Transcript SS_L4

4 star formation
Stellar Structure: TCD 2006: 4.1
massive interstellar clouds
The galactic interstellar medium (ISM) consists of a roughly uniform
gas with n~105 atoms/m3.
At current epoch, star formation takes place in massive interstellar
clouds.
Clouds have dimensions:
R~10 pc, n~5 109 atoms/m3, T~10 K.
Galaxy is pervaded by a magnetic field with field lines
approximately parallel to the galactic plane.
Field strongly tied to the ionized plasma in the ISM.
Small local perturbations in the field lead to local potential wells.
ISM condenses in these wells, increasing their strength as it
pulls the magnetic field with it  Rayleigh-Taylor instability.
Provides initial mechanism for the formation of dark clouds of
interstellar matter.
What causes these clouds to collapse to the point of forming stars?
Stellar Structure: TCD 2006: 4.2
Jeans’ criterion for gravitational collapse
To overcome internal pressure, cloud must be sufficiently compact that
gravity dominates: i.e. Egrav+Ekin <0.
4.1
Gravitational binding energy for a spherical cloud mass M, radius R is
E grav  
M
0
Gm
GM 2
dm   f
r
R
f depends on density distribution, f=3/5 for uniform density, increasing with
central condensation. We adopt f=1.
Since thermal energy per particle is (3/2)kT, total Ekin of cloud is
E kin 
3 M
kT
2 m H
where mH is the average mass of the cloud particles.
4.1 implies that a cloud of radius R can condense if either
M cloud
3kT
 MJ 
R or
2Gm H
 cloud
3
 J 
4M 2
MJ and J are the Jeans mass
and density.
Stellar Structure: TCD 2006: 4.3
 3kT 
 2Gm 
H 

3
4.2
contraction, fragmentation and formation
Initial contraction of a massive cloud proceeds providing it is not opposed
by increasing P. Releasing Egrav increases T (and hence P), but puts H2
molecules into excited rotational levels. De-excitation emits a photon at
28.2m. Cloud is transparent at 28.2m, so radiation cools cloud, allowing
contraction to proceed.
When cloud fragment reaches Jeans density, it will contract unopposed by
internal pressure, again providing the released Egrav is not converted to Ekin.
Possible while Egrav absorbed by dissociation of H2 (D=4.5eV) and
M
M


ionisation of H (I=1.6eV). Energy absorbed is 2m D m  I .
H
H
0
From initial radius, we can calculate the final radius from
 1
1
1  D

GM 
  
 I 


 R2 R1  m H  2
Stellar Structure: TCD 2006: 4.4
4.3
contraction, fragmentation and formation (2)
From initial radius, we can calculate the final radius from
 1
1
1  D

GM 
  
 I 


 R2 R1  m H  2
4.4
Example: The radius for a protostar of 1 solar mass is ~1015 m with a
Jeans density ~10-16 kg m-3. The dissociation+ionisation energy is 3 199 J.
The radius after gravitational contraction is ~1011 m. The timescale for this
contraction is tdyn~20,000 years.
Stellar Structure: TCD 2006: 4.5
approach to hydrostatic equilibrium
After ionisation is complete (all H  H+),
temperature and pressure rise,
contraction slows down  hydrostatic equilibrium.
<T> can be estimated using the Virial theorem.
Thermal kinetic energy is
Ekin~(M/mH)(3kT/2) = M/mH. 3kT.
Gravitational energy at end of collapse is
Egrav~-GM2/R2~-M/mH(D/2+I).
But 2Ekin+Egrav=0, so protostar approaches equilibrium at <T>
given by
k<T>~(D+2I)/12~2.6eV
Hence T~30,000 K, independent of M !
Stellar Structure: TCD 2006: 4.6
4.5
thermal contraction
Subsequent contraction governed by opacity.
Opacity controls the loss of radiation from the surface, and hence
the release of gravitational energy on a thermal timescale (tkin
~107 –108 years).
Virial theorem can again be used because the star remains close to
hydrostatic equilibrium.
Stellar Structure: TCD 2006: 4.7
minimum mass for nuclear reactions
If contraction not stopped by nuclear ignition, how far would it proceed?
When  so high that electron wavefunctions overlap, classical mechanics
breaks down, electrons become degenerate. Occurs when interelectron
spacing r approaches the de Broglie wavelength, r = B = h/mev.
Since (1/2)mev2 = (3/2)kT 
But  = mH / (4/3)r3
r = h / (3mekT)1/2
  = mH (mekT/h3)3/2
After this, further increase in  cannot affect T.
From the Virial theorem, we can derive <T>
kT 
GMm H
 Gm H M 2 / 3  1 / 3
3R
4.6
Substituting for ,
kTmax
 G 2 mH 8 / 3me  4 / 3
7
43

 M  Tmax  10 M K
2
h


For M~<0.08 solar masses, Tmax cannot trigger nuclear reactions.
Stellar Structure: TCD 2006: 4.8
4.7
stellar luminosity
We know Sun has been in hydrostatic equilibrium for 4.5 billion years,
implies Virial theorem useful. For example:
<P>~1014 Pa (104 atm), <>~103 kg m-3 (cf. water) <TI>~6.106 K.
We also know L=4R2 Teff4 (Stefan’s law: Eq. 1.4)
But Teff~6000 K ~ <TI>/1000.
Why are TI and Teff different ?
If star in equilibrium at temperature TI, it radiates as a blackbody with
L = 4R2 TI4 ~ 1012 L,
4.8
and photons would have mean energies kT~0.5 kEV (X-ray).
Fortunately, X-rays are trapped (absorbed and re-emitted) by ions, a
temperature gradient is established and radiant energy diffuses
towards the surface (Section 2).
Stellar Structure: TCD 2006: 4.9
stellar luminosity
If  is photon mean free path, then (Eq. 1.11) tdiff = R2 / c,
compared with the direct escape time (Eq. 1.12) tesc ~ R / c.
Diffusion slows down energy release rate by /R, so L is reduced by /R.
Hence Teff
 
 
R
1/ 4
TI
4.9
Since (3.20) relates TI to M and R, so from (1.4) and (4.9) we obtain:
L  4R 2Teff4  4R 2TI4

R
 GMmH   4   4
4
3
 4R 2 

G

m


M
H
35 k 4
 3kR  R
4
2
 L   M 3
 L
4.10
M3

Stellar Structure: TCD 2006: 4.10
4 star formation - review
Stars form from interstellar clouds, providing Jeans
criterion for gravitational collapse is satisfied.
Virial theorem used to determine radius after contraction to
point where opacity increases, timescale from earlier,
and average temperature.
Minimum electron separation used to estimate minimum
mass for nuclear ignition.
Photon mean free path used to estimate stellar luminosity.
Stellar Structure: TCD 2006: 4.11