Whilst playing around, I came up with the following proof of the Reynolds transport theorem. It’s by no means fully rigorous, but it is (at least, I hope) pretty intuitive.
(To-do: draw and upload a picture of the construction.)
Theorem statement
The Reynolds transport theorem is a theorem about how to take time derivatives inside moving volumes. Its form in terms of \(\int_\Omega \chi\, \rho\, {\mathrm d}{V}\) rather than \(\int_\Omega f\, \mathrm d V\) makes it much easier to use in fluid mechanics than its usual, Leibnizian, counterpart.
The statement is this: given an arbitrary function and an arbitrary volume, then the time derivative of an integral over that volume is given as follows, \[\frac{\mathrm d}{\mathrm d t} \, \int\limits_\Omega \chi\, \rho \, {\mathrm d}{V} = \int\limits_\Omega \frac{\mathrm D \chi}{\mathrm D t} \,\rho\, {\mathrm d}{V} + \oint\limits_{\partial \Omega} \chi \,\rho\left( {\vec{v}_{\text{b}}}-\vec{u} \right) \cdot{\mathrm d}{{\vec \Sigma}} \, ,\] where
- \(\chi(\vec x, t)\) is an arbitrary function into \(\mathbb{R}^n\) of space \(\vec x\) and time \(t\),
- \(\Omega = \Omega(t)\) is an arbitrary volume with boundary \(\partial \Omega\), immersed in a fluid,
- \(\rho\) is the density of that fluid, \(\vec u\) its velocity, and
- \(\vec{v}_{\text b}\) is the velocity of an infinitesimal element of \(\partial \Omega\).
Proof
First we start by proving the Leibniz integral theorem for generalised volumes. Consider a time-dependent volume \(\Omega(t)\) moving through a 3D fluid field with velocity \(\vec{u}\). Also, let \(\chi\) be some sufficiently smooth function \(\mathbb{R}^3 \times \mathbb{R} \to\mathbb{R}^n\) for any \(n\in\mathbb{N}\). Then, writing down the definition of a derivative and Taylor expanding inside the limit, \[\begin{align*} \frac{\mathrm d}{\mathrm d t} \!\int\limits_{\Omega(t)} \! \chi({\vec x},t) \, {\mathrm d}{V} &= \lim_{\delta t \to 0} \frac{1}{\delta t} \left\lbrace\, \int\limits_{~\Omega(t+\delta t)} \!\!\! \chi({\vec x},t+\delta t) \, {\mathrm d}{V} - \int\limits_{\Omega(t)} \chi({\vec x},t) \, {\mathrm d}{V} \right\rbrace \\ &= \lim_{\delta t \to 0} \frac{1}{\delta t} \left\lbrace\, \int\limits_{~\Omega(t + \delta t)} \!\!\! \chi({\vec x},t) + \frac{\partial \chi}{\partial t}({\vec x},t)\delta t + \mathrm{o}\left(\delta t\right) \, {\mathrm d}{V} - \int\limits_{\Omega(t)} \chi({\vec x},t) \, {\mathrm d}{V} \right\rbrace. \end{align*}\] For economy of writing, let’s ignore terms in \(\mathrm{o}\left(\delta t\right)\) in the following lines, as these will vanish anyway.
Now we want to expand the \(\Omega(t + \delta t)\) in the limits of integration. We achieve this by defining a vector \(\delta {\vec b}({\vec x},t)\) between every point \({\vec x}\) on \(\partial{\Omega}(t)\) and the corresponding point on \(\partial{\Omega}(t+\delta t)\) — so that the expression \[\oint_{\partial \Omega(t)} \delta {\vec b} \cdot {\mathrm d}{{\vec \Sigma}} = \oint_{\partial \Omega(t)} \det\left[\delta {\vec b} \middle| {\mathrm d}{{\vec \Sigma}_1} \middle| {\mathrm d}{{\vec \Sigma}_2}\right], \qquad \text{where}\ \ {\mathrm d}{{\vec \Sigma}_1} \times {\mathrm d}{{\vec \Sigma}_2} = {\mathrm d}{{\vec \Sigma}},\] is an approximation for the skin volume between two snapshots of \(\Omega\) — which then gives \[\begin{multline*} \frac{\mathrm d}{\mathrm d t} \!\int\limits_{\Omega(t)} \! \chi({\vec x},t) \, {\mathrm d}{V} \overset{\dots}{=} \lim_{\delta t \to 0} \frac{1}{\delta t} \left\lbrace\int_{\Omega(t)} \chi({\vec x},t) + \frac{\partial \chi}{\partial t}({\vec x},t)\, \delta t \, {\mathrm d}{V} - \int_{\Omega(t)} \chi({\vec x},t) \, {\mathrm d}{V} \right. \\ + \left. \oint_{\partial \Omega(t)} \left( \chi({\vec x},t) + \frac{\partial \chi}{\partial t}({\vec x},t) \, \delta t \right) \, \delta {\vec b} \cdot {\mathrm d}{{\vec \Sigma}} \right\rbrace. \end{multline*}\] But we can Taylor expand \(\delta {\vec b} = {\vec{v}_{\text{b}}}\, \delta t + \mathrm{o}\left(\delta t\right)\), where \({\vec{v}_{\text{b}}}\) is the velocity of a point on the boundary of \(\Omega\). Thus, simplifying: \[\begin{align*} \hphantom{\frac{\mathrm d}{\mathrm d t} \!\int\limits_{\Omega(t)} \! \chi({\vec x},t) \, {\mathrm d}{V}} &= \lim_{\delta t \to 0} \frac{1}{\delta t} \left\lbrace\int_{\Omega(t)} \frac{\partial \chi}{\partial t}({\vec x},t) \,\delta t \, {\mathrm d}{V} + \oint_{\partial \Omega(t)} \chi({\vec x},t) \, \delta t \, {\vec{v}_{\text{b}}} \cdot {\mathrm d}{{\vec \Sigma}} + \mathrm{o}\left(\delta t\right) \, {\mathrm d}{V} \right\rbrace \\ &= \int_{\Omega(t)} \frac{\partial \chi}{\partial t} \, {\mathrm d}{V} + \oint_{\partial \Omega(t)} \chi\, {\vec{v}_{\text{b}}} \cdot {\mathrm d}{{\vec \Sigma}}. \end{align*}\] This gives Leibniz’ theorem, \[\begin{equation*} \frac{\mathrm d}{\mathrm d t} \!\int\limits_{\Omega(t)} \! \chi \, {\mathrm d}{V} = \int\limits_{\Omega(t)} \frac{\partial \chi}{\partial t} \, {\mathrm d}{V} + \oint\limits_{\partial \Omega(t)} \chi \, {\vec{v}_{\text{b}}} \cdot {\mathrm d}{{\vec \Sigma}}. \end{equation*}\] Getting the Reynolds transform theorem is now a matter of applying conservation of mass \(\left.{\partial_t\,}\rho = - \nabla\cdot(\rho\vec{u}) \right.\) to the following integral, giving \[\begin{align*} \frac{\mathrm d}{\mathrm d t} \!\int\limits_{\Omega(t)} \! \rho \chi \, {\mathrm d}{V} &= \int\limits_{\Omega(t)} \frac{\partial(\rho \chi)}{\partial t} \, {\mathrm d}{V} + \oint\limits_{\partial \Omega(t)} \rho \chi {\vec{v}_{\text{b}}} \cdot {\mathrm d}{{\vec \Sigma}} \\ &= \int\limits_{\Omega(t)} \rho \frac{\partial \chi}{\partial t} - \chi \nabla\cdot(\rho\vec{u}) \, {\mathrm d}{V} + \oint \limits_{\partial \Omega(t)} \rho \chi {\vec{v}_{\text{b}}} \cdot {\mathrm d}{{\vec \Sigma}} \\ &= \int\limits_{\Omega(t)} \rho \frac{\partial \chi}{\partial t} + \rho\vec{u}\cdot\nabla \chi - \nabla\cdot(\rho \chi\vec{u}) \, {\mathrm d}{V} + \oint \limits_{\partial \Omega(t)} \rho \chi {\vec{v}_{\text{b}}} \cdot {\mathrm d}{{\vec \Sigma}} \\ &= \int\limits_{\Omega(t)} \rho \,\left( {\partial_t} + \vec{u}\cdot\nabla \right) \chi \, {\mathrm d}{V} - \oint \rho \chi\vec{u} \cdot {\mathrm d}{{\vec \Sigma}} + \oint\limits_{\partial \Omega(t)} \rho \chi {\vec{v}_{\text{b}}} \cdot {\mathrm d}{{\vec \Sigma}} \\ &= \int_\Omega \rho\, \frac{\mathrm D \chi}{\mathrm D t} \, {\mathrm d}{V} - \oint_{\partial \Omega} \rho \, \chi \, \left( \vec{u}-{\vec{v}_{\text{b}}} \right) \cdot{\mathrm d}{{\vec \Sigma}} \end{align*}\] as required. \(\square\)
Corollary: Rate of work
Given any fluid volume \(\Omega\), the rate of work done on the fluid is the change in its internal energy. Using the Reynolds transport theorem, inserting Navier-Stokes \(\left( \rho \frac{\mathrm D \vec{u}}{\mathrm D t} = \nabla\cdot\mathbb{s} \right)\), integrating by parts then collecting boundary terms: \[\begin{align*} \dot W &= \frac{\mathrm d}{\mathrm d t} \, \int_\Omega \frac12 \rho \left\lVert \vec{u} \right\rVert^2 \, {\mathrm d}{V} \\ &= \int_\Omega \rho \frac{\mathrm D \vec{u}}{\mathrm D t} \cdot \vec{u} \, {\mathrm d}{V} + \oint_{\partial \Omega} \frac12 \rho \left\lVert \vec{u} \right\rVert^2 \, \left( {\vec{v}_{\text{b}}}-\vec{u} \right) \cdot {\mathrm d}{{\vec \Sigma}} \\ &= \int_\Omega (\nabla\cdot\mathbb{s}) \cdot\vec{u} \, {\mathrm d}{V} + \oint_{\partial \Omega} \frac12 \rho \left\lVert \vec{u} \right\rVert^2 \, \left( {\vec{v}_{\text{b}}}-\vec{u} \right) \cdot {\mathrm d}{{\vec \Sigma}} \\ &= -\int_\Omega \nabla\vec{u} {\, {\cdot}{\cdot}\, } \mathbb{s} \, {\mathrm d}{V} + \oint_{\partial \Omega} \vec{u} \cdot \mathbb{s} \cdot {\mathrm d}{{\vec \Sigma}} + \oint_{\partial \Omega} \frac12 \rho \left\lVert \vec{u} \right\rVert^2 \, \left( {\vec{v}_{\text{b}}}-\vec{u} \right) \cdot {\mathrm d}{{\vec \Sigma}} \\ &= -\int_\Omega \nabla\vec{u} {\, {\cdot}{\cdot}\, } \mathbb{s} \, {\mathrm d}{V} + \oint_{\partial \Omega} \left( \vec{u} \cdot \vec{f} + j_{\text{kin}} \right) {\mathrm d}{\Sigma} \end{align*}\] where \(\mathfrak n\) is the (outward-pointing) normal, \(\mathbb s\) is the stress tensor (usually denoted \(\sigma\)), \(\vec{f} := \mathbb{s} \cdot {\mathfrak n}\) is the force applied on (?) a fluid boundary element, \(j_{\text{kin}} := \frac12 \rho \left\lVert \vec{u} \right\rVert^2 \, \left( {\vec{v}_{\text{b}}}-\vec{u} \right) \cdot {\mathfrak n}\) is the kinetic energy flux due to the displacement of the boundary, and \(\cdot\cdot\) denotes double tensor contraction, where \((\mathbb a {\,{\cdot}{\cdot}\,} \mathbb b)_{ik} := \mathbb{a}_{i\kern{-.1em}j} \, \mathbb{b}_{jk}\).
The body term can also be re-written with the strain rate \(\dot{ℽ} := \nabla \vec{u} + (\nabla \vec{u})^T\) as \(-\frac12 \int_{\Omega} \mathbb{s} {\,{\cdot}{\cdot}\,} \dot{ℽ} \, {\mathrm d}{V}\), which shows explicitly the notion of power as a product between stress and strain rate — an infinitesimal, continuum-mechanics version of \(\text{power} = \text{force} \cdot \text{velocity}\).
To-do
Add body force to corollary, by changing Navier-Stokes to \(\rho \frac{\mathrm D \vec{u}}{\mathrm D t} = \nabla\cdot\mathbb{s} + \rho \vec{f}\)
Write a second corollary about thermal energy balance, something like \[\dot{W}_\text{thermal} = \frac{\mathrm d}{\mathrm d t} \, \int_\Omega \rho e \, {\mathrm d}{V} \] where \[\rho \frac{\mathrm D e}{\mathrm D t} = \mathbb{s} {\,{\cdot}{\cdot}\,} \dot{ℽ} - \nabla \cdot \vec{q}\] — but for this to be satisfying want to find a good derivation that this is the right form to use.