1)

≈γ − 1γ + 1

, (9.74)

which implies that ρ1 and ρ2 are completely determined by oneanother; the behaviour of the shock must thus be entirely determi-ned by the explosion energy E and the surrounding matter densityρ1;

• if we now consider the shock at a time t when it has reached theradius R(t), the only quantity with the dimension of a length, whichcan be formed from E, t and ρ1 is(

Et2

ρ1

)1/5

; (9.75)

which makes us set

R(t) = R0

(Et2

ρ1

)1/5

(9.76)

with a dimension-less constant R0 which remains to be determined;

• the shock velocity is obviously

vs =dRdt

= R0

(Eρ1

)1/5 2t2/5−1

5=

25

Rt

; (9.77)

• we now use the jump conditions which we had obtained for thepiston in the tube; first, the velocity of the “piston” is, accordingto (9.67),

u =2vs

γ + 1, (9.78)


from which (9.66) yields for the pressure within (“behind”) theshock

P2 = P1γ(γ + 1)

2u2

c2s

= P12γv2

s

γ + 1

(γP1

ρ1

)−1

=2ρ1v

2s

γ + 1, (9.79)

where the sound speed (9.41) in the surrounding, unshocked medi-um was inserted;

• our earlier expression (9.74) for the density shows that the densityinside the shock remains constant, because ρ1 is constant; sincethe shock velocity (9.77) drops with time like

vs ∝ t2/5−1 = t−3/5 , (9.80)

the pressure inside the shock drops like

P2 ∝ t−6/5 , (9.81)

and the velocity of the gas behind the shock is

u ∝ vs ∝ t−3/5 ; (9.82)

• we can interpret these relations as follows: a shock wave drivenby the release of the energy E, which propagates outward with thetime-dependent radius R(t), collects material with mass

M ≈ ρ1R3 ; (9.83)

this material is accelerated from zero to a velocity ≈ R/t, such thatthe kinetic energy

ρ1R5

t2 (9.84)

is put into the collected material; the energy of the material behind(“within”) the shock is thus approximately

ρ1R2R3 ≈ ρ1R5

t2 ; (9.85)

equating this to the energy E, we immediately find

R =

(Et2

ρ1

)1/5

, (9.86)

i.e. the scaling relation (9.76) simply expresses energy conservati-on within the shock;

• we now know how the velocity, the radius, the pressure and thedensity at the shock; they are completely determined by the releaseof an amount of energy E into surrounding material with thedensity ρ1 whose energy can be neglected;


9.3.2 Similarity Solution

• the external radius of such an explosion is given by (9.86), whichsuggestes introducing

ξ ≡rR

(9.87)

as a dimension-less radial variable; we will now use ξ to expressthe radius in v(r, t), ρ(r, t) and P(r, t) and solve the hydrodynamicequations to determine the properties of the gas everywhere withinthe shock;

• the velocity at the shock is given by (9.76); imitating this behaviour,we put

v(ξ, t) =25

rtV(ξ) (9.88)

with a dimension-less function V(ξ) which needs to be determined;the gas velocity at the inner rim of the shock, given by (9.78),requires that V(ξ) satisfy the boundary condition

V(1) =2

γ + 1; (9.89)

• similarly, we use the ansatz

ρ = ρ1G(ξ) (9.90)

for the density and must, because of (9.74), satisfy the boundarycondition

G(1) =γ + 1γ − 1

(9.91)

for the as yet unknown function G(ξ);

• we finally express the pressure by the sound speed, using (9.42)together with (9.74), (9.77) and (9.79) to write

c2s =

γP2

ρ2=

2γρ1

(γ + 1)ρ2v2

s =2γ(γ − 1)(γ + 1)2

(25

Rt

)2

, (9.92)

which justifies the ansatz

c2s =

425

r2

t2 Z(ξ) , (9.93)

where Z(ξ) must satisfy the boundary condition

Z(1) =2γ(γ − 1)(γ + 1)2 ; (9.94)


• now, we can use energy conservation again to relate Z and V;the energy E within the shock must remain constant because noenergy can flow at supersonic velocities; since the solutions, ho-wever, have to be self-similar because they are expressed by thedimension-less radius ξ, energy must also be conserved within anyother sphere with ξ , 1, because ξ , 1 gives the shock positionat another than the considered instant of time; consequently, wecan set up the energy balance for any sphere and require that theenergy remain constant;

• per time interval dt, a sphere with radius r loses the energy

4πr2ρv

(v2

2+ w

)dt , (9.95)

while it gains energy by growing by an amount

4πr2vs dt = 4πr2(2r5t

)dt , (9.96)

incorporating the additional amount of energy

4πr2(2r5t

)dt

(ε +

v2

2

)ρ ; (9.97)

the energy balance then implies

v

(v2

2+ w

)=

2r5t

(v2

2+ ε

), (9.98)

and since the enthalpy is

w =ε + Pρ

=γ

γ − 1Pρ

=c2

s

γ − 1, (9.99)

the thermal energy per unit mass is

ε =ε

ρ=

1γ − 1

Pρ

=c2

s

γ(γ − 1); (9.100)

• this implies with (9.98)

v

(v2

2+

c2s

γ − 1

)=

2r5t

(v2

2+

c2s

γ(γ − 1)

), (9.101)

where we now insert the ansätze for v and c2s , (9.88) and (9.93);

the result can be written as

V(

Zγ − 1

+V2

2

)=

Zγ(γ − 1)

+V2

2, (9.102)

from which follows

Z =γ(γ − 1)(1 − V)V2

2(γV − 1); (9.103)


• the hydrodynamic equations which we now have to solve are ofcourse the continuity, Euler, and energy conservation equations;when specialised to our radially symmetric problem, they read

∂ρ

∂t+∂(ρv)∂r

+2ρvr

= 0

∂v

∂t+ v

∂v

∂r= −

1ρ

∂P∂r(

∂

∂t+ v

∂

∂r

)ln

(Pργ

)= 0 , (9.104)

where the conservation of entropy was used instead of the energyconservation equation;

• inserting G and V into the continuity equation and using∂

∂t=∂ξ

∂tddξ

= −2ξ5t

ddξ

(9.105)

yields after some straightforward manipulation

ξV ′ − (1 − V)ξG′

G+ 3V = 0 (9.106)

• noticing that

Pργ

=Pρρ1−γ =

c2s

γρ1−γ =

4r2ρ1−γ1

25γt2 ZG1−γ (9.107)

and dropping irrelevant constants, the logarithm in (9.104) can bere-written as

lnPργ

= ln(r2

t2 ZG1−γ), (9.108)

and substituting this into the entropy-conservation equation (9.104)yields

ξZ′

Z+ (1 − γ)

ξG′

G+

5 − 2V1 − V

= 0 ; (9.109)

• eliminating ξG′/G from the continuity equation (9.106), this latterequation becomes

ξZ′

Z+

1 − γ1 − V

(3V + ξV ′

)+

5 − 2V1 − V

= 0 , (9.110)

which is supplemented by (9.103), which implies

ξZ′

Z=

(2V−

11 − V

−γ

γV − 1

)ξV ′ ; (9.111)

taken together, (9.110) and (9.111) yield the single equation for V ,

ξV ′

V=γ(1 − 3γ)V2 + (8γ − 1)V − 5γ(γ + 1)V2 − 2(γ + 1)V + 2

; (9.112)

this ordinary first-order differential equation can directly be inte-grated in closed form after separating variables, using the boundarycondition (9.89); having found V(ξ), Z follows from (9.103), andG from (9.107);

Kapitel 10

Instabilities, Convection, HeatConduction, Turbulence

further reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapter 8; Pad-manabhan, “Theoretical Astro-physics, Vol. I: AstrophysicalProcesses”, sections 8.13–8.15;Landau, Lifshitz, “TheoreticalPhysics, Vol VI: Hydrodyna-mics”, chapter III

10.1 Rayleigh-Taylor Instability

• instabilities occur in many different forms in hydrodynamicalsystems; their analysis always proceeds according to the samescheme: one starts from an equilibrium configuration, perturbs itslightly, i.e. in linear approximation, decomposes the perturbationsinto plane waves, ∝ exp[i(ωt −~k · ~x)] and derives the dispersion re-lation ω(k); imaginary frequencies signal the onset of instabilities;

• we first study the situation in which two fluids with the densitiesρ1 and ρ2 are separated by a plane; we choose that plane to be thex-y plane and assume that gravity is directed into the negative zdirection with the acceleration g, corresponding to the gravitationalpotential Φ = gz; finally, the perturbation of the separating planebe described by a function ζ(x), i.e. perturbations are assumed tobe independent of y, without loss of generality;

• since ~v is assumed to be small, we can neglect the curl term inEuler’s equation (8.6) and assume that the velocity is the gradientof a velocity potential ψ; then, we can use Bernoulli’s law in theform (8.41), where we ignore the term ~v2/2 because it is of secondorder in ~v; thus,

∂ψ

∂t+ w + gz = 0 ; (10.1)

• for an incompressible fluid, γ → ∞ and the enthalpy turns into

w =γ

γ − 1Pρ→

Pρ

; (10.2)

multiplying (10.1) with ρ thus yields

ρ∂ψ

∂t+ P + ρgz = 0 ⇒ P = −ρgz − ρ

∂ψ

∂t; (10.3)

121

122KAPITEL 10. INSTABILITIES, CONVECTION, HEAT CONDUCTION, TURBULENCE

• pressure equilibrium at the boundary then implies

ρ1gζ + ρ1∂ψ

∂t

∣∣∣∣∣1

= ρ2gζ + ρ2∂ψ

∂t

∣∣∣∣∣2, (10.4)

which constrains the boundary by

ζ =1

g(ρ1 − ρ2)

(ρ2

∂ψ

∂t

∣∣∣∣∣2− ρ1

∂ψ

∂t

∣∣∣∣∣1

); (10.5)

since the velocity must be continuous across the boundary, thevelocity components in z direction must also agree on both sidesof the boundary, thus

∂ψ

∂z

∣∣∣∣∣2

=∂ψ

∂z

∣∣∣∣∣1

at z = ζ ; (10.6)

• finally, the velocity components in z direction can to first order beidentified with the displacement of the boundary,

∂ζ

∂t=∂ψ

∂z; (10.7)

differentiating the equation (10.5) for the boundary with respect tot, this yields

g(ρ1 − ρ2)∂ψ

∂z= ρ2

∂2ψ

∂t2

∣∣∣∣∣∣2

− ρ1∂2ψ

∂t2

∣∣∣∣∣∣1

(10.8)

• the incompressibility, ~∇·~v = 0, implies that the velocity potential ψsatisfies Laplace’s equation, ~∇2ψ = 0, and since ψ is independentof y by construction, we can set

ψ = f (z) cos(ωt − kx) ; (10.9)

the Laplace equation then demands

d2 fdz2 cos(ωt − kx) − f k2 cos(ωt − kx) = 0 , (10.10)

thus f satisfies an oscillator equation with the usual exponentialsolutions,

d2 fdz2 − k2 f = 0 ⇒ f ∝ e±kz ; (10.11)

• let now h1 and h2 be the heights of the layers, then the velocityneeds to vanish at both z = −h1 and z = h2; this specifies thesolution

ψ1 = A1 cosh [k(z + h1)] cos(kx − ωt) ,ψ2 = A2 cosh [k(z − h2)] cos(kx − ωt) ; (10.12)

10.1. RAYLEIGH-TAYLOR INSTABILITY 123

if we insert these solutions into the equation (10.8) for the pressurebalance, we find

g(ρ1 − ρ2)A1 sinh[k(z + h1)] k cos(kx − ωt) =

− ρ1A1 cosh[k(z + h1)]ω2 cos(kx − ωt)+ ρ2A2 cosh[k(z − h2)]ω2 cos(kx − ωt) ; (10.13)

from (10.13), we obtain the ratio A1/A2,

A2

A1=g(ρ1 − ρ2) k sinh[k(z + h1)] + ρ1 ω

2 cosh[k(z + h1)]ω2 ρ2 cosh[k(z − h2)]

,

(10.14)and a similar expression follows if ψ1 and ψ2 are swapped; equa-ting both yields the dispersion relation

ω2 =kg(ρ1 − ρ2)

ρ1 coth kh1 + ρ2 coth kh2; (10.15)

• if ω2 > 0 as required for a stable situation, ρ1 > ρ2 is obviouslynecessary, which means that the specifically heavier fluid mustlie below the specifically lighter fluid; in this case, perturbationspropagate as waves along the boundary between the fluids;

• if the density of the upper fluid is small compared to the lower, wecan approximate ρ2 = 0 and have

ω2 =kgρ1

ρ1 coth kh1= kg tanh kh1 , (10.16)

which gives the frequency of waves on a fluid under the influenceof gravity whose surface is being perturbed; in the limiting case oflow depth, h1 k−1, this simplifies to

ω2 ≈ k2gh1 , (10.17)

while for large depth, h1 k−1 and

ω2 ≈ kg ; (10.18)

these are the limiting cases of waves on shallow or deep water;

• if ρ2 > 0 and kh1 1 as well as kh2 1, we find

ω2 ≈ kgρ1 − ρ2

ρ1 + ρ2; (10.19)

this is the limiting case of very deep layers; in the limit of longwaves, kh1 1 and kh2 1, the dispersion relation becomes

ω2 ≈ k2 g(ρ1 − ρ2)h1h2

ρ1h2 + ρ2h1; (10.20)

• this linear stability analysis shows that ω becomes imaginary ifthe specifically heavier fluid lies on top of the specifically lighterone, ρ1 < ρ2, which is intuitively obvious;


10.2 Kelvin-Helmholtz Instability

• we now consider a situation in which one fluid flows with a velocity~v parallel to the surface of a fluid at rest, for instance like windover a lake;

• we choose the coordinate system such that the z axis is orthogonalto the boundary surface and direction of motion (i.e. the directionof ~v) points into the x direction;

• writing Euler’s equation in the form

d~vdt

= −~∇Pρ

(10.21)

and taking the divergence, we see that the pressure must satisfyLaplace’s equation

~∇2P = 0 (10.22)

if the fluid is incompressible, ~∇ ·~v = 0 and ρ = const.;

• as before, the ansatz δP = f (z) exp[i(kx − ωt)] leads to the oscilla-tor equation

d2 f (z)dz2 − k2 f (z) = 0 , (10.23)

with the solutions f (z) ∝ exp(±kz); the exponentially growingsolution ∝ exp(kz) is ruled out physically because it diverges atlarge distances from the boundary surface, and thus the pressureperturbation must behave as

δP2 ∝ ekzei(kx−ωt) (10.24)

above the boundary surface;

• to first order in the velocity perturbation δvz along the z axis,Euler’s equation reads

∂δvz

∂t+ v

∂δvz

∂x=

k δP2

ρ2, (10.25)

which, with δvz ∝ exp[i(kx − ωt)], implies

− iωδvz + ikvδvz =kδP2

ρ2, (10.26)

and thus

δvz =kδP2

iρ2(kv − ω); (10.27)

10.2. KELVIN-HELMHOLTZ INSTABILITY 125

• let now again ζ(x, t) be the boundary surface between the twofluids, then we must have

dζdt

= δvz =∂ζ

∂t+ v

∂ζ

∂x(10.28)

to linear order, and the ansatz ζ ∝ exp[i(kx − ωt)] implies

− iωζ + ikvζ = δvz = i(kv − ω)ζ , (10.29)

thus the pressure perturbation δP2 from (10.27) can be written as

δP2 = −ρ2ζ

k(kv − ω)2 ; (10.30)

• on the other side of the boundary surface, we have v = 0 and mustchoose the solution f (z) ∝ exp(kz); inserted into Euler’s equation,this yields

∂δvz

∂t= −

k δP1

ρ1⇒ δvz =

k δP1

iρ1ω=

ik δP1

ρ1ω; (10.31)

• the boundary surface then satisfies the equation

∂ζ

∂t= −iωζ = δvz , (10.32)

or

δP1 =ζρ1ω

2

k; (10.33)

• there must be pressure balance at the boundary, δP1 = δP2, there-fore

ρ1ω2 +ρ2(kv−ω)2 = 0 = (ρ1 +ρ2)ω2−2ρ2kω+ρ2k2ω2 , (10.34)

and this equation has the solutions

ω =2ρ2kv ±

√4ρ2

2k2v2 − 4(ρ1 + ρ2)ρ2k2v2

2(ρ1 + ρ2)

=kv

ρ1 + ρ2

(ρ2 ± i

√ρ1ρ2

); (10.35)

• for ρ1 , 0 , ρ2, ω always has an imaginary part, i.e. the pertur-bation grows; this so-called Kelvin-Helmholtz instability can bedamped by buoyancy forces, if ρ1 > ρ2; of course, a gravitationalfield must then be taken into consideration;

• the time scale for the perturbation to grow is obviously given by[=(ω)

]−1=

ρ1 + ρ2

kv√ρ1ρ2

; (10.36)

if ρ1 ρ2,

ω ≈ ±ikv√ρ2

ρ1(10.37)

follows, which holds for example for wind blowing over water;


10.3 Thermal Instability

• we assume that the system under consideration gains energy byheating and loses energy by cooling; the net cooling rate, i.e. thenet energy loss per unit time, be

L(ρ,T ) ; (10.38)

if the system is in thermal equilibrium, we require L(ρ,T ) = 0,which implicitly defines a relation between ρ and T ; for thermalbremsstrahlung, for example,

L(ρ,T ) = C ρ√

T − (heating) ; (10.39)

• this cooling functionL(ρ,T ) can adopt various forms, in particularbecause cooling processes are often related to thermal occupati-on numbers and atomic or molecular excitations; because of theBoltzmann factor, sometimes small temperature changes can giverise to large changes in occupation numbers, and the atomic ormolecular energy levels introduce discrete thresholds; the curveL(ρ,T ) = 0 thus typically looks as shown in the figure (to beinserted);

• at the same time, let the system be in pressure equilibrium withits surroundings, i.e. the pressure P be externally regulated; for anideal gas, we have

P ∝ ρT , (10.40)

such that pressure equilibrium may be represented by a straightinclined line; in this example, the curve P = const. intersects thecurve L = 0 in three points where both mechanical and thermalequilibrium are possible;

• if the system moves along the curve P = const., it gets out ofthermal equilibrium; the idea behind that is that mechanical isusually established much faster than thermal equilibrium suchthat a system more likely stays in mechanical than in thermalequilibrium if ρ and T are changed, and thus it remains on thecurve P = const.;

• if, at the point A, the density is increased, the system moves to-wards the lower right along the dashed line and thus into the regimewhere L > 0; here, the energy losses are larger than the energygain, the temperature decreases further, the density grows further,and the system moves further away from thermal equilibrium, it isthermally unstable;

• if the same happens in point B, the cooling function becomesnegative, L < 0, the system heats up and moves back into theequilibrium point;

10.3. THERMAL INSTABILITY 127

• we now consider a simple model for this thermal instability; thefundamental equations are

∂ρ

∂t+ ~∇ · (ρ~v) = 0

∂~v

∂t+~∇v2

2−~v × (~∇ ×~v) = −

~∇Pρ

T[∂s∂t

+ (~v · ~∇)s]

= −L(ρ,T ) , (10.41)

where the specific entropy s is

s = cv ln(

Pργ

)(10.42)

up to a constant which is irrelevant here; L is the net energy lossper unit mass, and the equation of state is

P =ρkTm

= νRT (10.43)

with the mol number ν and the gas constant R; the specific heatcapacities at constant volume and constant pressure are

cv =R

γ − 1(10.44)

and

cp = cv + R = R

(1

γ − 1+ 1

)=

γR

γ − 1= γcv (10.45)

• the equilibrium state has ρ = ρ0, T = T0, L(ρ0,T0) = 0 and ~v = 0;we perturb this state by small deviations δρ, δT , δ~v and linearisein these perturbations; this first yields

∂δρ

∂t+ ~∇ · (ρ0δ~v) = 0 ,

∂δ~v

∂t= −

~∇δPρ0

(10.46)

and, by substituting into the partial time derivative of the firstequation the divergence of ρ0 times the second equation,

∂2δρ

∂t2 + ~∇ ·

(ρ0∂δ~v

∂t

)=∂2δρ

∂t2 −~∇2δP = 0 (10.47)

we first allow perturbations with δP , 0 and ask later for theconditions for instability under constant pressure;


• up to the irrelevant constant, the entropy is

s = cv ln(

Pργ

)= cv (ln P − γ ln ρ)

= cv[ln(P0 + δP) − γ ln(ρ0 + δρ)

]= cv

ln

[P0

(1 +

δPP0

)]− γ ln

[ρ0

(1 +

δρ

ρ0

)]= cv

[ln P0 − γ ln ρ0 +

δPP0− γ

δρ

ρ0

]= s0 + cv

(δPP0− γ

δρ

ρ0

)= s0 + cv

δPP0− cp

δρ

ρ0; (10.48)

• this allows us to write the left-hand side of the entropy equation(10.41) as

(T0 + δT )(∂

∂t+ δ~v · ~∇

) (s0 + cv

δPP0− cp

δρ

ρ0

)(10.49)

= T0

(∂

∂t+ δ~v · ~∇

) (s0 + cv

δPP0− cp

δρ

ρ0

)+ δT

∂s0

∂t;

• equilibriums requires that

T0

(∂

∂t+ δ~v · ~∇

)s0 = 0 ⇒ T0

∂s0

∂t= 0 (10.50)

because of ~v0 = 0; this reduces the left-hand side of (10.41) to

T0∂

∂t

(cvδPP0− cp

δρ

ρ0

); (10.51)

• on the right-hand side, we have

− L(ρ0 + δρ,T0 + δT ) = −L(ρ0,T0)︸︷︷︸=0

−∂L

∂ρδρ −

∂L

∂TδT ; (10.52)

since ρ and T are furthermore connected through the equation-of-state, we can write

dL =

(∂L

∂ρ

)T

dρ +

(∂L

∂T

)ρ

dT

=

(∂L

∂ρ

)T

[(∂ρ

∂P

)T

dP +

(∂ρ

∂T

)P

dT]

+

(∂L

∂T

)ρ

dT

=

(∂L

∂T

)P

dTP +

(∂L

∂T

)ρ

dTρ ; (10.53)

at constant pressure or constant density ρ,

dTP = −Tdρρ, dTρ = −T

dPP

, (10.54)

10.3. THERMAL INSTABILITY 129

and thuscv

P0

∂δP∂t−

cv

ρ0

∂δρ

∂t=

(∂L

∂T

)P

δρ

ρ0−

(∂L

∂T

)ρ

δPP0

; (10.55)

• if we apply the Laplacian to this equation and use (10.47) for~∇2δP, we find

∂

∂t

[cv∂2δρ

∂t2 −P0

ρ0cp~∇

2δρ

]=(

∂L

∂T

)ρ

P0

ρ0

~∇2δρ −

(∂L

∂T

)P

∂2δρ

∂t2 ; (10.56)

dividing by cv, recalling thatcp

cv= γ , γ

P0

ρ0= c2

s , (10.57)

and introducing the abbreviations

Np ≡1cp

(∂L

∂T

)P, Nv ≡

1cv

(∂L

∂T

)ρ

, (10.58)

we find the relation∂

∂t

(∂2δρ

∂t− c2

s~∇2δρ

)= Npc2

s~∇2δρ − Nv

∂2δρ

∂t; (10.59)

• again, we expand the perturbations into plane waves, δρ ∝ exp[i(kx−ωt)], and thus transform (10.59) to

∂

∂t

[(c2

s k2 − ω2)δρ

]=

(Nvω

2 − Npc2s k2

)δρ (10.60)

which yields

(c2s k2 − ω2)iω = Nvω

2 − Npc2s k2 ; (10.61)

• this cubic dispersion relation is difficult to solve; in the limitingcase of small wave lengths, c2

s k2 ω2, we can approximate

iω ≈ −Np ⇒ ω ≈ iNp , (10.62)

which indicates an unstable solution if Np < 0, since then

δρ ∝ e−Npt (10.63)

grows exponentially; thermal instability thus sets in when(∂L

∂T

)P< 0 ; (10.64)

• in the opposite limiting case of very large wave length,

iω ≈ −Nv , (10.65)

i.e. instability then requires that(∂L

∂T

)ρ

< 0 ; (10.66)


10.4 Heat Conduction and Convection

10.4.1 Heat conduction

• a system can be in mechanical equilibrium, but out of thermalequilibrium; the most straightforward example is a star whichis being kept in mechanical equilibrium by the balance betweengravity and the pressure gradient, but nonetheless continuouslyradiates energy; the entropy equation reads in this case

ρTdsdt

= ~∇ · (κ~∇T ) + σi j∂vi

∂x j; (10.67)

• if the velocity gradient is too small to drive currents, the secondterm on the right-hand side can be neglected; from

dq = Tds and cp =

(dqdT

)P

(10.68)

followscpdT = Tds ⇒ ds = cpd ln T ; (10.69)

with that, the entropy equation (10.67) can be reduced to

ρcpdTdt

= κ~∇2T ordTdt

= χ~∇2T with χ ≡κ

ρcp; (10.70)

• in analogy to radiative energy transport, we define a conductiveopacity κcond through the conductive energy current

~Fcond = −c

3ρκcond

~∇(aT 4) ≡ −κ~∇T , (10.71)

from which we obtain the relation between heat conductivity κ andconductive opacity κcond,

κ =4caT 3

3ρκcond; (10.72)

• if both radiative and conductive energy transport is present, theeffective opacity is

1κeff

=1κrad

+1

κcond⇒ κeff =

κradκcond

κrad + κcond; (10.73)

• when we discussed the heat conductivity, we saw that it can bewritten as

κ =nv3

cvλ , (10.74)

10.4. HEAT CONDUCTION AND CONVECTION 131

where λ is the mead free path of the fluid particles; if we considerelectrons whose mean free path is determined by ions,

λ =1

niσ, (10.75)

where ni and σ are the number density and the scattering crosssection of the ions;

• typically, an electron will approach an ion up to a distance ri wherethe kinetic and potential energies equal,

mv2

2≈

Ze2

ri⇒ ri ≈

2Ze2

mv2 ; (10.76)

the cross section is then

σ ≈ πr2i , (10.77)

and we obtain the expression

κ =neve

3cv

m2ev

4e

4πniZ2e4 =13

(m2

e

4πZ2e4

) (ne

ni

)cvv

5e (10.78)

for the heat conductivity of electrons scattered by ions;

• in a thermal electron gas, cv = 3k/2 and

ve =3kTe

me, (10.79)

which yields the heat conductivity

κ =k2

(m2

e

4πZ2e4

) (ne

ni

) (3kTe

me

)5/2

(10.80)

for classical (non-degenerate) electrons; if we identify the Thom-son cross section σT, we can alternatively write

κ =3√

3kcZ2σT

(ne

ni

) (kTe

me

)5/2

(10.81)

with the obvious unit

[κ] =erg

cm s K; (10.82)

numerically,

κ ≈ 5.5 × 1012 Z−2(ne

ni

) (kTe

1 keV

)5/2

; (10.83)


10.4.2 Convection

• if the temperature gradient is too large, convection sets in: then,warm, rising bubbles cannot any more cool and return to theiroriginal locations, but continue to rise; we consider the situation,in which a volume V(P, s) characterised by the pressure P and thespecific entropy s rises against the gravitational force;

• we ignore again thermal compared to mechanical adaptation pro-cesses because they are typically slower and assume that the bubblewith volume V(P, s) rises by an amount ∆z, where its volume isV(P′, s); there, its buoyancy force is determined by the volumeV(P′, s′) which the bubble would adopt if it had the specific entro-py of its new environment; the situation is stable if the actualvolume V(P′, s) is smaller than the volume V(P′, s′), because thengravity will dominate the buoyancy force, and the bubble will thensink down again; we thus have the condition

V(P′, s′) > V(P′, s) (10.84)

for stability;

• withs′ = s +

dsdz

∣∣∣∣∣z∆z (10.85)

and because of

cpdT = Tds ⇒

(∂V∂s

)P

=Tcp

(∂V∂T

)P> 0 , (10.86)

we find

V(P′, s′) = V(P′, s) +

(∂V∂s

)P

∆s = V(P′, s) +

(∂V∂s

)P

dsdz

∣∣∣∣∣z∆z ;

(10.87)thus, the stability condition is satisfied if

dsdz

∣∣∣∣∣z> 0 (10.88)

• in an ideal gas with adiabatic index γ, the entropy is

s = cv ln[PT γ/(γ−1)

](10.89)

up to an irrelevant constant; thus

dsdz

=∂s∂P

dPdz

+∂s∂T

dTdz

= cv

[d ln P

dz+

γ

γ − 1d ln T

dz

]; (10.90)

the condition (10.88) then shows that the temperature gradientmust satisfy

−d ln Td ln z

<γ − 1γ

d ln Pd ln z

(10.91)

10.5. TURBULENCE 133

for the gas stratification to be stable against convection; the quan-tity

γ − 1γ≡ ∇ad (10.92)

is often called the adiabatic temperature gradient (“nabla adiaba-tic”); using this, we stability condition is written

−d ln Td ln P

≡ ∇ < ∇ad ; (10.93)

• when convection sets in, it is a very efficient means of transportingheat; viscosity hinders the convective energy transport;

10.5 Turbulence

• hydrodynamical flows with large Reynolds numbers turn out tobe highly unstable; for high viscosity (low Reynolds number),stable solutions of the Navier-Stokes equation exist which developinstabilities above a critical Reynolds number

R =uLν& Rcr ; (10.94)

• a full analysis of such instabilities is very difficult and in generalan unsoved problem; turbulence sets in, in the course of whichenergy is being transported from large to small scales until it isdissipated by the production of viscous heat on sufficiently smallscales;

• let λ be the size of an eddy and vλ the typical rotational velocityacross the eddy; let further ελ be the energy per unit mass in suchan eddy; then, the energy flow through an eddy of that size is

ε ≈

(v2λ

2

)︸︷︷︸

typical energy

(λ

vλ

)−1

︸︷︷︸time scale

≈v3λ

λ(10.95)

• energy is being fed into the turbulent cascade on the macroscopicscale L where the typical velocity is u; from there, the energy cas-cades through the turbulent eddies to progressively smaller scalesuntil it is finally viscously dissipated on a scale λv; in between,i.e. on scales λ satisfying

λs < λ < L , (10.96)

the energy flow ε must be independent of scale because the energycannot be accumulated anywhere; therefore, we conclude from


(10.95) that the typical eddy velocity must change with the eddyscale λ as

vλ ∝ λ1/3 (10.97)

or

vλ ≈ u(λ

L

)1/3

; (10.98)

the largest eddies thus carry the highest velocities, but the smallesthave the highest vorticity,

Ω ≈vλλ≈

u(λ2L)1/3 ; (10.99)

• to estimate the scale λv, we compare the viscous dissipation withthe specific energy flow ε; the viscous heating rate is approximately

hv ≈ ησ2 ≈ η

(vλλ

)2≈ η

(v3λ

λ

)2/3

λ−4/3 = ηε2/3λ−4/3 ; (10.100)

therefore, hv is negligibly small on large scales, but if the heatingrate becomes of order the energy flow rate,

hv ≈ ρε , (10.101)

viscous dissipation sets in; this happens on a length scale λv givenby

ηε2/3λ−4/3v ≈ ρε ⇒ λv =

(η

ρε1/3

)3/4

(10.102)

or, because of

ε ≈u3

L, (10.103)

the viscous scale can be transformed to

λv =

(ηL1/3

ρu

)3/4

= L(ν

uL

)3/4=

LR3/4 , (10.104)

where R is the Reynolds number on the scale L;

• finally, we consider the correlation function of the eddy velocityvλ, or rather its Fourier transform, the power spectrum; since

vλ ≈ (ελ)1/3 , (10.105)

the correlation function scales as

ξv ∝ (ελ)2/3 , (10.106)

while its Fourier transform Pv will then scale as

Pv ∝ λ3ξv ∝ k−3

(εk−1

)2/3∝ ε2/3k−11/3 ; (10.107)

the “power” per logarithmic k interval will thus scale as

k2Pv ∝ ε2/3k−5/3 , (10.108)

which is the Kolmogorov turbulence spectrum;

Kapitel 11

Collision-Less Plasmasfurther reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapters 28 and29; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophysi-cal Processes”, sections 9.1–9.4;Ishimaru, “Basic Principles ofPlasma Physics”, chapters 1–4

11.1 Basic Concepts

11.1.1 Shielding; the Debye length

• we would like to describe plasmas as fluids; this requires thatcollisions be random and short-ranged, such that equilibrium canbe locally established sufficiently fast; the fundamental differencebetween plasma physics and the hydrodynamics of neutral fluidsis the long-ranged Coulomb interaction between the particles;

• a plasma consists of electrons and ions of charge Ze; the existenceof two different types of charge allows shielding on a characteristiclength scale; to calculate it, we put into a plasma with chargedensity ±en± a point charge q; Poisson’s equation for the electro-static potential then reads

~∇2Φ = 4πe(n+ − n−) + 4πqδD(~x) (11.1)

if the charge is placed at the coordinate origin;

• in thermal equilibrium, the charge densities are

n± ∝ exp(∓eΦ

kT

), (11.2)

so that we can write

n+ − n− = n[exp

(eΦ

kT

)− exp

(−

eΦ

kT

)], (11.3)

with the mean particle number density n;

• sufficiently far from the central charge, the argument of the expo-nential

eΦ

kT 1 (11.4)

135

136 KAPITEL 11. COLLISION-LESS PLASMAS

so that we can approximate

n+ − n− ≈ n(2eΦ

kT

)(11.5)

and write Poisson’s equation as

~∇2Φ ≈8πe2n

kTΦ + 4πqδD(~x) , (11.6)

which has the solution

Φ =qr

exp− √2r

λD

, (11.7)

which contains the Debye length defined by

λD ≡

(kT

4πe2n

)1/2

= 6.9 cm(TK

)1/2 ( ncm−3

)−1/2

; (11.8)

this is the characteristic length scale for the shielding of a chargein a plasma;

• a plasma can be considered ideal if it contains sufficiently manyparticles within the Debye length, since then the interaction energyis small compared to the thermal energy; to see that, we compare

e2

rwith kT ; (11.9)

the first expression is the mean potential energy of the interactingcharges, the second is a measure for their thermal energy; r is themean separation of the particles, given by

4π3

r3n ≈ 1 ⇒ r ≈(

34πn

)1/3

; (11.10)

we thus obtain from the comparison between the two energies

e2

kT

(4πn

3

)1/3

=4πne2

3kT

(3

4πn

)2/3

=13

r2

λ2D

1 (11.11)

for λD r;

11.1.2 The plasma frequency

• the mean thermal velocity of the electrons is given by

〈v2〉 ≈kTme

, (11.12)

11.2. THE DIELECTRIC TENSOR 137

which means that an electron passes the Debye length in the time

tD ≈λD√〈v2〉≈

√kT

4πne2

me

kT=

√me

4πne2 ; (11.13)

this is the time sale on which the thermal motion of the electronscan compensate charge displacements by shielding;

• the time tD can be transformed into a characteristic frequency forplasma oscillations,

ωp ≈1tD

=

√4πne2

me≈ 5.6 × 104 Hz

( ncm−3

)1/2, (11.14)

which is the frequency with which charge inhomogeneities canoscillate against each other; with λD and ωp, we now have twoessential parameters for describing plasmas at hand;

11.2 The Dielectric Tensor

11.2.1 Polarisation and dielectric displacement

• external electric fields ~E polarise media, i.e. they induce in thesemedia a charge displacement which can be described by a polarisedcharge density ρpol; the polarisation ~P is defined by

~∇ · ~P = −ρpol ; (11.15)

the Maxwell equation in vacuum,

~∇ · ~E = 4πρ , (11.16)

changes to

~∇ · ~E = 4π(ρ + ρpol) = 4πρ − 4π~∇ · ~P ; (11.17)

the dielectric displacement, ~D ≡ ~E +4π~P, is defined as an auxiliaryfield which satisfies

~∇ · ~D = 4πρ ; (11.18)

• due to charge conservation, the charge density must satisfy

∂ρpol

∂t+ ~∇ · ~jpol = 0 ; (11.19)

if we substitute (11.15) here, we find

~∇ ·

−∂~P∂t

+ ~jpol

= 0 or ~jpol =∂~P∂t

; (11.20)


• the Maxwell equation

~∇ × ~B =1c∂~E∂t

+4πc

(~j + ~jpol

)=

1c

∂~E∂t

+ 4π∂~P∂t

+4πc~j (11.21)

can then be written as

~∇ × ~B =1c∂~D∂t

+4πc~j (11.22)

where ~j is the external (unpolarised) current;

11.2.2 Structure of the dielectric tensor

• we now need a relation between the external field ~E and the pola-risation ~P; for sufficiently weak fields, we assume this relation tobe linear and write for the Fourier-transformed quantities

Di(ω,~k) = ε ij(ω,~k) E j(ω,~k) , (11.23)

defining the dielectric tensor ε ij;

• due to the Fourier convolution theorem, the multiplication (11.23)in Fourier space corresponds to a convolution in real space; thismeans that ~D(~x, t) can also be influenced by fields at earlier timesand other places, ~E(~x − δ~x, t − δt);

• since the fields must remain real in real space, the dielectric tensormust satisfy the symmetry relation

ε ij(−ω,−~k) = ε∗ij (ω,~k) (11.24)

in Fourier space;

• the principal-axis directions of the tensor ε ij can only depend on

the vector ~k, so that we can start from the ansatz

ε ij(ω,~k) = Aδi

j +kik j

k2 B (11.25)

with functions A and B depending on (ω,~k); obviously,(kik j

k2

)k j = ki (11.26)

is parallel to ~k, while(δi

j −kik j

k2

)k j = ki − ki = 0 (11.27)

11.3. DISPERSION RELATIONS 139

and thus perpendicular to ~k; therefore, we split the tensor ε ij into

two parts,

ε ij =

(δi

j −kik j

k2

)εt +

kik j

k2 εl , (11.28)

which we call the transversal and the longitudinal components ofε i

j, with A = εt and B = εl − εt according to the ansatz (11.25);

• of course, εt and εl are generally functions of ω and k which alsoneed to satisfy the symmetry condition (11.24),

εl,t(−ω, k) = ε∗l,t(ω, k) ; (11.29)

• we have neglected in this derivation of the form of the dielectrictensor ε i

j that preferred macroscopic directions may exist in theplasma, e.g. due to magnetic fields ordered on large scales; if theyexist, they must also be built into the dielectric tensor;

• in the limit of long waves, |~k| → 0, the dielectric tensor tends to

ε ij → δi

jεt , (11.30)

and thus turns into the “normal” scalar dielectricity;

11.3 Dispersion Relations

11.3.1 General form of the dispersion relations

• the dielectric tensor determines which kinds of wave can propagatethrough the plasma; we shall now derive the dispersion relationsbetween the wave vectors ~k and the frequencies ω of such possiblewaves;

• if the fields are decomposed into Fourier modes ∝ exp[i(~k ·~x−ωt)],Maxwell’s equations in the plasma read

~k × ~E =ω~Bc

, ~k × ~B = −ω~Dc

,

~k · ~D = 0 , ~k · ~B = 0 , (11.31)

if we neglect external charge densities and currents;

• combining the curl of the first equation (11.31) with the secondyields

~k × (~k × ~E) =ω

c~k × ~B = −

ω2

c2~D , (11.32)


and if we expand the double vector product, we can write thedisplacement vector as

ω2

c2~D = k2 ~E − ~k(~k · ~E) ; (11.33)

we now use the dielectric tensor to write Di = ε ijE

j and find,written in components

ω2

c2 εijE

j = k jk jEi − kik jE j , (11.34)

or (δi

j −kik j

k2 −ω2

c2k2 εij

)E j = 0 ; (11.35)

• the expression in brackets can be represented by a matrix; equa-tion (11.35) has non-trivial solutions ~E , 0 if and only if thedeterminant of that matrix vanishes,

det(δi

j −kik j

k2 −ω2

c2k2 εij

)= 0 (11.36)

11.3.2 Transversal and longitudinal waves

• this condition defines a relation between ~k and ω which must besatisfied for waves allowed to propagate through the plasma; wenow insert ε i

j from (11.28) and find

detδi

j −kik j

k2 −ω2

c2k2

[(δi

j −kik j

k2

)εt +

kik j

k2 εl

]= 0 , (11.37)

or

det[(δi

j −kik j

k2

) (1 − εt

ω2

c2k2

)−ω2

c2

kik j

k4 εl

]= 0 ; (11.38)

• for transversal waves, kiEi = 0, and the corresponding termsdisappear from the matrix equation (11.35); what remains is(

1 − εtω2

c2k2

)Ei = 0 , (11.39)

which implies the dispersion relation

ω2 =c2k2

εt; (11.40)

if, conversely, ~E is parallel to ~k, terms proportional to(δi

j −kik j

k2

)(11.41)

11.4. LONGITUDINAL WAVES 141

do not contribute, and the dispersion relation is reduced to

ω2

c2

kik jE j

k4 εl = 0 , (11.42)

which generally demands that εl = 0; in order to understand thiscondition, we need to determine the form of εl;

11.4 Longitudinal Waves

11.4.1 The longitudinal dielectricity

• in order to determine the form of εl, we return to the collision-lessBoltzmann equation; we neglect the motion of the ions because oftheir lower velocities and write

f = f0 + δ f , (11.43)

i.e. we expect that sufficiently weak fields ~E and ~B will change thephase-space distribution function only little away from a homoge-neous and isotropic distribution function f0; to first approximation,Boltzmann’s equation then reads

∂δ f∂t

+~v · ~∇δ f − e(~E +

~v

c× ~B

)·∂ f0

∂~p= 0 ; (11.44)

for an isotropic distribution f0, we must further have

∂ f0

∂~p‖ ~v (11.45)

because no other direction is defined, thus

(~v × ~B) ·∂ f0

∂~p= 0 , (11.46)

and Boltzmann’s equation in linear approximation shrinks to

∂δ f∂t

+~v · ~∇δ f = e~E ·∂ f0

∂~p; (11.47)

• decomposing ~E and the perturbation δ f into plane waves∝ exp[i(~k·~x − ωt)], (11.47) yields

− iωδ f + i~v · ~kδ f = e~E∂ f0

∂~p, (11.48)

which can be solved for δ f ,

δ f =e~E

i(~k ·~v − ω)·∂ f0

∂~p; (11.49)


• charge and current densities are exclusively caused by δ f becausef0 is a homogeneous, isotropic and stationary equilibrium distribu-tion; therefore,

ρ = −e∫

d3 p δ f , ~j = −e∫

d3 p~vδ f ; (11.50)

these quantities are then also proportional to the phase factorexp[i(~k ·~x−ωt)], and the polarisation equations (11.15) and (11.20)can be written

i~k · ~P = −ρ , −iω~P = ~j ; (11.51)

with δ f from (11.49), we find

i~k · ~P = e2∫

d3 p ~E ·∂ f0

∂~p1

i(~k ·~v − ω); (11.52)

• this integral has a pole at ω = ~k ·~v and is therefore ill-defined; weremedy that by writing

1

i(~k ·~v − ω)→

1

i(~k ·~v − ω − iδ)(11.53)

which has no poles any more on the real axis, and later sending δto zero;

• for longitudinal waves, we have 4π~P = ~D− ~E = (εl − 1) ~E and ~E =

E~k/k; inserting this into (11.52) allows us to write the longitudinalpart εl of the dielectricity as

εl = 1 −4πe2

k2

∫d3 p~k ·

∂ f0

∂~p1

(~k ·~v − ω − iδ); (11.54)

• if we now place the coordinate system such that ~k points into thepositive ~x direction, the integral can be split up; we then have

~k ·∂ f0

∂~p= k

d f0

dpx(11.55)

and ∫d3 p~k ·

∂ f0

∂~p1

i(~k ·~v − ω − iδ)

= k∫

dpxd f (px)

dpx

1i(kvx − ω − iδ)

, (11.56)

with the definition

f (px) ≡∫

dpydpz f0(~p) ; (11.57)

11.4. LONGITUDINAL WAVES 143

11.4.2 Landau Damping

• obviously, the longitudinal dielectricity εl has a real and an imagi-nary part; the latter implies dissipation of electrical energy, as weshall shortly see; to begin with, the dissipation follows from

Q =∂

∂t

~E2

8π

+ ~E · ~j =~E · ~E4π

+ ~E · ~P

=~E

4π

(~E + 4π~P

)=~E · ~D4π

, (11.58)

where the missing hats indicate that ~E and ~D are to be taken asfunctions of ~x and t here;

• we consider the contribution of individual Fourier modes (ω,~k) tothe dissipation Q, i.e. we set

~E = ~E ei(~k·~x−ωt) , ~D = ~D ei(~k·~x−ωt) , (11.59)

and thus Q will be the dissipation per Fourier mode (ω,~k);

• in order to arrive at a real quantity, we need to replace ~E by(~E + ~E∗)/2 to obtain the real part of the ~E field which is definedcomplex here, and the same for ~D;

• since, for longitudinal waves, ~D = εl ~E, we can write

~D = −iω2

(εl ~E − ε∗l ~E

∗), (11.60)

where the minus sign on the second term comes from the changein sign in the phase factor exp[i(~k · ~x − ωt)] due to the complexconjugation of the ~E field; inserting this into Q from (11.58) gives

Q = −iω

16π

(~E + ~E∗

) (εl ~E − ε∗l ~E

∗)

; (11.61)

averaging this over time eliminates the products ~E · ~E and ~E∗ · ~E∗

because they vary with the phase factor exp(−2iωt), while themixed terms become independent of time; thus,

Q = −iω

16π

(εl ~E · ~E∗ − ε∗l ~E

∗ · ~E)

= −iω

16π(εl − ε

∗l)| ~E|2 ; (11.62)

the remaining expression in brackets is

εl − ε∗l = 2i=εl , (11.63)

and so we findQ =

ω

8π=εl| ~E|2 ; (11.64)


• the imaginary part of εl follows from (11.54),

=1

~k ·~v − ω − iδ= =

~k ·~v − ω + iδ

(~k ·~v − ω)2 + δ2=

δ

(~k ·~v − ω)2 + δ2,

(11.65)and in the limit δ→ 0 this turns into a Dirac delta function,

δ

(~k ·~v − ω)2 + δ2→ πδD(~k ·~v − ω) ; (11.66)

therefore, the imaginary part of εl is

=εl = −4π2e2

k2

∫dpx k

d fdpx

δD(kvx − ω)

= −4π2e2m

k2

d fdpx

∣∣∣∣∣∣px=ωm/k

(11.67)

• the damping then turns into

Q = −| ~E|2πme2ω

k2

d fdpx

∣∣∣∣∣∣px=ωm/k

; (11.68)

this is Landau damping, which is caused by the fact that electronswhich are slightly faster than the wave are slowed down, electronswhich are slightly slower than the wave are accelerated, and sincethe velocity distribution is typically monotonically decreasing,more electrons need to be accelerated than decelerated, and thusthe wave loses energy;

11.5 Waves in a Thermal Plasma

11.5.1 Longitudinal and transversal dielectricities

• in a thermal plasma, f0 is the Maxwell distribution

f0 =n

(2πmkT )3/2 exp(−

p2

2mkT

)dpxdpydpz , (11.69)

therefore the integrated distribution f is

f (px) =n(2πmkT )

(2πmkT )3/2 exp(−

p2x

2mkT

)=

n e−p2x/(2mkT )

(2πmkT )1/2 , (11.70)

and its derivative is

d fdpx

=−npx e−p2

x/(2mkT )

√2π(mkT )3/2

; (11.71)

11.5. WAVES IN A THERMAL PLASMA 145

• with these expressions, the longitudinal dielectricity can be trans-formed to read

εl = 1 +4πe2

k

∫ ∞

−∞

dpxnpx

√2π(mkT )3/2

e−p2x/(2mkT )

kpxm − ω − iδ

; (11.72)

introducing the Debye length λD (11.8) and using vx = px/m, wefind

εl = 1 +1

kλ2D

∫ ∞

−∞

dvxmvx

√2πmkT

e−mv2x/(2kT )

kvx − ω − iδ, (11.73)

and with the mean thermal electron velocity

ve =

√kTm

and z ≡vx

ve, (11.74)

we get the expression

εl = 1+1

√2π(kλD)2

∫ ∞

−∞

dzze−z2

z − x − iy≡ 1+

1k2λ2

D

W(x) , (11.75)

where the abbreviations

x ≡ω

kveand y ≡

δ

kve(11.76)

were used and the function

W(x) ≡1√

2π

∫ ∞

−∞

dzz e−z2/2

z − x(11.77)

was defined for x ∈ C;

• this function W(x) is well-defined and analytic for x in the upperhalf of the complex plane, i.e. for =x > 0, and it can analyticallybe continued into the lower half-plane by integrating over a closed(rectifiable and positively oriented Jordan) curve C in C whoseinteriour encloses x;

• we thus choose the contour such that it runs along the real axisfrom −R to R, possibly including x in a small extension if =x ≤ 0,and closing along a half-circle in the upper complex plane fromR to −R; because of the steep exponential drop of the integrand,the integral along that half-circle will not contribute in the limitR→ ∞, and the integral along the closed contour C will equal theintegral along the real axis,∮

Cdz

z e−z2/2

z − x→

∫ ∞

−∞

dzz e−z2/2

z − x, (11.78)

with =z = 0, as required for evaluating (11.75);


• beginning with =x > 0, we can integrate over z along the real axis;for doing so, we first substitute the identity

1z − x

= i∫ ∞

0dt exp [−i(z − x)t] (11.79)

into (11.77), which yields

W(x) =i√

2π

∫ ∞

0dt

∫ ∞

−∞

dz ze−z2/2+i(z−x)t

=i√

2π

∫ ∞

0dt e−ixt

∫ ∞

−∞

dz ze−z2/2+izt ; (11.80)

carrying out the z integration first, and splitting the remainingexponential exp(−ixt) into trigonometric functions, yields

W(x) =

∫ ∞

0dt t (cos xt − i sin xt) e−t2/2

= 1 − xe−x2/2∫ x

0dy ey

2/2 + i√π

2xe−x2/2 ; (11.81)

• series expansions of W(x) are useful for practical calculations; forsmall x, |x| < 1,

W(x) ≈ 1− x2 +x4

3+ i

√π

2xe−x2/2 ≈ 1− x2 +

x4

3+ i

√π

2x (11.82)

while, for large x,

W(x) ≈ −1x2 −

3x4 + i

√π

2xe−x2/2 ; (11.83)

• substituting this back into (11.75) and expanding the abbreviations(11.76), we find the longitudinal dielectricity

εl = 1 −1

(kλD)2

[(kve)2

ω2 +3(kve)4

ω4

− i√πω√

2kve

e−ω2/(2k2v2

e )]

≈ 1 −(ωp

ω

)2(1 +

3(kve)2

ω2

)+ i

√πωω2

p√

2(kve)3e−ω

2/(2k2v2e ) ; (11.84)

in the limiting case ω kve; we have inserted the plasma frequen-cy ωp = ve/λD in the first term here;

11.5. WAVES IN A THERMAL PLASMA 147

• in the opposite limiting case ω kve,

εl = 1 +1

(kλD)2

[1 −

ω2

(kve)2 + i√πω√

2kve

(1 −

ω2

2k2v2e

)]≈ 1 +

(ωp

kve

)2 1 − (ω

kve

)2

+ i√πω√

2kve

; (11.85)

• a similar calculation leads to the transversal dielectricity

εt = 1 +ω2

p

ω2

[W

(ω

kve

)− 1

]; (11.86)

• if ions need to be taken into account in addition to the electrons,the dielectricities are summed according to

ε − 1 =∑

i=species(εi − 1) ; (11.87)

11.5.2 Dispersion Measure and Damping

• the dispersion relation for transversal waves was given by (11.40);in the high-frequency limit,

εt ≈ 1 −ω2

p

ω2 (11.88)

because then x 1 and F(x) ≈ −1, and thus the dispersionrelation becomes

ω2 − ω2p = k2c2 ⇒ ω2 = ω2

p + k2c2 ; (11.89)

• the group velocity of such transversal waves is

cg =∂ω

∂k=

kc2

ω=

√ω2 − ω2

pc

ω= c

√1 −

ω2p

ω2 ; (11.90)

the propagation time of such waves is this

∆tω =

∫dlcg≈

∫dlc

1 +ω2

p

2ω2

=Lc

+2πe2

mcω2

∫dl n , (11.91)

where the quantity ∫dl n ≡ DM (11.92)

is called the dispersion measure (which is the column density ofthe free electrons); here, we have approximated ω ωp, whichshould not be confused with the assumption ω kve!


• for ω < ωp, the dispersion relation implies

k =

i√ω2

p − ω2

c, (11.93)

i.e. the transversal waves are damped; ω kve and ω < ωp =

ve/λD are possible if both ve and λD are small, which is the casein sufficiently cold plasmas; in the ionosphere of the Earth, n ≈106 cm−3, thus

λD ≈ 0.11 cm (11.94)

andve ≈ 6.4 × 106 cm s−1 , (11.95)

thusωp ≈ 60 MHz , (11.96)

which corresponds to a wavelength of

λ ≈cωp≈ 5 m , (11.97)

where all numbers are given assuming T = 273 K;

• the dispersion relation for longitudinal waves requires εl = 0, aswas shown above [see (11.42)]; to lowest order in kve/ω,

εl ≈ 1 −ω2

p

ω2 ⇒ ω = ωp ; (11.98)

to next higher order, setting the real part of εl to zero,

<εl ≈ 1 −ω2

p

ω2

[1 +

3(kve)2

ω2

]!= 0 , (11.99)

impliesω2

p

ω2 +3ω4

pk2λ2D

ω4 − 1 = 0 ; (11.100)

this quadradatic equation in ω2 has the solutions

ω2 =ω2

p

2

[1 ±

√1 + 12k2λ2

D

](11.101)

orω ≈ ωp

(1 + 3k2λ2

D

); (11.102)

Kapitel 12

Magneto-Hydrodynamics

further reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapter 21 and27; Padmanabhan, “TheoreticalAstrophysics, Vol. I: Astrophysi-cal Processes”, section 9.6

12.1 The Magneto-Hydrodynamic Equations

12.1.1 Assumptions

• magneto-hydrodynamics is built upon several assumptions whichgo significantly beyond hydrodynamics; this begins the fact thatplasmas consist of ions and electrons which should be described astwo fluids which are coupled to each other; this is simplified by theassumption that ions and electrons may be coupled to each otherso tightly by the electrostatic attraction that they can be treated asa single fluid;

• secondly, the usual assumption of hydrodynamics plays a criticalrole that the mean free path of electrons and ions is very smallcompared to any other relevant scales;

• thirdly, there is usually a small drift velocity ~vdrift between theelectrons and ions,

vdrift = ~ve −~vi , (12.1)

which causes an electric current, which in turn induces a magneticfield;

• finally, it is assumed that the plasma flows non-relativistically,such that terms of higher than linear order in v/c can be neglected;

• we thus get to the following mathematical assumptions: in the restframe of the plasma, the charge density ρ′ be negligible comparedto the current density ~j′,

ρ′ |~j′|c

; (12.2)

149

150 KAPITEL 12. MAGNETO-HYDRODYNAMICS

• the current density be related to the electric field ~E′ through Ohm’slaw,

~j′ = σ~E′ , (12.3)

with the conductivityσ, which is assumed to be very high (“ideal”),such that weak fields can be responsible for significant currents;

• because of the non-relativistic velocities, we neglect the displace-ment current,

1c∂~E′

∂t

4π~j′

c; (12.4)

the equations of the magnetic field thus read

~∇ · ~B′ = 0 , ~∇ × ~B′ =4πc~j′ ; (12.5)

• the rest frame of the plasma and the observer’s laboratory frameare related by the Lorentz transformation, which can be written, tolowest order in v/c, as

ρ′ = ρ −~j ·~vc2 , ~j′ = ~j − ρ~v ; (12.6)

since we have assimed ρ′ ~j′/c, we also have ρ ~j/c, andbecause

~∇ · ~E′ = 4πρ′ , ~∇ × ~B′ =4πc~j′ , (12.7)

this also implies |~E| |~B|;

• thus, the assumptions of magneto-hydrodynamics lead to the con-ditions

ρ |~j|c, |~E| |~B| ,

∣∣∣∣∣∣∂~E∂t

∣∣∣∣∣∣ ∣∣∣∣∣∣~jc

∣∣∣∣∣∣ , |~v|

c 1 ; (12.8)

from here, we can now derive the equations of magneto-hydrodynamics;

12.1.2 The induction equation

• from Maxwell’s equation

1c∂~B∂t

= −~∇ × ~E (12.9)

and Ohm’s law

~j ≈ ~j′ = σ~E′ = σ

~E +~v × ~B

c

, (12.10)

12.1. THE MAGNETO-HYDRODYNAMIC EQUATIONS 151

we conclude

~E =~jσ−~v × ~B

c(12.11)

and∂~B∂t

= −c~∇ × ~jσ−~v × ~B

c

, (12.12)

but we also need to satisfy

~j =c

4π~∇ × ~B , (12.13)

and thus

∂~B∂t

= −~∇ ×

(c2

4πσ~∇ × ~B

)+ ~∇ × (~v × ~B)

= ~∇ × (~v × ~B) −c2

4πσ

[~∇(~∇ · ~B) − ~∇2~B

]= ~∇ × (~v × ~B) +

c2

4πσ~∇2~B , (12.14)

because of ~∇ · ~B = 0; this induction equation determines theevolution of the magnetic field embedded into a plasma flow withthe velocity ~v; moreover, we have assumed that σ is spatiallyconstant, ~∇σ = 0;

12.1.3 Euler’s equation

• we now need equations for the back reaction of the magnetic fieldon the flow of the plasma; first, there is the continuity equation forthe mass density ρ, which is of course unalteredly valid,

∂ρ

∂t+ ~∇ · (ρ~v) = 0 ; (12.15)

Euler’s equation which describes the conservation of momentumor, more precisely, the transport of the specific momentum density,must be modified because of the Lorentz equation; the Lorentzforce on a charge e is

ec

(~v × ~B) ; (12.16)

multiplying that with the number density of charges, the producte~v turns into the current density ~j, which is

~j =c

4π~∇ × ~B , (12.17)

and thus Euler’s equation is modified to read

ρd~vdt

= −~∇P +1

4π(~∇ × ~B) × ~B ; (12.18)


the Lorentz force density on the right-hand side can also be re-written,

14π

(~∇ × ~B) × ~B =1

4πε i

jk

(ε jl

m∂Bm

∂xl

)Bk

= −1

4π

(δilδkm − δ

imδ

lk

)Bk∂Bm

∂xl

=1

4π

(Bk∂Bi

∂xk − Bk∂Bk

∂xi

)=

14π

(~B · ~∇)~B −1

8π~∇~B2 ; (12.19)

the first term specifies how ~B changes along ~B, i.e. it quantifiesthe tension of the magnetic field lines, which obviously tend tobe as straight as possible; the second term is the gradient of themagnetic energy density and augments the pressure gradient inEuler’s equation;

• we had seen in normal, viscous hydrodynamics that Euler’s equa-tion can be written in the form

∂(ρvi)∂t

+∂T i j

∂x j = 0 , (12.20)

with the stress-energy tensor

T i j = ρviv j + Pδi j − σi j (12.21)

and the shear tensor

σi j = η

(∂vi

∂x j+∂v j

∂xi−

23δi j~∇ ·~v

)− ζδi j~∇ ·~v ; (12.22)

in presence of the magnetic field, this changes to

T i j → T i j −1

4π

BiB j −~B2

2δi j

, (12.23)

i.e. the stress-energy tensor of the viscous fluid is augmented bythe stress-energy tensor of the magnetic field;

• together with an equation of state P = P(ρ), the induction equation,the continuity equation and Euler’s equation determine the flow;these are eight equations for the eight unknowns ρ, P, ~v and ~B;if the magnetic field is known, the current follows from equation(12.17), and the electric field from (12.11);

12.1. THE MAGNETO-HYDRODYNAMIC EQUATIONS 153

12.1.4 Energy and entropy

• the entropy equation, which read

ρTdsdt

= σi j ∂vi

∂x j + ~∇ · (κ~∇T ) , (12.24)

needs to be augmented by the production of Ohmic heat; per unittime, the induction current ~j′ in the rest frame of the fluid dissipatesthe energy

~j′ · ~E′ =~j′2

σ≈~j2

σ=

c2

16π2σ

(~∇ × ~B

)2, (12.25)

which must be added to the right-hand side of the entropy equation,

ρTdsdt

= σi j ∂vi

∂x j + ~∇ · (κ~∇T ) +c2

16π2σ

(~∇ × ~B

)2; (12.26)

• if we need to express this by the specific energy density ε insteadof the specific entropy density s, we start from the energy con-servation equation of viscous hydrodynamics and augments thatanalogously; this yields

∂

∂t

ρ~v2

2+ ε +

~B2

8π

+ ~∇ · ~q = 0 (12.27)

with the extended energy current density vector

qi = ρ

(~v2

2+ w

)vi − κ

∂T∂xi− σi

jvj +

c4πε i

jkE jBk , (12.28)

which now contains the Poynting vector

~S =c

4π~E × ~B =

c4π

~jσ−~v × ~B

c

× ~B=

c2

16π2σ(~∇ × ~B) × ~B −

14π

(~v × ~B) × ~B , (12.29)

and this yields the energy current density

qi = ρ

(~v2

2+ w

)vi − κ

∂T∂xi− σi

jvj (12.30)

−c2

16π2σ

[~B × (~∇ × ~B)

]i+

14π

[~B × (~v × ~B)

]i;

• for incompressible flows with ~∇ ·~v = 0, these equations simplifysomewhat; then, we first have to satisfy

~∇ · ~B = 0 , ~∇ ·~v = 0 , (12.31)


next, the induction equation,

∂~B∂t

= ~∇ × (~v × ~B) +c2

4πσ~∇2~B

= ~v(~∇ · ~B) − ~B(~∇ ·~v) + (~B · ~∇)~v − (~v · ~∇)~B +c2

4πσ~∇2~B

= (~B · ~∇)~v − (~v · ~∇)~B +c2

4πσ~∇2~B , (12.32)

thus∂~B∂t

+ (~v · ~∇)~B = (~B · ~∇)~v +c2

4πσ~∇2~B , (12.33)

and Euler’s equation reads with ~∇ ·~v = 0

∂~v

∂t+ (~v · ~∇)~v = −

1ρ~∇

P +~B2

8π

+1

4πρ(~B · ~∇)~B + ν~∇2~v , (12.34)

where ν = η/ρ is the specific viscosity per unit mass; likewise, theviscosity tensor σi

j in the energy conservation equation simplifies;

12.1.5 Magnetic advection and diffusion

• two terms determine the temporal change of the magnetic field inthe induction equation,

~∇ × (~v × ~B) andc2

4πσ~∇2~B ; (12.35)

• the first term, ~∇× (~v× ~B), determines the transport of the magneticfield with the fluid flow; it is called advection term; its order-of-magnitude is

vBL, (12.36)

if L is a typical length scale of the flow;

• the second term, c2~∇2~B/(4πσ), determines the diffusion of themagnetic field due to the finite conductivity of the plasma; itsmagnitude is of order

c2

4πσBL2 , (12.37)

and it vanishes if the conductivity is ideally (infinitely) large;

• the ratio of the two orders-of-magnitude,

advectiondiffusion

=4πσc2

L2

BvBL

=4πσvL

c2 ≡ RM , (12.38)

12.2. GENERATION OF MAGNETIC FIELDS 155

is called the magnetic Reynolds number; obviously, the diffusioncan be neglected if RM 1, and the induction equation simplifiesto

∂~B∂t− ~∇ × (~v × ~B) = 0 (12.39)

in this case, there is no diffusion, and the magnetic field is “frozen”into the plasma; the physical reason for that is that, if the conduc-tivity is ideally high, σ → ∞, each motion of the magnetic fieldwith respect to the plasma immediately induces strong currentswhich counter-act their origin, i.e. the motion of the field; this isthe typical case in astrophysical plasmas;

• in the opposite limit, RM 1, which occurs of the conductivity issmall, the induction equation reads

∂~B∂t

=c2

4πσ~∇2~B ; (12.40)

the magnetic diffusion time scale therefore is

τdiff ≈ 4πσL2

c2 ; (12.41)

plasmas in the laboratory are typically characterised by RM 1,while astrophysical plasmas typically have RM 1;

12.2 Generation of Magnetic Fields

• the induction equation constains no source term, i.e. it only descri-bes how existing magnetic fields change, but if ~B = 0 initially, thisremains so; this is a consequence of the assumption that ions andelectrons are ideally (tightly) coupled to each other;

• if that is not the case, the motions of the electrons and the ions needto be considered separately, in particular with different velocities~ve and ~vi; then, the two separate Euler equations for the electronsand the ions are

nemed~ve

dt= −~∇Pe − nee

(~E +

~ve

c× ~B

)− neme~∇Φ ,

nimid~vi

dt= −~∇Pi − nie

(~E +

~vi

c× ~B

)− nimi~∇Φ ; (12.42)

we divide these equations by neme and nimi and subtract the second


from the first; this yields

d(~ve −~vi)dt

= −~∇Pe

neme+~∇Pi

nimi

−e

me

(~E +

~ve

c× ~B

)−

emi

(~E +

~vi

c× ~B

); (12.43)

• since the ion mass mi is much larger than the electron mass me, butne = ni ≡ n, equation (12.43) can be approximated by

d(~ve −~vi)dt

= −~∇Pe

nme−

eme

(~E +

~ve

c× ~B

); (12.44)

• this equation must be accomplished by a phenomenological colli-sion term through which different electron and ion velocities canbe justified or produced in the first place; introducing a collisiontime τ, we write

d(~ve −~vi)dt

= −~∇Pe

nme−

eme

(~E +

~ve

c× ~B

)−~ve −~vi

dt; (12.45)

• the current is

~j = eni~vi − ene~ve = en(~vi −~ve) , (12.46)

where we emphasise the implicit assumption of singly-chargedions;

• we now assume that the relative drift velocity is constant,

d(~vi −~ve)dt

= 0 , (12.47)

which also means that the current density is constant; in such asituation, the electric field following from (12.44) is

~E = −~∇Pe

ene−~ve

c× ~B −

me

eτ(~vi −~ve)

= −~∇Pe

ene−~ve

c× ~B +

me~jne2τ

; (12.48)

this implies

∂~B∂t

= −c~∇ × ~E = c~∇ ×

~∇Pe

ene+~ve

c× ~B −

me~jne2τ

=

ce~∇ ×

~∇Pe

ne+ ~∇ × (~ve × ~B) −

mece2τ

~∇ ×~jn

; (12.49)

12.3. AMBIPOLAR DIFFUSION 157

• now, we have

~∇ ×~∇Pe

ne= ~∇Pe ×

~∇nn2 (12.50)

because ~∇ × ~∇Pe vanishes identically, and

~∇ ×~jn

= ~j ×~∇nn2 +

~∇ × ~jn

=c

4πn~∇ × (~∇ × ~B) + ~j ×

~∇nn2 ; (12.51)

if the current is flowing along the gradient in the number densityof the electrons, the latter term vanishes identically, and we obtainthe modified induction equation

∂~B∂t

=c

en2 (~Pe × ~∇n) + ~∇ × (~ve × ~B) +mec2

4πne2τ~∇2~B ; (12.52)

inserting the definition

σ ≡ne2τ

me(12.53)

of the conductivity, this equation is identical to the previous formof the induction equation, except for the first term,

cen2 (~Pe × ~∇n) , (12.54)

whcih now appears as the source of the magnetic field; mecha-nisms like this for creating magnetic fields are called batterymechanisms;

12.3 Ambipolar Diffusion

12.3.1 Scattering cross section

• in a mixture of neutral particles and plasma, the magnetic fieldcan be thought of “frozen into” the plasma; collisions between theplasma and neutral particles then creates a friction force betweenthe plasma and the neutral fluid, which causes a diffusion of themagnetic field; this diffusion process is called “ambipolar”;

• in order to determine it, we first need a cross section σ for the col-lisions, or, more conveniently, the velocity-averaged cross section〈σv〉; we adopt two limiting cases for it;

• if v is very large, we may approximate the cross section by itsgeometrical value; if ri and rn are the effective radii of the ions andthe neutral particles, respectively, we have

σ = π(ri + rn)2 (12.55)


and thus〈σv〉 = 〈v〉σ = 〈v〉π(ri + rn)2 ; (12.56)

the relative velocity during the interaction is

v = |~vi −~vn| ; (12.57)

• if v is sufficiently small, the ion can polarise the neutral particle,by which the cross section is enlarged; the electric field of the ion,assumed singly charged, is

~Ei =Zer2 ~er , (12.58)

while it should be possible to approximate the electric field of thepolarised neutral particle as a dipole field,

~En = −2~pr3 , (12.59)

where ~p = α~Ei is the plarised dipole moment with a parameter α;

• the dipole field of the neutral particle exerts the force

~F = Ze~En = −2ZeZeαr5 ~er = −2α

Z2e2

r5 ~er (12.60)

on the ion, which has the potential

U =α

2Z2e2

r4 ; (12.61)

the motion of the ion past the neutral particle can be characterisedby the minimum separation r0, the impact parameter b and thevelocity v∞ at infinity;

• angular-momentum conservation requires

µv∞b = µr0v0 (12.62)

with the reduced mass

µ ≡mimn

mi + mn; (12.63)

in addition, energy conservation demands

µ

2v2∞ =

µ

2v2

0 −αZ2e2

2r40

; (12.64)

from that, and because of

v0 =bv∞

r, (12.65)

12.3. AMBIPOLAR DIFFUSION 159

we obtain a quartic equation for the minimum separation,

r40 = b2r2

0 −αZ2e2

µv2∞

, (12.66)

or

r20,± =

b2

2±

√b4

4−αZ2e2

µv2∞

; (12.67)

both roots are mathematically possible, but physically, only r20,+ is

relevant because in the limiting case of vanishing coupling α, theminimum radius must equal the impact parameter, r0 = b, sincethen the ion is not scattered at all; r0 will become minimal of theimpact parameter is

b0 =

(4αZ2e2

µv2∞

)1/4

; (12.68)

since the force between the ion and the neutral particle decreasesvery steeply with r, by far the strongest effect occurs for closeencounters; thus, we estimate the cross section as

σ = πb20 =

2πZev∞

√α

µ(12.69)

obviously, 〈σv∞〉 is independent of the asymptotic velocity v∞,

〈σv∞〉 = 2πZe√α

µ; (12.70)

12.3.2 Friction force; diffusion coefficient

• a single collision transfers the momentum

|∆~p| = µ|~vi −~ve| , (12.71)

and the scattering rate per volume is ninn〈σv∞〉, thus the momen-tum transfer per time and volume is

~fr = ninn〈σv∞〉µ(~vi −~ve) , (12.72)

which corresponds to a friction force;

• withρiρn = niminnmn = (mi + mn)ninnµ , (12.73)

this can be cast into the form

~fr = γρiρn (~vi −~vn) , (12.74)

where γ is the friction coefficient

γ ≡〈σv∞〉

mi + mn; (12.75)


• in the two limiting cases of very small or very large relative velo-city, we find

γ =π(ri + rn)2

mi + mn|~vi −~vn| or

γ =2πZe

mi + mn

√α

µ(12.76)

• if the Lorentz force and the friction force balance each other, therelative drift velocity between the ions and the neutral particles iscorrespondingly established; the Lorentz force density is

~fL =~j × ~B

c=

14π

(~∇ × ~B) × ~B , (12.77)

from which follows

γρiρn(~vi −~vn) =1

4π(~∇ × ~B) × ~B ; (12.78)

or

~vd ≡ ~vi −~vn =(~∇ × ~B) × ~B

4πγρiρn; (12.79)

this is of order

vd ≈B2

4πγρiρnL; (12.80)

• the magnetic field, which is thought to be “frozen” into the flowof the plasma, satisfies the equation

∂~B∂t

+ ~∇ × (~B ×~vi) = 0 ; (12.81)

in order to calculate the flow of the magnetic field relative to theneutral particles, we replace ~vi by ~vd, which leads to the equation

∂~B∂t

+ ~∇ ×

~B × (~∇ × ~B) × ~B4πγρiρn

= 0 ; (12.82)

the second term is of order

B3

4πγρiρnL2 , (12.83)

and corresponds to a diffusion coefficient

D ≈B2

4πγρiρn≈ vdL ; (12.84)

Kapitel 13

Waves in Magnetised Plasmas

further reading: Shu, “The Phy-sics of Astrophysics, Vol II:Gas Dynamics”, chapter 22; Pad-manabhan, “Theoretical Astro-physics, Vol. I: AstrophysicalProcesses”, sections 9.7; Ishima-ru, “Basic Principles of PlasmaPhysics”, chapter 5

13.1 Waves in magnetised cold plasmas

13.1.1 The dielectric tensor

• we study the propagation of electromagnetic waves in a plasma inwhich random particle motion is negligible, whose temperature isthus low, and which can thus be considered as cold;

• the equation of motion of an electron in such a plasma with externalmagnetic field ~B0 is then only determined by the Lorentz force,

md~vdt

= −e~E − e~v

c× ~B0 , (13.1)

where ~E is the internal electric field;

• we now assume that the spatial change of the electric field isnegligible and decompose ~v and ~E into harmonic contributionsvarying with time as exp(−iωt); then, each monochromatic velocitymode is determined by

− iω~v = −em

(~E +

~v

c× ~B0

), (13.2)

which implies the matrix equation(iωδi

j −e

mcε i

jkBk0

)v j =

em

Ei ; (13.3)

• the term in parentheses on the left-hand side can be representedby the matrix iω 0 0

0 iω 00 0 iω

− 0 ω3 −ω2

−ω3 0 ω1

ω2 −ω1 0

≡ M , (13.4)

161

162 KAPITEL 13. WAVES IN MAGNETISED PLASMAS

where we have identified the Larmor frequency

ωB ≡eB0

mc(13.5)

and used the definition

ωi ≡eB0

mcBi

0

B0= ωB

Bi0

B0; (13.6)

• the inverse of the matrix (13.4) is

M−1 = −1

det M(13.7) ω2 − ω2

1 −ω1ω2 − iωω3 −ω1ω3 − iωω2

−ω1ω2 − iωω3 ω2 − ω22 −ω1ω3 − iωω2

−ω1ω3 − iωω2 −ω1ω3 − iωω2 ω2 − ω23

,with

det M = −iω(ω2 − ω2B) , (13.8)

which can be written in components as(M j

k

)−1=

−iω(ω2 − ω2

B)

(ω2δ

jk − ω

jωk − iωε jklω

l)

; (13.9)

• this allows us to write the solution for the velocity as

~v =em

M−1 ~E , (13.10)

which yields, with the inverse matrix (13.9), the velocity com-ponents v j

v j =−ie

mω(ω2 − ω2B)

(ω2E j − ω jωkEk − iωε j

klEkωl

), (13.11)

or, written in vector notation,

~v =−ieω

m(ω2 − ω2B)

[~E −

ω2B

ω2 (~E · ~eB)~eB −iωB

ω~E × ~eB

], (13.12)

where ~eB ≡ ~B0/B0 is the unit vector in the direction of the externalmagnetic field ~B0;

• because of ~E + 4π~P = ~D, the polarisation is

~P =~D − ~E

4π; (13.13)

its temporal change is the current density

~j = −ene~v =∂~P∂t

= −iω~P = −iω~D − ~E

4π, (13.14)

again under the assumption of a harmonic dependence on time,∝ e−iωt;

13.1. WAVES IN MAGNETISED COLD PLASMAS 163

• therefore, we find for the dielectric displacement

~D =4πiω~j + ~E = ~E (13.15)

+ −4πe2neω

m(ω2 − ω2B)

[~E −

ω2B

ω2 (~E · ~eB)~eB −iωB

ω~E × ~eB

];

if we can write this in the form D j = εjk Ek, we can read off the

dielectric tensor; we thus obtain

εjk = δ

jk

[1 −

4πe2ne

m(ω2 − ω2B)

]+

4πe2ne

m(ω2 − ω2B)ω2

B

ω2 e jBeBk

+4πie2ne

m(ω2 − ω2B)ωB

ωε

jkle

lB , (13.16)

which we abbreviate as

εjk ≡ ε⊥δ

jk + (ε‖ − ε⊥)e j

BeBk + igε jkle

lB ; (13.17)

• if we identify the plasma frequency there,

ωp =

√4πe2ne

m, (13.18)

we obtain the expressions

ε⊥ = 1 −ω2

p

ω2 − ω2B

,

ε‖ = ε⊥ +ω2

p

ω2 − ω2B

ω2B

ω2

= 1 −ω2

p

ω2 − ω2B

(1 −

ω2B

ω2

)= 1 −

ω2p

ω2 ,

g =ω2

p

ω2 − ω2B

ωB

ω; (13.19)

13.1.2 Contribution by ions

• if ions need to be taken into consideration, these quantities changeaccording to

ε⊥ − 1 → (ε⊥ − 1)e + (ε⊥ − 1)i ,

ε‖ − 1 → (ε‖ − 1)e + (ε‖ − 1)i ,

g → ge + gi ; (13.20)

there, the plasma and Larmor frequencies of the electrons and theions will then have to be distinguished;


• the larmor frequency of the ions is

ωBi =ZeB0

mic=

Zme

miωBe ωBe ; (13.21)

the plasma frequency of the ions is given by

ω2pi =

4πZ2e2ni

mi=

4πZ2e2ne

mi=

Zme

miω2

pe , (13.22)

therefore its ratio to the electron plasma frequency is

ωpi

ωpe=

√Zme

mi 1 ; (13.23)

• thus, the contribution of the ions to the longitudinal dielectrici-ty ε‖ is negligible, but not necessarily their contribution to theperpendicular dielectricity ε⊥; the ratio∣∣∣∣∣∣∣∣∣∣

ω2pi

ω2−ω2Bi

ω2pe

ω2−ω2Be

∣∣∣∣∣∣∣∣∣∣ =

ω2pi

ω2pe

2 ∣∣∣∣∣∣ω2 − ω2Be

ω2 − ω2Bi

∣∣∣∣∣∣ ≈ 1 (13.24)

only if ∣∣∣∣∣∣ ω2 − ω2Be

ω2 − f 2ω2Be

∣∣∣∣∣∣ ≈ 1f,with f ≡

Zme

mi(13.25)

holds; this is the case if

− f (ω2 − ω2Be) ≈ ω

2 − f 2ω2Be (13.26)

is satisfied, or

ω2 ≈f − f 2

1 + fω2

Be ≈ f (1 − f )ω2Be ; (13.27)

similarly, the contributions of ions and electrons to g are compara-bly large;

13.1.3 General dispersion relation

• with the expressions (13.17) and (13.19), the general dispersionrelation

det(δ

jk −

k jkk

k2 −ω2

k2c2 εjk

)= 0 (13.28)

leads to the equation

A(kcω

)4

+ B(kcω

)2

+ C = 0 , (13.29)


with the coefficient

A ≡ ε jkk jkk

k2 = ε⊥ + (ε‖ − ε⊥)eB jk jeBkkk

k2 + igεi jkkik jek

B

k2 ; (13.30)

if θ denotes the angle between the magnetic field and the wave,~eB · ~k = cos θ, this implies

A = ε⊥ + (ε‖ − ε⊥) cos2 θ = ε⊥ sin2 θ + ε‖ cos2 θ , (13.31)

because

εi jkkik jek

B

k2 =~kk·

~kk × ~eB

= 0 ; (13.32)

similarly, we find

B = −ε⊥ε‖(1 + cos2 θ) − (ε2⊥ − g

2) sin2 θ (13.33)

andC = ε‖(ε2

⊥ − g2) , (13.34)

and, by definition, A equals the longitudinal dielectricity εl;

13.1.4 Wave propagation parallel to the magnetic field

• if ~k ‖ ~B, the angle θ = 0 and thus

A = ε‖ , B = −2ε⊥ε‖ , C = ε‖(ε2⊥ − g

2) (13.35)

the roots of the dispersion relation (13.29) are then(kcω

)2

=1

2ε‖

[2ε⊥ε‖ ±

√4ε2⊥ε

2‖− 4ε2

‖(ε2⊥ − g

2)]

= ε⊥ ± g = 1 −ω2

p

ω2 − ω2B

±ω2

p

ω2 − ω2B

ωB

ω

= 1 −ω2

p

ω2 − ω2B

(1 ∓

ωB

ω

)= 1 −

ω2p

(ω ± ωB)ω; (13.36)

• in order to determine what kind of waves are described by theserelations, we return to the equation (11.35)(

δij −

kik j

k2 −ω2

k2c2 εij

)E j = 0 ; (13.37)

for longitudinal waves, ~E ‖ ~k, we have

ε ijE

j = ε⊥Ei + (ε‖ − ε⊥)kik j

k2

Ek j

k+ igε i

jkEk j

kkk

k, (13.38)


since we had also assumed ~B ‖ ~k; this can obviously be reduced to

ε ijE

j = ε⊥Ei + (ε‖ − ε⊥)Ei = ε‖Ei , (13.39)

and, in addition, we have(δi

j −kik j

k2

)E j = Ei −

kik j

k2

Ek j

k= 0 ; (13.40)

thus, longitudinal waves do not satisfy the dispersion relation(13.37), which implies that (13.37) describes transversal waves;

• for transversal waves, E jk j = 0, and thus

Ei −ω2

k2c2

(ε⊥Ei + igε i

jkE j kk

k

)= 0 (13.41)

or, in vector notation,

~E −1

ε⊥ ± g

ε⊥ ~E + ig

~E × ~kk = 0 ; (13.42)

if we turn the z axis into the direction of ~k, the components Ex andEy of ~E read

Ex −1

ε⊥ ± g

(ε⊥Ex + igEy

)= 0 and

Ey −1

ε⊥ ± g

(ε⊥Ey − igEx

)= 0 , (13.43)

which is solved ifEx = ±iEy , (13.44)

characterising circularly polarised light;

13.1.5 Faraday rotation

• the two solutions (13.36) for the dispersion relation thus describethe propagation of left- and right-circular polarised transversalwaves which obey different dispersion relations; left- and right-circular polarised light thus propagates differently along the ma-gnetic field; the polarisation direction of linearly polarised light isthus rotated; this effect is called Faraday rotation;

• in the limit of ω ωp, we have

k2± =

ω2

c2

1 − ω2p

(ω ± ωB)ω

≈ ω2

c2

1 − ω2p

ω2

(1 ∓

ωB

ω

) , (13.45)


or

k± ≈ω

c

1 − ω2p

2ω2

(1 ∓

ωB

ω

) =ω

c−

ω2p

2ωc±ω2

pωB

2ω2c≡ k0 ± ∆k ; (13.46)

the first term k0 corresponds to the wave vector in the unmagnetisedmedium, which the second term ∆k causes a phase shift betweenleft- and right-circular polarised light, and therefore to a rotationof linear polarisation by the angle

ψ =

∫∆kdz =

∫ω2

pωB

2ω2cdz =

∫4πe2ne

meBmc

dz2ω2c

=2πe3

m2c2ω2

∫dz neB ; (13.47)

obviously, the Faraday rotation is proportional to ω−2 or, equiva-lently, to the squared wave length λ2; the expression∫

dz neB ≡ RM (13.48)

is called the rotation measure;

13.1.6 Wave propagation perpendicular to the magne-tic field

• in this case, ~k ⊥ ~B, thus θ = π/2, and the coefficients A, B and Cturn into

A = ε⊥ , B = −ε⊥ε‖ − ε2⊥ + g2 , C = ε‖(ε2

⊥ − g2) ; (13.49)

these imply the solutions for the dispersion relation(kcω

)2

=ε‖

2+

C2ε⊥ε‖

±1

2ε⊥

(ε⊥ε‖ −

Cε‖

)= ε‖ or

Cε⊥ε‖

= ε⊥ −g2

ε⊥, (13.50)

whereB = −ε⊥ε‖ −

Cε‖

(13.51)

was used;

• the first of these solutions is a dispersion relation which is obvious-ly independent of ~B; perpendicular to the magnetic field, wavepropagation is thus possible as in an unmagnetised plasma; thesewaves are transversal and polarised in the direction of ~B; the otherdispersion relation corresponds to waves with longitudinal andtransversal components;


13.2 Hydro-Magnetic Waves

13.2.1 Linearised perturbation equations

• now we consider, in a way very similar to the treatment of soundwaves in a neutral fluid, the propagation of waves in a magnetisedplasma; for simplicity, we assume that dissipation and heat conduc-tion are unimportant, ζ = η = κ = 0, and that the conductivity beinfinite, σ−1 = 0; then, the equations of magneto-hydrodynamicsread

~∇ · ~B = 0 ,∂~B∂t

= ~∇ × (~v × ~B) ,∂ρ

∂t+ ~∇ · (ρ~v) = 0 ,

∂~v

∂t+ (~v · ~∇)~v = −

~∇Pρ

+1

4πρ(~∇ × ~B) × ~B ; (13.52)

the energy conservation equation is not relevant for the followingconsiderations;

• we assume that an equilibrium solution exists,

~B0 , ρ0 , P0 , ~v0 = 0 , (13.53)

which is perturbed by small quantities

δ~B , δρ , δP , δ~v ; (13.54)

• in absence of dissipation, entropy is conserved along flow lines;we further assume isentropic flow, thus s = const. everywhere inthe flow;

• we now linearise the MHD equations and obtain, to first order inthe perturbations

~∇ · δ~B = 0 ,∂δ~B∂t

= ~∇ × (δ~v × ~B) ,

∂δρ

∂t+ ~∇ · (ρδ~v) = 0 ; (13.55)

with ρ = const. for the homogeneous equilibrium solution, the lastequation implies

∂δρ

∂t+ ρ~∇ · δ~v = 0 ; (13.56)

finally, Euler’s equation reads to first order in the perturbations

∂δ~v

∂t= −

~∇(P + δP)ρ + δρ

+1

4π(ρ + δρ)

[~∇ × (~B + δ~B)

]×

(~B + δ~B

)≈ −

~∇δPρ

+(~∇ × δ~B) × ~B

4πρ, (13.57)

13.2. HYDRO-MAGNETIC WAVES 169

again under the assumption that the equilbrium solution is homo-geneous, thus ~∇P = 0 = ~∇ × ~B;

• again, we decompose the perturbations Q into plane waves,

δQ ∝ exp[i(~k · ~x − ωt)

], (13.58)

and find

i~k · δ~B = 0 , −iωδ~B = i~k × (δ~v × ~B) (13.59)

for the magnetic field and

− iωδρ+ iρ~kδ~v = 0 , −iωδ~v = −i~kδPρ

+(i~k × δ~B) × ~B

4πρ(13.60)

for the velocity;

• the pressure perturbation can be related to the density perturbationas usual,

δP = c2sδρ , (13.61)

where cs is the sound speed in the neutral gas, and this allows usto write Euler’s equation as

− iωδ~v = −ic2

s

ρ~kδρ +

(i~k × δ~B) × ~B4πρ

; (13.62)

• without loss of generality, we can now rotate the coordinate framesuch that ~k points along the positive x axis and that ~B falls into thex-y plane; further, we denote the angle between ~k and ~B with ψ;

• first, the continuity equation yields

δρ = ρ~k · δ~vω

= ρkδvx

ω; (13.63)

the phase velocity of the wave is

ck ≡ω

k, (13.64)

and thus

δρ = ρδvx

ck; (13.65)

• the induction equation turns into

δBx = 0 , δBy =δvxBy − δvyBx

ck, δBz = −

δvzBx

ck(13.66)


(the latter because of δBx = 0); and Euler’s equation finally reads,also in components

−ωδvx +c2

s kδρρ

= −kByδBy

4πρ,

−ωδvy =kBxδBy

4πρ,

−ωδvz =kBxδBz

4πρ; (13.67)

• with (13.64) and (13.65), we obtain from here the two sets ofequations

ckδBz = −δvzBx , ckδvz = −BxδBz

4πρ(13.68)

and

ckδBy = δvxBy − δvyBx , ckδvy = −BxδBy

4πρ,

ckδvx −c2

s

ckδvx = δvx

(ck −

c2s

ck

)=

ByδBy

4πρ; (13.69)

• the first set (13.68) of equations contains only δBz and δvz; thesecond set (13.69) couples δvx and δBx to δvy and δBy; we can thusdistinguish waves which are purely transversal and are describedby (13.68), and other waves, which have longitudinal and trans-versal components; since the density perturbations are caused byvelocity perturbations δvx, only longitudinal waves are responsiblefor them;

13.2.2 Alfvén waves

• the first set (13.68) of equations, describing transversal waves,yields

ckδvz =δvzB2

x

4πρck⇒ ck =

ω

k=

Bx√4πρ

(13.70)

if we eliminate δBz between the first and the second equation;since Bx = ~B · ~k, this can also be written as

ω =~B · ~k√

4πρ≡ cAk cosψ , (13.71)

where the Alfvén velocity

cA =

(B2

4πρ

)1/2

(13.72)

was identified;

13.2. HYDRO-MAGNETIC WAVES 171

• the phase velocity of Alfvén waves is

cA cosψ , (13.73)

their group velocity is

∂ω

∂~k=

~B√4πρ

; (13.74)

accordingly, Alfvén waves travel along ~k with the phase velocitycA cosψ, which depends on the angle between ~k and ~B; furthermo-re, they are transversal and transport physical quantities (energyand such) along the magnetic field ~B, independent of ~k!

13.2.3 Slow and fast hydro-magnetic waves

• we now set δvz = 0 and consider waves which obey the second set(13.69) of equations; if we eliminate

δBy =δvxBy − δvyBx

ck(13.75)

from the second and third equations, we find

c2kδvy = −

Bx

4πρ

(δvxBy − δvyBx

),(

c2k − c2

s

)δvx =

By

4πρ

(δvxBy − δvyBx

); (13.76)

further eliminating

δvy = −BxByδvx

4πρ

(c2

k −B2

x

4πρ

)−1

= −BxByδvx

4πρc2k − B2

x(13.77)

from the second equation yields

(c2

k − c2s

)δvx =

B2yδvx

4πρ+

B2xB2

yδvx

4πρ(4πρc2k − B2

x); (13.78)

• δvx cancels on both sides, and we find

(c2

k − c2s

)=

B2y

4πρ

(1 +

B2x

4πρc2k − B2

x

)=

B2y

4πρc2

k

c2k − B2

x/(4πρ)⇒

(c2

k − c2s

) (c2

k −B2

x

4πρ

)=

c2k B2

y

4πρ; (13.79)


• this quadratic equation in c2k has the solutions

c2k,± =

c2s

2+

B2

8πρ±

√(c2

s

2+

B2

8πρ

)2

−B2

xc2s

4πρ(13.80)

=c2

s + c2A

2±

√(c2

s + c2A

2

)2

− c2s c2

A cos2 ψ ;

thus, a fast and a slow wave mode are possible;

• we first consider the case in which ~B is (almost) parallel to~k, henceψ 1 and cos2 ψ ≈ 1 − ψ2/2; then,

c2k =

ω2

k2 ≈12

c2s + c2

A ±

√(c2

s − c2A

)2

=12

(c2

s + c2A ± |c

2s − c2

A|)

=

c2

s orc2

A(13.81)

to first order in ψ; accordingly, the fast wave propagates with thefaster of the sound and the Alfvén speeds, the slow wave with theslower;

• in this approximation, By ≈ ψB, and the Alfvén mode has

δvx =c3

A

c2A − c2

sψ2 δB

B, δvy = −

BxδBy

4πρcA, (13.82)

which can be approximated as

δvx ≈ 0 , δvy ≈ cAδBB

; (13.83)

the acoustic mode has

ByδBy = 0 , δvy = −BxδBy

4πρcs≈ −

c2A

cs

δBB

; (13.84)

• if, finally, cos2 ψ = 0 or ~B ⊥ ~k, we find

c2k,± =

12

(c2

s + c2A ± c2

s + c2A

), (13.85)

thusck =

√c2

s + c2A (13.86)

for the fast and ck = 0 for the slow MHD wave;

Kapitel 14

Jeans Equations and JeansTheorem

further reading: Binney, Tremai-ne, “Galactic Dynamics”, secti-ons 4.1–4.414.1 Collision-less motion in a gravitational

field

14.1.1 Motion in a gravitational field

• particles in a gas or a fluid move almost unaccelerated until theymeet another particle, which forces them to change their state ofmotion abruptly; we had based our treatment of hydrodynamics onthe assumption that the collisions occur on much smaller lengthscales λ than those that characterise the extent of the entire system,L; in plasma physics, we had seen that the shielding of chargeson length scales λD allows a hydrodynamical treatment despitethe inifite range of electrostatic interactions, provided there aresufficiently many particles in a volume ≈ λ3

D; in these cases, theinteractions are effectively extremely short-ranged; likewise, wehad assumed in our treatment of local thermodynamical equilibri-um that the mean free path of the photons is much smaller thanthe characteristic dimensions of the system under consideration;

• studying the motion of many point masses such as stars in a gra-vitational field, we encounter a fundamental change: the forcesbetween the particles are long-ranged and cannot be shielded; asingle star in a galaxy, for instance, thus experiences not only theattraction of its nearest neighbours, but of a large fraction of allstars in the entire galaxy;

• let us consider now a two-dimensional system, such as a galacticdisk, which we shall assume to be infinitely extended for now andin whose centre we assume a star; the disk be randomly coveredby stars such that their mean number density is spatially constant;

173

174 KAPITEL 14. JEANS EQUATIONS AND JEANS THEOREM

• in a circular ring of radius r and width dr, we find

dN = 2πrdr n (14.1)

stars, whose gravitational force on the star in the centre is

dF = 2πrdr nGm2

r2 (14.2)

of the mass of all stars is assumed to be the same, for simplicity;

• of course, the directions of all forces cancel in the mean, but thecontribution of arbitrarily distant rings diverges logarithmically,∫

dF = 2πGnm2∫

drr

= 2πGnm2 ln r ; (14.3)

thus, the structure of the entire stellar system is important for thedynamics of the stars in the gravitational field;

• in the spirit of our distinction of microscopic and macroscopicforces, which we had made when introducing hydrodynamics, theforces in a system which is dominated by self-gravity are alsomacroscopic; therefore, the collision terms, which describe theinteraction on a microscopic scale, can be neglected here at leastto first order of approximation;

• thus, we begin our treatment of self-gravitating systems with thecollision-less Boltzmann equation,

d f (~x,~v, t)dt

=∂ f∂t

+ ~x · ~∇ f + ~v ·∂ f∂~v

= 0 ; (14.4)

14.1.2 The relaxation time scale

• before we turn to a detailed study of Eq. (14.4) in a gravitationalfield, we investigate approximately how the trajectory of a starthrough a galaxy which is composed of individual stars deviatesfrom the trajectory through a hypothetical, “smooth” galaxy;

• we consider the passage of a star by another star employing Born’sapproximation, i.e. we integrate the deflection along a straighttrajectory passing the deflecting star at an impact parameter b;the perpendicular force at the location x along the hypothetical,straight trajectory is

F⊥ = | − ~∇⊥Φ| =

∣∣∣∣∣∣− ∂∂bGm2

√b2 + x2

∣∣∣∣∣∣ =Gm2b

(b2 + x2)3/2 , (14.5)

where Φ is the gravitational potential;

14.1. COLLISION-LESS MOTION IN A GRAVITATIONAL FIELD175

• with x ≈ vt, we have

F⊥ ≈Gm2

b2

[1 +

(vtb

)2]−3/2

, (14.6)

and Newton’s second law mv⊥ = F⊥ thus implies

δv⊥ ≈Gmb2

∫ ∞

−∞

[1 +

(vtb

)2]−3/2

dt

=2Gm

bv

∫ ∞

0(1 + τ2)−3/2dτ =

2Gmbv

; (14.7)

• let N be the number of stars in the galaxy and R be its radius, thenthe fiducial test star experiences

δN = 2πbδb n = 2πbδbNπR2 =

2NR2 bδb (14.8)

such encounters with an impact parameter between b and b + δb;the mean quadratic velocity change is thus

δv2⊥ ≈

2NbδbR2

(2Gm

bv

)2

=8NG2m2

R2v2

δbb

; (14.9)

• integration this expression, we need to take into account that theassumption of Born’s approximation requires that

δv⊥ . v ⇒2Gm

bv. v ⇒ b & bmin =

Gmv2 , (14.10)

and thus we obtain

∆v2⊥ =

∫ ∞

bmin

δv2⊥ ≈ 8N

(2Gm

bv

)2

ln b|Rbmin≡ 8N

(2Gm

bv

)2

ln Λ ,

(14.11)where

ln Λ ≡ lnR

bmin= ln

Rv2

Gm; (14.12)

is the so-called Coulomb logarithm;

• a typical velocity for the stars in a galaxy of mass M = Nm is,according to the virial theorem,

v2 ≈GMm

R⇒ R ≈

GNmv2 ; (14.13)


∆v2⊥

v2 ≈8 ln Λ

N; (14.14)

this shows by what relative amount the star’s velocity is changedduring one passage through the galaxy;


• the Coulomb logarithm ln Λ follows from

ln Λ = lnR

bmin= ln

Rv2

Gm≈ ln N , (14.15)

i.e. the relative velocity change is approximated by

∆v2⊥

v2 ≈8 ln N

N; (14.16)

• after ncross passages through the galaxy, the total relative velocitychange will approximately be

ncross8 ln N

N; (14.17)

if this should be of order unity, the number of passages needs to be

ncross ≈N

8 ln N; (14.18)

one passage takes approximately the time

tcross ≈Rv, (14.19)

i.e. a complete velocity change needs the relaxation time

trelax ≈Rv

N8 ln N

; (14.20)

• in a galaxy, we have

tcross ≈10 kpc

200 km s−1 ≈ 5 × 107 yr , (14.21)

and N ≈ 1011, thus the relaxation time is

trelax ≈ 3 × 1016 yr , (14.22)

which is much more than the age of the Universe; this illustratesthat in many astrophysically relevant systems, the collision-lessBoltzmann equation can be used;

• in a globular cluster, on the other hand, N ≈ 105 and tcross ≈ 105 yr,and thus

trelax ≈ 108 yr , (14.23)

which is short compared to the life time of the globular cluster; insuch cases, therefore, collisions do play a role;

14.2. THE JEANS EQUATIONS 177

14.2 The Jeans Equations

14.2.1 Moments of Boltzmann’s equation

• we thus begin again with Boltzmann’s collision-less equation, inwhich the right-hand side is set to zero,

d fdt

= 0 (14.24)

consider f (~x,~v, t) as a function of position, velocity and time, andreplace the time derivative of the velocity according to Newton’ssecond law

~v =~Fm

= −~∇Φ (14.25)

to obtain∂ f∂t

+~v · ~∇ f − ~∇Φ ·∂ f∂~v

= 0 ; (14.26)

• as several times in the course of this lecture, we now form momentsof equation (14.26) by integrating over velocity space,

∂

∂t

∫d3v f +

∫d3v~v · ~∇ f − ~∇Φ ·

∫d3v

∂ f∂~v

= 0 ; (14.27)

the last term here leads to boundary terms which vanish under theassumption that there are no infinitely fast point masses,

f (~x,~v, t)→ 0 for |~v| → ∞ ; (14.28)

it is a divergence in velocity space to which Gauß’ theorem can beapplied;

• also, the gradient can be pulled out of the integral in the secondterm, and this yields

∂n∂t

+ ~∇ ·

∫d3v f~v = 0 ; (14.29)

the mean velocity is defined as

〈~v〉 =1n

∫d3v, f~v , (14.30)

and so we find the continuitiy equation for the point masses in theform

∂n∂t

+ ~∇ · (n〈~v〉) = 0 , (14.31)

as expected;


• the second moment of Boltzmann’s equation is taken by multiply-ing the equation with ~v prior to integration over velocity space; inthis way, we obtain, written in components

∂

∂t

∫d3v f v j +

∫d3v

∂ f∂xi v

iv j −∂Φ

∂xi

∫d3v

∂ f∂vi v

j = 0 ; (14.32)

partial integration of the third term yields∫d3v

∂ f∂vi v

j = −

∫d3v f

∂v j

∂vi = −nδ ji ; (14.33)

this allows us to write (14.32) as

∂(n〈v j〉)∂t

+∂(n〈viv j〉)

∂xi + n∂Φ

∂x j = 0 , (14.34)

where

〈viv j〉 ≡1n

∫d3v f viv j (14.35)

is the correlation matrix of the velocity components;

• we now multiply the continuity equation (14.31) with v j,

v j∂n∂t

+ v j∂(n〈vi〉)∂xi = 0 , (14.36)

subtract it from (14.34) and use the continuity equation to obtain

n∂〈v j〉

∂t− 〈v j〉

∂(n〈vi〉)∂xi +

∂(n〈viv j〉)∂xi = −n

∂Φ

∂x j ; (14.37)

• the velocity-correlation matrix, 〈viv j〉, can be re-written to read

〈viv j〉 = 〈(vi − 〈vi〉)(v j − 〈v j〉)〉 + 〈vi〉〈v j〉

≡ (σ2)i j + 〈vi〉〈v j〉 , (14.38)

with which we can cast (14.37) into the form

n∂〈v j〉

∂t− 〈vi〉〈v j〉

∂n∂xi − n〈v j〉

∂〈vi〉

∂xi

+((σ2)i j + 〈vi〉〈v j〉

) ∂n∂xi + n

∂

∂xi

((σ2)i j + 〈vi〉〈v j〉

)= −n

∂Φ

∂x j (14.39)

which we can reduce to

n∂〈v j〉

∂t+ n〈vi〉

∂〈v j〉

∂xi = −n∂Φ

∂x j −∂

∂xi

((σ2)i jn

); (14.40)

14.2. THE JEANS EQUATIONS 179

• for convenience, we now abbreviate

vi ≡ 〈vi〉 (14.41)

because only averaged velocities and no velocities of individualparticles remain, and we obtain the two equations

∂n∂t

+ ~∇ · (n~v) = 0 ,

∂vi

∂t+ v j ∂v

i

∂x j = −∂Φ

∂xi −∂

∂x j

((σ2)i jn

); (14.42)

these are the Jeans equations which were derived for the first timeby Maxwell, but first applied to stellar-dynamical problems bySir James Jeans; obviously, the second equation corresponds toEuler’s equation in ideal hydrodynamics, where the divergence ofthe tensor (σ2)i jn takes the role of the pressure gradient,

~∇Pρ

=1ρδi

j∂P∂x j →

∂

∂x j

((σ2)i jn

); (14.43)

14.2.2 Jeans equations in cylindrical and spherical coor-dinates

• it is useful for many applications to write the distribution functionf as a function not of cartesian, but of curvilinear coordinates, suchas cylindrical or spherical coordinates; for instance, in cylindricalcoordinates, we first have

x = r cos φ , y = r sin φ , (14.44)x = r cos φ − rφ sin φ , y = r sin φ + rφ cos φ

and thus in the plane perpendicular to the z axis

~v = r(

cos φsin φ

)+ rφ

(− sin φcos φ

)= r~er + rφ~eφ (14.45)

as well as

x = r cos φ − 2rφ sin φ − rφ sin φ − rφ2 cos φ ,y = r sin φ + 2rφ cos φ + rφ cos φ − rφ2 cos φ , (14.46)

and thus~a = (r − rφ2)~er + (2rφ + rφ)~eφ ; (14.47)

the gradient in cylindrical coordinates is

~∇ = ~er∂

∂r+~eφr∂

∂φ+ ~ez

∂

∂z, (14.48)


and we obtain

vr = r , vφ = rφ , vz = z ,

ar = r − rφ2 = −∂Φ

∂r⇒

vr = −∂Φ

∂r+ rφ2 = −

∂Φ

∂r+v2φ

r,

aφ = 2rφ + rφ = −1r∂Φ

∂φ⇒

vφ = rφ + rφ = −1r∂Φ

∂φ−vrvφ

r, az = z = vz = −

∂Φ

∂z;

• this implies the collision-less Boltzmann equation in cylindricalcoordinates,

d fdt

=∂ f∂t

+ vr∂ f∂r

+vφ

r∂ f∂φ

+ vz∂ f∂z

+

v2φ

r−∂Φ

∂r

∂ f∂vr−

(vrvφ

r+

1r∂Φ

∂φ

)∂ f∂vφ−∂Φ

∂z∂ f∂z

= 0 ; (14.49)

in the same way, we can transform Boltzmann’s equation to sphe-rical coordinates;

• from Boltzmann’s equation in spherical coordinates, we find afterintegration over vr and under the practically important assumption

〈vφ〉 = 0 = 〈vθ〉 (14.50)

the equation

d(nσ2r )

dr+

nr

[2σ2

r − (σ2θ + σ2

φ)]

= −ndΦ

dr, (14.51)

where σ2r,θ,φ are the velocity dispersions

σ2r,θ,φ ≡

1n

∫d3v v2

r,θ,φ ; (14.52)

14.2.3 Application: the mass of a galaxy

• as an application of the Jeans equations, we consider a galaxy inspherical coordinates whose polar and azimuthal velocity dispersi-ons are assumed to be equal,

σ2θ = σ2

φ ; (14.53)

we introduce the anisotropy parameter

β ≡ 1 −σ2θ

σ2r

= 1 −σ2φ

σ2r, (14.54)

which is typically β ≥ 0;

14.3. THE VIRIAL EQUATIONS 181

• the spherically-symmetric Jeans equation then reads

d(nσ2r )

dr+

2nβrσ2

r = −ndΦ

dr; (14.55)

for a spherically-symmetric system, we can write

dΦ

dr=

GM(r)r2 , (14.56)

and thus we find

GM(r)r2 = v2

c = −rn

[σ2

rdndr

+ ndσ2

r

dr+

2nβrσ2

r

]= −σ2

r

[d ln nd ln r

+d lnσ2

r

d ln r+ 2β

]; (14.57)

here, vc is the orbital velocity on a circular orbit with radius raround the centre of the galaxy; given an assumption for β, suchas β = 0, this equation allows us to determine the mass of a galaxy,for instance if the surface-brightness profile is used as a measurefor d ln n/d ln r and the profile of the radial velocity dispersion ismeasurable;

14.3 The Virial Equations

14.3.1 The tensor of potential energy

• we return to the Jeans equation in the form (14.34)

∂(n〈v j〉)∂t

+∂(n〈viv j〉)

∂xi + n∂Φ

∂x j= 0 ; (14.58)

multiplication with the particle mass m and the spatial coordinatesxk, followed by integration over d3x yields∫

d3x xk∂(ρ〈v j〉)∂t

= −

∫d3x xk∂(ρ〈viv j〉)

∂xi −

∫d3x xkρ

∂Φ

∂x j;

(14.59)the second term on the right-hand side is Chandrasekhar’s tensorof the potential energy,

W ij ≡

∫d3x xiρ

∂Φ

∂x j , (14.60)

whose trace is the system’s potential energy;


• the properties of W ij become clearer if we insert the potential

explicitly,

Φ = −G∫

d3x′ρ(~x′)|~x − ~x′|

, (14.61)

and use the gradient

∂Φ

∂x j = G∫

d3x′(x j − x′ j)ρ(~x′)|~x − ~x′|3

; (14.62)

then, Chandrasekhar’s tensor becomes

W ij = −G

∫d3x

∫d3x′ ρ(~x)ρ(~x′)

xi(x j − x′ j)|~x − ~x′|3

; (14.63)

we now swap ~x and ~x′ and change the order of integrations, obtain

W ij = +G

∫d3x

∫d3x′ ρ(~x)ρ(~x′)

x′i(x j − x′ j)|~x − ~x′|3

; (14.64)

and add this to the previous expression (14.63) to find

W ij = −

G2

∫d3x

∫d3x′ ρ(~x)ρ(~x′)

(xi − x′i)(x j − x′ j)|~x − ~x′|3

; (14.65)

first, this shows that W ij is manifestly symmetric, W i

j = W ji , and its

trace is the potential energy,

W ii = −

G2

∫d3x

∫d3x′

ρ(~x)ρ(~x′)|~x − ~x′|

=12

∫d3x ρ(~x)Φ(~x) ,

(14.66)as claimed;

14.3.2 The tensor virial theorem

• now we return to the first term on the right-hand side of the spatialintegral (14.58),∫

d3x xk∂(ρ〈viv j〉)∂xi =

∫d3x

∂(xkρ〈viv j〉)∂xi

−

∫d3x ρ〈viv j〉

∂xk

∂xi ; (14.67)

the first term on the right-hand side is a divergence and vanishesupon integration over a closed system; the second is the tensor ofthe kinetic energy, multiplied by −2,

Kij ≡

12

∫d3x ρ〈viv j〉 ; (14.68)

14.3. THE VIRIAL EQUATIONS 183

we insert the definition (14.38) of (σ2)ij into this tensor and obtain

Kij =

12

T ij +

12

Πij , (14.69)

where T ij and Πi

j are the tensors

T ij ≡

∫d3x ρviv j , Πi

j ≡

∫d3x ρ(σ2)i

j , (14.70)

where the mean velocity 〈vi〉 was written as vi again;

• obviously, the tensor T ij corresponds to the stress-energy tensor in

ideal hydrodynamics up to the pressure term, while Πij describes

the momentum transport by unordered motion and thus a form ofanisotropic pressure;

• on the left-hand side of the spatial integral (14.58), the term∫d3x xk∂(ρ〈v j〉)

∂t(14.71)

remains; we use its symmetry to write it as

12

∫d3x

[xk∂(ρ〈v j〉)

∂t+ x j∂(ρ〈vk〉)

∂t

](14.72)

or12∂

∂t

∫d3x ρ

(xkv j + x jvk

); (14.73)

the partial time derivative can again be replaced by a total timederivative because the convective derivative ~v · ~∇ vanishes whenapplied to the volume integral, and this finally yields

12

ddt

∫d3x ρ

(xiv j + x jv

i)

= T ij + Πi

j + W ij , (14.74)

where the symmetry of the three tensors T ij, Πi

j and W ij was used

again;

• from our earlier considerations on the tensor virial theorem, weknow that

ddt

∫d3x ρ

(xiv j + x jv

i)

=d2Ii

j

dt2 , (14.75)

where Iij is the tensor of second moments of the mass distribution,

Iij ≡

∫d3x ρxix j ; (14.76)

thus, we obtain the tensor virial theorem for collision-less systems,

d2Iij

dt2 = 2(T i

j + Πij + W i

j

); (14.77)


• taking the trace of this equation leads us back to the ordinary(scalar) virial theorem, if the mass distribution is static,

d2Iii

dt2 = 0 ⇒ T ii + Πi

i + W ii = 0 ; (14.78)

now, the trace of the sum

T ii + Πi

i = 2Kii =

∫d3x ρv2 (14.79)

is twice the kinetic energy, and W ii is the total potential energy, as

we saw before; thus2K = −W , (14.80)

which is the scalar virial theorem;

14.4 The Jeans Theorem

• an integral of motion in a gravitational field Φ is a quantity fwhich satisfies

d f (~x,~v)dt

= 0 , (14.81)

where ~x(t) and ~v(t) are characterising arbitrary orbits; by means ofthe equation of motion, ~v = −~∇Φ, this can be cast into the form

d f (~x,~v)dt

= ~x · ~∇ f + ~v ·∂ f∂~v

= ~v · ~∇ f − ~∇Φ ·∂ f∂~v

= 0 ; (14.82)

• in comparison to the collision-less Boltzmann equation, we seethat f is an integral of motion if and only if it is a stationarysolution of Boltzmann’s equation,

∂ f∂t

= 0 ; (14.83)

• this leads us to Jeans’ theorem: Any stationary solution of thecollision-less Boltzmann equation depends on the phase-spacecoordinates only through integrals of motion, and conversely anyfunction depending only on integrals of motion is a stationarysolution of the collision-less Boltzmann equation;

• the proof of the first statement has already been given; as to thesecond statement, let Ii, 1 ≤ i ≤ n be n integrals of motion andf (I1, I2, . . . , In) an arbitrary function thereof; then,

d fdt

=∂ f∂Ii

dIi

dt= 0 , (14.84)

and f solves the collision-less Boltzmann equation;

14.4. THE JEANS THEOREM 185

• an orbit in a gravitational potential always has six integrals of mo-tion; namely, let the orbit be specified by ~x(t) and ~v(t), then it canbe traced back to the initial point ~x0, ~v0 by means of the equationsof motion; these six numbers can be considered as integrals ofmotion;

• one distinguishes isolating and non-isolating integrals of motion;isolating integrals such as the energy E or the angular momentum~L constrain the orbits; an orbit passing through the point ~x0, ~v0

in phase space is part of a subspace S n of phase space definedby the 6 − n conditions I1 = const., . . . , I6−n = const.; an integralI(~x,~v) is isolating for this orbit if an n-dimensional subspace S n

exists in which no point comes arbitrarily close to the hypersurfaceI(~x,~v) = I(~x0,~v0);

• isolating integrals are extraordinarily important, non-isolating or-bits have no practical importance;

Kapitel 15

Equilibrium, Stability andDisks

further reading: Binney, Tremai-ne, “Galactic Dynamics”, secti-ons 4.5, 4.7, 5.1 and 5.315.1 The Isothermal Sphere

15.1.1 Phase-space distribution function

• spherical systems which are independent of time have orbits with atleast the four integrals of motion energy E and angular momentum~L; the Jeans theorem then tells us that any (non-negative) functionf (E, ~L) is a stationary solution of the collision-less Boltzmannequation and may thus represent a stable, self-gravitating system;

• the gravitational potential caused by a system with phase-spacedistribution function f is determined by Poisson’s equation,

~∇2Φ = 4πGρ = 4πGm∫

d3v f ; (15.1)

• if the system is isotropic also in velocity space, it cannot dependon the direction of angular momentum, which implies

f (E, ~L) = f (E, L) ; (15.2)

• using the Laplacian operator in spherical symmetry,

~∇2 =1r2

ddr

(r2 d

dr

), (15.3)

the equation for the gravitational potential

1r2

ddr

(r2 dΦ

dr

)= 4πGm

∫d3v f

(mv2

2+ mΦ,m|~x ×~v|

)(15.4)

follows as the fundamental equation for self-gravitating sphericalsystems in equilibrium;

187

188 KAPITEL 15. EQUILIBRIUM, STABILITY AND DISKS

• it is often convenient to re-scale the Potential Φ and the energy Eby subtracting a constant Φ0,

ψ ≡ −Φ + Φ0 , E ≡ −Em

+ Φ0 = ψ −v2

2; (15.5)

• we now consider a simple example, in which f does not dependon L,

f (E) =n1

(2πσ2)3/2 eE/σ2

=n1

(2πσ2)3/2 exp(ψ − v2/2σ2

), (15.6)

with a normalising constant n1;

• integration over all velocities yields

4πn1eψ/σ2

(2πσ2)3/2

∫ ∞

0dv v2e−v

2/(2σ2) =4πn1eψ/σ

2

(2πσ2)3/2

√π(2σ2)3/2

4

= n1eψ/σ2≡ n , (15.7)

where n is the number density of particles;

15.1.2 Isothermality

• Poisson’s equation for this system reads

1r2

ddr

(r2 dψ

dr

)= −4πGnm = −4πGmρ1eψ/σ

2, (15.8)

or, usingψ = σ2 ln

nn1

= σ2(ln n − ln n1) , (15.9)

we find an equation for the number density n,

1r2

ddr

(r2 dn

dr

)= −

4πGσ2 n , (15.10)

which can of course also be considered as an equation for thedensity ρ = nm;

• in (ideal) hydrodynamics, we had derived the equation

M(r) = −kTrmG

(d ln ρgas

d ln r+

d ln Td ln r

)(15.11)

for a hydrostatic, spherical system; if this is isothermal, dT/dr = 0,we first have

r2 d ln ρgas

dr= −

mGkT

∫ r

0dr′ r′2ρgas(r′) (15.12)

15.1. THE ISOTHERMAL SPHERE 189

if we consider the mass as only contributed by the gas (withoutany dark matter); differentiation with respect to r and division byr2 yields

1r2

ddr

(r2 d ln ρgas

dr

)= −

4πGmkT

ρgas ; (15.13)

this equation is identical with (15.10) which we have just derivedfrom the Jeans theorem, if we set

σ2 ≡kTm

; (15.14)

thus, the corresponding stellar-dynamical model is called the iso-thermal sphere;

• the mean-squared velocity in the isothermal sphere is

〈v2〉 =1n

∫d3v v2 f =

∫dv v4 exp

(−v2

2σ2

)∫

dv v2 exp(−v2

2σ2

)=

3√π

8

(1

2σ2

)−5/2

√π

4

(1

2σ2

)−3/2 =32· 2σ = 3σ2 ; (15.15)

any individual velocity component thus has the mean square

〈v2x〉 = 〈v2

y〉 = 〈v2z 〉 = σ2 ; (15.16)

15.1.3 Singular and non-singular solutions

• one solution of the equation (15.10) for the density of the isother-mal sphere follows from the ansatz

n = Cr−α ; (15.17)

on the left-hand side of the equation, it yields

−α

r2 , (15.18)

and thus the ansatz indeed is a solution if α = 2 and C = σ2/(2πG),

ρ = mn =σ2

2πGr2 ; (15.19)

this solution is called the singular isothermal sphere;

• another solution which avoids the central singularity can be foundnumerically; for doing so, we conveniently introduce the dimension-less quantities

x ≡rr0, y ≡

ρ

ρ0, (15.20)


where ρ0 is meant to be the finite central density; then, the equationfor the scaled density y is

ddx

(x2 d ln y

dx

)= −

4πGσ2 ρ0r2

0 yx2 ; (15.21)

• if we set the scale radius r0 to

r0 ≡

√9σ2

4πGρ0, (15.22)

the equation simplifies to

ddx

(x2 d ln y

dx

)= −9yx2 ; (15.23)

• this equation can be integrated with the boundary conditions

y(0) = 1 ,dydx

∣∣∣∣∣0

= 0 ; (15.24)

within x . 2, i.e. r . 2r0, the numerical result can be approximatedby

y(x) ≈1

(1 + x2)3/2 ; (15.25)

15.2 Equilibrium and Relaxation

• is there an equilibrium state of a self-gravitating system, whichcorresponds to an entropy maximum? the entropy

S ∝ −∫

phase spaced3xd3 p p ln p (15.26)

is maximised if and only if p is the distribution function of theisothermal sphere; however, the isothermal sphere has infinitemass and energy and can thus not be an exact description of athermodynamical equilibrium state; this implies that there is nothermodynamical equilibrium of a self-gravitating system, and thatself-gravitating systems cannot have stable final configurations,but at best long-lived transient states!

• if we populate a narrow region in phase space with N stars, theirorbits will have slightly different initial conditions; as time pro-ceeds, they will progressively evolve away from each other andthus occupy a growing part of phase space; this phase mixing cau-ses the averaged phase-space distribution f to decrease, because

15.3. STABILITY 191

the averaged phase-space density is progressively diluted; thus,the macroscopic entropy

S ∝ −∫

d3xd3v f ln f (15.27)

does indeed increase;

• this process of phase mixing is in fact hardly different from thethermodynamical trend to equilibrium; there, too, the increase ofentropy is caused by macroscopically averaging over processeswhich are otherwise reversible;

• if the potential is changed while the particles are moving throughit, energy can be transported from particles to others; if, for ex-ample, the system contracts while a star approaches its centre,the potential deepens and the star looses energy; other stars cangain considerable amounts of energy; this process is called violentrelaxation (Lynden-Bell);

15.3 Stability

15.3.1 Linear analysis and the Jeans swindle

• as before in hydrodynamics, we consider an equilibrium soluti-on f0, Φ0 of the coupled system of the collision-less Boltzmannequation and Poisson’s equation,

∂ f∂t

+~v · ~∇ f − ~∇Φ∂ f∂~v

= 0 ,

~∇2Φ = 4πGm∫

d3v f ; (15.28)

in equilibrium, ∂ f /∂t = 0;

• then, we perturb f0, Φ0 by small amounts δ f , δΦ and linearise theequations in δ f , δΦ:

∂δ f∂t

+~v · ~∇δ f − ~∇Φ0∂δ f∂~v− ~∇δΦ

∂ f0

∂~v= 0 ,

~∇2δΦ = 4πGm∫

d3v δ f ; (15.29)

• as an equilibrium solution, we adopt an infinitely extended, homo-geneous distribution f0, which implies a density ρ0 and a potentialΦ0 given by

~∇2Φ0 = 4πGρ0 ; (15.30)


due to the infinite extend and the homogeneity, we must have

~∇Φ0 = 0 , (15.31)

which is possible only if ρ0 = 0; we invoke the “Jeans swindle” toset Φ0 = 0; this is practically permissible if we study perturbationswhich are small in scale compared to possible scales in ρ0;

• with the “Jeans swindle”, the linearised equations read

∂δ f∂t

+~v · ~∇δ f − ~∇δΦ∂ f0

∂~v= 0 ,

~∇2δΦ = 4πGm∫

d3v δ f ; (15.32)

again, we decompose the solution into plane waves,

δ f = δ fv(~v)ei(~k·~x−ωt) , δΦ = δΦvei(~k·~x−ωt) (15.33)

and obtain as a condition for δ f to solve the equations

−iωδ fv + i~v · ~kδ fv − iδΦv~k ·

∂ f0

∂~v= 0

−k2δΦv = 4πGm∫

d3v δ fv ; (15.34)

• the first equation yields

δ fv = δΦv~k ·

∂ f0

∂~v

1~k ·~v − ω

(15.35)

which, when inserted into the second, yields

− k2δΦv = 4πGm∫

d3vδΦv

~k · ∂ f0∂~v

~k ·~v − ω; (15.36)

since δΦv does not depend on ~v, we find

1 +4πGm

k2

∫d3v

~k · ∂ f0∂~v

~k ·~v − ω= 0 (15.37)

• this correponds exactly to the longitudinal dielectricity εl fromplasma physics,

εl = 1 −4πe2

k2

∫d3 p

~k · ∂ f0∂~p

~k ·~v − ω(15.38)

and thus leads to Landau damping, exactly as in plasma physics;

15.3. STABILITY 193

15.3.2 Jeans length and Jeans mass

• the boundary between stable an unstable solutions is defined byω = 0; if we suppose a thermal system with a Maxwellian velocitydistribution,

f0(~v) =n0

(2πσ2)3/2 e−v2/(2σ2) , (15.39)

we have∂ f0

∂~v= − f0(~v)

~v

σ2 ; (15.40)

if ~k is chosen parallel to the positive x axis, the condition

1 −4πGmn0

k2σ2(2πσ2)3/2

∫d3v

kvxe−v2x/(2σ

2)

kvx − ω(15.41)

follows; for ω = 0, the integral is∫d3v e−v

2x/(2σ

2) = (2πσ2)3/2 , (15.42)

and we find

1 −4πGρ0

k2σ2 = 0 ⇒ k2(ω = 0) ≡ k2J =

4πGρ0

σ2 ; (15.43)

instability sets in for smaller k or wave lengths larger than λJ =

2π/kJ; the quantity

λJ ≡2πkJ

=2πσ√4πGρ0

=

√πσ√

Gρ0(15.44)

is called the Jeans length;

• the Jeans length defines the volume

λ3J =

( √πσ√

Gρ0

)3

(15.45)

and thus the mass

MJ ≈ ρ0λ3J ⇒ λJ ≈

(MJ

ρ0

)1/3

=

(GMλ2

J

πσ2

)1/3

⇒ λJ ≈GMπσ2 ; (15.46)

according to the virial theorem,

σ2 ≈GM

R⇒ R ≈

GMσ2 ; (15.47)

the radius of the system is thus comparable to the Jeans length;this means that the assumption of homogeneity on the scale ofthe Jeans length cannot be satisfied and that the nature of theinstability needs to be studied for each system in detail once itsgeometry is specified; nonetheless, the Jeans length defines anorder of magnitude estimate for the boundary between stabilityand instability;


15.4 The rigidly rotating disk

15.4.1 Equations for the two-dimensional system

• as an example for a rotating system with flat geometry, we consideran infinitely thin disk (with thickness zero) which is rigidly rotatingaround the z axis with an angular velocity ~Ω = Ω~ez; the disk thusfills the x-y plane and have a surface-mass density Σ0;

• we consider perturbations in the plane of the disk and neglectwarps or twists; furthermore, we transform into a co-rotating coor-dinate frame and study the disk in the (simpler) fluid approximati-on;

• then, the continuity equation, Euler’s and Poisson’s equations read

∂Σ

∂t+ ~∇ · (Σ~v) = 0 ,

∂~v

∂t+ (~v · ~∇)~v = −

~∇PΣ− ~∇Φ − 2~Ω ×~v + ~Ω2~r ,

~∇2Φ = 4πGΣδD(z) ; (15.48)

here, we had to take into account in Euler’s equation that Coriolisand centrifugal forces occur in the co-rotating coordinate frame;

• the physical quantities occuring here are two-dimensional,~v(x, y, t),Σ(x, y, t) and so on, and for the pressure we assume a barotropicequation-of-state,

P(x, y, t) = P[Σ(x, y, t)] ; (15.49)

• the unperturbed quantities are obviously ~v = 0 and Σ = Σ0 aswell as P0 = P(Σ0); this trivially satisfies the continuity equation,Euler’s equation reads

~∇Φ0 = Ω2~r , (15.50)

and Poisson’s equation is

~∇2Φ0 = 4πGΣ0δD(z) ; (15.51)

• since no direction can be preferred on a homogeneous disk, ~∇Φ0

must point along the z axis, which contradicts Euler’s equation;thus, there is no gravitational force yet to balance the centrifu-gal force; therefore, we assume that the disk is embedded into asurrounding gravitational field which compensates the centrifugalforce, such as the halo of a galaxy;

15.4. THE RIGIDLY ROTATING DISK 195

15.4.2 Analysis of perturbations

• we now perturb the disk by small amounts δΣ, δ~v and so on andlinearise the equations in the perturbations; this implies

∂δΣ

∂t+ Σ0~∇ · δ~v = 0 ,

∂δ~v

∂t= −

c2s

Σ0

~∇δΣ − ~∇δΦ − 2~Ω × δ~v ,

~∇2δΦ = 4πGδΣδD(z) , (15.52)

where the sound velocity was introduced as

c2s =

(dP(Σ)

dΣ

)Σ0

; (15.53)

• as usual, we decompose the perturbations into plane waves,

δ~v = δ~vAei(~k·~x−ωt) ,

δΦ = δΦAei(~k·~x−ωt) ,

δΣ = δΣAei(~k·~x−ωt) (15.54)

valid at z = 0, and turn the x axis into the direction of ~k;

• we first consider Poisson’s equation; for z = 0, we have

δΦ = δΦAei(kx−ωt) , (15.55)

while~∇δΦ = 0 (15.56)

must hold otherwise; this is achieved by

δΦ = δΦAei(kx−ωt)−|kz| , (15.57)

where k = kx can have either sign;

• we now integrate Poisson’s equation along the z direction from −ζto ζ and then take the limit ζ → 0; due to the continuity of

∂2δΦ

∂x2 and∂2δΦ

∂y2 (15.58)

at z = 0, these two terms disappear from ~∇2δΦ, but what remainsis

limζ→0

∫ ζ

−ζ

dz∂2δΦ

∂z2 = limζ→0

∂δΦ

∂z

∣∣∣∣∣ζ−ζ

!= 4πGδΣ ; (15.59)

however, we also have

limζ→0

∂δΦ

∂z

∣∣∣∣∣ζ−ζ

= −2|k|δΦAei(kx−ωt) , (15.60)


and so δΦA is related with δΣA through

− 2|k|δΦA = 4πGδΣA ; (15.61)

thus, we find

δΦ = −2πGδΣA

|k|ei(kx−ωt)−|kz| ; (15.62)

• putting these equations into the continuity and Euler’s equations,we find

iΣ0kδvAx = iωδΣA ,

−iωδvAx = −ic2

s

Σ0kδΣA + i

2πGk|k|

δΣA + 2ΩδvAy ,

−iωδvAy = −2ΩδvAx ; (15.63)

these are three linear equations for the three variables δΣA, δvAx

and δvAy; writing them in the form−iω ikΣ0 0

ik(

c2s

Σ0− 2πG|k|

)−iω −2Ω

0 2Ω −iω

δΣA

δvAx

δvAy

= 0 (15.64)

makes it immediately obvious that non-trivial solutions exist ifand only if the determinant of the matrix vanishes, hence

iω(ω2 − 4Ω2) + ikΣ0

[ik

(c2

s

Σ0−

2πG|k|

)(iω)

]= 0 ; (15.65)

this means that either ω = 0 or

ω2 = 4Ω2 + k2c2s − 2πG|k|Σ0 ; (15.66)

this dispersion relation describes the modes of the perturbed, ri-gidly rotating disk; the modes are stable for ω2 ≥ 0 and unstablefor ω2 < 0;

15.4.3 Toomre’s criterion

• if Ω = 0, which is boring for a rotating disk, the disk is unstable if

|k| < kJ ≡2πGΣ0

c2s

, (15.67)

where kJ is the Jeans wave number for the disk; if the sound speedcan be arbitrarily low, cs → 0, perturbations are unstable forarbitrarily large k; their growth rate is then

γ = 2πGΣ0|k| (15.68)

and the growth is exponential, ∝ eγt; obviously, small perturbationswith |k| → ∞, λ → 0, grow particularly violently, i.e. the cold,non-rotating disk fragments violently on small scales;

15.4. THE RIGIDLY ROTATING DISK 197

• this is not suppressed by rotation either; for cs → 0, ω2 < 0 for

|k| >2Ω2

πGΣ0, (15.69)

i.e. even then the instability sets in on the smallest scales;

• pressure and rotation are obviously not able individually to stabili-se the disk, since for Ω = 0, the disk is unstable for perturbationswith small, and for P = 0 = cs for large wave lengths; however,pressure and rotation can be stabilising if they act together, sincethen the dispersion relation has a minimum where

∂

∂k

[|k|2c2

s − 2πGΣ0|k| + 4Ω2 − ω2] !

= 0 , (15.70)

which yields

2|k|c2s = 2πGΣ0 ⇒ |k| =

πGΣ0

c2s

=kJ

2; (15.71)

obviously, the disk is stable if and only if ω2 ≥ 0 at this wavenumber because it is then positive for all wave numbers; thus, thecondition for global stability is

4Ω2 − 2πGΣ0πGΣ0

c2s

+(πGΣ0)2

c2s

= 4Ω2 −

(πGΣ0

c2s

)> 0 (15.72)

orcsΩ

GΣ0>π

2≈ 1.57 ; (15.73)

this is Toomre’s criterion, which can also be applied to collision-less systems (recall that we had adopted the fluid approximation!);then,

csΩ

GΣ0& 1.68 (15.74)

is the condition for stability;

Kapitel 16

Dynamical Friction,Fokker-Planck Approximation

further reading: Binney, Tremai-ne, “Galactic Dynamics”, secti-ons 7.1 and 8.316.1 Dynamical Friction

16.1.1 Deflection of point masses

• an interesting effect occurs if a mass M moves through an environ-ment of masses m which are homogeneously distributed aroundthe mass M; although the motion of the masses can be consideredcollision-less, a deceleration occurs which is called dynamcialfriction;

• let ~vm and ~vM be the velocities of one of the masses m and of themass M, respectively; ~xm and ~xM are their locations; further,

~r ≡ ~xm − ~xM (16.1)

is the distance vector from M to m, and

~v ≡ ~r = ~vm −~vM (16.2)

is the relative velocity of m with respect to M; the system of twopoint masses obeys the equation of motion around a fixed forcecentre of a single body with the reduced mass,( mM

m + M

)~r = −

GMmr2 ~er ≡ −

α

r2~er , (16.3)

where ~er is the unit vector in radial direction away from M;

• obviously, the change of ~v equals the difference of the changes in~vm and ~vM,

∆~v = ∆~vm − ∆~vM , (16.4)

199

200KAPITEL 16. DYNAMICAL FRICTION, FOKKER-PLANCK APPROXIMATION

and, furthermore, the velocity of the centre of mass is unchanged,

~X ≡ m~xm + M~xM ⇒ ~X = m~vm + M~vM , (16.5)

and thus∆ ~X = 0 ⇒ m∆~vm + M∆~vM = 0 ; (16.6)

consequently then, ∆~vm = ∆~v + ∆~vM, and thus

m(∆~v + ∆~vM) = −M∆~vM ⇒ ∆~vM = −m

m + M∆~v ; (16.7)

we shall now determine ∆~v;

• the fictitous particle with the reduced mass, Mm/(M + m), nowdescribes a hyperbolic orbit around the (resting) centre of force;from the Kepler problem of celestial mechanics, we know that thecomplete scattering angle is given by

sinθ

2=

1ε, (16.8)

where ε is the orbit’s eccentricity;

• this implies that the cosine of (half) the scattering angle is

cosθ

2=

√1 − sin2 θ

2=

√1 −

1ε2 =

1ε

√ε2 − 1 , (16.9)

and thustan

θ

2=

1√ε2 − 1

; (16.10)

• generally, the treatment of the Kepler problem shows that the ec-centricity is related to energy E and angular momentum ~L through

ε =

√1 +

2L2Eα2µ

; (16.11)

if the impact parameter is b, the angular momentum is

L =mM

m + Mbv ≡ µbv (16.12)

and the energy is

E =mM

m + Mv2

2≡µ

2v2 (16.13)

because we need to insert the reduced mass

µ ≡mM

m + M(16.14)

16.1. DYNAMICAL FRICTION 201

for the mass of the fictitous particle whose motion we study; since

α ≡ GMm = Gµ(m + M) , (16.15)

the eccentricity is

ε =

√1 +

2µv2(bµv)2

2µ[Gµ(M + m)]2 =

√1 +

b2v4

G2(M + m)2 , (16.16)

and thus the tangent of (half) the scattering angle is

tanθ

2=

1√b2v4

G2(M+m)2

=G(M + m)

bv2 ; (16.17)

16.1.2 Velocity changes

• because of energy conservation, the velocity of the fictitous mass isv also for t → ∞, and because of the deflection by the angle θ, thevelocity components parallel and perpendicular to the asymptoticincoming direction are

v⊥ = −v sin θ = −v2 tan θ/2

1 + tan2 θ/2(16.18)

= −v2G(M + m)

bv2

1

1 +[

G(M+m)bv2

]2

= −2G(M + m)

bvb2v4

b2v4 + G2(M + m)2

= −2bv3

G(M + m)

1 +

[bv2

G(M + m)

]2−1

;

the velocity change of the mass M perpendicular to its initialdirection of motion is thus

∆vM⊥ = −m

M + mv⊥ (16.19)

=2bmv3

G(M + m)2

1 +

[bv2

G(M + m)

]2−1

;

• in parallel direction, we have

v‖ = v cos θ ⇒ ∆v‖ = v(cos θ − 1) = −v(1 − cos θ) (16.20)

or

∆v‖ = −v

(1 −

1 − tan2 θ/21 + tan2 θ/2

)= −v

2 tan2 θ/21 + tan2 θ/2

=−2v

1 + tan−2 θ/2=

−2v

1 +[

bv2

G(M+m)

]2 ; (16.21)


the velocity change in parallel direction is thus

∆vM‖ =m

M + mv‖ =

2mvM + m

1

1 +[

bv2

G(M+m)

]2 ; (16.22)

• if the mass M is moving through a homogeneous “sea” of massesm, all perpendicular deflections must cancel, while the parallel ve-locity changes must add up; therefore, the mass M will experiencea steady deceleration;

16.1.3 Chandrasekhar’s formula

• let f (~vm) be the phase-space density of the stars with mass m whichconstitute the “sea” of masses; then, the rate at which the masse Mencounters collisions with stars with an impact parameter betweenb and b + db is

2πbdb v f (~vm)d3vm ; (16.23)

these collisions change the velocity of M by

d~vM

dt= ~v f (~vm)d3vm (16.24)

×

∫ bmax

0db 2πb

2mvm + M

1 +

[bv2

G(M + m)

]2−1

where bmax is the maximum possible impact parameter, whichcould be defined by the physical size of the cloud of stars m;

• the integral∫ bmax

0db

b

1 +[

bv2

G(M+m)

]2 =

∫ bmax

0db

b1 + ab2

=1

2a

∫ 1+ab2max

1

dββ, (16.25)

where β ≡ 1 + ab2 was set; thus,∫ bmax

0db

b

1 +[

bv2

G(M+m)

]2 =12

[G(M + m)

v2

]2

(16.26)

× ln[1 +

b2maxv

4

G2(M + m)2

],

where

a ≡[

v2

G(M + m)

]2

(16.27)

16.1. DYNAMICAL FRICTION 203

was used; let further

Λ ≡bmaxv

2

G(M + m), (16.28)

then the rate of change for the velocity of M by encounters withstars of mass m and velocity ~vm is

d~vM

dt= ~v f (~vm)d3vm

2πmvm + M

[G(M + m)

v2

]2

ln(1 + Λ2)

= 2πG2 ln(1 + Λ2)m(M + m)

× f (~vm)d3vm~vm −~vM

|~vm −~vM |3, (16.29)

where we have inserted ~v = ~vm −~vM;

• the quantity Λ is typically Λ 1, thus

ln(1 + Λ2) ≈ ln Λ2 = 2 ln Λ ; (16.30)

hence, we replace from now on ln(1 + Λ2) ≈ 2 ln Λ; typical valuesfor this so-called Coulomb logarithm are

5 . ln Λ . 20 ; (16.31)

• for stars which are isotropically distributed in velocity space, theintegral over velocity space is determined by∫ ∞

0dvm v

2m f (vm)

~vm −~vM

|~vm −~vM |3

= −~vM

v3M

∫ vM

0dvm v

2m f (vm) , (16.32)

in perfect analogy to the gravitational field of a collection of masspoints with mean mass density ρ;

• together, this implies Chandrasekhar’s formula for dynamical fric-tion,

d~vM

dt= −16π2 ln ΛG2m(M + m)

~vM

v3M

∫ vM

0dvm v

2m f (vm) ; (16.33)

• if vM is small compared to the typical velocity of the stars m, theremaining integral can be approximated by∫ vM

0dvm v

2m f (vm) ≈

v3M

3f (0) ; (16.34)

then,

d~vM

dt= −

16π2G2

3ln Λm(M + m) f (0)~vM ; (16.35)

the friction is then proportional to ~vM like for Stokes-type friction;


• for sufficiently large vM, the integral converges to a constant, whichimplies that the friction force becomes proportional to v−2

M ;

• for a Maxwellian velocity distribution,

f (vm) =n0

(2πσ2)3/2 e−v2/(2σ2) , (16.36)

the friction force becomes

d~vM

dt= −

4πG2 ln Λ(M + m)ρ0

v3M

[erf(X) −

2X√π

e−X2

]~vM , (16.37)

where X ≡ vM/(√

2σ) scales the velocity vM; here, we have setρ0 ≡ mn0;

• if M m, (M + m) ≈ M, and the friction only depends on thedensity ρ0 of the scatteres stars, but not on their mass any more,

d~vM

dt= −

4πG2 ln Λρ0Mv3

M

[erf(X) −

2X√π

e−X2

]~vM ; (16.38)

similarly, in this case the friction force os proportional to M2

because the deceleration is proportional to M;

16.2 Fokker-Planck Approximation

16.2.1 The master equation

• so far, we have considered the collision-less Boltzmann equation,

d fdt

= 0 ; (16.39)

in presence of collisions, it has to be augmented by collision termson the right-hand side,

d fdt

= C[ f ] , (16.40)

which can be written as a functional of the phase-space distributionfunction f ; explicitly, this equation reads

∂ f∂t

+ ~x · ~∇ f − ~∇Φ ·∂ f∂~v

= C[ f ] ; (16.41)

• collisions transport particles from one position ~w in phase space tanother position ~w+∆~w; within a time interval ∆t, this may happenwith a probability

ψ(~w,∆~w)d6∆~w∆t ; (16.42)

like in hydrodynamics, we distinguish between long-ranged, “smooth”forces short-ranged collisions;

16.2. FOKKER-PLANCK APPROXIMATION 205

• the losses at ~w are

∂ f (~w)∂t

∣∣∣∣∣∣−

= − f (~w)∫

d6∆~wψ(~w,∆~w) , (16.43)

while the gain is

∂ f (~w)∂t

∣∣∣∣∣∣+

= − f (~w − ∆~w)∫

d6∆~wψ(~w − ∆~w,∆~w) ; (16.44)

their sum is the total change

∂ f (~w)∂t

=∂ f (~w)∂t

∣∣∣∣∣∣+

+∂ f (~w)∂t

∣∣∣∣∣∣−

= C[ f ] , (16.45)

which yields the so-called master equation

d fdt

= (16.46)∫d6∆~w

[ψ(~w − ∆~w,∆~w) f (~w − ∆~w) − ψ(~w,∆~w) f (~w)

];

16.2.2 The Fokker-Planck equation

• most collisions will change the velocity only by a small amount;we had seen in the beginning that

δv⊥ ≈2Gm

bv=

Gmv2

2vb

= 2vbmin

b, (16.47)

and thusδv⊥v≈

bmin

b; (16.48)

the relative velocity change thus decreases like b−1 while the num-ber of collisions increases with b proportional to b2; thus mostcollisions cause only small velocity changes;

• we can use that to simplify the master equation; for small δv, ∆~wis also small, and the first term under the integral in the masterequation can be expanded into a Taylor series,

ψ(~w − ∆~w,∆~w) f (~w − ∆~w) ≈ ψ(~w,∆~w) f (~w) −∂

∂wi

[ψ(~w,∆~w) f (~w)

]∆wi +

12

∂2

∂wi∂w j

[ψ(~w,∆~w) f (~w)

]∆wi∆w j , (16.49)

if we stop after the second order; now, we can integrate over ∆~w,which yields the scattering term

C[ f ] = −∂

∂wi

[f (~w)

∫d6∆w∆wiψ(~w,∆~w)

](16.50)

+12

∂2

∂wi∂w j

[f (~w)

∫d6∆w∆wi∆w jψ(~w,∆~w)

];


• we introduce the diffusion coefficients in phase space,

D(∆wi) ≡∫

d6∆w∆wiψ(~w,∆~w) (16.51)

andD(∆wi∆w j) ≡

∫d6∆w∆wi∆w jψ(~w,∆~w) (16.52)

which enable us to bring the master equation into the Fokker-Planck form

∂ f∂t

+ ~x · ~∇ f − ~∇Φ ·∂ f∂~v

= −∂

∂∆wi

[f (~w)D(∆wi)

](16.53)

−∂2

∂∆wi∂∆w j

[f (~w)D(∆wi∆w j)

];

• the great advantage of the Fokker-Planck approach is that thediffusion approximation in phase space depends only on the localphase space coordinates ~w of a test particle, such that the integro-differential master equation turns into a pure differential equation;the diffusion coefficients can now further be approximated byseveral simplifying assumptions;

theoretical astrophysics - uni-heidelberg.de · theoretical astrophysics matthias bartelmann ......

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