Neutron stars

the best studied supernovae to date, and through Earth-based detections of (a mere) 20 neutrinos from the supernova, confirmed many of the theories about how supernovae behave.

● The SNR Cas A (see Figure 7.2) had for a long time shown no evidence for a compact stellar remnant, but new images taken in X-rays seem to reveal one. If this is correct, was Cas A a type Ia or type II supernova?

❍ Type II. Type Ia completely disrupt the stars, leaving a brilliant ‘fireball’ but no compact remnant.

(c) (a)

(d) (b)

Figure 7.2 The appearance of

a supernova remnant (SNR) depends greatly on the

waveband observed. Cassiopeia A (known as Cas A), the remnant of a ∼ 300 year-old supernova in the Galaxy, is inconspicuous at optical wavelengths (a), where many stars in the field radiate with similarly brightness, but in infrared (b), radio (c), and X-ray wavelengths (d) it is the only source visible. The X-ray image, which is the most recent, shows a probable stellar remnant for the first time, and also shows that the gaseous remnant consists of two shock fronts, an inner one caused by the collision between the supernova ejecta with the circumstellar shell, which is heated to ∼ 107_{K, and}

an outer shock caused possibly by a sonic boom running ahead of the expanding ejecta.

7.2 Neutron stars

Core-collapse supernovae leave behind one of two remnants: either a neutron star or a black hole. These two classes of object will now be examined in turn. A newly formed neutron star will have a temperature of (1011_–1012_{) K but it will}

quickly cool, by emission of neutrinos, to around 109 _{K on a timescale of a day,}

and to around 108_{K within a hundred years. It will also have a mass of order} ∼ M&contained within a sphere only ∼ 10 km in radius, hence it will have a

density of a few ×1017_{kg m}−3_{. These extreme temperatures and densities lead to}

some extraordinary consequences for the neutron star’s properties.

7.2.1 Neutron star composition

For normal matter, the most stable configuration, with the most negative binding energy per nucleon, is nuclei around iron-56. When the density approaches 1014_{kg m}−3_{, the most stable nuclei are actually far more neutron-rich,}

namely those such as nickel-78 and iron-76. At still higher densities, above 4 × 1014_{kg m}−3_{, a phenomenon called neutron drip occurs, whereby neutrons}

‘leak out’ of nuclei and an equilibrium mixture of nuclei, neutrons and electrons exists. When the density exceeds that of normal nuclear matter (at about

2 × 1017_{kg m}−3_{), the nuclei begin to merge and a dense gas of protons, neutrons}

and electrons is produced. Under these conditions there are complicated and uncertain interactions between nuclei, so the equation of state for material with these properties is not known.

Despite this, a reasonable idea of the composition of a neutron star can be gained by considering a crude model in which the neutron star is treated as an ideal gas of degenerate electrons, protons and neutrons. The normal β−_{-decay of free}

neutrons is blocked in this situation because of the Pauli exclusion principle: there are no available quantum states left for the protons and electrons to occupy, so the neutrons are not able to decay. In particular, neutrons cannot decay if the Fermi energy of the neutrons is less than the sum of the Fermi energies of the electrons and protons, but they can decay if the Fermi energy of the neutrons is greater than that of the protons plus electrons. An equilibrium will therefore exist when:

EF(n) = EF(p) + EF(e).

This is equivalent to a relationship between the chemical potentials of the neutrons, protons and electrons.

Now, the neutrons and protons can be considered to be non-relativistic, hence their Fermi energies are:

EF(n) = mnc2+p 2 F(n) 2mn EF(p) = mpc2+p 2 F(p) 2mp

respectively, where mnand mpare the masses of the neutron and proton, and pF(n) and pF(p) are their Fermi momenta.

Conversely, the electrons are much less massive then the protons and neutrons, so they can be considered to be ultra-relativistic. Their Fermi energy is:

EF(e) = pF(e)c

where pF(e) is the electron’s Fermi momentum.

Now, since the Fermi momentum is pF = [3n/8π]1/3h, we can write the

equilibrium condition as:

mnc2+ , 3nn 8π -_2/3 h2 2mn = mpc 2₊,3np 8π -_2/3 h2 2mp + , 3ne 8π -_1/3 hc.

7.2 Neutron stars

Since the gas will be neutral, np = ne. Furthermore, the mass energy difference

between the neutron and proton is mnc2− mpc2= 1.3 MeV, so we can rearrange

this as: , 3np 8π -_1/3 hc + , 3np 8π -_2/3 h2 2mp − , 3nn 8π -_2/3 h2 2mn = 1.3 MeV. (7.1)

This equation can be solved numerically to find the ratio of neutrons to protons at a given density. For instance, at a density of 2.5 × 1017_{kg m}−3_{, the ratio is}

nn/np ∼ 200.

Worked Example 7.3

For a density of 2.5 × 1017_{kg m}−3_{, verify that Equation 7.1 predicts a}

neutron-to-proton ratio of about 200 to 1.

Solution

The number density and mass density are related by n = ρX/m, where X is the mass fraction. Since the neutron star is dominated by neutrons,

Xn ∼ 1 and we have nn ∼ 2.5 × 1017kg m−3/ 1.675 × 10−27kg ∼ 1.49 × 1044_m−3_.

Hence the stated ratio implies np≈ 7.45 × 1041m−3.

So, the first term on the left-hand side of Equation 7.1 becomes: , 3 × 7.45 × 1041_m−3 8π -1/3 × (6.626 × 10−34J s) × (2.998 × 108 m s−1) = 8.87 × 10−12J.

Similarly, the second term on the left-hand side of Equation 7.1 becomes: ,

3 × 7.45 × 1041_m−3

8π

-2/3 _{(6.626 × 10−34J s)2}

2 × 1.673 × 10−27_kg = 0.26 × 10−12J

and the third term on the left-hand side of Equation 7.1 becomes: ,

3 × 1.49 × 1044_m−3

8π

-2/3 _{(6.626 × 10−34J s)2}

2 × 1.675 × 10−27_kg = 8.93 × 10−12J.

So, the entire left-hand side of Equation 7.1 is (8.87+ 0.26− 8.93) × 10−12_J

= 0.20 × 10−12_{J or (0.20 × 10}−12_{/1.602 × 10}−19_{) eV which is 1.25 MeV.}

This is close enough to the 1.3 MeV energy difference between the neutron and proton, given the approximate ratio of 200 : 1 which we used as the starting point for the calculation.

So, as long as there is one proton and one electron for every 200 neutrons, the neutrons are prevented from decaying, and the neutron star remains supported, essentially by the pressure of degenerate neutrons. We see, then, that neutron stars are aptly named, and why they are not, for example, called nucleon stars.

7.2.2 The radius of a neutron star

When deriving equations to describe neutron stars we should be wary of

oversimplifying the assumptions about conditions inside a neutron star. In reality, Newtonian gravity should be replaced by Einstein’s general-relativistic treatment, because for a neutron star of mass M and radius RNS, a neutron’s gravitational

potential energy, GMmn/RNS, is comparable to its rest-mass energy, mnc2.

● Calculate the ratio of a neutron’s gravitational potential energy to its rest-mass energy, inside a neutron star of mass 1.4 M&and radius 10 km.

❍ The ratio is GMmn/RNSmnc2 = GM/RNSc2. This is equal to (6.673 ×

10−11_{N m}2 _kg−2_{× 1.4 × 1.99 × 10}30_kg)/104_{m × (2.998 × 10}8_{m s}−1₎2₌

0.2.

The momenta of neutrons also approach the relativistic limits, requiring special relativity. Nevertheless, the non-relativistic approximation can lead to useful insights into the size of the star, and an understanding of the dominant physics. In particular, it is interesting to compare calculations of the radius of a white dwarf, supported by degenerate electrons, and a neutron star, supported by degenerate neutrons. The following question takes you through such a calculation. (You may find it helpful to revise your study of white dwarfs as you complete this exercise.)

Exercise 7.3 Set up two columns on a sheet of paper, one for a white dwarf and