Spectroscopy Lecture 2 I. Atomic excitation and ionization II. Radiation Terms III. Absorption and emission coefficients IV. Einstein coefficients V. Black Body radiation
I. Atomic excitation and ionization E>0 I n 3 E=0 qualitative energy level diagram χ 2 1 E= I Mechanisms for populating and depopulating the levels in stellar atmospheres: radiative collisional spontaneous transitions
H, He relatively hard to ionize hot stars you see absorption lines of Hydrogen Metals relatively easy for first ionization
I. Atomic excitation and ionization The fraction of atoms (or ions) excited to the nth level is: N n = constant g n exp( χ n /kt) statistical weight Boltzmann factor Statistical weight is 2J+1 where J is the inner quantum number (Moore 1945) 1. For hydrogen g n =2n 2 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 : N n N m = g n Δχ exp ( ) g m kt Δχ = χ n χ m
I. Atomic excitation and ionization The number of atoms in level n as fraction of all atoms of the same species: N n N = g 1 χ n g ( n exp ) kt χ 2 + g ( 2 exp ) kt χ 3 + g ( 3 exp ) kt +... = g n χ n exp ( ) u(t) kt χ i u(t) = Σ g i exp ( ) kt Partition Function N n N = g n u(t) 10 θχ n θ= log e/kt = 5040/T
From Allen s Astrophysical Quantities Θ = 5040/T Y = stage of ionization. Y = 1 is neutral, Y = 2 is first ion.
I. Atomic excitation and ionization If we are comparing the population of the rth level with the ground level: log N r N 1 = 5040 χ + log T g r g 1
I. Atomic excitation and ionization Example: Compare relative populations between ground state and n=2 for Hydrogen g 1 = 2, g 2 =2n 2 =8 Temp. (K) θ=5040/t N 2 /N 1 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 N 2 /N 1 10000 20000 40000 60000
I. Atomic excitation and ionization : Saha Eq. For collisionally dominated gas: N 1 N P e = 3( 5 ( 2πm ) 2 kt ) 2 h 3 2u 1 (T) exp ( ) u 0 (T) kt I N 1 N u 1 u 0 = Ratio of ions to neutrals = Ratio of ionic to neutral partition function m = mass of electron, h = Planck s constant, P e = electron pressure
I. Saha Equation Numerically: log N 1 N P e = 5040 I + 2.5 log T + log 0.1762 T u 1 u 0 or N 1 N = Φ(T) P e Φ(T) = 0.65 u 1 u 0 T 2 5 10 5040I/kT
I. Saha Equation Example: What fraction of calcium atoms are singly ionized in Sirius? Stellar Parameters: T = 10000 K P e = 300 dynes cm 2 Atomic Parameters: Ca I = 6.11 ev log 2u 1 /u o = 0.18 log N 1 /N 0 = 4.14 no neutral Ca
I. Saha Equation Maybe it is doubly ionized: Second ionization potential for Ca = 11.87 ev u 1 = 1.0 log 2u 2 /u 1 = 0.25 log N 2 /N 1 = 0.82 N 2 /N 1 = 6.6 N 1 /(N 1 +N 2 ) = 0.13 In Sirius 13% of the Ca is singly ionized and the remainder is doubly ionized because of the low ionization potential of Ca.
From Lawrence Allen s The Atmospheres of the Sun and Stars T 25000 10000 6300 4200 The number of hydrogen atoms in the second level capable of producing Balmer lines reaches its maximum at Teff 10000 K
Behavior of the Balmer lines (H β ) Ionization theory thus explains the behavior of the Balmer lines along the spectral sequence.
How can a T=40000 star ionize Hydrogen? I = 13.6 ev = 2.2 x 10 11 ergs E = kt T = 160.000 K So a star has to have an effective temperature to of 160.000 K to ionize hydrogen? Answer later.
Predicted behavior according to Ionization Theory Ionization theory s achievement was the intepretation of the spectral sequence as a temperature sequence Observed behavior according to Ionization Theory
II. Radiation Terms: Specific intensity Consider a radiating surface: Normal θ Observer ΔA I ν = lim ΔE ν cos θ ΔA Δω Δt Δν Δω I ν = de ν cos θ da dω dt dν
II. Specific intensity Can also use wavelength interval: I ν dν = I λ dλ Note: the two spectral distributions (ν,λ) have different shape for the same spectrum For solar spectrum: I λ = max at 4500 Ang I ν = max at 8000 Ang c=λν dν = (c/λ 2 ) dλ Equal intervals in λ correspond to different intervals in ν. With increasing λ, a constant dλ corresponds to a smaller and smaller dν
II. Radiation Terms: Mean intensity I. The mean intensity is the directional average of the specific intensity: J ν = 1 4π I ν dω Circle indicates the integration is done over whole unit sphere on the point of interest
II. Radiation Terms: Flux Flux is a measure of the net energy across an area ΔA, in time Δt and in spectral range Δν Flux has directional information: +F ν -F ν ΔA
II. Radiation Terms: Flux F ν = lim Σ ΔE ν ΔA Δt Δν = de ν da dt dν = I ν cos θ dω Recall: I ν = de ν cos θ da dω dt dν F ν = I ν cos θ dω
II. Radiation Terms: Flux Looking at a point on the boundary of a radiating sphere 0 2π F ν = dφ 0 π I ν cos θ sin θ dθ 2π = dφ 0 2π + dφ 0 0 π/2 π π/2 I ν cos θ sin θ dθ = 0 I ν cos θ sin θ dθ Outgoing flux Incoming flux For stars flux is positive
II. Radiation Astronomical Example of Negative Flux: Close Binary system: Cool star (K0IV) Hot Spot Hot star (DAQ3)
II. Radiation Terms: Flux If there is no azimuthal (φ) dependence F ν = 2π π/2 I ν sin θ cos θ dθ 0 Simple case: if I ν is independent of direction: F ν = π Iν ( sin θ cos θ dθ = 1/2 ) Note: I ν is independent of distance, but F ν obeys the standard inverse square law
Flux radiating through a sphere of radius d is just F = L/4πd 2 L = 4πR 2 I ν (=σt 4 ) d
Energy received ~ I ν ΔA 1 /r 2 Source Detector element F r ΔA 1 Source image
Energy received ~ I ν ΔA 2 /4r 2 but ΔA 2 =4ΔA 1 = I ν ΔA 1 /r 2 F 2r ΔA 2
Energy received ~ I ν ΔA 3 /100r 2 but ΔA 3 = area of source Source image F ΔA 3 10r Detector element Since the image source size is smaller than our detector element, we are now measuring the flux The Sun is the only star for which we measure the specific intensity
II. K-integral and radiation pressure K ν = 1 4π I ν cos2 θ dω 2π π dω = 0 0 +1 sin θ dθ dφ = 2π 1 dµ µ = cos θ K ν = 1 2 +1 1 µ 2 dµ
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 E ν cosθ dt da θ component of momentum normal to wall per unit area per time = pressure
II. K-integral and radiation pressure P ν dν dω = 2I ν c cos 2 θ dν dω P ν = P ν dν = 4π c 0 +1 I ν cos 2 θ dν dω/c I ν (µ) µ 2 dµ = 2π I ν (µ) µ 2 dµ c -1 +1 P ν = 4π c K ν
II. K-integral and radiation pressure Special Case: I ν is indepedent of direction +1 P ν = 2π I ν (µ) µ 2 dµ c -1 P ν = 4π I ν 3c Total radiation pressure: P ν = 4π 3c 0 I ν dν = 4σ 3c T 4 For Blackbody radiation
Radiation pressure is a significant contribution to the total pressure only in very hot stars.
II. Moments of radiation J ν = 1 4π I ν dω +1 J ν = 1 I ν (µ) dµ 2 1 H ν = 1 2 +1 1 I(µ) µ dµ Mean intensity Flux = 4πH K ν = 1 2 +1 1 I(µ) µ 2 dµ Radiation pressure µ = cos θ
III. The absorption coefficient I ν I ν + di ν dx κ ν κ ν is the absorption coefficient/unit mass [ ] = cm 2 /gm. κ ν comes from true absorption (photon destroyed) or from scattering (removed from solid angle) di ν = κ ν ρ I ν dx
III. Optical depth L I ν I ν + di ν κ ν The radiation sees neither κ ν ρ or dx, but a the combination of the two over some path length L. τ ν = o L κ ν ρ dx Optical depth Units: cm 2 gm gm cm 3 cm
III. Optical depth Optically thick case: τ >> 1 => a photon does not travel far before it gets absorbed Optically thin case: τ << 1 => a photon can travel a long distance before it gets absorbed
Optically thin τ < 1 τ 1 Optically thick τ > 1 Luca Sebben
III. Simple solution to radiative transfer equation I ν I ν + diν di ν = κ ν ρ I ν dx dx κ ν di ν = I ν dτ I ν = I νο e τ Optically thin e τ = 1-τ I ν = I νo (1-τ)
III. The emission coefficient I ν I ν + diν dx j ν di ν = j ν ρ I ν dx j ν is the emission coefficient/unit mass [ ] = erg/(s rad 2 Hz gm) j ν 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 I ν. This is commonly referred to as the source function: S ν = j ν /κ ν The physics of calculating the source function S ν can be complicated. Let s consider the simple cases of scattering and absorption
III. The Source Function: Pure isotropic scattering dω dj ν to observer isotropic scattering The scattered radiation to the observer is the sum of all contributions from all increments of the solid angle like dω. 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 dω is proportional to dω and the absorbed energy κ ν I ν. This is isotropically re-radiated: dj ν = κ ν I ν dω/4π j ν = κ ν I ν dω/4π S ν = j ν κ ν = I ν dω/4π = J ν 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: S ν = 2hν 3 c 2 1 exp(hν/kt) 1
III. The Source Function: Scattering + Pure absorption j ν = κ νs I ν +κ νa B ν (T) S ν = j ν /κ ν where κ ν = κ νs + κ ν A S ν = κ ν S κ νs + κ ν A J ν + κ ν A κ νs + κ ν A B ν (Τ) 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 hν. The probability that an atom will emit its quantum energy in a time dt, solid angle dω is A ul. A ul is the Einstein probability coefficient for spontaneous emission.
IV. Einstein Coefficients If there are N u excited atoms per unit volume the contribution to the spontaneous emission is: j ν ρ = N u A ul hν 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 dω is B ul I ν dt dω. B ul is the Einstein probability coefficient for stimulated emission. True absorption is defined in the same way and the proportionality constant denoted B lu. κ ν ρi ν = N l B lu I ν hν N u B ul I ν hν The amount of reduction in absorption due to the second term is only a few percent in the visible spectrum.
IV. Einstein Coefficients N u True absorption, dependent on I ν N l B lu I ν A ul +B ul I ν hν Spontaneous emission, independent of I ν Negative absorption, dependent on I ν Principle of detailed balance: N u [A lu + B ul I ν ] = N l B ul I ν
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 I λ = c4 λ 5 F(c/λT) I ν = ν 3 F(ν/T) F is a function that is tabulated by measurements. This scaling relation was discovered by Wien in 1893 I ν = I λ = 2kTν 2 c 2 2πckT λ 4 Rayleigh-Jeans approximation for low frequencies
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. I ν = 2kTν 2 c 2 but as ν, I ν This is the ultraviolet catastrophe of classical physics
V. Black body radiation: Planck s Radiation Law Derive using a two level atom: N n N m = g n hν exp ( ) g m kt Number of spontaneous emissions: N u A ul Rate of stimulated emission: N u B ul I ν Absorption: N l B ul I ν
V. Black body radiation: Planck s Radiation Law In radiative equilibrium collisionally induced transitions cancel (as many up as down) N u A ul + N u B ul I ν = N l B lu I ν I ν = A ul B lu (N l /N u ) B ul I ν = A ul (g l /g u )B lu exp(hν/kt) B ul
V. Black body radiation: Planck s Radiation Law This must revert to Raleigh-Jeans relation for small ν Expand the exponential for small values (e x = 1+x) I ν A ul (g l /g u )B lu B ul + (g l /g u )B lu hν/kt hν/kt << 1 this can only equal 2kTν 2 /c 2 if B ul = B lu g l /g u A ul = 2hν 3 c 2 B ul Note: if you know one Einstein coeffiecient you know them all
V. Black body radiation: Planck s Radiation Law I ν = 2hν 3 c 2 1 (exp(hν/kt) 1) I λ = 2hc 2 λ 5 1 (exp(hc/λkt) 1)
V. Black body radiation: Planck s Radiation Law Maximum I λ = λt = 0.5099 cm K Maximum I ν = 5.8789 10 10 Hz K I ν = 2kTν 2 c 2 I λ = 2πckT λ 4 Rayleigh-Jeans approximation ν 0 I ν = I λ = 2πhν 3 c 2 2πhc 2 λ 5 e hν/kt e hc/kλt Wien approximation ν
V. Black body radiation: Stefan Boltzman Law In our black body chamber escaping radiation is isotropic and no significant radiation is entering the hole, therefore F ν = πi ν F ν dν = π 0 0 2hν 3 c 2 = 2π c 2 ( kt h ) 0 1 (exp(hν/kt) 1) 4 x 3 e x 1 dx dν x=hν/kt Integral = π 4 /15 2π 5 k 4 F ν dν = T 4 = σt 4 15h3 c 2 0
V. Note on Einstein Coefficients and BB radiation In the spectral region where hν/kt >> 1 spontaneous emissions are more important than induced emissions In the ultraviolet region of the spectrum replace I ν by Wien s law: B ul I ν = B ul 2hν 3 c 2 e hν/kt = A ul e hν/kt << A ul Induced emissions can be neglected in comparison to spontaneous emissions
V. Note on Einstein Coefficients and BB radiation In the spectral region where hν/kt << 1 negative absorption (induced emissions) are more important than spontaneous emissions In the far infrared region of the spectrum replace I ν by Rayleigh-Jeans law: B ul I ν = B ul 2νkT = c 2 c 2 A ul 2hν 3 2νkT c 2 = A ul kt hν >>A ul The number of negative absorptions is greater than the spontaneous emissions
log I ~ 4 log λ U B I J K T (K) 40000 20000 10000 5000 3000 1500 1000 750 500 log I ~ 5 log λ 1/λ
How can a T=40000 star ionize Hydrogen? 1. Blackbody has a distribution of energies and some photons have the energy to ionize hydrogen 2. The thermal velocities have a Maxwell Boltzmann distribution and some particles have the thermal energy to ionize Hydrogen. Fraction of total particles T=1000 K T=6000 K T=40000 K
I λ T = 6000 K I ν
V. Black body radiation: Photon Distribution Law N ν = 2hν 2 c 2 1 (exp(hν/kt) 1) N λ = 2hc 2 λ 4 1 (exp(hc/λkt) 1) Detectors detect N, not I!
Temperature of the Sun is almost a black body Steven Spangler, Univ. of Iowa
But the corona has a much higher temperature The 1905 book The Sun by Abbott commented on the unidentified green and red lines in eclipse spectra Red and green lines are FeX and FeXIV, indicating temperatures of 1-2 million K
kt 262 ev T 3 x 106 K