Continuum mechanics and continuity eq

[Niles]

Homework Statement

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?

 

[Chestermiller]

Homework Statement

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?

Your last equation should read:
$$\nabla \cdot u = -\frac{1}{\rho}(\frac{\partial \rho}{\partial t} + u\cdot \nabla\rho)$$
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
$$\frac{Dρ}{Dt}=\frac{\partial \rho}{\partial t} + u\cdot \nabla\rho$$

Chet