What does the Laplacian of a fluid velocity potential mean?

Question

I was studying something related to fluid mechanics and then I found that $\nabla^2 \Phi = 0$ where $\Phi$ is the fluid velocity potential ($\vec{V}=\nabla \Phi$). So I was wondering what does it mean that the laplacian of the fluid velocity potential is equal to zero (everywhere) and also in general.

I know the gradient of a scalar function tells us about the direction in which the function increases and the divergence of a field about how it spreads out in space. Then I could guess that the Laplacian is equal to zero because the fluid could be uniform and thus the velocity field doesn't spread out.

Is this a correct assumption?

Answer 1

One of the consequences of $\nabla\cdot V = 0$ is the incompressibility of the fluid. It is perfectly possible for air, for example, that some portions of space are temporary being depleted $\nabla\cdot V > 0$ or filled $\nabla\cdot V < 0$.

edit (Jan,$15^{th}$, 2020): if V can be expressed as a gradient of some scalar, its rotational (only for x component) is:

$\nabla$ X $V$ = $\nabla$ X $ (\nabla \phi) =$

$\frac{\partial (\nabla \phi)_z}{\partial y} - \frac{\partial (\nabla \phi)_y}{\partial z} = \frac{\partial (\partial \phi/\partial z)}{\partial y} - \frac{\partial (\partial \phi/\partial y)}{\partial z} = \frac{\partial^2\phi}{\partial z \partial y} - \frac{\partial^2\phi}{\partial y \partial z} = 0$

The same for the other components.

Resuming: $\nabla\cdot V = 0$ means incompressibility, but the additional condition of $\nabla^2 \phi = 0\;$, (what implies $V = \nabla \phi)\;$means the fluid has rotational = 0 everywhere.