Why is this the "no bubbles" condition?

Question

I'm reading through the "Waves" textbook from the Berkley series. In section 7.3, water waves are described. First, the condition that water is incompressible is derived as follows (assuming $\rho$ is constant):

$$\frac{\partial \rho}{\partial t} +\nabla \cdot (\rho \vec{v})=0 \\\to \nabla \cdot \vec{v} = \nabla \cdot \frac{\partial \vec{\psi}}{\partial t} = \frac{\partial}{\partial t}(\nabla \cdot \vec{\psi})=0 \\\to \nabla \cdot \vec{\psi} = \mathrm{constant} $$

This condition is reinforced by claiming that, if the constant is not $0$, then the surface integral over some small sphere of $\psi$ is not $0$ and "that could only mean that there are bubbles".

Can somebody explain to me how having no bubbles implies that $\nabla \cdot \vec{\psi} =0$?

Answer 1

I think what the author might be implying is that if the surface integral of $\vec{\psi} \cdot \vec{n}$ ($\vec{n}$ denoting the unit normal) over the small sphere is non-zero, the sphere has expanded (positive integral) or been compressed (negative integral) in comparison to the equilibrium position, which for an incompressible substance can only be realized through bubbles.

In equilibrium the displacement field is $\vec{\psi} = 0$ and thus $\nabla \cdot \vec{\psi} = 0$. In consequence, from $\partial_t \left(\nabla \cdot \vec{\psi}\right) = 0$ you obtain $\nabla \cdot \vec{\psi} = 0$ at any time for any motion that with a sufficiently regular motion can be transformed from or to the equilibrium position.