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SYLLABUS 2009
Hotter, less dense
Colder, denser
ISM: average properties
•  Mean density of ISM (in our Galaxy) ~ 1 particle/cm3
but is HIGHLY CLUMPED: filling factor ~3%.
•  The GAS is mainly:
Hydrogen (90%)
Helium (10%)
metals (C,N,O, ... Fe) <1% by number
•  Dust grains : gas atoms ratio = approx 1 : 1012
•  Dust grains have a power-law size distribution
radii ~ 5×10-9 - 2×10-7 m (smaller grains are more abundant)
A global model:
the 3+1 phases of the ISM
•  Cold, neutral medium (CNM)
n ~ 1-103 /cm3, T < 100 K, volume fraction: ~1-5%
•  Hot, ionized medium (HIM)
n ~ 10-4-10-2 /cm3, T ~ 106-107 K, volume fraction: ~30-70%
•  Warm interface media
  Warm ionized medium (WIM)
n ~ 0.01 /cm3, T ~ 1000 K, volume fraction: ~20-50%
hydrogen ionization fraction (XH) ~ 70%
  Warm neutral medium (WNM)
n ~ 0.1-10 /cm3, T ~ 1000-5000 K, volume fraction: ~10-20%
hydrogen ionization fraction (XH) ~ 10%
⊕  Dark neutral (molecular) clouds
n ~ 103-106 /cm3, T ~ 10-50 K, volume fraction: <1%
photoionized nebulae:
HII regions
Lagoon nebula
Rosette
Nebula
2 O-stars
Photoionisation: the removal of electrons from gas atoms by photons.
H + (ionizing photon)  p + eRecombination: Attraction between protons and electrons leads to the
recapture of the electron.
p + e-  H + (ionizing photon)
in summary
For a typical O-star (e.g. the O6V star in the Trapezium)
S = 1049 s-1
Typical nebula, n = 108 m-3; Te = 104 K
At r = 1 pc, J = 8.4×1014 m2 s-1
giving X = 0.9999999
i.e. the HII region is almost fully ionised!
We can calculate this for a range of r-values
 X as a function of r
M43
Real examples
1.  no. of ionising photons
-  Hotter star
-  More hot stars
2.  amount of gas (gas density)
IC 1274
Thermal balance in ionized nebulae
Photoionization adds kinetic energy (KE) into the H cloud since it creates hot, fast
photoelectrons.
T can increase indefinitely though… Energy must be lost somehow
Energy can be lost through photons: H recombination means KE of e-  photon which
can escape
Balance heating and cooling rates:
Predict temperatures of T = 30 000 – 60 000 K
These are much higher than observations
Typical nebula gas temperature has Te ~ 10 000 K.
 There must be other ways to cool the gas
What if we relax our assumption of a pure H nebula?
Metals
•  Real nebulae contain more than just H
•  Metals (heavy elements, C, N, O, Ne, S, Ar, Cl, etc.) are found in proportion
to Hydrogen of ~10-4 – 10-8
H recombination = ALLOWED transitions
high probability of occurring (~109 per sec)
Collisionally excited lines (CELs)
n=3
n=2
n=1
Forbidden line cooling
transition rates very low: ~10-3 - 100 per sec
(compared to H recombination rates: 109 per sec)
Therefore photons are v. likely to escape nebula before being absorbed
 can ignore absorption
They can remove a lot of heat from the nebula, thus solving our problem
Common forbidden lines:
Optical: [OIII] 4959,5007 Å, [NII] 6548,6584 Å, [SII] 6717,6731 Å
Infrared: [OIII] 52,88 µm, [NIII] 57 µm
•  density
•  temperature
•  element abundances
diffuse clouds
eagle nebula dark cloud
Snake nebula
Temp regulation within Diffuse Clouds
Diffuse clouds: HII region-type cloud, but located far from any ionization
source. All ionising photons have been absorbed by other material.
But diffuse clouds do not have temp of 0 K - how are they heated?
Photoionisation exists but is inefficient because:
•  available species (e.g. C I) not abundant
•  small range of photon energies (11.3-13.6 eV = 912-1110 Å)
•  max KE obtainable is 2.3 eV - quite low so not much heating
Remember DUST? (really smoke - small, sub-micron sized grains)
These dust grains give out (photo)electrons when struck by photons of
sufficent energy
Because of the photoelectric effect
(Call them
photoelectrons
because they are
produced by
photons)
Einstein's explanation of the photoelectric effect won him the Nobel Prize (in Physics) of 1921.
Absorption line spectroscopy
Q: How can we understand the gas if we can’t see it?
A: use a bright background source in order to see foreground ISM in
‘silhouette’
UV resonsance transitions
n = 1  n’
resonance transition
 resonance line
4
3
2
1
UV: 1200-3300 Å
FUV: 912-1200 Å
EUV: 100-912 Å
The spectral line equivalent width (EW):
definition
W
What is the equivalent width of a rectangular line profile
with area = the real line profile
I
0
A=I×W
I = intensity level of the continuum
W = equivalent width of the absorption line -- units of length (Å).
(identical for emission lines as well)
Abundance determination: the method
1.  observe high-resolution spectrum of a suitable bright source
use a space telescope (e.g. IUE, HST, FUSE).
2.  Identify set of resonance lines of elements/ions from comparison with a
theoretical line list.
3.  Measure the EW of the lines.
4.  Use curve of growth to get the column density of each species.
5.  Determine column density of Hydrogen (using same technique - use
Lyman resonance lines) ➙ get abundances of species relative to
Hydrogen.
Why are ISM abundances
for many elements so much
smaller than their solar
values?
The Sun is a normal star
born out of the ISM…
Reminder: do you know
what the Cosmos is mostly
made of?
H, He, O, C, Ne, N etc.
Dust in the ISM
Dust particles are extremely small (1-10 microns)
Particles are irregularly shaped
Composed of silicates, carbon, ices (water, methane, ammonia etc) and/or iron
compounds.
Effects of the Dust:
•  On Elemental Abundances: We saw how elements are depleted from the GAS phase
of the ISM sticking on to dust grains thus lowering the gas phase abundances of
certain (mostly refractory) elements.
•  On E/M radiation: dust grains absorb or scatter blue light more than red light
removing it from the line of sight to the observer. Distant objects thus appear
reddened.
Dust in the ISM: Where is it made?
Hourglass Nebula
SN1987A
Interstellar extinction law
2200 Å
bump: very broad (~400 Å), still unidentified!
(possibly absorption from Graphite grains)
Evidence for dust:
•  Extinction (absorption of light)
•  inc. Dark clouds
•  Reddening (scattering of light)
•  inc. Reflection nebulae
•  Direct observation (IR imaging and spectr.)
•  Abundances, depletion
HI (neutral atomic H) spin-flip
spin = magnetic orientation
photon emitted when spin changes (aligned  opposite)
wavelength = 21 cm (1.42 GHz)
prob of spontaneous flip = 1 in few million yrs, (highly forbidden)
but there is a lot of H!
Molecules, found in cold regions, typically emit cm or mm-wave radiation
➙ observable with radio-telescopes (large antennas).
The most abundant molecule, H2, has very few transitions - in fact v. difficult to
observe at any wavelength!
On the other hand, CO has a strong line at 2.6 microns - v. convenient!
The CO/H2 ratio in the Galaxy is about 10-4 (1:10000) and remains ~constant.
CO observations can therefore be used to trace molecular hydrogen.
star formation
Pressure vs. Gravity
Hydrostatic Equilibrium:
GRAVITY (inward) = PRESSURE (outward)
(all main sequence stars are in HE)
However, if gravity < pressure 
if gravity > pressure 
In order to get the highest gravity, we need lots of mass (density)
Therefore, in order to get the lowest gas pressure, we need the lowest T.
Stars are born in very cold clouds (~10-100 K).
 cold dense regions = dark clouds
Dark Clouds
Barnard
objects:
Bok globules
9 cm3
•  size
~410
n ~10
-10pc
•  Tmass
~ 103-10
~ 10-100
K. 4 M
•  size ~ 1 pc
Is there a minimum required mass for star-formation?
Use eqn. of HE: can show that there is a minimum mass for collapse to
happen (for a given T and ρ).
This is called the JEANS’ MASS (MJ):
T 3/ 2
MJ ~
ρ
For typical dark cloud: MJ = 105 M
BUT: Typical stellar masses are only 1–10 M
As cloud collapses, ρ↑and the smaller the minimum mass, MJ, becomes.
the cloud
breaks up into smaller ‘parcels’ before star
Thus after grav. collapse
begins…
formation starts
➙ FRAGMENTATION
frag & IMF
Massive molecular clouds fragment into many small clumps -- stellar nursery
Molecular cloud: gas mass = hundreds to a few millions of M
size: 10-100 pc
density: ~103 atoms/cm3
Star formation process thought to be quite inefficient (~30%)
Fragmentation ➙ range of star masses
The spread of stellar masses can be defined by the initial mass
function:
N(M) = M-(1+x)
N(M) = no. of stars of mass M
M = star mass
x = constant (= slope)
Salpeter (1955) found x = 1.35
for lower and upper mass limits of 0.1 and 125 M
PMS tracks
T-Tauri stars
•  age ~ 106 yr
•  mid to low mass; M ~ 3 M
•  surrounded by thin, hot gas, apparent from spectral emission lines.
•  material is being ejected at ~100 km/s in jets  stars are losing mass.
•  mass-loss rate of ~10-7 M/yr
•   after reaching the MS (in ~107 yr) the protostar has lost ~1 M of
material
•  this is partially lost from system, and partially incorporated into a debris
disk
•  debris disk is what forms a planetary system
Proplyds II
HH34
Herbig-Haro object
HH
model
Herbig-Haro objects: important signposts of starformation.
Outflows/jets ejected at few 100s km/s
impact the surrounding ISM and ionize it, making the
jets visible.
The jet creates bow-shocks, where the supersonic flow
collides with the stationary ambient gas.
In reality, the jet ‘lights up’ several dense knots along its
path: these knots possibly mark previous ejection
episodes from the protostar. The knots emit strongly in
forbidden emission lines.
Ejection episodes last 10 000-100 000 yrs
HH objects refer to these dense lit-up knots, not the jet
itself.
magnetic jets
Interactions between the jets/accretion disk/protostar result in energy
loss, and the protostar spins down.
This may help explain the slower rotation rate of MS stars compared to
protostars of equal mass.
Lack of knowledge of SF processes  empirical classification of
protostars
Class 0: Gas infall – no
protostar yet. Emission in the
far-IR/mm.
Class I: protostar embedded in
core; formation of outflows
illuminating the surrounding
dusty envelope. Shorter
wavelength emission results.
Class II: The star becomes
visible emitting in the UV/Opt;
the disk emits in the IR
(T-Tauri phase)
Class III: the infalling envelope
clears, the outflow stops and
the disk becomes optically
thin. Debris-disks follow, and
planetary formation.
Triggered Star Formation in Giant
Clouds
RCW79
Pismis 24
END
model exam q1
4 Spectral types O and B. Upper wavelength limit for ionizing H = 912 Å = 13.6 eV.
[2]
Photoionization occurs when photons with energy higher than 13.6 eV impact a neutral hydrogen atom and remove the
electron from the atom. This becomes unbound, and carries away kinetic energy corresponding to the difference between
the photon's original energy and 13.6 eV.
Recombination occurs when a free electron is captured by an ionized hydrogen atom. Radiation is emitted during
recombination, as the electron cascades down the energy levels.
[3]
These are the recombination lines and forbidden (also called collisionally excited) lines. Hydrogen and helium produce
most recombination lines, whereas metals produce most forbidden lines in typical nebular conditions.
[2]
model exam q2
11 Class 0: Gas infall – no protostar yet. Emission in the far-IR/mm.
Class I: protostar embedded in core; formation of outflows illuminating the surrounding dusty
envelope. Shorter wavelength emission results.
Class II: The star becomes visible emitting in the UV/Opt; the disk emits in the IR. Class III:
the infalling envelope clears, the outflow stops and the disk becomes optically thin. Debrisdisks follow, and planetary formation.
[5]
T-T stars are associated with Class II sources.
Proplyds (PROtoPLanetarY DiskS) have been identified in the Orion HII region by HST. They
are young stars, in the 1-3 solar mass range surrounded by a circumstellar disk and a dusty
cocoon; they are being exposed to the hard radiation field of the Trapezium cluster and are
undergoing mass loss because of this. Their fate is uncertain.
[4]