Lecture 28 8.284 Spring 2006

Galaxy Disks: rotation and epicyclic motion

1. Last time, we discussed how you measure the mass of an elliptical galaxy. You measurethe width of the line and apply the radial Jeans equation, making some assumptions aboutanisotropy.

2. Galaxy disks are much easier: they are mostly ordered, with very little randomness. Onceagain, you use a spectrograph.

spectroscopic slit

Hβ 4959 5007 OIII NII H NII 6563

an absorbtion line, maybe

increasing λ

Ionized gas� HII regions and di�use (ambient) starlight.

r

vc

most of the light

big bulge systems

small bulge systems

r-½flat!

(non-Keplerian)

looks very much like an isothermal sphere

the Milky Way is very typical

vc ∼ 220km/s

(a) As you might expect, vc increases as L increases. Empirical Tully-Fisher relation: L ∼ v4c .Can be used as standard candles.

(b) The corresponding Faber-Jackson relation for ellipticals is L ∼ σ4.

3. Of course, disks are not in�nitely cool� there's a velocity dispersion tensor with compenentsσ2rr, σ2θθ, and σ2zz in polar coordinates.

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Lecture 28 8.284 Spring 2006

(a) Now you know how to integrate an orbit numerically, but it turns out that if you assumethat the deviations from circularity are small, you can treat the problem analytically. Aperturbation approach.

(b) Let's go to a coordinate system in which a particle in circular motion would be at rest.

perturbed position

xy

r

r 0θ

Ω0t

Let

Ω(r) =vc(r)

r

~r = ~ro + xx̂ + yŷ

x ≈ r − ro

y ≈ ro(θ − Ωot)

4. Now you know that in an inertial frame, the equation of motion is given by

~̈r = −~∇Φ

(a) In the absence of a potential, ~̇r = constant.(b) This is not so in an accelerating frame. In a frame rotating with an angular velocity of

~Ωo,

~̈r = −~∇Φ− 2~Ωo × ~v︸ ︷︷ ︸Coriolisforce

− ~Ωo × (~Ωo × ~r)︸ ︷︷ ︸centrifugalforce

The Coriolis force explains why storms rotate counterclockwise in the northern hemi-sphere. The centrifugal force explains why the Earth bulges at the equator.

(c) Both of these forces are �ctitious, in the sense that we've chosen a non-intertial coordinatesystem.

(d) This is developed in section 7.2 of Symon's �Classical Mechanics� and in appendix 1.D.3of Binney & Tremain.

5. We will substitute our expression for ~r into this equation, but �rst notice that

~∇Φ = v2c

r= Ω2~r ≈

Ω2oro + Ω2odx︸ ︷︷ ︸Ω2o~r

+2roΩodΩdr

x

x̂2

Lecture 28 8.284 Spring 2006

where I've expanded the product in a 2D Taylor series around ro.

• One vector equation:

ẍx̂ + ÿŷ ≈ −Ω2orx̂− 2roΩodΩdr

xx̂− 2Ωoẋŷ + 2Ωoẏx̂︸ ︷︷ ︸Coriolis

+Ω2o~r

• 2 scalar equations: x̂:

ẍ− 2Ωoẏ = 4ΩoAox

Here, �Oort's A� is de�ned as

Ao ≡(−r

2dΩdr

)ro

ŷ:

ÿ + 2Ωoẋ = 0

• Anyone care to suggest a solution? Small motion about an equilibrium position. Try:

x = xo sin(κt + φ)

y = yo cos(κt + φ)

These give:

−κ2xo(sinκt + φ) + 2yoΩo sin(κt + φ) = 4ΩoAoxo sin(κt + φ)

and

−κ2yo cos(κt + φ) + 2κxo cos(κt + φ) = 0

yoxo

=2Ωoκ

κ2 = 4Ωo(Ωo −Ao)

6. What does motion look like? There are parametric equations for an ellipse in cartesian coor-dinates. Motion is said to be �retrograde.�

x

y

Ω

guiding center

κt=π/2

κt=0 κt=π

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Lecture 28 8.284 Spring 2006

7. This may be somewhat familiar. Kepler, when he introduced ellipses as the orbits of the orbitsof the planets, was trying to avoid the use of epicycles. Here we are taking what looks like agiant step backwards and reintroducing them!

(a) This is called the �epicyclic approximation� and κ is called the �epicyclic� frequency.

8. Case I: Galaxy vc is constant;

Ω(r) =vcr

A = −r2

d

dr

vcr

=12

vcr

=12Ω

κ2 = 4Ω(Ω− 12Ω) = 2Ω2

κ =√

2Ω

(a) Leads to an orbit that isn't closed� a rosette.(b) The epicycle is complete before the guiding center completes an orbit. This is what you

found.(c) yoxo =

√2� a bit squashed.

9. Case II: Kepler

Ω =(

GM

r3

)1/2

A(r) = −12r32

(GM)1/2

r3/2=

34Ω

κ2 = 4Ω(Ω−A) = Ω2

κ = Ω

(a) Here, one epicycle per guiding orbit; orbits close on themselves.(b) yoxo =

2Ωκ = 2� fairly �at: that's where Hipparchus went wrong. You can't completely

avoid ellipses!

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Lecture 28 8.284 Spring 2006

10. If you average over an orbit, you have

< ẏ2 >orbit< ẋ2 >orbit

=y2ox2o

=4Ω2

κ2

guiding center trails

guidingcenter leads

But that's not the whole story. When we measure velocity dispersion at a point, we're averagingover stars.(

1− AΩ

)=

< σ2yy >stars

< σ2xx >stars

x2oy2o

=< σ2θθ >stars< σ2rr >stars

=κ2

4Ω2

(This comes from second moments of collisionless Boltzmann)Case I:

< σ2θθ >stars< σ2rr >stars

=12

We measure a larger σ in the radial direction, then in the tangential direction.

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