What do we notice when we look at the night sky?

(1) Stars have different brightnesses (fluxes).  
This is a function of 

 -the intrinsic power (luminosity) of the stars 
 -their distance from us

(2) Stars have different colors.  The color depends 
on the surface temperature of the star; that is, 
how hot the surface is that is sending photons on 
their way toward us


Review: How is the continuum of a star formed? 
----------------------------------------------
	
Photons diffuse through a very hot, dense interior
 
The atoms in the interior are hitting one another 
so hard that much of the interior is completely ionized

All the atoms and free electrons are jostling against 
each other and interacting with the photons

Eventually, the photons reach a region sufficiently
low in density that they can escape, and the result 
is a very smeared spectrum---a smooth continuum

Recall that stars behave roughly as “blackbodies”:  
they are self-luminous, and the shape of the emitted 
spectrum depends mostly on their surface temperature


Thermal Radiation
-----------------

Wavelength of the peak is a measure of the star’s 
surface temperature. As the temperature increases, the 
peak wavelength decreases 

Recall:  temperature is a measure of how quickly the 
particles are jostling around

The higher the temperature, the greater the average 
speed of the particles

Wavelength of the peak determines the color of the star

If a star peaks at red wavelengths, then it basically 
looks red

Our eyes perceive as white the distribution of wavelengths 
that peaks in the green but has non-zero contributions 
elsewhere

Hotter stars also have higher intensity curves than cooler 
stars, which is characteristic of objects whose spectra 	
depend only on their temperature


Temperature and Balmer Lines
----------------------------

Now, in the outermost parts of the star, the atoms are 
reasonably cool and not moving all that fast; they are 
either neutral or missing only 1 or 2 electrons

These cool atoms, which still retain some of their 
electrons, can absorb the hot continuum produced from 
within the star and absorb that continuum at specific 
wavelengths

As the photons that make up the continuum spectrum fly 
outwards, some are absorbed by atoms in the cooler outer 
layers of the stellar atmosphere, called the photosphere 

Consequence:  absorption lines get imprinted on the 
continuum spectrum

Dark lines show where atoms in the photosphere have 
“stolen” the incident photons at specific wavelengths

There does not exist a simple relationship between the 
strength of a line and the abundance of its species

Rather, the strength of a line is determined by the
surface temperature of the star

Why?  Well, we know that an atom can only absorb light 
if the energy of the incoming photon exactly matches the 
difference between the atom’s current energy level and a 
higher energy level

Let’s suppose there are a lot of hydrogen atoms, and 
their electrons are all sitting in the 2nd energy level

Now a beam of white light is shone through the gas of 
hydrogen atoms

Let's focus on the incoming green photons  

A fair fraction of the electrons will absorb green 
photons and jump to the 4th energy level

This transition will remove green photons (486 nm) 
from the star's spectrum

The result is an absorption line at this wavelength, 
which we call the "Balmer beta" or "hydrogen beta" line

However, if a hydrogen atom is NOT in the proper n=2 
state, then this absorption will not occur 

For example, if the atom is in its ground state, n=1, 
then any photon with less than 10.2 eV will pass right 
through it unaffected 

Likewise, if the hydrogen atom initially is in a state 
higher than n=2, it will again be unable to absorb the 
green photon

So, in order to produce an absorption line, two 
conditions must be met: 

(1) there must be lots of the proper species (for ex, 
neutral hydrogen atoms) 

(2) the atoms must be in the proper initial energy level 
(for example, n=2 to produce Balmer absorption lines) 

Thus, we ought to see very strong Balmer absorption 
lines in the spectrum of a star if a large fraction 
of its hydrogen atoms are in the n=2 energy state

But what determines how many atoms are in the n=2 energy 
state? 

As the temperature rises in a gas, the atoms move more 
rapidly and collide more violently

So, as the temperature rises, atoms are . . .

more likely to be excited from the ground state n=1 
to the state n=2

. . but they are ALSO more likely to be ionized, and 
ionized hydrogen does not produce discrete absorption lines

The temperature required to ionize most of the hydrogen 
atoms is surprisingly low:  between 8000 and 12,000 Kelvin, 
the gas changes from mostly neutral to mostly ionized

So, producing strong Balmer lines is like producing 
good porridge

If the star’s outer atmosphere is TOO COLD, then all the 
hydrogen atoms are in the ground state n=1, from which 
they cannot absorb Balmer photons

If the atmosphere is TOO HOT, then most of the hydrogen 
atoms lose their electrons (they are ionized) and cannot 
bsorb any photons

We need the temperature to be JUST RIGHT to place the 
maximum fraction of hydrogen atoms in the right state 
to absorb Balmer photons


Spectral Classification of Stars
--------------------------------

We said at the beginning of the lecture that the surface 
temperature of a star is reasonably well determined by the 
peak in its spectrum

BUT, we now see that it can be even more precisely determined 
by looking at what absorption lines are present in the star’s 
spectrum!

This means that we can classify stars according to their 
spectral features . . .

. . . and end up with a classification by surface temperature! 
(see Table 13.1 in your book)

O stars have T > 30,000 K
B stars have 11,000 < T < 30,000 K
A stars have 7,500 < T < 11,000 K
F stars have 6,000 < T < 7,500 K
G stars have 5,000 < T < 6,000 K
K stars have 3,500 < T < 5,000 K
M stars have T < 3,500 K

Our Sun is a G star, but by far the majority of stars in our 
Milky Way galaxy are M stars

The O stars have spectra with weak Balmer lines of hydrogen 
and lines of ionized helium

The A stars (~10,000 K) have the strongest Balmer lines

The Balmer lines fade in the F stars, and metal lines appear

Q: Why are the lines weak in O stars? 
A: The temperature is so high that most hydrogen atoms are 
ionized; they cannot absorb any photons

Q: Why are the lines strong in A stars? 
A: The temperature is just right to put the largest fraction 
of hydrogen atoms into the n=2 state, from which they can 
bsorb Balmer photons 

Q: Why are the lines weak in K stars? 
A: The temperature is so low that almost all hydrogen atoms 
are in the ground state n=1, and very few hydrogen atoms are 
in the n=2 state. Thus, the atoms cannot absorb Balmer photons


Temperature vs Luminosity
-------------------------

Surface temperature and luminosity are two of the fundamental 
properties we use to study the nature of stars

Are all combinations of luminosities and surface temperatures 
possible?

Maybe there is something in the physics of stars that does 
not allow any arbitrary combination of the two?

Around 1910, two astronomers, Ejnar Hertzprung and Henry 
Russell, independently plotted the derived luminosities vs 
surface temperatures for the nearest few hundred stars whose 
parallaxes had been measured

Now when astronomers determine the

LUMINOSITY (from the apparent brightness and the distance)
and SURFACE TEMPERATURE (from the spectrum)

for many stars and plot L against T, they call it a
Hertzsprung-Russell diagram

Stars do not scatter randomly across the H-R diagram

Most stars fall in a relatively tight diagonal, from upper 
left (luminous, hot O stars) to lower right (low luminosity, 
cool stars), the “main sequence”

Some stars do not fall on the main sequence
(giant and supergiant stars, white dwarfs)

Compare a star with a surface temperature much like the Sun 
with a star that is 100 times more luminous

How could a star have the same surface temperature as our 
Sun but be much more luminous?  It has to be a LOT bigger!  

The Stefan-Boltzmann law tells us that per unit surface area, 
the amount of energy emitted is proportional to T^4

But if the stars have the same surface temperature, then the 
amount of energy emitted per unit area is the same

So, the more luminous star has to have a bigger surface 
area (and hence radius) than the less luminous star!

Ditto for the supergiants; if we take one with the same 
surface temp as the Sun, it can be 1000-100000 times more 
luminous, which means it must have a really huge surface area

Just how huge are supergiants?  If the Earth were orbiting 
Betelgeuse at our current distance from the Sun, it would 
be inside the star!  The star is, in fact, greater in size 
than Jupiter’s orbit

We will see later that Betelgeuse is on its way to a violent, 
explosive death that will disintegrate most of the star, 
leaving a very weird object at the center

Now let’s consider the other end of the scale:  white dwarfs

They can have temperatures hotter than the Sun, so per unit 
area, they emit MORE than the Sun emits

However, they are only 1/100th to 1/1000th as luminous as 
the Sun, which means they must have a smaller surface area 
(and hence radius)

White dwarfs are only about the size of Earth

We will see in future lectures that these various classes 
of stars are in different stages of their evolution

Main sequence stars are in their prime

Giants, supergiants, and white dwarfs are evolved stars, 
close to their deaths

Our own Sun near the end of its life will become a giant 
and will then end up as a white dwarf that slowly cools 
to oblivion


Spectroscopic Distances
-----------------------

We can use the H-R diagram to infer approximate distances 
to stars:

First, determine the spectral type of the star, which gives 
the temperature

Then, find the luminosity from the H-R diagram

Now, measure the flux of the star, and use that, the 
luminosity, and the inverse square-law to find the distance

How does one deal with the fact that there is a wide range 
of luminosities for a given spectral type (temp)?

The star’s spectrum can tell us (recall that the strength 
of the dark lines relate to a star’s temperature)

For gases at the same temperature, the rate of collisions 
depends on the density of the gases:  in a denser gas, 
collisions are more common

The more frequent collisions in a main sequence star makes 
certain absorption lines appear broader than in giants, 
giving the star’s size (and hence luminosity)


Main Sequence
-------------

We understand the main sequence as stars of different masses 
fusing hydrogen into helium

Higher mass means a larger inward force of gravity

To balance this, need a large central pressure, which means
a high central temperature

A high temperature means faster nuclear reactions, generating 
more luminosity

With higher mass, the surface temperature increases and the 
luminosity increases rapidly

Roughly, the luminosity of a main sequence star scales as 
the fourth power of the mass:  L proportional to M^4


Main Sequence Lifetimes
-----------------------

More massive stars have larger reservoir of fuel (hydrogen) 
for nuclear fusion

BUT they burn it much much faster

This means that massive stars have shorter lives
than low mass stars

Quantitatively, the fuel supply is proportional to the mass, 
and the lifetime is proportional to the fuel supply divided by
the luminosity, which means the lifetime is proportional
to 1/(M^3)

For example, a 2 Solar mass star lives for roughly 1/(2^3)
(one eighth) as long as the Sun

Some numbers from computer models of stellar evolution:
 
0.25 Solar masses: 1000 billion (a trillion!) years

1 Solar mass: 10 billion years

10 Solar masses: 10 million years

Short lifetimes of massive stars are crucial to 
the existence of the Earth - it allow products of 
nuclear burning to be recycled into the Galaxy 
and form new generations of stars