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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.

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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 $$\frac{1}{V}\left(\frac{\partial V}{\partial t}+v\centerdot \nabla V\right)=\frac{1}{V}\frac{DV}{Dt}=\nabla \centerdot v$$

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  • $\begingroup$ That is a very nice explanation, thanks a lot! I was wondering if the divergence term should have a negative sign next to it since it was positive on the RHS? Or is the sign arbitrary in this case? $\endgroup$
    – STOI
    Commented Jan 17, 2021 at 15:36
  • $\begingroup$ No. $\ln{\rho}=-\ln{V}$ $\endgroup$
    – Chet Miller
    Commented Jan 17, 2021 at 15:44
  • $\begingroup$ 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. $\endgroup$
    – STOI
    Commented Jan 17, 2021 at 18:58
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    $\begingroup$ No. You're missing a minus sign. $$\frac{D\ln{\rho}}{Dt}=-\frac{D\ln{V}}{Dt}$$. Please review differentiation of a quotient. $\endgroup$
    – Chet Miller
    Commented Jan 17, 2021 at 23:00
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    $\begingroup$ $d(1/V)=-dV/V^2$ so $Vd(1/V)=-\frac{dV}{V}$ so $\frac{1}{\rho}d\rho=-\frac{1}{V}dV$ $\endgroup$
    – Chet Miller
    Commented Jan 18, 2021 at 0:36

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