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


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.

  • 1
    $\begingroup$ Isn't this problem completely answered in Frankel's book on geometry for physicists ? In the chapter about Lie derivative. It is also in a paper by Flanders : jstor.org/stable/2319163 in a quicker way. The complete demonstration use differential forms and your notations are really ugly to that respect. $\endgroup$
    – FraSchelle
    Apr 20, 2016 at 13:10
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    $\begingroup$ Sorry, complete reference to Frankel's book is dx.doi.org/10.1017/CBO9781139061377 section 4.3.b of the third edition version. $\endgroup$
    – FraSchelle
    Apr 20, 2016 at 13:11
  • $\begingroup$ Thanks for the references. As for what is ugly or not, I think that one day after much hard work I will understand some of differential geometry, but I don't ever expect to understand aesthetics. $\endgroup$
    – Sean D
    Apr 20, 2016 at 14:07

1 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).


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