The protons a star cannot afford

The pressure that ionises squeezed hydrogen until its atoms had no room for a bound electron, and found the freed electrons crowded into a degenerate Fermi sea. That was the end of the density axis as far as atoms are concerned. It is not the end of the axis. Keep squeezing and the next thing to give way is the proton itself. The electrons crowd so tightly that it becomes cheaper for matter to absorb them into its protons, making neutrons, than to keep adding them to the top of the sea.

The textbook sentence is that in a neutron star gravity forces electrons into protons. That is true as far as it goes, and it leaves out the thing that decides what happens. Gravity supplies the density. What converts the matter, how far the conversion goes and why it stops are settled by the same condition that settles every chemical equilibrium in this subject: the chemical potentials on the two sides of a reaction must be equal. The reaction here is beta decay, and running it with the potentials of three crowded Fermi gases explains why a neutron star is made of neutrons. It also explains why a free neutron, which lives fifteen minutes on its own, lives for ever inside one.

One reaction, and the balance it has to reach #

A free neutron decays into a proton, an electron and an antineutrino, releasing 0.782 MeV of kinetic energy. The reverse reaction, a proton capturing an electron and emitting a neutrino, is allowed too, and in a laboratory it needs an electron that brings at least that 0.782 MeV with it. The neutrinos leave. At the densities considered here the matter is transparent to them, so they carry away energy but hold no population and have no chemical potential to contribute. The reaction that cannot go all the way found that equilibrium in any reaction means the chemical potentials, summed with their stoichiometric signs, balance. For beta decay that is one line:

$$\mu_n = \mu_p + \mu_e .$$

Add the condition that the star is electrically neutral, so there is one electron for every proton, and the composition at any density follows. Each species is a cold Fermi gas, and at zero temperature its chemical potential is simply the energy of the last particle added, the top of its sea, rest energy included. For a gas of density $n$ the top of the sea sits at a momentum $\hbar k$ with $k^3 = 3\pi^2 n$, and the energy there is $\sqrt{(\hbar c k)^2 + m^2c^4}$.

The figure solves that condition at every density from a white dwarf’s to several times that of an atomic nucleus. At low density there is no contest. A neutron costs 1.293 MeV more rest energy than a proton, a neutron would have to come from a proton and an electron together, and the electrons, filling a shallow sea, cannot supply the difference. The matter is all protons and electrons.

At $1.2\times10^{7}$ grams per cubic centimetre the top of the electron sea reaches exactly that difference. From there on, an electron arriving at the top of the sea has more energy than it would take to convert a proton into a neutron, so it is swallowed, and the proton fraction collapses. It falls by more than three powers of ten in the next five powers of ten of density. Its lowest value, 0.014 per cent, comes where the electrons have become highly relativistic while the neutrons, far heavier, are still gentle.

At nuclear density, $2.7\times10^{14}$ g/cm³, the fraction is 0.49 per cent. Beyond it the fraction turns and rises slowly, because the neutron sea is now deep enough to be expensive in its turn. It rises towards one ninth, a limit it approaches from below and never reaches. That limit reappears at the end of this essay as the most consequential number in it.

What the figure computes is the ideal case. Nucleons attract one another, and in a real star below about half nuclear density the matter is not a uniform gas of free nucleons at all but a lattice of neutron-rich nuclei in a sea of electrons and, deeper down, of free neutrons too. The dashed stretch of the curve marks that region honestly: the chemical-potential balance still holds there, but the nuclei take their own share of the protons, and the numbers change. What the free-nucleon curve gets right, and what survives every refinement, is the order of events. First the protons go, and then a slow recovery sets in that is capped by a geometry of momenta.

Three seas that have to end at one level #

Why so few protons? The balance says that a neutron at the top of its sea must have exactly the energy of a proton at the top of its sea plus an electron at the top of its. The seas have very different depths for the same number of particles, because their masses differ. A sea’s depth is its Fermi energy, and for the same density a light particle’s Fermi energy is far larger than a heavy one’s. At a given momentum the kinetic energy of a non-relativistic particle falls as its mass rises. An electron’s rest mass is 1,836 times smaller than a proton’s, and once it is relativistic its energy is simply $\hbar c k$, with no mass in it to slow the climb.

The figure draws the three seas as bars on one energy scale, measured from a proton at rest, and stacks the electron’s bar on top of the proton’s. The stacking is the reaction: a neutron that decays must produce a proton and an electron together. The dashed line is the level all three must share at equilibrium.

At nuclear density the neutron sea runs from its rest energy, 1.29 MeV above a proton’s, up to 57.9 MeV. The proton sea, holding only half a per cent as many particles, barely rises: its top is 1.68 MeV above rest. The electron sea holds exactly as many particles as the proton sea, and its top is 56.2 MeV up, because electrons are light and relativistic.

That is the whole explanation. A proton by itself is cheap. What makes it unaffordable is the electron that neutrality forces it to bring, and that electron has to be placed at the top of a sea already 56 MeV deep. A neutron, by contrast, only has to be placed at the top of its own sea. The composition adjusts until adding either costs the same, and with free nucleons that happens when protons are one nucleon in two hundred.

The three columns also show why the equilibrium moves with density. At half nuclear density the neutrons fill to 37.4 MeV and the electrons need 36.6 MeV; at four times nuclear density the neutrons fill to 137 MeV and the electrons need 128 MeV. The electron bar grows as the cube root of the proton number, and the neutron bar, once relativistic effects set in, eventually grows the same way. When both are ultra-relativistic, the momenta alone decide the balance. The proton’s own sea stays thin throughout, a few MeV at most, which is the visible form of the one-in-two-hundred fraction.

This is the same kind of accounting as in the product doping cannot move, where adding donors to silicon suppresses its holes because a reaction couples their populations. The coupling here runs the other way. Fixing the electron count to the proton count means the sea that is expensive to fill starves the sea that is cheap to fill, and it is the proton population that pays.

The minimum of one curve #

A chemical potential is a slope: the change in total energy for one more particle of a given kind. So the same equilibrium can be found without any chemical potentials at all. Fix the density and ask which composition has the lowest energy.

The figure plots the energy per nucleon of the three gases together, at fixed total nucleon density, as the share of protons is varied. To the right of the minimum each extra proton brings an electron at the top of an ever deeper sea, and the energy climbs steeply. To the left the protons are converted back into neutrons, which then have to be stacked above the neutron sea, and the energy rises again, more gently. The minimum lies at 0.49 per cent at nuclear density and at 1.5 per cent at four times that. In both cases the minimum sits exactly where the chemical-potential balance put it, which is a check on the arithmetic as much as a restatement of it: the slope of this curve, per proton converted, is $\mu_p + \mu_e - \mu_n$, and it vanishes at the bottom.

Two things about the curve are worth reading off, and neither shows in the first figure. The minimum is shallow. Pure neutron matter, with no protons at all, costs only 72 keV a nucleon more than the equilibrium at nuclear density, against a neutron Fermi energy of about 57 MeV. The matter does not strongly prefer its protons; it tolerates them. And the curve is asymmetric. Adding protons beyond the equilibrium costs far more than removing them, because electrons are expensive and neutrons are not.

The shallowness has a consequence. Real nucleon–nucleon forces add an energy, the symmetry energy, which penalises an excess of either kind of nucleon over the other. Its size at nuclear density is about 30 MeV a nucleon for pure neutron matter, which is hundreds of times the 72 keV by which the free gas prefers its equilibrium. So the symmetry energy moves the minimum a long way. Estimates with realistic forces put the proton fraction at nuclear density near 4 to 5 per cent, ten times the free-gas value. The free gas gets the structure of the answer right and its size wrong by a factor that depends on nuclear physics still being measured.

This is the standard habit of equilibrium thermodynamics, which the reaction that cannot go all the way applied to a flask and which applies unchanged to a star: equilibrium is a minimum, and the chemical potentials are its slopes. What differs here is only where the curvature comes from. In a flask it is the entropy of mixing, whose slope is infinite at the pure ends. At zero temperature there is no entropy, and the curvature comes entirely from the exclusion principle, which makes every added particle more expensive than the last.

A decay with nowhere to put its electron #

The same balance explains the most striking fact about these stars: their neutrons are stable. A free neutron has a mean life of about 878 seconds. A neutron star is a kilometre-sized object made mostly of neutrons that has existed for millions of years. Nothing has to be added to beta decay to reconcile the two. The decay has to put its electron somewhere, and in a crowded star there is nowhere to put it.

The electron from a neutron’s decay comes out with a spread of kinetic energies from zero to the 0.782 MeV end point, shared with the antineutrino in proportions fixed by the available phase space. The figure draws that spectrum and marks the tops of three electron seas. No two electrons can occupy the same state, so every decay whose electron would land below the sea’s top is forbidden. Only the part of the spectrum above the marked line survives.

The lifetime lengthens in proportion to how much survives. With the sea filled to 0.1 MeV, a density of $1.7\times10^{35}$ electrons per cubic metre, 89 per cent of the decays remain open and the neutron lives 992 seconds. With it filled to 0.3 MeV, 48 per cent remain and the lifetime nearly doubles. At 0.6 MeV only 4.2 per cent remain and the neutron lives six hours. When the sea’s top reaches the end point, at $7.4\times10^{36}$ electrons per cubic metre, no decay is open at all. That is the same density at which the first figure began converting protons into neutrons, and it has to be. The point where a neutron stops decaying and the point where a proton starts capturing are the same balance, approached from its two sides.

At nuclear density the electron sea is 56 MeV deep, seventy times the decay’s end point. A neutron there is as stable as a proton. It would decay only if an electron were removed from the sea, which is how a star that is disturbed, by compression or by cooling, relaxes back to equilibrium through the modified processes described below. The same blocking suppresses collisions in any degenerate Fermi system, the effect the collisions the exclusion principle forbids worked out for electrons in a metal. Here it suppresses a decay rather than a scattering, and it does so completely rather than as a power of the temperature.

The density at which each nucleus gives way #

Before matter reaches nuclear density it passes through a long stage in which it is made of nuclei, and the same balance applies to each of them. A nucleus with $Z$ protons can capture an electron and become the nucleus with one proton fewer and one neutron more. It does so when the electrons’ Fermi energy reaches the threshold for that capture. The threshold is the mass difference between the two nuclei, which the tables supply, and it has nothing to do with gravity.

The figure lists seven. Hydrogen converts first, at $1.2\times10^{7}$ g/cm³, because a free proton needs only 0.782 MeV. Among the rest the order is set by how unfavourable the neutron-richer neighbour is. Iron-56 already carries four more neutrons than protons, its neighbour manganese-56 is only 3.7 MeV heavier, and it captures at $1.1\times10^{9}$ g/cm³. Magnesium-24 and neon-20 follow at $3\times10^{9}$ and $6\times10^{9}$. Oxygen and carbon, whose neighbours are much heavier, hold out to $2\times10^{10}$ and $4\times10^{10}$. Helium-4 is the extreme case: its neighbour with one more neutron does not exist as a bound nucleus, the capture must break it apart, and it needs electrons of 20.6 MeV, reached only at $1.4\times10^{11}$ g/cm³. The tightly bound nucleus is the first to give way, and the one with the most symmetric neighbours is the last. The order has nothing to do with binding per nucleon and everything to do with the neighbours.

These numbers decide the fate of real stars. The core of a star of eight to ten solar masses ends its life made of oxygen, neon and magnesium, with its weight held up by the pressure of degenerate electrons, the pressure the mass no cold matter can hold up found to have a limit. As the core grows, its central density crosses the capture thresholds of magnesium and neon at a few times $10^{9}$ g/cm³. Each capture removes an electron from the sea, the pressure that was holding the core up falls, the core contracts, the density rises, and more captures follow. The collapse that results is an electron-capture supernova, and its trigger is not a temperature or a fuel running out but a Fermi energy crossing a table of mass differences.

The same thresholds are the reason the chemical environment can change a half-life in the laboratory, a much gentler effect worked out in the half-life that chemistry can change: a nucleus that decays by capturing one of its own electrons decays faster or slower as its electrons’ density at the nucleus changes. In a white dwarf the electrons are supplied by the whole star, and the effect is not a correction of a part in a thousand but a switch.

The ninth that decides how fast a star cools #

Now the limit the first figure approached and never reached. At very high density all three species are ultra-relativistic, so each chemical potential is simply $\hbar c$ times its Fermi momentum. The balance then reads $k_n = k_p + k_e$, and neutrality makes $k_e = k_p$, so the neutron Fermi momentum is exactly twice the proton’s. Densities go as the cube of the Fermi momentum, so there are eight neutrons for every proton, and the proton fraction is one ninth. At any finite density the masses still count for something, so the fraction stays just below.

That arithmetic looks like a curiosity until it is read as a triangle. A neutron at the top of its sea decaying into a proton and an electron at the tops of theirs must conserve momentum as well as energy. The antineutrino’s momentum, a few times $kT$, is negligible. So the three Fermi momenta must be able to form a triangle: $k_n \le k_p + k_e$. In the free gas, with the fraction below one ninth, $k_n$ is always slightly greater than $2k_p$, the triangle never closes, and the direct process is forbidden. Neither the decay of a neutron at the top of its sea nor the capture that reverses it can proceed with every particle near its Fermi surface.

This matters because the direct pair of reactions, called the direct Urca process, is the fastest way a neutron star can lose heat. Each cycle of decay and capture emits a neutrino, which escapes, and a star in which it is allowed cools in years rather than in hundreds of thousands of years. In the free gas it is forbidden at every density, by the geometry above, and a star must cool by slower processes in which a bystander nucleon takes up the spare momentum. With realistic nuclear forces the symmetry energy can push the proton fraction above the threshold, which for matter containing muons as well as electrons moves to between 11 and 15 per cent. Whether it does, and at what density, depends on how the symmetry energy grows with density, which is one of the least well-measured quantities in nuclear physics.

Observed neutron stars are hotter or colder than slow cooling alone would allow. The coldest known young ones, and the rapid cooling seen in some neutron stars after accretion stops, have been read as evidence that the fast process switches on in the most massive stars, whose centres are densest. So a number that appeared as the limit of an ideal-gas calculation turns out to be the boundary of the fastest cooling mechanism a neutron star has. The triangle and the one-ninth limit are the same fact, one written in momenta and the other in populations.

What the free gas leaves out #

Every number drawn here comes from three ideal Fermi gases at zero temperature, and three omissions are large enough to change the numbers, though not the argument.

The nuclear forces. Nucleons attract at a femtometre and repel more closely, and the energy of nuclear matter depends strongly on how unequal its neutron and proton numbers are. That symmetry energy is what raises the proton fraction at nuclear density from 0.5 per cent to about 5. Its dependence on density, which decides the proton fraction deeper in, is constrained by experiments that measure neutron skins on lead nuclei and by the observed radii of neutron stars, and the constraints still disagree at the level that matters for the Urca threshold.

The crust. Below half nuclear density matter is not uniform. It is a lattice of nuclei that grow steadily more neutron-rich with depth, then nuclei immersed in a gas of free neutrons once the density passes $4\times10^{11}$ g/cm³ and the neutron drips out of them. Deeper still they may take on exotic rod and slab shapes before dissolving into uniform matter. The capture thresholds drawn above describe the first steps of that sequence for single kinds of nuclei. A real crust is a mixture set by its history.

Other particles. At a few times nuclear density the electron’s chemical potential passes the muon’s rest energy, 106 MeV, and muons appear in the same equilibrium, sharing the negative charge. Deeper still, the neutron’s chemical potential may exceed the energy needed to make hyperons or to free quarks. Each new species opens a new channel in the same balance of chemical potentials, and each lowers the pressure, which is why the maximum mass of a neutron star is a test of the internal composition. None of this is in the figures, and the true composition of the centre of a neutron star is not known.

What survives all three is the principle the figures illustrate. Wherever a reaction can turn one set of particles into another, and the matter is cold and crowded, the composition is fixed by where the tops of the Fermi seas balance. A proton is rare in a neutron star for the same reason a hole is rare in n-type silicon and a free electron is rare in cold hydrogen gas: the equilibrium has found that making it would cost more than it can be paid for.

Still open: how many protons the core of a neutron star holds #

The proton fraction in the core of a neutron star has never been measured, and it may never be measured directly. It is inferred from the symmetry energy, whose value near nuclear density is known to about ten per cent and whose rise with density is uncertain by a factor of two or more. Laboratory measurements of how much further the neutrons extend than the protons in a lead nucleus gave results in some tension with each other and with astronomical constraints, and gravitational-wave observations of merging neutron stars measure a combination of the pressure and the radius that depends on the same physics. Whether the most massive neutron stars cross the one-ninth threshold for fast cooling, and whether the cold young stars that seem to require it are showing the effect or something else, such as a superfluid that suppresses the slow processes more than expected, is argued with each new measurement of a cooling curve.

The habit worth carrying away is to read an equilibrium as a statement about the most expensive participant. When a reaction couples the populations of several species, the composition is set by the one whose chemical potential rises fastest as it is added, and the others adjust to spare it. In a neutron star that is the electron, dragged into existence by every proton and stacked on top of a sea tens of megaelectronvolts deep. The star is neutrons not because gravity likes neutrons but because it has found the cheapest way to avoid making electrons.