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Hydrostatic Equilibrium

Intuition

A fluid at rest is a tug-of-war that ends in a draw: gravity pulls every parcel down, and the pressure difference between a parcel's bottom and top pushes it up. Equilibrium means these balance exactly — which forces pressure to increase with depth at precisely the rate needed to carry the weight of everything above.

Mathematical formulation

Consider a thin slab of fluid of area \(A\) and thickness \(\Delta y\) at depth \(d\). Balancing pressure forces against weight \(W = \rho g A\,\Delta y\) gives, in the limit:

\[\frac{dP}{dy} = \rho g \quad (y \text{ measured downward})\]

For an incompressible liquid (\(\rho\) constant), integrate from the surface (\(P(0)=P_0\)):

\[\boxed{\,P(y) = P_0 + \rho g y\,}\]

The compressible atmosphere: three barometric formulas

For a gas, \(\rho\) depends on \(P\) through the ideal-gas law \(P = \rho R T\), so the balance \(dP/dh = -\rho g\) (now \(h\) measured upward) becomes \(\frac{dP}{P} = -\frac{g}{RT}\,dh\), and the answer depends on the temperature profile:

Isothermal (\(T\) constant):

\[P(h) = P_0 \exp\!\left(-\frac{gh}{RT}\right)\]

Linear lapse rate (\(T = T_0 - \lambda h\)):

\[P(h) = P_0\left(1 - \frac{\lambda h}{T_0}\right)^{g/R\lambda}, \qquad \text{valid while } T_0 > \lambda h\]

Adiabatic (\(P/\rho^\gamma\) constant):

\[P(h) = P_0\left(1 - \frac{gh}{RT_0}\,\frac{\gamma - 1}{\gamma}\right)^{\gamma/(\gamma-1)}\]

All three agree to first order in \(h\) — near the ground, pressure falls off linearly either way.

Worked example: Superman with a straw

How high can anyone — even Superman — drink water through a straw?

Sucking creates low pressure at the top; the atmosphere pushes the water up. The best possible "suck" is a perfect vacuum, \(P_\text{top} = 0\):

\[P_\text{atm} = P_\text{top} + \rho g h \implies h_\text{max} = \frac{P_\text{atm}}{\rho g} = \frac{101.3\times 10^3}{1000 \times 9.8} \approx 10.3\ \text{m}\]

Lung power is irrelevant beyond this: the atmosphere, not the drinker, does the lifting.

Physical interpretation

\(dP/dh = -\rho g\) says pressure is the integrated weight per unit area of fluid above. That reading explains at a glance why mountain air is thin, why dams thicken toward the base, and why buoyancy exists at all: a submerged body's bottom feels more pressure than its top.

Common mistakes

  • Using the incompressible formula for tall gas columns. For air over kilometres, density varies — use a barometric formula.
  • Forgetting that only depth matters. The pressure at the bottom of a wide lake and a narrow tube of equal depth is identical (the hydrostatic paradox).
  • Sign errors: decide whether your coordinate increases up or down before integrating.

Knowledge graph position

Prerequisites: Pressure. Leads to: Buoyancy, Hydrostatic force on surfaces, Fluid in rigid-body motion.

Quiz

Q1 (computational). At what depth in fresh water does the absolute pressure double from its surface value?

Answer

Need \(\rho g h = P_\text{atm}\): \(h = 101.3\times10^3/(1000\times 9.8) \approx 10.3\) m — the same number as the straw limit, and not a coincidence.

Q2 (conceptual). Two barometric formulas (isothermal and adiabatic) give different pressures at 10 km. Which physical assumption differs?

Answer

The temperature profile: isothermal assumes \(T\) constant with height; adiabatic assumes parcels exchange no heat, so \(T\) falls with altitude. Reality (troposphere) sits between, closer to a constant lapse rate.

Q3 (multiple choice). In hydrostatic equilibrium, the pressure gradient vector \(\nabla P\) points:

  • (a) opposite to gravity — upward
  • (b) along gravity — downward
  • (c) horizontally
  • (d) it vanishes
Answer

(b). \(\nabla P = \rho \mathbf{g}\): pressure increases in the direction gravity points (downward).