1. Ge/Ay133 Disk Structure and Spectral Energy Distributions (SEDs)

2. Remember, disks are small: 1 A.U. = 7 milli- arcseconds for D=140 pc (Taurus, Ophichus) Angular momentum budget: For a MMSN mass profile Keplerian disk of radius R, J~(3 x 10 20 cm 2 /s)(R/100 AU) 1/2

3. With single telescopes, even HST, only outer zones resolved. Even worse at longer wavelengths, as we’ll see next time.

4. Imaging surveys/unresolved photometry rather easier. What do such surveys tell us? R=MIPS1 G=IRAC2 B=IRAC1 in Serpens VV Ser Blue=main sequence star Red =protostar/AGB star Spitzer

5. Characterizing large disk samples? SED Models: IR disk surface within several 0.1 – several tens of AU (sub)mm disk surface at large radii, disk interior. What determines disk properties (radius, flaring, T)? G.J. van Zadelhoff 2002 Chiang & Goldreich 1997 HH 30

6. Two different limits: Accretion-dominated vs. Passive How hot do accretion disks get? At the very least, infalling material must dissipate an energy of order (GM * /R disk ) per unit mass. Balance this against thermal radiation: (GM * /R disk )(M disk )~2  R disk 2  T 4 ·  where  accretion timescale. Numerically, T~(500 K)(1 AU/R disk ) 3/4 for  10 5 yr and a solar mass star. Thus, more massive and/or faster evolving disks are hotter. Notice the different vertical temperature structure for more realistic models.

7. Passive Disks: The SHAPE is critical If the central stellar mass dominates, the “vertical” acceleration at a distance R and height above the midplane z is g=g eff ≈ (GM * /R 3 )·z =   ·z where  =(v K /R)=the Keplerian angular vel. For an ideal gas, the sound speed c is simply c=(RT) 1/2 (R=ideal gas constant). From the equation of hydrostatic equilibrium dP/dz=-  g eff =-   z=-(P/RT)   z=-P(  /c) 2 z For a vertically isothermal disk this gives P = P 0 exp(-z 2 /H 2 ), Where the scale height H=c/  For a sound speed of 1 km/s and  R~10 km/s, H/R=(sound speed)/(Keplerian vel.)~0.1. More quantitative models give (a=R):

8. The simplest model: Blackbody disks The first disk SED models assumed a flat disk geometry and that the disk radiated as a perfect blackbody. In this limit, the long wavelength tail has F   Only a few disks obeyed this relationship, most showed much larger fluxes at longer wavelengths. The solution, as first recognized by Kenyon & Hartmann (1987, ApJ, 323, 714), was that actual disks are flared as derived in the slides above. The increased flaring with distance permits the disk to intercept more light from the star and re-radiate it in the far-IR.

9. Somewhat more realistic: Two layer disk models Chiang & Goldreich 1997, ApJ, 490, 368 While still not fully self-consistent, a rather better two-layer disk model was developed by Eugene Chiang for his thesis. The basic idea is that the stellar light is absorbed by the surface layers of the disk that are optically thin to re- radiated infrared energy from grains. This approach w/flared disks provides a good fit to most observed SEDs.

10. Next step: Include grain model opacities Predicts silicate emission bands for the SiO stretching and bending modes at 10/18  m. The grains in disk surfaces are not perfect black- or greybodies, but instead have wavelength dependent emissivities.

11. Grain Emission/Growth in Disk Surface Layers 10  m band20  m band Models Data Kessler-Silacci et al. 2006, ApJ

12. How unique are these models? Nearly all are degenerate for <30  m.

13. How bad are these model degeneracies? Can be broken with resolved images at longer wavelengths, as we’ll see next time. Bad! Note that for these two limiting cases the disk size differs by a factor of two, and the masses by a factor of nearly six!