lecture 25 stability of molecular clouds
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Lecture 24 Stability of Molecular Clouds
1. Stability of Cloud Cores 2. Collapse and Fragmentation of Clouds3. Applying the Virial Theorem
ReferencesOrigins of Stars & Planetary Systems eds. Lada & Kylafis
http://cfa-www.harvard.edu/creteMyers, “Physical Conditions in Molecular Clouds”Lada, “The Formation of Low-Mass StarsShu 2, Ch. 12 & Stahler, Appendix D for Virial Theorem
1. Stability of Molecular Cloud CoresStart with the molecular cloud core properties from Lec. 23 and focus on the items relating to virial equilibrium.
1. associated with star formation2. elongated (aspect ratio ~ 2:1)3. dynamics may be dominated by thermal or turbulent motion 4. temperature: T ~ 10 – 20 K5. size: R ~ 0.1 pc6. ionization: xe ~ 10-7
7. size-line width relation R ~ σ p , p = 0.5 +/- 0.2
8. approximate virial equilibrium9. mass spectrum: similar to GMCs.
non-thermal vs. thermal line widthsconsistency with virial equilibrium
Core Correlations (Myers 1999)
∆v2/R
N
For pure thermal support, GM/R ~ kT/m, m = 2.3 mH), andR ≈ 0.1 (M / M ) (10 K / T) pc.
Core Stability
Virial equilibrium seems an appropriate state from which cores proceed to make stars. However, as a significant fraction of cores are observed to have embedded protostars (observed by IRAS), they can’t be completely quiescent.
We are therefore concerned with the issue of core stability,e.g., why they are stable (if they are) and how they becomede-stabilized and collapse under gravity to form stars.
Although Myer’s data presentation suggests that randommotion (thermal and/or non-thermal) may stabilize coresagainst gravitational collapse, we should first consider other possibilities, in particular rotation and magnetic fields.
Effects of Rotation
R
grad v
J/M
β
Molecular cloud cores have modest velocity gradients,
0.4 - 3 km s-1 pc-1
and angular speedsΩ ~ 10-14 - 10-13 rad s-1.
β = Erot / Egrav ~ 0.02
Cores do not appear to be supported by rotation.
NB Not all of the observedgradients correspond to theoverall rotation of the core Goodman et al. ApJ 406 528 1993
Magnetic Pressure• Magnetic fields are difficult to
measure in cloud cores – OH absorption
measurements are hard because of the small chance of a background radio source
• Example of dark cloud B1|B| cos θ ≈ -19 ± 4 µG
OH probes nH ≈ 103 cm-3
• B2 /8π ~ 3 x 105 K cm-3
– Comparable to the thermal pressure of a 10 K core with n ~ 104 cm-3
1667 MHz
1665 MHz
Crutcher 1993 A
pJ 407 175
Stokes V
Stokes I
Recall Lecture 13 and radio measurements of the Zeemaneffect.
2. Collapse and FragmentationHow do cores condense out of the lower density regions of GMCs?
The conventional wisdom is local gravitational instabilitiesin a globally stable GMC via the classical Jeans instability.
Consider a uniform isothermal gas in hydrostatic equilibrium with gravity balanced by the pressure gradient. Now suppose a small spherical region of size r is perturbed
ρ → Xρ where X > 1
The over-density Xρ generates an outward pressure force per unit mass:
Simple Stability Criterion
00
22
02
2
or
ifout nsGravity wi .~/~
massunit per force onalgravtitati inwardan toleads alsodensity increased The
/~/|~|
ρρ
ρ
ρ
Gcr
Gcr
rGXrGMF
rXcpF
G
p
>>
∇
ρ0
Xρ0
rThe right hand side is essentially the Jeans length:
0J ρ
πλG
c=
The Jeans Length & Mass
( ) ( ) 2/13-42
2/3sun
0
32/1
0
2/32
03
0J
cm10/)H(K10/5.7
is valuenumerical whose
mass Jeans ingcorrespond aget we
length Jeans theFrom
−
−
=
∝⎟⎟⎠
⎞⎜⎜⎝
⎛==
=
nTMM
cGcM
Gc
J
JJ ρρπρλ
ρπλ
Length scales > λJ or masses > MJ are unstable against collapse.
Relevance of the Jeans AnalysisMany assumptions have been made implicitly in the above analysis that violate the basic fact that molecular clouds are not static but are turbulent on large scales.
Turbulence provides an effective pressure support, and its supersonic motions must be dissipated by shocks before the collapse can proceed: Is it enough to replace the sound velocity c in the Jeans formulae by
As the gas condenses the Jeans mass decreases,which suggests fragmentation (Hoyle 1953 ApJ 118 513), a collapse cascade into smaller and smaller masses.
2turb
2 σ+=mkTc
Does Fragmentation Occur• The dispersion relation for a perturbation δρ ~ ei(ωt - kx)
in the Jeans problem isω2 = c2 k2 - 4πGρ0
or ω2 = c2 (k2 - kJ
2), kJ2= 4π Gρ0 / c2
• k2 - kJ2 < 0 makes ω imaginary:
– Exponential growth occurs for k < kJ– Growth rate -iω increases monotonically with
decreasing k, which implies that • Longest wavelength perturbations (largest mass)
grow the fastest • The fast collapse of the largest scales suggests
that fragmentation is unlikely(Larson 1985 MNRAS 214 379)
Swindled by Jeans?Consider an alternate model, a thin sheet of surface density ∑.The dispersion relation is
ω2 = c2 k2 - 2π G ∑ |k|or
ω2 = c2 ( k2 - kc |k| ) Exponential growth occurs for
k < kc = 2π G ∑ / c2
with growth rate-i ω =( 2π G ∑ k - c2 k2 )1/2
which is maximum at kf = kc /2= π G ∑ / c2
and gives a preferred mass,Mf ~ (2π / kf )2 ∑ = 4 c4 / G2 ∑
This is a different result than the previous Jeans analysis for anisotropic 3-d medium. The maximum growth rate now occurs on an intermediate scale that is smaller than predicted by Jeans, possibly favoring fragmentation.
Preferred Length & Mass• Apply the thin sheet model to the Taurus dark cloud.
– T = 10 K, c = 0.19 km/s• λc ~ 0.05 pc
– A(V) ≈ 5 mag. or ∑ ≈ 0.032 g cm-2
• λf ~ 0.1 pc, Mf ~ 2 M• tf = λf /(π c) ~ 2 x 105 yr
• More realistic analyses tend to show that, when thermal pressure provides the dominant support against gravity, there is a minimum length & mass scale which can grow– If the cloud is non-uniform there is a preferred scale
• Typically ~ few times the characteristic length of the background, e.g., the scale height, H =c2 /πG∑, for a gaseous equilibrium sheet (Larson 1985 MNRAS 214 379)
3.The Virial Theorem and StabilityA. General Considerations
The derivation of virial theorem starts with the equation ofmotion for the macroscopic velocity of the fluid after averaging over the random thermal velocities. The mathematics is given by Shu2 Ch. 24 and Stahler Appendix D.
( )BBBT
BBBT
TPDt
vD
ijjiij
rrrtr
trrrr
∇⋅+∇−=⋅∇
−=
⋅∇+∇−∇−=
ππ
πδπ
φρρ
41
81
8/4/
2
2
The perspective of the Virial Theorem is that these equations contain too much information. Hence, take moments, specifically the first moment:
Virial Theorem: LHS AnalysisTake the dot product with r and integrate the equation of motion over a finite volume, and analyze the right and left hand sides separately.
The LHS side yields two terms since:
( ) ⎟⎠⎞
⎜⎝⎛ ⋅+⋅=⋅
⋅−⋅=⋅
••••
••
∫∫∫
rrrrrrDtD
dVvvdVrrDtDdVrr
VVV
rrrrrr
rrrrrr
2
and
)()(21
2
2
2
2
ρρρ
V
KVV
EDt
IDdVvdVrDtD 2
21
21
finally and
2
222
2
2
−=−⎟⎟⎠
⎞⎜⎜⎝
⎛∫∫ ρρ The 1st term vanishes
for a static cloud.
Virial Theorem: RHS Analysis• The RHS yields volume & surface terms
3PV∫ dV +
B2
8πV∫ dV −
r r ⋅
r ∇ φ( )
V∫ dm
− P +B2
8π
⎛
⎝ ⎜
⎞
⎠ ⎟
S∫ r
r ⋅ dr S + 1
4πr r ⋅
r B ( )
S∫
r B ⋅ d
r S ( )
• The 3rd term becomes the gravitational energy.• The 1st and 4th terms measure the difference between theexternal and mean internal pressures.
• The other terms are transformed into magnetic pressure and tension.
The Virial Theorem
( )
RHS. in the sexpression eapproximatinsert then and31
31
32
pressure external for the solvecan wecloud, static aFor 81)(
41
8
where
32
Kext
22
extK2
2
MWEVPVP
BrSdrBBSdBdVM
MWVPPEDt
ID
++=⟩⟨−
⋅−⋅⋅+=
++−⟩⟨+=
∫∫ ∫rrrrrr
πππ
Pext becomes a function of the global variables of thevolume V and the gas inside. If there is no rotation,EK = 0.
Simplest Application of the Virial Theorem
RGMMcPR
22
ext3
5334 −=π
Consider a spherical, isothermal, un-magnetized & non-turbulent cloud in equilibrium of mass M and radius R
where we have used P = ρc2. If we now also ignore gravity,we check that the external and internal pressures are the same. In the presence of gravity, the second term serves to lower the external pressure needed to confine the gasInside volume V.
Next, divide by 4πR3 and plot the RHS vs. R
Simplest Application of the Virial Theorem
4
2
3
2
ext1
2031
43
RGM
RMcP
ππ−=
• There is a minimum radius below which this model cloud cannot support itself against gravity.
• States along the left segment, where P increases with R, are unstable.
P
RRcr
P ~ 1/R3
Stable and Unstable Virial Equlibria
P
RRmax
A B
For a given external pressure, there are two equilibria, but only one is stable.
Pext
Squeeze the cloud at B (decrease R)– Requires more pressure to confine it: re-expands (stable)
Squeeze the cloud at A (decrease R)– Requires less pressure to confine it: contracts (unstable)
The Critical PressureDifferentiating the external pressure for fixed M and c
4
2
3
2
ext1
2031
43
RGM
RMcP
ππ−=
gives the critical radius,
2cr cGMR ≈
The corresponding density and mass are
cr
3
cr3
6
cr ρρ cM
Mc
≈≈
This last result is essentially the same as gotten from theJeans analysis.
Fragmentation vs. Stability(or Jeans vs. Virial)
• Both methods yield a similar characteristic “Jeans” mass– The difference is in the initial conditions
• Cloud scenario assumes a self-gravitating cloud core in hydrostatic equilibrium
– The cloud may or may not be close to the critical condition for collapse
• In the gravitational fragmentation picture there is no “cloud”– It cannot be distinguished from the background until it
has started to collapse Both perspectives may be applicable in different parts of GMCs
• There are pressure confined clumps in GMCs– Not strongly self-gravitating
• Cores make stars– Must be self-gravitating– If fragmentation produces cores these must be collapsing
and it is too late to apply virial equilibrium
Fragmentation vs. Stability• A very relevant question is how quickly cores
form (relative to the dynamical time scale) – NH3 cores do seem to be close to virial equilibrium but
they could be contracting (or expanding) slowly• Statistics of cores with and without stars should
reveal there ages– If cores are stable, there should be many more cores
without stars than with young stars– Comparison of NH3 cores and IRAS sources suggests
~ 1/4 of cores have young stellar objects (Wood et al. ApJS 95 457 1994), but Jijina et al. give a largerfraction
• The corresponding life time is < 1 Myr (Onishi et al. 1996 ApJ 465 815)
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