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Luminosity of stars The intrinsic luminosity of a star is its emitted power, [energy][time]-1. Apparent luminosities of stars are different due to: - different distances - different intrinsic luminosities Is it possible to quantify the apparent luminosity? A scale of magnitudes from 1 to 6 was...

Luminosity of stars The intrinsic luminosity of a star is its emitted power, [energy][time]-1. Apparent luminosities of stars are different due to: - different distances - different intrinsic luminosities Is it possible to quantify the apparent luminosity? A scale of magnitudes from 1 to 6 was already introduced around 150 b.C. (Ipparcus, Tolomeus): - mag1: the brightest stars; - mag6: the faintest stars, just visible at naked eye Since the human perception of light brightness is logarithmic, we introduce a logarithmic scale of luminosities: 𝑓1 = 100(𝑚2 −𝑚1 )/5 𝑓2 where fi is the flux (luminosity) and mi the corresponding magnitude. If the flux increases, the magnitude decreases. The faintest stars have highest magnitudes. Previous formula can be inverted: 𝑓2 𝑚2 − 𝑚1 = −2.5 log10 ( ) 𝑓1 1 -------------------------------------------------------- Examples: 1) Find the luminosity ratio between the faintest star detectable at naked eye and the faintest one visible using a powerful telescope. Ans. ≈ 108 2) A flux ratio is 106. Find the corresponding difference in magnitudes. Ans. 15 ------------------------------------------------------ Stars and, more generally, astronomic objects, emit not only in the visible but in other regions of the electromagnetic spectrum. Relationship between frequency ν and wavelength λ: λ ν =V where V is the velocity of the wave. In a vacuum (EM waves) V = c = 299792458 m/s ≈ 3 108 m/s. Region Wavelength Frequency (Hz) Radio > 10 cm < 3 109 Microwaves 0.1 mm – 10 cm 3 109 – 3 1012 Infrared 700 nm – 0.1 mm 3 1012 – 4.3 1014 Visible 400 – 700 nm 4.3 1014 – 7.5 1014 Ultraviolet 10 – 400 nm 7.5 1014 – 3 1016 X-rays 0.1 – 10 nm 3 1016 – 3 1018 Gamma rays < 0.1 nm > 3 1018 2 Visible light is only a tiny portion of the visible spectrum, but very important for us (our eye is sensitive only to visible light; most of the Sun’s emission is in the visible). A few definitions: - Flux: energy/time/surface (W m-2) - Luminosity: flux integrated over a surface (W) - Flux density: flux per frequency (or wavelength) interval (W m-2Hz-1). Especially in radio astronomy it is expressed in Jansky: 1 Jy = 10-26 Wm-2 Hz-1. The amount of energy with frequency in the range [ν, ν+dν] entering in the solid angle dω in time dt is: dEν = Iν cos θ dA dν dω dt - Iν is the specific intensity (or surface brilliance) of the radiation at the frequency ν in the direction of the solid angle dω. Its dimension is Wm−2 Hz−1 sterad−1. - Intensity (or total intensity) is obtained by integrating Iν over all the frequencies: ∞ 𝐼 = ∫ 𝐼𝜈 𝑑𝜈 0 - Spectrum: distribution of the flux density as a function of frequency/wavelength Continuous spectrum determines the color of a star, which depends on the temperature. Blackbody radiation is the base of the relationship color-temperature. Detection of a spectrum allows us to deduce the temperature of a star. 3 Short summary of the blackbody (BB) radiation. BB is an ideal source at thermodynamic equilibrium, absorbing all the energy incident on it. The spectrum of the radiation emitted by a BB depends only on the temperature: By increasing T, the irradiated flux increases as T 4 (Stefan-Boltzmann law) and the wavelength corresponding to the peak of the spectrum decreases: λmaxT = 0.29 cm K (Wien’s law). Connection with astronomy: continuous spectrum of a star is well approximated by a BB: -------------------------------------------------------- Examples. - Find the surface temperature of a star whose continuous peak is centered on the visible (500 nm). Ans. 5800 K - Find the wavelength corresponding to the maximum emission of the Earth. 4 Ans. ≈ 10 μm ---------------------------------------------------------- Intrinsic luminosity of a star. If we approximate the flux f to a BB, f =σT4 and: L =4πR2 σT4 (if the surface through which the flux is passing is assumed to be spherical). --------------------------------------------------------- Example. The luminosity of the Sun (Sun radius: 7 108 m; surface T = 5.8 103 K; σ =5.7 10-8 W m- 2 -4 K ). Ans. 4 1026 W. -------------------------------------------------------- Planck’s law: 2ℎ𝜈 3 1 𝑢(𝜈, 𝑇) = 2 ℎ𝜈/𝑘𝑇 𝑐 𝑒 −1 h = Planck constant, 6.63 10-34 J s. We can express the Planck’s law in terms of the wavelength, using the fact that: u(v, T)dv = u(λ, T)dλ: 2ℎ𝑐 2 1 𝑢(𝜆, 𝑇) = 5 ℎ𝑐/𝜆𝑘𝑇 𝜆 𝑒 −1 The ‘quantum’ of energy is the photon. Its energy: E = hv =hc/λ is often expressed in electronvolts (eV). 1 eV = 1.6 10-19 J. Region Wavelength Frequency (Hz) Energy (eV) Radio > 10 cm < 3 109 < 10-5 Microwaves 0.1 mm – 10 cm 3 109 – 3 1012 10-5 – 0.01 Infrared 700 nm – 0.1 mm 3 1012 – 4.3 1014 0.01 – 2 Visible 400 – 700 nm 4.3 1014 – 7.5 1014 2–3 Ultraviolet 10 – 400 nm 7.5 1014 – 3 1016 3 – 100 X-rays 0.1 – 10 nm 3 1016 – 3 1018 100–104 (0.1–10 keV) Gamma rays < 0.1 nm > 3 1018 > 104 (> 10 keV) 5 Let’s come back to the relationship color-temperature. In principle, by measuring the wavelength corresponding to the peak of the BB (which approximates the continuous spectrum of the star) we could deduce the surface temperature of the star, from the Wien’s law. In practice, it is not straightforward to measure the whole BB spectrum. Sometimes, it is impossible, when the maximum falls in the UV region. But by measuring the flux ratio r at two wavelengths the temperature is deduced: 𝑢(𝜆1 , 𝑇) 𝜆2 5 𝑒 ℎ𝑐/𝜆1 𝑘𝑇 − 1 𝑟= = ( ) ℎ𝑐/𝜆 𝑘𝑇 𝑢(𝜆2 , 𝑇) 𝜆1 𝑒 2 −1 Every quantity is known except T. Luminosity and distance. Apparent luminosity l depends on: - intrinsic luminosity L - distance - interstellar absorption (we neglect it in our discussion, for the sake of simplicity. See notes on interstellar medium). At a distance d the relationship between l and L is: L = 4πd2l 6  In order to calculate the distance between the observer and the star it is used the parallax. We want to measure d. The usual method is the parallax. The star is in R. T is the Earth, S the Sun (we assume a circular orbit). Parallax Pa(t) is the angle under which the star sees the Earth-Sun distance during the year. When TŜR = 90°, Pa = Pmax. Method: the star is observed at Pmax and 6 months later. The basis of the red triangle is known (3 108 km, 2AU), 2Pmax is measured as the displacement of the star R with respect to very far stars (‘fixed’ stars on the sky) and: tan(Pmax) ≈ Pmax =1AU/d and d is obtained. 7 Parsec (pc): distance corresponding to Pmax = 1”. 1 pc = 1 AU/tan(1”) = 3.09 1016 m = 3.26 ly (light-year). Although limited (ground-based telescopes measure up to 100 pc) the method is the only one model-independent. Exploited by spatial missions, e.g.: - Hipparcos (HIgh Precision PARallax COllecting Satellite). Started in 1989, ended in 1993. Devoted to the measurement of near stars. Resulted in ‘Millennium Star Atlas’, comprising 2·106 stars (up to magnitude 11) and 104 non-stellar objects. - GAIA (Global Astrometric Interferometer for Astrophysics). Started in 2013, developed by ESA, continuation of Hipparcos. The main goal is an accurate 3D map of the Milky Way (about 109 stars will be examined, up to magnitude 20) through monitoring each target up to 100 times to get high precision distances and changes in luminosity. Other goals: discovery of brown dwarfs, exoplanets, comets, asteroids in our solar system and new tests on general relativity. Duration: initially 5 years, has been prolonged up to 2022. Absolute magnitude. It is the magnitude of a star at a distance of 10 pc. Let’s assume that two identical stars are at distances d and d0 = 10 pc The ratio of their apparent luminosities: l/l0 = (d0/d)2 Since: 𝑓2 𝑚2 − 𝑚1 = −2.5 log10 ( ) 𝑓1 with m1 = M, m2 = m, it follows that: 𝑀 = 𝑚 + 5 − 5 log10 𝑑 Therefore: m (apparent magnitude) is measured; if either M or d is known, the other parameter can be obtained. 8 --------------------------------------------------------- Example. Betelgeuse has m = 0.45 and M = - 5.15. What is its distance from us? Ans. 132 pc. Example. Absolute magnitude of the Sun (m = -26.8, d = 1 AU = 4.85 10-6 pc). Ans. 4.8 --------------------------------------------------------- Reference: Karttunen, Fundamental Astronomy, ch. 4 http://www.cosmos.esa.int/web/hipparcos/millennium-star-atlas http://www.cosmos.esa.int/web/gaia Keywords: luminosity, magnitude, spectrum, parallax 9 Stars and spectral lines We have seen that the spectrum of a star consists of a continuous as well as of discrete lines. Continuous is approximated by BB radiation. Where do lines come from? * Dark lines were observed at the beginning of XIX century in the solar spectrum: Analogous lines were found in laboratory by Kirchhoff (1859) and associated to the presence of a chemical element. In fact, electromagnetic radiation coming from a BB and passing through a gas gives rise to lines, whose number and position depend on the gas composition. Hot gas under low pressure produces an emission spectrum consisting of such discrete lines. If the same gas is cooled down and observed against a source of white light (which has a continuous spectrum), the same lines are seen as dark absorption lines. Investigations on stars revealed lines analogous to those in the Sun, but not identical! Different stars may have different lines, coming from stellar atmospheres. Origin of the spectral lines – the case of hydrogen 1 Hydrogen gas shows a series of lines in the visible range, firstly discovered by Balmer. Empirically: 1 1 1 = 𝑅 (4 − 𝑛2 ) , with n = 3, 4,5, …. 𝜆 R = Rydberg constant (109677,55 cm-1). ---------------------------------------------------- Example. The wavelength corresponding to the first line (Hα, n = 3) of the series. Ans. 656,5 nm. Example. The minimum wavelength of the Balmer series. Ans. 364,7 nm. -------------------------------------------------- Spectral lines can be understood within the atomic model by Bohr, which is based on two postulates. The electron orbits the proton in a circular orbit. Bohr’s first postulate says that the angular momentum of the electron must be a multiple of h/2π, ℏ (′ℎ 𝑏𝑎𝑟 ′ ): mvr = nh/2π. (m: electron mass, v: its velocity, r: radius of the orbit). Bohr’s second postulate says that an electron moving in an allowed orbit around a nucleus does not radiate. Radiation is emitted only when the electron jumps from a higher energy state to a lower one. The emitted quantum has an energy hν, (according to Planck hypothesis, radiation is exchanged in terms of quanta) equal to the energy difference of these states: hν = En2 − En1. 2 As a consequence of the two postulates, the radius of the orbit results: 𝜖0 ℎ 2 2 𝑟𝑛 = 𝑛 𝜋𝑚𝑒 2 and the energy of an electron in the orbit of quantum number n: 𝑚𝑒 4 1 𝐸𝑛 = − 2 2 2 8𝜀0 ℎ 𝑛 For the ground state (n = 1), we get E1= −2.18×10−18 J= −13.6 eV. The energy of the quantum 1 1 emitted in the transition En2 → En1 is therefore: ℎ𝜈 = 𝐶 (𝑛2 − 𝑛2 ). In terms of the wavelength λ this 1 2 1 1 1 𝑚𝑒 4 can be expressed as = 𝑅( − ) where R = numerically corresponds to the Rydberg 𝜆 𝑛12 𝑛22 8𝜀02 ℎ3 𝑐 constant. The Bohr model allows us to frame theoretically the Balmer formula. ---------------------------------------------------------- Example. Radius of the first hydrogen orbit (n = 1). Ans. 0.053 nm ---------------------------------------------------------- After Balmer, other series were discovered for the hydrogen, in other regions of the electromagnetic spectrum. In fact, if the excited electron returns to its ground state (En →E1), we get the Lyman series, which is in the ultraviolet. The other series with specific names are the Paschen series (n1 = 3), Bracket series (n1 = 4) and Pfund series (n1 = 5), all in the infrared part of the spectrum. All the lines can be obtained by generalizing the Balmer formula: 1 1 1 = 𝑅 ( 2 − 2) 𝜆 𝑛1 𝑛2 and are explained by the Bohr model. 3 Previous considerations given for hydrogen can be generalized to every atom or molecule. Electromagnetic radiation is emitted or absorbed when an electron inside an atom or a molecule moves from one energy level to another. If the energy of the atom decreases by an amount ΔE, the atom emits (= radiates) a photon, whose frequency ν is obtained from the equation ΔE = hν where h is the Planck constant. An energy level of an atom refers to an energy level of its electrons. The energy E of an electron cannot take arbitrary values; the energy levels are quantized. An atom can emit or absorb radiation only at certain frequencies νif corresponding to energy differences between some initial and final states i and f : |Ei− Ef| = hνif. This gives rise to the line spectrum, specific for each element. At low temperatures most atoms are in their lowest energy state, the ground state. Higher energy levels are excitation states. Usually the excited atom will return to a lower state very rapidly, radiating a photon (spontaneous emission); a typical lifetime of an excited state might be 10−8 s. On the other hand, downward transitions can also be induced by radiation. Suppose our atom has ‘swallowed’ a photon and becomes excited. Another photon, whose frequency ν corresponds to some possible downward transition from the excited state, can now stimulate the atom, causing it to 4 jump to a lower state, emitting a photon with the same frequency ν. This is called induced or stimulated emission. Photons emitted spontaneously leave the atom randomly in all directions with random phases: the radiation is isotropic and incoherent. Induced radiation, on the other hand, is coherent; it propagates in the same direction as (and in phase with) the inducing radiation. If an electron with energy E < 0 receives more energy than |E|, it will leave the atom, which becomes an ion. Unlike in excitation all values of energy (E > 0) are now possible. The extraneous part of the absorbed energy goes to the kinetic energy of the liberated electron. When an electron scatters from a nucleus or an ion without being captured, the electromagnetic interaction can change the kinetic energy of the electron producing free–free radiation. In a very hot gas (T > 106 K) hydrogen is fully ionized, and the free–free radiation is the most important source of emission. It is then usually called thermal bremsstrahlung. Electromagnetic radiation is transverse wave motion; the electric and magnetic fields oscillate perpendicular to each other and also perpendicular to the direction of propagation. The light of an ordinary incandescent lamp has a random distribution of electric fields vibrating in all directions. If the directions of electric fields in the plane perpendicular to the direction of propagation are not evenly distributed, the radiation is polarized. The direction of polarization of linearly polarized light means the plane determined by the electric vector and the direction of the light ray. 5 If the electric vector describes a circle, the radiation is circularly polarized. If the amplitude of the electric field varies at the same time, the polarization is elliptic. Circularly polarized light Elliptically polarized light Scattering is an absorption followed by an instantaneous emission at the same wavelength but usually in a new (random) direction. The light coming from the sky is sunlight scattered from atmospheric molecules. Scattered light is generally partially polarized, the degree of polarization being highest in the direction perpendicular to the direction of the original radiation. 6  Concerning the spectral lines from stars, they are classified in spectral classes according to the temperature of the surface of the star. At the beginning (1900) lines were classified according to their intensity, but this last is not monotonically correlated to the temperature (see below): strongest lines are found at intermediate temperatures. Classes were ordered according to the following letters: O B A F G K M (Oh, Be A Fine Girl, Kiss Me!) from the highest (O) to the lowest (M) temperatures at the surface of a star. Most recent scheme is the following: C O−B−A−F−G−K−M−L−T. S The spectral classes C and S represent parallel branches to types G–M, differing in their surface chemical composition. The most recent addition are the spectral classes L and T continuing the sequence beyond M, representing brown dwarfs. The spectral classes are divided into subclasses denoted by the numbers 0... 9 (e.g. B3, G7…). (additional notations are Q for novae, P for planetary nebulae and W for Wolf–Rayet stars, but we will not enter into these topics). Important lines are the hydrogen Balmer lines, the lines of neutral helium, the iron lines, the H and K doublet of ionized calcium (at 396.8 and 393.3 nm), the G band due to the CH molecule and some metals around 431 nm, the neutral calcium line at 422.7 nm and the lines of titanium oxide (TiO). *A few examples of the classification scheme: 7 - O: Blue stars, surface temperature 20,000–35,000 K. Spectrum with lines from multiply ionized atoms, e. g. He II, CIII, NIII, OIII, SiV. He I visible, HI lines weak. - G: Yellow stars like the Sun, surface temperature about 5500 K. The HI lines weak, H and K lines (ionized calcium) very strong, strongest at G0. Metal lines getting stronger. Band due to CH group clearly visible. CN lines seen in giant stars. - C: Carbon stars. Very red stars, surface temperature about 3000 K. Strong molecular bands, e. g. C2, CN and CH. Of course, in hot stars molecules dissociate into atoms; thus, the absorption bands of molecules appear only in the spectra of cool stars of late spectral types. By convention, each line is indicated by: the element; the number n1; the number n2 appearing in the formula for the wavelength of the transition. If: n2 = n1 + 1, α n2 = n1 + 2, β n2 = n1 + 3, γ, …. Example: He2α Balmer series: H2α, H2β, H2γ, … Lyman series: H1α, H1β, H1γ, … also named Lyα, Lyβ, Lyγ,… Furthermore, a neutral atom (e.g. H) is indicated as H or HI. A ionized H is indicated as H+ or HII or HII. For multiple ionizations: O++, OIII, O(III), OIII are equivalent symbols. Why strengths of the lines change with temperature? Radiation comes from an enormous number of atoms. If the stellar atmosphere contains a lot of hydrogen, can we expect to find Balmer lines? Only if many H atoms are on the first excited state. If all the H are on the ground state, no lines corresponding to the Balmer series will appear. Therefore, intensity of a line depends on the number of atoms in the initial state. ‘Population’ of a state = number of atoms per unit volume initially in that state. ‘Excitation’: process which changes the population of a state. Efficiency of excitation depends on the temperature which is related to the average kinetic energy: 1 3 𝑚𝑣 2 = 𝑘𝑇 2 2 Higher T → higher kinetic energy per collision and higher collision rate. Population statistics is governed by the Boltzmann law: 8  𝑛𝑖 𝑔𝑖 −(𝐸 −𝐸 )/𝑘𝑇 = 𝑒 𝑖 𝑗 𝑛𝑗 𝑔𝑗 gi : statistical weight of the population ni. At sufficiently high temperature the electron is expelled from the atom (ionization) and cannot produce lines any longer! Therefore, at low temperatures, all H is neutral, most in the ground state: H2α is very weak. At intermediate T: most H is still neutral, population in the first excited state increases: H2α becomes more intense. At high T: many ionized H, population in first excited level decreases and so for the intensity of H2α. *Another example: the neutral helium lines at 402.6 nm and 447.2 nm. These are only seen in the spectra of hot stars. Indeed, the lines are due to absorption by excited atoms, and a high temperature is required to produce any appreciable excitation. As the stellar temperature increases (A→B), more atoms are in the required excited state, and the strength of the helium lines increases. When the temperature becomes even higher (B→O), helium begins to be ionized, and the strength of the neutral helium lines begins to decrease. 9 Position of the lines may be complicated by the Doppler effect: wavelength (or frequency) changes due to the relative source-observer motion. Shift concerns only the component of the velocity along the line source-observer (radial velocity vr). 𝑣𝑟 = (𝑣𝑜 𝑐𝑜𝑠𝜑 + 𝑣𝑠 𝑐𝑜𝑠𝜃) If λ is the observed wavelength, λ0 the wavelenght emitted at rest, Δ𝜆 𝜆 − 𝜆0 𝑣𝑟 Δ𝜈 = =− =− 𝜆0 𝜆0 𝑐 𝜈0 vr < 0, Δλ > 0, red shift vr > 0, Δλ < 0, blue shift 10  𝑣𝑟 To be noted: Δ𝜆 = − 𝜆0, which means that for the same vr the shift depends on the rest 𝑐 wavelength: it is not the same for all wavelengths. Reference: H. Karttunen, Fundamental Astronomy, ch. 5 G. R. Fowles, Introduction to modern optics, Dover Publications, ch. 2, 7, 8 Keywords: Spectral lines, Balmer formula, Bohr’s atomic model, star spectral classification 11 The Hertzsprung-Russell diagram Around 1910, Ejnar Hertzsprung (DK) and Henry Norris Russell (USA) independently studied the relation between the absolute magnitudes and the spectral types of stars of our Galaxy. The diagram showing these two variables is now known as the Hertzsprung–Russell diagram or simply the HR diagram. It has turned out to be an important aid in studies of stellar evolution. The x-axis represents the temperature while the y-axis shows the corresponding luminosity of a star. It has to be noted that the diagram is in double logarithmic scale and the x-axis is reversed. In view of the fact that stellar radii, luminosities and surface temperatures vary widely, one might have expected the stars to be uniformly distributed in the HR diagram. However, it is found that most stars are located along a roughly diagonal curve called the main sequence. The Sun is situated about the middle of the main sequence. We know that, in BB approximation: 𝐿 = 4𝜋𝑅 2 𝜎𝑇 4 (see Stars and spectral lines_2). It follows that for a given luminosity and temperature we can determine the radius of a star. Not all the stars are on the main sequence. White dwarfs with the same T of those on the main sequence have luminosity even 10-5 times lower. On the other hand, giants and supergiants are more luminous than the stars of the main sequence, the temperature being the same. Concerning the size: 𝐿 𝑅=√ 4𝜋𝜎𝑇 4 1 Therefore, stars with the same temperature will be greater (smaller) than those of the main sequence according to their luminosity; whence the names giants and dwarfs. ------------------------------------------------------------ Example. Determine the size of a white dwarf of spectral class G (the same of the Sun). 𝑅𝑊𝐷 𝐿𝑊𝐷 Since TWD = TS we have =√. From the diagram we see that LWD = 10-4LS. Therefore 𝑅𝑆 𝐿𝑆 RWD = 10-2RS = 10-2(7 105) km = 7000 km, about the size of the Earth. ------------------------------------------------------------ The diagram can be used to estimate the distance of a star. ----------------------------------------------------------- Example. A star with apparent magnitude m = 8 belongs to spectral class B0. By excluding it is a dwarf, determine its distance d from us. Remember the relationship between apparent and absolute magnitude: 𝑑 𝑚 − 𝑀 = 5 log10 10𝑝𝑐 Its absolute magnitude: 𝐿 𝑀 − 𝑀𝑆 = −2.5 log10 𝐿𝑆 From the HR diagram we read its luminosity relative to the Sun: 104. Therefore: M = - 5 (MS = 5, still from HR). Now we calculate: m – M = 13 from which we obtain: d =10pc 1013/5 ≈ 4 kpc. ---------------------------------------------------------------------- 2 How is it possible to discriminate among stars belonging to the same spectral class but with different luminosity? A spectroscopic line is broadened by different reasons. There is always a natural broadening, due to the Heisenberg principle: ∆𝐸∆𝑡 ≈ ℎ ΔE is the energy broadening and Δt is the lifetime of the excited state. It can be translated in terms of a frequency: ∆𝐸 ∆𝜈 ≈ ℎ Another cause of broadening are collisions: collision frequency increases the width of a line. Indeed, collisions among excited atoms in a gas can induce the release of photons, by reducing the lifetime of the excited state, Δt. Higher pressure → higher collision frequency → the line broadens. Therefore, pressure in the external layers of the stars produces broadening of the lines. Giants, having bigger sizes, have lower pressure and the lines are narrow. On the contrary, dwarfs have higher pressures and lines appear broadened. Stars of the main sequence are in the middle. We have seen how to get insight on luminosity, size, surface temperature, chemical composition (spectral lines). Is it possible to measure the mass of a star? To this purpose we exploit binary systems (about 50% of the star in our galaxy are actually binary stars). 3 There are 5 different kinds of binaries: - visual: both the members are resolved (very rare); - binaries with composite spectrum: the spectrum shows the presence of two different spectral classes overlapped; - spectroscopic binaries: spectral lines oscillate with time around an average value, due to Doppler shift generated by the motion along the orbit of a star; - astrometric binaries: the motion of the only visible component shows oscillations due to the presence of the twin star; - eclipse binaries: light shows periodic oscillations due to the reciprocal eclipse of the twin stars (only if the plane of the orbit is along the direction of the observer). 4 In the following we give a simplified treatment on how to get information on the mass. Let us suppose for simplicity circular orbits: We expect a Doppler shift: 𝑣𝑟 = 𝑣𝑠 𝑐𝑜𝑠𝜃 = 𝑣𝑠 cos⁡(𝜔𝑡) and Δ𝜆 𝑣𝑠 = cos⁡(𝜔𝑡) 𝜆0 𝑐 Generally, the orbit is inclined: Then, the previous formula becomes: Δ𝜆 𝑣𝑠 = cos(𝜔𝑡) sin(𝑖) 𝜆0 𝑐 Let us apply Newton mechanics: 𝑚1 𝑟2 By definition of center of mass: m1r1 = m2r2 and = 𝑚2 𝑟1 Furthermore, the two stars have the same period: 2𝜋𝑟1 2𝜋𝑟2 𝑃1 = = 𝑃2 = =𝑃 𝑣1 𝑣2 5 Whence: 𝑟2 𝑣 𝑚 = 𝑣2 = 𝑚1 (A) 𝑟1 1 2 Concerning the star S1, let us identify the gravitational force with the centripetal force: 𝑚1 𝑚2 𝑣12 𝐺 = 𝑚1 (𝑟1 + 𝑟2 )2 𝑟1 By introducing the common period P: 𝑚2 4𝜋 2 𝑟1 𝐺 = (𝑟1 + 𝑟2 )2 𝑃2 𝑟 𝑚1 +𝑚2 In terms of the distance 𝑅 = 𝑟1 + 𝑟2 = 𝑟1 (1 + 𝑟2 ) = 𝑟1 : 1 𝑚2 𝑚2 4𝜋 2 𝑚2 𝐺 2= 2 𝑅 𝑅 𝑃 𝑚1 + 𝑚2 or: 4𝜋 2 𝑅 3 = (𝑚1 + 𝑚2 )𝑃2 𝐺 But 𝑃 𝑅 = 𝑟1 + 𝑟2 = (𝑣 + 𝑣2 ) 2𝜋 1 Therefore: 𝑃 (𝑣1 + 𝑣2 )3 = 𝑚1 + 𝑚2 (B) 2𝜋𝐺 In conclusion: from the measurement of the Doppler shift we obtain the velocities; we measure the period and we get the sum of the masses (eq. B). From eq. (A) we also get the ratio of the masses and we can estimate the mass of each of the two stars. From the analyses of binary stars we know a great number of stellar masses. For the stars of the main sequence there is a correlation between mass and spectral class: 6 The mass of a star determines where this last is found in the main sequence. It is possible to find an empirical correlation between mass and luminosity (for the main sequence): Very roughly we can write: 𝐿 𝑀 𝛼 =( ) 𝐿𝑆 𝑀𝑆 with α ≈ 3.5, but the exponent is different for the various ranges of masses. References: H. Karttunen, Fundamental Astronomy, ch. 5, 9 Keywords: Hertzsprung-Russell diagram, line broadening, binary stars 7 Energy and elements produced by a star We know that nuclear processes occurring inside a star are responsible of its energy production. *We can exclude other possible processes. In fact: 1) Gravitational energy. Due to the enormous mass of a star a gravitational potential energy U is associated to it. 𝑀(𝑟)𝑑𝑀 16 𝑑𝑈 = −𝐺 = − 𝜋 2 𝐺𝜌2 𝑟 4 𝑑𝑟 𝑟 3 𝑅 16 2 2 1 5 3 𝑀2 𝑈 = ∫ 𝑑𝑈 = − 𝜋 𝐺𝜌 𝑅 = − 𝐺 3 5 5 𝑅 0 By inserting the data for the Sun we find: US = 2 1041 J. ‘Kelvin time’ is defined as the ratio between the gravitational energy of a star and the rate to which energy is lost by irradiation: 𝐸 𝐸 3 𝑀2 𝑡𝐾 = = = 𝐺 𝑑𝐸/𝑑𝑡 𝐿 5 𝑅𝐿 For the Sun L = 4 1026 W and we find tK =2 107 y. Since the geological age of the Earth is about 4.5 10 9 y, this excludes gravitational energy as a source for the stars. 2) Chemical energy. It could be due to disruption of molecular bonds. Order of magnitude of these last: ε = 1 eV = 1.6 𝑀𝑆 10-19 J. Total chemical energy available, in principle: 𝐸𝐶 = 𝜀 and we get about 2 1038 J, which would be 𝑚𝑝𝑟𝑜𝑡𝑜𝑛 irradiated in about 104 y. 3) Nuclear energy. Order of magnitude of the bonding energy among nucleons: 8 MeV, that is, about 106 times the chemical energy associated to molecular bonds. This gives about 1010 y, which is the correct order of magnitude for the lifetime of a star. However, we will see that such a lifetime can vary by orders of magnitude according to different kinds of stars. Nuclear interaction is ‘strong’: it is the most intense attractive short-range force among nucleons (protons and neutrons). In a nucleus, protons are repelled each other by coulomb repulsions, but nuclear force is stronger and binds the particles. Presence of neutrons increases the binding energy. 1 For a proton inside a nucleus, nuclear and electromagnetic forces are equilibrated: A proton at the border of the nucleus feels the nuclear force only by neighboring nucleons and the electromagnetic force by all the other protons. In a nucleus containing a small number of nucleons (light nucleus) the electromagnetic force is not sufficient to expel the proton: Electromagnetic force becomes intense in a heavy nucleus and a proton: or an alpha particle: can be emitted from the nucleus. In some cases, a gain of energy occurs when the (heavy) nucleus can be broken in two (or more) fragments (fission process): 2 Energy is delivered, such as in nuclear plants or in nuclear weapons. The binding energy per nucleon, typically about 8 MeV, depends on the mass number A (= total number of nucleons belonging to the nucleus): This justifies the fission process: the heavy nucleus is broken in lighter nuclei which are more stable. The most stable nuclei have mass around 60 (Fe). In the stars the number of intermediate nuclei is a very tiny fraction. Most of nuclei have A very low (H, He,...) and fusion process (combination of lighter nuclei to form a heavier nucleus) is energetically favored. Problem: 2 protons at distance > 10-13 m repel due to electromagnetic interaction; strong interaction is negligible. How is it possible to melt two protons? Let us consider one proton at rest, the other one is very far. Its energy is only kinetic, EK and must 1 𝑒2 overcome the coulomb potential energy at distance d = 10-15 m: 𝐸𝐾 = 4𝜋𝜀. If this kinetic energy 0 𝑑 is due to thermal energy: EK = kT we find: T ≈ 2 1010 K. In the core of the Sun T = 1.5 107 K. The process seems impossible. However, protons have a distribution of energy: even if the average temperature is ‘only’ 15 MK, a fraction of protons have a much higher energy (the tail of the distribution). 3 Since the number of protons in the Sun is enormous (1057) there is a sufficient number of them to give rise to a fusion reaction. But this explanation does not justify the amount of energy released by nuclear reactions inside a star. There is also another quantum effect, the tunnel effect, which contributes to the reaction. We remind that, in accordance with quantum mechanics, a proton is described by a wave whose motion is governed by the Schrödinger equation (not reported here for simplicity) and whose wavelength λ is related to the momentum p of the particle by the De Broglie relation: λ = h/p. The figure below describes the (qualitative) behavior of the wave function (in red) of a proton with energy E at a given time, in the presence of a Coulomb potential barrier (continuous black line) representing the repulsion due to a second proton in the origin. At large distance from the maximum of the repulsive potential barrier the wavefunction is represented by a sinusoid, corresponding to a free particle. By approaching the other proton, the wavelength increases due to a loss of momentum; indeed, the kinetic energy is spent to contrast the repulsive potential. At the distance r0 the potential energy equals the initial kinetic energy. At lower distances, that is inside the barrier, the solution of the Schrödinger equation shows that the amplitude of the wavefunction decreases exponentially but does not reach zero. Since the square modulus of the wavefunction gives the probability to find the particle, this shows that the incoming 4 proton can penetrate the barrier even if its energy is lower than the repulsive potential, contrasting the classical result according to which the particle should stop and reverse the motion. Due to this effect (tunnel effect) two protons can react even if their energies are not classically sufficient to overcome the repulsive potential. Thanks to the tunnel effect, reactions inside the stars can proceed and energy is produced. The proton-proton chains PP I 1) Two protons combine to produce deuterium; a positron and a neutrino are produced, too: p + p → d + e+ + νe (in addition: e+ + e- → γ , with energy 1.02 MeV) this reaction is very slow: lifetime 1.4 1010 y. Q value: 1.44 MeV. miu is neutrino Note that two protons cannot form a stable nucleus. In fact, the deuteron, the produced nucleus, is formed by a proton and a neutron, which means that during the very brief time interval when the two protons are within a distance of ≈ 10-15 m of each other, one of the protons must become transformed into a neutron. The duration of this time interval is about 2 ×10−21 sec. A free proton cannot transform into a neutron since this last has a higher mass of about 1.29 MeV. But inside a nucleus, in the presence of the strong force, with an attractive energy of 30–40 MeV, the transformation of a proton into a neutron becomes possible, “picking up” 1.29 MeV. The transformation of a proton into a neutron inside the nucleus occurs because of the operation of the weak force. The corresponding probability is very small: the weak interaction dominates the process in regulating the slowness of the whole chain. 2) deuterium reacts with another proton producing helium-3 (plus a gamma ray): d + p → 3He + γ Fast reaction: 1.5 s. Q value: 5.49 MeV 3) Two 3He interact and 4He (the usual helium) is produced, together with two protons; 3 He + 3He → 4He + p + p Lifetime 1 My, Q value: 12.86 MeV. 5  miu is neutrino Globally, the energy available for the star is: Ea = 2·(1.44-0.26+5.49)+12.86 MeV = 26.2 MeV. 0.26 MeV is the energy carried away by the neutrino ν, which is lost (neutrino interactions with matter are negligible). The fraction of mass transformed in energy is: Ea/(4mpc2) = 0.7% (the number of protons taking part to the reaction is 6 but 2 are returned at the end). PP I dominates at T = 10 MK. At higher temperatures two other reactions occur: PP II and PP III. PPII starts as PPI, but instead of the third reaction, we find: 3’) 3He + 4He → 7Be + γ Lifetime: 1 My; Q value: 1.59 MeV 4’) 7Be + e- → 7Li + νe Lifetime: 0.4 y; Q value: 0.06 MeV, neutrino energy: 0.86 MeV 5’) 7Li + 1H → 8Be* + γ Lifetime: 10 min; Q value: 17.35 MeV 6’) 8Be* → 2 4He Lifetime: 10-16 s. 6 PPIII differs from PPII starting from 4’: 4’’) 7Be + 1H → 8B + γ Lifetime: 66 y, Q value: 0.13 MeV 5’’) 8B → 8Be* + e+ + νe Lifetime: 1 s; Q value: 10.8 Mev; energy carried away be the neutrino: 7.2 MeV 6’’) 8Be* → 2 4He Lifetime: 10-16 s. The pp chain is the most important source of energy for stars with masses up to 1.5 MS. For the Sun the pp I branch produces 91% of He. Nevertheless, the other two branches are very important for basic research due to the production of neutrinos and related problem. We note the different energies carried by the neutrinos in the various chains. We will come back on this important point when we will discuss about the ‘neutrino problem’ for the Sun. We have to discuss three other points: i) how does H (that is, protons) in the stars come from? ii) do other processes exist for stars more massive than the Sun, and with higher core temperatures? iii) in the fusion reactions heavier nuclei are produced starting from lighter ones. But the pp chain produces only He. How are the other nuclei (and therefore the elements) produced? According to the Big Bang hypothesis, when matter started to exist hydrogen and helium were formed in a percentage 3/1 (cosmological nucleosynthesis). These elements entered in the composition of the first stars (actual stars in formation contain also other elements). For stars more massive of the Sun and whose core temperature > 16 MK another cycle has to be considered: the CNO cycle (Carbon Nitrogen Oxygen). This cycle forms He, too, but starting from carbon: 7 12 C acts as a catalyst. In more detail: Energy delivered in the star: 26.7 MeV – 1.7 MeV = 25 MeV. Contribution of each proton: 25/4 MeV = 6.25 MeV. It must be pointed out that CNO and pp cycles are not alternative: in each star both contribute to the production of energy. Each of them dominates in a suitable temperature range: 8 In the Sun the production of energy is due to: PP I 56% PP II 40% PP III 0.05% CNO 3.2% Obvious question: where does 12C come from? In stars more massive of the Sun the core temperature can reach 108 K: this makes possible for heavier nuclei to interact each other (in spite of the higher Coulomb repulsion). Carbon is formed (in more massive stars than the Sun) by means of the ‘3 alpha process’: a) 4He + 4He → 8Be + γ 8 Be rapidly decays (10-16 s) in two alpha particles (4He nuclei); nevertheless, it can react also with another alpha: 9 b) 8Be + 4He → 12C* → 12C + γ Excited state of carbon decays to the ground state by emitting one γ with energy 7.7 MeV. This excited state was predicted by F. Hoyle in 1954 (the same who coined the word ‘Big Bang’) and discovered experimentally by Fowler et al (Nobel prize in 1983). 12 C allows not only the CNO cycle to occur, but also the formation of heavier elements (under suitable temperature regimes) through various reactions, such as the ‘alpha reactions’: Each element produced is the starting point of the successive reaction, which is less and less probable. This justifies the production of the elements up to Fe. Elements can be produced also through direct interaction of two identical nuclei: Carbon burning (requires T > 600 MK): Oxygen burning (T > 1.5 GK are required): 10 Iron is the last element which can be built through a direct fusion reaction. How is it possible to form heavier nuclei (and therefore elements)? Free neutrons can be captured by a nucleus: (Z, N) + n → (Z, N + 1) The next step depends on the rapidity of neutron capture with respect to a beta decay (that is, a decay with emission of an electron): In the s-process (slow) the new nucleus undergoes a beta decay: (Z, N + 1) → (Z + 1, N) + e- + ν In the r-process (rapid) the nucleus is able to capture another neutron: (Z, N + 1) + n → (Z, N + 2) The r-process can be iterated and the nucleus increases its mass number A (= N + Z) until a beta decay breaks the chain. These processes require the presence of neutrons, which are generally available in non-stationary conditions (e.g., supernovae explosions). A few quite rare isotopes (about 40) cannot be produced by neutron capture. Rather, they are formed through the p-process: at temperatures > 1 GK a photon can decay in an electron-positron pair: γ → e+ + e- the positron can be annihilated or enters in the reaction: e+ + (Z, A) → (Z + 1, A) + 𝜈̅ Another possible reaction is: (Z, A) + p → (Z + 1, A +1) + γ This way all the elements can be formed. References: H. Karttunen, Fundamental Astronomy, ch. 10, 11 Keywords: nuclear interaction, nuclear fission, nuclear fusion, pp chain, CNO cycle, nucleosynthesis 11 Hydrostatic equilibrium and equation of state for a star Let us consider a spherical mass: The shell of radius r and thickness dr is attracted toward the center. It does not collapse, due to a pressure difference between external and internal layers of the shell. The gravity force acting on the small cylinder in the right figure is given by: 𝑀(𝑟)𝑑𝑚 𝑀(𝑟)𝜌(𝑟)𝑑𝑟𝑑𝐴 𝑑𝐹𝐺 = −𝐺 2 = −𝐺 𝑟 𝑟2 The force due to the pressure unbalance is given by: 𝑑𝐹𝑃 = 𝑃(𝑟)𝑑𝐴 − 𝑃(𝑟 + 𝑑𝑟)𝑑𝐴 = −𝑑𝑃𝑑𝐴 At equilibrium: dFG + dFP = 0 which implies: 𝑑𝑃 𝑀(𝑟) = −𝐺 𝜌(𝑟) equation of hydrostatic equilibrium 𝑑𝑟 𝑟2 The denser the fluid, the higher the pressure gradient. ----------------------------------------------- Example. Approximate pressure in the core of the Sun. We consider the Sun as a single shell with thickness R: 𝑑𝑃 𝑃𝐶 𝑀 ≈ ≈𝐺 2 𝑑𝑟 𝑅 𝑅 𝑀 < 𝜌 >≈ 𝑅3 and 𝑀2 𝑃𝐶 ≈ 𝐺 4 𝑅 By inserting the numerical values, we find about 1015 Pa. 1 -------------------------------------------------------------------- Equation of state. The approximation of an ideal gas is generally good for a non-relativistic plasma: PV = NkT where N is the total number of particles, k the Boltzmann constant. Since: 𝑁 𝜌= 𝑉 where is the average mass of the particles composing the plasma, we have the equation: 𝜌 𝑃= 𝑘𝑇 𝑚𝑝 +𝑚𝑒 𝑚𝑝 In the case of the Sun, most of the particles are protons and electrons: < 𝑚 >≅ ≅ 2 2 ---------------------------------------------------- Example. Temperature in the core of the Sun 𝑀 Using the previous equation, with 𝜌 = 4/3𝜋𝑅3 : 𝑚𝑝 4/3𝜋𝑅 3 𝑃 𝑇= 2 𝑘𝑀 which gives 4.4 107 K The calculation is rather rough; more accurate models give 1.5 107 K. ------------------------------------------------------ Example. Dependence of T on M and ρ 𝑑𝑃 𝑀(𝑟) 𝑀 = −𝐺 2 𝜌(𝑟) → 𝑃 ∝ 𝜌 𝑑𝑟 𝑟 𝑅 𝑀 1/3 𝑅∝( ) 𝜌 𝑀 −1/3 𝑃 ∝ 𝑀𝜌 ( ) = 𝑀2/3 𝜌4/3 ∝ 𝜌𝑇 𝜌 Therefore: 𝑇 ∝ 𝑀2/3 𝜌1/3 ------------------------------------------------------ 2 From the last proportionality we can deduce that: - The nucleus is hotter in more massive stars; - If a star contracts, its density increases and T increases as 𝜌1/3 - The temperature being the same, a star with smaller mass has a higher density: 𝜌 ∝ 𝑀−2 References: H. Karttunen, Fundamental astronomy, ch. 10 Keywords: hydrostatic equilibrium, pressure, equation of state 3 Degenerate gas and pressure of degenerate particles Let us consider a gas containing ne free electrons per unit volume; each electron occupies a volume: 𝑑𝑉 ≅ (∆𝑥)3 ≅ 1/𝑛𝑒 ∆𝑥 ≅ 𝑛𝑒 −1/3 Due to the Heisenberg principle: ℎ ∆𝑥∆𝑝𝑥 > =ℏ 2𝜋 the momentum of the electrons must satisfy the inequality: ℏ 1/3 ∆𝑝𝑥 > ≅ ℏ𝑛𝑒 ∆𝑥 This momentum generates a pressure, not to be confused with the thermal pressure. It is called pressure of the degenerate electrons and it is a purely quantum effect. Under which conditions such a pressure becomes important with respect to the usual pressure? For a gas at temperature T: 1 𝑝𝑥2 𝑚𝑒 𝑣𝑥2 = ≈ 𝑘𝑇 → 𝑝𝑥 ≈ √𝑚𝑒 𝑘𝑇 2 2𝑚𝑒 Let us impose that such a ‘thermal’ momentum be equal to the ‘quantum’ momentum: 1/3 ∆𝑝𝑥 = 𝑝𝑥 → √𝑚𝑒 𝑘𝑇 ≈ ℏ𝑛𝑒 Then, the approximate density to which the pressure of the degenerate electrons becomes important is: 3 𝑚𝑒 𝑘𝑇 2 𝑛𝑄 ≈ ( 2 ) ∝ 𝑇 3/2 ℏ which is called ‘density of quantum states’. Let us (approximately) calculate the pressure of the degenerate electrons: 1 1 − ∆𝑝𝑥 ∆𝑥 ≅ 𝑛𝑒 3 ∆𝑝𝑥 ≈ ℏ𝑛𝑒 3 𝑣𝑥 = (𝑣𝑥 ≪ 𝑐) 𝑚𝑒 1 The number dN of particles colliding against a wall with surface dA in a time interval dt: 𝑑𝑁 = (𝑑𝐴 ∙ 𝑣𝑥 𝑑𝑡)𝑛𝑒 Number of collisions N’ per unit time and surface: 𝑑𝑁 𝑁′ = = 𝑛𝑒 𝑣𝑥 𝑑𝐴𝑑𝑡 Each particle transfers a momentum ∆𝑝𝑥. Then, the pressure: 𝑃 = 𝑁 ′ ∆𝑝𝑥 = 𝑛𝑒 𝑣𝑥 ∆𝑝𝑥 = 𝑛𝑒 (∆𝑝𝑥 )2 /𝑚𝑒 Pressure on the surface` 𝑃 ≈ 𝑛 1 2 3 𝑒 (ℏ𝑛𝑒 ) 1 = ℏ2 53 𝑛 ≈ 0.025 ℎ2 5/3 𝑛 𝑚𝑒 𝑚𝑒 𝑒 𝑚𝑒 𝑒 A more rigorous calculation, based on quantum statistics, gives a very similar result, apart from a numerical factor: 1 3 2/3 ℎ2 5/3 ℎ2 5/3 𝑃= ( ) 𝑛𝑒 ≈ 0.05 𝑛 5 8𝜋 𝑚𝑒 𝑚𝑒 𝑒 Total density of the gas (electrons+protons+neutrons): 𝜌 𝜌 = 𝑍𝑚𝑝 𝑛𝑍 + (𝐴 − 𝑍)𝑚𝑛 𝑛𝑍 + 𝑚𝑒 𝑛𝑒 ≅ 𝑛𝑍 𝐴𝑚𝑝 → 𝑛𝑍 ≅ 𝐴𝑚𝑝 where we neglected the mass of the electrons and we equated the mass of the neutron to that of the proton. For a plasma, which is a highly ionized gas: 𝜌 𝑛𝑒 ≅ 𝑍𝑛𝑍 ≅ 𝑍 𝐴𝑚𝑝 The corresponding degenerate pressure: 2  5/3 1 3 2/3 ℎ2 5/3 ℎ2 𝑍 5/3 𝜌 𝑃= ( ) 𝑛𝑒 ≈ 0.05 ( ) ( ) 5 8𝜋 𝑚𝑒 𝑚𝑒 𝐴 𝑚𝑝 Main features of such a pressure: - it does not depend on the temperature - it depends on the inverse of the mass of the particle: 1/me. This means that the main contribution comes from the electrons. They are the first particles which introduce a degenerate pressure. - it depends on the density as ρ5/3. This means that for high density and low temperature (that is, low thermal pressure) we can find completely degenerate stars: the white dwarfs. ----------------------------------------------------- Example. Are electrons in the Sun degenerate? We have to compare the density of free electrons, ne, to the density of quantum states nQ, at which the degeneration becomes important. 3 4𝜋 2 𝑚𝑒 𝑘𝑇 2 𝑍 𝜌 𝜌 𝑛𝑄 ≈ ( ) ; 𝑛𝑒 ≈ ≈ 0.5 ℎ2 𝐴 𝑚𝑝 𝑚𝑝 Where we assume Z/A = 0.5 per solar nucleus, composed by He. Average values: ρ ≈ 1.4 g cm-3; T = 4.5 MK which give ne ≈ 4.2 1023 cm-3; nQ ≈ 3.6 1026 cm-3 ≈850ne That is, the gas is non-degenerate. In the core? ρ ≈ 150 g cm-3; T = 15 MK ne ≈ 4.6 1025 cm-3; nQ ≈ 2.2 1027 cm-3 ≈ 48ne Still non-degenerate, but the effect is now higher. 3 -------------------------------------------------------- Temperature and density determine the kind of pressure (hydrostatic or degenerate) which dominates inside a star. This is important in the case of a contracting star. From the equation of state: 2 1 𝑇 ∝ 𝑀3 𝜌3 (see lecture: Equation of hydrostatic equilibrium_5) which can be written: 1 2 𝑙𝑜𝑔𝑇𝑖𝑑𝑒𝑎𝑙 𝑔𝑎𝑠 = 𝑙𝑜𝑔𝜌 + 𝑙𝑜𝑔𝑀 + 𝐶1 3 3 A contracting star will evolve by following a straight line (slope 1/3) in the direction of increasing densities. On the other hand, in a degenerate gas: 3 2 4𝜋2 𝑚𝑒 𝑘𝑇 2 2 𝑛𝑄 ≈ ( ) → 𝑇𝑑𝑒𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒 𝑔𝑎𝑠 ∝ 𝜌3 → 𝑙𝑜𝑔𝑇𝑑𝑒𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒 𝑔𝑎𝑠 = 3 𝑙𝑜𝑔𝜌+C2 ℎ2 The slope of this straight line is 2/3. By starting from a non-degeneration condition, the two lines intersect at a density and temperature depending on the mass of the star: 11 10 9 8 log T (K) 7 6 5 4 3 degenerate gas ideal gas 2 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -3 log  (g cm ) In this state the collapse is blocked and the star does not contract anymore; rather, it starts to cool. In other words, degeneration prevents continuous collapse of a star (not always, only for some ranges of masses). 4 Stars having masses below a certain limit become degenerate before reaching a temperature to which nuclear fusion reactions start to ignite: the stars cool indefinitely. Such a threshold of mass is about 0.08 MS; the minimum mass of a star has to be higher than this limit, otherwise we speak of ‘brown dwarfs’. In the above discussion we supposed v 2.25 MS the core density is lower and electrons are not degenerate. Three alpha reactions take place but pressure (’normal’, non-degenerate pressure) increases due to energy production, this involves an expansion which prevents an acceleration of the reactions. Helium burning is not explosive. After a transient, for Sun-like stars the core burns He producing C and O, the envelope burns H. T increases and the star is located in a horizontal arm of the HR diagram. Precise position depends on the mass lost during the giant phase. 5 When also He is exhausted, the core made of carbon and oxygen contracts, and two concentric shells contain He and H which burn. Envelope dilates and the star evolves with constant luminosity until the Hayashi limit. When T is constant, luminosity increases and the star evolves along the asymptotic branch up to become a red supergiant. Since the temperature of the core is not sufficient to start the burning of carbon, the star loses all the envelope which forms a planetary nebula, while the core, very hot and degenerate, forms a white dwarf, without giving rise to nuclear reactions. The material of the nebula is available for a next generation of stars. 6 White dwarfs are sustained by the pressure of the degenerate electrons. Their typical radius: 𝐿 ∝ 𝑅 2 𝑇 4 → 𝑅 ∝ 𝐿1/2 𝑇 −2 𝐿𝑊𝐷 ≈ 10−4 𝐿𝑆𝑢𝑛 ; 𝑇𝑊𝐷 ≈ 𝑇𝑆𝑢𝑛 𝑅𝑊𝐷 ≈ 10−2 → 𝑅𝑊𝐷 ≈ 7000 𝑘𝑚 𝑅𝑆𝑢𝑛 7 Their density (supposing a mass equivalent to our Sun): 2 ∙ 1030 𝑘𝑔 𝜌𝑊𝐷 = ≈ 109 𝑘𝑔/𝑚3 (4𝜋/3)(7 ∙ 106 𝑚)3 (for comparison, the density of lead is ≈ 104 kg/m3). Pressure: 5/3 𝑍 5/3 𝜌 2 1 𝑃 ≈ 0.05ℎ ( ) ( ) ≈ 3 ∙ 1021 𝑃𝑎 𝐴 𝑚𝑝 𝑚𝑒 Relationship between radius and mass of a white dwarf: 5/3 𝑍 5/3 𝜌 1 Degenerate pressure: 𝑃 ≈ 0.05ℎ2 (𝐴) (𝑚 ) 𝑝 𝑚𝑒 𝐺𝑀2 𝑑𝑃 𝐺𝑀(𝑟) Pressure needed to gravitational equilibrium: 𝑃 ≈ ← =− 𝜌(𝑟) 𝑅4 𝑑𝑟 𝑟2 By equating, and considering that 𝜌 ≈ 𝑀/4𝑅 3 : 5/3 𝐺𝑀2 2 1 1 𝑀 5/3 ≈ 0.05ℎ ( ) ( ) 𝑅4 𝑚𝑝 𝑚𝑒 4𝑅 3 5/3 ℎ2 𝑍 5/3 1 It follows: 𝑀1/3 𝑅 ≈ 0.05 (𝐺𝑚 ) (𝐴) (4𝑚 ) 𝑒 𝑝 Therefore, if the mass of a white dwarf is known (e.g. it belongs to a binary system), its radius is determined. Previous relation points out that the bigger the mass, the lower the radius: 𝑅 ∝ 𝑀 −1/3. The surface temperature of a white dwarf is initially 50 kK, then cools up to a few kK. Cooling is very slow. Indeed, in a simple model, if we equate the luminosity to the energy loss: 𝐸𝑡ℎ 𝐿 = 4𝜋𝑅 2 𝜎𝑇 4 = 𝑑𝑡 𝑀 𝑑𝐸𝑡ℎ 𝑀 𝑑𝑇 𝐸𝑡ℎ ≈ 𝑘𝑇 ; ≈ 𝑘 𝑚𝑝 𝑑𝑡 𝑚𝑝 𝑑𝑡 𝑀 𝑑𝑇 4𝜋𝑅 2 𝜎𝑇 4 = 𝑘 𝑚𝑝 𝑑𝑡 𝑀 𝑘 1 𝑑𝑡 = 𝑑𝑇 𝑚𝑝 4𝜋𝑅 𝜎 𝑇 4 2 1 𝑘𝑀 |𝑡𝑊𝐷 | ≈ 𝑇 −3 3 4𝜋𝑅 2 𝜎𝑚𝑝 8 Since: 1 𝑇 ∝ 𝑡 −3 ; 𝐿(𝑡) ∝ 𝑅 2 𝑇 4 ∝ 𝑡 −4/3 More detailed models give 𝐿(𝑡) ∝ 𝑀𝑡 −7/5 Anyway, the important conclusion is that we can use measurements of luminosity of white dwarfs to estimate the age of stellar populations. Evolution of stars with M > 8 MS. In these massive stars the temperature of the core is sufficient to ignite reactions able to burn carbon (C) and heavier elements. When a burning is finished (e.g. C) the core contracts, with corresponding increase of pressure and temperature. Consequently, radiation pressure increases and pushes outside surface layers: the envelope expands, with decrease of the surface temperature: in the HR diagram the star moves along a horizontal line toward right (decreasing temperatures). On the contrary, when a new burning begins (e.g. O after C) the core expands and the envelope contracts, with increase of the surface temperature: the star moves toward left. In other words, the star moves alternatively toward right and left, with luminosity (about) constant: Every time that an alpha reaction is terminated, the core is surrounded by layers containing nuclei of the previous reactions; the star assumes an onion-like structure: 9 In the end the core contains 56Fe and 58Ni. Nuclear reactions cease. Indeed, remember that for heavier nuclei the energy per nucleon decreases (lecture Energy produced by a star_4, page 3): Density in the core is extremely high: 1012 kg m-3, nothing to do with ordinary matter. Reactions occur more and more rapidly: Until the mass of the core is lower than the Chandrasekhar limit, the pressure of the degenerate electrons keeps the star in equilibrium. Otherwise, core collapses and a supernova occurs (type II supernova). During the core collapse (in times of the order of 0.1 s) two processes occur: 10 - photodisintegration of the nuclei: gravitational energy delivered during the collapse gives rise to so energetic thermal photons to be able to break (almost all) the nuclei in single nucleons (protons and neutrons); - reaction between protons and electrons: e- + p → n + ν and the core contains almost neutrons. Star explodes and most of the mass is expelled. The mechanism is not clearly understood, but a few elements are established: - collapse of the core may be stopped by the degenerate pressure of neutrons; - a shockwave occurs by the neutrinos produced during the formation of neutrons. The wave crosses the various layers of the star and the temperature increases up to 5 GK: conditions for nucleosynthesis of nuclei heavier than Fe are produced in few seconds. - The energy delivered gives rise to a tremendous explosion: most of the material is expelled and only an ultra compact core is left (neutron star or black hole). Why do we call these stars type II supernova? There are other kinds of supernovae, in particular type Ia supernova. Some binary systems are formed by a red giant and a white dwarf. The two stars are sufficiently near that the external envelope of the giant feels the gravitational field of the dwarf. Material is transferred to this last: Mass of the white dwarf increases and, when the Chandrasekhar limit is overcome, equilibrium in the core (due to degenerate pressure) is broken and the carbon core collapses giving rise to explosive burning of carbon. The result is an explosion with complete disruption of the star (that is, without formation of a neutron star). This supernova Ia can be distinguished by the type II since the spectra 11 as well as the luminosities vs. time are different: Spectra of supernovae Ia do not contain hydrogen. Importance of supernovae: - enrich the universe of metals and heavy elements - heat the interstellar medium - allow us to detect extra solar neutrinos They are so luminous that they can be seen by naked eye. SN1054 (Type II, magnitude -5) was reported in China, Japan, Europe and North America; SN1006 (Type 1a, magnitude -7.5) was seen in China, Egypt, Switzerland, Japan, Iraq, America. SN1987A reached a magnitude 3; an increase in the neutrinos background was observed, compatible with the expected energy emitted during the explosion. The number in the code is the year of reported discovery. 12 References: H. Karttunen, Fundamental Astronomy, ch. 11 Keywords: protostars, Hayashi track, red giant, white dwarf, supernova, Chandrasekhar limit 13 Neutron stars and black holes A neutron star is formed (almost) exclusively by neutrons. The degenerate pressure prevents further collapse: 1 3 2/3 ℎ2 5/3 𝑃𝑛 = ( ) 𝑛 5 8𝜋 𝑚𝑛 𝑛 where mn = neutron mass, nn = its density. Note that the degenerate pressure of neutrons is less of that of electrons, by a factor 𝑚𝑛 /𝑚𝑒 = 1836 when the density is the same. Also in a white dwarf degenerate pressure (due to electrons) guarantees equilibrium; the ratio between the two pressures is: 5/3 𝑃𝑁𝑆 𝜌𝑁𝑆 5/3 𝑚𝑒 2 ∙ 1017 =( ) ( )=( ) (1/1836) ≈ 1.2 ∙ 1010 𝑃𝑊𝐷 𝜌𝑊𝐷 𝑚𝑛 2 ∙ 109 A few properties of a neutron star: Radius: Radius-mass relationship for white dwarfs: 5/3 𝑍 5/3 𝜌 1 𝑀2 degenerate pressure: 𝑃 = 0.05ℎ2 (𝐴) (𝑚 ) (lecture 6); hydrostatic pressure: 𝑃 = 𝐺 𝑅4 𝑝 𝑚𝑒 (Lecture 5). By equating the two expressions: 5/3 1/3 0.05ℎ2 𝑍 5/3 1 1 𝑀𝑊𝐷 𝑅𝑊𝐷 = ( ) ( ) 𝐺 𝐴 𝑊𝐷 𝑚𝑒 𝑚𝑝 For a neutron star: 5/3 1/3 0.05ℎ2 𝑍 5/3 1 1 𝑀𝑁𝑆 𝑅𝑁𝑆 = ( ) ( ) 𝐺 𝐴 𝑁𝑆 𝑚𝑛 𝑚𝑝 Also in this case the smaller objects are the most massive ones; if masses are the same: 1  5/3 (𝑍/𝐴)𝑁𝑆 𝑚𝑒 1 𝑅𝑁𝑆 =[ ] 𝑅𝑊𝐷 ≈ (3.2) ( )𝑅 ≈ 1.7 ∙ 10−3 𝑅𝑊𝐷 (𝑍/𝐴)𝑊𝐷 𝑚𝑛 1836 𝑊𝐷 Z represents the particles contributing to the degenerate pressure; A the particles contributing to the density. For a white dwarf, Z (electrons) = ½ A (protons + neutrons). For a neutron star, Z = A Starting from a star having M = 1.5 MS: white dwarf radius: 𝑅𝑊𝐷 = 10−2 𝑅𝑆𝑢𝑛 ≈ 7 ∙ 103 km For a neutron star the radius: RNS ≈ 12 km (just an order of magnitude) Density: Again, assuming 1.5 MS, R = 15 km: 1.5∙(2∙1030 𝑘𝑔) 𝜌 = (4𝜋/3)(1.5∙104 𝑚)3 = 2 ∙ 1017 kg/m3 For a comparison, the density of a nucleon (considered as a classical spherical particle): 1.7∙10−27 𝑘𝑔 𝜌 = (4𝜋/3)(10−15 𝑚)3 = 4 ∙ 1017 kg/m3 The density of a neutron star is like that of nuclear matter. In a first approximation it can be considered as a single nucleus with A ≈ 1057. Pressure: Degenerate pressure of a neutron star: 1 3 2/3 ℎ2 5/3 𝑃𝑛 = ( ) 𝑛 ; 𝜌 = 𝑛𝑛 𝑚𝑛 5 8𝜋 𝑚𝑛 𝑛 5/3 𝑛𝑛 −5/3 −8/3 = (𝜌5/3 𝑚𝑛 )𝑚𝑛−1 = 𝜌5/3 𝑚𝑛 𝑚𝑛 Therefore: 1 3 2/3 −8/3 5/3 𝑃𝑛 = 5 (8𝜋) ℎ 2 𝑚𝑛 𝜌 = 1036 Pa 2 Rotation. A neutron star is an object rotating with a very high angular velocity. Indeed, by applying the conservation of angular momentum: 2 𝐿 = 𝐼𝜔 = 𝑀𝑅 2 𝜔 5 If the initial rotation is comparable to that of the Sun (period about 1 month, typical for stars of the main sequence): (𝑅 2 𝜔)𝑏𝑒𝑓𝑜𝑟𝑒 = (𝑅 2 𝜔)𝑎𝑓𝑡𝑒𝑟 ↔ (𝑅 2 𝜔)𝑆𝑢𝑛 = (𝑅 2 𝜔)𝑁𝑆 2 𝑅𝑆𝑢𝑛 𝜔𝑁𝑆 = ( 2 ) 𝜔𝑆𝑢𝑛 𝑅𝑁𝑆 2 7 ∙ 108 𝑚 𝜔𝑁𝑆 =[ ] 𝜔𝑆𝑢𝑛 = 2 ∙ 109 𝜔𝑆𝑢𝑛 1.5 ∙ 104 𝑚 The period: 𝑃𝑁𝑆 = 𝑃𝑆𝑢𝑛 (𝜔𝑆𝑢𝑛 /𝜔𝑁𝑆 ) = (2.6 ∙ 106 𝑠)/(2 ∙ 109 ) ≈ 1.3 ∙ 10−3 s A neutron star can rotate with a period of the order of ms. Magnetic field. 1 If the magnetic flux, Φ𝐵 = 𝜋𝑅 2 𝐵 , is conserved, 𝐵 ∝ 𝑅2. Starting with a magnetic flux comparable 𝑅 2 to that of the Sun, the magnetic field of a neutron star increases by a factor: (𝑅 𝑆 ) ≈ 2 109. It 𝑁𝑆 6 would result a magnetic field of 10 T (for comparison, magnetic fields produced by superconductors ≈ 10 T). In fact, internal phenomena may increase the magnetic field by orders of magnitude. Acceleration of gravity at the surface: 𝐺𝑀 𝑔𝑁𝑆 = = 8.9 ∙ 1011 𝑚𝑠 −2 ≈ 1011 𝑔𝐸𝑎𝑟𝑡ℎ 𝑅2 Neutron stars were predicted around 1930. Today, we have detailed models of their structure; an example: 3 Do neutron stars really exist? In 1967 Jocelyn Bell-Burner, an Antony Hewish’s graduate student, discovered a radio source emitting pulses with irregular amplitude, but with extremely regular period (1.3 s). Today we know about 1000 pulsating stars (pulsars), whose period changes with time (generally increases by 0.1 – 001% per year). Of the possible candidates, that is, normal stars, white dwarfs and neutron stars, only these last are able to explain the behavior of the pulsars. Indeed, the origin of the radio pulses can be understood if the magnetic field of a neutron star is tilted at an angle of 45° − 90° with respect to the rotation axis. Then there will be a magnetosphere around the star, where the particles are tied to the magnetic field and rotate with it In this figure the radiating relativistic particles rotating around the neutron star by means of its magnetic field emit radiation in a narrow cone in their direction of motion (an accelerated charged particle emits radiation). Radiation from the point P arrives to the observer located in the direction 4 of the arrow. As the star rotates, this cone sweeps around like a lighthouse beam, and is seen as rapid pulses. Why does the period change with time? Kinetic energy decreases owing to emission of pulses as well as to gravitational waves. 1 For a rotating sphere: 𝐸 = 2 𝐼𝜔2 The rate of energy loss with time: 𝑑𝐸 𝑑𝜔 𝐸̇ = = 𝐼𝜔 𝑑𝑡 𝑑𝑡 There is a connection with the change of the period: 𝐸̇ 2 𝑑𝜔 𝜔̇ = 2 𝐼𝜔 =2 𝐸 𝐼𝜔 𝑑𝑡 𝜔 𝑑𝜔 2𝜋 𝑑𝑃 𝜔̇ 𝑃̇ =− 2 → =− 𝑑𝑡 𝑃 𝑑𝑡 𝜔 𝑃 By comparing the previous equations we get: 𝐸̇ 𝑃̇ = −2 𝐸 𝑃 The measurement of the increase of the period gives an estimate of the energy lost by the neutron star. This energy feeds the nebula remnant of the supernova explosion. ---------------------------------------------------------------------- Example. Neutron star with M = 2MS, P = 0.1 s, R = 15 km. The measured rate of change of the period: dP/dt: 3 10-6 s/y 1 1 2𝜋 2 1 2𝜋 2 Kinetic energy: 𝐸 = 2 𝐼𝜔2 = 5 𝑀𝑅 2 ( 𝑃 ) = 5 (4 ∙ 1030 𝑘𝑔)(1.5 ∙ 104 𝑚)2 (0.1 𝑠) = 7 ∙ 1041 J Rate of energy loss: 𝐸̇ 𝑃̇ 𝑑𝑃 𝑃̇ 10−13 = −2 𝑃 ; = 3 ∙ 10−6 s/y = 10-13 (1 y ≈ 3·107 s) ; = = 10−12 s-1 𝐸 𝑑𝑡 𝑃 0.1 𝑠 𝐸̇ Therefore: 𝐸 = −2 ∙ 10−12 s-1 𝐸̇ = (−2 ∙ 10−12 𝑠 −1 )(7 ∙ 1041 𝐽) = −1.4 ∙ 1030 W 5 which is sufficient to feed the energy emitted from the nebula (e.g. the luminosity of the Crab nebula L = 1031 W). 𝐸 1 Lifetime: 𝑡 = 𝐸̇ = (2∙10−12 𝑠−1 ) = 5 ∙ 1011 𝑠 ≈ 1.7 ∙ 104 𝑦 In addition to the steady slowing down of the rotation, sometimes small sudden jumps in the period are observed. These are interpreted as a sign of rapid mass movements in the neutron star crust (“starquakes”) or in its surroundings. Black holes. If the mass of the core is > MTOV = 3 MS (TOV limit, Tolman-Oppenheimer-Volkoff) the degenerate pressure of the neutrons is no longer sufficient to prevent the collapse of the core. We obtain a singularity in the sense that we do not know any process able to stop the collapse. We have a black hole. If we concentrate a mass M within a critical radius RC no information can emerge from the region internal to the sphere of radius RC. The radius is found by imposing that the escape velocity be equal to c: 1 2 𝐺𝑀𝑚 𝐺𝑀 1 2𝐺𝑀 𝑚𝑣𝑒𝑠𝑐𝑎𝑝𝑒 − =0 → > 2 𝑐2 ; 𝑟𝑆 = 2 𝑟 𝑣𝑒𝑠𝑐𝑎𝑝𝑒 =𝑐 𝑟 𝑐2 The critical radius is called Schwarzschild radius. Correct derivation of rS requires to solve the equations of general relativity. ---------------------------------------------- Example. (hypothetical) Schwarzschild radius of Earth and Sun. Ans. 0.71 cm; 2.95 km. ------------------------------------------------ Average density of a BH: 𝑀 3𝑀 𝑐 6 3𝑐 6 1 𝜌= = = ∝ 𝑀−2 𝑉 4𝜋 8𝐺 3 𝑀3 32𝜋𝐺 3 𝑀2 The higher the mass, the lower the density. Since the gravitational force is directed towards the centre of the hole and depends on the distance, different parts of a falling body feel a gravitational pull that is different in magnitude and direction. The tidal forces become extremely large near a black hole so that any material falling into the hole 6 will be torn apart. All atoms and elementary particles are destroyed near the central point, and the final state of matter is unknown to present day physics. It is improbable that a black hole could have a significant net charge. Rotation, on the other hand, is typical to stars, and thus black holes, too, should rotate. Since the angular momentum is conserved, stars collapsed to black holes must rotate very fast. In 1963 Roy Kerr found a solution of the field equations for a rotating black hole. In addition to the event horizon a rotating hole has another limiting surface, an ellipsoidal static limit. Objects inside the static limit cannot be kept stationary by any force, but they must orbit the hole. However, it is possible to escape from the region between the static limit and the event horizon, called the ergosphere. In fact, it is possible to utilize the rotational energy of a black hole by dropping an object to the ergosphere in such a way that part of the object falls into the hole and another part is slung out. The outcoming part may then have considerably more kinetic energy than the original object. A way in which a black hole could be directly observed is by means of the radiation from gas falling into it. For example, if a black hole is part of a binary system, gas streaming from the companion will settle into a disc around the hole. Matter at the inner edge of the disc will fall into the hole. The accreting gas will lose a considerable part of its energy (up to 40% of the rest mass) as radiation, which should be observable in the X-ray region. Some rapidly and irregularly varying X-ray sources of the right kind have been discovered. The first strong evidence for black hole in an X-ray binary was for Cygnus X-1. 7 Its luminosity varies on the time scale of 0.001 s, which means that the emitting region must be only 0.001 light-seconds or a few hundred kilometres in size. Only neutron stars and black holes are small and dense enough to give rise to such high-energy processes. Cygnus X-1 is the smaller component of the double system HDE226868. The larger component is an optically visible supergiant with a mass 20–25 MS. The mass of the unseen component has been calculated to be 10– 15 MS. If this is correct, the mass of the secondary component is much larger than the upper limit for a neutron star, and thus it has to be a black hole. Today many of such systems are known, where the compact component has a mass larger than 3 MS, and therefore is probably a black hole. Many frightening stories about black holes have been invented. It should therefore be stressed that BH obey the same dynamical laws as other stars – they are not lurking in the darkness of space to attack innocent passers-by. If hypothetically the Sun became a black hole, the planets would continue in their orbits as if nothing had happened. So far, we have discussed only black holes with masses in the range of stellar masses. There is however no upper limit to the mass of a black hole. Many active phenomena in the nuclei of galaxies can be explained with supermassive black holes with masses of millions or thousands of millions solar masses. Neutron stars as celestial GPS. On July 2017 the NICER (Neutron star Interior Composition ExploreR) project was installed onboard the ISS to measure the size of pulsars in order to understand their internal composition. To this purpose, an X-ray telescope was used, SEXTANT (Station Explorer for X-ray Timing And Navigation Technology), with a secondary goal to demonstrate that a spacecraft could orient itself using pulsars, just as the regular ticking of atomic clocks on satellites is exploited for GPS. Indeed, millisecond pulsars are highly stable ‘clocks’, with long term stability comparable to laboratory 8 atomic clocks. For these pulsars, a simple physical model with a small number of parameters can predict the arrival time of pulses to microsecond accuracy over months or years SEXTANT timed X-ray flashes coming from five pulsars, one of which is the closest and brightest known millisecond pulsar. The apparatus watched each of the signals for about 5–15 minutes before rotating autonomously to look at the next. By measuring tiny changes in the signals’ arrival time as the experiment orbited Earth, NICER could independently calculate its own position in space, with an accuracy of 7 km in two days of measurements. Presently, spacecraft must communicate with Earth regularly to confirm their position. But such communication — through systems such as NASA's Deep Space Network — is time-consuming, expensive, and more difficult the farther from Earth a probe travels. Pulsar navigation might work well for spacecraft in the outer Solar System because it could free probes to do many navigation- related tasks without waiting for instructions. NICER used 56 small X-ray telescopes for its study, but a few such telescopes could be sufficient for establishing the position of the spacecraft during navigation. Each instrument might weigh as little as 5 kilograms, making it relatively inexpensive to add to space missions. A view of the NICER X-ray Timing Instrument without its protective blanketing shows a collection of 56 close-packed sunshades-the white and black cylinders in the foreground-that protect the X-ray optics (not visible here), as well as some of the 56 X-ray detector enclosures, on the gold-colored plate, onto which X-rays from the sky are focused (image credit: NASA, Keith Gendreau). The NICER mission at work aboard the International Space Station (image credit: NASA/GSFC) References: H. Karttunen, Fundamental astronomy, ch. 14 Keywords: neutron stars, pulsars, black hole, Schwarzschild radius, NICER. 9 The Interstellar medium A star is born out of gas and dust that exists between the stars, known as the interstellar medium (ISM). Much of that material may be returned to the ISM through stellar winds and explosive events. Subsequent generations of stars can then form from this processed material. In this sense the evolution of stars is a cyclic process. It is important to study the nature of the ISM, in order to understand the formation of stars but also since it impacts our observations of everything, from relatively nearby stars to the most remote galaxies and quasars. The ISM is an enormous and complex environment that provides an important laboratory for testing our understanding of astrophysics. Indeed, the dynamics of the ISM involve turbulent gas motions, shocks, and galactic magnetic fields that lace through interstellar space. Modelling the ISM ultimately requires detailed solutions to the equations of magnetohydrodynamics. The production and destruction of dust grains and complex molecules requires a detailed understanding of chemistry in an environment not reproducible in a terrestrial laboratory. Interstellar extinction The stars located behind intervening dust clouds are obscured. This obscuration, referred to as interstellar extinction, is due to the summative effects of scattering and absorption of starlight. Given the effect that extinction can have on the apparent magnitude of a star, the distance modulus equation (see the lecture Luminosity of stars_1, p.8): 𝑘𝑑 𝑚 − 𝑀 = 5𝑙𝑜𝑔10 𝑑 − 5 = 5𝑙𝑜𝑔10 ( ) 10 𝑝𝑐 must be modified appropriately. In a given wavelength band centred on λ, we now have 𝑚𝜆 = 𝑀𝜆 + 5𝑙𝑜𝑔10 𝑑 − 5 + 𝐴𝜆 where d is the distance in pc and Aλ > 0 represents the number of magnitudes of interstellar extinction present along the line of sight. If Aλ is large enough, a star that would otherwise be visible to the naked eye or through a telescope could no longer be detected. Aλ must be related to the optical depth of the material, measured back along the line of sight. The optical depth τλ is the natural logarithm of the ratio of incident to transmitted radiant flux (intensity) through a material. Therefore 1  𝐼𝜆 = 𝐼𝜆,0 𝑒𝑥𝑝(−𝜏𝜆 ) where Iλ,0 is the intensity in the absence of interstellar extinction. Combining this with the equation relating the apparent magnitude to the radiant flux: 𝑓 𝑚1 − 𝑚2 = −2.5𝑙𝑜𝑔10 (𝑓1 ) 2 we can now relate the optical depth to the change in apparent magnitude due to extinction, giving 𝑚𝜆 − 𝑚𝜆,0 = −2.5𝑙𝑜𝑔10 (𝑒 −𝜏𝜆 ) = 2.5𝜏𝜆 𝑙𝑜𝑔10 𝑒 = 1.086𝜏𝜆 But the change in apparent magnitude is just Aλ, so Aλ = 1.086τλ. The change in magnitude due to extinction is approximately equal to the optical depth along the line of sight. It depends on the amount of interstellar dust that the light passes through. The optical depth through the cloud is given by 𝑠 𝜏𝜆 = ∫ 𝑛𝑑 (𝑠 ′ )𝜎𝜆 𝑑𝑠′ 0 where nd (s’) is the number density of scattering dust grains and σλ is the scattering cross section. If σλ is constant along the line of sight, then where Nd, the dust grain column density, is the number of scattering dust particles in a thin cylinder with a cross section of 1 m2 stretching from the observer to the star. This shows that the amount of extinction depends on the amount of interstellar dust that the light passes through. * If we assume spherical dust particles with radius a, then the geometrical cross section that a particle presents to a passing photon is just σg = πa2. We may now define the dimensionless extinction coefficient Qλ to be 2  𝜎𝜆 𝑄𝜆 = 𝜎𝑠 When the wavelength of the light is on the order of the size of the dust grains, then Qλ ∼ a/λ, implying that 𝑎3 𝜎𝜆 ∝ (𝜆 ≳ 𝑎) 𝜆 In the limit that λ becomes very large relative to a, Qλ goes to zero. On the other hand, if λ becomes very small relative to a, it can be shown that Qλ approaches a constant, independent of λ so that 𝜎𝜆 ∝ 𝑎2 (𝜆