The Balancing Act — Hydrostatic Equilibrium

Section 3 of 7

Hydrostatic Equilibrium

Part 2: The Equation of Hydrostatic Equilibrium

Setting up the force balance

Generated shell-force diagram with a rectangular gas shell labeled by density rho, area A, and thickness dr, plus arrows and equations for the inner pressure force, outer pressure force, and inward gravitational force. — Figure 5The hydrostatic equation comes directly from shell bookkeeping: pressure on the inner face pushes outward, pressure on the outer face pushes inward, and gravity pulls the shell inward.ASTR 201 (generated)

Consider a thin shell of gas at radius $r$ inside a star, with thickness $dr$ , cross-sectional area $A$ , and density $\rho(r)$ . Its mass is $dm = \rho(r)A\,dr$ . If the inward and outward forces do not cancel, the shell accelerates — which is exactly why this equation matters: hydrostatic equilibrium is the condition for a star to remain nearly static instead of beginning a rapid global readjustment.

Two forces act on the shell in the radial direction. The pressure on the inner face pushes outward, $F_{\text{in}} = P(r)A$ ; the pressure on the outer face pushes inward, $F_{\text{out}} = P(r+dr)A$ . The net pressure force is

$F_P = P(r)A - P(r+dr)A.$

Using the first-order Taylor expansion $P(r+dr) \approx P(r) + \frac{dP}{dr}\,dr$ , this becomes

$F_P \approx -A\frac{dP}{dr}\,dr.$

Because pressure decreases outward, $dP/dr < 0$ , so the pressure-gradient force points outward. Gravity pulls the shell inward with $F_g = -g(r)\,dm = -\frac{G M(r)\rho(r)}{r^2}A\,dr$ , using $g(r) = G M(r)/r^2$ . For hydrostatic equilibrium the net force vanishes, $F_P + F_g = 0$ :

$-A\frac{dP}{dr}\,dr - \frac{G M(r)\rho(r)}{r^2}A\,dr = 0.$

Dividing through by $A\,dr$ and substituting $g(r)$ gives the equation of hydrostatic equilibrium.

Read as a sentence: at every radius inside the star, the pressure must decrease outward at exactly the rate needed to support the weight of the overlying gas. Since $\rho > 0$ and $g > 0$ , the right-hand side is negative, so $dP/dr$ must also be negative — pressure is highest at the center and falls to nearly zero at the surface. This is a local force-balance law; solving for the full structure $P(r)$ , $\rho(r)$ , $M(r)$ needs additional equations (the subject of Reading 5).

Quick check

Classify each statement as a local law or a global scaling estimate, then say what physical question each one answers:

$dP/dr = -\rho g$
$P_c \sim G M^2/R^4$

Observable

The Sun holds its size for billions of years

The Sun keeps nearly the same size for many billions of years, far longer than its $\sim 50$ -minute dynamical timescale.

Model

Force balance on a thin shell

Treat the interior as a stack of thin shells and require the pressure-gradient force to balance gravity on each one.

Inference

Each shell satisfies dP/dr = -ρg

Each shell must satisfy $dP/dr = -\rho g$ . If that balance is violated, the shell accelerates inward or outward instead of staying in place.

Quick check

Hydrostatic equilibrium says $dP/dr < 0$ everywhere inside a star. What would happen if $dP/dr > 0$ at some radius (pressure increasing outward)? What if $dP/dr = 0$ (uniform pressure)?