Download 3.8 Case study: 21 cm line in the interstellar medium

3.8
Case study: 21 cm line in the interstellar medium
Due to the coupling between the magnetic moments of the electron and the proton, the basic state
of the hydrogen atom splits into two states (the hyperfine structure) with the energy difference
∆E ∼ µe µp /a3H . Substituting µe = e~/2me c and µp = 2.8e~/mp c yields
∆E ∼ 1.4α4
me
me c2 ∼ 10−6 eV.
mp
(3.56)
The corresponding photon wavelength is ∼ 1 m. A precise quantum calculation gives the wavelength of radiation emitted in the hyperfine transition of hydrogen to be λ = 21 cm; the frequency
is ν = 1400 MHz. Observations of this line is one of the basic tools for studying neutral hydrogen
in the interstellar medium.
The classical analog of this emission is the radiation of the electron magnetic moment precessing
in the magnetic field of the nucleus. The radiation power could be estimated as
P =
2 µ̈2
2 ω 4 µ2
=
.
3 c3
3 c3
(3.57)
This yields an estimate for the Einstein coefficient
A=
P
2 ω 3 e2 ~
π λ2C c
=
=
α
= 2 · 10−15 s−1 .
~ω
3 c3 4m2e c2
3 λ3
(3.58)
A precise quantum calculation yields A = 2.9 · 10−15 s−1 .
The absorption cross section is now obtained from eq. (3.50), where one has to substitute
gi /gi0 = 3. The radiation transfer equation in the line is given by eq. (3.54). In the radio band,
one can conveniently use the brightness temperature (2.12) instead of the intensity. Moreover, one
can conveniently characterize population of the levels by the spin temperature:
g2 −hν/kTs
hν
N2
= e
=3 1−
.
(3.59)
N1
g1
kTs
Then the radiation transfer equation is reduced to a simple form
where
dTB
= κν (Ts − TB ),
ds
(3.60)
1
hν
.
κν = σν N1 − N2 = σν N1
3
kTs
(3.61)
Note that the absorption coefficient, κν , is not equal to σν N1 ; this is because in the Rayleigh-Jeans
range, the absorption and induced emission nearly cancel each other (see sect. 3.3).
The spin temperature is determined by radiative transitions between the hyperfine structure
levels as well as by collisions between the atoms. The master equation (3.16) is straightforwardly
generalized to take into account collisional transitions:
dN1
= A [N2 (1 + n) − 3nN1 ] + C21 N2 − C12 N1 .
dt
16
(3.62)
where Cii0 Ni is the rate of the collisional transitions i → i0 . The coefficients Cii0 depend on the
density and the temperature of the gas. In this context, the temperature of the gas is called the
kinetic temperature.
In the thermal equilibrium, the spin temperature must be equal to the kinetic temperature,
which yields the relation
hν
hν
C12
= 3 exp −
=3 1−
;
(3.63)
C21
kTk
kTk
Now one can write the master equation via the spin, kinetic and brightness temperatures as
dN1
hν
hν
TB
= 3N1 A 1 −
+ C21
−
.
(3.64)
dt
Ts
kTk kTs
In the steady state, the lhs vanishes, which yields finally
B
1 + CA21 kT
Ts
hν
=
;
k
Tk
1 + CA21 kT
hν
(3.65)
The collisional cross section is roughly equal to the geometric cross section therefore one can
estimate the collision rate coefficient as C12 ∼ π(2aH )2 vT N .
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4.1
Ionization and recombination
Saha’s equation
Let us consider the ionization state of an ion with the charge Z in thermal equilibrium. In this
case, the populations of any two energy levels are related by the Boltzmann formula
gn −(En −Em )/kT
Nn
=
e
;
Nm
gm
(4.1)
where gi is the statistical weight of the level i. In the continuum, the number of states is ∼
V (me vT /h)3 ∼ V (me kT )3/2 /h3 therefore the ratio of densities of free and bound electrons (the
last is equal to the density of ions in the Z state of ionization) is estimated as
(me kT )3/2 −IZ /kT
Ne
∼
Ve
.
NZ
gZ h3
(4.2)
Rigorous consideration shows that V should be chosen to be the volume per an ion in the Z + 1th
−1
, and provides the exact coefficient thus yielding Saha’s equation
state of ionization, V = NZ+1
Ne NZ+1
2gZ+1 (2πme kT )3/2 −IZ /kT
=
e
.
NZ
gZ
h3
(4.3)
Since the number of the states in the continuum, (2πme kT )3/2 /(N h3 ), is typically very large,
ionization occurs not at T ∼ IZ but at significantly lower temperatures, T ∼ IZ /10. Since the
energy of the first excited level is typically of the order of the ionization energy, the population
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of the excited levels becomes comparable with the population of the basic state only at T ∼ IH ,
when the gas is practically totally ionized. Therefore the population of excited levels is small in
any case. Specifically for hydrogen, one can take N1 = Ne = N − N0 , where N is the total number
density of atoms. Introducing the degree of ionization α = Ne /N , one can write Saha’s equation
in the form
(2πme kT )3/2 −IH /kT
α2
=
e
.
(4.4)
1−α
h3 N
4.2
Radiative recombination and ionization (bound-free transitions)
An electron radiates when it passes by an ion therefore if it looses enough energy, it could remain
bound to the ion. This process is called the radiative recombination. In this case, we are interested
in relative close encounters such that the particle approaches the ion at the minimal distance
r0 b; which implies Ze2 /b me v 2 . Here we consider ions of an arbitrary charge Ze. This
distance, as well as the corresponding velocity, v0 , could be found from the conservation of energy
and angular momentum,
me v 2
me v02 Ze2
−
=
;
v0 r0 = vb.
(4.5)
2
r0
2
It follows from r0 b that v0 v; then one easily finds
r0 =
me v 2 b2
;
2Ze2
v0 =
2Ze2
.
me vb
(4.6)
When the electron closely approaches the ion, it radiates at the frequencies of the order of
ν∼
v0
2Z 2 e4
.
=
2πr0
πm2e v 3 b3
(4.7)
At the quantum language, one can say that the electron emits a photon of the frequency ν and
falls onto a bound level. The energy conservation implies
me v 2 Z 2 IH
+
= hν.
2
n2
(4.8)
We used the Rydberg formula for the energy levels because our quasiclassical approach is applicable
to the excited states, which are hydrogen-like. The electrons with the impact parameters from b
to b + δb fall onto the levels with the quantum numbers from n to n + δn. The relation between
δb and δn is obtained by differentiating Eq. (4.8) with account of (4.7):
IH
3Z 2 he4
δb.
δn
=
n3
πm2e v 3 b4
(4.9)
The electron radiates with the rate (3.17) during the time of the closest approach, τ ∼ 2r0 /v0 ∼
1/ν, therefore the probability of the radiative recombination is estimated as
w ∼ Aτ ∼
4π v0 2
α
.
3
c
18
(4.10)
The recombination rate onto the levels from n to n + δn is given by the electron flux through
the area 2πbdb multiplied by the recombination probability w. Therefore the recombination cross
section may be presented as
db
(4.11)
σnrec = 2πb w.
dn
Making use of Eqs. (4.6), (4.9) and (4.10), one gets finally
1
(nhνn )2
128π 4 Z 4 e10
8 σT
√
√
.
(4.12)
=
3 3 me c3 h4 v 2 ν n3
3 α me v 2 (hνn + me v 2 /2)
√
Here the correct coefficient 128π 4 /3 3 is written whereas our simple estimate gives 64π 4 /9. The
ionization threshold from the level n is νn = IH Z 2 /hn2 . Even though the expression (4.12) was
obtained in quasi-classical regime, n 1, it provides a good estimate even for the recombination
onto the basic state, n = 1.
The recombination rate of electrons with the velocity v onto the level n of hydrogen atoms in
the unit volume is written as
Z ∞
dNn
= βn N+ Ne ;
βn =
vσnrec fe (v)dv,
(4.13)
dt
0
σnrec =
where fe (v) is the electron distribution function. The averaging over the Maxwell distribution
could be done approximately just by substituting the thermal electron velocity for v. This yields,
with account of (4.8)
r
IH Z 2
σT c
.
(4.14)
βn ≈
2
kT
(1 + IkT Zn 2 )n
H
ion
is the radiation
The cross-section for photo-ionization from the level n is defined such that Iν σν,n
energy absorbed by the atom in the state n per unit frequency and unit time. Making use of the
detailed balance principle, one can express this cross-section via the recombination cross-section
(4.12). In the thermodynamical equilibrium, the recombination and ionization rates are equal for
any electron velocity,
Iν ion
N+ Ne σnrec vfe (v)(1 + nν )dv = 4π σν,n
Nn dν.
(4.15)
hν
Here v and ν are related by the energy conservation (4.8). Substituting the Plank distribution for
Iν and nν , the Maxwell distribution for fe ,
m 3/2
me v 2
e
2
exp −
,
(4.16)
fe (v) = 4πv
2πkT
2kT
and making use of Saha’s equation, one gets the so called Milne relation
ion
σν,n
=
g+ m2e v 2 c2 rec
σ .
gn h2 ν 2 n
(4.17)
For the hydrogen-like ions, gn = 2n2 g+ . Then one can write
8 σT n νn 3
64π 4 Z 4 e10 me
ion
√
= √
σν,n
=
3 3 ch6 ν 3 n5
3 α3 Z 2 ν
19
(4.18)
For ionization from the ground state of the hydrogen atom, this yields σ ≈ 10−17 cm2 near the
threshold.
The total photo-ionization rate is presented as
Z ∞
Iν bf
dNZ+1
= αNZ ;
α=
σZ dνdΩ.
(4.19)
dt
IH /h hν
In star’s interiors, where the temperature significantly exceeds the ionization energy of hydrogen, T IH /k = 1.6 · 105 K, the bound-free absorption occurs mostly on ions of heavy elements.
Formally, the absorption coefficient may be found by summing over all elements in all ionization
states at the given temperature:
X
=
σZbf NA,Z
(4.20)
κbf
ν
A,Z
where NA,Z,n is the number density of atoms of the element A in the ionization state Z. Taking
into account that the most abundant heavy elements (C, O, Si etc.) are fully ionized at T & 106
K, one can assume that the absorption is mostly determined by the hydrogen-like ions of heavy
elements. According to Saha’s equation (4.3), the number density of these ions is proportional to
Ne NA , where Ne is the free electron density, NA the density of heavy nuclei. Taking into account
that for hydrogen, A/Z = 1 whereas of other elements, A/Z = 2, the electron density is presented
as Ne = 21 (1 + x)ρ/mp , where x is the hydrogen abundance by mass. Then the Rosseland mean for
2
the bound-free absorption may be presented as κbf
R = (1 + x)zρ f (T ), where z is the abundance
by mass of heavy elements (a few per cent for typical astrophysical plasmas). The widely used fit
is (cf. (3.39))
ρ2
25
cm−1 .
(4.21)
κbf
=
4
·
10
z(1
+
x)
R
T 3.5
4.3
HII regions
Very hot, luminous stars emit enough photons with the energy > 13.6 eV to ionize the surrounding
interstellar gas; then an HII region is formed. Since the surface temperature of even the hottest
stars is well below IH /k = 160, 000 K, one can estimate the number of ionizing photons emitted
per second from the Wien approximation
2
Z ∞
2
2πν 2 −hν/kT∗
πIH R∗
kT∗ −IH /kT∗
15 IH
L∗ −IH /kT∗
2
e
dν
=
8
e
=
e
.
(4.22)
Ṅ = 4πR∗
2
3
4
c
h
π (kT∗ )3
IH /h c
For a star with the temperature 4 · 104 K and the luminosity 1039 erg/s, one gets Ṅ ≈ 1049 s−1 .
The radius of the ionized region (the Strömgren radius) is determined from the balance of
ionization and recombination,
4
Ṅ = πRS3 βN+ Ne .
(4.23)
3
Where β is the net recombination rate. Within the HII zone, the hydrogen is totally ionized
therefore N+ = Ne = N . Then eq. (4.23) yields
RS =
3 Ṅ
4π N 2 β
20
!1/3
.
(4.24)
The recombination onto the ground level is accompanied by the emission of a photon hν > IH ,
which immediately ionizes another atom, therefore the net recombination rate is determined by
the recombination on the excited levels,
β=
∞
X
βn .
(4.25)
n=2
Taking into account that βn decreases as 1/n at n < n∗ ≡ IH /kT and as 1/n3 at n < n∗ , one
roughly gets
r
r
n∗
X
IH
n∗
IH
ln
∼ σT c
.
(4.26)
β≈
βn ≈ σ T c
kT
2
kT
2
Substituting this estimate into eq. (4.24) yields RS = 60N −2/3 pc.
The net recombination is accompanied by the emission of photons with the total energy IH .
Therefore the total power of the recombination emission per unit volume is estimated as
r
IH
IH N 2 .
(4.27)
J = βIH N 2 ∼ σT c
kT
Comparing with eq. (3.36), one sees that the radiation cooling of hot plasmas is dominated by
the recombination at kT < IH and by the free-free emission at kT > IH .
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