Hydrostatic pressure ($dp/dz=-{\rho }_{f}g$)

Hydrostatic pressure is the pressure distribution in a resting fluid, balancing its body force from gravity. Take $z$ upward and a uniform downward gravitational field $\mathbf{g}=-g{\mathbf{e}}_z$, where $g>0$. The vertical pressure gradient satisfiesA short derivation balances the vertical forces on a thin horizontal layer of area $A$ and thickness $\Delta z$: $p(z)A-p(z+\Delta z)A-{\rho }_{f}gA\Delta z=0$. Dividing by $A\Delta z$ and taking the derivative gives the equation above. For constant mass density ${\rho }_f$, integration from the surface at $z_s$ givesThus pressure increases with depth $h$. For the stipulated water with mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ and gravitational acceleration $9.81\mathrm{m}{\mathrm{s}}^{-2}$, depths $0.20\mathrm{m}$ and $0.30\mathrm{m}$ give gauge pressures $1962\mathrm{Pa}$ and $2943\mathrm{Pa}$. Their difference produces the buoyancy of the example cube. ^[ref]

For a fixed symmetric triangular channel with depth $H$ and area $A=k{H}^2$, the hydrostatic first moment is $I={\int }_A(H-z)dA=k{H}^3/3=AH/3$. Since $dI/dH=A$, the depth-averaged pressure gradient is $g{I}_x/A=g{H}_x$. A mean geometric depth $H/2$ cannot replace $H$ in that gradient without an extra factor two. This distinction prevents confusion between pressure head, centroid depth, and hydraulic radius.

Gauge pressure is pressure measured relative to the surrounding atmospheric pressure; absolute pressure is instead measured relative to an ideal vacuum. At an open surface $p_{\mathrm{surface}}=p_{\mathrm{atmosphere}}$, so the constant-mass density hydrostatic pressure law becomes $p_g={\rho }_{f}gh$. Here $p_g=0$ at the surface denotes equality with atmospheric pressure. It does not denote zero absolute pressure. ^[ref]

The hydrostatic approximation neglects vertical acceleration relative to gravity and vertical pressure gradients. It reduces vertical momentum balance to $p_z=-\rho g$.

Vertical pressure gradient balances gravitational force per unit volume. Subtracting the resting hydrostatic state and using kinematic pressure gives $p_z^{\prime }=b^{\prime }$ for a Boussinesq approximation buoyancy anomaly. The approximation neglects vertical acceleration; this diagnostic relation is what couples buoyancy perturbation to the vertical pressure profile.

For an ideal gas with constant composition, set $\mathcal{R}=k_B/\bar{m}$, where $\bar{m}$ is the mean particle mass. Constant downward gravitational acceleration $g$ and $T=T_0-\alpha z>0$ give $p^{\prime }=-(g/\mathcal{R}T)p$ by hydrostatic equilibrium. Integration gives $p/p_0=(T/T_0{)}^{g/(\mathcal{R}\alpha )}$, and the ideal gas law gives the displayed mass density ratio. The isothermal limit $\alpha \to 0$ is $\rho /{\rho }_0=\exp [-gz/(\mathcal{R}{T}_0)]$.