star formation and pressure - lick observatorywoosley/lectures_winter2016/lecture11.16.pdf · star...
TRANSCRIPT
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Star Formation and Pressure
http://apod.nasa.gov/apod/astropix.html
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Component Fractional
volume Scale
Height (pc) Temperature Density State of
Hydrogen Observational
Technique
Cold dense Molecular
Clouds
< 1% but ~40% of mass
70 - 300 10 - 100 102 - 106 H2 Radio and infrared
(molecules)
Warm Neutral Medium (WNM)
30-70% volume about 50% of mass
300 - 1000 100-
10000 0.2 – 50 H I 21 cm
Coronal Gas (Hot Ionized
Medium)
30 – 70% but <5% of mass
1000 - 3000 106 - 107 10-4 - 10-2 H II
metals also ionized
x-ray ultraviolet
ionized metals recombination
THE THREE COMPONENT INTERSTELLAR MEDIUM
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In which of these components can star formation take place? A necessary condition is a region of gas that has greater gravitational binding energy than internal energy. (The force pulling the region together must be greater than the pressure pushing it apart.) Since internal energy increases with the amount of mass that is present while binding energy increases as M2, there is a critical mass that is bound.
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Ignore factor of 2 in the Virial Theorem. The
clouds we are envisioning have not reached
equilibrium.
≈ KE
≈3
5
GM2
R≈ (Number of particles) (
3
2kT )
≈M
mH
3
2kT (if made of pure hydrogen)
=NA M3
2kT (NA is Avogadro's Number, 6.02 ×1023)
This can be solved for the "Jean's Mass", MJ
3
5
GMJ
2
R=
3
2NAMJkT
MJ =5NAkTR
2G
Clouds of gas with radius R and temperature T that have a mass
bigger than this are unstable to gravitational collapse
The Jean’s Mass
"
"
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KE ∝ M
Ω ∝M2
MJeans
Energy
Mass
BOUND
UNBOUND
For masses larger than the Jean’s Mass gravitational binding energy exceeds internal energy
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It is easier to measure densities and temperatures rather than
radii, so the equation on the previous page can be transformed using
R =3M
4πρ⎛⎝⎜
⎞⎠⎟
1/3
MJ=
5NAkTR
2G=
5NAkT
2G
3MJ
4πρ⎛⎝⎜
⎞⎠⎟
1/3
MJ
2 /3=
5NAk
2G
3
4π⎛⎝⎜
⎞⎠⎟
1/3T
3
ρ⎛⎝⎜
⎞⎠⎟
1/3
MJ=
5NAk
2G
⎛⎝⎜
⎞⎠⎟
3/23
4π⎛⎝⎜
⎞⎠⎟
1/2T
3
ρ⎛⎝⎜
⎞⎠⎟
1/2
= 8.5×1022 gm T3/2
ρ1/2
⎛⎝⎜
⎞⎠⎟= 4.2×10−11 T
3/2
ρ1/2
⎛⎝⎜
⎞⎠⎟
M
assume sphere, constant density M =
4
3πR3 ρ
previous page
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It is more frequent that one finds the density in this contextexpressed as atoms/cm3 rather than gm/cm3.If n= ρNA (strictly true only for H I), then
MJ = 8.5×1022T 3/2NA1/2
n1/2 gm
MJ = 34 T3/2
n1/2 M
where n is the density in atoms cm-3. By this criterion, only molecular clouds and possibly portions of the coldest neutral medium (depending on mass) are unstable to collapse.
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Component Fractional
volume Scale
Height (pc) Temperature Density State of
Hydrogen Observational
Technique
Cold dense Molecular
Clouds
< 1% but ~40% of mass
70 - 300 10 - 100 102 - 106 H2 Radio and infrared
(molecules)
Warm Neutral Medium (WNM)
30-70% volume about 50% of mass
300 - 1000 100-
10000 0.2 – 50 H I 21 cm
Coronal Gas (Hot Ionized
Medium)
30 – 70% but <5% of mass
1000 - 3000 106 - 107 10-4 - 10-2 H II
metals also ionized
x-ray ultraviolet
ionized metals recombination
THE THREE COMPONENT INTERSTELLAR MEDIUM
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Example: Molecular cloud; T = 20 K, n = 104 atoms cm-3
MJ= 34
T3/2
n1/2
= 3420( )
3/2
104( )1/2
= 3489.4
100
= 30 M
Any cloud with this temperature and density and a mass over 30 solar masses is unstable to collapse
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How long does the collapse take?
vesc =2GM
Rτ ff ≈
Rvesc
= R3
2GMbut, ρ, the density, is given by
ρ=3M
4πR3 ⇒R3
M= 3
4πρso,
τ ff ≈3
8πGρ≈ 1300seconds/ ρ
but ρ≈n / NA , so
τ ff ≈ 30 million years/ n
where n is the number of atoms per cubic cm.
Denser regions collapse faster
3 million years if n = 102 atoms/cm3
Three million years is also the lifetime
of the shortest lived stars
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Complications: Rotation Magnetic fields
Fragmentation
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Star formation is inefficient. Even of the collapsing gas only 10 – 20 % of the gas ends up in the star, and overall an even smaller fraction of the cloud collapses to protostars.
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Krumholz et al (2000)
Formation of a massive binary
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http://www.astro.ex.ac.uk/people/mbate/Cluster/cluster3d.html
Collapse of a 500 solar mass cloud 2.6 ly across 285,000 years
Mathew Bate et al
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Power of observing in the infrared
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Spitzer - Orion
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LH 95 A Stellar
Nursery in the Large Magellanic
Cloud (HST)
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The star formation region N11B in the LMC taken by WFPC2 on the NASA/ESA Hubble Space Telescope
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T-Tauri discovered by John Hind in 1852 as a 10th magnitude star. A faint nebula was subsequently discovered nearby (“Hind’s nebula”). Both the star and nebula had variable brightness. The nebula was a “reflection” nebula, shining from the reflected light of T-Tauri.
T-Tauri – in Taurus close to the Pleaides
T-Tauri Stars
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By 1861 the nebula disappeared from view and by 1890 T-Tauri itself had faded to 14th magnitude, about the limit of telescopes then. A faint nebula at the site of T-Tauri itself was observed at that time, Over the next 10 – 20 years, T-Tauri brightened back to 10th magnitude and its local nebula became invisible against the glare. T-Tauri has remained at about 10th magnitude since (but varies).
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T-Tauri - about 400 ly away at the edge of a molecular cloud. FOV here is 4 ly at the distance of T-Tauri http://apod.nasa.gov/apod/ap071213.htm
T-Tauri
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T-Tauri Stars • Short lived phase in life of stars under 2 solar masses. Heavier stars evolve quicker and start burning by the time the star is visible. Above 2 solar masses the objects evolve rapidly and are rarely seen - “Herbig Ae and Be stars”. • Accretion disks and jets are common features • Emission and absorption lines. High sunspot and magnetic activity . • Powered by gravitational contraction, not nuclear burning. In a Kelvin-Helmholtz phase • May be forming planetary systems • High lithium abundance • Embedded in dense, dusty regions • Can be highly variable. Larger luminosity than main sequence stars of same temperature implies larger radii
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Because of rotational support matter hangs up in the equatorial plane forming an “accretion disk”. Matter first rains down on the poles, but then later reverses direction in a strong collimated outflow called a “jet”.
When the star first becomes visible it may still be surrounded by the gas and dust from which it formed. Often jets are seen.
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Protoplanetary disks orbit over half of T-Tauri stars. This shows 5 such stars in the constellation Orion. Picture using HST - field is about 0.14 ly across
http://en.wikipedia.org/wiki/T_Tauri_star
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T-Tauri Star – Drawing showing accretion disk
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30” west of the brightest point in Hind’s nebula is a disk-jet system, Herbig-Haro 30. At the center of this is probably another T-Tauri like star.
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Protostars start off with very large radii because they begin as contracting clouds of gas. They additionally have high luminosities because they are fully convective (more about this later) and able to transport the energy released by gravitational contraction efficiently to their surface. Most of the time is spent close to the main sequence.
ZAMS Sun (30 My)
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Stellar Interiors - Kinds of Pressure
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Pressure is force per unit area
Pressure = Force
Area
Mass 100 gm Area 100 cm2
Area 10 cm2
P=mg
A=
(100)(980)
100=
(100)(980)
10
980dyne cm−2 9800 dyne cm−2
g =GMearth
Rearth2
= 980 cm s−2 F=mg
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http://intro.chem.okstate.edu/1314f00/laboratory/glp.htm
Gas Pressure
More particles -> more pressure Faster particles -> more pressure Heavier particles at the same speed -> more pressure Particles exert pressure on on another, not just on the walls
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P ≈2(mv)(nv) = 2nmv2
Each particle delivers a “kick” = 2 m vx where vx is the change in x-velocity
Approximate this with a group of particles n in one cubic cm all moving to the right with vx = v. The particle flux then = n times v and each particle imparts momentum of roughly mv
*mv
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For a gas, pressure is defined as
P = 13
dndpv p dp∫
where (dn/dp) * dp is the number density (per cm3) of particles having momentum between p and p+dp, and v is their speed. The 1/3 is from an integral over angles. Pressure thus has units
1
cm3
cm
s
gm cm
s=gmcm
cm2s2=dyne
cm2
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Approximation:
suppose momentum p (and therefore v) is constant Then
1
3
dndp
v pdp ≈ 1
3v p dn
dpdp= 1
3v pn∫∫
where n is the total number density of particlesper cubic cm. When one integrates over a distribution of momenta, the 1/3 out front may change.
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IDEAL GAS PRESSURE
• Due to the thermal motion of particles such as electrons, ions, molecules, etc. Particles only interact during their collisions. Particles moving
slower than “c” and not “degenerate”
P ≈ 13
nmv 2 ≈ 13
n 3kT( ) = nkT
but 12
m vrandom2 = 3
2k T
So
P = n k T
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IDEAL GAS PRESSURE
But what is n? The number of particles per cm3 For a given density, n depends upon the composition. E.g. for pure neutral atomic hydrogen, H I, the number of atoms in 1 gram is Advogadro’s number, NA = 6.02 x 1023 atoms per mole. Note that NA= 1/mH where mH is the mass of the hydrogen atom.
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For H I then
Pideal (HI) = ρ NAkT = 8.31×107 ρT dyne cm−2
but what if the hydrogen were ionized? Then there would beone electron for every proton. The electron, though lighter, wouldmove faster and also contribute nekT to the pressure. The total pressure would then be twice as great
Pideal (H II) = 2ρ NAkT = 1.66×108 ρT dyne cm−2
But what if the gas were fully ionized, 75% H II and 25% HeIII like the interior of a recently born star?
In general Pideal = Fρ NAkT = 8.31×107 F ρT dyne cm−2
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It is not difficult to evaluate F but it can be tedious so here is F for various compositions you may encounter
Composition F H I 1 H II 2
ionized 75% H, 25% He 1.69 ionized carbon 0.583 ionized oxygen 0.563 50% C; 50% O 0.573
ionized 35%H; 65% He 1.19
“star stuff”
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Radiation T4
Ideal T
Degenerate electrons 5/3 4/3
* sun
Most main sequence stars have pressures that are dominantly due to ideal gas pressure
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Each pair of electrons occupies a cell
of size ~ (x)3, but x = h/p
Δx ⋅ p= h
Number of cells in volume V = V
Δx( )3 =Vλ 3 =
Vp3
h3
Number of electrons, N, in volume V = 2Vp3
h3
Number of electrons per unit volume ne =NV
= 2 p3
h3
So, pF ne h3
2
⎛
⎝⎜⎞
⎠⎟
1/3
DEGENERACY PRESSURE Pressure due entirely to quantum mechanics and the wavelike nature of the electron. Suppose one packs as many electrons with momentum p into a volume, V, as are quantum mechanically allowed by the wavelength of the electron
This is commonly called the “Fermi Momentum”
http://en.wikipedia.org/wiki/Pauli_exclusion_principle http://en.wikipedia.org/wiki/Electron_degeneracy_pressure
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Now the pressure
P 13
ne pF v= 13
ne pF
mvm
=ne pF
2
3m
ne
3mne h3
2
⎛
⎝⎜⎞
⎠⎟
2/3
Pdeg h2 ne
5/3
3i 22/3 me
DEGENERACY PRESSURE
The contribution of electrons, when present, is much larger than from neutrons or protons because of the 1/m
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pF=m
ev≈
neh3
2
⎛
⎝⎜⎞
⎠⎟
1/3
more accurately 3
8πn
eh3⎛
⎝⎜⎞⎠⎟
1/3
v=1
me
ne
3h3
8π⎛
⎝⎜⎞
⎠⎟⎛
⎝⎜
⎞
⎠⎟
1/3
ne≈
1
2ρ N
A
v=3ρ N
Ah3
16π me
3
⎛
⎝⎜⎞
⎠⎟
1/3
≈ 2 x 1010 ρ106 gm cm-3
⎛⎝⎜
⎞⎠⎟
1/3
cm s-1
At around 107 gm cm-3 the electrons will move close to the speed of light.
As ne goes up the speed of each electron rises
for elements other than H
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"non − relativistic"degeneracy pressure = PNRD
PNRD nepF
2
3me=ne
neh3
2⎛⎝⎜
⎞⎠⎟
2/3
3m
= h2
3 i 22/3mene
5/3 =0.210 h2
mene
5/3
A more accurate calculation gives
PNRD = 120
3π
⎛⎝⎜
⎞⎠⎟
2/3 h2
mene
5/3 = 0.0485 h2
mene
5/3
http://scienceworld.wolfram.com/physics/ElectronDegeneracyPressure.html
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Note that the degeneracy pressure depends only on the density and not on the temperature
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Once the electrons move near the speed of light, the pressure does not increase as rapidly with density as before.
13
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So to get momentum flux just divide energy flux by c.
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THE “PRESSURE” OF SUNLIGHT
From the sun, at the earth’s orbit (1AU), we receive a flux of radiation
φ =L
4πd 2=
L
4π (AU )2
= 1.37 × 106 erg cm−2 s−1
This corresponds to a momentum flux, or pressure
of
P =φ
c=
(1.37×106 )
(3.00×1010 )
erg (s)
cm2 s (cm)
= 4.57×10−5 dyne
cm2since (dyne)(cm)= erg
(1 square meter (104 cm2) would be accelerated 0.46 cm/s2 if it weighed 1 gm; would reach c in about 1000 years)
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1997 Comet Hale Bop
ion tail pushed back by
solar wind
dust tail accelerated by
radiation pressure
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Radiation T4
Ideal T
Degenerate electrons 5/3 4/3
* sun
Most main sequence stars have pressures that are dominantly ideal gas pressure