Download Stellar Relaxation Times

Stellar Relaxation Times
In order to appreciate the parameters of the Galactic disk, it is
necessary to understand the concept of “relaxation time”. How long
will it take for a star to gravitationally scatter enough so that it loses
information about its origin.
One way of parameterizing this is through energy exchange: when
does the kinetic energy exchanged during stellar encounters equal the
star’s original kinetic energy? In other words,
2
TE ⇒ ∑ ( ΔE ) ≈ E
Alternatively, one can define the relaxation time as the time it takes
a star to lose all memory of its original trajectory. In this case
€
TD ⇒ ∑ sin 2ϕ ≈ 1
The two values are closely related. Since its easier to derive the
latter quantity, we’ll use it as the definition of relaxation time.
€
Stellar Relaxation Times
Assume that
•  all deflections are two-body encounters
•  each encounter is statistically independent of all previous
encounters
•  close encounters are insignificant compared to long-range
encounters, so that during each encounter, |∆E| ≪ E.
Under these assumptions, all the deflections are small (sin ϕ ≪ 1), so
we can use the Born approximation, in which vinit ~ vfinal ~ v. Now
let’s consider a star’s gravitational encounter with another object.
s
e
r
b
M
v
v
Stellar Relaxation Times
s
e
r
b
M
v
v
For a single encounter, the deflection angle, ϕ, can be computed as a
function of the initial impact parameter, b, by
∞
dv⊥
1 ∞
F⊥ = m
⇒ v⊥ = ∫ dv⊥ = ∫ F⊥ dt
dt
m −∞
−∞
From the geometry of the encounter
€
⎛ b ⎞ ⎛ G M m ⎞ ⎛ b ⎞
F⊥ = F sin θ = F ⎜ ⎟ = ⎜ 2 ⎟ ⎜ ⎟
⎝ r ⎠ ⎝ r ⎠ ⎝ r ⎠
and from the Born approximation, v|| dt = v dt = ds ⟶!dt = ds / v, so
€
Stellar Relaxation Times
s
e
b
r
M
v
v
1 ∞
2
v⊥ = ∫ F⊥ dt =
m −∞
m
∞
∫
0
⎛ G M m ⎞ ⎛ b ⎞ ⎛ 1 ⎞
⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ds
⎝ r ⎠ ⎝ r ⎠ ⎝ v ⎠
Since r = (s2 + b2)½
€
2G M
v⊥ =
v
or, letting x = s/b,
€
∞
∫
0
b
2G M
ds =
2
2 3/2
v
(s + b )
∞
∫
0
ds /b
2 3/2
(1+ (s /b) )
Stellar Relaxation Times
s
e
b
r
M
v
v
2G M
v⊥ =
vb
∞
∫
0
dx
2 3/2
(1+ x )
2G M
x
=
⋅
vb (1+ x 2 )1/ 2
∞
0
2G M
=
vb
for small deflections, tan φ ≈ φ ≈ v⊥/v, so the deflection angle as a
function of impact parameter, b, is
€
2G M
ϕ= 2
v b
Stellar Relaxation Times
v dt
Now, let’s sum over all possible collisions. The number of collisions
per time dt depends on the impact parameter, the distance a star
travels in dt, and the density of stars in the stellar system, N, i.e.,
N coll = (2π b db) ⋅ (v dt ) ⋅ N
Consequently,
TD b max
∑ sin ϕ ≈ ∑ϕ
2
2
=1 =
∫ ∫ (2π b db)(vdt)N ϕ (b)
2
0 b min
€
TD b max
=
∫
0
⎛ 2G M ⎞ 2
∫ (2π b db) (vdt ) N ⎜⎝ v 2b ⎟⎠
b min
2
2
b max
8π G M N
db
=
T
D ∫
v3
b min b
⎛ bmax ⎞
8π G 2 M 2 N
=
TD ln ⎜
⎟
3
v
b
⎝ min ⎠
Stellar Relaxation Times
v dt
The only parameter still needed is bmax/bmin, and since this enters in as
the ln, the exact numbers chosen aren’t very important. One can start
with the obvious fact that no deflection angle can be larger than π, so
2G M
ϕ= 2
=π
v bmin
⇒ bmin
2G M
=
π v2
Conversely, no impact parameter can be greater than the mean stellar
distance, i.e.,
⎛
⎞1/ 3
€
1
N=
3
(4 /3)π bmax
⇒ bmax
3
= ⎜
⎟
4
π
N
⎝
⎠
So
€
⎛ 8π G 2 M 2 N ⎞
⎧ bmax π v 2 ⎫
⎬ = 1
⎜
⎟ TD ln ⎨
3
v
⎝
⎠
⎩ 2G M ⎭
€
Stellar Relaxation Times
Our simple dynamical relaxation time is therefore
⎧ bmax v 2π ⎫
⎛ v 3 ⎞ ⎧
⎛ bmax v 2 ⎞⎫
v3
9
⎬ = 2.1 × 10 ⎜ 2 ⎟ ln ⎨ 365 ⎜
TD =
ln⎨
⎟⎬ yr
2
2
8π G M N ⎩ 2G M ⎭
⎝ M N ⎠ ⎩
⎝ M ⎠⎭
with v in km/s, M in M, and N in stars/pc3. A more rigorous
derivation by Chandrasekhar gives
⎧ bmax v 2 ⎫
⎧ bmax v 2 ⎫
v3
v3
⎬ and TE =
⎬
TD =
ln ⎨
ln ⎨
2
2
2
2
8π G M NH( χ) ⎩ 2G M ⎭
32π G M NG( χ ) ⎩ 2G M ⎭
where H(χ) and G(χ) are factors of order unity that depend on the
stellar distribution function. Finally, Ostriker & Davidson (1968)
give an improved recursive expression for relaxation time:
⎧ v 3TP ⎫
v3
⎬
TP =
ln ⎨
2
2
8π G M N ⎩ 2G M ⎭
Star Clusters
In the solar neighborhood, the Sun is moving ~ 20 km/s with
respect to the surrounding stars. A density of ~ 1 star pc-3, then
implies a relaxation time of ~ 1014 yr. The Sun’s orbit about the
center of the Galaxy is not gravitationally affected by other stars.
On the other hand, giant molecular clouds have masses that are
~ 108 M. Although the number density of clouds is lower, it’s not
1016 times lower! The masses of these clouds are therefore high
enough to scatter stars out of their circular orbits, to produce σR,
σθ, and σz. The result is an increase in scale height with population
age.
Note that because all the objects are (approximately) in a plane, one
would expect σR > σz. This is what is seen.
Star Clusters
The Milky Way currently has two types of star clusters:
•  Open clusters: young systems containing ~ 103 stars
•  Globular clusters: very old systems containing > 105 stars
Our Milky Way no longer seems to make very massive star
clusters.
The H and χ Persei open clusters
M67 open cluster
Star Clusters
The Milky Way currently has two types of star clusters:
•  Open clusters: young systems containing ~ 103 stars
•  Globular clusters: very old systems containing > 105 stars
Our Milky Way no longer seems to make very massive star
clusters.
Pleides open cluster
Praesepe open cluster
Star Clusters
The Milky Way currently has two types of star clusters:
•  Open clusters: young systems containing ~ 103 stars
•  Globular clusters: very old systems containing > 105 stars
Our Milky Way no longer seems to make very massive star
clusters.
NGC 6649 open cluster
M11 open cluster
Star Clusters
The Milky Way currently has two types of star clusters:
•  Open clusters: young systems containing ~ 103 stars
•  Globular clusters: very old systems containing > 105 stars
Our Milky Way no longer seems to make very massive star
clusters.
M13 globular cluster
M3 globular cluster
Star Clusters
The Milky Way currently has two types of star clusters:
•  Open clusters: young systems containing ~ 103 stars
•  Globular clusters: very old systems containing > 105 stars
Our Milky Way no longer seems to make very massive star
clusters.
47 Tuc globular cluster
M15 globular cluster
Star Clusters
The Milky Way currently has two types of star clusters:
•  Open clusters: young systems containing ~ 103 stars
•  Globular clusters: very old systems containing > 105 stars
Our Milky Way no longer seems to make very massive star
clusters.
Open clusters have typical half-light diameters of ~ 1 pc and
velocity dispersions of ~ 5 km/s; globular clusters have halfdiameters of ~ 20 pc and σ ~ 20 km/s. These numbers imply
relaxation times of < 1 Gyr, hence stellar encounters are nonnegligible. The stars are exchanging energy!
Isothermal Spheres
In globular clusters, where stars have had plenty of time to
exchange energy, the stars approach an equipartition state, where
2E 1/ 2
−( E +Ω(r)) / kT
N(E)dE = N 0 1/ 2
e
3/2
π ( kT )
where
1
3
2
m v = kT
2
2
In other words, the stars distribute their velocities in a Maxwellian
fashion, with a different characteristic velocity at any position in
€
the cluster’s potential, Ω(r). Thus
⎛ m ⎞ 3 / 2
⎧ mv 2 Ω(r) ⎫
2
⎬ dv
N(v)dv = N 0 ⎜
−
⎟ 4 π v exp ⎨ −
kT ⎭
⎩ 2kT
⎝ 2π kT ⎠
or, more clearly,
⎛ m ⎞ 3 / 2
⎧ mv 2 ⎫
⎧ Ω(r) ⎫
2
⎬ dv and N(r) = N 0 exp ⎨ −
⎬
N(v)dv
= N(r) ⎜
⎟ 4 π v exp ⎨ −
€
⎩ 2kT ⎭
⎩ k T ⎭
⎝ 2π kT ⎠
€
Isothermal Spheres
Now let’s make a simple cluster where all the stars have the same
mass. In that case
ρ(r) = N(r)⋅ m = ρ 0 e −Ω / kT
⇒ ln ρ (r) = ln ρ 0 −
Ω(r)
kT
⇒
dΩ
d ln ρ
= −k T
dr
dr
If we throw this into the spherically symmetric Poisson equation, we
obtain an equation for the structure of this “isothermal” cluster.
1 d ⎛ 2 dΩ ⎞
⎜ r
⎟ = 4 π Gρ ⇒
2
r dr ⎝ dr ⎠
d ⎛ 2 d ln ρ ⎞
4π G 2
r ρ
⎜ r
⎟ = −
dr ⎝
dr ⎠
kT
One solution to this equation is a simple inverse square law. If we
substitute velocity dispersion for temperature using ⟨v2⟩ = 3kT/m, and
€define σ2 = ⟨v2⟩/3 as the line-of-sight velocity dispersion, then
σ2
ρ(r) =
2π Gr 2
Of course, this solution has an infinite central density. To fix this,
we can constrain the central density to be finite, and numerically
solve for the density
€ distribution.
Isothermal Spheres
The parameter r0 is the
“King radius”, or the
“core radius”, where
the projected mass
density (Σ) has dropped
by a factor of ~ 2.
Because an isothermal
sphere falls off as 1/r2
at large radii, the
distribution implies a
rotation speed that is
independent of radius,
and a total mass that is
infinite.
⎛ 3 v 2 ⎞1/ 2 ⎛ 9σ 2 ⎞1/ 2
⎟⎟ = ⎜
r0 = ⎜⎜
⎟
⎝ 4 π Gρ 0 ⎠
⎝ 4 π Gρ 0 ⎠
Isothermal Spheres
For
1/r2
distributions,
M(r) =
rr
rt
2
4
π
r
ρ(r) dr ∝
∫
∫ K dr = K r
0
t
0
where K is some constant. As rt ⟶∞, M(r) becomes infinite. Also
GM(r) v c2
€ 2 =
r
r
M(r) K r
⇒ v ∝
∝
=K
r
r
2
c
More specifically, if you keep track of the constants, vc = √2 σ.
Note that
€ the form of the isothermal sphere is numerical, but in the
central regions, r < 2 r0, a simple approximation (good to a couple of
percent) is
ρ(r) =
ρ0
{1+ (r /r ) }
2
0
3/2
and Σ(R) =
Σ0
1+ ( R /R0 )
2
Isothermal Spheres
z
The true density distribution and the projected
density distribution are easily related.
∞
∞
r
Σ(R) = 2 ∫ 0 ρ(z) dz = 2 ∫ R ρ (r)⋅
dr
2
2 1/ 2
(r − R )
If we adopt ρ(r) =
ρ0
{1+ (r /r ) }
2
R
r
then
3/2
0
∞
Σ(R) = 2 ρ 0 ∫
R
€
∞
r
{
1+ ( r /r0 )
2
r2 − R2
Setting η = 2 2
r0 + r
2
2 ρ 0 r03
Σ(R) = 2 2
r0 + r
∞
∫
0
}
3/2
(r
2
−R
2 1/ 2
)
dr = 2r ρ 0 ∫
3
0
R
r
(r
2
0
+r
2 3/2
) (r
then gives
dη
2 3/2
(1+ η )
=
2 ρ 0 r0
1+ ( r /r0 )
2
⋅
η
2 1/ 2
(1+ η )
=
2 ρ 0 r0
1+ ( r /r0 )
2
2
−R
2 1/ 2
)
dr
Isothermal Spheres
Note: isothermal spheres have many applications in astrophysics,
including x-ray emission from galaxy clusters, dark matter distributions
around galaxies, and, of course, the dynamics of star clusters.
Isothermal Spheres
Globular clusters are not exactly isothermal spheres because
•  Energy exchange continually populates the high-velocity tail of
the Maxwellian distribution. These stars leave the cluster (i.e.,
evaporate), decreasing the cluster mass and potential, facilitating
further evaporation.
escape velocity ⟶
Isothermal Spheres
Globular clusters are not exactly isothermal spheres because
•  Energy exchange continually populates the high-velocity tail of
the Maxwellian distribution. These stars leave the cluster (i.e.,
evaporate), decreasing the cluster mass and potential, facilitating
further evaporation.
•  Not all stars have the same mass:
heavier particles sink to the
cluster center, while less massive
systems migrate outwards. This
leads towards a “core collapse”.
Positions of milli-second pulsars
Isothermal Spheres
Globular clusters are not exactly isothermal spheres because
•  Energy exchange continually populates the high-velocity tail of
the Maxwellian distribution. These stars leave the cluster (i.e.,
evaporate), decreasing the cluster mass and potential, facilitating
further evaporation.
•  Not all stars have the same mass:
heavier particles sink to the
cluster center, while less massive
systems migrate outwards. This
leads towards a “core collapse”.
•  Binary stars take energy from the
cluster, creating harder binaries
and ejecting 3rd bodies. This
energy prevents core collapse.
Isothermal Spheres
Globular clusters are not exactly isothermal spheres because
•  Energy exchange continually populates the high-velocity tail of
the Maxwellian distribution. These stars leave the cluster (i.e.,
evaporate), decreasing the cluster mass and potential, facilitating
further evaporation.
•  Not all stars have the same mass:
heavier particles sink to the
cluster center, while less massive
systems migrate outwards. This
leads towards a “core collapse”.
•  Binary stars take energy from the
cluster, creating harder binaries
and ejecting 3rd bodies. This
energy prevents core collapse.
• The Galactic tidal field will truncate the cluster.
Tidal Truncation
The cluster’s motion through the Galaxy will have an effect on the
system’s structure. Consider a star at a distance r from a cluster’s
center. If the system has a galactocentric distance R, then the
Galaxy will exert a force on the star
G MG
2G MG
FG = − 2
⇒ dFG =
dr
3
R
R
Equating this to the cluster’s own gravitational force yields
€
2G MG
GMC
r
=
R3 t
rt2
⎛ MC ⎞1/ 3
⇒ rt = R ⎜
⎟
⎝ 2MG ⎠
Actually, when one includes centripetal force and the fact that the
cluster’s orbit is likely elliptical with eccentricity, ε, the equation
becomes
€
⎧ MC
⎫1/ 3
⎬
rt = R p ⎨
⎩ MG ( 3 + ε ) ⎭
Lowered Isothermal Models
[King 1962, AJ, 67, 274]
To compensate for the effect of
tides (and to fix the problem of
infinite mass), King (1966)
artificially truncated the isothermal
distribution at low energies using a
tidal energy. The result is a series
of models defined by the ratio of
the tidal radius to the core radius,
c = log10(rt/r0), or, alternatively, by
the ratio of the central potential to
the velocity dispersion, Ω(0)/σ2.
ρ(E)dE =
ρ1
e
{
(2πσ )
−E /σ 2
2 3/2
1 2
where E = v + Ω
2
−e
−E t /σ 2
} dE
Ω(0)/σ2 =
Ω(0)/σ2 =
Lowered Isothermal Models
[King 1962, AJ, 67, 274]
Note that isothermal (and lowered isothermal) distributions are
numerical. But a useful approximation for the projected surface
brightness is
⎧
⎪
1
Σ(R) = Σ 0 ⎨
⎪ 1+ (R /Rc ) 2
⎩
[
€
1/ 2
1
−
⎫2
⎪
1/ 2 ⎬
⎪
⎭
] [1+ (R /R ) ]
2
t
c