Optical Depth & Radiative transfer

Transcription

1 University of Naples Federico II, Academic Year Istituzioni di Astrofisica, read by prof. Massimo Capaccioli Lecture 8 Optical Depth & Radiative transfer

2 Learning outcomes The student will : learn what opacity is and how it links to the mean free path and to optical length; study the random walk; examine which are the sources of opacity; recap the Balmer and Lyman spectral series and see a general form for opacity; scrutinize the various sources of broadening; evaluate the transfer equation and its approximate solutions; recap some properties of the Sun.

3 Mean free path How far, in an ideal gas with number density n, does an atom move before interacting with another? 2a0 = d The effective collisional area 2 is: A = π d ( ) 2 2 2a a Collisional cross section: σ = π = 4π 0 0 Mean free path: l = 1 nσ the unit volume must be equal to the number n of individual volumes made by V = l σ ind

4 Opacity Opacity is the ability of material to absorb photons ( inverse of conductivity). Opacity κ photons of 2 m kg is cross-section per h: di = κ a specific wavelengt a unit mass of material for absorbing ρi ds. for a given wavelenght it is proportional to incoming flux I to density of scattering material ρ to effectiveness in scattering κ and to length of the path ds. I I di How does the intensity of radiation depend on opacity and distance travelled through a homogeneous ( d κ ds = dρ ds 0) medium? κρs =,0 =,0 I I e I e s l 1, where l = is the mean free path. κ ρ After the photon has traveled 1 by a factor e = one mean free path, its intensity has decreased

5 Opacity The following cartoon visualizes the process of attenuation of the radiation: Suppose that a beam of light with intensity I o runs into a slab of material. As the rays move through the material, some may be absorbed... or scattered... so that only a fraction I reaches the other side.

6 Opacity: two different cases with the same result thin but very effective/dense The slab might be relatively narrow, yet block a considerable fraction of the light, if its atoms are densely packed and efficient at interacting with light. thicher but less effective/dense On the other hand, a much wider slab of material, with a lower opacity and/or density might block exactly the same fraction of light.

7 Example: the atmosphere of the Sun 23 3 In the Sun's photosphere n m. 4 3 Assuming pure hydrogen, the density is ρ = nmh kg m. At the wavelength of visible light ( 500 nm ), the opacity in this region 2 1 of the atmosphere, is: κ = m kg. The photon mean free path is: l 1 = 150 km. κ ρ So photons can travel a very long way before the intensity decreases appreciably. The atmosphere is not in Local Thermodynamical Equilibrium (LTE): at a given place in the atmosphere there are photons which originated somewhere else with a different temperature.

8 Example: the atmosphere of the Earth The 3 density of Earth's atmosphere at sea lev el is 1.2 kg m. ρ What would the photon mean free path be if the atmosphere had the same opacity as the Sun? The answer is an increadibly small l = 31 m. So it would be murky... The high opacity in the Sun (as we will see) is that the high temperature leads to many free electrons which are able to absorb photons.

9 Local Thermodynamic Equilibrium Thermodinamic equilibrium requires that a system is in: - thermal equilibrium ( dt dt = dt dt 0 for the temperature T ), - mechanical equilibrium ( net force on each particle 0), - radiative equilibrium ( F 0 for the monochromatic vector flux density of radiation at each point in a region of space), F - chemical equilibrium ( invariance with time of concentration of elements). I n Global Thermodynamic Equilibrium the scale invariant parameters the temperature T ) do not vary throughout the system, while (e.g. in Local Thermodynamic Equilibrium they do vary but slowly enough, so that they remain roughly constant in the surroundings of each point.

10 Optical depth The optical depth is a unitless quantity that relates to how much light is absorbed along its path s: dτ = κ ρds = ds l. l = 1 κ ρ is the mean free path It is a distance coordinate s in units of the mean free path l. The difference in optical depth between the initial and final position of a light ray is then given by: τ = τ τ = κ ρds., f,0 0 Note that the optical depth is greater at the initial position than at the final, i. e. the optical depth decreases as the photon moves toward the observer. s

11 One mean free path = one optical depth The optical depth has several meanings. Consider a light ray making its way through some scattering medium. Suppose the mean free path of the light be small compared to the thickness of the material. Light bounces many times before finally escaping. Note the location of the last bounce. If we send a second light ray into the medium, it will follow a different path and a third ray will again scatter randomly before finding its way out. The location of the final bounce tends to be roughly one mean free path, or one optical depth, from the outer edge of the medium. If we turn the situation around and ask: How far can we see into the slab from the outside? The answer is: about one mean free path or, equivalently, about one optical depth.

12 Optical depth For a stellar atmosphere we can set τ = 0 at the surface. Thus, for a photon which traveled a distance s through the stellar atmosphere, the initial optical depth is: τ = 0 τ = s 0 κ ρds, and the intensity of the ligh t ra y can be expressed as: s κ ρds 0 I = I e = I e,0,0 τ.

13 Optical depth: another way to see it If ρ and κ are approximately constant over the photon's path: 0 1 where l = is the mean free path. κ ρ s τ = κ ρds κ ρs = s l The optical depth can be thought o f as the number of mean free paths a photon takes between its initial position and the surface.,

14 Local Thermodynamic Equilibrium (LTE) Local Thermal Equilibrium (LTE) holds if the distance over which both matter and radiation can travel between interactions is much smaller than the distance over which temperature changes. Compare the mean free path of a hydrogen atom in the Solar photosphere 23 3 (where the temperature gradient is 8.7 K km and n m ) to the temperature scale height. 11 The Bohr radius of hydrogen is a0 = m ( 0.53 Angstrom). This value can be computed in terms of other physical constants: πε a where is the permittivity of free space and 2 4 0ħ ħ 0 = = ε 2 0 mee mecα α is the fine structure constant.

15 Random walk As a photon moves through a dense gas, it interacts with atoms being scattered, absorbed, and re-emitted. If each scattering process is random, how far does the photon move (on average)? s N l = N = ( s l) 2 l N 1 Example of eight random walks in one dimension starting at 0. l 3 l N l 1 l 2 s cfr. the beautiful site:

16 Random walk Three cases of a random walk in 3D.

17 Random walk The optical depth can be thought of as the number of mean free paths a photon takes between its initial position and the surface. τ = s l. The number of interactions a photon has on the way to the surface is: N 2 s 2 = = τ. l Thus a photon can escape to the surface when τ 1 (since then s = l ), and the number of steps 2 ( N s ). required increases as the square of the distance travelled A more careful calculation ( coming soon) will show that the average optical depth from which photons originate is 2 3. τ

18 Optical depth: example Using the numbers we used before ( Lecture 7) for the solar photosphere, approximately how far into the Sun does it correspond to an optical depth of τ = 1? The photon mean free path is: 1 l = 150 km ρ κ The photons we see represent the conditions very close to the Sun's surface: , R. =

19 Sources of opacity The opacity is determined by the physical processes that can remove photons from a beam of radiation. In general this includes true absorption processes as well as scattering. 1. Bound-bound transitions 2. Bound-free transitions 3. Free-free absorption 4. Electron scattering (why not proton?)

20 Photon-electron interactions bo und-bound bound-free free-free 1 hν = m v v 2 ( ) 1 2 mv 2 χ ion Excitation energy hν = χ 3 χ 2 hν = χ 2 hν = χ + ion 1 2 mv 2 n = 3 n = 2 0 n = 1

21 Johann Jakob Balmer ( ) The Balmer jump The energy of an electron in the n = 2 orbit 13.6 of a hydrogen atom is: E = = 3.40 ev 2 2 So the atom can be photoionized by any photon with wavelength: hc = nm χ This gives rise to in stellar spectra. 2 the Balmer jump Note that the jump is not present in hotter stars, where most of the hydrogen is already ionized. Flux (arbitrary units) [nm] T e = T e = T e = 4350 K 5770 K 44, 500K

22 The Balmer jump The sudden decrease in the intensity of the continuous spectrum at the limit of the Balmer series of Hydrogen at 3646 Å, representing the energy absorbed when electrons originally in the second energy level are ionized.

23 The Lyman limit Theodore Lyman III ( ) Similarly, the energy of an electron in the n = 1 orbit of a hydrogen atom is: E = 13.6 ev So the atom can be photoionized by any photon with wavelength: 91.2 nm This is a very efficient process, and it takes very little neutral hydrogen to absorb all photons with this energy. For stars more than a few pc away, there is sufficient neutral hydrogen between them and us to absorb almost all the light at these UV wavelengths. We therefore know little about the far-uv flux of hot stars.

24 Typical sources of opacity In most stellar atmospheres, the primary source of continuum opacity is the photoionization of H ions. In cool stars, molecules can form. These complex arrangements of atoms can make many bound-bound and bound-free transitions which greatly increase the opacity

25 Rosseland mean opacity The total opacity is the sum of all the sources. It depends on the wavelength of the light being absorbed, as well as the composition, density and temperature of the gas. A weighted average over all wavelengths gives the Rosseland mean opacity κ : = kν Iν dν κ I Svein Rosseland ( ) at the Niels Bohr Institute

26 Rosseland mean opacity i i i Opacity increases with density at fixed T; At fixed density and low temperatures, opacity rises steeply as T increases; After the peak, the opacity decreases T 3.5 as κ ; log ρ = 4 log ρ = 6 log ρ = 2 log ρ = 0 i i The bu mp at T ~ 40,000 K is due to the second ionization of He; At the highest temperatures, the opacity is due to electron scattering. HeII

27 Absorption lines and Kirchoff s laws Photons that we see originate (on average) from an optical depth of 2/ 3. Recall the optical depth is related to the opacity along the photon path by: τ s = 0 κ ρds Gustav Robert Kirchhoff ( ) If the opacity is much higher/lower at some wavelength 0 (for example due to bound-bound absorption), the photons we see will originate from higher/lower layers. 0 Low opacity High opacity Low opacity

28 Absorption lines and Kirchoff s laws κ ( ) r > r 2 1 r T F ( ) r 1

29 Absorption lines and Kirchoff s laws Absorption does not matter if the temperature of the gas is uniform: the luminosity radiated from all points in the gas is the same, so it does not matter where the photon originates from. However, if the temperature is lower near the surface, the luminosity will 4 be much lower: L T. Thus, at the same spot the luminosity will be lowest at wavelengths where the opacity is the highes t: this gives rise to absorption lines. 0 T Low opacity High opacity Low opacity

30 Emission lines What would be the result if the star s temperature increased outward? In this case, light from regions of high opacity would arise from higher in the atmosphere, where the temperature (and therefore luminosity) is higher. We would therefore see an emission line. 0 T Low opacity High opacity Low opacity

31 The Solar Atmosphere The solar atmosphere extends thousands of km above the photosphere (from which the optical radiation is emitted). T ~ 10 6 K The temperature increases above the photosphere (in the chromosphere and corona). We can therefore see emission lines from these regions. T ~ 25,000 K T ~ 5,770 K HeI emission from the chromosphere (~ 20,000 K). Core T ~ 10 7 K

32 Line profiles The region near the central wavelength is the core of the line. The sides which join up with the continuum are the wings. wings deeper into the atmosphere The area of the line is related to the amount of absorption. The central regions of the line must be formed at higher (and cooler) regions of the atmosphere. The wings probe deeper into the atmosphere. W F F c = Fc core d higher & cooler regions

33 Recap on broadening Absorption lines are due to bound-bound atomic transitions of fixed energy. Why are the lines not infinitely narrow? Natural broadening Pressure or collisional broadening Doppler broadening

34 Natural broadening From Heisenberg's uncertainty principle: the electron in an excited state is only there for a short time, so its energy cannot have a precise value. Since energy levels are " fuzzy", atoms can absorb photons with slightly different energy, with the probability of absorption declining as the difference in the photon's energy from the "true" energy of the transition increases. The FWHM of natural of t is given by: o broadening for a transition with an average waiting time 2 c 1 E0 t = ħ 2 h ν t = ħ 2 h t = ħ 0 1/ 2 =. πc t o Natural broadening is usually very small; the typical value 4 is = 2 10 A. 1 2

35 Doppler broadening Two components contribute to the intrinsic Doppler broadening of spectral lines: 1. thermal broadening; 2. turbulence - the dreaded microturbulence! Thermal b roadening is controlled by the thermal velocity distribution 2 dn(v r ) m mv r (and the shape of the line profile): = exp dvr, NTotal 2π kt 2kT where v is the line-of-sight velocity compone nt. r 2 3 The Doppler w idth associated with the velocity v ( variance v = 2kT m) is: v 2 0 2kT 7 T D , µ = = = c c m and is the wavelength of the line center.

36 Thermal broadening light profile Combining these we get the thermal broadening line profile: 2 2 ( ν ν 0 ) 2 mc ν ν 2kT I c m = e I ν 2π kt total. At line center, ν = ν, and this reduces to: 0 I I ν total c = ν m 2π kt. Where the line reaches half its maximum depth, the total width is: 2 = kT ln 2. c m

37 Voigt profile The absorption line profile is due to the Doppler and damping (pressure and natural) broadening. The resulting profile is known as the Voigt profile. Woldemar Voigt ( )

38 Pressure broadening: example The atmospheres of supergiant stars are ~1 million times less dense than in the Sun. So the pressure broadening is weaker by this factor, which is why more luminous stars have much narrower lines (the cores are still dominated by Doppler broadening). But be careful in evaluating by eye the broading. This is done properly for lines with the same depth.

39 Emission coefficient The emission coefficient is the opposite of the opacity: it quantifies processes which increase the intensity of radiation: di = j ds The emission coefficient j has units of W m str kg. Thus, accounting for both processes of emission and absorption: ρ di = κ ρi ds + j ρds.

40 The source function The intensity of radiation is therefore determined by the of the emission coefficient and the opacity: relative importance 1 di di j = = I = I S κ ρ ds dτ κ dτ = κ ρds = ds l where we have defined the source j function: S =. κ 3 1 The source function has units of intensity, W m sr. As the ratio of two inverse processes (emission and absorption), the source function is relatively insensitive to the detailed properties of the stellar material.

41 Radiative transfer This is the time independent radiative transfer equation: 1 di di I S κ ρ ds = dτ = For a system in thermodynamic equilibrium ( e.g. a black body), every process of absorption is perfectly balanced by an inverse process of emission., Since the intensity I is equal to the black body function B and therefore di throughout the box: = 0, then ds S = I = B. constant

42 Radiative transfer: general solution di dτ = I S,,0 τ,0 τ I ( 0) I e S e d τ,,0 0 τ = + i.e. the final intensity is the initial intensity, reduced by absorption, plus the emission at every point along the path, also reduced by absorption.

43 Example: homogeneous medium,0,0 I (0) I e τ τ = + S e d,0 τ 0 τ Imagine a beam of light with I,0 at s = 0 entering a volume of gas of constant density, opacity, and source function. τ,0 τ ( ) I (0) = I e + S 1 e,0,0 In the limit of high optical dep th ) (0),0 ( e = 0 : I = S, τ,0,0,0,0,0 τ 1 ( e = 1 τ ) I ( I + S τ τ : 0).

44 Approximate solutions Approximation #1: Plane-parallel atmospheres. We can define a vertical optical depth such that: di I S dτ =, v 0 τ ( z) = κ ρdz where dz = ds cosθ z Light ray i. e. τ ( z) = τ cosθ, v and the transfer equation becomes di cosθ d τ, v = I S.

45 Approximate solutions Approximation #2: Gray atmospheres. Integrating the intensity and source function over all wavelengths, we get the following simplified transfer equation: di I S dτ = di cosθ = d τ Integrating over all solid angles, v I S d dτ v I cosθ dω = IdΩ S dω df dτ rad v ( I S ) = 4π where F is the radiative flux through unit area. rad

46 The photon wind di cosθ d τ v = I S In a spherically symmetric star with the origin at the centre: dp dr rad κρ = F c rad So the net radiative flux (i.e. movement of photons through the star) is by differences in the radiation pressure. driven

47 Approximate solutions Approximation #3: An atmosphere in radiative equilibrium. Since: F rad = σt 4 e σ 4 Prad = Te τ v + C c

48 The Eddington approximation To determine the temperature structure of the atmosphere, we need to establish the temperature dependence of the radiation pressure to solve: 1 c 2 Since: Prad = I cos θdω, σ 4 Prad = Te τ v + C c we need to assume something about the angular distribution of the intensity.

49 2π π φ = 0 θ = 0 2π π The Eddington approximation 1 1 I = I sinθdθdφ Iout I 4π = + 2 ( ) ( ) F = I cosθ sinθdθdφ = π I I rad out in φ = 0 θ = 0 2π π 1 2 4π rad = cos sin c θ θ θ φ = 3c φ = 0 θ = 0 P I d d I σ 4 Prad = Te τ v + C c in 4π σ 4 I = Te τ v + C 3c c 3σ 4 2 I = Te τ v + 4π 3 4π 2 Fsurface = I τ v + = π S( τ v = 2 / 3) 3 3 This is the Eddington-Barbier relation: the surface flux is determined by the value of the soure function at a vertical optical depth of 2/ 3. 1

50 The Eddington approximation dfrad dprad 1 = 0 = 4 π ( I S ) = F dτ dτ c v v rad I 3σ 4 2 = Te τ v + 4π 3 Approximation #5: Local thermodynamic equilibrium. In this case the source function is equal to the black body function. Recall: df dτ rad v 4 σt = 0 = 4 π ( I S ) I = B = π so I = S T = T τ + 4 e v 2 3

51 The Eddington approximation Note that T = T e when t v = 2/3. Thus the photons that we see (that give us the luminosity we use to define the effective temperature) originate at an optical depth of 2/3, not T = T e τ v T e

52 Limb darkening The solar disk is darker at the edge (limb) than at the centre. The light rays that we see from the edge of the Sun must originate from higher in the atmosphere; otherwise they would have to travel through a greater optical depth to reach us.

53 Limb darkening,0 τ,0 τ I (0) I e S e d,0 0 τ = + τ Now we will again assume a plane-parallel atmosphere, but we will not assume a gray atmosphere, LTE, or the Eddington approximat ion. We need to find the intensity as a function of angle, θ. Recall the definition of the vertical optical depth: τ, v( z) = τ cosθ sec θτ ( z) = τ, v Light ray So we can rewrite the general solution as: τ secθ v,,0 τ v,,0 secθ τ v, secθ I I e S secθe d = +,0 v, 0 τ

54 Limb darkening σ 4 I = T e + 4π I ( θ ) 2 + 3cosθ = I ( θ = 0) 5 ( 2 3cosθ ) We can compare this with observations, and the agreement is pretty good. This does not prove that our numerous assumptions are correct, but does show that they do not produce a result that is in strong conflict with the data.

55 The Sun It is a self-luminous ball of gas held together by its own gravity and powered by thermonuclear fusion in its core. Our Sun is a typical star among the various stars in the Galaxy, average in mass, size and temperature. It is a dwarf star with a radius of 10 9 Earth radii and a mass of Earth masses. The Sun's lifetime is about 10 billion years, meaning that after this time the hydrogen in its core will be depleted. The Sun will then evolve into a red giant, consuming Mercury, Venus and the Earth in its expanded envelope. The Sun is currently 5 billion years old. The most outstanding characteristic of the Sun is the fact that it emits huge quantities of electro-magnetic radiation of all wavelengths.

56 The Sun The total output of the Sun, called the solar constant, is ergs/sec. Only ergs/sec strikes the Earth (since this latter is small in angular size), but the amount of energy reaching the Earth in 30 minutes is more than the power generated by all of the human civilization. This energy is what powers the atmosphere and our oceans (storms, wind, currents, rainfall, etc.). The energy emitted by the Sun is divided into 40% visible light, 50% IR, 9% UV and 1% x-ray, radio, etc. The light we see is emitted from the surface of the Sun, the photosphere. The Sun below the photosphere is opaque and hidden.

57 The Sun is divided into six regions based on the physical characteristics of these regions. The boundaries are not sharp. The Solar structure fusion core - region of energy generation radiation shell - the region where energy transport is by radiation flow convection shell - the region where energy transport is by convection cells photosphere - the surface where photons are emitted chromosphere - the atmosphere of the Sun corona - the superhot region where the solar wind originates

58 The Solar structure The radii and temperatures of these regions are the following: region radius temperature fusion core 0.3 solar radii K radiation shell solar radii K radiation shell solar radii K photosphere 100 km 6000 K chromosphere 2000 km 30,000 K corona 10 6 km K The Sun rotates differentially since it is not a solid. The solar equator completes one rotation in days. The angular velocity depends on latitude φ as: ω = sinφ The poles complete one rotation in 36 days.