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I'm currently learning on substantial derivatives in fluid mechanics and kind of understand how partial derivatives $\frac{(\partial\rho)}{(\partial t)}$ and substantial derivatives $\frac{(D\rho)}{(Dt)}$ differ. However, I can't get the intuition from the vector notation of the two. What would be the good way to interpret the difference between two notations and apply it to partial and substantial derivative's difference. Is $\rho$ independent of coordinate in $\rho(\nabla\cdot v)$?

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2 Answers 2

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What's the difference between $\nabla\bullet(\rho v)$ and $\rho(\nabla\bullet v)$ as a physical intuition?

Physically, we are often interested in $\rho \vec v$, where $\vec v$ is the fluid velocity field (not the particle velocity) when we want to know how much mass is flowing out of some given volume $V$ with a boundary $\partial V$.

We calculate this as an integral over area $d^2\vec A$ as: $$ \oint_{\partial V} \rho \vec v \cdot{d^2\vec A} \;, $$ which can be turned into a volume integral: $$ \int_V \vec \nabla \cdot \left(\rho\vec v\right)d^3r\;, $$ where you see that it is the whole thing $\rho \vec v$ that is being differentiated in the integrand: $\vec \nabla \cdot (\rho \vec v)$, which is generally different from $\rho\vec \nabla \cdot \vec v$ when $\rho$ depends on the spatial coordinates.

For example, the change in mass of a fixed volume $V$ has to come from two places: (1) any explicit change in the density with time; (2) flow of mass out of the volume with time. So we have: $$ \frac{dM}{dt} = \int d^3r \frac{\partial \rho}{\partial t} + \int d^3r \vec \nabla \cdot \left(\rho \vec v\right)=0\;, $$ and we set this to zero since mass (usually) can't be created out of nothing...


More generally, we are interested in $\rho \vec v$ since we often are interested in quantities per unit mass and the flow of those is usually written as: $$ \theta \rho \vec v\;, $$ where $\theta$ is the amount of whatever per unit mass. (e.g., if you are interested in the mass itself, $\theta=1$. e.g., if you are interested in the momentum $\theta =\vec v$.)

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  • $\begingroup$ Thanks i can pretty much understand now. If im right, $\nabla\bullet\rho v$ and $\rho\nabla\bullet v$ is same as 0 for incompressible, but what would two values be when it is steady state or non-steady state? $\endgroup$
    – Lime nut
    Nov 28, 2023 at 10:12
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    $\begingroup$ "Steady state" just means that the quantities like $\rho$ and $\vec v$ do not explicitly depend on time. It doesn't really tell you anything directly about their spatial dependence. But, for example, in steady state you know that $\frac{\partial \rho}{\partial t}=0$, so using the continuity equation you can then also say that $\vec \nabla \cdot (\rho\vec v) = 0$ $\endgroup$
    – hft
    Nov 28, 2023 at 11:15
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In general, using the product rule for the divergence operator $\nabla \cdot$, we have

$\nabla \cdot (\rho \vec v) = \rho \nabla \cdot \vec v + (\nabla \rho) \cdot \vec v$

We can therefore see that a necessary and sufficient condition for

$\nabla \cdot (\rho \vec v) = \rho \nabla \cdot \vec v$

is that

$ (\nabla \rho) \cdot \vec v=0$

One way (although not the only way) to achieve this is to have $\nabla \rho = 0$, which means that the fluid's density is independent of location i.e. the fluid is homogenous.

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  • $\begingroup$ Nice explanation! thanks. Besides, I was taught that substantial derivatives are a tracking of the physical quantity by change of time while following the particle itself. Would there be a way to relate this definition to a vector notation $\rho\nabla\bullet v$? I just can't come up with the idea. $\endgroup$
    – Lime nut
    Nov 28, 2023 at 12:27

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