10.1: Hydrostatic Equilibrium

Before describing the geostrophic balance, let’s first consider the simplest solution of the momentum equation, the solution for an ocean at rest. It gives the hydrostatic pressure within the ocean. To obtain the solution, we assume the fluid is stationary:

$$u = v = w = 0; \nonumber$$

the fluid remains stationary:

$$\frac{du}{dt} = \frac{dv}{dt} = \frac{dw}{dt} = 0; \nonumber$$

and there is no friction:

$$f_{x} = f_{y} = f_{z} = 0. \nonumber$$

With these assumptions the momentum equation $(7.6.5-7)$ becomes:

$$\frac{1}{\rho} \frac{\partial p}{\partial x} = 0; \quad\quad \frac{1}{\rho} \frac{\partial p}{\partial y} = 0; \quad\quad \frac{1}{\rho} \frac{\partial p}{\partial z} = -g (\varphi, z) \nonumber$$

where I have explicitly noted that gravity $g$ is a function of latitude $\varphi$ and height $z$. I will show later why I have kept this explicit.

Equation $\PageIndex{4}$ require surfaces of constant pressure to be level surface (see Section 3.4). A surface of constant pressure is an isobaric surface. The last equation can be integrated to obtain the pressure at any depth h. Recalling that ρ is a function of depth for an ocean at rest.

$$p = \int_{-h}^{0} g(\varphi, z) p(z) \ dz \nonumber$$

For many purposes, $g$ and $\rho$ are constant, and $p = \rho gh$. Later, I will show that $(\PageIndex{5})$ applies with an accuracy of about one part per million even if the ocean is not at rest.

The SI unit for pressure is the pascal $(\text{Pa})$. A bar is another unit of pressure. One bar is exactly $10^{5} \ \text{Pa}$ (table $\PageIndex{1}$). Because the depth in meters and pressure in decibars are almost the same numerically, oceanographers prefer to state pressure in decibars.

Table $\PageIndex{1}$. Units of Pressure.

$1 \ \text{pascal (Pa)}$	=	$1 \ \text{N/m}^{2} = 1 \text{kg} \cdot \text{s}^{-2} \cdot \text{m}^{-1}$
$1 \ \text{bar}$	=	$10^{5} \ \text{Pa}$
$1 \ \text{decibar}$	=	$10^{4} \ \text{Pa}$
$1 \ \text{millibar}$	=	$100 \ \text{Pa}$