Post on 25-Dec-2015
Astronomical Constants
• CGS units used throughout lecture (cm,erg,s...)• AU = Astronomical Unit
= distance earth - sun = 1.49x1013 cm
• pc = Parsec = 3.26 lightyear = 3.09x1018 cm • ’’ = Arcsec = 4.8x10-6
radian• Definition: 1’’ at 1 pc = 1 AU
• M = Mass of sun = 1.99x1033 gram
• M = Mass of Earth = 5.97x1027 gram
• L = 3.85x1033 erg/s
Radiative transfer
Basic radiation quantity: intensity
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I(Ω,ν ) =erg
s cm2Hz ster
Definition of mean intensity
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J(ν ) =1
4πI(Ω,ν )dΩ
4π∫ =
erg
s cm2Hz
Definition of flux:
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F(ν ) =1
4πI(Ω,ν )ΩdΩ
4π∫ =
erg
s cm2Hz
Radiative transferPlanck function:
In dense isothermal medium, the radiation field is in thermodynamic equilibrium. The intensity of such an equilibrium radiation field is:
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Iν = Bν (T) ≡2hν 3 /c 2
[exp(hν /kT) −1](Planck function)
Wien Rayleigh-Jeans
In Rayleigh-Jeans limit (h<<kT) this becomes a power law:
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Iν = Bν (T) ≡2kTν 2
c 2€
2
Radiative transferBlackbody emission:
An opaque surface of a given temperature emits a flux according to the following formula:
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Fν = π Bν (T)
Integrated over all frequencies (i.e. total emitted energy):
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F ≡ Fν dν0
∞
∫ = π Bν (T)dν0
∞
∫If you work this out you get:
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F =σ T 4
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σ=5.67 ×10−5 erg/cm2/K4 /s
Radiative transfer
In vaccuum: intensity is constant along a ray
Example: a star
Non-vacuum: emission and absorption change intensity:
A B€
FA =rB2
rA2FB
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ΩA =rB2
rA2ΩB
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F = IΩ
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I = const
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dI
ds= ρ κ S − ρ κ I
Emission Extinction
(s is path length)
Radiative transfer
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dIνds
= ρ κ ν (Sν − Iν )
Radiative transfer equation again:
Over length scales larger than 1/ intensity I tends to approach source function S.
Photon mean free path:
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lfree,ν =1
ρ κ ν
Optical depth of a cloud of size L:
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τ =L
lfree,ν= Lρκ ν
In case of local thermodynamic equilibrium: S is Planck function:
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Sν = Bν (T)
Radiative transfer
Observed flux from single-temperature slab:
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Iνobs = Iν
0e−τν + (1− e−τν ) Bν (T)
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≈τ Bν (T)
for
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τ <<1
€
Iν0 = 0and
Radiative transfer
Emission/absorption lines:
Hot surface layer
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τ ≤1
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τ >>1
Flux
Cool surface layer
Flux
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Iνobs = Iν
0e−τν + (1− e−τν ) Bν (T)
Difficulty of dust radiative transfer• If temperature of dust is given (ignoring scattering for the
moment), then radiative transfer is a mere integral along a ray: i.e. easy.
• Problem: dust temperature is affected by radiation, even the radiation it emits itself.
• Therefore: must solve radiative transfer and thermal balance simultaneously.
• Difficulty: each point in cloud can heat (and receive heat from) each other point.
Rotating molecules
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≡Bh J(J +1)€
E rot =J 2
2IClassical case:
Quantum case:
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E rot =h2
2IJ(J +1)
I is the moment of inertia
J is the rotational quantum number:J = 0,1,2,3...
B has the dimension of frequency (Hertz)
Rotating molecules
Dipole radiative transition: JJ-1:
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ΔE = Bh J(J +1) − (J −1)J[ ]
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=2Bh J
Quadrupole radiative transition: JJ-2:
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ΔE = Bh J(J +1) − (J −2)(J −1)[ ]
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=2Bh (2J −1)
Transition energies linear in J
Rotating molecules
Carbon-monoxide (CO):
J = 10 =2.6 mmJ = 21 =1.3 mmJ = 32 =0.87 mm
CO: I = 1.46E-39
Ground based millimeter dishes. CO emission is brightest molecular rotational emission from space. Often used!
Plateau de Bure James Clerck Maxwell Telescope
Rotating molecules
Molecular hydrogen (H2)
H2: I = 4.7E-41 J = 20 =28 mJ = 31 =17 mJ = 42 =12 m
Need space telescopes (atmosphere not transparent). Emission is very weak. Only rarely detected.
Due to symmetry: only quadrupole transitions:
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Δ J = ±2
Spitzer Space Telescope
Rotating molecules
Molecules with 2 or 3 moments of inertia:
H
H
H
NH
H
O
“Symmetric top”: 2 different moments of inertia, e.g. NH3:
“Asymmetric top”: 3 different moments of inertia, e.g. H2O:
These molecules do not have only J, but also additional quantum numbers.
Water is notorious: very strong transitions + the presence of masers (both nasty for radiative transfer codes)
Rotating moleculesSymmetric top: NH3
H
H
H
N
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E rot ≡ Bh J(J +1) + (A − B)hK 2
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A ≅ 2.02 ×1011Hz
B ≅ 3.20 ×1011Hz
Quadrupole ΔK0 transitions are slow (10-9 s) but possible (along backbone)
backbone
Radiative J to J-1 transitions with ΔK=0 are rapid (~10-1...-2 s). But ΔK0 transitions are ‘forbidden’ (do not exist as dipole transitions).
After book by Stahler & Palla
Isotopes
• Molecules with atomic isotopes have slightly different moments of inertia, hence different line positions.
• Molecules with atoms with non-standard isotopes have lower abundance, hence lines are less optically thick.
• Examples: – [12CO]/[13CO] ~ 30...130– [13CO]/[C18O] ~ 10...40
Vibrating molecules: case of H2
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Evib ≡hν 0 v + 12( )
Atomic bonds are flexible: distance between atoms in a molecule can oscillate.
Vibrational frequency for H2:
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0 ≅1.247 ×1014 Hz
Vibrating molecule can also rotate. Sum of rot + vib energy:
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E tot = Evib + E rot
Selection rules for H2-rovib transitions: from v to any v’, but ΔJ=-2,0,2 (quadrupole transitions).Quadrupole transitions are weak: H2 difficult to detect...
Vibrating molecules: case of CO
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Evib = hν 0 v + 12( )
For CO same mechanism as for H2:
Vibrational frequency for CO:
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0 ≅ 6.4 ×1013 Hz
Often v=0,1,2 of importance
Selection rules for CO-rovib transitions: from v to any v’, but ΔJ = -1,0,1.
Δv=-1 is called fundamental 4.7 m
Δv=-2 is called first overtone 2.4 m
Vibrating molecules: case of CO
J to J-1: R-branch transitions: during vibrational transition also downward rotational transition: more energy release. Lines blueward of Q-lines, bluer for higher J.
J to J: Q-branch transitions: Pure vibrational transitions (no change in J). All lines roughly at same position.
Transitions from v to v-1 or v-2 etc:
J to J+1: P-branch transitions: during vibrational transition also upward rotational transition: less energy release. Lines redward of Q-lines, redder for higher J.
R Q P
Vibrating molecules: case of CO
• “Band head”:– Rotational moment of intertia for v=1 slightly larger than
for v=0– Therefore rotational energy levels of v=1 slightly less
than for v=0– R branch (blue branch): distance between lines
decreases for increasing J. Eventually lines reach minimum wavelength (band head) and go back to longer wavelengths.
Calvet et al. 1991
2.294 2.302 [m]
Band head CO first overtone 2-0
Photodissociation of molecules
First and second electronic excited state
Electronic ground state
Ultraviolet photons can excite an atom in the molecule to a higher electronic state.
The decay of this state can release energy in vibrational continuum, which destroys the molecule.
Formation of molecules: H2
Due to low radiative efficiency, H+H cannot form H2 in gas phase (energy cannot be lost). Main formation process is on dust grain surfaces:
Binding energy of H = 0.04 eV due to unpaired e-.
In lattice fault: 0.1 eV, so it stays there.
Once two H meet, they form H2, which has no unpaired e-. So H2 is realeased.
Many other molecules also formed in this way (e.g. H2O, though water stays frozen onto dust grain: ice mantel).