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Gas Content of Galaxies 4 Gas content of galaxies Interstellar space is very vast: the nearest star is about 4 light years away, but is not empty, filled with gas and dust. Light produced by stars may be absorbed and reemitted by the gas and dust, and so the observed spectrum of a galaxy may be different from that of its stellar populaion. 4.1 Main components of interstellar medium • cold (∼ 1000K) neutral gas, mainly atomic H in the disk, ∼ 90%; • cold (∼ 100K) molecular gas, mainly H2, in disk, ∼ 10% • dust grains, mainly in disk, ∼ 1%. • warm (∼ 10, 000K) ionized gas near hot stars • hot (106K) gas near supernova remnants • other components: magnetic fields, ∼ 10−6 Gauss; Cosmic rays: very high energy particles (electrons and protons). The total amount of gas in the Milky Way is about 1/5 of the total stellar mass. 1 2 Fig. 4.1. The images of the Milky Way in various wavebands. From top to bottom: radio continuum at 408 MHz; 21 cm lines of atomic hydrogen; radio continuum at 2.5 GHz; CO lines of molecular hydrogen; infrared (mainly dust), mid-infrared, near infrared, optical (mainly stars), X-ray (hot gas associated with supernova remnants), and gamma ray. [Courtesy of Astrophysics Data Facility at the NASA Goddard Space Flight Center] 4.2 Emission and absorption of interstellar gas Here is the electronic energy level of a hydrogen atom 4.2.1 Basic structure of an atom (2π)2m e4 E = − e n 2n2h2 νn = c/λn = |En|/h En0 − En ν 0 = nn h ch λnn0 = En0 − En −18 E1 = −2.17 × 10 Joule = −13.6eV Gas content of galaxies 3 Fig. 4.2. The electronic energy level for hydrogen atom 1eV = 1.60 × 10−19Joule 15 ν1 = |E1|/h = 3.3 × 10 Hz which is the Lyman-limit frequency λ1 = 91.2nm = 912A˚ E2 = E1/4; E3 = E1/9 Thus Lyman α: 3 E12 = E2 − E1 = − E1 , λ12 = 121.6nm. 4 Balmer (H) α: 5 E23 = E3 − E2 = −E1/9+ E1/4= − E1 , λ12 = 656nm. 36 4 4.2.2 Ionization, recombination and de-excitation Hydrogen as an example: • Photo-ionization H+ γ → H+ + e− Absorbing a photon with energy above 13.6 eV (912A);˚ surplus energy heats the gas • Photo-excitation: H(ground − state) + γ → H(excited − state) Absorbing a photon with energy equal to the energy difference of the two level, e.g. Lyman α (1216A˚). This selective absorption produces an absorption line at this wavelength. • Recombination: the capture of an electron by an ion: H+ + e− → H+ γ The capture is usually in an excited state. • De-excitation: an atom in an excited state can make a transition to a lower-energy state either due to spontaneous transition or due to the stimulated transition by collisions with a particle (electron and ion) or a photon. This kind of transitions can produce emission lines. For a given atom, all the processes can be calculated through quantum mechanics. 4.3 Detecting neutral hydrogen by absorption Neutral hydrogen can absorb UV photons from a background source. At the temperture of a few hundred K, almost all hydrogen atoms are in the ground level. Absorbing UV photons from the source, re-emitted photons in random direction, producing absorption lines, mainly in Lyman series (i.e. Lyman alpha, beta etc.) Shortcoming of this approach: The line absorptions are in the UV band, difficult to observe from the ground. You also need to have background sources. 4.4 21 cm emissions from neutral gas At a temperature T ∼ 10 - 100 K, all hydrogen atoms must be at the ground state Reminder, to excite a hydrogen atom from ground state to the first exited state, the energy is ∆E = E2 − E1 = 10.02eV, This energy corresponds to a temperture T =∆E/k ∼ 105K Gas content of galaxies 5 Fig. 4.3. The transition of spin state of hydrogen atom at ground state produce 21 cm emission. No emission is expected? Not quite!! 4.4.1 Radio observation at 21 cm In 1945, van de Hulst proposed to observe HI at a radio waveband at 21 cm. Physical principle: hyperfine structure of the ground state • The coupling of the spin of the proton (spin 1/2) and that of the electron (spin 1/2) • Two states of total spin: F = 0, and F = 1 (three degenerate states with quantum number, −1, 0, 1. • Energy: 6 × 10−6eV, between F = 0 and F = 1. • Wavelength: 21 cm: in radio band • Frequency: 1.43 GHz • Corresponds to a temperture of about 0.06 K ! • The lifetime of F = 1 state, τ = 11 milion years. Rate of collisional excitation and de-excitation ∝ n2. For typical ISM, n ∼ 106m−3, the rate is about 400 per year. Thus collisional excitation and de-excitation can establish thermodynam- ical equilibrium: n ∝ ge−E/kT . 6 Fig. 4.4. The observed 21 cm line profile showing two components. n2 g2 = e−hνkT ≈ 3 , n1 g1 independent of temperature! Thus the strength of 21 cm emission provides a direct measure of the total amount of neutral hydrogen gas. If density is lower than ∼ 104m−3, collision may not be able to establish thermodynamical equilibrium, then n2/n1 may not be equal to the equilib- rium value. In this case, one can define a ‘spin’ temperature: n2 = 3e−hν/kTspin . n1 4.4.2 Line profiles Line broadening due to Doppler’s effect caused by random motions of gas particles: • Cold component: 100 K • Warm component: 10,000 K 4.4.3 21 emission from the Milky Way For optically thin case, the total flux is directly proportional to the gas column density NHI (number of hydrogen atoms per unit area in the galaxy) Gas content of galaxies 7 Fig. 4.5. The production of a Stromgren sphere. Thus, 21 cm flux map tells us directly the HI gas distribution Total HI mass can be obtained by integrating NHI over the entire galaxy The 21 cm map of the Milky Way is given in Figure 4.1. 9 Total HI mass of the Milky Way is about 5 × 10 M . 4.5 HII regions HII denotes ionized hydrogen: HI denotes neutral hydrogen • O, B stars which produce a lot of UV photons that can ionize the gas through photoionizarion: H+ γ → H+ + e− • Producing an ionized sphere called Stromgren sphere 4.5.1 The radius of a Stromgren sphere (i) Assuming the medium is pure hydrogen gas of constant density (ii) Let N? be the number of ionizing photons with hν > 13.6eV produced by the star per unit time (iii) R: the number of recombinations of protons and electrons into hy- drogen atoms per unit time per unit volume 8 3 R(4πr /3) = N? where r is the radius of the Stromgren sphere. Since R must be proportional to npne, and since np = ne (why?), we write 2 R = αnpne = αne where α is the recombination coefficient which depends on the temperture of the gas. For HII regions, T ∼ 104K, we have α ≈ 3 × 10−19m3sec−1 Thus 1/3 3N? r = 2 4παne 7 −3 49 −1 Example: HI region: ne = 10 m ; N? = 3 × 10 sec for a O5V star, and r ∼ 60 light years 4.5.2 Emissions from HII regions Ionized hydrogen (proton) can capture an electron in a high energy level, and then cascade to the ground level, producing various emission lines; Lyman alpha emission lines; H-alpha emission; H-beta emission etc. Thus, if a cloud contains enough hydrogen, each ionizing photon will eventually be degraded into a Lyman alpha photon, plus a Balmer photon, plus perhaps some lower-energy photons. Since the total number of ionizing photons is determined by the number of young stars that produce them, the observations of H-alpha emissions (6563A˚: in optical) may tell us the star formation rate in a galaxy. Lyman-alpha not as good, because it is in UV, and so you need space- telescope to observe it, and because it is strongly affected by dust (see be- low). 4.5.3 Molecular gas Mainly H2 gas, but direct detection of H2 difficult in radio emission, because the energies of most transitions are not in the radio band. Observations usually made with CO lines and OH lines in the radio Molecular clouds are very dense, ∼ 108 – 1014 per m3 (but still only about 10−15 of the atmosphere), and are the places where stars are forming. 4.5.4 Interstellar Dust • Composition: (1) Grains of silicates, like beach sands Gas content of galaxies 9 Fig. 4.6. The cascading emission lines from hydrogen from n = 4. (2) Soot, exhaust of a diesel engine • Origin: Produced in the outer atmosphere of red giant stars in the late stage of evolution Dust extinction Dust grains can absorb and scatter light, causing a light source to become dimmer. Suppose the average absorption (scattering) cross section of a dust grain is σ, and the number density of dust grain is nd. Then the intensity will decrease according to ∆I = −Iσnd∆x , i.e. dI = −Iσn . dx d This gives: −τ I = I0e , where x x τ = σnddx = σ nddx Z0 Z0 10 Fig. 4.7. The observed reddening curve, Aλ/AJ . x is the optical depth, and 0 nddx is the column density of dust grains. The change in magnitudeR due to dust extinction is ∆m = m − m0 = −2.5 log(I) + 2.5 log(I0) = 2.5τ . Reddening by dust Extinction is ‘selective’, affecting bluer light more (the reason why the sun is redder at sunset).
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