# Velocity potential

{{ safesubst:#invoke:Unsubst||$N=Refimprove |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} A velocity potential is a scalar potential used in potential flow theory. It was introduced by Joseph-Louis Lagrange in 1788.[1]

It is used in continuum mechanics, when a continuum occupies a simply-connected region and is irrotational. In such a case,

${\displaystyle \nabla \times \mathbf {u} =0,}$

where ${\displaystyle \mathbf {u} }$ denotes the flow velocity. As a result, ${\displaystyle \mathbf {u} }$ can be represented as the gradient of a scalar function ${\displaystyle \Phi \;}$:

${\displaystyle \mathbf {u} =\nabla \Phi \;}$,

${\displaystyle \Phi \;}$ is known as a velocity potential for ${\displaystyle \mathbf {u} }$.

A velocity potential is not unique. If ${\displaystyle a\;}$ is a constant, or a function solely of the temporal variable, then ${\displaystyle \Phi +a(t)\;}$ is also a velocity potential for ${\displaystyle \mathbf {u} \;}$. Conversely, if ${\displaystyle \Psi \;}$ is a velocity potential for ${\displaystyle \mathbf {u} \;}$ then ${\displaystyle \Psi =\Phi +b\;}$ for some constant, or a function solely of the temporal variable ${\displaystyle b(t)\;}$. In other words, velocity potentials are unique up to a constant, or a function solely of the temporal variable.

If a velocity potential satisfies Laplace equation, the flow is incompressible ; one can check this statement by, for instance, developing ${\displaystyle \nabla \times (\nabla \times \mathbf {u} )}$ and using, thanks to the Clairaut-Schwarz's theorem, the commutation between the gradient and the laplacian operators.

Unlike a stream function, a velocity potential can exist in three-dimensional flow.

## Usage in Acoustics

In theoretical acoustics, it is often desirable to work with the acoustic wave equation of the velocity potential ${\displaystyle \Phi \;}$ instead of pressure ${\displaystyle p\;}$ and/or particle velocity ${\displaystyle \mathbf {u} \;}$.

${\displaystyle \nabla ^{2}\Phi -{1 \over c^{2}}{\partial ^{2}\Phi \over \partial t^{2}}=0}$

Solving the wave equation for either ${\displaystyle p\;}$ field or ${\displaystyle \mathbf {u} \;}$ field doesn't necessarily provide a simple answer for the other field. On the other hand, when ${\displaystyle \Phi \;}$ is solved for, not only is ${\displaystyle \mathbf {u} \;}$ found as given above, but ${\displaystyle p\;}$ is also easily found as

${\displaystyle p=-\rho {\partial \over \partial t}\Phi }$.

## Notes

1. {{#invoke:citation/CS1|citation |CitationClass=book }}