Continuity Equation in Differential Geometry

Question

I'm looking for a derivation of the mass continuity that applies in general on symplectic manifolds. In particular the "the amount of change in the mass in a volume is just amount that flows in or out" heuristic argument is less formal than I'd like. I've found an answer here that gives a good idea of how the proof should work.

Below I try to reproduce a derivation from here.

If we have a manifold $M$ with a diffeomorphic flow $\phi_t$ and volume-form $\mathrm d\omega$, then a sub-region $D$ of our manifold has mass $$M\left(D,t\right)=\int_{D}\rho_{t}\mathrm{d}\omega$$ which shouldn't change as it flows along $$ \int_{\phi_{t}D}\rho_{t}\mathrm{d}\omega=\int_{D}\rho_{0}\mathrm{d}\omega $$ and now we change variables and introduce the pullback $\phi^*$ $$ \int_{D}\phi_{t}^{*}\rho_{t}\mathrm{d}\omega =\int_{D}\rho_{0}\mathrm{d}\omega$$ and so, taking the time derivative, $$ \partial_t\int_{D}\phi_{t}^{*}\rho_{t}\mathrm{d}\omega = \int_{D}\partial_t\left(\phi_{t}^{*}\rho_{t}\mathrm{d}\omega\right) =\int_{D}\partial_t\left(\rho_{0}\mathrm{d}\omega\right)=0$$ The next (and crucial step) is to transform the integrand as follows,

$$ \partial_t\left(\phi_{t}^{*}\rho_{t}\mathrm{d}\omega\right)= \phi_{t}^{*}\left(\it\unicode{xA3}_X\left(\rho_{t}\mathrm{d}\omega\right)\right)+\phi_{t}^{*}\left(\left(\partial_{t}\rho_{t}\right)\mathrm{d}\omega\right)$$

Where$\it\unicode{xA3}$ is the Lie derivative along the vector field $X$ induced by $\phi$. This is the step I don't understand (mathematically and physically). After that one simply undoes the pullback and says the integrand must be zero since the integral is zero over arbitrary domains and hence

$$\it\unicode{xA3}_X\left(\rho_{t}\mathrm{d}\omega\right)+\left(\partial_{t}\rho_{t}\right)\mathrm{d}\omega=0$$

and if the fluid is incompressible then $\partial_t\rho=\it\unicode{xA3}_X\rho_{t}=0$ and so

$$\it\unicode{xA3}_X\mathrm d\omega=0$$

If anyone can shed light on the problematic step above, or provide a different derivation in the same spirit it would be much appreciated.

Answer 1

$$ \partial_t\left(\phi_{t}^{*}\rho_{t}\mathrm{d}\omega\right)= \phi_{t}^{*}\left(\it\unicode{xA3}_X\left(\rho_{t}\mathrm{d}\omega\right)\right)+\phi_{t}^{*}\left(\left(\partial_{t}\rho_{t}\right)\mathrm{d}\omega\right)\tag 1$$ This is quite easy, the fundamental reason for the result above is that

$\quad\quad$ the function $\phi_{t}^{*}\left(\rho_{t}\mathrm{d}\omega\right)$ has two independent temporal dependences due to the nature of $\rho$.

(a) One is due to the parametric dependence of $\rho$ on time, as $\rho=\rho(t,x)$.

(b) The other is due to the pullback $\phi_{t}^{*}$ which only acts in the spatial variables $x$.

Notice that, instead, $\omega$ depends on $t$ only through the spatial variables under the action of $\phi_t$, since it has no explicit dependences on $t$.

The second ingredient is the definition of Lie derivative with respect to the vector field $X$ generating the one-parameter group of (symplectic) diffeomorphisms $\{\phi_t\}_{t \in \mathbb R}$: If $\Xi$ is a tensor field, we have $$\partial_t \phi_{t}^{*} \Xi = \phi_{t}^{*} {\it\unicode{xA3}_X} \Xi\:.$$

Let us come to (1). Using the elementary Leibniz rule for $\partial_t$ taking the two dependences on $t$ into account, we immediately have, $$\partial_t\left(\phi_{t}^{*}\rho_{t}\mathrm{d}\omega\right)= \left.\partial_t\left(\phi_{t}^{*}\rho_{\tau}\mathrm{d}\omega\right)\right|_{\tau=t} + \left.\phi_{t}^{*}\left(\left(\partial_\tau\rho_{\tau}\right)\mathrm{d}\omega \right)\right|_{\tau=t}$$ that is $$\partial_t\left(\phi_{t}^{*}\rho_{t}\mathrm{d}\omega\right)= \left.\phi_{t}^{*}\left(\it\unicode{xA3}_X\left(\rho_{\tau}\mathrm{d}\omega\right)\right)\right|_{\tau=t} + \left.\phi_{t}^{*}\left(\left(\partial_\tau\rho_{\tau}\right)\mathrm{d}\omega \right)\right|_{\tau=t} $$ which is (1).