Stellar StructureRadiative Transfer
Amount of outgoing radiation energy passing per second along a bundle of light rays from a small area of the emitting body …
In stellar photosphere, with a spherical symmetry, the intensity at a position is a function of from the stellar surface, in the direction of angle , and normal to the gas layer.
2
Radiation of frequency between and Δ at point normal to the small area in the direction around ′
Radiation from each point of the area into the solid angle about ′
Envelope of the cones for a circular area
Bundles of radiation rays passing through into the solid angle about ′
3
Specific Intensity or simply “intensity”, or “brightness”, is the amount of radiation energy per unit frequency interval at per unit time interval per unit area per unit solid angle passing into the specified direction at a position .
= lim Δ→0 Δt→0 Δ→0 Δ→0
Δ
Δ Δt Δ Δ cos
In cgs unit, [ergs s−1 cm−2 sr−1 Hz−1]
Because Δ → 0, the energy does not diverge. The intensity is independent of the distance from the source (i.e., light ray).
4
[area/radius2]= Τm2 m2, dimensionless The entire sky = 4
1 sr = rad2 =
https://upload.wikimedia.org/wikipedia/commons/9/9b/Steradian_V2.svg
5
In general, the outward or inward radiation flux passing through Δ in the direction (projection) is
ν + = outward
hemisphere , cos d, = 0, Τ 2
or
ν − = − inward
hemisphere , cos d, = Τ 2 to
So the net flux is ν + − ν
− = sphere , cos d
Outward flux diminishes with increasing distance from the source, because the maximum solid angle decreases.
6
ω = sin
An integral over a sphere is then
=0
ν + = outward
= =0
2 =0
Τ 2 cos sin = 2 [
1
= 7
= cos = (, , , , )
Mean Intensity
= 1
1
Total Radiation = 0
∞
Flux = cos [ergs s−1cm−2 Hz−1] Total Flux =
Energy Density
= 1
Total Energy Density = = 4 = 4B

Estimate (1) the bolometric flux (2) bolometric intensity, and
(3) mean bolometric intensity, emitted at the solar surface.
What is the bolometric flux received from the Sun at the
Earth? Compare this with the “solar constant”.
Novotny
flux/magnitude
• A galaxy or the central part of a globular cluster
an extended source
e.g., 18.2 mag/sq arcsec
In a dark site, sky~20-21 mag/sq arcsec
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• Intensity: Flux per solid angle
• Radiance: For an extended source; flux per solid angle per projected area
• Candela (cd): Intensity/brightness in a specific direction
• Lumen (lm): For visible light; luminous flux; 1 lm = 1 cd sr
• Lux (lx): Illuminance for visible light; lumen per area;
1 lx= 1 lm m−2 = 1 cd sr m−2 = 10 −14.18−mV
2.5 ; i.e., 0 mag = 2.06 lx
• (Spectral) irradiance: [W m−2 nm−1] Total solar irradiance = 1361~1362 W m−2
• Jansky: spectral irradiance;
1 Jy = 10−26 W m−2 Hz−1 = 10−26 kg s−2; 1 Jy = 10 23.9−AB
2.5
×Watt: Historically the power assumption of incandescent lightbulb; no longer used for brightness
http://www.stjarnhimlen.se/comp/radfaq.html 11
Radiation Pressure
Each quantum of energy, = , there is associated a momentum Τ
Radiation pressure net rate of momentum transfer (cf. gas pressure)
Radiation passing per second, through a unit area, at an angle with the normal, in a solid angle , is cos → Momentum transfer = ( cos /) cos
normal to the surface ∴ = 2
cos2 d
For isotropic radiation, = 4
3 = Τ 3 Τ4 3
projection of the area
13
Exercise
In the solar photosphere, = 5700 K, ≈ 2.6 dynes cm−2, gas ≈ 8000 dynes cm−2
In the core of the Sun, T = 1.5 × 107K, ≈ 1014 dynes cm−2, gas ≈ 2 × 1017 dynes cm−2
Radiation pressure plays a very minor role.
14
=
For a sphere of uniform brightness an isotropic source, =
Flux on the surface

2 0
= 1 − cos2 = sin2 =

2
= B
2
1
Energy density , = 4
=
8
Τ −1
Total Energy = , , = 4 (Stefan-Boltzmann law)
In terms of wavelength,
5
1
|d | =c
→ max ≈ 2900 [μmK] (Wien’s displacement law)
Solar photosphere, ~6000 K max~0.5 μm = 500 nm (visible)
Solar corona, ~ 106 K max~3 × 10−3 μm = 3 nm (X rays)
Dark clouds, ~ 20 K max~150 μm (FIR)
Note max max ≠ max
≈ 5.88 × 1010 [Hz K−1] (Wien’s displacement law)
19
When Τ 1, (low freq. or high temperature, valid in all radio regimes in astronomical situations)
T ≈ 2 3
2

2 Τ− (Wien approximation)
Because ∝ T, in radio astronomy brightness temperature Also antenna temperature, noise temperature … even if the radiation is not thermal. 20
‘Rayleigh-Jeans catastrophe’
Brightness Temperature ≡
The temperature with which, at a certain frequency, the intensity of a source equals to that of the blackbody. is a nonlinear function of intensity.
≡
In the Rayleigh-Jeans regime, i.e., in radio astronomy, = 2
22
A convenient unit [K] rather than [erg s−1cm−2sr−1 Hz−1 ]
Antenna Temperature = 2
Other temperatures related to the Planck spectrum …
Color Temperature ↔ shape (slope between two freq.)
Effective Temperature The total flux of a source equals to that of a
blackbody of the temperature
= cos ≡ eff 4
For a star (spherically symmetric), luminosity = 4 ∗
2 eff 4
1. What is the surface temperature of the Moon?
2. What is the color temperature of the Moon? Of the Sun?
Ciocca & Wang (2013)
26
Probability Distribution = the distribution of speeds for a gas at a certain temperature (Maxwell-Boltzmann distribution)


2
3 =
2
= 1
28
Doppler shift a distribution of the line of sight velocities
29
This is the fraction of particles in the speed interval , + d in the line of sight (radial) component
d
most probable speed = max of : highest probability
= 2
∞
∞ 2
Exercise
1. How fast does a proton typically move in the solar
photosphere? How about an electron?
2. How about in an H II region?
3. How fast does a hydrogen molecule move in a typical
interstellar dark cloud?
32
33
The 3-sigma rule: 68-95-99.7; e.g., a 3-σ outlier is 3 chances out of 1000 in a normal distribution.
68.27, 95.45, 99.73%
= 1
2 exp −
− 0 2
Full width at half maximum
34
= =
: number of moles R: ideal gas constant = 8.314 J/(K mol) : Boltzmann constant = = Τ
: Avogardro constant= 6.02 × 1023 mol−1
: number of total particles : volume number density
A combination of 3 laws Boyle’s law ∝ Τ1 Charles’s law ∝ T Avogadro’s law ∝
= =

This equation is valid if interaction is negligible, i.e., if density is low OK in normal stars in the low-density upper layers or even in the deep hot regions. 35
Mean molecular weight (per particle) = Τ (average mass per particle, in unit of amu)
In a fully ionized gas (e.g., in stellar interior), = 1/2 (H) … 2 particles per H 2 = particles of H mass
= 4/3 (He) … 3 particles per 4 H
≈ 2 (metals) … particles per 2 H
1
[
particles
= Τ4 (6 + + 2) for a fully ionized gas
For the solar composition, = 0.747, = 0.236, = 0.017 ≈ 0.6
Recent revision = 0.0152 (Caffau+11) 36
Mean molecular weight per electron
Sometimes is used = equivalent H per electron, e.g., when electrons provide the main gas pressure in the degenerate state.
=
1+
Note that and must be continuous as a function of depth inside a star, but and therefore can be discontinuous.
37
Exercise
(1) an H I cloud;
(2) an all He gas (completely ionized versus neutral);
(3) a molecular cloud;
38
: heat capacity ratio = adiabatic index = Laplace’s coefficient = isentropic (adiabatic and reversible) expansion factor
40
2
for the translational (3-d) kinetic energy of an ideal monatomic gas.
For a diatomic gas av = 0
∞ 2
Bradt
3 rms =
+2
3 rms
e.g., dry air, = 1.403 (0 o C), = 1.400 (20
o C)
o C)
o C)
Heat capacity: heat supplied to increase one degree in temperature; and [joule K−1] = ΤΔ Δ
Specific heat capacity (=per unit mass), (at constant pressure) or (at constant volume)
Gas Thermodynamics
42
43
Note
When a gas cell rises in the atmosphere, it expands and cools almost adiabatically, i.e., with little heat exchange (air is a poor heat conductor), but keeps the same pressure and temperature, balanced with the surroundings.
Boltzmann Formula/Equation
Population ratio between two excited states (of an -times ionized species) 2
1 =
2
1 − Τ21
: number density of the particles in the -th energy state
: statistical weight of the -th energy state = degeneracy of the level = number of states with different quantum
numbers but with the same energy
21: difference in excitation energies (wrt to ground state) =
In general, it really should be or
for -times ionization.
45
When Zeeman splitting (by magnetic field) is neglected, all projections of the angular momentum are degenerate in energy, so
= 2 + 1 for -times ionization.
: angular momentum of the state
For the hydrogen atom (a pure Τ1 potential)
= − 13.6
where = 1, 2, 3, … is the principal quantum number.
For rotating linear molecules (e.g., CO) have = 2 + 1, where = 0, 1, 2, … is the angular momentum quantum number.
For each , there are 2 + 1 possible z-components, = −, − − 1 , … 0, 1, … − 1 ,
46
Exercise
Calculate the wavelength of the H line (transition between =3 and =2 levels).
Adopting a solar photospheric temperature of = 5800 K, what is the ratio of the population of the first excited state to the ground state of H? To all states?
For the H atom, the ground “state” has two quantum states of the same energy −13.6 eV, 1 = 2.
For the first excited states 2 = 8.
47
+1
2 Τ3 2
3 − Τ
: number density of the particles in the -th ionized state : number density of free electrons
, : partition functions of the ionized species, and of the electron = sum of the statistical weights of all bound states, each weighted by the Boltzmann factor
= σ , − , very often 1dominates; = 2
: ionization potential from the ionization stage to + 1 48
Clayton
Numerically,
[K] and [eV]
For hydrogen, neutral 1 ≈ 1 = 2 (i.e., most of the neutral H is in the ground state), and ionized 2 = 1 (just the proton). For neutral ground state of H, =13.6 eV
Sometimes the electron pressure is used instead of the number density , via = , for which ≈ 1 dyne cm−2
for cool stars, and ≈ 1000 dyne cm−2 for hotter stars. Carroll & Ostlie
49
Numerically,
+1 2
1 Pa = 1 N
cm2
50
51
“Metals” supply plenty electrons. Given a temperature, e.g., calcium ( = 6.11 eV) will lose electrons at a rate greater than will hydrogen ( = 13.6 eV).
The recapture (recombination) rate of ions depends on the electron density.
I II III IV
H 13.60
Ionization energies 52
Heated gas of density and volume =dA ds
Radiation in all directions, = ′
′ : the mass emission coefficient of the material
≡ ′ [erg cm−3 s−1 Hz−1 sr−1 ] is the monochromatic
emission coefficient = emission coefficient
is directional =
For isotropic emission, the emissivity is [erg g−1 s−1 Hz−1] = Τ4 55
Consider radiation through a slab of thickness , the intensity is reduced by an amount
= − ′
Define = ′ cm−1 , where is material density,
′ cm2 gm−1 : mass absorption coefficient (opacity coefficient)
cm−1 : absorption coefficient
= − ………….. (1)
Dividing (1) by and integrating ln = − + const
= 0 −
0 is the incident beam
s
we get =
0 −
=0.1, 90.5%
=0.5, 60.6%
=1.0, 36.8%
determines the fraction of the intensity from that layer that reaches the surface; e.g., from a layer of =2, a fraction of −2 ≈ 0.14 reaches the surface.
The apparent “surface” of a star (photosphere) ≈ 1, Τ1 = 37% radiation emerges from there.
57
Optical thickness: 1 optically thick = opaque 1 optically thin = transparent
= Energy emitted
= Energy absorbed (need something to absorb from
When abs and
sca are independent of , the opacities are gray.
Why is the sky blue? Why is a cloudy sky gray?
≈ 1 “surface”
Radiative Transfer Equation … governs how specific intensity varies with emission and absorption by a medium
= − +
If there is scattering radiation in and out of the solid angle an integrodifferential equation solution is complicated
= 0
is the source function.
This equation is used more often, because is a simpler function of physical quantities, and is more intuitive (dimensionless). 60
(2) = 0 (absorption only)
= 0 exp − 0
′ ′
integrated along the line of sight.
(1) = 0 (emission only)
= 0 + 0
′ ′
integrated along the line of sight.
62

= 0 − + 0

"
If = const (not valid in ISM but OK in
stellar atmosphere)
1 − −
LTE → T = 2 3
2
1
Τ − 1,
where ≡ Τc2 2 3 (dimensionless) is called the photon occupation number = number of photons per mode of polarization
Thermodynamic equilibrium = no net flows of matter or of energy into a system
Two systems in thermal equilibrium when T temperature the same
Two systems in mechanical equilibrium when P pressure the same
Two systems in diffusive equilibrium when chemical potentials the same
64

= − + In LTE, Τ = 0 → = Τ
and =
= (Planck-Kirchhoff law) cf Kirchhoff ’s circuit law
Kirchhoff: Τjν κν in TE depends only on (Planck law not unknown yet) good absorbers are also good emitters. Valid only for thermal emission; e.g., not for scattering
Finally, the solution is = 0 − + 1 − − , Under the assumptions of (1) LTE, and (2) = const
65
66
In the Rayleigh-Jeans regime, ∝ , and ∝T
= 0 − + 1 − −
If background is zero, i.e., 0 = 0, and dropping ,
(i) 1 → (measuring only the “surface”)
(ii) 1 → (measuring the entire medium)
= 0 − + 1 − −
67
What we actually measure is the flux density,
= source [erg s−1 cm−2 Hz−1] cf. Jansky
Integrating over the solid angle subtended by the source,
= source
1 − − ≈ 1 − −
Point sources Jy
Extended sources Jy sr−1
= 0 − + 1 − −
68
69
• Spontaneous emission
occurrence rate density of incoming photons of the
same , polarization, and direction of propagation
• Collisional deexcitation no emission of photons 73
Einstein Coefficients
Einstein (1917)
(induced) emission (Stimulated) absorption
2 1 + 1 + 22 + 1 + 2 = 2 − 1 /
2 21: # of spontaneous radiative transitions during
or then unit different
2 21 or 1 12 : # of (stimulated) or radiative transitions during when irradiated with
75
https://einstein.manhattanrarebooks.com/pages/books/17 /albert-einstein/zur-quantentheorie-der-strahlung-on-the- quantum-theory-of-radiation
https://einstein.manhattanrarebooks.com/pages/books/17/albert-einstein/zur-quantentheorie-der-strahlung-on-the-quantum-theory-of-radiation
Transition Probability
Considering a 2-level system, we calculate the emission arising from the transition:
[erg s−1 cm−3 sr−1 Hz−1]
Δ = 0
EmissionAbsorption
2
1
= [erg s−1cm−3 sr−1] volume emissivity
×
00
76
Once an atom is excited, there is a finite probability within t, 2,1 to jump spontaneously from level 2 to level 1 (deexcitation).
The total number of downward transitions 2 → 1 is 2 2,1 , where 2 is the number of atoms (population) in level 2 per unit volume.
[−]: Einstein coefficient for spontaneous transition = probability per unit time
Τ1 21 [s]: lifetime staying in level 2 (i.e., − remaining excited)
= 0
Allowed transitions (via an electric dipole) satisfying selection rules
1. Parity change 2. L = 0, ±1, = 0 → 0 forbidden 3. Δ = 0, ±1, = 0 → 0 forbidden 4. Only one electron with = ±1 5. = 0 (Spin not changed)
A forbidden transition is one that fails to fulfill at least one of the selection rules 1 to 4. It may arise from a magnetic dipole or an electric quadrupole transition.
Spectroscopic Notation Ionization State
II --- singly ionized atom, e.g., H II H+
III – doubly ionized atom, e.g., O III O++
….. and so on….e.g., Fe XXIII Peculiar Spectra
e (emission lines), p (peculiar, affected by magnetic fields), m (anomalous metal abundances), e.g., B5 Ve
• Forbidden Lines (a pair of square brackets), e.g., [O III], [N II]
• Semi-forbidden Lines (a single bracket), e.g., [OII
• Allowed (regular) Lines (no bracket), e.g., C IV 79
Some examples:
[O III] 21 = 0.021 s−1, 21 = 5007
21 = 0.0281 s−1, 21 = 4959
32 = 1.60 s−1, 32 = 4364
[S II] 21 = 4.7 × 10−5 s−1, 21 = 6716
H I 21 cm hyperfine line 21 ≈ 10−15 s−1, the probability is extremely low.
80
4 × B
An accelerated charge emits a total power (Larmor formula)
Power emitted proportional to the square of the charge and square of the acceleration
Radiation max perpendicular to acceleration; none along the acceleration Radiation is polarized

: The Einstein coefficient for absorption transition = probability per unit time be excited from level 1
to 2 by absorption of a photon; is the energy density of the radiation field
There is the corresponding Einstein coefficient for stimulated emission, for which gives the transition probability of the stimulated emission, i.,e., a photon incident on am atom on the upper state induces a transition to the lower state, producing a new photon of the same frequency, before spontaneous transition takes place.
82
In TE, the number of downward transitions (per unit time per unit volume)
= the number of upward transitions, i.e.,
2 21 + 2 21 = 1 12
Solving for ,
= Τ21 21
Τ1 2 Τ12 21 − 1
But in TE, Τ1 2 is governed by Boltzmann equation, so
= Τ21 21
Τ1 2 exp Τ0 Τ12 21 − 1
Also in TE, = and varies little within , → 0 83
= 4
c
Comparing to the Planck function,
T = 2 3
2 21
In fact, these detailed balance relations are related to the atomic properties, so should be independent, i.e., regardless of TE or not.
(exponent must equal to the exponent)
84
To relate the absorption and emission coefficients and Einstein coefficients, recall
= 0
= 0

= −