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Gravitational Dynamics
Gravitational Dynamics can be applied to:
•
•
•
•
•
•
Two body systems:binary stars
Planetary Systems
Stellar Clusters:open & globular
Galactic Structure:nuclei/bulge/disk/halo
Clusters of Galaxies
The universe:large scale structure
Syllabus
• Phase Space Fluid f(x,v)
– Eqn of motion
– Poisson’s equation
• Stellar Orbits
– Integrals of motion (E,J)
– Jeans Theorem
• Spherical Equilibrium
– Virial Theorem
– Jeans Equation
• Interacting Systems
– TidesSatellitesStreams
– Relaxationcollisions
How to model motions of 1010stars
in a galaxy?
• Direct N-body approach (as in simulations)
– At time t particles have (mi,xi,yi,zi,vxi,vyi,vzi),
i=1,2,...,N (feasible for N<<106).
• Statistical or fluid approach (N very large)
– At time t particles have a spatial density
distribution n(x,y,z)*m, e.g., uniform,
– at each point have a velocity distribution
G(vx,vy,vz), e.g., a 3D Gaussian.
N-body Potential and Force
• In N-body system with mass m1…mN,
the gravitational acceleration g(r) and
potential φ(r) at position r is given by:
r12
N



G  m  mi  rˆ12
F  mg (r )     2  m r
i 1
r  Ri
G  m  mi


m

(
r
)


mi

 
r  Ri
i 1
N
r
Ri
Eq. of Motion in N-body
• Newton’s law: a point mass m at position r
moving with a velocity dr/dt with Potential
energy Φ(r) =mφ(r) experiences a Force
F=mg , accelerates with following Eq. of
Motion:



d  dr (t )  F   r (r )
 


dt  dt  m
m
Orbits defined by EoM & Gravity
• Solve for a complete prescription of history
of a particle r(t)
• E.g., if G=0  F=0, Φ(r)=cst,  dxi/dt =
vxi=ci  xi(t) =ci t +x0, likewise for yi,zi(t)
– E.g., relativistic neutrinos in universe go
straight lines
• Repeat for all N particles.
•  N-body system fully described
Example: Force field of two-body
system in Cartesian coordinates
2

G  mi

 (r )     , where Ri  (0,0,i )  a, mi  m
i 1 r  Ri
Sketch the configurat ion, sketch equal potential contours
 ( x, y , z )  ?
 
  

g (r )  ( g x , g y , g z )   (r )  ( , , )
x y z
 
g (r )  ( g x2  g y2  g z2 )  ?
sketch field lines. at what positions is force  0?
Example: 4-body problem
• Four point masses Gm=1 at rest
(x,y,z)=(0,1,0),(0,-1,0),(-1,0,0),(1,0,0).
What is the initial total energy?
• Integrate EoM by brutal force with time
step=1 to find the positions/velocities at
time t=1. i.e., use straight-orbit V=V0+gt,
R=R0+V0t+gt2/2. What is the new total
energy?
Star clusters differ from air:
• Size doesn’t matter:
– size of stars<<distance between them
– stars collide far less frequently than
molecules in air.
• Inhomogeneous
• In a Gravitational Potential φ(r)
• Spectacularly rich in structure because φ(r)
is non-linear function of r
Why Potential φ(r) ?
• More convenient to work with force,
potential per unit mass. e.g. KE½v2
• Potential φ(r) is scaler, function of r only,
– Easier to work with than force (vector, 3
components)
– Simply relates to orbital energy E= φ(r) +½v2
nd
2
Lec
Example: energy per unit mass
• The orbital energy of a star is given by:
1 2

E  v   (r , t )
2


dE  dv dr

v
  
dt
dt dt
t
0

dv
 
since
 dt
0 for static potential.
and dr  v
dt
So orbital Energy is Conserved in a
static potential.
Example: Energy is conserved
• The orbital energy of a star is given by:
1 2

E  v   (r , t )
2


dE  dv dr

v
  
dt
dt dt
t

dv
 
0 since
 dt
0 for static potential.
and dr  v
dt
So orbital Energy is Conserved in a
static potential.
rd
3
• Animation of GC
formation
Lec
A fluid element: Potential & Gravity
• For large N or a continuous fluid, the gravity dg and
potential dφ due to a small mass element dM is calculated
by replacing mi with dM:
r12
r
R
dM
d3R

G  dM  rˆ12
dg     2
r  Ri
G  dM
d    
r R
Potential in a galaxy
• Replace a summation over all N-body particles with the
integration:
N
 dM  m
i
i 1
RRi
• Remember dM=ρ(R)d3R for average density ρ(R) in small
volume d3R
• So the equation for the gravitational force becomes:
3


G ( R)dR
F / m  g (r )   r , with  (r )     
r R
Poisson’s Equation
• Relates potential with density
   4G (r )
2
• Proof hints:
 Gm
 
    4Gm (r  R)
r R
 
4G (r )   4G (r  R)  ( R)dR 3
2
Poisson’s Equation
• Poissons equation relates the potential to the
density of matter generating the potential.
• It is given by:
 
      g  4G (r )
Gauss’s Theorem
• Gauss’s theorem is obtained by integrating
poisson’s equation:

2
   (r )dV   4G (r )dV  4GM
V
V


   (r )dV    (r ).ds
2
V
S

   (r ).ds  4GM
S
• i.e. the integral ,over any closed surface, of the
normal component of the gradient of the potential
is equal to 4G times the Mass enclosed within
that surface.
Laplacian in various coordinates
Cartesians :
2
2
2



2  2  2  2
x
y
z
Cylindrica l :
2
2
1


1




2 
R
 2

 2
2
R R  R  R 
z
Spherical :
2
1


1


1





2
2  2
r

 2
 sin 
 2 2
r r  r  r sin   
  r sin   2
th
4
• Potential,density,orbits
Lec
From Gravitational Force to Potential
r
 (r )   g dr

d
g    
dr
From Potential to Density
Use Poisson’s Equation
The integrated form of
Poisson’s equation is
given by:
1

 2
4G

Gd r 

 (r )     
r  r
3
More on Spherical Systems
• Newton proved 2 results which enable us to
calculate the potential of any spherical system
very easily.
• NEWTONS 1st THEOREM:A body that is inside a
spherical shell of matter experiences no net
gravitational force from that shell
• NEWTONS 2nd THEOREM:The gravitational
force on a body that lies outside a closed spherical
shell of matter is the same as it would be if all the
matter were concentrated at its centre.
From Spherical Density to Mass
M(R  dr)  M(R)  dM
4 3
2
dM   (r)d  r   4r  (r) dr
3

dM
dM
 (r ) 

2
4
4r dr

3
d  r 
3

4 3
M ( R)   d  r 
3

M(r+dr)
M(r)
Poisson’s eq. in Spherical systems
• Poisson’s eq. in a spherical potential with no θ or Φ
dependence is:
1   2  
 2
r
  4G (r )
r r  r 
2
Proof of Poissons Equation
• Consider a spherical distribution of mass of
density ρ(r).
g
g
GM (r )
r2

   g (r )dr
r
since   0 at  and is  0 at r
r

GM (r )
 
dr
2
r
r

Mass Enclosed   4r 2  (r )dr
r
• Take d/dr and multiply r2 

d
2
2
r
  gr  GM (r )  G  4r  (r )dr
dr
2

• Take d/dr and divide r2
1   2   1 
1 
2
GM   4G (r )
r g  2
r
 2
2
r r  r  r r
r r

2
    .g  4G


Sun escapes if Galactic potential
well is made shallower
Solar system accelerates weakly
in MW
• 200km/s circulation
g(R0 =8kpc)~0.8a0,
a0=1.2 10-8 cm2 s-1
Merely gn ~0.5 a0 from
all stars/gas
• Obs. g(R=20 R0)
~20 gn
~0.02 a0
• g-gn ~ (0-1)a0
• “GM” ~ R if weak!
Motivates
– M(R) dark particles
– G(R) (MOND)
Circular Velocity
• CIRCULAR VELOCITY= the speed of a test
particle in a circular orbit at radius r.
2
cir
v
GM (r )
g 
  
2
r
r
2
vcir
r
 M (r ) 
G
For a point mass:
GM
vc (r ) 
r
For a homogeneous sphere
4G
4 3
vc (r ) 
r since M(r)  r 
3
3
Escape Velocity
• ESCAPE VELOCITY= velocity required in order
for an object to escape from a gravitational
potential well and arrive at  with zero KE.
1 2
1 2
 (r )   ()  vesc   vesc
2
2
 vesc (r )   2 (r )
-ve
• It is the velocity for which the kinetic energy
balances potential.
Tutorial Question 1:
Singular Isothermal Sphere
GM 0
• Has Potential Beyond ro:  (r )   r
r
2
 (r )  v0 ln  o
• And Inside r<r0
ro
• Prove that the potential AND gravity is continuous
at r=ro if
0  GM 0 / r0  v02
• Prove density drops sharply
to 0 beyond r0, and
2
V0
inside r0
 (r ) 
4Gr 2
• Integrate density to prove total mass=M0
• What is circular and escape velocities at r=r0?
• Draw Log-log diagrams of M(r), Vesc(r), Vcir(r),
Phi(r), rho(r), g(r) for V0=200km/s, r0=100kpc.
Tutorial Question 2: Isochrone Potential
• Prove G is approximately 4 x 10-3 (km/s)2pc/Msun.
• Given an ISOCHRONE POTENTIAL
GM
 (r )  
b  (b  r )
2
2
• For M=105 Msun, b=1pc, show the central escape velocity
= (GM/b)1/2 ~ 20km/s.
• Argue why M must be the total mass. What fraction of the
total mass is inside radius r=b=1pc? Calculate the local
Vcir(b) and Vesc(b) and acceleration g(b). What is your
unit of g? Draw log-log diagram of Vcir(r).
• What is the central density in Msun pc-3? Compare with
average density inside r=1pc. (Answer in BT, p38)
Example:Single Isothermal Sphere Model
• For a SINGLE ISOTHERMAL SPHERE (SIS)
the line of sight velocity dispersion is constant.
This also results in the circular velocity being
constant (proof later).
• The potential and density are given by:
 (r )  V ln r 
2
c
2
c
V
 (r ) 
4Gr 2
Proof: Density
CircularVe locity_ vc(r)  const  vo
2
vc r
M (r ) 
r
G
dv 
GM vc 2
1 

  r ( r )  2 
   r (  2 )
dt
r
r

Log()
vc
dM
 (r ) 

-2
2
r
 4 3  4Gr
d  r 
n=-2
3

Log(r)
Proof: Potential
v
r
2
 (r )    rdr   dr  vc ln  
r



r
r
2
c
We redefine the zero of potential
  v ln( r )  constant
2
c
If the SIS extends to a radius ro then the mass and
density distribution look like this:

M
ro
r
ro
r
GM (ro )
 
r
• Beyond ro:
• We choose the constant so that the potential is
continuous at r=ro.
r
 (r )  v ln  o
ro
2
c
GM o
 (r )  v ln r  v ln ro 
ro
2
c
2
c
r
 r-1

logarithmic
So:
GM o
Inside_Sph ere :  (r )  vc ln r  vc ln ro 
ro
2
GM o
Outside_Sp here :  (r )  
r
2
Plummer Model
• PLUMMER MODEL=the special case of the
gravitational potential of a galaxy. This is a
spherically symmetric potential of the form:
 
GM
r a
2
2
• Corresponding to a density:
3M  r 2 
1  2 

3 
4a  a 

5
2
which can be proved using poisson’s equation.
• The potential of the plummer model looks like
this:

r

GM o
a
g    GM o r (r 2  a 2 )

3
2
 0 for r  0
 r  0 is minimum of potential

GM o 1 2
2GM o
 vesc  vesc 
a
2
a
• Since, the potential is spherically symmetric g is
also given by: g   GM
r2


GM
 2  GM o r r 2  a 2
r

 M  M or r  a
3
2


3
2

3

2 2
dM (r )
• The density can then be obtained from:   dV
• dM is found from the equation for M above and
dV=4r2dr.
5

2
2


3
M
r
• This gives
(as before from



1

4a 3  a 2 
Poisson’s)
Isochrone Potential
• We might expect that a spherical galaxy has
roughly constant  near its centre and it falls to 0
at sufficiently large radii.
• i.e.   r 2  constant
(for small r)
  r -1
(for large r)
• A potential of this form is the ISOCHRONE
POTENTIAL.
 (r )  
GM
b  (b  r )
2
2
th
5
• orbits
Lec
Stellar Orbits
• Once we have solved for the gravitational
potential (Poisson’s eq.) of a system we
want to know: How do stars move in
gravitational potentials?
• Neglect stellar encounters
• use smoothed potential due to system or
galaxy as a whole
Motions in spherical potential
Equation of motion
dx
v
dt
dv

 g  
dt
If no gravity
x(t )  v 0t  x 0
v (t )  v 0
If spherical

gr  
r
Conserved if spherical
1
E  v 2   (r )
2
L  J  x  v  rvt  nˆ
Proof: Angular Momentum is Conserved
  
• L  r v

 dL d r  v  dr   dv    dv

 v  r 
 v v  r 
dt
dt
dt
dt
dt
Since

0

then the force is in the r direction.
both cross products on the RHS = 0.
So Angular Momentum L is Conserved in Spherical
Isotropic Self Gravitating Equilibrium Systems.
Alternatively: =r×F & F only has components in the r

direction=0 so
L
t
0
In static spherical potentials: star
moves in a plane (r,)
• central force field
• angular momentum
g  grˆ
r  r  L
dL d
 r  r   r  r  r  r  r  g  0
dt
dt
• equations of motion are
– radial acceleration:
– tangential acceleration:
r  r 2  g (r )
2r  r2  0
r 2  constant  L
Orbits in Spherical Potentials
• The motion of a star in a centrally directed field of
force is greatly simplified by the familiar law of
conservation (WHY?) of angular momentum.
  dr
Lr
 const
dt
area swept
2 d
r
2
dt
unit time
Keplers 3rd law
pericentre
apocentre
Energy Conservation
(WHY?)

1  dr  1  d 
2
E   (r )      r

2  dt  2  dt 
 2
L
1  dr 
  (r )  2   
2r
2  dt 
2
eff
dr
  2 E  2eff (r )
dt
2
Radial Oscillation in an Effective
potential
• Argue: The total velocity of the star is slowest at
apocentre due to the conservation of energy
• Argue: The azimuthal velocity is slowest at
apocentre due to conservation of angular
momentum.
th
6
• Phase Space
Lec
dr
 0 at the PERICENTRE and APOCENTRE
•
dt
L2
• There are two roots for 2 E  2 (r )  2
r
• One of them is the pericentre and the other is the
apocentre.
• The RADIAL PERIOD Tr is the time required for
the star to travel from apocentre to pericentre and
back.
• To determine Tr we use:
dr
L2
  2E     2
dt
r
• The two possible signs arise because the star
moves alternately in and out.
ra
dr
 Tr  2 
2
L
rp
2 E  2  2
r
• In travelling from apocentre to pericentre and
back, the azimuthal angle  increases by an
amount:
r
r d
r L
2
d
dt
r
  2 
dr  2 
dr  2 
dr
a
rp
a
dr
rp
a
dr
dt
rp
dr
dt
2
Tr
• The AZIMUTHAL PERIOD is T 

2
• In general  will not be a rational number.
Hence the orbit will not be closed.
• A typical orbit resembles a rosette and eventually
passes through every point in the annulus between
the circle of radius rp and ra.
2
• Orbits will only be closed if
is an integer.

Examples: homogeneous sphere
• potential of the form   12  2 r 2  constant
• using x=r cos and y = r sin
• equations of motion are then:
x   2 x;
y   2 y
x  A cos(t   x );
– spherical harmonic oscillator
y  B cos(t   y )
• Periods in x and y are the same
so every
orbit is closed ellipses centred on the centre
of attraction.
homogeneous sphere cont.
A
B
A
t=0
B
• orbit is ellipse
• define t=0 with x=A, y=0
  x  0,  y   / 2
t  Pr / 2  x  0 if
P  2Pr
 
Pr
Radial orbit in homogeneous sphere
d 2r
GM ( r )
4G


r
2
2
dt
r
3
– equation for a harmonic oscillator
angular frequency 2/P
P
3
t dyn 
4

16G
Altenative equations in spherical
potential
• Let
u
1
r
2E
 dr  1 2 (r )

const

 2  2  2
L2
L
 r dθ  r
2 E  du 
2 (1 / u )
2
 2 
 u 
L
L2
 d 
 d 
Apply 

2
du


2
0
d  du 
GM (1 / u )

 u 
d  d 
L2
Kepler potential
g (r )  GM / r 2
• Equation of motion becomes:
d 2u
GM

u

d 2
L2
– solution: u linear function of cos(theta):
a 1  e 2 
r   
1  e cos 
with P  Pr and thus   2
• Galaxies are more centrally condensed than
a uniform sphere, and more extended than a
point mass, so     2
Pr  P
Tutorial Question 3: Show in
Isochrone potential
 (r )   GM
b  (b
2
r )
2
1
2


L
Pr 
, and   1 
3

 2 E  2
L2  4GMb
– radial period depends on E, not L
2GM
• Argue
    2

,
but  2
for
– this occurs for large r, almost Kepler


1
2


L2  4GMb
st
1
Tutorial
g
M
2
vesc
(r)
(E)
(r)
th
7
Lec
Tidal Stripping
• TIDAL RADIUS:Radius within which a particle
is bound to the satellite rather than the host
system.
• Consider a satellite of mass Ms inside radius R is
moving in a spherical potential (r) made from a
point mass M.
R
r
M(r)
• The condition for a particle to be bound to the
satellite rather than the host system is:
GM
GM GM s
 2  2
2
( r  R)
r
R
Differential (tidal) force on the
particle due to the host galaxy
Force on particle
due to satellite
GM s
GM
U 2
2
r
R
2
2R
 R
If R  r then U  1 -   1 
 .........
r
 r
GM
k 2
r
 R  GM s
   2 ,k  2
R
r
GM GM s
k 3 
r
R3
• Generally, fudge factor k=14, bigger for radial
orbits, bigger for point-like mass.
• Therefore Tidal Radius is (ambiguously defined as):
 ms 

RT (t )  r (t )
 kM (r ) 
• The tidal radius is smallest at pericentre where r is
smallest. Often tidal radius is only defined when
r(t)=pericentre.
• As a satellite losses mass, its tidal radius shrinks.
1
3
The meaning of tidal radius
• The inequality can be written in terms of the
mean densities.
ms ( r ) M ( r )

4 3 4 3
R
r
3
3
• The less dense part of the satellite is torn
out of the system, into tidal tails.
Size and Density of a BH
• A black hole has a finite (schwarzschild)
radius Rbh=2 G Mbh/c2 ~ 2au (Mbh/108Msun)
– verify this! What is the mass of 1cm BH?
• A BH has a density (3/4Pi) Mbh/Rbh3, hence
smallest holes are densest.
– Compare density of 108Msun BH with Sun (or
water) and a giant star (10Rsun).
Growth of a BH by capturing objects
in its Loss Cone
• A small BH on orbit with pericentre rp<Rbh is lost
(as a whole) in the bigger BH.
– The final process is at relativistic speed. Newtonian
theory is not adequate
• (Nearly radial) orbits with angular momentum
J<Jlc =2*c*Rbh =4GMbh/c2 enters `loss cone` (lc)
• When two BHs merger, the new BH has a mass
somewhat less than the sum, due to gravitational
radiation.
Tidal disruption near giant BH
• A giant star has low density than the giant
BH, is tidally disrupted first.
• The disruption happens at radius rdis > Rbh ,
Mbh/rdis3 ~ M*/R*3
• Giant star is shreded.
• Part of the tidal tail feeds into the BH, part
goes out.
Adiabatic Compression due to
growing BH
•
•
•
•
A star circulating a BH at radius r has
a velocity v=(GMbh/r)1/2,
an angular momentum J = r v =(GMbh r)1/2,
As BH grows, Potential and Orbital Energy
E changes (t)
• But J conserved (no torque!), still circular!
• So Ji = (GMi ri)1/2 =Jf =(GMf rf )1/2
• Shrink rf/ri = Mi/Mf < 1, orbit compressed!
Adiabatic Invariance
• Suppose we have a sequence of potentials p(|r|)
that depends continuously on the parameter P(t).
• P(t) varies slowly with time.
• For each fixed P we would assume that the orbits
supported by p(r) are regular and thus phase
space is filled by arrays of nested tori on which
phase points of individual stars move.
 P ( t )
• Suppose
0
t
• The orbit energy of a test particle will change.
• Suppose  Pi ( r )   Pf ( r )
Initial
Final
• The angular momentum J is still conserved
because rF=0.
    
J  ri  vi  rf  v f
• In general, two stellar phase points that started out
on the same torus will move onto two different
tori.
• However, if potential is changed very slowly
compared to all characteristic times associated
with the motion on each torus, all phase points that
are initially on a given torus will be equally
affected by the variation in Potential
• Any two stars that are on a common orbit will still
be on a common orbit after the variation in Pot is
complete. This is ADIABATIC INVARIANCE.
th
8
• Phase Space
Lec
Stellar interactions
• When are interactions important?
• Consider a system of N stars of mass m
• evaluate deflection of star as it crosses
system
X=vt
v
Fperp

r
b
• consider en encounter with star of mass m at
3
2
2
a distance b:Gm
Gm b
Gm   vt  
g 
r
2
cos  
x
2
b
2

3

2
b
2
1    
  b  
Stellar interactions cont.
• the change in the velocity vperp is then
– using s = vt / b
Gm
2Gm


v   g dt 
1

s
ds

bv 
bv



3
2  2

v  g  t 
Gm 2b

2
b
v
• Or using impulse approximation:
– where gperp is the force at closest approach and
– the duration of the interaction can be estimated
as :
t = 2 b / v
Number of encounters with impact
parameter b - b + b
• let system diameter be: 2R
• Star surface density ~ N/R2/Pi
• the number encounteringN
N   b  b   b 2 
2
R 2
2bbN 2N

 2 bb
2
R
R
v  0
b
b+b
change in kinetic energy
• but suming over squares (vperp2) is > 0
 2Gm  2 N
• hence
bb
v   

 bv  R
2
2

2
v 
2
2
 Gm  db
N
8

 Rv  b
 
b min
b max
• now consider encounters over all b
– then
bmin
GNm
Gm R
2
 2  , where we have used v 
R
N
v
– but @ b=0, 1/b is infinte!
– need to replace lower limit with some b
Relaxation time
• hence v2 changes by v2 each time it crosses
2 is:
the system where
v
2
R
 Gm 
v  8 N 
N
 ln , where  
bmin
 Rv 
2
Orbit deflected when v2 ~ v2
– after nrelax times across the system
nrelax
v2
N
 2
v
8 ln 
and thus the relaxation time is:
t relax  nrelaxtcross 
N
N
tcross 
tcross
8 ln 
8 ln N
Relaxation time cont.
• collisionless approx. only for t < trelax !
• mass segregation occurs on relaxation
timescale KE  12 mv2  constant  v 2  m 1
– also referred to as equipartition
– where kinetic energy is mass independent
– Hence the massive stars, with lower specific
energy sink to the centre of the gravitational
potential.
• globular cluster, N=105, R=10 pc
– tcross ~ 2 R / v ~ 105 years
– trelax ~ 108 years << age of cluster:
relaxed
• galaxy, N=1011, R=15 kpc
– tcross ~ 108 years
– trelax ~ 1015 years >> age of galaxy:
collisionless
•
cluster of galaxie: trelax ~ age
Dynamical Friction
• DYNAMICAL FRICTION slows a satellite on its
orbit causing it to spiral towards the centre of the
parent galaxy.
• As the satellite moves through a sea of stars I.e.
the individual stars in the parent galaxy the
satellites gravity alters the trajectory of the stars,
building up a slight density enhancement of stars
behind the satellite
• The gravity from the wake pulls backwards on the
satellites motion, slowing it down a little
• The satellite loses angular momentum and slowly
spirals inwards.
• This effect is referred to as “dynamical friction”
because it acts like a frictional or viscous force,
but it’s pure gravity.
• More massive satellites feel a greater friction since they
can alter trajectories more and build up a more massive
wake behind them.
• Dynamical friction is stronger in higher density regions
since there are more stars to contribute to the wake so the
wake is more massive.
• For low v the dynamical friction increases as v increases
since the build up of a wake depends on the speed of the
satellite being large enough so that it can scatter stars
preferentially behind it (if it’s not moving, it scatters as
many stars in front as it does behind).
• However, at high speeds the frictional force v-2, since the
ability to scatter drops as the velocity increases.
• Note: both stars and dark matter contribute to dynamical
friction
• The dynamical friction acting on a satellite of
mass M moving at vs kms-1 in a sea of particles of
density mXn(r) with gaussian velocity distribution
 vs2 
 


f (r , v )   (r ) exp  
2 
 2 

1
2 

3
 ( r )  n( r ) m
• Only stars moving slower than M contribute to the
force. This is usually called the Chandrasekhar
Dynamical Friction Formula.
• For an isotropic distribution of stellar
v
velocities this is:
f (vm )vm2 dvm
M
dvM
 16 2 ln G 2 ( M  m)
dt

0

3
M
VM
• For a sufficiently large vM, the integral
converges to a definite limit and the
frictional force therefore falls like vM-2.
• For sufficiently small vM we may replace
f(vM) by f(0) , define friction timescale by:
dvM
16 2
v

ln G 2 f (0)( M  m)vM   M
dt
3
t fric
Friction & tide: effects on satellite
orbit
• When there is dynamical friction there is a drag
force which dissipates angular momentum. The
decay is faster at pericentre resulting in the
staircase-like decline of J(t).
• As the satellite moves inward the tidal force
becomes greater so the tidal radius decreases and
the mass will decay.
th
9
• Phase Space
Lec
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
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
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 is defined by:
mdN=f(x,v)d3xd3v
where dN is the number of particles per unit
volume with a given range of velocities.
• The mass distribution function is given by
  3  3
f(x,v).
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 coordinates d3x=4πr2dr
• The total momentum is given by:

   3  3
Ptotal   v mdN   f ( x , v )v d xd v

3
 v dN  v n( x)d x
• The mean velocity is given by:
v

3
 v f ( x, v ) d v


3
 v f ( x, v ) d v
mn( x)
 f ( x, v ) d v

v dN


 dN
3
• Example:molecules in a room:
 vx2  v y2  vx2 

m  n o exp  
2

2



f(x, v) 
( 2  )3


v2 
2



exp

4

v
dv
3
3
2



 v dN   vfd xd v  0  2 

2
3
3
 dN  fd xd v  exp   v 2 4v 2 dv
 2 
0
These are gamma functions
• Gamma Functions:

(n)   e x dx
x
n 1
0
(n)  (n  1)(n  1)
1
   
 2
DF and its moments
d 3xA (x)  dM  A  d 3x  Af ( x, v)d 3 v
d 3x  dM  d 3x  f ( x, v)d 3 v
1 2

  l k
2
2
For : A( x, v)  Vx V y x y , Vx , VxVy, (Vx  Vy  Vz )    x , x  v

2
2

 d   sin d  d  2  2 , r  x 
0
x2  y2  z 2
0
2
d 3x  dxdydz  r drdΩ  4πr 2 dr (if spherical)
d 3 v  dv x dv y dv z  v 2 dvdΩ  4πv 2 dv (if spherical)

2
2
V  v 
vx
 v2
2[ E   ( x )]
y  vz 
AdM
A  A  
 dM

  , n 

ρ
m
mass - weighted average,
Additive equations
A B  A  B
f  f1  f 2
  1   2
st
1
Tutorial
g
M
2
vesc
(r)
(E)
(r)
th
10
• orbits
Lec
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
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
th
11
• orbits
Lec
DF & Integrals of motion
• If some quantity I(x,v) is conserved i.e.
dI ( x, v )
0
dt
• We know that the phase space density is conserved
df
i.e
0
dt
• Therefore it is likely that f(x,v) depends on (x,v)
through the function I(x,v), so f=f(I(x,v)).
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)
Analogy
•
•
•
•
•
•
DF(x,v) Analogous to density(x,y,z),
DF(E,L,Lz) analogous to density(r,theta,phi),
E(x,v) analogous to r(x,y,z)
Integrals analogous to spherical coordinates
Isotropic F(E) analogous to spherical density(r)
Normalization dM=f(E)*dx^3*dv^3=
density(r)*dv^3
Spherical Equilibrium System
• Described by potential φ(r)
• SPHERICAL: density ρ(r) depends on modulus of r.


  r ,   r 
  ,0,0   (0,  ,0)   (0,0,  )


 x  0  xy  0
• EQUILIBRIUM:Properties do not evolve with time.
f


0
0
0
t
t
t
Anisotropic DF f(E,L,Lz).
• Energy is conserved as:

0
t
• Angular Momentum Vector is conserved as:

0

• DF depends on Velocity Direction through L=r X v
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
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
d (  r2 ) / dr  (2 r2   t2 ) / r  d / dr
• To prove the above Jeans eq.
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 )
   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
Velocity dispersions and masses in
spherical systems
• 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
2
2
v  v
galactic centre
2
2



r

1

v
/
v

r
• Define anisotropy


th
12
• Phase Space
Lec
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.
Total Mass of spherical systems
• E.g. 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
  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 gam=3.5, Isotro 
2
2
M

4
.
5
(
v

v
r
 )r / G
0
Mass of the Milky Way
We find
d  vr
dr  r 

2
 2 v 2
GM r 
r



 r  1
r  2

2
2
d  vr r   1 vr r GM r 
  
 cst
dr  r  
r
r


Drop first term, solutions
v r
GM r 
(2      1) r  
 cst

r
r
2
vr  v 
2
2
GM
1

4.5r
r
Scalar Virial Theorem
• The Scalar Virial Theorem states that the kinetic
1
energy of a system with mass M is just K  M  v 2 
2
where <v2> is the mean-squared speed of the
system’s stars.
• Hence the virial theorem states that
W
GM
 v 

M
rg

 v 2   r . 
2
 2K  W  0
Virial

dv
 
• Equation of motion:
dt

T
T
1
d
v
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.
• 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
th
13
• orbits
Lec
Spherical Isotropic Self Gravitating Equilibrium
Systems
• ISOTROPIC:The distribution function only
depends on the modulus of the velocity rather than
the direction.

f v 
  
2
x
2
y

 vx v y  0
Note:the tangential
direction has  and 
components
2
z
1 2
   tangential
2
2
r
Isotropic DF f(E)
• In a static potential the energy of an particle is
conserved.
1 2
mE  mv  m (r )
2
dE
0
dt
Note:E=energy per
unit mass
• So,if we write f as a function of E then it will
agree with the statement:
df ( x, v)
0
dt
For incompressible fluids
• So:
f ( x, v)  f ( E ( x, v)) ,

0
• E=cst since
t
1 2

E  v   ( x)
2
• For a bound equilibrium system f(E) is positive
everywhere (can be zero) and is monotonically
decreasing.
• 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.
Eddington Formulae
• EDDINGTON FORMULAE:These can be used to
get the density as a function of r from the energy
density distribution function f(E).
0
(1)  (r )  4  f ( E ) 2( E   ) dE


f ( E )dE
( 2) 
 8 

(E   )

0
1 d d  d
(3) f ( E ) 

2
dE
8
E
 d
0
Proof of the 1st Eddington Formula
 (r )   f ( E )d 3v
4 3
  f ( E )d  v 
3

1 2
E  v   (r )
2
4 3
2
d v  d  v   4v dv
3

3
 v  2( E   (r ))
3
 4 32

2
so  (r )   f ( E )d   2 ( E   ) 
3

8 2 3

f ( E )( E   ) dE

3 2  (r )
0
 4 2
0
f ( E )( E   )


1
2
1
2
dE
(r )
0
  (r )  4  f ( E ) 2( E   ) dE

So, from a given distribution function we can
compute the spherical density.
Relation Between Pressure Gradient and
Gravitational Force
• Pressure is given by:
P  
1
 
3
1
 
3
 v 
2
1
2 3

  f ( E )v d v
3
3
2 4
3
f ( E )v d  v 
3

3
4


2
f ( E )v d   2E   2 
3

2
1 4

2
3 3
3
2

1
3
f ( E )v E   2 dE
2
2
1
4
2  f ( E )( E   ) 2 v 2 dE
3
but v 2  2( E   )
0
3
4
    2 2  f ( E )E   2 dE
3
 (r )
2
1
d
2
3
 

2
  4 2  dEf ( E ) E   2   
dr
3
2
 r 

  
  (r )      (r )
r
 r 
0
since  (r )  4  f ( E ) 2E   dE

• So, this gives:
d (  2 )
d
   (r )
dr
dr
Note: 2=P
• This relates the pressure gradient to the
gravitational force. This is the JEANS
EQUATION.

d
    dr
dr
r
2
Note:-ve sign has gone since we reversed
the limits.
So, gravity, potential, density and Mass are all related and
can be calculated from each other by several different
methods.
g
M
2
vesc
(r)
(E)
(r)
nd
2
Tutorial
g
M
2
vesc
(r)
(E)
(r)
Proof: Situation where Vc2=const is a Singular
Isothermal Sphere
• From Previously: dP d (  2 )
d

   (r )
dr
dr
dr
• Conservation of momentum gives:

u

 P  g
t
1 
 g   P

• 


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.
Finding the normalising constant for the phase
space density
• If we assume the phase space density is given by:
 vx2  v y2  vz2 
  n  constant
f ( x, v)  exp  
2


2



 Ek  1
 exp  - 2   2  constant
   r
 E k  2 2 ln r 
  constant
 exp  2



 E   

 exp  - k 2   constant   constant
  

• 
 E 
f ( E )  exp   2   constant
  
• We can then find the normalizing constant so that (r) is
reproduced.
0
2
c
 (r )  4  f ( E ) 2E   dE
V
 (r ) 
2
4Gr

• Note: you want to integrate f(E) over all energies that the
star can have I.e. only energies above the potential

 f ( E )dE
E (r )
• We are integrating over stars of different velocities ranging
from 0 to .


v 0
f ( E ) d 3v
• One way is to stick the velocity into the
distribution function:
• Using
1 2


  ( x)  v 
2 4v 2 dv
 ( x)  m  f o exp  
2



0




2
v
the substitution x  2 gives:
2
 ( x)  2

3
2 2
  ( x) 
f o exp   2 
  
• Now  can also be found from poissons equation.
Substituting in  from before gives:
 (r )  2

3
2 2

for
vc2
2
• Equating the r terms gives:
vc2
 2
1


r
2
r
vc  2
as before.
th
14
• orbits
Lec
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
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
Orbits in Axisymmetric Potentials
• We employ a cylindrical coordinate system (R,,z)
centred on the galactic nucleus and align the z axis
with the galaxies axis of symmetry.
• Stars confined to the equatorial plane have no way
of perceiving that the potential is not spherically
symmetric.
• Their orbits will be identical to those in
spherical potentials.
• R of a star on such an orbit oscillates around some
mean value as the star revolves around the centre
forming a rosette figure.
Reducing the Study of Orbits to a 2D Problem
• This is done by exploiting the conservation of the
z component of angular momentum.
• Let the potential which we assume to be
symmetric about the plane z=0, be (R,z).
• The general equation of motion of the star is then:

d 2r
  ( R, z )
2
dt
Motion in the
meridional plane
• The acceleration in cylindrical coordinates is:

R R  
R

2

z
z

• The component of angular momentum about the zaxis is conserved.
d 2 2
2

LZ  R   ( R  )  0
dt
2
• If  has no dependence on  then the azimuthal
angular momentum is conserved (rF=0).


1 2
R  R 2 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
Motions in Meridional Plane
• EoM in (R,z)
–:
d2r
  ( R, z )
2
dt
r  R Rˆ  z zˆ
 ˆ 
 
R
zˆ
R
z
– in cylindrical coords:
R  R 2    ;
R
– and Lz conserved

z  
z
 
d 2




2 R  R  0 
R 0
dt
• 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.
• This (non uniformly) rotating plane with cartesian
coordinates (R,z) is often called the
MERIDIONAL PLANE.
• eff(R,z) is called the EFFECTIVE POTENTIAL.
effective potential eff(R,z)
• coupled equations for oscillations in R,z
directions
• use Lz to replace  by Lz / R 2
 eff
 eff


z  
R
;
R
z
2
Lz
 eff   ( R, z ) 
2
2R
• reduced to motion in meridional plane (R,z)
• So 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 orbits are bound between two radii (where the
effective potential equals the total energy) and
oscillates in the z direction.
Example: Logarithmic potential
• oblate galaxy with Vcirc ~ V0 =100km/s


( R, z)  v0 ln R 2  2 z 2  cst
1
2
2
• Draw contours of the corresponding Selfgravitating Density to show it is unphysical.
• Plot effective potential contours for
Lz=100kpckm/s.
• orbits with E=Φ(1kpc,0), what is maximum
z-height?
• What is Rg of a circular orbit with
E= Φ(1kpc,0)?
th
15
• orbits
Lec
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
plane then become:
.


eff
  
R
R
eff
z  
z
R 2  J z
• 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  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.
Minimum in eff
• The minimum in eff has a simple physical
significance. It occurs where:
eff
 L2z
0

 3
R
R R
eff
0
z
(1)
(2)
• (2) is satisfied anywhere in the equatorial plane
z=0.
• (1) is satisfied at radius Rg where
L2z
  
 3  Rg 2
 
 R  Rr ,0  Rg
• This is the condition for a circular orbit of angular
speed 
conditions for a circular orbit at Rg
• minimum in effective potential at R,z = Rg,0
 eff
  
0

R
 R  R
2
g
,0
Lz
 3  Rg 2
Rg
with angular speed 
• circular orbit with angular momentum Lz
• If the energy of the orbit is reduced the two points
between which the orbit is bound eventually
become one.
• You then get no radial oscillation.
• You have circular orbits in the plane of the galaxy.
• This is one of the closed orbits in an axisymmetric
potential and has the property that.
eff
(Minimum in effective
0
potential.)
r R Rcircular
eff
1  Jz
  ( R, z )  
2 R
eff  J z2

 

 3 0
R
R R

2
Gravitational force
in radial direction
Centrifugal
Force
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 :
x   2 x,
 – vertical frequency :
 eff
– with
and
 ( R, z )  Lz
2
2




3
L
z

 2  

;
4
2 
 R  ( Rg , 0 ) Rg
z   2 z
2
2R 2
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
th
16
• orbits
Lec
JEANS EQUATION for
oblate rotator
: a steady-state axisymmetrical system in which 2 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
st
1
Tutorial
g
M
2
vesc
(r)
(E)
(r)
• Example:
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
th
17
• orbits
Lec
Orbits in Planar Non-Axisymmetrical
Potentials
• Here the angular momentum is not exactly
conserved.
• There are two main types of orbit
– BOX ORBITS
– LOOP ORBITS
LOOP ORBITS
• Star rotates in a fixed sense about the centre of the
potential while oscillating in radius
• Star orbits between allowed radii given by its energy.
• There are two periods associated with the orbit:
– Period of the radial oscillation
– Period of the star going around 2
• The energy is generally conserved and determines the
outer radius of the orbit.
• The inner radius is determined by the angular momentum.
apocentre
pericentre
Box Orbits
• Have no particular sense of circulation about the
centre.
• They are the sum of independent harmonic
motions parallel to the x and y axes.
• In a logarithmic potential of the form
1 2  2
y2 
2
l ( x, y)  vo ln  Rc  x  2 
2
q 

box orbits will occur when R<<Rc and loop orbits
will occur when R>>Rc.
Orbits in 2D elliptical potentials
bars in nulcear regions of disc galaxies: SBs
• E.g., non-rotating logarithmic potential
2
 2

y
2
 L x, y   v0 ln  Rc  x 
q 1
2 ,
q 

– equipotential ellipses constant axial ratios, q
– for small R << Rc, expand:
– potential of a 2-D harmonic oscillator (same as
for an homogenous ellipsoid)
2
2
v


2
2
y
2
0
1
 L  2 v0 ln Rc 
x 
2
2 
q
2 Rc 

1
2
2
non-rotating potentials cont.
• Box Orbits
– Within core, harmonic oscillators:
2
2
v0 x
v0 y
 L
 L

  2  x, and   2 2  y
x
y
Rc
Rc q
 x  v0 R ; and  y  v0 qR
c
c

x
/  y  n / m
– unless frequencies are commensurable:
– independent oscillations, stars eventually pass
close to every point inside a rectangle:
• For large R, R >> Rc,
– numerical integrations are required
• if launched with a tangential velocity
•
LOOP Orbits
• orbits rotate in fixed sense about centre of
potential
– oscillate in radius between Rmin and Rmax
– never approach centre!
– fills in annulus, of width determined by Lz,
Loop and Box orbits
• motions in plane y=0 and with vY > 0
– with given energy E
loop orbits, anticlockwise
vx
box orbits
0
loop orbits, clockwise
x
0
– loop orbits are in annuli around “closed loop
orbit”
» equivalent to circular orbit in axisymmetric potential
– box orbits
• outermost curve denotes y=0 and vY=0; “closed
box orbit”
• motion simply parallel to x-axis
• relative proportion of loop versus box orbits
• More box orbits when energy is lower
– all orbits with E < Phi(Rc) are box orbits
• More box orbits when increasingly nonaxisymmetric
• all orbits are loop orbits in axisymmetric potentials
– always rotate in fixed sense (conserves L)
– loop orbits become more elongated
– less epicyclic motion necessary to fill in central
hole
• becomes a box orbit
Orbits in 3-D Triaxial Potentials
– E.g., Elliptical Galaxies
• For given Energy have
• Within core of potential: (box orbits)
– 3-D harmonic oscillator
– adopt long-axis orbit as parent to family
– all axial orbits stable
• Outside core : potentially 3 axial orbits and
3 loop orbits
– one about each axis, stable?
– short and middle-axis (box) orbits are unstable
– middle-axis tube (loop) axis unstable
• have one box orbit (long axis) and two loop
orbits (long-axis and short-axis) closed
orbits and hence parents of non-closed
families
• disc stars in spirals are short-axis tube orbits
th
18
• orbits
Lec
2-D rotating potentials
• non-axisymmetric potentials generally
rotate!
• frame of reference (x,y,z) in which L is
static rotates at angular velocity b = b ez
• EoM in this rotating cordinate system are:
r    b  b  r   2b  r 
centrifugal coriolis
• Effective potential: eff    12 b 2 R2
r   eff  2b  r 
• as used in the binary stars section
Five Lagrange points
• as in Roche-lobe potential
 eff
 eff
 0,
and
0
x
y
– L3, potential minimum at
central stationary point
L5
– L1 and L2, saddle
L2
points : unstable
– L4 and L5, potential
L1
L3
maxima: stable for
L4
certain 
– Annulus bounded by L1,2,4,5 : stars appear to be
stationary, called Co-Rotation Radius
Co-Rotation Radius
Co-Rotation:   b
– where angular velocity of a star is the angular
velocity of the rotating potential
• stars stay in same place relative to potential
• For R>> RCo-Rotation, all closed orbits nearly
circular
• Barred potential spins much faster than stars
– so asymmetry is averaged out
• For R < RCo-Rotation, most heavily populated
orbits parented by long-axis (box) orbits
– aligned with potential
Lindblad Resonances
• barred potential
• in polar coordinates (R,) in frame rotating
with potential where =0 is long axis
• loop orbits as circular with small epicyclic
oscillationsR(t )  R0  R1 (t ), and  (t )   0  1 (t )
then put :  ( R, )   0 ( R )  1 ( R, )
with
1
• for a barred potential:
0
 1
1 ( R, )   b cosm 
– m=2 ensures a barred pot.
• small motions R1(t) are harmonic oscillator,
frequency 0, driven at frequency m(0-b)
Lindblad Resonances cont.
• At Co-Rotation, no oscillations
• At Lindblad resonances, m(0-b)= ± 0,
and star encounters successive crests of the
potential at a frequency that coincides with
natural frequency of radial oscillations.
• + sign: star overtakes potential --- Inner LR
• - sign: potential overtakes star --- Outer LR
outer LR
• e.g. MW:
bar
inner LR
Lagrange Points
• There is a point between two bodies where a particle can
belong to either one of them.
• This point is known as the LAGRANGE POINT.
• A small body orbiting at this point would remain in the
orbital plane of the two massive bodies.
• The Lagrange points mark positions where the
gravitational pull of the two large masses precisely cancels
the centripetal acceleration required to rotate with them.
• At the lagrange points:
eff
x
0
eff
y
0
Effective Force of Gravity
• A particle will experience gravity due to the
galaxy and the satellite along with a centrifugal
force and a coriolis term.
• The effective force of gravity is given by:


    
g eff  g  2  v    (  r )
Coriolis term
Centrifugal
term
• The acceleration of the particle is given by:
r  eff  2  r    (  r)
Jakobi’s Energy
• The JAKOBI’S ENERGY is given by:
1  2
E J  eff  r
2
• Jakobi’s energy is conserved because
dE J deff d  1  2 

  r 
dt
dt
dt  2 
 
 r .eff  rr

 

     
 r eff  eff  r .(2  r )  r (  (  r ))

     
 r .(2  r )  r (  (  r ))


 

  
  r is  to r  r .(  r )  0

   
  
  
  r  r so   (  r ) is  to r  r .(  (  r ))  0
dE J

0
dt
• For an orbit in the plane (r perpendicular to )

 1 2 2
eff (r )   (r )   r
2



 (r )   g (r )   s (r )
GM g GM s
    
r
r  rs
• EJ is also known as the Jakobi Integral.
• Since v2 is always positive a star whose
Jakobi integral takes the value EJ will never
tresspass into a region where eff(x)>EJ.
• Consequently the surface eff(x)>EJ, which
we call the zero velocity surface for stars of
Jakobi Integral EJ, forms an impenetrable
wall for such stars.
• By taking a Taylor Expansion r=rs, you end up
with:
1

3
1


m
 m 3
  
rJ   D 
 D
 3M 
 M 3  m 

 
M 
• Where we call the radius rJ the JAKOBI LIMIT of
the mass m.
• This provides a crude estimate of the tidal radius
rt.
th
19
• orbits
Lec
The Jeans Equations
• The DF (phase space density f) is a function
of 7 variables and hence generally difficult
3
 f  f 

f
to solve
  vi

0
t i 1  xi xi vi 
• Can gain insights by taking moments of the
equation. f 3
f 3  f 3
 t d
v   vi
xi
d v
xi
 v
d v0
i
• where integrate over all possible velocities
– where the summation over subscripts is implicit
Jeans equations cont.
f 3
f 3
 f 3
 t d v   vi x i d v  xi  vi d v  0
– first term: velocity doesn’t depend on time,
hence we can take partial w.r.t. time outside
– second term, vi does not depend on xi, so we
can take partial w.r.t xi outside
– third term: apply divergence theorem so that
3
2


Fd
x

F

d
S


V
S
 f 3

2

d
v

f
d
S


x i vi
x i S
First Jeans equation
n   fd 3 v
• define spatial density of stars n(x)
1
v i   fvi d 3 v
n
• and the mean stellar velocity v(x)
n  nv i 
0

t
x i
• then our first (zeroth) moment equation
becomes
nd
2
Jeans equation
• multiply the CBE by vj and then integrate
over all velocities

f 3 
f 3
3
f
v
d
v

v
v
d
v

v
d v0
j
i j
j



t
xi
xi
vi
• We get





nv j 
nvi v j   n
0
t
xi
x j
vi v j 
1
n
v v
i
j
fd 3 v
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 some sort of
pressure force
Jeans Equation
• Compact form, s=x, y, z, R, r, …


 n ss 2

as  

ns
s
• e.g., oblate spheroid, s=[R,phi,z],
– Isotropic rotator,
a=[-Vrot^2/R, 0, 0],
sigma=sigma_s
– Tangential anisotropic (b<0),
a=[b*sigma^2)/R, 0, 0],
sigma=sigma_R=sigma_z=sigma_phi/(1-b),
• e.g., non-rotating sphere, s=[r,th,phi],
a=[-2*b*sigma_r^2/r, 0, 0],
sigma_th=sigma_phi=(1-b)*sigma_r
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
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
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