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A short course in
Stellar Evolution
Dr. Maura McLaughlin
West Virginia University
[email protected]
July 8 2008
Pulsar Search Collaboratory
Outline
• What is a star? (5 minutes)
• Stellar properties (20 minutes)
• The HR diagram (20 minutes)
• The life cycle of a star (20 minutes)
• Supernovae (10 minutes)
• Compact objects (20 minutes)
The Amazing Power of Starlight
By analyzing the light received from a star,
what can we learn?
1. Brightness
2. Distance
3. Temperature
4. Composition
5. Size
6. Masses
Brightness
Apparent magnitude scale
m1 - m2 = -2.5log10(I1/I2)
Magn. Diff.
Intensity Ratio
1
2.5
2
2.5*2.5 = 6.25
…
…
5
(2.5)5 = 100
Betelgeuse
Magnitude = 0.41 mag
Rigel
Magnitude = 0.14 mag
For a magnitude difference of 0.41 –
0.14 = 0.27, we find an intensity ratio
of (2.5)0.27 = 1.28
Distance
1 parsec (3.23 light years)
is an angle of 1 arcsecond
Distance in parsecs =
1 / (Angle in arcseconds)
Betelgeuse
Magnitude = 0.45 mag
Absolute magnitude an
object would have at 10 pc
M = m - 5(log10D - 1)
Rigel
Magnitude = 0.18 mag
Distance
Back to our example of
Betelgeuse and Rigel:
Betelgeuse
Rigel
mV
0.41
0.14
MV
-5.5
-6.8
d
152 pc
244 pc
Difference in absolute magnitudes:
6.8 – 5.5 = 1.3
=> Luminosity ratio = (2.512)1.3 = 3.3
Betelgeuse
Rigel
Question
An astronomer measures a stellar parallax
to be 0.25 milli-arcseconds. The
distance to the star is therefore…
(a) 4 parsecs
(b) 4 kilo-parsecs
(c) 4 Mega-parsecs
(d) 4 Giga-parsecs
Luminosities
Luminosity is the amount of energy a body
radiates per unit time. For two objects of
Betelgeuse
the same apparent brightness,
Magnitude = 0.45 mag
Inverse
square law
2
Rigel
L1/L2=(D
D
)
1/ 2
Magnitude = 0.18 mag
Question
The Sun has an apparent magnitude of
-26.7 as measured from Earth. What
would the apparent magnitude be as
measured from Saturn (10 AU)
(a) -26.7
(b) -31.7
(c) -21.7
(d) -16.7
Temperatures
The light from a star is usually
concentrated in a rather narrow
range of wavelengths.
The spectrum of a star’s light is
approximately a thermal spectrum
called a blackbody spectrum.
A perfect blackbody emitter
would not reflect any radiation.
Thus the name ‘blackbody’.
Betelgeuse
Magnitude = 0.45 mag
Rigel
Magnitude = 0.18 mag
Temperatures
1. The hotter an object is, the more luminous it is.
Intensity is proportional to T4.
Betelgeuse
Magnitude = 0.45 mag
2. The peak of the black body spectrum shifts
Rigel
towards shorter wavelengths when the temperature
Magnitude = 0.18 mag
increases.  Wien’s displacement
law:
lmax ≈ 3,000,000 nm / TK
(where TK is the temperature in Kelvin).
Temperatures
The color of a star is
measured by comparing its
brightness in two different
wavelength bands.
The bluer a star appears, the
smaller the color index B – V.
The hotter a star is, the
smaller its color index B – V.
B band
V band
Question
Orion
Betelgeuse
Which star has a smaller
color index??
Rigel
Compositions
Spectral lines are the
absorption of photons at
discrete energy levels.
Temperatures too high
produce higher energy
photons which ionize all
hydrogen.
Temperatures too low do
not produce enough
excitation into higher
energy states.
Compositions
Spectral Classes
Spectral Classes
Sun is a G2 star!!
Mnemonics to remember
the spectral sequence:
Oh
Oh
Only
Be
Boy,
Bad
A
An
Astronomers
Fine
F
Forget
Girl/Guy
Grade
Generally
Kiss
Kills
Known
Me
Me
Mnemonics
Sizes
We already know that hotter stars are brighter.
But brightness also increases with size!
A
B
Quantitatively:
Star B will be
brighter than star
A, even if they have
the same
temperature.
L = 4 p R2 s T4
Surface area of the star
Surface flux due to a
blackbody spectrum
Question
Polaris has just about the same spectral type
(and thus surface temperature) as our sun,
but it is 10,000 times brighter than our sun.
How much larger is it than our Sun?
Masses
Recall Kepler’s 3rd Law:
Py2 = aAU3
Valid for the solar system: star
with 1 solar mass in the center.
We find almost the same law for
binary stars with masses MA and
MB different from 1 solar mass:
3
a
____
AU
M A + MB =
Py2
(MA and MB in units of solar masses)
Examples:
a) Binary system with period of P = 32
years and separation of a = 16 AU:
3
16
____
MA + M B =
= 4 solar masses.
2
32
b) Any binary system with a combination of
period P and separation a that obeys
Kepler’s 3rd Law must have a total mass of 1
solar mass.
c) Since we also know MA/MB = rA/rB, we can
solve for masses.
What can we learn about the Sun?
• Average star
• Spectral type G2
• Only appears so bright because it is so close.
• Absolute visual magnitude = 4.83 (magnitude if it were
at a distance of 32.6 light years)
• 109 times Earth’s diameter
• 333,000 times Earth’s mass
• Consists entirely of gas (av. density = 1.4 g/cm3)
• Central temperature = 15 million K
• Surface temperature = 5800 K
Organizing the Family of Stars: The
Hertzsprung-Russell Diagram
Example: useful way to understand populations
Organizing the Family of Stars: The
Hertzsprung-Russell Diagram
We know:
Stars have different temperatures, different
luminosities, and different sizes.
To bring some order into that zoo of different types
of stars: organize them in a diagram of
Absolute mag.
or
Luminosity
Luminosity versus Temperature (or spectral type)
Hertzsprung-Russell Diagram
Spectral type: O
Temperature
B
A
F
G
K
M
The Hertzsprung Russell Diagram
Most stars are
found along the
main sequence
The Hertzsprung Russell Diagram
Same
temperature,
but much
brighter
than MS
stars
 Must be
much larger
Same temp., but
fainter → Dwarfs
 Giant
Stars
Radii of Stars in the HR Diagram
100 times smaller
than the Sun
Masses of
Stars in the
HertzsprungRussell Diagram
The higher a star’s mass,
the more luminous
(brighter) it is:
L ~ M3.5
High-mass stars have
much shorter lives
than low-mass stars:
tlife ~ M-2.5
Sun: ~ 10 billion yr.
10 Msun: ~ 30 million yr.
0.1 Msun: ~ 3 trillion yr.
A Census of the Stars
Faint, red dwarfs
(low mass) are the
most common
stars.
Bright, hot, blue
main-sequence
stars (high-mass)
are very rare.
Giants and
supergiants are
extremely rare.
Life Cycle of a Star
?
The Contraction of a Protostar
Evidence of Star Formation
Nebula around S
Monocerotis (Foxfur)
Contains many massive,
very young stars,
including T Tauri stars:
strongly variable; bright
in the infrared.
Energy generation in the Sun:
Nuclear Fusion
Basic reaction:
4 1H  4He + energy
4 protons have more mass
than 4He.
 Energy gain = Dm*c2
What reaction rate does
Sun need to resist its
own gravity?
Need large proton speed to
overcome Coulomb barrier
(electromagnetic repulsion
between protons).
T ≥ 107 K =
10 million K
Hydrostatic
Equilibrium
Outward pressure force
must exactly balance the
weight of all layers above
everywhere in the star.
This condition uniquely
determines the interior
structure of the star.
This is why we find stable stars
on such a narrow strip (main
sequence) in the HR diagram.
Question
Why are high temperatures needed for nuclear fusion?
a) electrons must be moving quickly to change orbitals
b) the high temperature is necessary to fully cook the new
atom
c) protons need high speeds to overcome repulsive force
d) protons need high speeds to overcome gravity
Masses of Main-Sequence Stars
Stars with
masses greater
than 100 solar
masses are
unstable.
Stars with masses
less than 0.08 solar
masses cannot fuse
hydrogen in their
cores - brown dwarfs.
The Life of Main-Sequence Stars
Stars gradually
exhaust their
hydrogen fuel.
In this process of
aging, they are
gradually becoming
brighter, evolving
off the zero-age
main sequence.
Death line
The Lifetimes of Stars on
the Main Sequence
Expansion onto the Giant Branch
Once Hydrogen in the core is
completely converted into He,
H burning continues in shell.
Lots of energy produced ->
star expands to form Red
Giant.
Giants and supergiants are 101000 times larger than the
Sun and 10 - 106 times less
dense.
Expansion cools the star and
makes it more luminous.
Sun will expand beyond Earth’s orbit!
Question
After Hydrogen burning in the core of the
Sun ends, its SURFACE temperature will
________ and its size will __________.
(A) increase; decrease
(B) decrease; increase
(C) increase; increase
(D) decrease; decrease
Red Giant Evolution
H-burning shell keeps
dumping He onto the core.
He core gets denser and
hotter until the next
stage of nuclear burning
can begin in the core:
He fusion through the
“triple-alpha process”:
4He
+ 4He  8Be + g
+ 4He  12C + g
The onset of this
process is termed the
8Be
helium flash
Hyades Star Cluster
Open cluster in constellation Taurus. About 100 stars.
HR Diagram of Hyades
High-mass stars
evolved onto the
giant branch
Turn-off point
Low-mass stars
still on the main
sequence
Estimating
the Age of
a Cluster
The lower on
the MS the
turn-off
point, the
older the
cluster.
Endpoints of stellar evolution…
Depends almost completely on its mass.
Let’s start with the least massive stars
and go to the most!
Red Dwarfs (0.08 - 0.4 solar masses)
Stars with less
than ~ 0.4 solar
masses are
completely
convective.
Hydrogen and
helium remain
well mixed
throughout the
entire star.
No phase of shell “burning” with expansion to giant.
Star not hot enough to ignite He burning.
Could live for 100 billion yrs or more! Universe only 14
billion years old so can’t test this though…
Sunlike Stars
Sunlike stars
(~ 0.4 – 8
solar masses)
develop a
helium core.
 Expansion to red giant during H burning shell phase
 Ignition of He burning in the He core
 Formation of a degenerate C,O core
The Fate of our Sun
and the End of Earth
• Sun will expand to a red
giant in ~ 5 billion years
• Expands to ~ Earth’s orbit
• Earth will then be
incinerated!
• Sun may form a planetary
nebula (but uncertain)
• Sun’s C,O core will become
a white dwarf
Low luminosity; high temperature => White dwarfs are found
in the lower center/left of the H-R diagram.
White Dwarfs
Degenerate stellar remnant (C,O and degenerate electrons)
Matter in the He core has
no energy source left.
 Not enough thermal
pressure to resist and
balance gravity
Pressure in degenerate
core is due to the fact
that electrons can not be
packed arbitrarily close
together and have small
energies.
White Dwarfs
Extremely dense:
1 teaspoon of white dwarf material: mass ≈ 16 tons!!!
Chunk of white dwarf material the size of a
beach ball would outweigh an ocean liner!
white dwarfs:
Mass ~ Msun
Temp. ~ 25,000 K
Luminosity ~ 0.01 Lsun
The Chandrasekhar Limit
The more massive a white dwarf, the smaller it is.
 Pressure becomes larger, until electron degeneracy
pressure can no longer hold up against gravity.
WDs with more than ~ 1.4 solar
masses can not exist!
The Deaths of Massive Stars:
Final stages of fusion in high-mass stars (> 8 Msun), leading
to the formation of an iron core, happen very rapidly: Si
burning lasts only for ~ 1 day.
The Deaths of Massive Stars:
Iron core ultimately collapses, triggering an explosion
that destroys the star:
Supernova
The Famous Supernova of 1987:
Supernova 1987A
Before
At maximum
Unusual supernova in the Large
Magellanic Cloud in Feb. 1987
Where do we get elements
heavier than Iron???
Only in Supernovae!!! The huge heat generated in the
explosion makes elements like Gold, Silver and Platinum.
Cas A Supernova Remnant
Only 300 years old -> youngest known remnant in MW!
Cas A Supernova Remnant
What is that?!
Only 300 years old -> youngest known remnant in MW!
Neutron Stars
A supernova The central core will
Pressure becomes so high
collapse into a
explosion of
that electrons and
compact object
an M > 8 Msun
protons combine to form
supported by
star blows
stable neutrons
away its outer neutron degeneracy
throughout the object.
pressure.
layers.
Typical size: R ~ 10 km
Mass: M ~ 1.4 – 3 Msun
Density: r ~ 1014 g/cm3
 Piece of neutron
star matter of the
size of a sugar cube
has a mass of ~ 100
million tons!!!
Black Holes
Just like white dwarfs
(Chandrasekhar limit: 1.4
Msun), there is a mass
limit for neutron stars:
Neutron stars can
not exist with
masses > 3 Msun
We know of no
mechanism to halt the
collapse of a compact
object with > 3 Msun.
It will collapse a singularity: => A black hole!
Escape Velocity
Velocity needed to
escape Earth’s gravity
from the surface:
vesc
vesx ≈ 11.6 km/s.
Now, gravitational force
decreases with distance
(~ 1/d2) => Starting out
high above the surface =>
lower escape velocity.
If you could compress
Earth to a smaller radius
=> higher escape velocity
from the surface.
vesc
vesc
The Schwarzschild Radius
=> There is a limiting radius
where the escape velocity
reaches the speed of light, c:
___
Rs = 2GM
c2
Vesc = c
G = gravitational constant
M = mass
Rs is called the Schwarzschild radius.
Schwarzschild Radius
and Event Horizon
No object can travel
faster than the speed
of light.
=> Nothing (not even
light) can escape from
inside the
Schwarzschild radius.
 We have no way of
finding out what’s
happening inside the
Schwarzschild radius.
 “Event horizon”
What is the Schwarzchild radius of a typical human being?
Question
What would happen to the Earth’s orbit if the Sun were
replaced by a 1 solar mass black hole?
A) Nothing at all.
B) We would slowly (on a timespan of years) get
sucked into the black hole.
C) We would immediately (on a timespan of
seconds) get sucked into the black hole.
D) Our orbit would become larger because of the
smaller object at the center.
Observing Black Holes
No light can escape a black hole
=> Black holes can not be observed directly.
If an invisible compact
object is part of a
binary, we can
estimate its mass from
the orbital period and
radial velocity.
Mass > 3 Msun
=> Black hole!
Compact object
with > 3 Msun must
be a black hole!
Review - Compact Objects
Sunlike stars (masses 0.4 - 8 solar masses)
- become white dwarfs (R ~ 10,000 km)
- maximum mass limit of 1.4 solar masses
- supported by electron degeneracy pressure
- teaspoon of WD material would weigh as much as 100 elephants
More massive stars (roughly 8-10 solar masses)
- become neutron stars (R ~ 10 km)
- maximum mass limit of 2-3 solar masses
- supported by neutron degeneracy pressure
- teaspoon of NS star material would weigh as much as a
supertanker
Most massive stars (roughly > 10 solar masses)
- become black holes (Rs ~ 10 km)
- no maximum mass limit
- too massive to be supported by degeneracy pressure