Download Lecture 18

ASTR 200 : Lecture 18
Solar Stars : Pre- and Postmain-sequence evolution
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Midterm exam
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One week from today. In lecture, October 28
Will contain short answer questions and problems
Formula sheet will be provided
No calculators
Will cover lectures and associated textbook readings
listed on the course web site up to and including
TODAY. Friday Oct 21.
• A previous year's midterm has been posted, but it is a
different format and course topic order. No solution
set will be provided.
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Collapse phase of stars
• We have seen that cold interstellar clouds of order 0.1
pc in scale collapse to form stars
• The free-fall time was ~105 years, and the clouds
must span 0.01-100 solar masses, with smaller
masses being much more common
• The contracting
cloud forms a disk,
with a central
condensation called
a protostar.
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Collapse phase : energy source
• As the clouds collapse to smaller sizes, they heat up
due to Kelvin Helmhotz heating
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If
it
was
all
hydrogen,
(M/m
)kT
~
GM
/R
•
p
– mp is proton mass, M and R are the cloud's mass & radius
– T ∝M /R , although half potential energy--> rotation
• So as the cloud shrinks by a
factor of 10, T rises by a
factor ~10
• 10K when R~10,000 au,
to T~10,000K at 1 au
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- (BUT, a lot of the heat is
radiated away DURING the
contraction, so T of central
protostar is more ~3000 K)
Protostars : `reasonably hot' and big
• With T~3000 K, where is the peak of the blackbody?
– Wein's law: peak~ 1000 nm ~ 1 micron....infrared
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• As the solar mass protostar's T reaches 3000 K, it is
also shrinking, with a radius ~0.1 au = 20 solar radii
• Convince yourself
that this implies a
luminosity of
about 30 solar
luminosities (!)
• Right: protostars
glowing strongly
in IR in clouds
On the HR diagram, stars move
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• Low mass stars
contract initially
and then reach
main sequence
after 0.1-1 Myr
• 1 solar mass star
reaches ignites
core fusion and
settles to internal
equilibrium
where surface
T~6000 K
After ~10 Gyr of stability, the star
consumes the H in its core into He
• For a solar mass star, the
post main-sequence
evolution results in the star
leaving the main sequence
during the phase where it
burns He in core to carbon
• During this phase, becomes
giant briefly, with radius
100 solar radii and slightly
cooler surface T
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• With masses < about 3 solar
masses, the stars mass is
insufficient to cause a core
T high enough to burn
carbon
This is not going to
be.....good....
In ~5 Gyr the Sun will leave
the main sequence and reach a radius of about 1 au
The Earth will then be at roughly the solar surface of T~3000 K, and vaporize
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This is not going to
be.....good....
In ~5 Gyr the Sun will leave
the main sequence and reach a radius of about 1 au
The Earth will then be at roughly the solar surface of T~3000 K, and vaporize
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The C core contracts, but won't ignite
• Instabilities late in the red giant's life push the less
dense outer layers (beyond the core) of the star away
• In the meantime, the core
is contracting. This
contraction releases yet
more gravitational
energy, so much in fact
that the outward pressure
pushes all the layers
outside of the core out
into space, making a
planetary nebula
– (has nothing to do
with planets!)
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So if there is no fusion, there is no
energy generation in the core...
• Gas pressure gets less and less as the very hot core
radiates away its heat...why doesn't the star
collapse to a point (a 'singularity', or black hole?)
– Because another pressure, besides gas pressure or
radiation pressure, takes over
• As the star contracts, it becomes so dense that
electrons begin to exert a new form of 'pressure'
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Degenerate electron pressure
• These electrons exert pressure at the rate they transfer
momentum to a unit surface area.
– For one electron, this is (p/3) v/V, where v is the electron
speed v=(p/me) if non-relativistic electrons
– The 'number of states available' is V/h3, multiplied by 2
because two electrons can occupy each momentum state
p
2 V p pv
8π
2
4
P= 3 ∫0
4 π p dp= 3 ∫0 p dp
3V
h
3 h me
f
f
• Integrating, one finds
2 /3
8π
1 3
5
P=
pf = ( π )
3
20
15 h m e
14
2
h 5/3
ne
me
Equation of state: Degenerate matter
• The electrons thus exert a `quantum mechanical' pressure. If
this pressure exceeds the gas pressure (from the ideal gas law)
we say that the electrons are `degenerate'
• The number density ne of electrons is proportional to the
number of protons (n e ∝ρ ) so the (non-relativistic) equation of
5/3
state for degenerate matter has :
P∝ρ
– Different proportionality than ideal gas law: P∝ρT
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• The degenerate equation of state has NO temperature
dependence! Degenerate electrons create pressure even if the
temperate cools arbitrarily low...
• With this equation of state, and hydrostatic equilibrium, we
have two equations and two unkowns (P and) so we can
solve for the structure of the star
Radius of a white dwarf: pressure
• Hydrostatic equilibrium told us that an estimate of the central
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pressure of the object will be:
3 GM
P( 0)≈
4
4π R
• Now the degenerate electron pressure was:
2 /3
1 3
Pe = ( π )
20
2
h 5/3
ne
me
• We need ne . How many electrons are there per unit volume?
A white dwarf if composed mostly of carbon (and sometimes
oxygen); both give 1 electron per 2 mH when ionized, so
M
3M
n e≃
=
2m H V 8 π m H R3
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Radius of a white dwarf: Mass
• Equating the central pressure to be electron degeneracy, and
solving for R gives
2
−1 /3
M
RWD ≃0.24
(
)
2
G memH mH
ℏ
• This is a factor of 4 too low. A more accurate analysis gives
2
−1/ 3
M
RWD ≃0.95
(
)
2
G me mH mH
ℏ
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• Amazingly, once the total mass M is set, then given the mass
of the electron and proton, the white dwarf's size is determined
by Planck's constant
• Note that R depends inversely on the cube root of the mass
– Increase M by 8x, and the radius DROPS by a factor 2!
Mass-radius relation
• Evaluating all the constants yields
−1/ 3
M
RWD ≃0.009 (
)
M sun
R sun
• For reference, 0.009 Rsun ~ Rearth !
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• The mean electron number density ne ~ 10 electrons/m
• Of course, the mass is in the protons and neutrons, not
electrons
– 1 solar mass in 1 earth radius is about 2 tonnes/cm3
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The Sirius A and B system
• 1844: German astronomer F. Bessel realized that
Sirius A (brightest star in sky!) must have a
companion, based on wobble in its proper motion
• 1862: American astronomy A. Clarke directly
detects the faint Sirius B
• Surface temperatures are not too
different, but LA ~ 10,000 LB
Modern HST picture
A
– Thus RA ~ 100 RB
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• The small luminosity limits how
quickly the WD can radiate its
leftover heat
B
White Dwarf cooling
• Once they shed their planetary nebula,
the central star is initially very hot,
with T>40,000 K
• However, it still obeys the usual
luminosity relation and radiates into
space
• As T drops, the total
luminosity does too, and so
the WD gets cooler and less
luminous
• Takes billions of years
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Central WD
White Dwarves → Black dwarves
• There is no end to this process, the white dwarves will
eventually cool to nearly zero absolute temperature
• Such a star is
coined a
`black dwarf'
• However, none
exist yet, because
even the very
oldest stars have
not had time to
cool their surface
T<1000 K
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White dwarf cooling
Age (Gyr) after planetary nebula