---
title: "Bernoulli's theorem"
firstLetter: "B"
publishDate: 2021-04-02
categories:
- Physics
- Fluid mechanics
- Fluid dynamics

date: 2021-04-02T15:05:08+02:00
draft: false
markup: pandoc
---

# Bernoulli's theorem

For inviscid fluids, **Bernuilli's theorem** states
that an increase in flow velocity $\va{v}$ is paired
with a decrease in pressure $p$ and/or potential energy.
For a qualitative argument, look no further than
one of the [Euler equations](/know/concept/euler-equations/),
with a [material derivative](/know/concept/material-derivative/):

$$\begin{aligned}
    \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
    = \va{g} - \frac{\nabla p}{\rho}
\end{aligned}$$

Assuming that $\va{v}$ and $\va{g}$ are constant in $t$,
it becomes clear that a higher $\va{v}$ requires a lower $p$:

$$\begin{aligned}
    \frac{1}{2} \nabla \va{v}^2
    = \va{g} - \frac{\nabla p}{\rho}
\end{aligned}$$


## Simple form

For an incompressible fluid
with a time-independent velocity field $\va{v}$ (i.e. **steady flow**),
Bernoulli's theorem formally states that the
**Bernoulli head** $H$ is constant along a streamline:

$$\begin{aligned}
    \boxed{
        H
        = \frac{1}{2} \va{v}^2 + \Phi + \frac{p}{\rho}
    }
\end{aligned}$$

Where $\Phi$ is the gravitational potential, such that $\va{g} = - \nabla \Phi$.
To prove this theorem, we take the material derivative of $H$:

$$\begin{aligned}
    \frac{\mathrm{D} H}{\mathrm{D} t}
    &= \va{v} \cdot \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    + \frac{\mathrm{D} \Phi}{\mathrm{D} t}
    + \frac{1}{\rho} \frac{\mathrm{D} p}{\mathrm{D} t}
\end{aligned}$$

In the first term we insert the Euler equation,
and in the other two we expand the derivatives:

$$\begin{aligned}
    \frac{\mathrm{D} H}{\mathrm{D} t}
    &= \va{v} \cdot \Big( \va{g} - \frac{\nabla p}{\rho} \Big)
    + \Big( \pdv{\Phi}{t} + (\va{v} \cdot \nabla) \Phi \Big)
    + \frac{1}{\rho} \Big( \pdv{p}{t} + (\va{v} \cdot \nabla) p \Big)
    \\
    &= \pdv{\Phi}{t} + \frac{1}{\rho} \pdv{p}{t}
    + \va{v} \cdot \big( \va{g} + \nabla \Phi \big) + \va{v} \cdot \Big( \frac{\nabla p}{\rho} - \frac{\nabla p}{\rho} \Big)
\end{aligned}$$

Using the fact that $\va{g} = - \nabla \Phi$,
we are left with the following equation:

$$\begin{aligned}
    \frac{\mathrm{D} H}{\mathrm{D} t}
    &= \pdv{\Phi}{t} + \frac{1}{\rho} \pdv{p}{t}
\end{aligned}$$

Assuming that the flow is steady, both derivatives vanish,
leading us to the conclusion that $H$ is conserved along the streamline.

In fact, there exists **Bernoulli's stronger theorem**,
which states that $H$ is constant *everywhere* in regions with
zero [vorticity](/know/concept/vorticity/) $\va{\omega} = 0$.
For a proof, see the derivation of $\va{\omega}$'s equation of motion.


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