# 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)?

Edit: I wrote this answer a long time ago. Coming back, I realised there were a lot of things that could have been much clearer. I hope it's better now.

There is a connection, indeed. 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 $$\left.\frac{d}{d\tau}\right|_0\int_{\chi_\tau[\Omega]}\omega = \int_\Omega L_X\omega \tag{1}$$ 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". To formalize this, take a $$k$$-dimensional manifold $$\Omega$$ and an embedding $$i:I\times\Omega\to I\times M$$ where the first component is the identity. Then the submanifold $$\Omega_t=i[\{t\}\times \Omega]\subseteq \{t\}\times M$$ is the "position of $$\Omega$$ inside $$M$$ at time $$t$$".

Now we want to get the "spacetime velocity" of $$\Omega$$, which will be a vector field defined on the submanifold $$i[I\times \Omega]$$ (the "worldsheet" of $$\Omega$$). To do this, observe that for every $$x\in\Omega$$ we get a smooth curve $$\gamma_x:I\to I\times M$$ given by $$\gamma_x(t)=i(t,x)$$. The vectors tangent to this family of curves give us a vector field $$X$$ defined by $$\tilde X_{i(t,x)}=\gamma'_x(t)$$ 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 $$\tilde X$$ on $$i[I\times \Omega]$$. Its spacelike part $$V=\text{proj}_{TM}X$$ is the vector field that appears in Reynold's transport theorem, and we have $$X=\frac{\partial}{\partial t}+V$$.

Now a time-dependent $$k$$-form $$\omega$$ on $$M$$ can be regarded as a $$k$$-form on $$I\times M$$ such that $$\iota_{v}\omega=0$$ for any "purely timelike" vector $$v\in T(I\times\{x\})$$. Note that we may (and will) always choose charts where the projection $$t:I\times M\to I$$ is a coordinate, so that $$\omega$$ has no $$dt$$ component.

Now we have all the pieces. Let $$\chi$$ be the one-parameter family of diffeomorphisms induced by $$X$$ (we only need $$\chi$$ to be defined in a neighbourhood of $$\Omega_0$$). Applying formula (1) to the submanifold $$\Omega_0$$, we have

$$\left.\frac{d}{dt}\right|_0\int_{\chi_t[\Omega_0]}\omega = \int_{\Omega_0} L_X\omega .$$

Since $$X$$ at time $$t$$ is the velocity of $$\Omega_t$$, we have $$\chi_t[\Omega_0]=\Omega_t$$. Also, by Cartan's magic formula, we have $$L_X\omega = d(\iota_X\omega) + \iota_X(d\omega)$$. Hence

$$\left.\frac{d}{dt}\right|_0\int_{\Omega_t}\omega = \int_{\Omega_0} d(\iota_X\omega) + \int_{\Omega_0}\iota_X(d\omega) \tag{2}$$

Now lets treat the two integrals on the right hand side of (2) separately. For the first integral, since $$V$$ is the spatial part of $$X$$, we have $$X=\frac{\partial}{\partial t}+V$$ and hence $$\iota_X\omega=\iota_V\omega$$. On the other hand, since the tangent space of $$\Omega_0$$ is contained in $$\{0\}\times M$$, it is annihilated by $$dt$$. These two observations give the first two equalities in $$\int_{\Omega_0} d(\iota_X\omega)=\int_{\Omega_0} d(\iota_V\omega)=\int_{\Omega_0} d_M(\iota_V\omega)=\int_{\partial\Omega_0}\iota_V\omega \tag{3}$$ and the third one comes from Stokes' theorem.

Now let's look at the second integral on the right hand side of (2). 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 since $$X=\frac{\partial}{\partial t}+V$$, 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}\partial_t\omega_Idx^I+\iota_V(d_M\omega) \\ &= \int_{\Omega_0}\dot\omega+\int_{\Omega_0}\iota_V(d_M\omega) \end{aligned} \tag{4} The the crossed terms of the interior product vanish because they match $$dt$$ with $$V$$ and purely spatial basis forms with $$\frac{\partial}{\partial t}$$.

At last, plugging (3) and (4) into (2), 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.

• Misto sheaf, what books did you use to study these topics? Jun 30, 2021 at 9:34
• In case you didn't see the message, ping again @Jackozee Hakkiuz Jul 3, 2021 at 20:54
• @Buraian ooh, sorry. Do you mean integration of differential forms? I went superficially through many books, but I'd say the ones that gave me the most insight were O'Neills Elementary differential geometry, Loring Tu's An introduction to manifolds, Kobayashi and Nomizu's Foundations of differential geometry and Lee's Smooth manifolds. Again I didn't read the whole books, I just looked for the topics I was interested in at the moment. Jul 3, 2021 at 22:07
• I think it may be worth documenting the recommendations some where which may be found easily, because, from my understanding you have learned this all from scratch (technically everyone does, but I don't think so you follow a set 'course' ) @Jackozee Hakkiuz Jul 3, 2021 at 22:15
• @Buraian there are many threads in this site about differential geometry books. You just need to put "differential geometry books" or "differential forms books" in the search bar. Or did you have something else in mind? Jul 3, 2021 at 22:27

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.