week 1: chapter 8 in carroll and ostlie

Introduction to

Astrophysics

You should already know the basics - e.g.,:

•! Laws of Gravity

•! Kepler’s Laws

•! Stellar magnitude scale

•! Wien’s Law

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!


OBAFGKM

List everything you know about these letters!

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?


Maxwell-Boltzmann velocity distribution: gives the fraction of particles per unit volume with speeds between v and v + dv:

nvdv = n

m

2!kT"

# $

%

& '

3

2

e

(mv 2

2kT 4!v 2dv

nv

= !n /!v = total number density

(particles per unit volume)

n

m = particle mass

= Boltzmann’s constant

k

T = temperature

of gas [K]


Area under the curve (shaded region) gives the fraction of particles with velocity between v and v + dv

First HW assignment: write a program to generate this plot, showing vmp, vRMS, and shading the area between v and v+dv.


In this distribution, useful quantities can be defined: the most probably velocity, vMP, the mean speed, <v>, and the root-mean-square speed, vRMS:


vMP

=2kT

m

vRMS

= v2

=3kT

m

v =8kT

!m

vMP

< v < vRMS

Why is the distribution asymmetric?

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.

Orbitals of higher energy are less likely to be occupied by

electrons.

Let sb and sa represent specific sets of quantum numbers.

Ratio of the probability that a system is in state sb to the probability that a system is in state sa is given by:


P(sb)

P(sa)

= e!(E

b!E

a) / kT

T is the (common) temperature of the two systems.

k is Boltzmanns constant

is the Boltzmann factor

e!E / kT

At low temperature, the

Boltzmann factor goes

to zero: P(sb) is much

less than P(sa)

At high temperature,

the Boltzmann factor

approaches 1 and a

significant population

occupies the higher

energy state.


P(Eb )

P(Ea )=gb

gae!(Eb !Ea ) / kT =

gb

gae!"# / kT

Energy levels can be degenerate, with more than one quantum state having the same energy (simplification).

If sa and sb are degenerate, then Ea ! Eb, but sa ! sb

gb is the number of states with energy, Eb, the statistical weight of the energy level.

gn = 2n2

, where n is the the orbital energy level


n shell l orb angular momentum

ml magnetic quant number

Electrons to fill

Total per shell (gn)

Shell type

1 0 (s) 0 2 2 One lobe

2 0 (s) 0 2 One lobe

1(p) -1,0,1 6 8 Two lobes

3 0 (s) 0 2 One lobe

1 (p) -1,0,1 6 Two lobes

2 (d) -2,-1,0,1,2 10 18 Three lobes

l = n-1

m = -l…0…+l

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(Eb )

P(Ea )=gb

gae!(Eb !Ea ) / kT

1=2(2

2)

2(12)e![(!3.4 )!(!13.6)] / kT

1= 4e!10.2 / kT

ln(0.25) = !10.2eV /kT

ln(4) =10.2 /kT

T =10.2eV

k * ln(4)= 85400K


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(Eb )

P(Ea )=gb

gae!(Eb !Ea ) / kT

1=2(2

2)

2(12)e![(!3.4 )!(!13.6)] / kT

1= 4e!10.2 / kT

ln(0.25) = !10.2eV /kT

ln(4) =10.2 /kT

T =10.2eV

k * ln(4)= 85400K

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?

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.


The Boltzmann equation: the ratio of the number of electrons in each energy level is the same as the ratio of

the probabilities.

(Emission)

Nb

Na

=gb

gae!(Eb !Ea ) / kT

In order to calculate this ratio, need to also account for the fraction of

ionized electrons.


Step 1: the partition function

Calculate the number of ways that an atom can arrange electrons with the same energy.

(what will the independent variables be?)

Z = g j

j=1

!

" e#(E j #E1 / kT )

gn = 2n2

But in general, and for your homework, use pre-calculated partition functions from handout.


Ni+1

Ni

=2Z

i+1

neZi

2! mekT

h2

"

# $

%

& '

3 / 2

e()

i/ kT

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.

Step 2: the Saha equation

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,….


Ni+1

Ni

=2Z

i+1

neZi

2! mekT

h2

"

# $

%

& '

3 / 2

e()

i/ kT

Pe

= nekT

Ni+1

Ni

=2kTZ

i+1

PeZi

2! mekT

h2

"

# $

%

& '

3 / 2

e()

i/ kT

**Remember, these are ionization stages from different energy levels

Re-write the Saha eqn in terms of electron pressure rather than density.


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

N1

N0

Pe

=2" m

e( )3 / 2

kT( )5 / 2

h3

2Z1(T)

Z0(T)e#$1 / kT

logN1

N0

Pe

%

& '

(

) * =

#5040T

$1 + 2.5logT + logZ1

Z0

# 0.1762

N1

N0

=+(T)Pe

+(T) = 0.665Z1

Z0

T5 / 210#5040$1 / kT

+(T) =1.2020 ,109Z1

Z0

T5 / 210

#! $1

Change from base e to base 10

See handout from Gray “Observational Astrophysics”


Plotting the Saha equation

Important hints for your code: ;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

Hydrogen particles

Pe = 20 N m-2

50% ionization at 9500K

pe=200 ;dyn cm-2 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?


Step 3: combining the Boltzmann and Saha equations

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.)

N2

Ntotal

=N2

N1

+ N2

!

" #

$

% &

NI

Ntotal

!

" #

$

% & =

N2/N

1

1+ N2/N

1

!

" #

$

% &

1

1+ NII/N

I

!

" #

$

% &

From the old Boltzmann equation

Fraction remaining after

Saha ionization

Re-written in terms of ratios

that can be calculated


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)'

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).


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!


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.


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.

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


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


What is the ratio of doubly to singly ionized Ca atoms?

NIII

NII

=!(T)Pe

!(T) =1.202 "109ZIII

ZII

#

$ %

&

' ( T

(5 / 2)10

)* +

0.002 Fraction of CaIII to CaII atoms is small


Dependence of spectral line strength on effective temperature

Chapter 9 Stellar Atmospheres: Radiation fields

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.

dA is a patch on the radiating surface of the star

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!

I!

="I

"!=

E!d!

d! dt dAcos# d$

Let’s jump ahead just a bit and look at some definitions.

Chapter 9: Stellar Atmospheres

I! =1

4"I!# d$ =

1

4"I! sin% d%% = 0

"

#&= 0

2"

# d&

I! = I!

For an isotropic field (same intensity in all directions)

isotropic


I!

= B!

Blackbody radiation is isotropic. For blackbody radiation:

Specific energy density with wavelength between ! and !+d! is defined as:

u!d! =1

cI"!d! d#

=1

cI!$ = 0

%

"&= 0

2%

" d! sin$ d$ d&

=4%c

I! d!


I!

= B!

For isotropic radiation:

u!d! =

4"

cI!d!

For blackbody radiation:

B!(T) =

2hc2

!5

1

ehc /! k T

"1

u!d! =

8hc

!5

1

ehc /! k T

"1d!

Energy density in blackbody radiation for a characteristic wavelength:


I!

= B!

Of course:

u = u!d!0

"

# = u$ d$0

"

#

For blackbody radiation,

u =4!

cB"(T) d"0

#

$ =4% T

4

c= aT

4

a = 4% /c = 7.565767 &10'16Jm'3K

'4


Specific Radiative Flux:

F!d! = I!" d!cos# d# d$

= I! cos# sin# d# d%# = 0

&

"%= 0

2&

"

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!


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?


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.

http://tauceti.sfsu.edu/learning/

•! 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)

Chapter 8: The H-R Diagram

Stefan-Boltzmann Law:

L = 4!R2"T

4

R =1

Te

2

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.


Recall:

dist(pc) =1

parallax (arc sec)

dist(pc) =10(m!M +5)/ 5

Distance modulus

Trigonometric dist


Download an IDL structure containing the Hipparcos catalog:

http://www.physics.sfsu.edu/~fischer/data/hip.dat

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 VMAG FLOAT 9.10000 PRLAX FLOAT 0.00354000 RA_MOTION FLOAT -0.00520000 DEC_MOTION FLOAT -0.00188000 PRLAXERR FLOAT 0.00139000 B_V FLOAT 0.482000 EB_V FLOAT 0.0250000 HD STRING '224700' SPTYPE STRING 'F5 ' SPTYPSRC STRING 'S'

Given this information, how would you calculate distance to a star?

hip.prlax is the parallax and 1./hip.prlax = distance

pro junk ; an example of how to start a program

vel=findgen(400)*100. nv=fltarr(400) temp=8000. kb=1.38e-23 m=1.67e-27

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

stop

end


Download an IDL structure containing the Hipparcos catalog (Hmwk problem 5):

http://www.physics.sfsu.edu/~fischer/data/hip.dat

idl>x=where(1./hip.prlax lt 100.) ;stars closer than 100 pc

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

Plot HR Diagram:

Xrange: 0.0 < B-V < 1.5 Yrange: 15 < V < -5

idl> plot, BMV, absmag, xra=[0., 1.5], yra=[15, -5], ps=3

Of course, you’ll write a program

to do this!


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

less than 200 pc, select stars brighter than V=9.

Main sequence

subgiants

giants

How big are stars?


Earth Mars Jupiter Saturn Uranus Neptune Sun


How big are stars?

Sun Sirius Pollux Arcturus


How big are stars?

Arcturus Rigel Aldebaran Betelguese Antares

.

Sun


Stellar mass is one of the most fundamental parameters. More massive stars have stronger gravitational pressure, more fusion rxn’s, higher temperatures, greater luminosity.

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!

•! 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?


! =Msun

4

3" Rsun

3

=1.4gcm#3

! =MSirius

4

3" RSirius

3

= 0.76gcm#3

! =MBet

4

3" RBet

3

=10#11gcm

#3

The Sun (G2V):

Sirius (A1V):

Betelgeuse (M2I):

Average density of stars:

Betelgeuse has an average density that is 100,000

times less dense than the air we breathe!

week 1: chapter 8 in carroll and ostlie

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