The formation of stars and planets Day 1, Topic 2: Radiation physics Lecture by: C.P. Dullemond
Astronomical Constants CGS units used throughout lecture (cm,erg,s...) AU = Astronomical Unit = distance earth - sun = 1.49x10 13 cm pc = Parsec = 3.26 lightyear = 3.09x10 18 cm = Arcsec = 4.8x10-6 radian Definition: 1 at 1 pc = 1 AU M = Mass of sun = 1.99x10 33 gram M = Mass of Earth = 5.97x10 27 gram L = 3.85x10 33 erg/s
Radiative transfer Basic radiation quantity: intensity I(",#) = Definition of mean intensity erg s cm 2 Hz ster J(") = 1 4# % I($,")d$ = 4 # erg s cm 2 Hz Definition of flux: F(") = 1 4# % I($,")$d$ = 4 # erg s cm 2 Hz
Planck function: Radiative transfer In dense isothermal medium, the radiation field is in thermodynamic equilibrium. The intensity of such an equilibrium radiation field is: I " = B " (T) # In Rayleigh-Jeans limit (hν<<kt) this becomes a power law: I " = B " (T) # 2kT" 2 2h" 3 /c 2 [exp(h" /kt) $1] c 2 Wien (Planck function) " 2 Rayleigh-Jeans
Blackbody emission: Radiative transfer An opaque surface of a given temperature emits a flux according to the following formula: F " = # B " (T) Integrated over all frequencies (i.e. total emitted energy): If you work this out you get: $ F " % F # d# = & % B # (T)d# 0 $ 0 F = "T 4 "= 5.67 #10 $5 erg/cm 2 /K 4 /s
Radiative transfer In vaccuum: intensity is constant along a ray Example: a star F A = r 2 B r F 2 B A " A = r 2 B r " 2 B A F = I " A B I = const Non-vacuum: emission and absorption change intensity: di ds = "# S $ "# I (s is path length) Emission Extinction
Radiative transfer Radiative transfer equation again: di " ds = #$ (S %I ) " " " Over length scales larger than 1/ρκ ν intensity I tends to approach source function S. Photon mean free path: l free," = 1 #$ " Optical depth of a cloud of size L: In case of local thermodynamic equilibrium: S is Planck function: " # = L l free,# = L$% # S " = B " (T)
Radiative transfer
Radiative transfer Observed flux from single-temperature slab: I " obs = I " 0 e #$ " + (1# e#$ " ) B " (T) " # $ B $ (T) for " # <<1 and I 0 " = 0
Emission/absorption lines: Radiative transfer " # $1 " # >>1 Hot surface layer Cool surface layer Flux λ Flux λ I " obs = I " 0 e #$ " + (1# e#$ " ) B " (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.
Heating: Thermal balance of dust grains Optically thin case: Q + = " a 2 % F # $ # d# a = radius of grain ε ν = absorption efficiency (=1 for perfect black sphere) Cooling: Q " = 4# a 2 & # B $ (T)% $ d$ " # = $ a2 % # m Thermal balance: 4" $ a 2 B % " B # (T)$ # d# = " a 2 % " (T)# " d" = 1 $ F " # " F d" # $ # d# %
Optically thick case Additional radiation field: diffuse infrared radiation from the grains di " d J " d = 1 4# I " d % I " d d$ Intensity obeys tranfer equation along all possible rays: ( d) ds = #$ " B " (T) % I " Thermal balance: ( 1 $ B " (T)# " d" = % F " e&' " d + $ * + J ) " -#, " d"
Dust opacities. Example: silicate Opacity of amorphous olivine (silicate) for different grain sizes
Crystalline vs. amorphous silicates Bouwman et al.
Rotating molecules I is the moment of inertia Classical case: Quantum case: E rot = J 2 2I E rot = h2 2I J(J +1) " Bh J(J +1) J is the rotational quantum number: J = 0,1,2,3... B has the dimension of frequency (Hertz)
Rotating molecules Dipole radiative transition: J J-1: "E = Bh J(J +1) # (J #1)J [ ] = 2Bh J Quadrupole radiative transition: J J-2: "E = Bh J(J +1) # (J # 2)(J #1) [ ] = 2Bh (2J "1) Transition energies linear in J
Carbon-monoxide (CO): CO: I = 1.46E-39 Rotating molecules J = 1 0 J = 2 1 J = 3 2 λ=2.6 mm λ=1.3 mm λ=0.87 mm Ground based millimeter dishes. CO emission is brightest molecular rotational emission from space. Often used! Plateau de Bure James Clerck Maxwell Telescope
Molecular hydrogen (H 2 ) Rotating molecules H 2 : I = 4.7E-41 J = 2 0 λ=28 µm Due to symmetry: only J = 3 1 λ=17 µm quadrupole transitions: " J = ±2 J = 4 2 λ=12 µm Need space telescopes (atmosphere not transparent). Emission is very weak. Only rarely detected. Spitzer Space Telescope
Rotating molecules Carbon-monoxide (CO) and Molecular hydrogen (H 2 )
Rotating molecules Molecules with 2 or 3 moments of inertia: Symmetric top : 2 different moments of inertia, e.g. NH 3 : Asymmetric top : 3 different moments of inertia, e.g. H 2 O: H H N H H O H 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 molecules H Symmetric top: NH 3 E rot " Bh J(J +1) + (A # B)hK 2 H N H A " 2.02 #10 11 Hz B " 3.20 #10 11 Hz Radiative J to J-1 transitions with ΔK=0 are rapid (~10-1...-2 s). But ΔK 0 transitions are forbidden (do not exist as dipole transitions). backbone Quadrupole ΔK 0 transitions are slow (10-9 s) but possible (along backbone) 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: [ 12 CO]/[ 13 CO] ~ 30...130 [ 13 CO]/[C 18 O] ~ 10...40
Vibrating molecules: case of H 2 Atomic bonds are flexible: distance between atoms in a molecule can oscillate. E vib "h# ( 0 v + 1 ) 2 Vibrational frequency for H 2 : " 0 #1.247 $10 14 Hz Vibrating molecule can also rotate. Sum of rot + vib energy: E tot = E vib + E rot Selection rules for H2-rovib transitions: from v to any v, but ΔJ=-2,0,2 (quadrupole transitions). Quadrupole transitions are weak: H 2 difficult to detect...
Vibrating molecules: case of H 2 Transitions from v to v-1 or v-2 etc: J to J: J to J-2: Q-branch transitions: Pure vibrational transitions (no change in J). All lines roughly at same position. S-branch transitions: during vibrational transition also downward rotational transition: more energy release. Lines blueward of Q-lines, bluer for higher J. Example: S(1): v=1-0, J=3-1 2.12 micron J to J+2: O-branch transitions: during vibrational transition also upward rotational transition: less energy release. Lines redward of Q-lines, redder for higher J. S Q O
Vibrating molecules: case of CO For CO same mechanism as for H 2 : E vib = h" ( 0 v + 1 ) 2 Vibrational frequency for CO: " 0 # 6.4 $10 13 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 λ 4.7 µm Δv=-2 is called first overtone λ 2.4 µm
Vibrating molecules: case of CO Transitions from v to v-1 or v-2 etc: J to J: J to J-1: Q-branch transitions: Pure vibrational transitions (no change in J). All lines roughly at same position. 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+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. Band head CO first overtone 2-0 2.294 2.302 λ [µm] Calvet et al. 1991
Overview of location of molecular lines
Freeze-out of gases: Ices Form ice coatings on dust grains Rotational lines disappear (molecules are in a solid) Expect each bond (C-C, C-H, C-O etc) to produce a wide vibr. band (like typical dust features), because molecules can exchange energy and momentum. Various ices studied: CO (<20 K), CO 2 (<70..80 K), H 2 O (<90 K), etc. (Note: evaporation temperatures depend on various factors). Exist far away from star (must be cold enough). But ice bands in near-/mid-ir: ices too cold for emission: ice bands only observed in absorption!
CO ice Example: solid and gas-phase CO CO ice+gas CO gas From lecture Ewine van Dishoeck
A two-level molecule LTE level population: n u n d = g u g d e "#E / kt Emissivity and absorptivity: j " = h" 4# n u A ud $(") "# $ = h$ 4% (n d B du & n u B ud )'($) u d "E = h# n u is population of upper level n d is population of lower level di " ds = j " # $% " I " extinction stimulated emission Einstein relations: A ud = 2hv 3 B c 2 ud B ud g u = B du g d
A two-level molecule Collisional excitation: LTE / Non-LTE Rate of change of population of upper level: "n u "t = n d B du J # n u A ud # n u B ud J +n d C du " n u C ud = 0 Absorbtion Collision rates satisfy: Stimulated emission Spontaneous emission Collisional de-excitation Collisional excitation Statistical equilibrium equation C du ~ N H2 " du "#E / kt C du g d = C ud g u e
A two-level molecule Critical number density N of the gas defined such that: n u A ud + (n u B ud " n d B du ) J + n u C ud " n d C du = 0 Dominates when N<N crit. We call this: non-lte. Dominates when N>N crit. We call this: local thermodynamic equilibrium (LTE)
Photodissociation of molecules Ultraviolet photons can excite an atom in the molecule to a higher electronic state. First and second electronic excited state The decay of this state can release energy in vibrational continuum, which destroys the molecule. Electronic ground state
Formation of molecules: H 2 Due to low radiative efficiency, H+H cannot form H 2 in gas phase (energy cannot be lost). Main formation process is on dust grain surfaces: In lattice fault: 0.1 ev, so it stays there. Once two H meet, they form H 2, which has no unpaired e -. So H 2 is realeased. Binding energy of H = 0.04 ev due to unpaired e -. Many other molecules also formed in this way (e.g. H 2 O, though water stays frozen onto dust grain: ice mantel).
Formation of molecules: H 2 If sufficient free electrons are present, H 2 can also form in the gas phase, until all free electrons are used up: H + e - H - +hν H - + H H 2 + e - But these reactions are rare due to limited electron abundance! Main formation is by grain surface reactions.