Fundamental Astrophysics Course Lecture 1 - Overview
• An Introduc on to Stars • Observed Features of Star Forma on • Overview of the Star Forma on Sequence • The InterStellar Medium • Mean Molecular Weight • The Angular Momentum Issue • The Virial Theorem • Linearized Hydrodynamical Equa ons
10:10 pm An Introduc on to Stars
10:10 pm Star field image
10:10 pm The Sun is the best studied star
• Stellar interiors not directly observable. • Solar neutrinos emi ed in core and detectable. • Helioseismology - vibra ons of solar surface can be used to probe density structure • Models of stellar interiors constructed and predic ons of these models are tested by comparison with observed proper es of individual stars.
10:10 pm The building blocks in the universe:stars
corona photosphere
Stellar core
chromosphere 10:10 pm Star clusters
Star clusters are very useful to understand stars and their evolu on: • Stars all at same distance • Stars are dynamically bound • Stars have same age • Stars have same chemical composi on
Can contain 103 – 106 stars
NGC3603 from Hubble Space Telescope
10:10 pm Globular clusters are more massive star clusters
10:10 pm Observable properties of stars
Basic parameters to compare theory and observations: • Stellar mass (M) • Luminosity (L) ∞ – The total energy radiated per second i.e. power (in W): L = L dλ ∫0 λ – If F is the flux (W·m−2) and L is the luminosity (W) or where F is the flux (erg·s−1·cm−2) and L is the luminosity (erg·s−1) and D the distance to the star: L F = 2 • Radius (R) 4πD €
• Effective temperature (Te) – The temperature of a black body of the same radius as the star that would radiate the €same amount of energy. Thus 2 4 L= 4πR σ Te where σ is the Stefan-Boltzmann constant (5.67× 10-8 Wm-2K-4)
10:10 pm A few fundamental values
30 • Solar mass (M) = 1.989x10 kg ~ 2x1030 kg 26 • Solar luminosity (L) = 3.839x10 W (3.939x1026W accoun ng for solar neutrinos) 8 8 • Solar radius (R) = 6.955x10 m ~ 7x10 m (432 000 miles) 2 4 • L= 4 π R σ Te => Te4 = 1.114x1015 => Te (Sun) = 5 777 K
10:10 pm σ = 5.67x10-8 W m-2 K44 Stellar radii • Angular diameter of sun at distance of 10pc: -9 -3 θ= 2 R / 10pc = 5× 10 radians = 10 arcsec • Compare with Hubble resolu on of ~0.05 arcsec ⇒ very difficult to measure R directly
• Radii of ~ 1000 stars measured with techniques such as interferometry and eclipsing binaries.
10:10 pm • 1 parsec = 206 265 AU = 3.26 light-year = 3.0856x1016 m = 1.92x1013 miles • Θ ~ tanθ for small angles
• Θ = 2R / 10 pc = (2 x 7x108) / (10 x 3.0856x1016) = 4.537x10-9rd = 9x10-4 arcsec
10:10 pm The Hertzsprung-Russell diagram
M, R, L and Te do not vary independently. Two major rela onships:
1. L with T 2. L with M
The first is known as the Hertzsprung- Russell (HR) diagram or its observed counterpart, the colour-magnitude diagram (CMD).
10:10 pm Colour-Magnitude Diagrams (CMD)
2 3 Measuring accurate Te for ~10 or 10 stars is an intensive task – spectra needed and model atmospheres
Magnitudes of stars are measured at different wavelengths: standard system is UBVRI Band U B V R I λ/nm 365 445 551 658 806 W/nm 66 94 88 138 149
10:10 pm Other filters exist over the en re electromagne c spectrum and not only in the op cal range
10:10 pm Magnitudes and Colours
(γ Orionis)
(α Orionis)
10:10 pm Magnitudes and Colours Various calibra ons* can be used to provide the colour rela on:
B-V = f (Te) or any color = mag1 – mag2 = f (Te) rela on
•Note that the observed (B - V) must be corrected for the interstellar ex nc on to get
(B - V)0
10:10 pm Extinction
•Ex nc on: Absorp on and sca ering of electromagne c radia on by dust and gas between the observed object and the observer •The ex nc on due to dust is not equally effec ve at all wavelengths. The shorter the wavelength, the higher the ex nc on - blue light is affected more strongly than red light.
•Therefore , stars behind a lot of dust look redder than they really are. This is called interstellar reddening.
10:10 pm UV and Dust Emission in the Far-IR at all scales
UV/op cal Far-infrared
10:10 pm UV and Dust Emission in the Far-IR at all scales
10:10 pm Absolute magnitude and bolometric magnitude
• Absolute Magnitude M (≠ from mass) defined as apparent magnitude of a star if it were placed at a distance of 10 pc m – M = 5 log(d/10) – 5 where d is in pc
• Magnitudes are measured in some wavelength band e.g. UBV. To compare with theory it is more useful to determine bolometric magnitude – defined as absolute magnitude that would be measured by a detector sensitive to all wavelengths. We define the bolometric correction to be:
BC = Mbol – Mv Bolometric luminosity is then: Mbol – Mbol = -2.5 log L/L
10:10 pm Bolometric corrections for Main-Sequence Stars
From Allen’s Astrophysical Quan es (4th edi on)
10:10 pm 2. The Mass-luminosity relation
For main-sequence stars, there is a Mass-luminosity relation.
L ∝ Mn Where n ~ 3.5. Slope changes at extremes, less steep for low and high mass stars.
This implies that the main- sequence (MS) on the HRD is a function of mass i.e. from bottom to top of main-sequence, stars increase in mass We must understand the M-L rela on and L-Te rela on theore cally. Models must reproduce observa ons
10:10 pm Over a wide mass range
10:10 pm Age and metallicity
There are two other fundamental properties of stars that we can measure: age and chemical composition
Composition is parameterised with X, Y, Z ≡ mass fraction of H, He and all other elements
e.g. X = 0.747 ; Y = 0.236 ; Z = 0.017 Note: Z often referred to as metallicity (while z is redshift)
Ideally, astronomers would prefer studying stars of same age and chemical composition to keep these parameters constant and determine how models reproduce the other observables => star clusters.
10:10 pm Globular cluster example Selec on of Open clusters Age (yrs)
2.0x106 branch Horizontal 6.5x106
2.8x107 Turn off
1.6x108
1.9x109
• In clusters, t and Z must be same for all stars 7.1x109 • Hence differences must be due to Minit • Stellar evolution assumes that the 10 differences in cluster stars are due mainly to 2.9x10 initial mass: Minit • Cluster HR (or colour-magnitude) diagrams are quite similar: the age defines the overall appearance
10:10 pm Globular vs. Open clusters
Globular Open
• MS turn-off points in similar • MS turn off point varies position. Giant branch joining massively, faintest is MS consistent with globulars • Horizontal branch from giant • Maximum luminosity of MS
branch to above the MS turn- stars can get to Mv ≈ -10 off point • Very massive stars found in
• Old stellar populations (at these clusters least in the MW)
The differences are interpreted due to age. Open clusters lie in the disk of the Milky Way and have large range of ages. In the Milky Way, globulars are all ancient, with the oldest tracing the earliest stages of the forma on of Milky Way (~ 12× 109 yrs)
10:10 pm Summary • Definition : a star burn hydrogen in its core • Characteristics:
o 0.08 < M / Msun < a few 100 -4 6 o 10 < L / Lsun < 10 o L ~ M3.5 (strong dependence) o Lifetime τ = M / L ~ M / M3.5 = M-2.5 • Four fundamental observables used to parameterise stars and compare with
models M, R, L, Te • M and R can be measured directly in small numbers of stars • Age and chemical composition also shape the HR diagram • Stellar clusters are crucial laboratories to study stars and their evolution: all stars at same distance, same t, and initial Z
• We will develop models to attempt to reproduce the M, R, L, Te relationships and understand HR diagrams
10:10 pm Mass func ons • The stellar masses are one of the most important factor in determining their evolu on, so when studying a stellar popula on, we are interested in es ma ng their masses. • An important informa on is the number of stars per unit mass which is called a mass func on. • We define the mass func on Φ(M) such that Φ(M) dM is the number of stars with masses between M and M + dM.
• With this defini on, the total number of stars with masses between Mlow
and Mup is: Mup N(M ,M ) = Φ(M)dM low up ∫Mlow • By deriving both sides, we get: dN = Φ(M) € dM • Φ is the deriva ve of the number of stars with respect to mass, i.e. the number of stars dN within some mass interval dM. € 10:10 pm Total mass of stars between Mlow and Mup
• If we are interested in the total mass of the stars between Mlow and Mup in a given system rather than the number of such stars, we must integrate Φ mes the mass per star. Thus the total mass of stars with masses between Mlow and Mup is: Mup M (M ,M ) = MΦ(M)dM * low up ∫Mlow • or equivalently: y = ln(m) dM dy/dm = 1/m * = MΦ(M) = ξ(M) dy = dm/m dM d[ln(m)] = dm/m € • ξ (M) gives the number of star per ln(M), rather than per number in mass. • Physical explana on:
– Suppose that€ Φ(M) = Cte constant: there are as many stars from 1 − 2 M as there are from 2 − 3 M as there are from 3 − 4 M , etc. – Instead suppose that ξ(M) = Cte: there are equal numbers of stars in intervals that cover an equal range in logarithm, so there would be the same number from 0.1 − 1 M , from 1 − 10 M , from 10 − 100 M , etc.
10:10 pm Star Forma on Rate (SFR)
SFR = the stellar mass formed in an object (e.g. a galaxy) per unit me:
In the Milky Way, it amounts to ~ 2 M / yr But it could be up to several 1000 M / yr in high-redshift starbursts.
• Instantaneous SFR: limΨ(t) ⎯dt ⎯→⎯ 0→ SFRinst
• SFR averaged over 10 millions years: SFR10Myrs • SFR averaged over 100 millions years: SFR100Myrs €
10:10 pm • The distribu on of stellar masses (IMF) might be invariant [hot topic].
• If we examine two popula ons of different sizes, then Φ(M) = dN/dM will be different because they have different numbers of stars. However, they may have the same frac on of their stars in a given mass range.
• So, it is generally common to normalize Φ or ξ so that the integral is unity, i.e. to compute a normaliza on factor for Φ or ξ such that the integrals are equal to 1.
• When the mass is normalized, Φ(M) dM and ξ(M) dM give the frac on of stars (frac on by number for Φ and frac on by mass for ξ ) with masses between M and M + dM.
with mlow= 0.1M and mup = 120M In LMC at 165 000 l.y. •R136a1 : 265 M (?) •R136a2 : 195 M (?) ηCarinae : 120 M
10:10 pm The Salpeter IMF
If we ignore the low-mass fla ening of the IMF below ~ 1M, we might assume that the same slope holds over the range Mmin = 0.1M to Mmax = 120M.
-2.35 This is known as the Salpeter IMF: ϕ(M) = ϕ0 M The normaliza on ϕ0 (in units of M ) is evaluated by requiring that the integral equals to 1: M max M max A A 1 = Φ(M)dM = AM −2.35dM = (M −1.35 − M −1.35) = (M −1.35 − M −1.35) ∫M min ∫M min −1.35 max min 1.35 min max 1.35 A = −1.35 −1.35 = 0.060 with Mmin = 0.1MΘ and 120MΘ Mmin − Mmax € In a similar way, we have in terms of mass: € M max M max B B 1 = ξ(M)dM = BM −1.35dM = (M −0.35 − M −0.35) = (M −0.35 − M −0.35) ∫M min ∫M min −0.35 max min 0.35 min max 0.35 B = −0.35 −0.35 = 0.17 with Mmin = 0.1MΘ and 120MΘ Mmin − Mmax 10:10 pm €
€ -2.35 -1.35 So, we will have: ϕ(M) = 0.060 M and ξ(M) = 0.17 M for M = 0.1 – 120M We can es mate the frac on of stars by number and by mass in a given mass range. For instance, what is the frac on of stars by number (1) and by mass (2) more massive than the Sun in a new-born popula on ?
120 120 1) f (> M ) = 0.060Φ(M)dM = 0.060M −2.35dM = N Θ ∫1 ∫1 0.060 f (> M ) = (1−1.35 −120−1.35 ) = 0.045 N Θ 1.35
120 120 2) f (> M ) = 0.17MΦ(M)dM = 0.17M −1.35dM = M Θ ∫1 ∫1
€ 0.17 −0.35 −0.35 f (> M ) = (1 −120 ) = 0.40 M Θ 0.35 Only 4.5% of the stars are more massive that the Sun but, the mass of these stars amounts to 40% of the new-born stars. €
10:10 pm Two remarks
The total stellar mass of a system is computed from the following integral: M max M max M max M = MΦ(M)dM = AMM −2.35dM = AM −1.35dM tot ∫M min ∫M min ∫M min A A M = (M −0.35 − M −0.35 ) = (M −0.35 − M −0.35 ) tot −0.35 max min 0.35 min max It shows that mostmost of the of the stellarstellar mass lies in mass lies in low-masslow-mass stars. stars
3.5 € The total luminosity (assuming L = C M ) is computed as follows: M max M max M max L = L(M)Φ(M)dM = CM 3.5 AM −2.35dM = ACM1.15dM tot ∫M min ∫M min ∫M min AC L = (M 2.15 − M 2.15) tot 2.15 min max Which shows that the total the total stellar luminosityluminosity isis dominated by the by the most massive stars. massive stars
€
10:10 pm How to evaluate the IMF ? The first IMF was determined by Salpeter (1955) and the following methodology is used:
(1) Es mate the luminosity func on in a volume: dN/dMV where MV = absolute magnitude. (2) Es mate the mass-luminosity: dMV / dln(m) -> PDMF (Present-Day Mass Func on):
The PDMF is the mass distribu on observed today (t0 = age universe today) Then, we must correct the PDMF for the star forma on history and dead stars For m ≤ 0.9 M: life me > age of the galaxy (t0 - tform, where tform = forma on galaxy).
For m > 2 M: stars have a very short life me (τm) compared to the age of the galaxy.
In the intermediate range 0.9 < m ≤ 2 M: compu ng the IMF is more complex ...
10:10 pm Typical masses for a Salpeter IMF
• Median mass: 0.17 M
• Mean mass: 0.57 M
• 5% of the stellar mass produces 95% of the luminosity (m > 25 M)
• 50 % of the luminosity mass at m* = 72 M 1% of the gas mass converted into stars
(m>72 M) produces 50 % of the luminosity
10:10 pm Other more realis c IMFs (mainly for faint stars)
IMF Scalo (1998): ξ(m) = m-α IMF Kroupa 2001: ξ(m) = m-α
• α = -0.2 ± 0.3 for 0.08 ≤ m/M < 1 M [α= 0.3 for m/M < 0.08] • α = -1.7 ± 0.5 for 1 ≤ m/M < 10 M α= 1.3 for 0.08 ≤ m/M <0.5 • α = -1.3 ± 0.5 for 10 ≤ m/M < 100 M α= 2.3 for 0.5 ≤ m/M
10:10 pm Sample Ini al Mass Func ons of Stars
⎧ −2.35 ΦSalp (M) ∝ M , for 0.1MΘ < M <100MΘ Φtop −heavy (M) ∝ ⎨ −1 ⎩ e.g. M , for 100MΘ < M < 500MΘ 39 10:10 pm
€ Variability of the IMF ? An important if not crucial Marks et al. (2012) ques on is:
• Is the Ini al Mass Func on universal or does it vary with the environment, the element abundances (metallicity), the redshi , the star forma on rate, the Jeans mass, … ?
10:10 pm Why is the IMF of major importance ? Several reasons than range from a basic be er understanding of the star forma on process to wider Xtragalac c / cosmological consequences:
• Changing the IMF => changing the stellar mass (M*) of the stars in a system (e.g. a galaxy)
• Changing M* => changing the star forma on rate (SFR = M* / Δt) and therefore all parameters based on SFR: o Cosmic SFR density = density of star forma on per unit volume of the universe at a func on of reds it
o Specific SFR = SFR / M* => very important parameter that provides an es mate of the star forma on ac vity in galaxies Parcours astrophysique: o φ of Galaxies by Buat & Ilbert o φ of Cosmic Structure by Cattaneo & Adami
10:10 pm Cosmological implica ons of a stellar ini al mass func on that varies with the Jeans mass in galaxies Evolu on of cosmic SFR density.
The purple solid line shows model form from Hopkins et al. (2010), assuming a Kroupa IMF With a varying IMF model, the SFR density decreases, as is shown by the red dashed line The blue dash–do ed line shows the Wilkins et al. (2008) SFR density that would be necessary to match the observed present-day stellar Narayanan & Davé (2012) mass density
Vincole o et al. (2012) Kistler et al. (2009)
10:10 pm IMF of nonrota ng primordial stars (Z = 0). We dis nguish 4 regimes of ini al mass:
• LOW-MASS STARS < 10 M⊙ => WHITE DWARFS
• MASSIVE STARS [10, 100 M⊙] => NEUTRON STARS
• VERY MASSIVE STARS [100, 1000 M⊙] => BLACK HOLES (DELAYED OR DIRECT) BUT [140, 260 M⊙] SINGLE PULSATION DISRUPT THE STAR
• SUPERMASSIVE STARS >1000 M⊙ (?) => COMPLETE COLLAPSE TO BLACK HOLE
10:10 pm 10:10 pm Giant molecular clouds (GMC) Star formation takes place in cold, dense gas clouds: The molecular clouds. Stars form in groups or clusters. The largest GMC in Orion is about 1000 light years away. Hot young stars (25-50 million year old) ionize their surroundings and are therefore easily visible. ! main characteristics: - molecular hydrogen - ~1% of dust (Si and C) - organic and non-organic molecules ammoniac NH3, formaldehyde H2CO, acetylene HC3N - ~ 105-106 solar masses - ~ 1000 atoms per cm3 - clumpy: 103-104 solar masses clumps - T ! 10 K -3 ⇒ Wien law λmax = 2.9x10 / T ~ 290 μm ! structure ⇒ Observation in Far-IR (Herschel, ALMA, PdB)
- emission of the H2 molecule but difficult because small dipole moment. Molecules with larger dipole moment (CO) are better but do not represent the mass... - dust emission (radio band). Interpretation difficult since many parameter unknown (departure from LTE, opacity, dust temperature, etc.) - IR cameras on large telescopes measure the absorption on thousands of Bok globule: Barnard 68 background stars → map the dust distribution in the cloud
Bok globules Example characteristics: - 410 light years away - diameter ~ 12'500 AU - T ~ 16 K - ~ 2.1 solar masses 10:10 pm - in gravitational equilibrium -12 with Pbound=12.5 " 10 Pa Observed Features of Star Forma on
10:10 pm Star formation: observations - Herbig-Haro objects HH-30 - first discovery by Herbig and Haro in 1951. - patchy nebulous structures at an extended jet (0.1-0.3 ly=6000-18000 AU) - formed through collision of jet launched by young stars colliding with ISM
- frequent in star forming regions knots - super-sonic flows: 200 to 400 km/s - 50% bipolar, knots in jets, variable on year timescales collimation - mass loss: 10-10 to 10-6 solar mass / year by B fields - hot bow shocks with molecular emission lines (CO, OH, H2O) - Not older than a few 100 000 years DG Tau HR 4796 - T-Tauri stars - pre-main sequence precursors of F, G, K, M stars - energy generation mostly by contraction - IR excess (protoplanetary disk) - variable (irregular, accretion phenomena), active - emission lines (Ca, H, Balmer lines, Li) - disk + bipolar outflows - age: 1 to 10 million years - 2 classes: - classical T-Tauri CTTS (with disks, strong IR excess) 1206 HERNA´ NDEZ ET AL. Vol. 686 Hernandez et al. 2008 - weak-line T-Tauri WTTS This could indicate that the low disk presence observed in the Velorum cluster is abnormal for its evolutionary stage, and envi- (without disks, weak/no IR excess) ronmental effects, such as strong stellar winds and/or strong ra- diation fields from the Velorum system, could provide the physical mechanism for the low disk frequency. Moreover, 75% - Disks of the disk-bearing stars in the Velorum cluster show IRAC SED slopes smaller than the median values of other 5 Myr old - mass: 2 ! 10-3 to 10-1 solar masses stellar groups plotted in Figure 10 suggesting that the disks of the Velorum cluster have a higher degree of dust settling. In par- - no correlation mass vs. age ticular, the median IRAC SED slopes for the disk population of NGC 2362, the k Orionis cluster, the OB1b subassociation and - no disks around stars older than 10 Myrs the Velorum cluster are 1.72, 1.60, 1.70, and 1.82, re- - temperature: " 50-300 K at 1 AU spectively (with a typical errorÀ ofÀ 0.06). À À Studying the young (2–3 Myr) open cluster NGC 2244, Balog T ! r q with q = -1/2 to -3/4 et al. (2007) showed that high-mass stars (O-type stars) can affect the primordial disks of lower mass members only if they are - outer radius by about 100 AU within 0.5 pc of the high-mass star. We find similar results for the Velorum cluster. Using the photometric members with - sometimes gap in the center V J > 3:5( M2 or later), we find a disk frequency of 4% 3%À at a projected distance of 0.25–1.0 pc; inside 0.25 pc theÆ - UV excess at the inner boundary (accretion shock) central objects of the cluster contaminate the optical photometry used to select photometric candidates ( 3). The closest primor- dial disk is located at projected distancex of 0.5 pc. At larger Fig. 11.—Fraction of stars with near-infrared disk emission as a function of projected distance (>1.0 pc), the disk frequency is larger (8% the age of the stellar group. Open circles represent the disk frequency for stars in 2%). This suggests that the relative fast dispersion of disks in theÆ the T Tauri (TTS) mass range ( K5 or later), derived using JHKL observations: Velorum cluster is produced by the strong radiation fields and NGC 2024 and Trapezium ( Haisch et al. 2001), and Chameleon I (Go´mez & Kenyon 2001). Solid symbols represent the disk frequency calculated for stars strong stellar winds from the central objects. This result must still in the TTS mass range using Spitzer data (see text for references). be considered tentative given the small number of stars with disks, which results in large errors in disk frequencies. Alternatively, Figure 11 indicates that the disk frequency drops introduced by theoretical evolutionary tracks, Figure 10 shows rapidly at 5 Myr and the relative low frequency of disks ob- that the Velorum cluster is in a similar evolutionary stage as the served in the Velorum cluster could be explained if the cluster other stellar groups with ages normally quoted as 5 Myr: the is slightly older than 5 Myr. Additional studies of photometric k Orionis cluster (Barrado y Navascue´setal.2007;Dolan& members presented in Table 1 are necessary to explain the disk Mathieu 2002), the cluster NGC 2362 (Dahm & Hillenbrand population found in Velorum cluster. 2007), and the Orion OB1b subassociation (Bricen˜o et al. 2007; Herna´ndez et al. 2007b). We estimate an error of 1.5 Myr com- 6. CONCLUSIONS paring the standard deviation in the color V J for the member We have used the IRAC and MIPS instruments on board the sample (see 3) to the standard deviationÀ obtained using the Spitzer Space Telescope to conduct a study of disks around the theoretical isochronesx from Siess et al. (2000) with a reference Velorum cluster. Since the central object is a binary system age of 5 Myr. consisting of the closest known Wolf-Rayet star and a high-mass Figure 11 shows the disk frequencies of late-type stars (stars K O star, a strong UV radiation field and stellar winds are present in middle or later) with near-infrared disk emission in different stel- the cluster. Using optical photometry of X-ray sources (2XMM) lar groups, as a function of age (Herna´ndez et al. 2005, 2007a; and members confirmed by spectroscopy (R. D. Jeffries et al. Haisch et al. 2001). Using the number of members with V J > 2008, in preparation), 579 photometric candidates were selected 2:0( K5 or later) expected in our photometric sample (Fig.À 3, as possible members of the cluster. The level of contamination dashed histogram) and the disk detected in 4.3, we calculated a by nonmembers depends on the V J color range, showing the primordial disk frequency of 6% 2%for thex Velorum cluster. highest level of contamination (À68%) at V J 1:5–3.5, We include recent Spitzer resultsÆ for the young stellar clusters where the field giant branch crosses the young stellarÀ population.¼ NGC 1333 (Gutermuth et al. 2008), NGC 2068/71(Flaherty & Combining optical, 2MASS, and Spitzer data we have detected Muzerolle 2008), Taurus (Hartmann et al. 2005), NGC 7129 infrared excess in 29 stars. One of the infrared excess stars is a (Gutermuth et al. 2004), Chameleons (Megeath et al. 2005), Be star. We report five debris disks around A-type stars, five debris Tr 37 and NGC 7160 (Sicilia-Aguilar et al. 2006), IC 348 (Lada disks around solar-type stars (spectral type range F to early K) et al. 2006), NGC 2244 (Balog et al. 2007), NGC 2264 (Cieza and one solar-type star with infrared excess produced by a very & Baliber 2007), Ori (Herna´ndez et al. 2007a), NGC 2362 massive debris disks or by a primordial disk with a high degree (Dahm & Hillenbrand 2007), k Ori (Barrado y Navascue´s et al. of dust settling. Seventeen disk-bearing low-mass stars (K5 or 2007), Upper Scorpius (Carpenter et al. 2006), and Orion OB1b later) were found in the cluster with a range of disk properties. and 25 Ori (Herna´ndez et al. 2007b). We classified these objects in three classes using the infrared ex- The disk frequencies decrease toward older ages with a time- cess at 5.8 m and the SED slope 8:0 24 : nine Class II stars, scale for primordial disk dissipation of 5 Myr. It is apparent seven evolved disks star, and one½ pretransitionalÀ½ candidate. We that the disk frequency found in the Velorum cluster is lower found that 76% of the stars bearing primordial disks have color than that found in young stellar populations with similar ages, V J > 4( later than M3.5), indicating a mass-dependent and comparable to the disk frequency in older stellar groups. timescaleÀ for disk dissipation in the Velorum cluster, similar Quick Overview of the Star Forma on Sequence
10:11 pm Young stellar objects (YSO)
Classification of YSO according to evolutionary stage based upon spectral and physical characteristics. 0) GMC *Dust is a strong absorber in VIS and UV->heats up->radiates thermally at ~300 K ->IR excess 1) radiation (over what is expected from the star) at 10-100 microns.
Protostar
2) Main accretion phase The cloud collapse is very rapid!
3) Late accretion phase Giant planets ? * (direct instability)
PMS star Classical T Tauri Optically thick disk Planet formation 4) Giant planets (core accr.)
Weak line T Tauri 5) Optically thin disk
10:11 pm Rocky planets Star Forma on Sequence in brief
5 • Jeans instability => gas cloud collapse begins (~ 10 Msun) • Isothermal collapse (free-fall me 108 years) • Fragmenta on of the cloud of gas • Center of cloud becomes op call thick: adiaba c compression (~ 10-13 g/cm3) => thermodynamically isolated i.e., no heat transfered to the surroundings • First hydrosta c core forms: ~ 170K (first equilibrium phase)
• H2 dissocia on (T ~ 2 000K) and second core collapse -3 • Ionized H in second core, dynamically stable again (10 Msun, 20 000K, second equilibrium phase) • Pre-Main Sequence contrac on • Zero Age Main Sequence (ZAMS): luminosity produced by H
fusion, minimum mass 0.08 Msun
10:12 pm Overview star formation sequence: Nomenclature •Protostar •optically thick stellar core •forms during the end of the adiabatic contraction phase and grows during the accretion phase. -5 •large accretion rates ~10 Msun/yr
•Pre Main Sequence Star •visible in the optical -7 •small accretion rates ~10 Msun/yr •energy generation mainly via contraction (via protoplanetary disk)
•ZAMS: Zero age main sequence •PMS contraction => center heats up •~3~3 10 Mio6 K: K: H burning ignites •contraction stops, energy production mainly via fusion
10:15 pm Composition of the interstellar medium
Nuclear abundances The atomic composition of the interstellar medium is the result of Big Bang nucleosynthesis (H, He, some Li and Be) followed by enrichment due to processes connected with stellar evolution. Galactic chemical evolution can be summarized by the following drawing: GALAXY
gas stars Since there is no compelling evidence of strong fractionation at the time of the formation of the solar system, we can suppose that the infall enriched gas chemical composition of the solar system reflects the composition of the interstellar matter at 8.5 kpc of the galactic center 4.6 billion wind remnants: years ago. neutron stars white dwarfs galaxy
closed box models (if no exchange) Arnett 1996
The "closed box" model, as introduced by Tinsley (1974), is characterized by a SFR ψ(t), total mass Mt, and gas mass Mg(t) in the form of gas, in the instantaneous recycling approxima on (IRA). The system is closed, with MNote:g(0) = Mt. The evolu on of gas mass and metal content is regulated by the following expressions: •abundances decrease on average with increasing nucleon number and •nucleons with even numbers are more wherestable R isand the returnedhence frac on of more abundantgas defined as: …………………………. •nucleons with magic numbers of protons (2,8,20) are also more abundant and P is the primary metal produc on factor defined as: …………….. •ironZ (A=56) is the nuclei with the largest Big bang 10:15 pm binding energy nucleosynt. stellar nucleosynt. supernova nucleosynt. (neutron capt.) The Mean Molecular Weight
5:00 pm The Mean Molecular Weight [kg][m2][s-2][K-1][kg-1][kg][m3] [kg][m3][kg][m2][s-2][K-1][kg-1] • For an ideal gas with n par cles per unit volume, ℜ ℜρ molecular weight μ, the equa on of state (EoS) is: P = nkT = ρT ⇒ µ = µ nk -23 with ρ = n μ mu (the Boltzmann constant k = 1.38 x 10 J / K, the universal gas constant 7 -23 R = k / mu = 8.31 x 10 erg / K / g and the atomic mass unit mu = 1 amu = 1.66053 x 10 kg). Note that the molecular weight μ is dimensionless € • The plasma in stars could be taken as one single (complex) component if all the components follow the ideal gas law
• The chemical composi on can be described by specifying all Xi, the weight frac ons of each nuclei with a molecular weight μi and charge Zi. Assuming ni nuclei per unit volume and a density per nuclei ρi with Xi = ρi / ρ
ρi ρ X i ni = = µimu mu µi 10:03 am
€ The Mean Molecular Weight
• To es mate the total pressure P, we account for all the par al ones: ⎛ ⎞ P = Pe + ∑Pi = nekT + ∑nikT = ⎜ ne + ∑ni ⎟ kT i i ⎝ i ⎠
with Pe is due to free electrons and Pi are the par al pressures due to each nuclei. • The contribu on of € ionized atoms of element i
(i.e. the nucleus plus Zi free electrons) :
n = ne + ∑ni =∑Zini + ∑ni = ∑(Zi +1)ni i i i i • So: ρ X i P = nkT = ∑(Zi +1)nikT = ∑(Zi +1) kT = i i mu µi
k X i X i (1+ Zi ) € P = ∑(Zi +1) ρT = ℜ∑ ρT mu i µi i µi
10:03 am
€ Then Mean Molecular Weight
X (1+ Z ) • So from: P = ℜ∑ i i ρT i µi
ℜ • And, by analogy with: P = nkT = ρT € µ
• We define the mean molecular weight: € X (1+ Z ) µ−1 = ∑ i i i µi
10:03 am € X (1+ Z ) The Mean Molecular Weight µ−1 = ∑ i i i µi
• What is the objec ve of introducing the mean molecular weight ? € • Doing so means that we can handle a mixture of ideal gases as a single ideal gas. The idea is to use the mean molecular weight in replacement to the classical molecular weight.
−1 X i • For a neutral gas, no free electrons => 1 + Zi -> 1: µneutral = ∑ i µi
• If we wish to define the mean molecular weight for free electrons:
−1 X iZi µe = ∑€ i µi
5:42 pm € The Mean Molecular Weight
For instance • Mass frac ons:
o X = mass of H / total mass with μatom = 1 and μmolecule = 2 o Y = mass of He / total mass with μ = 4
o Z = mass of metals / total mass with <μZ> = 15.5 (in average for the Sun)
−1 X i X Y Z X Y Z Then: µneutral = ∑ = + + = + + i µi µH µHe µZ 1 4 15.5 The proto-solar mass frac ons are X = 0.7110, Y = 0.2741 and Z = 0.0149 (e.g. Lodders 2003). 0.7110 0.2741 0.0149 µ −1 = + + = 0.7805 € So, we have for atomic H: atom 1 4 15.5
µatom =1.281 0.7110 0.2741 0.0149 And for molecular H: µ −1 = + + = 0.4250 molecular 2 4 15.5
5:46 pm € µmolecular = 2.353
€ The Mean Molecular Weight
• For a fully ionized gas, we have approximately the same mass but more par cles (electrons). • That translates into a lower mean molecular weight: X (1+ Z ) X × (1+1) Y × 3 1+ Z µ−1 = ∑ i i = + + × Z i µi 1 4 µZ 1+ Z • The same number of protons and neutrons in a nucleus means: ≈ 0.5 µZ • € And therefore (remember: X+Y+Z = 1): 3Y 1 µ−1 = 2X + + (1− X −Y) 4 2 € 1 2 2 2 µ = = = = = 0.612 3/2X +Y /4 +1/2 3X +Y /2 +1 3 × 0.7110 + 0.2741/2 +1 3.270
10:38 pm € The Angular Momentum Issue
10:38 pm • Let’s assume a cloud of gas (mass M, ini al
radius R0, ini al angular velocity ω0). 2 • Ini al Kine c energy KE0 = ½ I ω0 where I is 2 the rota onal iner a [similar to ½ M v0 linear) 2 2 • v0 = ω0R0 and KE0 = ½ M ω0 R0 2 • Ini al gravita onal energy* Ω0 = 3/5 GM /R0 • The conserva on of the angular momentum 2 2 L = I ω0 = M ω0 R0 = M ω R at any me • When the cloud collapses, R and ω to 2 keep L constant and KE = ½ I ω un l KE = Ω0
10:42 pm 2 2 2 • KE / Ω = (1/2 Mω R ) / (3/5 GM /R) 2 2 2 = (1/2 Mω0 R0 ) / (3/5 GM /R0) x (R0 / R)
= (KE0 / Ω0) x (R0 / R) • At equilibrium, KE / Ω ~ 1 • => the amount of contrac on to reach the equilibrium for a cloud of gas en rely depends
on KE0 / Ω0
10:42 pm *Gravita onal poten al energy of a uniform gaseous sphere of mass M and radius R • V = 4πr2dr • M = 4πr2ρ(r)dr • dU = -GM(r)dM/r • For each shell: M(r) = 4/3 πr3<ρ> • dM = 4πr2<ρ> dr • dU = -GM(r)dM/r = -G[4/3 πr3<ρ>][4πr2<ρ>]/r dr • dU = -16/3 Gπ2 <ρ>2 r4 dr 2 2 R 4 • ∫ dU = -16/3 Gπ <ρ> ∫0 (r dr) • ∫ dU = -16/3 Gπ2<ρ>2 R5/5 with <ρ> = M/(4/3 πR3) • U = -3/5 GM2/R
7:18 pm The Theproblem problem of the of angularangular momentum momentum Structure formation in the universe (galaxies, stars, planets, ...) one must get rid of angular momentum,For the structure (galaxies, stars, as structure formation isplanets associated) to form with in the size decrease.Universe, one must get rid of angular momentum. Structure forma on is associated with a decrease in size.
grav. only
grav. “centripetal force” (ang. momentum)
Disk formation How to shrink the disk? One possibility is to have the angular momentum redistributed, e.g. give all L to a small One possibility is to get the angular momentum redistributed. Give all L to a small fraction of frac on of the mass. In the Keplerian poten al, the orbital frequency Ω from Kepler’s 3rd mass, accrete the rest. law for circular orbits is given as: In the Keplerian potential, the orbital frequency2 3 is 2given as2 3 3 3 4π R ⎛ 2π ⎞ 4π R GM − − T 2 = ⇒ ⎜ ⎟ = ⇒ ω = = GM × R 2 = ω × R 2 so for circularGM orbits⎝ ω ⎠ GM R3 0 2 2 -3/2 1/2 whichSo, for meanscircular that orbits also a, L = smallmRv mass, = mR butω = mR at a largeω0 x R distance, = mω can0R contain significant L. Wewhich thus needmeans to that segregate a small mass mass, if andat angular large radius, momentum.can contain a significant L. We thus need to segregate€ mass and angular momentum at large radius to form structures. We recall that the Sun contains 99.96% of the mass, but only 0.6% of the angular momentum. Note: the Sun contains 99.96% of the mass of the Solar system but 0.6% of the angular momentum ! The problem of angular momentum II
Angular momentum is an enemy of star formation. To form a star, a gas cloud must get rid of a large amount of angular momentum.
•average density: -of the ISM: ~10-24 g/cm3 -of a star: ~ 1 g/cm3 •to form a star, the volume must therefore shrink by 24 orders of magnitude and therefore in radius by about 8 orders. If angular momentum is conserved, the cloud must rotate 10-8 times slower than young stars. The latter typically rotate with 50-200 km/s which implies a cloud rotation speed of 0.05 to 0.1 cm/s i.e. mm/s! a) Minimal rotation from galactic differential rotation Differential rotation of the galaxy leads to R [CM = Center of Mass] angular momentum relative to the CM at decoupling of the cloud. v1 r=radius cloud (e.g. 20 pc~) 4 Mio AU) R=galactocentric distance of the vCM r r cloud (Sun: ca 8 kpc) R>>r
!v="cr angular frequency " = 2#/T vrot(R)= " R
v1: rotational velocity around the galaxy center of a gas parcel at R+r Galac cto GC Center vCM: rotational velocity around the galaxy center of a gas parcel at R 7:10 pm (at the mass center of the gas cloud) The problem of angular momentum III
If the cloud collapse, the velocity difference must be transformed into cloud rotation
(uv)’ = u’v + uv’
Examples •Solid body rotation prograde •Keplerian rotation
7:09 pm retrograde The problem of angular momentum IV
b) Measuring galactic rotation (subtract solar velocities) Ro GC Sun l R d l !0R0 From simple geometry α Star
l=galactic longitude !R α
We further assume d< (w. Taylor) Inserting this in the expressions for the velocities yields with the Oort constant with the Oort constant All quantities measurable from radial velocity (vrad), astrometry (proper motion μ, parallax d). A=14.5 ± 1.5 km/s kpc-1 Numerical values: 7:06 pm B=-12 ± 3 km/s kpc-1 R GC 0 Sun l 1 r 2 R d Star 1 2 2 2 R = r + (R0 – d cos l) 2 d2 = (d cos l)2 + r2 => r2 = d2 - (d cos l)2 1 2 2 2 2 2 2 2 2 2 2 & => R = d – d cos l + R0 + d cos l - 2 R0 d cos l = d + R0 - 2 R0 d cos l With (d << R & d << R0) 2 2 2 2 2 2 2 2 1 = d /R + R0 /R - 2 R0/R d cos l ~ R0 /R - 2 R0/R d cos l with R ≠ 0 2 2 2 2 R = R0 - 2 R0d cos l => R - R0 = - 2 R0d cos l = (R – R0) (R + R0) ~ 2 R0 (R - R0) 2 R0 (R - R0) = - 2 R0d cos l => R – R0 = - d cos l 4:32 pm The problem of angular momentum V c) Resulting cloud rotation Recalling that the cloud rotation is given by We obtain by inserting the expressions for the Oort constants: For a typical GMC, R~20 pc, and the rotation velocity is therefore vrot ~ 50 m/s. This is a factor 104 to 105 larger than the velocity deduced from stellar rotation assuming conservation of angular momentum. There must be mechanism to loose angular momentum. Loss mechanism: • magnetic fields (but needs a certain ionisation, problem at very low T) • accretion disk (friction) 7:06 pm The Virial Theorem 7:09 pm • We define a « control volume » as a non-par cular h p://www.av8n.com/physics/euler-flow.htm and representa ve volume represented by the box in the figure. t0 • At some ini al me t0 (top) and later at some me t1 (bo om). t1 • One part of fluid is shown in blue, another one in red and the rest in black to « see » the flow. • Qualita vely, we can note: – The fluid is flowing le -to-right. – The density depends on posi on: The red par cles are denser than the blue par cles. – The density depends on me: The density in the control volume decreases as me passes from t0 to t1. – The velocity depends on posi on: during the me interval in ques on, the right edge of the blue parcel moves a rela vely short distance, while the right edge of the red parcel moves a rela vely long distance. – The fluid is conserved. The total number of red and blue par cles does not change. (Some par cles that flow out of the frame of the diagram are not accounted for, but all par cles (red, blue, or black) that affect the control volume are accounted for). 7:14 pm Conserva on of mass • Suppose we have a fluid with local density ρ (t, x, y, x) and local velocity v (t, x, y, z). Let’s consider a control volume V (not necessarily small, not necessarily rectangular) with surface S. The total mass in this volume is M = ∫ ρ dV (1) • The rate-of-change of this mass is: ∂M / ∂t = ∫ ∂ρ / ∂t dV (2) • The only way such a change can occur is when par cles flow across the boundary, so: ∂M / ∂t = ∫ ρ v · dS (3) • Using Gauss’ theorem, we change the surface integral into a volume one: ∂M / ∂t = − ∫ ∇ · (ρ v) dV (4) • Eq. 2 and Eq. 4 are equal for any volume V and the integrands must be equal. This gives us an expression for the local conserva on of mass: ∂ρ / ∂t + ∇ · (ρ v) = 0 (5) 7:48 pm Conserva on of mass ∂ρ / ∂t + ∇ · (ρ v) = 0 (5) • At the right edge of the control volume, the fluid is rela vely dense and has a rela vely high velocity, causing a large ou low of fluid from the control volume (the red par cles). • At the le edge, the density is rela vely low and the velocity is rela vely low, causing a small inflow of fluid. • The x-deriva ve of ρvx is posi ve. • This is the crucial contribu on to ∇· (ρ v) • The other two contribu ons vanish in this example. We have a net ou low of fluid, which causes a decrease in the density of the fluid in the control volume, in accordance with Eq. 5. • Eq. 5 is called a con nuity equa on and it is one way to express the conserva on of mass 7:48 pm Conserva on of momentum • We can go through the same process for momentum instead of mass. We use Π to represent momentum, to avoid conflict with P which represents the pressure. The total momentum in the control volume is: Πi = ∫ ρ vi dV (6) • where the index i runs over the three components of the momentum. The rate of change: ∂Πi / ∂t = ∫ ∂(ρvi) / ∂t dV (7) • In the absence of other forces i.e. no gravity and no pressure (e.g. a fluid of non-interac ng dust par cles) and under the assum on of no viscosity, the only way a momentum-change can occur is by momentum flowing across the surface: ∂Πi / ∂t = ∫ (ρ vi) v · dS = ∫ (ρ vi) vj djS (8) • We are expressing dot products using the Einstein summa on conven on, i.e. implied summa on over repeated dummy indices, such as j in the previous expression. 8:01 pm Conserva on of momentum • We can change the surface integral into a volume integral using Gauss’ theorem: ∂Πi / ∂t = − ∫∇j (ρ vi vj) dV (9) • Eq. 7 and Eq. 9 are equal for any volume V and so the integrands must be equal. This gives us an expression for the local conserva on of momentum: ∂Πi / ∂t = ∂ (ρvi) / ∂t = − ∇j (ρ vi vj) (10) • We can understand this equa on as follows: each component of the momentum- density ρ vi (for each i separately) obeys a local conserva on law. There are strong parallels between Eq. 5 and Eq. 10. • Note that the ∇j operator on the right-hand side is differen a ng two veloci es (vi and vj) only one of which undergoes dot-product summa on (namely summa on over j). Using vector-component nota on (such as ∇j vj) is a bit less elegant than using pure vector nota on (such as ∇·v) but in this case it makes things clearer. • We now add the effect of pressure. It contributes a force on the par cles in the control volume: Fi = ∫ P diS = −∫∇i P dV (11) • A uniform gravita onal field contributes another force: Fi = ∫ ρgi dV (12) 8:00 pm Conserva on of momentum • These forces change the momentum, by Newton’s second law: dΠi’ / dt = Fi (13) • Note the tricky nota on: we write d/dt rather than ∂/∂t, and Π’ rather than Π, because Newton’s laws apply to par cles, not to the control volume itself. The rate-of-change of Π, the momentum in the control volume, contains the Newtonian contribu ons (Eq. 11 and Eq. 12 via Eq. 13) plus the flow contribu ons (Eq. 10). • Combining all the contribu ons, we obtain the main result, the equa on of mo on: ∂(ρvi) / ∂t + ∇j (ρ vi vj) = −∇i P + ρgi (14) • This way of expressing the equa on of mo on has many advantages. It has a somewhat elegant symmetrical form. It expresses conserva on of momentum in a way that is strongly analogous to conserva on of mass. This 3-D expression can be generalized to give an expression that is valid in (3+1)-D space me. 8:05 pm Conserva on of momentum • However, one some mes encounters other ways of expressing the same equa on of mo on. Expand the le -hand side, we get: ρ ∂(vi) / ∂t + vi ∂(ρ) / ∂t + vi ∇j (ρvj) + ρvj ∇j (vi) = −∇i P + ρ gi (15) • where the second and third terms cancel because of conserva on of mass (Eq. 5), leaving us with ρ ∂(vi) / ∂t + ρvj ∇j (vi) = −∇i P + ρgi (16) • Then, if we convert from component nota on to vector nota on, we get a version of Euler’s equa on, ρ ∂v / ∂t + ρ (v ·∇) v = −∇P + ρg (17) • If we divide by ρ, we obtain the tradi onal Euler’s equa on: ∂v / ∂t + (v ·∇) v = −∇P / ρ + g (18) 8:05 pm Conserva on of momentum ∂v / ∂t + (v ·∇) v = −∇P / ρ + g (18) • g = - ∇Φ where Φ is the grav. Poten al ∂v / ∂t + (v ·∇) v = −∇P / ρ -∇Φ (19) Also called: Euler Equa on 8:32 pm Euler equations a) Continuity equation Temporal change of fluid mass in a volume V must be due to in/outflow Gauss Mass Flow Divergence So for an i.e. infinitesimal volume: b) Momentum equation Control volume V of fluid. Momentum content (i=1,2,3): Temporal change: In the absence of gravity and pressure (and viscosity), the only way of changing the momentum is to transport it in our out: Gauss En u lisant la (usingconven on de Einsteinsche Summenkonvention everywhere)nota on d’Einstein therefore8:06 pm Euler equations II Additionally, pressure and gravity are acting on the particles in the control volume: Gauss Newtons law apply to particles, not control volumes. The momentum change in the control volume thus contains the flow and the Newtonian contributions Using the continuity equation we can also write this as where the second part on the LHS reads as We also note as it must be. 10:10 pm The virial theorem Let’s assume that a star is an ensemble of par cles. Once i-1 par cles form the star, we add the par cle i which contributes to the gravita onal poten al energy Ω: And therefore Ω is: The iner al momentum I of a body formed by par cles of mass mi : And: Let’s define p as: We can write: We have with Ec the total kine c energy 10:10 pm The virial theorem Moreover: Or: And: Finally: If the only force is gravita on: 10:10 pm The virial theorem Which gives: With d d d p dr p r p r i r p i 2E we have: ∑( i i ) = ∑ ( i i ) = ∑ i +∑ i = Ω + c dt i i dt i dt i dt And finally, we get the virial theorem: In €numerous cases (slow evolu on, quasi-sta c), the second deriva ve of the iner al momentum is zero and we have: This theorem applies to stars but also to globular clusters and even galaxies. 10:10 pm The virial theorem • However, Fi represents all the forces ac ng on the par cles, internal or external to the system: • Assuming an ideal gas with internal forces that produce an internal pressure pint = (γ-1)ρu with the specific heat ra o equal to γ = cp / cv [cp: heat capacity at constant pressure and cv: heat capacity at constant volume]. u is the internal energy density and pS the constant surface pressure : ∑Fi ri = 3(γ −1) ∫ ρudV + ∫ −pS n ext .rdS = 3(γ −1)Utherm − pS ∫ ∇r.dV i S V ∂x ∂y ∂z ∑Fi ri = 3(γ −1)Utherm − 3pSV ∇r = + + =1+1+1 = 3 i ∂x ∂y ∂z € −pS ∫ ∇r.dV = −3pS ∫ dV = −3pSV V V 2 € d I € • Finally: = 2E c + Ω − 3PSV + 3(γ −1)Utherm dt 10:10 pm € € The virial theorem d 2I • From the previous equa on: = 2E + Ω − 3P V + 3(γ −1)U dt c S therm • Let’s assume a sta c configura on with no external pressure: d 2I = 2E + Ω + 3(γ −1)U dt € c therm • In absence of any internal energy, we find the same equa on as before: € 2E c + Ω = 0 • In the absence of any bulk kine c energy (e.g. not rota on): € Ω + 3(γ −1)Utherm = 0 • Let’s define: E tot = Ω +Utherm • Then: and Ω = E tot −Utherm Utherm = E tot − Ω € € 10:10 pm € The virial theorem • So from: Ω + 3(γ −1)Utherm = 0 E −U + 3(γ −1)U = 0 • We have: tot therm therm € E tot = (1− 3γ + 3)Utherm = (4 − 3γ)Utherm E tot = (4 − 3γ)Utherm • And: Ω + 3(γ −1)(E tot − Ω) = 0 € Ω + 3γEtot − 3E tot − 3γΩ + 3Ω 3(γ −1)E tot = −Ω − 3Ω + 3γΩ (3γ − 4) E = Ω tot 3(γ −1) • Which is an example of a non-rota ng star in hydrosta c equilibrium€ . 10:10 pm f is the number of degrees of freedom: The virial theorem f = 3 for mono-atomic • Remarks: and f = 5 for diatomic – γ > 4/3 => Etot < 0 and the system is bound. For instance for a mono-atomic gas, γ = (f+2) / f = 5/3 1 E = Ω = −U i.e. if the tot 2 therm constrac on is slow and quasi-sta c, half of the poten al energy is transformed in internal energy (heat) and the rest € is lost in radia on. The system is not stable. – γ = 4/3 => Etot = 0 i.e. the gas is radia on dominated and the total energy is independant of the radius. – γ < 4/3 => Etot > 0 i.e the system is not bound and the system is stable. 10:10 pm Linearized hydrodynamic equa ons 10:10 pm Linearized hydrodynamic equations The basic equations for an inviscid, self-gravitating fluid are: 1) Continuity equation (mass conservation) 2) Euler equation (momentum conservation) 3) Poisson equation (Laplace operator) + an equation of state. Often taken as isothermal gas, so p=p(ρ) cf . Otherwise, one must also consider an energy conservation equation. We linearize this system of equations, assuming that we know a time dependent solution to this set of equations which we call We now consider small, time dependent perturbations to these quantities and write the new variables: where ε << 1. For the mass conserva on: ∂ρ / ∂t + ∇ · (ρ v) = 0 ∂[ρ0(x)+ερ1(x,t)] / ∂t + ∇ · [ρ0(x)+ερ1(x,t) . v0(x)+εv1(x,t)] = 0 2 ∂ρ0(x) / ∂t + ∂[ερ1(x,t)] / ∂t + ∇ · [ρ0(x).v0(x) + ερ1(x,t).v0(x) + ρ0(x).εv1(x,t) + ε ρ1(x,t)v1(x,t)] = 0 2 ∂ρ0(x) / ∂t + ∂[ερ1(x,t)] / ∂t + ∇ · [ρ0(x).v0(x) + ερ1(x,t).v0(x) + ερ0(x).εv1(x,t) + ε ρ1(x,t)v1(x,t)] = 0 ε ∂[ρ1(x,t)] / ∂t + ε∇ · [ρ1(x,t).v0(x)] + ε∇ · [ ρ0(x).v1(x,t)] = 0 ∂ρ1 / ∂t + ∇ · ρ1.v0 + ∇ · ρ0 . v1 = 0 10:12 pm 1 =1− x + o(x) with x <<1 1 x For the momentum conserva on: + ∂v / ∂t + ( v .∇) · v = -1 / ρ ∇. p - € ∇ϕ ∂[v0(x)+εv1(x,t)] / ∂t + (v0(x)+εv1(x,t) .∇) · [v0(x)+εv1(x,t)] = -1 / [ρ0(x)+ερ1(x,t)] [∇. [p0(x)+εp1 (x,t)]] - ∇. [ϕ0(x)+εϕ1(x,t)] ∂v0 / ∂t + ε ∂v1 / ∂t + (v0+εv1 .∇) · v0 + ε(v0+εv1 .∇) · v1 = -1 / (ρ0 + ερ1) [∇. p0 + ε∇. p1) - ∇. ϕ0 - ε∇. ϕ1 2 ∂v0 / ∂t + ε ∂v1 / ∂t + (v0 .∇) · v0 + ε(v1 .∇) · v0 + ε(v0 .∇) · v1 + ε (v1 .∇) · v1 = - [1/ρ0 / (1 + ερ1/ρ0)] ∇. P0 – [1/ρ0 / (1 + ερ1/ρ0)] ε∇. P1 - ∇. ϕ0 - ε∇. ϕ1 2 ∂v0 / ∂t + ε ∂v1 / ∂t + (v0 .∇) · v0 + ε(v1 .∇) · v0 + ε(v0 .∇) · v1 + ε (v1 .∇) · v1 - [1/ρ0 * (1 - ερ1/ ρ0)] ∇. P0 – [1/ρ0 * (1 - ερ1/ρ0)] ε∇. P1 - ∇. ϕ0 - ε∇. ϕ1 2 ∂v0 / ∂t + ε ∂v1 / ∂t + (v0 .∇) · v0 + ε(v1 .∇) · v0 + ε(v0 .∇) · v1 + ε (v1 .∇) · v1 = -1/ρ0 ∇. P0 + 2 2 2 ερ1/ρ0 ∇. p0 - 1/ρ0 ε∇. P1 + ε ρ1/ρ0 ∇. p1 - ∇. ϕ0 - ε∇. ϕ1 2 ε ∂v1 / ∂t + ε(v1 .∇) · v0 + ε(v0 .∇) · v1 = ερ1/ρ0 ∇. p0 + - ε/ρ0 ∇. P1- ε∇. ϕ1 ∇ ∇ 2 ∇ ∇ ∇ 10:12 pm ∂v1 / ∂t + (v1 . ) · v0 + (v0 . ) · v1c = ρ1/ρ0 . P0 - 1/ρ0 . p1 - . ϕ1 Linearized hydrodynamic equations II Plugging back these variables in the equations listed above, taking into account that the time independent base solution solves the equations, and neglecting terms higher than first order in the perturbation ε, we get 1) lin. continuity equation 2) lin. momentum equation 3) lin. Poisson equation Note that all quantities with the subscript 0 are know constants, therefore these linearized equations do not contain any non-linear terms (as e.g. in the original equations) any more. We recall that these linearized equations are valid in the limit of very small perturbations. They allow to discuss the stability of a configuration and the early growth of perturbations but loose their validity as soon as the perturbations leave the linear regime. 10:10 pm Application: sound waves a) non-self gravitating fluid Let us consider a fluid of constant density and temperature (and thus also pressure), initially at rest: The linearized equations then simplify to: 1) ∂ρ1 / ∂t + ∇ · ρ1.v0 + ∇ · ρ0 . v1 = 0 2) 2 ∂v1 / ∂t + (v1 .∇) · v0 + (v0 .∇) · v1 = ρ1/ρ0 ∇. p0 - 1/ρ0 ∇. p1 - ∇. ϕ1 Let us further assume an equation of state of the type p=p(ρ): => Inserting this into eq. 2 yields: Take the time derivative of this expression (keeping in mind that all 0 quantities are constants): 10:21 pm Application: sound waves II Inserting eq. 1 in this expression, using div grad = laplace, and with We recognize the standard wave equation indicating that these small perturbations propagate as a wave called a sound wave through the medium at a speed c called the sound speed. Stability analysis Formally, we can try a solution of the type Inserting this into the wave equation, we get the so called dispersion relation for a sound wave in a non self-gravitating medium: We see, as expected, that the solution is always oscillatory in nature without any sign of instability (ω is always real). Application: sound waves III b) self-gravitating fluid Here again we consider a uniform fluid of constant density and temperature initially at rest. Taking the time derivative of the lin. continuity equation, and the divergence of the lin. momentum equation, and combining this with the lin. Poisson equation yields now for the density perturbation: Stability analysis Assuming a similar plane wave solution as above gives the new dispersion relation We see that the new term due to gravity can make instabilities possible, since if ω2<0, the perturbations will grow exponentially in time. Note that for a given density, the instabilities occur at small wave number k, i.e. at large wavelengths. 10:10 pm Further reading R. Kippenhahn & A. Weigert Stellar structure and evolution, Springer Verlag, Berlin, 3rd Ed. 1994 S. Stahler, F. Palla The formation of stars, Wiley-VCH, Weinheim, 2004 10:10 pm Quetions? 10:10 pm