The Stellar Graveyard 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 elec- tron 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.