Trouble with the derivation of the Reynolds Transport Theorem

I have been trying to reconcile two different forms of the Reynolds Transport Theorem (RTT) that I have seen in textbooks.

The first form comes from a finite volume method cfd textbook. It directly relates the rate of change of a property within a material volume (MV) to some integrals of an associated control volume (CV). I did not understand how they arrived at this form and was trying to derive it myself from the more well know form of RTT (described in paragraph below).

The second RTT form I think is much more common, particularly from a math perspective. $$ \label(1)~~~~~~~~~~~~\frac{d}{dt} \int_{V(t)} F ~dV = \int_{V(t)} \frac{\partial F}{\partial t} ~dV ~+ \int_{A(t)} F\mathbf b \cdot \mathbf n ~dV $$

where V is an arbitrary volume bounded by surface A. F is the (scalar in this case) field of interest, b is the velocity of the surface A and n is the normal vector for the surface A. All of these quantities are in general variable, both spatially and temporally.

My understanding is that this second form of the RTT (eqn 2) can be applied to both a material volume (MV) as well as any arbitrary control volume (CV) (this is done in some derivations of the continuity equation I have been looking at.

So now we have both

$$ 2)~~~~~~~~~~~~ \frac{d}{dt} \int_{MV(t)} F ~dV = \int_{MV(t)} \frac{\partial F}{\partial t} ~dV ~+ \int_{A_{MV}(t)} F\mathbf u \cdot \mathbf n ~dV $$

(In the case of the material volume analysis the velocity of the surface A (i.e. b) is equal to the fluid velocity u)

And

$$ 3)~~~~~~~~~~~~\frac{d}{dt} \int_{CV(t)} F ~dV = \int_{CV(t)} \frac{\partial F}{\partial t} ~dV ~+ \int_{A_{CV}(t)} F\mathbf b \cdot \mathbf n ~dV $$

Now borrowing again from the derivation of the continuity equation and assuming that the two volumes CV and MV are instantaneously coincident we can draw some conclusions. Obviously the first terms of eqns 2 and 3 are not, in general, equal even with CV and MV coincident.

However if MV is the same as CV even temporarily then the second terms from each equation should be equal

$$ 4)~~~~~~~~~\int_{MV(t)} \frac{\partial F}{\partial t} ~dV = \int_{CV(t)} \frac{\partial F}{\partial t} ~dV $$

Additionally the third term of equation 2 can be rewritten in terms of an integral over CV rather than MV.

$$ 5)~~~~~~~ \int_{A_{MV}(t)} F\mathbf u \cdot \mathbf n ~dV = \int_{A_{CV}(t)} F\mathbf u \cdot \mathbf n ~dV$$

In an effort to derive the first form of the RTT described earlier I substituted eqns. 4 and 5 into 2 to arrive at

$$ 6)~~~~~~~~~~~~ \frac{d}{dt} \int_{MV(t)} F ~dV = \int_{CV(t)} \frac{\partial F}{\partial t} ~dV ~+ \int_{A_{CV}(t)} F\mathbf u \cdot \mathbf n ~dV $$

So now we have the rate of change of the field inside the material volume described by integrals over the control volume but I feel like it is absolute non-sense for several reasons. The most evident reason being that there is now no reliance at all on b which was never assumed to be zero and should therefore play a role I think in this final equation.

So my question FINALLY is what did I do wrong here? Equating the integrals (MV and CV) when the volumes coincide is right out of the textbook derivation of the continuity equation so I feel okay about that.

The apparent paradox is resolved once you realize that $\mathbb{u}$ and $\mathbb{b}$ are not necessarily the same vector fields when $CV(t)$ is an arbitrary moving control volume and $MV(t)$ is a material control volume -- even if they coincide at some instant of time.

Consider how the surface velocity arises in the derivation of the Reynolds Transport Theorem. An arbitrary moving control volume $V(t) \subset \mathbb{R}^3$ is the image of a fixed region under some family of smooth maps

$$\mathbb{\xi}(\cdot,t) : \mathbb{R}^3 \to \mathbb{R}^3$$

where, given some fixed initial region $V(0)$, we have

$$\mathbb{\xi}(\cdot,t) : V(0) \mapsto V(t).$$

If $F$ is an integrable scalar field, then with a change of integration variable $\mathbb{x} = \mathbb{\xi}(\mathbb{x}_0,t),$ we have

$$\int_{V(t)} F(\mathbb{x},t) \, d\mathbb{x} = \int_{V(0)} F(\mathbb{\xi}(\mathbb{x}_0,t),t) \, \left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right| \, d\mathbb{x}_0, $$

where the Jacobian determinant $\left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right|$ is positive for sufficiently small $t$ if the map $\mathbb{\xi}$ is continuous. Taking the derivative with respect to $t$, we can interchange the derivative and integral on the RHS over the fixed region $V(0),$ to obtain

$$\begin{align}\frac{d}{dt}\int_{V(t)} F(\mathbb{x},t) \, d\mathbb{x} &= \int_{V(0)} \frac{\partial}{\partial t} \left\{F(\mathbb{\xi}(\mathbb{x}_0,t),t) \, \left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right| \right\} \, d\mathbb{x}_0 \\ &= \int_{V(0)} \left\{ \frac{\partial}{\partial t} F(\mathbb{\xi}(\mathbb{x}_0,t),t) + \nabla F(\mathbb{\xi}(\mathbb{x}_0,t),t) \cdot \mathbb{\xi}_t(\mathbb{x}_0,t)\right\} \left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right| \, d\mathbb{x}_0 + \int_{V(0)}F(\mathbb{\xi}(\mathbb{x}_0,t),t) \frac{\partial}{\partial t}\left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right| \, d \mathbb{x}_0. \end{align}$$

Using some matrix calculus we can obtain

$$\frac{\partial}{\partial t}\left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right| = \nabla \cdot \mathbb{\xi}_t(\mathbb{x}_0,t)\left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right|. $$

Hence,

$$\begin{align}\frac{d}{dt}\int_{V(t)} F(\mathbb{x},t) \, d\mathbb{x} &= \int_{V(0)} \left\{ \frac{\partial F}{\partial t} + \nabla F \cdot \mathbb{\xi}_t + F \nabla \cdot \mathbb{\xi}_t \right \}\left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right|\, d\mathbb{x}_0 \\ &= \int_{V(0)} \left\{ \frac{\partial F}{\partial t} + \nabla \cdot \left(F \mathbb{\xi}_t \right) \right \}\left| \frac{\partial \mathbb{x}}{\partial \mathbb{x}_0}\right|\, d\mathbb{x}_0\end{align}.$$

Changing the variable of integration back to $\mathbb{x}$ we obtain

$$\frac{d}{dt}\int_{V(t)} F(\mathbb{x},t) \, d\mathbb{x} = \int_{V(t)} \left\{ \frac{\partial F}{\partial t} + \nabla \cdot \left(F \mathbb{b} \right) \right \}\, d\mathbb{x},$$

where

$$\mathbb{b}(\mathbb{x},t) = \mathbb{\xi}_t(\mathbb{\xi}^{-1}(\mathbb{x},t),t)$$

is the "velocity field" corresponding to the map defining the moving control volume.

An application of the divergence theorem yields

$$\frac{d}{dt}\int_{V(t)} F(\mathbb{x},t) \, d\mathbb{x} = \int_{V(t)} \frac{\partial F}{\partial t} \, d\mathbb{x} + \int_{A(t)} F \mathbb{b} \cdot \mathbb{n} \, dS.$$

A material control volume is one which always contains the same fluid particles and the field $\mathbb{b}$ coincides with the Eulerian fluid velocity field $\mathbb{u}.$