Superfluidity of neutron stars

What Are Neutron Stars?

Nick: please MAKE SURE to add: (1) some pictures, and (2) references and links with more detailed information...

Stellar Evolution

Neutron stars are one of the many possible ends to the life of a star. All stars begin as space dust, usually in a giant cloud. Should perturbations in the density of cloud permit enough mass to be in a certain radius, the cloud will collapse. As the dust free-falls the dust will form a disk, most of which will end up in the star and the rest will form planets, moons, asteroids, and meteoroids. Then assuming that the dust cloud was massive enough, the core of the condensed cloud will ignite forming a main sequence star. This phase of stellar life is characterized by hydrogen burning via the pp-chain or CNO-cycle. Some stars (those of less than 0.7 solar masses) finish their lives here as the core contraction after the red giant phase cannot raise the temperature high enough to burn helium. These very low mass stars will become white dwarfs. The rest will then move on to helium burning in the phase known as the horizontal branch. After the helium shell burning the next mass cutoff occurs. Any star less than 8 solar masses will then turn into white dwarf as the core will not become hot enough to produce carbon. The stars that are still massive enough will then begin burning helium to carbon and oxygen, carbon to neon, neon and oxygen to silicon, and silicon to nickel (nickel will then beta decay to cobalt and cobalt then decays to iron). Once the star produces nickel it is doomed as it cannot gain any more energy by fusing elements together. Of course mass limits the exact stopping stage, but whatever the cause, when the burning does cease these stars explode.

The end state of any star occurs suddenly. Stars maintain a balance between the inward gravitational pressure and the outward energy pressure. But the star eventually cannot burn past a certain stage. The core of the star shrinks rapidly until the atoms squeeze next to each other as the electron clouds are what is holding the atoms apart. The outward pressure of this state, the white dwarf, is known as the electron degeneracy pressure. For larger mass stars, the core shrinks so rapidly that there is too much pressure it is then favorable for the protons and electrons to react to form neutrons. The reverberation from this rapid compression is the classical cause of the supernova explosion. This way the neutrons become degenerate and the star becomes a neutron star. If the mass then shrinks below the Schwarzschild radius, the star becomes a black hole. The Schwarzschild radius is given by:

$r_{s}=(3M/M_{\odot })km$

where M is the mass of the star and the M with the dot is the mass of the sun. Interestingly, after the red supergiant phase (after helium burning has ceased) the time it takes for the energy from the core to make it to the surface is 10,000 years. However, the rest of the star's life will last less than 1000 years. This means that the surface of the star has no idea that the star is dead until it explodes.

There is another method of neutron star formation. This method occurs only in binary star systems. After one of the two stars has become a white dwarf it may strip mass from its partner. After the white dwarf has stolen enough mass to reach the Chandrasekhar limit of 1.4 solar masses the white dwarf explodes leaving behind a neutron star (or black hole if the accretion is fast enough).

Properties of the Neutron Star

Magnetism

Neutron stars are formed entirely of neutrons which are fermions. Neutrons like all particles have a magnetic spin associated with them despite being uncharged.

caption

Extreme Densities

As mentioned before, a star with eight times the mass of the sun is eligible to be a neutron star. The neutron star itself will not retain this mass as much of it will be blown away in the explosion. A white dwarf (which the sun will one day be) is usually about one hundredth the size of the sun, or about the size of the earth. Neutron stars are obviously much smaller than this. One can use a simple calculation to approximate the size of this star. From statistical physics one has the following relations for a gas:

The particle density: $n=\int _{0}^{\infty }{\omega (E)f(E)dE}\;$

The pressure: $P={1 \over 3}\int _{0}^{\infty }{pv\omega (E)f(E)dE}\;$

where omega is the state density, f is the probability distribution, p is the momentum, and v is the velocity. These values for a non-relativistic degenerate Fermi gas are:

$\omega (E)=4\pi {\frac {(2m)^{3/2}}{h^{3}}}E^{1/2}\;$

$f(E)={\frac {1}{e^{\frac {E-\mu }{k_{B}T}}+1}}\;$

$p={\sqrt {2mE}}\;$

$v={\sqrt {\frac {2E}{m}}}\;$

Then plugging these in to the equations for n and P.

$n=4\pi {\frac {(2m)^{3/2}}{h^{3}}}\int _{0}^{\infty }{{\frac {E^{1/2}}{e^{\frac {E-\mu }{k_{B}T}}+1}}dE}\;$

$P={\frac {8\pi }{3}}{\frac {(2m)^{3/2}}{h^{3}}}\int _{0}^{\infty }{{\frac {E^{3/2}}{e^{\frac {E-\mu }{k_{B}T}}+1}}dE}\;$

Then one can let $x={\frac {E}{k_{B}T}}\;$ one gets:

$n=4\pi {\frac {(2mk_{B}T)^{3/2}}{h^{3}}}\int _{0}^{\infty }{{\frac {x^{1/2}}{e^{x-{\frac {\mu }{k_{B}T}}}+1}}dx}\;$

$P={\frac {8\pi }{3}}{\frac {(2m)^{3/2}(k_{B}T)^{5/2}}{h^{3}}}\int _{0}^{\infty }{{\frac {x^{3/2}}{e^{x-{\frac {\mu }{k_{B}T}}}+1}}dx}\;$

These integrals are known as Fermi integrals which can be looked up in tables. But the general form is:

$F_{n}(\alpha )=\int _{0}^{\infty }{{\frac {x^{n}}{e^{x+\alpha }+1}}dx}\;$

Therefore the two integrals become:

$n=4\pi {\frac {(2mk_{B}T)^{3/2}}{h^{3}}}F_{1/2}({\frac {-\mu }{k_{B}T}})\;$

$P={\frac {8\pi }{3}}{\frac {(2m)^{3/2}(k_{B}T)^{5/2}}{h^{3}}}F_{3/2}({\frac {-\mu }{k_{B}T}})\;$

Then one can substitute n into P and get:

$P={\frac {2}{3}}nk_{B}T{\frac {F_{3/2}({\frac {-\mu }{k_{B}T}})}{F_{1/2}({\frac {-\mu }{k_{B}T}})}}\;$

Now one can take the limit as the argument of the Fermi function goes to negative infinity to represent the extreme degeneracy and get:

$\lim _{{\frac {-\mu }{k_{B}T}}\to -\infty }F_{3/2}({\frac {-\mu }{k_{B}T}})={\frac {2}{5}}({\frac {\mu }{k_{B}T}})^{5/2}$

$\lim _{{\frac {-\mu }{k_{B}T}}\to -\infty }F_{1/2}({\frac {-\mu }{k_{B}T}})={\frac {2}{3}}({\frac {\mu }{k_{B}T}})^{3/2}$

Which makes n and P:

$n=4\pi {\frac {(2mk_{B}T)^{3/2}}{h^{3}}}({\frac {2}{3}}({\frac {\mu }{k_{B}T}})^{3/2})\;$

$P={\frac {2}{3}}nk_{B}T({\frac {3}{5}}{\frac {\mu }{k_{B}T}})\;$

Then solving for $\mu /(k_{B}T)$ in the n equation and substituting into P:

${\frac {\mu }{k_{B}T}}=({\frac {3}{8\pi }}n)^{2/3}{\frac {h^{2}}{2mk_{B}T}}\;$

$P={\frac {h^{2}}{20m}}({\frac {3}{\pi }})^{2/3}n^{5/3}\;$

This is the equation for the pressure of an extremely degenerate Fermi gas. Pairing this with the equation for Hydrostatic Equilibrium one can get the radius of the star. The Hydrostatic Equilibrium equation is:

${\frac {\partial P}{\partial R}}=-{\frac {GM\rho }{R^{2}}}\;$

Here G is the gravitational constant, M is the mass of the star, and rho is the mass density. Integrating with the assumption of constant mass and density one gets:

$P={\frac {GM\rho }{R}}$

Now one needs to equate the two and solve. However to make this a little easier one can do the following to the density variables n and rho.

$n={\frac {\rho }{m_{N}}}\;$

$\rho ={\frac {3M}{4\pi R^{3}}}$

Though this may seem strange, this changes the density of particles to the mass density. Then makes the assumption of uniform density. Then solving one gets:

$R={\frac {h^{2}}{20}}({\frac {9}{4\pi ^{2}}})^{2/3}({\frac {1}{m_{N}}})^{8/3}{\frac {1}{GM^{1/3}}}$

From here one can input numbers. So a 1.5 solar mass star will have a radius of about 12 km. This is a simple calculation made with some approximations. The calculation however is a good judge of the order of magnitude of the true result. From this one can see that the entire sun can be reduced to the size of Tallahassee if it became a neutron star. As another order of magnitude result, the mass density of the star would be on the order of 4E14 g/cm^3. So one teaspoon of this would weigh roughly 4.5E12 lbs. Of course these calculations are crude but give a general idea of the order of magnitude of how dense this stellar body is.