Categories: Fluid mechanics, Fluid statics, Physics.

Hydrostatic pressure

The pressure pp inside a fluid at rest, the so-called hydrostatic pressure, is an important quantity. Here we will properly define it, and derive the equilibrium condition for the fluid to be at rest, both with and without an arbitrary gravity field.

Without gravity

Inside the fluid, we can imagine small arbitrary partition surfaces, with normal vector n^\vu{n} and area dS\dd{S}, yielding the following vector element dS\dd{\va{S}}:

dS=n^dS\begin{aligned} \dd{\va{S}} = \vu{n} \dd{S} \end{aligned}

The orientation of these surfaces does not matter. The pressure p(r)p(\va{r}) is defined as the force-per-area of these tiny surface elements:

dF=p(r)dS\begin{aligned} \dd{\va{F}} = - p(\va{r}) \dd{\va{S}} \end{aligned}

The negative sign is there because a positive pressure is conventionally defined to push from the positive (normal) side of dS\dd{\va{S}} to the negative side. The total force F\va{F} on a larger surface inside the fluid is then given by the surface integral over many adjacent dS\dd{\va{S}}:

F=Sp(r)dS\begin{aligned} \va{F} = - \int_S p(\va{r}) \dd{\va{S}} \end{aligned}

If we now consider a closed surface, which encloses a “blob” of the fluid, then we can use the divergence theorem to get a volume integral:

F=SpdS=VpdV\begin{aligned} \va{F} = - \oint_S p \dd{\va{S}} = - \int_V \nabla p \dd{V} \end{aligned}

Since the total force on the blob is simply the sum of the forces dF\dd{\va{F}} on all its constituent volume elements dV\dd{V}, we arrive at the following relation:

dF=pdV\begin{aligned} \boxed{ \dd{\va{F}} = - \nabla p \dd{V} } \end{aligned}

If the fluid is at rest, then all forces on the blob cancel out (otherwise it would move). Since we are currently neglecting all forces other than pressure, this is equivalent to demanding that dF=0\dd{\va{F}} = 0, which implies that p=0\nabla p = 0, i.e. the pressure is constant.

p=0\begin{aligned} \boxed{ \nabla p = 0 } \end{aligned}

With gravity

If we include gravity, then, in addition to the pressure’s contact force Fp\va{F}_p from earlier, there is also a body force Fg\va{F}_g acting on the arbitrary blob VV of fluid enclosed by SS:

Fg=VρgdV\begin{aligned} \va{F}_g = \int_V \rho \va{g} \dd{V} \end{aligned}

Where ρ\rho is the fluid’s density (which need not be constant) and g\va{g} is the gravity field given in units of force-per-mass. For a fluid at rest, these forces must cancel out:

F=Fg+Fp=VρgpdV=0\begin{aligned} \va{F} = \va{F}_g + \va{F}_p = \int_V \rho \va{g} - \nabla p \dd{V} = 0 \end{aligned}

Since this a single integral over an arbitrary volume, it implies that every point of the fluid must locally satisfy the following equilibrium condition:

p=ρg\begin{aligned} \boxed{ \nabla p = \rho \va{g} } \end{aligned}

On Earth (or another body with strong gravity), it is reasonable to treat g\va{g} as only pointing in the downward zz-direction, in which case the above condition turns into:

p=ρg0z\begin{aligned} p = \rho g_0 z \end{aligned}

Where g0g_0 is the magnitude of the zz-component of g\va{g}. We can generalize the equilibrium condition by treating the gravity field as the gradient of the gravitational potential Φ\Phi:

g(r)=Φ(r)\begin{aligned} \va{g}(\va{r}) = - \nabla \Phi(\va{r}) \end{aligned}

With this, the equilibrium condition is turned into the following equation:

Φ+pρ=0\begin{aligned} \boxed{ \nabla \Phi + \frac{\nabla p}{\rho} = 0 } \end{aligned}

In practice, the density ρ\rho of the fluid may be a function of the pressure pp (compressibility) and/or temperature TT (thermal expansion). We will tackle the first complication, but neglect the second, i.e. we assume that the temperature is equal across the fluid.

We then define the pressure potential w(p)w(p) as the indefinite integral of the density:

w(p)1ρ(p)dp\begin{aligned} w(p) \equiv \int \frac{1}{\rho(p)} \dd{p} \end{aligned}

Using this, we can rewrite the equilibrium condition as a single gradient like so:

0=Φ+pρ=Φ+dwdpp=(Φ+w(p))\begin{aligned} 0 = \nabla \Phi + \frac{\nabla p}{\rho} = \nabla \Phi + \dv{w}{p} \nabla p = \nabla \Big( \Phi + w(p) \Big) \end{aligned}

From this, let us now define the effective gravitational potential Φ\Phi^* as follows:

ΦΦ+w(p)\begin{aligned} \Phi^* \equiv \Phi + w(p) \end{aligned}

This results in the cleanest form yet of the equilibrium condition, namely:

Φ=0\begin{aligned} \boxed{ \nabla \Phi^* = 0 } \end{aligned}

At every point in the fluid, despite pp being variable, the force that is applied by the pressure must have the same magnitude in all directions at that point. This statement is known as Pascal’s law, and is due to the fact that all forces must cancel out for an arbitrary blob:

F=Fg+Fp=0\begin{aligned} \va{F} = \va{F}_g + \va{F}_p = 0 \end{aligned}

Let the blob be a cube with side aa. Now, Fp\va{F}_p is a contact force, meaning it acts on the surface, and is thus proportional to a2a^2, however, Fg\va{F}_g is a body force, meaning it acts on the volume, and is thus proportional to a3a^3. Since we are considering a point in the fluid, aa is infinitesimally small, so that Fp\va{F}_p dominates Fg\va{F}_g. Consequently, at equilibrium, Fp\va{F}_p must cancel out by itself, which means that the pressure is the same in all directions.


  1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.