Transcript Gravitational Dynamics

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: 5-body rectangle problem
• Four point masses m=3,4,5 at rest of three
vertices of a P-triangle, integrate with time
step=1 and ½ find the positions at time t=1.
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
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: Force field of two-body
system in Cartesian coordinates
2

G  mi

 (r )     , where Ri  (0,0,i )  a, mi  i  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: 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
0

dv
 
since
 dt
0 for static potential.
and dr  v
dt
So orbital Energy is Conserved in a
static potential.
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
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.
Poisson’s Equation
• Poissons equation relates the potential to the
density of matter generating the potential.
• It is given by:
 
      g  4G (r )
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
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 uniform distribution of mass of density
ρ.
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


From Density to Mass
M(R  dr)  M(R)  dM
4 3
2
dM   (r)d  r   4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)
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 )      2
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.
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.
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 

2 
4a  a 

5
3
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 3 r 2  a


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
3


3
M
r
• This gives
(as before from



1

4a 2  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
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 of angular momentum.
  dr
Lr
 const
dt
area swept
2 d
r
2
dt
unit time
Keplers 3rd law
pericentre
apocentre
2
 2
1  dr  1  d 
E     r
   (r )
2  dt  2  dt 
2
1  dr  1 L2
   
  (r )
2
2  dt  2 r
eff
2E
 dr  1 2 (r )
 const   2   2  2
2
L
L
 r dθ  r
1
u
r
2 E  du 
2 (1 / u )
2
 2 
 u 
2
L
d

L


2
dr
L2
  2 E  2 (r )  2
dt
r
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:
d
L
d
d
r2
dt
  2 
dr  2 
dr  2  dr  2 
dr
dr
dr
dr
dr
rp
rp
rp
rp
dt
dt
ra
ra
ra
ra
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.

?????????????Graphs
• The velocity of the star is slower at apocentre due to the
conservation of angular momentum.
• ??????????????????????????
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):
 v x2  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:
 v x2  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
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!
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)
EoMJeans eq.
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
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
Proof: Energy is conserved
• The total energy of an orbit is given by:
1 2

E  v   (r , t )
2


dE  dv dr

v
  
dt
dt dt
t
0

dv
 
since
 dt
and dr  v
0 for equilibrium
systems.
dt
So Energy is Conserved.
Linear Momentum
• The change in linear momentum with time is

given by: dP
dt
 
• Since   0 P is not conserved.
• However, when you sum up over all stars, linear
momentum is conserved.
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

 v x 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 ))
so  (r )  
3
 4 32

2
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
1 4 2
3

2  f ( E )v 2 E   2 dE
3 3
2
3
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)
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
c
2
c
2
v v
  
4G 2r
vc2 vc2

4Gr 2 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:
 v x2  v y2  v z2 
  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
Vc2
 (r ) 
4Gr 2
 (r )  4  f ( E ) 2E   dE

• 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.
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
• 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
• 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.
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 
• 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 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   ( 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.
• 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.
• 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
Can generally be ignored since the
gravitational force is only in the z
direction close to the 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
• The orbits are bound between two radii (where the
effective potential equals the total energy) and
oscillates in the z direction.
• ????????graph of z vs R??????????
• 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
• The following figure shows the total angular momentum
along the orbit.
• ??????????????????????????
• Angular momentum is not quite conserved but almost.
• This is because the star picks up angular momentum as it
goes towards the plane and vice versa.
• 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 precess about this axis.
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.
JEANS EQUATION
• The 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  v z2  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
    v  v
2
2
2
rot
• 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
• Example: suppose  ( R, z )  A ( R, z )
2n
P    
2
2n2

n 1
n
(since 
 (  2 )
 ( A 2 n  2 )


z
z
but 
2n2


2 ( n 1)

2 n 1
1  A


z

 ( )
2 n 1
 n 1 
2n

 n 

2n2
)

1
2n


A n 1 
 2
z
A 2 n
 (n  1) 2

z
 (n  1) 2 
so 

z
z
Integrate from  to z and assume  ()  0 and
the velocity dispersion   0 at z  .
 ( R, z )  B ( R )
2
 
n 1
Tidal Stripping
• TIDAL RADIUS:Radius within which a particle
is bound to the satellite rather than the host
system.
• Consider a satellite of mass profile ms(R) moving
in a spherical potential (r) made from a mass
profile M(r).
R
r
V(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
GM s
 2
2
R
R
2
r 1  
r

GM
2
2R
 R
If R  r then 1 -   1 
 .........
r
 r
GM
 2
r
 R  GM s
  2
R
r
Force on particle
due to satellite
GM R GM s
k 2
 2
k  1  3 3 for keplerian
• So,
r r
R
• Therefore the formula for Tidal Radius is:
 ms 

RT  r 
 kM (r ) 
1
3
• The tidal radius is smallest at pericentre where r is
smallest.
• The inequality can be written in terms of the mean
ms (r ) M (r )
densities.
4 3
R
3

4 3
r
3
• The satellite mass is not constant as the tidal radius
changes.
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.
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 ms moving at vs kms-1 in a sea of particles
with gaussian velocity distribution
 vs2 
 


f (r , v )   (r ) exp  
2 
 2 
is
vs
 vs 


t fric
 t 

1
2 

3
 ( r )  n( r ) m
• For an isotropic distribution of stellar
velocities this is:
v
M
dvM
 16 2 ln G 2 m( M  m)
dt

0
f (vm )vm2 dvm

3
M
VM
• i.e. only stars moving slower than M contribute to the force.
This is usually called the Chandrasekhar Dynamical
Friction Formula.
• For sufficiently small vM we may replace f(vM) by f(0) to
give:
dvM
16 2

ln G 2 f (0)m( M  m)vM
dt
3
• For a sufficiently large vM, the integral converges to a
definite limit and the frictional force therefore falls like
vM-2.
• 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 appearance.
• ??????????????????????????????????
• As the satellite moves inward the tidal force
becomes greater so the tidal radius decreases and
the mass will decay.
• ???????????????????????????????????????
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
 v 
2
W

GM
rg
M

2
 v   r . 
 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.
 v x v x  x x 
 v x v y  x y  etc
 v 2  v x2    v y2    v z2 

 r . 
d
2
 r
 vcir
 ( spherical )
dr
2
 v 2  vcir

• 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
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
• E.g. circular orbit in a flat potential:
2
 Pi ,cir  Let P(t)=Vcir
 Pi ,cir 

rf  ri 

 T f  Ti 
P 
2

r
P 
 f ,cir  T 
f ,cir 

vcir
• So, if Vcir increases by a factor of 2 then the radius
of the orbit would decrease by a factor of 2 and
the period would decrease by a factor of 4.
• E.g. in a potential  p  p 2 (t ) ln r  const
• In general the orbital eccentricity is conserved e.g.
a circular orbit will stay circular.
d
v r
dr
2
rp
2
 vcir 
r
2
cir
so v
2
cir
p
 vcir  p
2
• In general, two stellar phase points that started out
on the same torus will move onto two different
tori.
• However, if P 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 P
• Any two stars that are on a common orbit will still
be on a common orbit after the variation in P is
complete. This is ADIABATIC INVARIANCE.