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Derivation

Equation of Stellar Hydrostatic Equilibrium

D-332 Home PU-308 Threads force · energy Depends on newtonian-gravitation, poisson-equation-gravity
Statement

For a self-gravitating, non-rotating, non-magnetic fluid sphere in mechanical equilibrium, the outward force per unit volume from the radial pressure gradient exactly balances the inward gravitational body force. This yields the equation of stellar hydrostatic equilibrium, \( \frac{dP}{dr} = -\frac{G\,m(r)\,\rho(r)}{r^2} \), closed by the mass-continuity relation \( \frac{dm}{dr} = 4\pi r^2 \rho(r) \), where \(m(r)\) is the mass enclosed within radius \(r\).

Why it matters

This is the single most important structural equation in stellar astrophysics. Every static stellar model — from the Sun to a white dwarf — is built on it, because it fixes how pressure must rise inward to hold the star up against its own weight. Together with the mass equation it forms two of the four coupled equations of stellar structure, and it alone already sets the central-pressure and virial scales of a star before any thermodynamics or nuclear physics is added.

It also underpins order-of-magnitude astrophysics: combined with an equation of state it predicts central temperatures hot enough for fusion, and its breakdown — when gravity wins — is precisely the onset of dynamical collapse in supernovae and star formation.

Assumptions
Static equilibrium (no bulk acceleration).If dropped, the momentum equation retains \(\rho\,\ddot r\), giving the full Euler equation; the star is then collapsing, pulsating, or expanding rather than in balance.
Spherical symmetry.If dropped, \(P\) and \(\rho\) depend on angle; the single ODE becomes a genuine 3D PDE and gravity need not be radial.
Fluid supports no static shear stress (isotropic pressure).If dropped, a solid crust or anisotropic pressure adds tangential stress terms; relevant for neutron-star crusts.
Newtonian gravity.If dropped, strong-field compact objects require the Tolman–Oppenheimer–Volkoff equation with relativistic corrections.
Negligible rotation and magnetic fields.If dropped, centrifugal and Lorentz body forces enter, breaking spherical symmetry and flattening the figure of the star.
Gravity sourced only by the star's own mass (self-gravity, no external tides).If dropped, a companion's tidal field adds a position-dependent term and the equilibrium figure is distorted.
Derivation
1
\[ \rho\,\frac{D\vec v}{Dt} = -\nabla P + \rho\,\vec g \]
Euler's equation for an inviscid fluid: mass density times acceleration equals the surface pressure force per volume plus the gravitational body force per volume. A
2
\[ 0 = -\nabla P + \rho\,\vec g \]
Impose static equilibrium: every fluid element has zero velocity and zero acceleration, so \(\frac{D\vec v}{Dt}=0\). This is the hydrostatic condition \(\nabla P = \rho\,\vec g\). A
3
\[ \vec g = -\nabla \Phi, \qquad P = P(r),\ \ \Phi = \Phi(r) \]
Write gravity as the gradient of a potential and impose spherical symmetry, so all fields depend on \(r\) alone and gradients are purely radial: \(\nabla P = \frac{dP}{dr}\hat r\). B
4
\[ \frac{dP}{dr}\,\hat r = -\rho\,\frac{d\Phi}{dr}\,\hat r \;\Longrightarrow\; \frac{dP}{dr} = -\rho\,\frac{d\Phi}{dr} \]
Project the vector balance onto \(\hat r\); the angular components vanish identically under spherical symmetry. A
5
\[ \nabla^2 \Phi = 4\pi G \rho \;\Longrightarrow\; \frac{1}{r^2}\frac{d}{dr}\!\left(r^2 \frac{d\Phi}{dr}\right) = 4\pi G \rho \]
Poisson's equation for the gravitational potential, written with the spherically-symmetric radial Laplacian. This is the assumed prior result poisson-equation-gravity. C
6
\[ \frac{d}{dr}\!\left(r^2 \frac{d\Phi}{dr}\right) = 4\pi G \rho\, r^2 \;\Longrightarrow\; r^2 \frac{d\Phi}{dr} = \int_0^r 4\pi G \rho(r')\,r'^2\,dr' \]
Multiply through by \(r^2\) and integrate from the centre outward; the lower limit contributes nothing because the integrand vanishes at \(r'=0\) and regularity forces \(r^2 d\Phi/dr \to 0\) there. C
7
\[ m(r) \equiv \int_0^r 4\pi \rho(r')\,r'^2\,dr' \;\Longrightarrow\; \frac{dm}{dr} = 4\pi r^2 \rho \]
Define the enclosed mass as the volume integral of density over concentric shells of volume \(4\pi r'^2\,dr'\); differentiating recovers the mass-continuity equation. A
8
\[ r^2 \frac{d\Phi}{dr} = G\,m(r) \;\Longrightarrow\; \frac{d\Phi}{dr} = \frac{G\,m(r)}{r^2} \]
Substitute the enclosed-mass definition into Step 6. This is Newton's shell theorem in differential form: the field at \(r\) depends only on the mass interior to it (prior result newtonian-gravitation). B
9
\[ \frac{dP}{dr} = -\rho\,\frac{d\Phi}{dr} = -\frac{G\,m(r)\,\rho}{r^2} \]
Insert the field of Step 8 into the radial hydrostatic balance of Step 4. The minus sign encodes that pressure must increase inward to support the overlying weight. A
Result
\[ \boxed{\ \frac{dP}{dr} = -\frac{G\,m(r)\,\rho(r)}{r^2}\,,\qquad \frac{dm}{dr} = 4\pi r^2 \rho(r)\ } \]

Reading. At every radius the outward push from the pressure decreasing with height (\(dP/dr<0\)) exactly cancels the inward pull of the weight of a unit-volume fluid element, \(g\rho = Gm\rho/r^2\). The mass equation says the enclosed mass grows by one thin shell's worth per step outward. Two coupled first-order ODEs in the two unknowns \(P(r)\) and \(m(r)\); a third relation (an equation of state, \(P=P(\rho,\dots)\)) is needed to close the system.

Units check. Left side: \([P]/[r] = \mathrm{Pa\,m^{-1}} = \mathrm{N\,m^{-3}}\). Right side: \([G][m][\rho]/[r]^2 = (\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{kg})(\mathrm{kg\,m^{-3}})/\mathrm{m^2} = \mathrm{kg\,m^{-2}\,s^{-2}} = \mathrm{N\,m^{-3}}\). Both sides are a force per unit volume. ✓

Limiting cases
  • Constant density (incompressible sphere). With \(\rho=\text{const}\), \(m(r)=\tfrac{4}{3}\pi r^3\rho\) and the equation integrates in closed form to \(P(r)=P_c\big[1-(r/R)^2\big]\) with \(P_c=\tfrac{3GM^2}{8\pi R^4}\).
  • Thin-shell / near-surface limit. For \(r\to R\) with \(m\to M\) and \(g\approx\text{const}\), the equation reduces to the plane-parallel atmosphere \(dP/dz=-g\rho\), giving an exponential scale height \(H=P/(g\rho)\).
  • Order-of-magnitude estimate. Replacing derivatives by ratios, \(P_c \sim GM^2/R^4\), the central-pressure scaling used throughout astrophysics.
  • Polytrope \(P=K\rho^{1+1/n}\). Combining with Poisson's equation reduces the pair to the dimensionless Lane–Emden equation for \(\theta(\xi)\).
Breaks when
  • Dynamical (non-static) conditions. If the pressure support is insufficient or suddenly removed — e.g. core collapse when electron degeneracy pressure fails above the Chandrasekhar mass — the neglected \(\rho\,\ddot r\) term dominates and the star collapses on a free-fall timescale \(t_\text{ff}\sim(G\bar\rho)^{-1/2}\). Hydrostatic equilibrium simply does not hold.
  • Strong gravity. For neutron stars and near black holes, \(GM/(Rc^2)\) is of order unity; the Newtonian equation must be replaced by the general-relativistic Tolman–Oppenheimer–Volkoff equation, which adds pressure and energy-density source terms and a metric factor that make gravity effectively stronger.
  • Rapid rotation or strong magnetic fields. Centrifugal and Lorentz body forces break spherical symmetry; equilibrium then requires the full vector balance and the star becomes oblate (e.g. rapidly rotating early-type stars).
  • Radiation-pressure-dominated instability. When radiation pressure approaches the point that the effective adiabatic index falls below \(4/3\), no stable hydrostatic configuration exists — very massive stars are dynamically unstable.
Failure modes
  • Sign error on the gradient. Writing \(dP/dr = +Gm\rho/r^2\) — forgetting that pressure rises inward, so the derivative with respect to increasing \(r\) is negative.
  • Using total mass \(M\) instead of enclosed mass \(m(r)\). Only the interior mass exerts a net field (shell theorem); the exterior shells contribute nothing.
  • Treating \(g\) as constant throughout the star. Valid only in the thin near-surface layer, not in the deep interior where \(g=Gm(r)/r^2\) varies strongly.
  • Forgetting the closure relation. Trying to integrate \(dP/dr\) without an equation of state linking \(P\) and \(\rho\) — the system is underdetermined with two equations and three unknowns.
  • Dropping the \(4\pi\) in the mass equation by conflating \(dm/dr\) with a surface density rather than a shell of volume \(4\pi r^2 dr\).
  • Applying it to a collapsing or pulsating object, where the acceleration term is not negligible and the "equilibrium" premise is false.
Discussion

The equation is a statement of Newton's second law for a fluid at rest: the buoyant force arising from the pressure difference across a fluid element balances its weight. Its power lies in decoupling mechanics from thermodynamics — the balance of forces holds regardless of why the pressure exists, whether from ideal gas, radiation, or electron degeneracy. This is why the same equation governs a main-sequence star, a red giant, and a white dwarf; only the closure relation changes.

Multiplying by \(4\pi r^3\) and integrating over the whole star yields the virial theorem \(3\int_0^R P\,dV = -E_\text{grav}\), tying the volume-averaged pressure directly to the gravitational binding energy. This connection to the energy thread shows that a star's total thermal energy is fixed, up to a factor, by its gravity — the origin of the negative heat capacity that makes stars heat up as they radiate and contract.

The mass equation and the pressure equation are the first two of the four equations of stellar structure; the remaining two govern energy transport and energy generation. Because the pressure equation contains no time derivatives, it enforces the crucial physical fact that stars evolve quasi-statically: the hydrostatic (dynamical) timescale is seconds to hours, vastly shorter than the thermal (Kelvin–Helmholtz) or nuclear timescales, so the star remains in near-perfect force balance throughout its life while its thermodynamic state slowly changes.

More deeply, hydrostatic equilibrium is the extremum condition for the total energy at fixed entropy: a self-gravitating fluid configuration is in equilibrium precisely when the first variation of \(E = E_\text{int} + E_\text{grav}\) vanishes under mass-conserving perturbations, and it is stable when the second variation is positive. This variational viewpoint unifies the force balance derived here with the stability analysis that gives, for example, the \(\gamma > 4/3\) criterion and the onset of gravitational instability in the Jeans problem, where the same balance between pressure and self-gravity determines whether a gas cloud is supported or collapses to form a star.

Common misconceptions. The pressure gradient does not "cause" support in isolation — it is sourced by gravity through the requirement of balance; remove gravity and the pressure gradient would simply drive expansion. A second misconception is that the equation determines the star uniquely: it does not, until an equation of state and boundary conditions (\(m(0)=0\), \(P(R)=0\)) are supplied.

Worked examples
1
Central pressure of the Sun (uniform-density estimate). Estimate \(P_c\) for the Sun treated as an incompressible sphere, and comment on the true central value. B
\[ \frac{dP}{dr} = -\frac{G m(r)\rho}{r^2}, \quad \rho=\text{const}=\frac{3M}{4\pi R^3}, \quad m(r)=\frac{4}{3}\pi r^3 \rho \]
Substitute constant density and integrate from surface (\(P(R)=0\)) inward.
\[ \frac{dP}{dr} = -\frac{4}{3}\pi G \rho^2 r \;\Longrightarrow\; P_c = \int_0^R \frac{4}{3}\pi G \rho^2 r\,dr = \frac{2}{3}\pi G \rho^2 R^2 = \frac{3 G M^2}{8\pi R^4} \]
Now insert numbers: \(G=6.674\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}\), \(M_\odot=1.989\times10^{30}\,\mathrm{kg}\), \(R_\odot=6.96\times10^{8}\,\mathrm{m}\).
\[ P_c = \frac{3(6.674\times10^{-11})(1.989\times10^{30})^2}{8\pi (6.96\times10^{8})^4} \approx 1.34\times10^{14}\ \mathrm{Pa} \]
\[ P_c \approx 1.3\times10^{14}\ \mathrm{Pa} \approx 1.3\times10^{9}\ \mathrm{atm} \]

Reading. The uniform-density model underestimates the true solar central pressure (\(\sim 2.5\times10^{16}\,\mathrm{Pa}\)) by two orders of magnitude, because real stars are strongly centrally concentrated; still, it captures the correct scaling \(P_c\propto GM^2/R^4\).

Units check. \((\mathrm{m^3 kg^{-1} s^{-2}})(\mathrm{kg^2})/\mathrm{m^4} = \mathrm{kg\,m^{-1}\,s^{-2}} = \mathrm{Pa}\). ✓

2
Pressure scale height of the Earth's isothermal atmosphere. Use the near-surface (constant-\(g\)) limit with the ideal-gas law to find the height over which pressure falls by a factor \(e\). B
\[ \frac{dP}{dz} = -g\rho, \qquad P = \frac{\rho k_B T}{\mu m_H} \;\Longrightarrow\; \rho = \frac{\mu m_H P}{k_B T} \]
Eliminate density and separate variables, treating \(T\) and \(g\) as constant.
\[ \frac{dP}{dz} = -\frac{\mu m_H g}{k_B T}\,P \;\Longrightarrow\; P(z) = P_0\, e^{-z/H}, \quad H = \frac{k_B T}{\mu m_H g} \]
Insert values for air: \(k_B=1.381\times10^{-23}\,\mathrm{J\,K^{-1}}\), \(T=288\,\mathrm{K}\), mean molecular weight \(\mu=28.96\), \(m_H=1.66\times10^{-27}\,\mathrm{kg}\), \(g=9.81\,\mathrm{m\,s^{-2}}\).
\[ H = \frac{(1.381\times10^{-23})(288)}{(28.96)(1.66\times10^{-27})(9.81)} \approx 8.4\times10^{3}\ \mathrm{m} \]
\[ H \approx 8.4\ \mathrm{km} \]

Reading. Atmospheric pressure drops by a factor \(e\) roughly every 8.4 km, matching the observed value — the plane-parallel limit of the stellar equation applied to a planetary atmosphere.

Units check. \([k_B T]/([m][g]) = \mathrm{J}/(\mathrm{kg\cdot m\,s^{-2}}) = \mathrm{(kg\,m^2 s^{-2})}/(\mathrm{kg\,m\,s^{-2}}) = \mathrm{m}\). ✓

Problems
  1. (A) Starting from \(dP/dr=-Gm\rho/r^2\), show that the weight per unit area of a column from radius \(r\) to the surface equals the pressure at \(r\) (assume \(g\) constant).
    Solution With \(g\) constant, \(dP=-g\rho\,dr\). Integrate from \(r\) (pressure \(P\)) to \(R\) (pressure \(0\)): \(0-P=-g\int_r^R\rho\,dr'\), so \(P=g\int_r^R \rho\,dr' = g\,\sigma\), where \(\sigma=\int_r^R\rho\,dr'\) is the column mass per unit area. Thus \(P=g\sigma\) is precisely the weight per unit area of the overlying column.
  2. (A) The Sun's mean density is \(\bar\rho=1408\,\mathrm{kg\,m^{-3}}\). Estimate the free-fall (dynamical) timescale \(t_\text{ff}\approx(G\bar\rho)^{-1/2}\) and compare to a human lifetime.
    Solution \(t_\text{ff}=(G\bar\rho)^{-1/2}=\big[(6.674\times10^{-11})(1408)\big]^{-1/2}=(9.40\times10^{-8})^{-1/2}\,\mathrm{s}\approx 3.26\times10^{3}\,\mathrm{s}\approx 54\,\mathrm{min}\). This is roughly an hour — vastly shorter than a human lifetime, confirming that the Sun re-establishes hydrostatic balance almost instantaneously compared to its evolutionary timescales.
  3. (B) For a uniform-density sphere, derive \(P(r)=P_c[1-(r/R)^2]\) and identify \(P_c\).
    Solution With \(\rho=\text{const}\), \(m(r)=\tfrac{4}{3}\pi r^3\rho\), so \(dP/dr=-\tfrac{4}{3}\pi G\rho^2 r\). Integrate from \(r\) to \(R\) with \(P(R)=0\): \(P(r)=\tfrac{2}{3}\pi G\rho^2(R^2-r^2)\). Factor out \(R^2\): \(P(r)=\tfrac{2}{3}\pi G\rho^2 R^2[1-(r/R)^2]\), so \(P_c=\tfrac{2}{3}\pi G\rho^2 R^2=\tfrac{3GM^2}{8\pi R^4}\) using \(\rho=3M/4\pi R^3\).
  4. (B) Using the estimate \(P_c\sim GM^2/R^4\) and the ideal-gas law \(P_c=\rho_c k_B T_c/(\mu m_H)\) with \(\rho_c\sim M/R^3\) and \(\mu=0.6\), estimate the Sun's central temperature.
    Solution Set \(GM^2/R^4 \sim (M/R^3)k_B T_c/(\mu m_H)\), giving \(T_c \sim \mu m_H GM/(k_B R)\). Numerically: \(T_c\sim (0.6)(1.66\times10^{-27})(6.674\times10^{-11})(1.989\times10^{30})/[(1.381\times10^{-23})(6.96\times10^{8})]\). Numerator \(=1.32\times10^{-7}\); denominator \(=9.61\times10^{-15}\); \(T_c\sim1.4\times10^{7}\,\mathrm{K}\) — the right order of magnitude for hydrogen fusion (true value \(\approx1.5\times10^7\,\mathrm{K}\)).
  5. (C) Combine \(dP/dr=-Gm\rho/r^2\) and \(dm/dr=4\pi r^2\rho\) with a polytropic relation \(P=K\rho^{1+1/n}\) to obtain the Lane–Emden equation.
    Solution From the two structure equations, \(\frac{1}{r^2}\frac{d}{dr}\!\left(\frac{r^2}{\rho}\frac{dP}{dr}\right)=-4\pi G\rho\). Write \(\rho=\rho_c\theta^n\) so \(P=K\rho_c^{1+1/n}\theta^{n+1}\). Then \(\frac{1}{\rho}\frac{dP}{dr}=(n+1)K\rho_c^{1/n}\frac{d\theta}{dr}\). Define \(r=\alpha\xi\) with \(\alpha^2=\frac{(n+1)K\rho_c^{1/n-1}}{4\pi G}\). Substituting gives the dimensionless Lane–Emden equation \(\frac{1}{\xi^2}\frac{d}{d\xi}\!\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^n\), with \(\theta(0)=1\) and \(\theta'(0)=0\). Its solutions \(\theta(\xi)\) are the standard polytropic stellar models.