Transcript Lecture 5

Lecture 5
•
•
•
•
Summary of galaxy classification
Example problems
HW review
Introduction to emission line spectroscopy:
review of atomic structure
Next topic: AGN
• Before we start this topic, need to review some basics on
some of the most fundamental probes of AGN: emission
lines
• AGN emit high energy photons with energies more than
sufficient to ionize hydrogen
• We will therefore concentrate on only this “phase” of the
ISM : HII regions
Useful References
Gaseous Nebulae:
•
•
•
Shu (pg 216)
Online notes
Osterbrock
AGN:
•
•
•
•
C&O Ch 26
Introduction to Active Galactic Nuclei by Peterson(first chapter available free
online; see reading references on class webpage)
Active Galactic Nuclei by Krolik (more advanced graduate-level textook)
Quasars and Active Galactic Nuclei - An Introduction , by A.K.Kembhavi and
J.V.Narlikar (slightly more advanced graduate-level textbook)
The Interstellar Medium
• Perfect vacuum exists nowhere in space
• This “stuff” between stars ~ 10-15% of visible mass of Galaxy
• 99% is gas, 1% dust
• So what? Why study it?
• It affects starlight, or other radiation sources
• Interesting physics in itself - probes chemical evolution and
starformation history of all galaxies. For our next topic, it is
very important
Physics of ISM
 rmean ~ 0.3 cm-3
 rmolecular cloud ~ 106 cm-3
 ratmosphere ~ 21018 cm-3
Many “phases” of ISM
Thermodynamic equilibrium is not generally the case.
Emission-line spectroscopy from
Ionized Gas
• Emission-lines are a powerful probe of the underlying
abundances, temperature, density, and ionization
states of AGN
• Rules governing transitions between different states
of an atom are governed by quantum mechanics
• Let us review some basic atomic physics
Spectroscopic notation
•
Solution of radial part of
Four quantum numbers describe any electron in an atom:
Schrodinger equation
– n= principal quantum number = 1, 2, 3 …
– l = orbital angular momentum quantum number = 0, 1, … n-1
Solution of
angular part of
– ml = magnetic quantum number = -l, -(l –1), … , 0, 1, …, (l –1), l ((2l+1 values)
Schrodinger
– ms= spin quantum number = +/- ½
equation
•
No two electrons can have the same quantum numbers. This is the “Pauli exclusion
principle” so there are 2n2 possible states for an electron with a given quantum number n
•
We can describe location of an electron as being in “orbitals”, where for historical reasons,
the terminology below is used:
l=
0
1
2
3
4
Letter
used
s
p
d
f
g
Spectroscopic notation
e.g., Oxygen has 8 electrons. What is its electronic configuration? What
about Neon which has 10 electrons?
Spectroscopic notation
•
•
For hydrogen, we can use the Bohr model of the atom and find that the
allowed energies for a hydrogen atom are:
– En = -13.6eV (1/n2)
And the frequency of a transition between two levels is:
–  = RH x (1/n2low - 1/n2high)
– RH = Rydberg’s constant = 109677.5 cm-1 for hydrogen
•
Similar relations work for other hydrogenic atoms (He II, Li III, etc …)
Spectroscopic notation
•
For an atom, we don’t specify the quantum numbers of every electron, instead we sum the individual
l together vectorially to find the total angular momentum:
– L=li
– L = 0 are called S states, L= 1 are called P states, L= 3 are D states, etc!
•
Similarly one sums the total electron spin vectorially:
– S =  msi
–
•
•
•
•
For an atom with an even number of electrons, S is an integer, with an odd number it’s +1/2,
+3/2, etc. S = 0 and L= 0 for a closed shell.
LS coupling assumption (good for light atoms) implies that the total angular momentum J is just the
(vector) sum of L+S
The magnetic quantum number M takes on values of J, J-1, …, 0, …-J-1, -J
An atomic level is described as: n2S+1LJ, where 2S+1 is called the “multiplicity”
(inner full shells sum to zero)
For a term with total angular momentum J, there are (2J + 1) separate states. Because the atom
doesn’t care how it is oriented in space, all these states are degenerate in energy (except in the
presence of a magnetic or electric field)
Write down the possible term symbols for Li, which has 3 electrons
Example Configuration:
Selection Rules
Not all transitions between levels occur with equal probability:
•
•
•
•
For several reasons (angular momentum must be conserved so since a photon takes away 1
unit of angular momentum), there are the following rules governing transitions:
– l = +/- 1
– L=-1,0,+1
– J=-1,0,+1, unless J=0, then J=+/-1
– S=0
– M=-1,0,1
Permitted lines are lines whose where the rules above are obeyed and the transition
probability is high; can radiate via dipole radiation
Forbidden lines are correspond to the case when the rules above are not obeyed. They can
still occur but with low transition probability (via 2-photon process or require quadrupole or
octopole transitions)
Forbidden lines are given a symbol of square brackets, e.g. [OI]
Multiplets
•
•
•
•
•
While in hydrogen the energy levels depend only on the quantum number n, in other atoms
transitions arising from a specific nL term (with a number of values of J) to another term can
give rise to multiplets
One example of a multiplet is the [OIII]4959,5007 multiplet which arises from the 1D2 to 3P2
and 3P1 levels
This is called fine structure
There is also hyperfine structure due to the alignment of the proton spin and the electron spin
which gives rise to the 21 cm hydrogen continuum line
Also Zeeman splitting in a magnetic field (which breaks the degeneracy of the separate J
states)
[OIII] Ground state lines
Transition Rates
•
•
•
•
•
•
For any possible transition between two states m and n, there exist Einstein transition
coefficients: Amn, Bmn, and Bnm
– Amn = spontaneous transition rate per unit time per atom
– Bmn = stimulated emission coefficient (a photon with an energy equivalent to the
transition from n to m can prompt – stimulate – an electron decay)
– Bnm = absorption coefficient
These coefficients are related:
– gmBmn = gnBnm where gm is the statistical weight of the level
– Amn = (h3/c2)( gn/gm) Bnm
Transitions with high A values (~108 per second) are “permitted”, if A= 108 per second, then
an electron will take, on average, 10-8 seconds to decay
Also note that the larger the energy difference of a transition, the higher the A value as A ~ 3
“Forbidden” transitions have low A values, A < 1 per second
On Earth, the density is so high that collisional de-excitation dominates. You can see allowed
transitions but forbidden lines are not typically seen. In the ISM, densities can be low enough
that forbidden transitions can actually occur before collisional de-excitation.
Populating Atomic Levels
• An electron may change levels for a number of reasons:
–
–
–
–
–
Photo-excitation from a lower level (very unimportant in the ISM)
Collisional excitation (must overcome the Boltzmann factor)
Collisional de-excitation from higher level
Cascade from a higher level
Recombination to an excited state
• In the ISM, where the density is extremely low, collisional
excitations (or de-excitations) are only important for
levels with no permitted decays
Cross section for photo-ionization
• Hydrogen requires 13.6 eV of energy to ionize, which
corresponds to =912 Å, = 3.29 x 1015 Hz. At this
energy, the atom’s cross section to photoionization is
A0=6.30 x 10-18 cm2.
• a= A0 (0/)3
• Extreme UV light with wavelengths just shortward of
912 Å are absorbed very rapidly by neutral hydrogen
in the interstellar medium
• At shorter wavelengths (say the soft x-ray) absorption
by the ISM is much less important, the ISM is
transparent in the hard x-ray and gamma-ray bands
Cross section for photo-ionization of hydrogen
Photo-ionization of heavy elements
Energies needed to ionize metals:
Element
II
III
IV
V
Hydrogen
13.6eV
Helium
24.6
54.4
Carbon
11.3
24.4
47.9
64.5
Nitrogen
14.5
29.6
47.4
77.7
Oxygen
13.6
35.1
54.9
77.4
Neon
21.6
41.0
63.5
97.1
Note that when hydrogen is ionized,
so is oxygen!
Recombination Coefficient
• Rate of recombination
a  Te-1/2
Ionization Balance
• The rate of photo-ionizations occurring in a nebula is
where  is the optical depth (i.e., the attenuation factor), and a is
the ionization cross-section.
Meanwhile, the rate of recombinations is
In a steady state nebula, the rate of photo-ionizations equals the
rate of recombinations, so
Ionization Balance
For the case of a constant density nebula, this simplifies. First,
multiply through by 4  r2
Now integrate both sides of the equation over radius. If R is the
radius of the ionized region, the
Ionization Balance
For a constant density nebula, integrating the right side of the
equation is easy:
For the left-hand side, we can get rid of the radius by noting that
the attenuation is due to the absorption of neutral hydrogen along
the path of the photon. In other words,
Ionization Balance
By definition, at the edge of the nebula, all the ionizing photons
have been absorbed, so the optical depth at that point is infinite.
So
The last integral is equal to one. Therefore, if we define Q(H0) as
the number of photons capable of ionizing hydrogen, i.e.,
we get
where we have assumed Ne = Np = N(H). This gives the radius of
the Strömgren Sphere.
Forbidden Lines from Ionized Gas: “2-Level Atom”
2
E21
1
Upward transitions can occur
through:
Downward transitions can occur
through:
• photoionization from a lower
level
• Spontaneous decay (rate
determined by A21)
• collisional excitation (inelastic
collisions with electrons)
• collisional de-excitation (elastic
collisions with electrons)
In equilibrium, upward transition rate from level 1
balances downard transition rate to level 1
This means:
Einstein A-coefficient
n e n 1 q12 (T)  n 2A21  n e n 2q 21(T)
[1]
Collisional de-excitation coefficient
Increases with ne
Increases with n1
Collisional excitation coefficient
The collisional coefficients are related through the Boltzmann factor
g 2 - EkT21
q12 (T)  q 21 (T) e
g1
Where g2, g1, are the statistical weights of the two levels (=2J+1)
[2]
The collisional de-excitation coefficient is given approximately by:
q 21  v 21
Where the thermal speed of the electrons is given by:
1
3
kT
2
m e v  kT ; v 
2
2
me
And an elastic cross section given approximately by
2
h
 12 
m e kT
q 21 
h2
3
2
m e kT 
1
2
A more detailed calculation yields:
8.63 10-6 12
q 21 
g2 T
Where 12 comes from quantum
mechanical calculations
Using Equations [1] and [2], we get that the level populations are related by:
n1n e q 21 (T) g 2 - EkT2 1
n2 
e
A 21  n e q 21 (T) g1
[3]
The limiting cases are:
n eq 21(T)  A21
g 2 - EkT2 1
n 2  n1 e
g1
Boltzmann distr.
A 21
n e 
 n critical
q 21 (T)
This defines the critical density for a transition; at densities above the critical
density, transitions from the upper to lower level are more likely to occur
through collisions rather than spontaneous decay.
How do we use this information?
The luminosity of a line (per unit volume) caused by a downward transition
from 2 -> 1 is:
L21  n 2 A 21hν
Using equation [3], this becomes:
n1n e q 21 (T)A 21h g 2 - EkT21
L 21 
e
A 21  n e q 21 (T) g1
At low densities:
g 2 - EkT2 1
L 21  n1n e q 21 (T)h 
e
 n2
g1
At high densities:
g 2 - EkT2 1
L 21  n1A 21h
e n
g1
The transition between these two regimes occurs at the critical density:
A 21
n e 
 n critical
q 21 (T)
This is a useful diagnostic - if we don't see certain lines of an ion, it may be
because the density is too high for their radiation to occur. This happens
when the mean time between collisions is comparable to or less than the
radiative lifetime A21-1. (This is why we don’t see forbidden lines on Earth)
For permitted transitions (electric dipole radiation), the transitions rates
occur at A ~ 108 sec-1 and corresponding critical densities of 1015 cm-3.
For forbidden lines, the A coefficients are much smaller so the critical
densities are much lower - comparable to densities in typical nebulae.
Critical densities for some common transitions:
Identification Wavelength (Å) log Ncr (cm -3) Identification Wavelength (Å) log Ncr (cm -3)
[C III
1909
9 [Fe VII]
5721.1
7.6
[O II]
3726.1
3.5 [N II]
5754.6
7.5
[O II]
3728.8
2.8 [Fe VII]
6086.9
7.6
[Fe VII]
3760.3
7.6 [O I]
6300.3
6.3
[Ne III]
3868.8
7 [S III]
6312.1
7.2
[Ne III]
3967.5
7 [Fe X]
6374.6
9.7
[S II]
4068.6
6.4 [N II]
6583.4
4.9
[O III]
4363.2
7.5 [S II]
6716.4
3.2
[Ar IV]
4711.3
4.4 [S II]
6730.8
3.6
[Ar IV]
4740
5.6 [Ar III]
7135.4
6.7
[O III]
5006.9
5.8 [O II]
7319.9
6.8
Appenzeller and Östreicher 1988 (AJ 95, 45) and Table 3.11 of Osterbrock
Many of these lines are seen in gaseous nebulae,
and the strength of the lines depends on the
density of the gas
So, summarizing, because these lines are excited primarily by collisions
with electrons, the observed line strengths can be used to measure the
density, temperature, and then composition of the gas!!
To measure density:
Use line ratios from lines with different critical densities (but are close
together in energy to eliminate temperature dependence ) one in the lowdensity limit (L a n2) and one in the high-density limit (L a n).
In this example, the [SII] at 6716Å
comes from a 4S3/2-2D5/2 transition
and has a critical density of 1.5 x
103 cm-3. The 6731Å line comes
from the 4S3/2-2D3/2 transition with
critical density 3.9 x 103 cm-3. Note
that the line ratio is sensitivite to
density between about 100 to
10,000 cm-3.
To measure temperature:
Again use single ion (independent of abundance and ionization) but pick
lines that have large differences in excitation potential.
The most-used example is [O III] with lines at 4363 and 4959+5007 Å.
Once these line ratios are used to constrain density and temperature,
abundances can be derived!
The Nature of the Ionizing Source:
Since the ions observed are ionized by photons, and the ionization
potential of metals varies, line ratios of lines from various ions can be
used to infer the spectral shape of the ionizing radiation field. Here
are the ionization potentials (eV) for a few common ions:
Element
H
He
C
N
O
Ne
I
II
13.6
24.6
11.3
14.5
13.6
21.6
III
54.4
24.4
29.6
35.1
41
IV
47.9
47.4
54.9
63.5
64.5
77.7
77.4
97.1
Since even the most massive (hottest) stars produce very few
photons beyond ~ 50 eV, the presence of lines from highly ionized
species indicated the presence of some source of hard photons other
than stars. What could this be?
Next topic:AGN