Download Topic 2 Chemical Composition of Stars

Topic 2
Evidence of abundances
from spectral lines
Chemical Composition of Stars
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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
'
0.01 ≤ )Tλ (z) =
(
∞
∫k
z
*
(z)
ρ
(z)dz
, ≤ "few"
λ
+
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
Black Body Radiation
Cool stars (top) peak emission is red
Hot stars (bot.) peak emission is blue
2hν 3
1
I(ν ) = 2 hν / kT
c e
−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
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HOTTER
! 
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
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
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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.
Redshift of spectral lines in the
optical spectrum from a
supercluster of distant galaxies (right)
compared to that of the Sun (left)
Reminder: the key equations
for the Doppler effect are:
Δf =
fv
c
f' = f ±
for radiation travelling
at the speed of light
fv
c
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
Equivalent Width
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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 lead to spectral line broadening,
they are classified in different ways, for example:
• 
The effect on the line width (profile)
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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
Due to thermal motion
of atoms along the
line of sight
Gaussian profile
Natural
Due to Heisenberg
Uncertainty Principle
Lorentzian profile
Pressure
Emitted Radiation is
affected by nearby
particles
Lorentzian profile
Doppler Broadening
Due to random thermal
!  Mathematically, from MB
motion of atoms in a gas
the most probable speed is
vmp = 2kT /m
!  In thermal equilibrium the
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 wavelengths of the
width as
2 λ 2kT
Δλ =
photons emitted or absorbed €
c
m
are thus Doppler shifted
!  Taking into account
different atoms’ directions
For hydrogen atoms in the
€ is refined to
this
Sun’s photosphere (T=5770K)
2 λ 2kT (ln2)
(Δλ)1/ 2 =
Δλ ≈ 0.427Å for Hα
c
m
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Natural Broadening
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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:
ΔE =
! 
!
Δt
With a full calculation the uncertainty in the
photon’s wavelength thus becomes
€
(Δλ)1/ 2 =
λ2 1
πc Δt 0
A typical value for natural
broadening is
(Δλ)1/2 = 2.4 x 10-4 Å
where (Δλ)1/2 is the full width at half-maximum
and Δt0 is the average waiting time for a specific
transition to occur
€
Pressure Broadening
Pressure Broadening is due to
the nearby presence of particles
and fields which affect the
emitted radiation
Linear Stark broadening
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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
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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
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Neutral atoms with momentary
dipoles perturb each other
Energy shift goes with E3
Common in cooler stars
Zeeman Broadening
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Lines split by magnetic fields
Amount of splitting is
proportional to λ2
Effect varies (Normal,
Anomalous, Paschen-Back)
Other sources of broadening
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Large scale turbulent or rotational motion of large
masses of gases (important in giants and
supergiants). Modifies the previous Doppler width:
'
2 λ $ 2kT
2
+ vturb ) ln2
(Δλ)1/ 2 =
&
(
c % m
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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
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Comparison of Doppler and
Natural Broadening
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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
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Consider a smooth continuum spectrum passing through a
uniform gaseous column that absorbs light at preferential
wavelengths
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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
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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
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The figure opposite compares a
standard Doppler and Damping
profile with both curves
normalised to the same area
1
Damping
Doppler
0.5
λ0
Curve of Growth
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For optically
very thick lines
Wλ ∝
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€
For optically
thick lines
W λ ∝ ln( fN)
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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
fN
For optically
thin lines
W λ ∝ fN
€
Extracting chemical abundances
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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
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Boltzmann equation
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Saha equation
for atoms of an element
Predicts the ratio of particle
in a specific ionisation
densities for 2 different
state gives the ratio of the
ionisation levels
3/2
χ
number of atoms Nb/Na
− i
N i+1 2kTZ i+1 # 2πme kT &
=
%
( e kT
with energies Eb,Ea in
2
'
Ni
Pe Z i $ h
different excitation states
N b gb −( E b −E a )/ kT
= e g accounts for
N a ga
degenerate states
€
where: Z is the partition function,
χ is the ionization energy and
Pe is the electron pressure
Case study: Na lines in the Sun
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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
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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:
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For 3302.38Å line:
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For 5889.97Å line:
# fN a λ &
log10 %
( = 13.20
$ 5000Å '
# fN a λ &
log10 %
( = 14.83
$ 5000Å '
€
Total no. of ground state atoms
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Next determine Na using :
# fN a λ &
# fλ &
log10 N a = log10 %
( − log10 %
(
$ 5000Å '
$ 5000Å '
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€
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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
state and ionised). Note: T = 5800 K, Pe = 10-4 N cm-2
Total number of neutral atoms
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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 b gb −( E b −E a )/ kT
= e
N a ga
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Assume ga = gb and (Eb-Ea)= hc (photon energy) to give:
Nb
N
= 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
€
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Finally we need to determine the total number of Na
atoms per unit area in all stages of ionisation
Now include all ionisation levels
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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
N II 2kTZ II # 2πme kT &
=
%
( e kT
2
'
NI
Pe Z I $ h
with ZI = 2.4, ZII = 1.0 and χI = 5.14 eV (ionisation
energy of neutral sodium) gives
NII = 2480 NI
€
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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
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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
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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)
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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
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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
2 λ0
c
2kT(ln2)
Since the Doppler shift is ten times
v = 20
≈ 213 km.s−1
m
the Doppler width: €
The Doppler width due to thermal broadening is given by: Δλw =
€
€
2kT(ln2)
m