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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
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
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 )  rp 

 kM (r ) 
• The tidal radius is smallest at pericentre where r is
smallest. Often tidal radius is only defined when
r(t)=pericentre rp.
• 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.
M s (r ) M (r )

4
4 3
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/c 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.
st
1
Tutorial
g
M
2
vesc
(r)
(E)
(r)