Topic 2

Evidence of abundances from spectral lines

Chemical Composition of Stars !  The chemical composition of

stars can be investigated from looking at star light

!  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

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 density €

0.01≤ Tλ (z) = kλ (z)ρ(z)dzz

∞

∫'

( )

*

+ , ≤ "few"

Black Body Radiation

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

Cool stars (top) peak emission is red Hot stars (bot.) peak emission is blue

€

I(ν) =2hν 3

c 21

ehν / kT −1

Spectral Lines: Origin and Appearance

Continuous Spectrum

Emission Spectrum

Absorption Spectrum

Absorption Spectra !  The spectra that are

observed depend very much on the temperature and pressure in the stellar atmosphere

!  The hotter the temperature the greater the mean thermal energy •  Break up of molecules •  Ionization of atoms •  Those atoms with the highest

binding energy are only seen in their ionized state in the hottest stars

HO

TTER

We will discuss Binding 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. wavelength appropriate

•  The amount of the particular element that is present•  The strength of the transition probability or cross-section for

absorption

Interpreting Spectral Lines !  Spectral lines can

give us information on what elements/ions are present in a star’s photosphere as well as the abundance of that particular element or ion.

!  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 equivalent width of the observed

spectral line 3.  Interpreting the resulting line width due to broadening

mechanisms

Reminder: the key equations for the Doppler effect are:

for radiation travelling at the speed of light .

€

Δf =fvc

€

f ' = f ±fvc

Redshift of spectral lines in the optical spectrum from a

supercluster of distant galaxies (right) compared to that of the Sun (left)

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)

€

Wλ =FC − FλFC

$

% &

'

( ) ∫ dλ

Main Sources of Line Broadening !  There are many mechanisms that can lead 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

Natural Due to Heisenberg

Uncertainty Principle Lorentzian profile

Pressure Emitted Radiation is affected by nearby

particles Lorentzian profile

Doppler Due to thermal motion

of atoms along the line of sight

Gaussian profile

Local effects Non-Local effects

•  Opacity Broadening •  Rotational Broadening

Doppler Broadening !  Due to random thermal

motion of atoms in a gas !  In thermal equilibrium the

gas atoms are moving with speeds described by the Maxwell-Boltzmann (MB) distribution

!  The wavelengths of the photons emitted or absorbed are thus Doppler shifted

!  Mathematically, from MB the most probable speed is

!  Using the (non-rel.) Doppler shift formula:

!  This gives the spectral line width as

!  Taking into account different atoms’ directions this is refined to

€

vmp = 2kT /m

€

Δλ /λ = ± v /c

€

Δλ =2λc

2kTm

€

Δλ( )1/ 2 =2λc

2kT ln2( )m

For hydrogen atoms in the Sun’s photosphere (T=5770K)

Δλ ≈ 0.427Å for Hα

Natural Broadening !  Natural Broadening is due to the Heisenberg

Uncertainty Principle - a direct result of Quantum Mechanics

!  When an electron is moved to an excited state 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:

!  With a full calculation the uncertainty in the photon’s wavelength thus becomes

where (Δλ)1/2 is the full width at half-maximum and Δt0 is the average waiting time for a specific transition to occur

€

ΔE =!Δt

€

Δλ( )1/ 2 =λ2

πc1Δt0

A typical value for natural broadening is

(Δλ)1/2 = 2.4 x 10-4 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)

proportional to E field strength •  Strongly affects HI lines in hot

stars !  Quadratic Stark Broadening

•  Atoms/ions perturbed by passing electrons

•  Broadening proportional to the square of the E field strength

•  Affects most lines in hot stars

!  Van Der Waals Broadening •  Neutral atoms with momentary

dipoles perturb each other •  Energy shift goes with E3 •  Common in cooler stars

!  Zeeman Broadening •  Lines split by magnetic fields •  Amount of splitting is

proportional to λ2

•  Effect varies (Normal, 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:

!  Collisional broadening due to the orbitals of an atom 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

€

Δλ( )1/ 2 =2λc

2kTm

+ vturb2$

% &

'

( ) ln2

Comparison of Doppler and Natural Broadening

!  Each line increases N by 10 !  The top figure shows the

absorption line profile for Doppler broadening 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

Voigt Profile and Curve of Growth

!  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.

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

!  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 and Damping profiles result in something known as the Voigt profile

!  The figure opposite compares a standard Doppler and Damping profile with both curves normalised to the same area

λ0 0.5

1

Damping Doppler

Curve of Growth !  For optically

very thick lines

!  For optically thick lines

!  For optically thin lines

€

Wλ ∝ fN€

Wλ ∝ ln( fN)€

Wλ ∝ fN

Note that this is a special form of the Curve of Growth where information from several spectral lines

(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 for atoms of an element in a specific ionisation state gives the ratio of the number of atoms Nb/Na with energies Eb,Ea in different excitation states

!  Saha equation Predicts the ratio of particle densities for 2 different ionisation levels

where: Z is the partition function, χ is the ionization energy and Pe is the electron pressure

€

Ni+1

Ni

=2kTZi+1

PeZi

2πmekTh2

#

$ %

&

' ( 3 / 2

e−χ ikT

€

Nb

Na

=gbgae−(Eb −Ea ) / kT

g accounts for degenerate states

Case study: Na lines in the Sun !  Using the data below we will predict the total number of

Sodium 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:

!  For 3302.38Å line:

!  For 5889.97Å line:

€

log10fNaλ5000Å#

$ %

&

' ( =13.20

log10fNaλ5000Å#

$ %

&

' ( =14.83

Total no. of ground state atoms !  Next determine Na using :

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

€

log10 Na = log10fNaλ5000Å#

$ %

&

' ( − log10

fλ5000Å#

$ %

&

' (

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

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:

!  Assume ga = gb and (Eb-Ea)= hc (photon energy) to give:

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

€

Nb

Na

=gbgae−(Eb −Ea ) / kT

€

Nb

Na

= 5.45 ×10−4 (3302.38 Å),Nb

Na

=1.48 ×10−2(5889.97 Å)

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:

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

€

NII

NI

=2kTZIIPeZI

2πmekTh2

#

$ %

&

' ( 3 / 2

e−χ I

kT

Solar and local Galactic abundances !  Photometric absorption lines provide lots of information.

Isotopic ratios agree well with terrestrial observations !  Also emission lines from solar chromosphere, 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 nucleosynthesis)

•  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.The Doppler shift is given by

The Doppler width due to thermal broadening is given by:

Since the Doppler shift is ten timesthe Doppler width:

€

Δλs = λ' − λ0 = λ0vc

€

Δλw =2λ0c

2kT(ln2)m

€

v = 20 2kT(ln2)m

≈ 213 km.s−1