Hydrostatic equilibrium and mass continuity
Construct the mechanical stellar equations and separate a slowly evolving hydrostatic configuration from thermal equilibrium.
On this page
The question for today
What you’ll understand
- Derive the enclosed-mass and pressure-gradient equations with consistent signs.
- State the dynamical and boundary assumptions behind a spherical stellar model.
- Distinguish local nuclear heating from luminosity and gravitational energy release.
A star can change its composition and radius while remaining very close to mechanical equilibrium. The essential comparison is between its evolution time and the time required for gravity and pressure to rearrange the material. We begin with Newtonian gravity, spherical symmetry, negligible magnetic stress, and sufficiently slow rotation that centrifugal support can be ignored. These assumptions describe much of ordinary stellar evolution, but fail during rapid collapse, violent shocks, or strong rotational deformation. Our numerical sphere below is an analytic toy, not a calibrated star.
Count shells before balancing forces
Define m(r) as the mass enclosed by radius r, and rho as mass per volume. A thin spherical shell has volume equal to its area times its thickness. This gives mass continuity without requiring a constant density profile. Shells outside a chosen radius exert no net gravitational force there under spherical symmetry. The inward gravitational acceleration therefore depends on enclosed mass, rather than the total stellar mass everywhere. Near a regular center, enclosed mass grows as the cube of radius.
The outward pressure force on a small element is minus the pressure gradient divided by density. Setting its acceleration to zero requires pressure to fall outward. Here P is total isotropic material plus radiation pressure when radiation is trapped sufficiently well to be treated locally. Adding a separate radiative force while also retaining its same contribution in total pressure would double count support. An equation of state must still relate pressure, density, temperature, and composition; the two mechanical equations alone do not close a stellar model.
Equilibrium is a controlled approximation
The characteristic dynamical time follows by comparing a displacement of order the stellar radius with an acceleration of order its surface gravity. Hydrostatic evolution is appropriate when structural changes occur much more slowly than this time. It does not mean every fluid velocity vanishes: small convective motions can coexist with an averaged hydrostatic background. Strong pulsations require the acceleration term in the momentum equation, and relativistically compact objects require a different gravitational structure equation.
At the center, radius and transported luminosity vanish while pressure and temperature remain finite. At the exterior, an atmosphere supplies pressure and temperature conditions at a specified matching depth. Zero surface pressure is a useful mathematical idealization for our uniform-density sphere, but is not an exact photospheric condition. A genuine stellar calculation is a boundary-value problem constrained at both ends. Choosing central values arbitrarily and integrating outward need not produce the desired total mass or atmospheric match.
Mechanical balance does not enforce energy balance
Let L(m) be the outward luminosity transported through a mass shell, excluding freely escaping neutrinos. Its derivative measures local net heating after allowing for changing internal energy and compression work. Nuclear reactions can supply heat, neutrinos can remove it, and contraction can release stored gravitational and thermal energy. Hydrostatic equilibrium removes acceleration from the momentum equation; it does not remove these thermal time derivatives. A contracting object can shine even when its local nuclear heating is negligible.
We define specific internal energy u without nuclear rest-mass energy and put the associated reaction energy in epsilon nuclear. Neutrino losses must be counted once using the same convention. At fixed composition, the thermal derivative combination is minus temperature times the specific-entropy derivative; changing composition requires consistent chemical terms or their inclusion in the reaction-energy convention. Thermal equilibrium sets the derivative combination to zero. Temperature transport and abundance evolution then supply the remaining equations needed to follow the star through time.
Work it through
Integrate a uniform-density toy
Take constant density ρ₀ = 10³ kg m⁻³, radius R = 10⁶ m, G = 6 × 10⁻¹¹ m³ kg⁻¹ s⁻², and P(R)=0. Find total mass, central pressure, and pressure at r=R/2.
Integrating shell mass gives , about 4.19 × 10²¹ kg. The assumed density is uniform rather than inferred from an equation of state.
Substituting the enclosed mass into hydrostatic balance makes the pressure gradient proportional to minus r. Integrating inward from the specified surface gives .
At half the radius, the pressure is three quarters of its central value, so . It remains positive and decreases toward the surface.
M ≈ 4.19 × 10²¹ kg, Pc ≈ 1.26 × 10⁸ Pa, and P(R/2) ≈ 9.42 × 10⁷ Pa for this mechanical toy.
The living notebook
Explore the model
Interactive lab · function
Integrating a shell-area power law
Reset and set BOTH Coefficient a to 1 and Position x to 1; predict y=1 and area=1/3. Then Reset and set BOTH a to 1 and x to 2; predict y=4 and area=8/3. Interpreting x as scaled radius and y as a scaled shell-mass derivative illustrates the cubic mass integral for constant density. This quadratic instrument has no gravity, pressure, atmosphere, or stellar structure solver.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forSpherical shell geometry turns a quadratic mass derivative into a cubic enclosed mass.
Pause & reason
Check your understanding
Your turn
Try it yourself
At one shell of a different hypothetical model, let m=5 × 10²⁹ kg, r=10⁷ m, ρ=10³ kg m⁻³, and G=6 × 10⁻¹¹ SI. Find dP/dr. Estimate the pressure change over an outward distance of 100 m, treating the gradient as locally constant.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
First compute the positive inward acceleration Gm/r², then multiply by minus density.
Compare with the solution
The acceleration magnitude is . Therefore the gradient is −3 × 10⁸ Pa m⁻¹, and the outward pressure change is −3 × 10¹⁰ Pa. The negative sign is essential: lower exterior pressure leaves an outward pressure force balancing inward gravity. The finite change uses a local approximation and presumes the shell properties vary little across the chosen distance.
Keep the idea
Take it with you
- Spherical mass continuity and hydrostatic balance require an equation of state and boundary conditions.
- Hydrostatic evolution can include convection and slow contraction when dynamical accelerations are small.
- Energy conservation distinguishes nuclear heating, neutrino loss, compression, and changing internal energy.
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