The Reynolds Transport Theorem for Control Volumes
Matthew Louis / September 2025 (503 Words, 3 Minutes)
The Reynolds transport theorem is an all-around important relation for deriving integral conservation laws in fluid mechanics. Despite that, it’s almost completely unrelated to fluid mechanics, and is essentially a 3 dimensional generalization of the Leibnitz rule for differentiation under the integral sign. I made this post to clear up a very specific point of confusion that I myself, as well as some stack exchange users (see this post and this one), about how the velocity of the integration domain comes into play. The Wikipedia article states the equation in the following (general) form
\[\begin{equation}\label{eq:general-rtt} \frac{d}{dt} \int_{\Omega(t)} \boldsymbol{f}dV = \int_{\Omega(t)} \frac{\partial \boldsymbol{f}}{\partial t}dV + \int_{\partial \Omega(t)} (\boldsymbol{v}_b\cdot \boldsymbol{n})\boldsymbol{f}dA \end{equation}\]where $\boldsymbol{f}$ is an arbitrary vector (tensor) field, $\Omega(t)$ a time-dependent region with boundary $\partial \Omega(t)$, and $\boldsymbol{v}_b$ the velocity of the area element of the boundary region. Now, if the region $\Omega(t)$ is fixed, then $\boldsymbol{v}_b=0$ and we just get
\[\frac{d}{dt} \int_{\Omega} \boldsymbol{f}dV = \int_{\Omega} \frac{\partial \boldsymbol{f}}{\partial t}dV\]i.e. the plain old Leibnitz integral rule, which we ought to expect. This, however, is not the form often used in fluid mechanics textbooks, where even for a static control volume, there are still surface terms. This doesn’t mean it’s wrong either, though it is fair to say that this form is less convenient. So how do we get the form used in fluid mechanics from this one? Well, we first note the following fact, letting $\boldsymbol{f} = \rho \overline{\boldsymbol{f}}$:
\[\begin{equation}\label{eq:mat-derivative-and-integral} \frac{d}{dt}\int_{\Omega(t)}\boldsymbol{f}dV = \int_{\Omega(t)}\rho\frac{D\overline{\boldsymbol{f}}}{Dt}dV \end{equation}\]which assumes mass is conserved (continuity equation). And, by definition of the material derivative
\[\rho \frac{D\overline{\boldsymbol{f}}}{Dt} = \rho \frac{\partial \overline{\boldsymbol{f}}}{\partial t} + \rho \left(\boldsymbol{v}\cdot \nabla \overline{\boldsymbol{f}}\right)\]where $\boldsymbol{v}$ is the fluid velocity. Note that
\[\begin{equation*} \begin{split} \rho \left(\boldsymbol{v}\cdot \nabla \overline{\boldsymbol{f}}\right) &= \nabla \cdot (\rho \boldsymbol{v}\overline{\boldsymbol{f}}) - \overline{\boldsymbol{f}}\nabla\cdot (\rho \boldsymbol{v})\\ \rho \frac{\partial \overline{\boldsymbol{f}}}{\partial t} &= \frac{\partial (\rho \overline{\boldsymbol{f}})}{\partial t} - \overline{\boldsymbol{f}}\frac{\partial \rho}{\partial t} \end{split} \end{equation*}\]and we have
\[\begin{equation*} \begin{split} \rho \frac{D\overline{\boldsymbol{f}}}{Dt} = \frac{\partial (\rho \overline{\boldsymbol{f}})}{\partial t} +\nabla \cdot (\rho \boldsymbol{v}\overline{\boldsymbol{f}}) - \overline{\boldsymbol{f}}\left(\frac{\partial \rho}{\partial t} + \nabla\cdot (\rho \boldsymbol{v})\right) \end{split} \end{equation*}\]And if $\rho$ is a physical density field, then the second term vanishes by conservation of mass, leaving
\[\begin{equation}\label{eq:identity} \rho \frac{D\overline{\boldsymbol{f}}}{Dt} = \frac{\partial (\rho \overline{\boldsymbol{f}})}{\partial t} +\nabla \cdot (\rho \boldsymbol{v}\overline{\boldsymbol{f}}) \end{equation}\]Now, integrating both sides over the volume $\Omega(t)$ and using Eq. \ref{eq:general-rtt}, we get
\[\begin{equation*} \begin{split} \int_{\Omega(t)}\rho \frac{D\overline{\boldsymbol{f}}}{Dt}dV &= \int_{\Omega(t)}\frac{\partial (\rho \overline{\boldsymbol{f}})}{\partial t}dV +\int_{\Omega(t)}\nabla \cdot (\rho \boldsymbol{v}\overline{\boldsymbol{f}})dV\\ &= \left(\frac{d}{dt}\int_{\Omega(t)} \rho \overline{\boldsymbol{f}}dV - \int_{\partial \Omega(t)}\rho \overline{\boldsymbol{f}}(\boldsymbol{v}_b\cdot \boldsymbol{n})dA\right)+\int_{\Omega(t)}\nabla \cdot (\rho \boldsymbol{v}\overline{\boldsymbol{f}})dV\\ &= \frac{d}{dt}\int_{\Omega(t)} \rho \overline{\boldsymbol{f}}dV + \int_{\partial \Omega(t)}\rho \overline{\boldsymbol{f}}((\boldsymbol{v}-\boldsymbol{v}_b)\cdot \boldsymbol{n})dA \end{split} \end{equation*}\]And the left hand side can easily be related to, for example, the momentum conservation equation
\[\rho\frac{D\boldsymbol{v}}{Dt} = \nabla \cdot \boldsymbol{\sigma} + \rho \boldsymbol{f}\](where $\boldsymbol{f}$ above is a generic body force, not to be confused with the $\boldsymbol{f}$ of Eq. \ref{eq:general-rtt}) to give
\[\begin{equation*} \begin{split} \int_{\Omega(t)} (\nabla \cdot \boldsymbol{\sigma} + \rho \boldsymbol{f})dV&= \frac{d}{dt}\int_{\Omega(t)} \rho {\boldsymbol{v}}dV + \int_{\partial \Omega(t)}\rho {\boldsymbol{v}}((\boldsymbol{v}-\boldsymbol{v}_b)\cdot \boldsymbol{n})dA\\ \implies \frac{d}{dt}\int_{\Omega(t)} \rho {\boldsymbol{v}}dV &= \int_{\Omega(t)} (\nabla \cdot \boldsymbol{\sigma} + \rho \boldsymbol{f})dV - \int_{\partial \Omega(t)}\rho {\boldsymbol{v}}((\boldsymbol{v}-\boldsymbol{v}_b)\cdot \boldsymbol{n})dA \end{split} \end{equation*}\]and the left hand side is clearly the total momentum source in $\Omega(t)$, and we can use the divergence theorem to get the typical surface stress terms. Likewise, if $\overline{\boldsymbol{f}}=1$, then $\frac{D\overline{\boldsymbol{f}}}{Dt}=0$, and the entire lefthand side is zero, giving the following mass conservation law
\[\frac{d}{dt}\int_{\Omega(t)} \rho dV = - \int_{\partial \Omega(t)}\rho((\boldsymbol{v}-\boldsymbol{v}_b)\cdot \boldsymbol{n})dA\]which makes physical sense if we consider a (stationary $\boldsymbol{v}_b=0$) box with a mass jet entering in the left with velocity $\boldsymbol{v}_1$ (with cross sectional area $A$) and exiting with velocity $\boldsymbol{v}_2$ (area $A$). If $\boldsymbol{v}_2>\boldsymbol{v}_1$, then the rate of change of the total mass in the control volume is negative, and vice versa.
Conclusion
The following two statements of the Reynolds Transport Theorem are interchangable (assuming conservation of mass)
\[\frac{d}{dt} \int_{\Omega(t)} \rho\overline{\boldsymbol{f}}dV = \int_{\Omega(t)} \frac{\partial (\rho\overline{\boldsymbol{f}})}{\partial t}dV + \int_{\partial \Omega(t)} (\boldsymbol{v}_b\cdot \boldsymbol{n})\rho\overline{\boldsymbol{f}}dA\] \[\frac{d}{dt} \int_{\Omega(t)} \rho\overline{\boldsymbol{f}}dV = \int_{\Omega(t)} \rho \frac{D\overline{\boldsymbol{f}}}{Dt}dV - \int_{\partial \Omega(t)} ((\boldsymbol{v}-\boldsymbol{v}_b)\cdot \boldsymbol{n})\rho \overline{\boldsymbol{f}}dA\]Note: If you look at the above two equations, and rearrange Eq. \ref{eq:identity} to get
\[\frac{\partial(\rho\overline{\boldsymbol{f}})}{\partial t} = \rho \frac{D\overline{\boldsymbol{f}}}{Dt} - \nabla \cdot (\rho \boldsymbol{v}\overline{\boldsymbol{f}})\]you can immediately derive the second form from the first.
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