Neutron Stars - uni-tuebingen

Neutron Stars

Kostas Kokkotas

December 17, 2012

Kostas Kokkotas Neutron Stars

Bibliography

I Black Holes, White Dwarfs, and Neutron Stars S.L.Shapiro and S.A.Teukolsky, see Chapters 2, 3, 8 and 9.

I GRAVITY J.B. Hartle see Chapter 24

I Duncan R. Lorimer, ”Binary and Millisecond Pulsars at the NewMillennium”, Living Rev. Relativity 4, (2001),http://www.livingreviews.org/lrr-2001-5

I D. R. Lorimer & M. Kramer; Handbook of Pulsar Astronomy; CambridgeObserving Handbooks for Research Astronomers, 2004


Neutron Stars: A Laboratory for Theoretical Physics

I A neutron star is the collapsed core of a massive star left behind after asupernova explosion. The original massive star contained between 8 and25 M. (More massive stars collapse into black holes.) The remnantneutron star compresses at least 1.4M into a sphere only about 10-16km across.

I This material is crushed together so tightly that gravity overcomes therepulsive force between negatively charged electrons and positivelycharged protons. The resulting structure of the star is complex, with asolid crystalline crust about one kilometer thick encasing a core ofsuperfluid neutrons and superconducting protons. Above the crust existsboth an ocean and atmosphere of much less dense material.

Figure: Neutron Star Sructure


Pauli Principle

An isolated WD or NS cools down to zero temperature and it is the pressureassociated with matter at T = 0 that supports these stars against gravitationalcollapse.A simple but very important example of nonthermal source of pressure is theFermi pressure arising from the Pauli exclusion principle.

The Pauli exclusion principle:

I restricts the quantum states allowed to half-integral spin particles (e, p,n) known as fermions

I prohibits any two fermions from occupying the same quantum state.

I is crucial for the structure of atoms and their chemical properties

I has the consequence that in the lowest-energy state of an atom, theelectrons are not in the lowest-energy level near the nucleus; instead theyare arranged in higher-energy-level shells.

I it supports the outer electrons in an atom against the attractive electricforces of nucleus.


Electron degeneracy pressure

I Electron degeneracy pressure is a force caused by the Pauli exclusionprinciple, which states that two electrons cannot occupy the samequantum state at the same time. This force often sets a limit to howmuch matter can be squeezed together.

I A material subjected to ever increasing pressure will become ever morecompressed, and for electrons within it, the uncertainty in positionmeasurements, ∆x , becomes ever smaller. Then, as dictated by theHeisenberg’s uncertainty principle, ∆x ∆p ≥ ~

2, the uncertainty in the

momenta of the electrons, ∆p, becomes larger. Thus, no matter how lowthe temperature drops, the electrons must be traveling at this”Heisenberg speed”, contributing to the pressure.

I When the pressure due to the “Heisenberg speed” exceeds that of thepressure from the thermal motions of the electrons, the electrons arereferred to as degenerate, and the material is termed degenerate matter.

I Electron degeneracy pressure will halt the gravitational collapse of a star ifits mass is below the Chandrasekhar Limit (1.44 M). This is thepressure that prevents a white dwarf star from collapsing.


Maximum Mass of White Dwarfs I

White Dwarfs support themselves against gravity by the pressure of electronsarising from the Pauli exclusion principle.

I This pressure is called Fermi pressure and the correspondingcompressional energy is called Fermi energy.

I A rough estimate of the maximum mass that can be supported againstgravity by Fermi pressure can be made by studying the Fermi energy of aspherical configuration of radius R consisting of N electrons and Nprotons (electrically neutral).

I The heavier protons supply most of the mass and the lighter electronssupply most of the pressure.

I Since electrons exclude each other, we can think of each of them asoccupying a volume of characterisc size λ ∼ R/N1/3

I The Fermi momentum of electrons is

pF ∼ ~/λ ∼ N1/3~/R (1)

I If the sphere is compressed, R shrinks, pF rises, the Fermi energy of theelectrons rises and work has to be done to make the compression.


Maximum Mass of White Dwarfs II

If the compression has been carried out to the point that electrons becomerelativistic with energies E = [p2

F + m2ec

2]1/2 c ≈ pF c. Then the total energy inthis approximation is:

EF ∼ N · (pF c) ∼ N4/3~c/R . (2)

The protons supply most of the gravitational energy EG , which is roughly

EG ∼ −G(mpN)2/R (3)

Both EF and EG vary as 1/R. For large N the total energy will be negative andit will be energetically favorable for the configuration to collapse.The critical N at which gravitational collapse is favored is

Ncrit ∼(~c/Gm2

p

)3/2

∼ 2.2× 1057 (4)

and the critical mass is1

Mcrit ∼ mpNcrit ∼ 1.85M! (5)

The exact solution of the maximum mass is called the Chandrasekhar mass and

it is about 1.4M!1mp = 1.673 × 10−24gr and M = 1.989 × 1033gr


Radius of White Dwarfs

The equilibrium masses associated with masses M approaching Mcrit isdetermined by the onset of the relativistic degeneracy:

EF = pF c ≥ mc2 (6)

where m refers to either electrons or neutrons. Then by using equations (1)and (4) this condition gives

R ≤ ~mc

(~cGm2

p

)1/2

≈ 5.03× 108cm for m = me

2.74× 105cm for m = mn(7)

Thus there are two distinct regimes of collapse:

I one for densities above white dwarf values

I and another for densities above nuclear densities

but in both cases M ∼ M. 2

2Note: mp = 1.242 × 10−52cm, h = 2.612 × 10−66cm2 & me = 6.764 × 10−56cm.Kostas Kokkotas Neutron Stars

Maximum Mass of White Dwarfs III

An ideal fermion gas is described by Fermi-Dirac statistics (if bosons thenBose-Einstein statistics) with distribution function (µ: chemical potential)

f (E) =1

e(E−µ)/kT + 1(8)

I For low particle densities and high temperatures it reduces toMaxwell-Boltzmann distribution f (E) = e(µ−E)/kT with f (E) 1.

I For completely degenerate fermions (T → 0 i.e. µ/kT →∞) µ becomesthe Fermi energy, EF = [p2

F c2 + m2

ec4]1/2 and

f (E) = 1 E ≤ EF

0 E > EF(9)

That means that there is only one electron per allowed quantum statewith momentum p up to a maximum value the pF (Fermi momentum).

The number density of electrons will be:

ne =2

~3

∫ pF

0

4πp2dp =8π

3~3p3F =

1

3π2λex3 (10)

where x = pF/mec is the dimensionless Fermi momentum and λe = ~/mec is

the electron Compton wavelength.


Maximum Mass of White Dwarfs IV

The pressure will be Pe ≈ 13ne < pv > where v = pc2/E is the velocity and

the energy density εe ≈ neE or analytically (why?):

Pe =1

3

2

~3

∫ pF

0

p2c2

(p2c2 + m2ec4)1/2

4πp2dp =mec

2

λ3eφ(x)

= 1.42× 1025φ(x) dyne/cm2 (11)

εe =2

h3

∫ pF

0

(p2c2 + m2

ec4)1/2

4πp2dp =mec

2

λ3eχ(x) (12)

where

φ(x) =1

8π2

x(1 + x2)1/2(2x2/3− 1) + ln

[x + (1 + x2)1/2

]→ x5

15π2 x 1x4

12π2 x 1

χ(x) =1

8π2

x(1 + x2)1/2(1 + 2x2)− ln

[x + (1 + x2)1/2

]→ x3

3π2 x 1x4

4π2 x 1


Maximum Mass of White Dwarfs V

Even if degenerate electrons contribute most of the pressure, the density isusually dominated by the rest-mass of the ions. The density is

ρ0 =∑i

nimi ≡ nmB =nemB

Ye= µemune = 0.974× 106µex

3 g/cm3 (13)

where mi is the mass of ion species i , mB is the mean baryon density, Ye is themean number of electrons per baryon, mu = 1.66× 10−24g is the atomic massunit and µe = mB/muYe is the mean molecular weight per electron.Equations (11) and (13) provide in a parametric form the ideal degenerateequation of state (EoS) P = P(ρ0) 3.If we use a polytropic form for the (EoS) P = KρΓ

0 where K and Γ areconstants we get two limiting cases:

I Non-relativistic electrons ρ 106g/cm3 ,x 1

Γ =5

3, K =

32/3π4/3

5

~2

mem5/3u µ

5/3e

= 1.004× 1013µ−5/3e cgs (14)

I Extremely relativistic electrons ρ 106g/cm3, x 1

Γ =4

3, K =

31/3π2/3

4

~cm

4/3u µ

4/3e

= 1.24× 1015µ−4/3e cgs (15)

3ρ = ρ0 + εe/c2 is the total energy density but the last term εe/c2 is negligible.Kostas Kokkotas Neutron Stars

The Tolman-Oppenheimer-Volkov (TOV) Solution

The energy momentum tensor for a perfect fluid is:

Tµν = (ρ+ p)uµuν − gµνp where uµ =dxµ

ds

Then the law for conservation of energy and momentum leads to:

Tµν;ν = gµν ∂νp + [(p + ρ)uµuν ];ν + (ρ+ p)Γµνλu

µuλ = 0 (16)

For a spherically symmetric and static solution

ds2 = eν(r)dt2 − eλ(r)dr 2 − r 2(dθ2 + sin2 θdφ2

)the fluid velocity is: uµ = (e−ν/2, 0, 0, 0) and thus (16) becomes

Tµν;ν = ∂νpg

µν + (ρ+ p)Γµ00u0u0 = 0 (17)

where Γµ00 = − 12gµνg00,ν = − 1

2eνν′


TOV - II

...multiplying with gµλ we get

∂λp = −1

2(ρ+ p) ∂λν ⇒ dp

dr= −1

2(ρ+ p)

dν

dr(18)

which is the relativistic version of the equations for hydrodynamical equilibrium,since in the Newtonian limit g00 = eν ≈ 1 + 2U and ρ p which leads to:

∂p

∂r= −ρ∂U

∂r(19)

Still we need to find a way, via Einstein’s equations

Rµν = −8π

(Tµν −

1

2gµνT

λλ

)(20)

to estimate ν(r) in the same way that we need to solve Poisson equation toestimate the gravitaional potential U.


TOV - III

θθ : 1− e−λ[1 +

r

2(ν′ − λ′)

]= −4πr 2(ρ− p) (21)

rr :ν′′

2+

(ν′)2

4− ν′λ′

4− λ′

r= 4πeλ(ρ− p) (22)

tt :ν′′

2+

(ν′)2

4− ν′λ′

4+ν′

r= −4πeλ (3ρ+ p) (23)

which leads to the following ODE rλ′ + eλ − 1 = 8πeλr 2ρ or(re−λ

)′= 1− 8πρr 2 leading to

eλ(r) = (1− 2M/r)−1 where M(r) = 4π

∫ r

0

ρ(r ′)r ′2dr ′ (24)

which together with (18) give

dp

dr= −(ρ+ p)

M + 4πr 3p

r(r − 2M)or

dν

dr= 2

M + 4πr 3p

r(r − 2M). (25)


TOV - IV : A uniform density star

For the special case ρ = const there is an analytic solution. For example themass function will become:

M(r) =4

3πr 3ρ for r ≤ R and M(r) =

4

3πR3ρ for r ≥ R

The we get:p

ρ=

(1− 2Mr 2/R2)1/2 − (1− 2M/R)1/2

3(1− 2M/R)1/2 − (1− 2Mr 2/R3)1/2. (26)

But substituting the above relations in (18) we get an analytic solution for g00

i.e.

eν/2 =3

2

(1− 2M

R

)1/2

− 1

2

(1− 2Mr 2

R3

)1/2

. (27)

Maximum allowed mass when p(r = 0)→∞ :

M

R=

4

9


Spherical Relativistic Stars

The equations of structure for a spherical relativistic star are:

dm

dr= 4πr 2ρ (28)

dP

dr= −(ρ+ p)

m + 4πr 3P

r(r − 2m)(29)

dν

dr= − 1

ρ+ p

dP

dr(30)

together with and EoS P = P(ρ) and an equation for the metric function λ(r)i.e. e−λ(r) = 1− 2m/r . If we move from geometrical units back to the cgs units

m→ G

c2m, ρ→ G

c2ρ, P → G

c4P and ν(r)→ 2

c2Φ (31)

we get the following equations describing a spherical star in hydrostaticequilibrium in the non-relativitic limit

dm

dr= 4πr 2ρ (32)

dP

dr= −ρGm

r 2(33)

dΦ

dr= −1

ρ

dP

dr=

Gm

r 2(34)


Polytropes I

The ideal Fermi gas EoS reduces to simple polytropic form P = KρΓ0 in the

limiting cases of extreme non-relativistic (Γ = 5/3) and ultrarelativisticelectrons (Γ = 4/3). Equilibrium configurations with such an EoS are calledpolytropes and herey are quite simple to analyze.The hydrostatic equilibrium equations (32) and (33) can be combined to give

1

r 2

d

dr

(r 2

ρ

dP

dr

)= −4πGρ (35)

if we substitute P = KρΓ0 , and we write Γ = 1 + 1/n (n is called polytropic

index) we can get a dimensionless form of this equation the so calledLane-Emden equation

1

ξ2

d

dξ

(ξ2 dθ

dξ

)= −θn (36)

where we have made the following substitutions

ρ = ρcθn, r = aξ, a =

[(n + 1)Kρ

1/n−1c

4πG

]1/2

(37)

where ρc = ρ(r = 0) is the central density. While the boundary conditions are

θ(0) = 1 , and θ′(0) = 0 . (38)


Polytropes II

Lane-Emden equation can be easily integrated numerically starting at ξ = 0with the BC (38). For n < 5 the solution decreases monotonically and has azero at finite value ξ = ξ1 i.e. θ(ξ1) = 0 which corresponds to the surface ofthe star, where p = ρ = 0. Thus we get:

R = aξ1 =

[(n + 1)K

4πG

]1/2

ρ(1−n)/2nc ξ1 (39)

M =

∫ R

0

4πr 2ρdr = ... = 4π

[(n + 1)K

4πG

]3/2

ρ(3−n)/2nc ξ2

1 |θ′(ξ1)| (40)

= 4πR(3−n)/(1−n)

[(n + 1)K

4πG

]n/(n−1)

ξ(3−n)/(1−n)1 ξ2

1 |θ′(ξ1)|. (41)

For the special cases that we are interested:

Γ =5

3, n =

3

2, ξ1 = 3.65375, ξ2

1 |θ′(ξ1)| = 2.71406 (42)

Γ =4

3, n = 3, ξ1 = 6.89685, ξ2

1 |θ′(ξ1)| = 2.01824 (43)


Polytropes II

Figure: θ(ξ) for n = 0, 1, 2, 3, 4 and 5


Maximum Mass of White Dwarfs : Chandrasekhar limit

For low-density white dwarfs Γ = 5/3 we get:

R = 1.122× 104

(ρc

106 g/cm3

)−1/6 (µe

2

)−5/6

km (44)

M = 0.5

(ρc

106 g/cm3

)1/2 (µe

2

)−5/2

M = 0.7

(R

104 km

)−3 (µe

2

)−5

M (45)

For the high-denity case Γ = 4/3 we get:

R = 3.35× 104

(ρc

106 g/cm3

)−1/3 (µe

2

)−2/3

km (46)

M = 1.457

(2

µe

)2

M (47)

Note that M is independent of ρc and R in the ultra-relativistic limit. Thus asρc →∞, the electrons become more and more relativistic throughout the star,then R → 0 and the mass asymptotically will be given by equation (47).

The mass limit (47) is called Chandrasekhar limit (MCh = 1.457M) and is the

maximum possible gas of a white dwarf. For cold perfect gas the dependence

of MCh on composition is contained entirely in µe = mB/muYe = A/Z which is

the mean molecular weight per electron, for He : A = 4 &Z = 2.


Neutron Stars I

The above results can be easily scaled to determine the equation of state of anarbritrary species i of ideal fermions. For pure neutrons we get:

εn =mnc

2

λ3nχ(xn) = 1.625× 1038χ(xn) (48)

where xn = pF/mnc.

ρ0 = mnnn =mn

λ3n

1

3π2x3n = 6.107× 1015x3

ng/cm3 (49)

If we use a polytropic form for the equation of state (EoS) P = KρΓ0 we get for

the two limiting cases:

I Non-relativistic neutrons ρ 6× 1015g/cm3, x 1

Γ =5

3, K =

32/3π4/3

5

~2

m8/3n

= 5.38× 109cgs (50)

I Extremely relativistic neutrons ρ 6× 1015g/cm3 , x 1

Γ =4

3, K =

31/3π2/3

4

~cm

4/3n

= 1.23× 1015cgs (51)

Notice that in this case the mass density ρ = εn/c2 is due entirely to neutrons

and greatly exceeds ρ0 whenever the neutrons are extremely relativistic.


Neutron Stars II

Low density neutron stas with the ideal neutron gas equation of state can beapproximated by n = 3/2 Newtonian polytropes. From eqns (39) & (41) weget:

R = 14.64

(ρc

1015g/cm3

)−1/6

km (52)

M = 1.102

(ρc

1015g/cm3

)1/2

M (53)

Thus there is no minimum neutron star mass i.e. M → 0, ρc → 0 and R →∞.

In nature, of course, neutrons become unstable to β-decay at low densities.


Neutron Stars III : Nuclear Density

In neutron stars the density is higher than the nuclear density!The maximum value for the momentum of a neutron will be ∆p ∼ mpc. Thenthe uncertainty principle ∆p ∆x ∼ ~ suggests that

∆x ∼ ~mpc

∼ 10−13cm (54)

actually the exact value is ∆x ∼ 1.8× 10−13cm i.e. the neutrons “touch” eachother.The number density will be

n ∼ 1

∆x3≈ 1.7× 1038cm−3 (55)

and the density will be

ρn = mpn ≈ 2.8× 1014g/cm3 (56)

In neutron stars the average density is higher than this value.


Neutron Stars IV : Composition

I NS’s surface is believed to consist of a very thin fluid/gas atmosphere.

I Beneath that, the outer crust will typically consist of a lattice of 5626Fe.

The pressure in this region is mainly provided by the iron lattice.

I At higher densities more and more electrons will become free and thepressure will increasingly be due to the degenerate electron gas. In theregion were ρ 106 g/cm3 the free electrons are non-relativistic and thecompressibility index

γ = (ρ+ p)/pdp/dρ

takes the value γ = 5/3.

I As we go further towards the center of the star the electrons becomeincreasingly relativistic and γ → 4/3.

I At a density of about 106g/cm3 the electrons and protons startrecombining in inverse β-decay e− + p → n + νe thus forming increasinglyneutron rich nuclei.

I At ρ ∼ 3× 1011g/cm3 the nuclei gets unstable and the free neutrons areemitted. This is called the ”neutron drip” region and here the EoS willbecome very soft (low compressibility index).


Neutron Stars IV : Composition

I As we move towards the center, the matter consists mostly of freeneutrons and small abundances of free protons, electrons and muons. Thepressure will be provided mainly by the degenerate neutrons and thecompressibility index will decrease from the non-relativistic value 5/3 tothe relativistic 4/3.

I The neutrons become superfluid at some point in the neutron star.

I If no phase transitions occur the EoS will approach p → ρ/3

I The actual EoS in these ultra high densities is highly unknown e.g. theremight exists a quark-gluon plasma, kaon condensates and other exoticstates of matter etc


Neutron Stars V

Figure: Schematic cross section of a neutron star, showing the outer crustconsisting of a lattice of nuclei with free electrons, the inner crust which alsocontains a gas of neutrons, the nuclear “pasta” phases, the liquid outer core,and the possibilities of higher-mass baryons, Bose-Einstein condensates ofmesons, and possible quark matter in the inner core.


Neutron Stars VI

Figure: Neutron star interior as predicted by theory


Neutron Stars VII

Figure: Mass-radius curve for HWW EoS configurations. The curve isparametrized by central density measured in gr/cm3. At the extrema (max ormin) the configuration becomes unstable/stable for increasing central density.


Neutron Stars VIII

Figure: Mass-radius relationship of neutron stars and strange stars. Thestrange stars may be enveloped in a crust of ordinary nuclear matter whosedensity is below neutron drip density.


Neutron Stars: History

I Walter Baade and Fritz Zwicky from CalTech first predicted theexistence of neutron stars in 1934, but they were not discovereduntil over 30 years later.

I Pulsars are a special category of spinning neutron stars, discoveredin 1967 by Jocelyn Bell, an astronomy graduate student workingwith Prof. Antony Hewish at Cambridge. Pulsars derive their namefrom ”pulsating radio sources” because they were first observed atradio wave frequencies. Hewish won the 1974 Nobel Prize in Physicsalong with Sir Martin Ryle for their ”pioneering discoveries in radioastrophysics.” Hewish was cited for his ”decisive role in thediscovery of pulsars.”


Neutron Stars: Supernovae

A supernova explosion is usually associated with the formation of black holesand neutron stars.

I Young stars are hydrogen, and the nuclear reaction converts hydrogen tohelium with energy left over. The left over energy is the star’sradiation–heat and light.

I When most of the hydrogen has been converted to helium, a new nuclearreaction begins that converts the helium to carbon, with the left overenergy released as radiation.

I This process continues converting the carbon to oxygen to silicon to iron.Nuclear fusion stops at iron.

I A very old star has layers of different elements. The outer layers ofhydrogen, helium, carbon, and silicon are still burning around the ironcore, building it up.

I Eventually, the massive iron core succumbs to gravity and it collapses toform a NS. The outer layers of the star fall in and bounce off the neutroncore which creates a shock wave that blows the outer layer outward.

I This is the supernova explosion.


Neutron Stars: Supernovae (sequence of events)

I Within about 0.1 sec, the core collapses.

I After about 0.5 sec, the collapsing envelope interacts with the outwardshock. Neutrinos are emitted.

I Within 2 hours, the envelope of the star is explosively ejected. When thephotons reach the surface of the star, it brightens by 8 orders of mag.

I Over a period of months, the expanding remnant emits X-rays, visiblelight and radio waves in a decreasing fashion.


Neutron Stars: Modelling

I Stars more massive than ∼ 8M end in core collapse ( 90% are starswith masses ∼ 8− 20M ).

I Most of the material is ejected

I If M > 20M more than 10 % falls back and pushes the PNS abovethe maximum NS mass leading to the formation of BHs (type IIcollapsars).

I If M > 40M no supernova is launched and the star collapses toform a BH (type I collapsars)

I Formation rate: 1-2 per century / galaxy

I 5-40% of them produce BHs through the fall back material

I Limited knowledge of the rotation rate! Initial periods probably< 20ms. Maybe about 10% of pulsars are born spinning withmillisecond periods.


Neutron Stars: Maximum Mass I

The mass of a stable neutron star can be estimated by using the followingassumptions (Rhoades & Rufinni 1974):

1. GR is the correct theory of gravity and Tolman-Oppenheimer-Volkovequations describe the equilibrium structure.

2. The EoS satisfies the “microscopic stability” condition dP/dρ ≤ 0. If thiscondition is violated small elements of matter will spontaneously collapse.

3. The EoS should satisfy the causality condition dP/dρ ≤ c2 i.e. the speedof sound is less than the speed of light.

4. The EoS below some critical density ρ0 is known

The most fundamental from the above statements is the 3rd one which leadsinto P = P0 + (ρ− ρ0)c2 for ρ ≤ ρ0.Rhoades & Rufinni adopted ρ0 = 4.6× 1014g/cm3 and one of the very first EoS(the Harisson-Wakano) and they got

Mmax ≈ 3.2

(4.6× 1014g/cm3

ρ0

)1/2

M (57)


Neutron Stars: Maximum Mass II

Actually for uniform density stars one can proves that the maximum mass canbe derived by studying the stability. That is by extremising the energy (for afixed number of baryons):∣∣∣∣∂M∂χ

∣∣∣∣A

= 0 and

∣∣∣∣∂2M

∂χ2

∣∣∣∣A

> 0 (58)

where M is the stellar mass, A is the total baryon number and sin2 χ = 2M/R.These condition imply that there is a critical adiabatic index

Γ ≡ ∂ lnP

∂ ln n=(

1 +ρ

P

) dP

dρ> Γc(χ) = (ζ+1)

1 +

(3ζ + 1

2

[ζ + 1

6ζtan2 χ− 1

](59)

where

ζ(χ) ≡ 6 cosχ

9 cosχ− 2 sin3 χ/(χ− sinχ cosχ)− 1

in the Newtonian limit

Γc =4

3+

89

105

M

R(60)

and the maximum mass is : Mmax ∼ 3.6M.


Neutron Stars: Maximum Mass III

Figure: The stability domain in the uniform-density approximation. The curveseparating the stable and unstable regions is the function Γc given by eqn (59).The dashed line shows the adiabatic index of a free neutron gas.


Neutron Stars: Maximum Rotation

The shortest period for a rotating star occurs when it is rotating at break-upvelocity called also Kepler limit:

Ω2R ∼ GM

R2(61)

where Ω is the angular velocity of rotation. This can be written also in terms ofthe mean density ρ as:

Ω2 ∼ Gρ (62)

the universal relation between dynamical time-scale and density for aself-gravitating system.

I For mean densities of 108g/cm3 (white dwarfs) this relation gives a period:

P =2π

Ω≥ 1 sec

I For mean densities of 1014g/cm3 (neutron star) this relation gives aperiod:

P =2π

Ω≥ 1 msec


Neutron Stars: Magnetic Field

During the collapse if we assume that the initial angular momentum andmagnetic flux are conserved we will end up in a neutron star with msec periodand magnetic field of the order of 1012 G.

I Conservation of angular momentum

MR2Ω = MR2NSΩNS ⇒ PNS = P

(RNS

R

)2

∼ 0.5ms (63)

for P ∼ 1 month and R/RNS ∼ 7× 105.

I Conservation of magnetic flux

R2B = R2NSBNS ⇒ BNS = B

(R

RNS

)2

∼ 5× 1012Gauss (64)

for initial B ∼ 100 Gauss.


Neutron Stars: Pulsars I

I From our earthly vantage point, pulsars appear to pulse with light witheach rotation.

I Their light, like a lighthouse beam, sweeps across the Earth.

I Some pulsars emit visible light, X-rays, and even gamma-rays.

I All pulsars are NS, but (so far as we know) not all NS are pulsars.


Neutron Stars: Pulsar Observation


Neutron Stars: Pulsars - The magnetic dipole model

The pulsar emission is derived from the kinetic energy of a rotating NS.In the oblique rotator model, it is assumed that the NS rotates uniformly invacuo at a frequency Ω and possess a magnetic dipole moment ~m oriented atan angle α to the rotation axis.A dipole magnetic field at the pole of the star BP is related to ~m by

|~m| =1

2BPR

3 R is the radius of the star (65)

The radiation at infinity due to the time-varying dipole moment will be:

E = − 2

3c3| ~m|2 = ... = −B2

PR6 sin2 α

6c3Ω4 (66)

The energy carried away by the radiation originates from the rotational kineticenergy of the NS E = IΩ2/2, where I is the moment of inertia. Thus

E = IΩΩ (67)

i.e. the pulsar slows down. The characteristic age T of pulsar is:

T = −(

Ω

Ω

)0

=6Ic3

B2PR

6 sin2 αΩ20

(68)


Neutron Stars: Pulsars - The magnetic dipole mode

By integration we get the initial angular velocity

Ωi = Ω0

(1 + 2

Ω2i

Ω20

t

T

)1/2

(69)

where t is the present age of the pulsar given by:

t =T

2

(1− Ω2

0

Ω2i

)≈ T

2(70)

For the Crab pulsar this method predicts 1260 yr while the true is 947 yr.The magnetic field strength can be estimated

BP =

√6c3I

2πR3 sin2 α

√PP ≈ 5.2× 1019

√PP Gauss (71)

For a typical NS (M = 1.4M, R = 12km and I = 1045g/cm2) we get

E = 1.8× 1049 erg , E = 4.6× 1038 erg/s (72)

from the above estimation we get an approximate value for the magnetic fieldstrength of the Crab pulsar.

BP = 4.4× 1012G (73)

which is the same as the predicted value earlier.


Neutron Stars: Pulsars - P vs P Diagram


Neutron Stars: Pulsars - P vs P Diagram


Neutron Stars: Pulsars - Braking index

If the pulsar slowdown follows a power-law then we can write

Ω ≈ Ωn (74)

the parameter n is called the braking index and for the magnetic dipole modeln = 3.In general we can define

n ≡ ΩΩ

Ω2(75)

which suggests that the braking index can be measured directly from the pulsar

frequency and its derivative.


Neutron Stars: Pulsars and Gravitational Radiation

Slight deformations of a pulsar (eg deformed ellipsoid) can be sources of GWsand account for their slow down. The energy loss in GW is:

EGW = −32

5

G

c5I 2ε2Ω6 (76)

Small ellipticities can generate GW emission. Magnetic fields of the order of1015-1016 Gauss can generate such ellipticities.The deceleration law for constant ε is

EGW = IΩΩ ∼ Ω6 (77)

By integration we get the initial angular velocity

Ωi = Ω0

(1 + 4

Ω4i

Ω40

t

TGW

)1/4

(78)

where t is the present age of the pulsar given by:

t =TGW

4

(1− Ω4

0

Ω4i

)≈ TGW

4(79)

For the Crab pulsar this method predicts 620 yr while the true is 947 yr i.e.gravitational radiation cannot alone be responsible for the Crab slowdown.

A combination of gravitational and magnetic dipole radiation can be found

which may give both the correct age and the observed pulsar deceleration rate.Kostas Kokkotas Neutron Stars

Neutron Stars: Pulsars - The Aligned Rotator


Neutron Stars: Pulsars - The Aligned Rotator

I Goldreich-Julian(1969) suggested that strong E-fields parallel to thesurface will rip off charged particles creating a dense magnetosphere

I The charged particles in the magnetosphere within the light cylinder, mustcorotate with the pulsar.

I The light cylinder extends out to the distance at which the corotationvelocity attains the speed of light. The corotation radius is thus given by

Rc ≡c

Ω= 5× 104 P km (P : is the pulsar period in sec) (80)

I As the particles approach Rc in the equatorial plane, they become highlyrelativistic.

I Whether or not the dipole field is aligned with the rotation axis plasmanear the pulsar can in principle carry away sufficient angular momentumand energy to supply the necessary and energy to supply the necessarybraking torque.

I The energy loss will be dictated again by E = −B2PR

6

c3 Ω4 which leads againto a breaking index n = 3.

I Particles accelerate along the B-field lines. Relativistic e− emit highenergy EM radiation (γ-rays) which may disintegrate γ → e− + e+ andthis goes on!


Neutron Stars: Pulsars II

Pulsars are divided into two main categories, isolated pulsars and binarypulsars.

I Isolated pulsars (which include most radio pulsars) produce radiationprimarily through their rotation, as they gradually slow down andcool off. Their light is generated by electrons caught in the pulsar’sstrong magnetic field, concentrated and emitted near the magneticpoles. Thus, much of an isolated pulsar’s visible energy is funneledfrom the rotating magnetic pole region. The precise location of thebeam is debated.

I Pulsars in Binarys are those in orbit with a companion star, mostoften hydrogen-burning stars like our Sun. Such a pulsar can pullover, or accrete, matter from its stellar companion as their orbitsbring these objects in close contact with each other. The violentaccretion process can heat the gas being transferred and produceX-rays. The X-rays from this matter, under the influence of magneticfields, can also appear to pulsate at the pulsar’s rotation rate.


Neutron Stars: Significant Pulsars

I The first radio pulsar, CP 1919 (now known as PSR 1919+21), with apulse period of 1.337 seconds and a pulse width of 0.04 second, wasdiscovered in 1967

I The first binary pulsar, PSR 1913+16, whose orbit is decaying at theexact rate predicted due to the emission of gravitational radiation bygeneral relativity

I The first millisecond pulsar, PSR B1937+21

I The brightest millisecond pulsar, PSR J0437-4715

I The first X-ray pulsar, Cen X-3

I The first accreting millisecond X-ray pulsar, SAX J1808.4-3658

I The first extrasolar planets to be discovered orbit the pulsar PSRB1257+12

I The first double pulsar binary system, PSR J0737?3039

I The magnetar SGR 1806-20 produced the largest burst of energy in theGalaxy ever experimentally recorded on 27 December 2004

I PSR J1748-2446ad, at 716 Hz, the pulsar with the highest rotation speed.

I PSR J0108-1431, the closest pulsar to the Earth. It lies in the direction ofthe constellation Cetus, at a distance of about 86 parsecs.


Neutron Stars: X-ray binaries

I Binary pulsars are often called X-ray binaries. The flow of matter from thestellar companion, the accretion disk, also glows in X-rays owing to itshigh temperature. While some X-ray binaries are steady X-ray sources,others are bright only for a few weeks or months at a time, lying dormantfor years between outbursts. The X-ray sky is thus highly variable, unlikethe visible night sky.

I X-ray binaries were first discovered by R. Giacconi and collaborators in theearly ’60s, with their binary nature established in the early ’70s byGiacconi and others. Giacconi was awarded the 2002 Nobel Prize inPhysics for ”pioneering discoveries in astrophysics, which have led to thediscovery of cosmic X-ray sources.”


Neutron Stars: Cannibals

I The process of accretion can speed up the spin of a binary pulsar,since the high-velocity accreting material hits the pulsar at a grazingangle, constantly spinning it faster.

I Rotation rates for such accretion-powered pulsars can reachhundreds of Hz, nearly kHz (millisecond pulsars)

I The transfer of material onto the neutron star also slowlycannibalizes the stellar companion so that, over millions of years,enough matter is accreted away to completely whittle away a oncehealthy star.

I There are pulsars orbiting stars with as little as 10 Jupiter masses.One well-known example is the pulsar B1957+20, known as the”Black Widow” pulsar, which is emitting a stream of high-energyradiation that will soon blow away its now feeble companion so thatno trace of this star remains.


Neutron Stars: Millisecond Pulsars

I There are many known isolated millisecond pulsars detected in theradio-wave regime. The fastest known is the pulsar PSR J1748-2446ad,spinning at an astonishing 716 Hz (2nd 649Hz and 3rd at 588 Hz).

I It is believed that these isolated millisecond radio pulsars were once X-raybinary millisecond pulsars that accreted material, spun up, andcannibalized their companions.

I This link was first confirmed in 1998 when scientists using the Rossi X-RayTiming Explorer found that the X-ray binary millisecond pulsar SAXJ1808.4-3658 was accreting matter from a companion with only 0.05M.


Neutron Stars: Population

I The Galaxy is thought to contain about 100,000 pulsars, yet about twothousand are known.

I Only a small fraction of these are millisecond pulsars.

I Isolated pulsars are hard to find because they are dim.

I X-ray binary millisecond pulsars are discovered when they flare up in arare, sporadic accretion event that lasts only a few weeks. In many X-raybinaries, however, the accreted material on the neutron star is occasionallyconsumed in a violent thermonuclear explosion that sweeps across thesurface, generating a bright burst of X-rays lasting a few seconds.

I Using the Rossi Explorer in 1996 discovered rapid X-ray flickering duringsome of these explosions, called ”X-ray burst oscillations,” and suggestedthat this flickering might offer another way to measure the spin rates ofneutron stars. An advantage of this approach is that burst oscillations arebright and fairly common.


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