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+---
+title: "Hydrostatic pressure"
+firstLetter: "H"
+publishDate: 2021-03-12
+categories:
+- Physics
+- Fluid mechanics
+
+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:
+
+$$\begin{aligned}
+    0
+    = \nabla \Phi + \frac{\nabla p}{\rho}
+    = \nabla \Phi + \dv{w}{p} \nabla p
+    = \nabla \Big( \Phi + w(p) \Big)
+\end{aligned}$$
+
+By defining the **effective gravitational potential** $\Phi^* = \Phi + w(p)$,
+we get the cleanest form yet of the equilibrium condition,
+which states that $\Phi^*$ must be a constant:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \Phi^*
+        = 0
+    }
+\end{aligned}$$
+
+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.