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Oxford Master Course in Mathematical and Theoretical Physics Astrophysical Fluid Dynamics S. Balbus 1
Transcript of Oxford Master Course in Mathematical and Theoretical ... · Oxford Master Course in Mathematical...
Oxford Master Course in Mathematicaland Theoretical Physics
Astrophysical Fluid Dynamics
S. Balbus
1
Contents
1 Fundamentals 5
1.1 Opening Comment . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Governing Equations . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Mass Conservation . . . . . . . . . . . . . . . . . . . . 6
1.2.2 Newtonian Dynamics . . . . . . . . . . . . . . . . . . . 6
1.2.3 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2.4 Energetics . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 The vector “v dot grad v” . . . . . . . . . . . . . . . . . . . . 12
1.4 Rotating Frames . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.4.1 The Taylor-Proudman Theorem . . . . . . . . . . . . . 14
1.5 The Indirect Potential . . . . . . . . . . . . . . . . . . . . . . 14
1.6 Local Equations in Discs and Stars . . . . . . . . . . . . . . . 16
1.7 Manipulating the Fluid Equations . . . . . . . . . . . . . . . . 17
1.8 The Conservation of Vorticity . . . . . . . . . . . . . . . . . . 19
2 Magnetohydrodynamics (MHD) 23
2.1 Magnetic Forces . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2 Induction Equation . . . . . . . . . . . . . . . . . . . . . . . . 24
2.3 Self-consistency . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.4 MHD Fundamentals . . . . . . . . . . . . . . . . . . . . . . . 26
3 Gravity 37
3.1 Legendre Exapansion . . . . . . . . . . . . . . . . . . . . . . . 37
3.2 Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.3 Poisson Equation . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.4 Gravitational Tidal Forces. . . . . . . . . . . . . . . . . . . . . 40
3.5 The Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . 43
3.6 The Lane-Emden Equation . . . . . . . . . . . . . . . . . . . . 45
3.7 Solution properties of the Lane-Emden Equation . . . . . . . . 47
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3.8 Polytrope Masses . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.9 The Gravitational Potential Energy of a Polytrope. . . . . . . 49
3.10 E↵ects of Rotation . . . . . . . . . . . . . . . . . . . . . . . . 50
3.10.1 Example 1: Rotating Liquid . . . . . . . . . . . . . . . 51
3.10.2 Example 2: Sub-Keplerian Disks . . . . . . . . . . . . 51
3.11 Self-gravity and Rotation . . . . . . . . . . . . . . . . . . . . . 52
4 Waves and Instabilities 55
4.1 Small Perturbations . . . . . . . . . . . . . . . . . . . . . . . . 55
4.2 Some Useful Relationships Between �, �, and ⇠. . . . . . . . . 57
4.3 Variational Derivation of the Equation of Motion . . . . . . . 58
4.4 Sound Waves and Shock Waves in One Dimension . . . . . . . 61
4.4.1 Linear Waves . . . . . . . . . . . . . . . . . . . . . . . 61
4.4.2 Harmonic Solutions . . . . . . . . . . . . . . . . . . . . 62
4.5 MHD Waves: Fast, Slow, Alfven . . . . . . . . . . . . . . . . . 64
4.6 The Rankine-Hugoniot Jump Conditions for a Shock Wave . . 66
4.7 The Propagation of Strong Shocks . . . . . . . . . . . . . . . . 68
4.7.1 Formulating the problem . . . . . . . . . . . . . . . . . 68
4.7.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.8 The E↵ects of Rotation on Sound Waves . . . . . . . . . . . . 74
4.8.1 The Epicyclic Frequency . . . . . . . . . . . . . . . . . 74
4.8.2 Density Waves . . . . . . . . . . . . . . . . . . . . . . 75
4.8.3 Inertial Waves . . . . . . . . . . . . . . . . . . . . . . . 78
4.9 Angular Momentum Transport . . . . . . . . . . . . . . . . . . 78
5 Classic Instabilities 80
5.1 The Jeans Instability . . . . . . . . . . . . . . . . . . . . . . . 80
5.2 Sound Waves and Gravity in Cosmology . . . . . . . . . . . . 82
5.2.1 The Bonnor-Lifschitz Equation . . . . . . . . . . . . . 82
5.2.2 Adiabatic Solutions of the Bonnor-Lifschitz Equation . 86
5.3 Rayleigh-Taylor Instability . . . . . . . . . . . . . . . . . . . . 87
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5.4 The Kelvin-Helmholtz Instability . . . . . . . . . . . . . . . . 90
5.5 Stability of Continuous Shear Flow . . . . . . . . . . . . . . . 92
5.5.1 Analysis of the inflection point criterion . . . . . . . . 92
5.6 Rotational Instability . . . . . . . . . . . . . . . . . . . . . . . 93
5.6.1 The Rayleigh Criterion . . . . . . . . . . . . . . . . . . 93
5.7 The Magnetorotational Instability . . . . . . . . . . . . . . . . 95
5.8 Thermal Instability . . . . . . . . . . . . . . . . . . . . . . . . 98
6 Classical Flow Solutions 102
6.1 Accretion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
6.2 Spherical Accretion: The Bondi Problem . . . . . . . . . . . . 102
6.3 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
6.4 The Parker Wind Problem . . . . . . . . . . . . . . . . . . . . 106
6.5 Disk Accretion: Hydrodynamical Model . . . . . . . . . . . . 107
6.5.1 Turbulent Model . . . . . . . . . . . . . . . . . . . . . 107
6.5.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . 108
6.6 MHD Theory of Accretion Disk Turbulence . . . . . . . . . . . 111
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Most of the fundamental ideas
of science are essentially simple,
and may, as a rule, be expressed
in a language comprehensible to
everyone.
— Albert Einstein
1 Fundamentals
1.1 Opening Comment
We arose from the gas cloud that collapsed to form our solar system, andwhen, in the distant future, the sun enters its red giant phase, we shall allreturn to gas and dust. Little wonder, then, that the topic of astrophysicalgasdynamics holds our fascination.
The behaviour of a gas subject to large-scale gravitational and magneticforces is enormously rich and full of surprises. One of my goals in giving thiscourse is to try to give you, the student encountering the topic for the firsttime, a sense of both the generality and the depth of the problems we arestruggling with. Truly, there isn’t an area of modern astrophysics that is nottouched in some way by the dynamical behaviour of gases. Astrophysicalgas dynamics may be the most fundamental domain of astrophysics. It isimpossible to understand star formation, stellar structure, planet formation,accretion discs, or anything in the early universe without a detailed knowl-edge of gas dynamics. It is an excellent way to begin a study of theoreticalastrophysics.
1.2 Governing Equations
Although the ultimate fundamental objects are the atomic particles thatcomprise our gas, we shall work in the limit in which the matter is regarded asa nearly continuous fluid. The fact that this is not exactly a continuous fluidmanifests itself in many ways, the most important of which is the equationof state of our gas, which depends upon the notion of atomic collisions. Butmore subtle e↵ects are also present, like viscosity and thermal conduction,both of which are a consequence of finite mean free paths.
One of the most interesting features of astrophysical gases is that they arealmost always magnetized. This allows modes of behaviour that are absent in
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an ordinary gas (e.g. shear waves), sometimes with profound consequences,especially in rotating systems. The dynamics of magnetized gases is knownas magnetohydrodynamics, or MHD for short, and we will have much to sayabout this topic. The ohmic resistivity of magnetized gas is another exampleof a collisional process involving individual particles, in this case chargedparticles.
I shall assume that the reader is familiar with the basic equations ofstandard hydrodynamics. If not, (s)he may review a standard textbook (myfavorite is Elementary Fluid Dynamics by D.J. Acheson), or the set of ex-tensive notes I have prepared for my course Hydrodynamics, Instability andTurbulence. Let us begin with a very brief review.
1.2.1 Mass Conservation
The statement of mass conservation is expressed by the equation:
@⇢
@t+r·(⇢v) = 0, (1)
Here ⇢ is the mas density and v is the velocity field. The content of thisequation is simply that if there is net mass influx into or mass outflux froma fixed volume, the mass within that volume must change accordingly. Ifthe flow is divergence free, the density of an individual fluid element remainsconstant.
1.2.2 Newtonian Dynamics
Our second fundamental equation is a statement of Newton’s second law ofmotion, that applied forces cause acceleration in a fluid. The accelerationrefers to an individual element of fluid, hence the time derivative is expressedas a total, or Lagrangian derivative, following the path of the element:
⇢Dv
Dt= ⇢
"@v
@t+ (v·r)v
#
= F (2)
where the right side is the sum of the forces on the fluid element.
The most fundamental force acting on a fluid is the pressure. We shallalmost always be working with an ideal gas in this course, and the pressureis then given by the ideal gas equation of state
P =⇢kT
µ(3)
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where T is the temperature in Kelvins, k is the Boltzmann constant 1.38 ⇥10�23 J K�1, and µ is the mass per particle. The quantity kT/µ has dimen-sions of velocity squared, and arises often enough that it is given its ownname:
c2S
⌘ kT
µ(4)
where the subscript S refers to “sound” for reasons that will become clearlater. c
S
is the “isothermal sound speed.”
The pressure arises from the kinetic energy of the gas particles themselves,which must never be confused with fluid elements. A fluid element is smallenough that it has uniquely defined dynamic and thermodynamic attributes(e.g. density and pressure), but large enough to contain a vast number ofatoms. A fluid element has a well-defined entropy for example, an atom doesnot.
There is a very simple relationship between the pressure P and internalenergy density E of an ideal gas:
E =P
� � 1. (5)
Here � is the adiabatic index of the gas. It is equal to 5/3 for single particles,and 7/5 for diatomic molecules.
A pressure exerts a force only if it is not spatially uniform. For example,the pressure force in the x direction on a slab of thickness dx and area dy dzis
[P (x� dx/2, y, z, t)� P (x+ dx/2, y, z, t)]dy dz = �@P
@xdV (6)
There is nothing special about the x direction, so the force per unit volumefrom a pressure is more generally �rP dV .
Other forces can be added on as needed. One force of obvious importancein astrophysics is gravity. The Newtonian gravitational acceleration g canalways be derived from a potential function
g = �r� (7)
If the field is from an external source, then � is a given function of r andt, otherwise it must be computed along with the evolution of the fluid itself.We shall discuss the problems of self-gravity later in the course.
Another force that we must consider arises from the presence of a mag-netic field. Magnetic fields allow a gas to behave in ways not allowed whenthe field vanishes, and the additional degrees of freedom imparted to a gasmean that magnetic forces can be very important even when the field appears
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to be weak! To calculate the magnetic force per unit volume exerted by amagnetic field, start with the Maxwell equation
r⇥B = µ0
J + ✏0
µ0
@E
@t(8)
The e↵ects of the displacement current are negligible for nonrelativisitic flu-ids, since they involve time delays associated with light propagation. Hence,the current density is determined by the magnetic field geometry:
J = (1/µ0
)r⇥B (9)
The Lorentz force per unit volume is J⇥B, assuming that the gas is every-where locally neutral.
In the absence of dissipational processes, the equation of motion for amagnetized gas is therefore
⇢@v
@t+ (⇢v·r)v = �rP � ⇢r�+
1
µ0
(r⇥B)⇥B (10)
1.2.3 Viscosity
The fact that there is a finite distance between collisions of the mass particlesof gas creates internal so-called viscous stresses in the flow. As a result, avelocity gradient tends to relax and become erased with time if undisturbed.To leading order, the forces arising from the velocity gradients are found tobe linear in these gradients. This is reasonable, rather like the leading termin some sort of a Taylor expansion.
To represent this, we introduce index notation i, j, k which take on vectorCartesian components x, y, z. A repeated index is summed over. Hencer · vis denoted @v
i
/@xi
. The viscous stress tensor �ij
, dimensions of momentumper unit second per unit area, takes the form first introduced by Stokes,
�ij
= ⌘
@v
i
@xj
+@v
j
@xi
� 2
3�ij
@vk
@xk
!
(11)
where ⌘ is a scalar that may be a function of density and pressure (morebelow). This is the most general linear superposition of velocity gradientsthat vanishes for solid body rotation (no shear) and for uniform expansion(no preferred direction for momentum flow). As an Exercise, you shouldprove this.
Notice that �ij
= �ji
, i.e., the i component of the momentum beingtrasnported in the j direction is physically equivalent to the j component of
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of the momentum being transported in the i direction. The momentum being“carried” and the momentum doing the “transporting” are interchangeable.
Notice as well that the �ij
tensor has a zero trace: the sum of its diagonalelements �
ii
= 0. This is an indication that the deformations of the flowcaused by the shear that a↵ects the viscous stress are associated with nochange in volume. There can be a viscous stress associated with the velocitydivergence itself under unusual circumstances, but we will not pursue this inthis course (see e.g. Landau & Lifschitz, Fluid Mechanics).
The parameter ⌘ is known as the dynamic viscosity coe�cient, with di-mensions of mass per length per time. The viscosity is often ignorable inmany applications, but is needed when small scales are important for dis-sipation. This may be the case for turbulent flow. In fact, astrophysicistslove to represent the whole of turbulent flow phenomenologically with a fake“turbulent viscosity parameter,” as a way to account for the enhanced trans-port turbulence often produces. Avoid joining this crowd, but if you must,at least beware of drawing detailed mathematical conclusions this way.
The equation of motion with viscosity may be written, in a mixed vector-index notation as
⇢@v
@t+ (⇢v·r)v = �rP � ⇢r�+
1
µ0
(r⇥B)⇥B +@�
ij
@xj
(12)
where i is the component of all boldface vector quantities being selected.
In the limit of no gravity, no magnetic field, constant density and constant⌘, we obtain the so-called Navier-Stokes equation:
@v
@t+ (v·r)v = �rP 0 + ⌫r2
v (13)
where P 0 = P/⇢ and ⌫ = ⌘/⇢ is the kinematic viscosity. As an Exercise, youshould prove this. If you can further prove that this equation, together withr·v = 0, has no solutions developing singularities from nonsingular initialconditions (or much to everyone’s surprise find such a singular solution), youcan win $1,000,000 from the Clay Mathematics Institute1. Without viscositythis statement is false; there are singular solutions that develop from the“inviscid” equations, as we will see later in the course.
1.2.4 Energetics
The thermal energy behaviour of the gas is described by the internal energyloss equation, which is most conveniently expressed in terms of the entropy
1http://www.claymath.org/millenium-problems/navierstokes-equation
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per particle. The entropy is defined up to an (unimportant) additive con-stant, and is given by
s =S
N=
k
� � 1lnP⇢�� (14)
where N is the number of particles, � is the adiabatic index (equal to 1+2/fwhere f is the number of degrees of freedom of a particle).
Exercise. Derive the above expression from dE = �PdV + TdS, P =⇢kT/m = (� � 1)E , E = EV .
The entropy of a fluid element is conserved unless there is a loss or gainfrom radiative processes, or the gas is heated by dissipation. If n is thenumber of particles per unit volume, then
nTDs
Dt=
P
� � 1
D lnP⇢��
Dt= volume heating rate ⌘ Q (15)
If there are no radiative losses or gains and no dissipation, as is often thecase when the fluid motions are too rapid for heat to escape, the fluid is saidto be adiabatic and the right side of the above is zero. Note that the internalthermal energy is not conserved in an adiabatic fluid because of compressionor expansion. As an exercise, the reader should show that c2
S
satisfies theequation
⇢D
Dt
c2S
� � 1= �Pr·v (16)
for an adiabatic gas. (Use the entropy and mass conservation equations.)The temperature of a fluid element, like the density, remains fixed only if themotions are incompressible.
In the presence of “bulk” radiative losses, meaning that the photons caneasily escape, a typical form for Q might be
Q = n�� n2⇤(T ) ⌘ �⇢L (17)
where � is an external heating rate, and the ⇤ term represents the e↵ectof collisional losses (hence the n2 dependence from binary collisions). In anearly influential model of the interstellar medium, � was the heating ratedue to cosmic rays and n2⇤ the volumetric losses due to the excitation of CIIlines. More generally, one uses the net loss function per unit mass L as ameasure of the departure from adiabatic behaviour. Typically, L is writtenas a function of ⇢ and T (e.g. thermal bremsstrahlung losses are proportionalto ⇢2T 1/2), but any two thermodynamic variables will do.
To apply the energy equation to stellar interiors, the nature of radiativeenergy losses must be carefully assessed. If active convection is not taking
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place, the dominant mode of energy loss is the transport of the radiationenergy density. This may seem odd at first glance, because the energy densityof a stellar interior is usually completely dominated by the thermal energyof the matter particles; only in the most massive stars does the radiationenergy density become comparable. The reason that energy transport fromradiation is more e↵ective is that the mean free path for photon scattering ishuge compared with any collision mean free path associated with the matter.In other words, the radiation energy leaks out much more rapidly from astellar interior, and is thence radiated at the stellar surface very much likea blackbody: the energy loss per unit area of surface is �T 4, where � isthe Stefan-Boltzmann constant and T is the surface temperature. (� =5.67⇥ 10�8 J s�1 m�2 K�4.)
A stellar interior is about as close to a perfect thermodynamic equilibriumthat one can imagine, but it is not exactly perfect. It is, after all, hotternear the core than near the surface. It is this temperature gradient thatis responsible for the drift of radiation out of the interior plasma into thesurrounding space. We expect that the flux of energy caused by this gradientshould be of order c�r(aT 4), where � is the mean free path for photonscattering. In fact, the precise radiative flux F is given by
F = �c�
3r(aT 4) (18)
the factor of 3 being little more than a directional average. (See MartinSchwarzschild’s classic text, The Structure and Evolution of the Stars, for acareful derivation.)
The mean free path is given by elementary kinetic theory as
� =1
n�(19)
where n is the total of scattering particles and � is an average cross sectionper particle (not to be confused with Stefan-Boltzmann constant!) It isconvenient to work with the mass-related quantities. With µ being a massper particle, we define the mass density ⇢ = µn and the average cross sectionper particle = �/µ, and write
F = �4acT 3
3⇢rT (20)
is called the opacity, generally a complicated function of the temperature,density and and abundances. At typical stellar temperatures, an approxima-tion known as a Kramers law in which
/ ⇢
T 3.5
(21)
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is often used (see e.g. Principles of Stellar Evolution, by D. Clayton, orgood old Schwarzschild). At very high temperature however, the scatteringis dominated by electron scattering, and is a constant. Only the mostmassive stars come into this regime.
Exercise. Evaluate this constant for the case (a) of a purely hydrogenic gas;(b) a gas in which helium is 10% of the total number of baryons. Answers:(a) 0.397 g cm�2; (b) 0.6435 g cm�2. Note the convenient cgs units! (DATA:Proton mass = 1.6726⇥ 10�24g, electron cross section = 6.6526⇥ 10�25 cm2,helium mass = 6.64466⇥ 10�24g, electron mass = 9.109⇥ 10�28g.)
Exercise. In a plasma, the ordinary nonradiative thermal heat is transportedby electron Coulomb collisions at a flux (energy per area per time) of F
C
=��T 5/2rT , where � in cgs units is about 6⇥ 10�7. (See Spitzer, Physics ofFully Ionised Gases.) Compare the radiative and Coulomb heat fluxes for aplasma at 1 gram per cm3 and T = 106K.
In the presence of di↵usive radiative losses (or any other sort of di↵usivelosses), the entropy equation becomes simply
nTDs
Dt=
P
� � 1
D lnP⇢��
Dt= �r·F (22)
with F given by (20). In radiative equilibrium, the left hand side of thisequation is identically zero, so that in a spherical star, the flux magnitude Fis proportional to 1/r2. In terms of the star’s lumimosity L,
F = �4acT 3
3⇢
dT
dr=
L
4⇡r2. (23)
1.3 The vector “v dot grad v”
The vector (v·r)v is more complicated than it appears. In Cartesian coor-dinates, matters are simple: the x component is just (v·r)v
x
, and similar fory, z. But in cylindrical coordinates (R,�, z), the radial R component (say)of this vector is NOT (v·r)v
R
. Rather, we must take care to write
(v·r)v = v·r(vR
e
R
+ v�
e
�
+ vz
e
z
) (24)
where the e
i
are unit vectors in their respective directions. In Cartesiancoordinates, these unit vectors would be constant, but in any other coordinatesystem they change with position. You should be able to show that
@eR
@�= e
�
,@e
�
@�= �e
R
, (25)
12
X
φ
Y
X
θ
φ
r
Y
RZ
Z
Spherical Coordinates Cylindrical Coordinates
Figure 1: Spherical (r, ✓,�) and cylindrical coordinates (R,�, z).
and that there are no other unit vector derivatives in cylindrical coordinates.Thus, the radial component of (v·r)v is
v·rvR
� v2�
R, (26)
and the azimuthal component is
v·rv�
+vR
v�
R(27)
The extra terms are related to centripetal and Coriolis forces, though morework is needed to extract the latter...a piece of it still remains in the gradientterm!
We will use both spherical and cylindrical coordinates throughout thiscourse, as shown in figure [1].
1.4 Rotating Frames
It is often useful to work in a frame rotating at a constant angular velocity⌦, perhaps the frame in which an orbiting planet appears at rest around itsstar. The same rule that applies to ordinary point mechanics applies here aswell: add
�2⌦⇥ v +R⌦2
e
R
(28)
to the applied forces operating on a fluid element (the right side of the equa-tion). The first term is the Coriolis force, the second is the centrifugal force,⌦ is in the vertical direction, and all velocities are measured relative to therotating frame of reference.
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1.4.1 The Taylor-Proudman Theorem
Suppose that in the rotating frame the fluid velocity v is much less thanR⌦. Suppose further that the density is constant and the pressure term(1/⇢)rP is an exact gradient. Then, the sum of the enthalpy, gravity, andcentrifugal terms is expressible as a gradient, say rH, and the steady-statefluid equation is simply
2⌦⇥ v = rH (29)
Since ⌦ lies along the z axis, the left hand side has no z component, soneither does the right. But that means that H is independent of z. Thenthe radial and azimuthal velocity components are independent of z as well,that is, the flow is constant on cylinders! In the case of constant density, themass conservation equation is r·v = 0. But the above equations of motionimply
@vx
@x+
@vy
@y= 0 (30)
so that @vz
/@z = 0, and vz
is also independent of z. If there is no vz
atthe upper or lower boundaries, v
z
vanishes everywhere, and the motion istwo dimensional! The fact that small motions in rotating systems are oftenindependent of height is called the Taylor-Proudman theorem, and you willoften hear “Taylor Columns” referred to in fluid mechanics talks. Now younow where they come from.
It may seem that our arguments go through even if the density is notconstant when P = P (⇢), since it is still possible to form an H function. Butnow mass conservation cannot be so trivially satisfied, because of the v·r⇢term. With only three quantities (v
x
, vy
, ⇢) we cannot satisfy four equations,so there must be a v
z
present, and the flow is no longer so simple.
1.5 The Indirect Potential
A common problem of interest is a gaseous disc around a dominant centralbody M⇤ in which a secondary body M is embedded (or otherwise exerts anexternal gravitational force). With M⇤ as the origin, the acceleration on afluid element at location R is NOT
d2R
dt2= �GM⇤R
R3
� GMr
r3. (NO!) (31)
(See figure [2] for vector definitions.) This is the acceleration that would bemeasured in an inertial frame. The right side of this equation is actuallyd2X/dt2, the acceleration at the same location from the point of view ofthe center-of-mass of M⇤ and M . Sitting on top of the central star, one is
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R"r"
M"CM""x"
X"
X*" XM"
P"
RM""="XM"")"X*"
M*"
Figure 2: Central star M⇤ surrounded by a disc with a perturbing mass M . CMdenotes the center-of-mass; all X vectors are measured relative to this point. Thee↵ect of working in the convenient but noninertial frame centered on M⇤ can beincluded by adding an “indirect” potential to the equations of motion. See text.
no longer in an inertial frame: the star itself is accelerating. The correctacceleration in the star’s frame is given by adding a term �GMR
M
/R3
M
to the right side, i.e. minus the acceleration of the star itself. The correctequation is
d2R
dt2= �GM⇤R
R3
� GMr
r3� GMR
M
R3
M
(YES!) (32)
The last term is known as the indirect term, derivable from the indirectpotential �GM/R
M
. The total perturbing potential due to M is then
� = �GM
r� GM
RM
= � GM
|R�RM
| �GM
RM
(33)
Note that if RM
� r, then the indirect term may be neglected; it is importantwhen R
M
is comparable to r.
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1.6 Local Equations in Discs and Stars
It is possible to simplify the dynamics in a disc or star by working in a smallneighbourhood around a point. Often, this is all that is necessary to revealthe critical dynamics of a flow, and simplifying the problem mathematicallyallows for deeper physical understanding.
Consider the case of a Keplerian disc, gas flow in a central potential�GM/r. In a frame rotating at the angular velocity ⌦(R
0
) ⌘ ⌦0
, whereR
0
is particular cylindrical location, the added rotational force terms are�2⌦0 ⇥ v plus R⌦2
0
e
R
. The radial force due to gravity may be written as�R⌦2(R), where ⌦(R) is the angular velocity of the gas as a function ofR. At R = R
0
there is a force balance, but at a slightly di↵erent locationR = R
0
+ x (x small), there is an imbalance given by �xd⌦2/d lnR = 3⌦2
0
x:
(R0
+ x)[⌦2(R0
)� ⌦2(R0
+ x)] ' �xd⌦2
d lnR.
Dropping the “0” subscript, the local equation of motion in a Keplerian discis given by
Dv
Dt+ (v · r)v + 2⌦ ⇥ v = �1
⇢rP + 3⌦2xe
R
� z⌦2
e
z
(34)
Exercise. Where did the �z⌦2 term come from?
In taking the Lagrangian derivative, the 1/R curvature terms may beignored, since we are working in a very small neighbourhood of a patch ofdisc, in essence taking the limit of R ! 1 while ⌦ remains finite. Then wemay use ordinary Cartesian coordinates with dx, dy, dz replacing dR,Rd�, dz:
Dvx
Dt� 2⌦v
y
= �1
⇢
@P
@x+ 3⌦2x (35)
Dvy
Dt+ 2⌦v
x
= �1
⇢
@P
@y(36)
Dvz
Dt= �1
⇢
@P
@z� ⌦2z (37)
Local forces (perhaps magnetic) may be added to the right side at will. Theseequations are sometimes referred to as the Hill system.
Exercise. The indirect potential should be ignored if we wish to study thee↵ects of an embedded point mass in a disc in the local approximation. Why?
A similar, but somewhat simpler reduction is used for spherical stars orplanets. Here, the centrifgual terms in R⌦2 are generally negligible. Now,
16
we may identify dx, dy, dz with rd✓, r sin ✓d�, dr. The “�–plane” equationsfor motion on a spherical surface are
Dvx
Dt� fv
y
= �1
⇢
@P
@x(38)
Dvy
Dt+ fv
x
= �1
⇢
@P
@y(39)
where f = 2⌦ cos ✓ is the “Coriolis parameter.” Note that f = 0 at theequator.
Exercise. Derive these equations.
An interesting two-dimensional oceanographic application is to add a tidalpotential force on the right and use P = ⇢g⇣, where ⇢ is the (here constant)density of water and ⇣ = ⇣(x, y, t) is the (varying) height of the sea, relativeto the equilibrium sea level.
Exercise. Show that the equation of mass conservation for this problem is
@⇣
@t+
@(hvx
)
@x+
@(hvy
)
@y= 0 (40)
where h is the total height of the sea. (Note that to leading order, h can bereplaced by undisturbed sea depth h
0
(x, y). In what sense do I mean ”leadingorder?”) The set of three equations, with the additional tidal forcing terms,are known as the Laplace tidal equations.
1.7 Manipulating the Fluid Equations
For a particular problem, working in cylindrical or spherical cooridinates isoften the most convenient, but for proving general theorems or identities,Cartesian coordinates are usually the simplest to use. There is a formalismthat makes working with the fluid equations much easier in this case.
As in our brief introductory discussion of viscosity, we will let the indexi, j, or k will represent Cartesian component x, y, or z. Hence v
i
means theith component of v, which may any of the three depending upon what valuei is chosen. So v
i
is a way to write v. The gradient operator r is written @i
,in a way that should be self-explanatory.
As before, if an index appears twice, it is understood that it is to besummed over all the values x, y, and z. Hence
A · B = Ai
Bi
= Ax
Bx
+ Ay
By
+ Az
Bz
, (41)
17
and(v·r)v = (v
i
@i
)vj
(42)
In this last example i is a dummy index, and the vector component is repre-sented by j. The dynamical equation of motion in this notation is
⇢[@t
+ (vi
@i
)]vj
= �@j
P � ⇢@j
� (43)
Sometimes the “rot” (or “curl”) operator is needed. For this, we intoducethe Levi-Civita symbol ✏ijk. It is defined as follows:
• If any of the i, j, or k are equal to one another, then ✏ijk = 0.
• If ijk = 123, 231, or 312, the so-called even permutations of 123, then✏ijk = +1.
• If ijk = 132, 213, or 321, the so-called odd permutations of 123, then✏ijk = �1.
The reader should be able to convince him(her)self that
r ⇥ A = ✏ijk@i
Aj
(44)
Here, the vector component is represented by the index k. Don’t forget tosum over i and j! ✏ijk is used in the ordinary cross product as well:
A ⇥ B = ✏ijkAi
Bj
. (45)
Notice thatA · (B ⇥ C) = ✏ijkA
k
Bi
Cj
(46)
which proves that any even permutation of the vectors on the left side ofthe equation must give the same value, and an odd rearrangement gives thesame value with the opposite sign.
A double cross product looks complicated:
A ⇥ (B ⇥ C) = ✏lkmAl
(✏ijkBi
Cj
) = ✏mlk✏ijkAl
Bi
Cj
. (47)
The last equality follows because mlk is an even permutation of lkm. Thislooks unpleasant, but fortunately there is an identity that saves us:
✏mlk✏ijk = �mi
�lj
� �mj
�li
(48)
where �ij
is the Kronecker delta function (equal to zero if i and j are di↵erent,unity if they are the same). The proof of this is left as an exercise for the
18
reader, who should be convinced after a few simple explicit examples. Withthis identity, our double cross product becomes
A ⇥ (B ⇥ C) = Bm
Aj
Cj
� Cm
Ai
Bi
= B(A · C)�C(A · B). (49)
Our final example is to derive an expression for
A⇥(r⇥B) = ✏ijkAi
(✏lmj@l
Bm
) = ✏kij✏lmj(Ai
@l
Bm
) (50)
Using our identity (48), this becomes
(�kl
�im
� �km
�il
)(Ai
@l
Bm
) = Ai
@k
Bi
� Ai
@i
Bk
= Ai
@k
Bi
� (A·r)B (51)
One consequence of this result is a representation of Ai
@k
Bi
in any coordinatesystem:
Ai
@k
Bi
= A⇥(r⇥B) + (A·r)B (52)
Another particular important application of (51) is to the Lorentz force ex-pression
(r⇥B)⇥B = �1
2rB2 + (B·r)B (53)
The first term on the right side has the form of a magnetic pressure gradient;the second behaves like a tension force. It depends on the derivative of Balong its length, and if the magnitude of B remains fixed, the force mustbe perpendicular to B itself. The e↵ect of this tension force is profound,allowing a magnetized gas to support shear waves (known as Alfven waves)that ordinarily do not exist in a fluid.
1.8 The Conservation of Vorticity
A quantity of great interest to fluid dynamicists is the vorticity, ! = r⇥v.We will see that it is intimately related to the angular-momentum-like cir-culation element v·dl integrated around a closed fluid loop. Under someinteresting circumstances, this circulation integral is conserved.
We start with the following identity, which follows immediately from theresults of the previous section:
v ⇥ (r ⇥ v) =1
2rv2 � (v·r)v (54)
Using this result to replace (v·r)v in the inviscid, umagnetised dynamicalequation of motion results in
@v
@t+
1
2rv2 � v⇥! = �1
⇢rP �r�. (55)
19
Exercise. If P = P (⇢) and dH = dP/⇢, show that
v2
2+H + �
is constant along a velocity flow streamline. This is called the Bernoulliconstant, B. B need not be the same constant on every streamline!
If we take the curl of equation (55), and remember that the curl of thegradient vanishes, we find
@!
@t�r⇥(v⇥!) =
1
⇢2(r⇢⇥rP ) (56)
Let us once again consider the case where either ⇢ is constant, or when P isa function only of ⇢. Then the right hand side vanishes, and:
@!
@t�r⇥(v⇥!) = 0. (57)
To interpret this physically, consider a closed curve, a loop, frozen into thefluid. The integral
Rv · dl around the loop is by Stokes’ theorem
R! · dA,
taken over an area that is bounded by the loop. This is the vorticity flux.How does the vorticity flux change with time as the fluid evolves? The answeris contained within (57).
We consider more generally an equation of the form
@A
@t= v ⇥ (r ⇥ A)+r� (58)
where � is a potential function. The curl of this equation leads back toequation (57) for the special case A = v, but we retain generality here.Expanding the double cross product on the right and regrouping leads to
DAi
Dt= v
j
@i
Aj
+ @i
� (59)
where D/Dt is the usual Lagrangian derivative and we have switched toindex notation. Next, consider the change in the line integral of the vectorfield A over a closed loop moving with the fluid:
D
Dt
ZA · dl =
Z "DA
Dt·dl +A·Ddl
Dt
#
(60)
20
Notice the strange derivative of the line element dl! The Lagrangian deriva-tive of the line element as it moves through the fluid is just the di↵erencebetween the velocity field at the two endpoints of the segment dl:
Ddl
Dt= (dl·r)v (61)
orDdl
j
Dt= dl
i
@i
vj
(62)
From equation (59)
DA
Dt·dl = dl
i
vj
@i
Aj
+ dli
@i
�, (63)
and we have just seen that
A·Ddl
Dt= A
j
dli
@i
vj
. (64)
Adding these last two equations gives
D
Dt(A · dl) = dl
i
@i
(�+ vj
Aj
) (65)
In equation (60) we thus have a perfect gradient integrated over a closedcurve, hence the integral must vanish. The line integral
RA · dl is conserved
with the fluid. In particular when A = v, the velocity circulation integralalong with the vorticity flux surface integral are conserved in the Lagrangiansense, moving with the fluid. We shall see very soon that the same is truefor the magnetic vector potential and the magnetic flux.
The fact that the integralRv · dl around any closed curve embedded in
the fluid remains constant as the fluid flows is known as vorticity conserva-tion. Another way to say the same thing is that the field lines of vorticity !
are “frozen” into the fluid. Once again, this is not a general fluid result, butdepends upon the fact that either ⇢ is constant or that P and ⇢ are directlyfunctionally related. If the pressure gradient can push with the same forcealong a surface of varying inertial response, vorticity will surely be generated.
With the help of our ✏ijk✏lmk identity and just a little work, it is straight-forward to show that
@!
@t�r⇥(v⇥!) = 0 (66)
is the same as
@!
@t+ (v · r)! = +(! · r)v � !r · v (67)
21
Mass conservation may be written
D ln ⇢
Dt= �r · v, (68)
so that our equation becomes
D!
Dt� !
D ln ⇢
Dt= (! · r)v, (69)
orD
Dt
!
⇢
!
=1
⇢(! · r)v (70)
In strictly two-dimensional flow, this is a very powerful constraint. Then! has only a z component, and the right side must vanish. The density ⇢may be replaced by the surface density ⌃ on the left side of the equation:multiply by ⇢2, expand the D/Dt derivative, integrate over height, and refoldthe D/Dt derivative. We find that
D
Dt
✓!
⌃
◆= 0 (71)
This is known as the conservation of potential vorticity. In this case, potentialvorticity (PV in the parlance) labels fluid elements. PV is extremely usefulin the study of two-dimensional turbulence, and in studying long wavelengthwave propagation in planetary atmospheres.
Exercise. Consider rotational flow, with the velocity v having only a �component v
�
. In general, v�
could depend upon R and z, but show that ifvorticity conservation holds, under steady conditions v
�
cannot depend uponz. This is known as von Zeippel’s theorem.
22
The nation that controls mag-netism will control the universe.
— Dick Tracy, created by ChesterGould
2 Magnetohydrodynamics (MHD)
2.1 Magnetic Forces
Astrophysical gases are almost always at least partially ionized. This is nottoo surprising: a glass of distilled water is ionized at the level of one part in107, and salty sea water is much more ionized: it is a very good conductor. Amedium can be almost entirely neutral and still behave like a good conductor.All but the coolest and densest astrophysical gases (e.g., protostellar discs)are electrodynamically active.
The Lorentz force per unit volume in the gas is
F = ⇢e
E + J ⇥ B (72)
where ⇢e
is the charge density, E is the electric field, J is the current density,and B is the magnetic field. The gases of interest here are all electricallyneutral, so that ⇢
e
= 0. This means that the only part of the Lorentz forcethat a↵ects the gas is the magnetic part.
We have already encountered the Lorentz force in our discussion of theequation of motion for a magnetized gas:
J ⇥ B =1
µ0
(r ⇥ B) ⇥ B (73)
With our ✏ijk✏lmk identity, we have already seen that
(r ⇥ B) ⇥ B = �1
2rB2 + (B · r)B (74)
Thus, the dynamical equation of motion for a magnetized gas is
⇢Dv
Dt= �r
P +B2
2µ0
!
� ⇢r�+
B
µ0
·r!
B (75)
23
The first magnetic term on the right clearly behaves like a sort of pressure.Magnetic fields lines of force do not like to be squeezed any more than gasmolecules do.
The (B · r)B term is less obvious. It corresponds to a sort of magnetictension. Notice that it vanishes when the magnetic field does not changealong its own direction. On the other hand, when there are such changes,the resulting force acts in the direction of restoring the field line back toan unstretched position. In fact, this can be made quantitative: there isa magnetic analogue to waves propagating along an ordinary string that isunder tension. In the case of “magnetic strings,” these waves are calledAlfven waves.
2.2 Induction Equation
Having introduced the magnetic field, we need to know how it evolves asthe flow changes. The magnetic field adds one more variable to our problem(well, three actually, since there are three components of B), so we need moreequations. The motion of the gas causes the charged particles to move, theions and electrons respond di↵erently to the applied forces, currents form,these currents in turn generate new fields that a↵ect the currents all overthat change the fields ... Help. It seems like a complicated mess!
Fortunately there is a great simplifying principle to save us: in a perfectconductor, the electric field vanishes. Actually, what we should say is thatin the rest frame of the conductor, the electric field locally vanishes. In aframe in which the conductor (in our case a fluid element of conducting gas)moves, the total Lorentz force (not the electric field) must vanish. In otherwords,
E + v ⇥ B = 0. (76)So even though we have assumed conditions for charge neutrality, there mustbe an electric field: �v⇥B. But if the divergence of this electric field doesnot vanish, then there must be a local charge density, and charge neutralitycannot hold, which looks like a contradiction. Well, maybe it just turns outthat r·(v⇥B) = 0. Guess what? The divergence doesn’t vanish. In amoment, we’ll come back and explain why this is not really a contradictionafter all, but for the time being let us nervously continue.
Faraday’s law of induction is
@B
@t= �r ⇥ E (77)
and with E = �v ⇥ B, this becomes
@B
@t= r ⇥ (v ⇥ B) (78)
24
This is the extra set of equations we need to determine the magnetic field. Byknowing how the spatial gradients of B are behaving, we may compute howthe field evolves in time, thanks to the powerful constraint that the Lorentzforce on the charge carriers must vanish.
2.3 Self-consistency
Why don’t we have a contradiction with the fact that r · E is not zero? Theanswer is that while not zero, it is in fact small. Small?? That answer is notgood enough. How small? Well, of order v2/c2 (c is the speed of light), which,as we will see, is precisely of the same order as the displacement current thatwas also neglected.
To estimate r · (v ⇥ B), assume that any magnetic field gradients areas large as they can be (of order µ
0
J), and that J is also as large as it canbe, of order the ion charge density times v, ⇢
i
v (it could be smaller since itis proportional to the di↵erence between ion and electron velocities). Then
r·(v⇥B) ⇠ vµ0
J ⇠ ⇢i
✏0
v2
c2. (79)
That answer, that the divergence of the electric field is of order v2/c2 timesthe ion charge density, is good enough. Not only is it permitted to ignore thedivergence of the electric field, it is required! We have already not includedthe displacement current, and this too is a correction of order v2/c2. In thiscase, if L is a characteristic length and @/@t ⇠ v/L, then
✏0
µ0
@E
@t⇠ ✏
0
µ0
vE
L⇠ ✏
0
µ0
v2B
L⇠ ✏
0
µ2
0
v2J (80)
which is indeed of order (v2/c2)µ0
J . Corrections of order v2/c2 are relativis-tic, and we must ignore them to be self-consistently nonrelativistic!
Notice something quite remarkable: the magnetic field satisfies the sameequation as the vorticity. In particular, equation (78) can be recast in theform of equation (58), by “uncurling” it! That means everything we learnedabout vorticity, in particular that it is frozen in to the fluid, also holds forthe magnetic field. Magnetic flux,
RB · dA, is conserved as the area moves
with the fluid. But unlike the case of vorticity conservation, which dependedupon a restrictive relationship between P and ⇢, magnetic flux conservationdepends only upon there being no dissipation (i.e., electrical resistance) inthe gas. This is generally an excellent approximation.
25
A Summary of the Dissipationless Equationsof Motion
From now on, we shall drop the subscript “0” on µ0
, and write µ.
@⇢
@t+r·(⇢v) = 0 (81)
⇢
@
@t+ v·r
!
v = �r
P +B2
2µ
!
� ⇢r�+1
µ(B·r)B (82)
P
� � 1
@
@t+ v·r
!
lnP⇢�� = 0 (83)
@B
@t= r⇥(v⇥B) (84)
2.4 MHD Fundamentals
Now then, Dimitri ...— President Merkin Mu✏ey to Premier Kisso↵, Dr. Strangelove
In this section, a detailed derivation of the fundamental MHD equations ispresented. The discussion will be more technical here than in most of therest of the course, but it is very important to see how the basic governingequations of the subject arise, and much of this material is not so easy to findoutside of specialized treatments. I hope the reader will have the patienceto read carefully through this section, but it may be skipped the first timethrough without loss of continuity.
In astrophysics, we are very often interested in the MHD behaviour of agas that is almost entirely neutral but is still a good conductor. This mayseem like contradictory, since a neutral gas has no charge carriers, but thekey word is “almost.” Even a very small population of charge carriers willmake the gas magnetized and highly conducting, as we will shortly see.
A typical environment is a gas cloud consisting of neutral particles (pre-dominantly H
2
molecules), electrons, and ions. Each species (denoted bysubscript s) is separately conserved, and obeys the mass conservation equa-tion
@⇢s
@t+r·(⇢
s
v
s
) = 0 (85)
26
where ⇢s
is the mass density for species s and v
s
is the velocity. The symbolsof the flow quantities (e.g. v, ⇢, etc.) for the dominant neutral species willhenceforth be presented without subscripts.
So far, everything is simple. The dynamical equations become more cou-pled, however, since we need to include interactions between the di↵erentspecies. The dynamical equation for the neutral particles is
⇢@v
@t+ ⇢(v·r)v = �rP � ⇢r�� p
nI
� p
ne
(86)
where P is the pressure of the neutrals, � the gravitational potential andp
nI
(pne
) is the momentum exchange rate between the neutrals and the ions(electrons).
The ion equation is
⇢I
@vI
@t+ ⇢
I
(vI
·r)vI
= eZnI
(E + v
I
⇥ B)�rPI
� ⇢I
r�� p
In
(87)
and the electron equation is
⇢e
@ve
@t+ ⇢
e
(ve
·r)ve
= �ene
(E + v
e
⇥ B)�rPe
� ⇢e
r�� p
en
(88)
The subscript I (e) refers to the ions (electrons). When not in a subscript butused in an equation, e is the fundamental charge of a proton, i.e. it is alwayspositive. The electron charge is always �e. The momentum exchange ratep
In
is precisely �p
nI
, and the same holds for pen
. (Why?) The quantity Zis the mean charge per ion, n is a number density, and the fluid is neutral inbulk, eZn
I
= ene
.
The key point is that for the charge carriers, all terms proportional tothe mass densities ⇢
I
and ⇢e
are small compared with the Lorentz force andmomentum exchange rates. Hence, to a very good approximation,
0 = eZnI
(E + v
I
⇥ B)� p
In
(89)
0 = �ene
(E + v
e
⇥ B)� p
en
(90)
Adding these two equations and using bulk neutrality leads to ,
0 = eZnI
(vI
� v
e
) ⇥ B � p
In
� p
en
(91)
But eZnI
(vI
� v
e
) is just the current density J , so that
p
In
+ p
en
= J ⇥ B (92)
27
Using this in the neutral equation leads to
@v
@t+ ⇢(v·r)v = �rP � ⇢r�+ J ⇥ B (93)
Remarkably, the net Lorentz force appears unmodified in the equation forthe neutrals.
For a sparsely ionized fluid, departures from ideal MHD appear mainly inthe induction equation. To relate E and B it is best to use the electron forceequation, since the ions may be more closely locked to the neutrals. Thus
E = �v
e
⇥ B � p
en
ene
= �[v + (ve
� v
I
) + (vI
� v)] ⇥ B � p
en
ene
(94)
Now matters start to get very detailed. I present these details in thefollowing section, but for purposes of this course I view this material asentirely optional. Having made the details available to you, however, I feelfree to make a quick summary of the results, leaving it for you to read thenext section if you wish more explanation.
The term v
e
� v
I
is �J/ene
.
The term v
I
� v is related to p
In
by an equation of the form
p
In
= �⇢⇢I
(vI
� v)
where � is a coe�cient that may be calculated from knowledge of the interac-tion cross sections. (See equation (104).) But p
In
is simply related to J ⇥ B
from equation (92), because p
In
is in fact dominant over p
en
. Ultimately,the reason is that the ions are more massive than the electrons.
The final term proportional to p
en
represents ohmic dissipation. Wedenote the electrical conductivity by �
cond
.
Putting all of this together leads to the full induction equation:
@B
@t= r⇥
"
v⇥B � (r ⇥ B)⇥B
µ0
ene
+[(r ⇥ B) ⇥ B] ⇥ B
µ0
�⇢⇢I
� r ⇥ B
µo
�cond
#
(95)
The Details....
Let us examine matters a little more closely. (Please don’t worry about everylast detail. My purpose here is to give you a feeling for all that goes into acalculation like this, and to be able to understand the nomenclature that you
28
will encounter in the literature. You don’t have to become an expert in theminutiae of interstellar kinetic theory for this course!) p
nI
takes the form
p
nI
= nµnI
(v � v
I
)⌫nI
(96)
where n is the number density of neutrals, and µnI
is the reduced mass of anion–neutral particle pair,
µnI
⌘ mI
mn
mI
+mn
, (97)
mI
and mn
being the ion and neutral mass respectively. ⌫nI
is the collisionfrequency of a neutral with a population of ions,
⌫nI
= nI
h�nI
wnI
i. (98)
In equation (98), nI
is the number density of ions, �nI
is the cross sectionfor neutral-ion collisions, and w
nI
is the relative velocity between a neutralparticle and an ion. The angle brackets represent an average over all possiblerelative velocities in the thermal population of particles. Notice that equation(96) has the dimensions of a force per unit volume, and that it is proportionalto the velocity di↵erence between the species: if there is no di↵erence in theirmean velocities, two population of particles cannot exchange momentum.
Why does the reduced mass µnI
appear? Because the reduced mass alwaysappears in any interaction between two individual particles: in the center ofmass frame the equations reduce to a single particle equation with the particlemass equal to the reduced mass. In an elastic one-dimensional collision, forexample, if v is initial relative velocity of the two interacting particles, thenthe momentum exchange is 2µ
12
v, where µ12
is the reduced mass. (Showthis.)
For neutral-ion scattering, we may approximate the cross section �nI
tobe geometrical, which means that the quantity in angle brackets will be pro-portional to µ�1/2
nI
. The order of the subscripts has no particular significancein either the cross section �
nI
, reduced mass µnI
, or relative velocity wnI
. But⌫In
does di↵er from ⌫nI
: the former is proportional to the neutral density n,the latter to the ion density n
I
.
Putting all these definitions together gives
p
nI
= nnI
µnI
h�nI
wnI
i(v � v
I
) (99)
In accordance with Newton’s third law, this is symmetric with respect tothe interchange n $ I, except for a change in sign, p
nI
= �p
In
. All ofthese considerations hold, of course, for electron-neutral scattering as well.Explicitly, we have
p
ne
= nne
µne
h�ne
wne
i(v � v
e
) ' nne
me
h�ne
wne
i(v � v
e
). (100)
29
The gas is assumed to be locally neutral, so that ne
= Zni
where Z is thenumber of ionizations per ion particle. In a weakly ionized gas, Z = 1. Thereduced mass µ
ne
is very nearly equal to the electron mass me
. The collisionrates are given by (see Draine, Roberge, & Dalgarno 1983 ApJ 264, 485 foryet more details) (note, cgs units!):
h�nI
wnI
i = 1.9⇥ 10�9 cm3 s�1 (101)
h�ne
wne
i = 10�15 (128kT/9⇡me
)1/2 = 8.3⇥ 10�10T 1/2 cm3 s�1 (102)
The electron-neutral collision rate is just the ion geometric cross section timesan electron thermal velocity. (The peculiar factor of (128/9⇡)1/2 is a detail ofthe averaging procedure.) But the ion-neutral collision rate is temperatureindependent, much more beholden to long range induced dipole interactions,and significantly enhanced relative to a geometrical cross section assumption.Even if the ion-neutral rate were determined only by a geometrical crosssection, |p
nI
| would exceed |pne
| by a factor of order (me
/µnI
)1/2. In fact,the dipole enhancement of the ion-neutral cross section makes this factorlarger still2.
In the astrophysical literature, it is common to write the ion-neutral mo-mentum coupling in the form
p
In
= ⇢⇢I
�(vI
� v), (103)
where � is the so-called drag coe�cient,
� ⌘ h�nI
wnI
im
I
+mn
(104)
and we will use this notation from here on. Numerically, � = 3 ⇥ 1013 cm3
s�1 g�1 for astrophysical mixtures (Draine, Roberge, & Dalgarno 1983).
We come next to the ions and electrons. The dynamical equations for theions and electrons are
⇢I
@vI
@t+ ⇢
I
v
I
·rv
I
= �rPI
� ⇢I
r�+ ZenI
(E + v
I
⇥B)� p
In
(105)
and
⇢e
@ve
@t+ ⇢
e
v
e
·rv
e
= �rPe
� ⇢e
r�� ene
(E + v
e
⇥B)� p
en
, (106)
2I should be a little bit more careful. The statement that |pne| is larger than |pnI | bya factor of (me/µnI)
1/2 assumes that the velocity di↵erences v � ve and v � vI do notintroduce any mass dependencies, which is generally true.
30
respectively. e will always denote the positive charge of a proton, the absolutevalue of the electron charge, 1.602 ⇥ 10�19 Coulombs or 4.803 ⇥ 10�10 esu.3
For a weakly ionized gas, the Lorentz force and collisional terms dominatein each of the latter two equations. Comparison of the magnetic and inertialforces, for example, shows that the latter are smaller than the former bythe ratio of the proton or electron gyroperiod to a macroscopic flow crossingtime. Thus, to an excellent degree of approximation,
ZenI
(E + v
I
⇥B)� p
In
= 0, (107)
and�en
e
(E + v
e
⇥B)� p
en
= 0. (108)
The sum of these two equations gives
J⇥B = p
In
+ p
en
(109)
where charge neutrality ne
= ZnI
has been used, and we have introducedthe current density
J ⌘ ene
(vI
� v
e
). (110)
The equation for the neutrals becomes
⇢@v
@t+ ⇢v·rv = �rP � ⇢r�+ J⇥B (111)
Due to collisional coupling, the neutrals are subject to the magnetic Lorentzforce just as though they were a gas of charged particles. It is not themagnetic force per se that changes in a neutral gas. As well shall presentlysee, it is the inductive properties of the gas.
Let us return to the force balance equations for the electrons:
�ene
(E + v
e
⇥B)� p
en
= 0. (112)
After division by �ene
, this may be expanded to
E + [v + (ve
� v
I
) + (vI
� v)]⇥B +m
e
⌫en
e[(v
e
� v
I
) + (vI
� v)] = 0,
(113)where we have introduced the collision frequency of an electron in a popula-tion of neutrals:
⌫en
= nh�ne
wne
i. (114)
We have written the electron velocity v
e
in terms of the dominant neutralvelocity v and the key physical velocity di↵erences of our problem. It has
3Beware: esu units are still commonly used in the astrophysical literature! You shouldbecome comfortable with them.
31
already been noted that in equation (109), pen
is small compared with p
In
,provided that the velocity di↵erence |v
e
�v| is not much larger than |vI
�v|.As we argued earlier, the p
en
term in equation (109) is small relative to p
In
:
J⇥B ' p
In
= nnI
µnI(vI
� v)⌫nI
. (115)
It then follows that the final term in equation (113)
me
⌫en
e(v
I
� v),
which is proportional to J ⇥ B, becomes small compared with the thirdterm
(ve
� v
I
)⇥B,
which also proportional to J⇥B, by a factor of order (me
/µIn
)1/2. Thesesimplifications allow us to write the electron force balance equation as
E + v⇥B � J ⇥ B
ene
� J
�cond
+(J ⇥ B) ⇥ B
�⇢⇢I
= 0, (116)
where the electrical conductivity has been defined as
�cond
⌘ e2ne
me
⌫en
(117)
The associated resistivity ⌘ is
⌘ =1
µ0
�cond
, (118)
which has units of m2 s�1. Numerically (e.g. Blaes & Balbus 1994 ApJ, 421,163; Balbus & Terquem 2001, ApJ, 552, 235):
⌘ = 0.0234✓n
ne
◆T 1/2 m2 s�1 (119)
Equation (116) is a general form of Ohm’s law for a moving, multiple fluidsystem.
Next, we make use of two of Maxwell’s equations. The first is Faraday’sinduction law:
r⇥E = �@B
@t. (120)
32
We substitute E from equation (116) to obtain an equation for the self-induction of the magnetized fluid,
@B
@t= r⇥
"
v⇥B � J ⇥ B
ene
+(J ⇥ B) ⇥ B
�⇢⇢I
� J
�cond
#
(121)
It remains to relate the current density J to the magnetic field B. Thisis accomplished by the second Maxwell equation,
µ0
J = r⇥B +@E
@t(122)
The final term in the above is the displacement current, and it may beignored. Indeed, since we have not, and will not, use the “Gauss’s Law”equation
r·E = (e/✏0
)(ZnI
� ne
), (123)
we must not include the displacement current. In Appendix B, we showthat departures from charge neutrality in r·E and the displacement cur-rent are both small terms that contribute at the same order: v2/c2. Thesemust both be self-consistently neglected in nonrelativisitc MHD. (The finalMaxwell equation r·B = 0 adds nothing new. It is automatically satisfiedby equation (120), as long as the initial magnetic field satifies this divergencefree condition.) These considerations imply
J =1
µ0
r⇥B (124)
for use in equation (121).
To summarize, the fundamental equations of a weakly ionized fluid aremass conservation of the dominant neutrals (eq.[85])
@⇢
@t+r·(⇢v) = 0, (125)
the equation of motion (eq. [111] with [124])
⇢@v
@t+ ⇢v·rv = �rP � ⇢r�+
1
µ0
(r⇥B)⇥B, (126)
and the induction equation (eq. [121] with [118] and [124])
@B
@t= r⇥
"
v⇥B � (r ⇥ B)⇥B
µ0
ene
+[(r ⇥ B) ⇥ B] ⇥ B
µ0
�⇢⇢I
� r ⇥ B
µo
�cond
#
(127)
33
It is only natural that the reader should be a little taken aback by thesight of equation (127). Be assured that it is rarely, if ever, needed in fullgenerality: almost always one or more terms on the right side of the equationmay be discarded. When only the induction term v⇥B is important, we referto this regime as ideal MHD. The three remaining terms on the right are thenonideal MHD terms.
To get a better feel for the relative importance of the nonideal MHD termsin equation (121), we denote the terms on the right side of the equation,moving left to right, as I (induction), H (Hall), A (ambipolar di↵usion), andO (Ohmic resistivity). We will always be in a regime in which the presenceof the induction term is not in question. More interesting is the relativeimportance of the nonideal terms. The explicit dependence of A/H andO/H in terms of the fluid properties of a cosmic gas has been worked out byBalbus & Terquem (2001):
A
H= Z
9⇥ 1012 cm�3
n
!1/2 ✓
T
103 K
◆1/2
✓vA
cS
◆(128)
andO
H=✓
n
8⇥ 1017 cm�3
◆1/2
✓cS
vA
◆(129)
Here n is the total number density of all particles, T is the kinetic tempera-ture, v
A
is the so-called Alfven velocity (much more about this quantity willcome later!),
v
A
=Bpµ0
⇢(130)
and cS
is the isothermal speed of sound,
c2S
= 0.429kT
mp
(131)
where k is the Boltzmann constant and mp
the mass of the proton. The coef-ficient 0.429 corresponds to a mean mass per particle of 2.33m
p
, appropriateto a molecular gas with a 10% helium admixture.
As reassurance that the fully general nonideal MHD induction equation isnot needed for our purposes, note that equations (128) and (129) imply thatfor all three nonideal MHD terms to be comparable, T ⇠ 108 K! Obviouslythis is not a weakly ionized regime. In figure (2), we plot the domains ofrelative dominance of the nonideal MHD terms in the nT plane.
Our emphasis of the relative ordering of the nonideal terms in the in-duction equation should not obscure the fact that ideal MHD is often an
34
Figure 3: Parameter space for nonideal MHD. The curves correspond to thecase v
A
/cS
= 0.1. (From Kunz & Balbus 2004, MNRAS, 348, 355.)
35
excellent approximation, even when the ionization fraction is ⌧ 1. For ex-ample, the ratio of the ideal inductive term to the ohmic loss term is givenby the Lundquist number
` =vA
H
⌘(132)
where H is a characteristic gradient length scale. To orient ourselves, let usconsider the case of a protostellar disc and set H = 0.1R, where R is theradial location in the disc. (This would correspond to H being about thevertical thickness of the disc.) Then ` is given by
` ' 2.5(ne
/n)(vA
/cS
)Rcm
,
Rcm
being the radius in centimeters. In other words, the critical ionizationfraction at which ` = 1 is about
(ne
/n)crit
= 0.4(cS
/vA
)R�1
cm
⇠ 10�13(cS
/10vA
)
at R = 1 AU. The actual ionization fraction at this location may be aboveor below this during the course of the solar systems evolution, but the pointworth noting here is that R
cm
is a large number for a protostellar disc! Ion-ization fractions far, far below unity can render an astrophysical gas a nearperfect electrical conductor. It therefore makes a great deal of sense to beginby examining the behaviour of an ideal MHD fluid.
Exercise. Show that the Lorentz force may be written
J ⇥ B = @i
B
i
Bj
µ� �
ij
B2
2µ
!
⌘ @i
TL
ij
. (133)
Exercise. Show that the Newtonian self-gravity force may be written
�⇢r� = @i
� gi
gj
4G⇡+ �
ij
g2
8G⇡
!
⌘ @i
TN
ij
. (134)
where gi
= �@i
�. (Hint: @i
@i
� = 4⇡G⇢.)
Exercise. Show that the inertial terms in the equation of motion be written
⇢@t
vi
+ ⇢vj
@j
vi
+ @i
P = @t
(⇢vi
) + @i
(⇢vi
vj
+ �ij
P ) ⌘ @t
(⇢vi
) + @i
TR
ij
, (135)
which defines the Reynolds stress TR
ij
.
Exercise. Show that the equation of motion may be written
⇢@t
vi
+ @i
Tij
= 0, (136)
where Tij
= TL
ij
+TN
ij
+TR
ij
is the energy-momentum stress tensor. This formof the equation of motion is most readily generalized when relativity becomesimportant.
36
... I deduced that the forceswch keep the Planets in theirOrbs must [be] reciprocally as thesquares of their distances fromcentres about wch they revolve: &thereby compared the force requi-site to keep the Moon in her Orbwith the force of gravity at thesurface of the earth, & found themanswer pretty nearly.
— Sir Isaac Newton
We can lick gravity, but some-times the paperwork is over-whelming.
— Dr. Werner von Braun
3 Gravity
3.1 Legendre Exapansion
The gravitational field a distance r from a point massM located at the originis
g = �GMr
r3, (137)
or more generally,
g = �GM(r � r
0)
|r � r
0|3 (138)
if the point mass is located at r
0. This field is derivable from a potentialfunction �,
� = � GM
|r � r
0| , g = �r� (139)
Gravity is a linear theory, and the fields (and thus the potentials) from anextended mass distribution superpose. Therefore, in general,
� = �GMZ ⇢(r0)d3r0
|r � r
0| , (140)
37
with|r � r
0| = (r2 � 2rr0 cos ✓ + r02)1/2 (141)
where ✓ is the angle between r and r
0. For r � r0,
(r2 � 2rr0 cos ✓ + r02)�1/2 = r�1
1� 2r0
rcos ✓ +
r02
r2
!�1/2
(142)
We are very often interested in the potential at great distances from thesource, r � r0. The last two terms in r0/r are small, so we define
� = �2r0
rcos ✓ +
r02
r2⌧ 1. (143)
Then,
1� 2r0
rcos ✓ +
r02
r2
!�1/2
= r�1(1 + �)�1/2 = r�1
"
1� �
2+
3�2
8+ ...
#
(144)
Expanding � and retaining terms through order (r0/r)2, we find
r�1
1� 2r0
rcos ✓ +
r02
r2
!�1/2
=1
r
2
41 +
r0
r
!
cos ✓ +
r0
r
!2
1
2(3 cos2 ✓ � 1) + ...
3
5
(145)The expansion consists of powers of (r0/r) multiplied by a polynomial incos ✓. These latter are denoted P
l
(cos ✓) and are known as Legendre polyno-mials. Their properties are discussed very clearly in Jackson’s text, ClassicalElectrodynamics. The most important of these for our purposes is that thePl
are orthogonal when integrated over spherical solid angles:
ZPl
(cos ✓)Pl
0(cos ✓) d⌦ =4⇡
2l + 1�ll
0 (146)
Because of the symmetry between r and r
0, in general we must have
1
|r � r
0| =1
r>
1X
l=0
r<
r>
!l
Pl
(cos ✓), (147)
where r>
(r<
) is the greater (lesser) of r and r0.
38
g
dA*
dA
g dA*dA =
M
||
r
|g|||
Figure 4: Gauss’s law. The area element dA has projection dA⇤ parallel to�g. dA⇤ is thus a di↵erential area element of a sphere surrounding M atdistance r.
3.2 Gauss’s Law
A remarkable property of a 1/r2 force law is that
Zg · dA = �4⇡GM (148)
where the surface integral is over any volume containing a total mass M .To prove this, we show that it is true for a single point mass. Then bysuperposition, it is true for any distribution.
Note that |g · dA| is just the product of g with the area dA⇤, the projec-tion of dA parallel to �g. For a point mass, g is radial, and the projectedarea of dA⇤ is precisely that of a small portion of a sphere centered on themass point, at the same radius r as dA. Since this small spherical area isr2d⌦, where d⌦ is the solid angle subtended by the area at the point mass,and g = �GM/r2,
g · dA = �GM d⌦ (149)
This is independent of r, and the integral of the whole area just adds upthe total solid angle subtended by the volume enclosing the mass: 4⇡. Thisproves Gauss’s law for a point mass. By linear superposition, it is thereforetrue for any mass distribution interior to the surface.
39
3.3 Poisson Equation
We have shown Zg · dA = �4⇡G
Z⇢ dV (150)
for any mass distribution inside the volume V . Using the divergence theorem,Z
r · g dV = �4⇡GZ
⇢ dV (151)
and since g = �r�, Zr2� dV = 4⇡G
Z⇢ dV (152)
The volume is arbitrary, hence we conclude
r2� = 4⇡G⇢, (153)
which is known as the Poisson equation. It allows us to compute the gravita-tional potential, and hence the graviational forces, from a given distributionof mass. It must added to the fluid equations of motion to calculate theevolution of a self-gravitating system.
Exercise. Show that the Lorentz force may be written
J ⇥ B = @i
B
i
Bj
µ� �
ij
B2
2µ
!
(154)
and that for a self-gravitating gas the Newtonian gravitational force has avery similar form:
�⇢r� = �@i
gi
gj
4⇡G� �
ij
g2
8⇡G
!
(155)
where gi
= �ri
�, g2 = gi
gi
.
The quantities inside the @i
operators are known respectively as theMaxwell and gravitational stress tensors. They play a key role in momentumand energy transport in magnetic and self-gravitating systems.
3.4 Gravitational Tidal Forces.
As an illustration of how the expansion of the potential function can be used,let us calculate the height of the tides that are raised on the earth by the
40
r
s
Earth
θ
Moon
Figure 5: Geometry for tides raised on the earth by the moon.
moon—though our calculation will be completely general for any two bodyproblem apart from the numbers we use.
We define the z axis to be along the line joining the centers of the earthand the moon. The distance between the centers will be r, and a point onthe earth’s surface will be at a vector location r + s relative to the centerof the moon. Let s = (x, y, z) in Cartesian coordinates with origin at thecenter of the earth. Note that
1
|r + s| = (r2 + s2 + 2rs cos ✓)�1/2 (156)
so we need to keep track of the sign, which is di↵erent from our r, r0 expansion.We regard r as fixed, and calculate forces by taking the gradient with respectto x, y, and z. We have, with r � s,
� GMm
|r + s| = �GMm
r
"
1� s cos ✓
r+✓s
r
◆2
P2
(cos ✓) + ...
#
(157)
whereMm
is the mass of the moon. Di↵erentiating with respect to z = s cos ✓gives, to first approximation
�@�
@z= �GM
m
r2(158)
which looks familiar: it is the Newtonian force acting between the centersof the two bodies, directing along the line joining them. It is not the tidalforce, which comes in at the next level of approximation. The tidal potentialis:
� (tidal) = �GMm
s2
r3P2
(cos ✓) (159)
And the tidal force is, after carrying out the gradient operation,
g (tidal) = �r� =GM
m
r3(�x,�y, 2z) (160)
41
Tidal forces try to squeeze matter along the directions perpendicular to theline joining the bodies, and try to stretch matter along the direction parallelto this line. Note that we speak here only of the forces; the resulting dis-placements can be much more complex. Not only are they sensitive to localsurface features in the oceans, there are also time delays in the response ofthe displacement, due to the presence of dissipation.
Let us assume, however, that the new shape of the earth has adjustedso that the surface now follows an equipotential of the earth’s gravitationalfield plus that of the moon. Let �
1
be the potential function of the earth’sunperturbed spherical field. Let �
2
be the new potential function in thepresence of the moon’s potential, di↵ering slightly from �
1
at a given location.The tidal force causes a displacement of the original spherical equipotentialsurfaces by an amount ⇠, and this is what we wish to calculate. Let �
1
(s) = �be constant on a sphere of radius s. The new equipotential surface with this(constant) value of � is �
2
(r + ⇠), where ⇠ is the small displacement causedby the moon. It is this quantity that we wish to calculate. If an equipotentialsurface of �
2
has the same value as an equipotential surface of �1
, but onlyafter the surface has been displaced by ⇠, then
� = �2
(s+ ⇠) = �2
(s) + ⇠ · r�2
= �1
(s) (161)
where s is the radius of the earth. But at the same location s:
�2
(s)� �1
(s) = � (tidal), (162)
and to leading order we may replace �2
with �1
in the term proportional to⇠. Then
�⇠ · r�1
= � (tidal) (163)
which states the physically very sensible result that the work done againstthe gravitational force of the earth in distorting the surface is provided by theadditional tidal potential energy. Writing the potential functions explicitly:
⇠s
GMe
s2=
GMm
s2
r3P2
(cos ✓) (164)
or
⇠s
= sM
m
Me
✓s
r
◆3
P2
(cos ✓) (165)
This works out to be
⇠s
= 0.32P2
(cos ✓) meters (166)
for the earth-moon system. Notice how extremely sensitive the height of thetidal displacement is to the separation distance r. When the moon was afactor of 2 closer to the earth, as it is believed to have been on a timescaleof 109 years ago, the tidal forces were almost an order of magnitude larger.
42
3.5 The Virial Theorem
The Virial Theorem is one of the most useful theorems in astrophysical gas-dynamics. Basically, it is an integral form of the equation of motion in fullgenerality. When the dominant balance is between two forces, the theoremstates that the associated energies must be comparable in strength. We shalluse Cartesian index notation in our proof.
Begin with
⇢Dv
i
Dt= �@
i
P � @i
B2
2µ
!
� ⇢@i
�+B
j
µ@j
Bi
(167)
where
�(r) = �GZ ⇢(r0) d3r0
|r � r
0| (168)
is the gravitational potential the system. Note that
�@i
� = �GZ ⇢(r0) (r
i
� r0i
) d3r0
|r � r
0|3 (169)
Multiply the equation of motion by ri
and sum over i,
⇢ri
Dvi
Dt= �r
i
@i
P � ri
@i
B2
2µ
!
� ⇢ri
@i
�+ ri
Bj
µ@j
Bi
(170)
and then integrate over a fixed volume V . For the pressure integral,
�Z
ri
@i
P dV = �Z
@i
(ri
P ) dV + 3Z
P dV (171)
= �Z
Pr · dA+ 3Z
P dV
= �Z
Pr · dA+ 2Z
Utherm
dV (172)
where Utherm
= (3/2)P is the thermal energy density.
The integral involving the potential isZ
⇢ri
@�
@ri
d3r = GZ ⇢(r)⇢(r0)r
i
(ri
� r0i
)
|r � r
0|3 d3r d3r0 (173)
If we switch the labels r and r
0, we obtainZ
⇢ri
@�
@ri
d3r = GZ ⇢(r)⇢(r0)r0
i
(r0i
� ri
)
|r � r
0|3 d3r d3r0 (174)
43
Adding and dividing by 2:
Z⇢r
i
@�
@ri
d3r =G
2
Z ⇢(r)⇢(r0)
|r � r
0| d3r d3r0 ⌘ �V. (175)
i.e., this is just minus the gravitational potential energy V . (The factor of1/2 is present because each pair of interacting fluid elements occurs twice inthe integration, but should only be counted once.)
On to the magnetic integrals:Z
ri
@i
(B2/2µ) d3r =Z(B2/2µ)r · dA� 3
Z(B2/2µ) d3r, (176)
where we have integrated by parts and used the divergence theorem. And
Z ri
µ@j
(Bi
Bj
)d3r =Z(r · B)
B
µ·dA�
Z �ij
µB
i
Bj
d3r
=Z(r · B)
B
µ·dA�
Z B2
µd3r. (177)
The sum of all the terms on the right side of our equation is then
2Etherm
+ V +M �Z
P +B2
2µ
!
r · dA+1
µ
Z(r · B)B · dA (178)
where
Etherm
=Z
Utherm
d3r, M =Z B2
2µd3r (179)
are the total thermal and magnetic energies.
For the left side of the virial equation we start with the following identity:
Z⇢DQ
Dtd3r =
Z⇢
@Q
@t+ v·rQ
!
d3r =Z @(⇢Q)
@td3r +
Zr·(⇢vQ) d3r
(180)where the second equality follows from mass conservation. The last integralcan be converted to a surface integral of the flux ⇢vQ over a bounding area.If we choose the surface so that the velocity vanishes at this surface, thenthis integral vanishes, and we shall make this assumption. We then have:
Z⇢DQ
Dtd3r =
Z @(⇢Q)
@td3r =
d
dt
Z⇢Qd3r (181)
44
With this in hand, we perform the following manipulations:
Z⇢r
i
Dvi
Dtd3r =
Z⇢D(r
i
vi
)
Dtd3r �
Z⇢v
i
Dri
Dtd3r (182)
=d
dt
Z⇢r
i
vi
d3r �Z
⇢v2 d3r
=d
dt
Z⇢r
i
Dri
Dtd3r � 2KE
=d
dt
Z ⇢
2
Dr2
Dtd3r � 2KE
=d2
dt21
2
Z⇢r2d3r � 2KE
=1
2
d2I
dt2� 2KE (183)
where I isR⇢r2d3r and KE denotes the total kinetic energy of the fluid. The
Virial theorem is then:
1
2I = 2KE + 2E
therm
+ V +M �Z
P +B2
2µ
!
r · dA+1
µ
Z(r · B)B · dA
(184)where the velocity is assumed to vanish over the bounding surface. TheVirial theorem shows that when a dominant steady-state balance is presentbetween two e↵ects–pressure and gravity, say—the two associated energiesare comparable. In particular, for a star in hydrostatic equilibrium,
Etherm
= �1
2V, E
total
= Etherm
+ V =V
2(185)
since the pressure vanishes at the surface, and magnetic fields are generallynegligible for stellar hydrostatic equilibrium.
3.6 The Lane-Emden Equation
Consider the hydrostatic equilibrium of a star with a very simple equation ofstate: P = K⇢�. Special cases of interest include � = 1 (isothermal spheres),� = 6/5, 2 (elementary analytic solutions), � = 5/3, 4/3 (nonrelativistic,relativistic white dwarfs). Our fundamental equation
�r� =1
⇢rP (186)
45
becomes
�r2� = �4⇡G⇢ = r· 1
⇢rP
!
= r· 1
⇢rK⇢�
!
(187)
Assuming spherical geometry and carrying through the di↵erentiation,
K�
� � 1
1
r2d
drr2 d⇢��1
dr
!
= �4⇡G⇢ (188)
For the case of an isothermal gas,
K1
r2d
drr2 d ln ⇢
dr
!
= �4⇡G⇢ (189)
Next, we introduce the polytropic index,
n =1
� � 1(190)
and let the density ⇢ be written
⇢ = ⇢c
✓n (191)
where ⇢c
is the central density. If we further rescale the length r as r = a⇠,with a a length to be determined, our equation becomes
1
⇠2d
d⇠⇠2d✓
d⇠= �
"4⇡G⇢1�1/n
c
(� � 1)a2
K�
#
✓n (192)
Obviously, we should choose a so that the constant in square brackets isunity, or
a2 =K(n+ 1)⇢(1/n)�1
c
4⇡G(193)
The resulting equation:1
⇠2d
d⇠⇠2d✓
d⇠= �✓n (194)
is known as the Lane-Emden equation.
46
3.7 Solution properties of the Lane-Emden Equation
To solve the Lane-Emden equation, we need boundary conditions at theorigin ⇠ = 0. Let us return to the original first order equation of hydrostaticequilibrium. Since we are in spherical symmetry, this may be written
dP
dr= �GM(r)⇢
r(195)
where M(r) is the mass interior to radius r. We have made use of Gauss’stheorem, which assures us that the mass exterior to r contributes nothingto the force at r, and that the mass interior to r acts as though it wereconcentrated at r = 0. When r is very small, M(r) ' 4⇡⇢
c
r3/3, and thusdP/dr ! 0. But since P ⇠ ✓1+n, it follows immediately that the ⇠ gradientof ✓ must also vanish at the origin. Our two boundary conditions on ✓ at⇠ = 0 are
✓(0) = 1, ✓0(0) = 0. (196)
It is now a simple matter to show at all solutions to the Lande-Emdenequation must have the same asymptotic form as ⇠ ! 0:
✓(⇠) ! 1� ⇠2
6+ ... (197)
It is natural to ask whether there are solutions that extend to infinity, andif so, what their asymptotic form is. This can be done at once by lookingfor solutions of the ✓ = A⇠p, and solving for A and p by demanding self-consistency. We have
1
⇠2d
d⇠⇠2d✓
d⇠= p(p+ 1)A⇠p�2 = ✓n = �An⇠np (198)
This requires
p =2
1� n, An�1 =
2(n� 3)
(n� 1)2(199)
In fact, this is an exact solution, not just an approximate one. But it isnot the exact solution we want, because it does not satisfy the boundaryconditions at the origin. Moreover, there is a subtle flaw in our analysis.The case p = �1, n = 3 breaks down, because what we thought was theleading order term on the right side of the Lane-Emden equation actuallyvanishes. We cannot exclude a solution with asymptotic behaviour ✓ ⇠ 1/⇠on the basis of our analysis; we simply have to retain the next order terms!
In fact, notice that a function of the form (1 + ⇠2/3)�1/2 satisifes boththe boundary conditions at ⇠ = 0, and has the asymptotic form ✓ ⇠ A/⇠.
47
Amazingly enough, this turns out to be an exact solution of the equation,the only known analytic solution of a nonlinear form of the Lane-Emdenequation satisfying all the boundary conditions at ⇠ = 0. (The cases n = 0and n = 1 reduce to elementary cases, which you should solve explicitly.)The solution
✓ = (1 + ⇠2/3)�1/2 (200)
corresponds to n = 5 (show!). All solutions with n � 5 extend to infinity,those with n < 5 are finite.
Exercise. Expand equation (200) through terms of order 1/⇠3 to showthat it satisfies the Lande-Emden equation only to leading order, with n = 5.
3.8 Polytrope Masses
The mass interior to radius r is given by
M(r) = � r2
⇢G
dP
dr= �Kr2
G
✓1 +
1
n
◆⇢1/n
d ln ⇢
dr(201)
Translating this to ✓ and ⇠ variables gives:
M(r) =
"K3(n+ 1)3
4⇡G3
#1/2
⇢(3�n)/2n
c
⇠2�����d✓
d⇠
����� (202)
The total mass of a polytrope can be evaluated once the surface values of ⇠and ✓0(⇠) are known: no further integration is needed to determine the massonce ✓ is tabulated. Notice something remarkable: for n = 3, the mass is auniversal constant, independent of whatever is chosen for the central density⇢c
!
This is a result of tremendous significance for astrophysics. For whitedwarf stars, the Lande-Emden equation is not simply an approximation tohelp us understand more complex and realistic models. A polytropic equa-tion of state is in fact exact for a gas of degenerate electrons in either thenonrelativistic limit, in which case P ⇠ ⇢5/3 and n = 1.5, or in the extremerelativistic limit, in which case P ⇠ ⇢4/3 and n = 3. The pressure from thenuclei in the white dwarf, is nondegenerate, and negligible compared withthe degenerate electron pressure.
Imagine constructing white dwarfs by slowly increasing the value of thecentral density ⇢
c
. At first, when the the gas in nonrelativistic, the largerthe value of ⇢
c
, the larger the value of the stellar mass M . Note howeverthat the radius of the star, which is proportional to a, gets smaller andsmaller. At some point the electrons become so confined that they become a
48
relativisitic degenerate gas, and n ! 3. The mass of the star can increase nomore! This limiting mass is called the Chandrasekhar Mass, and it may beevaluated by calculating the value of ⇠2✓ = 2.01824 at the surface of an n = 3polytrope (notice that the n = 5 polytrope has the value 1.732 evaluated atthe “surface” ⇠ ! 1), and by using K = 4.935 ⇥ 109 in SI units for anextreme relativistic degenerate gas. One finds
MCh
= 1.44M� (203)
It is the existence of this upper mass limit, whose value depends upon thevery simple solution of the n = 3 Lane-Emden equation, that leads to theproduction of neutron stars and black holes.
3.9 The Gravitational Potential Energy of a Polytrope.
It is a remarkable fact that even though we cannot express the solution tothe Lane-Emden equation in a closed analytic form for arbitrary �, there isa very simple expression for the gravitational potential energy as a functionof mass M , radius R and of course �. The calculation is very igeneous.
The Potential Energy U is given by
U = �Z
R
0
GMr
r4⇡⇢r2dr (204)
where Mr
is shorthand for M(r), the mass within radius r. We then performa series of substitutions and integrations by parts:
U = �Z
R
0
GMr
⇢
r24⇡r3dr =
ZR
0
dP
dr4⇡r3 dr = �3
ZR
0
P
⇢
!
4⇡⇢r2 dr (205)
Continuing with dMr
/dr = 4⇡⇢r2
U = �3Z
R
0
P
⇢
! dM
r
dr
!
dr = 3Z
R
0
d
dr
P
⇢
!
Mr
dr (206)
Notice that we have yet to use anything but hydrostatic equilibrium by wayof physics! Now, for the first time, we use the fact that we have a polytrope.From P = K⇢� follows
d
P
⇢
!
=
� � 1
�
!dP
⇢,
49
and we then have
U = 3
� � 1
�
!ZR
0
Mr
⇢
dP
dr= �3
� � 1
�
!ZR
0
GM2
r
r2dr (207)
A last integration by parts gives
U = 3
� � 1
�
!ZR
0
GM2
r
d
dr
1
rdr = 3
� � 1
�
! GM2
R+ 2U
!
(208)
where we have written U for � R GMr
dMr
/r on the right. We now have asimple equation involving only U , GM2/R, and �! It’s solution is:
U = �3(� � 1)
5� � 6
GM2
R(209)
Notice the recovery of the well-known constant density solution,
U = �3
5
GM2
R(constant density)
in the limit � ! 1. We can also make sense of why the case n = 5 insection 3.7 marks the division between finite and and infinite extent of thepolytrope: it corresponds to the the gravitational potential energy becominginfinite.
3.10 E↵ects of Rotation
We conclude our discussion of polytropes by examining a few simple rotatingsystems. In the presence of rotation, the equation for hydrostatic equilibriumbecomes:
�1
⇢rP �r� = �R⌦2
e
R
(210)
If P = P (⇢), the vorticity conservation equation implies that ⌦ = ⌦(R).(This is von Zeippel’s theorem, proved in an earlier exercise.) Then
R⌦2(R) = r"Z
R
R0⌦2(R0) dR0#
⌘ r�⌦
(211)
which defines the velocity potential function �⌦
. We see that all terms arederivable from a potential function. The equation of hydrostatic equilibriumbecomes
H + �� �⌦
= C = CONSTANT, (212)where H is the enthalpy function
H ⌘Z dP
⇢=
�
� � 1
P
⇢(213)
50
3.10.1 Example 1: Rotating Liquid
As a first example, consider the case of a constant density, uniformly rotatingliquid in a constant vertical gravitational field g. (Constant density corre-sponds to � ! 1.) Then � = gz and �
⌦
= R2⌦2/2. If z0
is the surfaceheight of the liquid at the center R = 0, then C = gz
0
, assuming that thepressure vanishes at the liquid surface. The solution is
P
⇢= g(z
0
� z) +1
2R2⌦2, (214)
and the P = 0 surface of the liquid is given by
z = z0
+R2⌦2
2g(215)
The primary astronomical application of this equation is that it tells us howto make parabolic mirrors! Liquid glass is spun in huge ovens, and its surfaceacquires a perfectly parabolic shape. When silvered, the parabolic mirrorreflects all incoming parallel rays onto a single point, the focus. In this way,an image is formed. Even a small segment of a parabolic mirror has thisproperty. Very large telescopes can be constructed (and controled) by usingmany small mirrors.
3.10.2 Example 2: Sub-Keplerian Disks
The Keplerian rotation law of planetary motion is
v2�
=GM
R(216)
In a gas disc, this law is modified by gas pressure. But if we assume thatthe rotation law follows the same functional form (linearly proportional tocentral mass), the modification may be parameterized as
v2�
=GM cos �
R(217)
where � is a free parameter. Such a profile is said to be sub-Keplerian, withthe additional support provided by a pressure gradient. As before, v
�
dependsonly upon cylindrical radius R for a polytropic gas. With � = �GM/r and�
⌦
= �GM cos �/R, the integrated equation of hydrostatic equilibrium isnow:
�
� � 1
P
⇢� GM
r+
GM cos �
R= H1 (218)
51
where H1 is the (constant) enthalpy at R = 1. Recall that R = r sin ✓,where ✓ is the usual spherical angle measured from the positive z axis. Wedefine the latitude angle � = ⇡/2 � ✓. Then, equipotential surfaces (as wellas isobaric and isochoric4 surfaces) are given by the equation
cos �
cos�� 1
!
= Cr (219)
where the constant C, a combination of enthalpy terms, may be either pos-itive or negative. When C > 0, then � > � everywhere, and the surface isunbounded in spherical radius, extending to the cylinder
Rmax
= r cos� = (cos �)/C (220)
When C = 0, the surface is a cone � = �. Note that all surfaces approach thiscone as r ! 0. Since these are surfaces of constant density, this result gives usa geometrical interpretation of what was originally a dynamical parameter:� is the opening angle of the disc wedge. Finally, when C < 0, then � < �everywhere, and the surface is bounded, reaching a maximum radial extentof
rmax
= |C|�1 (1� cos �) (221)
at � = 0. These surfaces, which form the interior surfaces of constant densityin the disc, are shaped like drops of water.
Since gas can move freely along equipotential surfaces, the unbounded sur-faces suggest the possibility of subsonic winds and outflows from discs, witha small enough velocity to allow the assumption of hydrostatic equilibrium.Whether this occurs in reality depends upon how easily the escaping gas canacquire angular momentum to orbit at its asymptotic cylinder R = R
max
.
The bounded surfaces correspond to the body of the disc itself. Theirshape suggests that slow but significant radial mixing might occur in thedisc, if gas can flow along these surfaces and exchange angular momentumas needed. The protostellar disc that formed the solar system o↵ers someevidence that mixing has occurred because dust grains far from the sun seemto have been exposed to the higher temperatures of the inner regions. Thispossibility of radial mixing in discs is currently being explored by computersimulations.
Our results are summarized in figure 3.
3.11 Self-gravity and Rotation
The simplest problem with self-gravity is that of a uniformly rotating, con-stant density cylinder. An application of Gauss’s law shows that at a distance
4Isochoric means constant density.
52
1 2 3 4 5 6 7 8 9 100
1
2
3
4
5
6
7
8
9
x H∞
/(G M)
y H
∞/(
G M
)
isoenergy (density, pressure) plots for H∞>0, β = π/6
log10
(cs2/((γ−1)H
∞))
OUTFLOW
ρ =
0
−0.2
−0.1
0
0.1
Figure 6: Equipotential disc contours for the case when ⌦ is 0.93 of itsKeplerian value, corresponding to an opening wedge angle of � = 30�, markedby the line marked OUTFLOW. (The colour scale on the right gives contourinformation for c2
S
⌘ �P/⇢ as shown, but it may be ignored if you do not havea colour printer!) The constant H1 is the enthalpy at infinity, �/(��1) timesP1/⇢1. Open contours become very closely packed near the ⇢ = 0 boundary,are not drawn.
53
R from the rotation axis, the gravitational field is
gR
= �2Gµ(R)
R(222)
where µ(R) is the mass per unit length interior to R,
µ(R) =Z
R
0
2⇡⇢RdR = ⇡R2⇢. (223)
This givesgR
= �2⇡G⇢R (224)
and an associated potential of
� = ⇡G⇢R2. (225)
The potential integration constant is not important here, since it can be ab-sorbed into the integration constant of the hydrostatic equilibrium equation,which is
P
⇢+ ⇡G⇢R2 � 1
2R2⌦2 = CONSTANT =
P0
⇢(226)
where P0
is the central density. Our solution is then
P
⇢=
P0
⇢+
1
2R2⌦2
✓1� 2⇡G⇢
⌦2
◆(227)
When self-gravity dominates over rotation, the pressure goes to zero at afinite radius, R
P=0
:
[RP=0
]2 =2P
0
⇢(2⇡G⇢� ⌦2)(228)
On the other hand, when rotation dominates, we find that the pressure risessharply away from the center:
P
⇢=
P0
⇢+
1
2R2⌦2 (229)
There is interesting application of this simple formula, which is to the centralregion of a hurricane, where the pressure can be very, very low. (At largerotation velocities the earth’s Coriolis force is negligible; we also assume thatthe compressibility of air is unimportant.) The characteristic feature of ahurricane is a central low pressure “eye”, surrounded by an “eye wall,” inwhich the pressure rises sharply. This is reproduced in our quadraticallygrowing solution for the pressure profile. In the exterior region of the hur-ricane, which is not reproduced in the above formula, significant departuresfrom solid body rotation occur, and the pressure decreases outwards.
54
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