AST1100 Lecture Notes

Part 3EThe end state of stars

Questions to ponder before the lecture

1. When helium burning in the core of a star ceases due to lack of helium, what can possiblyhappen further? Clearly hydrostatic equilibrium is lost. The core consists mostly of carbon andoxygen after the helium burning. Can you imagine the next phases in the star’s life?

2. The answer to the previous question depends strongly on the mass of the star. Can you imaginewhy this should be different for a low mass star than for a high mass star?

3. If the core of the star in the end contains mainly iron, can the fusion processes continue? Whynot?

4. Can you imagine what could happen to the iron core when hydrostatic equilibrium is lost andiron cannot fuse to heavier elements?

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AST1100 Lecture Notes

Part 3EThe end state of stars

Before reading this section, you need to readpart 1G again, otherwise parts of this sectionmay be very difficult to understand. We will nowcontinue the discussion on stellar evolution fromthe previous lecture. The star has reached theasymptotic giant branch having a radius of up to1000 times the original radius. The core consistsof carbon and oxygen but the temperature is nothigh enough for these elements to fuse to heavierelements. Helium fuses to carbon and oxygen ina shell around the core. Hydrogen fuses to he-lium in another shell further out. In the outerparts of the star, the temperature is too low forfusion reactions to take place. In the low andmedium mass stars, convection has been trans-porting heavy elements from the core to the sur-face of the star allowing observers to study thecomposition of the core and test stellar evolution-ary theories by studying the composition of ele-ments on the stellar surface. The core is still con-tracting trying to reach a new hydrostatic equi-librium. The further evolution is now stronglydependent on the mass of the star.

1 Low mass stars

We will soon find out how we define low massstars, but for the moment we will only say that atypical low mass star is our Sun. The core of thestar, consisting mainly of carbon and oxygen con-tracts until the density of electrons is so high thatthe core becomes electron degenerate and the de-generation pressure works against gravity to keephydrodynamic equilibrium. In the more massive

’low mass stars’ nuclear fusion may to some extentburn these elements to heavier elements like neonand magnesium. But eventually the core tem-perature is not high enough for further nuclearreactions and the core remains electron degener-ate.

As the star contracts, the temperature in theouter parts of the star increases and the hydro-gen burning again becomes more efficient thanthe helium burning in lower shells. The heliumproduced in the upper shells ’falls’ down on lowershells where no helium burning takes place (he-lium burning takes place even further down inthe hotter areas). After a while, the density inthe lower helium rich shell becomes very highand partially degenerate. At a point, the tem-perature in the lower shell is high enough for anexplosive ignition of helium and a helium shellflash occurs, similar to the helium core flash de-scribed in lecture 20, but less energetic. The flashlifts the hydrogen burning shell to larger distancesfrom the center. Hydrogen burning ceases, thestar contracts until the temperature again is highenough for hydrogen burning. The whole processrepeats, the produced helium falls on to lower lay-ers which finally start burning helium in anotherhelium shell flash. The star is very unstable andthe repeated helium flashes result in huge masslosses from the star. The outer layers of the starare blown away in the helium flashes (this is oneof the theories describing the huge mass lossesthe star undergoes in this period). A huge cloudof gas and dust is remaining outside the core ofthe star. After a few millions years, all the outerlayers of the star have been blown off and only

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Fact sheet: Many factors influence the rate of stel-lar evolution, the evolutionary path, and the natureof the final remnant. By far the most important fac-tor is the initial mass of the star. This diagram illus-trates in a general way how stars of different massesevolve and whether the final remnant will be a whitedwarf, neutron star, or black hole. (Note that for thehighest mass stars it is also possible for the super-nova explosion to obliterate the star rather than pro-ducing a black hole. This alternative is not illustratedhere.) Stellar evolution gets even more complicatedwhen the star has a nearby companion. For example,excessive mass transfer from a companion star to awhite dwarf may cause the white dwarf to explode asa Type Ia supernova. (Figure: NASA/CXC/M.Weiss)

the degenerate carbon/oxygen core remains. Thisstar which now consist only of the remaining de-generate core is called a white dwarf. The sur-rounding cloud of gas which has been blown off iscalled a planetary nebula (these have nothing todo with planets)

Figure 1: Motion in the HR-diagram for the last stagesof stellar evolution. This is the path that a low massM < 8M� star follows. From the asymptotic giantbranch, the outer layers are blown off and the hotter innerparts become the new and hotter ’surface’. The effectivetemperature increases and the star moves from 1 to 2. Asthe layers were nuclear fusion takes place are blown off,the total luminosity decreases as the star is no more ca-pable of producing energy. The star therefore falls downfrom 2 to 3 to the white dwarf area in the HR-diagram.

As the star was blowing away the outer layers, thehotter inner parts of the star made up the surface.Thus, the surface temperature of the star was in-creasing, and the star was moving horizontally tothe left from the asymptotic giant branch in theHR-diagram. Finally when the layers producing

energy by nuclear fusion are blown off, the lumi-nosity of the star decreases dramatically and thestar ends up on the bottom of the HR-diagram asa white dwarf (see HR-diagram in figure 1). Thedegenerate white dwarf does not have any sourcesof energy production and will gradually cool off asthe heat is lost into space. The white dwarf willmove to the right in the HR-diagram becomingcooler and dimmer.

How large is a white dwarf star? We can use theequation of hydrostatic equilibrium to get an es-timate of the radius R assuming uniform densityof the white dwarf (look back at lecture 20 wherewe did a similar approximation in the equation ofhydrostatic equilibrium):

P

R≈ GM

R2

M

(4/3)πR3=

3GM2

4πR5(1)

The pressure P is now the degeneration pressure.Inserting the expression for the degeneration pres-sure from part 1G in this equation gives(

3

π

)2/3h2

20me

n5/3e =

3GM2

4πR4. (2)

The electron number density ne can be written interms of the total gas density as

ne = np =ρpmH

=Z

A

ρ

mH

.

Here Z is the number of protons per nucleus andA is the number of nucleons per nucleus. As thegas in total is neutral, the number of electronsin the gas equals the number np of protons. The

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Fact sheet: At around 215 pc away, the famous Helix Nebula in the constel-lation of Aquarius is one of the nearest planetary nebulae to Earth. The redring measures roughly 1 pc across and spans about one-half the diameter ofthe full Moon. A forest of thousands of comet-like filaments, embedded alongthe inner rim of the nebula, points back toward the central star, which is asmall, super-hot white dwarf. In general, a planetary nebula is an emissionnebula consisting of an expanding glowing shell of ionized gas ejected dur-ing the asymptotic giant branch phase of certain types of stars late in theirlife. They have nothing to do with planet formation, but got their name be-cause they look like planetary disks when viewed through a small telescope.Planetary nebulae have extremely complex and varied shapes, as revealed bymodern telescopes. (Figure: NASA/ESA)

number density, i.e., total number of protons pervolume in the gas equals the mass density ρp ofprotons divided by the hydrogen mass (basicallyequal to the proton mass). The mass density ofprotons in a gas equals the mass density of the gastimes the fraction of the mass in protons given byZ/A. A nucleus typically contains the same num-ber of neutrons as protons such that the totalnumber of nucleons is twice the number of pro-tons and Z/A = 0.5. We will use this number inthe calculations. Inserting this expression for ne

in the equation of hydrostatic equilibrium (equa-tion 2) we have(

3

π

)2/3h2

20me

(Z

AmH

)5/3M5/3

(4/3π)5/3R5=

3GM2

4πR4,

or

RWD ≈(

3

2π

)4/3h2

20meG

(Z

AmH

)5/3

M−1/3.

For M = M� (where M is the mass which re-mained in the degenerate core, the star originallyhad more mass which was blown off) the radiusof the white dwarf is similar to the radius of theEarth. A white dwarf is thus extremely dense,one solar mass compressed roughly to the size ofthe Earth. There is another interesting relation tobe extracted from this equation. Elevating bothsides of the equation to the third power and thenmultiplying the mass from the right to the leftside we have

MR3 ∝MV = constant,

where V is the volume of the white dwarf. Thus,if the mass of a white dwarf increases, the volume

decreases. A white dwarf gets smaller and smallerthe more mass it gets. It shrinks by the addi-tion of mass. This can be understood by lookingat the degenerate equation of state (look back topart 1G): When more mass is added to the whitedwarf, the gravitational inward forces increase.This has to be balanced by an increased pressure.From the equation of state we see that since thereis no temperature dependence, the only way toincrease the pressure is by increasing the density.The density is increased by shrinking the size. Soa white dwarf must shrink in order to increase thedensity and thereby the degeneration pressure inorder to sustain the gravitational forces from anincrease in mass.

Can a white dwarf shrink to zero size if we justadd enough mass? Clearly when the density in-creases, the Fermi energy increases and the energyof the most energetic electrons increases. Finallythe energy of the most energetic electrons will beso high that the velocity of these electrons will beclose to the speed of light. In this case, the rel-ativistic expression for the degeneration pressureis needed (see part 1G). We remember that therelativistic expression for the degeneration pres-sure went as P ∝ ρ4/3 instead of P ∝ ρ5/3 inthe non-relativistic case. Inserting the relativisticexpression in the equation of hydrostatic equilib-rium we thus expect to obtain a different relationbetween radius and mass. Inserting the relativis-tic expression in equation 1 instead we obtain(

3

π

)1/3hc

8

(Z

AmH

)4/3M4/3

(4/3π)4/3R4=

3GM2

4πR4.

We see that the radius cancels out of the equationand we are left with a number for the mass of the

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relativistic white dwarf:

M ≈ 3

16π2−3/2

(hc

G

)3/2(Z

AmH

)2

.

A more exact calculation taking into account non-uniform density would have given

MCh ≈√

3/2

2π

(hc

G

)3/2(Z

AmH

)2

≈ 1.4M�.

This is the Chandrasekhar mass MCh which givesthe upper limit of the mass of a white dwarf. For arelativistic degenerate gas the pressure can onlywithstand the gravitational forces from a maxi-mum mass of M = 1.4M�. If the mass increasesbeyond that, the white dwarf will collapse. Wewill discuss this further in the next section. Hav-ing both the typical mass (or actually the upperbound on the mass) and the typical radius of awhite dwarf we can find the typical density: asmall needle made of white-dwarf material wouldweight about 50 kg.

So we are now closer to the definition of a ’lowmass star’. A ’low mass star’ is a star which hasa mass low enough so that the remaining coreafter all mass losses has a mass less than theChandrasekhar mass 1.4M�. The final result ofstellar evolution for low mass stars is therefore awhite dwarf. It turns out that stars which haveup to about 8M� when they reach the main se-quence will have a core mass lower than the Chan-drasekhar limit. We will now discuss the fate ofstars with a main sequence mass larger than 8M�.

2 Intermediate and high massstars

For stars of mass M > 8M�, the evolution isdifferent. The higher mass makes the pressureand thereby the temperature in the core higherthan in the case of a low mass star. The forcesof gravity are larger and therefore the pressureneeds to by higher in order to maintain hydro-static equilibrium. The carbon-oxygen core con-tracts, but before it gets degenerate the temper-ature is high enough for these elements to fuseto heavier elements. This sequence of processes

which hydrogen and helium followed is repeatedfor heavier and heavier elements. When one ele-ment has been depleted, the core contracts untilthe temperature is high enough for the next fu-sion process to ignite while burning of the differ-ent elements takes place in shells around the core.This will continue until the core consists of iron5626Fe. We learned in the lecture on nuclear reac-tions that in order to produce elements heavierthan iron, energy needs to be added, no energyis released. For elements heavier than iron, themass per nucleon increases when the number ofnucleons in the core increases. This is why no en-ergy can be released in further fusion reactions.Nuclear processes which need energy input aredifficult to make happen: The quantum mechani-cal probability for a nucleus to tunnel through theColoumb barrier only to loose energy in the fusionprocess is very small. Thus, when the stellar coreconsists of iron, no more nuclear processes takeplace and the core starts contracting again. Atthis point, the star might look like figure 2. Thereare several layers of elements which resulted fromprevious nuclear fusion processes around the ironcore. Fusion processes are still taking place inthese shells.

Figure 2: The structure of a star just hours before a super-nova explosion. Energy production in the core has ceasedas it now consists of Fe—the final product of nuclear fu-sion. Different elements are still burning in layers aroundthe core.

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At this point the temperature in the core is ex-tremely high T ∼ 109−1010 K containing a densegas of high energy photons. The core continuescontracting and no more nuclear fusion processesare available to produce a pressure to withstandthe forces of gravity pushing the core to higherand higher densities. The energy of the photonsis getting so high that they start splitting the ironatoms by the photo disintegration process

5626Fe +0

0 γ → 1342He + 41

0n

and helium atoms are further split into single pro-tons and neutrons

42He +0

0 γ → 211p + 21

0n

reversing the processes which have been takingplace in the core of the star for a full stellar lifetime. To split iron nuclei or other nuclei lighterthan iron (with a few exceptions) requires energy.Again, by looking at the plot from the lecture onnuclear processes we see that nuclear fission pro-cesses only produce energy for elements heavierthan iron. For lighter elements, the mass per nu-cleon increases when a core is split and energy isneeded in the process. Thus, the photodisintegra-tion processes take thermal energy from the core,energy which would contribute to the gas pres-sure preventing a rapid gravitational collapse ofthe core. When this energy is now taken away inthe nuclear fission processes, the temperature andthereby the gas pressure goes down and the forcesacting against gravity are even smaller. The corecan now contract even faster. The result of thefast collapse is that the core is divided in twoparts, the inner core which is contracting and theouter core being in free fall towards the rapidlycontracting inner core.

The inner core becomes electron degenerate, butthe degeneration pressure is not sufficient to with-stand the weight from the mass around the core.But quantum physics absolutely forbids morethan one electron to occupy one quantum state inmomentum space, so how can the inner core con-tinue contracting? Nature has found a solution:electron capture. Electrons and the free protonswhich are now available after the splitting of thenuclei combine to form neutrons

11p +1

−1 e→10 n +0

0 ν (3)

The final result is an inner core consisting almostentirely of neutrons. The neutron core continuescollapsing until it becomes degenerate. This time,it is the neutrons and not the electrons which aredegenerate. The stellar core is now so dense thatall the quantum states of the neutrons are oc-cupied and the core cannot be compressed fur-ther. The neutron degeneration pressure with-stands the forces of gravity. Why can the neu-tron degeneration pressure withstand the forcesof gravity when the electron degeneration pres-sure could not? The answer can be found in theexpression for the degeneration pressure in part1G. Even if the neutron mass is much larger thanthe electron mass which should lead to a smallerdegeneration pressure, the density is now muchhigher than it was in the electron degenerate gas.The higher density makes the total degenerationpressure larger.

When the inner core consisting mainly of neu-trons becomes degenerate, the collapse is sud-denly stopped, the core bounces back and an en-ergetic shock wave is generated. This shock wavetravels outwards from the core but is blocked bythe massive and dense ’iron cap’, the outer core,which is in free fall towards the inner core. Theenergy of the shock wave heats the outer core totemperatures large enough for photon disintegra-tion and electron capture processes to take place.Thus, almost all the energy of the shock waveis absorbed in these energy demanding processes.So the huge amounts of energy carried outwardsby the shock wave could in principle have blownthe whole star apart, but the wave is blockedby the outer core where the energy is absorbedin photo disintegration and electron capture pro-cesses.

The final part of the story is truly remarkable.We see that the electron capture process releasesneutrinos (see equation 3). So parts of the en-ergy ’absorbed’ in this process is reemitted in theform of energetic neutrinos. A large amount ofelectron capture processes are now taking placeand computer simulations show that a neutrinosphere is created, an immensely dense wall of neu-trinos is traveling outwards. Normally, we do notneed to take neutrinos into account when study-ing processes in the stellar interiors because neu-

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trinos hardly interact with matter at all and justleaves the star directly without influencing thestar in any way. Now the outer core is extremelydense increasing the reaction probability of pro-cesses where neutrinos are involved. There is ahuge amount of energetic neutrinos trying to passthrough the very dense outer core. The combi-nation of extreme densities and extreme neutrinofluxes makes the ’impossible’ (or actually improb-able) possible: a large part of the neutrinos reactswith the matter in the outer core. About 5% ofthe energy in the neutrinos is heating the outerparts just enough to allow the shock wave to con-tinue outwards. The shock wave lifts the outerparts of the star away from the core, making thestar expand rapidly. In short time, most of thestar has been blown away to hundreds of AU awayfrom the remaining inner core. This is a super-nova explosion. The luminosity of the explosionis about 109L� which is the luminosity of a nor-mal galaxy. The total luminsity of the supernovais thus similar to the total luminosity of an entiregalaxy. And the energy released in photons is justa fraction of the energy released in neutrinos. Anenormous amount of energy is released over veryshort time scales. Where does the energy fromthis explosion originally come from? We will dis-cuss this in the exercises. Now we will look at thecorpse of the dead star.

3 The fate of intermediate andhigh mass stars

For stars with mass M < 25M�, the neutron de-generation pressure is high enough to withstandthe forces of gravity and thereby to maintain hy-drostatic equilibrium. Normally only 2 − 3M�have remained in the core, the rest of the starwas blown away in the supernova explosion. Theremaining 2−3M� star consists almost entirely ofneutrons produced in the electron capture processtaking place in the last seconds before the super-nova explosion. It is a neutron star. As you willshow in the exercises, the density of the neutronstar is similar to the density in atomic nuclei. Theneutron star is a huge atomic nucleus. In the ex-ercises you will also show that the typical radiusof a neutron star is a few kilometers. The mass of

2-3 Suns are compressed into a sphere with a ra-dius of a few kilometers. The density is such thatif you make a small needle out of materials froma neutron star it would weight about 106 tons.

It is thought that neutron stars have solid outercrusts comprised of heavy nuclei (Fe) and elec-trons. Interior to this crust the material is com-prised mostly of neutrons, with a small percent-age of protons and electrons as well. At a suf-ficiently deep level the neutron density may be-come high enough to give rise to exotic physi-cal phenomena such as super-fluidity and perhapseven a quark-gluon plasma where one could findfree quarks. To model the physics of the inte-rior of neutron stars, unknown particle physics isneeded. These neutron stars are therefore macro-scopic objects which can be used to understanddetails of microscopic physics.

For white dwarfs we found that there is a an up-per limit to the mass of the dwarf. Repeatingthese calculations reveals that there is a similarupper limit to the mass of neutron stars. De-pending somewhat on the less known physics inthe interiors of neutron stars, one has found thisupper limit to be somewhere between 2 and 3solar masses. If the collapsing core has a masslarger than about 3 solar masses, the gravitationalforces will be higher than the neutron degenera-tion pressure. In this case, no known physicalforces can withstand the forces of gravity and thecore continues to contract and becomes a blackhole. Using current theories of quantum physicsand gravitation, the core shrinks to an infinitelysmall point with infinite mass densities. Infiniteresults in physics is usually a sign of physical pro-cesses which are not well understood. When thecollapsing core becomes sufficiently small, boththe general theory of relativity for large massesas well as quantum physics for small scales areneeded at the same time. These theories are atthe moment not compatible and a more generaltheory is sought in order to understand what hap-pens at the center of the black hole.

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Fact sheet: This dramatic image shows the Crab Nebula, a 3.4 pc wide ex-panding supernova remnant in the constellation of Taurus. The supernovaexplosion was recorded in 1054 by Chinese, Japanese, Arab, and (possibly)Native American observers. The orange filaments are the tattered remainsof the star and consist mostly of hydrogen. The rapidly spinning and highlymagnetized neutron star embedded in the center of the nebula is the dynamopowering the nebula’s interior bluish glow. The blue light comes from electronswhirling at nearly the speed of light around magnetic field lines. The neutronstar, like a lighthouse, ejects twin beams of radiation from gamma-rays to radiowaves that appear to pulse 30 times a second due to the neutron star’s rotation.(Figure: NASA/ESA)

4 Pulsars

When the stellar core is contracting, the rota-tional velocity of the core increases because ofconservation of angular momentum. In order tomaintain the angular momentum when the radiusdecreases, the angular velocity needs to increase.In the exercises you will calculate the rotationalspeed of the Sun if it had been compressed to thesize of a neutron star. After the formation of theneutron star the rotational period is typically lessthan a second.

In 1967, Jocelyn Bell discovered a source of radioemission which emitted regular radio pulses. Thepulses where found to be extremely regular, withexactly the same period between each pulse. Theperiod between each pulse turned out to be aboutone second. Later, several of these regular radioemitters have been discovered, most of which witha period of less than a second. At first, no phys-ical explanation for the phenomenon was foundand the first radio emitter was called LGM-1 (Lit-tle Green Men). Later the name pulsar has beenadopted. Today about 1500 pulsars are known,the fastest is called the millisecond pulsar due tothe extremely short pulsation period.

Figure 3: A rotating neutron star: The magnetic axis isnot aligned with the rotation axis. Therefore the magneticfield lines are dragged around, producing synchrotron ra-diation as they accelerate electrons outside the neutronstar. The synchrotron radiation is directed in the direc-tion of the magnetic field lines. Every time the fields linessweep over our direction we see a pulse of radiation.

Today the leading theory trying to explain theregular radio pulses from pulsars is that the pul-sars are rotating neutron stars. In the exercisesyou will show that in order to explain pulsars interms of a rotating object, the radius of the ob-ject needs to be similar to the radius of a neutronstar. For larger object to rotate sufficiently fast,the outer parts of the object would need to rotatewith a velocity larger than the velocity of light.The physics behind the process leading to the ra-

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dio pulses of pulsars is still an an active field ofresearch and the details are not well understood.In short, the current theory says that the neutronstar has a strong magnetic field (which is createdduring the collapse of the stellar core) with theaxis of the magnetic field lines shifted with re-spect to the axis of rotation (see figure 3). Whenthe star rotates, the field lines are sweeping outa cone around the rotation axis. Electrons in ahot electron gas around the neutron star are ac-celerated in the strong magnetic field lines fromthe neutron star. When electrically charged par-ticles are accelerated, they emit electromagneticradiation, synchrotron radiation. The electronsonly feel the magnetic field when the magneticpoles (which are not aligned with the rotationaxis) of the neutron star points almost directlyin their direction. Thus, only the electrons inthe gas where the magnetic field lines are point-ing at the moment emit synchrotron radiation.This radiation is emitted radially outwards fromthe neutron star. If the Earth is along the line ofsuch an emission, observers will see a pulse of syn-chrotron radiation each time the magnetic polesof the neutron star points in the direction of theEarth. We will therefore receive a pulse of radia-tion once every rotation period.

5 Generations of stars

In the Big Bang, mainly hydrogen and heliumwere produced. We have learned that heavier ele-ments are produced in nuclear fusion processes inthe interior of stars. But fusion processes produceenergy only when the nuclei involved are lighterthan iron. Fusion processes which produce ele-ments heavier than iron need energy input andare therefore extremely difficult to make happen.How can it be that the Earth consists of largeamounts of elements heavier than iron? Humanbeings contain elements heavier than iron. Wheredid they come from?

We learned that in the final stages of the stel-lar evolutionary process for high mass stars, thetemperature in the core is very high and high en-ergy photons are able to split nuclei. At thesehigh temperatures, the iron nuclei around the corehave so high thermal energy that even nuclear

processes requiring energy may happen. Even theheavy nuclei have high enough energy to breakthe Coloumb barrier and fuse to heavier elements.All the elements in the universe heavier than ironhave been produced close to the core of a massivestar undergoing a supernova explosion. Whenthe shock wave blows off the material around thecore, these heavier elements are transported tothe interstellar material. We remember that astar started its life cycle as a cloud of interstellargas contracting due to its own weight. So the ele-ments produced in supernova explosions are beingused in the birth of another star. Parts of theseelements go into the planets which are formed ina disc around the protostar.

The first stars which formed in the universe arecalled population III stars. These stars containedno heavier elements (in astrophysics, all the heav-ier elements are called ’metals’, even if they arenot metals in the normal sense). These stars havenever been observed directly but theories for theevolution of the universe predict that they musthave existed. The next generation, produced inpart from the ’ashes’ of the the population IIIstars are called population II stars. These starshave small traces of metals but are generally alsometal poor stars. Finally, the population I starsis the latest generation of stars containing a non-zero abundance of metals. The Sun is a popu-lation I star. The exact details of stellar evolu-tion are different depending on whether it is apopulation I, II or III star: computer simulationsshow that the metal content (which is usually verysmall even in population I stars) plays an impor-tant role in stellar evolutionary processes.

6 Type Ia supernovae

We learned in previous lectures that supernovaecan be divided into type I and type II accord-ing to their spectra. The type I supernovae couldfurther be divided into type Ia, Ib and Ic. Thetype I supernovae did not show any hydrogen linesin their spectra whereas type II supernovae showstrong hydrogen lines. It is currently believedthat type II as well as type Ib and Ic are core col-lapse supernovae discussed above. Type Ib andIc do not have hydrogen lines because the outer

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Fact sheet: A pulsar is a highly magnetized, rotating neutron star that emitsa beam of intense electromagnetic radiation. The magnetic axis is not alignedwith the rotation axis, therefore the magnetic field lines are dragged around,producing synchrotron radiation as they accelerate electrons outside the neu-tron star. The synchrotron radiation is emitted in the direction of the mag-netic axis and can only be observed when the beam of emission is pointingtowards the Earth, much the way a lighthouse can only be seen when thelight is pointed in the direction of an observer. The precise periods of pulsarsmake them useful tools. E.g., observations of a pulsar in a binary neutron starsystem (PSR B1913+16) were used to indirectly confirm the existence of grav-itational waves, and the first exoplanets were discovered around a pulsar (PSRB1257+12). Certain types of pulsars rival atomic clocks in their accuracy inkeeping time. (Figure: NRAO )

hydrogen rich parts of the star have been blownoff before the supernova explosion.

A type Ia supernova is believed to be a completelydifferent phenomenon. There are several differenttheories trying to explain type Ia supernova, nonof which are understood well. In one of the mostpopular theories, a type Ia supernova occurs ina binary system: A white dwarf star and a mainsequence or giant star orbit a common center ofmass. When the two stars are sufficiently close,the white dwarf starts accreting material from theother star. Material from the other star is ac-creted in a shell on the surface of the degener-ate white dwarf. The temperature in the core ofthe white dwarf increases and nuclear fusion pro-cesses burning carbon and oxygen to heavier ele-ments ignite. Since the white dwarf is degenerate,a process similar to the helium flash occurs. Fu-sion processes start everywhere in the white dwarfat the same time and the white dwarf is com-pletely destroyed in the following explosion. Theexact details of these explosions are still studiedalong with other completely different theories incomputer models. Hopefully, in the future, thesemodels will be able to tell us exactly what hap-pens in type Ia supernova explosions. In fact,understanding type Ia supernovae is crucial forunderstanding the universe as a whole as they areone of the main objects used for determining thepresence of so-called dark energy making the uni-verse expand faster and faster. We also remem-ber that type Ia supernovae are used as standardcandles to measure distances. It turns out thatthe luminosity is usually almost the same in mosttype Ia explosions. If these explosions really are

white dwarf stars exploding, we can understandwhy the luminosity is almost the same in all su-pernovae of this kind: There is one common factorin all cases: the white dwarf stars usually have amass close to the Chandrasekhar mass of 1.4M�when exploding.

7 An Example: SK 69o202

Let us take as an example the blue giant thatexploded as a supernova of type ii in the LargeMagellenic Cloud in 1987; the star SK 69o202,later known as SN 1987a. The details of the star’slife have been obtained through computer simu-lations. This originally 20M� star’s life may besummarized as follows:

1. H→He fusion for a period of roughly 107 yrwith a core temperature Tc ≈ 40× 106 K, acentral density ρc ≈ 5 × 103 kg/m3, and aradius ≈ 6R�.

2. He→C, O fusion for a period of roughly106 yr with a core temperature Tc ≈ 170 ×106 K, a central density ρc ≈ 9× 105 kg/m3,and a radius ≈ 500R�. The core mass is6M�. The star is in this phase a red super-giant.

3. C→Ne, Na, Mg fusion for a period of roughly103 yr with a core temperature Tc ≈ 700 ×106 K, a central density ρc ≈ 1.5×108 kg/m3,and a radius ≈ 50R�. The core mass is 4M�.From this stage and onward the star losesmore energy through the emission of neutri-nos than from the emission of photons. In

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addition, from this point onwards the evolu-tion of the core is very rapid and the outerlayers do not have time to adjust to thechanges happening below: the star’s radiusremains unchanged.

4. Ne→O, Mg fusion for a period of some fewyears with a core temperature Tc ≈ 1.5 ×109 K, a central density ρc ≈ 1010 kg/m3.

5. O→S, Si fusion for a period of some few yearswith a core temperature Tc ≈ 2.1 × 109 K.The neutrino luminosity is at this stage 105

greater than the photon luminosity.

6. S, Si→“Fe” (actually a mix of Fe, Ni, Co)fusion for a period of a few days with a coretemperature Tc ≈ 3.5×109 K, a central den-sity ρc ≈ 1011 kg/m3. Si “melts” into α-particles, neutrons and protons which againare built into “Fe” nuclei. The core mass isnow roughly 1.4M�.

7. The core looses energy in the electron cap-ture process and starts contracting rapidly.The energy in the core collapse is releasedand the star explodes as a supernova.

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Fact sheet: For problem 3E.1: This image shows Sirius A, the brightest starin our nighttime sky, along with its faint, tiny stellar companion, the whitedwarf Sirius B, which is the tiny dot at lower left. Using the Hubble telescope’sSTIS spectrograph astronomers have been able to isolate the light from thewhite dwarf and disperse it into a spectrum. STIS measured that the light fromSirius B was stretched to longer, redder wavelengths due to the white dwarf’spowerful gravitational pull. Based on those measurements, astronomers havecalculated Sirius B’s mass at 98 percent that of our Sun (its diameter is only12 000 km). Analysis of the spectrum also yielded an estimate for its surfacetemperature: about 25 000 K. (Figure: NASA & ESA)

8 Exercises

Exercise 3E.1 In part 1G (equation 6) we foundthe condition for a gas to be degenerate in termsof the temperature T of the gas and the numberdensity ne of electrons (number of electrons pervolume). Look back to 1G and study this deriva-tion before doing this exercise. We will now tryto rewrite this expression into a condition on themass density ρ of the gas.

1. Assume that the gas is neutral, i.e. that thereis an equal number of protons and electrons.Show that this gives

ne =Zρ

AmH

,

where Z is the average number of protonsper nucleus, A is the average number of nu-cleons per nucleus, mH is the hydrogen massand ρ is the total mass density.

2. Find the expression for the condition for de-generacy in terms of the total mass densityρ instead of ne.

3. Find the minimum density a gas with tem-perature T = 109 K must have in order tobe degenerate. A typical atom in the gas hasthe same number of protons and neutrons.

4. If you compress the whole Sun into a spherewith radius R and uniform density until itbecomes degenerate, what would be the ra-diusR of the degenerate compressed Sun (as-sume the temperature T = 109 K for the fi-nal stages of the Sun’s life time)? This isbasically what will happen at the end of the

Sun’s life time. Gravitation will compressit until it becomes a degenerate white dwarfstar. A white dwarf star typically has a ra-dius similar to the radius of the Earth. Doesthis fit well with your result?

5. What about Earth? To which radius wouldyou need to press the Earth in order for itto become degenerate (assume again that thetemperature will reach T = 109 K when com-pressing the Earth)?

Exercise 3E.2

Where does the huge amount of energy released ina supernova explosion originally come from? Doesit come from nuclear processes or from other pro-cesses? Explain how the energy is released andhow the energy is transferred between differenttypes of energy until it is released as a huge fluxof photons and neutrinos. Read the text care-fully to understand the details in the processesand make a diagram of the energy flow.

Exercise 3E.3

In this exercise we will study the properties of aneutron star.

1. In the text we find an approximate expres-sion for the radius of a white dwarf star us-ing the equation of hydrostatic equilibriumand the degeneration pressure for electrons.Go back to the text and study how this wasdone in detail. Now, repeat the same ex-ercise but for a neutron degenerate neutronstar. What is the radius of a neutron star

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having 1.4M�? (The size of a neutron staris typically 10 km, your answer will not becorrect but should be roughly this order ofmagnitude)

2. In the following, use a neutron star radius of10 km which is more realistic. What is thetypical density of a neutron star? Expressthe result in the following way: To which ra-dius R would you need to compress Earth inorder to obtain densities similar to neutronstar densities?

3. Compare the density in the neutron star tothe densities in an atomic nucleus: Use thedensity you obtained in the previous ques-tion for the neutron star density. For thedensity of an atomic nucleus: Assume that aUranium atom has about 200 nucleons andhas a spherical nucleus with radius of about7 fm (1 fm= 10−15 m).

4. Express the radius of a 1.4M� neutron star(use numbers from previous questions) interms of the mass of the neutron star. Howclose are you to the Schwarzschild radiuswhen you are at the surface of a neutronstar? Is general relativity needed when mod-eling a neutron star?

5. The Sun rotates about its axis once every 25days. Use conservation of angular momen-tum to find the rotation period if the Sun iscompressed to

(a) a typical white dwarf star

(b) a typical neutron star

Compare your answer to numbers given inthe text for the rotational period of a neu-tron star. If you found a faster period, canyou find some possible reasons why the ob-served rotational period is slower? Hint 1:The angular momentum for a solid object isgiven by L = Iω where I is the moment ofinertia. Hint 2: Assume that the Sun is asolid sphere and remember that the momentof inertia of a solid sphere is (2/5)MR2

6. We will assume that we do not know whatpulsars are, but we suspect that they mightbe rotating objects. The larger the radius ofa rotating object, the faster the velocity of

an object on the surface of the rotating ob-ject. The fastest observed pulsar is the mil-lisecond pulsar. Assume that its rotationalperiod is 1 ms. What is the maximum radiusR that the object can have without havingobjects at the surface of the object movingfaster than light? What kind of astronom-ical objects could this possibly be? If youfind several possibilities, try to find reasonsto eliminate some of them.

Exercise 3E.4

We learned in the lectures on distances in the uni-verse that by observing the light curve of a super-nova one could find the luminosity and therebythe absolute magnitude in order to obtain the dis-tance. Here we will study a very simple model fora supernova in order to see if we can understandthis relation between light curve and luminosity.

1. We have seen that in a supernova explosion,the outer shells of a star is basically liftedaway from the central core. Assume that wecan model the supernova in this way: theshell of gas simply expands spherically veryfast outwards in all directions. It is equal tosaying that the radius R of the star increases.Assume that we use Doppler measurementsevery day after the explosion of the super-nova and find that the shell around the coreexpands with a constant velocity v. A time∆t after the explosion started we also mea-sure the effective temperature of the shell byspectroscopic measurements and find it tohave the temperature T . If we assume theshell to be a black body, show that the lumi-nosity of the supernova at this point is givenapproximately by

L = 4πσT 4v2∆t2.

Which assumptions did you make in order toarrive at this expression?

2. A supernova is observed in a distant galaxy.Nobody had so far managed to measure thedistance to this distant galaxy, but now therewas a supernova explosion there and thisallowed us to find the distance to the su-pernova and therefore also to the galaxy.

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You can now make all the assumptions fromthe previous question about the supernova.The supernova is observed every day afterthe explosion and the velocity of the shellis found to be moving at a constant veloc-ity of 9500 km/s. After 42 days, the ef-fective temperature is measured to be 6000K. Find the luminosity of the supernova af-ter 42 days expressed in solar luminosities

(L� = 3.8× 1026W)

3. Find the absolute magnitude of the super-nova after 42 days. (The absolute magnitudeof the Sun is M = 4.83).

4. The apparent magnitude of the supernova af-ter 42 days is m = 10. What is the distanceto the supernova? Express your answer inMpc.

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