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UMD ASTR 680 - Neutron Star Structure

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Shapiro and Teukolsky, Chapters 2, 8, 9Neutron Star StructureWe now enter the study of neutron stars. Like black holes, neutron stars are one ofthe three possible endpoints of stellar evolution (the remaining one being white dwarfs).Also like black holes, neutron stars are very compact objects, so GR is important in theirdescription. Unlike black holes, they have surfaces instead of horizons, so they are a lotmore complicated than black holes. We’ll start with an overall description of neutron stars,then discuss their high densities and strong magnetic fields.Summary of Neutron StarsA typical neutron star has a mass of 1.5 − 2 M¯and a radius of ∼ 10 km, give or takea few (couple?) of kilometers. It can have a spin frequency up to ∼1 kHz and a magneticfield up to perhaps 1015G or more. Its surface gravity is around 2 − 3 × 1014cm s−2, somountains of even perfect crystals can’t be higher than < 1 mm, meaning that these are thesmoothest surfaces in the universe. They have many types of behavior, including pulsing (inradio, IR, opt, UV, X-ray, and gamma-rays, but this is rarely all seen from a single object),glitching, accreting, and possibly gravitational wave emission. They are the best clocks inthe universe; the most stable are a million times more stable in the short term than thebest atomic clocks. Their cores are at several times nuclear density, and may be composedof exotic matter such as quark-gluon plasmas, strange matter, kaon condensates, or otherweird stuff. In their interior they are superconducting and superfluid, with transitiontemperatures around a billion degrees Kelvin. All these extremes mean that neutron starsare attractive to study for people who want to push the envelope of fundamental theoriesabout gravity, magnetic fields, and high-density matter.High densitiesLet’s start, then, with high densities. An essential new concept that is introducedin high densities is Fermi energy. The easiest way to think about this is in terms of theuncertainty principle,∆p∆x > ¯h . (1)If something is localized to a region of size ∆x, then its momentum must be at least ¯h/∆x.That means that in a dense environment, there is a momentum, and hence an energy,associated with the confinement. Therefore, squeezing something increases its total energy,and this Fermi energy acts as a pressure (sometimes called degeneracy pressure). Theexistence of this energy has a profound role in the structure of white dwarfs, and especiallyneutron stars. In fact, if degeneracy pressure dominates, then unlike normal stars, whichget larger as they get more massive, degenerate stars are smaller at higher masses. Inparticular, an approximate relation is that R ∼ M−1/3for a degenerate star.First, let’s get some basic numbers. If the energy and momentum are low, then theFermi energy EFis related to the Fermi momentum pF∼ ¯h/∆x by EF≈ p2F/2m, wherem is the rest mass of the particle. Since ∆x ∼ n−1/3, where n is the number density ofthe particle, in this nonrelativistic regime EF∼ n2/3. At some point, however, EF> mc2.Then EF∼ pFc, so EF∼ n1/3. For electrons, the crossover to relativistic Fermi energyhappens at a density ρ ∼ 106g cm−3, assuming a fully ionized plasma with two nucleonsper electron. For protons and neutrons the crossover is at about 6 × 1015g cm−3(it scalesas the particle’s mass cubed). The maximum density in neutron stars is no more than1015g cm−3, so for most of the mass electrons are highly relativistic but neutrons andprotons are at best mildly relativistic.Let’s now think about what that means. Ask class: suppose we have matter in whichelectrons, protons, and neutrons all have the same number density. For a low density,which has the highest Fermi energy? The electrons, since at low densities the Fermi energygoes like the inverse of the particle mass. Ask class: given what we said before, whatis the approximate value of the electron Fermi energy when ρ = 106g cm−3? That’s therelativistic transition, so EF≈ mec2≈ 0.5 MeV. Then at 107g cm−3the Fermi energyis about 1 MeV, and each factor of 10 doubles the Fermi energy since EF∼ n1/3in therelativistic regime. What that means is that the energetic “cost” of adding another electronto the system is not just mec2, as it would be normally, but is mec2+ EF. It thereforebecomes less and less favorable to have electrons around as the density increases.Now, in free space neutrons are unstable. This is because the sum of the massesof an electron and a proton is about 1.5 MeV short of the mass of a neutron, so it isenergetically favorable to decay. Ask class: what happens, though, at high density? Ifmp+ me+ EF> mn, then it is energetically favorable to combine a proton and an electroninto a neutron. Therefore, at higher densities matter becomes more and more neutron-rich.First, atoms get more neutrons, so you get nuclei such as120Rb, with 40 protons and 80neutrons. Then, at about 4 × 1011g cm−3it becomes favorable to have free neutronsfloating around, along with some nuclei (this is called “neutron drip” because the effect isthat neutrons drip out of the nuclei). At even higher densities, the matter is essentiallya smooth distribution of neutrons plus a ∼ 5 − 10% smattering of protons and electrons.At higher densities yet (here we’re talking about nearly 1015g cm−3), the neutron Fermienergy could become high enough that it is favorable to have other particles appear.It is currently unknown whether such particles will appear, and this is a focus of muchpresent-day research. If they do, it means that the energetic “cost” of going to higherdensity is less than it would be otherwise, since energy is released by the appearance ofother, exotic particles instead of more neutrons. In turn, this means that it is easier tocompress the star: squashing it a bit doesn’t raise the energy as much as you would havethought. Another way of saying this is that when a density-induced phase transition occurs(here, a transition to other types of particles), the equation of state is “soft”.Well, that means that it can’t support as much mass. That’s because as more massis added, the star compresses more and more, so its gravitational compression increases.If pressure doesn’t increase to compensate, in it goes and forms a black hole. What allthis means is that by measuring the mass and radius of a neutron star, or by establishingthe maximum mass of a neutron star,


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