# Volumetric Dilatation Rate, Material derivatives, and Divergence

in class we derived the following relationship: $$\frac{1}{V}\frac{dV}{dt}= \nabla \cdot \vec{v}$$ This was derived though the analysis of linear deformation for a fluid-volume, where: $$dV = dV_x +dV_y + dV_z$$ I understood the derived relation as: $$\frac{1}{V}V'(t) = \nabla \cdot \vec{v}$$ However, my professor recently told me that the $$d/dt$$ operator before V, stood for the material derivative and not the common derivative. I am very confused as to how is that the case, given that we did an infinitesimal analysis of linear deformation, in a way I could call analogous to any other infinitesimal analysis that results in the common derivative.

I also tried deriving the equation by taking the material derivative of $$V$$, and dividing by $$V$$: $$\frac{1}{V}\frac{DV}{Dt} = \frac{1}{V}\frac{\partial V}{\partial t} + \frac{1}{V}(\vec{v} \cdot gradV)$$

but I was unable to.

The continuity equation reads $$\frac{\partial \rho}{\partial t}+v\centerdot \nabla \rho+\rho \nabla \centerdot v=0$$where $$\rho$$ is the fluid density. Dividing this by $$\rho$$ gives $$\frac{1}{\rho}\left(\frac{\partial \rho}{\partial t}+v\centerdot \nabla \rho\right)+\nabla \centerdot v=0$$But, since the density is the inverse of the specific volume V, we have $$\frac{1}{V}\left(\frac{\partial V}{\partial t}+v\centerdot \nabla V\right)=\frac{1}{V}\frac{DV}{Dt}=\nabla \centerdot v$$
• No. $\ln{\rho}=-\ln{V}$ Jan 17, 2021 at 15:44
• So since $$1/\rho = \nu = V/m$$, then $$\frac{1}{\rho} (\frac{\partial \rho}{\partial t} + \vec{v} \cdot \nabla \rho)+ \nabla \cdot \vec{v} = V (\frac{\partial 1/V}{\partial t} + \vec{v} \cdot \nabla (1/V)) +\nabla \cdot \vec{v} = \frac{D(1/V)}{Dt} + \nabla \cdot \vec{v}= 0$$ and then do you integrate? Or am I doing this incorrectly.
• No. You're missing a minus sign. $$\frac{D\ln{\rho}}{Dt}=-\frac{D\ln{V}}{Dt}$$. Please review differentiation of a quotient. Jan 17, 2021 at 23:00
• $d(1/V)=-dV/V^2$ so $Vd(1/V)=-\frac{dV}{V}$ so $\frac{1}{\rho}d\rho=-\frac{1}{V}dV$ Jan 18, 2021 at 0:36