Upload
vandan
View
244
Download
4
Embed Size (px)
Citation preview
Protoplanetary disksPart I
Lecture 5
Lecture Universität Heidelberg WS 11/12Dr. C. Mordasini
Based partially on script of Prof. W. Benz Mentor Prof. T. Henning
Lecture 5 overview1. Observed features of protoplanetary disks
2. The minimum mass solar nebula
3. Disk structure
3.1 Vertical structure
3.2 Radial structure
3.3 Application for the MMSN
4. Stability of disks
4.1 Formation by direct collapse
4.2 Numerical simulations
1. Observed properties
General structure
Dullemond & Monnier 2010
Before direct images of disks were made, it was noted that some stars have peculiarities in their spectra which were attributed to the presence of material in orbit.
Spectral features
1) Infrared excess
Micrometer sized dust in orbit around the star, radiating in the IR. It is heated by viscous dissipation (liberated potential energy), and by re-radiated stellar irradiation. Excess over the photospheric (stellar) contribution.
Armitage 2007
Andrews+ 2010
The shape of the SED in the IR is used to classify sources (Class 0, I, II, III)
Spectral features II2) UV excess
UV excess emission above the photospheric level: hot spots on the star, shock-heated by accreting material hitting the stellar surface at high velocities.
Provides an estimate of the gas accretion rate.
Muzzerolle et al. 2010
UV& Vis spectrumsolid=observeddotted=star (model)dashed=accretion shock (model)
Derived accretion rates are between 10-11 to 10-6 Msun/yr in bursts. The rate seems to be higher for more massive stars.
Characteristic radii and masses
By doing such observations for many disks, one can derive rough statistics (still small numbers):
Observations in the submm wavelength allow to derive the mass and spatial distribution of cold dust at tens of AU.
Thermal continuum emission from cold dust at mm and submm wavelengths (Ophiuchus nebula).
Andrews et al. 2010
1246 ANDREWS ET AL. Vol. 723
(a) (b) (c) (d) (e)
Figure 3. Derived distributions of the disk structure parameters for the composite sample (combining the results presented here and in Paper I). From left to rightare the disk masses (Md), radial surface density gradients (! ), characteristic radii (Rc), scale-heights at 100 AU (H100), and the radial scale-height gradients ("). Thecontributions of the four disks with diminished millimeter emission in their central regions are hatched.
Figure 4. Comparison of the data with the best-fit disk structure models. The left panels show the SMA continuum image, corresponding model image, and imagedresiduals (data-model). Contours are drawn at the same 3# intervals in each panel. Cross hairs mark the disk centers and major axis position angle; their relativelengths represent the disk inclination. The right panels show the broadband SEDs and deprojected visibility profiles, with best-fit models overlaid in red. The inputstellar photospheres are shown as blue dashed curves.(A color version of this figure is available in the online journal.)
4 3 2 1 00.0
0.2
0.4
0.6
0.8
LogMM
Andrews et al. 2010
Gives total dust mass. Multiply by factor ~100 to get gas masses.
Disks have masses between 0.0001 to 0.1 Solar masses, following a lognormal distribution.
We will see later why disks clearly more massive than ~0.1 Mstar are not expected (grav. instability).
Characteristic radii and masses II
Disks have 1) characteristic radii between tens to hundreds of AU.
2) surface density power law exponents ~1
3) vertical scale heights at 100 AU of 5-20 AU.
4) alpha viscosity coefficients of ~10-3 to 10-1.
Note: the derivation of these results involves a considerable amount of assumptions and modeling...Andrews et al. 2010
Disk lifetimesThe disk lifetime (or more exactly, the lifetime of hot dust close to the star) can be derived by determining at which age of the star the IR excess disappears.
L-band (3.4 μm) photometry:- excess caused by μ-sized dust @ ~900K... ok to < 10 AU
Haisch et al. 2001, Fedele et al. 2010
NGC 2024
Trapezium
IC 348
NGC 2362
D. Fedele et al.: Accretion Timescale in PMS stars
Table 2. Adopted age, spectral type range, facc and fIRAC (when available) in Figs. 3 and 4.
Cluster Age Sp.T range facc fIRAC Age ref. facc ref. fIRAC ref.[Myr] [%] [%]
rho Oph 1 K0–M4 50 ± 16 M05 M05Taurus 1.5 K0–M4 59 ± 9 62 M05 M05 Ha05NGC 2068/71 2 K1–M5 61 ± 9 70 FM08 FM08 FM08Cha I 2 K0–M4 44 ± 8 52–64 Lu08 M05 Lu08IC348 2.5 K0–M4 33 ± 6 47 L06 M05 L06NGC 6231 3 K0–M3 15 ± 5 S07 this work! Ori 3 K4–M5 30 ± 17 35 C08 this work He07Upper Sco 5 K0–M4 7 ± 2 19 C06 M05 C06NGC 2362 5 K1–M4 5 ± 5 19 D07 D07 D07NGC 6531 7.5 K4–M4 8 ± 5 P01 this work" Cha 8 K4–M4 27 ± 19 50 S09 JA06 S09TWA 8 K3–M5 6 ± 6 D06 JA06NGC 2169 9 K5–M6 0+3 JE07 JE0725 Ori 10 K2–M5 6 ± 2 B07 B07NGC 7160 10 K0–M1 2 ± 2 4 SA06 SA05 SA06ASCC 58 10 K0–M5 0+5 K05 this work# Pic 12 K6–M4 0+13 ZS04 JA06NGC 2353 12 K0–M4 0+6 K05 this workCollinder 65 25 K0–M5 0+7 K05 this workTuc-Hor 27 K1–M3 0+8 ZS04 JA06NGC 6664 46 K0–M1 0+4 S82 this work
References. Schmidt (1982, S82), Park et al. (2001, P01), Hartmann et al. (2005, Ha05), Kharchenko et al. (2005, K05), Mohanty et al.(2005, M05), Sicilia-Aguilar et al. (2005, SA05), Carpenter et al. (2006, C06), Lada et al. (2006, L06), Jayawardhana et al. (2006, JA06),Sicilia-Aguilar et al. (2006, SA06), Dahm & Hillenbrand (2007, D07), Briceño et al. (2007, B07), Je!ries et al. (2007, JE07), Hernández et al.(2007, He07), Sana et al. (2007, S07), Caballero (2008, C08), Flaherty & Muzerolle (2008, FM08), Luhman et al. (2008, L08), Sicilia-Aguilaret al. (2009, S09), Zuckerman & Song (2004, ZS04).
Fig. 3. Accreting stars-frequency as a function of age. New data (basedon the VIMOS survey) are shown as (red) dots, literature data as (green)squares. Colored version is available in the electronic form.
He ! 5876 Å in emission (EW = !0.5 Å, –0.6 Å respectively).The evidence of large H$10% together with the He ! emission ismost likely due to ongoing mass accretion, and these two starsare classified as accreting stars. We estimate a fraction of accret-ing stars in NGC 6231 of 11/75 or 15 (±5%). We warn the readerthat this might be a lower limit to the actual fraction of accret-ing stars; further investigation is needed to disentagle the nature(accretion vs binarity/rapid rotation) of the systems with largeH$10% (>300 km s!1) but low EW [H$].
Fig. 4. facc (dots) versus fIRAC (squares) and exponential fit for facc (dot-ted line) and for fIRAC (dashed line).
NGC 6531
We identified 26 cluster members in NGC 6531 based on thepresence of H$ emission and presence of Li. 13 other sourcesshow presence of Li 6708 Å, but have H$ in absorption. As inthe case of NGC 6231, these might be cluster members with noor a reduced chromospheric activity. We measured the EW of Li6708 Å of these 13 sources and compared them with the typi-cal EW of the 26 stars in NGC 6531 showing also H$ emission
Page 5 of 7
On then studies the fraction of stars with excess as a function of stellar cluster age.
Notes: -mean lifetime ~3 Mrs-longest lifetime ~10 Myrs-similar lifetime for dust and gas
Jupiter must form within a few Myrs.
2. The minimum mass solar nebula
MMSNThe basic idea behind the minimum mass solar nebula (Weidenschilling 1977, Hayashi 1981) is to use the structure of the solar system as we observe it now to derive the structure of the disk from which the planets formed.
The solar system was formed from a gas cloud which must have had solar composition. The minimum mass of the nebula is found by replenishing the planet’s (observed) composition until they reflect solar abundance. Doing so yields the minimum possible mass hence the term MMSN.
The key ingredient in the analysis is the dust to gas ratio:
Note that not all high Z material condenses at temperatures in the nebula. The dust to gas ratio is a function temperature, and thus of the distance from the star.
In hot regions close to the star, only refractory “rocky” elements (iron, silicates) remain solid. In colder regions further away from the star both refractory and volatile material (ices of water, methane, ammonia) are condensed.
The two regions are devised by the so called ice line, at T~170 K, at about 2.7 AU. Probably it is no coincidence that this lies between Mars (terrestrial region) and Jupiter (giants).
Ice line
MMSN II
2)Hot region: T>170 K (a<aice )
which means a jump by a factor 4 at the iceline.
1)Cold region: T<170 K (a>aice )
This is clearly the minimum since it assumes a 100% efficiency in incorporating matter into planets.
Ruden 2000
Hayashi 1981
aice
We can estimate the minimum mass as follows:-high-Z in giant planets: ~80 Earth masses-high-Z in terrestrial planets: ~2 Earth masses
i.e. only 1.6 %.
Note: the position of the iceline is quite uncertain. It likely moves in time.
Solar system inventory
A more detailed analysis can be made by spreading each planet out to its nearest neighbors.Column density in the Minimum Mass Solar Nebula
Spread rock and ice in the solar system planets evenly over thedistance to the neighbouring planets
Assume rock and ice represent ≈1.8% of total material ⇒ originalgas contents
Σr(r) = 7 g cm−2 r
AU
−3/2for 0.35 < r/AU < 2.7
Σr+i(r) = 30 g cm−2 r
AU
−3/2for 2.7 < r/AU < 36
Σg(r) = 1700 g cm−2 r
AU
−3/2for 0.35 < r/AU < 36
Total mass of Minimum Mass Solar Nebula
M =
r1
r0
2πrΣr+i+g(r)dr ≈ 0.013M⊙
Anders Johansen (Lund) Protoplanetary discs 14 / 29
The total mass between 0.35 and 36 AU is then
Note -this assumes planets do not migrate.-lies in the domain of observed protoplanetary disk masses.-the power law slope is 1.5. This is steeper than for the observed disks (~1).-the actual mass was probably a few times higher (~3 to 4).
MMSN III
Column density in the Minimum Mass Solar Nebula
Spread rock and ice in the solar system planets evenly over thedistance to the neighbouring planets
Assume rock and ice represent ≈1.8% of total material ⇒ originalgas contents
Σr(r) = 7 g cm−2 r
AU
−3/2for 0.35 < r/AU < 2.7
Σr+i(r) = 30 g cm−2 r
AU
−3/2for 2.7 < r/AU < 36
Σg(r) = 1700 g cm−2 r
AU
−3/2for 0.35 < r/AU < 36
Total mass of Minimum Mass Solar Nebula
M =
r1
r0
2πrΣr+i+g(r)dr ≈ 0.013M⊙
Anders Johansen (Lund) Protoplanetary discs 14 / 29
The zone limits are simply the arithmetic mean between adjacent orbits.
Weidenschilling 1977r-3/2
Temperature?•Much more difficult to determine is the temperature in the solar nebula.•Two dominant energy sources:
-solar irradiation-viscous heating
•Simplest case: only solar irradiation in optically thin nebula. This is a reasonable assumption in the outer parts of the disk and/or at low accretion rates, but not in the inner parts and/or at high accretion rates.
for solar like stars on the main sequence where L goes as ~M4
Temperature in the solar nebula
Much more difficult to determine the temperature in the solar nebula
Several energy sources: solar irradiation, viscous heating, irradiationby nearby stars
Simplest case: only solar irradiation in optically thin nebula
F⊙ =L⊙4πr2
Pin = πinR2F⊙ Pout = 4πR2outσSBT 4eff
Teff =
F⊙4σSB
1/4
T = 280K r
AU
−1/2
Anders Johansen (Lund) Protoplanetary discs 15 / 29
Temperature in the solar nebula
Much more difficult to determine the temperature in the solar nebula
Several energy sources: solar irradiation, viscous heating, irradiationby nearby stars
Simplest case: only solar irradiation in optically thin nebula
F⊙ =L⊙4πr2
Pin = πinR2F⊙ Pout = 4πR2outσSBT 4eff
Teff =
F⊙4σSB
1/4
T = 280K r
AU
−1/2
Anders Johansen (Lund) Protoplanetary discs 15 / 29
Temperature in the solar nebula
Much more difficult to determine the temperature in the solar nebula
Several energy sources: solar irradiation, viscous heating, irradiationby nearby stars
Simplest case: only solar irradiation in optically thin nebula
F⊙ =L⊙4πr2
Pin = πinR2F⊙ Pout = 4πR2outσSBT 4eff
Teff =
F⊙4σSB
1/4
T = 280K r
AU
−1/2
Anders Johansen (Lund) Protoplanetary discs 15 / 29
Temperature in the solar nebula
Much more difficult to determine the temperature in the solar nebula
Several energy sources: solar irradiation, viscous heating, irradiationby nearby stars
Simplest case: only solar irradiation in optically thin nebula
F⊙ =L⊙4πr2
Pin = πinR2F⊙ Pout = 4πR2outσSBT 4eff
Teff =
F⊙4σSB
1/4
T = 280K r
AU
−1/2
Anders Johansen (Lund) Protoplanetary discs 15 / 29
A. Johansen
3. Disk structure
Disk structureThe rotation, density, temperature in the protoplanetary disk are very important for the formation of planets: They are the initial and boundary conditions of planet formation.
From what we have seen, protoplanetary disks are generally believed to have relatively small mass, typically a few percents or the mass of the star. We will see further down why very massive disks are not expected to exist (for a long time). In what follows, we assume that the disk is axisymmetric and adopt cylindrical coordinates r, ϕ, z.
3.1 Vertical structure
Vertical disk structureThe disk’s vertical hydrostatic equilibrium is given in good approximation by:
We have assumed that the disk is thin (z<<r) and that its own mass is negligible compared to the star’s mass.
1ρ
∂p
∂z= −GMs
r2
z
r
= −Ω2z
Ω =
GMs
r3
Ms: mass of the star
The equation of vertical hydrostatic equilibrium is then with
Vertical gravityRadial density structure solar nebula
Σ(r) = 1700 g cm−2r−1.5
What about the vertical structure?
⇒ Hydrostatic equilibrium between gravity and pressure
d
r
z
g
g
zg
r
The distance triangle and the gravity triangle are similar triangles ⇒gz/g = z/d
gz = gz
d= −GM
d2
z
d≈ −GM
r3z = −Ω2
Kz
Anders Johansen (Lund) Protoplanetary discs 16 / 29
α
α
A. Johansen
z<<d
We can obtain an order of estimate of the thickness of the disk H by recalling that the pressure is given by
p =k
µmH
ρT = c2ρ
1ρ
c2ρ
H= Ω2
H → H =c
Ω
then we can estimate from the hydrostatic equilibrium
Vertical disk structure II
Observations as well as theoretical considerations suggest that H/r is relatively small, typically H ≤ 0.1 r. This implies that c << vrot and therefore protoplanetary disks are said to be cold. In other words, disks are mainly rotationally supported, pressure gradients are only of secondary importance.
For a vertically isothermal disk we can solve the eq. of hydrostatic equilibrium:
Vertically isothermal structure
Replacing Ω by vrot/r we obtain H
r=
c
vrotH/r=h=aspect ratio of the disk
= midplane density
Integrating on both sides and
Vertical disk structure IIIBecause disks are thin, it is convenient to use vertically averaged quantities such as surface density Σ:
For a MMSN like this with H/r~0.05, we find a midplane density of about 10-9 g/cm3 at 1 AU.
3.2 Radial structure
Radial disk structure IIn the radial direction, the gravitational attraction by the star is counteracted by the pressure force (only for the gas) and the centrifugal force.
Fg Fp+Fc
In the radial direction, the combined hydrostatic and centrifugal equilibrium is given by
v2
r=
GMs
r2+
1ρ
∂p
∂r(→ vgas < vKep)
solids
As the pressure is decreasing towards the exterior, we have . We thus see fromthe equation that v<vkeper. The gas is going around the sun slower than the solids (dust, planets). This is a consequence of the additional pressure support.
These pressure gradients are however small, as already mentioned. We can estimate them as follows
v2
r≈ Ω2
r − c2
r= Ω2
r
1− c
2
r2Ω2
= Ω2
r
1− H
2
r2
Multiplication with r gives
Radial disk structure II
1ρ
∂p
∂r=
c2
r
∂lnp
∂lnrwe obtain for the velocity perturbation:
∆v =12
∂lnp
∂lnr
H
r
c
Since (H/r) << 1 and that the pressure derivative with density is never very big, we conclude that Δv << c. Since c itself is already considerably smaller than Ωr, we conclude that Δv << Ωr.
In other words, the departure from Keplerian speed is small and the disks are said to be nearly Keplerian. This is a property of thin accretion disks, since thin means a small pressure gradient, and thus a small pressure support.
Still, as we will see, has the difference between the speed of solids and the speed of the gas important implications for the aerodynamics in the nebula (e.g. dust drift).
This shows that the speed of gas is the same as the one of the solids i.e. the Keplerian velocity to a factor -(H/r)2. As H/r is already small, this is a very small quantity. Let us now write the velocity as v=Ωr - Δv where the first term is the Keplerian speed and the other a correction term. Inserting this in the equilibrium equation and recalling that
3.3 Application for the MMSN
With the results form the last chapters, we can set up a simple model based on the MMSN:
Application for the MMSN
•Mid-plane gas density varies from 10−9 g/cm3 in the terrestrial planet formation region down to 10−13 g/cm3 in the outer nebula•Blue line (not a straight line) shows location of z = H: The aspect ratio H/r increases with r, so solar nebula is said to be (slightly) “flaring”.•The most serious drawback of this model is the temperature structure (passive irradiation only). Still, this model is quite of used for first estimates.
Minimum Mass Solar Nebula density
Density contours in solar nebula:
10 20 30 40
r/AU
!4
!2
0
2
4
z/A
U
!16!15!14
!13
!13
!16!15
!14
!13
!13
Mid-plane gas density varies from 10−9 g/cm3 in the terrestrial planetformation region down to 10−13 g/cm3 in the outer nebula
Blue line shows location of z = H
Aspect ratio increases with r , so solar nebula is slightly flaring
Anders Johansen (Lund) Protoplanetary discs 21 / 29
Density contours
A. Johansen
4. Stability of disk
Stability of an uniformingly rotating sheetWe consider the stability of a fluid disk or sheet of zero thickness and with constant surface density Σ0 and temperature T. The sheet is located in the z=0 plane and is rotating with constant angular velocity Ω=Ωz. The equations governing the evolution of the sheet in the rotating frame of reference are:
Because the sheet is assumed to be isothermal, the vertically integrated pressure is given by:
p = p(Σ) = c2Σ
Note that since the sheet is uniform, the gradient of the potential cannot really lie in the (x,y) plane as indicated by the above equation. So the adopted initial conditions are not strictly speaking a solution to the set of equations. Remember the Jeans swindle....
In the unperturbed state, we assume an equilibrium solution given by:
Σ = Σ0; v = 0; p = p0 = c2Σ0 → ∇φ0 = Ω2(xex + yey); ∆φ0 = 4πGΣ0δ(z)(2)
(rot. frame!)
(1)∂Σ∂t
+∇ · (Σv) = 0
(2)∂v∂t
+ v ·∇v = − 1Σ∇p−∇φ− 2(Ω× v) + Ω2(xex + yey)
(3) ∆φ = 4πGΣδ(z) (mass is in the z plane)
Stability of an uniformingly rotating sheet IIWe now introduce small perturbations in the equilibrium quantities:
Σ(x, y, t) = Σ0 + Σ1(x, y, t); v(x, y, t) = v1(x, y, t); . . . ; 1
We keep only the terms linear in ε. We obtain the linearized equations for the evolution of the perturbations:
(4)∂Σ1
∂t+ Σ0∇ · (v1) = 0
(5)∂v1
∂t= − c2
Σ0∇Σ1 −∇φ1 − 2(Ω× v1)
(6) ∆φ1 = 4πGΣ1δ(z)
As in the stability analysis for the Jeans mass, we now look for solutions of the type:
Σ1(x, y, t) = Σae−i(k·r−ωt)
v1(x, y, t) = (vaxex + vayey)e−i(k·r−ωt)
φ1(x, y, t) = φ0e−i(k·r−ωt)
Five unknowns
Stability of an uniformingly rotating sheet IIITo simplify but without loss of generality, we chose the x-axis to be parallel to the propagation of the perturbation k. In other words, we chose: k = kex
Consider first the Poisson equation. For points outside the sheet, we must have ∆φ1 = 0whereas for points in the z=0 plane we have the solution given above. The only function that satisfies these constraints and that vanishes at infinity is given by:
φ1 =2πGΣa
|k| e−i(k·r−ωt)
This solution substituted back into the linearized equation yields:(7) − iωΣa = −ikΣ0vax
(8) − iωvax =c2ikΣa
Σ0+
2πGiΣak
|k| + 2Ωvay
−iωvay = −2Ωvax
This set of equations can be written in form of a matrix. It has a non trivial solution only when
ω2 = 4Ω2 − 2πGΣ0|k| + k2c2 ≥ 0
Dispersion relation for the uniformingly rotating sheet.
Stability of an uniformingly rotating sheet IV
k
ω2
unstable region
most unstable wave length
ω2 = 4Ω2 − 2πGΣ0|k| + k2c2 ≥ 0 Note: - long wavelengths (small k) are stabilized by rotation - short wavelength (large k) are stabilized by pressure
This result is equivalent to the one found for the rotating cloud.
Overall stability is achieved if ω(k) ≥ 0 everywhere, i.e. the minimum -determined by setting the derivative equal zero - must still be positive. This condition implies that the condition necessary for stability of the uniformly rotating sheet is given by the so called Toomre criterion
Q =2cΩ
πGΣ0> 1 stability criterion for the uniformly rotating sheet
cold, massive disks are instable
A very similar criterion can be derived for the differentially rotating sheet. In this case, we can show that stability is given by:
κ2 = rdΩdr
+ 4Ω2Q =2cκ
πGΣ0> 1 with the epicyclic frequency defined by:
The same criterion also applies for spiral galaxies.
QToomre from intuitive argumentsIn the last paragraph, we have derived Q from perturbation theory. An intuitive, order of magnitude derivation is found by comparing the involved energies.
Energies per unit mass:
Gravity
Rotation
Thermal
For instability, gravity must be dominant both over the rotational and the thermal energy:
Q =2cΩ
πGΣ0> 1
Plugging in the expression above yields
This corresponds, to numerical constants of order unity, to the same as the conditions we derived before, i.e.
4.1 Formation by direct collapse
In order to additionally allow fragmentation into bound clumps, the timescale τcool on which a gas parcel in the disk cools and thus contracts must be short compared to the shearing timescale, on which the clump would be disrupted otherwise, which is equal the orbital timescale τorb= 2 π/Ω. This means that
Formation by direct collapseThe Toomre criterium describes the stability of the disk agains the development of axi-symmetric radial rings. Hydrodynamical simulations show that disks become unstable to non-axisymmetric perturbations (spiral waves) already at slightly higher Q values of about Qcrit = 1.4 to 2.
Only if both criteria are fulfilled, the formation of self-gravitating, bound gas clumps can occur.
This also explains why no disk with a mass similar as the central mass exist (at least not for a long time).
Here, ξ is of order unity (Gammie, 2001). If this conditions is not fulfilled, only spiral waves develop leading to a gravoturbulent disk. The spiral waves efficiently transport angular momentum outwards, and therefore let matter spiral rapidly inwards to the star. This process liberates gravitational binding energy, which increases the temperature, and reduces the gas surface temperature, which both reduce Q, so that the disk evolves to a steady state of marginal instability only, without fragmentation.
Cooling criterion
The disk corresponding to the MMSN is therefore very stable. Planets do not form through direct collapse, at least not at 1 AU.
Note:
The more massive the disk, the smaller Q.
But it nowhere falls below 1, so fragmentation does not happen.
Formation by direct collapseFor thin disks we can write the sound speed as: c =
H
rvrot
For a MMSN disk we thus find at 1 AU
QToomre
In a more detailed disk model (alpha disk with irradiation), Q reaches minimum at about 30 AU. τcoolΩ has a peculiar form coming from the variation of the opacity with temperature.
QToomre
Mdisk=0.024 Msun Mdisk=0.1 Msun
Mordasini et al. 2011
τcoolΩ τcoolΩ
4.2 Numerical simulations
Numerical simulationsEarly simulations assumed a locally isothermal equation of state. This corresponds to an infinitely short cooling timescale, as it means that when the density increases, the temperature remains constant, as compressive work is immediately radiated away. This is of course not correct ...
Color-coded face-on density maps of a run with Qmin~1.3 (top) and Qmin~1.5 (bottom), at 200 (left panels) and 350 yr (right panels). The equation of state is isothermal but switched to adiabatic close to fragmentation. Brighter colors are for higher densities, and the disks are shown out to 20 AU.
We see that both disks develop strong spiral waves, but only the upper one fragments into bound clumps.
This happens very fast!
Mayer et al. 2004
Size: 20 AU Mdisk: 0.085 Msun
Numerical simulations II
Evolution of Q-profiles. Profiles at t = 0 (thin solid line), 160, (dashed line), and 240 yr (thick solid line). Fragmentation occurs between 160 and 240 yr in the model shown below (Mdisk=0.1 Msun, Qmin=1.38) while model shown in the upper panel (Mdisk=0.08 Msun, Qmin=1.65) develops only strong spiral arms.
Mayer et al. 2004
Numerical simulations III– 40 –
Fig. 4.— A time series of SPH particle positions for a disk of mass MD/M!=0.2 and initial minimumQmin=1.5 (simulation A2me). Spiral structure varies strongly over time. Length units are defined as1=10AU and time in units of the disk orbit period TD= 2!
!
R3D/GM!. With the assumed mass of the star
of 0.5 M" and the radius of the disk of 50 AU, TD! 500 yr.
The plot shows a time series of SPH particle positions for a disk of mass Mdisk/Mstar=0.2 and initial minimum Qmin=1.5. Spiral structure varies strongly over time. Length units are defined as 1=10 AU. With the assumed mass of the star of 0.5 M⊙ and the radius of the disk of 50 AU, TD≈ 500 yr.
Nelson, Benz et al. 2000
Uses realistic cooling by computing the structure of the disk and taking the photospheric temperature for computing the radiative losses.
In this simulation, T rises when the density rises. Spiral arms still develop but there is no evidence for local collapse. A correct treatment of the thermodynamics is essential for accurate dynamics.
To date, no general agreement exists on the feasibility of direct collapse as planetary formation mechanism.
Questions?