# Derivations

A velocity field that is a gradient of a scalar potential field is irrotational

It is a direct result from vector calculus that the curl of the gradient of a scalar is always zero, $\nabla \times \nabla \phi = 0$:

$\nabla \times \nabla \phi = \nabla \times (\frac{\partial \phi}{\partial x}\hat{x} + \frac{\partial \phi}{\partial y}\hat{y} + \frac{\partial \phi}{\partial z}\hat{z})$
$\nabla \times \nabla \phi = (\frac{\partial^{2} \phi}{\partial z \partial y} - \frac{\partial^{2} \phi}{\partial y \partial z})\hat{x} - (\frac{\partial^{2} \phi}{\partial z \partial x} - \frac{\partial^{2} \phi}{\partial x \partial z})\hat{y} + (\frac{\partial^{2} \phi}{\partial y \partial x} - \frac{\partial^{2} \phi}{\partial x \partial y})\hat{z} = 0$

Incompressible flow is divergence free

By definition, small fluid elements in an incompressible flow do not change in density. Therefore the material derivative of density is zero in an incompressible flow: $\frac{D \rho}{Dt} = 0$.

The differential statement of conservation of mass is:

$\frac{D \rho}{Dt} + \nabla \cdot \rho \vec{v} = 0$

The first term is zero for an incompressible flow, which leads to:

$\nabla \cdot \rho \vec{v} = 0$
$\rho(\nabla \cdot \vec{v}) = 0$
$\nabla \cdot \vec{v} = 0$

A divergence free velocity field is an equivalent statement of conservation of mass for an incompressible flow.

Bernoulli's equation is valid in the full fluid domain in potential flow

The Euler equations are a statement of conservation of momentum in an inviscid flow:

$\rho \frac{\partial \vec{v}}{\partial t} + \rho(\vec{v} \cdot \nabla)\vec{v} + \nabla p + \rho g \nabla z = 0$

The second term can be rewritten through a vector identity:

$(\vec{v} \cdot \nabla)\vec{v} = -\vec{v} \times \omega + \nabla (\frac{| \vec{v} |^{2} }{2})$

Vorticity is zero in irrotational flow, so Euler's equation simplifies to:

$\rho \frac{\partial \nabla \phi}{\partial t} + \rho \nabla(\frac{| \vec{v} | ^{2}}{2}) + \nabla p + \rho g \nabla z$
$\nabla(\rho \frac{\partial \phi}{\partial t} + \rho(\frac{| \vec{v} |^{2}}{2}) + p + \rho gz) = 0$
$\rho \frac{\partial \phi}{\partial t} + \rho(\frac{| \vec{v} |^{2}}{2}) + p + \rho gz = constant$

This is a statement of the unsteady version of Bernoulli's equation, which is valid not just along streamlines, but between points anywhere in the fluid domain.

A dipole flow element is a combination of a point source and sink

Combine the potentials of a source and sink, both strength $m$ and located on the x-axis separated by distance $2a$:

$\phi = \frac{m}{2\pi}[ln(\sqrt{(x - a)^{2} + y^{2}}) - ln(\sqrt{(x + a)^{2} + y^{2}})]$

Let the dipole strength, $\mu$, be equal to a combination of the sink/source strengths and the distance: $\mu = 2ma$. Take the limit as the source and sink get closer together and $\mu$ remains constant:

$\phi = \textrm{lim}_{a\rightarrow 0} \frac{\mu}{4\pi a}[ln(\sqrt{(x - a)^{2} + y^{2}}) - ln(\sqrt{(x + a)^{2} + y^{2}})]$
$\phi = \frac{\mu}{2\pi} \frac{x}{x^{2} + y^{2}}$

This same methodology can be used to prove a more general case of a dipole at any angle.