White Dwarves, Supernovae and Neutron Star: theory

First need to understand the

Fermi Distribution

Consider electrons in a 3-d infinite potential well. ψE(x,y,z) = (2/L)3/2sin(n₁πx/L)sin(n₂πy/L)sin(n₃πz/L)

E = π²ħ²/2mL² (n₁²+n₂²+n₃²)= ħ²k²/2m,  n1, 2, 3 = 1, 2, 3...

Degeneracy: same energy associated with sets of {n₁, n₂, n₃}. Only 2 spin 1/2 fermions allowed in each energy level. For ground state of N fermions, fill up lowest levels E_F Fermi Energy.

(n₁²+n₂²+n₃²) = 2mL²/π²ħ²E≡ n²

k² = π²/L²n² = (2π/λ)²

n = n₁i+n₂j+n₃k

n = √(n₁²+n₂²+n₃²)

The number of states --> counting points in octant with ni positive Each point in this "lattice" space occupies unit volume. So the number of "points" corresponding to a set of quantum numbers { n₁, n₂, n₃ } up to a given energy is given by the volume in the lattice space. # of points ≡ Np = (1/8)(4/3π n³)

# of fermions = 2N_p ≡ N (2 spin states for each level)

N = 1/3π (nF)³= 1/3π[2mL²/π²ħ²EF]3/2

Define fermion # density ρ ≡ N/L³

3ρ/π= [2mEF/π²ħ²]3/2 EF = ħ²π²/2m[3ρ/π]2/3

We can define the Fermi momentum (actually wave #)

kF = (2mEF)½ = (3π²ρ)1/3 = 2π/λF where

λF = 2.03ρ-1/3 ∼ 2d 

is the de Broglie λ ∼ inter-particle separation d

So N fermions fill energy levels (2/level) up to E_F. This corresponds to a radius r_f in lattice space.

Total energy: Integrate over the octant in lattice space; multiply by 2 for spins.

Etot= (2)(1/8)∫d³n E

= (2)(1/8)4π∫₀nF(n²dn)[(π²ħ²/2mL²)n²]

= ħ²π³/2mL²∫₀nFdn n⁴ = ħ²π³/10mL²(nF)5

But we had (n_F)³ = 3N/π so,

Et = ħ²π³/10mL²(3N/π)5/3 = ħ²π³/10m(3ρ/π)5/3L³

So with N fermions we have levels populated up to E_F. Holding N constant,

EF∼ V-1, Et ∼ V-2/3

Metals

Assume that only the valence electrons are free to move: (good approximation, since the rest are bound by 20eV - many keV). e.g.
what is E_F for potassium? (valence 1, density ρ ∼ 2000 kg m^-3, At. wt A = 39.
N ∼ 3x10²⁸ m^-3, so E_F ∼ 4 eV or T_F ∼ 400000 K. (this means that we would have to heat up a metal to T_F to get a significant # of electrons out of their ground state.). Why don't metals feel hot?

We can be more precise: Classical atoms form a Maxwell-Boltzmann distribution NMB(E) ∝ E½e-E/kBT It "can be shown that" the corresponding Fermi-Dirac distribution is NFB(E) ∝ E½ 1 + e(E-EF)/kBT (We've been working with the T = 0 distribution up until now)

White Dwarves

1. Gas pressure: P V = N R T or (more useful)

   P =  k ρT/m
   

   
   
2. Radiation pressure: photons form part of gas, but can be created and destroyed

   P = 1/₃ σc T4
   

   Which depends strongly on temp, but not pressure

3. Degeneracy Pressure: Electron pressure

   P = K ρ5/3/m₀ ≈ 1012 ρ5/3m₀ Pa
   

   Note this does not depend on temperature at all, but depends very strongly on pressure

Radiation pressure decreases and gravitational contraction resumes. Gravitational contraction restricted by opposition of overlapping wave functions - Pauli principle. Result is a white dwarf at ρ ∼ 10⁶ g/cm (i.e.: The degeneracy or Pauli pressure is balanced by the gravitational pressure.

Pdeg = ∂Etot/∂V = (ħ²π³)/15me(3ρ/π)5/3= (ħ²π³)/15me(3N/π)5/3V-5/3

Pgrav = -∂Egrav/∂V

The gravitational potential energy calculated by assuming constant density, ρ, and spherical shape.

M<r= (4/3π r³)ρ,  Mshell = (4π r² dr)ρ

dEgrav = -GMMshell/r = -G(4π)²/3ρ²r4 dr

Egrav= ∫₀RdEgrav= -(4π)²/15Gρ²R5 = -3/5GM²/R

Where R is the radius of the star and M is the mass of the star. Assume the star is made of N nucleons. M = Nm_n where m_n is the nucleon mass

Egrav = -3/5G(Nmn)²(4π/3)1/3V-1/3

Pgrav= -∂Egrav∂V = 1/5G(Nmn)²(4π/3)1/3V-4/3

Equate the two pressures:

ħ²π³/15me(3N/π)5/3V-5/3= 1/5G(Nmn)²(4π/3)1/3V-4/3

Assume ∼ equal numbers of neutrons and protons: Solving

R* =      ((3/4)⁴ π²)1/3 ħ²    N-1/3

          GMemn² 

M₀ = 2 x 1030 kg , N ∼ M/mn ∼ 1.2x1057, R*∼7x10³km, ρ ∼1.1x10⁹kg/m³

i.e. about the size of the earth. Note that this implies that the heavier the star, the smaller it is (R^* ∝ N^-1/3). Historically, this was a remarkable prediction: the size of a white dwarf depends on 4 numbers measured on earth. (But surely neutron stars are hot: how come we can use a zero-temp approx?

EF ∼ 300 keV ≈ kTF ⌉⇒ TF ∼4x109K
Tcore ∼ 108 K << TF !!!

E = (p²c²  + m²c⁴)½ ≈ p²/(2m) (non-rel) 
                                ≈ pc (ultra-rel)

Type I

Type 1 have a compact object (white dwarf) with a red giant, which expands and spills material onto companion, finally triggering catastrophic collapse. All type 1a seem to be the same (very important for later on!)

Drawing Credit: ST ScI, NASA

Material ⇒ Accretion disk onto compact object

Triggers explosive burning
⇒ shock wave compresses accretion disk
⇒ more burning
⇒ massive ejection of disk material
⇒ formation of ⁵⁴Fe (half-life 70 days!)
decay provides energy for slow light curve

Type II

Many in external galaxies:spectrum show ejected material has v - 10⁴ km s^-1

Agrees with models of core collapse of heavy (>10 M₀ ) star

Pre-collapse:

ρ ≈109 gm cm-3, Tc > 109 K

Supernova Theory

E = (p²c²  + m²c⁴)½ ≈ p²/(2m) (non-rel) 
                                ≈ pc (ultra-rel)

P = K ρ5/3/m₀ ⇒ K 'ρ4/3

Neutrons are also fermions so the Pauli pressure still holds the star for collapsing further. Degeneracy pressure for neutrons gives

R*n = (34π2/24 )1/3 ħ²/Gmn³ N-1/3

P = Kρ5/3   (Non-Rel. electron gas)
P = K'ρ4/3 (Rel. electron gas)=>
P = K"ρ5/3 (Neutron gas)

This creates a neutron star. e.g..

M = 1.5 M₀,  R = 11 km, ρ ∼ 4x1017 kg/m³, n = ρ/mP ≈ 1045
 (nuclear density)

i.e. about the size of Ottawa! Upper limit for neutron stars is about 3.5M₀: beyond that we get black holes. (upper limit is not so rigorous: can ignore electron interaction but not neutrons).

Collapse details

γ's56Fe ⇒ 20 p + 26 n i.e. heavy atoms transformed back to nucleon gas 

e- + p ⇒ n + ve  energetically preferable to absorb electron

The electrons are removed, so pressure drops: also positrons, neutrinos etc.. can be produced

γ (nucleus) -> e+ + e- (nucleus)

e+ + e- -> ν ν
(Z,A) -> e+ + ν + (Z-1,A)
(Z-1,A) + e- -> ν + (Z,A)

No electron pressure -> collapse Shock travels at c/5. Inner layers collapse very fast (core collapse time ≈ 10 minutes) At time t1, core is entirely converted to neutrons, becomes rigid again and "bounce" sets in Details messy ⇒ T = 1011K (typical binding energy/nucleon) on shock front ⇒ dissociated nuclei, ⇒ energy absorbed ⇒ shock weakened But weaker shock ⇒ t ≈ 109K triggers (e.g.) Si -> Fe burning, produces energy explosively ⇒ stronger shock

Energy conservation in core collapse

ΔU = GM²/R2 - GM²/R1 ≈ 2 x 1046 J

Best bet
⇒ 10⁴⁶ J light & KE in ejected shell
⇒ 10⁴⁶ J neutrinos
Neutrino flash should be easily detectable

Period

L₂ = L₁, ω₂R₂²   =   ω₁R₁²

so

f₂ ≈ 107 /(3600x24) ≈ 120

i.e. neutron star should have period of P ≈ 1...10 ms

Magnetic Field

Solar B ∼100 G ∼ .01T, estimate core fields ∼ 10⁶T: What can we expect Neutron star fields to be?

Magnetic flux must be conserved Φ = ∫ B dA = πa² B = πa² B₂(R₂/R₁)² so B₂ = (R₁/R₂)²B₂

This would imply mag fields in neutron stars to be extremely high: magnetars have fields ∼ 10¹¹-10¹² T!

Neutrinos

Note that there are two different mechanisms:
Neutronisation (Fe => neutrons) which produces entirely ν̄ 's and lasts ≈ 1s

Cooling produces ν's in flavour indep. manner

Supernova Sn 1987a

Pulsar seen with f ≈ 2760 Hz

Is there a maximum speed of rotation?

Mass at equator will fly off if centrifugal force is too large
GMm/R² = mω²R so
ω = (G M/R³)^1/2 = (4/3πGρ)^1/2
We know density ≈ 2.4x10¹⁷kgm^-3
so ω ≈ 8200 or
ν ≈ 1400 Hz

Maybe not!

Neutrino observations

2 hours before the light arrived a pulse of neutrinos hit the various detectors running at the time (Kamiokande, Mt.. Blanc) Theoretically predicted but never seen before or since.4 labs live at the time

Mt Blanc, - 6 events, E ≈ 10 MeV @ 23.119 in 7 s Kamiokande, - 12 events 10<E<20 MeV @ 23.315 in 12s I.M.B., - 8 events E>20 MeV @ 23.316 in 6s Baksan, - Nothing General comments Mt blanc observations inconsistent with others (Δt≈4.2 hrs) Rough (ΔΘ ≈ 20°) alignment with L.M.C. Time precedes first optical observation by ~ 2 hrs

If 3 x 10⁵⁷ ν's emitted

In addition to providing reasonable confirmation of supernova theory, SN 1987a provides strong limits on some particle physics e.g. ν mass
travel time

t = L/c √(1-v²/c²) = v²/c² = p²/E²

Δτ'  =  Lm²/2c (1/ E₀² - 1/E1²)

Worst limit is to assume that all ν's emitted simultaneously ⇒ m < 33ev.

{However KK has low energy events first}

This gives a limit (open to a lot of squabbling) m < 10 eV which is better than any terrestial expt.

Nucleosynthesis

He⁴ + C12 -> 016 + γ>
C12 + C12 -> Mg24 + γ

Nuclear Abundances show oddities: E-E means even no. of protons, even no. of neutrons, e.g. C12 Why are E-E nuclei more common? B,Be,Li under-abundant Si,C,N,O,Fe overabundant How are heavier elements than Fe formed

1. Pairing interaction ⇒ EE nuclei being made stable
   p↑p↓n↑n↓ -> E-E: -> no "last" nucleon
   p↑ -> O-E
   n↑ -> E-O
   p↑n↓ -> O-O: -> "two last"

2. B,Be,Li only formed during pp cycle; not end-product

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3. C,N,O, formed by C cycle: more stable than lighter elements

4. To form heavier elements than iron exothermically is not possible

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   s-Process:

   
   slow production of heavy elements via neutrons
   H² + He³-> He⁴ + n
   O¹⁶ + O¹⁶ -> S³¹ + n

   This neutron can then be captured by heavier nucleus, which may then decay

   ---

   e.g. n capture from ¹¹⁵In gives ¹¹⁶In, which decays to ¹¹⁶Sn,
   which then neutron captures to give ¹¹⁷Sn,¹¹⁷Sn,¹¹⁸Sn,¹¹⁹Sn,¹²⁰Sn,¹²¹Sn⇒¹²¹Sb

   
   

   (Cd¹¹³ has lifetime of 10¹⁵ years)

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   r Process

   
   Sn¹²² cannot by formed by capture + decay, since Sn ¹²¹ is very short-lived
   Supernova ⇒ huge fluxes of neutrons
   ⇒ nuclei which have large excess of neutrons which decay back to equilibrium
   

   Mix of isotopes on earth indicates r-process contributed i.e. further evidence that we were part of supernova

   

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   That finishes stars: we want to move to a larger scale