Spectroscopy – Lecture 2 Atomic excitation and ionization Radiation Terms Absorption and emission coefficients Einstein coefficients Black Body radiation

· · I. Atomic excitation and ionization ∞ qualitative energy level diagram · n c 3 2 Mechanisms for populating and depopulating the levels in stellar atmospheres: radiative collisional spontaneous transitions 1 E=–I

I. Atomic excitation and ionization The fraction of atoms (or ions) excited to the nth level is: Nn = constant gn exp(–cn/kT) statistical weight Boltzmann factor Statistical weight is 2J+1 where J is the inner quantum number (Moore 1945)1. For hydrogen gn=2n2 1 Moore, C.E. 1945, A Multiplet Table of Astrophysical Interest, National Bureau of Standards

( ) I. Atomic excitation and ionization Ratio of populations in two levels m and n : Nn Nm = gn gm ( kT ) – exp Dc Dc = cn – cm

( ) ( ) ( ) ( ) ( ) ... I. Atomic excitation and ionization The number of atoms in level n as fraction of all atoms of the same species: gn ( kT ) – exp cn Nn N = + g2 ( kT ) – exp c2 + g3 ( kT ) – exp c3 ... + g1 = gn u(T) ( kT ) – exp cn u(T) = S ( kT ) – exp ci gi Partition Function Nn N = gn u(T) 10 –qcn q= log e/kT = 5040/T

I. Atomic excitation and ionization If we are comparing the population of the rth level with the ground level: gr Nr N1 –5040 log c + log = g1 T

I. Atomic excitation and ionization Example: Compare relative populations between ground state and n=2 for Hydrogen g1 = 2, g2=2n2 =8 Temp. (K) q=5040/T N2/N1 6000 0.840 0.00000001 8000 0.630 0.0000016 10000 0.504 0.00031 15000 0.336 0.00155 20000 0.252 0.01100 40000 0.126 0.209

( ) ) ( I. Atomic excitation and ionization : Saha Eq. For collisionally dominated gas: h3 ( 2pm ) 2 3 kT 5 N1 N Pe 2u1(T) u0(T) ( kT ) – exp I = N1 N = Ratio of ions to neutrals u1 u0 = Ratio of ionic to neutral partition function m = mass of electron, h = Planck´s constant, Pe = electron pressure

I. Saha Equation Numerically: N1 N T Pe u1 u0 log –5040 I + 2.5 log T = T Pe u1 u0 log –5040 I + 2.5 log T – 0.1762 or N1 N = F(T) Pe F(T) = 0.65 u1 u0 T 2 5 10–5040I/kT

I. Saha Equation Example: What fraction of calcium atoms are singly ionized in Sirius? T = 10000 K Pe = 300 dynes cm–2 Stellar Parameters: Ca I = 6.11 ev log 2u1/uo = 0.18 Atomic Parameters: log N1/N0 = 4.14 no neutral Ca

I. Saha Equation Maybe it is doubly ionized: Second ionization potential for Ca = 11.87 ev u1 = 1.0 log 2u2/u1 = –0.25 N2/N1 = 6.6 N1/(N1+N2) = 0.13 log N2/N1 = 0.82 13% Ca II 87% Ca III

T 25000 10000 6300 4200

II. Radiation Terms: Specific intensity Consider a radiating surface: Normal q Dw Observer DA In = DEn cos q DA Dw Dt Dn lim In = dEn cos q dA dw dt dn

Can also use wavelength interval: II. Specific intensity Can also use wavelength interval: Indn = Ildl Note: the two spectral distributions (n,l) have different shape for the same spectrum For solar spectrum: Il = max at 4500 Ang In = max at 8000 Ang c=ln dn = –(c/l2) dl Equal intervals in l correspond to different intervals in n. With increasing l, a constant dl corresponds to a smaller and smaller dn

∫ II. Radiation Terms: Mean intensity I. The mean intensity is the directional average of the specific intensity: 1 4p ∫ Jn = In dw Circle indicates the integration is done whole unit sphere on the point of interest

II. Radiation Terms: Flux Flux is a measure of the net energy across an area DA, in time Dt and in spectral range Dn Flux has directional information: +Fn DA -Fn

∫ ∫ S DEn ∫ II. Radiation Terms: Flux Fn = lim DA Dt Dn dEn = dA dt dn In cos q dw In = dEn cos q dA dw dt dn Recall: ∫ In cos q dw Fn =

∫ ∫ ∫ ∫ II. Radiation Terms: Flux Looking at a point on the boundary of a radiating sphere ∫ 2p ∫ p Fn = df In cos q sin q dq = df ∫ 2p p/2 In cos q sin q dq Outgoing flux + df ∫ 2p p/2 p In cos q sin q dq = 0 Incoming flux For stars flux is positive

Astronomical Example of Negative Flux: Close Binary system: II. Radiation Astronomical Example of Negative Flux: Close Binary system: Hot Spot Cool star (K0IV) Hot star (DAQ3) Negative and postive flux through this area

∫ (∫ sin q cos q dq = 1/2 ) II. Radiation Terms: Flux If there is no azimuthal (f) dependence p/2 ∫ Fn = 2p In sin q cos q dq Simple case: if In is independent of direction: Fn = p In (∫ sin q cos q dq = 1/2 )

Energy received ~ InDA1/r2 Detector element Source F DA1 area covered by solid angle q2 Detector subtends angle q on the sky r Source image

Energy received ~ InDA2/4r2 but DA2 =4DA1 = In DA1/r2 F 2r DA2

Energy received ~ InDA3/100r2 but DA3 = area of source Source image F DA3 Detector element 10r Since the image source size is smaller than our detector element, we are now measuring the flux

∫ ∫ ∫ ∫ II. K-integral and radiation pressure Kn = In cos2q dw 1 4p = 2p p sin q dq df ∫ –1 +1 2p dm m = cos q Kn = 1 2 m2 dm ∫ –1 +1

II. K-integral and radiation pressure This intergral is related to the radiation pressure. Radiation has momentum = energy/c. Consider photons hitting a solid wall Pressure= 2 c d En cosq dt dA q component of momentum normal to wall per unit area per time = pressure

∫ ∫ II. K-integral and radiation pressure Pn dn dw = cos2q dn dw 2In c In cos2q dn dw/c +1 Pn = 4p ∫ In (m) m2 dm = 2p -1 c 4p c Kn Pn =

∫ ∫ II. K-integral and radiation pressure Special Case: In is indepedent of direction Pn = 2p ∫ -1 +1 In (m) m2 dm c 3c Pn = 4p In Total radiation pressure: Pn = ∫  In dn 3c 4p 4s T4 For Blackbody radiation

∫ ∫ ∫ ∫ II. Moments of radiation Jn = 1 4p In dw Jn = 1 In (m) dm 2 –1 +1 In (m) dm 2 Mean intensity Hn = 1 2 I(m) m dm ∫ –1 +1 Flux = 4pH Kn = 1 2 I(m) m2 dm ∫ –1 +1 Radiation pressure

III. The absorption coefficient In + dIn In dx kn kn is the absorption coefficient/unit mass [ ] = cm2/gm. kn comes from true absorption (photon destroyed) or from scattering (removed from solid angle) dIn = –knr In dx

∫ III. Optical depth In + dIn In kn The radiation sees neither knr or dx, but a the combination of the two over some path length L. ∫ o L tn = knr dx Optical depth gm cm3 Units: cm2 gm cm

III. Optical depth Optically thick case: t >> 1 => a photon does not travel far before it gets absorbed Optically thin case: t << 1 => a photon can travel a long distance before it gets absorbed

III. Simple solution to radiative transfer equation In + dIn In dx kn dIn = – In dt In = Ino e–t Optically thin e–t = 1-t In = Ino(1-t)

III. The emission coefficient In + dIn In dx jn dIn = jnr In dx jn is the emission coefficient/unit mass [ ] = erg/(s rad2 Hz gm) jn comes from real emission (photon created) or from scattering of photons into the direction considered.

III. The Source Function The ratio of the emission to absorption coefficients have units of In. This is commonly referred to as the source function: Sn = jn/kn The physics of calculating the source function Sn can be complicated. Let´s consider the simple cases of scattering and absorption

• III. The Source Function: Pure isotropic scattering dw djn to observer • isotropic scattering The scattered radiation to the observer is the sum of all contributions from all increments of the solid angle like dw. Radiation is scattered in all directions, but only a fraction of the photons reach the observer

∫ ∫ III. The Source Function: Pure isotropic scattering The contribution to the emission from the solid angle dw is proportional to dw and the absorbed energy knIn. This is isotropically re-radiated: djn = knIn dw/4p ∫ jn = knIn dw/4p Sn = jn kn = ∫ In dw/4p = Jn The source function is the mean intensity

III. The Source Function: Pure absorption All photons are destroyed and new ones created with a distribution governed by the physical state of the material. Emission of a gas in thermodynamic equilibrium is governed by a black body radiator: 2hn3 1 Sn = c2 exp(hn/kT) – 1

III. The Source Function: Scattering + Pure absorption jn = knSIn +knABn(T) Sn = jn/kn where kn = knS+ knA Sn = knS knS+ knB Jn knA Bn (T) + Sum of two source functions weighted according to the relative strength of the absorption and scattering

IV. Einstein Coefficients When dealing with spectral lines the probabilities for spontaneous emission can be described in terms of atomic constants Consider the spontaneous transition between an upper level u and lower level l, separated by energy hn. The probability that an atom will emit its quantum energy in a time dt, solid angle dw is Aul. Aul is the Einstein probability coefficient for spontaneous emission.

IV. Einstein Coefficients If there are Nu excited atoms per unit volume the contribution to the spontaneous emission is: jnr = Nu Aul hn If a radiation field is present that has photons corresponding to the energy difference between levels l and u, then additional emission is induced. Each new photon shows phase coherence and a direction of propagation that is the same as the inducing photon. This process of stimulated emission is often called negative absorption.

IV. Einstein Coefficients The probability for stimulated emission producing a quantum in a time dt, solid angle dw is Bul In dt dw. Bul is the Einstein probability coefficient for stimulated emission. True absorption is defined in the same way and the proportionality constant denoted Blu. knrIn = NlBluInhn – NuBulInhn The amount of reduction in absorption due to the second term is only a few percent in the visible spectrum.

IV. Einstein Coefficients Nu hn BluIn True absorption, dependent on In Aul Spontaneous emission, independent of In Negative absorption, dependent on In +BulIn Nl Principle of detailed balance: Nu[Aul + BulIn] = NlBluIn

V. Black body radiation Detector Light enters a box that is a perfect absorber. If the container is heated walls will emit photons that are reabsorbed (thermodynamic equilibrium). A small fraction of the photons will escape through the hole.

V. Black body radiation: observed quantities Il = F(c/lT) l5 F is a function that is tabulated by measurements In = n3 F(n/T) I n= 2kTn2 c2 2pckT l4 Rayleigh-Jeans approximation for low frequencies I l=

V. Black body radiation: The Classical (Wrong) approach Lord Rayleigh and Jeans suggested that one could calculate the number of degrees of freedom of electromagnetic waves in a box at temperature T assuming each degree of freedom had a kinetic energy kT and potential energy Radiation energy density = number of degrees of freedom×energy per degree of freedom per unit volume. 2kTn2 c2 I n = but as n ∞, In  This is the „ultraviolet“ catastrophe of classical physics

( ) V. Black body radiation: Planck´s Radiation Law Derive using a two level atom: Nn Nm = gn gm ( kT ) – exp hn Number of spontaneous emissions: NuAul Rate of stimulated emission: NuBulIn Absorption: NlBluIn

NuAul + NuBulIn = NlBluIn V. Black body radiation: Planck´s Radiation Law In radiative equilibrium collisionally induced transitions cancel (as many up as down) NuAul + NuBulIn = NlBluIn In = Aul Blu(Nl/Nu) – Bul In = Aul (gl/gu)Bluexp(hn/kT) – Bul

V. Black body radiation: Planck´s Radiation Law This must revert to Raleigh-Jeans relation for small n Expand the exponential for small values (ex = 1+x) In ≈ Aul (gl/gu)Blu – Bul + (gl/gu)Bluhn/kT hn/kT << 1 this can only equal 2kTn2/c2 if Bul = Blugl/gu Aul = 2hn3 c2 Bul Note: if you know one Einstein coeffiecient you know them all

V. Black body radiation: Planck´s Radiation Law In = 1 (exp(hn/kT) – 1) 2hn3 c2 Il = 1 (exp(hc/lkT) – 1) 2hc2 l5

V. Black body radiation: Planck´s Radiation Law Maximum Il = lT = 0.5099 cm K Maximum In = 5.8789×1010 Hz K I n = 2kTn 2 c2 Rayleigh-Jeans approximation n→ 0 2pckT l4 I l= I n = 2phn 3 c2 e–hn/kT I l = 2phc2 l5 e–hc/klT Wien approximation n→∞

∫ ∫ ∫ ( ) V. Black body radiation: Stefan-Boltzmann Law In our black body chamber escaping radiation is isotropic and no significant radiation is entering the hole, therefore Fn = pIn Fn dn = p ∫ ∞ 1 (exp(hn/kT) – 1) 2hn3 c2 dn 4 x3 = 2p c2 ( kT h ) ∫ ∞ ex–1 dx x=hn/kT ∞ 2p5k4 Fn ∫ dn = 15h3c2 T4 = sT4 Integral = p4/15

V. Note on Einstein Coefficients and BB radiation In the spectral region where hn/kT >> 1 spontaneous emissions are more important than induced emissions In the ultraviolet region of the spectrum replace In by Wien´s law: BulIn = Bul 2hn3 c2 e–hn/kT = Aul e–hn/kT << Aul Induced emissions can be neglected in comparison to spontaneous emissions

V. Note on Einstein Coefficients and BB radiation In the spectral region where hn/kT << 1 negative absorption (induced emissions) is more important than spontaneous emissions In the far infrared region of the spectrum replace In by Rayleigh-Jeans law: BulIn = Bul 2nkT c2 = Aul c2 2hn3 2nkT = Aul kT hn >>Aul The number of negative absorptions is greater than the spontaneous emissions

log I ~ –4 log l log I ~ –5 log l – 1/l I U B J K T (K) 40000 20000 10000 5000 log I ~ –5 log l – 1/l 3000 1500 1000 750 500

T = 6000 K Il In

V. Black body radiation: Photon Distribution Law Nn = 1 (exp(hn/kT) – 1) 2hn2 c2 Nl = 1 (exp(hc/lkT) – 1) 2hc2 l4 Detectors detect N, not I !

Il Nl