8. Galactic Dynamics Aim: Understand equilibrium of galaxies 1 ...

8. 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 of
equilibria ? [like stars] ?
Equilibrium of stars as comparison:
• gas pressure versus gravity
• hydrostatic equilibrium, thermal
equilibrium
• timescales corresponding to the above,
plus nuclear burning
• structure completely determined by
mass plus age (and metallicity)
• very little influence from “initial
conditions”
Galaxies:
Much harder!
1

Galaxy dynamics: assume ensemble of point
masses, moving under influence of their own
gravity
Point masses can be: stars, mini BHs,
planets, very light elementary particles ...
Overview
8.1 What are the equations of motion ?
8.2 Do stars collide ? (BT p. 3+4)
8.3 Virial Theorem
8.4 Application: Galaxy masses (BT p.
211-214)
8.5 Binding Energy and Formation of
Galaxies (BT p. 214)
8.6 Scaling Relations
8.7 Relevant Time scales (BT p. 187-190)
(Dynamical timescale, particle interaction
timescale, Relaxation time for large
systems)
2

8.1 Equations of motion
Assume a collection of masses mi at
location xi, and assume gravitational
interaction. Hence the force � Fi on particle i
is given by:
d 2
�Fi = mi
dt2�xi = �
j�=i
�xj − �xi
|�xj − �xi| 3Gmimj
• Only analytic solution for 2 point masses
• “Easy” to solve numerically (brute
force) - but slow for 10 11 particles. See for
example:
• Analytic approximations necessary for a
better understanding of solution
In the following we investigate properties of
gravitational systems without explicitly
solving the equations of motion.
3

8.2 Do stars collide?
Is it safe to ignore non-gravitational
interactions ?
Example: our galaxy
10 11 stars with a total mass of 5 × 10 10 M⊙
A disk of 20 kpc in size, and thickness of 1
kpc
The sun is 8 kpc from the center
Cross section of the sun σ = π ∗ (2R⊙) 2
with R⊙ = 7 × 10 10 cm
Typical orbital speed 200 km/s.
1 km/s ∼ 1 pc in 1 Myr.
One orbit: 250 Myr.
Age of the galaxy: 10 1 0 yr
Sun has made 30 revolutions and disk is in
quasi steady state
4

Calculate number of collisions between t1
and t1 + dt of stars coming in from the left,
in a galaxy with homogeneous number
density n.
• incoming star has velocity v
• suppose all stars have radius r∗
• incoming star will collide with stars in
cylinder 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
5

This gives a collision rate of 1.6×10 −26
sec −1 = 5×10 −19 yr −1 .
Note that the age of the Universe is
13.75 ± 0.11 Gyr ∗
So collisions are very rare indeed. Hence we
can ignore these collisions without too
much trouble.
∗ Komatsu, E., et.al.,
http://arXiv.org/abs/1001.4538
6

8.3 Virial Theorem:
Relation for global properties:
Kinetic energy and Potential energy.
Again consider our system of point masses
mi with positions �xi.
Construct �
i �pi�xi and differentiate w.r.t.
time:
d
dt
�
i
�pi�xi = d
dt
where I = �
inertia.
�
i
i mix 2 i
d�xi
mi
dt �xi = d
dt
= 1d
2
2I dt2 �
i
1
2
d
dt (mix 2 i )
, which is the moment of
However, using the chain rule, we can also
write:
d
dt
�
i
�pi�xi = �
i
d�pi
dt �xi + �
i
d�xi
�pi
dt
7

Then
�
i
d�xi
�pi
dt
= �
i
mi�v 2 i
with K the kinetic energy.
Since d�pi
dt = � Fi we have
1d2I
2 dt
2 = �
i
= 2K
�Fi�xi + 2K.
Now assume the galaxy is ’quasi-static’, i.e.
its properties change only slowly so that
d2I dt2 = 0. Then the equation above implies
.
K = − 1
2
�
i
�Fi�xi
8

Then assume that the force � Fi can be
written as a summation over ’pairwise
forces’ � Fij
Now realize that
�
i
�Fi = �
j,j�=i
�Fi�xi = �
i
�Fij
�
j�=i
�Fij�xi
This summation can be rewritten. We sum
the terms over the full area 0 < i ≤ N,
0 < j ≤ N, i �= j. However, we can also limit
the summation over just half this area:
0 < i ≤ N, i < j ≤ N, and add the (j, i) term
explicitly to the (i, j) term within the
summation.
9

Hence instead of summing over � Fij�xi, we
sum 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 the
summation is done, it does not use any
special property of the force field.
Because � Fij = − � Fji (forces are equal and
opposite for pairwise forces) the last term
can be rewritten
�
i
�Fi�xi = �
i
�
j>i
�Fij(�xi − �xj)
For gravitational force � Fij = − Gmimj
|�xi−�xj| 2
�xi−�xj
|�xi−�xj|
10

�
i
�Fi�xi = − �
which equals
= − �
i
�
j>i
i
�
j>i
Gmimj
|�xi − �xj|
Gmimj
|�xi − �xj| 3(�xi − �xj)(�xi − �xj)
= −1
2
�
i
�
j�=i
Gmimj
|�xi − �xj|
with W the total potential energy.
Therefore for a galaxy in quasi-static
equilibrium:
K = − 1
2 W,
= W
eq.(1)
which is the virial theorem for quasi-static
systems. The more general expression for
non-static systems is:
1d
2
2I = W + 2K.
dt2 11

8.4 Application: Galaxy masses
Consider a system with total mass M
Kinetic energy K = 1 2 M〈v2 〉 with
〈v 2 〉 = mean square speed of stars
(assumption: speed of star not correlated
with mass of star)
Define gravitational radius rg
GM 2
W = −
rg
Spitzer found for many systems that
rg = 2.5r h, where r h is the radius which
contains half the mass
Virial theorem implies:
M〈v 2 〉 =
GM 2
rg
12

or
M = 〈v 2 〉rgG −1
Hence, we can estimate the mass of
galaxies if we know the typical velocities in
the galaxy, and its size!
13

8.5 Binding Energy and Formation of
Galaxies
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 very
large homogeneous distribution
• A galaxy cannot just form from an
unbound, extended smooth distribution −>
E total = Estart ≈ 0, E gal = −K, so energy
must be lost or the structure keeps
oscillating:
Possible energy losses through
• Ejection of stars
• Radiation (before stars would form)
14

8.6. Scaling Relations
Consider a steady state galaxy with
particles mi at location xi(t).
Can the galaxy be rescaled to other physical
galaxies?
• scaled particle mass ˆmi = ammi
• scaled particle location ˆxi = axxi(att)
• am, ax, at are scaling parameters
In the ’rescaled’ galaxy we have
ˆ�Fi = ammi d2
dt 2ax�x(att) = amaxa 2 t � F i,orig
The gravitational force is equal to
ˆ�F G = �
j�=i
a 2 m
a 2 x
�F G,orig
ax�xj−ax�xi
|ax�xj−ax�xi| 3Gammiammj =
15

Equilibrium is satisfied if the two terms
above are equal
ˆ�Fi = ˆ� FG
Since we have equilibrium when all scaling
parameters are equal to 1, we obtain
or
amaxa 2 t = a2m a2 ,
x
am = a 3 xa 2 t .
Now how do velocities scale ?
ˆv = d d
ˆx =
dt dt axx(att) = axatvorig Hence, the scaling of the velocities satisfies:
av = ax at. We can write the new scaling
relation as
am = axa 2 v
Hence ANY galaxy can be scaled up like
this!
16

Notice that it is trivial to derive this from
the virial theorem - the expression for the
mass has exactly the same form.
As a consequence, if we have a model for a
galaxy with a certain mass and size, we can
make many more models, with arbitrary
mass, and arbitrary size.
17

8.7 Time scales
Dynamical timescale,
particle interaction timescale
Is gravitational force dominated by short or
long range encounters? (N.B. in a gas, only
short range forces are relevant).
In a galaxy, the situation is different.
Consider force with which stars in cone
attract star in apex of cone.
18

Force ∼ 1/r 2 , with r the distance from
apex. If ρ is almost constant, then the mass
in a shell with width dr increases as r 2 dr.
Hence differential force is constant at each
r, and we have to integrate all the way out
to obtain the total force.
Realistic densities decrease after some
radius, so that the force will be determined
by the density distribution on a galactic
scale (characterized by the half mass
radius).
19

Relaxation time
Short range encounters do not dominate →
Approximate force field with a smooth
density ρ(x) instead of point masses.
• Contrary of situation in gas: only
consider long range encounters (long range
∼ scale of the galaxy)
Assume all stars have mass m. Analyze
perturbations due to the fact that density is
not smooth, but consists of individual stars.
Simplify, and look first at single star-star
encounter.
What is the amount δv induced by one star
crossing another?
Estimate: straight line trajectory past
stationary perturber
20

The perpendicular force � F ⊥ gives
perturbation δ�v ⊥:
�F ⊥ = Gm2 cos θ
r 2
=
Newton: d
dt δ�v ⊥ = � F⊥ m
Gm
bv
δ�v ⊥ =
∞�
−∞
�
dt � F ⊥
m =
= Gm2cos θ
(b2 +x2 ) =
Gm2b (b2 +x2 ) 3/2
Gm 2
b 2 [1+(vt/b) 2 ] 3/2
�
ds
(1 + s2 Gm
=
) 3/2 bv
⇒
Gm
b 2 [1 + (vt/b) 2 ]
s
�
1 + s 2|∞ −∞
3/2dt =
= 2Gm
bv
In words: δv is roughly equal to acceleration
at closest approach, Gm/b 2 , times the
duration 2b/v.
21

Note: approximation fails when
δ�v ⊥ > v ⇒ b < Gm/v 2 ≡ b min
Galaxy has characteristic radius R.
Define crossing time tc as the time it takes
a star to move through the galaxy tc = R/v
Calculate number of perturbing encounters
per crossing time tc
In a crossing time, the star has 1
“encounter” with each other star in the
galaxy
The impact parameter of each encounter
can be derived by projecting each star onto
a plane perpendicular to the unperturbed
motion of the star
Hence “flatten” the galaxy in the plane
perpendicular to the motion of the star, and
22

assume that the stars are homogeneously
distributed in that plane, out to a radius R,
and no stars outside R. This is obviously a
simplifying assumption, but it is reasonably
accurate.
This can be used to derive the distribution
of impact parameters:
N stars in total in Galaxy, distributed over
total surface πR 2
stellar surface density / # per unit area:
N
πR 2
In a crossing time, the star has δn
encounters with impact parameter between
b and b + db. δn is given by the area of the
annulus 2πbdb times the density of stars on
the surface, which is N/(πR 2 ):
δn = N
πR22πbdb =
2N
R2 b db
23

Result: � δv ⊥ = 0 as the perturbations are
randomly distributed, and will not change
the average velocity
� δv 2 ⊥ = δv 2 δn =
= 8N
� �2 2Gm 2Nb
db
bv R2 � �2 Gm db
Rv
as each perturbation adds to � δv ⊥ by an
equal amount (2Gm/bv) 2 .
The encounters do not produce an average
perpendicular velocity, but they do produce
an average (perpendicular velocity) 2 .
Hence, on average, the stars still follow
their average path, but they tend to
“diffuse” around it.
b
24

The total increase in rms perpendicular
velocity can be calculated by integrating
over all impact parameters from bmin to
infinity:
Total rms increase:
∆v 2 =
R
�
b min
=8N( Gm
Rv )2 ln Λ
� δv 2 ⊥ db =
R
�
b min
8N
� Gm
Rv
�2
with ln Λ = Coulomb logarithm = ln R
b min
We can rewrite this equation. Recall:
b min ≡ Gm/v 2
From virial theorem:
v 2 = GM/R = GNm/R
db/b =
25

Hence: bmin = Gm/(GNm/R) = R/N
ln Λ = ln R/bmin = ln R
R/N
Furthermore from virial theorem:
GM 2
= ln N
R = Mv2 → GM
R = v2 → GNm
R
so that: 〈∆v2 ⊥ 〉
v 2
→ N = v2 R
Gm
=
8 ln N
N
= v2
This last number is the fractional change in
energy per crossing time. Hence we need
the inverse number of crossings N/(8 ln N)
to get 〈∆v2 ⊥ 〉 ∼ v2
26

The timescale t relax is defined as the time it
takes to deflect each star significantly by
two body encounters, and it is therefore
equal to
t relax = N
8 ln N tc ≈ 0.1N
ln N tc
with crossing time: tc ≡ R/v
27

Conclusions
• effect of point mass perturbations
decreases as N increases
• even for low N=50, 〈∆v 2 ⊥ 〉/v2 = 0.6,
hence deflections play a moderate role.
• for larger systems the effect of
encounters become even less important
Notice: one derives the same equation when
the exact formulas for the encounters are
used. Put in another way, the encounters
with b < bmin do not dominate.
28

Relaxation time for large systems
System N tc (yr) t relax (yr)
globular cluster 10 5 10 5 2 × 10 8
galaxy 10 11 10 8 10 17
galaxy cluster 10 3 10 9 3 × 10 10
Age of Universe 13.7 Gyr.
⇒ Galaxies are collisionless systems
• motion of a star accurately described by
single particle orbit in smooth gravitational
field of galaxy
• no need to solve N-body problem with
N = 10 11 (!)
29

Some more typical questions that can be
asked at exam: (not homework
assignments)
• What is the virial theorem ?
• How can we use it to scale galaxies ?
• Do galaxies have a main sequence like
stars ?
• How is relaxation time defined ? What
is the approximate expression ? What is the
typical relaxation time for a galaxy ? How
does the long relaxation time make it easier
to produce models for galaxies ?
30

8. Homework Assignments:
1) How empty are galaxies ? Calculate the
mean distance between stars. Take the
radius of the sun as typical radius for 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.
2) calculate the mass of our galaxy for two
assumed radii:
a.) 10 kpc (roughly the distance to the sun)
b.) 350 kpc (halfway out to Andromeda)
Take as a typical velocity the solar value of
200 km/s.
3) Show explicitly that the virial theorem
produces the same result for the scaling
relations.
4) Take typical sizes, velocities, number of
stars for globular clusters, galaxies, and
31

galaxy clusters. Calculate for each of the
systems, the average time between stellar
collisions, crossing times and relaxation
times.