Nebulae containing Heavy Elements

PHYS3710: The Interstellar Medium School of Physics, UNSW
Professor Michael Burton
4. Nebulae Containing Heavy Elements Cooling Lines Low abundance of heavy elements (e.g. C, N, O) but efficiently excited by electron
collisions
Excitation Potential ~ kT whereas for H, He >> kT
⇒Important Coolants
‘Forbidden’ Lines:
Oxygen is the most important
O
→ O+ occurs for hν > 13.6eV
O+ → O++ occurs for hν > 35.1eV
Diagram: Forbidden Lines of Oxygen
These are relatively slow Magnetic Dipole or Electric Quadrupole Transitions, with decay
rates A ~ 1 s-1
c.f. rapid Electric dipole transitions of permitted lines, with A ~ 108 s-1
Statistical Equilibrium Excitation occurs by electron collisions (recombination into excited states is negligible)
Diagram: 2-level Atom
Collisional excitation rate is
Collisional de-excitation rate is
Radiative Decay Rate
Rate
j → i:
Rate
i → j:
Cij cm+3s-1
Cji cm+3s-1
Aji s-1
ne nj Cji + nj Aji
ne ni Cij
In statistical equilibrium
Transitions up = Transitions down; i.e.
nenjCji + nj Aji = neniCij
(i < j)
(j > i)
(52)
⎧ E ⎫
In LTE:
where
−⎨ i ⎬
nI
g exp ⎩ kT ⎭
ni =
Z(T) i
nI is the number density of a particular ion
gi is the degeneracy of level i with energy Ei
−E / kT
Z(T) is the Partition Function = ∑ gn e n
(53)
(54)
n
[exercise: Sum over all states to demonstrate this]
g (E −E )/ kT
n
⇒ i = i e j i
nj
gj
It can be semi-empirically shown that the collisional de-excitation rate is given by
⎛ qij ⎞1/ 2
(i.e. with i > j)
Cij (Te ) ≅ ⎜ ⎟
⎝ Te ⎠
(55)
where qij is a constant, which depends on the ion and the transition.
⎛ 2π ⎞1/ 2  2 Ω(1,2)
C21 = ⎜ ⎟
We write this as, for instance,
⎝ kT ⎠ me3 / 2
g2
for collisional de-excitation from level 2 to level 1
8.629 ×10−6 Ω(1,2)
C21 =
cm3 s-1
0.5
T
g2
where Ω(1,2) is the collision strength between level 1 and level 2.
(56)
[exercise]
The postulate of detailed balance states that we can equate equilibrium between 2 levels by
just considering the just transitions between those 2 levels (without having to worry about
interlocking loops involving other levels). Justification? It works!
Then, suppose that collisions dominate over radiative decay
∴ n j C ji = n i Cij
or
Cij
n
g (E −E )/ kT
= j = j e i j
C ji
ni
gi
(57)
This relation only depends on atomic coefficients. Thus equation (57) must always apply!
Let us compare Radiative Decay vs. Collisional De-excitation
Balanced when
Aji = ne Cji
i.e. there exists a critical density ne=ncrit given by ncrit = Aji/Cji
(58)
For n > ncrit then collisions dominate and the level populations are given by the thermal
distribution for the temperature T.
For n < ncrit radiative decay occurs first, whenever an atom is collisionally excited. The level
populations are then sub-thermally excited.
The value of the density, in relation to the critical density, then determines the cooling rate of
the gas, as we illustrate below.
Examples for critical density:
n crit ≈ 1.6 10 4 , 3.1 10 3 cm−3
for [OII] 2 D3/2 , 2 D5/2
≈ 7.0 10 5 , 3.8 10 3 , 1.7 10 3 cm-3 for [OIII] 1D2 , 3 P2 , 3 P1 (T ~ 10 4 K)
Cooling Rate Cooling Rate is given by L ∝ n e n i < σvkT >
(energy per unit time per unit volume that is lost from the gas)
ne, ni are the number density of the colliding partners
<σv> is the volume swept out per unit time by an electron
<kT> is the energy lost per collision due to cooling (thermal energy transferred to the atom
which then radiates away).
There are two cases to consider:
Case (i)
n<< ncrit
Then for every collisional excitation we have a radiative decay
Thus Cooling ∝ Collisional Excitation Rate
i.e.
Lij = n e n i Cij (Te ) (E j − E i )
(59)
Then, in the low density limit virtually all the ions are in the ground state, so ni ~ nI
Let
For O:
nI = yI nz where yI fractional ionization,
nz number density of nuclei of element Z
nz = az n
az abundance of element Z relative to density of H (with n = ne)
az ≅ 6 ×10 -4
Then:
L
o+
(
2
D5 / 2 + 2D3 / 2 ) ≅ 1.1×10 -33 y
o+
⎛ −3.89 ×10 4 ⎞
n2
3
exp
⎜
⎟ J/m /s
Te1/ 2
T
⎝
⎠
e
(60)
Here the ΔT=38,900K comes from ΔE = kΔT = hν = hc / λ for λ = 3726/29Å
Example: Assume all cooling through O+ and y
o+
≈1
Balancing cooling (L) through O+ with heating (G) through photoionization:
L
o+
= G = N ionization Q = N recombination Q
= n 2 α B kT* with α B = 2 × 10 −10 Te−3 / 4 from eqn. (34)
Here Q is the energy injected per ionization and T* is the effective stellar temperature.
Thus
Te1/ 4 exp(−3.89 ×10 4 /Te ) = 2.5 ×10−6 T*
(exercise)
(61)
We then derive: (exercise):
T*
(K)
2 104
4 104
6 104
Te(O+)
(K)
7,450
8,500
9,300
Te (H only)
(K) (=2/3T* eqn. [42])
13,000
27,000
40,000
Including O++ lowers the temperature still further.
Note: a Thermostat is in operation, leading to small variation in Te despite a wide range in T*:
i.e. if T* ↑, so Nrec ↓ (∝1/T 0.75 ), so n H 0 ↓, so G ↓ which counteracts the increase in Q.
Case ii : n>> ncrit
Cooling ∝ population of excited level
= nj Aji (Ej - Ei)
⎧ E − E j ⎫
g
= j n i A ji exp⎨ i
⎬(E j − E i )
gi
⎩ kT ⎭
Diagram: Cooling Power Law with Density
Note dependence
(62)
on n for
on n2 for
n>> ncrit and
n<< ncrit
jν UL = n u AUL hνUL
Line Intensity
Where U is the upper, L is the lower level, and ν the frequency of the emitted photon.
Diagram: Line Intensity
Distribution of Heavy Elements Ionization Potentials (eV)
Species
I
II
III
H
13.6
He
24.6
54.4
N
14.5
29.6
47.4
O
13.6
35.1
54.9
C
11.3
24.3
47.9
Diagram:
Ionization Stratification
(63)
The Saha Equation (not examinable)
Gives the distribution of atomic species among various stages of ionization, when in thermal
equilibrium.
Generalised Boltzmann Law:
+
⎧ χ + 1/2me v 2 ⎫
g
dN x (v)
=
exp⎨− I
⎬
kT
go
Nx
⎩
⎭
where go
g
χi
v
(64)
statistical weight of species X
statistical weight of X+. ge
ionization potential
speed of particle
It can be shown (……..; see Rybicki and Lightman) that:
N (i)
j +1
N (i)
j
U j +1 (T) ⎛ 2πm e kT ⎞ 3 / 2
−χ
/ kT
ne = 2
⎟ exp j, j +1
⎜
2
⎠
U j (T) ⎝ h
(65)
linking the 2 ionization states j, (j+1) of element N(i) when in LTE, where
Uj is the partition function
χ j, j +1 = χ j +1 − χ j = the difference in ionization potentials.
Combine with
Conservation of Nuclei
∑N
(i)
j
Conservation of Charge n e =
= N (i)
∑∑ z
i
j
j
N
(i)
j
⎫
⎬
⎭
(66)
Radio Frequency Continuum of a Nebula Bremsstrahlung (free-free) radiation arises from the acceleration of electrons around positive
ions. Results in continuum emission.
emitted at Radio Frequencies
absorption within nebula important
Radiative Transfer:
dIυ
= ( jυ − α υ Iυ )
ds
and in thermodynamic equilibrium
Put T = Te :
Suppose
(i)
(ii)
(iii)
[eqn. 10]
jυ
= Bυ (T) Kirchoff's Law (eqn.18)
αυ
dIυ
= Bυ (Te ) − Iυ (eqn. 19) where dτ υ = α υ ds
dτ υ
Te constant,
τν is the optical depth across the nebula, and
Iν = 0 at τν = 0
Diagram: Radio Emission from an HII Region
We show below that:
Iν = Bν (Te )[1− e−τν ] :
dIν
∫ B (T ) − I
ν
e
∫ dτ
=
ν
ν
I
⇒ [−ln(Bν − Iν )] oν = τ ν − 0
⎛ Bν ⎞
∴ln⎜
⎟ = τ ν
⎝ Bν − Iν ⎠
∴ln[1− Iν /Bν ] = − τ ν
∴1− Iν /Bν = e−τν
∴ Iν = [1− e−τν ] Bν
(67)
Let us consider the two limiting forms:
⎤
(a) τ ν << 1 ⇒ Iν = Bν (Te ) τ ν ⎥
(b) τ ν >> 1 ⇒ Iν = Bν (Te ) ⎥
⎦
(68)
−2.1
1.35
2
For Bremsstrahlung radiation τ ν is given by Aν Te n e L
Where ε = ne2 L, the emission measure, measured in cm-6 pc
This is an observable quantity as we will see below.
Rayleigh-Jeans approximation (eqn. 20) ⇒ Bν (Te ) =
Hence
(69)
(70)
2 ν2
kTe
c2
Iν α ν −0.1 for τ ν << 1
Iν α ν 2 for τ ν >> 1
(71)
Diagram: Radio Continuum Spectrum of an HII Region
ν = ν o for τ ν o = 1. So put τ ν o = 1 in eqn. (69) to determine ν0, which then gives
emission measure, assuming that Te is known, by applying equation (70).
ε , the
If we can estimate L [e.g. ~ 1pc from the angular size of HII region], then we can derive ne,
the electron density. Since virtually all the gas is ionized this is also the density of the gas.

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