chapter 10 neutron stars and general...

Chapter 10

Neutron Stars and General Relativity

Neutron stars are relevant to our discussion of general relativity ontwo levels.

• They are of considerable intrinsic interest because theirquanti-tative description requires solution of the Einstein equations inthe presence of matter.

• In addition, they also explain the existence of pulsars andthesein turn provide the most stringent observational tests of generalrelativity.

231

232 CHAPTER 10. NEUTRON STARS AND GENERAL RELATIVITY

10.1 The Oppenheimer–Volkov Equations

Neutron stars have an average density of order1014−1015 g cm−3.

• This produces gravitational fields that are of moderate strengthby general relativity standards (enormous by Earth standards).

• Escape velocity at the surface is around13c− 1

2c.

• Thus a general relativistic treatment is necessary for their correctdescription.

• Unlike the vacuum Schwarzschild solution, we must now dealwith mass distributions and a finite stress–energy tensor.

• We shall, however, simplify by assuming astatic, sphericallysymmetric configurationfor the matter.

• Boundary condition:With these assumptions we may assumethe solutionoutsidethe neutron star to correspond to the Schwarzschildsolution, so the interior solution must match Schwarzschild atthe surface.

Thus, we consider the general solution of the Einsteinequations for the gravitational field produced by a static,spherical mass distribution that matches sto the exterior(Schwarzschild) solution at the surface of the sphericalmass distribution.

10.1. THE OPPENHEIMER–VOLKOV EQUATIONS 233

• The matter inside the star is a perfect fluid, with a stress energytensor given

T µν = (ε +P)uµuν −Pδ µ

ν ,

where for later convenience we’ve written tensors in mixed form.

• Spherical symmetry, with a line element of the general form

ds2 = −eσ(r)dt2+ eλ (r)dr2+ r2dθ2+ r2sin2θdϕ2,

implying non-vanishing metric components

g00(r) = −eσ(r) g11(r) = eλ (r) g22(r) = r2 g33(r,θ) = r2sin2θ .

Must match smoothly to the Schwarzschild metric at the surface.

• Assumed in equilibrium, soσ(r) andλ (r) are functions only ofr and not oft, andthe 4-velocity has no space components:

uµ = (e−σ/2,0,0,0) = (g−1/200 ,0,0,0).

Inserting these 4-velocity components, the stress–energytensortakes the diagonal form

T µν = (ε +P)uµuν −Pδ µ

ν =

−ε 0 0 0

0 −P 0 0

0 0 −P 0

0 0 0 −P


• For the vacuum Einstein equation we need only the Ricci tensorto construct the Einstein tensor, but in the general non-vacuumcase we need both the Ricci tensorRµν and the Ricci scalarR.It is convenient to express Einstein in mixed tensor form

Gνµ ≡ Rν

µ −12δ ν

µ R = 8πT νµ .

SinceT νµ is diagonal, only diagonal components ofGν

µ needed.

Because of the Bianchi identity

Gµν ;µ = 0

and the Einstein equations

Gµν = 8πT µ

ν ;µ

the stress–energy tensor obeys

T µν ;µ . = 0

This implies that we can choose to solve the equationT µ

ν ;µ = 0 in place of solving one of the Einstein equa-tions. In many cases this can lead to a faster solution thansolving all the Einstein equations directly.

We shall employ that strategy here, using twoEinstein equations and the constraint equa-tion T µ

ν ;µ = 0 in mixed-tensor form to obtaina solution.


The constraint equation has been solved in Exercise 8.19, where youwere asked to show that

T µν ;µ = 0 −→ P′+ 1

2(P+ρ)σ ′ = 0.

(primes denoting partial derivatives with respect tor) for a metric andstress–energy tensor

ds2 = −eσ(r)dt2+ eλ (r)dr2+ r2dθ2+ r2sin2θdϕ2,

T µν =

−ε 0 0 0

0 −P 0 0

0 0 −P 0

0 0 0 −P

We require two additional equations, with the simplest choices being

G00 = 8πT 0

0 G11 = 8πT 1

1

The Einstein tensorsG00 andG11 were derived for this metric in Ex-ercise 8.1. Using contraction with the metric tensor to raise an indexwe obtain from those results

G00 = g00G00 = −e−σ G00 = e−λ

(1r2 −

λ ′

r

)

−1r2

G11 = g11G11 = e−λ G11 = e−λ

(1r2 +

σ ′

r

)

−1r2.


From the preceding equations we find then that we mustsolve

−e−λ(

1r2 −

λ ′

r

)

+1r2 = 8πε(r)

e−λ(

1r2 −

σ ′

r

)

−1r2 = 8πP(r)

P′+ 12(P+ρ)σ ′ = 0.

To proceed we note that the first Einstein equation may be rewrittenas

G00 =

1r2

ddr

[

r(

1− e−λ)]

=2r2

dmdr

= 8πε,

where we have defined a new parameter

2m(r) ≡ r(1− e−λ).

At this pointm(r) is only a reparameterization of the met-ric coefficienteλ since, upon multiplying byeλ ,

eλ =r

r−2m(r)=

(

1−2m(r)

r

)−1

,

but m(r) will be interpreted below as the total mass–energy enclosed within the radiusr. With this interpre-tation we note thateλ is of the Schwarzschild form forroutside the spherical mass distribution of the star.


from the first Einstein equation

G00 =

2r2

dmdr

= 8πε → dm = 4πr2εdr,

and thusm(r) = 4π

∫ r

0ε(r)r2dr,

with an integration constantm(0) = 0 chosen on physical grounds.

Now consider the second Einstein equation

e−λ(

1r2 −

σ ′

r

)

−1r2 = 8πP(r)

Solving it forσ ′ = dσ/dr gives

dσdr

= eλ(

8πrP(r)+1r

)

−1r,

and substitution ofeλ =

rr−2m(r)

leads todσdr

=8πr3P(r)+2m(r)

r(r−2m(r)).

Therefore the preceding two equations may be used to de-fine the metric coefficientseσ andeλ in terms of the pa-rameterm(r) and the pressureP(r).


Finally, we may combine

P′+ 12(P+ρ)σ ′ = 0 and

dσdr

=8πr3P(r)+2m(r)

r(r−2m(r))

to give

dPdr

= −(P(r)+ ε(r))(4πr3P(r)+m(r))

r(r−2m(r)).

Collecting our results, we have obtained theOppenheimer–Volkov equationsfor the structure ofa static, spherical, gravitating perfect fluid

dPdr

=(P(r)+ ε(r))

(m(r)+4πr3P(r)

)

r2

(

1−2m(r)

r

) ,

m(r) = 4π∫ r

0ε(r)r2dr

wherem(r) is the total mass contained within a radiusr


dPdr

=(P(r)+ ε(r))

(m(r)+4πr3P(r)

)

r2

(

1−2m(r)

r

) ,

m(r) = 4π∫ r

0ε(r)r2dr

• Solution of these equations requires specification of an equationof state that relates the density to the pressure.

• They may then be integrated from the origin outward with initialconditionsm(r = 0) = 0 and an arbitrary choice for the centraldensityε(r = 0) until the pressureP(r) becomes zero.

• This defines the surface of the starr = R, with the mass of thestar given bym(R).

• For a given equation of state each choice ofε(0) will give auniqueR andm(R) when the equations are integrated.

• This defines a family of stars characterized by a specific equationof state and the value of a single parameter (the central density,or a quantity related to it like central pressure).


These equations represent the general relativistic (covariant) descrip-tion of hydrostatic equilibrium for a spherical, gravitating perfectfluid.

• The condition of hydrostatic equilibrium was built into the solu-tion through the assumption

uµ = (e−σ/2,0,0,0) = (g−1/200 ,0,0,0).

which constrains the fluid to be static since the 4-velocity has nonon-zero space components.

• They reduce to the Newtonian description of hydrostatic equi-librium in the limit of weak gravitational fields

However, the Oppenheimer–Volkov equations imply significant devi-ations from the Newtonian description in strong gravitational fieldssuch as those for neutron stars. To see this clearly, we may rewritethem in the form

4πr2dP(r) =−m(r)dm(r)

r2

×

(

1+P(r)ε(r)

)(

1+4πr3P(r)

m(r)

)(

1−2m(r)

r

)−1

dm(r) = 4πr2ε(r)dr.

These equations may be interpreted in the following way:


4πr2dP(r)︸︷︷︸

Force acting on shell

=−m(r)dm(r)

r2︸︷︷︸

Newtonian

×

1+P(r)ε(r)︸︷︷︸

GR

1+4πr3P(r)

m(r)︸︷︷︸

GR

1−2m(r)

r︸︷︷︸

GR

−1

dM(r) = 4πr2ε(r)dr︸︷︷︸

Mass–energy of shell

.

• The second equation gives the mass–energy of a shell lying be-tween radiir andr +dr.

• The left side of the first equation is the net force acting outwardon this shell.

• The first factor on the right side of the first equation is the attrac-tive Newtonian gravity acting on the shell because of the massinterior to it.


4πr2dP(r)︸︷︷︸

Force acting on shell

=−m(r)dm(r)

r2︸︷︷︸

Newtonian

×

1+P(r)ε(r)︸︷︷︸

GR

1+4πr3P(r)

m(r)︸︷︷︸

GR

1−2m(r)

r︸︷︷︸

GR

−1

dm(r) = 4πr2ε(r)dr︸︷︷︸

Shell mass–energy

.

• The last three factors on the right side of the first equation—thefactors on the second line—represent general relativity effectscausing deviation from Newtonian gravitation.

• Since all three factors on the second line of the first equation ex-ceed unity as the star becomes relativistic, in general relativitywe find that gravity is consistently stronger than in the corre-sponding Newtonian description of the same problem.

Gravity is enhanced by coupling to pressure in the gen-eral relativistic description.This will ultimately imply thatthere are fundamental limiting masses for strongly gravi-tating objects.

10.2. INTERPRETATION OF THE MASS PARAMETER 243

10.2 Interpretation of the Mass Parameter

The parameterm(r) entering the Oppenheimer–Volkov equations hasbeen interpreted provisionally as the total mass–energy enclosed withina radiusr. Let us now provide some more substantial justification forthis interpretation.

1. Outside a star of radiusR, m(r) becomes equal tom(R), whichis the mass that would be detected through Kepler’s law for theorbital motion if the star were a component of a well-separatedbinary system.

2. In the Newtonian limit, it is clear from

m(r) = 4π∫ r

0ε(r)r2dr,

that m(r) can be unambiguously interpreted as themass con-tained within the radiusr.

3. For relativistic starsm(r) may be consistently split into a con-tribution from arest massm0(r), an internal energyU(r), and agravitational energyΩ(r),

m(r) = m0(r)+U(r)+Ω(r),

as we now demonstrate. Formally we can split the energy den-sity ε into a contribution from the rest mass and one from inter-nal energy,

ε = µ0n+(ε −µ0n),

where the first term is the total rest mass ofn particles of averagemassµ0 and the second term in parentheses is the contributionof internal energy. The proper volume for a spherical shell ofthicknessdr is

dV = 4πr2√

detg11dr = 4πr2√

eλ dr = 4πr2(1−2m/r)−1/2dr.


Thus the total rest mass inside the radiusr is

m0(r) =∫ r

0µ0ndV = 4π

∫ r

0r2(1−2m/r)−1/2µ0ndr,

the total internal energy insider is

U(r) =∫ r

0(ρ −µ0n)dV = 4π

∫ r

0r2(1−2m/r)−1/2(ρ −µ0n)dr,

and the total mass–energy insider is

m(r) = 4π∫ r

0ε(r)r2dr.

Thus, the difference

Ω(r) = m(r)−m0(r)−U(r)

=−4π∫ r

0r2ρ

(

1− (1−2m/r)−1/2)

dr

must be the total gravitational energy insider.

These observations give us some confidence thatm(r) may indeed beinterpreted as the total mass–energy inside the coordinater.

10.3. SOME QUANTITATIVE ESTIMATES FOR NEUTRON STARS 245

10.3 Some Quantitative Estimates for Neutron Stars

Detailed properties of neutron stars require numerical so-lution of the Oppenheimer–Volkov equations with real-istic equations of state. However, many of their basicproperties can be estimated by employing these equations,or even Newtonian concepts, in simpler ways (see Exer-cises).

• The simple assumption that in a neutron star gravitypacks the neutrons down to their hard-core radius oforder 10−13 cm yields that

– The most massive neutron stars contain about3×1057 baryons (mostly neutrons) within a ra-dius of about 7 km, with a mass of about 2.3M⊙.

– This implies an average density> 1015 g cm−3

(several times nuclear matter density).

– This implies a (gravitational) binding energy∼100 MeV (order of magnitude larger than thebinding energy of nucleons in nuclear matter).

• The total gravitational binding energy is within an or-der of magnitude of the rest mass energy, and the es-cape velocity is∼ 50% of the speed of light. Both in-dicate that general relativistic effects are significant.


Although general relativity is important for the overallproperties of neutron stars, over a microscopic scale char-acteristic of nuclear and other sub-atomic interactions themetric is essentially constant.

• Thus the microphysics (nuclear and elementary par-ticle interactions) of the neutron star can be describedby quantum mechanics implemented in flat space-time (special relativistic quantum field theory).

• For neutron stars it is possible to decouple gravity(which governs the overall structure) from quantummechanics (which governs the microscopic proper-ties)

10.4. THE BINARY PULSAR 247

10.4 The Binary Pulsar

The Binary Pulsar PSR 1913+16 (also known as theHulse–Taylor pulsar) was discovered using the Arecibo305 meter radio antenna.

• It is about 5 kpc away, near the boundary of the con-stellations Aquila and Sagitta.

• This pulsar rotates 17 times a second, giving a pulsa-tion period of 59 milliseconds.

• It is in a binary system with another neutron star (nota pulsar), with a 7.75 hour period.

• The precise repetition frequency of the pulsar meansthat it is basically a very high quality clock orbit-ing in a binary system that feels very strong, time-varying gravitational effects.

−→ Precise tests of general relativity


Puls

es/S

econd

Time (hours)

Radia

l Velo

city

(km

/s)

16.93

16.94

16.95

16.96

0 2 4 6 7.75

0

-100

-200

-300PeriastronPeriastron

Receding

Approaching

Figure 10.1:Pulse rate and inferred radial velocity as a function of timefor theBinary Pulsar.

10.4.1 Periodic Variations

The repetition period for a pulsar is associated with the spin of thepulsar and is clock-like in its precision. Thus

• Variations in that period as observed from Earth must be associ-ated with orbital motion in the binary.

• These variations can be used to give very precise informationabout the orbit.

• When the pulsar is moving toward us, the repetition rate of thepulses as observed from Earth will be higher than when the pul-sar is moving away (Doppler effect), and this can be used tomeasure the radial velocity (see Fig. 10.1).


The pulse arrival times vary as the pulsar moves through its orbit

• It takes three seconds longer for the pulses to arrive from the farside of the orbit than from the near side.

• From this, the Binary Pulsar orbit can be inferred to be about amillion kilometers (three light seconds) further away fromEarthwhen on the far side of its orbit than when on the near side.


Periastron

Apastron

Orbital PlaneTilted by 45o

1.1 R

4.8 R

Figure 10.2:Binary Pulsar orbits.

10.4.2 Orbital Characteristics

The orbits determined for the binary system are shown in Fig.10.2.

• Each neutron star has a mass of about 1.4 solar masses.

• The orbits are very eccentric (eccentricity∼ 0.6.).

• The minimum separation (periastron) is about 1.1 solar radii.

• The maximum separation (apastron) is about 4.8 solar radii

• The orbital plane is inclined by about 45 degrees, as viewedfromEarth.


Centero f Mass

Orb i t 3

Orb i t 2

Orb i t 1

P1

P2

P3

Figure 10.3:Precession of the periastron.

• By Kepler’s laws, the radial velocity of the pulsar varies sub-stantially as it moves around its elliptical orbit, as illustrated inFig. 10.1 earlier.

• These orbits are not quite closed ellipses because of precessioneffects associated with general relativity.

– This causes the location of the periastron to shift a smallamount for each revolution (Fig. 10.3).

– The points P1, P2, and P3 are periastrons on three succes-sive orbits (with the amount of precession greatly exagger-ated for clarity


10.5 Precision Tests of General Relativity

The discovery and study of the Binary Pulsar was ofsuch fundamental importance that Taylor and Hulse wereawarded the Nobel Prize in Physics for their work (onlyNobel ever given for relativity). Chief among the reasonsfor this importance is that the Binary Pulsar has providedthe most stringent tests of general relativity available be-fore the discovery of the Double Pulsar.

10.5. PRECISION TESTS OF GENERAL RELATIVITY 253

Centero f Mass

Orb i t 3

Orb i t 2

Orb i t 1

P1

P2

P3

10.5.1 Precession of Orbits

Because spacetime is warped by the gravitational field in thevicinityof the pulsar, the orbit will precess with time.

• This is the same effect as the precession of the perihelion ofMercury, but it is much larger for the present case.

• The Binary Pulsar’s periastron advances by4.2 degrees per year,in accord with the predictions of general relativity.

• In a single day the orbit of the Binary Pulsar advances by asmuch as the orbit of Mercury advances in a century!

10.5.2 Time Dilation

• When the binary pulsar is near periastron, gravity is stronger andits velocity is higher and time should run slower.

• Conversely, near apastron the field is weaker and the velocitylower, so time should run faster.

• It does both, in the amount predicted by GR.


Figure 10.4:Shrinkage of the orbit of the Binary Pulsar because of gravitationalwave emission.

10.5.3 Emission of Gravitational Waves

The revolving pair of masses is predicted by general relativity to ra-diate gravitational waves, causing the orbit to shrink (Fig. 10.4).

• The time of periastron can be measured very precisely and isfound to be shifting.

• This shift corresponds to a decrease in the orbital period by 76millionths of a second per year.

• The corresponding decrease in the size of the orbit by about3.3millimeters per revolution.

10.5. PRECISION TESTS OF GENERAL RELATIVITY 255

1975 1980 1985 1990 1995 2000 2005-40

-35

-30

-25

-20

-15

-10

-5

0C

um

ula

tive

Pe

ria

str

on

Sh

ift (s

)

Year

General Relativity

The quantitative decrease in periastron time is illustrated by the datapoints in the above figure.

• Because the orbital period is short, the shift in periastron arrivaltime has accumulated to more than 30 seconds (earlier) sincediscovery.

• This decay of the size of the orbit is in agreement with the amountof energy that general relativity predicts should be leaving thesystem in the form of gravitational waves (dashed line in figure)

• Although gravitational waves have not yet been detected directly,precision measurements on the Binary Pulsar give strong indi-rect evidence for the correctness of this key prediction of generalrelavity.


10.6 Origin and Fate of the Binary Pulsar

Formation of a neutron star binary is not easy. One of twothings must happen

• A binary must form with two stars massive enough tobecome supernovae and produce neutron stars, andthe neutron stars thus formed must remain bound toeach other through the two supernova explosions.

• The neutron star binary must result from gravita-tional capture of one neutron star by another.

These are improbable events, but not impossible, and theexistence of the Binary Pulsar (and several similar sys-tems) demonstrates empirically that mechanisms exist forit to happen.

10.6. ORIGIN AND FATE OF THE BINARY PULSAR 257

Once a neutron star binary is formed its orbital motion radiates energyas gravitational waves, the orbits must shrink, and eventually the twoneutron starsmust merge.

• Because of the gravitational wave radiation and the correspond-ing shrinkage of the Binary Pulsar orbit (3.3 millimeters per rev-olution), merger is predicted in about 300 million years.

• The sum of the masses of the two neutron stars is likely abovethe critical mass to form a black hole. Therefore,the probablefate of the Binary Pulsar is merger and collapse to a rotating(Kerr) black hole.

• As two neutron stars in a binary approach each other they willrevolve faster (Kepler’s third law).

• This will cause them to emit gravitational radiation more rapidly,which will in turn cause the orbit to shrink even faster.

• Thus, near the end the merger of two neutron stars will pro-ceed rapidly in a positive-feedback runaway and will emit verystrong gravitational waves that may be detectable with current-generation gravitational wave detectors.

These considerations are valid for any binary star system, not just theBinary Pulsar, but the gravitational wave effects are more pronouncedfor binaries involving highly compact objects like neutronstars.


PSR J0737-3039B

To Earth

PSR J0737-3039A

Figure 10.5:Orbital configuration of the Double Pulsar.

10.7 The Double Pulsar

In 2003 a binary neutron star system (the Double Pulsar) was discov-ered in whichboth neutron stars were observed as pulsars in a verytight, partially eclipsing orbit (Fig. 10.5).

• The two neutron stars have masses of 1.3381±0.0007M⊙ (com-ponent A), and 1.2489±0.0007M⊙ (component B).

• They have spin periods of 22.7 ms (component A) and 2.77 s(component B).

• The orbit is slightly eccentric (ε = 0.088).

• The orbit has a mean radius of about 800,000 km (1.25R⊙).

• Thus the orbital period is only 147 minutes, with a mean orbitalvelocity of about 106 km h−1.

• The very fast orbital period and the exquisite timing associatedwith the pulsar clocks has allowed the Double Pulsar to give themost precise tests of general relativity to date.

chapter 10 neutron stars and general...

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