# Liouville's Theorem For Spacetime

Liouville's theorem states that the density of phase space governs the continuity equation. $$\frac{\partial\rho}{\partial t}+\sum_{i=1}^n\Big(\frac{\partial(\rho\dot{q_i})}{\partial q_i}+\frac{\partial(\rho\dot{p_i})}{\partial p_i}\Big)=0$$ This means that the number of states in your statistical system is conserved in time.

My question is, is there a similar structure to the density of spacetime $\sqrt{-g}$? Is the volume of spacetime conserved? If you want to think of the geometry of spacetime as physical, I feel like there must be something which could be interpreted as a continuity equation for the density of spacetime $\sqrt{-g}$, spacetime events shouldn't disappear for no reason!

Perhaps the continuity equation would read something like this: $$\big(\sqrt{-g}\frac{dx^\mu}{d\tau}\big)_{;\mu}=0$$ Where $x^\mu$ is somehow tracking a spacetime event's evolution. I am unable to find any results of this kind in a general relativity textbook, even though it seems like such an important question to me. What if this equation is nonzero? What makes it nonzero? The presence of matter?

Edit: I realize now that in order to define this continuity equation properly, it would need to involve the induced metric $h_{\mu\nu}$ of a set of space like surfaces, with a time like normal vector $n$ for each surface, and the equation would then read something like $$\big(\sqrt{h}\frac{dx^\mu}{dn}\big)_{;\mu}=0$$ The $x^{\mu}$'s should be geodesics of the full spacetime with initially velocities normal to one of these space like surfaces. For every point on this surface, there corresponds a geodesic, so the $x^{\mu}$'s depend on the location on the surface.

• There is no Liouville theorem for generic spacetimes. Think e.g. of gravitational lensing or the geodesic deviation equation. – Qmechanic Jun 24 '18 at 13:54
• Gravitational lensing occurs in the presence of matter, correct? Since the density of spacetime is not equivalent to the curvature, it could be that there is a loss of spacetime events in the matter and a curvature is set up outside the matter to compensate for the lack of uniformity of spacetime events. I apologize for speaking in such an imprecise way, I haven't fully mastered GR or differential geometry. – fewfew4 Jun 24 '18 at 14:01

For a general Riemannian or pseudo-Riemannian manifold with metric $g$, the covariant derivative with the Levi-Civita connection satisfies,

$$\nabla_\mu \sqrt{|g|} = 0$$

where $g = \det g_{\lambda\sigma}$. This can be proved from the metric compatibility condition, $\nabla_\mu g_{\lambda\sigma} = 0$ which the Levi-Civita connection ensures. In particular, it is obvious when expressing the determinant as,

$$g = \frac{1}{4!}\epsilon^{\alpha\beta\gamma\delta}\epsilon^{\mu\nu\lambda\sigma}g_{\alpha\mu}g_{\beta \nu}g_{\gamma\lambda}g_{\delta\sigma}.$$

The vanishing of the volume element under covariant differentiation is useful in manipulations with tensor densities, since these are constructed with factors of $\sqrt{|g|}$.

This isn't really a physical fact though, it's just a consequence of differential geometry. If we take your proposal,

$$\nabla_\mu(\dot x^\mu \sqrt{|g|}) = 0$$

since $\nabla_\mu \sqrt{|g|} = 0$, your suggestion is equivalent to, $\nabla_\mu \dot x^\mu = 0$.