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Material scientists have discovered a new fluid property called "radost" that is carried along with a fluid as it moves from one place to the next (just like a fluid's mass or momentum). Let $r(x,y,z,t)$ be the amount of radost/unit mass in a fluid. Let $\rho(x,y,z,t)$ be the mass density of the fluid. Let $\vec{v}(x,y,z,t)$ be the velocity vector of the fluid. Use the divergence theorem to derive a conservation law for radost.

We did an example like this in class, but for conserving mass, so it was a little different. What we ended up with was the following expression $$\frac{\partial \rho}{\partial t}+\nabla \cdot (\rho \vec{v})=0$$

We started by writing, $$dM=\rho\, dV$$ Thus, $$M=\int_V \rho\, dV$$ Then we applied the divergence theorem and that was basically it.

I'm just kind of confused how to start this one.

dfeuer
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Lefty
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2 Answers2

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Figured this out:

Use $dr=r\rho dV$ and continue as above.

Lefty
  • 55
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Given a point $\vec{r}$, divide the space in two pieces: 1) $V$ such that $\color{#0000ff}{\Large\vec{r} \in V}$ and 2) $\tilde{V}$. $V \cup \tilde{V}$ is the whole space. At any time $t$, the number of particles $N_{V}\left(t\right)$ in $V$ plus the number of particles $N_{\tilde{V}}\left(t\right)$ in $\tilde{V}$ is a constant of motion: $N_{V}\left(t\right) + N_{\tilde{V}}\left(t\right) = \mbox{constant}\,, \forall\ t$, such that

$$ {{\rm d}N_{V}\left(t\right) \over {\rm d}t} + {{\rm d}N_{\tilde{V}}\left(t\right) \over {\rm d}t} = 0\,, \qquad \forall\ t $$

$N_{V}\left(t\right)$ is given by $$ N_{V}\left(t\right) = \int_{V}\rho\left(\vec{R},t\right)\,{\rm d}V \qquad\Longrightarrow\qquad {{\rm d}N_{V}\left(t\right) \over {\rm d}t} = {\LARGE\int}_{\!\!\!\!\!\!V}\!\!\!\!\!\!{\partial\rho\left(\vec{R},t\right) \over \partial t}\,{\rm d}V $$

where $\rho\left(\vec{R},t\right)$ is, at time $t$, the particle density at point $\vec{R}$. In an infinitesimal ${\rm d}S$ surface element, $\mbox{at}\ \vec{R}$, of $V$ and during the time interval $\Delta t$, the number of particles that cross ${\rm d}S$ is given by $\Delta N_{\tilde{V}}\left(t\right) \equiv \rho\left(\vec{R}, t\right)\left\{\left[\vec{v}\left(\vec{R}, t\right) \Delta t\right]{\rm d}S\cos\left(\angle\right)\right\}\quad$. $\vec{v}\left(\vec{R}, t\right)$ is the flux velocity and $\angle$ is the angle between $\vec{v}\left(\vec{R}, t\right)$ and a normalized vector $\vec{n}$ perpendicular to ${\rm d}S$. $\left\vert\vec{n}\right\vert = 1$. Then ${\rm d}N_{\tilde{V}}\left(t\right) = \vec{\rm J}\left(\vec{R}, t\right)\cdot{\rm d}\vec{S}\Delta t$ where $\vec{\rm J}\left(\vec{R}, t\right) \equiv \rho\left(\vec{R}, t\right)\vec{v}\left(\vec{R}, t\right)$ and ${\rm d}\vec{S} \equiv {\rm d}S\,\,\vec{n}\quad$. $\vec{\rm J}\left(\vec{R}, t\right)$ is the $Current\ Density\ Vector$.

$$ {{\rm d}N_{\tilde{V}}\left(t\right) \over {\rm d}t} = \lim_{\Delta t \to 0}\int_{S}{\Delta N_{\tilde{V}}\left(t\right) \over \Delta t} = \int_{S}\vec{\rm J}\left(\vec{R}, t\right)\cdot{\rm d}\vec{S} = \int_{V}\nabla\cdot\vec{\rm J}\left(\vec{R}, t\right)\,{\rm d}V $$

where we have used the Gauss Divergence Theorem. Then

$$ {\LARGE\int}_{\!\!\!\!\!\!V}\!\left[% {\partial\rho\left(\vec{R},t\right) \over \partial t} + \nabla\cdot\vec{\rm J}\left(\vec{R}, t\right) \right]\,{\rm d}V = 0 $$

\begin{align} &\lim_{V \to 0}{1 \over V}\!\!\!{\LARGE\int}_{\!\!\!\!\!\!V}\!\!\!\!\left[% {\partial\rho\left(\vec{R},t\right) \over \partial t} + \nabla\cdot\vec{\rm J}\left(\vec{R}, t\right) \right]\,{\rm d}V = 0\ \Longrightarrow\ \color{#ff0000}{\large% {\partial\rho\left(\vec{r},t\right) \over \partial t} + \nabla\cdot\vec{\rm J}\left(\vec{r}, t\right) = 0\,\color{#0000ff}{;\forall\, \vec{r}, t}} \end{align}

Felix Marin
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