Introduction to
Astrophysics
You should already know the basics e.g.,:
•!
Laws of Gravity
•!
Kepler’s Laws
•!
Stellar magnitude scale
•!
Wien’s Law
Chapter 8: Spectral line formation
OBAFGKM
List everything you know about these letters!
This course will focus on fundamental physics of stars and star formation
with an emphasis on understanding the Sun.
Handout: syllabus, Homework 1
Week 1: Chapter 8 in Carroll and Ostlie
Chapter 8: Spectral line formation
OBAFGKM
List everything you know about
these letters!
Chapter 8: Spectral line formation
Statistical Mechanics: study of properties of large
number of particles. e.g., a gas can contain particles
with a range of speeds and energies. The gas as a
whole has well-defined properties (temperature,
pressure, density).
For gases in thermal equilibrium: velocity distribution
for atoms in the atmospheres of stars are given by
the Maxwell-Boltzmann function.
What do you think the independent variables for this
distribution will be?
Chapter 8: Spectral line formation
Chapter 8: Spectral line formation
Maxwell-Boltzmann velocity distribution: gives the
fraction of particles per unit volume with speeds
between v and v + dv:
3
Area under the curve
(shaded region) gives
the fraction of particles
with velocity between v
and v + dv
" m % 2 (mv
2
n v dv = n$
' e 2kT 4 !v dv
# 2!kT &
2
First HW assignment:
write a program to
generate this plot,
showing vmp, vRMS, and
shading the area
between v and v+dv.
n v = !n / !v
n = total number density
(particles per unit volume)
T = temperature
m = particle mass
of gas [K]
k = Boltzmann’s constant
Chapter 8: Spectral line formation
In this distribution, useful
quantities can be defined: the
most probably velocity, vMP, the
mean speed, <v>, and the rootmean-square speed, vRMS:
v MP =
v =
Why is there so much variation in the spectra of a star?
What determines the probability of line formation?
Atoms in a gas gain and lose energy when they collide. As a
result, the MB distribution produces a definite distribution of
electrons among the atomic orbitals.
2kT
m
Orbitals of higher energy are less likely to be occupied by
electrons.
8kT
!m
v RMS = v 2 =
Chapter 8: Spectral line formation
Why is the distribution asymmetric?
3kT
m
v MP < v < v RMS
Chapter 8: Spectral line formation
Chapter 8: Spectral line formation
Let sb and sa represent specific sets of
quantum numbers.
Energy levels can be degenerate, with more than one
quantum state having the same energy (simplification).
Ratio of the probability that a system is in
state sb to the probability that a system is
in state sa is given by:
If sa and sb are degenerate, then Ea ! Eb, but sa ! sb
At low temperature, the
Boltzmann factor goes
to zero: P(sb) is much
less than P(sa)
P(sb ) !(E b !E a )/ kT
=e
P(sa )
At high temperature,
the Boltzmann factor
approaches 1 and a
significant population
occupies the higher
energy state.
T is the (common) temperature of the
two systems.
k is Boltzmanns constant
e!E / kT
gb
is the number of states with energy, Eb, the
statistical weight of the energy level.
gn = 2n 2 , where n is the the orbital energy level
is the Boltzmann factor
Chapter 8: Spectral line formation
Chapter 8: Spectral line formation
n shell
l orb angular
momentum
ml magnetic
quant number
Electron
s to fill
Total per
shell (gn)
Shell type
1
0 (s)
0
2
2
One lobe
2
0 (s)
0
2
1(p)
-1,0,1
6
0 (s)
0
2
1 (p)
-1,0,1
6
2 (d)
-2,-1,0,1,2
10
3
P(E b ) gb !(E b !E a )/ kT gb !"# / kT
= e
= e
P(E a ) ga
ga
l = n-1
m = -l…0…+l
One lobe
8
Two lobes
One lobe
Two lobes
18
Three lobes
For n=1, what is the degeneracy, gn?
For n=2? For n=3?
For a gas of neutral H-atoms, at
what temperature will equal
numbers of atoms have electrons
in the ground state and the first
excited state?
P(E b ) gb !(E b !E a )/ kT
= e
P(E a ) ga
1=
2(2 2 ) ![(!3.4 )!(!13.6)] / kT
e
2(12 )
1 = 4e!10.2 / kT
ln(0.25) = !10.2eV /kT
ln(4) = 10.2 /kT
10.2eV
T=
= 85400K
k * ln(4)
Chapter 8: Spectral line formation
For a gas of neutral H-atoms, at
what temperature will equal
numbers of atoms have electrons
in the ground state and the first
excited state?
Chapter 8: Spectral line formation
P(E b ) gb !(E b !E a )/ kT
= e
P(E a ) ga
1=
But, if a temperature of 85000K puts
only half of the electrons in the n=1
state, then whey do the Balmer lines
(n=2) peak at the much lower
temperature of ~9500K?
2(2 2 ) ![(!3.4 )!(!13.6)] / kT
e
2(12 )
1 = 4e!10.2 / kT
ln(0.25) = !10.2eV /kT
ln(4) = 10.2 /kT
10.2eV
T=
= 85400K
k * ln(4)
The spectral lines become weaker in
stars withT > 9500K because atoms
have a MB energy distribution. A
significant fraction of atoms are in the
high energy tail of the distribution and
are ionized.
Chapter 8: Spectral line formation
Calculate the number of ways that an atom can
arrange electrons with the same energy.
(what will the independent variables be?)
Z = " g je
#( E j #E1 / kT )
j=1
gn = 2n
N b gb !( E b !E a )/ kT
= e
N a ga
In order to calculate this ratio, need
to also account for the fraction of
ionized electrons.
Chapter 8: Spectral line formation
Step 2: the Saha equation
Step 1: the partition function
!
(Emission)
The Boltzmann equation: the ratio of
the number of electrons in each
energy level is the same as the ratio of
the probabilities.
2
But in general, and for your homework, use precalculated partition functions from handout.
Right now, not talking about the fraction of atoms in different energy
levels. Saha eqn accounts for the fraction of electrons ionized from
each energy level: i, i+I,….
3/2
N i+1 2Z i+1 " 2! me kT % ( ) i / kT
=
$
' e
Ni
ne Zi # h 2 &
The ratio of the number of atoms in ionization stage i+1 to those
in stage i depends on the ratio of the partition functions. Like the
M-B distribution, there is also a dependence on temperature and
(ionization) energy. Also a dependence on the electron number
density, ne because when there are more electrons, the ions can
recombine.
Chapter 8: Spectral line formation
Chapter 8: Spectral line formation
Re-write the Saha eqn in terms of electron pressure rather than density.
3/2
N i+1 2Z i+1 " 2! me kT % ( ) i / kT
=
$
' e
2
Ni
ne Zi # h
&
Pe = n e kT
3/2
N i+1 2kTZ i+1 " 2! me kT % ( ) i / kT
=
$
' e
Ni
Pe Z i # h 2 &
Need to re-parameterize variables for your computer code.
k = Boltzmann constant = 1.38066 x 10-23 J/K = 8.617385 x 10-5 eV/K
Let ! = loge /kT
3/2
Change from
base e to
base 10
See handout
from Gray
“Observational
Astrophysics”
**Remember, these are ionization stages from different
energy levels
Chapter 8: Spectral line formation
;partition function for an array of temperatures
;temp=findgen(200)*100.+5000.
theta = 5040./temp(i)
case species of
'hydrogen': begin
logZ1 = 0.30103 + (0.00001)*alog10(theta)
chi1=13.6
logZ2 = 0.
end
5/2
Chapter 8: Spectral line formation
Step 3: combining the Boltzmann and Saha equations
Plotting the Saha equation
Important hints for your code:
(2" me ) (kT ) 2Z1(T) e# $1 / kT
N1
Pe =
N0
h3
Z 0 (T)
% N1 ( #5040
Z
log' Pe * =
$1 + 2.5logT + log 1 # 0.1762
T
Z0
& N0 )
N1 +(T)
=
N0
Pe
Z
+(T) = 0.665 1 T 5 / 210#5040 $1 / kT
Z0
Z
#! $
+(T) = 1.2020 ,10 9 1 T 5 / 210 1
Z0
Hydrogen particles
Pe = 20 N m-2
50% ionization at 9500K
cm-2
pe=200 ;dyn
phi1 = (-5040.*chi1/temp(i)) + 2.5*alog10(temp(i)) + logZ2 - logZ1 - 0.1762
NII_to_NI(i) = (10.^(phi1))/pe
; fraction of ionized to neutral species
fx(i) = NII_to_NI(i) / (1. + NII_to_NI(i)) ; Saha equation
What does the plot of the Saha eqn demonstrate?
Very narrow range of ionization for hydrogen - starts ionizing at
T = 8K and by T = 11K 100% is ionized!
What does this mean for the strength of H lines in stars?
The fraction of electrons in a higher energy state is that fraction
from the Boltzmann equation, times the fraction of atoms that are
not ionized. (If the atoms are ionized, then can’t have the electron
in a higher energy level.)
From the old
Boltzmann
equation
Fraction
remaining after
Saha
ionization
! N 2 $! N I $ ! N 2 /N1 $!
$
N2
1
=#
&#
&=#
&#
&
N total " N1 + N 2 %" N total % " 1+ N 2 /N1 %"1+ N II /N I %
Re-written in terms of ratios
that can be calculated
Chapter 8: Spectral line formation
What does this plot of the Boltzman
distribution demonstrate?
Again, a narrow ionization range of
about 3000K , starting at T=8000K.
Once the peak ionization is reached,
the fraction of electrons in the n=2
level begins to decline with increasing
temperature (lost to ionization).
Plotting the Boltzman equation
Important hints for your code:
; now the Boltzmann equation
N2_to_N1=(g2/g1)*10.^(-10.2*5040/temp(i))
N2_to_N(i) = (N2_to_N1/(1+N2_to_N1))*(1./(1.+NII_to_NI(i)))
plot,temp/1000.,n2_to_N*1000000., xtit='!6 Temperature K / 1000', $
ytit='!6 N!d2!n / N!dtotal!n (10!u-6!n)'
Chapter 8: Spectral line formation
Chapter 8: Spectral line formation
The narrow region inside a star
where hydrogen is partially
ionized is called the hydrogen
partial ionization zone and
has a characteristic
temperature of 10000K for a
wide range of stellar
parameters.
The Hydrogen Balmer lines attain maximum intensity at about 9500K,
not the higher temperature of 85,000K required to pump electrons up
to the n=2 energy level.
Even in cool stars, the Balmer transitions are important: while there
may not be many electrons pumped up to the n=2 (Balmer baseline)
state, there are an enormous number of hydrogen atoms in stars!
Chapter 8: Spectral line formation
The part of the Sun that we see is a thin outer layer,
called the photosphere. This layer has a characteristic
temperature of 5770K. It has about 500,000 H atoms
for each Ca atom and a pressure of about 1.5 Nm-2.
Estimate the relative strength of Balmer Hydrogen
absorption lines and those due to Ca II H and K lines.
But, stellar atmospheres have ~1 He atom for every 10 H atoms.
What affect will the presence of He have on the ionization temperature?
He donates 2 electrons when ionized, providing more electrons for
recombination for the ionized H atoms. Thus, a higher temperature is
required to achieve the same degree of H ionization.
Output from computer program (Hmwk problem 2):
The ratio of singly ionized hydrogen to neutrals is: 8.6e-05
The fraction of atoms in the n=1 level is: 8.6e-05
The ratio of singly ionized calcium to neutrals is: 973.821
The fraction of Ca atoms in the n=1 level is: 0.995104
Chapter 8: Spectral line formation
Chapter 8: Spectral line formation
NII / NI = 8.6e-05
N2 / Ntotal = 5.5e-09
Hardly any Hydrogen is ionized, only 1 atom in a billion is in the
n=2 level and capable of producing Balmer absorption.
Note: distribution for H has shifted to cooler temperatures - why?
NII / NI = 973.821
N2 / Ntotal = 0.0038
Ca is overwhelmingly ionized at 5770K. Most of the remaining Ca
is in the ground state.
500,000 hydrogen atoms for every calcium atom, but only
5e-9 are un-ionized and in the n=2 state: 2.4e3 hydrogen atoms
Essentially all Ca atoms are ionized (Ca II) and in the ground state,
so roughly 400 Ca atoms for Ca II H and K line formation for every 1
H atom available for Balmer line formation.
Ca II H line: spans about 1000 A
H!: spans about 100 A
Chapter 8: Spectral line formation
Chapter 8: Spectral line formation
What is the ratio of doubly to singly ionized
Ca atoms?
N III !(T)
=
N II
Pe
0.002
Fraction of CaIII to CaII
atoms is small
#Z &
)*
!(T) = 1.202 "10 9 % III (T (5 / 2)10
$ Z II '
+
Dependence of spectral line strength on effective temperature
Chapter 9 Stellar Atmospheres: Radiation fields
Chapter 9: Stellar Atmospheres
Let’s jump ahead just a bit and look at some definitions.
The light we see emerging from a star comes from the
outer layers. The temperature, density and composition
of the outer layers determines the features of the stellar
spectrum.
I! =
I! =
1
4"
#I
!
d$ =
1
4"
# #
2"
&= 0
"
I sin % d% d&
%=0 !
For an isotropic field (same intensity
in all directions)
"I
E ! d!
=
"! d! dt dAcos # d$
I! = I!
Specific intensity, I!!
Energy passing through a
solid angle from a point on
the surface, at a given time,
in a direction !, with
wavelength between ! and
!+d!
isotropic
dA is a patch on the radiating
surface of the star
Chapter 9: Stellar Atmospheres
Chapter 9: Stellar Atmospheres
Blackbody radiation is isotropic. For
blackbody radiation:
For isotropic radiation:
I! = B!
Specific energy density with
wavelength between ! and !+d! is
defined as:
u! d! =
1
c
"I
!
d! d#
1 2% %
" " I! d! sin$ d$ d&
c &= 0 $ = 0
4%
=
I! d!
c
u! d! =
4"
I! d!
c
For blackbody radiation:
I! = B!
B! (T) =
2hc 2
1
hc / ! k T
5
! e
"1
=
Energy density in blackbody
radiation for a characteristic
wavelength:
u! d! =
8hc
1
d!
hc / ! k T
5
! e
"1
Chapter 9: Stellar Atmospheres
Chapter 9: Stellar Atmospheres
Of course:
u=
Specific Radiative Flux:
#
"
0
u! d! =
For blackbody radiation,
u=
4!
c
$
#
0
#
"
0
=
I! = B!
B" (T) d" =
" I d! cos# d# d$
F! d! =
u$ d$
4% T
c
4
= aT 4
!
" "
2&
%= 0
&
I cos # sin # d# d%
#=0 !
This is the net energy with a wavelength between
!!and !+d! that passes each second through a
unit area in the direction of the z-axis.
Because of the factor cos!, oppositely directed rays
can cancel!
a = 4% /c = 7.565767 &10'16 Jm '3K '4
Chapter 9: Stellar Atmospheres
Both the radiative flux and the specific
intensity measure light received from a
celestial source.
When you point a photometer at a light
source, which of these are you measuring?
Chapter 9: Stellar Atmospheres
For a resolved source (e.g. observations of
the Sun from an orbiting satellite) you are
measuring specific intensity, I!, the amount
of energy passing through a solid angle
!min.
For an unresolved source (a distant star) it is
the radiative flux that is being measured.
The detector integrates the specific
intensity over all solid angles. This is the
definition of radiative flux. As the distance to
the source increases, the amount of energy
decreases as 1/r2.
Chapter 8: The H-R Diagram
http://tauceti.sfsu.edu/learning/
Stefan-Boltzmann Law:
•! Programming IDL for Astronomy (M. Perrin) a very good introduction
to IDL with important philosophy too!
•! The following 4 postscript files (also from C. Heiles, UCB) give you a
good introduction to IDL:
1) Basics
2) Data Types, Including Structures
3) Plotting
4) Color
•! IDL help page written by Eric Williams, former SFSU student
•! D. Fanning's IDL tips and tricks
•! IDL Astronomy Library
•! C. Markwardt's IDL library
•! idl> ? (this gives you the idl help manual)
L = 4 !R 2"T 4
R=
1
Te2
L
4 !"
If star A and star B have the same surface
temperature, but star A is 100 times more
luminous than the other, then how do the radii
of stars A and B compare?
Right, the radius of star A is 10 times larger.
Chapter 8: The H-R Diagram
Chapter 8: The H-R Diagram
Download an IDL structure containing the
Hipparcos catalog:
Recall:
Trigonometric dist
Distance modulus
http://www.physics.sfsu.edu/~fischer/data/hip.dat
dist( pc) =
1
parallax (arc sec)
dist( pc) = 10(m!M +5)/ 5
idl> restore,’hip.dat’
idl> help,/st,hip
** Structure <2659ba4>, 13 tags, length=92, data length=92, refs=1:
HIPNO
STRING '1'
RA
DOUBLE
6.1111111e-05
DEC
DOUBLE
1.0890000
Given this information, how would
VMAG
FLOAT
9.10000
you calculate distance to a star?
PRLAX
FLOAT
0.00354000
RA_MOTION
FLOAT
-0.00520000
hip.prlax is the parallax and 1./
DEC_MOTION
FLOAT
-0.00188000
hip.prlax = distance
PRLAXERR
FLOAT
0.00139000
B_V
FLOAT
0.482000
EB_V
FLOAT
0.0250000
HD
STRING '224700'
SPTYPE
STRING 'F5
'
SPTYPSRC
STRING 'S'
Chapter 8: The H-R Diagram
pro junk
; an example of how to start a program
Of course, you’ll
write a program
to do this!
Download an IDL structure containing the
Hipparcos catalog (Hmwk problem 5):
vel=findgen(400)*100.
nv=fltarr(400)
temp=8000.
kb=1.38e-23
m=1.67e-27
http://www.physics.sfsu.edu/~fischer/data/hip.dat
idl>x=where(1./hip.prlax lt 100.) ;stars closer than 100 pc
for i=0,399 do begin
a1=4.*!pi*vel(i)^2
a2=(m/(2.*!pi*kb*temp))^(1.5)
a3=exp((-m*vel(i)^2)/(2.*kb*temp))
nv(i)=a1*a2*a3
end ;for
idl> newhip=hip(x)
; stars closer than 100 pc
idl> dist=1./newhip.prlax
idl> absmag=newhip.vmag - 5.*alog10(dist) + 5.
idl> BMV=newhip.b_v
stop
Plot HR Diagram:
Xrange: 0.0 < B-V < 1.5
end
Yrange: 15 < V < -5
idl> plot, BMV, absmag, xra=[0., 1.5], yra=[15, -5], ps=3
Chapter 8: The H-R Diagram
Chapter 8: The H-R Diagram
For HW problem 5, you should create a plot like this (for the first part).
Then, make a second plot, but instead of selecting stars with distance
How big are stars?
less than 200 pc, select stars brighter than V=9.
giants
subgiants
Main
seque
nce
Earth Mars Jupiter
Saturn Uranus Neptune
Sun
Chapter 8: The H-R Diagram
Chapter 8: The H-R Diagram
How big are stars?
How big are stars?
.
Sun
Sun
Sirius
Pollux
Arcturus
Chapter 8: The H-R Diagram
Arcturus Rigel Aldebaran Betelguese
Antares
Chapter 8: The H-R Diagram
Average density of stars:
Stellar mass is one of the most fundamental parameters. More
massive stars have stronger gravitational pressure, more fusion
rxn’s, higher temperatures, greater luminosity.
The Sun (G2V):
! =
M sun
= 1.4gcm#3
4
3
" Rsun
3
! =
M Sirius
= 0.76gcm#3
4
3
" RSirius
3
! =
M Bet
= 10#11 gcm #3
4
3
" RBet
3
Appendix G (textbook) lists stellar masses and radii as a function of
spectral type.
With the stellar mass and radius, you can calculate density, but first
guess!
Sirius (A1V):
•! the density of rocky material (earth) is about 5 g cm-3
•! the density of water is 1 g cm-3
What do you think the density of stars might be?
Do you think high mass (main sequence) stars will have higher or
lower density than the lower mass stars?
Betelgeuse (M2I):
Betelgeuse has an average density that is 100,000
times less dense than the air we breathe!