= Velocity potential
{title2=$\mathbf u=\nabla\phi$}
{wiki}
Use a <velocity potential> to reconstruct an <irrotational flow> from one <scalar field>. A <velocity potential> $\phi(\mathbf x,t)$ satisfies
$$
\mathbf u=\nabla\phi.
$$
It has units $\mathrm{m^2\,s^{-1}}$ and is unchanged as a description of the <velocity field> when any function of time alone is added. For twice continuously differentiable $\phi$, the identity <curl of a gradient> proves zero <vorticity>. In <incompressible flow>, taking the <divergence> also gives $\nabla^2\phi=0$, the <Laplace equation>. This turns a three-component <velocity field> into a single <harmonic function> with appropriate <boundary conditions>. https://ocw.mit.edu/courses/2-016-hydrodynamics-13-012-fall-2005/c472432debcf6ee250209b68cf18cc12_2005reading4.pdf[A. H. Techet, MIT Hydrodynamics reading 4, pp. 1–3]{ref}
For <uniform flow> $\mathbf u=(U,0,0)$, take $\phi=Ux$. With $U=2\ \mathrm{m\,s^{-1}}$, the change from $x=0$ to $x=3\ \mathrm m$ is $\Delta\phi=6\ \mathrm{m^2\,s^{-1}}$. By the <fundamental theorem for line integrals>, the <line integral> of <velocity> between these points equals that change along any path in the domain; a <closed curve> gives zero <circulation>.
For the potential <line vortex>, choose a local angle branch in <polar coordinates> and set $\phi=K\theta$. Then $u_r=\partial_r\phi=0$ and $u_\theta=r^{-1}\partial_\theta\phi=K/r$. One counterclockwise turn changes the continued <velocity potential> by $2\pi K$, although the <velocity field> returns to its original value. With $K=1\ \mathrm{m^2\,s^{-1}}$, this change is $2\pi\ \mathrm{m^2\,s^{-1}}$, exactly the <circulation>. A full <annulus> therefore admits local angle branches and a multivalued <velocity potential>, but no global single-valued <velocity potential> when $K\ne0$. Cutting the <annulus> along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is <path independence>, not a failure of the local <gradient> formula.
Back to article page
