The formation of stars and planets. Day 1, Topic 2: Radiation physics. Lecture by: C.P. Dullemond

Transcription

1 The formation of stars and planets Day 1, Topic 2: Radiation physics Lecture by: C.P. Dullemond

2 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

3 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

4 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

5 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

6 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

7 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)

8 Radiative transfer

9 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

10 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)

11 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.

12 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# %

13 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"

14 Dust opacities. Example: silicate Opacity of amorphous olivine (silicate) for different grain sizes

15 Crystalline vs. amorphous silicates Bouwman et al.

16 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)

17 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

18 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

19 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

20 Rotating molecules Carbon-monoxide (CO) and Molecular hydrogen (H 2 )

21 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)

22 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 (~ 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

23 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] ~ [ 13 CO]/[C 18 O] ~

24 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...

25 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= 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

26 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

27 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

28 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 λ [µm] Calvet et al. 1991

29 Overview of location of molecular lines

30 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 (< 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!

31 CO ice Example: solid and gas-phase CO CO ice+gas CO gas From lecture Ewine van Dishoeck

32 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

33 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

34 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)

35 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

36 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).

37 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.