Why velocity potential?

(Aero-)Acoustics (among other parts of fluid-dynamics) loves the velocity potential $\varphi$ defined as $\vec{v} = \nabla \ \varphi$ with the condition of $\nabla \times \ \vec{v} = 0$. I am definitely not trying to undermine this concept, I intuitively like that as well, but why exactly is that so cool?

I am aware of:

• It's a scalar quantity with easy connections to acoustic pressure and velocity.
• It can be used as a descriptor of curl-free part of the field (using e.g. Helmholtz decomposition) when more complex fields are given.

If you would prefer more exact questions:

• Given $\nabla \times \ \vec{v} = 0$, what is the equation that can't be formulated 'equivalent elegant' for acoustic pressure? (Which is scalar as well...)
• Are there any typical operations, derivations etc. that are significantly harder to be formulated for pressure than for velocity potential?
• One reason to introduce it is that $\nabla\times\nabla\phi=0$ identically for any $\phi$; so if $\mathbf v=\nabla\phi$, then we're guaranteed that $\nabla\times\mathbf v=0$. Sep 23 '15 at 16:41
• That's right, but it's kind of tautology, isn't it? One way: we presume $v \equiv \nabla \phi$. Other way: $\nabla \times v \equiv 0$. Sep 23 '15 at 16:51
• Well it's only under certain circumstances that $\nabla\times\mathbf v=0$, so it's not really a tautology but a special case of the vorticity. Sep 23 '15 at 17:02

For (usually) simply connected regions whenever you have an irrotational field, that is $\textrm{curl}\,\mathbf {v}=0$, then it exists a function $f$ such that $\mathbf{v} = \textrm{grad}\,f$ in every point $x$ where the domains of definition make sense. The function $f$ is then said to be a potential for the field $\mathbf{v}$. Likewise, knowledge of the potential function in any point of the space allows to derive back the field, equivalently (again, provided the correct assumptions on connectivity of the regions to hold).