Overview 3. Galactic Dynamics

Aim: Understand equilibrium of galaxies

1. What are the dominant forces ?2. Can we define some kind of equilibrium?3. What are the relevant timescales?4. Do galaxies evolve along a sequence ofequilibria ? [like stars] ?

Equilibrium of stars as comparison:

• gas pressure versus gravity• hydrostatic equilibrium, thermalequilibrium• timescales corresponding to the above,plus nuclear burning• structure completely determined bymass plus age (and metallicity)• very little influence from “initialconditions”

Galaxies:

Much harder!1

Galaxy dynamics: assume ensemble of pointmasses, moving under influence of their owngravity

Point masses can be: stars, mini BHs,planets, very light elementary particles ...

Overview

3.1 What are the equations of motion ?3.2 Do stars collide ? (BT p. 2+3)3.3 Virial Theorem3.4 Application: Galaxy masses (BT p.360,361)3.5 Binding Energy and Formation ofGalaxies (BT 361)3.6 Scaling Relations3.6 Relevant Time scales (BT 1.2.1)(Dynamical timescale, particle interactiontimescale, Relaxation time for largesystems)

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3.1 Equations of motion

Assume a collection of masses mi atlocation xi, and assume gravitationalinteraction. Hence the force �Fi on particle i

is given by:

�Fi = mid2

dt2�xi =

�

j �=i

�xj − �xi

|�xj − �xi|3Gmimj

• Only analytic solution for 2 point masses• “Easy” to solve numerically (bruteforce) - but slow for 1011 particles. See forexample:lahmu.phyast.pitt.edu/˜ gardner/Galaxy/index.html

• Analytic approximations necessary for abetter understanding of solution

In the following we investigate properties ofgravitational systems without explicitlysolving the equations of motion.

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3.2 Do stars collide ? (BT 1 to 1.1)

Is it safe to ignore non-gravitationalinteractions ?

Example: our galaxy

1011 stars with a total mass of 5× 1010 M⊙

A disk of 20 kpc in size, and thickness of0.5 kpcThe sun is 8 kpc from the center

Cross section of the sun σ = π ∗ (2R⊙)2

with R⊙ = 7× 1010 cm

Typical orbital speed 200 km/s.1 km/s ∼ 1 pc in 1 Myr.One orbit: 250 Myr.Age of the galaxy: 109 yrSun has made 40 revolutions and disk is inquasi steady state

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Calculate number of collisions between t1and t1 + dt of stars coming in from the left,in galaxy with homogeneous number densityn.

• incoming star has velocity v

• suppose all stars have radius r∗• incoming star will collide with stars incylinder with volume V1:−> V1 = π(2r∗)2 × v × dt

• number of stars in V1 : N1 = nV1• number of collisions per unit time:

N1

dt=

nV1

dt= 4πr2∗nv

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This gives a collision rate of 1.6×10−26

sec−1 = 5×10−19 yr−1.

Note that the age of the Universe is13.75± 0.11 Gyr ∗

So collisions are very rare indeed. Hence wecan ignore these collisions without toomuch trouble.

∗Komatsu, E., et.al.,http://arXiv.org/abs/1001.4538

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3.3 Virial Theorem: (not in BT)

Relation for global properties:Kinetic energy and Potential energy.

Again consider our system of point massesmi with positions �xi.

Construct�

i �pi�xi and differentiate w.r.t.time:

d

dt

�

i

�pi�xi =d

dt

�

i

mid�xi

dt�xi =

d

dt

�

i

1

2

d

dt(mix

2i )

=1

2

d2I

dt2

where I =�

i mix2i , which is the moment of

inertia.

However, using the chain rule, we can alsowrite:

d

dt

�

i

�pi�xi =�

i

d�pi

dt�xi +

�

i

�pid�xi

dt

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Then

�

i

�pid�xi

dt=

�

i

mi�v2i = 2K

with K the kinetic energy.

Since d�pidt = �Fi we have

1

2

d2I

dt2=

�

i

�Fi�xi + 2K.

Now assume the galaxy is ’quasi-static’, i.e.its properties change only slowly so thatd2Idt2

= 0. Then the equation above implies

K = −1

2

�

i

�Fi�xi

.

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Then assume that the force �Fi can bewritten as a summation over ’pairwiseforces’ �Fij

�Fi =�

j,j �=i

�Fij

Now realize that

�

i

�Fi�xi =�

i

�

j �=i

�Fij�xi

This summation can be rewritten. We sumthe terms over the full area 0 < i ≤ N ,0 < j ≤ N , i �= j. However, we can also limitthe summation over just half this area:0 < i ≤ N , i < j ≤ N , and add the (j, i) termexplicitly to the (i, j) term within thesummation.

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Hence instead of summing over �Fij�xi, wesum over �Fij�xi + �Fji�xj

Hence�

i

�

j �=i

�Fij�xi =�

i

�

j>i

(�Fij�xi + �Fji�xj)

This is simply a change in how thesummation is done, it does not use anyspecial property of the force field.

Because �Fij = −�Fji (forces are equal andopposite for pairwise forces) the last termcan be rewritten

�

i

�Fi�xi =�

i

�

j>i

�Fij(�xi − �xj)

For gravitational force �Fij = − Gmimj

|�xi−�xj|2�xi−�xj|�xi−�xj|

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�

i

�Fi�xi = −�

i

�

j>i

Gmimj

|�xi − �xj|3(�xi − �xj)(�xi − �xj)

which equals

= −�

i

�

j>i

Gmimj

|�xi − �xj|= −

1

2

�

i

�

j �=i

Gmimj

|�xi − �xj|= W

with W the total potential energy.Therefore for a galaxy in quasi-staticequilibrium:

K = −1

2W,

which is the virial theorem for quasi-staticsystems. The more general expression fornon-static systems is:

1

2

d2I

dt2= W + 2K.

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Homework assignments:

1) derive the relation for the potentialenergy given above. Suppose each particlehas mass fmi, 0 ≤ f ≤ 1. Slowly add anextra mass midf to each particle frominfinity and calculate how much energy isreleased. Integrate f from 0 to 1.

2) How empty are galaxies ? Calculate thedistance between stars, and calculate theradius of a star. Take the ratio of the two.Compare this to the same ratio of galaxies:take the average distance between galaxies,and the radius of galaxies.

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3.4 Application: Galaxy masses (BT p.

360,361)

Consider a system with total mass M

Kinetic energy K = 12M�v2� with

�v2� = mean square speed of stars(assumption: speed of star not correlatedwith mass of star)

Define gravitational radius rg

W = −GM2

rg

Spitzer found for many systems thatrg = 2.5rh, where rh is the radius whichcontains half the mass

Virial theorem implies:

M�v2� =GM2

rg

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or

M = �v2�rgG−1

Hence, we can estimate the mass ofgalaxies if we know the typical velocities inthe galaxy, and its size!

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Homework Assignments:

3) calculate the mass of our galaxy for twoassumed radii:

a.) 10 kpc (roughly the distance to thesun)

b.) 350 kpc (halfway out to Andromeda)

Take as a typical velocity the solar value of200 km/s.

4) Give the correct expression for the totalkinetic energy if the velocities of stars AREcorrelated with their mass.

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3.5 Binding Energy and Formation ofGalaxies

The total energy E of a galaxy is

E = K + W = −K = 1/2W

• Bound galaxies have negative energy -cannot fall apart and dissolve into a verylarge homogeneous distribution• A galaxy cannot just form from anunbound, extended smooth distribution −>Etotal = Estart ≈ 0, Egal = −K, so energymust be lost or the structure keepsoscillating:

Possible energy losses through

• Ejection of stars

• Radiation (before stars would form)

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3.6. Scaling Relations

Consider a steady state galaxy withparticles mi at location xi(t).

Can the galaxy be rescaled to other physicalgalaxies?

• scaled particle mass m̂i = ammi

• scaled particle location x̂i = axxi(att)• am, ax, at are scaling parameters

In the ’rescaled’ galaxy we have

�̂Fi = ammid2

dt2ax�x(att) = amaxa2

t�Fi,orig

The gravitational force is equal to

�̂FG =�

j �=iax�xj−ax�xi

|ax�xj−ax�xi|3Gammiammj =

a2m

a2x

�FG,orig

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Equilibrium is satisfied if the two termsabove are equal

�̂Fi = �̂FG

Since we have equilibrium when all scalingparameters are equal to 1, we obtain

amaxa2t =

a2m

a2x

,

or

am = a3xa2

t .

Now how do velocities scale ?

v̂ =d

dtx̂ =

d

dtaxx(att) = axatvorig

Hence, the scaling of the velocities satisfies:av = ax at. We can write the new scalingrelation as

am = axa2v

Hence ANY galaxy can be scaled up likethis!

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Notice that it is trivial to derive this fromthe virial theorem - the expression for themass has exactly the same form.

As a consequence, if we have a model for agalaxy with a certain mass and size, we canmake many more models, with arbitrarymass, and arbitrary size.

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Homework Assignments:

5) Show explicitly that the virial theoremproduces the same result for the scalingrelations.

6) How does this compare to stars ? Whathappens to a star when we scale it in thesame way ? Which equilibrium ismaintained, which equilibrium is lost ?

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3.6 Time scales (BT 1.2.1)

Dynamical timescale,particle interaction timescale

Is gravitational force dominated by short orlong range encounters? (N.B. in a gas, onlyshort range forces are relevant).

In a galaxy, the situation is different.Consider force with which stars in coneattract star in apex of cone.

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Force ∼ 1/r2, with r the distance fromapex. If ρ is almost constant, then the massin a shell with width dr increases as r2dr.

Hence differential force is constant at eachr, and we have to integrate all the way outto obtain the total force.

Realistic densities decrease after someradius, so that the force will be determinedby the density distribution on a galacticscale (characterized by the half massradius).

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Relaxation time

Short range encounters do not dominate →Approximate force field with a smoothdensity ρ(x) instead of point masses.

• Contrary of situation in gas: onlyconsider long range encounters (long range∼ scale of the galaxy)

Assume all stars have mass m. Analyzeperturbations due to the fact that density isnot smooth, but consists of individual stars.Simplify, and look first at single star-starencounter.

What is the amount δv induced by one starcrossing another?

• Exact: BT §3.1(d): hyperbolic Keplerianencounter• Estimate: straight line trajectory paststationary perturber

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The perpendicular force �F⊥ givesperturbation δ�v⊥:

�F⊥=Gm2cos θ

r2=

Gm2cos θ

(b2+x2)=

Gm2b

(b2+x2)3/2

=Gm2

b2[1+(vt/b)2]3/2

Newton:d

dtδ�v⊥ =

�F⊥m

⇒

δ�v⊥=�

dt�F⊥m

=� Gm

b2[1 + (vt/b)2]3/2dt =

Gm

bv

∞�

−∞

ds

(1 + s2)3/2=

Gm

bv

s�

1 + s2|∞−∞ =

2Gm

bv

In words: δv is roughly equal to accelerationat closest approach, Gm/b2, times theduration 2b/v.

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Note: approximation fails when

δ�v⊥ > v ⇒ b < Gm/v2 ≡ bmin

Galaxy has characteristic radius R.

Define crossing time tc as the time it takesa star to move through the galaxy tc = R/v

Calculate number of perturbing encountersper crossing time tc

In a crossing time, the star has 1“encounter” with each other star in thegalaxy

The impact parameter of each encountercan be derived by projecting each star ontoa plane perpendicular to the unperturbedmotion of the star

Hence “flatten” the galaxy in the planeperpendicular to the motion of the star, and

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assume that the stars are homogeneouslydistributed in that plane, out to a radius R,and no stars outside R. This is obviously asimplifying assumption, but it is reasonablyaccurate.

This can be used to derive the distributionof impact parameters:

N stars in total in Galaxy, distributed overtotal surface πR2

stellar surface density / # per unit area:

N

πR2

In a crossing time, the star has δn

encounters with impact parameter betweenb and b + db. δn is given by the area of theannulus 2πbdb times the density of stars onthe surface, which is N/(πR2):

δn = NπR22πbdb = 2N

R2 b db

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Result:�

δv⊥ = 0 as the perturbations arerandomly distributed, and will not changethe average velocity

�δv2⊥ = δv2δn =

�2Gm

bv

�2 2Nb

R2 db

= 8N�

Gm

Rv

�2 db

b

as each perturbation adds to�

δv⊥ by anequal amount (2Gm/bv)2.

The encounters do not produce an averageperpendicular velocity, but they do producean average (perpendicular velocity)2.Hence, on average, the stars still followtheir average path, but they tend to“diffuse” around it.

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The total increase in rms perpendicularvelocity can be calculated by integratingover all impact parameters from bmin toinfinity:

Total rms increase:

∆v2 =R�

bmin

�δv2⊥ db =

R�

bmin

8N�

Gm

Rv

�2db/b =

=8N(Gm

Rv)2 lnΛ

with lnΛ = Coulomb logarithm = lnR

bmin

We can rewrite this equation. Recall:

bmin ≡ Gm/v2

From virial theorem:

v2 = GM/R = GNm/R

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Hence: bmin = Gm/(GNm/R) = R/N

lnΛ = lnR/bmin = lnR

R/N= lnN

Furthermore from virial theorem:

GM2

R= Mv2 →

GM

R= v2 →

GNm

R= v2

→ N =v2R

Gm

so that:�∆v2

⊥�v2 = 8 lnN

N

This last number is the fractional change inenergy per crossing time. Hence we needthe inverse number of crossings N/(8 lnN)to get �∆v2

⊥� ∼ v2

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The timescale trelax is defined as the time ittakes to deflect each star significantly bytwo body encounters, and it is thereforeequal to

trelax =N

8 lnNtc ≈

0.1N

lnNtc

with crossing time: tc ≡ R/v

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Conclusions

• effect of point mass perturbationsdecreases as N increases

• even for low N=50, �∆v2⊥�/v2 = 0.6,

hence deflections play a moderate role.

• for larger systems the effect ofencounters become even less important

Notice: one derives the same equation whenthe exact formulas for the encounters areused. Put in another way, the encounterswith b < bmin do not dominate.

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Relaxation time for large systems

System N tc (yr) trelax (yr)

globular cluster 105 105 2× 108

galaxy 1011 108 1017

galaxy cluster 103 109 3× 1010

Age of Universe 13.7 Gyr.

⇒ Galaxies are collisionless systems

• motion of a star accurately described bysingle particle orbit in smooth gravitationalfield of galaxy

• no need to solve N-body problem withN = 1011 (!)

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Homework Assignment:

1) What is the relaxation time for our solarsystem according to the formula above.Why is this calculation incorrect ?

2) Take typical sizes, velocities, number ofparticles for globular clusters, galaxies, andgalaxy clusters. Calculate the crossing timesand relaxation times.

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Some more typical questions that can beasked at exam: (not homeworkassignments)

• What is the virial theorem ?• How can we use it to scale galaxies ?• Do galaxies have a main sequence likestars ?• How is relaxation time defined ? Whatis the approximate expression ? What is thetypical relaxation time for a galaxy ? Howdoes the long relaxation time make it easierto produce models for galaxies ?

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