Dynamics of Astrophysical Discs Mathematical Tripos, Part IIIProfessor Gordon Ogilvie Lent Term 2019

[Lecture 1, 17.01.19]

[Introductory lecture – see separate document]

[Lecture 2, 22.01.19]

1. ORBITAL DYNAMICS

1.1. Orbits in an axisymmetric potential

Test particle in gravitational potential Φ of star, planet, black hole, galaxy, etc.

Cylindrical polar coordinates (r, φ, z) (radial, azimuthal, vertical).

Newtonian dynamics.

Assume Φ is

• axisymmetric:Φ = Φ(r, z),

• reflectionally symmetric:Φ(r,−z) = Φ(r, z).

Important special case: point-mass potential ⇒ Keplerian orbits:

Φ = − GM√r2 + z2

.

Lagrange’s equations of motion:d

dt

∂L

∂qi=∂L

∂qi.

Lagrangian for a particle of unit mass:

L = T − V =1

2

(r2 + r2φ2 + z2

)− Φ(r, z).

Conserved quantities:

• specific angular momentum

h =∂L

∂φ= r2φ,

• specific energy

ε =∑i

qi∂L

∂qi− L = T + V =

1

2

(r2 + r2φ2 + z2

)+ Φ(r, z).

1

Radial and vertical equations:

r = rφ2 − Φr,

z = −Φz.

Equivalent to

r = −Φeffr ,

z = −Φeffz ,

with effective potential

Φeff = Φ +h2

2r2.

Note that

ε =1

2

(r2 + z2

)+ Φeff .

Family of circular orbits in midplane (z = 0):

0 = −Φeffr (r, 0) = −Φr(r, 0) +

h2

r3,

0 = −Φeffz (r, 0) (X by reflectional symmetry).

Specific angular momentum hc(r), angular velocity Ωc(r) and specific energy εc(r) ofcircular orbits:

hc =√r3Φr(r, 0),

Ωc =hc

r2,

εc =h2

c

2r2+ Φ(r, 0).

Important relation:dεc

dhc

= Ωc.

Proof:

dεc

dr=hc

r2

dhc

dr− h2

c

r3+ Φr(r, 0)

= Ωcdhc

dr.

Keplerian case:

Φ(r, 0) = −GMr, hc =

√GMr, Ωc =

√GM

r3, εc = −GM

2r.

(See Example 1.1 for revision of Keplerian orbits.)

2

1.2. Oscillations and precession

Small departures from circular orbit of radius r in midplane:

δr = −κ2 δr,

δz = −ν2 δz,

withκ2 = Φeff

rr (r, 0), ν2 = Φeffzz (r, 0),

defining epicyclic frequency (radial frequency) κ(r) and vertical frequency ν(r).

Note that Φeffrz (r, 0) = 0 by reflectional symmetry.

Orbit is stable if κ2 > 0 and ν2 > 0, i.e. if orbit minimizes ε for given h.

We have

κ2 = Φrr(r, 0) +3h2

c

r4

=d

dr

(h2

c

r3

)+

3h2c

r4

=1

r3

dh2c

dr

= 4Ω2c + 2rΩc

dΩc

dr,

ν2 = Φzz(r, 0).

Keplerian orbits:κ = ν = Ω.

FIGURE

If κ ≈ Ω, an eccentric orbit precesses slowly. Minimum r (periapsis) occurs at timeintervals ∆t = 2π/κ, corresponding to

∆φ =2πΩ

κ

= 2π

(Ω

κ− 1

)+ 2π

= 2π

(Ω

κ− 1

)mod 2π.

Apsidal precession rate∆φ

∆t= Ω− κ.

Similarly, if ν ≈ Ω, an inclined orbit precesses slowly with nodal precession rate

Ω− ν.

3

1.3. Mechanics of accretion

Two particles in circular orbits in midplane.

Can energy be released by exchange of angular momentum?

Total angular momentum and energy

H = H1 +H2 = m1h1 +m2h2,

E = E1 + E2 = m1ε1 +m2ε2,

In an infinitesimal exchange:

dH = dH1 + dH2 = m1 dh1 +m2 dh2,

dE = dE1 + dE2 = m1Ω1 dh1 +m2Ω2 dh2,

If dH = 0 thendE = (Ω1 − Ω2) dH1.

In practice dΩ/dr < 0,so energy is released by transferring angular momentum outwards.

FIGURE

Generalize argument to allow for exchange of mass:

dM = dm1 + dm2 = 0,

dH = dH1 + dH2 = 0, dHi = mi dhi + hi dmi,

dEi = miΩi dhi + εi dmi

= Ωi dMi + (εi − hiΩi) dmi,

dE = (Ω1 − Ω2) dH1 + [(ε1 − h1Ω1)− (ε2 − h2Ω2)] dm1.

In practiced

dr(ε− hΩ) = −h dΩ

dr> 0,

so energy is released by transferring angular momentum outwards and mass inwards.

FIGURE

This is the physical basis of an accretion disc.

[Lecture 3, 24.01.19]

2. GLOBAL AND LOCAL VIEWS

In the global view we consider the full disc, usually in cylindrical polar coordinates (r, φ, z).

In the local view (local approximation, local model, shearing sheet, shearing box, Hill’sapproximation) we consider a small volume of the disc in the neighbourhood of an orbitingreference point, using local Cartesian coordinates (x, y, z) in the radial, azimuthal andvertical directions.

4

2.1. Local view of orbital dynamics

Select a reference point in a circular orbit of radius r0. The orbit has angular velocityΩ0 = Ω(r0), etc.

Introduce local coordinates (x, y, z):

r = r0 + x, φ = Ω0t+y

r0

, z = z.

Expand the Lagrangian for a particle of unit mass,

L =1

2

(r2 + r2φ2 + z2

)− Φ(r, z)

=1

2

(x2 + (r0 + x)2

(Ω0 +

y

r0

)2

+ z2

)− Φ(r0 + x, z),

to second order in (x, y, z):L = L0 + L1 + L2 + . . . ,

L0 =1

2r2

0Ω20 − Φ0 = constant,

L1 = r0Ω0y + (r0Ω20 − Φr0)x

= r0Ω0y,

L2 =1

2

(x2 + y2 + z2

)+ 2Ω0xy +

1

2Ω2

0x2 − 1

2Φrr0x

2 − 1

2Φzz0z

2.

L0 = constant and L1 =d

dt(r0Ω0y) generate no motion.

L2 is separable into horizontal and vertical parts:

L2 = Lh + Lv =1

2

(x2 + y2

)+ 2Ω0xy −

1

2(Φrr0 − Ω2

0)x2 +1

2z2 − 1

2Φzz0z

2.

Motion due to L2:

x = 2Ω0y + (Φrr0 − Ω20)x,

y + 2Ω0x = 0,

z = −Φzz0z.

Interpret (x, y, z) as Cartesian coordinates in a frame rotating with orbital angular velocityΩ0. Equation of motion includes Coriolis force and tidal potential

Φt =1

2(Φrr0 − Ω2

0)x2 +1

2Φzz0z

2.

Comes from expansion of gravitational plus centrifugal potential, Φ − 12Ω0r

2, to secondorder in (x, z).

5

Introduce the orbital shear parameter q(r):

q = −d ln Ω

d ln r

(=

3

2for Keplerian orbits

).

We found previously

κ2 = 4Ω2 + 2rΩdΩ

dr= 2(2− q)Ω2.

From radial force balance,

Φr(r, 0) = rΩ2 ⇒ Φrr(r, 0) = Ω2 − 2qΩ2,

so

Φt = −q0Ω20x

2 +1

2ν2

0z2.

In practice q > 0, so Φt has a saddle point at the origin.

Rewrite equation of motion in local view as

x− 2Ω0y = 2q0Ω20x,

y + 2Ω0x = 0,

z = −ν20z.

Conserved quantities:

py =∂L2

∂y= y + 2Ω0x,

εh =∑i

qi∂Lh

∂qi− Lh =

1

2

(x2 + y2

)− q0Ω2

0x2,

εv =∑i

qi∂Lv

∂qi− Lv =

1

2z2 +

1

2ν2

0z2.

These come from the expansion of h and ε in the local view. Consider the conservedquantities

h

r0

=r2φ

r0

=(r0 + x)2

r0

(Ω0 +

y

r0

)= constant + (y + 2Ω0x) + · · · ,

ε− Ω0h =1

2

(r2 + r2φ2 + z2

)+ Φ(r, z)− Ω0r

2φ

=1

2

(x2 + (r0 + x)2

(Ω0 +

y

r0

)2

+ z2

)+ Φ(r0 + x, z)− Ω0(r0 + x)2

(Ω0 +

y

r0

)= constant +

1

2

(x2 + y2 + z2

)− q0Ω2

0x2 +

1

2ν2

0z2 + · · ·

(exercise).

6

Local equivalent of circular orbital motion in midplane is

x = constant, y = −q0Ω0x, z = 0,

an orbital shear flow. Coriolis force balances tidal force.

FIGURE

General solution of local equation of motion: note that

x− 2Ω0(py − 2Ω0x) = 2q0Ω20x

x+ 2(2− q0)Ω20x = 2Ω0py

x+ κ20x = 2Ω0py,

so

x = x0 + Re(Ae−iκ0t

),

y = y0 − q0Ω0x0t+ Re

(2Ω0A

iκ0

e−iκ0t),

z = Re(B e−iν0t

),

for complex oscillation amplitudes A,B. These are the local representation of (slightly)eccentric and inclined orbits. The conserved quantities are

py = (2− q0)Ω0x0,

εh =1

2κ2

0

(|A|2 − 1

2q0x

20

),

εv =1

2ν2

0 |B|2.

[Lecture 4, 29.01.18]

Having derived local model, usually omit subscript zero on Ω, q, κ, ν, etc. These quantitiesare regarded as constants, evaluated at r0.

2.2. Symmetries of the local model

Symmetries inherited from global problem:

• Translational symmetry in y: y 7→ y + c

• Reflectional symmetry in z: z 7→ −z

Additional symmetries not present in global view:

• Translational symmetry in x (together with Galilean boost in y): x 7→ x + c, y 7→y − q0Ω0ct

7

• Rotational symmetry by π about z axis (cannot tell inside from outside): x 7→ −x,y 7→ −y

• Scale-invariance (no characteristic length-scale, because zoomed in to scales r0):x 7→ cx

• Separability of horizontal and vertical dimensions

Combination of translational symmetries in x and y means the local model is horizontallyhomogeneous.

FIGURE

3. EVOLUTION OF AN ACCRETION DISC

3.1. Conservation of mass and angular momentum

Evolution of an accretion disc is regulated by conservation of mass and angular momen-tum. These are embodied in 3D equations of fluid dynamics. Reduce to 1D form.

Mass conservation:

∂ρ

∂t+

1

r

∂

∂r(rρur) +

1

r

∂

∂φ(ρuφ) +

∂

∂z(ρuz) = 0.

Integrate ∫Cr

· dA =

∫ ∞−∞

∫ 2π

0

· r dφ dz

over cylinder Cr of radius r:∂M

∂t+∂F

∂r= 0

assuming no loss or gain through vertical boundaries. 1D mass density

M (r, t) =

∫Cr

ρ dA,

radial mass flux

F (r, t) =

∫Cr

ρur dA.

Angular momentum conservation:

Start from equation of motion in the form

∂u

∂t+ u · ∇u = −∇Φ +

1

ρ∇ · T,

where symmetric stress tensor T includes collective effects (pressure, viscosity, magneticfields, self-gravity, turbulence, etc.)

8

φ-component, multiplied by r: assume Φ is axisymmetric:

∂

∂t(ruφ) + u · ∇(ruφ) =

1

ρr

∂

∂r(r2Trφ) +

1

ρr

∂

∂φ(rTφφ) +

1

ρ

∂

∂z(rTφz) = 0.

Combine with mass conservation:

∂

∂t(ρruφ) +

1

r

∂

∂r[r2(ρuruφ − Trφ)] +

1

r

∂

∂φ[r(ρu2

φ − Tφφ)] +∂

∂z[r(ρuφuz − Tφz)] = 0.

Integrate over cylinder Cr of radius r. Assume that ruφ = h(r) from orbital dynamics(examine this assumption later):

∂

∂t(Mh) +

∂

∂r(Fh+ G ) = 0,

assuming no loss or gain through vertical boundaries. Internal torque

G (r, t) = −∫

Cr

rTrφ dA.

Since h depends only on r:∂M

∂t+∂F

∂r= 0,

∂M

∂th+

∂

∂r(Fh+ G ) = 0.

Eliminate M :

Fdh

dr+∂G

∂r= 0,

which determines F . So M evolves according to

∂M

∂t=

∂

∂r

[(dh

dr

)−1∂G

∂r

].

Interpretation:

• angular momentum determines orbital radius

• angular-momentum transport determines mass evolution

[Lecture 5, 31.01.18]

More usual notation:

M = 2πrΣ, F = 2πrΣur, G = −2πνΣr3dΩ

dr,

where surface density Σ(r, t), mean radial velocity ur(r, t) and mean effective kinematicviscosity ν(r, t) are defined by

Σ =

∫ ∞−∞〈ρ〉 dz,

9

Σur =

∫ ∞−∞〈ρur〉 dz,

νΣrdΩ

dr=

∫ ∞−∞〈Trφ〉 dz,

and

〈·〉 =1

2π

∫ 2π

0

· dφ

is an azimuthal average (if required).

Then obtain:∂Σ

∂t=

1

r

∂

∂r

[(dh

dr

)−1∂

∂r

(νΣr3dΩ

dr

)].

For a Keplerian disc:Ω ∝ r−3/2, h = r2Ω ∝ r1/2,

∂Σ

∂t=

3

r

∂

∂r

[r1/2 ∂

∂r

(r1/2νΣ

)],

a diffusion equation for surface density.

Interpretation:

FIGURE

3.2. Evolution of orbital energy

Recall dε = Ω dh for circular orbits.

Consider

∂

∂t(M ε) +

∂

∂r(Fε) = ε

(∂M

∂t+∂F

∂r

)+ F

dε

dr

= 0 + FΩdh

dr

= −Ω∂G

∂r,

so∂

∂t(M ε) +

∂

∂r(Fε+ G Ω) = G

dΩ

dr.

Here G Ω is a radial energy flux associated with the internal torque. The RHS is minusthe rate of dissipation of orbital energy per unit radius.

In more usual notation

∂

∂t(Σε) +

1

r

∂

∂r

(rΣurε− νΣr3dΩ

drΩ

)= −νΣr2

(dΩ

dr

)2

.

10

Assuming that this is converted into heat and lost locally by blackbody radiation, thesurface temperature Ts(r, t) of the disc is given by

2σT 4s = νΣr2

(dΩ

dr

)2

.

3.3. Boundary conditions

At a free boundary the torque vanishes: G = 0.

Inner boundary condition is most important because a spreading ring will reach r = 0 infinite time. IBC depends on nature of central object.

Black hole

• circular orbits are unstable close to event horizon

FIGURE

• e.g. non-rotating (Schwarzschild) BH:

Ω2 =GM

r3, κ2 = Ω2

(1− 6GM

c2r

),

so orbits unstable for r < rms =6GM

c2= 3 rS (marginally stable orbit).

• rapid inspiral from rms

• ur ↑↑ and Σ ↓↓ within rms, so effectively G ≈ 0 at r = rin ≈ rms

Star with negligible magnetic field

• disc may extend to stellar surface r = R∗

• star (supported mainly by pressure) usually rotates at only a fraction of Keplerianrate:

Ω∗ <

√GM

R3∗.

• Ω makes rapid adjustment from Keplerian to stellar in viscous boundary layer

FIGURE

• usual argument:dΩ

dr= 0 at r = rin ≈ R∗ so G = 0 there

11

Star with significant magnetic field

Disc is disrupted within magnetospheric radius, leading to a cavity and magnetosphericaccretion along field lines towards poles of star.

FIGURE

Star may exert torque on disc through magnetic fields. Outcome depends on stellarrotation, field strength, accretion rate.

Treatment for r rin

If mainly interested in scales r rin, we may let rin → 0.

To allow mass flux at origin but no torque:

F → constant, G → 0 as r → 0.

To allow torque at origin:G → constant as r → 0.

For a Keplerian disc:

F ∝ rΣur ∝ r1/2 ∂

∂r

(r1/2νΣ

), G ∝ r1/2νΣ.

First case (no torque): νΣ→ constant as r → 0.

Second case (torque): r1/2νΣ→ constant as r → 0.

3.4. Analysis of the diffusion equation

Consideration of angular momentum transport processes and local vertical structure ofdisc (later) leads to

ν = ν(r,Σ) (e.g. double power law).

• If ν = ν(r) only, diffusion equation is linear

• If ν = ν(r,Σ), diffusion equation is nonlinear

[Lecture 6, 05.02.19]

Linear case

Can be treated using Green’s function.

12

Let G(r, r0, t) be the solution M (r, t) = 2πrΣ(r, t) of the diffusion equation with theinitial condition M (r, 0) = δ(r − r0), i.e. a very narrow ring of radius r0 and unit mass.Then the solution for any initial condition M0(r) is

M (r, t) =

∫ ∞0

G(r, r0, t)M0(r0) dr0

by linear superposition.

Can calculate G(r, r0, t) in terms of Bessel functions for any power law ν ∝ ra and anyboundary conditions.

Easiest special case: suppose ν ∝ r.

Let

ν = Ar, y =

√4r

3A, g = r1/2νΣ = Ar3/2Σ

to obtain classical diffusion equation

∂g

∂t=∂2g

∂y2.

Spreading-ring solution with zero torque (g = 0) at inner boundary y = yin:

g =C√4πt

exp

[−(y − y0)2

4t

]− exp

[−(y + y0 − 2yin)2

4t

].

Initial condition

M = δ(r − r0) ⇒ g =Ar

3/20

2πr0

δ(y − y0)1√

3Ar0

(last factor coming from dy/dr at r = r0) gives 2πC =√A/3, so

G(r, r0, t) =1√

12πArt

exp

[−

(√r −√r0)2

3At

]− exp

[−

(√r +√r0 − 2

√rin)2

3At

].

The error function erf(z) and the complemetary error function erfc(z) are defined by

erf(z) =2√π

∫ z

0

e−t2

dt, erfc(z) = 1− erf(z).

FIGURE

Remaining mass in disc ∫ ∞rin

G(r, r0, t) dr = erf

(√r0 −

√rin√

3At

),

which declines ∝ t−1/2 for large t.

13

FIGURE

Remaining angular momentum (with h ∝√r)∫ ∞

rin

G(r, r0, t)h dr = h0 − hin erfc

(√r0 −

√rin√

3At

),

i.e. initial angular momentum minus that accreted through rin.

FIGURE

Nonlinear case

Consider ν ∝ raΣb and rin → 0.

The problem is then scale-free and admits special algebraic similarity solutions (see Exam-ples 1.5 and 1.6). These are generally attractors for solutions of the initial-value problem.Unlike the linear case, these solutions may have free boundaries beyond which the den-sity vanishes. If the torque vanishes at the origin, then the total angular momentum isconserved, while the total mass of the disc declines as a power-law in time.

3.5. Steady accretion discs

In a steady state, mass conservation gives

F = −M = constant,

where M is the mass accretion rate.

(Neglect the slow variation of the potential due to the increase in M .)

Angular momentum conservation gives

Fh+ G = constant.

If the inner boundary condition is G = 0 at r = rin, the solution is

G = M(h− hin), hin = h(rin).

For a Keplerian disc, we then have

νΣ =M

3π

(1−

√rin

r

).

If we know the function ν(r,Σ), this provides a complete solution for the disc.

From the expression for the surface temperature,

2σT 4s = νΣr2

(dΩ

dr

)2

,

14

in the case of a steady Keplerian disc we have

σT 4s =

3GMM

8πr3

(1−

√rin

r

).

FIGURE

Total luminosity of disc extending to rout =∞ is

Ldisc =

∫ ∞rin

3GMM

4πr3

(1−

√rin

r

)2πr dr =

1

2

GMM

rin

.

This is exactly equal to the rate at which orbital binding energy is transferred to the gasas it passes from to rin. But it is only half the potential energy released. In the case ofaccretion on to a stellar surface, the remaining energy is released in the boundary layer.

[Lecture 7, 07.02.19]

4. VERTICAL STRUCTURE

4.1. Hydrostatic equilibrium

Dominant force balance in z direction perpendicular to plane of disc:

0 = −ρ∂Φ

∂z− ∂p

∂z.

Expand Φ about midplane:

Φ(r, z) = Φ(r, 0) +1

2Φzz(r, 0)z2 + · · · ,

so, for a thin disc,∂Φ

∂z≈ ν2z,

where ν(r) is the vertical frequency (ν = Ω for Keplerian disc).

Equation of hydrostatic equilibrium:

∂p

∂z= −ρν2z.

FIGURE

Notes:

• Essentially an ODE in z at each r (and t).

15

• In the local approximation the ODE

dp

dz= −ρν2z

is exact for hydrostatic solutions independent of (x, y).

• In a self-gravitating disc the disc also contributes to the vertical gravity and affectsthe hydrostatic structure.

4.2. Hydrostatic models

If p and ρ are related in a known way, we can solve for the hydrostatic structure.

e.g. isothermal ideal gas: p ∝ ρ. Write p = c2sρ, where cs = constant is the isothermal

sound speed. Then the solution is a Gaussian:

p ∝ ρ ∝ exp

(− z2

2H2

),

with scaleheight

H =cs

ν

(=cs

Ωfor Keplerian

).

More general approach:

Define the surface density Σ, vertically integrated pressure P and scaleheight H by

Σ =

∫ρ dz, P =

∫p dz, ΣH2 =

∫ρz2 dz,

where integrals are over full vertical extent of disc.

Note that (with suitable boundary conditions)

P =

∫1 · p dz = [zp]−

∫zdp

dzdz = 0 +

∫zρν2z dz = ΣH2ν2.

Reduce to dimensionless form:

ρ(z) = ρ · ρ(z), p(z) = p · p(z),

with

ρ =Σ

H, p =

P

H, z =

z

H.

ρ and p are characteristic values of density and pressure.

Then dimensionless functions satisfy

dp

dz= −ρz

16

and normalization conditions∫ρ dz =

∫p dz =

∫ρz2 dz = 1.

Isothermal solution (p ∝ ρ):

ρ = p =1√2π

exp

(− z

2

2

).

Uniform solution (ρ = constant):

ρ =

1

2√

3, z2 < 3,

0, z2 > 3,, p =

1

4√

3(3− z2), z2 < 3,

0, z2 > 3.

Polytrope of index n (p ∝ ρ1+1/n, n > 0 not necessarily an integer):

ρ = Cρ

(1− z2

2n+ 3

)n, p = Cp

(1− z2

2n+ 3

)n+1

,

(valid for z2 < 2n+ 3 only, otherwise ρ = p = 0), with normalizing constants

Cρ =Γ(n+ 3

2)

Γ(n+ 1)

1√(2n+ 3)π

,

Cp =(n+ 3

2)

(n+ 1)Cρ =

Γ(n+ 52)

Γ(n+ 2)

1√(2n+ 3)π

.

Here

Γ(p) =

∫ ∞0

xp−1 e−x dx, p > 0

is the Gamma function, equal to (p− 1)! for integers p. Useful integral:∫ 1

−1

(1− x2)p dx =

√π Γ(p+ 1)

Γ(p+ 32), p > −1.

Can be shown that polytrope produces uniform disc in limit n → 0 and isothermal discin limit n→∞.

Isothermal disc extends formally to z =∞, while other models have surfaces and vacuumexteriors.

4.3. Order-of-magnitude estimates and time-scales

Simple scaling relations (∼), omitting numerical factors of order unity.

17

Important dimensionless parameter of a thin disc: aspect ratio

H

r 1.

From hydrostatic equilibrium,

p

H∼ ρΩ2H ⇒ cs ∼ ΩH,

where cs =√p/ρ is the isothermal sound speed.

Dimensions of dynamic viscosity

[ρν] = ML−1T−1

are the same of those of p/Ω. We write

ρν =αp

Ω,

where α is the dimensionless viscosity parameter. This is the alpha viscosity prescription.Then

ν =αc2

s

Ω∼ αcsH.

In kinetic theory of gases, kinematic viscosity is

ν ∼ v`,

where v is mean speed of molecules and ` is mean free path. This molecular viscosityis negligible for astrophysical discs. But a similar estimate can be made for the effective‘eddy viscosity’ of turbulence, if v is a typical turbulent velocity and ` is the correlationlength of the turbulence. For subsonic turbulence with v . cs and ` . H, we expect thatα . 1.

[Lecture 8, 12.02.19]

The stress is then

Trφ = ρνrdΩ

dr= −qαp.

The idea behind |Trφ| ∼ αp is that whatever physical process gives rise to the stress, itshould scale with the pressure – probably correct for local processes such as small-scaleturbulent motions.

Three important characteristic time-scales:

Dynamical time-scale (orbital motion, vertical hydrostatic equilibrium):

tdyn ∼1

Ω∼ H

cs

.

18

Viscous time-scale (radial motion, evolution of surface density):

tvisc ∼r2

ν∼ α−1

(H

r

)−2

tdyn.

Thermal time-scale (vertical thermal balance):

tth ∼P

νΣΩ2∼ c2

s

νΩ2∼ H2

ν∼ α−1tdyn.

For a thin disc with α < 1:tdyn < tth tvisc.

All three time-scales usually increase with r.

Mach number of orbital motion:

Ma ∼ rΩ

cs

∼(H

r

)−1

.

Typical accretion velocity

|ur| ∼ν

r∼ α

(H

r

)cs.

For a thin disc:|ur| cs rΩ,

so orbital motion is highly supersonic while accretion flow is highly subsonic.

Relative contribution of radial pressure gradient to radial component of the equation ofmotion:

∂p

∂r

/ρrΩ2 ∼ c2

s

r2Ω2∼(H

r

)2

.

Vertical variations of the radial gravitational acceleration are also of this order. Otherterms in the radial equation of motion, such as inertial terms associated with the radialmotion, are smaller still. Therefore

uφ = rΩ

[1 +O

(H

r

)2],

so treating the azimuthal fluid velocity as equal to the orbital velocity of a test particleis an excellent approximation for a thin disc. In general, the thin-disc approximationsinvolve fractional errors of O(H/r)2. A formal asymptotic treatment of thin discs ispossible, using as small parameter a characteristic value of (H/r)2.

4.4. Radiative models

Energy dissipated by viscosity is carried away by radiation.

Radiative diffusion (optically thick disc):

F = −16σT 3

3κρ∇T,

19

where κ is the (Rosseland mean) opacity.

Dominant balance in thermal energy equation (thin disc):

0 = ρν

(rdΩ

dr

)2

− ∂Fz∂z

.

Contributions from Fr and from radial advection are smaller by O(H/r)2.

Vertical structure of radiative Keplerian disc:

∂p

∂z= −ρΩ2z,

∂Fz∂z

=9

4ρνΩ2,

Fz = −16σT 3

3κρ

∂T

∂z,

together with:

• equation of state, e.g.

p =RρT

µ+

4σT 4

3c(ideal gas + radiation)

(where R is the gas constant and µ the mean molecular weight),

• opacity function κ(ρ, T ),

• boundary conditions, e.g. ‘zero BCs’ρ = p = T = 0 at z = ±zs (or match toatmopsheric model).

Analogous to radial structure of a star.

In local approximation, these are exact ODEs for equilibrium solutions.

Opacity often approximated by a power law, e.g. for ionized discs:

• Thomson opacity (electron scattering: hotter regions)

κ = constant ≈ 0.33 cm2 g−1,

• Kramers opacity (free-free/bound-free: cooler regions)

κ = CκρT−7/2, Cκ ≈ 4.5× 1024 cm5 g−2 K7/2.

In cooler discs, dust and molecules dominate.

4.5. Radiative Keplerian disc with alpha viscosity, ideal gas pressure andpower-law opacity

20

dp

dz= −ρΩ2z,

dFzdz

=9

4αpΩ,

Fz = −16σT 3

3κρ

dT

dz,

p =RρT

µ,

κ = CκρxT y.

Reduce to dimensionless form:

ρ(z) = ρ · ρ(z), p(z) = p · p(z), T (z) = T · T (z), Fz(z) = F · F (z),

with dimensionless vertical coordinate z = z/H and characteristic values

ρ =Σ

H, p =

P

H, T =

µ

R

p

ρ, F =

16σT 4

3τ,

where the characteristic optical thickness is

τ = κΣ = CκρxT yΣ.

Dimensionless equations:

dp

dz= −ρz,

dF

dz= λp,

dT

dz= −ρx+1T y−3F ,

p = ρT ,

with dimensionless eigenvalue

λ =9

4αΩ

pH

F=

27

64α

ΣH2Ω3τ

σT 4=

27

64αCκσ

(R

µ

)4−y

Σx+2H−x+2y−6Ω2y−5

and normalization conditions∫ρ dz =

∫p dz =

∫ρz2 dz = 1.

Zero BCs ρ = p = T = 0 at surfaces z = ±zs. Dimensionless cooling rate is

2Fs = λ.

21

[Lecture 9, 14.02.19]

Numerical solution for Thomson opacity (x = y = 0) gives λ = 3.213 and zs = 2.895.Solution is similar to a polytropic model with n ≈ 2.4.

Vertically integrated viscosity

νΣ =

∫ρν dz =

∫αp

Ωdz =

αP

Ω= αΣH2Ω.

We have foundHx−2y+6 ∝ Σx+2Ω2y−5,

soν ∝ H2Ω ∝ r−3(x+2y−4)/2(x−2y+6)Σ2(x+2)/(x−2y+6).

e.g. for Thomson opacity (x = y = 0),

ν ∝ rΣ2/3.

Heating rate per unit area

H =

∫9

4αpΩ dz =

9

4αPΩ.

22

Cooling rate per unit area

C = 2σT 4s = 2Fz

∣∣∣z=zs

= 2F Fs = λ16σT 4

3τ,

i.e.

T 4s =

8λ

3τT 4.

These two are equal in a thermal steady state.

5. VISCOUS AND THERMAL STABILITY

5.1. Viscous instability

Nonlinear diffusion equation for Keplerian disc:

∂Σ

∂t=

3

r

∂

∂r

[r1/2 ∂

∂r

(r1/2νΣ

)], ν = ν(r,Σ).

Linearize about any given solution Σ0(r, t):

Σ(r, t) = Σ0(r, t) + Σ′(r, t), |Σ′| Σ0,

(νΣ)′ =∂(νΣ)

∂ΣΣ′ = βνΣ′, β =

∂ ln(νΣ)

∂ ln Σ.

Linearized diffusion equation:

∂Σ′

∂t=

3

r

∂

∂r

[r1/2 ∂

∂r

(r1/2βνΣ′

)].

Unstable (antidiffusive) for β < 0: rapid growth on short length-scales, causing disc tobreak into rings. Astrophysical application may be complicated by thermal instability(see below).

5.2. Thermal instability

So far, assumed a balance between heating and cooling:

9

4νΣΩ2 = H = C = 2Fs.

Relax this assumption, but assume that α 1 so that

tdyn tth tvisc.

Consider behaviour on the timescale tth:

• disc is hydrostatic;

23

• surface density does not evolve.

By solving equations of vertical structure except thermal balance, can calculate H andC as functions of (Σ, νΣ). In fact H depends only on νΣ. Equation of thermal balanceH = C defines a curve in the (Σ, νΣ) plane.

Along the equilibrium curve, dH = dC and d(νΣ) = βν dΣ, where β = ∂ ln(νΣ)/∂ ln Σ:

dH

d(νΣ)d(νΣ) =

∂C

∂ΣdΣ +

∂C

∂(νΣ)d(νΣ)

dH

d(νΣ)=

1

βν

∂C

∂Σ+

∂C

∂(νΣ).

Internal energy content of disc per unit area ∼ P ∼ (Ω/α)νΣ. If some heat is added, νΣincreases but Σ is fixed on tth. Unstable if excess heating exceeds excess cooling, i.e. if

dH

d(νΣ)>

∂C

∂(νΣ)

i.e. if1

βν

∂C

∂Σ> 0.

In practice ∂C /∂Σ < 0 (because, at fixed νΣ, Σ ∝ 1/ν ∝ 1/(αT )), so thermal instabilityoccurs (like viscous instability) when β < 0. Thermal instability then dominates (shortertimescale).

5.3. Outbursts

Radiative disc with gas pressure and Thomson opacity has νΣ ∝ rΣ5/3 and is viscouslyand thermally stable.

For cooler discs undergoing H ionization, instability can occur.

‘S curve’ and limit cycle:

FIGURE

Note that, for steady accretion and r rin,

νΣ ≈ M

3π.

Explains outbursts in many cataclysmic variables, X-ray binaries and other systems.

24

[Lecture 10, 19.02.19]

6. INCOMPRESSIBLE HYDRODYNAMICS

6.1. Hydrodynamics of the shearing sheet

Recall local view of an astrophysical disc:

Linear shear flow u0 = −qΩx ey in a frame rotating with Ω = Ω ez.

Ω and q = −d ln Ω/d ln r are evaluated at reference radius r0.

FIGURE

Boundary conditions: either horizontally unbounded or apply (modified) periodic BCs(see later).

Possible treatments of vertical structure:

• ignore z completely (2D shearing sheet)

• neglect vertical gravity: homogeneous in z

• include vertical gravity: isothermal, polytropic, radiative, etc. models

6.2. Homogeneous incompressible fluid

3D system, unbounded or periodic in x, y, z.

Uniform kinematic viscosity ν.

Equation of motion

∂u

∂t+ u · ∇u + 2Ω× u = −∇Φt −

1

ρ∇p+ ν∇2u.

Incompressibility:∇ · u = 0.

Basic state: u = u0 = −qΩx ey, hydrostatic pressure p = p0(z).

Uniform viscous stress, but no divergence and so no accretion flow.

Introduce perturbations (not necessarily small):

u = u0 + v(x, t), p = p0 + p′(x, t).

Then∂v

∂t+ u0 · ∇v + v · ∇u0 + v · ∇v + 2Ω× v = −∇ψ + ν∇2v,

∇ · v = 0,

25

where ψ = p′/ρ. In components:(∂

∂t− qΩx ∂

∂y+ v · ∇

)vx − 2Ωvy = −∂ψ

∂x+ ν∇2vx,(

∂

∂t− qΩx ∂

∂y+ v · ∇

)vy + (2− q)Ωvx = −∂ψ

∂y+ ν∇2vy,(

∂

∂t− qΩx ∂

∂y+ v · ∇

)vz = −∂ψ

∂z+ ν∇2vz.

Shearing waves (after Kelvin/Thomson 1887):

Consider a plane-wave disturbance of the form

v(x, t) = Re v(t) exp[ik(t) · x] ,

ψ(x, t) = Reψ(t) exp[ik(t) · x]

,

with time-dependent wavevector k(t).

Then (∂

∂t− qΩx ∂

∂y

)v = Re

[dv

dt+

(idk

dt· x− qΩx iky

)v

]exp[ik(t) · x]

.

If we choosedk

dt= qΩky ex

then two terms cancel and we are left withdv

dt.

This meanskx = kx0 + qΩkyt, ky = constant, kz = constant.

Tilting of wavefronts by the shear flow:

FIGURE

Dual shear flow in Fourier space:

FIGURE

Furthermore

v · ∇v = Re[v eik·x

]· ∇Re

[v eik·x

]= Re

[k · v eik·x

]Re[iv eik·x

]= 0,

because ∇ · v = 0 implies ik · v = 0.

Special result for incompressible fluid. Nonlinearity doesn’t vanish for a superposition ofshearing waves.

26

Amplitude equations for shearing waves:

dvxdt− 2Ωvy = −ikxψ − νk2vx,

dvydt

+ (2− q)Ωvx = −ikyψ − νk2vy,

dvzdt

= −ikzψ − νk2vz,

ik · v = 0,

with k2 = |k|2.

Viscous terms are taken care of by a viscous decay factor

Eν(t) = exp

(−∫νk2 dt

)= exp

−ν[(k2x0 + k2

y + k2z)t+ qΩkx0kyt

2 +1

3q2Ω2k2

yt3

].

Faster than exponential decay if ky 6= 0.

Write v = Eν(t)v(t), ψ = Eν(t)ψ(t) to eliminate ν terms.

Eliminate variables in favour of vx to obtain (see Example 2.1)

d2

dt2(k2vx

)+ κ2k2

z vx = 0,

where κ2 = 2(2− q)Ω2 (square of epicyclic frequency in local approximation).

Summary of outcomes (see Example 2.1):

• Stable if κ2 > 0:

• |v|2 oscillates if ky = 0

• |v|2 decays algebraically if ky 6= 0

• Unstable if κ2 < 0:

• |v|2 grows exponentially if ky = 0

• |v|2 grows algebraically if ky 6= 0

When ν > 0, Eν kills off any algebraic growth for ky 6= 0. But axisymmetric disturbancesof sufficiently large scale grow exponentially.

Conclude that a rotating shear flow is linearly stable when κ2 > 0, unstable when κ2 < 0.

Agrees with stability of circular test-particle orbits.

Agrees with Rayleigh’s criterion for the linear stability of a cylindrical shear flow u =rΩ(r) eφ to axisymmetric perturbations.

The case κ2 = 0 (either non-rotating shear flow or one with uniform specific angularmomentum) is marginally Rayleigh-stable and allows algebraic growth in the absence ofviscosity.

27

[Lecture 11, 21.02.19]

6.3. 2D incompressible dynamics: vortices

2D perturbations in the plane of the disc satisfy(∂

∂t− Sx ∂

∂y+ v · ∇

)vx − 2Ωvy = −∂ψ

∂x+ ν∇2vx,(

∂

∂t− Sx ∂

∂y+ v · ∇

)vy + (2− q)Ωvx = −∂ψ

∂y+ ν∇2vy,

∂vx∂x

+∂vy∂y

= 0,

where we write the basic shear flow at u0 = −Sx ey with shear rate S = qΩ.

Introduce the streamfunction χ(x, y, t) such that

vx =∂χ

∂y, vy = −∂χ

∂x.

Instantaneous streamlines are curves χ = constant.

Vorticity perturbation

∇× v =

(∂vy∂x− ∂vx

∂y

)ez = (−∇2χ) ez = ζ ez.

Take curl of equation of motion to eliminate ψ: many terms cancel, leaving(∂

∂t− Sx ∂

∂y+ v · ∇

)ζ = ν∇2ζ, (1)

a nonlinear advection-diffusion equation to be solved in conjunction with Poisson’s equa-tion

∇2χ = −ζ.

Total absolute vorticity is (2Ω− S + ζ) ez, with contributions from background rotation,background shear and velocity perturbation.

Coriolis force drops out of 2D incompressible dynamics. Model is too constrained to allowepicyclic/inertial oscillations. It involves pure vortex dynamics with background shear.

Multiply equation (1) by ζ to obtain enstrophy equation(∂

∂t− Sx ∂

∂y+ v · ∇

)(1

2ζ2

)= νζ∇2ζ

=∇ · (νζ∇ζ)− ν|∇ζ|2.

Integrate over area, assuming suitable boundary conditions:

d

dt

∫1

2ζ2 dA = −

∫ν|∇ζ|2 dA.

28

So enstrophy decays.

To maintain vorticity perturbations in the presence of viscosity requires baroclinic or 3Deffects or other source terms.

Shearing-wave solutions of equation (1),

ζ(x, t) = Reζ(t) exp[ik(t) · x]

,

satisfydζ

dt= −νk2ζ .

The nonlinear term v ·∇ζ vanishes because ∇·v = 0 implies ik · v = 0. So the vorticityamplitude decays viscously:

ζ ∝ Eν(t).

Kinetic energy

∝ |v|2 ∝ k2|χ|2 ∝ k2

∣∣∣∣∣ ζk2

∣∣∣∣∣2

∝ E2ν

k2.

FIGURE

6.4. Dynamics of elliptical vortex patches

Set ν = 0. Can vorticity resist shear?

Vorticity equation reduces toDζ

Dt= 0.

Consider a uniform vortex patch defined by a closed contour, inside which ζ = ζ0, anon-zero constant, and outside which ζ = 0.

Vorticity perturbation ζ generates a velocity field v that (together with the backgroundshear −Sx ey) advects the contour. Do steady solutions exist?

Elliptical vortex patch (centred on origin WLOG; semi-axes a and b, inclined at angle θwith respect to y and x axes).

FIGURE

v induced by ζ0 causes ellipse to rotate with angular velocity

θ =abζ0

(a+ b)2

(see Example 2.3).

Background shear −Sx ey deforms ellipse according to

a

a= − b

b= S sin θ cos θ, θ =

S(b2 cos2 θ − a2 sin2 θ)

a2 − b2

29

(see Example 2.3).

Combine effects:

a

a= − b

b= S sin θ cos θ,

θ =S(b2 cos2 θ − a2 sin2 θ)

a2 − b2+

ab ζ0

(a+ b)2.

Area πab is conserved. Rewrite in terms of aspect ratio r = a/b:

r

r= 2S sin θ cos θ,

θ =S(cos2 θ − r2 sin2 θ)

r2 − 1+

r ζ0

(r + 1)2.

2D autonomous dynamical system.

Fixed points (equilibria) have θ = 0 WLOG (letting r < 1 if necessary) and

S

r2 − 1+

r ζ0

(r + 1)2= 0 ⇒ ζ0

S= − (r + 1)

r(r − 1).

FIGURE

Stability of fixed point θ = 0: linearized equations:

δr = 2Sr δθ,

δθ = S∂f

∂rδr,

where

f =1

r2 − 1+

r

(r + 1)2

ζ0

S

vanishes at equilibrium. After some algebra,

∂f

∂r=

2− (r + 1)2

r(r2 − 1)2.

So

δr = 2S2r∂f

∂rδr,

which implies instability for

∂f

∂r> 0, i.e. r <

√2− 1.

FIGURE

Other instabilities exist, e.g. elliptical instability (3D).

30

[Lecture 12, 26.02.19]

7. COMPRESSIBLE HYDRODYNAMICS: GRAVITATIONAL INSTABIL-ITY

2D compressible sheet: inviscid, self-gravitating.

Surface density Σ(x, y, t) =∫ρ dz, 2D pressure P (x, y, t) =

∫p dz.

Basic equations (only a model, not derivable exactly from 3D equations):

∂Σ

∂t+∇ · (Σu) = 0,

∂u

∂t+ u · ∇u + 2Ω× u = −∇Φt,m −∇Φd,m −

1

Σ∇P,

where the tidal potential in the midplane is Φt,m = −qΩ2x2, the disc potential Φd(x, y, z, t)satisfies Poisson’s equation

∇2Φd = 4πGΣ δ(z),

and its value in the midplane is Φd,m(x, y, t) = Φd(x, y, 0, t).

Assume a barotropic relation P = P (Σ) for simplicity.

Solve Poisson’s equation in the Fourier domain: let

Σ(kx, ky, t) =

∫ ∞−∞

∫ ∞−∞

Σ(x, y, t) e−ikxx e−ikyy dx dy,

etc., so that (−k2 +

∂2

∂z2

)Φd = 4πGΣ δ(z),

wherek =

√k2x + k2

y.

The relevant solution (decaying as |z| → ∞) for k 6= 0 is

Φd = −2πGΣ

ke−k|z|,

so that [∂Φd

∂z

]0+

0−

= 4πGΣ,

as required. So

Φd,m = −2πGΣ

k.

(The k = 0 component gives no horizontal force anyway.)

31

Conservation of potential vorticity / ‘vortensity’

Use the identity(∇× u)× u = u · ∇u−∇

(12|u|2)

to rewrite the equation of motion as

∂u

∂t+ [(2Ω +∇× u)× u] =∇(· · · ),

since P = P (Σ). Take the curl:

∂

∂t(∇× u) +∇× [(2Ω +∇× u)× u] = 0.

Use the identity

∇× (A×B) = B · ∇A−A · ∇B + A(∇ ·B)−B(∇ ·A)

to obtain (since the problem is 2D)(∂

∂t+ u · ∇

)(2Ω +∇× u) = −(2Ω +∇× u)(∇ · u)

= (2Ω +∇× u)1

Σ

(∂

∂t+ u · ∇

)Σ.

ThusDf

Dt= 0,

where

f =2Ω + (∇× u)z

Σ

is the potential vorticity or ‘vortensity’ (vorticity divided by surface density).

Unlike the incompressible 2D case, vortex dynamics is not the whole story. Vorticaldisturbances are coupled to acoustic ones.

Linear stability of a uniform 2D self-gravitating sheet

Basic state: Σ = constant, u = −qΩx ey.

Linearized equations for perturbations Σ′, v, etc.:(∂

∂t− qΩx ∂

∂y

)Σ′ + Σ∇ · v = 0,(

∂

∂t− qΩx ∂

∂y

)v − qΩvx ey + 2Ω× v = −∇Φ′d,m −

1

Σ∇P ′,

∇2Φ′d = 4πGΣ′ δ(z),

with

P ′ =dP

dΣΣ′ = v2

s Σ′,

32

where vs is the (adiabatic) sound speed.

Solutions are shearing waves:

Σ′(x, t) = Re

Σ′(t) exp [ik(t) · x],

etc. Amplitude equations:

dΣ′

dt+ Σ ik · v = 0,

dvxdt− 2Ωvy = −ikx

(Φ′d,m + v2

s

Σ′

Σ

),

dvydt

+ (2− q)Ωvx = −iky

(Φ′d,m + v2

s

Σ′

Σ

),

Φ′d,m = −2πGΣ′

k.

Vortensity perturbation

f ′ =ikxvy − ikyvx

Σ− (2− q)ΩΣ′

Σ2

satisfies df ′/dt, as expected (exercise).

Consider axisymmetric waves: ky = 0, kx = constant, k = |kx|.

Equations have constant coefficients, so assume amplitudes ∝ e−iωt:

− iωΣ′ + Σ ikxvx = 0,

− iωvx − 2Ωvy = −ikx(v2

s −2πGΣ

|kx|

)Σ′

Σ,

− iωvy + (2− q)Ωvx = 0.

Multiply second equation by iω and eliminate Σ′ and vy:

ω2vx − 2(2− q)Ω2vx = k2x

(v2

s −2πGΣ

|kx|

)vx.

Deduce dispersion relation for ‘density waves’:

ω2 = κ2 − 2πGΣ|kx|+ v2sk

2x. (1)

There is also a vortical solution with ω = 0 and vx = 0. This involves a sinusoidalazimuthal velocity perturbation vy(x) giving rise to a Coriolis force that is balanced bypressure and self-gravity. Such ‘zonal flows’ may be related to the rings observed inprotoplanetary discs.

33

The dispersion relation (1) has positive, stabilizing contributions from inertial forces (κ2)and acoustic forces (v2

sk2x), and a negative, destabilizing contribution from self-gravity. It

describes ‘acoustic–inertial waves’ that can potentially be destabilized by self-gravity.

[Lecture 13, 28.02.19]

Disc is unstable to axisymmetric disturbances if ω2 < 0 for some kx.

ω2 is minimized wrt |kx| when

0 = −2πGΣ + 2v2s |kx| ⇒ |kx| =

πGΣ

v2s

,

so

(ω2)min = κ2 − (πGΣ)2

v2s

= κ2

(1− 1

Q2

),

whereQ =

vsκ

πGΣ

is the Toomre stability parameter. (Product of stabilizing influences divided by destabi-lizing influence.)

Gravitational instability (GI) if Q < 1.

Q is:

• an inverse measure of self-gravity;

• a measure of temperature.

Dispersion relation for density waves:

FIGURE

If Q < 1, disc tends to form rings (axisymmetric instability, exponential growth).

If 1 < Q . 1.5, disc tends to form spiral waves or clumps (non-axisymmetric instability,transient growth).

Since Q ∝ vs ∝√T , thermostatic regulation is possible: instability produces motion that

is dissipated through shocks and viscosity, heating the disc.

Two possible outcomes of GI:

• fragmentation: formation of gravitationally bound objects (moonlets, planets, stars,etc.);

• gravitational turbulence: sustained activity of non-axisymmetric density waves.

34

Efficient cooling promotes fragmentation.

Thermal balance in a Keplerian disc with gravitational turbulence

Assume a cooling timescale τ = β/Ω, where β is dimensionless.

Cooling rate per unit area

C =P

γ − 1

1

τ.

Heating rate per unit area

H =9

4αPΩ.

Equating these gives a relation between α, β and γ:

9

4αβ(γ − 1) = 1.

Simulations of GI show fragmentation for β . 4 and turbulence for β & 4, implying amaximum α of about 1/(9(γ − 1)).

What is ν in a self-gravitating Keplerian disc with Q ≈ 1?

Q =vsΩ

πGΣ⇒ ν =

αc2s

Ω=αv2

s

γΩ=αQ2(πGΣ)2

γΩ3.

If β = constant then α = constant and ν ∝ Σ2r9/2.

8. SATELLITE–DISC INTERACTION

Common problem:

• orbiting companion, e.g. on circular orbit within disc

• gravitational (rather than hydrodynamic) interaction with disc

• perturbs orbital motions and excites waves

• exchanges of energy and angular momentum

• orbital evolution of satellite (e.g. planetary migration)

Test-particle dynamics in xy plane, in local approximation.

x− 2Ωy = 2qΩ2x− ∂Ψ

∂x,

y + 2Ωx = −∂Ψ

∂y.

Satellite on circular orbit at reference radius (xs = ys = 0). Satellite potential

Ψ = − GMs√x2 + y2

.

35

General solution in absence of Ψ (recall Section 2.1):

x = x0 + Re(Ae−iκt

),

y = y0 − qΩx0t+ Re

(2ΩA

iκe−iκt

).

Guiding centre (x0, y0 − qΩx0t) follows circular orbit. Complex epicyclic amplitude A.

To express x0 and A in terms of position and velocity:

x = x0 + Re(Ae−iκt

),

x = Re(−iκA e−iκt

)= κ Im

(Ae−iκt

),

x = −κ2 Re(Ae−iκt

).

So

Ae−iκt = Re(Ae−iκt

)+ i Im

(Ae−iκt

)= − x

κ2+ix

κ

A =

[−2Ω

κ2(y + qΩx) +

ix

κ

]eiκt,

x0 = x+x

κ2

= x+2Ω

κ2(y + qΩx)

=2Ω

κ2(y + 2Ωx).

[Lecture 14, 05.03.18]

Canonical y-momentum (per unit mass)

py = y + 2Ωx =κ2

2Ωx0 = constant.

Energy (per unit mass)

ε =1

2

(x2 + y2

)− qΩ2x2.

Now

κA =

[−2Ω

κ(y + qΩx) + ix

]eiκt

κ2|A|2 = x2 +4Ω2

κ2(y + qΩx)2

= 2ε+2qΩ2

κ2y2 +

8qΩ3

κ2yx+

8qΩ4

κ2x2

= 2ε+2qΩ2

κ2(y + 2Ωx)2,

36

so

ε =1

2κ2|A|2 − qΩ2

κ2p2y = constant.

In the presence of a satellite potential:

py = −∂Ψ

∂y,

ε+ Ψ = constant,

A =

[−2Ω

κ2(y + qΩx) +

ix

κ− 2iΩ

κ(y + qΩx)− x

]eiκt,

=

[−2Ω

κ2(y + 2Ωx) +

i

κ(x− 2Ωy − 2qΩ2x)

]eiκt,

=

(2Ω

κ2

∂Ψ

∂y− i

κ

∂Ψ

∂x

)eiκt.

Calculate ∆A using linear perturbation theory.

On the unperturbed circular orbit (A = 0):

x = x0 = constant, y = −qΩx0t.

Then, with Ψ = −GMs(x2 + y2)−1/2,

A =

(2Ω

κ2

∂Ψ

∂y− i

κ

∂Ψ

∂x

)eiκt

= GMs(x2 + y2)−3/2

(2Ωy

κ2− ix

κ

)eiκt

≈ −iGMs

κx20

(1 + q2Ω2t2)−3/2

(1− i2Ω

κqΩt

)eiκt,

giving

∆A =

∫ ∞−∞

A dt

= −iGMs

κx20

∫ ∞−∞

(1 + q2Ω2t2)−3/2

(cosκt+

2Ω

κqΩt sinκt

)dt.

Let

f(k) =

∫ ∞−∞

(1 + x2)−3/2 cos kx dx = 2kK1(k) (k > 0),

where K1 is a modified Bessel function. Then

∆A = −iC GMs

κqΩx20

,

where

C = f

(κ

qΩ

)− 2Ω

κf ′(κ

qΩ

)37

is a function of q only. For Keplerian orbits (κ/qΩ = 2/3), C ≈ 3.359.

So encounter with satellite excites an epicyclic oscillation at first order.

Long before and after the encounter, Ψ→ 0.

Since ε+ Ψ is exactly conserved, ∆ε = 0 in the encounter.

But

ε =1

2κ2|A|2 − qΩ2

κ2p2y,

so

∆(p2y) =

κ4

2qΩ2∆(|A|2).

Assume a circular orbit before the encounter:

A = 0, py =κ2

2Ωx0.

Then, after the encounter,

A ≈ −iC GMs

κqΩx20

,

∆(p2y) ≈ 2

κ2

2Ωx0 ∆py =

κ4

2qΩ2

(CGMs

κqΩx20

)2

,

so

∆py =(CGMs)

2

2q3Ω3x50

correct to second order.

Irreversibility/dissipation is implicit in assuming circular initial orbit.

FIGURE

Simplified version: ‘impulse approximation’:

FIGURE

∆v⊥ ≈GMs

x20

1

qΩ

∆(v2⊥) + ∆(v2

‖) = 0 (energy)(GMs

qΩx20

)2

+ 2qΩx0 ∆v‖ ≈ 0

∆v‖ ≈ −(GMs)

2

2q3Ω3x50

,

the same result but lacking the C2 factor.

38

y-force on disc per unit x at location x:

(CGMs)2

2q3Ω3x5· Σ · |qΩx| ∝ x−4 sgn(x).

Torque per unit radius is the same ×r0.

Satellite experiences equal and opposite torque.

Effect is of second order in Ms.

Similar result for density waves (response of a fluid disc).

x−4 divergence is moderated within |x| . H (or Hill radius).

Gravitational interaction is ‘repulsive’ !

FIGURE

One-sided torque leads to gap opening if Ms is large enough and ν small enough.

Asymmetry leads to net torque on satellite and to migration (usually inwards).

[Lecture 15, 07.03.19]

Now include periodic nature of y coordinate (Ly = 2πr0):

A =

(2Ω

κ2

∂Ψ

∂y− i

κ

∂Ψ

∂x

)eiκt

= F (t) eiκt

=∞∑

n=−∞

fn e−inωt eiκt,

with ω = 2π/T = qΩ|x0|/r0, where T = 2πr0/qΩ|x0| is the ‘synodic period’.

Add damping of epicyclic motion:

A =∞∑

n=−∞

fn e−inωt eiκt − γA.

The long-term response

A =∞∑

n=−∞

ifn e−inωt eiκt

(nω − κ) + iγ

is resonant at Lindblad resonances where

x

r0

=1

n

κ

qΩ.

In a Keplerian disc, these locations correspond to orbital commensurabilities

Ω

Ω0

=n

n− 1.

39

In a fluid disc, density waves are launched there. Wave emission resolves singularity inepicyclic response.

9. MAGNETIC FIELDS IN DISCS: MAGNETOROTATIONAL INSTABIL-ITY

9.1. Equations of magnetohydrodynamics (MHD)

Homogeneous, incompressible, inviscid fluid. Density ρ, electrical conductivity σ, mag-netic diffusivity η = 1/µ0σ. (Set µ0 = 4π for Gaussian units.)

Lorentz force per unit volume is divergence of Maxwell stress

M =BB

µ0

− |B|2

2µ0

I

(tension and pressure).

Equation of motion including Lorentz force:

∂u

∂t+ u · ∇u = −∇Φ− 1

ρ∇Π +

1

µ0ρB · ∇B.

Total (gas plus magnetic) pressure

Π = p+|B|2

2µ0

.

Induction equation∂B

∂t+ u · ∇B = B · ∇u + η∇2B.

Incompressibility and solenoidal conditions

∇ · u =∇ ·B = 0.

In rotating frame of local model, add Coriolis and centrifugal forces:

∂u

∂t+ u · ∇u + 2Ω× u = −∇Φt −

1

ρ∇Π +

1

µ0ρB · ∇B.

Rotation does not affect induction equation.

40

Write u = −qΩx ey + v, where v is departure from orbital motion. Look for solutionswith v,B,Π independent of x, y. Then ∂vz/∂z = ∂Bz/∂z = 0 and

DvxDt− 2Ωvy =

Bz

µ0ρ

∂Bx

∂z,

DvyDt

+ (2− q)Ωvx =Bz

µ0ρ

∂By

∂z,

∂vz∂t

= −ν2z − 1

ρ

∂Π

∂z,

DBx

Dt= Bz

∂vx∂z

+ η∂2Bx

∂z2,

DBy

Dt= −qΩBx +Bz

∂vy∂z

+ η∂2By

∂z2,

∂Bz

∂t= 0,

withD

Dt=

∂

∂t+ vz

∂

∂z.

Bz = constant is a conserved uniform vertical flux.

Disc extends over z− < z < z+ (possibly time-dependent), with full thickness z+ −z− = Σ/ρ = constant. Above and below disc, ρ = 0 and magnetic field is force-free:Bx = constant, By = constant. Let Bx = B+

x and By = B+y for z > z+, Bx = B−x and

By = B−y for z < z−. Constants B±x,y are determined by global considerations.

Integrate vz equation over disc, with BCs p = 0 at z = z±:

Z = −ν2Z − (Π+ − Π−)

Σ,

where Z = (z+ + z−)/2 (centre of mass) and Π± = |B+|2/2µ0.

Free oscillation about equilibrium position (Z = 0 if Π+ = Π−). Assume oscillation absentso vz = 0.

Reduces to linear problem

∂vx∂t− 2Ωvy =

Bz

µ0ρ

∂Bx

∂z,

∂vy∂t

+ (2− q)Ωvx =Bz

µ0ρ

∂By

∂z,

∂Bx

∂t= Bz

∂vx∂z

+ η∂2Bx

∂z2,

∂By

∂t+ qΩBx = Bz

∂vy∂z

+ η∂2By

∂z2.

[Lecture 16, 12.03.19]

41

9.2. Equilibrium solutions

In a steady state

−2Ωvy =Bz

µ0ρ

dBx

dz,

(2− q)Ωvx =Bz

µ0ρ

dBy

dz,

0 = Bzdvxdz

+ ηd2Bx

dz2,

qΩBx = Bzdvydz

+ ηd2By

dz2.

Eliminate to obtain (exercise)

d2Bx

dz2+K2Bx = 0, K2 =

2qΩ2v2az

v4az + η2κ2

,

where

vaz =Bz√µ0ρ

is the vertical Alfven velocity.

Solution is a combination of sin(Kz) and cos(Kz).

Usual situation: symmetrical BCs Bx = ±B+x and By = ±B+

y at z = ±z+. Relevantsolution is

Bx = B+x

sin(Kz)

sin(Kz+),

By = B+y

z

z+− (2− q)Ωη

v2az

B+x

[sin(Kz)

sin(Kz+)− z

z+

],

vx =v2

az

(2− q)Ωz+

B+y

Bz

− ηB+x

Bz

[K cos(Kz)

sin(Kz+)− 1

z+

],

vy = −v2az

2Ω

B+x

Bz

K cos(Kz)

sin(Kz+).

Interpretation:

• field bends in xz plane (close to parabola if Kz+ 1, i.e. strong field and/or highdiffusivity, otherwise more bendy)

• z-dependent departure from orbital motion (vy) with Coriolis and Lorentz forcebalancing

• isorotation (constant angular velocity) along field lines if η = 0 (ratio of two termsof RHS of y-component of induction equation is η2κ2/v4

az)

42

• mean radial velocity (accretion flow) if non-zero magnetic torque ∝ B+y Bz acting on

disc

Above the disc, assuming very low density, we have

Bx = B+x , By = B+

y , vx = constant, vy =B+x

Bz

qΩz + constant.

B acts as a rigid channel for the gas. Net acceleration ‖ B due to inertial forces andgravity is proportional to

B+x

Bz

2Ωvy −B+y

Bz

(2− q)Ωvx − ν2z =

(B+x

Bz

)2

2qΩ2z − ν2z + constant.

If the field is sufficiently inclined to the vertical, i.e.(B+x

Bz

)2

>ν2

2qΩ2,

then this net acceleration increases with z and will become positive at some height abovethe disc. A hydrostatic solution is then impossible and an outflow (jet/wind) is launchedalong the field lines. This is magnetocentrifugal acceleration. For a Keplerian disc (ν = Ω,q = 3/2), it requires i > 30, where i = arctan |B+

x /Bz| is the inclination.

9.3. Stability analysis

Consider perturbations to this equilibrium, of the form

δvx = Re(vx e

st+ikz),

etc., where s is the growth rate and k is the vertical wavenumber.

Equations require

svx − 2Ωvy =ikBz

µ0ρBx,

svy + (2− q)Ωvx =ikBz

µ0ρBy,

(s+ ηk2)Bx = ikBzvx,

(s+ ηk2)By + qΩBx = ikBzvy.

Multiply first two equations by ikBz and use last two to substitute for vx and vy:

ssηBx − 2Ω(sηBy + qΩBx

)= −ω2

aBx,

s(sηBy + qΩBx

)+ (2− q)ΩsηBx = −ω2

aBy,

where sη = s+ ηk2 and the Alfven frequency is

ωa = k · va =kBz√µ0ρ

.

43

Algebraic elimination leads to the magnetorotational dispersion relation

(ssη + ω2a)2 + s2

ηκ2 − 2qΩ2ω2

a = 0.

This gives marginal stability (s = 0) for k2 = K2 and instability for k2 < K2.

The BCs for modes are δBx = δBy = 0 at z = ±z+. This gives solutions in sin(kz) andcos(kz) (dispersion relation is even in k) with quantization

k =nπ

2z+, n = 1, 2, 3, . . . .

The n = 1 mode determines overall stability.

Ideal case of magnetorotational instability (MRI): η = 0, sη = s.

Quadratic for s2:s4 + (2ω2

a + κ2)s2 + ω2a(ω2

a − 2qΩ2) = 0,

with roots

s2 = −ω2a +

1

2κ2

(−1±

√1 +

16ω2aΩ2

κ4

).

Assume κ2 > 0, otherwise hydrodynamically unstable. + root maximized wrt ω2a when

0 =∂s

∂ω2a

= −1 +4Ω2

κ2

1√· · ·

√· · · = 4Ω2

κ2

ω2a = Ω2 − κ4

16Ω2= q

(1− q

4

)Ω2,

giving

(s2)max = −Ω2 +κ4

16Ω2− 1

2κ2 + 2Ω2

= Ω2

(1− κ2

4Ω2

)2

=

(1

2qΩ

)2

.

So maximum growth rate is half the orbital shear rate.

Fastest growing mode has δvx = δvy and δBx = −δBy, which maximizes the correlationsleading to outward angular momentum transport:

−Txy = ρ δvx δvy −δBx δBy

µ0

.

This shear stress also extracts energy from the orbital shear, allowing the perturbation togrow.

44

For ideal instability we require

ω2a(ω2

a − 2qΩ2) < 0.

The n = 1 mode k = π/2z+ is the last to be stabilized as Bz is increased. Disc is unstablefor

0 <vaz

Ωz+<

2√

2q

π,

which is about 1.1 for a Keplerian disc.

The MRI leads to sustained turbulence in discs that are sufficiently ionized (η smallenough). It leads to outward angular-momentum transport with typically α . 0.1, de-pending on the field strength and degree of ionization.

Please send any comments and corrections to gio10@cam.ac.uk

45