Transcript Dead Stars

Dead Stars
Some Later Stages of Low- Mass Stars
•Eventually, hydrogen at the core gets completely used up
–Core is now pure 4He “ash”
–It continues burning 1H  4He in a thin shell
–This star becomes a red giant
•After a while, the 4He becomes so
3 4He  12C + 
hot, it can undergo nuclear fusion
•And shortly thereafter, it can add one
12C + 4He  16O + 
4
16
more He to make O
•In the Sun, this is the last nuclear process that occurs
•The Sun eventually loses almost all its 4He and 1H and ends up as a burnt-out
carbon/oxygen star
•No longer able to produce new energy it loses heat slowly forever
•It becomes a white dwarf
White Dwarfs
•White dwarfs are made primarily of 12C mixed with 16O
•They are no longer undergoing nuclear fusion
– Therefore, they are not in a balance of energy lost vs. energy generated
•They are not held up by ideal gas pressure
•Instead, the Pauli exclusion principle creates pressure
– No two particles of the same type can be in the same quantum state
– Therefore, as you add more and more to a small volume, they must go
into higher energy states
•Each type of fermion creates its own degeneracy pressure
•Degeneracy pressure exists even at zero temperature
•It has nothing to do with interactions between particles
Degeneracy Pressure (1)
•We will do an order of magnitude calculation to find the degeneracy pressure
– Similar to what you did in a homework problem
1/3
3
L

V
•Consider first a single particle in a box of size V = L
•The energy of a particle in
such a box is given by
•The minimum energy is
•Write in terms of volume
•Pressure is the derivative of
energy with respect to volume
2
2
2
2
2
n

n

n
y
z 
2  x
2mL
3 2 2
3 2 2

E
2
2mV 2/3
2mL
2 2
E

P
mV 5/3
V
E
L
L
L
•Because of Pauli exclusion principle, each particle needs its own volume (sort of)
•The total volume is V  NV
5/3
2
2
2 2 5/3
tot
 N 


n
•Therefore we have
Pd 


Pd 
m  Vtot 
m
Degeneracy Pressure (2)
•Actual computation is more complicated
– Take into account two spin states
– Particles actually share a volume
– Make assumption of many particles
•Correct formula works out to
Pd 
Pd 
2
5m
 2 2 n5/3
3
m

2 23
n5/3
•Note that calculation is independent of temperature
– Cooling the star doesn’t change degeneracy pressure
•Effect is inversely proportional to mass
– Electrons contribute much more than protons, neutrons, or nuclei
•This effect has nothing to do with particle interactions
– Even neutrinos would have degeneracy pressure
Radius vs. Mass for White Dwarfs (1)
•Order of magnitude estimate for size of white dwarf
R
– Ignore factors of 2, , etc.
•As we did before, assume a star of radius R and mass M
•Divide the star in half
– Each half has mass ½ M ~ M and are separated by a distance R
GMM
•The force between the two halves is about
F
•Pressure is force over area
F
R2
P
•Area of circle is R2 ~ R2
A
GM 2
•This must match degeneracy pressure
P
4
R
•Number density is number of particles over volume
•Volume ~ R3
M
•Number of particles is mass divided by mass per particle N 

P
2
5m
3

2 23
n5/3 
2 5/3
n
m
N
  
m V 
2
5/3
2
5/3
N

mR5
2
M 5/3
P
m 5/3 R5
R
Radius vs. Mass for White Dwarfs (2)
2
M 5/3
P
m 5/3 R5
2
M 5/3 GM 2
GM 2
 4
P
5/3 5
4
m R
R
R
•Equate these two expressions
•Solve for R
R
2
Gm 5/3 M 1/3
M = mass of star
m = mass of particle
causing pressure
 = mass per degenerate particle
•For a white dwarf, the electron
m  9.109 1031 kg
causes the pressure
• is the mass per electron
12  12.000 u  6  2.0000 u
– Partly 12C, which would have
– Partly 16O, which would have 16  15.995 u  8  1.9994 u
•Clearly,   2 u  3.3211027 kg
•Write the mass in terms of the Sun’s mass
M  1.989 1030 kg   M M

Radius vs. Mass for White Dwarfs (3)
1.0546 10 J  s  1.989 10 kg   3.32110

R
 6.674 10 N  m kg 9.109 10 kg 
34
2
30
11
 1968 km   M M
2

2
1/3
27
31
kg 
5/3
 M 


M


1/3
1/3
•If you do all the differential equations R  8800 km  M M 1/3
properly, you actually get
– Compare, REarth = 6370 km
•Note that the more massive the white dwarf, the smaller the star
•This is still not accurate, because for masses > 0.5 MSun, the electrons become
relativisitic
•This causes a faster reduction in the size
•In fact for the Chandrasekhar mass, 1.4 MSun, the star becomes unstable and
undergoes catastrophic collapse
Radius vs. Mass for White Dwarfs (4)
•By the time the Sun dies, it will
have lost a significant fraction of
its mass
– Mf = 0.6 Msun
•Most white dwarfs are in the
range 0.5 – 1.3 MSun
White Dwarf Cooling and the H-R Diagram
•White dwarfs have a relatively small range of
radii and masses
– M = 0.6 – 1.3 Msun
– R = 3,000 – 10,000 km
•As they age, they cool, but their size doesn’t
change
2
L  R  T 

  
L
 R  T 
4
T 
 0.2  2  10  
T 
4
•Their luminosity drops as they age
•They end up in a small strip in lower left side
of H-R diagram
•Young, hot ones near the top
•Old, cool ones near the bottom
4
Later Stages of High-Mass Stars
•Stars lighter than about 8 solar masses eventually shed most of their mass and
end as carbon/oxygen white dwarfs with mass < 1.4 Msun
•Stars heavier than this instead go through a large variety of additional fusion
processes, ending with 56Fe
•They then undergo catastrophic core collapse
•The 56Fe disintegrates back into protons and neutrons
•The degeneracy pressure from the electrons is going crazy
–Electrons have enormous energy because of their low mass
•The electrons are then absorbed by electron capture to make neutrons
•With the electrons gone, the collapse speeds up
n0 + 
p+ + e•But eventually, the neutrons stop the collapse
•The energy released by this fall explodes the rest of the star
–A type Ia supernova
•But the ball of neutrons – a neutron star - remains
Radius vs. Mass for Neutron Stars (1)
•We can – approximately – use the same
formulas we did for white dwarfs
R
2
Gm 5/3 M 1/3
M = mass of star
m = mass of particle
causing pressure
 = mass per degenerate particle
•For the neutron star, the
particle causing pressure
  m  mn  1.675 1027 kg
is the neutron
•For a neutron star, this is also , the mass per neutron
M  1.989 1030 kg   M M
•As before, write it in terms of the Sun’s mass

•Assuming our formula is right, we then find
R
1.0546 10
34
J  s  1.989 10 kg 
2
30
11
2
2
27
6.674

10
N

m
kg
1.675

10


1/3
 M 

8/3 
kg   M 
1/3
 M 
  3.348 km  

M 
1/3
Radius vs. Mass for Neutron Stars (2)
R   3.3 km  M M

1/3
This value is not actually right, because:
• We didn’t actually solve the differential equations
– This increases the radius a bit
• The neutrons are strongly interacting, so ignoring interactions is wrong
• The neutrons are relativistic
– Makes the neutron star unstable if mass gets too large
• The gravity is so strong that you have to take general relativity into account
– See next section of course
• Because of all these effects, there is once again a maximum mass for neutron
stars
• The actual limit is not known very well
– Current estimates put it in the range 2.5 – 3.0 MSun
Radius vs. Mass for Neutron Stars (3)
20
15
10
5
0
•Stars in the range 8 – 30 Msun end up as neutron stars
•Final neutron star mass 1.4 – 3.0 Msun
•More massive stars end as black holes (next section)