10. Collisionless Boltzmann Equation
(BT 4.1 p. 190-193)
Consider a system with a large nunber of
stars.
At any t define the distribution function
(DF) or phase space density function:
f (~
x, ~v , t)d~
xd~v
as the # of stars in volume d~
x with
velocities in range d~v (centered on ~
x, ~v )∗.
at all ~
x, ~v : f ≥ 0.
We now have reduced 6N functions ~
xi, ~vi
into one 7-dimensional function.
If we derive the time evolution of f , we can
completely ignore individual particles !
∗ Note
R
that in BT
f (~
x, ~v , t)d~
xd~v = 1
f
is
normalised
so
that
1
(Notice that we can always rewrite f (~
x, ~v , t)
as a summation of δ functions. We would
then get back our “original” particles. It
shows that in some sense, one
7-dimentional function is “more complex”
than 6N 1-dimensional functions. But if we
take smooth distribution functions, they are
much simpler than the 6N 1-dimensional
functions.)
In order to derive the time evolution, first
define a new coordinate w:
~
w
~ ≡ (~
x, ~v ) ≡ (w1, w2, ...w6)
where w1 ≡ x1, w2 ≡ x2, ... w4 ≡ v1, etc.
Hence the star has coordinate w
~ in
phase-space.
The flow of the star is given by
~
w
~˙ = (~
x˙ , ~v˙ ) = (~v , −∇Φ).
The flow w
~˙ conserves stars. (Only direct
collisions would would have stars jump from
one point in phase space to another).
2
Hence we have the continuity equation:
6
X
∂f
∂(f ẇα)
+
= 0.
(1)
∂t
α=1 ∂wα
Why is this ? Integrate over any volume.
The first term gives the increase in number
of stars in the volume. The second term is
equal to:
Z
V
~ · (f w)
∇
~˙ =
Z
S
(f w)
~˙ · d2S.
This is the surface integral over the flow
out of the volume. Hence the equation
guarantees that stars are conserved (the
density can only increase if stars move into
the volume).
3
A special property of w
~˙ is
6
X
∂ ẇα
3
X
∂vi
∂ v˙i
=
+
∂w
∂x
∂vi
α
i
α=1
α=1
3
(∗) X
∂
=
−
α=1 ∂vi
∂Φ
∂xi
!
!
(∗∗)
= 0.
(*) By definition ∂vi/∂xi = 0 because xi and
vi are independent coordinates.
(**) This is also equal to zero because the
potential does not depend on vi.
Remember equation (1):
6
X
∂f
∂(f ẇα)
+
=0
∂t
∂w
α
α=1
or
6
X
∂f
∂f
∂ ẇα
+
· ẇα + f ·
= 0.
∂t
∂wα
α=1 ∂wα
The last term on the right is zero, as we
have seen above. Hence:
4
6
X
∂f
∂f
+
ẇα
= 0.
∂t
∂wα
α=1
We can write this as:
3
X
∂Φ ∂f
∂f
∂f
+
vi
−
=0
∂t
∂xi ∂vi
i=1 ∂xi
or
∂f
~ − ∇Φ
~ · ∂f = 0.
+ ~v · ∇f
∂t
∂~v
These equations are the Collisionless
Boltzmann Equation (CBE).
The CBE is sufficient to calculate the
evolution of any f with time.
5
Define the Lagrangian or convective
derivative:
6
X
df
∂f
∂f
≡
+
w˙α
.
dt
∂t
∂wα
α=1
Hence the CBE can be written as
df
= 0.
dt
In words: if you move along with a particle
with velocity wα, their phase space density
does not change or their mass/density is
conserved.
We say that the flow in phase-space is
incompressible (the density remains
conserved along a flow-line).
Note that encounters between stars cause a
change in ~v due to the additional
component in the general potential Φ. CBE
is then not valid.
6
A 1-D view∗
Focus on a small element of phase space at
x and vx with size dx by dvx:
In interval dt, net flow in x is :
vxdtdvx[f (x, vx, t)−f (x+dx, vx, t)] = −vxdtdvx
∂f
dx
∂x
the net flow due to the velocity gradient is
dx
dvx
dt[f (x, vx, t) − f (x, vx + dvx, t)] =
dt
−dxdt
dvx ∂f
dvx
dt ∂vx
the sum of these equals the net change to f
in the region, ie at x, vx of size dx dvx:
∗ see
also: http://www.astro.virginia.edu/class/
whittle/astr553/Topic08/Lecture 8.html
7
Courtesy:
http://www.astro.virginia.edu/class/
whittle/astr553/Topic08/Lecture 8.html
8
dxdvx
∂f
∂f
dvx ∂f
dt = −vxdtdvx dx − dxdt
dvx
∂t
∂x
dt ∂vx
or, dividing by dxdvxdt, we get:
∂f
dvx ∂f
∂f
+ vx
+
=0
∂t
∂x
dt ∂vx
x = − ∂Φ we get the one-D CBE:
using dv
dt
∂x
∂f
∂Φ ∂f
∂f
+ vx
−
=0
∂t
∂x
∂x ∂vx
Again, the phase space density (f) along the
star’s orbit is constant ie the flow is
”incompressible” in phase-space. If a region
gets more dense, σv will increase, if a region
expands, σv will decrease.
Example – an idealised marathon race with
all runners running at different but constant
speeds: start : n high, ∆v high; end : n
low, ∆v low
9
Equilibrium models
(BT 4.4.1, pages 220 - 221)
Remember that we defined the
Integral of motion as any function constant
along any orbit:
d
I[~
x(t), ~v (t)] = 0.
dt
In equilibrium models, we have by definition
∂f
∂t = 0. Furthermore, we have seen from
the CBE df /dt = 0.
Hence the distribution function for an
equilibrium model is an integral of motion !
10
As a consequence: Jeans theorem
(i) Any function of integral of motions
F (I1(~
x, ~v ), I2(~
x, ~v ), ...)
is also a solution of the time-independent
CBE.
(ii) Any integral of motion is a solution of
the time-independent CBE.
First part obvious: Since the CBE is a linear
equation, then functions of solutions are
themselves solutions.
This is extremely useful since it allows us to
construct legitimate DFs using integrals of
motion.
Second part can be proven as follows:
x + ∂I · d~v = ~
∂I = 0
dI/dt = ∇I · d~
v
·
∇I
−
∇Φ
·
dt
∂~v dt
∂~v
and this is identical to the time-independent
CBE df (I)/dt = 0 (see page 6)
11
A consequence of Jeans theorem:
A galaxy can be constructed by adding up
orbits. Along each orbit, the DF is constant.
This is another justification of
Schwarzschild’s method.
12
Spherical systems
(BT 4.4.2 p. 221, 222)
For a system with a spherical potential, the
classical integrals are energy E and angular
~
momentum L.
For system with a spherically symmetric
~ but only
DF, the DF can not depend on L,
on L2.
Not individual orbits are used to build
models, but sets of orbits !
13
Isotropic models
Distribution function f (ε)
Define
ψ = −Φ + Φ0
1 v2
ε = −E + Φ0 = ψ − 2
Relative potential :
Relative energy :
Choose Φ0 such that
f >0
for
ε>0
f =0
for
ε≤0
The density is the integral of distribution
function over all velocities:
ρ(r) =
=
Z v
max
0
Z
f (ε)d~v
f (ε)4πv 2dv
1 v 2 , dε = −vdv.
Substitute ε = ψ − 2
14
Since ε will run from 0 (for v = vmax) to ψ
(for v = 0):
ρ(r) = 4π
Zψ
q
f (ε) 2(ψ − ε) dε
0
We can make models by specifying f (ε). We
find a relation of the form ρ(r) = F (ψ). We
then have to find solutions of this equation,
which also satisfy the Poisson equation:
~ 2Φ(~
4πGρ(~
x) = ∇
x)
Solutions can be constructed. For example,
assume f () = γ . These give densities of
the form ρ = cnst × ψ (γ+1.5). These models
are analogous to gas polytropes.
15
What is “basic set of building blocks” for
models of type f (ε) ?
• fill in delta function f (ε) = δ(ε − ε0)
√ q
ρ = 4 2π ψ − ε0
maximum in the center, decreasing
outwards, to zero at ψ(r) = ε0.
These are NOT individual orbits, but
combinations of orbits.
16
Distribution function f (ε) from density
(BT p. 236-237)
As we saw on the previous pages, we can
calculate the density of a model from the
distribution by the integral equation:
ρ(r) = 4π
Zψ
q
f (ε) 2(ψ − ε) dε.
0
We now show how to derive f from ρ.
Write ρ as ρ(ψ). This is always possible
since ψ is a monotonic function of r. Then:
1 dρ
√
=
2π 2 dψ
Zψ
0
f (ε) dε
√
.
ψ−ε
This is an “Abel” integral equation, with
solution:
f (ε) =
Zε
d
1
√
{
2
2π 2 dε
0
dρ dψ
√
}.
dψ ε − ψ
Thus, f (ε) ≥ 0 if and only if {. . .} increases
monotonically with ε. This is true for
models that q
are more centrally concentrated
than ρ(r) = 1 − r2.
17
Intermezzo: Abel integral equation
Let
f (x) =
Z x
g(t)dt
0 (x − t)1/2
then
1 d t f (x)dx
g(t) =
.
π dt 0 (t − x)1/2
Z
Proof: substitute the first equation into the
right hand side of the second:
g(s)
1d t x
1
F (t) =
dsdx,
1/2
1/2
π dt 0 0 (x − s)
(t − x)
Z
Z
and interchange order of integration
1d t t
g(s)
1
F (t) =
dxds.
1/2
1/2
π dt 0 s (x − s)
(t − x)
Z
Z
Use
Z t
1
s (x − s)1/2 (t − x)1/2
dx = π
18
Hence
1d t
F (t) =
g(s)πds.
π dt 0
Z
It is easy to see that this results into
F (t) = g(t).
19
Velocity moments
Define velocity dispersions vr2 , vθ2 , vφ2 :
1
2
vi =
ρ
Z
f (~v )vi2d~v .
If f = f (ε), f is isotropic: at a given
location ~
x, the distribution in velocities is
the same in all directions:
2)
v
f = f (ε) = f (ψ(r) − 1
2
Hence vr2 = vθ2 = vφ2 ≡ σ 2.
We can write for such systems:
11
σ2 =
3ρ
4π
=
3ρ
Z
Z
f (~v )4π v 2dv
f (~v )(2(ψ − ε))1/2dε.
2 , dε = −vdv.
again using ε = ψ − 1
v
2
20
Isotropic models: Examples:
Plummer model
BT p 42, 43, 223-225
Potential: ψ = q
Mass: M (r) =
GM
r2 + a2
M r3
(r2 + a2)3/2
Density: ρ(r) =
3M a2
4π(r2 + a2)5/2
∼ polytrope with index
Plummer model =
n = 5:
ρ = Cψ 5
with
3a2
C=
4πG5M 4
Eddington formula:
64C 7/2
f (ε) = √
≥0
2
7 2π
21
Velocity dispersion:
GM
q
ψ
=
vr2 = 1
6
6 r2 + a2
Circular velocity:
GM (r)
vc2 =
=
r
GM r2
(r2 + a2)3/2
22
Singular isothermal sphere (BT p. 226-229)
The equation of hydrostatic support of an
isothermal gas:
dp
k T dρ
GM (r)
,
= B
= −ρ
2
dr
m dr
r
where kB is Boltzmann’s constant, and m is
the mass per particle.
Multiplying with (r2m/ρkB T ), differentiating
wrt r and using (dM/dr) = 4πr2ρ, we find:
d
d ln ρ
Gm
2
r
=−
4πr2ρ.
dr
dr
kB T
(2)
Now take:
f (ε) =
ρ1
(2πσ 2)
2
ε/σ .
e
3/2
23
1 v 2 , the local velocity
Since ε = ψ − 2
distribution is proportional to
2 /2σ 2
−v
f (~v ) ∝ e
.
This is a Gaussian distribution in velocity in
each dimension v1, v2, v3, with dispersion σ.
Hence the dispersion in 1 dimension is given
by σ.
Use fundamental integral equation
R
ρ = f d3~v to find
2
ψ/σ
ρ(r) = ρ1e
Substitute ψ = σ 2 ln(ρ/ρ1) in the Poisson
equation:
1 d 2 dψ
r
= 4πGρ
2
r dr dr
and obtain:
d 2 d ln ρ
4πG
(r
) = − 2 ρr2.
dr
dr
σ
(3)
This equation and equation (2) are identical
if we set:
24
k T
σ2 = B .
m
Hence the adopted name of an “isothermal
sphere”.
Now let’s try to solve equation (3).
Try solution of the form ρ = Cr−p. Then
4πG 2−p
−p = − 2 Cr
.
σ
This works if
p = 2,
σ2
C=
.
2πG
Thus
σ2
ρ(r) =
,
2
2πGr
ψ = −2σ 2 ln r.
vc2 = 2σ 2
vr2 = σ 2.
Also
25
10. Homework assignments
1. In the isotropic models (pages 14-16),
what kind of orbits manage to reach the
maximum distance to the center given by
ψ(r) = ε0. What is the velocity of stars at
this radius?
2. Calculate the exact relation between ρ
and ψ for polytropes, assuming f (ε) = εγ
3. Calculate the density
related to the
q
potential ψ = 1/ 1 + r2. Show that this
model is a polytrope [i.e., that we can
write ρ = cnst × ψ g ].
4. Derive the exact form of the distribution
function for the Plummer model given
earlier.
26
Typical questions for tentamen:
What is the CBE ?
From what assumption can we derive it ?
What does the CBE imply for the
Lagrangian derivative df /dt ?
How do we make equilibrium models ?
Explain the Jeans theorem.
27