---
title: "Hydrostatic pressure"
firstLetter: "H"
publishDate: 2021-03-12
categories:
- Physics
- Fluid mechanics
- Fluid statics

date: 2021-03-12T14:37:56+01:00
draft: false
markup: pandoc
---

# Hydrostatic pressure

The pressure $p$ 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 $\vu{n}$ and area $\dd{S}$,
yielding the following vector element $\dd{\va{S}}$:

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

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

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

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

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

If we now consider a *closed* surface,
which encloses a "blob" of the fluid,
then we can use Gauss' theorem to get a volume integral:

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

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

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

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 $\dd{\va{F}} = 0$,
which implies that $\nabla p = 0$, i.e. the pressure is constant.

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


## With gravity

If we include gravity, then,
in addition to the pressure's *contact force* $\va{F}_p$ from earlier,
there is also a *body force* $\va{F}_g$ acting on
the arbitrary blob $V$ of fluid enclosed by $S$:

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

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

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

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

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

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

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

Where $g_0$ is the magnitude of the $z$-component of $\va{g}$.
We can generalize the equilibrium condition by treating
the gravity field as the gradient of the gravitational potential $\Phi$:

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

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

$$\begin{aligned}
    \nabla \Phi + \frac{\nabla p}{\rho}
    = 0
\end{aligned}$$

In practice, the density $\rho$ of the fluid
may be a function of the pressure $p$ (compressibility)
and/or temperature $T$ (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** $w(p)$ as
the indefinite integral of the density:

$$\begin{aligned}
    w(p)
    = \int \frac{1}{\rho(p)} \dd{p}
\end{aligned}$$

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

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

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

$$\begin{aligned}
    \boxed{
        \Phi^* = \Phi + w(p)
    }
\end{aligned}$$

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

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

At every point in the fluid, despite $p$ 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:

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

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



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