We had introduced the Euler flow equations, and saw it gave us (in $n$ dimensions) $n$ equations in $n+2$ unknowns. This requires introducing another 2 equations for us to solve the equations of motion. Today we will do half the required job, we will introduce one additional equation: the conservation of mass, or continuity equation.
Euler's Derivation
A relatively clever derivation begins by examining a fluid parcel of "small volume" $V$ and density $\rho$, so small such that $\int_{V}\rho\,\D^{n}x\approx\rho V$. Since the particles in the fluid parcel remain in the parcel over time, we have its total mass be constant: \begin{equation} \frac{\mathrm{D}(\rho V)}{\mathrm{D} t} = 0. \end{equation} Since this fluid parcel is nonempty, $\rho V\gt0$, so taking the logarithm of the mass and then taking its time derivative gives us: \begin{equation} \frac{1}{\rho}\frac{\mathrm{D}\rho}{\mathrm{D} t} + \frac{1}{V}\frac{\mathrm{D} V}{\mathrm{D} t} = 0. \end{equation} Supposing we have the fluid parcel be described as a rectangle with edges $\delta x$, $\delta y$, $\delta z$, then we see \begin{equation} V = \delta x\,\delta y\,\delta z \end{equation} hence Taylor expanding the volume as a function of time gives us \begin{equation} V + \frac{\mathrm{D} V}{\mathrm{D} t}\delta t = \left[1 + \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right)\delta t\right]\delta x\,\delta y\,\delta z. \end{equation} Collecting these together give us the equation \begin{equation} \frac{\mathrm{D}\rho}{\mathrm{D} t} + \rho\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right) = 0. \end{equation} This is the continuity equation.
Puzzle 2. This derivation seems to use the Lagrangian description; in general, however, we would need to include a factor of the Jacobian to describe the volume of the fluid parcel. Work out this derivation. Cheaters may consult Childress.
Modern Textbook Derivation
Consider now a fixed cube in the fluid, whose side lengths are again $\delta x$, $\delta y$, $\delta z$. What amount of mass flows into the cube across the $\delta y\delta z$ face?
We find, assuming the flow moves away from the origin, \begin{equation} \begin{pmatrix}\mbox{mass which}\\ \mbox{flows in}\end{pmatrix} = \left(\rho - \frac{1}{2}\frac{\partial(\rho u)}{\partial x}\delta x\right)\delta y\,\delta z. \end{equation} Similarly, the amount flowing out \begin{equation} \begin{pmatrix}\mbox{mass which}\\ \mbox{flows out}\end{pmatrix} = \left(\rho + \frac{1}{2}\frac{\partial(\rho u)}{\partial x}\delta x\right)\delta y\,\delta z. \end{equation} We then have the net mass gained be $-\delta x\,\delta y\,\delta z\partial_{x}(\rho u)$. Doing this for the other faces, we get \begin{equation} \begin{pmatrix}\mbox{net mass}\\ \mbox{flows}\end{pmatrix} = \nabla\cdot(\rho\vec{u})\,\delta x\,\delta y\,\delta z. \end{equation} On the other hand, the mass flux varies over time as \begin{equation} \begin{pmatrix}\mbox{mass}\\ \mbox{flux}\end{pmatrix} = \frac{\partial}{\partial t}(\rho\,\delta x\,\delta y\,\delta z). \end{equation} Setting equals to equals, and simplifying, we find \begin{equation} \frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\vec{u})=0 \end{equation} which is another, equivalent, form of the continuity equation.
Exercise 3. Prove Eq (10) is equivalent to Eq (5).
References
I have profited most from Lamb's presentation, though Childress takes greater care when working through the derivation in the Lagrangian description.
- Leonhard Euler, "Principes généraux du mouvement des fluides". Mémoires de l'académie des sciences de Berlin 11 (1757) 274–315; English translation arXiv:0802.2383
- Horace Lamb, Hydrodynamics. Dover, sixth ed., 1932; §7 (archive)
- Stephen Childress, An Introduction to Theoretical Fluid Mechanics. AMS Press, 2009.
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