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Fluids can be characterized by the fact, that initially neighboring material fluid particles do not remain neighboring during the course of deformation.

The deformation gradient in continuum mechanics is defined via

$$\boldsymbol{F} = \dfrac{\partial \boldsymbol{\varphi}(\boldsymbol{X},t)}{\partial \boldsymbol{X}}$$

whereas $\boldsymbol{\varphi}$ is the function of motion and $\boldsymbol{X}$ is the position of a fluid particle in the reference configuration and $t$ is the time.

The Reynolds Transport Theorem (RTT) explains how to compute the material time derivative of a field quantity integrated over volume $\omega(t)$ via $$ \frac{d}{dt}\int_{\omega(t)} \chi dv = \int_{\omega(t)} \dfrac{d \chi}{d t} + \chi \text{div}(\boldsymbol{v}) dv ~. $$

Now my questions concerns the derivation of that formula:

Typically, we make an integral transformation from our spatial volume $\omega(t)$ to the (time) fixed volume $\Omega$ in the reference configuration. In order to do that we make use of the relation $$ dv = \det( \boldsymbol{F} ) dV = J dV~. $$ How is that allowed? If in fluid mechanics two neighboring particles do not stay neighbors, then the deformation gradient $\boldsymbol{F}$ will be non-continuous (in turbulent flows, this function will have jumps everywhere I suppose). Don't we need a continuous mapping $\boldsymbol{\varphi}(\boldsymbol{X},t)$ in order to apply these differential geometric relations?

P.S.: When we consider solid, I do not have a problem with the RTT since neighboring particles always stay neighbors and $\boldsymbol{\varphi}(\boldsymbol{X},t)$ is a smooth function (although it might be not possible to explicitly state that function).

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  • $\begingroup$ If in fluid mechanics two neighboring particles do not stay neighbors Do you actually mean "particle" here or do you mean a fluid parcel (which would consist of many particles but would be an "element" of a fluid)? In either case, I do not believe I have seen such a requirement before; can you provide a source where it is set to be a requirement? $\endgroup$
    – Kyle Kanos
    Commented Jan 15 at 17:08
  • $\begingroup$ @KyleKanos Thanks for your comment. I actually mean fluid particle in the sense, that one particle sits at one particular point in space. The fluid itself is then composed out of a continuous set of points that we can handle in a differential geometric manner (e.g. compute derivatives and so on). The fact, that two neighboring points do not remain neighbors during flow (or deformation how this process is addressed within a continuum mechanical setting) is not a requirement but a fact. Have a look at this very nice video: youtube.com/watch?v=UqBmdZ-BNig $\endgroup$
    – Schneitz
    Commented Jan 15 at 19:13
  • $\begingroup$ @KyleKanos A quick search for a source, stating that the mapping $\boldsymbol{\varphi}(\boldsymbol(X),t)$ must be $C^1$ can be found here: brown.edu/Departments/Engineering/Courses/En221/Notes/… -> Section 3.2. annotation (iii). In my understanding, the smoothness of the mapping between reference and spatial configuration is a crucial requirement in continuum mechanics. $\endgroup$
    – Schneitz
    Commented Jan 15 at 19:38
  • $\begingroup$ without sufficient regularity assumptions on your mapping $\varphi$, the integral $\int_{\omega(t)}\chi\,dV$ is not a differentiable function of $t$, and if you make $\varphi$ even worse then the map is not even a continuous function of $t$ so the question doesn’t really make sense. You can try to go down the analysis rabbit hole of considering weak-differentiability, or distributional derivatives (which can take jump discontinuities into account by means of a distributional Dirac delta as derivative), but is this really what you want to consider? $\endgroup$
    – peek-a-boo
    Commented Jan 15 at 22:48
  • $\begingroup$ @peek-a-boo Thanks for your comment! Your point is exactly what my concern is about. In fluid mechanics $\boldsymbol{\varphi}$ is continious in $t$ (one can always follow a single fluid particle) but not in $\boldsymbol{X}$ (in turbulent flow neighboring particles to not remain neighboring and hence $\boldsymbol{\varphi}$ must jump w.r.t. $\boldsymbol{X}$) . So how is it still okay to apply Reynolds transport theorem?? $\endgroup$
    – Schneitz
    Commented Jan 16 at 8:19

2 Answers 2

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Here you can find a (not 100% rigorous) proof of the Reynolds' transport theorem

https://basics.altervista.org/test/Math/time_derivatives_of_integrals/time_derivatives_of_integrals.html

Let's evaluate the time derivative of a volume integral, using the definition of derivative

\begin{equation} \dfrac{d}{dt} \int_{V(t)} f(\mathbf{r}, t) dV = \lim_{\Delta t \rightarrow 0} \left[ \frac{1}{\Delta t} \left( \int_{V(t+\Delta t)} f(\mathbf{r}, t+\Delta t) dV - \int_{V(t)} f(\mathbf{r}, t) dV \right) \right] \end{equation}

Now, let's expand the function $f(\mathbf{r},t+\Delta t)$ in series \begin{equation} f(\mathbf{r}, t + \Delta t) = f(\mathbf{r}, t) + \Delta t \, \dfrac{\partial}{\partial t} f(\mathbf{r},t) + o(\Delta t) \ , \end{equation} and the compose the volume $V(t+\Delta t)$ as the (algebraic) sum of the volume $V(t)$ and the difference of the volumes $\Delta V(t,t+\Delta t) = V(t+\Delta t) - V(t)$, so that the first integral reads \begin{equation} \int_{V(t+\Delta t)} = \int_{V(t)} + \int_{\Delta V(t,t+\Delta t)} \ . \end{equation}

Putting these two manipulations together in the first formula \begin{equation} \begin{aligned} \dfrac{d}{dt} \int_{V(t)} f(\mathbf{r}, t) dV & = \lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \int_{V(t)} \left( \Delta t \dfrac{\partial f}{\partial t}(\mathbf{r}, t) + o(\Delta t) \right) dV \\ &+ \lim_{\Delta t \rightarrow 0} \dfrac{1}{\Delta t} \int_{\Delta V(t,t+\Delta t)} \left( f(\mathbf{r}, t) + O(\Delta t) \right) dV \end{aligned}\end{equation}

  1. The first integral on the right-hand side is the volume integral of the time partial derivative \begin{equation} \lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \int_{V(t)} \left( \Delta t \dfrac{\partial f}{\partial t}(\mathbf{r}, t+\Delta t) + o(\Delta t) \right) dV = \int_{V(t)} \dfrac{\partial f}{\partial t} \end{equation}

  2. The second integral is the one that contains the motion of the boundary of the volume. The elementary $\Delta V$ can be computed with the dot product of the elementary displacement and the vector area \begin{equation} "d \Delta V" = \mathbf{v}_b \Delta t \cdot \mathbf{\hat{n}} dS \ , \end{equation} being $\mathbf{v}_b$ the velocity of the points of the boundary of the volume $V$. Evaluating the limit for $\Delta t \rightarrow 0$ using average value theorem, you can get \begin{equation}\begin{aligned} \lim_{\Delta t \rightarrow 0} \dfrac{1}{\Delta t} \int_{\Delta V(t,t+\Delta t)} \left( f(\mathbf{r}, t) + O(\Delta t) \right) dV & = \lim_{\Delta t \rightarrow 0} \dfrac{1}{\Delta t} \left[ \oint_{\partial V} f(\mathbf{r}, t) \Delta t \mathbf{v}_b(\mathbf{r},t) \cdot \mathbf{\hat{n}}(\mathbf{r},t) dS + o(\Delta t) \right] = \\ & = \oint_{\partial V} f(\mathbf{r}, t) \mathbf{v}_b(\mathbf{r}, t) \cdot \mathbf{\hat{n}}(\mathbf{r}, t) dS \ . \end{aligned} \end{equation}

Putting everything together, without explicitly writing the independent variables, we've got Reynolds' transport theorem

\begin{equation} \dfrac{d}{dt} \int_{V(t)} f = \int_{V(t)} \dfrac{\partial f}{\partial t} + \oint_{\partial V(t)} f \mathbf{v}_b \cdot \mathbf{\hat{n}} \ . \end{equation}

EDIT. Some details about integration: independent variables and domains enter image description here

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  • $\begingroup$ After looking over your suggestion, I am not quite sure about your step 2. I can relate to the statement, that $\Delta V = \int_{\partial V_t}\Delta t \boldsymbol{v}\cdot \boldsymbol{v} dS $ since the $\Delta $ infront of the $V$ correlates to the $\Delta$ infront of the $t$. But it is not clear to me why is it possible to write $\int_{\Delta V} \dots dV = \int_{\partial V_t} \Delta t \boldsymbol{v} \cdot \boldsymbol{n} dS$. I would rather write $\int_{\Delta V} \dots dV = \int_{ \boldsymbol{v} \cdot \boldsymbol{n} \partial V_t} \dots dV$. But then I am still stuck with the $dV$... $\endgroup$
    – Schneitz
    Commented Jan 16 at 15:24
  • $\begingroup$ I've just edited the answer with some details about integration $\endgroup$
    – basics
    Commented Jan 16 at 15:50
  • $\begingroup$ If you change the parametrization, aka the independent variables, you need to change the domain of integration to be consistent with the change of variables (since independence is one of the most important principle in physics and math: the results should not depend on an arbitrary choice of variables and parameters) $\endgroup$
    – basics
    Commented Jan 16 at 15:54
  • $\begingroup$ awesome. now it is clear. Many thanks again. I wonder why most text books use the transformation back to $V(0)$ and do not provide this more clean derivation. $\endgroup$
    – Schneitz
    Commented Jan 16 at 15:56
  • $\begingroup$ My answer is because many books and professors suck. Maybe it's a bit rude, but I haven't found better answers yet. I'm telling you that with an aerospace engineering background with focus on fluid, structure and control dynamics. I lost too much time to make things clear when I was studying, and this is the reason why I started collecting hand-written notes on the website I shared with you $\endgroup$
    – basics
    Commented Jan 16 at 15:59
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Yes you are correct, the application of the transport formula requires sufficient regularity of the field. The assumption of regularity may be hard to justify physically especially for turbulent flow.

Consider for example Onsager's conjecture. Starting from Euler's equation (Navier-Stokes with no friction), you would expect conservation of energy: $$ E = \frac12\int u^2d^3x $$ which you can derive from the transport equation assuming regularity. However, for weak solutions this reasoning may fail. Actually, it was proven (see for example Energy dissipation without viscosity in ideal hydrodynamics I. Fourier analysis and local energy transfer by Eyink) that Hölder 1/3-continuity suffices for conservation of energy, so the usual argument involving smoothness can be strengthened to a certain limit. Conversely, you can construct weak solutions that are $\mathcal C^\alpha$ with $\alpha<1/3$ that do not conserve energy (in particular wild solutions which initially start from rest and come back to rest after a finite time). While it is still an open question, the dissipation anomaly could be a result of the weak convergence of Navier-Stokes equation to such an irregular, dissipative solution of Euler.

Similarly, Helmholtz' theorem which states the conservation of circulation, can similarly be proven using the transport theorem. It too can be extended to non-smooth fields, up to $\mathcal C^{1/2}$, but for more irregular fields you can find counter-examples of non conservation.

Hope this helps.

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  • $\begingroup$ Thanks for your answer and hinting me to Onsager's conjecture which I was not aware of. However, I am not talking about the regularity of the field, which I called $\chi$, but the regularity of the mapping $\boldsymbol{\varphi}$. Obviously this regularity is violated already in frictionless flow link. So I think my question is very fundamental, why is it allowed to use RTT for non-smooth deformation mapps $\varphi$? $\endgroup$
    – Schneitz
    Commented Jan 16 at 14:21
  • $\begingroup$ For Onsager's conjecture, it is the regularity of the velocity field that is relevant, $u$, which in your Lagrangian setting corresponds to $\partial_t\phi\circ \phi^{-1}$. Your $\chi$ rather corresponds to the specific energy $u^2$. This is why it is concrete example of the transport theorem not working due to the (spatial) irregularity of $u$, i.e. your $\phi$. To make it work outside the smooth setting, you'll need more sophisticated mathematical tools like Payley-Littlewood etc which is detailed in the linked article. $\endgroup$
    – LPZ
    Commented Jan 16 at 14:30

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