The Stellar Graveyard

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White Dwarfs

• What is the fate of the Sun? As we have seen, stars like the Sun can evolve to the point where they fuse helium into heavier elements like carbon and oxygen, but beyond that the star simply is not massive enough to create heavier elements. What then is the fate of stars like the Sun?

• When helium fusion shuts down, Kelvin-Helmholtz contraction takes over and shrinks what is left of the star into a remnant that is supported entirely by electron degeneracy pressure.

• This object is called a white dwarf. What are the properties of this object? We turn to observations for help.

• The ﬁrst white dwarf ever discovered was in the 40 Eridani system (a triple star system), by William Herschel in 1783. The second white dwarf discovered was the companion to Sirius, predicted to exist as early as 1844, but not observed until 1862 by Alvan Clark.

Sirius A/B The components are called A (primary) and B (white dwarf). They have magnitudes MA = 1.42 and MB = 11.18 (mA = −1.46 and mB = 8.30). Stellar interferometry had measured the size of Sirius A to be RA = 1.71R⊙. The stars have luminosities of LA = 26L⊙ and LB = 0.024L⊙. The observed temperatures are TA = 9900 K and TB = 24, 800 K! What is the radius of Sirius B?

2 4 1 2 2 L R T R L / T L =4πR2σT 4 → A = A A → B = B A L R T R L T B B B A A B Putting in the numerical values from above gives

RB =0.0048RA =0.0084R⊙ =0.92R⊕

We have detailed observations of the binary character of Sirius. From its orbit, we can determine the masses to be:

MA =2.12M⊙ → MB =0.97M⊙

1 Astrophysics – Lecture • From the computations related to Sirius, we deduce two basic principles about the nature of white dwarfs:

⊲ White dwarfs are small only about the size of Earth. ⊲ White dwarfs have about the mass of the Sun. ⊲ Combining these two facts to compute the average density gives:

3 Mwd R⊙ −3 6 9 3 ρwd = ρ⊙ =0.96 (0.0084) ρ⊙ ≃ 2 × 10 ρ⊙ ≃ 2 × 10 kg/m M⊙ R wd • This is about 200, 000× denser than lead, or about 10 tons per teaspoon

◮ Twiddle Math — White Dwarf Pressure......

• If the electrons in the degenerate remnant were an ideal gas, then we could compute the pressure from the ideal gas law. We call this the thermal pressure: N P = kT = n kT ∼ n m v2 th V e e e th

where ne is the number density of electrons and vth is the usual thermal speed from microki- netics 1 2 kT / v ∼ th m e • Recall that we can understand the degeneracy pressure in the context of the uncertainty principle ~ ∆x∆p ∼ ~ → ∆p ∼ → ∆p ∼ ~n1/3 ∆x e I can opt to think in terms of the velocity v instead of the momentum p: 1/3 ~ne ∆p = me∆v → ∆v ∼ me

• The usual game we play in these quantum systems is to say that the average value is on order of the uncertainty.

• Our ﬁrst guess as to the pressure due to the degenerate electrons is to assume it has the same functional form as the thermal pressure 2 1/3 5/3 2 ~ne 2 ne Pdeg ∼ neme(∆v) ∼ neme ∼ ~ me ! me

• Note this has the predicted dependence on density from our previous statement that the degenerate remnant would be well approximated by a polytrope: P = κρ5/3

• We will say the electrons are degenerate when Pdeg >Pth

2 Astrophysics – Lecture ◮ Twiddle Math — White Dwarf Mass-Radius Relation ......

• Doing a precise derivation of the relationship between the mass and radius of a white dwarf is beyond the scope of what we would like to cover in this class. But we can make good back-of-the-envelope estimates by applying our physical intuition.

• The white dwarf is supported by degeneracy pressure. Since the star is in hydrostatic equilibrium (it isn’t collapsing!) the pressure must balance the force of gravity M 2 P ∼ G g R4

• The pressure support is provided by the degeneracy pressure we noted above, so we set these equal: Pg ∼ Pdeg

• Here it is convenient to get rid of the number density in favor of the mass density ρ, since we are interested in the mass. The base assumption is that the remnant is electrically neutral, so the number of protons is equal to the number of electrons. The number density of electrons then may be written

ρ ∼ M 1 ne = 3 mp mp R

• Using this in the pressure equality and solving for the radius R gives: 5 3 −1 3 M 2 M / 1 ~2 1 M / P ∼ P → G ∼ ~2 → R ∼ deg g R4 m m R5 G m m2 m p e e p p

• This could be converted to mass density as well. Expressing a density from the mass radius relationship gives: 3 3 5 M G m m ρ ∼ ∼ e p M 2 mr R3 ~6

• There are some remarkable things to note about this result

⊲ Note that M R3 = constant ⊲ As the mass of the star goes up, the volume it occupies goes down. As Barbara Ryden (Ohio State University) notes: this makes no sense! you don’t expect 40 pounds of manure to ﬁt in a 20 pound bag!

• What this seems to suggest is that if I continue to pile mass onto the white dwarf, I should be able to shrink its volume down to zero! This does not happen, of course, because physics saves me.

• When does physics save me? Look at the speed of the electrons when the density gets high 1 3 1 3 ~n / ~ ρ / ∆v ∼ e ∼ m m m e e p 3 Astrophysics – Lecture • When ρ ∼ 1 × 109 kg/m3, then ∆v ∼ c/3 — the particles are becoming relativistic!

• This implies that there is a density that becomes unphysical because ∆v c, implying that the star cannot be supported by electron degeneracy pressure in all circumstances.

◮ Twiddle Math — the Chandrasekhar Mass......

• The Chandrasekhar Mass (also goes by the moniker “Chandrasekhar Limit”) was ﬁrst worked out by Subrahmanyan Chandrasekhar in 1930 on a boat trip from India to England (as the folklore goes).

• Chandra solved the hydrostatic equation together with the equation of state for both the non-relativistic and relativistic Fermi gas, giving rise to the limiting mass for a stable white dwarf star.

• Let’s make a crude approximation of the limit, at the level of our twiddle math, by estimating the stellar properties when we think the non-relativistic treatment is starting to go bad.

• First, consider the conditions when ∆v ∼ c. From our uncertainty relationship

1/3 3 ~ne cme 37 −3 ∆v ∼ → ∆v ∼ c → ne ∼ ∼ 2 × 10 m me ~ • A typical white dwarf has one proton and one neutron for every electron, so the density from the uncertainty principle would be

3 3 ∼ ∼ c memp ∼ × 10 3 ρup 2mpne 2 ~3 6 10 kg/m

• If we compare this to the white dwarf density we twiddled withe non-relativistic case, we can ﬁnd the mass of a white dwarf where relativistic eﬀects are essential

3 3 5 1 2 G m m c3m3m ~3c3 / ρ ∼ ρ → e p M 2 ∼ e p → M ∼ ∼ 4 × 1030kg mr up ~6 ~3 G3m4 p

◮ Twiddle Math — the Chandrasekhar Mass in the Relativistic Limit ......

• So how do we handle relativistic eﬀects? It is clear that the transition to relativistic scenario must have important consequences for the structure of the star. When the energy is much greater than the rest energy, E ≫ mc2, we can in practice treat the particles as if they are massless particles, and use the fundamental behaviours we use to describe traditional massless particles (such as photons)

4 Astrophysics – Lecture • Of particular use here is we can calculate the relativistic pressure by noting that for massless particles the energy is given by E = pc and the radiative pressure is P = u/3, where u is the energy density. Applying these to our relativistic electron gas (“reg”) then

~c E ∼ (∆p)c ∼ (~n1/3)c → u ≡ E n ∼ ~cn4/3 → P ∼ n4/3 reg e reg reg e e reg 3 e

• Remembering that the number density of electrons may be written in terms of the mass density as ne ∼ ρ/mp, we have

4 3 ρ / ~c M 4/3 P ∼ ~c ∼ reg m 4/3 R4 p mp

• For the star to be in equilibrium, the hydrostatic equilibrium condition must apply — the relativistic pressure must balance the pressure due to the gravitational action of the star.

2 ~ 4/3 ~3 3 ∼ → M ∼ c M → ∼ c Pg Preg G 4 4/3 4 M 3 4 R mp R sG mp

• This gives a mass of M =1.86M⊙. A more detailed calculation, with added care to ﬁll in the twiddles as well as some more detailed physics of the nuclear properties of the material will give the famous Chandrasekhar limit, Mch =1.44M⊙.

• NB 1: None of our degenerate pressure estimates, Pdeg or Preg depend on temperature! The pressure of a completely degenerate electron gas is independent of the temperature. This means the mechanical structure of the star is eﬀectively decoupled from the thermal properties. This is not completely plausible; a formal treatment of the structure of white dwarfs requires one to only use a partially degenerate electron gas.

• NB 2: Note this is exactly the expression for M that we derived when we asked what the properties of the star would be at the transition from non-relativistic to relativistic regimes! This should not surprise us! We are working at the level of twiddles, where all numerical factors are tossed away in favor of keeping dimensionful quantities and constants. Because all the numbers we keep are the ones with units, the combination that will give us units of mass better be the same in both cases, because the same constants are involved!

5 Astrophysics – Lecture ◮ Super-Chandrasekhar Mass ......

• The mass-radius curve for white dwarfs is shown in the ﬁgure below. No white dwarf has ever been discovered with a mass in ex- cess of m =1.44M⊙.

• But one must ask if there is any way the Chandrasekhar limit could be exceeded?

• In 2003 the Supernova Legacy Sur- vey (http://cfht.hawaii.edu/SNLS/) detected a supernova known as the “Cham- pagne Supernova” (SN 2003fg, SNLS- 03D3bb). The supernova was classiﬁed as aberrant Ia.

• Type Ia supernova result from white dwarfs accreting matter from a companion until their exceed the Chandrasekhar mass, at which point the degenerate gas cannot support the star, and it collapses.

• The dynamics of SN 2003fg suggest the progenitor had a mass m 2M⊙!

• There are two possible explanations for this: (1) the white dwarf progenitor was rotating so rapidly that the centrifugal force in the frame of the electron gas reduced the pressure in the core; (2) two white dwarfs merged to make the supernova, momentarily exceeding the Chandrasekhar limit.

• You can read more about the SNLS result in their paper:

⊲ The type Ia supernova SNLS-03D3bb from a super-Chandrasekhar-mass white dwarf star, D. A. Howell et al., Nature 443, 208 [2006]

Massive Star Evolution

• For massive stars (the exact transition is unknown, and a matter of current research, but generally assumed to be M 7M⊙), burning in the core can continue past carbon. The repeating sequence of transition to shell burning as the star’s core undergoes increased heating leads to burning into heavier elements: H → He → C → Ne → O → Si → F e.

• The burning continues until a core of 56F e builds up. 56F e has the highest binding energy per nucleon of all the isotopes; that means that it is the most

6 Astrophysics – Lecture stable nucleus, and any reaction (be it ﬁssion or fusion) requires an input of energy (as opposed to energy being released).

• Remember once a carbon core forms in the star, it is supported by electron degeneracy pressure. This fact does not change as the star continues to evolve. By the time the iron core is forming, it is also supported by electron degeneracy pressure.

• The silicon burning shell rains iron ash down on the core, and its mass slowly increases until mcore = Mch =1.44M⊙.

• At this point, the core can no long be supported by electron degeneracy pressure, and it collapses. When the collapse occurs, it has a radius r ∼ 1R⊙, and an enormous density, ρ ∼ 1012 − 1013 kg/m3

Free Fall Time Suppose we want to know how long it takes the collapse to occur. When the degeneracy pressure vanishes, gravity becomes the dominant force, and we approximate all the dynamics as if gravity acted alone. Assume the core is a constant density sphere, ρ =3M/(4πR3). If we wanted to properly derive the free fall time, we should write the equation of motion for matter radially infalling under only the inﬂuence of gravity:

GM(r) 4πGρ a =r ¨ = g(r) → r¨ = − → r¨ = r r2 3

This takes a bit of work (a couple of pages), so let’s make a hand-wavy back-of-the- envelope calculation that will get us close to the right answer. Assume the acceleration g(r)= const, then appeal to the kinematic equations. Recall that the position equation is 1 1 x = x + v t + at2 → r = at2 o o 2 2 where we have assumed the particle has to fall a distance x = R and starts from rest. Using a =r ¨ from above, we can solve for the free fall time

2r 2R 3 t = = = a 4 πGρR 2πGρ r s 3 r This is remarkably close to the value we would obtain from a proper treatment

3 1/2 3π 1012kg/m t = → t =0.0664s ff 32Gρ ff ρ r ! 12 3 For a core of density ρ ∼ 10 kg/m , this gives a collapse timescale of tff ∼ 0.07 s.

7 Astrophysics – Lecture • As the core collapses, the density rises dramatically. The iron nuclei begin to disintegrate, and the core becomes a soup of free electrons, protons, and neutrons. The protons and the electrons begin to combine to form neutrons:

− p + e → n + νe

• The neutrinos carry an enormous amount of energy! A Chandrasekhar mass of iron has ∼ 1057 protons, so ∼ 1057 neutrinos are made during the collapse. The burst carries away an amount of energy equivalent to ∼ 1046 J1; compare this to the Sun’s total main sequence energy output of ∼ 1044 J.

◮ Twiddle Math — Neutron Stars ......

• The core has become a sphere of ∼ 2 × 1057 neutrons in free fall. What could possibly stop the collapse? Neutron degeneracy pressure!

• Neutrons, like electrons, are fermions and obey the Pauli Exclusion Principle. In analogy with the degenerate electron gas, the expected pressure for a degenerate neutron ﬂuid is

5/3 5/3 2 ne 2 nn Pdeg ∼ ~ → Pdnf ∼ ~ me mn

• In complete analogy with the white dwarf, we can derive a mass-radius relationship for the neutron star −1 3 ~2 1 M / R ∼ G m m2 m n p p • It is convenient to express this as a ratio with the result from the white dwarf

1/3 Rns ∼ me Mwd R m M wd n ns

• Choosing white dwarf values to be Mwd ∼ (1/2)1.4M⊙ ∼ 0.7M⊙ and Rwd ∼ 0.01R⊙ yields

−1/3 Mns Rns ∼ 3km 1.4M⊙

• In reality, we’ve glossed over some facts here in all our twiddling and handwaving. In particular, we’ve ignored some nuclear physics. We don’t know the exact equation of state

1I think BOB has an odd value for this on pg. 532, quoting a neutrino luminosity of 3×1038 W. Supernova explosions generate ∼ 1046 J of energy, and we know most of that is in neutrinos. For the BOB number to be right, the timescale would have to be t ∼ E/L ∼ 3 × 107 s; it does not take the core a year to collapse!

8 Astrophysics – Lecture in the core of a neutron star (this is still a matter of current research, both theoretical and observational).

• As the neutrons are pressed together, the strong nuclear force transitions from being attractive to being repulsive, effectively stiffening the star against having the neutrons being pushed arbitrarily close together. In addition, the gravitational field becomes strong enough that it cannot be treated as a Newtonian system; general relativity must be used to correctly describe the structure of the star. For a plausible model taking these factors into account

−1/3 Mns Rns ≃ 11km 1.4M⊙

Astrophysics Scaling Games You may have noticed some of the ways we’ve been scaling different expressions for numerical values. This is a standard astrophysics method for making formulae easy to evaluate. The method works like this ⊲ For each physical parameter in the formula, pick a canonical value typical of the system you are describing. ⊲ Multiply the formula by 1, formed by the ratio of the canonical value divided by the canonical value, all to the power of the parameter in your formula. ⊲ Repeat for all parameters in the formula. Factor out one piece of each of the canonical values, leaving behind (parameter/canonical value). Let’s do an example. Consider the Steffan-Boltzmann Law: L = 4πR2σT 4. Now scale this to solar values 2 4 2 4 2 R⊙ 4 T⊙ 2 4 R T L =4πσR T =4πσR⊙T⊙ R⊙ T⊙ R⊙ T⊙ Now multiply the known constants together, though you should be able to look at how we’ve written this and see that it is just going to yield a solar luminosity (by definition)

2 4 2 4 26 R T R T L =3.84 × 10 W =1L⊙ R⊙ T⊙ R⊙ T⊙

Now, if you know the scalings of your star in terms of solar normal values, it becomes very easy to estimate the expected luminosity. This is an extremely useful technique, which we will continue to use.

◮ Properties of Neutron Stars ......

• These are extremely compact objects with sun-like masses, which leads to the inescapable conclusion that their surface gravity must be enormous. One way of characterizing the

9 Astrophysics – Lecture surface gravity is to compute the escape velocity: 2/3 2GM Mns vesc = → vesc ∼ 0.6c R 1.4M⊙ r where we have used the mass-radius relation to express this in terms of the total mass of the 11 star. This corresponds to gns ∼ 10 g⊕.

• These objects are not stars in the normal sense — they aren’t actively fusing and producing energy. Can we see them?

• They are still eﬀectively blackbodies, for which we can compute a luminosity. The surface 6 temperatures are hot by stellar standards, Tns ∼ 10 K ∼ 170 T⊙. The radius for our −5 canonical model is Rns ∼ 11km ∼ 1.6 × 10 R⊙. Using our scaling formula we obtain 2 4 R T L = L⊙ ∼ 0.2L⊙ R⊙ T⊙ • Wow! They are apparently pretty luminous! Taking the eﬀective temperature and con- verting it to the peak emission wavelength using Wien’s Law gives λ ∼ 3 × 10−9 m, which is in the x-ray band. High energy telescopes such as Chandra and Fermi have observed and discovered many isolated neutron stars.

◮ Pulsars ......

• The ﬁrst detected neutron star was actually a member of a special class of neutron stars known as pulsars — bright radio sources that pulse on and oﬀ with a regular period P

• Jocelyn Bell (graduate student) and her advisor (Anthony Hewish) had set up an extended radio telescope known as a dipole array over 4.5 acres in England. In 1967, Bell noticed a recurring bit of “scruﬀ” at intervals of 23h 56m, the sidereal period of the Earth.

• The scruﬀ was resolved to be a series of regularly spaced radio pulses, 1.337 s apart.

• Such a regular astrophysical clock had never been detected or speculated on before. The discovery caused some consternation and, for a time, was dubbed LGM-1 — “Little Green Men 1”

• It was with great relief that other sources were soon discovered in other parts of the sky and were soon observed by other radio telescopes. But what were they?

• Many ideas were considered and discarded, but in 1968 Franco Pacinni and Tommy Gold suggested they could be rapidly rotating neutron stars. This idea was strengthened further by the discovery of two pulsars in supernova remnants — the Crab Pulsar and the Vela Pulsar.

• In what is generally regarded as a travesty by much of the astronomical community, Bell did not share in the 1974 Nobel Prize, which was awarded to Hewish and Ryle. By all accounts,

10 Astrophysics – Lecture Hewish was dismissive of her initial detection of the pulsar; it was Bell’s own initiative and persistence in continuing to observe the source that ultimately led to the conclusion that a discovery had been made (though they didn’t know what it was). This, more than anything else, has garnered much criticism that Bell did not share in the Nobel Prize.

11 Astrophysics – Lecture • By all accounts, Bell has been most gracious about this aﬀair. In 1977 in an after dinner speech, she commented on the matter:

⊲ There are several comments that I would like to make on this: First, demarcation disputes between supervisor and student are always difficult, probably impossible to resolve. Secondly, it is the supervisor who has the final responsibility for the success or failure of the project. We hear of cases where a supervisor blames his student for a failure, but we know that it is largely the fault of the supervisor. It seems only fair to me that he should benefit from the successes, too. Thirdly, I believe it would demean Nobel Prizes if they were awarded to research students, except in very exceptional cases, and I do not believe this is one of them. Finally, I am not myself upset about it – after all, I am in good company, am I not!

◮ Properties of Pulsars......

• The basic model of a pulsar is it is a rotating, magnetized neutron star. The pulsing signal is created by a lighthouse mechanism. Emis- sion of electromagnetic radiation arises from the poles of a strong dipole magnetic ﬁeld. As the “emission cones” centered on the magnetic poles sweep across the line of sight to the Earth, we see a pulse.

• The exact model of pulsar emission is a matter of debate and current research. The fundamental idea in most models is that the rotating magnetic field is a changing magnetic field which the Maxwell Equations tells us gives rise to a electric field. This electric field strips charged ions off the surface of the neutron star, carrying them into the magnetosphere where they spiral along the magnetic field lines onto the magnetic poles and through some interac- tion generates electromagnetic emissions.

• There are several generic properties of pulsars:

⊲ Pulsars are excellent clocks owing to their well deﬁned pulse periods. For example, PSR 1937+214 has a period of P = 0.00155780644887275 s, rivaling the accuracy of the best atomic clocks ⊲ Most pulsars have periods between 0.25 s and 0.2s. There is also a small population of millisecond pulsars, with periods centered around 4 ms. ⊲ The periods of pulsars increase gradually. Typical values for the period derivative are P˙ = dP/dt ≃ 10−15.

12 Astrophysics – Lecture • The fact that the pulsar is spinning down (the period is increasing) means the pulsar is losing energy. One can estimate a characteristic lifetime as the time it takes the spin to reach 0, given by τ = P/P˙ . For typical values of P and P˙ , this yields values τ ∼ 107 yrs.

◮ Pulsar Interior ......

• The exact nature of the pulsar interior is a matter of intense interest. The densities are extremely high — nuclear density 17 3 (ρnuc ∼ 4 × 10 kg/m ) and above. These are densities that are not obtainable in the laboratory, and may be probes of exotic matter states that cannot exist anywhere else in the Cosmos.

• Generally the NS is thought of as a thin crust (∼ 0.5 km thick), covering a dense interior superﬂuid of neutrons.

• Superﬂuids have interesting properties. Those of particular importance to us are:

⊲ Superﬂuids have zero viscosity — it will ﬂow without friction past a surface

⊲ Superﬂuids have extremely high thermal conductivity

13 Astrophysics – Lecture ⊲ Superﬂuids are irrotational: ∇ × v = 0 — the vorticity is zero.

• Note that the vorticity being zero does not mean that the trajectories taken by a superﬂuid element are not closed circles. Do not be misled by the phrase irrotational. Consider the velocity ﬁeld in the equatorial plane v v = o φˆ r

This represents a ﬂuid element travelling at constant speed vo/r in a closed circle of constant radius r. Taking the curl of this velocity ﬁeld (in either cylindrical or spherical coordinates) gives ∇ × v =0

• The most prominent consequence of this structure is that it is difficult to change the angular momentum of the star. A common model is that the superfluid is threaded by vortices of normal fluid, which are pinned to the crust and where the angular momentum is concentrated. Moving/changing the vortex structure changes the angular momentum of the star.

◮ Pulsar Glitches ......

• The slow, steady spindown of pulsars is occasionally interrupted by a glitch — a sudden spin-up of the pulsar! After the glitch, the pulsar slowly returns back to its original spin period.

• IfΩ=2π/P is the rotational frequency of the pulsar, the size of the glitch is is roughly ∆Ω/Ω ∼ 10−6.

• There is no agreed upon understanding of what causes glitches, though many models have been proposed and discussed in the literature (starquakes, crust cracking, vortex unpinning, . . . ).

14 Astrophysics – Lecture