A floor under the centre that assumes nothing

Almost every statement in this collection about the inside of a star is conditional. The centre of the Sun is at fifteen million kelvin if the opacity is right, if the composition is what the surface says, if the mixing length is what it was calibrated to. The conditions are usually reasonable and they are always there.

There is one exception, and it is worth knowing precisely because it is so cheap. From the single statement that the pressure gradient carries the weight, and from nothing else whatever, a lower bound on the central pressure follows:

$$P_c \ \ge\ \frac{GM^{2}}{8\pi R^{4}}.$$

No equation of state. No composition. No temperature. No assumption about how the density is distributed. Any body of mass $M$ and radius $R$ in hydrostatic equilibrium, made of anything, has at least that pressure at its centre.

The derivation, which is three lines #

Consider the quantity

$$\Psi(r) = P(r) + \frac{Gm(r)^{2}}{8\pi r^{4}},$$

where $m(r)$ is the mass inside radius $r$. Differentiate it. The pressure term gives $-Gm\rho/r^{2}$, from hydrostatic equilibrium. The second term gives $Gm\rho/r^{2}$ from the derivative of $m^{2}$ — using $dm/dr = 4\pi r^{2}\rho$ — and $-Gm^{2}/2\pi r^{5}$ from the derivative of $r^{-4}$. The first two cancel exactly, leaving

$$\frac{d\Psi}{dr} = -\frac{Gm^{2}}{2\pi r^{5}} < 0.$$

So $\Psi$ decreases outward, always. At the centre $m^{2}/r^{4}$ vanishes — $m$ goes as $r^{3}$ — so $\Psi(0) = P_c$. At the surface the pressure is zero and $\Psi(R) = GM^{2}/8\pi R^{4}$. A decreasing function is larger at the start than at the end, and the inequality is done.

What makes this work is the exact cancellation, and the cancellation is not a coincidence: $Gm^{2}/8\pi r^{4}$ is the gravitational self-energy density of the interior, and what has been constructed is a statement that pressure plus that energy density is monotone. It is the same kind of argument as the virial theorem, which also gets a hard result out of hydrostatic equilibrium by integrating against a well-chosen weight.

It is worth pausing on how little was used. Nothing in those three lines mentions what the body is made of, whether it is hot or cold, whether it is degenerate, whether it is rotating slowly or not at all, or whether it has a core. The only inputs are that gravity is Newtonian, that the body is spherically symmetric, that it is in equilibrium, and that the pressure at the surface is zero. Drop any one of them and the argument fails — a rotating body has an extra term, and a body with an atmosphere has a nonzero surface pressure — but none of them is a statement about matter.

That is a rare kind of result in this subject. Most of the quantitative statements a reader meets about stellar interiors are the output of a computation with a hundred inputs; this one is the output of an integration by parts.

Why it is so weak, and what the weakness measures #

For the Sun the bound gives $4.5\times10^{13}$ pascals. A solar model gives $2.3\times10^{16}$. The bound is five hundred times too small — true, and nearly useless.

That factor is not a defect of the derivation. It is a measurement of something. The bound is saturated only by a body with a very particular density profile, and a real star is far more centrally condensed than that. Most of the Sun’s central pressure comes not from the total weight above the centre but from the concentration of that weight close in, and the derivation deliberately knows nothing about concentration, because knowing about it would require an equation of state.

So the ratio of the true central pressure to the bound is a dimensionless measure of central condensation — a cousin of $\rho_c/\bar\rho$ and of the Love number, differently weighted. The Earth’s ratio is 6.3, Jupiter’s 10, the Sun’s 522. The ordering is exactly the ordering of the bodies’ central condensations, and it comes out of an inequality rather than a model.

There is a second reading of the same weakness, and it is the more useful one for thinking about planets. For the Earth the ratio is 6.3 and for the Sun it is 522, and the difference is not that one body is better understood than the other. It is that a rocky planet is nearly incompressible — its density varies by a factor of four from surface to centre — while a star’s varies by five orders of magnitude. A bound derived without any assumption about compressibility is naturally much tighter on the object that barely compresses. So the inequality is a good tool for planets and a poor one for stars, and it says so itself, through the width of its own bracket.

The other side #

The companion result costs one extra assumption — that the density does not increase outward, which is true of anything stable — and gives a ceiling:

$$P_c \ \le\ \frac{1}{2}\left(\frac{4\pi}{3}\right)^{1/3}G M^{2/3}\rho_c^{4/3}.$$

It needs a central density, which is a model output rather than an observation, so it is the softer of the two edges. But for the terrestrial bodies it is surprisingly tight: the Earth’s modelled central pressure is $3.6\times10^{11}$ pascals against a ceiling of $5.4\times10^{11}$, so the two bounds together confine the answer to within a factor of two. For Jupiter the pair gives $4.0\times10^{11}$ to $4.5\times10^{13}$, a factor of a hundred, which is the price of a body that compresses. And there is a hidden circularity to watch for: the ceiling needs $\rho_c$, and the only way anyone knows a giant planet’s central density is by building the interior model the bound was supposed to be independent of.

For a star the bracket is three and a half orders of magnitude wide, which is a fair summary of how much a stellar interior depends on details the inequalities decline to consider.

A worked reading, for one body #

It is worth putting a number through the whole chain once, because the arithmetic is short enough to follow and the result is more surprising than the algebra.

Take the Sun. $M = 1.989\times10^{30}$ kilograms, $R = 6.957\times10^{8}$ metres. Then $GM^{2} = 2.64\times10^{50}$ in SI units, $R^{4} = 2.34\times10^{35}$, and $8\pi = 25.13$. The quotient is $4.5\times10^{13}$ pascals, or about four hundred and fifty million atmospheres.

That is already an extraordinary statement to have obtained from two numbers a person can measure from the ground — a mass from the Earth’s orbit and a radius from an angular diameter and a distance. It says that the middle of the Sun is at a pressure four hundred million times the bottom of the ocean, and it says so without knowing that the Sun is made of hydrogen, without knowing its temperature, and without any theory of what holds it up beyond the requirement that something does.

Now do the same for a body of one Earth mass compressed to the size of a city — the shape of the calculation that made a collapsed star a serious proposition in the 1930s. The bound goes as $R^{-4}$, so shrinking a body by a factor of a thousand raises its floor by $10^{12}$. It is that steepness, and not any detail of the matter, that makes compact objects a different subject.

Where the bracket breaks #

The Newtonian ceiling is exceeded by a neutron star, and by a large factor.

A 1.4-solar-mass star twelve kilometres across, with a central density near $10^{18}$ kilograms per cubic metre, is given a ceiling of about $10^{34}$ pascals by the formula above. Every published equation of state puts its actual central pressure somewhere between $6\times10^{34}$ and $10^{35}$. The Newtonian bracket is violated, and the violation is not marginal.

The reason is that the ceiling assumes Newtonian hydrostatics. The relativistic equation — the Tolman–Oppenheimer–Volkoff equation — has three corrections to the Newtonian pressure gradient and all three make it steeper: the pressure itself gravitates, the energy in the pressure adds to the enclosed mass, and space is curved so that the distance between two shells exceeds the difference of their radial coordinates. Each is a factor greater than one at these compactnesses, and their product is of order three.

The lower bound, interestingly, survives. Since general relativity makes the pressure gradient steeper than Newtonian gravity does, the integrated pressure can only exceed what the Newtonian derivation demands — so $P_c \ge GM^{2}/8\pi R^{4}$ remains a valid floor even in the relativistic case, and it is not a weak one there. For the neutron star it gives $10^{33}$ pascals, which is a substantial fraction of $\rho c^{2}$: the pressure at which the energy stored in the pressure gravitates about as much as the rest mass does.

That is the sense in which an inequality with no physics in it says something about which physics is needed. Applied at a small enough radius, the Newtonian bracket breaks, and where it breaks is where Newtonian hydrostatics has stopped being the right equation. Nothing about the interior had to be assumed to find that out.

It is worth spelling out why the Newtonian result fails upward rather than downward, since a broken bound could in principle break either way. The ceiling is derived by asking what the largest central pressure is that a given amount of mass, distributed with density falling outward, can generate — an integral of $Gm\rho/r^{2}$ over the interior. Relativity does not add mass; it makes each gram weigh more, because pressure gravitates. So the same mass distribution needs more pressure to hold itself up, and the Newtonian answer, computed with gravity too weak, comes out too small. The bracket does not fail because the star is unusual. It fails because the theory used to build it is.

What can be checked, and what cannot #

The masses and radii in the figure are not all equally secure, and the difference matters for how much the bracket is worth.

The Earth’s mass and radius are known to eight and nine figures. Jupiter’s to six. The Sun’s mass is known through $GM_\odot$ to nine figures — the product, not the mass, because $G$ is the worst-measured constant in physics — and its radius to five.

A neutron star’s are a different matter. Masses are known well for the ones in binaries, where the orbit gives them, and a few are known to a per cent from pulsar timing. Radii are much worse: the best come from modelling the pulse profile of a hot spot on a rotating neutron star’s surface, which is a fit with many parameters, and from the tidal deformability measured in a merger. Twelve kilometres is a round number standing for something between eleven and thirteen, and since the bound goes as $R^{-4}$, that is a factor of two in the floor.

What rotation does to it #

The derivation assumed spherical symmetry, and it is worth being explicit about how a rotating body escapes the bound, because the escape is small for everything in the figure and is not small in general.

In a rotating body the pressure gradient carries the weight minus the centrifugal support, so the hydrostatic equation acquires a term and the exact cancellation in the derivation no longer happens. What survives is a modified inequality in which the effective gravity is reduced, and the floor is lowered by roughly the ratio of the centrifugal acceleration at the equator to the surface gravity.

For the Sun that ratio is about two parts in a hundred thousand, so the correction is invisible. For Jupiter it is nearly nine per cent, which is measurable and still small. For a rapidly rotating early-type star it can approach a half, and for a millisecond pulsar spinning near its break-up limit it is of order unity — at which point the bound is not merely weakened but no longer meaningful, because the body is not spherical and there is no single radius to put in the denominator.

The dependence on radius is what makes that last case severe. The bound goes as $R^{-4}$, and a rapidly rotating body has an equatorial radius substantially larger than its polar one — twenty per cent or more for the fastest rotators — so the choice of which radius to use moves the answer by a factor of two before any centrifugal correction is applied.

The honest summary is that the bound is a statement about non-rotating spheres and that almost everything it is usefully applied to is close enough to one. Where it is not — the fastest-rotating stars, and any body supported substantially by rotation rather than by pressure — the right generalisation exists and is considerably less clean, because the enclosed mass at a given radius is no longer well defined. An inequality that assumes nothing about matter still assumes a great deal about geometry, and the geometry is the assumption that fails first.

It is worth noticing that the same limitation applies to every other result in this anchor and is rarely stated there either. A polytrope is spherical, the Lane–Emden equation is an ordinary differential equation in one variable because of it, and the entire apparatus of stellar structure is built on an assumption that the fastest-rotating tenth of stars visibly break.

Why bother with a bound at all #

Three reasons, and the first is pedagogical rather than practical.

An inequality with no assumptions in it is a useful thing to hold in mind while reading results that have many. When a paper reports a central pressure, the bound is the sanity check that costs nothing: if the reported value is below $GM^{2}/8\pi R^{4}$, either the body is not in hydrostatic equilibrium or the calculation is wrong, and there is no third option.

Second, the bound scales exactly. It goes as $M^{2}/R^{4}$ with no residual dependence on anything, so it can be carried across regimes where every model would have to be rewritten — from a planet to a star to a degenerate remnant to a body at nuclear density — and the arithmetic does not change. Very little else in stellar physics has that property.

Third, and most usefully, the gap is a measurement. The ratio of a body’s true central pressure to its bound is a dimensionless number that summarises how centrally condensed it is, computable for anything with a model and comparable across objects with nothing else in common. It behaves like the moment-of-inertia factor — a single dimensionless summary of a whole interior — and like that quantity it is not invertible, but is worth a great deal as a common currency.

The bound is a comparison across objects, so it is worth drawing for a different set of them and beside the structure it deliberately refuses to assume.

Where the ladder goes #

This anchor’s later rungs are about the other inequalities that follow from hydrostatic equilibrium alone. There is a bound on the mean temperature of a star made of ideal gas, which follows from the virial theorem and says that any star of the Sun’s mass and radius must average several million kelvin whatever it is made of. There is a bound relating a body’s compactness to its maximum possible mass in general relativity — the Buchdahl limit, $2GM/Rc^{2} \le 8/9$ — which is derived the same way and is the reason a neutron star cannot be arbitrarily compact even with an infinitely stiff equation of state.

There is a practical rung too, and it belongs to planets rather than stars. Applied to an exoplanet with a measured mass and radius, the bound says what pressure its interior must reach, and therefore which laboratory equations of state are relevant to it — a super-Earth of five Earth masses reaches pressures no diamond anvil has produced, and knowing that before building a model is worth something. It is the same difficulty that makes one density consistent with many compositions so hard to escape: the interior is at conditions that must be extrapolated to.

And there is the thread that leads from all of them to the polytropes: supply the missing relation between pressure and density, and every inequality collapses to an equality. What the bounds show is exactly how much of a star’s structure is decided by mechanics and how much by matter, and the answer, in the case of the Sun, is a factor of five hundred.