Some Later Stages of Low- Mass 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 hot, it can undergo nuclear fusion And shortly thereafter, it can add one more 4He to make 16O 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 3 4He  12C +  12C + 4He  16O + 

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 Consider first a single particle in a box of size V = L3 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 Because of Pauli exclusion principle, each particle needs its own volume (sort of) The total volume is Therefore we have L

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 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 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 The force between the two halves is about Pressure is force over area Area of circle is R2 ~ R2 This must match degeneracy pressure Number density is number of particles over volume Volume ~ R3 Number of particles is mass divided by mass per particle R R

Radius vs. Mass for White Dwarfs (2) M = mass of star m = mass of particle causing pressure  = mass per degen- erate particle Equate these two expressions Solve for R For a white dwarf, the electron causes the pressure  is the mass per electron Partly 12C, which would have Partly 16O, which would have Clearly, Write the mass in terms of the Sun’s mass

Radius vs. Mass for White Dwarfs (3) If you do all the differential equations 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 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

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 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 + + p+ n0 e- 

Radius vs. Mass for Neutron Stars (1) We can – approximately – use the same formulas we did for white dwarfs For the neutron star, the particle causing pressure is the neutron For a neutron star, this is also , the mass per neutron As before, write it in terms of the Sun’s mass Assuming our formula is right, we then find M = mass of star m = mass of particle causing pressure  = mass per degen- erate particle

Radius vs. Mass for Neutron Stars (2) 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) 5 10 15 20 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)