# Incompressible flow condition intuition

The derivation of the continuity equation leads to $$\frac{D\rho}{Dt}=-\rho(\vec\nabla\cdot\vec u)$$ as shown in this wiki article.

If we assume that the flow is incompressible, it implies that $$\rho$$ is constant w.r.t time. Thus $$\vec\nabla\cdot\vec u=0$$

But how do we understand this equation? How is there a relation between the density of the fluid and its speed? I feel like I am missing a very obvious understanding of the divergence theorem, but I can't quite point it out.

An intuitive explanation comes from the meaning of the word divergence itself. If the divergence of the velocity is zero, it means that there is no (net) inflow or outflow of fluid from a given volume. If the fluid inside a given volume were to diverge, i.e. reduce in amount over time, then the density (amount per unit volume) would also decrease. Similarly for a negative divergence (net inflow into a given volume).

Thus, for the density to remain constant (incompressibility), the net inflow/outflow in any given volume must be zero. This is precisely $$\vec{\nabla}\cdot\vec{u}=0$$. Hope that helps.

• Initially, I thought about this, but here we are talking about the density of the fluid itself and not about the average density of the surface taken in the divergence theorem. That is $\rho=\text{mass of liquid/volume of liquid}\neq \text{mass of liquid/volume of imaginary surface used in divergence theorem}$ Commented Jul 10 at 21:54
• Look at it as the density of the fluid present inside the given volume - or equivalently, let the volume be that of a fluid parcel of infinitesimal size Commented Jul 11 at 9:54

As you have said it is derived from the continuity equation (which is one of Euler fluid equations) $$\frac{\partial \rho}{\partial t}+\nabla\cdot(\rho \mathbf u)=0$$ if the density of the fluid $$\rho$$ is just a constant (does not change in time and does not depend on position), then the equation reduces to $$\nabla\cdot \mathbf u =0$$ where $$\mathbf u$$ is the velocity field.

Note that the original equation says that no fluid is entering/escaping a differential volume. The second equation is saying the same for fluids of constant density.

How is there a relation between the density of the fluid and its speed?

The relation is the continuity equation, it relates density to velocity. For constant density, the fluid cannot "diverge", fluid cannot become more disperse or more concentrated in a given volume.

You ask"How is there a relation between the density of the fluid and its speed"? There is none $$\nabla u=0$$ says only that the velocity does not change for example in x direction .
• This isn't helpful as you're basically saying "I don't see $\rho$ in the equation, so there's no relation" while completely ignoring the fact that the equation was derived from one involving both velocity and density. Commented Jul 11 at 13:56
Density of a fluid is $$\rho(\mathbf{x},t),$$ expressed in spatial coordinates. A fluid is incompressible if its total derivative is zero, i.e. $$\frac{\mathrm{d}\rho(\mathbf{x},t)}{\mathrm{d}t}=0.$$ so that $$\frac{\mathrm{d}\rho}{\mathrm{d}t}=\frac{\partial\rho}{\partial t}+\frac{\partial\rho}{\partial x_i}\frac{\mathrm{d}x_i}{\mathrm{d}t}=\frac{\partial\rho}{\partial t}+\frac{\partial(\rho u_i)}{x_i}=\frac{\partial\rho}{\partial t}+\rho\frac{\partial u_i}{\partial x_i}=\frac{\partial\rho}{\partial t}+\rho\nabla\cdot\mathbf{u}=0.$$ If $$\rho$$ is expressed in material coordinates $$\mathbf{X}$$, then $$\partial_t\rho(\mathbf{X})=0$$. Hence, $$\nabla\cdot\mathbf{u}=0.$$