Hydrostatic Equilibrium and the Virial Theorem

A star is a self-gravitating ball of gas that neither collapses nor disperses on a dynamical timescale. Something holds it up. That something is a pressure gradient, and the statement that the gradient exactly cancels gravity at every radius is hydrostatic equilibrium, the first and most important of the stellar-structure equations. This lesson derives it from a force balance on a mass shell, extracts a rigorous lower bound on the central pressure, computes the gravitational potential energy of a star, and assembles both into the virial theorem. The virial theorem carries two consequences that shape everything a star does: its total energy is negative and bound, and its effective heat capacity is negative, so radiating energy away raises the internal temperature. The last section turns the virial energy into the Kelvin-Helmholtz timescale, the time gravitational contraction alone could sustain the solar luminosity.

The condition of hydrostatic equilibrium

Consider a thin spherical shell inside a star at radius $r$ , of thickness $d r$ and unit cross-sectional area, containing gas of mass density $ρ (r)$ . Two forces act on it along the radial direction. Gravity pulls it inward with the weight of the shell in the field of the mass $m (r)$ interior to it; only the interior mass matters, because a spherically symmetric shell exerts no net gravitational force on material inside it. Pressure pushes on both faces: the larger pressure on the inner face pushes outward, the smaller pressure on the outer face pushes inward, and the net pressure force is set by the gradient.

The mass of the shell element (per unit area) is $ρ d r$ , so its weight per unit area is $- G m (r) ρ d r / r^{2}$ , directed inward. The net outward pressure force per unit area is $P (r) - P (r + d r) = - d P$ . In equilibrium the two cancel:

- d P - \frac{G m ( r ) ρ}{r ^{2}} d r = 0,

which rearranges to the equation of hydrostatic equilibrium,

\frac{d P}{d r} = - \frac{G m ( r ) ρ ( r )}{r ^{2}} .

The pressure gradient is negative everywhere: pressure decreases outward, falling from a maximum at the center to essentially zero at the surface.¹ Written against the interior mass instead of the radius, using $d m = 4 π r^{2} ρ d r$ (mass conservation, derived in the next lesson), it takes the Lagrangian form

\frac{d P}{d m} = - \frac{G m}{4 π r ^{4}},

which is the form used to build numerical models, where $m$ rather than $r$ is the independent variable.

The pressure gradient across a shell of thickness dr supports its weight in the gravitational field of the interior mass m(r).

Central pressure: a rigorous lower bound

The hydrostatic equation alone, without knowing the density profile, already bounds the central pressure from below. Start from the Lagrangian form and integrate from the center ( $m = 0$ , $P = P_{c}$ ) to the surface ( $m = M$ , $P \approx 0$ ):

P_{c} = 0 \int M \frac{G m}{4 π r ^{4}} d m .

Inside the star every mass shell lies at a radius $r < R$ , the total radius, so $1/ r^{4} > 1/ R^{4}$ for every shell. Replacing $r$ by $R$ therefore only decreases the integrand:

P_{c} > 0 \int M \frac{G m}{4 π R ^{4}} d m = \frac{G}{4 π R ^{4}} 0 \int M m d m = \frac{G M ^{2}}{8 π R ^{4}} .

This is a strict inequality that uses no equation of state and no assumption about the internal structure beyond spherical symmetry and monotone density. For the Sun, $M_{⊙} = 1.989 \times 1 0^{30} kg$ and $R_{⊙} = 6.96 \times 1 0^{8} m$ give a bound

P_{c} > \frac{( 6.674 \times 1 0 ^{- 11} ) ( 1.989 \times 1 0 ^{30} ) ^{2}}{8 π ( 6.96 \times 1 0 ^{8} ) ^{4}} \approx 4.5 \times 1 0^{13} Pa .

The true central pressure of a solar model is about $2.5 \times 1 0^{16} Pa$ , roughly five hundred times the bound. The gap is expected: replacing $r$ by $R$ everywhere is a crude underestimate, because most of the mass sits at radii well inside $R$ where the integrand is far larger. The value of the bound is that it is exact given only mechanical equilibrium, and it already tells us the interior is at a pressure enormously larger than anything in a laboratory.

Replacing every shell radius r by the full radius R shrinks the integrand, so the true central pressure exceeds the closed-form lower bound.

Gravitational potential energy

Assembling a star from dispersed matter releases gravitational energy. The gravitational potential energy $U$ is the work done against gravity to disperse the star to infinity, taken with a negative sign, and it is built up shell by shell. Bringing a shell of mass $d m$ from infinity to radius $r$ , where the interior mass is $m (r)$ , releases $- G m d m / r$ . Summing over all shells,

U = - 0 \int M \frac{G m}{r} d m .

Evaluating the integral needs the density profile. For the simplest case, a sphere of uniform density $ρ = 3 M / (4 π R^{3})$ , the interior mass is $m (r) = M (r / R)^{3}$ , so $d m = 3 M r^{2} d r / R^{3}$ , and

U = - 0 \int R \frac{G M ( r / R ) ^{3}}{r} \frac{3 M r ^{2}}{R ^{3}} d r = - \frac{3 G M ^{2}}{R ^{6}} 0 \int R r^{4} d r = - \frac{3}{5} \frac{G M ^{2}}{R} .

Real stars are centrally concentrated, so their density rises toward the center and the magnitude of $U$ is larger than the uniform-sphere value. The general result for a polytrope of index $n$ (derived in the equation-of-state lesson) is

U = - \frac{3}{5 - n} \frac{G M ^{2}}{R},

which reduces to the uniform sphere at $n = 0$ and grows as $n$ increases toward the $n = 5$ limit of infinite central concentration. A useful summary is that $U = - α G M^{2} / R$ with a dimensionless structure factor $α$ of order unity ( $α = 3/5$ uniform, $α \approx 1.5$ for a Sun-like $n = 3$ model).

The virial theorem

Hydrostatic equilibrium and the gravitational energy combine into an exact relation between the total thermal energy and the gravitational energy. Multiply the Lagrangian hydrostatic equation by the volume $V (r) = \frac{4}{3} π r^{3}$ interior to $r$ and integrate over the star. It is cleaner to multiply the Eulerian form by $4 π r^{3}$ and integrate from $0$ to $R$ :

0 \int R 4 π r^{3} \frac{d P}{d r} d r = - 0 \int R \frac{G m}{r ^{2}} ρ 4 π r^{3} d r = - 0 \int R \frac{G m}{r} d m = U,

where the right-hand side used $d m = 4 π r^{2} ρ d r$ and reproduced the potential-energy integral. Integrate the left-hand side by parts, with the boundary term vanishing because $P (R) = 0$ at the surface and $r^{3} = 0$ at the center:

0 \int R 4 π r^{3} \frac{d P}{d r} d r = [4 π r^{3} P]_{0}^{R} - 0 \int R 12 π r^{2} P d r = - 3 0 \int R P (4 π r^{2} d r) = - 3 0 \int R P d V .

Equating the two evaluations gives the mechanical form of the virial theorem,

- 3 0 \int R P d V = U, equivalently 3 \int P d V = - U .

The step to a thermal statement uses the equation of state. For a nonrelativistic ideal gas the pressure and the internal thermal energy density $u$ are related by $P = \frac{2}{3} u$ , since a monatomic gas stores $\frac{3}{2} k T$ of kinetic energy per particle and exerts $nk T$ of pressure. Then $\int P d V = \frac{2}{3} \int u d V = \frac{2}{3} K$ , where $K$ is the total thermal (kinetic) energy of the star. Substituting,

3 \cdot \frac{2}{3} K = - U ⟹ 2 K + U = 0.

This is the virial theorem in its most useful form for stellar physics.²

Two facts follow immediately. The total energy $E = K + U$ equals $- K$ , so it is negative: the star is gravitationally bound and it takes a positive energy input to disperse it. And $E = \frac{1}{2} U$ , so the total energy is half the gravitational energy, the other half having gone into thermal motion.

Of the gravitational energy released by contraction, half becomes thermal energy and half is radiated; the total energy sits at U/2, below zero.

Negative heat capacity and the thermostat

The relation $E = - K$ is the source of the most counterintuitive property of stars. Suppose the star slowly radiates energy from its surface at luminosity $L$ , so that its total energy decreases, $d E / d t = - L < 0$ . Since $E = - K$ , the thermal energy must increase: $d K / d t = + L$ . The star gets hotter as it loses energy. The gravitational reservoir supplies the difference: from $U = 2 E$ , losing $d E$ releases $d U = 2 d E$ of gravitational energy, half of which replaces the radiated energy and half of which goes into raising $K$ .

Because the thermal energy scales with the mean internal temperature, this means a self-gravitating star has an effective negative heat capacity: extract heat and its temperature rises; add heat and it cools and expands. Formally, writing the mean temperature $\overset{ˉ}{T}$ through $K = \frac{3}{2} \frac{M}{μ m _{H}} k \overset{ˉ}{T}$ for a gas of mean molecular weight $μ$ , the total energy is $E = - K \propto - \overset{ˉ}{T}$ , so

C_{⋆} \equiv \frac{d E}{d T ˉ} < 0.

This is the thermostat that stabilizes nuclear burning. If a fusion-powered core generates energy slightly faster than it is radiated, the extra heat expands and cools the core, throttling the temperature-sensitive reaction rate back down; if it generates too slowly, the core contracts and heats, speeding the reactions up. The negative heat capacity is why a normal (nondegenerate) star burns stably rather than exploding. The same mechanism fails when the core becomes degenerate and the pressure decouples from the temperature, which is what makes the helium flash in low-mass stars and thermonuclear supernovae runaway events.

The Kelvin-Helmholtz timescale

Before nuclear fusion was known, Kelvin and Helmholtz proposed that the Sun shines by slow gravitational contraction, converting the released gravitational energy into radiation. The virial theorem sets the timescale. If contraction is the only power source, the luminosity is supplied by the decrease in total energy, and the available energy is $∣ E ∣ = \frac{1}{2} ∣ U ∣ \approx \frac{1}{2} G M^{2} / R$ up to the structure factor. The Kelvin-Helmholtz (thermal) timescale is this reservoir divided by the luminosity,

t_{KH} = \frac{∣ E ∣}{L} \approx \frac{G M ^{2}}{2 R L} .

For the Sun,

t_{KH} \approx \frac{( 6.674 \times 1 0 ^{- 11} ) ( 1.989 \times 1 0 ^{30} ) ^{2}}{2 ( 6.96 \times 1 0 ^{8} ) ( 3.828 \times 1 0 ^{26} )} \approx 5 \times 1 0^{14} s \approx 1.6 \times 1 0^{7} yr,

about thirty million years. This is the fatal problem with the contraction hypothesis: the geological and fossil record demanded an age far greater than $1 0^{7}$ years, so gravitational contraction cannot be the Sun's long-term power source. The resolution is nuclear fusion. Hydrogen burning converts a fraction $\approx 0.007$ of the rest mass to energy, and only the innermost $\sim 10%$ of the Sun's mass reaches core temperatures, giving a nuclear timescale

t_{nuc} \approx \frac{0.1 \times 0.007 M _{⊙} c ^{2}}{L _{⊙}} \approx \frac{( 7 \times 1 0 ^{- 4} ) ( 1.989 \times 1 0 ^{30} ) ( 2.998 \times 1 0 ^{8} ) ^{2}}{3.828 \times 1 0 ^{26}} \approx 3 \times 1 0^{17} s \approx 1 0^{10} yr .

Nuclear burning outlasts contraction by a factor of roughly a thousand, matching the Sun's known age of $4.6 \times 1 0^{9}$ years with room to spare. The Kelvin-Helmholtz timescale is not obsolete, though: it governs the pre-main-sequence contraction of protostars, the readjustment of a star between nuclear burning stages, and the cooling of white dwarfs, wherever gravitation rather than fusion sets the pace.

The Sun's three characteristic timescales span thirteen orders of magnitude, from the free-fall dynamical time to the nuclear-burning lifetime.

Summary

Hydrostatic equilibrium, $d P / d r = - G m ρ / r^{2}$ , is the mechanical backbone of every static star and yields a rigorous central-pressure bound $P_{c} > G M^{2} / (8 π R^{4})$ . The gravitational energy $U = - α G M^{2} / R$ combines with it through the virial theorem $2 K + U = 0$ , forcing the total energy $E = \frac{1}{2} U$ to be negative and giving the star a negative heat capacity that stabilizes nuclear burning. The Kelvin-Helmholtz time $t_{KH} \approx G M^{2} / (2 R L) \sim 3 \times 1 0^{7} yr$ for the Sun is far too short to be the Sun's power source, which fixed the need for fusion, but it still governs contraction phases throughout stellar life. The next lesson promotes hydrostatic equilibrium into the full set of coupled structure equations, adding mass conservation, energy generation, and energy transport.

Carroll & Ostlie, §10.1 — Hydrostatic Equilibrium: the pressure-gradient support of a mass shell and the central-pressure estimate. ↩
Carroll & Ostlie, §10.4 — The Virial Theorem: the derivation $2 K + U = 0$ , the negative-heat-capacity consequence, and the Kelvin-Helmholtz timescale. ↩