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th
11
• Phase Space
Lec
Collisionless Systems
• We showed collisions or deflections are rare
• Collisionless: stellar motions under
influence of mean gravitational potential!
• Rational:
• Gravity is a long-distance force, decreases
as r-2
– as opposed to the statistical mechanics of
molecules in a box
Collisionless Systems
• stars move under influence of a smooth
gravitational potential
– determined by overall structure of system
• Statistical treatment of motions
– collisionless Boltzman equation
– Jeans equations
• provide link between theoretical models
(potentials) and observable quantities.
• instead of following individual orbits
• study motions as a function of position in
system
• Use CBE, Jeans eqs. to determine mass
distributions and total masses
Fluid approach:Phase Space Density
PHASE SPACE DENSITY:Number of stars
per unit volume per unit velocity volume
f(x,v) (all called Distribution Function DF).
number of stars  m
Nm
f(x, v) 
 3
space volume  velocity volume pc (kms 1 )3
The total number of particles per unit
volume is given by:
  
m  n( x ) 


    
f ( x , v )dvx dv y dvz
• E.g., air particles with Gaussian velocity
(rms velocity = σ in x,y,z directions):
 vx2  v y2  vx2 

m  n o exp  
2

2



f(x, v) 
( 2  )3
• The distribution function f(x,v) is defined
by:
mdN=f(x,v)d3xd3v
where dN is the number of particles per unit
volume with a given range of velocities.
  3  3
mdN  f ( x , v )d xd v
• The total mass is then given by the integral
of the mass distribution function over space
and velocity volume:
  3 3 
3
M total    ( x)d x   f ( x , v )d v d x
• Note:in spherical symmetry d3x=4πr2dr,
• for isotropic systems d3v=4πv2dv
• The total momentum is given by:

   3  3
Ptotal   v mdN   f ( x , v )v d xd v
• Example:mean speed of air molecules in a
box of dx3 :
 v2  v2  v2 
y
m  n o exp   x
2
2


f(x, v) 
( 2  )3

x
 v2
exp  
3
3
2

vdN
vfd
xd
v
2




0


2
3
3

 dN  fd xd v  exp   v 2
 2
0
These are gamma functions




3
4

v
dv



2
 4 v dv

• Gamma Functions:

(n)   e x dx
x
n 1
0
(n)  (n  1)(n  1)
1
   
 2
How to calculate

2
0
0
 d    sin  d
3
dx
and
3
dv
2
2
2
d


2

2

,
r

x

x

y

z

d 3 x  dxdydz  r 2 drdΩ  4πr 2 dr (if spherical)
d 3 v  dvx dv y dvz  v 2 dvdΩ  4πv 2 dv (if isotropic)
V  v  vx2  v y2  vz2  2[ E   ( x)]
DF and its moments
d 3 x A(x)  dM  A  d 3 x  Af ( x, v )d 3 v
d 3 x  dM  d 3 x  f ( x, v)d 3 v
1
For : A(x,v )  Vx , VxVy, (Vx 2  Vy 2  Vz 2 )    x  , x  v
2
AdM

A  A 
 dM
   ,

mass-weighted average,
Additive: subcomponents add up
to the total gravitational mass
A B  A  B
f  f1  f 2
  1   2
  
Full Notes online
• http://www-star.st-and.ac.uk/~hz4/gravdyn/
GraviDynFinal3.ppt
GraviDynFinal3.pdf
Liouvilles Theorem
We previously introduced the concept of phase space
density. The concept of phase space density is useful
because it has the nice property that it is incompessible for
collisionless systems.
A COLLISIONLESS SYSTEM is one where there are no
collisions. All the constituent particles move under the
influence of the mean potential generated by all the other
particles.
INCOMPRESSIBLE means that the phase-space density
doesn’t change with time.
Consider Nstar identical particles moving in a small bundle
through spacetime on neighbouring paths. If you measure
the bundles volume in phase space (Vol=Δx Δ p) as a
function of a parameter λ (e.g., time t) along the central
path of the bundle. It can be shown that:
dVol
dNstar
 0,
 0,
d
d
px
' LIOUVILLES THEOREM'
px
x
x
It can be seen that the region of phase space occupied by
the particle deforms but maintains its area. The same is
true for y-py and z-pz. This is equivalent to saying that the
phase space density f=Nstars/Vol is constant. df/dt=0!
motions in phase-space
• Flow of points in phase space corresponding
to stars moving along their orbits.
• phase space coords: ( x, v)  w  (w1 , w2 ,..., w6 )
w  ( x, v)  (v,)
• and the velocity of the flow is then:
– where wdot is the 6-D vector related to w as
the 3-D velocity vector v
relates to x
• stars are conserved in this flow, with no
encounters, stars do not jump from one
point to another in phase space.
• they drift slowly through phase space
• In the COMBINED potential of stars and
dark matter
fluid analogy
• regard stars as making up a fluid in phase
space with a phase space density
f (x, v,t)  f (w,t)
•
• assume that f is a smooth function,
continuous and differentiable
– good for N >105
• as in a fluid, we have a continuity equation
• fluid in box of volume V, density , and
velocity v,
the change in mass is
then: dM   

dt  
V
d 3 x    v  d 2 S

t 
S
3
2


F
d
x

F

d
S


V
S
        v d 3 x  0
 t

V
– Used the divergence theorem
continuity equation
• must hold for any volume V, hence:

t      v   0
• in same manner, density of stars in phase
space obeys a continuity equation:
6
 
f
  fw

0
t


1
w
If we integrate over a volume of phase space
V, then 1st term is the rate of change of the
stars in V, while 2nd term is the rate of
outflow/inflow of stars from/into V.
0
3
3
 vi vi 

w

 




vi  i 1 vi
 1 w
i 1  xi
6
  
 x   0
 i
Collisionless Boltzmann
Equation
• Hence, we can simplify the continuity
equation to the CBE:
6
f
f
  w 
0
t  1 w
3
 f  f 
f
  vi

0

t i 1  xi xi vi 
• Vector form f
f
 v  f   
0
t
v
• in the event of stellar encounters, no longer
collisionless
• require additional terms to rhs of equation
CBE cont.
• can define a Lagrangian derivative
• Lagrangian flows are where the coordinates
travel along with the motions (flow)
– hence x= x0 = constant for a given star
• then we have:
• and
d 
  w  6
dt t
6
df f
f

  w 
0
dt t  1 w
incompressible flow
• example of incompressible flow
• idealised marathon race: each runner runs at
constant speed
• At start: the number density of runners is
large, but they travel at wide variety of
speeds
• At finish: the number density is low, but at
any given time the runners going past have
nearly the same speed
DF & Integrals of motion
• If some quantity I(x,v) is conserved i.e.
dI ( x, v )
0
dt
• Assume f(x,v) depends on (x,v) through the
function I(x,v), so f=f(I(x,v)).
• Such phase space density is incompressible, i.e
df
0
dt
Jeans theorem
• For most stellar systems the DF depends on (x,v)
through generally three integrals of motion
(conserved quantities), Ii(x,v),i=1..3  f(x,v) =
f(I1(x,v), I2(x,v), I3(x,v))
• E.g., in Spherical Equilibrium, f is a function of
energy E(x,v) and ang. mom. vector L(x,v).’s
amplitude and z-component
 
f ( x, v)  f ( E, || L ||, L  zˆ)
3D Analogy of 6D Phase space
• If DF(x,v) is analogous to density(x,y,z),
• Then DF(E,L,Lz) is ~ density(r,theta,phi),
• Integrals analogous to spherical coordinates
– E(x,v) analogous to r(x,y,z)
• Isotropic DF(E) ~ spherical density(r)
– Normalization dM=f(E)dx3dv3 ~ dM=density(r)dr3
– Have non-self-gravitating subcomponents: DF1+DF2,
like rho1+rho2 to make up total gravity.
th
12
• Phase Space
Lec
Tensor Virial
Theorem

• Equation of motion: dv  
dt
T
T
1
dv
1


dt

r
  dtr .



T 0  dt  T 0


d (rv )  dr

v
dt
dt
T
T
T
(rv )
1 
1


  v vdt   dtr .
T 0 T 0
T 0


 v v  r  
This is Tensor Virial Theorem
• E.g.
 vx vx  x x 
 vx v y  x y  etc
 v 2  vx2    v y2    vz2 

 r . 
d
2
 r
 vcir

dr
2
 v 2  vcir

( spherical )
• So the time averaged value of v2 is equal to the
time averaged value of the circular velocity
squared.
Scalar Virial Theorem
• the kinetic energy of a system with mass M is just
where <v2> is the mean-squared speed of the
system’s stars.
1
K  M  v2 
2
• Hence the virial theorem states that
W
GM
 v 

M
rg

 v 2   r . 
2
 2K  W  0
Virial
Stress Tensor
n ij 2
• describes a pressure which is anisotropic
– not the same in all directions
• and we can refer to a “pressure supported”

system


n   P
x
• the tensor is symmetric.
• can chose a set of orthogonal axes such that
the tensor is diagonal  ij 2   ii 2 ij
• Velocity ellipsoid with semi-major axes
11 ,  22 , 33
given by
2
ij
i
Subcomponents in Spherical
Equilibrium Potential
• Described by spherical potential φ(r)
• SPHERICAL subcomponent density ρ(r) depends on
modulus of r.
  r  ,  r 
  r , 0, 0    (0, r , 0)   (0, 0, r )
 x  0
 xy  0
• EQUILIBRIUM:Properties do not evolve with time.
f


0
0
0
t
t
t
• In a spherical potential
r 2  x2  y2  z 2
d (r 2 )  d ( y 2 )
rdr  ydy
dr y


dy r
 (r )
r  (r )
x
x
y
y r
y d (r )
x
r dr

 xy r
r
So <xy>=0 since the average value of xy will be
zero.
<vxvy>=0
Spherical Isotropic f(E) Equilibrium Systems
• ISOTROPIC:The distribution function f(E) only
depends on the modulus of the velocity rather than
the direction.
f  E  , E  v / 2   (r )
2
   
2
2
x
 vx v y  0
Note:the tangential
direction has  and 
components
2
y
2
z
1 2
    tangential
2
2
r
Anisotropic DF f(E,L) in spherical
potential.
• Energy E is conserved as:

0
t
• Angular Momentum Vector L is conserved as:

0

• DF depends on Velocity Direction through L=r X v
• Hence anisotropic
1 2
2
 r   tangential
2
e.g., f(E,L) is an incompressible fluid
• The total energy of an orbit is given by:
1 2

E  v   (r , t )
2
df ( E , L) f ( E , L) dE f ( E , L) dL


 00
dt
E
dt
L
dt
0 for static potential,
0 for spherical potential
So f(E,L) constant along orbit or flow
• spherical Jeans eq. of a tracer density rho(r)
• d (  r2 ) / dr  (2 r2   t2 ) / r  d / dr
• Proof :
vr  2 E  2  L2 / r 2
2

vr dvr / dr   d / dr  L2 / r 3

dv 3  d (vt2 )dvr  d (L2 )dE / vr r 2
fdv3  f ( E , L)d (L2 )dE / vr r 2
 d / dr  L / r r
2
3
2
fdv3  f ( E , L)d (L2 )dE  (dvr / dr )
Jeans eq. Proof cont.
   fdv3   f ( E , L)d L2 dE /( vr r 2 )
 t2   f ( E , L)d L2 dE * L2 /( r 4 vr )
 r2 r 2 
 d L   f ( E , L)dEv
2
L 0
r
vr2  0
 
  f ( E , L)d L dE (d / rdr ) / v
  f ( E , L)d L dE L / r  /v
d (  r2 r 2 ) / rdr   f ( E , L)d L2 dE *(dvr / dr ) / r
2
2
2
4
r
rd (  r2 ) / dr  2  r2   rd / dr   t2
r
• SELF GRAVITATING:The masses are kept
together by their mutual gravity.
• In non-self gravitating systems the density
that creates the potential is not equal to the
density of stars. e.g a black hole with stars
orbiting about it is NOT self gravitating.
th
13
• Phase Space
Lec
Velocity dispersions of a
subcomponent
in spherical potential
• For a spherically symmetric system we have




d
n
2
2
2
2
n vr 
2vr  v  v
dr
r
• a non-rotating galaxy has
  n ddr
– and the velocity ellipsoids are spheroids with
their symmetry axes pointing towards the
galactic centre
v 2  v 2
2
2



r

1

v
/
v

r
• Define anisotropy


Spherical mass profile from velocity
dispersions.
• Get M(r) or Vcir from:
 
1 d
2  vr
d
GM r 
2
n vr 


n dr
r
dr
r2
vcirc
2
2
2


d GM r 
d ln vr
2 d ln n
r

  vr 

 2 
dr
r
d ln r
 d ln r

• RHS: observations of dispersion and 
as a function of radius r for a stellar
population.
• Isotropic Spherical system, β=0
d (  2 )
d
   (r )
dr
dr
Note: 2=P
• This is the isotropic JEANS EQUATION, relating
the pressure gradient to the gravitational force.


d
P      dr    | g |dr
dr
r
r
2
Above Solution to Isotropic Jeans Eq:
negative sign has gone since we reversed
the limits.
Hydrostatic equilibrium  Isotropic
spherical Jeans equation
dP d (  2 )
d

   (r )
dr
dr
dr
• Conservation of momentum gives:
0  P   g
1
 g   P

Tutorial
g
M
2
vesc
(r)
(E)
(r)
Tutorial Question 3
• Question: Show dispersion sigma is constant in potential Phi=V02ln(r).
What might be the reason that this model is called Singular Isothermal
Sphere?
• 


1 d

 2   r
 dr
vc2
 r 
r
r  ro
r
2
c
r
2
2
v
v
v
1
  2     dr    c 2 c dr
r
4G r r


At r  ro , P   2  0
2
2
v
v
c
  2  c
4G 2r 2
vc2 vc2

4Gr 2 2

• 
2
v
 2   c
2
2
v
 2  c
2
vc

2
• Since the circular velocity is independent of radius
then so is the velocity dispersionIsothermal.
Flattened Disks
• Here the potential is of the form (R,z).
• No longer spherically symmetric.
• Now it is Axisymmetric
1       2 
 ( R, z )   ( R, z ) 
R
 2 
R
4G  R  R  z 

gr  
R

gz  
z
Question 4: Oblate Log. potential
• oblate galaxy with Vcirc ~ V0 =100km/s
 ( R, z )  12 v0 2 ln  R 2  2 z 2   C0
• Draw contours of the corresponding Selfgravitating Density to show it is unphysical.
• Let Lz=1kpc*V0 , E=0.55*V02 +C0, Plot
effective potential contours in RZ plane to
show it is an epicycle orbit.
• Taylor expand the potential near (R,z)=(1,0)
to find epicycle frequencies and the
approximate z-height and peri-apo range.
Orbits in Axisymmetric
Potentials (disk galaxies)
z
y
x

R2=x2+y2
R
• cylindrical (R,,z) symmetry z-axis
• stars in equatorial plane: same motions as in
spherically symmetric potential
– non-closed rosette orbits
• stars moving out of plane
– can be reduced to 2-D problem in (R,z)
– conservation of z-angular momentum, L
• Angular momentum about the z-axis is conserved,
toque(rF=0) if  no dependence on .
d 2 2
2

LZ  R   ( R  )  0
dt
2
• Energy is also conserved (no time-dependence)
1 2
R  R22  z 2   (R, z)  const
2
Specific energy density in 3D


• Eliminating in the energy equation using
conservation of angular momentum gives:
2
1 2
J
( R  z 2 )   ( R, z )  z 2  E
2
2R
eff
Total Angular momentum almost
conserved
• These orbits can be thought of as being
planar with more or less fixed eccentricity.
• The approximate orbital planes have a fixed
inclination to the z axis but they process
about this axis.
• star picks up angular momentum as it goes
towards the plane and returns it as it leaves.
Orbital energy
• Energy of orbit is (per unit mass)
E 
1
2

1
2

1
2


 2  R
R
R
R
2
 z 2
2
 z 2



2

 z 2  
2
Lz


2
2R
  eff
• effective potential is the gravitational
potential energy plus the specific kinetic
energy associated with motion in  direction
• orbit bound within E   eff
• The angular momentum barrier for an orbit of
energy E is given by eff ( R, z )  E
• The effective potential cannot be greater than the
energy of the orbit.
R 2  z 2  2 E  2 ( R, z )
eff
0
• The equations of motion in the 2D meridional
(RZ)plane then become:
.


   eff
R
R
eff
z  
z
R 2  J z
• Thus, the 3D motion of a star in an axisymmetric
potential (R,z) can be reduced to the motion of a
star in a plane (Rz).
• This (non uniformly) rotating plane with cartesian
coordinates (R,z) is often called the
MERIDIONAL PLANE.
• eff(R,z) is called the EFFECTIVE POTENTIAL.
• The orbits are bound between two radii (where the
effective potential equals the total energy) and
oscillates in the z direction.
• The minimum in eff occurs at the radius at which
a circular orbit has angular momentum Lz.
• The value of eff at the minimum is the energy of
this circular orbit.
eff
J z2
2R 2
R
E

Rcir
• The effective potential is the sum of the
gravitational potential energy of the orbiting star
and the kinetic energy associated with its motion
in the  direction (rotation).
• Any difference between eff and E is simply
kinetic energy of the motion in the (R,z) plane.
• Since the kinetic energy is non negative, the orbit
is restricted to the area of the meridional plane
satisfying E - eff
. (R,z)>= 0
• The curve bounding this area is called the ZERO
VELOCITY CURVE since the orbit can only
reach this curve if its velocity is instantaneously
zero.
Nearly circular orbits: epicycles
• In disk galaxies, many stars (disk stars) are
on nearly-circular orbits
 eff
R    eff ;
• EoM:
z  
R
• x=R-Rg
z
 eff  eff

0
R
z
at R  Rg , z  0
– expand in Taylor series about (x,z)=(0,0)
2
2






 eff
2
eff
1
1

 eff  2 
x  2 
2
2

R

z

( R ,0)

– then
  2 x2 / 2   2 z 2 / 2
g


z2 
 ( Rg , 0 )
• When the star is close to z=0 the effective
potential can be expanded to give
eff
1  2 2
eff ( R, z )  eff ( R,0) 
z
z
2
z
2 z
Zero, changes sign above/below z=0
equatorial plane.
1 2 2
eff ( R, z )  eff ( R,0)   z  .......
2
z   2 z
So, the orbit is oscillating
in the z direction.
2
epicyclic approximation
• ignore all higher / cross terms:
• EoM: harmonic oscillators
 – epiclyclic frequency :
R   2 R ,
 – vertical frequency :
 eff   ( R, z ) 
– with
2
2


3
L


z
2  

;
2 
4
Rg
  R ( Rg ,0)
and
Lz 2
2R2
z   2 z
epicycles cont.
• using the circular frequency , given by
1   
Lz
2
 ( R)  
 4

R  R  ( R ,0 ) R
2
2



3
L


– so that  2     z   R 2   3 2
R  R  R 4
R
 


 R
 2   4 2 
 R
R
disk galaxy:  ~ constant near centre
g
– so  ~ 2
  ~ declines with R,
Vrot
» slower than Keplerian R-3/2
» lower limit is  ~ 
in general  <   2
R
Example:Oort’s constants near Sun
  
A   R
 ;
 R  R0
1
2
 1 

B   2 R
 
 R
 R0
– where R0 is the galacto-centric distance
• then 2 = -4A(A-B) + 4(A-B)2 = -4B(A-B) =
-4B0
• Obs. A = 14.5 km/s /kpc and B=-12 km/s
/kpc

B
0
0
2
A B
 1.3  0.2
the sun makes 1.3 oscillations in the radial
direction per azimuthal (2) orbit
– epicyclic approximation not valid for z-motions
when |z|>300 pc
General Jeans Equations
• CBE of the phase space density f is eq. of 7
variables and hence generally difficult to
3
  f   f 

f
solve
  vi

0
 t i 1   xi  xi  vi 
• Gain insights by taking moments of the CBE
f 3
f
  f 3
3
  t Ud v   vi  xi Ud v   xi   vi Ud v  0
• where integrate over all possible velocities
– U=1, vj, vjvk
1st Jeans (continuity) equation
• define spatial density of stars n(x)
n   fd 3 v
• and the mean stellar velocity v(x)
1
v i   fvi d 3 v
n
• then our zeroth moment equation becomes
n  nv i 
0

x i
t
3rd Jeans Equation
 

2
 vj

n

 vj

ij
n
 nvi
 n

t
xi
xi
xi

v
1
 v   v    P
t

similar to the Euler equation for a fluid flow:
– last term of RHS represents pressure force
JEANS EQUATION for
oblate rotator
: a steady-state axisymmetrical system in which
ASSUME ij2 is isotropic and the only streaming
motion is azimuthal rotation:
1  (  ) 


 z
z
2
1  (  ) v rot 



 R
R
R
2
2
• The velocity dispersions in this case are given by:
2
 2 ( R, z )  vr2  vz2  v2  vrot
  2
since
   (v  vrot )  v  2v vrot  v
2
2
2
2
rot
but v  v rot since apart from v rot it' s isotropic
2
  2  v2  vrot
• If we know the forms of (R,z) and (R,z) then at
any radius R we may integrate the Jeans equation
in the z direction to obtain 2.
Obtaining 2


 ( R, z )    dz
 z z
2
1
Inserting this into the jeans equation in the R
direction gives:

v
2
rot
 R 

R


dz

R  R z z
th
20
• orbits
Lec
Applications of the Jeans
Equations
• I. The mass density in the solar
neighbourhood
• Using velocity and density distribution
perpendicular to the Galactic disc
– cylindrical coordinates.
– Ignore R dependence
E.g.: Total Mass of spherical Milky
WAY
• Motions of globular clusters and satellite
galaxies around 100kpc of MW
– Need n(r), vr2,  to find M(r), including dark
halo
• Several attempts all suffer from problem of
small numbers N ~ 15
• For the isotropic case, Little and Tremaine
TOTAL mass of 2.4 (+1.3, -0,7) 1011 Msol
• 3 times the disc need DM
Power-law model of Milky Way
  0,  v 2  vr 2
• Isotropic orbits:
  1, v 2  0
• Radial orbits
• If we assume a power law for the density
distribution


n  r , M (r )  r
– E.g. Flat rotation a=1, Self-grav gamma=2,
Radial anis.  0.
– E.g., Point mass a=0, Tracer gamma=3.5, Isotro
2
2
M

4
.
5
(
v

v
r
 )r / G
0
Mass of the Milky Way: point-mass
potential model
We find
GM
d
r
 vcirc 2  vr 2 p
r
dr
 vr 2  1/ r
 d ln n d ln vr 2

p  

 2     1  2
 d ln r d ln r

For  =3.5, and isotropic tracer  =0, we have
p  4.5
M  pvr 2 r / G  pv2 r / G
Vertical Jeans equation
• Small z/R in the solar neighbourhood, R~8.5
kpc, |z|< 1kpc, R-dependence neglected.
• Hence, reduces to vertical hydrostatic eq.:




2
nv z   n
z
z
mass density in solar
neighbourhood
• Drop R, theta in Poisson’s equation in
cylindrical coordinates:
1     1  2  2
 

 4G
R

R R  R  R  2 z 2
2
 2
 4G
2
 z
local mass density  = 0
Finally
-
 
 1 
2 
n
v
z  / 4G  

z  n z

• all quantities on the LHS are, in principle,
determinable from observations. RHS
Known as the Oort limit.
• Uncertain due to double differentiation!
local mass density
• Don’t need to calculate for all stars
– just a well defined population (ie G stars, BDs
etc)
– test particles (don’t need all the mass to test
potential)
• Procedure
– determine the number density n, and the mean
square vertical velocity, vz2, the variance of the
square of the velocity dispersion in the solar
neighbourhood.
local mass density
• > 1000 stars required
• Oort : 0 = 0.15 Msol pc-3
• K dwarf stars (Kuijken and Gilmore 1989)
– MNRAS 239, 651
• Dynamical mass density of 0 = 0.11 Msol
pc-3
• also done with F stars (Knude 1994)
• Observed mass density of stars plus
interstellar gas within a 20 pc radius is 0 =
0.10 Msol pc-3
• can get better estimate of surface density
• out to 700 pc
S ~ 90 Msol pc-2
• from rotation curve Srot ~ 200 Msol pc-2
• Question 5:
In potential  ( R, z )  0.5v02 ln( R 2  2 z 2 )
 v02 (1  ( R 2  z 2 ) / 1kpc2 ) 1/ 2 ,
due to dark halo (1st term) and stars (2nd term),
assume V0  100km / s. calculate total mass of stars
and star density  s ( R, z ).
What is the dark halo density on equator (R, z)  (1kpc,0)?
0
Calculate stellar  s (1kpc,0)   
2

 s  (1kpc, z )
z
2
Show isotropic rotator have unphysical vrot
(1kpc,0)
dz
Helpful Math/Approximations
(To be shown at AS4021 exam)
• Convenient Units
• Gravitational Constant
• Laplacian operator in
various coordinates
• Phase Space Density
f(x,v) relation with the
mass in a small position
cube and velocity cube
1km/s 
1kpc
1pc

1Myr 1Gyr
G  4  10  3 pc (km/s) 2 M - 1
sun
G  4  10  6 kpc (km/s) 2 M - 1
sun
     2   2   2 (rectangul ar)
z
y
x
 R - 1 ( R )   2  R - 2 2 (cylindric al)

z
R
R
2
 (r 2 )  (sin  )

 
r  
(spherical )
 r
2
2
2
2
r sin 
r
r sin 
dM  f ( x, v)dx 3dv3
th
21
• orbits
Lec: MOND