In reality, stars are not spherical & time-steady.

Next we’ll look at a few departures from that ideal equilibrium:• rotation (non-spherical; usually time-steady)

• tidal distortion (non-spherical)• pulsation (not time-steady! may be non-spherical)

First, let’s look at general non-spherical versions of the stellar structureequations.

Hydrostatic equilibrium:

dP

dr= −

GMr

r2ρ is really... ∇P = −ρ∇Φ

If only gravity is present, the potential Φg is relatively simple to write.

For a point-mass (or all stellar mass concentrated at the center),

Φg = −GM∗

r

and for a spherically symmetric distribution, you can (kind of) replace M∗ with

Mr, the mass interior to the local spherical radius.

Of course, the fully general description is to use the gravitational Poissonequation that relates Φg and density,

∇2Φg = 4πGρ

and this is actually the non-spherical version of mass conservation(dMr/dr = 4πr2ρ).

In integral form, Poisson’s equation is

Φg(r) = −G

∫

d3r′ρ(r′)

|r− r′|

If you’re taking Jackson E&M, you’ll learn more than you ever cared to aboutequations like this.

For now, let’s go back to the simple assumption of a fully centrally condensedstar, so

ρ(r′) = M∗ δ(r′) , which gives us back Φg = −GM∗

r.

5.1

However, if there are other sources of perturbing acceleration in the system,hydrostatic equilibrium is:

∇P = −ρ∇Φg + ρa

and in cases where a can be written as the gradient of a potential,

∇P = −ρ∇Φeff .

Let’s think about uniform (rigid) rotation with a fixed angular velocity Ω(radians per second).

In a rotating frame, the outwardcentrifugal force is ⊥ to the rotationaxis.

vφ = xΩ

|acen| =v2φx

= Ω2x = Ω2r sin θ

Note: There’s no Coriolis force, since v in the rotating frame is zero.

In spherical coordinates, a is given by

ar = Ω2r sin2 θ

aθ = Ω2r sin θ cos θ

and it’s not hard to prove that this vector acceleration is the gradient of a

potential

Φrot = −Ω2r2

2sin2 θ a = −∇Φrot =

(

−∂Φrot

∂r

)

er +

(

−1

r

∂Φrot

∂θ

)

eθ

(True when Ω = constant.)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Thus, we’ll use Φeff = Φgrav +Φrot.

5.2

Some of the other stellar structure equations are:

Radiative energy transport:

dT

dr= −

3κρ

4acT 3

Lr

4πr2becomes... ∇T = −

3κρ

4acT 3Frad

Energy conservation:

dLr

dr= 4πr2ρ ǫ becomes... ∇ · Frad = ρ ǫ

Why? Write out the divergence of the flux in gory detail...

∇ · Frad =1

r2∂

∂r

(

r2Lr

4πr2

)

=1

4πr2∂Lr

∂r= ρǫ .

Convective energy transport? Still ∼locally okay, but we’ll talk about itsqualitative changes later. (Blobs don’t rise radially...)

For gravity + rotation, let’s write the “effective potential” version ofhydrostatic equilibrium.

∇P = −ρ∇Φeff ≡ ρgeff ,

and Φeff = −GM∗

r−

Ω2r2

2sin2 θ .

Rotation makes Φeff a function of both r & θ.

Consider surfaces of constant Φeff = equipotential surfaces(i.e., “level surfaces”).

Parallel to these surfaces, ∇Φeff = 0, so no forces exist.

All forces are normal to equipotential surfaces (i.e., ∇Φeff points normal tothese surfaces, and thus ∇P does, too).

For non-rotating stars, level surfaces = concentric spheres.

What do level surfaces look like for rotating stars? This will tell us the SHAPE

of a rotating star!

5.3

The Roche Problem: i.e., what is R∗(θ) for a given (constant) Ω ?

Let’s start by looking at one easy limiting case: the equatorial plane (θ = π/2),where acen points radially outward.

The radial component of hydro. equilib.1

ρ

∂P

∂r= −

GM∗

r2+Ω2r

There’s a key radius r at which the right-hand side = 0, r3 =GM∗

Ω2

Inside that radius, gravity wins. Outside, the centrifugal force wins.

For a known equatorial stellar radius, we can solve for the critical/breakup

velocity,

Ωcrit =

√

GM∗

R3eq

for Ω < Ωcrit , a star can hold together (i.e., at surface, gravity wins)

Ω > Ωcrit , the star breaks apart! (i.e., at surface, centrifug. force wins)

For context, the present-day Sun’s rotation period Π is ∼27 days, and

Ω =2π

Π≈ 3× 10−6 rad/s , Ωcrit ≈ 6× 10−4 rad/s , ω ≡

Ω

Ωcrit

≈ 0.004 .

Keeping in mind that ω ≤ 1 is needed for a “closed” star,

ω ≪ 1 means that rotational distortion is a tiny departure from sphericalsymmetry.

We can find R∗(θ) by recognizing thatif a given equipotential surface has a constant Φeff,

then we can equate Φeff for any two points along it.

Φeff(r, θ) = Φeff(r, 0) (right-hand side: pole)

Defining the polar radius as Rp, this equation is

−GM∗

R∗−

Ω2R2∗

2sin2 θ = −

GM∗

Rp

.

5.4

Multiply by −R∗/GM∗, and define a dimensionless oblateness factorx(θ) = R∗(θ)/Rp, and we get

1− x+x3

2

(Ω2

GM∗/R3p

)

sin2 θ = 0 (cubic polynomial for x)

You’ve probably seen several ways of obtaining a closed-form solution of acubic equation. Some are nasty.

George Collins (in 1963) found an elegant trigonometric solution. You canprove to yourself that...

x(ω, θ) =3

ω sin θcos

[π + cos−1(ω sin θ)

3

]

satisfies the cubic, where we recall

ω =Ω

Ωcrit

, and Ωcrit =

√

8GM∗

27R3p

...is rigorously true if Rp is fixed.

The trigonometric solution has physically relevant solutions for 0 ≤ ω ≤ 1.

Note that at breakup speed, Req =3

2Rp , so that Ωcrit =

√

GM∗

R3eq

which makes this definition of Ωcrit consistent with the one derived earlier.

5.5

Caveats: This is true for a centrally condensed star (i.e., most of its M∗ at thecenter), and for Ω = constant.

For “stiffer” equations of state (lower polytropic n) and/or differentially

rotating stars, Req/Rp can be larger.

Examples from Stoeckly (1965, ApJ, 142, 208):

Recently, Lock & Stewart (2017, JGR, 122, 950) proposed that terrestrial

planets can accrete into torus-like configurations while partially molten...

Ω = constant deep inside, but outer regions are accreted later, and they tend

to conserve angular momentum (Ω ∝ 1/r2).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

For stars, let’s stay with the rigidly rotating Roche model.

Handout: Illustration of the shapes of Roche equipotential surfaces.

5.6

Meridional cross sections of Roche equipotential surfaces, plotted for a uniform distribution of ω values between 0 and

1.05, at increments of 0.05. The dotted curve is a representative post-breakup surface (ω = 1.05).

Color representation of the bolometric flux emergent from a B2 main-sequence star rotating at 0, 60%, 80%, and 99% of

its critical “breakup” rotation speed (left to right) and obeying von Zeipel (1924) gravity darkening.

Teff dependence of the gravity darkening exponent, calculated from Claret’s (2004, A&A, 424, 919) evolutionary grids.

Each point shows a star between 0.8 and 125 M⊙ at some point along its evolutionary sequence. Symbol color denotes

stellar luminosity, spanning the minimum (dark violet, L∗/L⊙ = 0.2) and maximum (red, L∗/L⊙ ≈ 106) values found

in the models.

5.7

Can we observe this? Yes!

Oblateness:

• interferometry shows departures from circular shapes

• eclipsing binary light curves show it (although tidal forces also distort

equipotentials...)

• exoplanet transit light curves (Barnes et al. 2013, ApJ, 774, 53)

Rotation itself: For some stars, spectral lines are broadened.

The inclination i = angle between rotation axis and the line-of-sight to theobserver. i = 0: pole-on ... i = 90: equator-on.

For some other stars, there’s starspot photometry.

observable quantity: Π =2π

Ωitself, not projected by sin i .

(It’s getting precise enough to probe differential rotation, by seeing differentrecurrence periods for low/high latitude spots.)

Also, the Rossiter-McLaughlin effect...

5.8

In eclipsing or transiting systems, spectra give the radial velocity of thecomponent that’s being partially occulted:

(Observer is at bottom.)

Note: The star goes in the opposite direction as the orbiting occulter.This helps diagnose exoplanet spin-orbit alignment (or misalignment).

With a star rotating so fast that it’s oblate, do its other properties (ρ, T , P )

vary versus θ, too?

Yes: Let’s repeat the 1924 work of Hugo von Zeipel:

Gravity Darkening

For stars in which rotation can be derived from a potential, we know that

∇P = −ρ∇Φeff .

Vectors normal to equipotential surfaces are parallel to vectors normal tosurfaces of constant pressure.

i.e., P = constant along equipotential surfaces.

Thus, we can (kind of) go back to a 1D model:dP

dn= −ρ

dΦeff

dn

where n is a radius-like position vector that always points normal to the localequipotential surface.

5.9

Since it’s all nice and 1D, we can write ρ = −dP

dΦeff

and, in this model, ρ is a function only of Φeff , too.

Stars in which

const. P (isobaric)const. ρ (isopycnic)

surfaces coincide are called barotropic.

Stars in which that’s not true are called baroclinic.

If it’s an ideal gas, then P ∝ ρT , so that T = T (Φeff only), also.

i.e., ∇P , ∇Φeff, and ∇T vectors are all parallel to one another.

Note: even though P , ρ, T , and Φeff are constant on equipotentials,

|∇Φeff|, |∇P |, etc., aren’t necessarily constant!

For a radiative interior, the energy transport equation is

∇T = −3κρ

4acT 3︸ ︷︷ ︸

Frad

The scalar factor in front of Frad is just a function of ρ and T , so it’s constant

on equipotentials, too. Following Collins Chapter 7,

Frad ∝ ∇T ∝dT

dnn ∝

dT

dΦeff

dΦeff

dnn

Thus, Frad = f(Φeff)∇Φeff

|Frad| = f(Φeff) |geff | . (1)

5.10

This is “von Zeipel gravity darkening.”

As one goes from pole to equator in a rotating (radiative) star, the emergentflux decreases.

Note that in our usual way of writing things,

Frad =L∗

4πR2∗

= σT 4eff

and Teff is a function of θ (hot poles, cold equator).

This doesn’t contradict the earlier idea that T = constant on level surfaces.

(The visible “photosphere” may not correspond to a level surface!)

How do we solve for Teff(θ)?

The general way of defining luminosity is L∗ =

∮

dA Frad

where the integral is taken over the full distorted area, and∮

dA Frad ∝

∮

dA geff .

Usually, we assume L∗ is independent of the rotation rate, so if we know it, we

can write

Frad(θ) = σ[Teff(θ)]4 =

L∗

∮dA geff

geff(θ)

5.11

The quantity in is the “von Zeipel constant” (depends on Ω, but not θ).

Teff ∝ gβeff where β = the “von Zeipel exponent.”

There’s a cottage industry in trying to determine β for all types of rotatingstars.

β = 0.25 for radiative interiors.

β ≈ 0–0.1 for late-type stars with convective envelopes.

(Convective mixing redistributes energy in latitude.) (see handout)

Observers infer β from interferometry & eclipsing binaries.

There are some problems with all the approximations we had to make thus far.

von Zeipel himself ended up deriving something absurd: now called the“von Zeipel paradox.”

Let’s look at energy conservation in more detail: ∇ · Frad = ρ ǫ .

If Frad = f(Φeff)∇Φeff, then

∇ · Frad = ∇ · (f ∇Φeff) = f ∇2Φeff + ∇f · ∇Φeff

Let’s work out these two terms.

5.12

For our constant-Ω Roche model, one can show that

∇2Φeff = ∇2

(

−GM∗

r−

Ω2r2

2sin2 θ

)

= −2Ω2

The gravity term goes away, and ∇2Φ = constant.

Recall, however, that the full Poisson equation for a self-gravitating systemwas given by

∇2Φgrav = 4πGρ

Our Roche model assumes ρ = M∗ δ(r′), which is zero away from the core.

Thus, the generalized Poisson equation, with both gravity and centrifugal

terms, should be

∇2Φeff = 4πGρ− 2Ω2 (i.e., true for r = 0 and r 6= 0) .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

This was just the first term in ∇ · Frad.

The full energy conservation is: ρ ǫ = f (4πGρ− 2Ω2) +∇f · ∇Φeff .

Left side: a function of ρ and T only, so it is constant along equipotentialsurfaces.

1st term on right side: above is also true.

2nd term on right side: we know that ∇Φeff = −geff is NOT constant alongequipotential surfaces!

(e.g., follow the surface of a critically rotating star, from pole to equator)

Thus, to make the whole energy conservation equation true, this 2nd term

must be equal to 0

To do that, we must set ∇f = 0.i.e., make f(Φeff) not only constant on equipotential surfaces, but constant

everywhere inside the star.

5.13

von Zeipel thus solved for ǫ...

ǫ = f

(

4πG−2Ω2

ρ

)

= constant ×

(

1−Ω2

2πGρ

)

Wait, what? We know now that ǫ is defined by nuclear physics.

It shouldn’t depend so “naively” on Ω, nor should it go negative for fastenough rotation, nor should it be rigidly constant when Ω = 0.

This absurd result is “von Zeipel’s paradox.”

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Resolving the paradox:

In reality, a rotating star

• isn’t in strict radiative equilibrium,

• doesn’t have Ω = constant,

• so there’s no exact Φeff , and

• surfaces of ρ, P , and T don’t all coincide with equipotentials.

What happens when there’s a component of ∇P along an equipotentialsurface? A gas pressure gradient drives a force → gas flows!

Meridional circulation:

Eddington & Sweet derived themagnitude of vcirc. We won’t deriveit...

vcirc ≈

(Ω

Ωcrit

)2L∗R

2∗

GM2∗

For a more complete derivation, see Collins Chapter 7.

5.14

It’s also straightforward to show that the circulation period is

tcirc ∼R∗

vcirc∼

tK−H

(Ω/Ωcrit)2

Makes sense that it’s of the same order as Kelvin-Helmholtz, since that’s thetime over which thermal energy diffuses out of the star. (Here, it’s diffusing

around the star.)

For the Sun, vcirc ∼ 10−14 km/s.This means that tcirc ∼ 1012 years, ≫ main sequence lifetime!

This is a tiny effect, but it’s still enough to

• fix the paradox, and

• make our earlier results on gravity darkening still “close to correct.”

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The above was focused mainly on massive stars with radiative interiors.

The Sun (and other low-mass stars) show different things...

Helioseismology has allowed us to probe the Sun’s

• interior differential rotation (Ω 6= constant), and

• meridional circulation:

opposite sense as Eddington-Sweet (rises at equator)faster, with vcirc ∼ 10−2 km/s (but still only ∼1% of surface Vrot).

Contours show constant Ω.

Radiative interior is ∼rigid.

Convection zone:equatorial Π ∼ 25 days,

polar Π ∼ 35 days.

∆Ω /Ω ≈ 0.3

5.15

What maintains it? Still the topic of active research. Definitely has a lot to dowith the solar dynamo.

Magnetic fields are transported by rising/falling convection blobs.

Tachocline: the strong shear layer at base of convection zone, at which thedynamo activity appears to be concentrated.

Recent work suggests there may be two meridional circulation cells in theconvective zone (e.g., Zhao et al. 2013, ApJ, 774, L29).

Also, g-mode helioseismology (Fossat et al. 2017, A&A, 604, A40) suggests theinner core may be rotating several times more rapidly than the radiative zoneat r ∼ 0.5R⊙. (Some are skeptical...)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Other stars seem to show some ∆Ω, too, with pole-to-equator behavior both

“solar” and “anti-solar.”

There’s also radial differential rotation: later, we’ll see how the cores and

envelopes of RED GIANTS are decoupled and rotate at different rates.

However, slightly massive MS stars with convective cores (M∗ ∼ 1.6M⊙) have

radiative envelopes that rotate quite uniformly (arXiv:1507.01140). Theremust be efficient angular momentum transport in the interiors.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Summary of rotation speeds for various kinds of stars:

• Early-type (O,B,A): Veq ∼ 100–300 km/s. (Ω/Ωcrit can be > 0.5)

• Late-type (G,K,M): Veq ∼ 1–10 km/s.

• Young low-mass stars: Veq ∼ 10 to a few 100 km/s.

We’ll talk more about rotational evolution later, too.

5.16

Digression on the solar dynamo

For the present-day Sun: in just a few rotations, the equator “laps” the higherlatitudes. The interior gets twisted up... we’ll see soon how this affects the

magnetic field.

The meridional circulation is slower; it takes ∼decades to transport plasma

over distances of order R⊙.

In the convection zone, the plasma is in motion...

• in the r direction (rising/falling blobs),

• in both r and θ (meridional circulation), &

• in the φ direction (differential rotation).

If any of those motions give rise to currents, then they can influence B viaAmpere’s law. Recall the MHD induction equation:

∂B

∂t= ∇× (v ×B) + DB∇

2B

where 1st term shows how the field is frozen-in to the flow, and the 2nd term

describes magnetic diffusion (i.e., gradual spreading and dissipation of thefield, due to resistivity & particle collisions).

Given a tiny “seed field,” one can imagine the right set of motions (v) and

fields (B) that would make the first term contribute to a net amplification(∂B/∂t > 0).

5.17

Over long times, the 1st term generates B, and the 2nd term destroys it...dynamic equilibrium = dynamo!

When we see magnetic fields in the universe, we can compare the object’s age

to the time it would take for DB to destroy any “fossil field” from the object’sbirth;

∂B

∂t≈ DB∇

2B i.e.,B

t∼ DB

B

ℓ2 tdiffuse ∼

ℓ2

DB

where ℓ ≈ system size, and DB is a function of plasma n, T , & composition.

If the system’s age ≫ tdiffuse then a dynamo must be there to maintain the fieldwe see.

This seems to be the case for the Sun, but what’s the v?

It’s still a topic of active research, but I’ll take you through the overall process

that most believe is going on: the αΩ dynamo.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Start with observations:

But what’s going on with the magnetic field over the 11 (or 22) year cycle?

5.18

George Ellery Hale noticed several patterns... “Hale’s law(s)”

• The B-field reorganizes itself on a roughly 11-year cycle... but not exactclockwork repetition; it’s 11 ±1 or 2.

• Sunspot minimum: “dipole” field, aligned with rotation axis.

• Sunspot maximum: most of the field is concentrated into sunspot pairs

(one +, one −) with the leading/following polarities of the opposite sensein the N/S hemispheres.

• Polarities “flip” every ∼11 years, so one full cycle is really ∼22 years.

Joy’s law:

Soon after Hale’s original work, AlfredJoy noticed that most sunspot pairs

tilt towards the equator.

The leading spot tends to be at lowerlatitude than the following spot.

[no matter their polarities!]

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Next, can the so-called αΩ dynamo theory explain all these trends?

5.19

(1) The Ω effect:

Start with a poloidal field (i.e., Br and Bθ components only; field lines stay inplanes of constant φ.

(Example: the solar minimum ∼dipole field.)

Differential rotation stretches the field lines by carrying plasma at equator tohigher φ faster than the plasma nearer the poles.

Poloidal field gets turned into toroidal field: all Bφ.

(2) The α effect:

To maintain a cyclic regeneration of the Sun’s field, the toroidal field should beturned back into poloidal (preferably with stronger field strength) so the

process can keep going indefinitely.

This is the step with the most controversy.

One thing we know, however, is that there are many ways for small “blobs” torise up from the base of the convection zone, and carry magnetic fields with

them. (high β)

• Are the blobs just the turbulent convective blobs themselves (i.e., theyrise because they’re hot and in pressure equilibrium)?

• Or are they “knots” of unusually strong B, which in total P equilibrium

have lower density (high Pmag, low Pgas) and thus rise buoyantly, andstretch the field into loops?

5.20

When the field is mostly toroidal, the magnetic field gets pushed up by a risingblob, and forms a sunspot pair.

If Bφ > 0, the leading spot has negative

polarity.

If Bφ < 0, the leading spot has positive

polarity.

The observed Hale polarity law is consistent with this, e.g.,

In 1996, north polar cap is +−→ Ω effect makes Bφ < 0 in north hemisphere

−→ At next solar maximum, the leading spots in north are +

But we haven’t examined how the toroidal field is converted back to poloidal!

No matter what the rising blobs are, their motions are affected by the

Coriolis force, acor = −2Ω× v.

The rising speeds are slow... the dominant motion is due to differential

rotation (i.e., vφ > 0 near equator in a rotating frame).

If vφ was the only component, this pretty much explains Joy’s law (tilts).

acor,θ = 2Ω vφ cos θ .

5.21

Babcock & Leighton noticed that tilted sunspot pairs eventually DECAY (i.e.,the fields diffuse away, DB 6= 0). The leading spots are nearest the equator,and they cancel out with their trans-equatorial counterparts.

However, the trailing spots are nearer to the poles, and diffusion may beenough to “ooze” their polarity up to the poles → i.e., making new poloidal

field. It agrees with Hale’s polarity laws, and we sort of see it:

That’s one idea.

Parker had an alternate idea. He noted that there are other components of vthat are affected by the Coriolis force.

Deep down (near base of convection zone), the “blobs” probably have strong

fields; i.e., high Pmag & thus low Pgas, if in total pressure equilibrium.

Low-pressure regions are sites of converging flow, and the Coriolis force turns

those into cyclonic flow (see right side of above cartoon).

5.22

Sense of twist is similar to hurricanes on Earth (counter-clockwise in north,clockwise in south). Again, the B-field gets dragged along:

Because the sign of Bφ is different in north & south, the new poloidal fields

line up in the opposite direction as the original poloidal field.

Eventually, the fragmented bits of poloidal field merge together & migrate to

the poles, to produce a new ∼dipolar poloidal field.

This process occurs with roughly ∼11-year periodicity, so the full “solar cycle”

(to get back to the same poloidal polarity) is really ∼22 years in duration.

Both ideas are summarized in this cartoon:

(Upper branch: Parker’s cyclonic blobs; lower branch: Babcock/Leighton

diffusing fields)

5.23

Note: There is still no complete theory of the dynamo! These basic steps mustbe going on, but we still don’t know why the period is ∼10 years, and not 1 or100.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Still, this process does enhance the field magnitude, and thus is a true dynamo.

Schematically, the field is morphed as follows:

• The field is stretched by the bulk flows (i.e., kinetic energy is transferred

to magnetic energy).

• New magnetic energy is redirected in space by twisting and folding thefield, which further amplifies it.

• Magnetic diffusion smears out the localized twists and folds, thus leadingto merging of the field on large spatial scales.

To maintain a steady solar cycle, all that buildup must be balanced by losses.That’s accomplished by some combination of

• Diffusivity, which drains magnetic field from the system in addition tofacilitating the merging above.

• The Sun also “sheds” magnetic fields up through its surface in the form of

highly twisted plasmoids (e.g., prominences, coronal mass ejections).

5.24

Tidal Distortion

The 2nd type of non-spherical distortion we’ll be discussing.

Distortion caused by gravitational attraction from an external mass.

Usually we need to consider onlyM2’s mass, not its shape.

M1 is deformed, and the deformation moves as M2 orbits or passes by.

Changes in shape vs. time cause tidal torque on M1...

• It slows M1’s rotation rate.

• If the rotation rate slows down to the point of tidal locking (i.e.,

synchronous rotation), then the torque can cause...

– eccentricity e → 0 (circularization)

– an increase in D (angular momentum loss from system)

• Also, tidal torques can cause internal energy losses: tidal heating(e.g., volcanism on Io).

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Basic idea: Look at the differential acceleration between 2 extreme points on

M1:

The mean, zero-order accel. on

M1 due to M2 is

g ≈ −GM2

D2

But consider ∆g = gnear − gfar.

If R ≪ D, then differences are differentials:

∆g ≈

(dg

dr

)

∆r , g = −GM2

r2,

dg

dr=

2GM2

r3, so ∆g ≈

2GM2

D3∆r

and ∆r ≈ 2R, or maybe just R if the differential is taken between the center ofM1 and either extreme point.

5.25

However, to really figure out how a star or planet is distorted by a companion,let’s look at the force in more detail.

The full potential felt by a point at (r, θ) is

Φeff = −GM1

r−

GM2

d

This is the Roche point-mass approximation, and will lead to the classical

Roche equipotential surfaces if computed exactly.

Here, however, let’s expand d in terms of other known quantities;

e.g., the “law of cosines” for triangles,

d2 = D2 + r2 − 2rD cos θ .

We really want1

d=

1

D

[

1 +( r

D

)2

− 2( r

D

)

cos θ

]−1/2

and for r ≪ D, let’s expand using the binomial formula,

(1 + ǫ)α ≈ 1 + αǫ+α(α− 1)

2!ǫ2 + · · ·

and keep all terms up to (r/D)2 consistently. For small quantity ,

(1 + )−1/2 ≈ 1−

2+

3

8

2

and we get

1

d=

1

D

[

1−1

2

( r

D

)2

+( r

D

)

cos θ +3

2

( r

D

)2

cos2 θ + · · ·

]

=1

D

1 +

( r

D

)

cos θ︸︷︷︸

P1

+( r

D

)2(3

2cos2 θ −

1

2

)

︸ ︷︷ ︸

P2

+ · · ·

5.26

Collins (chapter 7) shows how this expansion keeps going in terms ofhigher-order Legendre polynomials.

Let’s look at each term in the expansion.

Eventually we want to know about the acceleration due to the potential termfrom M2...

a = −∇Φeff,2 = ∇

(GM2

d

)

.

Zero-order term: The Φ term is a constant, so a = 0.

First-order term:

ar = −∂Φ2

∂r=

GM2

D2cos θ

aθ = −1

r

∂Φ2

∂θ= −

GM2

D2sin θ

This is straightforward

gravitational attraction ofthe whole body.

If M1 is in orbit around M2, then this is cancelled out if we go into the

co-orbiting reference frame.

Second-order term: Here’s the true tidal distortion:

Φ2 = −GM2r

2

2D3(3 cos2 θ − 1)

So, like in the Roche oblateness problem, let’s compare Φeff at different valuesof θ in order to derive the equilibrium shape of M1.

Φeff(r, θ) = Φeff(r, 0) (right-hand side: pole)

−GM1

R∗−

GM2R2∗

2D3(3 cos2 θ − 1) = −

GM1

Rp

−GM2R

2p

D3

5.27

Simplify by specifying the equator on the left side (θ = π/2), so that we’reeventually solving for xe = Req/Rpol.

Working it through like in the single-star rotation problem, we get a similar

cubic equation,

(Q

2

)

x3e + (1 +Q)xe − 1 = 0 where Q ≡

M2

M1

(Rp

D

)3

.

There are analytic solutions, but in a lot of cases we care about weak tidaleffects, in which Q ≪ 1.

In that case, xe ≈1

1 +Qisn’t so bad.

Note that xe < 1. The star is prolate (stretched out along its θ = 0 axis).

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It’s useful to estimate the so-called “bulge height” due to tidal deformation.

Compare it to a sphere of equivalentvolume.

Also assume it’s a prolate spheroid, with

V = (4π/3)abc = (4π/3)RpR2eq

= (4π/3)R3px

2e

Equate it to the sphere’s equivalent volume V0 = (4π/3)R30.

Thus, R30 =

R3p

(1 +Q)2 Rp = (1 +Q)2/3R0 .

The bulge height is defined as

∆r ≡ Rp −R0 =[

(1 +Q)2/3 − 1]

R0

5.28

For Q ≪ 1, (1 +Q)2/3 ≈ 1 +2

3Q+ · · ·

and...∆r

R0

≈2

3Q ≈

2

3

M2

M1

(R0

D

)3

(using R0 in the definition of Q).

We’ve made a lot of assumptions. Not only Q ≪ 1, but we also assumed theinfinite series of Legendre polynomials can be cut off at P2.

This assumption is essentially that r ≪ D, and so it’s similar to Q ≪ 1 aslimiting us to “weak tides.”

What happens when the tides are strong?

Well, M1 certainly won’t hold together if the tidal force along the line ofcenters (i.e., at r = Rp and θ = 0) is stronger than its own self-gravity!

That occurs forGM1

R2p

≈2GM2Rp

D3

i.e., a critical value of Q =M2

M1

(Rp

D

)3

=1

2.

If Q exceeds that value, it’s unlikely that M1 will remain a single, centrallycondensed body.

The above is the “L1 Lagrange point.” For a synchronously rotating binary

system, the Roche equipotentials include a centrifugal term, which changes theabove force-balance a bit.

In the limit of r ≪ D, the force-balancing Lagrange point radii (on either sideof the centroid of M1) turn out to be

r

D=

(1

3

M1

M2

)1/3

, or exactly Qcrit =1

3.

This is also often called the Hill radius.

5.29

This critical point is sometimes written in terms of the minimum distance Dthat M2 can have before its tidal forces break up M1.

i.e., Dcrit = R1Q−1/3crit

(M2

M1

)1/3

and numerical models give values of Qcrit between about 0.07 and 0.3,depending on the internal structure of M1.

All of the above is true if the equipotentials are allowed to “float freely,”

i.e., if the fluid inside the star/planet is perfectly elastic.

For rocky planets & moons, it’s not so elastic.

Earlier we defined the compressive strength S. Here, let’s define a shearstrength (or shear rigidity) S⊥. It also has units of pressure, and we’llestimate its value below.

Planetary scientists define a so-called “tidal Love number,”

kT ≡3/2

1 + [S⊥/(ρgR)]

For elastic fluids S⊥ is negligible (i.e., there’s no resistance to deformation), sokT ≈ 3/2.

For rigid materials, S⊥ is large, so kT ≪ 3/2.

A more general way to write the bulge height is

∆r

R0

≈4

9kT

M2

M1

(R0

D

)3

.

Deriving the shear strength will tell us more about tidal heating.

When there’s resistance to deformation, the tidal energy still has to gosomewhere... friction will dissipate it as heat.

The shear strength is formally defined as

S⊥ =F⊥/A

r/R0

=applied shear stress

relative sheared displacement.

5.30

In other words, in order to shear a rigid body over a distance r, one needs toapply a transverse force

F⊥ = AS⊥r

R0

.

The work done by applying this force gives the amount of energy expended:

∆E =

∫ r

0

dr′ F⊥ = =A

2S⊥

r2

R0

.

What is the area A? Very roughly, if the force is acting over the whole planet,

then A ≈ πR20.

Neglecting order-unity constants, ∆E ≈ S⊥R0 r2

≈ S⊥R0 (∆r)2

In a binary system, we can estimate the average power released over oneorbit, with ∆t = the period.

(Over one orbit, the tidal bulge gets swept through the whole planet.)

Thus, let’s try Ltide =∆E

∆t≈

S⊥(∆r)2R0

∆t

and if S⊥ scales with ρgR0 ∼M2

1

R40

(recall central pressure scaling)

and if we use the bulge height approximation above (scaling out constants like

kT), we get

Ltide ∝M2

2 R50

D6∆t(M1 drops out!).

If we work all this out for Io’s molten core, we first would look up S⊥ ∼ 1011

dynes/cm2. This is several orders of magnitude bigger than the compressivestrength of a rock or iron core.

For Io, we would get

∆r

R0

≈ 10−3 and Ltide ≈ 1025 erg/s .

This is about 10−9L⊙. Tiny, but since the mass of Io is about 10−8M⊙, that’s

not too shabby.

Problem: Observationally, Io emits only Ltide ≈ 1021 erg/s.

5.31

In reality, all that ∆E work isn’t all going into heating.

In many systems, most of the tidal work goes into changing the angularmomentum of the planet, or of the orbit.

However, if the orbit is elliptical, there’s a net change in the magnitude of thetidal work done over the orbit.

That component is more likely to go directly into heating.

For an orbit with ellipticity e, roughly speaking the relative sheared

displacement

isn’t∆r

R0

, instead it’s ∼ e∆r

R0

.

Thus, one must multiply our Ltide above by a factor of e2.

For Io, e ≈ 0.0043, so Ltide is reduced to about 3× 1020 erg/s. Much closer to

the observed value of ∼1021.

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For more information...

Modern rheological models tend to produce higher tidal dissipation thanolder classical models (e.g., arXiv:1707.06701).

Do exoplanets undergo plate tectonics? Evaluating their tidal stresses is key(e.g., arXiv:1711.09898).

5.32