The formation of stars and planets

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 = 1.49x1013 cm • pc = Parsec = 3.26lightyear = 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

3. Basic radiation quantity: intensity Definition of mean intensity Definition of flux: Radiative transfer

4. Radiative transfer Planck function: In dense isothermal medium, the radiation field is in thermodynamic equilibrium. The intensity of such an equilibrium radiation field is: (Planck function) In Rayleigh-Jeans limit (h<<kT) this becomes a power law: Wien Rayleigh-Jeans

5. Radiative transfer Blackbody emission: An opaque surface of a given temperature emits a flux according to the following formula: Integrated over all frequencies (i.e. total emitted energy): If you work this out you get:

6. Radiative transfer In vaccuum: intensity is constant along a ray Example: a star A B Non-vacuum: emission and absorption change intensity: (s is path length) Emission Extinction

7. Radiative transfer equation again: In case of local thermodynamic equilibrium: S is Planck function: Photon mean free path: Optical depth of a cloud of size L: Radiative transfer Over length scales larger than 1/ intensity I tends to approach source function S.

8. Radiative transfer

9. for and Radiative transfer Observed flux from single-temperature slab:

10. Flux  Radiative transfer Emission/absorption lines: Hot surface layer Cool surface layer Flux 

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

13. Rotating molecules Classical case: Quantum case: I is the moment of inertia J is the rotational quantum number: J = 0,1,2,3... B has the dimension of frequency (Hertz)

14. Rotating molecules Dipole radiative transition: JJ-1: Quadrupole radiative transition: JJ-2: Transition energies linear in J

15. Rotating molecules Carbon-monoxide (CO): J = 10 =2.6 mm J = 21 =1.3 mm J = 32 =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

16. Due to symmetry: only quadrupole transitions: Rotating molecules Molecular hydrogen (H2) H2: I = 4.7E-41 J = 20 =28 m J = 31 =17 m J = 42 =12 m Need space telescopes (atmosphere not transparent). Emission is very weak. Only rarely detected. Spitzer Space Telescope

17. Rotating molecules Carbon-monoxide (CO) and Molecular hydrogen (H2)

18. H N O H H H H Rotating molecules Molecules with 2 or 3 moments of inertia: “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)

19. H N H Radiative J to J-1 transitions with K=0 are rapid (~10-1...-2 s). But K0 transitions are ‘forbidden’ (do not exist as dipole transitions). H backbone Rotating molecules Symmetric top: NH3 Quadrupole K0 transitions are slow (10-9 s) but possible (along backbone) After book by Stahler & Palla

20. 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

21. Atomic bonds are flexible: distance between atoms in a molecule can oscillate. Vibrational frequency for H2: Vibrating molecule can also rotate. Sum of rot + vib energy: Vibrating molecules: case of H2 Selection rules for H2-rovib transitions: from v to any v’, but J=-2,0,2 (quadrupole transitions). Quadrupole transitions are weak: H2 difficult to detect...

22. For CO same mechanism as for H2: Vibrational frequency for CO: v=-1 is called fundamental   4.7 m v=-2 is called first overtone   2.4 m Vibrating molecules: case of CO Often v=0,1,2 of importance Selection rules for CO-rovib transitions: from v to any v’, but J = -1,0,1.

23. Vibrating molecules: case of CO Transitions from v to v-1 or v-2 etc: J to J: Q-branch transitions: Pure vibrational transitions (no change in J). All lines roughly at same position. 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+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

24. Band head CO first overtone 2-0 2.294 2.302  [m] Calvet et al. 1991 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.

25. Overview of location of molecular lines

26. Photodissociation of molecules First and second electronic excited 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. Electronic ground state

27. 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: Once two H meet, they form H2, which has no unpaired e-. So H2 is realeased. In lattice fault: 0.1 eV, so it stays there. Binding energy of H = 0.04 eV due to unpaired e-. Many other molecules also formed in this way (e.g. H2O, though water stays frozen onto dust grain: ice mantel).