# Reynolds transport theorem: link with the Lie derivative?

In this Wikipedia article (see "Higher dimensions") there seems to be a connection between the Reynolds transport theorem (here) and the Lie derivative:

$$\frac{d}{dt}\int_{\Omega(t)}\omega=\int_{\Omega(t)} i_{\vec{\textbf v}}(d\omega)+\int_{\partial \Omega(t)} i_{\vec{\textbf v}} \omega+\int_{\Omega(t)}\dot{\omega} \qquad(1)$$

were $\omega$ is a p-form and the domain is time-varying. The symbol $i_X$ denotes the contraction with the vector field $X$.

By using the Cartan magic formula $L_{X} \omega = d (i_X \omega) + i_X (d \omega)$, it seems that

$$\frac{d}{dt}\int_{\Omega(t)}\omega=\int_{\Omega(t)} i_{\vec{\textbf v}}(d\omega)+\int_{\Omega(t)} d(i_{\vec{\textbf v}} \omega)+\int_{\Omega(t)}\dot{\omega} \, ,$$

namely

$$\frac{d}{dt}\int_{\Omega(t)}\omega = \int_{\Omega(t)} L_{\vec{\textbf v}}\omega+\int_{\Omega(t)}\dot{\omega} \, .$$

It is clear that the Reynolds transport theorem is equivalent to the conservation of mass in hydrodynamics:

$$\frac{d}{dt}\int_{\Omega(t)}\rho = \int_{\Omega(t)} div(\rho \vec{\bf{v}}) +\int_{\Omega(t)}\dot{\rho} \, .$$

What is the relation between the two formulations? Clearly $\omega =\rho$ is not the correct thing to do: we can only integrate 3-forms and $\rho$ is a 0-form (the domain $\Omega(t)$ is three dimensional). How to derive the mass conservation from equation (1)?

There is indeed a connection. Recall the definition of the Lie derivative for a $$k$$-form $$\omega$$ along a vector field $$X$$. If $$\chi$$ is the one-parameter family of diffeomorphisms defined (at least locally) by $$X$$, then $$L_X\omega = \left.\frac{d}{d\tau}\right|_0(\chi_{\tau}^{*}\omega)$$ Integrating over some $$k$$-submanifold $$\Omega$$, we get $$\int_\Omega L_X\omega = \int_{\Omega}\left.\frac{d}{d\tau}\right|_0(\chi_{\tau}^{*}\omega) = \left.\frac{d}{d\tau}\right|_0\int_{\Omega}\chi_{\tau}^{*}\omega = \left.\frac{d}{d\tau}\right|_0\int_{\chi_\tau[\Omega]}\omega$$ Note that we first used the Leibniz integral rule in one variable, and then the fact that diffeomorphisms preserve the integral. I claim Reynold's transport theorem is a special case of the formula $$\left.\frac{d}{d\tau}\right|_0\int_{\chi_\tau[\Omega]}\omega = \int_\Omega L_X\omega$$ we just obtained.

Say your space is an $$n$$-dimensional manifold $$M$$, and consider an interval $$I\subseteq \mathbb R$$ (say with $$0\in I$$) which for us will represent time. In the context of Reynold's transport theorem you have a submanifold $$\Omega$$ of $$M$$ (say of dimension $$k$$) which "varies (smoothly) over time". We can interpret this as having a fixed manifold $$\Omega$$ and a family of embeddings $$i_t:\Omega\to M$$. Using these embeddings, we can trace the "worldsheet" of $$\Omega$$ inside the "spacetime" $$I\times M$$. Define another family of embeddings $$j_t:\Omega\to I\times M$$ given by $$j_t(x)=(t,i_t(x)),$$ and define $$\Omega_t:=j_t[\Omega]\subseteq I\times M$$. Now we want to get the "spacetime velocity" of $$\Omega$$, which will be a vector field defined on $$\Theta:=\bigcup_{t\in I}\Omega_t$$. To do this, we observe that the embeddings $$j_t$$ are smooth, so they define a family of smooth curves $$\gamma_x:I\to I\times M$$ given for every $$x\in\Omega$$ by $$\gamma_x(t)=(t,i_t(x))$$. The vectors tangent to this family of curves give us a vector field $$Y$$ defined by $$Y_{(t,i_t(x))}=\gamma'_x(t)$$ for every $$(t,i_t(x))\in\Theta=\bigcup_{t\in I}\Omega_t$$. This is the spacetime velocity we wanted. Using the extension lemma for vector fields on submanifolds, we get a vector field $$X$$ on $$I\times M$$ which agrees with $$M$$ on $$\Theta$$ (we actually need to show that $$\Theta$$ is a submanifold, but I think this follows from the smoothness of all the maps).

Similarly, if you had a $$k$$-form $$\bar{\omega}$$ on $$M$$ which depends on time $$t\in I$$, then it can be regarded as a $$k$$-form $$\omega$$ on $$M\times I$$ by defining $$\omega_{(t,m)}(v_{(t,m)}) := \bar{\omega(t)_m}(\text{proj}_{TM}(v_{(t,m)}))$$ for every tangent vector $$v_{(t,m)}\in T_{(t,m)}(I\times M)$$.

Now applying the formula we first derived to the submanifold $$\Omega_0$$, we have $$\left.\frac{d}{dt}\right|_0\int_{\chi_t[\Omega_0]}\omega = \int_{\Omega_0} L_X\omega$$ where $$\chi$$ is the one-parameter family of diffeomorphisms induced by $$X$$ (we might need to show that $$\chi$$ is actually defined for all $$t\in I$$, but I think this is true at least in a neighbourhood of $$\Omega_0$$). Since $$X$$ extends $$Y$$, I think one can show that $$\chi_t[\Omega_0]=\Omega_t$$. Also, by Cartan's magic formula, we have $$L_X\omega = d(\iota_X\omega) + \iota_X(d\omega)$$. Hence

$$\left.\frac{d}{dt}\right|_0\int_{\Omega_t}\omega = \int_{\Omega_0} d(\iota_X\omega) + \int_{\Omega_0}\iota_X(d\omega)$$

When integrating the right hand side we can take charts consisting of a chart on $$I$$ times a chart on $$M$$, so $$\omega$$ has no $$dt$$ component because we defined it using $$\bar\omega$$. This way, letting $$V=\text{proj}_{TM}Y$$ we have $$\iota_Y\omega=\iota_V\omega$$ and even $$\iota_X\omega=\iota_Y\omega=\iota_V\omega$$ on $$\Omega_0$$. On the other hand, since $$\Omega_0$$ is contained in $$\{0\}\times M\subseteq I\times M$$ we have $$dt|_{\Omega_0}=0$$. These observations give the first two equalities in $$\int_{\Omega_0} d(\iota_X\omega)=\int_{\Omega_0} d(\iota_V\omega)=\int_{\Omega_0} d_M(\iota_V\omega)=\int_{\partial\Omega_0}\iota_V\omega$$ and the third one comes from Stokes' theorem.

Similarly, over $$\Omega_0$$ we have $$X=Y=\frac{\partial}{\partial t}+V$$, and since $$\omega=\omega_Idx^I$$ (where $$I$$ is a multiindex where $$t$$ does not appear) we have \begin{aligned} d\omega &=\partial_t\omega_Idt\wedge dx^I+\sum_i \partial_i\omega_Idx^i\wedge dx^I \\ &=\partial_t\omega_Idt\wedge dx^I+d_M\omega \end{aligned} and the second integral on the right hand side becomes \begin{aligned} \int_{\Omega_0}\iota_X(d\omega) &= \int_{\Omega_0}\iota_{\frac{\partial}{\partial t}+V}(\partial_t\omega_Idt\wedge dx^I+d_M\omega) \\ &= \int_{\Omega_0}\left[\iota_{\frac{\partial}{\partial t}}(\partial_t\omega_Idt\wedge dx^I)+\iota_{\frac{\partial}{\partial t}}(d_M\omega)+\iota_V(\partial_t\omega_Idt\wedge dx^I)+\iota_V(d_M\omega)\right] \\ &= \int_{\Omega_0}\left[\partial_t\omega_Idx^I+\iota_V(d_M\omega)\right] \\ &= \int_{\Omega_0}\dot\omega+\int_{\Omega_0}\iota_V(d_M\omega) \end{aligned} (note the second and third terms vanish because they pair "spacelike" basis forms with "timelike" basis vectors and viceversa).

Putting al this together, we have $$\left.\frac{d}{dt}\right|_0\int_{\Omega_t}\omega = \int_{\partial\Omega_0}\iota_V\omega+\int_{\Omega_0}\dot\omega+\int_{\Omega_0}\iota_V(d_M\omega)$$ as desired.

Instead of $\rho$, think about the three form $\rho ~\mathrm{d}x$. (Like I tell my calculus students, you should always write the integration element in an integral.) Here $\mathrm{d}x$ is the standard volume form.

By definition of the divergence of a vector field you have that $$\mathrm{div}(\rho \vec{v}) ~\mathrm{d}x = \mathrm{d}( \iota_{\rho\vec{v}} ~\mathrm{d}x ) = \mathrm{d} (\iota_{\vec{v}} \rho ~\mathrm{d}x ) = L_{\vec{v}} \left(\rho~\mathrm{d}x\right)$$ which is what you expect.