4
M. C. Miller
¯
h = 1.05457 × 10 −27 erg s is the reduced Planck constant. Done more precisely
and assuming isotropic matter we find that this minimum is the Fermi momentum
p F = (3π 2 ¯
h
3 n) 1/3 . The Fermi energy adds to the rest-mass energy via E tot =
(m 2 c 4 + p 2
F c 2 ) 1/2 = mc 2 + E F , and is E F ≈ p 2
F /2m for p F mc and E F ≈ p F c
for p F mc, where c = 2.99792458 × 10 10 cm s −1 is the speed of light.
Matter is degenerate when E F > kT , and strongly degenerate when E F kT ,
where k = 1.38065 × 10 −16 erg K −1 is the Boltzmann constant and T is the
temperature. As a result, electrons (with their low masses m e = 9.109382 ×
10 −28 g = 0.510999 MeV/c 2 ) become degenerate at much lower densities than
do neutrons or protons. Matter dominated by nuclei heavier than hydrogen has
∼2 baryons per electron, and hence electrons become relativistically degenerate
(p F = m e c) at a density of ≈2 × 10 6 g cm −3 . In addition, above ≈2.5 × 10 7 g cm −3
the total energy of electrons becomes larger than m n c 2 −m p c 2 = 1.294 MeV, where
m n = 1.674927 × 10 −24 g = 939.566 MeV/c 2 and m p = 1.672622 × 10 −24 g =
938.272 MeV/c 2 are respectively the rest masses of neutrons and protons. As a
result, at these and greater densities electrons and protons can undergo inverse beta
decay e − + p → n + ν e . At higher densities the ratio of neutrons to protons in
nuclei increases, and then at the “neutron drip” density ρ nd ≈ 4.3 × 10 11 g cm −3
neutrons are stable outside nuclei. The neutron drip density is derived in detail in,
e.g., [22], but a good estimate can be obtained by a simple argument. Neutrons
can drip out of the nucleus when the total electron energy per nucleon equals
the nuclear binding energy per nucleon (as described in [22] there are small
corrections due to lattice energy and the nonzero energy of neutron continuum
states). The binding energy per nucleon is ∼8 MeV (see, e.g., Table 3 of [22]).
At high densities, Z/A ∼ 0.3 in contrast to the Z/A ∼ 0.5 common at lower
densities. Thus the condition on the electron Fermi energy for neutron drip is
E F,e ≈ 8(A/Z) MeV. The density at which this happens is therefore approximately
2 × 10 6 g cm −3 (0.5/0.3)[(8 MeV/0.3)/0.5 MeV] 3 ≈ 4 × 10 11 g cm −3 , where we
extrapolate from the density at which the electron Fermi energy becomes relativistic.
At infinite density, equilibrium matter consisting of just neutrons, protons, and
electrons would have eight times as many neutrons as protons (and electrons,
because charge balance has to be maintained). To see this, note that at infinite density
all species are ultrarelativistic and their chemical potentials are thus dominated by
their Fermi energies. Charge balance means that n p = n e , so equilibrium implies
E F,n = E F,p + E F,e −
= 2E F,p
n
1/3
n = 2n
1/3
p
n n = 8n p .
(1.1)
For a neutron star with a canonical mass M = 1.4 M (where M =
1.989 × 10 33 g is the mass of the Sun) and radius R = 10 km that for simplicity
we will treat as made entirely of free neutrons, the average number density is
n = (M/m n )/(4πR 3 /3) = 4×10 38 cm −3 . This implies p F = 2.4×10 −14 g cm s −1
and thus E F ≈ 2 × 10 −4 erg= 100 MeV. This corresponds to a temperature of
M. C. Miller
¯
h = 1.05457 × 10 −27 erg s is the reduced Planck constant. Done more precisely
and assuming isotropic matter we find that this minimum is the Fermi momentum
p F = (3π 2 ¯
h
3 n) 1/3 . The Fermi energy adds to the rest-mass energy via E tot =
(m 2 c 4 + p 2
F c 2 ) 1/2 = mc 2 + E F , and is E F ≈ p 2
F /2m for p F mc and E F ≈ p F c
for p F mc, where c = 2.99792458 × 10 10 cm s −1 is the speed of light.
Matter is degenerate when E F > kT , and strongly degenerate when E F kT ,
where k = 1.38065 × 10 −16 erg K −1 is the Boltzmann constant and T is the
temperature. As a result, electrons (with their low masses m e = 9.109382 ×
10 −28 g = 0.510999 MeV/c 2 ) become degenerate at much lower densities than
do neutrons or protons. Matter dominated by nuclei heavier than hydrogen has
∼2 baryons per electron, and hence electrons become relativistically degenerate
(p F = m e c) at a density of ≈2 × 10 6 g cm −3 . In addition, above ≈2.5 × 10 7 g cm −3
the total energy of electrons becomes larger than m n c 2 −m p c 2 = 1.294 MeV, where
m n = 1.674927 × 10 −24 g = 939.566 MeV/c 2 and m p = 1.672622 × 10 −24 g =
938.272 MeV/c 2 are respectively the rest masses of neutrons and protons. As a
result, at these and greater densities electrons and protons can undergo inverse beta
decay e − + p → n + ν e . At higher densities the ratio of neutrons to protons in
nuclei increases, and then at the “neutron drip” density ρ nd ≈ 4.3 × 10 11 g cm −3
neutrons are stable outside nuclei. The neutron drip density is derived in detail in,
e.g., [22], but a good estimate can be obtained by a simple argument. Neutrons
can drip out of the nucleus when the total electron energy per nucleon equals
the nuclear binding energy per nucleon (as described in [22] there are small
corrections due to lattice energy and the nonzero energy of neutron continuum
states). The binding energy per nucleon is ∼8 MeV (see, e.g., Table 3 of [22]).
At high densities, Z/A ∼ 0.3 in contrast to the Z/A ∼ 0.5 common at lower
densities. Thus the condition on the electron Fermi energy for neutron drip is
E F,e ≈ 8(A/Z) MeV. The density at which this happens is therefore approximately
2 × 10 6 g cm −3 (0.5/0.3)[(8 MeV/0.3)/0.5 MeV] 3 ≈ 4 × 10 11 g cm −3 , where we
extrapolate from the density at which the electron Fermi energy becomes relativistic.
At infinite density, equilibrium matter consisting of just neutrons, protons, and
electrons would have eight times as many neutrons as protons (and electrons,
because charge balance has to be maintained). To see this, note that at infinite density
all species are ultrarelativistic and their chemical potentials are thus dominated by
their Fermi energies. Charge balance means that n p = n e , so equilibrium implies
E F,n = E F,p + E F,e −
= 2E F,p
n
1/3
n = 2n
1/3
p
n n = 8n p .
(1.1)
For a neutron star with a canonical mass M = 1.4 M (where M =
1.989 × 10 33 g is the mass of the Sun) and radius R = 10 km that for simplicity
we will treat as made entirely of free neutrons, the average number density is
n = (M/m n )/(4πR 3 /3) = 4×10 38 cm −3 . This implies p F = 2.4×10 −14 g cm s −1
and thus E F ≈ 2 × 10 −4 erg= 100 MeV. This corresponds to a temperature of