Cooling Neutron Stars. What we actually see.

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Cooling Neutron Stars What we actually see.

The Equilibrium We discussed the equilibrium in neutron star cores through this reaction (direct Urca). nëp + e à + ö e ö n = ö p + ö e + ö öe Does the reaction actually occur? E n = E p + E e + E öe ö n + kt ù ö p à kt + ö e à kt + E öe E öe ø kt

Why? People assumed that the first neutron stars to be discovered would be glowing embers of a supernova explosion, slowly releasing the heat generated by the cataclysm that created them. How does a neutron star cool? How bright is its surface?

Conservation Laws Energy: E n = E p + E e + E öe ö n + kt ù ö p à kt + ö e à kt + E öe E ø kt öe p n = p p + p e + p öe p n ô p p + p e + p öe kt p F,n ô p F,p + p F,e + c Momentum: Remember that the neutron is a bit above the Fermi sea and the others are a bit below.

Can momentum be conserved? If the fraction of protons equals or exceeds 1/8 the number of neutrons, momentum can be conserved. Low densities and high densities, but not nuclear densities. ð ñ ú 1/3 x n ù 0.4 ún

Where is Urca?

What happens at nuclear densities? Modified Urca reaction: n + n 0 ëp + n 0 + e à + ö e p n + p n 0 = p p + p n 0 + p e + p öe p n à p n 0 ô p p + p n 0 + p e + p öe 0 ô p F,p + p F,n + p F,e + c This last inequality is indeed true! Modified Urca is a go! kt

What about quark matter? In quark matter, the quarks are relativistic (at least the ups and downs) so dëu + e à + ö e can occur and produce neutrinos.

Reaction rates Fermi s Golden Rule: õ if = 2ù M ~ if 2 ú f Transition probability Matrix element for the interaction Density of final states Energy loss rate ï ø E õ if f i n Energy of neutrino Fraction of particles in initial state

Getting the scaling! (1) The density of states of the outgoing neutrino is (kt) 2 and its energy is kt. The number of states for the outgoing proton and electron are: 4ùp 2 dp F kt g g =4ùp h 3 F ökt h 3 c 2 de The fraction of initial particles near the Fermi energy. 3ökT p F c 2

Getting the scaling! (2) Let s drop all dimensionless numbers to make it easier. Direct Urca: Quark Urca: ï (kt) 3 p F,e ö e (kt)p F,p ö p (kt) p 2 (kt) 6 p 2 F,n c2 p F,e p F,p ö e ö p ö n ï (kt) 6 n ö n kt F,n c2

Getting the scaling! (3) Modified Urca: ï (kt) 3 p F,e ö e (kt)p F,p ö p (kt)p F,n ö n (kt) ö n kt p 2 F,n c2 2 p F,e p F,p p F,n ö e ö p ö 3 (kt) 8 n T 8 p 4 F,n c4

The Results Quark Urca: L quark ù (1.3 â 10 44 erg s à1 ) M Mì T 6 9 Modified Urca: ð ñ L Urca ù (5.3 â 10 39 erg s à1 M ú nuc 1/3 ) Mì ú T 8 9

Photon Cooling (1) The luminosity from the surface of a neutron star L í =4ùR 2 ût 4 e =7 â 10 36 erg s à1 ð ñ R 2 10km T 4 e,7 It is this radiation that we may observe. The photons diffuse through the envelope of the star. You can define a conductivity: dt F = à ôà dr

Photon Cooling (2) In astrophysics is customary to use the opacity. 3 ôú L dt = à dr 4ac Let s combine this with the hydrostatic equilibrium equation: dp Gm(r)ú = à dr r 2 dp 4ac T 3 4ùGM 4aT 3 û T L Edd = = dt 3 ô L 3 m p ô L T 3 4ùr 2

Photon Cooling (3) Combine this with the EOS and opacity. P = ú kt, and ô = ô ömu 0 út à7/2 Yields the following: dp 4a = dt 3 û T m p ô 0 L Edd L P 2 4 = 17 4 17 ò ó k 2 út = öm u 4a 3 4a 3 T 13/2 4a = ú 3 û T m p ô 0 û T m p ô 0 L Edd L L Edd L û T m p ô 0 k öm u k öm u L Edd L T 17/2 k T 17/2 öm u T 15/2 P

Photon Cooling (4) Rearranging: ú 2 4 = 17 To keep things simple let s assume that the temperature where the electrons become degenerate is the core temperature. At this point, T F = T. 4a 3 m p ô 0 I have assumed that the electrons are not relativistic: k T «m e c 2, T «5.93 x 10 9 K. û T L Edd L öm u T 13/2 k p 2F kt F = 2me,ú = 8ù 3h p 3 (ö 3 F em u )

Photon Cooling (5) ð Substituting ñ the equation for ρ in terms of T F : 8ù 2 (2me kt) 3 (ö e m u ) 2 4 4a û T L Edd = 17 3 L k 3h 3 L =7 â 10 à32 1 Z(1 + X) L Edd ö ö 2 e L í =9.3â10 6 erg s à1 1 Z(1 + X) m p ô 0 ò ó T 7/2 1K ö ö 2 e ò ó T 7/2 1K ò ó =1.6â10 5 erg s à1 T 7/2 for iron 1K ò ó T e,7 = T 7/8 R à1/2 9 10 km öm u T 13/2

Heat Capacity Let s estimate the heat capacity of ða neutron star. C v = x n ù 0.4 ù 2 (x 2 +1) 1/2 kt Nk mc 2 Integrating up from zero temperature U n =31 ò ó ú à2/3 únuc x 2 ð ñ 1/3 ú ún M =6â 10 47 erg Mì M m n ò k 2 T 2 ó m n c ò ó 2 ú à2/3 T 2 ú 9 nuc ñ

Thermal Evolution (1) We can now determine how the inside of the star cools. du = à (L + L í ) dt These equations look a lot like the ones for the spin. If we assume that only one neutrino or photon process dominates, they are integrable.

Thermal Evolution (2) Modified Urca: Quark Urca: Photons: ð t ' 1 yr ð t ' 1 hr ú únuc ú únuc t ' 5000 yr ñ ú à1/3 h i û T à6 T 6 1 à 9 T9,i 9 ñ ú à1/3 h i û T à4 T 4 1 à 9 T9,i únuc 9 ð ñ ú à2/3 T à3/2 9 ú h i û T 3/2 1 à 9 T9,i

Thermal Evolution (3)

Thermal Evolution (4)

Young Neutron Stars (1)

Young Neutron Stars (2)

Young Neutron Stars (3) On the P-P-dot diagram. 10 15 10 3 Things to notice: most red dots (isolated radio pulsars): 10 11-13 G, 10 5-8 yr 10 14 10 6 most PSRs in binaries have short periods Many young PSRs have high-energy emission or are AXPs (no radio and thermal x-ray) 10 10 10 12 10 9

Anomalous X-ray Pulsars Young isolated neutron stars (often in SNRs): consistent spin down with glitches, periods of several seconds, thermal spectra in X-rays, L ~ 10 34 erg/s, really faint in optical inferred B ~ 10 15 G too bright to be standard cooling, too faint for standard accretion Accretion from tiny disk?

Powering the AXPs Magnetic fields play a dominant and dynamic role. Electron conduction, Field decay, Early on neutrinos dominate the cooling; later photons do.

Soft Gamma Repeaters Young isolated neutron stars (sometimes in SNRs): consistent spin down, periods of several seconds, thermal spectra in X-rays, L ~ 10 34 erg/s, really faint in optical inferred B ~ 10 15 G too bright to be standard cooling, too faint for standard accretion

Plus they burst! Bursts last a few tenths of a second and radiate as much energy as the sun does in a year. Soft compared to GRBs. Biggest explosions that don t destroy the source. Magnetic stress builds in the crust until it fractures and the field rearranges itself locally leading to hard X-ray burst.

Some bursts are really big! March 5, 1979: SGR 0526-66 August 27, 1998: SGR 1900+14 The entire crust is disrupted leading to large-scale reconnection like a solar flare.

What is special about these objects? The energy of an electron in a magnetic field is quantized: ð ñ 1 eb E = n + 2 ~ω g where ω g = me c We can define a characteristic magnetic field for an electron. ~ω g = ~ eb me = m c e c 2 so B = e ~ m 2 e c3 ù 4.4 â 10 13 G

What happens? (1) First, what doesn t happen: If you had an electric field this strong (about 10 18 V/m), you would pull electron-positron pairs out of the vacuum (Klein paradox).

What happens? (2) Such a strong magnetic field is not unstable, but it does funny things to light. The index of refraction depends on the polarization of the light and its intensity. Photons can split or produce pairs.

Looking at the surface (1) Gravity distorts our view of neutron stars.

Looking at the surface (2) ] The magnetic field magnifies one polarization but not the other. The poles are magnified more than the equatorial regions.

Electromagnetic Shocks The phase velocity of light depends on the strength of the wave. The peaks go slower than the troughs, so the wave steepens as it travels near the neutron stars.

Formation of a Fireball

Neutron Star Energetics Some typical energies for isolated neutron stars: Rest-Mass Energy Gravitational Energy M Mc 2 =1.8 â10 54 M ì GM 0.6 2 R =1.6â10 53 Spin Energy 1 IΩ 2 =2â10 46 P à2 2 0 Thermal Energy Magnetic Energy 12 erg ð ñ M 2 M R à1 erg ì 6 erg U n =6â10 47 M T 2 M erg ì 9 1 R 3 B 2 =8â10 46 R 3 6 B2 erg 15