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Topic 2

Evidence of abundances from spectral lines

Chemical Composition of

! The chemical composition of stars can be investigated from looking at ! The continuous thermal spectrum is often punctuated with absorption lines ! The study of such absorption lines yields a lot of information on the chemical composition of the star’s photosphere

! The photosphere is defined as having an optical depth Tλ given by

' ∞ * 0.01≤ ) Tλ (z) = ∫ kλ (z)ρ(z)dz, ≤ "few" ( z +

where Tλ is the number of mean free paths of radiation experienced before the light escapes from the star, z is the geometric height above some arbitrary layer in the star, €k is the absorption co-efficient or opacity and ρ is the mass Black Body Radiation Cool stars (top) peak emission is red Hot stars (bot.) peak emission is blue

2hν 3 1 I(ν) = c 2 ehν / kT −1

! In€ general absorption and emission lines are superimposed on a black body spectrum

Spectral Lines: Origin and Appearance

Continuous Spectrum

Emission Spectrum

Absorption Spectrum Absorption Spectra

! The spectra that are observed depend very much on the and pressure in the stellar atmosphere ! The hotter the temperature the greater the mean thermal energy • Break up of

• Ionization of • Those atoms with the highest HOTTER binding energy are only seen in their ionized state in the We will discuss Binding hottest stars Energy in depth later

Will an Absorption Line be seen?

! There are a number of factors to consider here • For example, the atmosphere of the hottest stars is opaque • (Highly) ionized atoms give absorption spectra in the UV which is absorbed by the Earth’s atmosphere • Principal factors in an absorption line’s intensity are: • The extent to which the element is in the right state of ionization or excitation to produce lines which can be observed – i.e. appropriate • The amount of the particular element that is present • The strength of the transition probability or cross-section for absorption Interpreting Spectral Lines

Redshift of spectral lines in the ! Spectral lines can optical spectrum from a give us information on supercluster of distant galaxies (right) what elements/ compared to that of the (left) are present in a star’s Reminder: the key equations photosphere as well for the are: fv fv as the abundance of Δf = f ' = f ± c c that particular element for radiation travelling or . at the speed of light .

! This involves a number of€ steps € 1. Possibly correcting the observed spectral line for Doppler effect due to the star’s relative velocity 2. Calculating the of the observed spectral line 3. Interpreting the resulting line width due to broadening mechanisms

Equivalent Width

! The first step is to calculate the Equivalent Width

! The equivalent width (Wλ) of a spectral line is defined as that which, at 100% absorption, occupies the same area as the observed line ! The relationship between the EW and the effective number of absorbing atoms is known as the curve of growth (later)

$ ' FC − Fλ Wλ = ∫ & ) dλ % FC (

€ Main Sources of Line Broadening

! There are many mechanisms that can to spectral line broadening, they are classified in different ways, for example: • The effect on the line width (profile) • Doppler or Gaussian • Damping or Lorentzian • Microscopic or macroscopic in origin, also known as intrinsic or extrinsic or local and non-local

Local effects Non-Local effects

• Opacity Broadening • Rotational Broadening

Doppler Natural Pressure Due to thermal motion Emitted Radiation is Due to Heisenberg of atoms along the affected by nearby line of sight particles Lorentzian profile Gaussian profile Lorentzian profile

Doppler Broadening

! Due to random thermal ! Mathematically, from MB motion of atoms in a gas the most probable speed is v 2kT /m ! In thermal equilibrium the mp = gas atoms are moving with ! Using the (non-rel.) Doppler speeds described by the shift formula: Maxwell-Boltzmann (MB) Δλ /λ = ± v /c € distribution ! This gives the spectral line ! The of the width as 2λ 2kT emitted or absorbed Δλ = € c m are thus Doppler shifted ! Taking into account different atoms’ directions For atoms in the this€ is refined to Sun’s photosphere (T=5770K) 2λ 2kT(ln2) 0.427Å for H (Δλ)1/ 2 = Δλ ≈ α c m

€ Natural Broadening

! Natural Broadening is due to the Heisenberg Uncertainty Principle - a direct result of ! When an is moved to an it occupies this state for a short period of time Δt and so the energy of this state, E cannot have a precise value ! The uncertainty associated with E is called ΔE and is given by Heisenberg’s famous formula: ! ΔE = Δt ! With a full calculation the uncertainty in the ’s wavelength thus becomes 2 € λ 1 (Δλ)1/ 2 = πc Δt0 A typical value for natural where (Δλ)1/2 is the full width at half-maximum and Δt0 is the average waiting time for a specific broadening is transition to occur -4 € (Δλ)1/2 = 2.4 x 10 Å

Pressure Broadening

! Pressure Broadening is due to the nearby presence of particles and fields which affect the emitted radiation ! Linear Stark broadening • Only affects atoms/ions with permanent EDM • Broadening (energy shift) ! Van Der Waals Broadening proportional to E field strength • Neutral atoms with momentary • Strongly affects HI lines in hot dipoles perturb each other stars • Energy shift goes with E3 • Common in cooler stars ! Quadratic Stark Broadening ! Zeeman Broadening • Atoms/ions perturbed by Lines split by magnetic fields passing • • Amount of splitting is • Broadening proportional to the proportional to λ2 square of the E field strength • Effect varies (Normal, • Affects most lines in hot stars Anomalous, Paschen-Back) Other sources of broadening

! Large scale turbulent or rotational motion of large masses of gases (important in giants and supergiants). Modifies the previous Doppler width:

2λ $ 2kT 2 ' (Δλ)1/ 2 = & + vturb ) ln2 c % m ( ! Collisional broadening due to the orbitals of an being perturbed in a collision with a neutral atom€ ! Opacity broadening due to absorption and possible re-emission of photons between the star and the Earth

Comparison of Doppler and Natural Broadening

! Each line increases N by 10 ! The top figure shows the absorption line profile for only. Note how the basic Gaussian shape is kept and when the number of atoms N is large the total absorption increases very slowly ! The bottom figure shows the absorption line profile for natural width only. As N increases very strong damplng wings dominate the absorption line and Curve of Growth

! Consider a smooth continuum spectrum passing through a uniform gaseous column that absorbs light at preferential wavelengths

! For moderate optical depths only the Gaussian (Doppler) broadening is significant. Here increasing N results in a proportional increase in EW. This line is said to be optically thin.

! At higher optical depths EW varies only very slowly with column density. This is because the Gaussian broadening “bottoms out” − increasing N doesn’t increase the area of the absorption line very much. This is an optically thick line.

Voigt Profile and Curve of Growth

! At very high optical depths the Lorentzian damping becomes dominant and EW grows with N but is not linearly proportional to N

! The combination of the Doppler 1 and Damping profiles result in something known as the Voigt profile

! The figure opposite compares a standard Doppler and Damping Damping profile with both curves Doppler 0.5 normalised to the same area λ0 Curve of Growth

! For optically very thick lines

Wλ ∝ fN

! For optically thick lines

€ Wλ ∝ ln( fN)

! For optically thin lines Note that this is a special form of the Curve of Growth where information from several spectral lines Wλ ∝ fN (from the same initial orbital) has been€ combined

€ Extracting chemical abundances

! Once a curve of growth exists for a particular star then by using an equivalent width, along with the Boltzmann and Saha equations the total number of atoms of a particular element lying above the photosphere can be determined

! Boltzmann equation ! Saha equation

for atoms of an element Predicts the ratio of particle in a specific ionisation for 2 different state gives the ratio of the ionisation levels 3 / 2 χ number of atoms N /N # & − i b a Ni+1 2kTZi+1 2πmekT kT with energies in = % 2 ( e Eb,Ea N P Z $ h ' different excitation states i e i N g where: Z is the partition function, b = b e−(Eb −Ea )/ kT χ is the ionization energy and g accounts for N g Pe is the electron pressure a a degenerate states €

€ Case study: Na lines in the Sun

! Using the data below we will predict the total number of atoms per unit area above the Sun’s photosphere:

Transition λ (Å) W (Å) f log10(W/λ) log10[f(λ/5000Å)] 3s-4p 3302.38 0.088 0.0214 -4.58 -1.85 3s-3p 5889.97 0.730 0.645 -3.90 -0.12

! Since both transitions start from the ground state orbital of

neutral Na (Na I) then Na (the number of sodium atoms per unit area) is the same in both cases. Use the values of λ(Å) above along with the curve of growth (back 2 slides) to get:

# fNa λ & ! For 3302.38Å line: log % ( =13.20 10$ 5000Å'

# fNa λ & ! For 5889.97Å line: log % ( =14.83 10$ 5000Å'

€ Total no. of ground state atoms

! Next determine Na using : # fN λ & # fλ & log N = log % a ( − log % ( 10 a 10$ 5000Å' 10$ 5000Å' to give: ! For 3302.38Å line: ! For 5889.97Å line: € (take 15.0 as the average value) ! so there are 1015 Na I ground state atoms per cm2 in the Solar photosphere

! However we want the total number of Na atoms (both ground -4 -2 state and ionised). Note: T = 5800 K, Pe = 10 N cm Total number of neutral atoms

! First we need to calculate how many Na I atoms there are in the different excitation states corresponding to the 2 absorption lines. For this we use the Boltzmann equation: N g b = b e−(Eb −Ea )/ kT Na ga ! Assume ga = gb and (Eb-Ea)= hc (photon energy) to give: N N b = 5.45 ×10−4 (3302.38 Å), b =1.48 ×10−2 (5889.97 Å) Na Na so we conclude€ that most of the neutral Na I atoms are in the ground state

€ ! Finally we need to determine the total number of Na atoms per unit area in all stages of ionisation

Now include all ionisation levels

! Finally, using the Saha equation we can calculate how many singly ionized Na atoms (Na II) there are for each Na I atom: 3 / 2 χ # & − I NII 2kTZII 2πmekT kT = % 2 ( e NI PeZI $ h '

with ZI = 2.4, ZII = 1.0 and χI = 5.14 eV (ionisation energy of neutral sodium) gives

€ NII = 2480 NI

! Therefore the total number of sodium atoms per unit area above the Sun’s photosphere is 2.48×1018 cm-2 Solar and local Galactic abundances

! Photometric absorption lines provide lots of information. Isotopic ratios agree well with terrestrial observations ! Also emission lines from solar , prominences and corona ! Radioactive decays (e.g. 40K) must be taken into account when interpreting terrestrial isotope ratios (e.g. 40Ar/36Ar) ! Meteorite composition (excluding volatiles) is in good agreement with Solar absorption lines especially CC1 meteorites which are used to ‘fill in the gaps’ where photometric abundances are poorly measured/unknown ! The local abundance curve is typical for nearby stars and ISM

Abundance Variations

! Outside of the Solar System abundances can vary due to different factors which include: • Point in the star’s evolution − for example, some advanced stars can lose the “envelope” of gas around them (we will look more closely at this later in the course) • Mixing of surface layers • Certain stars are Chemically Peculiar (CP) and exhibit dramatic over/ under abundances of elements which can occur in spots due to gravitational settling and/or radiative levitation modified by magnetic fields (nothing to do with ) • Population effects (see below) where a star’s composition reflects that of the local ISM at the time of formation (enriched or not by earlier nucleosynthesis)

! Population effects: • Population I stars: young, circular orbits close to the plane of the galaxy, low velocities, metal rich • Population II stars: old, in galactic halo, elongated/inclined orbits, high velocities, metal poor Elemental Abundance Summary

! Putting all this information together allow us to build up a picture of Universal Elemental Abundances ! The main features are: ! Most stars are 98% H and He (by mass) ! There is a deep minimum in the abundance curve corresponding to Li, Be, B ! This is followed by a peak around C,N,O,Ne ! Then there is a further decline until the next peak at Fe

! Abundances generally diminish until A = 100 and Z = 45 after which the curve flattens out ! Some stars have considerably lower content of heavier metals (pop II vs pop I) ! Heavy element content varies with the position in the Galaxy ! The interstellar medium is mainly H, either neutral (HI) or ionised (HII)

Solved Problem 1

! The wavelength and equivalent width of a given spectral line are 493 nm and 0.052 nm. What is the frequency of the line and the equivalent width in terms of frequency?

! The frequency is directly obtained from ν = c/λ while the relationship between widths in these two domains can be obtained by differentiating this expression − since the equivalent width covers such a small wavelength range. The frequency is 6.085 x 1014 s-1 and dν = -c dλ/λ2 where the negative sign which just gives the relative direction of the incremental change can be ignored in this case. Putting the numbers in gives the equivalent width in terms of frequency as 6.42 x 1010 s-1.

It is easy to see from this simple example that the ratio of width to frequency is the same as the ratio of width to wavelength. Solved Problem 2

! The observed spectrum of a given star contains extremely broad hydrogen emission lines centred approximately at their natural frequency and relatively sharp hydrogen absorption lines in the short wavelength wing of the emission lines.

Explain how the observed spectrum is produced, assuming that the star is a nova surrounded by an expanding shell of material which is mostly hydrogen at 104 K.

If the absorption lines are shifted from their natural wavelength by 10 times their width, estimate the velocity of the expanding shell.

! The emission occurs in all directions from the expanding shell which has a broad spread of velocities with respect to the direction of the observer. The absorption occurs along the line of sight and involves the nearside of the shell which is moving towards the observer. ' v The Doppler shift is given by Δλs = λ − λ0 = λ0 c 2λ 2kT(ln2) The Doppler width due to thermal broadening is given by: Δ λ = 0 w c m Since the Doppler shift is ten times 2kT(ln2) 1 v = 20 ≈ 213 km.s− the Doppler width: € m €