# Continuum mechanics and continuity eq

1. Dec 12, 2013

### Niles

1. The problem statement, all variables and given/known data
Hi

I can't follow the derivaton in this link. It is the following equality they have in the beginning, which I don't understand:
$$\nabla \cdot u = \frac{1}{\rho}\frac{d\rho}{dt}$$
Following the very first equation on the page, I believe it should be
$$\nabla \cdot u = -\frac{1}{\rho}(\frac{d\rho}{dt} + u\cdot \nabla\rho)$$
Do you agree with me? Or is there something I am missing here?

2. Dec 12, 2013

### Staff: Mentor

$$\nabla \cdot u = -\frac{1}{\rho}(\frac{\partial \rho}{\partial t} + u\cdot \nabla\rho)$$
rather than dρ/dt. The quantity $\frac{\partial \rho}{\partial t} + u\cdot \nabla\rho$ is called the material derivative of ρ, and represents physically the time rate of change of the density measured by an observer who is traveling with the fluid. The material derivative is often represented using capital D's, so that
$$\frac{Dρ}{Dt}=\frac{\partial \rho}{\partial t} + u\cdot \nabla\rho$$